MECHANICS: NEWTONIAN, CLASSICAL, RELATIVISTIC: THEORY, PROBLEMS, APPLICATIONS Prasun Kumar Nayak
Asian Books Private Limited
MECHANICS: NEWTONIAN, CLASSICAL, RELATIVISTIC THEORY, PROBLEMS, APPLICATIONS
By
Prasun Kumar Nayak Lecturer and Head. Department of Mathematics, Bankura Christian CoI\ege, Bankura - W.B.
,tSlAII
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PREFACE Mechanics plays an important role in the spheres of Applied Mathematics, Physics, and Engineering due to its inherent viabilities. The aim of this book is to give a rigorous and thorough analysis and applications of various aspects of Mechanics. Also, the present book has been designed in a lucid and coherent manner so that the Honours and Postgraduate students of various Universities may reap consIderable benefit out of it. I have chosen the topics with great care and have tried to present them systematically. The author expresses his sincere gratitude to his great teachers Prof. P.K. Chowdhury and Prof. P.R Chowdhury, Department of Applied Mathematics, Calcutta University, who taught him this course at PG level in 1992-1994. Author is thankful to his friends and colleagues, specially, Dr. G.B. Sural, Dr. A.P. Sinha of Bankura ChrIstian College, for their great help and valuable suggestions in the preparation of the book. Author also extends his thanks to Dr. Madhumangal Pal, Dept. of Applied Mathematics, Vidyasagar University, Dr. RRN. Bajpai, Principal, Bankura christian College, for their effective encouragement. This book could not have been completed without the loving support and encouragement of the parents, wife (Mousumi) and son (Bubai). I extend my thanks to my students, relatives, family members for their constant encouragement and support. Critical evaluation and suggestions for the improvement of the book will be appreciated and greatly acknowledged.
Prasun Kumar N ayak Bankura Christian College, Bankura, India, 722 101.
~
I
( About the Author) Prasun Kumar Nayak, a graduate (1992) of Calcutta University
(with Honours in Mathematics from R.K. Mission Residential College, Narendrapur), received his M.Sc. in Applied Mathematics (1994), from the same University. He is currently a voluntary research scholar at Department of Applied Mathematics, Vidyasagar University. His research interests are in Fuzzy Game Theory. At present, he has been serving as a Lecturer and Head of the Department of Mathematics, in Bankura Christian College, Bankura (West Bengal), a highly reputed constituent college (NAAC accredited B++) to the University ofBurdwan, India.
Contents Preface Bibliography 1 Mechanics of Particles 1.1 The Particle. . . . . 1.2 Newton's Laws of Motion 1.3 Conservative force field. . 1.4 Mechanics of a Particle. . 1.4.1 Conservation of linear momentum 1.4.2 Conservation of angular momentum 1.4.3 Conservation of energy. . 1.4.4 Equation of motion . . . . 1.5 Mechanics of System of Particles 1.5.1 Centre of mass . . . 1.5.2 Linear momentum .. 1.5.3 Angular Momentum . 1.5.4 Energy of the system 1.5.5 Kinetic energy . 1.5.6 Rigid body . . . . . 1.6 Motion of variable mass . . 1.7 Nonrealistic rocket motion . 1. 7.1 Single-stage rocket . 1.7.2 Multistage rockets . 1.8 Limitations of Newton's Law 1.9 Virial Theorem . 1.10 Exercise . . . . . . . . . 2
Generalised Co-ordinates 2.1 Constraints . . . . . . . . . . . . . 2.1.1 Holonomic Constraints . . . 2.1.2 Non-Holonomic Constraints 2.1.3 Scleronomic Constraints 2.1.4 Rheonomic Constraints 2.1.5 Bilateral Constraints . . 2.1.6 Unilateral Constraints . 2.1.7 Conservative Constraints (\'ii)
1
3 4 4 6 8 8 9 10 12 15 15 16 18 21 23 24 24 26 26 30 32 33 35
37 37 38 38 39 39 39 39 40,
(viii) 2.1.8 Dissipative Constraints 2.2 Generalised .Co-ordinates . . . . 2.2.1 Notation for GC . . . . 2.2.2 Transformation Equations 2.2.3 Configuration Space 2.2.4 Degrees of Freedom 2.3 Generalised Displacement 2.4 Generalised Velocity . . . . 2.5 Generalized Acceleration .. 2.6 Generalized Kinetic Energy 2.7 Generalised Momentum 2.8 Generalised Force . . . . . 2.9 Generalized Potential .. . 2.10 Principle of Virtual Work 2.10.1 Virtual Displacement. 2.10.2 Virtual Work . . . . . 2.10.3 Principle of Virtual Work 2.10.4 D'Alembert's Principle. 2.11 Exerci se . . . . . . .
CONTENTS
40 40 42 42 42 43 43
44 45 45
49 50 51 52 52 52 52 53 55
3 .Variational Principle 3.1 Properties of Functionals . . . . . . . . . . . . 3.2 'Euler's Equation . . . . . . . . . . . . . . . . . 3.3 Functionals depend on higher order derivatives 3.4 Functionals on several independent variables. 3.5 Variational problems with moving boundaries 3.6 Parametric representation 3.7 Exercise . . . . . .
57 58 60 66
4
77
Lagrangian Mechanics 4.1 Lagrange's equation of motion. 4.2 Dissipative system 4.3 Non-Holonomic System . . . . 4.4 Cyclic Co-ordinates. . . . . . . 4.4.1 Systems with ignorable co-ordinates 4.4.2 Separation of variables . . . . . 4.4.3 Liouville's class of Lagrangians . 4.5 Conservation Theorems . . . . . . . . . 4.5.1 Conservation theorem of energy . 4.5.2 Conservation theorem of linear momentum 4.5.3 Conservation theorem of angular momentum 4.5.4 Condition of invariance 4.6 Exercise . . . . . . . . . . . . . . . . . . . . . . . . .
68 71
74 75
77 89
91 95 96 99 100 104 104
105 106
109 116
CONTENTS
(ix)
5 Hamiltonian Mechanics 5.1 Phase Space . . . . . . . . . . . 5.2 Hamiltonian Function . . . . . 5.3 Hamilton's Equation of Motion 5.4 Hamilton's Principle . . . . . . 5.4.1 Modified Hamilton's principle. 5.5 Action Principles . . . . . . . . . . . . 5.5.1 Action as function of co-ordinates 5.5.2 Principle of Least Action .. 5.5.3 Second Form of Least Action Maupertui's Principle . . . . 5.5.4 5.6 Symmetry, Conservation Laws . . . . 5.6.1 Conservation of linear momentum 5.6.2 Conservation of angular momentum 5.6.3 Conservation of energy 5.6.4 Noether's Theorem 5.7 Conclusion 5.8 Exercise.......
119 119 120 123 133 135 139 139 140 143 146 149 150 151 151 153 154 154
6 Transformation Theory 6.1 Poisson Brackets .. 6.1.1 Properties of PB 6.1.2 Basic PBs . . . . 6.2 Lagrange Bracket . . . . 6.2.1 Angular Momentum and PB relations 6.3 Point Transformation . . . 6.4 Legendre Transformations 6.5 Canonical Transformation 6.6 Generating Functions. 6.6.1 First Form 6.6.2 Second Form 6.6.3 Third Form 6.6.4 Fourth Form 6.6.5 Cyclic relations 6.6.6 Examples of CT 6.7 Condition for Canonicality . 6.7.1 Exact Differential Form 6.7.2 Poisson Bracket Form 6.7.3 Lagrange Bracket Form 6.7.4 Bilinear Covariant Form 6.8 Properties of CT . . . . . . . . 6.9 Equation of Motion . . . . . . . 6.10 Infinitesimal Contact Transformation. 6.11 Finite Transformations . . . . . . . . .
157 157 158 161 162 163 165 165 166 168 170 174 175 176 178 178 182 182 184 186 186 189 196 201 204
(x)
CONTENTS 6.12 Liouville's Theorem 6.13 Exercise . . . . . . .
205 208
7 Hamilton-Jacobi Theory 7.1 Hamilton-Jacobi Equation . . . . . . . . . . . . . 7.2 Hamilton principal function . . . . . . . . . . . . 7.2.1 Connection with canonical transformation 7.2.2 Significance of S . . . . . . 7.3 Hamilton's characteristic function. 7.3.1 Significance of W . 7.4 Separation of Variables. 7.5 Action-Angle Variables.
211 211 212 213 214 218 220 222 228
8
239 239 241 242 243 246 247 248 264
Reference Frames 8.1 Translation motions 8.2 Rotating frames of reference . . . . . . . . 8.2.1 Rate of change of arbitrary vector 8.2.2 Velocity and Acceleration 8.2.3 Equation of motion .. 8.3 Lagrangian and Hamiltonian 8.4 Motion of rotating earth 8.5 Exercises
9 Rigid Body Motion 9.1 Moments, Products of Inertia . . . . . . . . 9.1.1 Axes theorem for moment of inertia 9.1.2 Radius of gyration. . . . . . . 9.1.3 MI of some symmetrical bodies 9.2 Independent Co-ordinates . . . . 9.2.1 Generalised Co-ordinates 9.2.2 Eulerian angles . . . . . . 9.3 Angular Velocity . . . . . . . . . 9.3.1 Precession and Nutation. 9.4 Angular Momentum 9.5 Kinetic Energy . . . . . . . 9.6 Euler's Theorem . . . . . . 9.7 Euler's Equation of Motion 9.8 Force/Torque-free motion . 9.8.1 Poinsot's geometrical representation 9.8.2 Polhode and Herpolhode 9.8.3 Body and Space cone 9.8.4 Binet Ellipsoid . . . 9.9 Motion of Symmetrical Top 9.9.1 Integrals of motion.
265 265 266 267 267 268 268 268 271 272 272 273 276 276 278 282 285 287 288 290 292
CONTENTS
(xi)
9.9.2 Precession without Nutation 9.9.3 Nutational Motion . . . . 9.9.4 Steady motion . . . . . . 9.9.5 Stability of steady motion 9.10 Exercise .
294 296 298 299 304
10 Central Force Field 10.1 Central force . . . . . . . . . 10.1.1 Angular momentum. 10.2 Centre of Mass . . . . . . . . 10.3 Two-Body Problem . . . . . . 10.4 Lagrange's equation of motion. 10.4.1 Expression for total energy 10.4.2 Potential energy . . . 10.4.3 Integrals of motion .. 10.5 General equation of the orbit 10.5.1 Equation of orbit . . . 10.5.2 Stability of orbits. . . 10.6 Motion under arbitrary potential field 10.7 Kepler's Law . . . . 10.8 Unbounded Motion . . . . 10.9 Scattering . . . . . . . . . 10.9.1 Elastic Scattering. 10.9.2 In-elastic scattering 10.9.3 Reaction . . . . 1O.10Direct/linear collisions .. . 10.10.1 Loss of KE . . . . . 10.10.2 Characteristic of direct collision . 10. 11 Oblique collisions. 10.11.1 Loss of KE 10. 12Exercises . . . . .
305 305 306 307 308 313 314 315 316 321 323 323 326 328 329 330 330 333 333 334 334 335 336 336 337
11 Theory of Oscillations 11.1 Equilibrium . . . . . . 11.1.1 Stable equilibrium .. 11.1.2 Unstable equilibrium. 11.2 Oscillatory problems . . . . . 11.2.1 Simple pendulum .. 11.2.2 Damped harmonic oscillation 11.2.3 Forced harmonic oscillation 11.3 Small oscillation . . . . . . . . . . . 11.3.1 Generalised co-ordinates . . . 11.3.2 Kinetic and Potential energy 11.4 Lagrange's Equation of Motion . . .
339 339 339 340 340 340 342 343 345 345 345 348
CONTENTS
11.4.1 The Eigenfunctions and Eigenfrequencies . . . 11.4.2 Normal co-ordinates, Frequencies of Vibration 11.4.3 Oscillation under constraint . . . . . . . . 11.4.4 Stationary property of the normal modes 11.5 Exercises 12 Relativistic Mechanics 12.1 Inertial frame of reference 12.2 Newtonian relativity .. 12.3 Galilean transformation . 12.4 Galilean invariance . . . 12.4.1 Invariance of space 12.4.2 Invariance of Time interval 12.4.3 Invariance of Velocity . . . 12.4.4 Invariance of acceleration . 12.4.5 Illvariance of Newton's Law 12.4.6 Conservation law of linear momentum 12.4.7 Conservation law of Energy 12.5 Special Theory of Relativity .. . 12.6 Lorentz Transformation . . . . . 12.6.1 Relativity of simultaneity 12.6.2 Length contraction. 12.6.3 Time dilation . . . . . . 12.7 Addition of velocities. . . . . . 12.8 Transformation of acceleration 12.9 Variation of mass with velocity. 12.lOMass-energy relation . . . . . . . 12.11Momentum and energy . . . . . . 12.12Transformation of energy and momentum 12.13Relativistic force . . . . . 12.14Equation of motion . . . . 12.14.1 Relativistic rocket 12.15Inelastic collision .. 12.16 Minkowski space .. 12.16.1 Proper time. 12.16.2 Four velocity 12.16.3Four momentum 12.16.4Four force . . . . 12.16.5 Kinetic energy . 12.17The relativistic Lagrangian 12.18Exercise . . . . . . . . . . .
(XIl)
348 351 361 362 370 373 373 373 374 375 376 376 376 377 377
378 378 380 381 384 385 '388 391 393 395 397 400 401 404 405 405 406 407 409 409 410 411 412 412 413
Bibliography [1] A. Sommerfield, Classical Mechanics, Academic Press, New York, 1964. [2] A. S. Gupta, Calculus of Variations with Applications, Prentice-Hall of India Private Limited, New Delhi, 2005. [3] A. S. Ramsey, Dynamics (part-2), CBS Publishers and Distributors. [4] C. R. Mondal, Classical Mechanics, Prentice-Hall of India Private Limited. [5] Gupta, Kumar and Sharma, Classical Mechanics, Pragati Prakashan. [6] E. T. Whittaker, A treatise on the Analytical Dynamics of Particles and Rigid bodies, 4/e, Dovar Publication.
[7] H. Goldstein, Classical Mechanics, Addision-Weseley Publishing Company. [8] J. Synge and B. Griffith, Principle of Mechanics, 2/e, McGraw Hill, New York, 1949. Prentice-Hall of India Private Limited. [9] L. Greiner, Classical Mechanics, System of Particles and Hamiltonian Dynamics, Springer International Edition. [10] N. C. Rana and P. S. Joag, Classical Mechanics, TMH Publishing Company Limited. [11] P. K. Nayak, Numerical Analysis (Theory and Applications), Asian Books Pvt. Limited. [12] R. G. Takwale and P. S. Puranik, Introduction to Classical Mechanics, Tata McGraw-Hill Publishing Company Limited.
1
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Chapter 1
Mechanics of Particles Mechanics can be defined as that science which describes and predicts the conditions of rest or motion of material bodies around us under the action of forces where a body is said to be at rest, when it does not change its position with reference to the surrounding objects, but when it changes its position, we say that it is in motion. The topic of mechanics has three major sections 1. mechanics of rigid bodies 2. mechanics of deformable bodies
:t mechanics of fluids The mechanics of rigid bodies may be subclassified into three subsections (i) Kinematics: It deals with the all possible motion of material systems with reference to the agency which causes motion and gives the geometrical description of the body. Here we are to describe the motion of the material bodies in terms of quantities such as displacement, velocity, acceleration etc. (ii) Dynamics: It deals the causes of motion which determine, among all possible motions, which motion will actually take place in any particular system. Here we introduce the concept of force. (iii) Statics: It deals with the system of forces which actually gives no motion (rest) to the system relative to some given frame of reference. For equilibrium of the system the resultant effect of the forces should be zero. There are many fields of study in mechanics, such as classical, relativistic, quantum, continuum, space mechanics, and so on. Classical mechanics which is primarily supposed as the study of motions of physical objects such as motion of celestial bodies, is now considered as the part of mechanics dealing with the objects neither too big so that there exists a close agreement between theory and experiment, nor too small interacting object so that systems are considered on an atomic scale. 3
CHAPTER 1. MECHANICS OF PARTICLES
4
1.1
The Particle
By a term 'particle', we mean a point body having some mass or a body of some mass but having negligible physical dimensions in comparison with the dimensions of the other bodies involved in the systems under consideration. For example, in the planetary motion, the planet which rounds the sun can be treated as a particle. In atomic structure, au atom which can go too close to another charged body, such a molecule, may not be treated as a particle. Thus by a particle, we mean, an object whose size, internal structure, rotational and vibrational motions are negligible for the problem with which we are concerned.
1.2
Newton's Laws of Motion
Newton's laws of motion are the basis of the development of classical mechanics. These laws were formulated by Newton by considering mass, length and time as three fundamental quantities, which are invariant and independent of each other. Newton's laws of motion are stated in the following form: (i) Every body continues to be in its state of rest or of uniform motion in a straight line unless it is compelled to change that state by external forces acting on it (Law of inertia). (ii) The time rate of change of momentmn of the particle is proportional to the external force and is in the direction of the force (Law of causality). (iii) To every action there is always an equal and opposite reaction. This law prescribes the general nature of forces of reaction in relation to the forces of action (Law of reciprocity). Note that, these laws of motion has meaning only with respect to a particular coordinate system. They are valid in a class of reference frames, called inertial frames. Thus these laws successfully explain the motion of celestial bodies, such as planets, stars, etc. From the laws we have the following discussions ; Result 1.2.1. The first law of motion introduces two important concepts Concept of inertia (ii) Concept of Force.
(i)
Deduction 1.2.1. Inertia: The first law gives the concept of state of rest or of uniform motion that are actually kinematical concepts. The term inertia means the tendency of a body to continue, i.e., to remain in a state of rest or of uniform motion for ever in absence of any external force. The property of a body to remain in the state of rest or of uniform motion when no external force acts on it is known as inertia and a frame of reference in which such a state is observed is known as inertial frame of reference. In fact, we can turn around and define the inertial frames to be those frames in which Newton's laws of motion are valid.
1.2. NEWTON'S LAWS OF MOTION
5
Deduction 1.2.2. Force: The first law of motion gives us the qualitative definition of 'force'. A force is something which changes or tends to change the state of rest or of uniform motion of a body in a straight line. Result 1.2.2. Measurement of force: The second law of motion gives us a quantitative definition of a force, i.e., it provides us with a measure of the applied ---+ force. Let a particle of mass m be moving with velocity 11. Let F be the applied force acting on the particle in the direction of motion. Thus the velocity changes, and produces acceleration 0:. Thus from the Newton's second law we get, ---+
F
---+
<X
dp ---+ dt; p = dp
----> m v = change momentum d
---+
---->
d7f
=> F = k dt = k dt (m v ) = km dt = km a
where k is the constant of proportionality. Let us define, the unit of force as that force which acting on a body of unit mass produces a unit acceleration. Thus, ---+ m = 1,1 F 1= 1, so, k = 1. Thus the expression of the second law becomes ----> d7f ---+ F =km- =1na. (1.1) dt Since the second law demands the priori knowledge of the total force that a body experiences, this total force must also include all the forces of reactions that it experiences. F is known as the imprelised force and m 0: is called the effective force. Now we have the following discussions: ---+
---+
(i) From the second law, it follows that when external force F = 0 ,i.e., if there is no impressed force, momentum m 7f is a constant, i.e., the body will continue to be in the state of uniform motion. We can consider the state of rest as a special case of the second law.
(ii) It follows from the equation that, 0: = ~ F, i.c., accelera.tion may be defined as the force per unit mass. ----> ----> (iii) The equation F = m a can be used to measure the mass of a body. Thus, ---+ let the same force F act on two bodies having ml and m2 so as to produce acceleration 0: 1 and 0: 2 respectively. Therefore, ----> 1
ml 1a 1
1 ml = m2 1----> a 2 => - = m2
10:2 1 1----> I' al
If, now, m2 is known or taken as unit mass, ml can be calculated by measuring :~~:. The mass of a body measured in the manner is called the inertial mass.
(iv) The second law states that the change of momentum produced by a force in a body is in the direction of the straight line in which the force acts. Thus, if two or more forces act on a body at the same time, then each produces an acceleration proportional to it in its own line of action independent of the motion of the body. This is the principle of the physical independent of forces.
6
CHAPTER 1. MECHANICS OF PARTICLES (v) Another way of measuring the mass of a body is the weighting it, i.e., by comparing the gravitational force acting on it with that on a standard mass. In this procedure, we make use of the fact that the gravitational field the weight of a body is exactly equal to the gravitational force acting on it. In -t --t this case, the equation F = m 71 becomes the weight W = m g, where 9 is the acceleration due to gravity. The mass of a body measured in this manner is called the gravitatio~al mass.
Deduction 1.2.3. The third law states that the mutual actions of any two bodies are equal and oppositely directed along the same straight line. This law can be applied only to the two isolated particles exerting forces on each other when forces due to all other particles are completely absent. The forces of action and reaction or of mutual interaction between the two particles (not in the same body) do so along the liue joining the two particles.
Ex 1.2.1. If (X, Y) be the component of forces acting in 2 dimension on a particles of unit mass, then the Newton's equation can be written as fx[(Y -X~)/~1-2X = o. SOLUTION: The given particle have unit mass, i.e., m second law of motion, we have,
1.3
= 1, hence by Newton's
Conservative force field
Conservative forces have a natural occurrence in classical mechanics. Typically, conservative forces are such that the work done by them, as the system moves from one configuration to another, depends only upon the initial and final co-ordinates of the particles. Gravitational and electrostatic fields are the common examples of conservative fields. This forces may be distinguished by anyone of the following equivalent features : (i) The work done by the force is independent of path.
1.3. CONSERVATIVE FORCE FIELD
7
(ii) Around any closed path in the force field, the work done is 0, i.e.,
f
---t
F .dr:> = O.
---t
(iii) If (Fx, FYI F z ) are the cartesian components of the force F, then Fxdx+Fydy+ Fzdz is an exact differential. ---t
---t
---t
---t
---t
(iv) F is only a function of position and \l x F = 0, i.e., one can write F as the gradient of some scalar point function. Since the gradient points towards the direction of increasing potential and forces cause the system to move towards ---t ---t the lower potential, so F = - \lV. Scalar point function V introduced here is called potential energy of the particle at that point. Thus a conservative force is also central in nature. (v) A potential energy function V exists that has a definite value at every point. (vi) At every point T
+V
=constant for a particle, where T is the kinetic energy. ---t
---t
Note that, in the equation F = - \lV, the potential energy V introduced makes it ---t --; --; clear that, it is not unique, as F = - \l V = - \l (V + C), where C is a coristant, so the zero level of V is arbitrary. Hence the absolute value of the PE has no meanin.g. But we can calculate the difference of PE by: B
WAB =
B
J
J
A
A -.
F.drt = -
VV.drt = VA - VB
where WAR is the work done by the force F in displacing the particle from position A to position B.
F
3
Ex 1.3.1. Prove that = (2xy + z3)i + x 2 + 3xz2k is a conservative force field. Find the scalar potential function V. Find also the work done in moving an object in this field from (1, -2, 1) to (3,1,4). SOLUTION: For the given force field, we have ---t
---t
k
J
\l x F -
..JL ..JL..JL ax a~ az . 2xy + z3 X 3xz 2 = (0 - O)i - (3z 2 - 3z 2)3 + (2x - 2x)k
= O.
Thus the given force field is conservative. Thus :3 a scalar potential function, V such ---t ---t that F = - \lV, Le.,
(2xy
+z =}
3
2
2
av
av
av
)i
+ x j + 3:z;z k = - [ax - i + !:lj + !:lk] uy uz
av ax
= -2xy - z3; V = _x 2y - xz 3 + 'lh(y, z)
A
av
!:l =
uy
av az
A
A
A
_x 2; V = _x 2y + 'ljJ2(X, z)
= -3xz2; V = -xz 3 + 1/J3(:t, y)
A
A
CHAPTER 1. MECHANICS OF PARTICLES
8
Thus the scalar potential function is Vex, y, z) = -x 2 y-xz 3 +c, where c is independent of x, y, z. Now the work done in moving an object in this field from A(I, -2, 1) to B(3, 1,4) is WAB = VA - VB = 202units.
1.4
Mechanics of a Particle
Let P be the position of the particle whose position vector is ~ with respect to some convenient origin O. The path of the particle may be described by some parameter l. If l depend on time t, then
~ = ~(t) = x(t)i + yet)] The velocity
+ z(t)k.
v is the time rate of change of the displacement of a particle, given by d--:t ~ v =-= r dt
----+
which is the vector tangent to the path l at PI and is directed in the sense in which the variable t increases along the path. The acceleration vector is similarly defined as the time rate of change of velocity vector
a
dv d2~ a =-=-2-'
----+
dt
dt
Newton's second law of motion is basic law of mechanics, which states that --+() F e
where
a = d;X
dv ----+ = -d (----+) Tn v = ' f n - = 'm a
dt
dt
(1.2)
is acceleration produced in the motion of the particle on applying --+
the external force F(e). The reference frame in which equation (1.2) is valid is called an inertial or Galilean system. Many of the important conclusions of mechanics can be expressed in the form of conservation theorems, which indicates under what conditions various mechanical quantities are constant in time. Now we derive the following conservation theorems for a particle in motion using Newtonian mechanics.
1.4.1
Conservation of linear momentum
Let Ii be the linear momentum of a body and p(e) be the external force applied on the particle. If the particle is of mass m, which is constant in time and is moving with velocity then
v,
-+
p
----+
= mv
d~ = m dt '
(1.3) --+
with the dimension [.MLT-Ij. According to Newton's second law of motion,F(e) = ->.
¥,: = Ii,
where
--+ F(e)
is the applied external force in the motion of the particle.
1.4. MECHANICS OF A PARTICLE
9
Suppose that there is no external force acting on the particle, then ----7
F(e)
dp
---+
= dt = 0 ::::}
---+
P
---+
= m v = a constant.
Thus if no external force acts on the free particle, then its linear momentum of the particle remains constant. It is the law of conservation of linear momentum. Also, if the mass of the particle is assumed to remain constant, it will continue to move along a straight line. This corresponds to a statement of Newton's first law of motion.
1.4.2
Conservation of angular momentum
A moment of momentum is angular momentum. Consider a particle P, whose position vector at any instant is r?, with respect to the base point O. If P is the ---+ linear momentum of the particle at a given instant, then its angular momentum L and the moment of force or torque -;;J about the point 0 is defined as ---+
---+
---+
L=rxPi
---+
---+
---+
(1.4)
T=TXF
---+
where P is the total force, experienced by the particle. Angular momentum has the dimension [AI L 2 T- 1]. The time rate of change of angular momentum L is given by ---+
dL dt
=
---+
d ---+ ---+ ) dt (r x p ---+
---+
=
---+
dr ---+ dt x p
---+
+
dP r x dt
---+
dp
=vxp+rxdt
=
---+
r
dp
x -
[as11 and
dt
p
are parallel, so11 x
p
---+
= 0]. ---+
Now, we are to show that, the time rate of change of angular momentum L is ----7
defined as torque. If an external force F(e) acts on the particle, then the external torque acting on the particle is defined as ----+
T(e)
= r?
----7
x p(e)
= r?
x
d---+
dL
dt
dt
--.!!... = -
=
. L.
Let us suppose that no external torque acts on the particle, then we have () T e
---+
=
dL dt
---+
---+
= 0 ::::} L = a constant vector.
Thus, if no external torque acts on a particle, then its angular momentum is conserved. It is the law of conservation of angular momentum. Planets moving around the sun is an example of this conservation law. Ex 1.4.1. The position vector of a particle of mass m 1rw'lring u:ndeT the influence of a force is given by r' = A sin wti + B cos wt] . Find the expressions for linear momentum and force. Also find the angular momentum particle about the origin.
CHAPTER 1. MECHANICS OF PARTICLES
10
= Asinwti + Bcoswt3.
SOLUTION: Here given that -:t at time t is given by --7
v
The velocity of the particle
d-:t = dt d [A' = dt smwti~ + B coswt]~'l = w[Acoswti - Bsinwt31.
Therefore, the linear momentum of the particle is given by
p = ml1 = mw[A cos wti -
Bsinwt31.
-+
The force F acting on the particle is given by
dp
-+
~
d
= dt = dt[mw(Acoswti -
F
~
Bsinwt.i)l
= -mw 2[Asinwti + Bcoswt31
= -mw 2-:t.
Now, the angular momentum of the particle about origin is,
L = 1" x p = [Asinwti + Bcoswt31 x mw[Acoswti ]
~
BSinwt31
k
Asinwt Bcoswt 0 mwAcoswt -mwAsinwt 0
= -mwAB[sin2 wt + cos 2 wt]k = -mwABk. 1.4.3
Conservation of energy
Energy conservation has most wide applicably, encompassing all branches of sciences equally. Consider that a particle of mass m is moved from a point A to B under --->
the action of an external force F(e). Then, work done by the force in moving the particle from A to B is given by
. =. j B
WAB
j' d .d-:t =. dt (m l1).d-:t B
--->
F(e)
A
A B
=m
J A
J B
dl1 -+ dt.d r
=m
A
J-+-+ B
d-:t d v.Tt -+
=m
v.d v
d-:t -+ as-= v dt
A
where v A, v n are the velocities at point A and n respectively, AND the scalar quantities TA, Tu are the kinetic energies of the particles in the positions A and B respectively. This relation is true for non-relativistic motions only.
1.4. MECHANICS OF A PARTICLE
11
(i) If TA > TB, WAB < 0, i.e., work is done by the particle against the force and the KE has decreased. The work done against dissipative forces like the frictional force always negative. (ii) If TB > TA, WAB > 0, i.e., work done by the force on the particle and the KE of the particle has increased.
Suppose that the external force acting on the particle is conservative in nature. Therefore, work done in moving the particle from the point A to B can also be written as B
WAB = ]
~.d7 =
A
B
]
-
~~ dr =
VA - VB
A
where VA and VB are the values of the potential energy of the particles at points A and B respectively. Thus WAB
= TB - TA = VA - VB =;. TA + VA = TB + TB
=;. T
+V =
constant .
Thus, if a particle moves under the action of conservative force, then the total energy of the particle is conserved, and has the dimension [ML 2T-2]. It is law of conservation of energy for a particle. A particle moving under the gravitational force is an example of this conservation law. However, if an given system frictional or other dissipative forces are present, then F .d7 is always positive and the closed path integral is non-zero. Such a system is called a non-conservative system. Ex 1.4.2. Find the period of oscillation for the potential V
= Vo tan 2 ax.
SOLUTION: Consider the conservation of energy in one dimension as
~mx2 + V(x) or, t =
= E
Vfin] 2:
dx
JE _ V(x)
+
constant.
For a moving particle, KE is always a positive definite quantity and as such there is the constraint V(x) < E. The period T of the oscillation is given by the time that the particle takes to travel from A to B and back. Thus, b
T
=
v'2m] JE-V(x)' dx a
CHAPTER 1. MECHANICS OF PARTICLES
12
where a, b are the roots of the equation V(x) = E, assuming E to be given. For the given potential V(x) = Vo tan 2 ax, the period of oscillation is
I a
T= v'2m
dx ; a= ~tan-l Vo tan 2 ax a
. VE -a
[E
YVa
a
= 2 ~J
YE+""Vo o Vk
cos ax dx; sin2 ax
k;2
=
2 -
E
E
+ Va
2 ~. -l[VE+Vo. ] 7r ~ = ;y E+""Vo sm E smaa =;y E+""Vo'
1.4.4
Equation of motion
The study of motion of a particle involves the relationship of space and time. When we consider the motion of a particle in an inertial frame, then we make the following basic assumptions: (i) The inertial frame is homogeneous: This assumption ensures that the equation of motion of the particle is independent of position of the origin of the coordinate system and orientation of the co-ordinate axes. (ii) The inertial space is isotropic: Under this assumption it follows that the physical properties are the same in all directions in the frame. (iii) The inertial frame time is homogeneous: This assumption ensures that a free particle which moves with a certain velocity in the inertial frame during a certain interval, does not move with a diffe'rent constant velocity during a later or an earlier time interval, Generally, the force acting on a particle may depend on its position, velocity and time and hence the equation of motion of the particle can be written in the form (1.5) This second order ordinary differential equation in space co-ordinates (7) and alter integrating it twice, we will get the equation of trajectory of the particle. There will be two constants of integration introduced and these will be determined by knowing the initial conditions of the particle. Taking the initial condition 7(to) = 11 0 , integrating equation (1.5) with respect to time, we get t
m,
t
J ~d' r
,f =
to
J--+P(--+ l' , ~ T ,t )d' ,t to
t
'* m[~ -
'Vo] =
JF to
(-;;7, ~,t)dt'.
(1.6)
13
1.4. MECHANICS OF A PARTICLE t~
.
J F (7, 7, t )dt'
The right hand integral
represents the impulse imparted to the
to
particle during time interval (t - to) and is equal to the change in the momentum of the particle. Integrating equation (1.6) again with respect to time, we get
J 'J ~F (~~ t
~ ( - to ) + m 1 r - ~ r a= ~ vat
t
dt
r , r ,t )d" t.
to
(1.7)
to
If the force Fe?, 7, t) is given, performing the integration in (1.7), we get the explicit form of the equation of the trajectory of the particle. For some given simple forms of the force F(7, 7, t) we shall discuss the nature of the trajectory.
(i) Let the given force F(7, ~
·ma.
7, t)
ticular, if F = mg. when considered constant.
g,
be of the form F(7, 7, t) = In parthe acceleration due to earth's gravity can be
~~~
~~~
~
(ii) Let the given force F ( .,. , 'r , t) be of the form F ( .,. , .,. , t) = F (t). For exam. pIe F (7, 7, t) be of the form F = F a sin wt which is experienced by charged particles in alternating fields. ~
~
~
~~~
~~~
~~
(iii) Let the given force F( r , r, t) be of the form F( r, r, t) = F( r). For exam. pIe F (7, 7, t) be of the form F = ~ 7 which is experienced by gravitational or electrostatic fields, under inverse square law. ~
~
~~~
~~~
~~
(iv) Let the given force F ( r , r ,t) be of the form F ( r , r ,t) = F ( r). This type of frictional force is experienced by a particle moving in a viscous medium. Ex 1.4.3. Forced Motion: For simplicity, let us consider an one dimensional motion along the x-axis. Let a time dependent force of the type F = Fa sin wt be acting on the particle. The equation of motion is
.. x with the initial condition t
=?
Fa. = -smwt m
= 0, x = Xa, v = Va.
:i: = Va
+
x = Xa
F + [va + -]t 'mw
Fa [1 - coswt];
mw
Fa
--2
'fnw
Thus we get integrating, sinwt.;
integrating
This problem is of interest in connection with scattering of electromagnetic radiation by free electrons such as those appearing in the ionosphere. Ex 1.4.4. Linear Harmonic Oscillator:
14
CHAPTER 1. MECHANICS OF PARTICLES
A block slides on a frictionless horizontal table. At equilibrium position, the spring is relaxed and exerts force on the body. When the body is displaced to the right, the direction of the force exerted by the spring on the body points to the left, and when the body is displaced to the left the direction of the force points to the right (restoring force). In each case, the magnitude of the force is proportional to the magnitude of the displacement of the body. When the mass is displaced through
Figure 1.1: Linear harmonic oscillator the distance x, the restoring force set up in the stretched spnng is F(x) The equation of motion can be written as
-kx.
(I2x dv F(x) = m dt 2 = m dt x =?
x
I F(x)dx = m I Xu
t
~~ dx = m I ~~ vdt to
Xo
t
X
~ I(-kx)d:r = m I :t(~v2)dt to
Xu
~
1
2
--kx + 2
1 .) 1 2 1 2 -k.1:6 = -mv - -mvo 2
2
2'
Let the mass be released initially from rest at a distance Xo from the origin. Therefore, the initial conditions are x = :1:0, v = Vo = 0 at t = O. Thus,
dx /'kV 2 2 v=-=V .!::.... x-x dt mO. ~
I
dx
CE
yf x 5-
x2
-
~A -
Xll
~ X
=
Xo
.' sm(wt + 2'7T )
m
It
dt .
0
= :1:0 cos wt;
w=~.
Deduction 1.4.1. In the above section, we have derived three immutable conservation laws of mechanics from Newton's laws of motion. These laws are the results of the first integral of equation of motion, and are thus called the first integrals of equations of motion. The range of validity of the conservation laws is quite broad, extending even to the domain of relativistic motions, where Newton's laws are replaced by laws of relativistic mechanics. These laws prove to be very powerful tools in solving mechanical problems in the following respects: (i) Conservation laws a.re independent of the details of the trajectory and often, of
1.5. MECHANICS OF SYSTEM OF PARTICLES
15
details of the particular force. These laws assure us many times that some aspects of motion are impossible and must left out. (ii) Conservation laws have an intimate relation with invariance. Their failure in certain cases may result in the discovery of new and not yet understood phenomena.
1.5
Mechanics of System of Particles
Here we extend the considerations of the previous section to a system of more than one particles. In mechanics, many time it is simpler to consider the motion of a body by dividing the body into large number of discrete points and by analyzing the motion of the latter. Hence, 'a system of particles' is a very useful concept. A system of particles is called a dynamical system.
1.5.1
Centre of mass
Consider a system of n particles, the ith particle of the system having mass mi and let its position vector with respect to some arbitrary base point is r\. The point G whose position vector is given by ----->
OG
~
n ~
~ - n - - L..J m 1 r i = m 1 i=l
1
=R=
2::
n 1~
~
M L..J mi r' 1 i=l
i=l
is called the centre of mass of the system of particles, where M is the total mass of all the particles. Physically, the centre of mass of the system can be thought of as a weighted average position of the system of particles. If the mass is continuously distributed, then the CM of bodies of simple geometric shape can be calculated by replacing summation by integration.
Theorem 1.5.1. The position of the CM is independent on the origin chosen. PROOF: To prove, the position of the centre of mass does not depend on the origin -----> ~ ~ ~ chosen, let us take some other point 0' as origin. Let 00' = a. If 0 P = r' ~, then
With 0' as origin, the centre of mass G', by definition, will be given by,
L
n
L
n
'i=l
i=l
~ 1 ~, 1 (~ o G = -M rn' '/' . = -M m' r' + ~) a il il 1 = !vI
L
~
'II
mi
1
ri
+M
1=1
----->
= OG
+~ Q =
~
n
L
mi
Q
i=l
----->
~
~
OG + 0 0 = 0 G.
This means that, G' and G are the same point. So, G, the centre of mass, is a fixed point whatever the co-ordinate system is used. Now
16
CHAPTER 1. MECHANICS OF PAR'UCLES
(i) If we take a co-ordinate system with the centre of mass (G) as origin (0'), then n
2::::
m~ T7:'
=
~
O. Here,
T7:'
is the position vector of the ith particle with centre
i=l
of mass as the origin. Thus, we may define the centre of mass of a system of particles as 'that point about which the vector sum of the first moments of all the masses vanishes.' (ii) A 'body' is a system of particles in which the mass is continuously distributed. If a body or any non-continuous system of particles occupies a region at all points of which the acceleration 9 due to gravity is the same, the centre of mass and centre of gravity coincide. In the absence of gravity, it is idle to speak of centre of gravity. It is better to speak of centre of mass.
1.5.2
Linear momentum
Consider a system of n particles, the ith particle of the system having mass mi and moving with velocities v\, 11 2 .... ,11 n respectively. T7~ is the position vector of the ith particle with respect to an arbitrary origin 0 and T7~ is the position vector of the same particle with respect to 0', the centre of mass, and R is the position vector of CM with respect to O. Then ~
.
where
Vi = T7 ~
.,
and v~.
= T7 i are the velocities of the ith particle with respect to 0 ~ ~ and R = V is the velocity of the CM with respect to O. The ---+
and 0' respectively total linear momentum P of the system of particles is given by
~ P = ""' ~ ---+ Pi = ""' ~ m~ ~ r i = ""' ~ mil ~ R + ~, r il d ""'~, ~ =M ~ V + dt ~ m'i r i = M V; ~
""'
---+,
as ~ mi r ~
=
~ 0
~
where we have assumed that mass of the individual particle does not change with time. The particles of the system exert force on each other due to mutual interaction (action and reaction). These forces are called inertial forces. In addition, the particles of the system may be acted upon by external forces due to sources outside the system. Thus, any particle will be under the action of the following two forces : Internal Force: If represents the interacting force on the ith particle due to the lh particle, then total inertial force on the ith particle will be ,
F;.
~
--+
F(t)",=LF J1.; j=1,2, ... ,n. J
1.5. MECHANICS OF SYSTEM OF PARTICLES
17
Since a particle does not exert any force on itself, it follows that in the above summation j i= i. Therefore, the total inertial force on the ith particle due to all other particles should be expressed as
where the symbol L.j will imply that i = 1,2, ... , n, but j i= i. For example, if the particles are charged, the Coulomb forces are inertial forces. ~
External force: Let
F(e)
be the external force acting on the system, so that the
~
external force F(e)i on the ith particle. If Pl is the linear momentum of the ith particle, then according to Newton's second law of motion, the equation of motion for the ith particle is ~ l
~
or,
p(e)
d-t
---+
p(e)
+ p(i) =-.!!....:.. dt l
-
d-t
+ LF.1'i = %t
(1.8)
l
J
or,
L# + LFJi = L d~l; i,J .
For the system.
The equation (1.8) gives the equation of motion for the ith particle of the system. Now if F(t) be the total external force on the system, then
The factor ~ appears in the above equation, because in summing over both i and j, the pair of the ith and the lh particles appears twice, once for the ith particle due to lh and secondly for the lh particle due to the ith. Since both are same and should appear only once, the factor ~ has been included. Therefore, -t(t)
F
1~ -. -t ~ dPi +2~[PJi+PljJ=~&.
(1.9)
According to Newton's third law of motion, mutual action and reaction forces are -t -t equal and opposite ( such as gravitational field, electrostatic field), so P ji + p lJ = 0, i.e., the force of self-interaction is zero. This assumption is sometimes referred to as weak law of action and reaction. Thus, the simplified form of the equation (UJ) is (1.10) This equation shows that in the case of a system of n particles, under the action of external forces, the centre of mass of the system behaves like a particle whose
18
CHAPTER 1. MECHANICS OF PARTICLES
mass is equal to the total mass of the system and is acted upon the total external force. The motion of the eM of the system is independent of the inertial forces that exist among the particles of the system. Therefore, in an inertial reference frame the time rate of change of the total linear momentum of a system is equal to the total external force acting on the system. If the total external force acting on the ---t ---t system is zero, i.e., P (t) = 0, then the above equation becomes
dp t dt
~
L..,
---t
= 0
=>
~ ---t
a constant vector.
L.., P i =
i
i
Thus, if a system of particles is acted upon by no net external force, then the total linear momentum of the system is conserved. It is the law of conservation of linear momentum of a system of particles.
1.5.3
Angular Momentum
If -:;7 i be the position vector of the 'i th particle, then the torque acting on the particle is given by ~ ---t 'f' . t
x P (e)
L p.
-
+ ---t 'f' . X t
ith
d---t
Pi - ---t 'r t. x --' dt'
---t
(1.11)
Jt -
j
--->
The first factor -:ti x p(e) represents the torque on the ith particle due to external force acting on it, while the second factor -:ti x ",£Fji represents the torque on the j
particle due to internal forces acting on it due to all other particles of the system. If the internal forces between two particles in addition to being equal and opposite also lie along the line joining the particles - a condition known as the strong law of ith
--->
action and reaction, then all of the cross product vanish. Denoting -:tt x p(e) = the torque on the
ith
-n(e) ,
particle due to external force, the equation (1.11) becomes, ---t (e)
T.i
+ ---tr i
X
~ ---tp L..,
ji
=
---t
r i
X
dp i
dt
j
or,
L
-n(e)
+ I:[-:t i
x
I:F
ji ]
=
I:[-:t
i
x d!t]
j
or,
I:
-n(e)
+ L[-:t i
x
F ji ]
= 2)-:ti x d!i].
tJ
1
Now, the first factor of the (1.12)
(1.12)
i
"'£ T:;(e) 'i
system and the second factor ",£[-:;7i x
= -:f(e) , total external torque on the
Fji ] = -;P(i) , the sum of the torques due to
i,j
internal forces acting on the system. Thus, the equation (1.12) may be written as
-;pee) + -;P(i) = ~[-:ti t
x d!i].
(1.13)
1.5. MECHANICS OF SYSTEM OF PARTICLES
19
-
The vector 7i - 7 j is a vector directed from the particle j to i. Since the force F ji is directed from particle i to j, the vectors 7 i - 7 j and Fji are parallel to each other and so -(') T ~ = O. Therefore, the equation (1.13) reduces to
-
-
- --
Let L ~ = 7i X Pi be the angular momentum of the ith particle, then the angular momentum of the system is ~ L = 2::: L ~ = 2:::[ l' i x p iJ. Thus, i
If the total external torque
-
7(e)
dL = -0 dt
i
acting on the system of particles is zero, then =}
-L
= a constant vector .
Thus, if total external torque acting on a system of particles is zero, then the angular momentum is conserved. It is the law of conservation of angular momentum. Theorem 1.5.2. Prove that if no external force acts on a system of particles, then the centre of mass of the system moves with constant velocity. PROOF: Let us consider a system of n particles of masses ml, m2, ... , mn with position vectors 7 1 , 7 2 , ... , 7 n respectively. If 11,1 = ml + m2 + ... + m n , then the
-
position vector R of the centre of mass of the system is given by
CHAPTER 1. MECHANICS OF PARTICLES
20 n
=?
= Lmt~t =
MV
t=1 -+
=?
where
dJ}
V is the
M
dV dt
n
LP\ i=1 n-+
n
= ~[LPi] = L dt
i=1
i=1
dp't dt
linear velocity of the centre of mass and
dr'
=
~i is the
linear velocity of the 'ith particle of the system. If F(t) is the total external force on the system, then according to Newton's law of motion, we get,
-+
--+
dX
where P = is the acceleration of the centre of mass of the system. If the total -+ -+ external force acting on the system is zero, i.e.,F(tot) = 0, then we have -+
dV -+ M dt = 0
-+
=?
V = a constant vector.
Thus, if no external force acts on a system of particles, then the centre of mass of the system moves with constant velocity. Theorem 1.5.3. Prove that angular momentum of a system of particles about the origin is equal to the sum of the angular momentum of the centre of mass of the system about the origin and the angular momentum of the constituting particles about the centre of mass. PROOF: Let us consider a system of n particles of masses ml, m2, ... , mn with position vectors -;;:'1, -;;:'2, ... , -;;:' n respectively. Let -;;:'~ be the position vector of the ith particle with respect to the centre of mass of the system, then -;;:' t = Ii + -;;:'~, -+ -+ -+ -+ where R. be the position vector of the centre of mass. Also, 11 i = V + v~, where
1/ t
-+
--+
= d
It'
and V
--+
= dd~
are the respective velocities of the
;t
ith
particle and the centre
d--+'
of mass, while 1/~ = is the velocity of the ith particle with respect to the centre of mass. The angular momentum of the system of particles is given by, 1
-+
L
-+ =~ L.J r i
-+~"""' -+
x mi v i
=
L.J (R
i
= ~-+ L.J R ~-+
-+
+ ~-+ L.J R
-+
=
x mt V
Now,L.J R x mi V
-+
~ -+
-+-+/
x m t (V
+
v i)
t
x mt ~
-+/ Vi
+ ~-+/ L.J r i x
-+
R x (L.J mt)V
i
+ -+/ r i)
-+
=
-+
mi V
+ ~-+/ L.J r i
-+/
x mi v i'1.14)
-+
R x MV
i
-+/
-+
~
/
d ~
-+
-+
L.J R x mi Vi = R x (L.... mi Vt ) = R x dt L.... mi( r i
t
-+
-+ i -
R)
i
-+
-+
= R x [M R - M R]
=
-+
~
0 as L.... m't -;;:' t
-+
= MR.
1.5. MECHANICS OF SYSTEM OF PARTICLES ~ -+1
~
---+
-+1
~ r i x mi V = ~ (mi r
i)
X
21
~
-+
-+
V = ~ mi ( r
i
-+
R)
i -
X
-+
V
i
-+
-+
= [M R - M R]
-+
X
V
-+
= O.
Substituting the values of the above factors in the equation (1.14), we have, -+
-+
L = R
-+
X
MV
-+1 +~ ~ r i
-+1
X
Pi'
i
-+
-+
The first factor R x M V represents the angular momentum of the centre of mass of the system about the origin, while the second factor L: T7~ x p~ represents the 1.
angular momentum of the constituting particles about the centre of mass. In case, -+ -+ the centre of mass of the system of particles is at rest, i.e., V = 0, or moves parallel -+ -+ -+ -+ -+ -+-+ to its position vector R, then R x MV = 0, so that L = L: r ~ x p~. Hence the t
angular momentum of the particles of the system about the centre of mass is same as the angular momentum of the particles of the system about the origin. Hence, the angular momentum of a system of particles is independent of the choice of reference point, in case the centre of mass is at rest or moves parallel to its position vector.
1.5.4
Energy of the system
The total force acting on a particle of a system is equal to the sum of the external force acting on it and the internal forces due to other particles of the system on it. If denotes the total force on the ith particle of the system, then
Pi
p. - p(ext) + ~p .. i ~
2 -
)2'
j
In case, the internal structure of the system does not change, the effect of total -+ -+ r force F 2 on the ill. particle will be to impart acceleration, i.e., F i = m 2 It i. Let us suppose that, the system of particles is moved from the configuration 1 to 2. Then, amount of work done is given by 2
W 12
=L t
JPi. dT7 1
1.
(1.15)
CHAPTER 1. MECHANICS OF PARTICLES
22
where Tl and T2 are the total KE of the system in configuration 1 and 2 respectively. A system of particles possesses following two types of potential energies. (i) Due to the interactions of the particles with each other, the system possesses internal potential energy. The internal potential energy of the ith particle due to its interaction with /h particle is denoted by ~~int), so that its internal potential energy due to interaction with all the particles of the system will be
v;,(int)
= L~~int). Since J
internal forces are conservative in nature, the internal forces on the to all other particles can be expressed as " ' F .. = _ ~ y(int) = _ "'~ y(int) ~
J1
1
~
t
1
J1
ith
particle due
•
j
J
The script i on the del operator indicates that derivatives implied by V -operator is to be carried out with respect to the co-ordinates of the ith particle. (ii) Due to the actions of external forces on the particles of the system, the system possesses external potential energy. The external potential energy of the ith particle due to its external force acting on it is denoted by v;,(ext). In case, the external forces acting on the particles of the system are also conservative, then ~F (ext) t
= _ ri .V(ext) v t . 1
Now, amount of work done in moving the system of particles from configuration 1 to 2 is given by
1 I: 1F~ext) 2
W12
= 2;= I
2
F 1.dT't
= I: 1[F~ext) + I:Fjt].dT'1
1
=
t
The first term
J
1
I
2
1 2
.dT'i
+ I:
1
F Jt .dT'1'
t,J 1
ext ~ L J2 --> F; ) .d T .,. represents change in potential energy of the system 1
t
_2~
due to external forces acting on it, while the second term
J F ji.dT't
L t,j 1
represents
the change in potential energy due to inertial forces between the particles of the system. Now,
I: .l F~ext) .dT'i = - 2;= .l ~ v;,(ext) .dT' 2
2
i
t
1
1
2
Further, "'1~ ~ F ji.d ~ r i = I,J 1
i
= V1(ext)
1
1"'1 "2 [~ 2
~
I,J 1
r i F ji. d ~
+~ F ij.d r J]. --J
-
V2(ext).
1.5. MECHANICS OF SYSTEM OF PARTICLES
23
The factor ~ appears in the equation, because in summing over both i and j, the pair of the ith and lh particles appears twice, once for the ith particle due to the yth and secondly for the lh particle due to ith. Since both are same and should appear only once, the factor ~ has been included. Now, ~~int), so the above j{ = equation becomes,
F
Vi
2
= -~L:
/ Vi~~int).(d7i
- d7 j )
t,J 1
--
_~ ~ / 2~
2
Vy(mt) d-7 .. _ ji . r t
tJ -
V(int) _ V;(int) 1
2·
i,j 1
+ [Vj(int) - V;(mt)] =? T+ V = Constant.
=? WI2 = [V?xt) - V;(ext)] =? TI
+ VI = T2 + V2
= VI - V2
Thus, if the external and internal forces acting on a system of particles are conservative, then the total energy of the system of particles is conserved. It is the law of conservation of energy for a system of particle system.
1.5.5
Kinetic energy
Here we are to show that, the KE of a system of particles is equal to the sum of the KE of the motion of the centre of mass and the KE of motion of the particles with -7 -7 respect the centre of mass. Let V be the velocity of the centre of mass and v~, the -7 velocity of the ith particle with respect to the centre of mass, then 7J i = V + 7J~. The KE of a system of particle with reference to the origin of the co-ordinate frame is given by
CHAPTER 1. MECHANICS OF PARTICLES
24
1L: m-v-. n
1 2 = -MV +-2 2
12
t
t
i=l
Like angular momentum, we find the kinetic energy is also of two terms. Thus, the KE of a system of particles is equal to the sum of the KE of the motion of the mass particle Ai, i.e., centre of mass and the KE of motiol). of the particles with respect the centre of mass.
1.5.6
Rigid body
A system of particles is called rigid body, if the following conditions are satisfied (i) The distance between any two neighbouring particles is infinitesimally small. (ii) The distance between any two particles does not change due to any external force. (iii) As each particle exerts action on its neighbouring particle in contact with it, so equal reaction is introduced in the opposite direction. Then the resultant of all such actions and reactions must be zero.
1.6
Motion of variable mass
So far we have discussed the motion of bodies whose mass remain constant throughout the motion. In this section, we shall discuss the motion of the bodies whose mass changes during motion. Let a body of variable mass m( t) moving with a velocity v at time t receive an increment 8m of mass in an interval of time Ot. Let u be the component of velocity of this increment 8m in the direction of v just before union with m. Let F be the external impressed force acting on m, which may be due to gravitational and atmospheric resistance in the same direction and let v + 8v be the velocity of the body at time t + 6t. Hence the change in momentum in time ot is
= (m + om)(v + 8v) - (mv + uom) = m8v + vOm + 8m.8v - u8m. During the time interval 8t, the change in linear momentum of the main body Can be written as
.
mOv + 'v8m + 8m.ov - u8m
6t->O
Ot
= hm - - - - - - - - - - = m
dv dt
dm
+ vyt -
dm d dm uYt = dt(mv) - uYt.
Following the Newton's second law of motion, applied force will balance the rate of change of linear momentum of the system. Hence,
d dm dt(mv) -1J.Yt
=F
(1.16)
1.6. MOTION OF VARIABLE MASS
25
which is the equation of motion for variable mass. Below are some important examples of varying mass. Ex 1.6.1. A spherical raindrop of radius acms falls through a vertical height h, receiving throughout the motion an accumulation of condensed vapor at the rate of I< grammes per square cm. per second, no vertical force but gravity acting; show that when it reaches the ground its radius will be I< ~[1
+
VI +
2\a;'2]'
SOLUTION: Let the radius and the mass of the raindrop be respectively rand M, when it has fallen through a distance x in time t. If p be the density /c.c, then, 4 37fr3p.
=
M
Initially, t = 0, x = 0, r = a. Since, the vapor accumulates at the rate of kgm./sp.cm./sec., so, by the given condition, dM 2 dr dt = p.47fr .k =;. dt = k =;.
Using the relation, ftCMv) d dx dt(M dt)
T
+ a;
= kt
as t = 0, r = a.
= P, we have
=
d4 3 dx Mg =;. dt[3 7fr p dt]
=;.
~(r3dx) dt
=;. r3
= g1'2 =
dt
~; =
gp
ft r 3dr k
:k 1'4 -
43
= 37fr
.~:4;
dt
as r = a,
~; =
0
9 4 4 9 a4 d1' 4kT 3 (r - a ) = 4k2 (r - 1'3) dt 9 1'2 a4 ga 2 =;. x = 4k 2 [2' + 2r 2 ]- 4k 2 ; as l' = a,x = 0 g a4 g r2 - a 2 2 =;> x = 8k 2 [r2 + r2 - 2a ] = 8k 2 ( )2. r
dx
=;. dt
=
When x = h, we have,
{2h
2
2
r - 2ky T.r - a = 0 =;.
T
=
~[2k fih ± 2
YT
,-----
2 8k h
g
+ 4a 2 = k fih[1 +
YT
1+
ga 2 2hk 2 ]
where we are taking the positive sign. Ex 1.6.2. A mass M is fastened to a chain of mass m per unit length coiled upon (J, 1'0'//,.'111, hO'f"izontal plane, who8e coefficient of friction is p,. The mass is projected from the coil with velocity V. Show that it will be brought to rest in a distance M[(1 m
+ 3mV2)1/3 -1]. 2111/1,9
CHAPTER 1. MECHANICS OF PARTICLES
26
SOLUTION: Let x be the chain uncoiled at time t, then a mass of (M + mx) is moving with velocity v(say) and a mass of [M + m(x + 8x)] is moving with velocity v + 8v at time t + 8t. Hence, [1\1 + m(x
+ 8x)](v + 8v)
- (M + mx)v = change in momentum
= impulse of the acting forces = -/L[M + mx]g8t. In the limit, 8t
~
0, we have,
(M
dv
d
or, dt[(M +mx)v] or,2(M
dx
+ mx) dt + m dt v =
= -J.L(M +mx)g
+ mx)v d~ [(M + mx)v] = or,[(M + mx)v]
2
-211.(M + mx)2g
2J.Lg = --(M + mx) 3 + Cj 3m
where C is a constant of integration. Initially, x 21.J'l~J3 = C. Hence,
Now, let the velocity v
1. 7
=
°
when x
-J.L(M + mx)g
= Xo,
=
0, v
= V,
integrating,
so that C
= M2V2 +
so we have,
Nonrealistic rocket motion
A rocket fixed from the earth will always be affected by the gravitational pull of the earth. Other celestial objects are at great distances from the rocket and the effect of such objects on the motion of the rocket can be ignored. To simplify the problem tltill further, we are considering the free flight of the rocket based on the motion of a body of variable mass, the thrust being created by ejecting a part of the mass of the body, neglecting the effect of rotation and gravitational pull of the earth.
1.7.1
Single-stage rocket
Suppose that, a rocket (propelled by burning fuel) having a mass M(t) at the instant t and moving at a velocity v throws away a fuel mass dM' at a velocity u. It should be borne in mind that 1\1 and d1\1' are relativistic masses and the velocities v and u are
1.7. NONREALISTIC ROCKET MOTION
27
measured relative to the inertial reference frame in which the motion is considered (and not relative to the rocket). Then the law of conservation mass gives dM+dM'=O.
Obviously, dAf < 0 as the mass of the rocket decreases. To derive the equation of motion we are to use the conservation law of momentum. At the instant t, the total momentum of the system is Mv, while at the instant t + dt, it is given by (!vI + dJll) x (v + dv) + udM'. The momentum conservation law for this isolated system can be written in the form Mv
= (M + dM) x (v + dv) + udM' => Mdv
+ vdM + udM' = O.
The term dvdlvI has been neglected, as it is an infinitesimal of the second order of smallness. Taking dlvI + dM' = 0, into consideration, the equation of motion is dM !vIdv+vdM-udM , =o=> -d ( Mv ) = u dt dt
(1.17)
which is valid for both relativistic and non-relativistic cases. If the velocities are small, they can be added by using the formula from classical mechanics, and hence u can be written as u = u' + v, where u' is the velocity of the ejected mass relative to the rocket. Hence, d ( ) dt M v
=
(' u
) dM
+ v dt => M
dv dt
, dM
= u dt'
(1.18)
where 111 = Mp + Mf(t) + Ms; Mp denotes the mass of the payload, Mf denotes the mass of the fuel and Ms the mass of the structure or casing of the rocket. This is the equation describing the motion of rockets at non-relativistic velocities in the absence of external forces. If the rocket is subjected to the action of a force F, then the equation of motion takes the form M dv,.._ F dt -
+u
' dM dt .
(1.19)
Let M f + !vIs = Mo. Initially, let Mf(O) = EMo, so that Ms = (1 - E)Jl10. Here (1 - E) is called structural factor and the parameter E is the ratio of the initial fuel mass to the initial rocket mass. Obviously, 0 < E < 1 and it is preferable to have E, nearly equal to 1, in the sense that the structure should be as light as possible, such that we shall be able of carry more fuel. In practice, E lies in (0.7,0.9). Once the fuel is ignited, at any time, we have M = Mp + Mf(t) + (1 - E)Mo.
Deduction 1.7.1. Tsiolkovsky formula: Let us consider the acceleration of a rocket moving in a straight line, assuming the velocity of the ejected gases to be constant relative to the rocket. In absence of all external forces (including gravity and air resistance), the equation of motion of the rocket is M dv dt
=
_u,dM. dt
CHAPTER 1. MECHANICS OF PARTICLES
28
The negative sign on the RHS is due to the fact that in the case of acceleration, the exhaust velocity u' is directed against the velocity v. In other words, v, increases with the decrease of lvI. Let v and M t be the velocity and mass of the system at instant t respectively and vo, Mo those at t = 0 (before acceleration). In this case, v
/
Mt
dv
Vo
=
-u' / -dM => v M
t M = Vo - u ,log -.
Mo
Mo
Hence the change in speed of the rocket of the rocket in any interval of time is independent of ~'1, i.e., how the mass is released and depends only on the exhaust velocity 'u' and on the friction of mass exhausted during that time interval. When ~f~ = e ~ 2.72, then v - Vo = u' and ~~ = e 2 ~ 7.4, then v - Vo = -2u'. However, it is impossible to construct a rocket with mass ratio 7.4 : 1. Also
which is the Tsiolkovsky formula. The first form expresses the change in the velocity of the rocket, when it changes mass from Mo to M t , while the second formula expresses the mass of the rocket whose velocity changes from Vo to v. Deduction 1. 7.2. Characteristic velocity: Suppose that it is required to change the velocity of a rocket (acceleration, deceleration, change in direction of flight). In the reference frame, in which the rocket is at rest at the given instant of time, the problem can be reduced to imparting a certain velocity v to the rocket in a direction which would ensure the desired maneuver. The fuel consumption in this maneuver for a rocket flying outside the gravitational field can be taken into account with the help of the formula M = Moe-(v-vo)/u', with Vo = 0, where Mo is the mass of the rocket before maneuver. The velocity v required to be imparted to the rocket is called characteristic velocity of the maneuver. Deduction 1.7.3. Let us suppose that the fuel is burnt and ejected at constant rate d~f = -/-L and it lasts for time T. The quantity /-LU' is called the relative force. If the direction of u' is opposite to that of v, the rocket is accelerated. If the two directions coincide, the rocket is retarded. For any other relation between u' and v, the rocket's velocity changes not only in magnitude but also in direction. Let Ms be the mass of the structure and that of the fuel as t = 0 be Mf' then Mo = Ms + Mf. Hence at an instant t, the mass of the vehicle-fuel system can be written as
Suppose at t
= T(burn
out), the entire fuel is burnt, in which Mf(T)
T = ~~o. Hence the rocket mass at any time t in (0, T) as M(t) = Mp
=
0 so that
+ Mo -
/-Lt.
1.7. NONREALISTIC ROCKET MOTION
29
Using this value of M t , from equation (1.18), we have,
dv, J.Lt ) dt = -u J.L
+ Mo -
(Mp
'J t
v
J
dv = -u J.L
:!:::}
Mp
0
va
,
=> ~} = Vo + u log[
dt
+ Mf(O) + Ms -
Mp
J.Lt
+ Mf(O) + Ms - J.Lt + Ms 1
A1p + A1f(0)
= vo + u' log[1 - M J.L M tl p+ 0
(1.20)
which is the velocity acquired by the rocket in time t. From this equation, it is clear that, the performance of the rocket depends on (i) u', the exhaust velocity, (ii) (1 - e), the structural factor, (ii'i) ~ = Mf(~rtMs = mass nitio. Also, ,
v = Vo
t
x
=>
+ u + log[1 -
J
elx = Vo
o
t
J+ J elt
_
=> x - vot
log[1 -
u'
0
t
eMo
+ Mf(O) + Ms Tl
Mp
M:~~o ;ldt
0
_ T(Mp
+ Mo) u'[(I _
1I
elViO
M
eMo p
t ) 1 ~(1 _ c:Mo t) og e M 1I T p + lViO
+ lViO T 1I
+ 11
which is the distance covered by the rocket in time t.
Deduction 1.7.4. The most important problem is to attain the maximum velocity with the minimum consumption of fuel, i.e., for the minimum difference between Mo and M I,. However, the velocities of ejected gases are restricted. Therefore, the rocket attains maximum velocity at t = T, when all its entire fuel is burnt out and the velocity V max caused by the expenditure of the entire propellent is given by, V max
'[
eMo
= Vo + u log 1 - Mp + Mf(O) + Ms = Vo
[ + u'log 1-
1
fMO M M .1
p+
0
This shows that, the larger the value of the ratio ~, the greater will be the maximum velocity attained by the rocket. It is this logarithmic dependance on mass that requires enormous rockets to deliver small payloads.
Deduction 1.7.5. The rocket is moving vertically upward if the constant gravitational pull of the earth (neglecting air resistance) on it is assumed to be constant,
30
CHAPTER 1. MECHANICS OF PARTICLES
then the equation of motion of the single stage rocket can be written as
A1
dv = dt
_u
,dM _ Mg dt t
v
=* JdV = -UIJ.LJ 0
110
M: p+
=* v = Vo + u I log [1 Now , at t = T =
EMo 11
v
'
I
0 -
t
gJdt
J.Lt
0
J.Lt] 111, - gt. Mp+ 0
we have ,
= 'U -
= 'U I log[l -
'00
eNlo geMO ] - -Mp+Mo 1-"
which gives the velocity increment caused by the expenditure of the entire propellant. Also,
v = Vo
+ u I log [1 t
h
NIp
J.Lt] + Mo - gt
t
t
=* J dx = Vo J dt + u ' J log[l - Mp ~ Mo ]dt - 9 J tdt o
0
0
0
J.Lt J.Lt =* h = vot - -gt - -(Mp + Mo)[(l - 11'[ A1,) log(l - M 111, )]. 2 J.L p+ 0 p+ 0 1
u'
2
which is the expression for the height attained by the rocket at time t. Hence the change speed of a rocket in the gravitational field does depend (in construct to that in free space) on how the mass is released. The shorter the burn time, greater the change in velocity. Hence for greater gain in speed, the rocket fuel should burn as quickly &"l possible in the region of gravitational field (in absence of atmosphere). Also at burn out, T = E~O, hence at burn out, the height attainable by a single stage rocket is 2
voeMo 1g eM6 H -- - - - - -u ' [111,0 (1 J.L 2 2J.L2 J.L
1.7.2
-
eMo] . e ) + M]log [1 p Mp + Mo
Multistage rockets
Normally, the rockets are expected to carry some load, called the payload. Payload may be the load of a satellite which is to be placed in an orbit around the earth, or of bombs in the case of missiles. The payload and the body of the r~cket have a fixed mass. All the load carried by a rocket is not useful right up to the end of the flight. Ratio ~:, has practical limit and it cannot be increased indefinitely. Hence it is expedient to get rid of such a load as soon as possible. A single rocket (one stage rocket) will not attain high velocities that are required, and for this purpose, the multistage rockets are used.
1.7. NONREALISTIC ROCKET MOTION
31
Ex 1.7.1. Find the optimal parameters for a two-stage rocket in order to impart a given velocity to a given payload of mass 500kg, the final velocity 8km/ s, and the velocity of ejected gas 2km/s. Assume that according to the constructional requirements, the mass of the fuel in this stage. SOLUTION: Let MI and M2 be the masses of the fuel in the first and second stages of the rocket respectively and M be the mass of the payload. If EMI and EM2 be the masses of the structure of the empty rockets in the first and second stages respectively, then the total masses of the two stages are (1 + E)MI and (1 + e)M2' The total initial mass mo of the rocket is
rno
=
M + (1 + E)(MI + M 2);
e = 0.1.
Now, if VI and V2 be the speeds of the rocket at the end of consumption of fuel of the first and second stages respectively, then I
VI -
Vo
mo
= u log - ;
V2 -
ml
VI = 1l
I
log
lvI + (1 ~I
11'
+ E)M2 + EM 2
where the negative signs are dropped to represent the numerical value only. The initial velocity Vo = 0 and let ~7 = a, then we get,
M
mo -
=}
+ (1 + E)M2 M +cA12
mI'
[AI + EMI + (1 + E)AI2][M + cA12J M + (1 +E)M2
Also, for minimum value of mo, we have ~~2 C)(MI + M2)' Thus,
= 0,
-e
a
-
-a
= moe
i.e., ~~;
=
-1 as mo
= M + (1 +
[E~ + (1 + E)](AI + EM2) + [M + EMI + (1 + E)AI2Jc M + (1 + E)M2 _ [M + eMI + (1 + E)M2J(M + EM - 2)(1 + E) = 0 [M + (1 + e)M2J2 M2 =} MI = M2 + (1 + e) ;) . Using the relation mo = M + (1 + E)(MI + M2), we get e
a
= M+(I+E)(MI +M2) M+(I+E)M2+EMI
M+(I+E)M2 M+EM2
X ---'---'----
[M + 2(1 + E)M2 + (1 + E)2%!-][M + (1 + E)M2J 1If2
[M + (1 + 2E)M2 + E(I + e):xf][M + EM2J Now, a
= ~ = 4,e = O.I,M = 500kg,
tcM M i][500 + l.IM2J
[500 + 2.2M2 + e [500 + I.2M2 + 4 _
=}
lOe4
so
~'cici J[500 + 0.IM2J
11 M 5 0+0.1 2
= [500 + 2.2M2 + 0.00242MiJ [ 0
e4
----J. 500 + l.IM2
32
CHAPTER 1. MECHANICS OF PARTICLES
The mass ma = 500+2.2M2+O.00242Mi has its minimum value when the expression within the parentheses has its maximum value as a function of M 2 . The extremum condition for the expression within the parentheses gives the equality, e2
=
2 500 + 1.1Ah =? M2 = 500(e - 1) = 8846.795kg 500 + 0.IM2 1.1 - 0.le 2
+ 0.1) (8~~~)2 = 181039.6998kg. mo = 500 + 1.01 x (8847 + 781040) = 209375.7kg.
Ni} = 8847 + (1
The mass of the two stages and that at take-off are (1
+ e)Ml =
199144kg.; (1
+ e)M2 = 9732kg.; ma =
209376kg.
For a single stage rocket, v - Va = u/log[M.t~~~;f], where M f is the mass of the fuel. Using Va = 0 and the above data, we see that for V = 8km/ sec, 500 + 1.1Mf 8 log 500 + O.IMf = "2 =? Mf = -6147kg. But the amount of fuel cannot be negative. Thus, a single stage rocket cannot achieve above final speed. Also, M < < M f so, the mass ratio Mr is given by Mr
::::::J
M
+ (1 + e)Mf + eMf
M
::::::J
1+e
= 1+~
E
e
i.e., mass ratio in a rocket cannot be increased at will. Here e and with this mass, a rocket can attain a speed V
= 0.1 and so Mr = 11
1
= u 10g(1 + -) = 2.083km/ sec. c
which is much less than 8km/ sec. This can be increased further by constructing a lighter empty (e-smaller) rocket, but technically it is impossible to construct a rocket with e < 0.1.
1.8
Limitations of Newton's Law
The limitations for applicability of Newton's law are, (i) These laws are only valid in inertial frame. For non-inertial frame, the transformed v~rsion of.Newton's Laws are made because of the presence of pseudo force in that frame.
cit particle on which Newton's law is required to apply, particle velocity be close to that of light in vacuum, the by Einstein's relativity, mass will change with velocity.
(ii) The velocity
1.9.
VIRIAL THEOREM
33
But Newton's law only is applicable for stationary mass. Also in setting up the differential equations of motion we usually choose particular co-ordinate system from symmetry and boundary conditions of the particular problem which helps to simplify it. However, in Newtonian formulation, Newton's equations of motion no longer maintain their simple form form in all co-ordinate systems and in fact writing down of fiuch equations in different co-ordinate systems is not straight forward. Vector form (independent Cartesian Spherical Cylindrical of co-ordinate form) (x,y,z) (r,e, =1= m¢; Fz =mz Frj> =1= m¢; Fz =1= mz Such a dependence on the co-ordmate system IS, however, undesIrable and we should be able to write the equations of motion without any specific reference to the coordinate system used. After Newton, these developments in mechanics were contributed by D'Alembet, Lagrange, Euler, Poinsot, Einstein and others.
me
1.9
Virial Theorem
Virial theorem was introduced by Rudolf Clausius (1870). Consider a system of n interacting particles that has achieved dynamical equilibrium through some kind of central force interactions. (i) Let be the applied forces on the ith particle of mass m~ with position vector -:t i. The virial of the system is defined as
Pi
which is connected with the total potential energy of the system. (ii) Let T be the total translation kinetic energy of the system, then d--+ - 2 ~ m~ dt . dt . n
--+
T-~"
.drt~
i=l
The virial theorem states that, for an n body system, which is bounded in size and velocity, the average value of the KE over a large time interval equals minus half times the average value of the PE of the system and is equal to minus times the total energy of the n body system. For this, taking the time derivative of the quantity n
G
= I: p~.ri,
which is a bounded function, we get,
1,=1
dG
It
n
n
n
i=l n
i=l
i=l
i=l
dt = LP~ri + LPiTi = LPirt + L = LFiri + 2T = W + 2T; i=l
miTiTi ~-:ti
as
mi--a:i2 =
~
Pi
_
= Fi
CHAPTER 1. MECHANICS OF PARTICLES
34
=?
d n -["'p-r'] dt L.J t 1. = W
+ 2T.
i=l
Thus the time average of a quantity over the time interval r is obtained by integrating both sides with respect to t from 0 to r, and diving by r, we get
J
JdI
T n T
1 :;:
d [L.JPirt] '" dt
o
= :;:1
dG dt
=W +2T
0
t=l
=?
W
-
+ 2T =
1 -[G(r) - G(O)], r
where Wand T stand for the time averaged values of Wand T respectively. If the motion is periodic with r is chosen as period, then G(r) - G(O) = O. For a stable system, -Y;\, the distance of the body and the velocity of that body are bounded and then also G(r) - G(O) ~ 0 as r ~ 00. When the motion is not periodic, the co-ordinates and velocities for all particles remain finite and hence there is an upper bound of G. Taking r sufficiently long, then G(r) - G(O) >::::J O. Thus in both cases,
dJ:',
TV
+ 2T = 0;
if r --;
00
which is the well-known virial theorem and -~ W is called virial of Claussius. The time scale over which the process of virialisation takes place is called the virial time scale of the system. This theorem has wide applications in Stellar dynamics and forms the basis for statistical calculations of stellar clusters. Ex 1.9.1. Consider a system in which the total forces acting on the particles consist of conservat'ive forces PI and frictional forces Ii proportional to the velocity. Show that for such a system the virial theorem holds in the form T = -~ 2: P:r t , providing t
the motion reaches a steady state and is not allowed to die down as a result of the frictional forces.
SOLUTION: Here the forces are the sum of non-frictional forces (conservative) PI and the frictional forces ft, proportional to the velocity. Thus the virial W = 2: Pir t t
depends only on the conservative forces PI and have no contribution of it- Further it is assumed that the motion of the system must not be allowed to die down as a result of the frictional forces. The KE T of the system is T
1
1
2 = -2 '" L.J m'v t t = 2 '" L.J m'v'v' t t t i
t
Here given that the frictional force
Ii
=
CVi; C
= some constant and
Vi
= velocity.
1.10. EXERCISE
35
From Newton's fundamental equations of motion, we have,
Pi = F: + I! = F: + CV! => 2:P!r! = 2: F:ri + C 2: VITi =>
2: Pir! = 2: FIT! + 2: V!T!; C
Taking time average
To maintain the motion, energy will constantly involved so that all time averages of the quantity which is a total time derivative of a bounded function zero, we have,
1.10
Exercise
Ex 1.10.1. State Newton's laws of motion. What are inertial frames? Give examples of inertial and non-inertial frames. Ex 1.10.2. Define angular momentum of a system of particles. Prove that the rate of change of linear momentum of a system is equal to the mass x acceleration of the mass particle at the centroid. Ex 1.10.3. Prove that, the total angular momentum of a system about a fixed origin o is (i) the total angular momentum of the mass particle at the centroid about 0, and (ii) the angular momentum of the system about the centroid for motion related to the centroid. Ex 1.10.4. Show that the angular momentum is conserved in motion under central force. Ex 1.10.5. Prove that if the centre of mass of a system is at rest or moves parallel to its position vector, the angular momentum of the system of particles is independent of of the choice of the reference point. Ex 1.10.6. Find the period of oscillation for the potential V(x)
= -Vosech2 ax.
Ans: T= ~Jj. Ex 1.10.7. Prove that the centre of mass of a system of particles moves as if the external force on the system on the constituting particles were concentrated at the centre of mass. Ex 1.10.8. The KE of a system is the sum of (i) the KE of the mass particle at the centroid and (ii) the KE of the system for motion relative to the centroid.
36
CHAPTER 1. MECHANICS OF PARTICLES
Ex 1.10.9. The rate of change of angular momentum of a system ahout a fixed point is equal to the SU'll of the moments of the external forces about the same points. Ex 1.10.10. Show that the KE of a system of particles is equal to the sum of the KE of the motion of the centre of mass and the KE of motion of the particles with respect the centre of mass. Ex 1.10.11. A rocket of total mass 1.11 x 105 with fuel 8.70 x 104 Kg is to be launched vertically. It is designed to consume fuel at a constant rate 820Kg/ s. What should be the minimum exhaust speed for lift off at launch? Vr = 1.327km/ s Ex 1.10.12. A two stage rocket is to be built capable of accelerating a lOOK 9 payload to a velocity of 6000m/ s in free flight in empty space. If the exhaust velocity of the fuel used be 1500m/ s and an empty rocket (without fuel or payload) weights 10 percent as much as the fuel it can carry, then find the optimum choice of masses for the two s.tages so that the total take-off weight is minimum. Show that a single 36, 196kg, 1769kg. stage rocket can never attain such a speed.
Chapter 2
Generalised Co-ordinates A particular co-ordinate system chosen from the symmetry of the physical system helps to simplify the problem. For example, when the equation of motions are expressed through vectors, these equations do not depend on any specific co-ordinate system and the physical significance of the equation is also clearly seen. Such a dependence on the co-ordinate system, however, is undesirable and we should be able to write the equations of motion Without any specific reference to the co-ordinate system used. Here we shall define 'neralised co-ordinates which are independent of the co-ordinate system and may not be the usual spatial co-ordinates.
2.1
Constraints
We know motion of a system can not proceed arbitrarily and is restricted to occur alollg some specified path, or on a surface (plane or curved) arbitrarily oriented in space. Any restriction (geometrical conditions) which must be satisfied by the coordinates, on the freedom of motion of a dynamical system, in some way, is known as constraint and the force responsible is called constraint force or force of constraint. (i) A constraint is analytically written in the form
f(7 k, -t k, t) = 0 where 7
k,
(2.1)
-t k are position vector and velocity of the kth particle respectively,
t is time. This type of constraint is called differential or kinematical constraint.
(ii) A system without any constraint is called free system.
-t
(iii) A finite or geometric constraint is free from k and can be written in the form f(7 k, t) = O. If the finite constraint does not contain the time t explicitly, i.e., = 0, then the constraint is called stationary, otherwise it is non-stationary.
¥t
n
(iv) The general form of the constraint can be written as
k=l
lk' D are functions of t and
.
L lk 7 k + D
-t k, lk are not simultaneously zero. 37
= 0, where
CHAPTER 2. GENERALISED CO-ORDINATES
38
Physically, constraint motion is realized by the forces which arise when the object in motion, is in contact with the constraining surfaces or curves. They reduce the number of co-ordinates needed to simplify the configuration of a system and therefore simplify the mathematical description of the problem. A motion along a specified path is the simple example of the constraint motion. In a rigid body, the constraints on the motion of the particle are the distance Tij remain unchanged. The constraints may be classified as follows, for all dynamical systems, according to different types of constraints and forces to which the dynamical system may be subjected.
2.1.1
Holonomic Constraints
If the constraints involved in a system be such that the condition of constraints can be expressed as equations connecting the co-ordinates of the particles ( and possibly the time) having the form
(2.2) then the constraints are said to be holonomic and the system is called connected holonomic system. In this equation, j denotes the jth constraint. We know, the distance between any two particles with position vectors Ti and Tj of a rigid body remains unchanged, Le., (Ti - TJ )2 = Cfj. This equation of constraint is an example of a holonomic constraint of rigid body. It may be pointed out that, sometimes ." the constraints on a system are specified by a restriction on the velocities of the particles, rather than the co-ordinates and still the constraints are holonomic. This is true due to the fact that the velocity equation is integrable with respect to time to yield a resultant relation of the form (2.2), which is free from velocities. Ex 2.1.1. Prove that the constraint qlql holonomic.
+ q2q2 + qlq2 + q2ql = k,
k
= constant,
is
SOLUTION: The given constraint can be written as 1d
2
2
d
2dt [ql + q2J + dt (qlq2) = k => [q~ + q~J + 2qlq2 = 2kt + c => [ql + q2]2 = 2kt + c. Hence the given constraint qlql + q2q2 + qlq2 + q2ql = k is holonomic. 2.1.2
Non-Holonomic Constraints
Constraints that can not be expressed in the type (2.2) is called non-holonomic constraints. It may be in the form of an inequality. Some of the constraints of a system may be expressed as (2.3)
For example, the walls of a gas container constitute a non-holonomic constraint. The constraint of a particle placed on the surface of a sphere, under the effect of
2.1. CONSTRAINTS
39
gravity, which can be expressed as r2 - a 2 2: 0; a = radius of the sphere, is also non-holonomic. There are, however, dynamical systems in which equations among velocities cannot be integrated to give the relation between the co-ordinates. Such dynamical systems with non-integrable differential constraint are then taken as nonholonomic constraints.
2.1.3
Scleronomic Constraints
If the constraints are explicitly independent of time, i.e., finite stationary constraint, then it is called scleronomic. In other words, a scleronomic system is one which has only 'fixed' constraints. For example, the constraint in the case of a pendulum with fixed support is scleronomic constraint.
2.1.4
Rheonomic Constraints
If the constraints contain time as an explicit variable, i.e., non-stationary, then it is called rheonomous. In other words, a rheonomic system has 'moving' constraints. For example, consider, a simple pendulum whose length changes with time, its constraint expression is time dependent and this assigned motion is rheonomic. Similarly, if the radius of the sphere is changing with time.
2.1.5
Bilateral Constraints
When the equations of constraint are in . . he form of equalities, then it is called bilateral constraints. The general form of the bilateral constraint can be written as
(2.4) At any point on the constraint surface both the forward and backward motions are possible. In the bilateral constraint of motion on the surface, the force of constraint may be either direction normal to the surface.
2.1.6
Unilateral Constraints
When some of the constraints are expressed in the form of inequalities, then it is called unilateral constraints. At some points no forward motion is possible. The general form of the unilateral constraint can be written as : (2.5)
In the unilateral constraint of motion, the force of constraint acts only one direction which is the outward normal to the surface.
40
2.1. 7
CHAPTER 2. GENERALISED CO-ORDINATES
Conservative Constraints
When the total mechanical energy of the system is conserved, while performing the constraint motion, is called conservative constraint. In this case constraint forces do not do any work. A conservative system is subjected to potential type of forces. For example, gravitational force field is conservative.
2.1.8
Dissipative Constraints
When the total mechanical energy of the system is not conserved, and the constraint forces do not work, then it is called dissipative constraint. In this case constraint forces do not do any work. For example, Lorentz force (force on a charged particle moving in an electromagnetic field) is a conservative force, because, the potential energy (V) of the charged particle in an electromagnetic field is velocity dependent. Result 2.1.1. When we are to solve mechanical problems, the constraints introduce the following two types of difficulties 1. Non-independence of Co-ordinates: Since the co-ordinates qi of the particles of a system are connected by the equations of constraints, they are no longer all independent. Hence the equations of motion for the particles of the system -t
F(ext)
+ ""' ~
-t
F .. tJ -
d-tp. dt
-_.
J()#i) -t
-t
cannot be solved independent of all others, where F(ext) is an external force, F ij is the internal force on the ith particle due to the lh particle, Pi is the linear momentum of the ith particle. 2. Forces of Constraints : Constraints on a system are always associated with the forces exerted on the system by which they impose restrictions on the motion of the system. Such forces are called the forces of constraints. They are among the unknowns of the problem and cannot be specified directly and must be obtained from the solution we seek. The problems involving holonomic constraints allow elimination of dependent coordinates (by introducing proper set of generalised co-ordinates) and hence a formal solution. So we shall overcome the first difficulty, we assume that any constraint, if present, is holonomic. We shall overcome the second difficulty by formulating the laws of mechanics in such a way that the forces of constraints disappear and in that case a problem can be solved with only known applied forces. This is done by using the fact that "the virtual work done by the forces of constraint is zero" .
2.2
Generalised Co-ordinates
The cartesian co-ordinate system is not only possible system of co-ordinates required to define the position of mathematical system. In fact, the choice of the co-ordinates is perfectly arbitrary and is determined on the basis of the problem concerned. Now,
2.2. GENERALISED CO-ORDINATES
41
we may think a wide variety of possible co-ordinate system, any set of parameters which give an unambiguous representation of the configuration of the system of coordinates in more general sense. The minimum number of independent co-ordinates required to completely specify the configuration of the dynamical system at any given time is called the generalised co-ordinates of the system. The generalised co-ordinates may be any parameters, they are not necessarily positional ( i.e., rectangular cartesian, curvilinear, etc) coordinates in the conventional orthogonal system, dimensionless parameters, angles, axes, moments or their combinations or any set of suitable parameters. Following are some examples of generalised co-ordinates
Ex 2.2.1. A dynamical system be a motion of a simple pendulum (of length l) oscillating in a vertical plane; the suitable generalised co-ordinate is (), the angular displacement from the vertical line through point of suspension. Similarly, for a spherical pendulum, the usual polar angle () and ¢> are suitable generalised coordinates. Ex 2.2.2. A particle on the surface of the sphere; generalised co-ordinates are (), ¢> where (), ¢> are the polar co-ordinates on the ~urface. Ex 2.2.3. A rod lying on a plane surface; generalised co-ordinates are x, y, () where (x, y) are the co-ordinates of one end of the rod and () is the angle between x-axis and the rod. While choosing a suitable set of generalised co-ordinates in a mathematical problem, the following points are to be noted: (i) The values of the selected generalised co-ordinates will specify the configuration of the system completely. (ii) They can be varied arbitrarily and independently without violating the constraints which may be imposed on the system. (iii) There is no uniqueness in the choice of generalised co-ordinates, the different sets so chosen will be functionally related. Then our choice should fall on a set of co-ordinates that will give us a reasonable mathematical simplification of the problem. Unlike, cartesian co-ordinates, generalised co-ordinates belonging to any particular particle are not divided into groups of three that can be associated together to form vectors. They are associated with a system as a whole, so that the configuration of the system can be described by a minimum number of independent co-ordinates. Number of independent variables being minimized, constraints are automatically obeyed. So one should not introduce the forces of constraints in treatment with generalised co-ordinates.
CHAPTER 2. GENERALISED CD-ORDINATES
42
2.2.1
Notation for GC
A set of n generalised co-ordinates are represented by ql, q2, ... , qn or simply qj j (j = 1,2, ... , n). When we are to describe a specific problem, the symbols ql, q2,···, qn correspond to co-ordinates that we choose to describe the motion. For example, (i) When a particle moves in a plane, it may be described by cartesian co-ordinates x, y, z, then ql = x, q2 = y, q3 = z. (ii) When the problem involves some spherical symmetry, it is suitable to use spherical co-ordinates ql = r = (x2 +y2 +z2)1/2j q2 = () = cot- 1 (x2+=2)1/2 j q3 =
4> = tan-l~. with greater advantage.
2.2.2
Transformation Equations
Consider, a system of N -point masses each with position co-ordinates -r'i for the ith particle. Suppose there are k constraints among the 3N position co-ordinates so the number of generalised co-ordinates will be reduced accordingly. Let the number of independent co-ordinates n = 3N - k, which are known as generalised co-ordinates and are denoted by ql, q2, .. . , qn. If the old co-ordinates -r'i contain the set of mathematical equations (2.6) expressing the old co-ordinates -r'i; i = 1,2, ... , N as function of generalised coordinates ql, q2, ... , qn and time t are called transformation equations for the set of T7i variables to the qj set. Also the generalised co-ordinates qi can be expressed as the function of old co-ordinates T7 1, -r' 2, ... , -r' Nand t by the equation qi
=
qi ( -+ r 1, -+ r 2, ... , r+ N, ) t .
The necessary and sufficient condition that the transformation form a set of coordinates qj to the set T7 i is effective is that the Jacobian determinant J of the equation (2.6) be different from zero at all points. For the solution of a I!lechanical problem, first we set up the equation of motion in terms of generalised co-ordinates qj and then integrate the simple equations of motion to obtain time dependence. Then the function -r'(t) can be obtained from (2.6).
2.2.3
Configuration Space
The span of the set of all independent co-ordinates qij i = 1,2, ... , n in n-dimensional Euclidean space ~n is known as configuration space of the dynamical system. Any point in the configuration space describes all constitutes of the system. The point is called a system point. At any given time t, the system is located at some point in its configuration space, and as the system moves, its co-ordinates will change and the system point in the configuration space will describe a curve giving the trijectory of
2.3. GENERALISED DISPLACEMENT
43
the system. Thus, in case of a system of N -particles, the 3N dimensional space in the absence of constraints or n dimensional space, when the constraints are present, is called the configuration space.
2.2.4
Degrees of Freedom
The minimum number of independent variables, required to describe the configuration uniquely of a given dynamical system which are compatible with the given constraints is called the degrees of freedom of the system. If the dynamical system of N particles is connected by k independent constraints then, the degrees of freedom will n = 3N - k in three dimensional space. (i) Holonomic systems are characterized by the fact that, the number of generalised co-ordinates require to specify the configuration of the system is equal to the number of degrees of freedom. (ii) For non-holonomic system, the number of generalised co-ordinates require is greater than the number of degrees of freedom of the dynamical system. These coordinates are called quasi-genemlised co-ordinates. The number of degrees of freedom is also defined as the number of independent ways, a mechanical system can execute, without violating any constraints imposed on it. For example, let us consider, the motion of a particle, constrained to move on the surface of a sphere of radius a. The centre of the sphere coincides with the origin of the cartesian co-ordinate system, then x 2 + y2 + z2 = a 2. Let us introduce ql = ~,q2 = ~, then q3 = Jl - q? - q~, so that q3 is not an independent co-ordinate. So number of degrees of freedom = 2.
2.3
Generalised Displacement
When a body is moved from one position to another, the change position is called a displacement of the body. Consider, a dynamical system of N particles of masses mi(i = 1,2, ... , N), having position vectors 7i with respect to some origin 0, in which constraints are also present. In the N system constraints, let the number of generalized co-ordinates be n = 3N -k. Then for the ith particle, the transformation equation 7i = 7 i(qt, q2, .. ·, qnj t)j i = 1,2, ... , N gives,
6-+ r i
n
a-+
ri = "LJ" a.
j=l
6qj
a-+ ri + 7ft
6tj
.
t
= 1,2, ... ,N.
(2.7)
%
Any real displacement d7 is always associated with the change in time dt. A virtual displacement 67 i is assumed to take place at fixed instant t, so variation with respect to time is absent, i.e., 8t = o. Hence from equation (2.7) the virtual displacement 67 is given by (2.8)
44
CHAPTER 2. GENERALISED CO-ORDINATES n
fJqJ
=L
ai:?fJr\; detlaijl =j; O.
)=1
8qj are called the generalised displacements for the ith particle. The equation (2.8)
gives the virtual displacement of the ith particle of the system in terms of generalised displacements fJqj. Since qj'S need not have the dimension of length, so generalised displacements fJqj's do not necessarily have the dimensions of linear displacement. But, because of equation (2.8), ~l fJqj must have always the dimensions of linear displacement. If qj is the angle co-ordinate, then fJqj is the angular displacement.
2.4
Generalised Velocity
Here we obtain an expression for the velocity of a particle in terms of the generalised co-ordinates. Velocity may be described in terms of time derivative qj of the generalized co-ordinate qj, is called generalized velocity. The N holonomic, time-dependent constraint T'i = T'i(q1, q2, .. ·, qn; t); i = 1,2, ... , N gives, --+
d ri dt
n!l--+!l
~Urt
- - = L..J - J=l
aqJ
u%
-at
!l--+
Uri +-= at
n!l--+
~uri.
L..J - - qj . aq)' )=1
!l--+
Uri +at
(2.9)
a:;t,
appears only when there are constraints depending where f3iJ = ~~' and -;y = on time or in some cases where it is convenient to introduce moving co-ordinate --+ axes. For the scleromic constraint (-;y = 0), the generalised velocity qj are given by
q] =
n
.
L:: 13;/ T'i.
It is easy to express velocities in generalised co-ordinates because
.1=1
from the equation (2.9), we see that, the velocity components of the ith particle is the linear functions of the generalized n velocity components qj no matter how the generalised co-ordinates are defined. Since q] need not have the dimensions of length, qj do not necessarily have the dimensions of linear velocity. But because of equation (2.9), ~~i qj must always have the dimension of linear velocity. Deduction 2.4.1. Velocity in cartesian co-ordinates: In cartesian co-ordinates, the generalised co-ordinates are q1 = X, q2 = y, q3 = z. The orthogonal unit vectors are i,J, k. Any arbitrary displacement dT' of a point may be resolved into three mutually perpendicular displacements: dxi, dYJ and dzk. Thus, dT' = dxi
=>
-+ V
d--+ r
= -
dt
+ dYJ + dZk .": = x~ + Yl + zk .~
.A
2.5. GENERALIZED ACCELERATION
45
Deduction 2.4.2. Velocity in spherical polar co-ordinates: In spherical polar co-ordinates, the generalised co-ordinates are ql = r, q2 = (), q3 = ¢. The orthogonal unit vectors are r, 0, ¢. Any arbitrary displacement d--;;7 of a point may be resolved into three mutually perpendicular displacements : drr, rdOO and r sin ¢d¢¢. Thus, ~
~ ::::} V
2.5
+ rdf)f) + r sin ¢d¢¢ A
d r = drr
=
d--;;7 dt
A
.
A
•
A
= 'H + rOf) + r sin ¢¢¢
Generalized Acceleration
Consider a system of N particles with --;;7i as the position vector for the ith particle. For acceleration, differentiate the expression of (2.9) with respect to t, we get
(2.10)
Thus we see that, the cartesian components of acceleration are not linear functions of compon'ents of generalised accelerations ijj alone, but depend quadratically and linearly on the generalised velocity components qj.
2.6
Generalized Kinetic Energy
We consider a dynamical system in which the ith particle has co-ordinate -;;;"'i and mass mi' Let us assume constraints are rhenomic. Now we write down an expression for kinetic energy in terms of generalized velocities. The kinetic energy T of a system of N free particles for a generally rhenomic system is given by ,
46
CHAPTER 2. GENERALISED CD-ORDINATES
n
=
n
LL
n
O!jkqjqk
+L
j=lk=l
where O!·k = ! 3
(3jqj
+ "y = T(2) + T(l) + T(O)
(2.11)
j=l
f m·~r-i ~T\.
2 i:::;:l
1
qj
q,,'
(3. = 3
f m·~r-i
i=l
1
qj
alL
t'
"y
=!
f
2 i:::;:l
m·(aJii)2 '
are definite
functions of the ~'s and t and hence of q and t, called the coefficients of inertia. Thus the KE in terms of generalised velocities has three distinct terms : contains terms homogeneous quadratic in generalised velocities and this fact is indicated by the superscripts 2 in T(2). This form is known as canonical form. Since the matrix [O!jkJ is symmetric and positive definite, so T(2) is nonnegative. If this term is free from cross terms, i.e., ~~i ~:' = 0; j =fi k, the generalised co-ordinate system in qk's is referred to as an orthogonal system.
(i)
T(2)
(ii)
T(l)
(iii)
T(O)
a-+
-+
is a linear combination of velocity qj and vanish when ~ = 0, i.e., when ~ i does not contain the time explicitly. This situation occurs when moving co-ordinate system are not involved. Also T(l) is either positive or negative. a-+
is independent of generalized velocities. It will vanish when ~ = From the expression we see that T(O) is non-negative.
-+
o.
Thus the KE is in general made up of a term that is independent of the generalised velocities, a term linear, and a term that is quadratic in them. When the constraints are sc1eronomous, i.e., the transformation equation (2.6) is independent of time, then a;;. = 0, 'i = 1,2, ... , N. Hence (3j = 0 and "y = 0 but O!jk =fi 0 so that T(O) = T(l) = 0 and T = T(2). In this case T(2) ~ a hom~eneous quadratic form of generalized velocities or a canonical form. But ~;.i and ~;ki, by defillition, represent -+
-+
J
two base vectors b j and b k at ~i. When the constraints are scleronomous, then n
from the expression KE
T
=
n
E E O!jkqjqk of the system, we have, j=lk=l
47
2.6. GENERALIZED KINETIC ENERGY =
n
n
n
n
n
n
j=l
k=l
;=1
k=l
k=l
j=l
L L Ojkt5jlqk + L L Ojk t5klQj = L OlA:Qk + L OjlQ;
naT
nn
nn
or, Lril[j7" = LLOlkrikql 1=1 q, 1=1 k=l n
+ LLOjlqjql. 1=1 ;=1
n
= 2 L L olkQkQ, = 2T.
--
1=1 k=l
In orthogonal (q) system, b;. b k = Ojj -::/: k and consequently, 0jk = 0 when j n
In this case, T
=/: k.
= !m E Ojj Q1j=l
Deduction 2.6.1. KE in cartesian co-ordinates: For the orthogonal cartesian co-ordinate system, the generalised co-ordinates are q1 = X, q2 = y, q3 = z. The transformation equation is -:t = xi + YJ + zk which does not conta.in time explicitly, so that ~ = O. The system being orthogonal, so
87 87
A
87 87
A
A
A
87 87
A
A
8x' 8y =i.j =OJ 8y' 8z =j.k=Oj 8z' 8x =k.i=O.
Therefore, the KE is given by
T =
1
3
" Ojj qj·2 = 2m. '~
·2 Ollq1
+ 022q2·2 + 033q3·2
j=l
1 a-:t a7 1 1 = 2m ax . 8x = 2mi .i = 2m . A
where,
all
Similarly, 022 = 033 =
A
~m.
Thus the KE of a particle of mass m is T = !m[x2 + iP + z2].
Deduction 2.6.2. KE in polar co-ordinates: In plane polar co-ordinate (r,6) system, we have x = rcos6,y = rsin6, where (x,y) is the orthogonal cartesian co-ordinate system. Hence, Therefore, the KE is given by 2
1 m '~ " Ojj qj ·2 = oUq1 ·2 · 2 Ql = r, Q2 T = 2 + 022Q2j
=6
j=l
where, au = Similarly, 022
8Y2 21m [( 8x2 8r) + (ar) 1=
1 2 .2 1 2m[cos 6 + sm 6] = 2m .
= ~m[( ::)2 + (:: )2] = ~m[r2 sin2 0 + r2 cos2 OJ = ~mr2.
Thus, the KE of a particle of mass m is T = !m[f2 + r 292J. Similarly in spherical polar co-ordinate system, the KE is T = !m[f2 + r 292 + r2 sin2 6~2J.
CHAPTER 2. GENERALISED CO-ORDINATES
48
Ex 2.6.1. KE of double pendulum: A double pendulum in a uniform gravitational field consists of two bobs of masses ml and m2 at ends of two weightless rods of length II and b. The second pendulum i~ ~uspended from the first bob of the first pendlllum, which is fixed to a rigid support. Let us assume that the pendulums move in the xy-plane only. Since the pendulum is constraint to move in the vertical plane, it has two degrees of freedom. The generalised co-ordinates selected are the angles ql = (h, q2 = fh made
h~O
x
y
Figure 2.1: Double pendulum by the two rods respectively, (which are actually angular displacements) with the vertical. Taking 0 as origin, let the co-ordinates of PI and P2 be (Xl, Yl) and (X2, Y2) respectively. Now from geometry, Xl
=
h sint'h;
Yl = h cosfh
X2 = h sin 01 + l2 sin 02; Y2 = h cos 01 + l2 cos 02· Thus the KE can be written as T
1~
1'2
1
..
1
·2
= 2 L..J CXjk iwik = 2CX(hOl fh + 2[CX0102 + CX0201](h02 + 2CX020202 ),k
OXI 2 0Yl 2 OX2 2 0Y2 2 whereCXOI0l=m1[(001) +(00 ) ]+m 2[(001) +(001)] 1 = ml [l~ cos 2 01 + l~ sin2 01] + m2[(h cos 01)2 + (-ll sinOl)2] = (ml + m2)l~. OXI OXI 0Yl 0Yl OX2 OX2 0Y2 OY2 CX0102 = ml [00 1 002 + 001 002] + m2 [00 1 002 + 001 002] = m1 x 0 + m2[it cos 0112 cos 02 + (-it sin 01)( -l2 sin 02)] = md1l2coS(01 - (h) = CY0201· OXI 2 0Yl 2 OX2 2
0Y2 2 CX0202=ml[(002) +(00 ) ]+m 2[(00) +(002)] 2 = m1 x 0 + m2[(l2cOS02)2 + (-l2sin02)2] = m2l~. Thus the kinetic energy T for the double pendulum is given by
1 T = 2[ml
2'2
+ m2]l101 + m2hl2 COS(OI -
..
1
2·2
O2)01 02 + 2m2l202 .
49
2.7. GENERALISED MOMENTUM
Deduction 2.6.3. If we are to use the plane polar co-ordinates (r,O) as generalised co-ordinates, then the KE (T) is T = ~m[r2 + r 2(P]. The associated generalised momenta are given by Pr
aT . = ar = mr =
linear momentum; Po
aT . = aiJ = mr 2 0 =
angular momentum.
Ex 2.6.2. Find the KE for a free particle in paraboloidal co-ordinates ((, ry, ¢). SOLUTION: Using the relation between cartesian co-ordinate (x, y, z) and parabolic co-ordinate system ((, ''7, ¢), we get
x =
~ i;
~ry cos ¢;
y =
~ry sin ¢;
= ~ry cos ¢ + ~iJ cos ¢
y = ~ry sin ¢
- ~ry sin ¢¢
+ (iJ sin ¢ + f.ry cos ¢¢;
i = ~~ - ryiJ
Thus the KE T for a free particle is T
= !m[i;2 + y2 + i 2] 2
= ~m[(~ry cos ¢ + ~iJ cos ¢ - ~ry sin ¢¢)2 + (~ry sin ¢
+ ~'r,sin¢ + ~rycos¢¢)2 + (~~ - ryiJ)2] = !m[~2(ry2 + e) + iJ2(ry2 + e) + ery2¢2] 2
= ~m[(ry2 + e)(iJ2 + ~2) + ery2¢2].
2.7
Generalised Momentum
The momentum associated with generalized co-ordinate qk is similarly defined and is called the generalized momentum Pk associated with the co-ordinate qk as
aT
Pk = aqk = mkq.. k,
T
1·2
= K .E = -2 "m;J7· ~ , ~
(2.12)
so that the generalised components of momentum can readily be obtained from the generalised expression of the KE of the system. Differentiate the equation (2.11) with respect to qk, we get
a---tri a---trio ",,1 mi = ~ ~ 2 T·a N
Pk
11.
qj
i=l j=l
q)
qk
N
a---t a---t
" ri ri + ~ mia·fit· i=l
(2.13)
qk
The first term is a linear function of generalised velocities. Last term of (2.13) will again be absent if the generalized system is stationary. It is a linear function of generalized velocities. Pk need not always have dimensions of linear momentum. The product of any generalised momentum and the associated/conjugate co-ordinate must always have the dimension of angular momentum or of the dynamic variable 'action'.
CHAPTER 2. GENERALISED CD-ORDINATES
50
Deduction 2.7.1. Let us consider a particle, defined by (r, e) generalised coordinates in which Fr, Fo are the components of force. Then the generalised force associated with rand e co-ordinates are respectively as
where F = FrT + FoB and 7 = rT. Here Qr is the component of force in r direction and Qo is the torque acting on the particle to increase e.
2.8
Generalised Force
Consider a N particle system with ith particle having the position co-ordinates 7 i, there are k number of holonomic constraints, then n = 3N - k is the number of generalised co-ordinates ql, q2, ... , qn.Here we consider, the amount of work done 6W by the force L F i on the system in terms of arbitrary small displacement 67 i i
of the dynamical system. Then the work done for arbitrary small displacemet:lt is
(2.14) Qj are called the components of the generalised force associated with the co-ordinates qj and possibly on time t and also determines the work done in a virtual displacement in which qj alone changes. Also, qj need not have dimensions of length, generalised, force Qj need not have dimensions of force, but Q j 6qj must have always dimensions of work. For this reason the generalised force need not always have the dimensions force. As Qj is the coefficient of 6qk in (2.10), so its name generalised force. We will make use of generalised force to obtain Lagrange's equation of motion using D' Alembert's principle. If qj be an angle, the 6qj measures the angular rotation of certain part of the system about certain axis, Qj would be torque corresponding to that angular co-ordinate qj. If the system is conservative, then, if V be the potential energy of the system in any configuration space, we have 6W = -6V. In this case, V = V(ql, q2,.··, qn). Therefore
If the dynamical system be holonomic, dql, dq2,"" dqn are independent and so we get Qk = - g~ k = 1,2, ... , n.
;
2.9. GENERALIZED POTENTIAL
2.9
51
Generalized Potential
If the forces acting on the system are derivable from scalar potential energy function V of r+ i or qj but not 7 i or qj, the work done by the force on the system is an arbitrary displacement dr+ i of the system as N 8V 8V 8V &W = -&T = - 2:(-dXi + -&Yi + -dZi) i=l 8Xi 8Yi 8 Zi N
8V = 2:(8V 8xi 8qk i=l 8 Xi 8qk
(2.15)
+ 8V 8Yi + 8V 8zi) aYi 8qk
aZi aqk
N 8-+ r = - 2:(Fi. 8 i) - Qk.
i=l
qk
Thus, the generalized forces are given by the potential energy function Qj = -~, in this sense, the definition of Qj as generalized force is a natural one. When the system is not conservative, we define the generalized force Qj associated with a co-ordinate qj by
8U Qj = - 8qj
d 8U
+ dt(8qj)
where the generalised potential energy function U = U(qk, qk, t), which is a function of position co-ordinates q and generalised velocities, may be called a 'velocity dependent potential' or simply 'generalized potential' since it gives rise to generalized force Qj. We can come across such a velocity dependent potential when we consider the electromagnetic forces on moving charges. Ex 2.9.1. The Lorentz force experienced by a charged particle in an electromagnetic field is given by -+
aA -+ -+ -+ F=e[E+v xB; E=-"V4>- at; B="VxA.
-+
-+
-+
-+
-+
-+
-+
-+
-+-+
Here e is the charge, 4> = 4>( r ,t), A = A ( r ,t) are the scalar and vector electromagnetic potential respectively. In such a case, the velocity dependent potential U is given by -+-+
U = e4> - e( v . A). As the electromagnetic force is gyroscopic in nature, so it can be incorporated in the suitably generalised potential U as defined above. Ex 2.9.2. Let us consider the motion of a particle subject to Weber's electrodynamic law of at.tract.ion t.o a fixed point, t.he force per unit mass acting on t.he particle being F = ..!.[1- f2 - 2rf] r2
c2
52
CHAPTER 2. GENERALISED CO-ORDINATES
where T is the distance of the particle from the centre of force. In this case, the velocity dependent potential function U(r, r, t) is given by U = ~[1 + ~l. This is another example of velocity dependent potential.
2.10
Principle of Virtual Work
The word virtual means possible or permissible. Here we shall discuss one important feature of constraint forces.
2.10.1
Virtual Displacement
Any imaginary displacement which is consistent with the constraint relation at a given instant is called a virtual displacement. The system is subjected to an virtual (infinitesimal) displacement with the forces and constraints imposed on the system at the given instant t. If the transformation equation be (2.6), the virtual displacement 8r\ is linked with the virtual displacement 8qj by (2.16)
as there is no passage of time in a virtual displacement. This change in the configuration of the system is not associated with a change in time t, i.e., the displacement takes place instantaneously.
2.10.2
Virtual Work
The work done by any force on a particle due to its virtual displacement is called virtual work. If the sum of virtual works done by reactions caused by constraint be zero, then the constraints are called ideal constraint.
2.10.3
Principle of Virtual Work
Let us consider a dynamical system of N -particles which is in static equilibrium and in which time-dependent holonomic constraints are also present. Then the total --t --t force on each particle vanishes, i.e., F i = O. Now let us give a virtual displacement to the system which refers to a change in the configuration of the system as a result of any infinitesimal change of the co-ordinates 8-:t i consistent with forces and constraints imposed on the system at the given instant. Then the virtual work of the iLl! particle is F .. 8-:t. and so, the total virtual work done on such system due to the arbitrary virtual displacement must identically vanish 8W
= """' ~ --t F i.8 --t r • = o.
(2.17)
2.10. PRINCIPLE OF VIRTUAL WORK Let us divide the force
F
t
53
into the applied force F~a) and the force of constraint
f 1.,.l.e., -Fi = -w Fi + -f i, then
We now consider the system for which the virtual work of tbe forces of constraints is zero (i.e., L ft.8-:ti = 0). An example of such a system can be had in mind if t
we consider that the forces of constraints being perpendicular to the surface while virtual displacement tangential to it, then virtual work done by forces of constraints will be zero. Thus the condition of static equilibrium reduces to (2.18)
This condition for equilibrium of a system is known as the principle of virtual work and which is applicable to the statics and indeed to mechanics of a system of particles. Thus the principle states that, a system of particles is in equilibrium, if the total virtual work done by the applied forces of constraint is zero.
2.10.4
D' Alembert's Principle
D' Alembert's principle is based on the principle of virtual work. To interpret the static equilibrium of the system, D'Alembert adopted an idea of a reversed force. He conceived that a system will remain in equilibrium under the action of a force . ( which is the linear equal to the actual force F i plus reversed effective force Iii momentum of the particle according to Newton's law of motion). The equation of motion of the ith particle in a dynamical system is
-
Itt
appears as an effective force called the reversed force of inertia on the where particle, supplementing the already existing externally applied force F i . Thus the principle of virtual work takes the form ith
L)Ft -lti)' 8-:ti = 0 or,
L)F(a) t
or, I:(F~a)
-
~) Pi .
8-r i + I:.f i· 8-:ti = 0
-lti)' 8-:ti = O.
CHAPTER 2. GENERALISED CO-ORDINATES
54
It is called D'Alembert's principle in which forces of constraints disappear from the problem. Since forces of constraints no longer appear, D'Alembert's principle may be expressed as
(2.19)
Fi
where means the applied force on the ith. particle of the system. Thus the equation (2.19) implies that the sum of the virtual work done by the external force and the reverse effective force is zero, which is known as D'Alemberl's principle. Since the virtual displacement 8-;;:\ are not.independent of each other, but are related by the constraints, so the coefficients of 8~ i can not be set equal to zero and as a result, D' Alembert's principle can not be used to obtain the equations of motion for the particles of the system. Thus our required task to transform the equation (2.19)into an expression involving virtual displacements of independent generalised co-ordinates. Ex 2.10.1. Plane pendulum:
Let us take the plane pendulum problem, where the generalised co-ordinate is chosen as O. So the work done by the applied force is (-mg sin 0) (180), where 80 is the virtual displacement that 0 undergoes. Also, the acceleration of the bob is Ie. Thus from D'Alembert's principle, we have,
(-mgsinO)(IMJ) - mle(180) = 0 9 9 '* 0.. = -(T)sinO ~ -TO;
80
-I 0,0 small.
This is the usual equation of the SHM. Next suppose that, the length of the string changes with time. HereJ;he only change in the above equation is that the pendulum
s
0
mg
Figure 2.2: Simple pendulum bob has a component of acceleration (Ie + 2iiJ) in the 0 direction. Hence the work done by the inertial force is m[le + 2iiJ]180. So the modified form of the equation is
(-mg sin 0)(180) - m[le + 2iO]180 =}
d
=0
2'
dt[ml 0] = -mglsinO.
The LHS is the rate of change of the angular momentum of the bob about the point of support which, in the absence of gravity, remains constant even when the length of the pendulum is changing with time.
2.11. EXERCISE
55
Ex 2.10.2. A bead of mass, Tn is constrained to move on a rod in xy-plane. The rod i8 made to rotate with constant angular velocity w in the xy-plane. The reaction of the rod on the bead is R, the force of constraint. Prove that the virtual work done by R is zero. SOLUTION: Let the rod OA be moving with constant angular velocity w around O. Let P(r,O) be the position of the bead at any instant t, then
x = r cos (); y = rsin(}; () = wt. If the bead of mass Tn is confined with the rod, then x = r cos wt, y = r sin wt. Hence for arbitrary displacement, 8s, in which bead stays with the rod and time varies, Le., 8s is a real displacement, then 8s = (ox, 8y), where
8x = 6r cos wt - rw&t sin wt &y = &ysinwt + rwcoswt&t.
In the real displacement, &s = ac. Therefore, we easily see R&s =I- 0, as R has component along ac. Hence for real displacement force of consistent does do work. Consider now the virtual displacement, in which time-variation is not taking place, Le., &t = 0. In this case, &x
= &rcoswt;
&y
= &rsinwt.
In this virtual displacement case &s is along &r therefore, R.8s = 0, as R is orthogonal to rod, and r is the radius vector along the rod. Hence for virtual displacement, the virtual work done by forces of constraint is zero.
Exercise
2.11
Ex 2.11.1. What are generalised co-ordinates of a dynamical system? Define holonomic and non-holonomic dynamical system. Write short note with examples on unilateral and bilateral constraints.
Ex 2.11.2. For a scleronomous dynamical system of n dJ., prove that the KE is a homogeneous quadratic function of generalised velocities. Ex 2.11.3. For a scleronomous holonomic dynamical system of n dJ., moving under conservative force field, prove that KE +PE =constant.
Ex 2.11.4. For a scleronomous holonomic dynamical system of n dJ., prove that 'It
L qk k=l
zt -
L = constant, is the integrals of motion, where qi'S are the n generalised
"
co-ordinates, L= Lagrangian. Ex 2.11.5. A body subject to gravity moves on a smooth sphere along a vertical great circle. Find its angular acceleration assuming that the great circle is at rest in an inertial frame.
56
CHAPTER 2. GENERALISED CO-ORDINATES
Ex 2.11.6. A body is sliding down an inclined plane which is moving horizontally with constant velocity. Find the position of the body as the function of time. Ex 2.11.7. Show that for a hoop on a plane, the constraint of rolling without slipping i:::; not holonomic. Write down the constraint in differential form.
Chapter 3
Variational Principle Calculus of variations is one of the most important mathematical tools of great scientific significance used in Mathematicians, Physics, Engineers. The generalization of the elementary theory of maxima ana minima of function of a single or several variables is the origin of the calculus of variations. The basic object of calculus of variation is to explore for finding extrema or stationary values of a quantity whose values are determined by one or more functions, known as functional. The domain of a functional is the set of admissible functions, rather than a region of a co-ordinate space. For example, (i) A functional is the length L between two given points (xo, YO) and (Xl, Y1) on the curve y = y(x) which is Xl
L[y(x)J
=
J[l + (~~)2P/2dx Xo
where we are to determine the curve y = y(x). (ii) The functional is the area A of a surface bounded by a given curve C as A is determined by the choice of the surface
-JJ
az 2 +(ay)J az 2 1/2 dxdy [l+(ax)
A[z(x,y)J-
D
where D is the projection of the area bounded by the curve C on the xy-plane. Several physical laws can be deducted from concise mathematical principles to the effect that a certain functional in a given process attains a extrema. Here Xl
we are to find the stationary values of the functionals like
J F(x, y(x), y'(x))dx, Xo
Xl
Xl
J F(x, y(x), y'(x), . .. ,y(n) (x))dx, J F(x, Y1 (X), ... ,Yn(x), yi (x), . .. , y~(x))dx and Xo
57
CHAPTER 3. VARIATIONAL PRINCIPLE
58
J J F(x, y, z(x, y),
Zr,
zy)dx in which the function F b given, and the functions
D
y(x), yt(x), ... , Yn(:r), z(x, y) are the arguments of the functionals. Eor example, moment of inertia, the co-ordinates of the CG of a curve of surface are functionals, as their values are determined by the choice of curve or surface.
3.1
Properties of Functionals
A variable quantity I[y(x)] depend on a function y(x) is a functional if to each function y(x) belonging to a certain class of functions, there is a definite value of I. Exoperty 3.1.1. [Variation] : Let y(x) and Yl(X) be two functions belonging to a certain class. Then the difference 81/ of the argument y(x), defined by 8y = y(.r.) - Y1 (x) is called increment or variation. Property 3.1.2. [Closeness ]: The function y(x) and Yl(X) are closed, if IY(x) Yl(x)1 is small for all x for which y(x) and Yl(X) are defined. WIlen it occurs, we say, y(x) is closed to Yl(X) in the sense of zero-order proximity. However, with this definition, the functional
J X2
I[y(x)]
=
F[x, y(x), y'(x)]dx,
Xl
which occurs in many applications, is seldom continuous due to the presence of y'(x) in the argument of I. This necessities the extension of the concept of closeness of the curves y = y(x) and y = JJ1(x) such that both Iy(x) - Y1(x)1 and Iy'(x) - y~(x)1 are small for all values of x for which y( x) and Yl (x) are close in the sense of first order proximity. In general the curves Y = y(x) a-ad Y Yl(X) are close of the nth order proximity, if Iy(x) - Yl(x)l. IY'(x) - y~ (x)I ... , Iy(n)(x) - yin) (x) I are small for values of x for which these functions are defined.
=
Property 3.1.3. [Continuity]: If a small change in y(x) results in a small change in the functional I[y(x)], then the functional I[y(x)] is said to be continuous. This definition is, however, somewhat imprecise since we have not specified what we mean by 'a small change in y(x)'. The functional I[y(x)] is said to be continuous at Y = yo(x), in the sense of kth order proximity if given any positive number E, :l a 8> 0 such that,
II[y(x)]- I[yo(x)]1 <
E
< 8, Iy'(x) - yb(x)1 < 8, ... , Iy(k)(x) - y~k)(x)1 < 8.
for IY(x) - yo(x)1
Property 3.1.4. [Linear functional]: Let l[y(x)] be defined in the normed linear space !vI of the functions y(x). This functional is said to be linear, if it satisfies (i) I[cy(x)]
= cI[y(x)];
where c is an arbitrary constant,
3.1. PROPERTIES OF FUNCTIONALS
59
b
For example, /[y(:r)J
= J[y'(x) + 5y(x)Jdx
defined in the space C1[a, bJ.
a
Property 3.1.5. [Variation in functional J: If the increment 61 in the functional I can be written in the form
61
= I[y(x) + £5y(x)j- I[y(x)J = L[y(x), £5yj + f3(y(x), £5y) max /£5y/,
where L[y(x), £5yj is a functional linear in £5y not in y and max /£5y/ is the maximum value of /£5y/ and f3(y(x), £5y) ~ 0 as max /£5y/ ~ 0, then L[y(x), £5yj is called the variation of the functional I and is denoted by M. An alternative definition of the variation of a functional I is also given. Consider the functional I[y(x) + ll'£5yJ, for fixed y and 5y and different value of the parameter ll'. Now,
61
'*
= I[y(x) + ll'£5yJ - I[y(x)J = L[y(x), ll'£5yJ + f3[y(x), ll'£5yJlll'l max l£5yl
lim 61 = lim 61 = lim L[y(x), ll'£5yj 6ll' 01->0 a 01->0
+ f3[y(x) , ll'£5yjlll'/ max /£5yl a
6a->0
=
lim L[y(x), ll'£5yJ a
a->O
+ lin
f3[y(x) , ll'£5yj/ll'l max l£5y/ a
a->O
= L[y(x), 5yJ = M;
as L is linear.
Thus, the variation of the functional I[y(x)J is given as
a
all'I[y(x)
+ ll'£5yj
at a
= O.
Property 3.1.6. [Optimal value of the functionalj: A functional I[y(x)J attains a maximum on a curve y = yo(x) if the values of I on any curve close to y = yo(x) for which 61 = I[y(x)J - I[Yo(x)J ~ O. Similarly, if 61 = I[y(x)J - I[yo(x)J ~ 0, the minimum of I[y(x)J attained on the curve y = yo(x). Theorem 3.1.1. If a functional I[y(x)J attains its optimal value on y = yo(x) such that the domain of definition belong to a certain class of functions, then at y = yo(x),£5I = O. The calculus of variations explores methods that permit finding maximum or minimum values of the functionals. Such problems are known as variable problems. Several physical laws can be deducted from concise mathematical principles to the dfect that a certain functional in a given process attains a optimal. In mechanics, we have the principle of least action, the principle of conservation of linear momentum, Fermat's principle in optics, principle of Castigliano in the theory of elasticity are some examples of variational problems.
60
3.2
CHAPTER 3. VARIATIONAL PRINCIPLE
Euler's Equation
Let us assume that J(x, y, y') is thrice differentiable. Then the necessary condition for the extremum of the functional b
I[y(x)] = j J(x, y(x), y'(x))dx; y(a) = Yl, y(b) = Y2
(3.1)
a
where Yl and Y2 are specified at the fixed boundary points a and b, is that its variation must vanish, assuming that the admissible curves on which an extremum is achieved, admits of continuous first order derivatives. PROOF: Let y = y(x) be the curve which extremizes the functional I[y(x)] in (3.1) such that y(x) is twice differentiable and satisfies the boundary conditions y(a) = Yl, y(b) = Y2. Let y = y(x) be an admissible curve close to y(x) such that both y(x) and y(x) can be included in a one-parameter family of curves
y(x, a) = y(x)
+ a[y(x) -
y(x)].
(3.2)
For a = 0, y = y(x) and for a = 1, y = y(x). The difference y(x) - y(x) is the variation 6y of the function y(x). Now on the curves of the family (3.2), the functional (3.1) reduces to a function of a, say 'ljJ(a). But the extremizing curve corresponds to a = 0 so that 'ljJ(a) is extremized for a = 0, i.e., [~]Q=o = 0, where b
'ljJ(a) = I[y(x)
+ a6y] =
j J(x, y + a6y, y' + a6y')dx a
b
= j J(x, y(x, a), y'(x, a))dx Q.
b
'* 'ljJ'(a)
= j [Jy(x, y(x, a), y'(x, a))8y + Jyl(X, y(x, a), y'(x, a))8y']dx a
b
'* ./[Jy(x, y(x), y'(x))8y + Jyl(X, y(x), y'(x))6y']dx = 0 a b
b
'* j[Jy(x, y(x), y'(X))6y]dx + j a
Jyl d~ (8y)dx = 0; as d~ linear
a b
b
[J.y (x, y(x), y'(x))8y]dx + [Jy8y]~ - j [:x Jyl ]8ydX = 0
'* j a
a
b
'* j a
[Jy(x, y(x), y'(x)) -
d~Jyl]8YdX =
0 as [8Y]a
= 0 = [8Y]b'
(3.3)
3.2. EULER'S EQUATION
61
In view of assumptions made on J(x,y(x),y'(x)) and the extremizing curve y(x), it follow that the integrand in (3.3) on the curve y(x) is a given continuous function, while 8y is an arbitrary continuous function subject to [8Y]a = 0 = [8Y]b. Thus we must have
Jy(x, y(x), y'(x)) -
d~JYI =
0
(3.4)
on the extremizing curve. This equation is known as Euler's equation. If J(x, y, y') does not contain x explicitly so Jxy' = O. Using d~Jyl = Jxy' + JyylY' + JylylY" from (3.4), we get (3.5) which is the second order equation in y(x). The two arbitrary constants appearing in the solution are determined from the boundary conditions yea) = Y1, y(b) = Y2. Now we have the following discussions: (i) The existence of the solution (3.5) can not be taken for granted and even if solution exists, it may not unique. However in many problems, the existence of a solution is evident from the geometrical or physical nature of the problem. Hence in such cases, if the solution is unique, then this solution gives the solution gives the solution of the variational problem. (ii) When the function J in (3.1) depends on y and y' only, Euler's equation becomes Jy - JyylY' - Jylyly" = O. Now,
-d [J - Y'f] yl dx
= fyy' + fy'Y " - Y"fyl - Y'2fyy' - f ylyly'" Y = y'[Jy - y' Jyyl - JylyIY"] = 0 =} J - y'Jyl = C1; where C 1 is a constant.
(iii) Invariance of Euler's equation: Suppose that a functional I[y(x)]
=
b
J F(x, y, y')dx is transformed by replacing the independent variable or by a sia
multaneous replacement of the Jesired function and the independent variable. Then, the extremals of the functional are found from Euler equation for the transformed integrand. Herein lies the principle of the invariance of Euler's equation. Let x
J
=
x( u, v), y
F(x, y, y')dx
= =
=
J J
y( u, v) and
IXu Xv I =I- 0, then Yu Yv
F[x(u, v), y(u, v), Yu
Xu
+ yv'V~ ](xu + xvv~)du + xv'Vu
62
CHAPTER 3. VARIATIONAL PRINCIPLE
Now, the extremals of the original fUllctional are determined from Euler's equation for J rp( u, v, v~Jdu as given by
1
Ex 3.2.1. Test for an extremum the functional I[y(x)]
J(xy
o
+ y2
- 2y2Y')dx;
subject to y(O) = 1, y(l) = 2. 1
SOLUTION: To examine the extremum of the functional I[y(x)]
J(xy
o
+ y2
-
2y 2 Y')d:r, the Euler's equation is d Fy - -FyI dx =}
x
+ 2y -
= OJ
= xy + y2
F
4yy' - dd (_2y2)
- 2y2y'
=0
X
=} X
+ 2y =
0
=}
X
-"2'
Y=
Clearly, this extremal can not satisfy the boundary conditions y(O) = 1, y(l) Thus, an extremum cannot be achieved in the class of continuous functions.
=
2.
1
Ex 3.2.2. Find the extremum of the functional I[y(x)] the condition y(O)
= J(y'2 + 12xy)dx
satisfying
o
= 0, y(l) = O.
[GATE'05] 1
SOLUTION: To examine the extremum of the functional I[y(x)]
o
the Euler's equation is d Fy - dx FyI =}
= J(y'2 + 12xy)dx,
12x - 2y"
= 0;
= 0 =}
= y'2 + 12xy y(x) = x 3 + C 1 x + C2, F
where C1, C 2 are constants, to be determined by the condition y(O) = 0, y(l) = O. Using these conditions, we have, C1 = C 2 = O. Thus, y = x 3 is the only curve on which extremum can be attained. Ex 3.2.3. Shortest distance between two points: In the problem of geodesics, it is rf'ql~;red to determine the line of shortest length connecting two given points (X1,]/1) and (.'1:2, Y2) on a surface S . .Suppose, the length of an arc in a plane is ds = dx2 + dy2. This is a typical variational problem with a constraint as here we are rC)quired to minimize the arc length I joining two points (.E1, yd and (X2, Y2) on S given by the functional,
vi
J JJ X2
I
=
X2
ds
=
J X2
dx 2 + dy2
=
:1'1
3.2. EULER'S EQUATION
63
*
The condition that the curve be the shortest path is that I be a minimum, i.e., 6I = O. Therefore, taking y' = and f = + y,2, we get,
VI
8f _ ~(8f) = 0 8y dx 8y' =}
d -d [ x
y'
VI + y,2
=}
Y
,
] = 0 =}
y'
VI + y'2
= C( constant)
dy C ( = A say); where A dx VI - C2 =;=? Y = Ax + B; Integrating,
=-
C v 1 - C2
= -./r==~
where A and B are constants. So, the straight line has only been proved to be extremum path, since the second derivative of y with respect to x gives a positive constant quantity. The constants of integration A and B are determined by the conditions that the curve passes through the points through the two points (Xl, Yl) and (X2. Y2). Ex 3.2.4. The Brachistochrone problem: The Brachistochror~e problem was first introduced by Johann Bernoulli [1696]. Here we are to find the curve joining given points A and B not on the same vertical A
Figure 3.1: The Brachistochrone problem line which is traversed by a particle moving under gravity from A to B (in absence of any resistance) in the shortest time. Fix the origin at A with x-axis horizontal and y-axis vertically downward. The speed of the particle in any position (x, y) is (2gy) 1/2, g being the acceleration due to gravity. Hence the time taken by the particle in moving from A(O, 0) to B(Xl' Yl) is given by ds
dt
= v =} t =
=}
I
ds v
= IX' o
1 t[y(x)] = yI2g
V1+
Y[ dx
2gy
+ JXIPJ ~dx 1
y2
o
which does not contain x explicitly. So f(y, y') = Jl~Yr, with the boundary conditions on yare y(O) = 0; Y(Xl) = Yl. Although the integral is improper, it is
64
CHAPTER 3. VARIATIONAL PRINCIPLE
convergent. Since the integrand f(y, y') gral of Euler's equation is given by
f -
y
,
f y' =
q
=
V yr is independent of x, a first intel
:
J1+ yl
=}
11r;; -21 (1 + y)12 1/2 .2y1 vy y(l + y/2) = Constant = C
=}
y = 2[1- cos2tj; y'
=}
dy dx = ,.
=}
-- -
Y
C
Y
=}
x
=
C2
= C[l
= Cl
= cott
- cos 2tjdt
+ ~ [2t -
sin2tj.
Putting 2t = tl and remembering y(O) = 0 from the boundary condition we find that C2 = O. Thus x = ~ [il - sin tIl, y = ~ [1 - cos tIl represent a family of cycloids with ~ as the radius of the generating (rolling) circle. Further C is determined from the fact that the cycloid passes through (Xl, Yl). Ex 3.2.5. Minimal surface area: In the problem of minimal surface of revolution, a curve y = Y(X) 2': 0 is rotated
Figure 3.2: Minimum surface revolution about the x axis through an angle 27r. Here determination of the curve y = y(x) for which the surface of revolution is minimum, constitute a variational problem. Let we form a surface of revolution by taking some curve passing between two fixed end points (Xl, Yl) and (X2' Y2) and revolving it about the X axis. Then we are to find the curve for which the surface area is a minimum, subject to the condition y(Xl) and y(X2) is fixed. The surface area of a strip of the surface is 27ryds. Hence the resulting surface bounded by the planes X = Xl a1. J X = X2 has the area,
A
=
X2
J
27ryds
=
J
27ryV1 + y/2dx; y'
X2
= 27r
J
YV1
X2
+ y/2dx = 27r
J
f(y, y')dx
dy dx
3.2. EULER'S EQUATION
65
YVI
where the functional f(y, V') = + y/2 does not contain x explicitly. For extremum of the integral, using first integral Euler's equation, we have
f =?
Y
VI
constant = c y' + y/2 - y' Y ---,:.====;;: y' fu'
=
VI + y/2
=?
dy
=?
=
vx
c c2
2 -
y
VI + y/2 -= c
---;==::::::;:
dx CI
X -
Y = c cos h(--) c
which is the required curve, called two parameters catenary, where c and CI are constants. The constants are determined from the conditions, that the given curve passes through the given points. The minimum surface is called catenoid. Ex 3.2.6. Find the curve, using invariance of Euler's equation, for which the funclog 2
tional I[y(x)]
=
J [p-Xy/2 -
e Xy2]dx be extremized?
o
log 2
SOLUTION:
J [e-
For extremum of the functional" I[y(x)]
Euler's equation, we have
y /2 - e X y2]dx using
o.
af- -d( af) _ -
ay
=?
X
o
dx
y" - y'
ay' - ,
+ e 2x y = O.
The solution of the nonlinear differential equation poses some problem. Let us give a transformation, x = log 11., 1} = v, then the above functional reduces to 2
h[v(u)]
= /[v /2
2 - v ]du.
1 2
=
For extremum of the functional h[v(u)]
f[v /2 - v 2]d11. using Euler's equation, we 1
have v" + v = 0, readily gives rise to the solution v required solution for extremals is given by, y
= C1 cos(e + C 2 sin(e X
)
=
C 1 cos U
+ C2 sin u.
Thus the
X ).
Ex 3.2.7. Prove that the sphere is the solid figure of the revolution which, for a given surface area, has maximum volume. SOLUTION:
The surface area is given by, a
a
S = /27r1}dS = / 27rYVI
o
0
+ y/2dx
CHAPTER 3. VARIATIONAL PRINCIPLE
66
a
and the volume of the solid so formed is V
= I7ry 2dx. Here, we are to maximize o
V subject to fixed S. Taking 1 = 7ry2 and 9 = 27rYVl auxiliary function H(x, y, y') as
+ ).9(·r, y, y') = 7ry2 + 27ry).Vl + y,2 j ). =
+ y,2,
we are to consider an
H(x, y, y') = 1(.r, y, y')
Lagrangian parameter.
Now, H has to satisfy Euler's equation, where H does not contain x explicitly. Hence,
BH = constant = c( say) By'
H - y' -
=}
try
2
+
27r).y
Jl + y,2
=
c.
It passes through 0(0,0) and A(a, 0), for which c = 0, and so, 2),
y+
VI + y,2
=o=}y =} X
V4).2 - y2
,
=-'-----
=k-
Y
V4).2 -
,.--"..-~
y2.
When x = 0, y = 0 then k = 2),. Thus,
which is a circle with centre (2).,0) and radius 2),. Hence the figure formed by the revolution of the given arc is a sphere.
3.3 'Functionals depend on higher order derivatives Let us now consider the extremum of a functional of the form
J b
1(x, y(x), y'(x), ... , y(n) (x))dx
(3.6)
a
where we assume that the functional 1 is (n + 2) times differentiable with respect to all its arguments. The boundary conditions are _ '() _ , (n-l)( ) _ (n-l) y ( a ) - yo, Y a - Yo,···, Y a - Yo ' ... , y (n-l)(b) -_ Yl(n-l) y (b) -- Yl, Y'(b) - Yl,
We further assume that the extremum of the above functional is attained on a curve y = y(x) which is differentiable 2n times and any admissible function is also 2n
3.3. FUNCTIONALS DEPEND ON HIGHER ORDER DERIVATIVES times differentiable. It is clear that y one- parameter family of curves
y(x, a) = y(x)
67
= y(x) and y = y(x) can be embedded in
+ a[y(x) -
y(x)l.
(3.7)
On the curves of the above family, the functional becomes a function of a, say '0 (a). Since the extremizing curve corresponds to a = 0, we must have '0'(0) = O. Thus, b
I[y
+ m5Yl =
j[f(x, y(x, a), y'(x, a), ... , y(n)(x, a))ldx a b
::::}
[~~la=o ::::} [d~ j
f(x, y(x, a), y'(x, a), ... , y(n)(x, a))dxla=o = 0
a b
::::} j[JyJy + fylJy' + ... + fy(n) Jy(n)ldx = 0; by chain rule. (3.8) a b
b
Now, j fylJy'dx =
[JyIJyl~ - j[d~fy/lJYdx
a
a b
j[d~fY/1JYdx;
= -
Using boundary condition
a
b
b
j fyllJy" dx = j fyll a
d~2 (Jy)dx;
as linear operator
a
a
b
2
= j[ d 2 fylllJydx; dx
Using boundary conditions
a
and so on. Thus from (3.8), we find that on the extremizing curve, b
d [Jy dXfyl j
d2
+ dx2fyll
- ...
dn
+ (-It dxnfy(n)ldxJy = 0
a
for an arbitrary choice of Jy. Due to conditions of continuity imposed on f, the first factor in the forgoing integral is a continuous function of x in [0" bl. Thus the function y = y(x), which extremizes I satisfies d d2 dn (3.9) fy - dXfyl + dx 2 fyll - ... + (_l)n dxnfy(n) = 0,
68
CHAPTER 3. VARIATIONAL PRINCIPLE
which is known as Euler-poisson equation. Clearly the solution of this differential equation of order 2n involves 2n arbitrary constants, which are determined by using boundary conditions.
Ex 3.3.1. Determine the extremal ofthefunctionaZ I[y(x)]
/
J[~I1,y"2+py]dx subject
=
-/
to y( -Z)
= y'( -Z) = 0 =
y(l)
= y'(l). /
SOLUTION:
Here we are to find the extremum of the functional I[y(x)] = f[~p,y"2+ -/
py]dx, where we assume that F = ~p,y"2 + py is to be differentiable 4 times with respect to all of its arguments. For the Euler's-Poission equation gives d d2 , dy " d 2y fy - dX fyl + dx 2fyll = O;y = dx'y = dx2 2
=? =?
p,y
P - 0 + dd 2 ( l.Ly")
(w)
x
+ P = 0 =? Y =
= 0;
p"
P
4
---x 24p,
p constants
•
3 C2 2 + -x + -x + C3 X + C4 ' 6 2
CI
where the constants CI, C2, C3, C4 can be obtained from the prescribed boundary conditions y( -l) = y'( -Z) = 0 = y(Z) = y'(Z). Using the boundary conditions, we have, _~Z4_
24p,
L
Z3
6fL
_LZ4 24p, _LZ3 6fL
3+
CI Z
6
+ CI Z2 _ 2
+ (;1 Z3 + 6
C2Z2_C3Z+C4=0
2
C2 Z
+ C3 = 0
2 c2 Z2 2
+ C3 Z+ C4 = 0
+ CI Z2 + C2 Z+ C3 = O. 2
2
t&,
Solving these equations, we get, CI = 0,C3 = 0, C2 = solution satisfying the prescribed boundary conditions is,
C4
= -2~:'
Thus the
y = _~[x4 - 2Z 2 x 2 + Z4]. 24p, This vibrational problem arises in finding the axis of the flexibly bent cylindrical beam clamped at the ends. If the beam is homogeneous, then p and p, are constants.
3.4
Functionals on several independent variables
We now consider the problem of determining the extrema of functionals involving multiple integrals leading to one of more partiai differential equations. For example, consider the problem of finding the extremum of the functional
J[u(x,y)]
=
JJ
F(x,y,u,ux,uy)dxdy
D
(3.10)
3.4. FUNCTIONALS ON SEVERAL INDEPENDENT VARIABLES
69
over a region of integration D by determining u which is continuous and has continuous derivatives up to second order, and takes on prescribed values on the boundary of D. We further assume that F is thrice differentiable. Let the extremizing surface be u = u(x, y) so that an admissible one parameter surface is
u(x, y, a) = u(x, y)
+ a1](x, y)
where 17(.7:, y) = 0 on the boundary of D. Then the necessary condition for an extremum is 8.1 = [taJ(u + CY.1])]a=O = O. Now,
.1[u + a1](x, y)] = j j F(x, y, u + a1], U x + a1]x, uy + a1]y)dxdy D
=?
[~~la=o =
j j[FU1] + Fux1]x
+ Fuy1]y]dxdy =
(3.11)
O.
D
This may be transformed by integration by parts. We assume that the boundary of D admits of a tangent which turns continuously piecewise. Then by virtue of. Green's theorem, we have
r
j j[17xFu-r:+1]Fuy]dXdy= j j[:x(Fux 1]:+ n D
~(Fuy1])-17{:xFux+ :xFux}]dxdy
= j 1][Fu xdy - Fuydx] - j j 1]{:xFux r D
/'f' [Fu -
=.
a Fu x - ay a FUy ]1]dxdy + ax
j 1][Fux dy - Fuydxjr
r
D
= 0;
+ !Fu",}]dxdy
by (3.11).
Since 1] = 0 on r and the above relation holds for any arbitrary continuously differentiable function 1], it follows from the above relation that,
a
a
Fu - axFux - &yFu~ = O.
(3.12)
This second order partial differential equation known as Euler-Ostrogradsky equation. In general, if we consider the functional
.1[u(x, y)] = j j F(x, y, u, U x , u y, U xx , u xy , uyy)dxdy, D
we get the equation for extremals as (3.13)
70
CHAPTER 3. VARIATIONAL PRINCIPLE
Ex 3.4.1. Transverse vibration of a string: Let u(x, t) be the transverse displacement of the string, where 0 the KE T is T
=
~
x
~
l. Then
I
J p( %~ )2dx, where p is the density per unit length.
~
The potential
o energy V is given by,
J I
V = measured by the extension = k
[ds - dx]; k = constant
o au 1 + (-)2dx - dx] ax
o I
I
lau2 }dx - dx] = -kJau = k [{l + -(-) (-) 2dx. 2 ax 2 ax J o 0 Applying the Hamilton's principle, we get,
For the Euler-Ostrogradsky equation, the functional is F(x, t, u, Ux , Ut) = ~p(~~)2_ ~ (~~.)2. Thus,
If the string is homogeneous, p is constant, so, this equation becomes,
ku xx
- pUtt =
0
=} Utt
2 2 k = c U xx ; where, c = -
p
which is well-known one-dimensional wave equation. Thus it follows from EulerOstrogradsky equation for the functional that the equation for extremals is Utt = c2 ,uxx , which is the Laplace's equation. Along with the boundary condition, this problem is the famous Dirichlet's problem of partial differential equations. X2
Ex 3.4.2. Find the extremum of the functional I =
0: y(¥) = -1 and z(O) = 0; z(¥) = 1.
J [y,2 + z,2 + 2yz]dx with yeO) =
3.5. VARIATIONAL PROBLEMS WITH MOVING BOUNDARIES
71
X2
SOLUTION: For the extremum of the functional I
= J[y12 + Zl2 + 2yz]dx,
the Euler's
Xl
equation gives z" - y = 0 and y" - z y(iv) - y = O. The solutions are
=
O. These two equations gives an equation
Using the initial conditions, we get C 1 + C 2 - C3 = 0 Cle~ + C2e-~ + C4 = -1; Cle~ + C2e-~ - C 4 = 1 CI = C2 = 0, C4 = -1, C3 = O. CI
+ C2 + C3 = 0;
'*
Thus, the curves y = - sin x; z = sin x extremizes the functional I.
3.5
Variational problems with moving boundaries
Let us consider a functional of the form
J X2
I[y(x)]
=
F(x, y, y')dx; y'
dy dx'
Xl
So far, we have considered extremals of the above functional when the boundary points Xl and X2 are fixed. Let us now consider the variational problem, when one or both the boundary points Cc.ll move. This implies that the class of admissible curves is extended because in addition to the comparison curves with boundary points, we have to consider curves with variable boundary points. It is evident that, if on the curve y = y(x), an extremum is attained in a problem with moving boundary points, then the extremum is definitely attained on the narrower class of curves with fixed boundary points. Hence the curves on which extremum of the above functional is attained in a moving boundary value problem are solutions of Euler's equation
d Fy - dXFyl
= O.
(3.14)
Thus these curves must be extremals. In the variational problem with fixed boundary points, the two arbitrary constants in the solution of Euler's equation are determined by the two boundary conditions at Xl and X2. But in a moving boundary problem such conditions are missing and the arbitrary constants in the solution of Euler's equation have to be obtained from the vanishing of the variation 8I, which is the necessary condition for the extremum. For the sake of simplicity, we assllme that one of the boundary points (Xl, YI) say, is fixed, while the other point (X2' Y2) can move and goes to the point (X2 +
CHAPTER 3. VARIATIONAL PRINCIPLE
72
OX2, Y2 + 0Y2). Since the extremum is attained only on extremals, i.e., on solutions Y = y(x, G l , C2) of Euler's equations, from now on we consider the values of I on such curves. Hence l[y(x, C1, G2)J becomes a function of C1 and C2 and the limits of integration. We shall assume the admissible curves y = y( x) and y = y( x) + oy( x) close if \oy\ and \<5y'\ are both small. The extremals passing through (Xl, Y1) from a pencil of extremals y = y(x, C1). The functional l[y(x, G1)J on the curves of the pencil is a function of C1 and X2. If the curves of the pencil do not intersect, the l[y(x, C1)J is a single valued function of (X2' Y2). Thus the fixing of the point (X2' Y2) determines the extremal uniquely. Let us find the variation on l[y(x, C 1 )J when the boundary point moves from (X2' Y2) to (X2 + <5x2, Y2 + <5Y2). Thus, x2+6x2 61 = j F(x, y + Oy, y'
X2 + <5y')dx - j F(x, y, y')dx
~ 7[F( x, y + iJy, y' + iJy') -
F(x, y, y')]dx
x + x7 ,
F(x, y + iJy, y' + iJy')dx
Now by MVT, the second term can be written as
x2+6x2 F(x, y + <5y, y' + oy')dx = [FJx2+l1.6x2.0X2; where 0 <
j
e < 1.
Xl
Due to continuity of F, we may write [FJx2+l1 6x2 = [FJx2 +E.', where E.' -+ 0 as <5x2 -+ 0 and OY2 -+ O. Also, by Taylor's theorem we transform the first term OIl the RHS as,
X2 ./ [F(x, y + <5y, y' + oy') - F(x, y, y')Jdx Xl
.£2
= j[Fy(x, y, y')<5y + FyI (x, y, y')<5y'Jdx + K; K an infinitesimal Xl .£2
= [FylOyJ~; + j[Fy -
d~FylJ<5ydx + K.
Xl
X2
Since the values of I are taken only on extremals, the term J[Fy - d~FylJOydx in 'Xl
the above expression vanishes and so the above expression becomes [FyloY]X2' as [<5yJxl = O. Note that, [OY]X2 is not equal to <5Y2. Also, [<5Y]X2 = OY2 - y'(X2).OX2.
3..5. VARIATIONAL PROBLEMS WITH MOVING BOUNDARIES
73
Therefore,
J X2
61 =
[F(x, y
+ 8y, y' + 8y') -
F(x, y, y')ldx
= Fy,lx2·[8Y2 - y'(x2). 8x 21 = [F - y'FlI,llx2·8x2 + FlI,lx2.8Y2. Since 8X2 and 8Y2 are independent ( as (X2' Y2) move freely), the necessary condition of the extremum n = 0 then gives (3.15)
When the boundary point (X2' Y2) moves along the curve Y2 = ¢(X2), the condition n = 0 reduces to (3.16)
which provides the condition at the free boundary. Ex 3.5.1. Find the extremum of the functional 1
=
X2
J [y,2 + z,2 + 2yzldx with y(O) = 0
Xl
and z(O) = 0 and the point (X2' Y2, Z2) moves over the fixed plane x = X2. X2
SOLUTION: For the extremum of the functional 1
= J [y,2 + z,2 + 2yzldx, the Euler's Xl
equation gives z" - y = 0 and y" - z = O. These two equations gives an equation y(iv) - y = o. The solutions are
= C1 cosh x + C2 sinh x + C3 cos x + C4 sinx z = C1 cosh x + C2 sinh x - C3 cos x - C4 sinx. y
The constants C1, C2, C3 and C4 are determined by the given conditions y(O) = 0 and z(O) = 0, which give C1 = C3 = O. Further, the condition at the moving boundary point (X2' Y2, Z2) can be derived from the condition that 6Y2 = 0, (since X2 is fixed) as [Fy'lx=x2 = 0 and [Fz,lx=x2 = 0 =? y'(X2) = 0; and Z'(X2) = o.
Thus the above solutions lead to,
If, cos X2 f:. 0, then C2 = C4 = 0 and, therefore, an extremum is attained on y = 0, z = O. But if cos X2 = 0, then C 2 = 0 and C4 remains arbitrary. In this case, the extremal is y = C4 sin x, z = -C4 sin x.
74
CHAPTER 3. VARIATIONAL PRINCIPLE
Ex 3.5.2. Find the shortest distance between the parabola y = x 2 and the straight line x - y = 5. SOLUTION: Here the problem is to find the extremum of the functional I[y(x)] = X2
1
f [1 + y'2] 2 dx, subject to the condition that the left end of the extremal moves along Xl
y = x 2 , while the right moves along x - y = 5. Thus the condition (3.16) gives,
[(1
+ y'2)~ + (2x -
y')(1 + y'2)-~]X=Xl = 0
Since the general solution of Euler's equation for the above functional is y = C 1 x + C 2 ; C1, C 2 are constants, it follows that y' = C1. Further, both the end points lie on the extremal y = C1X + C2, so we must have,
C1X1
+ C2 = xi;
C1X2
+ C2 = X2 -
5.
Putting the value of y' by C1 in the transversality conditions and solving the resulting equations, we get,
C1
3
1
23
= -1,C2 = 4,x 1 = 2,x2 = s·
Thus the required extremal is y = -x given parabola and the straight line is 23/3
. [1 ./
+~
and the shortest distance between the
+ 1]~dx =
19v'2. 8
1/2
3.6
Parametric representation
Let the equation of the curve be x = x(t), y = y(t); to ~ t ~ t1. Let I = f F(t, x(t), y(t), x(t), y(t))dt and linear functional depend on the path but not the parameter t (not time). Assume t is not present explicitly in F and F is homogeneous form with respect to x(t) and y(t), then 1=
J
Introduce a new parameter r =
F(t, x(t), y(t), x(t), y(t))dt. ~(t); ~ =1=
0, we have,
..
dr F(x(r),y(r),XTe,YTe)T
J J~F(x(r),y(r),xT,ih)d;; =J as~ =
1=
due to homogeneity of F
=
F(x(r),y(r),xT,ih)dr;
T.
3.7. EXERCISE
75
To optimize, we use Euler's equation as
3.7
Exercise
Ex 3.7.1. Find the extremals of the functionals I[y(x)]
(1+u 2 )
= J +dx. y = sinh(Clx + C'!). Xl
Xo 1f
Ex 3.7.2. Find the extremals of the function,als I[y(x)] to the boundary condition y(O)
2"
= J[ yl2 -y2+2xy]dx, subject o
= 0, y(~) = O.
y Xl
Ex 3.7.3. Find the exremals of the functional I[y{x)] = J[2xy
=x-
~sinx.
+ yIll2]dx.
xo
Ex 3.7.4. Establish Euler-Lagrange differential equation for the determination of b
the curve y
= f(x), which makes the integral J f(x, y,'y')dx stationary, the symbols a
having their usual meaning. Ex 3.7.5. Apply variational principle to find the equation of motion of a onedimensional harmonic oscillator. Ex 3.7.6. Show that the path followed by a particle in sliding from the another in the absence of friction in the shortest time as a cycloid.
po~nt
to
Ex 3.7.7. Use variational principle to prove that the shortest distance between two points in a plane is a straight line. Ex 3.7.8. Minimize the cost integral I = ~ condition x(O)
2
J i; 2dt
o
where x - x(t) satisfies the
= 1, ±(O) = 1, x(2) = 0, ±(2) = O.
Ex 3.7.9. A particle of mass m falls in a given distance Zo in time to = the distance travelled in time t is given by z = at +
bt2 ,
[f¥ and
where the constants a, bare
to
such that the time to is always the same. Show that values of the coefficients only when a = 0 and b = ~.
J Ldt is an extremum for real
o
Hints: The Lagrangian is L = !m1?+mgy. The Lagrange's equation gives, y-g Using y
= at +
get a = 0, as
bt2 ,
we get b =
[f¥ =1= o.
= O.
~. Put to = [f¥ in the equation Yo = ato + bt~ and
"This page is Intentionally Left Blank"
Chapter 4
Lagrangian Mechanics A close inspection of Newton's second law of motion reveals that the number of available equations are fewer than the number of unknpwns, the constraint forces being not known beforehand. Here we are going to set up Lagrange's equations of motion, a great advantage of which is that the number of unknowns equals the number of degrees of freedom. This approach opens up a new procedure of handling particle dynamics; the main difference with its Newtonian counterpart lies in the fact that the energies of the system are addressed to rather than the forces themselves.
4.1
Lagrange's equation of motion
Here, we are to derive the Lagrange's equation of motion for both conservative and non-conservative systems from D'Alembert's principle. Consider a simple dynamical system consisting of n-particles. The co-ordinate transformation equations in terms of generalised co-ordinates is 7i = 7 i (q1, q2, .. . , qn, t); i = 1,2, ... , N. Now we consider an infinitesimal virtual displacement 157 1, at a particular instant t. This displacement 157t can be connected with t5qi as 15 ~ ri
a~
n
a~
a~
n
ri ri '"" ri = '"" ~ T t5 qj + {itt5t = ~ T t5 qj; q]
j=l
(4.1)
q]
.1=1
a;;;.
as t5t = 0, for virtual displacement. Now if the 'i th particle is acted upon by . the force F i and (- T1) is the reversed effective force, then from the D'Alembert's principle, we have ~
..
.
~.
L(Fi-Pi).t5ri=0;p=mi7i i
a7 r ui q '- 0 aq]
]-
(4.2) j
]
77
CHAPTER 4. LAGRANGIAN MECHANICS
78
where QJ
= 2:= Ft. ~~'
are the components of generalised force. Now we wish to
t
express the
2:= 2:= Itt' i
~~!5qj in terms of virtual displacements of the generalised
J
co-ordinates. Now,
since -; i is assumed to have continuous second order derivatives. Further
This shows that the order of differentiation with respect to t and qj can be interchanged. This result is called interchange of d and 5. Therefore differentiating the transformation equation we get
(4.4)
This important result is called cancellation of dots and are widely used in Lagrangian method. Hence using (4.3), from equation (4.4) we get,
4.1. LAGRANGE'S EQUATION OF MOTION where T =
2: ~miv;
79
represents the kinetic energy of the dynamical system. Thus
t
the D'Alemberts principle (4.2) becomes
d aT aT "Q.8q:" [ - ( - . )--]8q-=O ~ J J ~& aq aq- J )
)
)
)
d aT aT or, "[-d (-a' ) - -a - QJ]8qj ~ t q' q-J j J
= O.
(4.5)
Due to constraints, 8r'i were not all independent. Since the constraints are assumed to be holonomic, the virtual displacements 8qj are all independent of each other and hence to satisfy the equation (4.5), the coefficient of each displacement 8qj should separately vanish, i.e.,
d aT aT -d (-a' ) - -a t q) qj
= Qj;
j
= 1,2, ... , n.
(4.6)
These n linear, second order differential equations are called Lagrange's equations for a holonomic system and are valid for both conservative and nonconservative forces (except sliding, friction, etc). The general solution of the ODE (4.6) involves 2n arbitrary constants of integration. If we know the state of the dynamical system (generalised co-ordinates and velocities) at some instant of time, then the 2n constants can be determined, and hence the motion of the dynamical system is specified for all time. These equations are sometimes written as
d aT aT dt(aq)-2 =Qj; j=1,2, ... ,n aqJ
(4.7)
which is known as Nielsen form of the Lagrange equation. From the above derivations, we have the following discussions (i) We have derived Lagrange's equation via D'Alembert's principle in which we have explicitly used Newton's laws of motion. So they do not represent a new physical theory but an equivalent way of expressing the same law of motion. Also both the laws are valid only in inertial frames. (ii) As mentioned earlier, the Lagrangian formulation of mechanics is not a new theory, it is only alternative and equivalent formulation. The Newtonian and Lagrangian equations of motion lead to ordinary second-order differential equations of motion of the system which, when solved, describe the nature of motion. However, the equations are obtained from different considerations that cause difference in the approach to the two formulations. (iii) In the Newtonian approach, we are concerned with the applied forces acting on a system, the result of which to accelerate the system. The forces are due to external agencies and they represent the cause of the acceleration produced in the body. The resulting accelerated motion is the effect of the force.
CHAPTER 4. LAGRANGIAN MECHANICS
80
(iv) In the Lagrangian approach, we consider the scalars KE and PE, and so of the system and so the concept of the vector quantities like, force, momentum, torque, acceleration, etc., do not enter into the discussion directly. Scalars being far more easier to deal with Lagrange's method greatly simplifies the problem compared to the Newtonian formulation which involve vector quantities. These are the basic differences between the two formulations.
Deduction 4.1.1. Conservative System: Now we fit this problem to conservative system so that above set of equations may be identified as Lagrange's equations of --; motion. For a conservative system, forces F i are derivable from scalar potential function V, of position only but not generalised velocities, i.e.,
Hence equation (4.6), we get d ( aT ) aT '-- av dt aq] - aqj - - aqj
a
d aT or, - ( - . ) - - ( T - V) dt aq] aqj
=0
d [ !-lf)• ( T - V )] - -a f) ( T - V ) . of . or, -d = 0;. V IS not a functIOn qj. t uq] qj d DL DL or, dt [aq) 1- 8qj = 0
(4.8)
where L = T - V is known as Lagrangian for conservative system. The equation (4.8) is known as Lagrange's equation of motion for conservative, holonomic system.
Ex 4.1.1. Simple pendulum in uniform gravitational field Consider a point mass m attached to an extensible string of length I and suspended from S. Let m be displaced to the position P, from the mean position 0 of the point mass, so that the string makes an angle () with the vertical. Let 8 be the angular displacement from the vertical SO at any time t. The dynamical system consists of a single particle of mass m and generalised co-ordinate 8. The kinetic and potential energy of the particle at P is given by
T
= ~m1202
and V
=
-mg(l - I cos 8)
where the point 0 is taken as zero level of the potential energy. The given system is conservative and hence we have L = T - V = !m1202 + mgl cos 8. Hence the
4.1. LAGRANGE'S EQUATION OF MOTION
81
Lagrange's equation of motion gives d aL dt
aL _ 0
-(-) - -
ae
ae -
d 2' (ml e) + mgl sin e = 0 =? dt·
=? -
e.. + -9 sin e = 0 l
which is the required equation of motion of the simple pendulum. When () is very small, then sin () ~ () and jj = i.e., the motion executed by the simple pendulum
-1 (),
is SHM with the time period T
= 27r
If. Note that while ~; stands for the angular
momentum of the mass about the point of support, ~i represents the torque. Ex 4.1.2. Double pendulum in uniform gravitational field The double pendulum in an uniform gravitational field consists of two masses ml and m2 at ends of two weightless rods of lengths hand 12, one of them is fixed to a rigid support. Since the pendulum is constrained to move in the vertical plane, it has two degrees of freedom. The generalised co-ordinates selected are the angles ()I, ()2 made by the two rods respectively with the vertical. Taking 0 as the origin, let co-ordinates of PI and P2 be (Xl, YI) and (X2, Y2) respectively. Now, Xl = h sinfh; X2 The
J( E(T)
TI VI T2
=h
sin (h
YI
+ 12 sin e2;
= h cosfh
Y2 = h cos ()l
+ 12 cos ()2.
and P E(V) for the masses ml and m2 are given by
1 2 2 1 . 2 . 2 1 2 '2 =2 -ml(xi +:z.h) = -ml[(hcoS()le l ) + (-llsinele l ) ] = --mlll()l 2 2 = -mlgh cos ()l 122 . '2 . '2 = 2m2(X2 + ]12) = (h cos ()l()l + l2 C0S ()2()2) + (-h sin ele l - l2 sin ()2()2) 1
2 '2
2 '2
= 2m2[lle1 + l2 e2 + 2hl2 COS(()1 V2 = -m2g[h cos e l
. . ()2)e1()2]
+ l2 cos e2]
Therefore, the Lagrangian (L) is given by L
= (TI + T 2) - (Vi + V2) 1
2 '2
1
2 '2
= 2(ml + m2)lle l + 2m2l2()2 + m2hl2 cos(el + (ml + m2)gh COS()1 + m2gl2 COS()2.
. . - ()2)e1()2
The equations of motion are derived from the Lagrange's equations as k
= 1,2
82
CHAPTER 4. LAGRANGIAN MECHANICS
= m2 = m
Particularly, when ml ..
and h
..
= 12 = 1,
the equations becomes g
'2
2e 1 + e2cos(e1 - e2) + e2sin(e 1 - e2) + 2 sine 1 = 0 T ..
..
2e2+ el COS(el - e2) -
g
'2
e 1 sin(e1
-
e2) + Tsine2 = o.
Further, for small oscillation, sin e ~ e and cos e ~ 1. Neglecting the second and higher order terms, the equations become ..
..
g
....
g
2e 1 + e2+ 2Te1 = 0; e1 + e2 + Te2 = o. Deduction 4.1.2. Cartesian Co-ordinate System: Let P(x, y, z) be the position of a particle of mass m at any time t in a space curve r. Let (Fx, F y , F z ) be the force components acting on the particle in the increasing direction of x, y and z respectively. Here the dynamical system consists of a single particle of mass m and generalised co-ordinates of the system are x, y, z. If Qx, Qy, Qz be the generalised force components, then
= QxJx + QyJy + QzJz = FxJx + FyJy + FzJz Qr. = F Qy = Fy; Qz = F z .
JW :::::}
T ;
At any time t, the Kinetic Energy T of the system is given by
= ~m[x2 + ii + i 2] 2 aT aT aT a T . aT -8 = -8 = -8 = 0 and -8' = mx; ay = x y z x T
:::::}
. -=mz aT .
my;
ai
Now, using the Lagrange's equations of motion we have
These are the equations of motion of a particle of mass m in cartesian co-ordinates from Lagrange's equation. We shall now consider some simple applications of the above procedure. Ex 4.1.3. Atwood's Machine
Atwood's machine is a simple, holonomic, conservative, mechanical system with one degree of freedom if the pulley is frictionless, constraints and scleronomous. It consists of two masses ml and m2 tied together by means of light inextensible string
4.1. LAGRANGE'S EQUATION OF MOTION
83
of constant length I. The string passes round a light frictionless pulley of radius r and two masses hang on the two sides of the pulley. Here the tension in the string and the reaction of the pulley are the forces of constraint and the two applied forces are mIg and m2g. We see from the figure that, only variable x is independent, since length I of the string is constant. The kinetic energy (T) and the potential energy (V) of the system are given by T = 2mIx 1 ·2
and V
1 (i - x. )2 = 2 1 (mi + m2 ) x·2 ; + 2m2
= -mIYX -
as i = 0
m2g(l- x)
where the point 0 is considered as the zero level of the potential energy. Hence, the
Figure 4.1: Atwood's Machine Lagrangian function (L) is given by
For x as the generalised co-ordinate, the Lagrange equation of motion of the system may be written as
which gives the acceleration of the motion of the system. Solving, we get, x = Xo
1 2
+ Vo t + - (
mi - m2 mi +m2
)gt
2
which gives the displacement of mi towards 0 at any time t. Thus if mi > m2, mi will move downward with acceleration m 1 +-m 2 g. Obviously, mi = m2 indicates ml m2 equilibrium and then either mi and m2 remain at rest or move with uniform speed in opposite direction.
84
CHAPTER 4. LAGRANGIAN MECHANICS
.Ex 4.1.4. Linear Harmonic Oscillator A body of mass m is connected to a spring of negligible mass and rests on a frictionless horizontal floor. The spring has a spring constant k. When the body is displaced from its normal equilibrium position by x, the potentIal energy of the spring becomes
J X
V=-
J
--+ --+ F.dr
=-
1 2 (-kx)dx=2kx.
o
The mass m being springless, i.e., rigid, it does not contribute to V. The spring being assumed massless it also does not contribute to the KE of the system. Therefore, KE of the system is the KE of the body which is given by T = !mx 2 . For x as the generalised co-ordinate, the Lagrangian is given by
1 2
1 2 2 8L aL. -;:} = -kx; -8' = mx. .2
L = T - V = -mx - -kx =}
x
IIX
Therefore, Lagrange's equation for one dimensional oscillator is
mx+kx
= O.
Thus the motion executed by the mass m attached to the spring is S.H.M. and the period of oscillatory motion is T = 211" JIfI. Ex 4.1.5. A particle of mass m is projected with initial velocity u at an angle Q with the horizontal. the resistance of the air may be neglected. Use Lagrange's equations to describe the motion of pmjectile. Let a particle of mass AI is projected from the fixed point 0, with a velocityu at an angle Q with the horizonal. Let P(x, y) be the position of the particle at any time t. Let us consider the co-ordinates x and y as the generalised co-ordinates. At P. the kinetic and potential energies are given by SOLUTION:
T
= ~m(i:2 + Ii); V = mgy =} L = ~m(i:2 + Ii) - mgy.
Since x and yare generalised co-ordinates, the Lagrange's equations of motion for the projectile are
~(f)L) _ 8 Lx = 0', dt ax
=}
a
=}
aL
- ay = 0;
= 0; dd (my) + mg = 0 t x = 0; jj + g = O.
d (mx) - 0
dZ
d (aL) dt ay
4.1. LAGRANGE'S EQUATION OF MOTION
85
Now the initial conditions are t = 0; x = 0; i: = ucosa; Applying the initial conditions, the solution is given by X
= ( U cos a ) t;
=
y
.) ( 7L sm a t -
12
- gt =? Y = xtana -
2
y
=
0;
iJ = usina.
9 2 x. 2 2 2u cos a
e,
Deduction 4.1.3. Spherical Polar Co-ordinate System: Let P(r, ¢) be the position of a particle at any time t of mass m in spherical polar co-ordinate System (r, e, ¢). Let F r , Fe, F be the components of force acting on the particle in the increasing direction of r, e and ¢ respectively. Here the dynamical system consists of a single particle of mass m and generalised co-ordinates of the system are r, e, ¢. 1£ Q.r, Qe, Q be the generalised force components, then
8W = Qr 8r + Qe8e + Q¢8¢ and 8W = Frdr + rFede + rsineF¢d¢ =? Qr = Fr; Qe = rFe; Q¢ = rsineF¢. At any time t, the Kinetic Energy T of the system is given by T =
~rn['r2 + '1'2(;2 + T2¢2 sin 2 e].
Now, using the Lagrange's equations of motion we have
d [aa~] - aaT r dt '1' d aT
-d [-.]t
ae
d aT
-d [-.]t a¢
= Qr
aT -a e=
Qe
+ rJ} sin2 e) = Fr
=?
dd (mr) - m(rfp t
=?
-d (mr e) - mr ¢ sin e cos e = rFe t
d
2·
2 ·2
aT d 2 2 . -a = Q¢ =? -d (mr sin e) = rF¢sme ¢ t
or, m(f - riP - r¢2 sin2 e)
= Fr
m[! (r 2iJ - r2¢2 sin e cos e)] = rFe
d
.
2
m dt (r¢sin e)
= rF¢ sine.
These are the equations of motion of a particle of mass m in spherical polar coordinates from Lagrange's equation. Ex 4.1.6. Spherical Pendulum In a spherical pendulum, bob moves through the entire space about the point of suspension. The bob of such pendulum moves on the surface of the sphere whose radius is equal to the length of the pendulum. The position of the bob is located The distance x of the bob from the origin will be the radius of the sphere on which the particle or bob moves. The force in this case will not be central, but is constant in the vertical direction. The kinetic energy of the pendulum is T = ~mI2(iJ2+sin2 e¢2)
CHAPTER 4. LAGRANGIAN MECHANICS
86 z
~------y
p
:Figure 4.2: Spherical Pendulum and by spherical co-ordinate 9, e/>. the potential energy of the bob due to gravity relative to the horizontal plane z = 0 is V = mgl cos 9. Hence the Lagrangian function L for the dynamical system is given by
L= T- V =
aL aiJ
= mr
~ml2(92 + sin2 (J~2) -
mgl cos
e
2 '2 ej aL ae == mr sin ecos ee/> + mgr sin e
2'
a~ = mr2 sin2 e~j ae/>
a~ = o. ae/>
The equations of motion corresponding to the co-ordinates 9 and e/> are
d aL aL 2" 2 '2 dt ( ao ) - a(} = 0 => mr e - mr sin 9 cos 9e/> L d (a~) _ aa = 0 => mr2dd (sin2 e~) = O. dt ae/> e/> t
mgr sin 9 = 0
We shall discuss the nature of the solution in the next chapter.
Deduction 4.1.4. Non-Conservative System: If potentials are velocity dependent, called generalised potentials, then though the system is not conservative, yet the above form of Lagrange's equation can be retained provided Qj, the components of generalised force, are obtained from a function U(qj, qj) such that
au
Qj = - aqj
d au
(4.9)
+ dt ( aqj ).
Hence from equation (4.6) we get
d aT aT au d au dt ( aq'J) - aq·.1 = - aqJ + dt ( aq'J) or,
~[~(T dt aqj d aL
or, -d (-a' ) t qj
U)]-
aL
-a qJ
~(T aqj
= 0
U) = 0 (4.10)
4.1. LAGRANGE'S EQUATION OF MOTION
87
which is the same form of the equation (4.8) in which L = T - U is the Lagrangian. U is known as the generalised velocity dependent potential and we across such a potential when we consider the electromagnetic forces on moving charges.
Ex 4.1.7. Velocity-Dependent Potential of Electromagnetic Field Here we derive the Lagrange's equation of motion for the non-conservative system. Consider a charged particle of mass m and charge q is moving in electromagnetic field. The Maxwell's equations are --+ a B --+ V'xE+-=O' --+
at
--+
--+
--+
--+
' a--+ E
V'.E = cO
--+--+
= J-loj + J-lOC07ft; A
V' x B
--+
P
--+--+
--+
V'.B
= O.
--+
The quantity cO E = D and B = J-loH are respectively called electric displacement --+ and the magnetic induction respectively. From this equation, we observe that V' x --+ --+ E I- 0, so E can not be expressed as a gradient of a scalar potential function. But, --+ --+ --+ --+ --+ --+ --+ as V'. B = 0, B can be expressed as B = V' x A, where A is called a magnetic vector potential. --+
--+
V' x E
+
a--+ B at
--+
=
--+
--+
+
=> V' x [E
--+
--+
0 => V' x E
a--+A
at]
=
a
--+
+ at (V'
--+
--+
0 => E
--+
x A)
a--+A
=
+ 7ft = -
--+
0
--+
V'
where is the scalar potential functions. Now the Lorentz electromagnetic field --+ force F on a charged particle (in cgs Gaussian system) having charge q and moving in an electromagnetic field is given by --+ --+ 1 --+ --+ F = q[E + -(V x B)]; c = Velocity of light
c
a--+
1 A 1 --+ --+ --+ or, F = q [- V' - ~ at - ~ (V x V' x A)]. --+
--+
--+
--+
in which the magnetic force %(V x B) is gyroscopic in nature. The power associated --+
--+ --+
with these forces F is zero, i.e., F. V = O. To write more convenient form, let us write the x component of various terms on right hand side. Now,
(V x V x
A)x
= Vy(aA y _ aAx) _ Vz(a.Ax _ aAz) =
ax Vx aA x ax
a
az B.T + 11: aAy + Vz aA z _ Vx aA x _ 11: aAy _ v~ aA x y ax ax ax y ay ~ az ay
--+ --+
= ax ( V . A) Al
dAx _ aA x so, dt - at
+
v
V. V' Ax
aAx x ax
dAx _ aA x at
=> dt
--+ --+
+
11: aAx y ay
= --+v ~A .v
x
+
v z
aA x az
CHAPTER 4. LAGRANGIAN MECHANICS
88
Now, as the velocity V does not contain x explicitly, we have ~ = o. Also as %~7: = 0, we have (V.A) = Ax. Thus, consider the generalised co-ordinate x co~ponents of the va;ious terms, the Lorentz's force is given by
dt
PI;
a<1> 1 aAx = q[- ax - 7;Tt a<1>
1 -.
+ 7;(V
1 aAx
-; x
v
-;
x A)x]
1 a -; -;
1-; -;
1-;-;
= q[- ax - 7;Tt + 7; v.v Ax + 7;ax(V.A) - 7; V. 6A x a
1 dAx 7;Tt 1-;-; 1d = q [- ax (<1> - 7; V . A) - 7; dt {8V (V. A ) }] x 1-;-; 1d 1-;-; = q[--, (<1> - - V.A) - --{-(<1> - - V.A)}]; as <1> ax c edt avx c au d au = - ax + dt(DV)· 1-; -;
= q[- ax (<1> - 7; V .A) -
a
a -;-; .
a
a
= <1>(x,y,z,t)
-;
where the generalised potential energy is U = q<1> - ~ A .11. This shows that, when a charged particle moves in electromagnetic field, the force acting on the particle is derivable from a potential dependent on velocity of the particle. Therefore the force in generalised co-ordinates is given by
Such a particle system in electromagnetic field is called non-conservative one and the equations of motions in generalised co-ordinates is given by d DT
dt(~) -
a8TqJ =Qj
uqJ d 8T aT au or, dt ( 8qj) - 8qj = - 8qj
.:!.- (8(T -
or
, dt
8q]
d au
+ dt ( 8qj )
U)) _ 8(T - U) 8qJ
d 8L t qj
=0
8L
or'-d(-a.)--a =0. qj
Thus the Lagrangian for the charged particle moving in an Electromagnetic field is, L
=T
- U
=T
- q<1>
q-; -;
+ -c A . v
which shows that the form of Lagrange's equation remains same even if the system is non-conservative, where U is the generalised velocity and time dependent potential.
Deduction 4.1.5. Explicit form of Lagrange Equation :We shall now show how the second derivatives of the co-ordinates with respect to the time can be
4.2.
DISSIPATIVE SYSTEM
89
found explicitly from Lagrange equation. Let the configuration of the dynamical system considered be specified by co-ordinates qI, q2, . .. , qn, we shall suppose that the configuration can be completely specified in terms of these co-ordinates alone, without t, so that the kinetic energy of the system is a homogeneous quadratic function of IiI, 1i2, .. , , tin. Also from the previous section, we see that, this is always the case when the constraints are independent of time, but not in general when the constraints have forced motions. Suppose the Kinetic Energy is
where akl = alk, and where the coefficients akl are known functions of q1. q2,· .. , qn' The Lagrangian equations of motion for the system are
d aT dt
aT (-a') - -a. = Qj; j qJ
= 1,2, ... ,n.
qJ
j = 1,2, ... , n
= 1,2, ... , n
j
where
e
(4.11)
~) is the Christoffel's symbol of second kind and is defined by the relation
J
•
These equations, being linear in the accelerations, can be solved for the quantities = laijl be the determinant and Ars be the minor of ars in 6.. Multiply the n equations of (4.11) by Ali, A 2i , ... , Ani respectively, and add them we get iis. Let 6.
These n equations of iiI, ii2, . .. , iin which are give explicitly as a functions of (iI, tI2, ... , tin; ql, q2, ... ,qn can be regarded as explicit form of Lagrange equation of motion.
4.2
Dissipative system
When the forces acting on the dynamical system are derivable from a potential, Lagrange's equation of motion is given by (4.6), where L contains only conservative forces and Qj includes the forces that are not derivable from a potential. The forces
CHAPTER 4. LAGRANGIAN MECHANICS
90
Qj are dissipative, i.e .. to energy of the system is not conserved. This type of force occurs in the situation when frictional forces are present, and that the frictional force is proportional to the velocity of the particle. Thus the x component of such force may be written as
where kx is the :1' component of the frictional force per unit velocity. Frictional forces of this type may be derived in terms of a function'S, known as Rayleigh's dissipative function, and can be written as
which covers all the particles of the dynamical system. Since the frictional forces ---> -+ -+ -+ F f can be derived in terms of the function'S, we have, F f = \1 v'S where \1 v = ~ -aa + 3aaVv + kaaVz is t.he velocit.y• differential operator. For physical int.erpretation lJ r of Rayleign!s dissipative function'S, let us calculate the work done by the system against friction, which is given by dWf
-+ -+-+ = - ---> F f.d r = - F f. v dt = [kxv; + kyv~ + kzv~ldt
dW:r _ 2 2 2 _ => dt - kxvx + kyv y + kzv z - 2'S. Thus the rate of energy of dissipation due to friction is equal to twice Rayleigh's dissipation function. The component of generalised force Qj arising as a result of force of friction is given by n
~-+
a-+r i
n
~-+
a-+r i
Qj = Lt F if'--a-: = - Lt \1 v'S'--a-: ! % ! qJ _
~ -+
0.<
a-;j:>! _
a'S
- - ~ V' v'S· aqj - - aqj .
Hence the Lagrange's equation of motion becomes
Therefore, to derive the equations of motion, if the frictional forces are acting on the syst.em, we lllUst be specified two scalar functions - the Lagrangian L and the Rayleigh's dissipative function'S. Ex 4.2.1. Viscous force F retarding a sphere in a fluid of co-efficient of viscosity 'TJ -+ is given by Stokes law F = -6'Jr'f]'r 11, where 11 is the terminal velocity of the sphere. Find thf Rayleigh dissipation function.
4.3. NON-HOLONOMIC SYSTEM
91
SOLUTION: Let z be measured from the ground. Let Fz be the dissipative force, then Fz = -kzv z ; kz = 67rT7J. Therefore, the Rayleigh's dissipative function;} is given by
1
2
;} = -'2kz1Jz; k z =
67rT7J
where V z = i is the velocity along z axis. Let p be the density of the sphere and M be the mass of the sphere, then M = ~ 7rT3 p. The potential energy (V) is V = - M 9 z, and so the Lagrangian (L) is L = ~1\1v~ - Algz. Thus the lagrange's equation of motion is given by
-d [oL - - -oL J + -a;} = 0 ::::} Mooz dt
4.3
oi
8z
oi
67r7JTZ.
+ M 9 = 0.
Non-Holonomic System
Let q1. q2, ... , qn be the generalised co-ordinates of a non-holonomic bilateral dynamical systems having N number of particles and k be the number of holonomic constraints. When the constraints acting on the system are non-holonomic, the generalised co-ordinates are not independent of each other. Let for such dynamical system have k' non-holonomic constraint relations, with degrees of freedom = n, so the number of quasi-generalised co-ordinates required is n+k' where n = 3N -~~-k'. Let the differentials dqi of the generalised co-ordinates are connected by m( < 11.) relations of constraints as n
I: alkdqk + alt dt = OJ
I
= 1,2, ... ,m.
(4.13)
k=l
where alk's are functions of qk's and t. These equations are the equations of the constraints for a non-holonomic system. The virtual displacements tSqk at any instant t satisfy the following equations of constraints n
I: alk
tSqk = OJ I = 1,2, ... ,m
(4.14)
k=l
as Hamilton's principle does not involve variation of time. These equations can be used to reduce the number of virtual displacements to only that of the independent ones. In the case of non-holonomic system, the virtual displacements tSqk's are not all independent, but they are connected by m relations (4.14), so that (n - m) of 6qk's are independent. To eliminate these extra dependent virtual displacements we shall use Lagrange's method of undetermined multipliers. Multiply (4.14)by m constants ),1, ),2,· ·.,),m respectively, sum over I and finally integrate it we get (4.15)
CHAPTER 4. LAGRANGIAN MECHANICS
92
where the unknowns Ak'S are called Lagrange multipliers which may be, in general, functions of time. Now the Hamilton's principle for conservative system we get
L
=T - V =
Lagrangian
(4.16)
All the qk's are not independent but are connected by m equations (4.14). However, the first (n - m) of these co-ordinates may be chosen independently, the last m ones being fixed by eqnations (4.14). Since AI'S can be chosen arbitrary so let us choose Ak'S (k = 1,2, ... ,m) such that 8L
d
8L
'Tn
-8qk - -(-. ) + '" Al dt 8qk ~
alk
= 0;
k
= n-m+ 1, ... ,n
(4.17)
1=1
which are the nature of the equations of motion for the last m of the qk variables. Having found these Ak'S from equation (4.17), we can write equation (4.16) as
(4.18)
where £5'11. £5'12, ... ,£5qn-m are independent. Thus the expression in the third bracket must vanish, therefore
a8Lqk - dtd (88Lqk. ) + L Al 'TIL
alk
= 0;
k
= 1,2, ... , n
- m
E4.19)
1=1
Combination of equations (4.18),(4.19) are then finally the complete set of Lagrange's equations for non-holonomic system. Although there are (n - m) + m = n equations in all, we have (n + m) unknowns, namely n generalised co-ordinates qk's and m constants AI. Thus the additional equations are given by the constraints (4.13)as n
L alk
qk
+ alt = 0;
l
= 1,2, ... , m.
(4.20)
k=l
Therefore equations (4.19) and (4.20)together constitute (n + m) equations for (n + m) unknowns. Physical significance of A :Suppose, we want to remove the constraints on the system, but instead. applied external forces Q~ in such a manner as to keep the motion of the system unchanged. The equations of motion would likewise remain the
4.3. NON-HOLONOMIC SYSTEM
93
same. Clearly, these extra applied forces must be equal to the forces of constraint, for they are the forces applied to the system so as to satisfy the condition of constraint. Under the influence of these forces Q~ the equations of motion are
But these must be identical with the equations
d 8L 8L -d (-a' ) - -a t
qk
Hence we can identify
qk
=
L
),lalk;
l
k
= 1,2, ... , n.
L ),lalk with Q~, the generalised forces of the constraints.
In
l
this type of problems, we really do not eliminate the force of constraints from the formulation and they are obtained as apart of the answer. Result 4.3.1. The above equations(4.19) include holonomic constraints as a special case. Let the generalised co-ordinates ql, q2, ... ,qn be connected by the m relations fl(qI, q2,"" qn, t) = 0;
1 = 1,2, ... ,m
where m < n. Differentiating, we get 1 = 1,2, ... ,m
which is of the form of equation (4.20). Thus the coefficients are given by alk = ~t~; au = 1 • Thus the Lagrange's multiplier method can also be used for holonomic constraints when (i) it is inconvenient to reduce all the generalised co-ordinates qk's are independent or (ii) we might wish to obtain the forces of constraints.
1t
Ex 4.3.1. Particle In a Sphere Let a particle of mass m, move on the surface of a frictionless surface of a smooth sphere of radius l, under the action of gravity. Let the origin of the co-ordinates be at the centre of the sphere and let the z axis be vertically upwards. Here co-ordinates of any point on the sphere will be (r, 0, ¢) of which r is constant as the particle is constrained to move on the surface of the sphere of radius r = l. The equation of constraint is r - I = 0, where r is the radial distance of the particle. Thus the coefficient in the equation of constraint are, a r = 1, ao = 0 = ar/J. Let us suppose that the particle is initially at rest and let it slide down along the surface. The particle will obviously move in a vertical plane which we shall take for convenience ¢ = O. The Lagrangian for the particle is,
L=
~[r2 + r 2lP]- mgz = ~12(P -
mg1' cos O.
CHAPTER 4. LAGRANGIAN MECHANICS
94 Using the results r
= 1, r = f = 0,
-mliP
the equations of motion are
+ mg cos (J =
Aj ml 2 (j - mgl sin (J = O.
The undetermined multiplier A is dependent on (J, in general. Also, differentiating the first equation containing A with respect to time t, and using (j = Tsin (J from the second equation, we get, ... . -2ml(}8 - mg sin (J(J
'* -3mgsin8 =
dA' d(Je
=
dA· d88
'* A(e) =
3mg cos (J
+c
where c is the constant of integration, determined by the condition 8 = 0, A = mg, the force constraint at the top of the sphere. Thus A(8)
= 3mgcos 8 -
2mg ~ 0
as the particle will move on the surface as long as the surface pushes it outward. The equality sign holds, when cos(Jc = ~, i.e., at an angle 8c = cos- 1 ~ the particle flies off the surface.
Ex 4.3.2. Cylinder Rolling on Inclined Plane Consider a cylinder rolling without slipping on a inclined plane of length l, which makes an angle 0: with the horizontal. We wish to find the acceleration of the cylinder and the frictional force of constraint. Let the initial position of the cylinder of radius
o x
Q
B~------'------'A
'Figure 4.3: Cylinder rolling, without slipping r be 0 and roll down the plane along the line of greatest slope without slipping. Let P he the position when it has moved through a distance x down the inclined plane. Let C be the centre of the cylinder and LQCP = (J. Here the generalised co-ordinates are x and (J and they are connected by a relation x = r(J. Then, the equation of the constraint is rd() = dx, as the friction is sufficient to prevent the slipping. The constraint force is the force of friction. Since there is only one equation of constraint, only one Lagrangian multiplier ,is required. Now, the coefficients in the equation of the constraints are aT = Tj ax = -1. Due to the translation and rotational motion, the kinetic energy is made up by the two components. Thus the
4.4. CYCLIC CO-ORDINATES
95
kinetic and potential energy of the cylinder is 1
T= -mi:
2
1'2 1 2 1 2 '2 + -I() = -mi: + -mr () 2 2 4
2 V = mgh = mg(l- x) sina =?
~mi:2 + lmr 2(P -
L= T - V =
mg(l- x) sina.
The Lagrange's equation for () is given by
~(a~) _ aL = a A
a() a() ~mr20 - rA =
()
dt
=?
=?
2 I" "2mx
-
O· as rO = x ' Imx ". /\'=0 =?' /\ = "2
The Lagrange's equation for x is given by
d aL
aL
*
mgSina~mx = 0
dt (ai:) - ax = axA * mx - mg sin a + A = 0
mx "
=? x
2.
= 3gs111a =?
()"
2 . = 3rgsllla.
Hence the frictional force of constraint A is img sin a.
4.4
Cyclic Co-ordinates
If the Lagrangian L of a dynamical system does not contain a co-ordinate explicitly, then that co-ordinate is called an ignorable or cyclic co-ordinate. Thus, some of the co-ordinates ql, q2, ... , qk are not exactly contained in L, although the corresponding velocities ql, q2, ... , qk are not so centained. The Lagrangian L is written as a function of qj and q]. If. anyone co-ordinate say qj {j~. an ignorable ?o-~~dina1;e, then ~ = 0, where may not~e zero. We have seen that the motIOn of a conservative, holonomic dynamical system with n degrees of freedom, for which the co-ordinates are ql, q2, .. . , qn and the kinetic potential L, is determined by the differential equations
gt
d 8L 8L -d (-8' ) - -8 = 0 t
qj
qj
=?
d 8L 8L -d (-8' ) = 0 =? -8' t
qj
qj
= Pj =
Constant = /3j.
This Pl is known as the conjugate momentum for q]. So, th«:: Lagrangian of a dynamical system does not contain a co-ordinate qj, then the corresponding conjugate momentum is conserved. The presence of ignorable co-ordinates is the most frequently occurring reason for the solubility of particular problems by quadratures.
96
4.4.1
CHAPTER 4. LAGRANGIAN MECHANICS
Systems with ignorable
co~ordinates
Let a dynamical system be represented by n generalised co-ordinates ql, q2, ... , qn' Suppose ql, q2, .. ·, qs(s < n) be cyclic. We define a function R, the Routhian as s
R
= R(ql, ... ,qn, Pl,· .. , Ps, tis+1, ... ,tin, t) = LPitii - L. i=1
n 8L n 8L 8L Now, dR= Lqidpi + LPid,qi - L ~dqi - L ~dqi - Ft dt s
s
~1
=1.
~1
~
~1
~
n 8L· 8L n = Ltitdpi - L ~dtii - a;dt - LPidqi t=1 t=s+1 ~ i=1 n 8R n 8R s 8R 8R dt . Also, dR = L ~dqi + L ~dtii + L a.dpi + t=1 qt t=s+1 qt i=1 Pt s
m
(4.21) (4.22)
The infinitesimal quantities occurring on the above expressions are arbitrary and independent. Comparing, (4.21) and (4.22), we have
8R . 8R . i = 1,2, ... , s -8- = -Pi; -8- = q.i; qi Pi 8R 8L 8R 8L - - ; - - - - 8 ; i=s+1,s+2, ... ,n. 8tii 8qi qi
(4.23) (4.24)
Now the Lagrange's equation of motion gives,
(4.25) in which the Routhian function has replaced the Lagrangian function. Now, the Routhian R is a function only of the variables qs+1, qs+2, .. . , qn; til, ti2, . .. , tin, and the constants {31, {32, ... ,{3s. Thus this is a new Lagrangian system of equations, which we can regard as defining a new dynamical problem with only (n - k) degrees of freedom, the new co-ordinates being qs+1, qs+2, ... , qn, and the new Lagrangian being R. When the variables qs+1, qs+2, ... , qn have been obtained in terms of t by solving this new dynamical problem, the remainder of the original co-ordinates namely ql, Q2, ... ,qs can be obtained from the equations qj
=-
J;~
dt; j
= 1,2, ... ,s.
Thus, a dynamical problem with n degrees of freedom, which has k ignorable coordinates, can be reduced to dynamical problem which has only (n - s) degrees of freedom. This process is called ignoration of co-ordinates.
4.4. CYCLIC CO-ORDINATES
97
Ex 4.4.1. Kepler's Problem for planetary motion: Let us now apply Routhian procedure of ignoration of co-ordinates to the Kepler's planetary problem. In plane polar co-ordinates (r, (J), the Lagrangian function L of such system is given by
where V (r) = - ~ is the potential. Here the Lagrangian L does not contain explicitly, so (J is ignorable co-ordinate and so Po
oL = mr 2'(J = constant = [( = -. of}
(J
say)
where [ is the constant magnitude of the angular momentum. Also, it follows that :it 2 0) = O. But we know that 2 0 is just the areal velocity- the area swept out by the radius vector per unit time. This says that, the well-known Kepler's second law of planetary motion: the radius vector sweeps out equal areas in equal times. We now set up the Routhian function as,
(!r
!r
r
Note that R contains rand
only. Now, the equation of motion is given by
.!!:..(oR) _ oR
=0
dt oqi Oqi d. '2 =? - (mr) - mr(J dt
d
=?
+ -r2k = 0 [2
-d [mr2 t
+ 2V + -2] =0 mr
which is the equation of a particle moving under central force. Also, mr2+2V + ~~2 = 2T + 2V, so that the constant of the above equation is nothing but the two times of total energy E. Its first integral gives the conservation of total energy E as 1,2
-mr
2
[2 e E + -+V = onstant = mr2
J r
=? t -
-
ro
dr [2(E - V - 6)]1/2 m
2mr
where ro is the initial value of 'r. This equation gives t as a function of l' and the constants E, [ and roo However, it may be inverted, at least formally, to give r as a
98
CHAPTER 4. LAGRANGIAN MECHANICS
function of t and the constants. Once the solution of r is obtained, the solution 0 follows immediately as t
T
o- 00 = 1 j'
=
dt
mr2
o
j' mr2 . /1. [E - V - L ] ldr
.
Ym
TO
2r1lT'<
U
ldu
J =- J =-
UO
V2m[E + ku -
l~~]
u
uo
~
+
/[2mE
Y!2
=> 0 - 0' = cos- 1 (
u -
. 12mE
Y!2
du m2k2 _ (u _ mk)2]
--y:r
= [cos- 1 (
u. /2mE
V !2
j2
j2
+ m2k2
--y:r
)]U uo
!!f# I ).
+ --y:r m 2k2
The quantity 0' can be determined using the initial condition. Thus we can identify four constants of integration I.E, ro, 00 . Also, the above equation can be written as U
=
mk
r
2mE
m 2k 2
+ ------z:r- + [ 4 cos( 0 -
l2/mk
, 0)
2El2 1 + mk2 cos(O - 0').
or'-r- = 1 +
Now, the general equation of a conic with one focus at the origin is ~ = 1 + ccos(O - 0'), where e is the eccentricity of the conic section. Comparing we get,
VI
e= + ~S· The nature of the orbit depends on the magnitude of e according to the following scheme: Values of E Values of e Nature of Orbit Hyperbola E>O e>1 Parabola e=1 E=O Ellipse E
Ex 4.4.2. In a dynamical system with two degrees of freedom, the kinetic and poten·2 tial energies are T = ~ [a;£q~ + q~] j V = c + dq~ where a, b, c, d are give constants. Find the complete solution. SOLUTION: Since The kinetic and potential energy do not contain the co-ordinate ql explicitly, so ql is an ignorable co-ordinate. The kinetic potential or the Lagrangian L of the system is given by
1
'2
ql 2 a+bq~
L= -[
+ q2.2] -
2.
c ~ dq2'
4.4. CYCLIC CO-ORDINATES
99
The integral corresponding to the ignorable co-ordinate q1 is given by
aL
{3,
where {3 is a constant, whose value is determined by the initial conditions of the motion. The Lagrangian of the new dynamical system obtained by ignoring the co-ordinate q1 is
which is the Routhian function and the problem is now reduced to the solution of the single equation d (aa~) dt q2 q2
'*
- aaR = 0 '* ih + (2d + b(32)q2 = 0 =
q2 A sin{(2d + b(32)1/2t + E},
where A and E are constants of integration, to be determined by the initial conditions of motion. This equation gives the required expression of the co-ordinate q2 in terms of time. Now the value of the co-ordinate ql then can be deducted from the equation ql
= {3 = {3 = {3
J + bq~)dt J+ J+ (a
b(A sin{ (2d + b(32)1/2t + d )2Jdt
[a
1 1 2bA2) - "2bA2 cos2{(2d + b(32)1/2t + €}Jdt
[(a
1
= (3[(a + 2 bA2 )t -
bA 2 4(2d + b(32)1/2 sin2{(2d + b(32)1/2t +
dJ
and so the dynamical problem is completely solved.
4.4.2
Separation of variables
A class of dynamical equations which are obviously soluble by quadratures is constituted by the equations of those systems for which the kinetic energy and the potential energy are of the form
= ~[V1(q1)
for a system having n degrees of freedom, where VI, V2,··., Vn ; WI, W2, .. . , Wn are arbitrary functions of their respective arguments, so that the kinetic potential L breaks up into sum of parts, each of which involves only one of the variables. Let the
100
CHAPTER 4. LAGRANGIAN MECHANICS
forces are conservative in nature. A great advantage with the problems of Liouville'R type is that these admit separation of variables and hence can be solved completely. For in this case the Lagrange's equation of motion are
:t[l)r(qr).Qr l=} 1)1'
=}
=}
J[
t=
.. (qr ) ·qr
~V~(qr).Q; = -w~(qr);
r
= 1,2, ... , n
+ 1, (q1' )'2 ·qr = -w'r (qr) 21)7'
~ V1' (qr ) q; + Wr (qr) = a r ;
vr(qr) ( )ll/2dqr 2er - 2wr qr
+ /31';
Integrating
r = 1,2, ... , n
(4.26)
where a r , /31' are new constants of integration. The equations (4.26) constitute the solution of the problem. Ex 4.4.3. In a dynamical system with two degrees of freedom the KE is T
=
'2
2(a~bq2) +
~qiqi, and the potential energy is V = c + dq2, where a, b, c, d are constants. Show that the values of q2 in terms of the time is given by the equation of the form (q2 - k)(q2 + 2k)2 = h(t - to?, where h, k and to are constants. SOLUTION:
The KE and PE for the dynamical system are T
=
'2
2(a~bq2)
+ ~q~qi,
and V = r:+ciq2' One can easily identify Vl(ql) = 2(a~bq2);'V2(q2) = q~;Wl(ql) = c; W2(q2) = dq2. Now from the equation ~vr(qr)q; + wr(qr) = a r for a = 2 gives
~V2 (q2)q~ + W2 (q2) = C2 =}
1 2.2 -2q2q2
+ dq2
.
= c2
JJfJ vI2d J(d -
=}
_1_
C2
2
q2
q2 dq2
v'2d
=}
=}
= =
..;=-2C-2----=2d......q-2 q2 -t + to
q2
z 2)dz
=t-
to;
dC2 -
q2
= Z2
c2 ~ 1 C2 3/2 d =}dVd-q2-"3[d-q2l =..j2(t-to) =}
C2
2C2
1
2
d2
d)[3J + "3q2l = -2(t - to) =} (q2 - k)(q2 + 2k)2 = h(t - to)2
(q2 -
2
where the constants k and h are respectively given by k
4.4.3
= cd
and h
=-
3f .
Liouville's class of Lagrangians
The Liouville's class of Lagrangians is the one in which the KE and PE are has the special form. A great advantage with the problem of Liouville's type is that these
4.4. CYCLIC CO-ORDINATES
101
admit separation of variables and hence can be solved completely. For a· class of dynamical systems, this idea is extended by Liouville, who shows that all dynamical problems for which the KE and PE can respectively be put in the forms
where U r , V r , Wr are functions of qr only, 1" = 1,2, ... , n for a dynamical system having n degrees of freedom. For this, let us take
where qi, q~, ... , q~ are new variables, we can replace all the functions VI (qI), V2 (q2), ... , Vn (qn) by unity; we shall suppose this done, so that the KE and PE take the form
T = !u[qr 2 where u
+ q§ + ... + q;];
V = ![WI(qI) + W2(q2) u
= UI (qI) + U2(q2) + ... + Un(qn).
~u[qr + q~ + ... + q~] =
+ ... + wn(qn)]
Using the integral of energy of the system,
h - V; h = constant,
the Lagrange's equation of motion for the co-ordinate qi is given by
where 11 is a constant of integration. We obtain similar equations for each of the co-ordinates (qI, q2, ... ,qn); the corresponding constants (tI, 12, ... , In) must satisfy the relation 11 + 12 + ... + ,n = 0 in virtue of the integral of the system.
!(qr +
Ex 4.4.4. For a dynamical system, the KE (T) and the PE (V) are T = q~)(q? + q~); V = ::2.":2+1 • Solve completely the problem using Liouville's approach. ql
q2
CHAPTER 4. LAGRANGIAN MECHANICS
102
SOLUTION: Consider a class of dynamical problems, in which the KE and PE has the special form T = ~ ( + qi)( cit + cii); V = ~+1 . One can easily identify Ul (ql) =
qr
ql
fJ2
qr,U2(q2) = qi;Vl(ql) = 1,v2(q2) = l;Wl(ql) = 1;w2(q2) = O. Taking U = Ul(ql) U2(q2), the equation of ql can be written w
+
Similar ly, the equation for q2 can be written as ·2 = hq22 + /2, "21 [2 ql + q22J q2
h were /1
+ /2 =
O.
Therefore, the ratio of the above equations give,
=>
cil y'hqr - 1 + /1 ci2 = y'hq~ + /2
J
dql y'hqr - 1 + /1 -
=> cos
-1
J
dq2
y'hq~ + /2
+
Constant
QlvfJh -1 Q2vfh [ ~] - cos [. r;:;;; J = Constant 1 - /1 v/1
= 11' -
/(
say )
I I # , #,
=> cos- aql - cos- bQ2 = /; a = :::} cos- 1 [aQlbQ2 => 1 - b2q22 - a 2 Ql2 :::} a 2qr + b2qi
-; b = /1
-
/1
V(1-
+ aQf)(l - bQi)] = 11' - /. + a 2b2Ql2Q22 = ( - cos / - a bQl)Q22 + 2abQlQ2 cos/ = sin2 /.
Thus the problem is completely solved and the relation between ql and q2 is given by a2qr + /)2qi + 2abQlq2 cos/ = sin2 /, where a = ~; b = ~, / are constants of Y-::rJ. Y"il integration. Ex 4.4.5. A particle moves in a plane under the action of two Newtonian centres of attraction at the points (c, 0) and (-c, 0), the attractions being ~ and f.z respectively, where r, r' being the distances from (c, 0) and (-c, 0) respectively. Show that the problem is of Liouville's type and hence solve it. SOLUTION: Here the KE (T) of the particle whose rectangular co-ordinates are (x, y) is T = ~mv2 = r;: [::i:2 + 1?J and its potential energy (V) is J-L'
y'(x + c)2
+ y2
4.4. CYCLIC CO-ORDINATES
103
Let us take the new variables ql, q2 defined by the quantities r' r = ql - q2. Therefore the potential energy is given by
(qf - c2 )(c2 (qf - c2 )(c2
+ qlq2)2 + [qlql(C2 -
-
q~)(qlq2
-
c2(qf - c2 )(c2 - q~) q~)[qfq~ + qfq~] + qfqf(c2 - q~) c2 (qf - c2 )(c2 - q~)
=
ql
+ q2
and
q~) - q2q2(qf - c 2 )j2
+ q~q~(qf -
c2 )
,c
=}
T
=
c2(qf - q~)[(qf - c2)q~ + 2 - q~)qf] c2(qf - c2 )(c2 - q~) 1 '2 '2 -2 (qf - q~) [ 2 ql 2 + 2 q2 2]' c - q2 ql - c
Thus the given problem is the Liouville's class of Lagrangian. Comparing with equation (4.27), w~ get, Ul(ql) = qf, U2(q2) = -q~; Vl(ql) = ~,V2(q2) = h; Wl(ql) = ql C C q2 -(fL + fL')ql, W2(q2) = -(fL - fL')q2. Therefore,
~u2qf = hUl(ql) -
Wl(ql) +"11
=}
"21 (2 ql -
·2 = h ql2 + ( fL + fL ') ql q22) ql
. '1ar Iy, SImI
"21 (2 ql -
·2 q22) q2
= - I1q22 + ( fL -
+ "11
fL ') q2
+ "12·
Dividing, the the equations and integrating, we get,
The reader may simplify this problem as similar of the previous problem.
104
CHAPTER 4. LAGRANGIAN MECHANICS
Ex 4.4.6. Show that the dynamical system for which 2T = r1r2(fr ~. + can be expressed of Liouville's types. Hence solve it.
/2
+ f~)
and V
=
Hints: Put r1 = f]1 +f]2 and r2 = f]1-f]2· Then T = (f]r-f]~)(qr+q~); and V = ~. ql q2 Comparing with equation (4.27), we get, 11.1 = f]r,11.2 = -q~,v1 = 1,v2 = 1,W1 2q1. W2 = O. The solution is of the type,
=
. -l(ql - l/h) -sm . h- 1 ( -q2- ) ="'f. sm
v'Il
a
4.5
Conservation Theorems
Lagrange equations of motion are second order differential equations. To solve them completely for a system having k degrees of freedom, k such differential equations have to be integrated 2k times and likewise 2k constants of integration will be involved. These 2k constants of integration can be determined, if k initial values of C/k and qJ,; are known. It is not always possible to integrate every equation of motion. However, sufficient information about the physical nature of the motion of the system can be obtained without solving these equations completely through the constants of motion. The conservation theorems for a system in motion provide the knowledge of the constants of motion. As we shall see, the conservation theorems are the relations in the form of the first order differential equations. As such equations can be readily solved, the solution of the mechanical problem by employing conservation theorems becomes much easier. Before discussing Lagrangian formulation of conservation theorems, let us first discuHs about cyclic co-ordinates.
4.5.1
Conservation theorem of energy
Consider a conservative system, then the potential energy V is a function of generalised co-ordinates only, so that aay = 0 and ~ exist. Thus, . qk aqk
av = ~ (T _ V) = 8qk
8qk =
8T _ 8V 8qk 8qk
8 (~1 mkqk .2) a~ "2 qk
k
-
0
. = mkqk·
Also, let the constraints do not change with time so that the co-ordinate transformation equation is r?i = r?i(ql, q2, ... ,qn). Its Lagrangian L is a function of f]1, ... , f]n; q1, ... ,qn; t. If the Lagrangian does not contain the time explicitly, then
or,
L dL
=
L(ql, q2,· .. ,'qn; q1, Q2, . .. , qn) ~ 8L. ~ 8L ..
di= ~aqk+ ~a-qk k
qk
k
qk
d ( 8L ) . ~ 8L.. =~ ~ dt a- qk + ~ a-qk; k
qk
k
qk
U' L . smg agrange equatIOn
4.5. CONSERVATION THEOREMS
105
' " d (8L . )
= L...J dt ~qk k
Lk
~(8~ qk dt 8qk
'
qk
L) = 0 =>
~(8~ qk dt 8qk
L) = 0
d
=> dt (mkqkAk - L) = 0 => mkq~ - L = Constant = .1. The constant of motion .1, is one of the first integrals of Illotion and is, called .1acobi's integral of the system, and it represents the total energy of the system. Since !mkq~ = T and L = T - V, we have 2T - (T - V) = constant
=> T + V
= constant.
Therefore, if L of a conservative system does not contain time explicitly, then the total energy of the system is conserved. It is the conservation theorem of energy.
4.5.2
Conservation theorem of linear momentum
Let qk be the generalised co-ordinate and dqk represents the translation of the dynamical system in some given direction. Lagrange's equation for the motion of a dynamical system are given by
Let us consider the system is conservative. Then, its potential energy V is function of the generalised co-ordinates qk only so that g~ = 0 and ~ exists. Since the KE T depends upon velocities (which involve time derivative of displacement) so T of the dynamical system is independent of qk i.e., ~ = 0 and also unaffected by a shift of origin. Thus the Lagrange's equation of motion becomes, d
8T
8V
.
8V
-(-. ) + -8qk = 0 => Pk = -8qk - = Qk dt 8qk where Qk represents the components of the generalised force along the direction of translation of qk and Pk is the component of linear momentum along the same direction. Now, we interpret the equation Pk = Qk. The generalised force is Qk = - t aa- represents the unit vector along the direction of translation "~ F i. aTqki. Here, ~ aqk ~
of generalised co-ordinate qk. If we denote it as
n,
then
-
the component of the total force along the direction of n
-
along the direction of translation of the co-ordinate qk.
CHAPTER 4. LAGRANGIAN MECHANICS
106
~-.
~
= L-J P t· n
= -. p.n ~
i
so that Pk is the component of the total linear momentum along the direction of translation of the co-ordinate qk. Thus the equation Pk = Q k represents the equation of linear motion of the system in generalised co-ordinates. Now suppose that the generalised co-ordinate qk is cyclic. Then,
Qk =
a=v .Pk = --a 0 qk =}
0 =>- Pk
=
constant.
Thus, if a given component of the total applied force on the system is zero, then the corresponding component of the linear momentum is conserved. It is the conservation theorem of linear momentum.
4.5.3
Conservation theorem of angular momentum
To ptove the conservation theorem of angular momentum, consider that the cyclic
(,)
N
o
o Figure 4.4: Angular momentum
-F"""-
co-ordinate qk is such that dqk corresponds to the rotation of the system of particles about some axis. Since T cannot contain' qk and V is independent of qk, it can be again proved that (as before) Pk = Qk. Since dqk corresponds to rotation of the system about some axis, we shall show that now the generalised momentum Pk represents the component of the angular momentum and the generalised force Qk represents the component of torque along the direction of rotation of the generalised
4.5. CONSERVATION THEOREMS
107
co-ordinate qk. Th~gure shows th~ictorial representation of the rotation of the position vector OP = ~i(qk) to OQ = ~i(qk + dqk) through an infinitesimal rotation dqk, so that PQ = d~i and LPNQ = dqk. Keeping the magnitude of vector constant, we find
-
Id"7 i l = =>
(NP) x dqk
= TisinfJdqk
. (J I d~il dqk = Tism .
n be the unit vector along ON. It follows that dd7qk is perpendicular to both ~i and n and is given by ~;:i = n x ~i' Now, the generalised force Qk is given by Let
!
=
L n."1\; r\ =
torque acting on the ith particle about the axis of rotation
i A,,",~
= n. L..J
T i
A~
= n.
T
i
where T = is the total torque. It shows that Qk is the component of the total torque acting on the system along the unit vector n, i.e., the axis of rotation. Again, from the relation Pk = ltr-, we get, oqk
~ = fI,. ""'~ L..J Li = n.L ~
where L is the total angul~mentum. Thus Pk is the component of the total angular momentum of the system along the unit vectors n, Le., the axis of rotation.Therefore we say that the equation Pk = Qk is the equation for .rotational motion of the system of generalised co-ordinates. Now, suppose that the generalised co-ordinate qk is cyclic, so, Qk =
_aOVqk = 0 => Pk = 0 => Pk =
Constant.
Thus, if a given component of the total torque on the system is zero, the corresponding component of the angular momentum is conserved. It is the conservation theorem of angular momentum.
CHAPTER 4. LAGRANGIAN MECHANICS
108
Ex 4.5.1. A particle of mass m moves on a plane in the field of force given by -+ F = - kr- cos (H, where k is a constant and f is the radial unit vector. Will the angular momentum conserved? SOLUTION: Let P(r, 0) be the position (in polar co-ordinates) of the particle of mass m in the given plane. The KE of the given particle is given by
Since the given force is only in the radial direction, so Qo = O. Thus from the Lagrange's equation of motion, in terms of the ignorable co-ordinate 0 is d aT dt [ao d
~ dt [mr
aT
- ao] = Qo
2'
2'
0] = 0 ~ mr () = constant.
Now the quantity mr 2 0 is the angular momentum of the particle of mass m about the origin and so is conserved. Also the Lagrange's equation of motion in generalised co-ordinate r, where Qr = Fr = -kr cos (), is
aT d aT dt [ar - ar] ~
mr - mr0
2
= Qr + kr cos () = 0
which is the differential equation of the orbit of the particle.
Theorem 4.5.1. If L is a Lagrangian for a system of n degrees of freedom satisfying Lagrange's equations, then L' = L + clj; (ql, ... , qn, t) also satisfies Lagrange's equations where F is arbitrary, but differentiable, function of its arguments. PROOF: Here we are to show that Lagrange's equation of motion does not change, if we add to the Lagrangian a total time derivative of any arbitrary function F(ql' ... , qn, t). Let L(qk, tik, t) be the Lagrangian of a system and F(qk' t) be arbitrary differentiable function. Then L + clj; is also satisfies Lagrange's equation of motion, for
!!:..[8(d~/dt) _ ~(dF) dt
8qk
= !!:..[-t(8F
8qi dt
dt 8qk 8t
+ 8F til)]- ~[8F + 8F tid 8ql
=~(8F)+~(8F)iIl_ 8t 8qk
8qi 8t 2
8ql 2
8 F _ 8 F iII 8qkat 8qk8ql
8ql 8qk 4 (.) 8F 8qi =OasF.,Fqj so-8' =Oand- =0. qj 8 qj
The arbitrary function F(qk' t) is called the Gauge function for the Lagrangian. The transformation L' = L + of the Lagrangian, which keeps the dynamical equations of motion unchanged, is known as Gauge transformations. This introduces
!Jft
4.5. CONSERVATION THEOREMS
109
an arbitrariness in the form of the Lagrangian that can preserve the same form of the equation of motion written in terms of L. So if L = L(qk' qk, t) can be produce a set of equations of motion through Lagrange equation of motion, then any other Lagrangian L' = L + dF~tt) formed from any arbitrary choice of F(qk' t) can be plugged in the same Lagrange equations of motion replacing the original L.. The explicit forms of the equations of motion in qk and t would not be different for tb.ese two different Lagrangians. However, the generalised momenta and Jacobi integral corresponding to L' = L + dF / dt are
aL'
aL'
a aF
aF
aF
P~=-a' ·· a qk qk =-a' qk +-a' qk [-at+-a·qzj=Pk+ql I • aL' I . aF aF aF J = qk-. - L =.J + qk- - - = J - - . aqk aqk at at
Therefore, under Lagrange Gauge transformation, the canonical momenta and the Jacobi are also changed. The momenta change due to explicit spatial variation of the Gauge function F and the energy-like Jacobi integral changes due to the explicit time variation of F. In this process the new gyroscopic potential energy and its ordinary energy is also modified.
4.5.4
Condition of invariance
Equations of motions are said to be invariant when these are same for the old and new variables corresponding to some suitable transformations of co-ordinates and velocities. Such a transformations are called invariance transformations. Let us define a inverse point transformation q: = q~(ql' q2,.' ., qn; t) with q~ forming the Lagrangian L'(q', q', t) which is
L'(q', q', t) = L[q(q', t), q(q', q', t), t] which is the condition of invariance for the transformation defined by q~ = q~(ql' q2, ... , qn; t). For example, same equations of motion are obtained if L' differs from L by a total time derivative, say ~f, where F = F(q, t), i.e.,
L'( qI , q. /,t) = L( qI , q",t)
dF (ql,···, qn, t ) . + dt
At the infinitesimal level, keeping first order terms in 8q and 8q, we have,
L(q, q, t) = L(q + 8q, q + 8q, t) . = L(q, q, t)
n aL
+ :t (8F) aL.
d
ql
t
+ I)-a, 8qi + -a', 8qi] + -d (8F) i=l
ql
CHAPTER 4. LAGRANGIAN MECHANICS
110 n
8L
d 8L
d 8L
d
d 8L dt 8q·t
d n 8L dt L...J 8q·t i=l
=> "[-5q' - -(-)5q·] + -(5F) = 0 L...J a· t + -(-5q·) dt a·· t dt 8·. 1. dt ~1 ~ ~. ~ 8L L...J 8q·t i=l n
=> " [ - - -(-. )]5qi+-["-. 5qi+5F] =0 d
=> dt
8L_ [L: ~oqi + 5F] = 0; qt n
by Lagrange equation
i=l
=>
t
~~ 5qi + 5F =
Constant in motion.
qt
i=l
Thus associated with an infinitesimal invariance transformation there is a constant of motion. Also the constancy of angular momentum from the rotational invariance of n
the Lagrangian may be readily derived from
2: g~ 5qi + 5F = i=l q,
Constant in motion,
for the general setting of the physical system with n particles. Ex 4.5.2. The point of support of a plane pendulum of mass m and length I is moving along horizontal axis with constant velocity V. Write down the Lagrange's equation. Let at an instant t, the co-ordinate of the point of support of the plane pendulum be (0, Yo), where x axis is taking vertical. Let P(x, y) be the position of the mass m, which makes an angle 8 with the vertical. Thus, SOLUTION:
x
= Leos 8, y = Yo + l sin 8; yO = V t.
Thus the KE and PE are given by,
T
= ~m[x2 + :z?] = ~m[1202 + V2 + 2Vlcos80];
V
= mgl(l- cos 8).
Using Lagrangian L = T - V, the Lagrange's equation of motion for the generalised co-ordinate 8 is given by,
d 8L dt ( 80 ) For
=
8L
..
g.
DO => 8 + I sm 8 = o.
s~all 8, the above equation becomes jj + r8 = 0, the time period is T = 27rJI.
This is the same result as finds in simple pendulum. The term mVI cos 80 did not contribute because, /t(VI sin 8) = ftF(t). Hence one can add a term which is a total time derivative to the Lagrangian L. Ex 4.5.3. Discuss the problem of damped oscillator described by the Lagrangian L = e"Yt [~mq2 - ~ kq2];" m, k, positive constants by considering the different cases. SOLUTION:
For the Lagrangian, L = e"Yt[~mq2 - ~kq2], the equation of motion is
!!...(8~) _ dt 8q
8L oq
= 0 => !!...(e"Ytmq + e"Ytkq) = 0 dt .
k 0 => q.. +,q. + -q =
m
4.5. CONSERVATION THEOREMS
111
which is the differential equation for the damped oscillator. Let us take the trial solution of the form q = eQt for the linear second order differential equation, we have the auxiliary the equation
Three cases may arise
(i) Let
~<
#; then the solution can be written as
where A, B are constants and ,B =
q
= e-~ [A cos,Bt + B sin,Bt] ,
-ie, so that ,B is real.
Let q = e-~
e, then
Thus the Lagrange's equation of motion is transformed to
e+ w e= 0; ..
2
2
w
k
,2
= -m --. 4
#; (iii) Let ~ > #; (ii) Let
~=
#.
~< then the solution can be written as q = qoe-¥, q(t = 0) = qo. then, the solution can be written as q = e-~ [Ce Ot + De- Ot ],
It resembles the form of a simple harmonic motion for
where C, D are constants and e is real. Since e > ~ > 0, we have to set C = - to avoid an explosive term. Now, it is not possible to obtain the Lagrange's equation for the equation of the damped oscillator simply by adding the Lagrangian of the simple harmonic motion, namely Ls = ~m(? - ~mw2q2, a potential term corresponding to the damping force (-,mtj), i.e., (-,mqq). The resulting Lagrangian L' = ~mq2 - ~mw2q2 - ,mqq gives ij + w 2 q = 0 instead of (i). The additional term -,mqq is actually the total derivative ~f; F = - ~,mq2. It is therefore not surprising that both Ls and L' lead to the same equation of motion. The correct Lagrangian for the damped oscillator must contain an explicit time dependence in the form of an overall exponential factor as given in our problem.
Ex 4.5.4. A uniform rod, of mass 3m and length 21, has its middle point fixed and a mass m attached at one extremity. The rod when in horizontal position is set rotating about a vertical axis through its center with an angular velocity equal to J2ng/1. Show that the heavy end of the rod will fall till the inclination of the rod to the vertical is cos- 1 [v'n 2 + 1 - n]. SOLUTION: Let 0 be the middle point of the uniform rod L1vI(= 21) attached at the point L. Initially let the rod lies along OX in the plane of the paper. On the
CHAPTER 4. LAGRANGIAN MECHANICS
112 M
~~---------------x
y
Figure 4.5
°
rod M L, take a point P such that P = r, the small element = dr. Further, at any time t, let the plane through M L and the vertical have turned through an angle
= r sin () cos
or,
z
= r cos ()
iJ = rO cos () sin
Vfl = :i;2 + iJ2 + z2 = r2(02 = [2(0 2 + ¢2 sin 2 ()).
+ ¢2 sin2 ())
vi
Now, the mass of the element PQ Tl
1 = 2 dm V2p
= 327dr = dm(say)
and therefore the KE(T1 ) is
=
~ 3m dr r2(02 + ¢2 sin2 ())
=
3'; (02
2 2Z
+ ¢2 sin2 ())r 2dr.
Hence the K E of the uniform rod is T2
= 3';,
J 1
(0 2 + ¢2 sin2 ())
r 2dr
=;
(0 2 + ¢2 sin2 (J)Z2.
-I
Therefore the total K E(T) of the system is given by
+ KE of the particle = m (02 + ¢2 sin2 {J)Z2 + m (0 2 + ¢2 sin 2 {J)Z2
T = KE of the rod (T1 )
2
= m12(02
+ ¢2 sin2 (J).
2
(T2)
4.5. CONSERVATION THEOREMS
113
Also, the work function is given by W = mgl cos () Now, the Lagrange's <jJ equation is
d aT
where 01 is a constant.
aw
aT
=
dt ( a(p) - u<jJ
+ C1, a<jJ
or, :t (2ml2(psin2 ()) = 0 => (psin2 () = C2 where O2 is a integration constant determined from the initial condition () = ~ and ¢ = J2ngjl. Thus C 2 = J2ngjl and consequently ¢sin 2 () = J2ngjl. Using the Lagrange's () equation A.(U) - aT = oW = 0 we have
!
dt ae
Be
7JO
'
(2ml20) - 2ml2¢2 sin () cos ()
=
-mgl sin ()
=> 210 - 2l2¢2 sin () cos () = -g sin () => 210 - 4ngcot(}cosec2(} = -g sin () => 210 2 + 4ngcot 2B = 2g cos B + 03; Integrating where the integration constant 0 3 is determined by the initial condition () position) and 0 = O. Thus C3 = 0 and therefore 2102 + 4ngcot2() = 2g cos (). The rod will fall down till iJ position () is given by,
=
O. Also as cos ()
=
2n cos 2 B - cos Bsin 2
or, cos or, B = Since
0 comes out to be positive,
()
=
~ (horizonta
0 gives the initial position, the
=0
+ 2ncos(} -1 = 0 cos- 1 [Jn 2 + 1 - n].
2
()
hence from this position, the rod will rise again.
Ex 4.5.5. A bead slides without friction on a wire which is rotating with angular velocity w in the force free space. Write down the Lagrange's equation of motion and find the radial velocity and acceleration.
SOLUTION: Let us consider the polar co-ordinate (r, ()) so that x = r cos B, y = r' sin B. As the wire is rotating with constant angular velocity, we have iJ = w(given) so that () = wt. Now the KE and PE are respectively T
= ~m[f2 + r 2iJ2];
L
=T
=>
- V
V
= mgy = mgr sin ()
= ~m[f2 + r 2iJ2]_ mgr sin 0
The Lagrange's equation of motion for rand () are given by
~(aL) _ aL _ dt af
=> f =
'2
rB - gsinB;
ar -
d 2' dt(mr 0)
= 0,
o.
~(aL) _ aL -
, dt ao 2'
or,mr ()
=
aB -
o.
,
constant.
CHAPTER 4. LAGRANGIAN MECHANICS
114
These two equations give the required equations of motion. For the force free space, the kinetic energy (T) is conserved, so that
= ~m[r2 + r 2w2] => r =
T
r = riJ2
Also,
[r 2w2 +
constant]~.
= rw 2 -
- 9 sin e => f
9 sin wt.
Ex 4.5.6. A bead slides on a wire in the form of a cycloid described by the equations x = a( 0 - sin 0),11 = 0.(1 + cos 0); 0 ::; 0 ::; 271", where a is the radius of the generating (;l,rde.
Neglecting fnctwn between bead and w'ire .find the equation of mot'ion.
Let P(x, y) be the position of the bead of mass m at any time t on the cycloid x = a( e - sin e), y = a(l + cos e); 0 ::; e ::; 271". The KE of the bead is SOLUTION:
T =
~m[x2 + I?]
=
~m[a2(1
2 2 2 '2 = ma [1 - cose]O .
- cos 0)2
+ a2 sin2 e]iJ2
Taking x-axis as the reference level, the potential energy of the bead is given by
v = mgy = mga(l -
cos e),
where y is measured upwards, opposite to mg. Thus the Lagrangian of the motion can be written as L
=T
- V
= ma 2[1- cose]02
- mga(l- cos e)
in which we see that 0 is the only generalised co-ordinate. Thus the Lagrange's equation of motion for the generalised co-ordinate e, i.e., [~~] ~~ = 0 gives
1t
-
d [2ma2iJ(1 - cos e)] - [ma 2iJ2 sin e + mga sin e] = 0 d,f .. 1 '2 9 => (1 - cosO)e + -2 sinee - - sine = 0 2a '2
or jj ,
+ ~ cot ~ 2
2
-
JL cot ~ = 0 2a
2
'
which is the desired differential equation of motion. Further let, U = cos(~), then 2 2 Thus the differential equation du - _12 sin(f!.)iJ and (llj'Id u - _12 sin(f!.)jj - 14 cos(f!.)iJ dO2 2 2 . of motion becomes, d 2 ,u d02
9
=-
4a u.
This differential equation represents a simple harmonic motion. The solution of this equation gives,
4.5. CONSERVATION THEOREMS
115
where C1 and C2 are constants. This shows that cos(~) returns to its original value
/f-,
after a time of 271" which is the required period of one oscillation between cusps of the wire. As this period is same as that of the simple pendulum of length 4a, hence this system is also referred to as Cycloidal pendulum. Ex 4.5.7. Consider a dynamical system with T = !(qr + q~), V = f(q1 - q2), where f is a given function. By choosing suitable new co-ordinates Q1 and Q2, reduce the problem of determining the motion to the evaluation of an integral involving the function f. Determine q1, q2 as functions of t if f (x) = x 2 . SOLUTION: Let us choose the new co-ordinate Q1 and Q2 as Q1 = q1 - q2, and Q2 = q1 + q2· Therefore, the old co-ordinates in terms of new co-ordinates are q1 = !(Q1 + Q2) and q2 = !(Q2 - Q1). The KE and PE in terms of new variables Q1 and Q2 are
and so t~e Lagrangian L becomes L for Q2 gives,
=
HQr+Q~)- f(Q1). The Lagrange's equation
where A and B are constants of integration. The Lagrange's equation for Q1 gives,
Now, given that, f(x)
= x 2 , so the above equation gives,
Thus we get, two equations, q1 these equations give
+ q2 =
At + B and q1 - q2
=
ke it . The solution of
116
4.6
CHAPTER 4. LAGRANGIAN MECHANICS
Exercise
Ex 4.6.1. Find the Lagrangian of the Sun-Earth system.
GATE -1997
Ex 4.6.2. The Lagrangian for the Kepler's problem is L = ![f 2 + r 2e2] + ~; f.L > 0 where Cr, (7) denote the polar co-ordinates and the mass of the particle is unity. The angular momentum of the particle about the centre of attraction is a constant. GAT E - 2001 Ex 4.6.3. A rod of mass m and length l is free to move in any direction on a frictionless inclined plane. The angle of inclination of the plane with respect to ground is O. How many degrees of freedom are there for the rod? Write down the equations of motion for the rod. What are the conserved quantities for the motion of the rod? Ex 4.6.4. The lagrangian L for a motion of a particle of unit mass is L = !(i;+y2 + i 2 ) - V + i:A + yB + iC; where V, A, B, C are functions of x, y, z. Show that the equations of motion are x = -~~ +y[~~ - ~:l-i[~~ - ~~] and two similar equations. Ex 4.6.5. Deduce the Lagrange's equations of motion of a dynamical system of n degrees of freedom specified by n generalised co-ordinates qk{k = 1,2, ... , n) and moving under a conservative field of forces. Ex 4.6.6. A bead is able to slide along a smooth wire in the shape of a parabola z = CT2. The parabola is rotating with constant angular velocity w about its vertical symmetry-axis. Write down the equation of motion of the bead. Ex 4.6.7. Use Lagrange's equations to find an expression for the angular acceleration of a pendulum bob moving in ~ vertical plane. Also find the tension in the string which supports the bob. Ex 4.6.8. A particle of mass m, moves on a frictionless horizontal plane {xy)plane. It is attached to one end of each of two identical massless springs A and B. Each spring is of upstretched length l and spring constant k.' The other ends of A and B are fixed at (0, l) and (~, ~) respectively. The equilibrium position of the particle is at (0,0). Obtain the Lagrangian of the particle. Assuming that x « l, y « l, simplify L, retaining terms up to order x 2 ,y2 and xy. GATE -1991
Ex 4.6.9. An artificial satellite moves in the gravitational field of the earth of mass M. Regarding the satellite as a point particle of mass m, obtain the Lagrange's equations of motion for the satellite. Hence find expressions for the canonical momenta Pr , Po, Pcp. 'Which of these is constant in time. GATE - 1992 Ex 4.6.10. A thin, uniform rod of mass m and length R slides with its ends on a smooth vertical circle of radius R. Set up the Lagrangian of the system. Find the period of small oscillations in terms of given quantities. GATE - 1994 Ex 4.6.11. A thin uniform rod of mass M and length l is oscillating in a vertical plane about a fixed horizontal axis through one of its ends. Write Lagrange's equation of motion of the rod and determine its time period for small oscillation. GATE - 1995
4.6. EXERCISE
117
Ex 4.6.12. A rod of mass m and length l is free to move in any direction on a frictionless inclined plane. The angle of inclination of the plane with respect to ground is (). How many degrees of freedom are there for the rod? Write down the Lagrange's equation of motion. What are the conserved quantities for the motion of the rod? GATE -1997 Ex 4.6.13. Consider the lagrangian L = !q log(q2) - )..q + qf(q); where f(q) an arbitrary function and ).. > O. Write down the Lagrange's equation of motion. calculate the Hamiltonian of the system. Is total energy of a constant of motion. Does the Lagrangian transform covariantly under the transformation q --+ aq for a a real constant. GATE - 2001 Ex 4.6.14. A particle of mass m is constrained to move on the plane curve xy = c(c> 0) under gravity (y-axis vertical). Find the Lagrange's equation of motion for the particle. GATE - 2001 Ex 4.6.15. A particle of mass m is constraint to move on the surface of a sphere of radius R. The sphere is resting on the ground. Set up the Lagrangian for the particle. What are the constants of motion for the particle. GATE - 2000 Ex 4.6.16. Consider the motion of the particle with Lagrangian L = ~(:i:+y2 + z2) as viewed from the rotating co-ordinate system x' = x cos () + y sin (), y' = - x sin () + y cos (), z' = z where () = ()(t) is some function of time. Find the new Lagrangian L' in terms of those co-ordinates. Ex 4.6.17. Prove that a dynamical system having n dJ. and k( < n) ignorable co-ordinates can be reduced to one (n - k) rJ. Ex 4.6.18. Define cyclic co-ordinates. If all the co-ordinates of a dynamical system are cyclic, prove that the co-ordinates can be found by integration. In particular, if the system is scleronomous, show that the co-ordinates are linear functions of time. '2
Ex 4.6.19. The kinetic potential of a dynamical system is L = aq;!+b +!q~+2q~+cq21 where a, b, c are given constants. Find the q2 in terms of elliptic functions. Ex 4.6.20. The kinetic potential of a dynamical system is L = ~f(q2)qr+h~-1P(q2). Show that, the relation between the variables ql and q2< is given by a differential equation (~) ~ = 0, where q~ = ~, and where L' is defined by the equation,
d:l
-
1
L' = [2h - 21P(Q2)] 2 [f(q2)
1
+ q~]2.
"This page is Intentionally Left Blank"
Chapter 5
Hamiltonian Mechanics The study of classical mechanics based on Lagrangian dynamics is not only simpler but also advantageous in solving complex problems. The study of classical theory of fields is extremely suitable via, the approach. For a holonornie dynamical system, described by a set of n generalised co-ordinates, the Lagrange's equations of motion are the second order differential equations. Here we shall derive an alternative formula of dynamics in terms of Hamiltonian, consists of two first order differential equations for a system with n degrees of freedom, in generalised co-ordinates and momenta. We shall assume that, the constraints are holonomic and the forces are monogenic, i.e., they are derivable from potentials which depend either on position or are velocity dependent. This Hamilton formulation is the basic foundation of statistical and quantum mechanics, field theories, that are outside classical mechanics, Hamilton-Jacobi theory for complete classical solution of dynamical problems, that is within classical mechanics.
5.1
Phase Space
Let us consider a system of n particles. The configuration space is not adequate to represent the history of the system. If only 3n momenta are used as axes, we will get the momenta space. Any point in the momenta space will describe the state of motion of the whole system and the locus of the point is called Hodogmph. The space traced by the dynamical system during its motion, formed by 6n coordinates; 3n generalised co-ordinates ql, q2,"" q3n and 3n momentum components PI, P2, ... ,P3n is called phase space. The extended phase space or state space is (2n+ 1) dimensional space where one more dimension is added to the phase space to include the parameter time. We know, only three initial values (only of co-ordinates) for the particle and hence specification of the path followed by the system is not possible. Hence there are infinite number of possible paths through any point in the configuration space. The specification of a point on the path in the phase space need six initial values for each particle, so that the state of the system is defined completely. Therefore, there is less arbitrariness about the path in phase space as
119
CHAPTER 5. HAMILTONIAN MECHANICS
120
compared to the path in configuration space. Thus, there is unique path in phase space, which refers to actual dynamical path.
5.2
Hamiltonian Function
Let Pk = 8!- be the conjugate momentum corresponding to the generalised co• I' . 0 f varIa . bl ordinate Pk.8qk The pair (qk,Pk) is called the canomca conjugate paIr es. Consider a Lagrangian L = L(qk, qk, t), which is a function of positiWl'-co-ordinates qk, the generalised velocities qk and time t of a dynamical system. Then, dL -dt,
~ 8L .
=~ k
8L .. 8L aqk qk + ~ ~ ;y-qk + at k qk '
~ dt d ( 8L ) . ~ ~qk 8L .. 8L =~ ;y- qk + ~ + at; k
qk
k
d 8L.
qk
8L ..
d or, dt
8L . [2:k ;y-qk qk
L]
~ d
8L
= 2:[dt (-8' )qk + ~qk] + at k qk qk
b L 'E t' y agrange s qua IOn
=
8L.
8L
~ dt (~qk) + at k qk
8L
= -at
d
or, dt[2:Pkqk - L] =
8L
-at;
Pk
=
8L 8qk
k
or, dd H(q,p, t)
t
=-
L 88 ;
H
t
= 2:Pkqk -
L.
k
The function H = H(q,p, t) = LPkqk - L is called precisely Hamiltonian or Hamilk
ton's function. Normally, the Hamiltonian for each problem must be constructed via Lagrangian formulation. Thus H consists of two quantities :
(i) L Pkqk is the part that is homogeneous in qk in the second degree and k
(ii) L is a part of Lagrangian independent of generalised velocities. The quantities qk and Pk form a set of independent variables in Hamiltonian formalism, whereas qk and qk form the independent set in Lagrangian formulation. We shall use the relation H(q,p, t) = LPkqk - L to obtain Hamilton's equations of motion. k
It is at once implied from the above relation that if L does not contain the time t explicitly, then H is reduced to a constant in time. For the dynamics of microscopic systems like atoms, molecules etc., the forces involved are known definite forces exerted by microscopic particles on one another and can be taken as the applied forces on the system. Consequently, the concept of constraints lose its meaning. Thus, for microscopic systems, Hamiltonian dynamics enjoys wide application. 2x 2 Ex 5.2.1. For an anharmonic oscillator the Lagrangian L(x, ±) = !±2 ax
3
+ f3x±2.
!w
Find the corresponding Hamiltonian.
5.2. HAMILTONIAN FUNCTION
121
=
From the given Lagrangian L(x, x) get the conjugate momentum as SOLUTION:
oL P = ox
.
=x+
2{3" xx => x
!x 2 - !w 2x 2 - ax 3
+ (3xx 2, we
P
= 1 + 2{3x'
Therefore the Hamiltonian H is given by H(x,p) = xx - L(x, x) = xx -
~x2 + ~w2x2 + ax3 -
px = 1 + 2{3x
p2 2(1 + 2{3x)
2px - p2 - 2(1 + 2{3x)
+ 2w
1
2 2
x
+
(3xx 2
1 223 2w x + ax
+ ax
3
.
Ex 5.2.2. Weber's electrodynamic law of attraction: Let us consider the motion of a particle subject to Weber's electrodynamic law of attraction to a fixed point, the force per unit mass acting on the particle being
where r is the distance of the particle from the centre of force. In this case, the velocity dependent potential function V(r,r,t) is given by V = ~[1 + ~J. Here we construct the Hamiltonian H for velocity dependent potentials. Now the Lagrangian for the system is
The conjugate momenta corresponding to the generalised co-ordinate r is given by pr
oL 2r => r. = Pr [m = n-:= mr. - -2 u1' 're
2
-2
re
J- 1 .
Thus, the one dimension Hamiltonian H for the velocity dependent potential is
1 ·2 + -1 [1 +r 2J Prr. - -mr 2 r c2 2 -1 1 2 2 -2 1 [ p ; 2_2 = pr2[m - -J - -mp [m - -J + -r 1 + -(m - -) J. rc2 2 r rc2 c2 rc2 1 1 '2 Now, T + V = -2mi·2 + -[1 + r 2J r c 1 2 2 2 1 2 2J = -mPr[m + -[1 + -c2 m - ) - ¥= H. 2 rc2 r rc2 H
= Prr. - L =
-r
Hence the given system is not conservative.
p; (
122
CHAPTER 5. HAMILTONIAN MECHANICS
Deduction 5.2.1. We have seen that for a conservative, scleronomic system the n
quantity
2: qi g!-q, i=1
L is a constant and that if the KE is a homogeneous quadratic
function of the velocities, the total energy is constant as well, i.e,
Therefore the Hamiltonian H can also be interpreted as generalised energy. Deduction 5.2.2. For the kinetic energy function homogeneous in the second degree of the generalised velocities as given by T = ~ 2: 2: o,ijqilj, the Lagrangian L can i
j
be written as L
=~
L L aijqiqj - V(q) i
J
where q = q(qI, q2,' .. , qn) and aij = aji. In matrix notation it can be written as L =
~qT Aq -
V(q).
Let aij1 =1= (aij) -1, in general, be the elements of A-I, then from Lagrange's equation of motion we have, n n n 8 8V d ~ .1 1~ ~ aij .. - [ ~ak'q' = - ~~--q'q' - dt .1=1 J J 2 i=1 .1=1 aqk t J 8qk
8L ::::::} Pj = a'. =
%
L aji(q)qj; n
inverting and qi = aijlpj
i=1
. 1 (-1 ) 8 a ij (-1 ) av ::::::} Pk = -2 ais Ps -8 ajmPm - - a qk qk = ~pT A-I aA A- 1p _ av 2 aqk 8qk 1 1 8A8V = __ pT _ _ p _ _ . 2 aqk 8qk
Thus the Hamiltonian H can be defined as H(q,p) -
aH·
DH
aqk ;
apk'
qk =
=
~pT A-lp
+V
with Pk =
Ex 5.2.3. Write down Hamiltonian in spherical polar co-ordinates for the Lagrangian . 2 L = ~m r+ + ~; J.L constant.
123
5.3. HAMILTON'S EQUATION OF MOTION SOLUTION: In spherical polar co-ordinates the Lagrangian is given by
I. TA f..L = -q q+-
2
r
where the matrices A and A-I are given by
A
=
m 0 0 ) 0 mr2 0 ; ( o 0 mr2 sin2 ()
A-I
=
(
m 1. 0 0 ~
o
0
0
)
0
mr
I
.
mr 2 sin 2 8
Thus the Hamiltonian H becomes
;:pr)
(
5.3
f..L
- ~
Hamilton's Equation of Motion
We have already introduced the Hamiltonian function H, which is related to the Lagrangian L by H(q,p, t) = EPkqk - L(q, q, t). The Hamiltonian H of a dynamical k
system is function of generalised co-ordinates qk, generalised momenta Pk and the time t. Here we want to describe the motion of the system in terms of equation of motion involving the Hamiltonian H, by transforming the set of variables (qk, qk) to a new set (qk,Pk). So,
CHAPTER 5. HAMILTONIAN MECHANICS
124
Comparing the coefficients of dPk, dqk, dt, we obtain,
. qk
aH
= apk;
. -Pk
aH
aL - at
= aqk;
=
aH at'
(5.1)
These 2n set of first order differential equations of motion (5.1), are known as H amilton's canonical Equations of motion or simply Hamilton's equation, due to William Hamilton[1835]. The 2n constants of integration can be evaluated in terms of initial conditions. Equations (5.1) predict that generalised momenta and generalised co-ordinates are dynamically equivalent sets of variables because their roles can be trivially interchanged just by making a change of sign. The Hamiltonian methods are not particularly superior to Lagrangian techniques for the direct solution of the mechanical problems. Rather, the usefulness of the Hamiltonian viewpoint lies in providing a framework for theoretical extensions of many areas of Physics. This gives an elegant and most useful method in the theory of classical mechanics. Ex 5.3.1. Generalised Harmonic Oscillator: For the generalised harmonic oscillator, the Lagrangian L is given by,
L = -1,2 q - -1 [X - -y2] q2 - Y. -qq 2Z 2 Z Z where X, Y, Z are time dependent. The conjugate momenta is given by
aL 1. Y P=aq=zq-zq; =*q=pZ+Yq. Thus the Hamiltonian H is given by, H = pq. - L 1.2
= Z1 [.q -
Y] .
1'2
q q - 2Z q
+"21 [X -
0
zy2] q2 + Y. zqq
y2 2
1
= 2Z q + "2 [X - Z]q 1
= 2Z[PZ + Yq]
2
1
+ "2 [X -
y2 2
1
ZJq = "2[Zp
2
2
+ 2Ypq+Xq J.
5.3. HAMILTON'S EQUATION OF MOTION
125
The Hamilton's canonical equation of motion gives,
. 8H . 8H p=--=-(Yp+Xq); q=-=(Zp+Yq] 8q 8p
~ (~) = (-J -:) (~) ~ x(t) = Ax(t) where x(t)
=
(~)
and A
=
(-J -:). The eigenvalues of
A are given by,
1-YZ- A Y-~A = 0 1
~ A = ±JY2 - XZ = A1,A2' The linearly independent eigenvectors corresponding to the distinct eigenvalues AI: A2 are give by (Y + Al Z)T and (Y + A2 z)T respectively. Thus the solution is given by x(t)
= C! (Y ~ AI) eA1t + c2 (Y ~ A2)
eA2t.
Deduction 5.3.1. Physical significance of H : Let us consider the Hamiltonian H = H(q,p, t). Now, the total time derivative of H is given by,
dH
"
dt = ~ k =dH ~ dt
aH . aH Pk + at aaHqk qk. + "~k a:Pk
"~Pkqk .. k
8L
. . -8L +" ~qkPk k
at
an
= -at = 8t; from (5.1).
If the Lagrangian of a system does not involve time t, then ~f
dd~
= 0
~H
= 0 and consequently
= LPkqk - L = Constant = E.
Thus, quite generally for the generalised co-ordinates and its canonical conjugate the Hamiltonian function H(qk,Pk) may be interpreted as the energy of the system. This gives a physical meaning of the Hamilton's function. Therefore, if the time t is cyclic in L, then t is also cyclic in the Hamiltonian H and so the Hamiltonian H is constant in motion. In this case H = E is the energy integral for conservative systems, for which H = T + V = E, otherwise in general H # H(t) and L # L(t), there will exist a constant of motion, called the Jacobi '8 integral given by H = constant = J
which need not be identical with the actual energy E.
126
CHAPTER 5. HAMILTONIAN MECHANICS
Deduction 5.3.2. Now let us consider the conservative system. For a conservative system, potential energy V is a function of generalised co-ordinates qk only and is independent of the generalised velocities, i.e., aay = 0 and the conjugate momentum qk .IS gIven . by Pk = ~. aT [lqk
Deduction 5.3.3. If the Hamiltonian H does not involve a particular co-ordinate qk, then from (5.1) we get, 8H . = 0 =? Pk = 0 =? Pk = constant . 8qk
Such co-ordinate qk is called cyclic or ignorable co-ordinate. Thus, a cyclic coordinate in the Lagrangian will be absent in the Hamiltonian H. Conversely, if a generalised co-ordinate does not occur in H, the conjugate momentum is conserved. Ex 5.3.2. The Hamiltonian of a dynamical system is given by H = qp2 - qp + bp, where b is a constant. Solve the problem. SOLUTION: The Hamiltonian of the given dynamical system is H = qp2 - qp + bp. Hamilton canonical equatkms of motion give . 8H d'p = 8H q=-an 8p 8q =? q = (2p - 1)q + band _
p=
p2 - p.
From the last equation, we have,
dp p2 _ P
= -dt =?
1
[p -
1 P _ 1ldp = dt
log -p- = t + k; integrating p-1 e t +k 1 t +k =? P = et +k _ 1 = 2"[1 + coth(-2-)]' =?
From the first equation, we have,
. (t+k) q = qcoth -2-
+ b =?
=? qe-[J coth(~)dtJ =?
=
dq (t+k) dt - qcoth -2b
t+k = b qcosech 2(-2-)
J J
e-[J coth(~)dtJdt + c
t+k cosech2(-2-)dt
t+ k = -2bcoth(-2-) =?
=b
+c
+c
t+k t+k t+k q = cSinh 2(-2-) - 2bcosh(-2-)sinh(-2-)'
5.3. HAMILTON'S EQUATION OF MOTION
127
Ex 5.3.3. The Hamiltonian 'of a dynamical system of two degrees of freedom is given by H = qlPI - q2P2 - aqf + bq~, where a, b are a constants. (i) Solve the problem. (ii) Also show that the quantities H = P1 ~:q1 and F2 = qlq2 are constants of the motion. Are there any other independent algebraic constants of the motion'? SOLUTION: (i)" The Hamiltonian of the given dynamical system is H = qlPI - q2P2 aqf + bq~. Hamilton canonical equations of motion give
.
aH
.
aH
i=1,2.
qi=-a andpi=--a; Pi
=}
qi
liI = ql; q2 = -q2; PI = -(pI - 2aql); P2 = P2 - 2bq2·
The first canonical equation gives,
-dql = dt =} ql = et+ c1 • ql
Similarly, q2 we get
= e- HC2 , where CI
=}
and
C2
-dPI + PI = dt
2ae t+ C1
Ple t
e2HC1 dt
= 2a
J
are constants. Also using PI
=
-(pI - 2aql),
+ C2 =} PI = 2aeH C! + eC2 - t .
Similarly, using h = P2 - 2bq2, we get P2 = -2bte t +C2 + eHC3 • Thus the expressions for ql, q2, PI, P2 give the solution of the dynamical problem of two degrees of freedom. (ii) Using the Hamilton's canonical equation of motion, we have ~[PI - aql] = q2(PI - aliI) - Q2(PI - aql) q2 q~ dt
-
=}
FI
PI - aql
+ PI -
aql
q2
= -PI
+ 2aql -
= PI -
aql
q2
=
aql q2
. ; as q2 = -q2
+ PI -
aql = 0
constant.
Similarly, it can be shown that P2-q1bq2 is also constant of motion. Also
=
_..!!.
Q2 q2
_P1-:Q1 = PI - 2aql q2 q2 ql
Thus FI and F2 are independent integrals of motion.
i= o.
~~~~--~~~~
128
CHAPTER 5. HAMILTONIAN MECHANICS
Ex 5.3.4. A Hamiltonian of one degree of freedom has the form H = ~ - bqpe- at + b;q2e- ot [a + be- nt ] + where a,b,a and k are constants. Find a Lagrangian corresponding to this Hamiltonian.
kf
SOLUTION:
From the Hamilton's canonical equation of motion, we have,
.
8H 2p ab = - bqe- at + _(2p)e- at [a 8p 2a 2 = E - bqe- nt + abpe-at[a + be-nt]
q= -
+ be-at]
a
+ bqe- ot = p[-1 + abe-at(a + be- nt )] a q + bqe- ot q + bqe- nt =*p= =~-~~ + abe-at(a + be-ot) D =* q
where D = ~ + abe-ot(a Hamiltonian, we get,
+ be-ot).
From the relation between Lagrangian and
q= 8H 8p q[q + bqe- nt ] p2 -at kq2 = D - ?:D + bqpe - 2 nt ot q2 + bqqe- _ 1 [q + bqe- ]2 + bqpDe- ot
L=pq-H;
_
•
2
kq2
-
D 2 q2 _ b2q2e-2ot _ 2bqqe- at + 2bqpe- at D kq2 2D 2 ii + 2bqe- at [q + bqe- at ] - b2q2e-2od kq2 = - - ; pD = q+bqe- at 2D 2 q2 + b2q2e-2at + 2bqqe-at kq2
2q2
+ 2bqqe- at _
2D
(q
+ bqe- at )2 2D
kq2 -
1
2 - '2
2 (q + bqe- at )2 ~ + abe-at(a + be-at)
which is the required Lagrangian for the given Hamiltonian.
Deduction 5.3.4. Cylindrical Co-ordinates: Here we are to use the cylindrical co-ordinates (r, e, z), which are taken as generalised co-ordinates, we are to write the Hamilton's equations of motion for a particle of mass m moving in a force field of potential V(r,e,z). Using the transformation equations x = rcose; y = r sin e; z = z between cartesian and cylindrical co-ordinates we get T
= ~m[±2 + ii + z2]
L
=T
2
- V
=
~m[r2 + r 2iP + z2] 2
= ~m[1~2 + r 2iP + z2]_ VCr, e, z)
129
5.3. HAMILTON'S EQUATION OF MOTION
which does not contain the time explicitly. The conjugate momenta are then becomes
=
Pr
m a'r
= mr;
Po
=
m aiJ
= mr
2'
e; pz
=
m aZ
.
= mz.
where Pr is the component of linear momentum along the line through the origin in xy-plane, pz is the third component of linear momentum of the particle, PIb is the third component of angular momentum of the particle. Since L does not contain the time explicitly, the Hamiltonian H is given by H
= T + V = ~m[f2 + r 2iJ2 + z2J + V(r, e, z) 1
P;
= -2 m [-2 m
p~
P;
+ 22 + 2"J + V(r, e, z) mr m
= -2 [p~ + P~r + P;J + V(r, e, z). m 1
We also calculate H from the relation H = Pr'r + poiJ + pzz - L, and get the same result, Thus the Hamilton's canonical equation of motion gives · aH P~ av . aH PI' p r = - - = -3- - - · r = - = ar mr ar ' apr m · __ aH __ av. iJ _ aH _ ~ PO ae ae ' - apo - mr2 · aH av Pz=--=--; az az
. aH pz z=-=-. apz m
When V is not a function of e, then Po = 0, i.e., Po = constant. If Po = 0 and V is independent of r, then Pr is a constant. Potential functions that are independent of e are said to be possesses cylindrical symmetry. Deduction 5.3.5. Spherical Co-ordinates: Here we are to use the spherical polar co-ordinates (1", () , ¢ ), which are taken as generalised co-ordinates, we are to write the Hamilton's equations of motion for a particle of mass m moving in a force field of potential V(r, e, ¢). Using the transformation equations x = r sin e cos ¢; y = r sin esin ¢; z = r cos e between cartesian and spherical polar co-ordinates we get, T
= ~m[x2 + 'Ii + z2J = ~m[f2 + 1'21;2 + r2 sin2e~2J 2
=> L = T - V =
2
~m[f2 + r 2iJ2 + r2 sin2e~2J -
V(r, e, ¢).
Since the Lagrangian L does not contain the time explicitly, the conjugate momenta are then becomes aL . aL 2. aL 2. 2 . Pr = af = mr; Po = aiJ = m1' e; PIb = a~ = mr sm e¢.
CHAPTER 5. HAMILTONIAN MECHANICS
130 The Hamiltonian H is given by
= Pri· + poe + P
H
p~
L
p~
P~
= -2 22 + 2mr 2sm · 2 m + -mr
The Hamilton's canonical equations of motion
e + V( r" e 'I'A.) .
qi = Wi; ap, Pi = -
BaH; q, i
=
1,2,3 gives
· aH p~ P~ av. aH Pr Pr=--=--+ -_. r=-=a1' m".3 mr 3 sin2 e ar' aPr m · aH p~cose av . aH Po Po=--= -_. e=-=ae m1,2 sin3 e a(} , apo mr2 · aH P
=-
av l a¢; 'I'
=
aH ap
=
P
Ex 5.3.5. Simple pendulum: Consider a point mass m attached to an extensible string of length 1 and suspended from a point 8 which is constraint to move along a straight horizontal line. The pendulum oscillates in the vertical plane. Let m be displaced to the position P, from the mean position 0 of the point mass, so that the string makes an angle () with the vertical. Let () be the angular displacement from the vertical 80 at any time t. The dynamical system consists of a single particle of mass m and generalised co-ordinate e. The kinetic and potential energy of the particle at Pis,
T
= ~ml2e2
and V
=
-mg(l-lcose)
where the point 0 is taken as zero level of the potential energy and 9 is the constant acceleration due to gravity. The given system is conservative and hence we have
L
=T
- V
= ~ml2(P + mgl(1 -
cos(})
Now, we are to derive the equation of motion by using Lagrange's equations. The Lagrange's equation of motion gives
.!!:.-(a~)_ dt ae
aL ae
=O=>fj+~sin(}=O 1 .. 9 => () + -()
1
= 0;
e small .
Thus in configuration space it represents the SHM motion with period 211"
Po
=
i:JL
ao
= ml H
2··
.
e, I.e., ()
=
~
~ T + V = Poe H(po, (})
=
.
.
, so the Hamlltoman becomes, L
= m 2l 2e2 - [~m2l2e2 -
1 2 Po 2 2ml [ml 2 ]
mgl(1- cos e)]
+ mgl(1- cos(}) = E
JIg. Since
5.3. HAMILTON'S EQUATION OF MOTION
131
where E is the total energy. Thus the Hamilton's canonical equation becomes,
. -Po
[ . 0 = - 8H 80 = mg sm .
These are the two first order total differential equations. Combining, we get,
dpo dO
=
m 2g[3 sinO 2 2 3 Po ~ Po + m g[ cos () ~ p~
= C( constant)
+ m2g[302 = C;
0
= small
which is the equation of the ellipse in co-ordinate Po and 0, on phase space. Ex 5.3.6. Kepler's problem: For the planetary problem, the KE T and PE V are respectively given by, T
= ~m[r2 + r2{P];
L
=T -
V
-t;.
= ~m[r2 + r2fj2] + t;.
The conjugate momenta are given by, Pr Hamiltonian H is given by, H(r, O,Pr,PO)
V =
=
ff = mr;
Po
=~=
mr 20. Thus the
= T + V = ~m[7~2 + r 202]_ !!:. 2
- _1_(P2 - 2m r
r
+ p~]_ !!:.
r·
1'2
The Hamilton's canonical equation of motion gives,
.
8H
8H
P~
J-t
- _. P - - - = - - - 8Pr - m' r 81' m1' 3 r2
.
aH
() = -
apo
Now,
Pr.
r -
Po = 0 ~
d
dt (m1' ..
Po
.
= --2; Po = mr 2·
o. 2·
e) = 0 ~ m1' 0 = constant = [( say) [2
f..L
mr 3
r2
~mr=----.
We shall discuss the nature of the solution in the next chapter. Ex 5.3.7. Motion of a particle sliding on a parabolic wire : Consider the sliding of a particle on a wire bent to a form of a parabola. Ignoring the friction, we see that, the particle is acted upon by gravity only. Let x be the generalised co-ordinate and the parabola be given in the form y = ~ x 2 . Thus, 1 . 2 .2 1 L = 2m(x + y ) - mgy = 2m(1
+ x 2 )x. 2 -
mg 2
TX .
132
CHAPTER 5. HAMILTONIAN MECHANICS -
The conjugate momenta p is given by,
8L
P = -8' X
This relation shows that p H
.
= px -
(
= m 1+x
i= mi:.
L
2).. x
x
P
= m(l+x2) .
Hence the Hamiltonian H is given by, P
= p m(1 + x 2 )
p2
=}
-
mg X 2 21 m(1 + x 2) x·2 + T
mg
2m(1 +x2)
+ Tx2 = T+ V.
The .damilton's canonical equation of motion gives,
Ex 5.3.8. The Hamiltonian of a system having two degrees of freedom is H = [Prqi + p~qi - 2aql], where a is a constant. Show that, ql varies sinusoidally with q2.
!
SOLUTION: Since the Hamiltonian H does not contain the time explicitly, so H is constant, say ~. Hence,
+ p~Qr - 2aql = A =} Plqr = A + 2aql - p~Qr p!Qi
J
Also, as q2 is a cyclic co-ordinate, so the corresponding momenta From Hamilton's canonical equation, we get,
Therefore,
q2
varies sinusoidally with
Q2.
P2
is constant.
133
5.4. HAMILTON'S PRINCIPLE
5.4
Hamilton's Principle
Hamilton's principle is a powerful technique for solving a wide variety of problems. Hamilton's principle states that, the motion of a conservative, holonomic dynamical system from time tl to t2 takes place along such a path (among of all possible paths) that the line integral of the Lagrangian of the system between the limits tl and t2 is extremum (stationary) for the path (consistent with constraints, if any) of the motion. Mathematically
J t2
1=
J t2
L(q, q, t)dt = extreme;
or, 6I = 8
tl
L(q,q, t)dt = 0
tl
where 8 is variation symbol, that does not include time variation. Unlike all other techniques, this one does not start with a differential equation, rather it starts with an integral which is then optimized against some possible variations of the path. This principle helps to distinguish the actual path of the system from the infinite number of neighboring possible paths. Deduction : Consider a conservative, holonomic dynamical system (defined by n
Q
p
Figure 5.1: Different paths generalised co-ordinates), which moves from initial configuration P of the system at time tl to the configuration Q at time t2. Let P RQ be the actual path, while P R' Q and P R" Q be the two neighboring varying paths out of infinite number of possible paths between the end points P and Q. For the deduction of the principle, the following two conditions must be satisfied by the system (i) All paths are traversed in the same time, i.e., at time tl, the system must be at point P and at time t2, it must be at Q, i.e., 8t must be zero at the end points P and Q. (ii) For all paths whether dynamical or varied, have the same termini. Since the points P and Q are fixed in configuration space, the co-ordinate variation 8-:;:' must be zero at the end points P and Q.
-
-->
Suppose that, the various particles of the system are acted upon by forces. If F i is .. the force on the ith particle, then F i = mi 7 i . From D' Alembert's principle,
"'L.)
-
Yi - ~ P J8 r
t
=
0; ~ Pi =
mi ~ r i
CHAPTER 5. HAMILTONIAN MECHANICS
134
(5.2)
Let there be little variation from the actual path and 7~ be the position vector of the ith particle on this neighbouring path, such that ~I
~
r i:= r i
+ () r
~~.
i,
~~
Le., Uri
=
~I
~
r i -
r i
where 8 can be treated as an operator which produces a small change in 7 connected with time. Thus,
i,
not
2°
~ ~ ""' [ d (--; ~) 1 ---; 2 or, ""' ~ F i.8 r ~ - ~ mi dt r i. 8 r i r i J=0 i
i
(5.3)
or,
~
Now, ~ F i.87=\
=
8W, the work done by the forces acting on the system during
i
. 2
the infinitesimal displacement 87=\ and ~ ~mi7i = T, the total kinetic energy of i
the system. Therefore, the equation becomes ,,",d( --; ~) 8(W + T) = ~ dt mi r i· 8 r i i
i.e., the time of motion for the system along every path, whether actual or varied, is the same. It is possible only when we imagine the varied paths in configuration space to be build up by a succession of virtual displacements from the actual path of motion. Integrating the above equation, we have t2
J
8(W + T)
tl
t2
=
J~ ! tl
~
(mi ?i. 87 i)dt
5.4. HAMILTON'S PRINCIPLE t2
135
t2
or, 8 j(W +T) tl
= !2;d(mi-:ti .87i )j tl
as 8[(W +T)dt]
t
"'" miTt.o d7 ~ = OJ j( W + T )dt = 'L..J t2
or, 0
i
1I
= 8(W +T)dt
l' i
~i
as r
= ~O.
i
For a conservative system, if V is potential energy of the system, then oW = -8V, and so, ~
~
oj(T-V)dt=O:::::} 8 j 1I
~
Ldt =
o:::::}
b
j Ldt= extremum b
which is the Hamilton's principle. This principle is stated in a general form independent of any co-ordinate system and hence is useful iIi. non-mechanical systems and fields. This principle can also be used for holonomic systems as well as systems having infinite degrees of freedom.
5.4.1
Modified Hamilton's principle
The Hamiltonian for a dynamical system is given by the expression H(q,p, t) t2
2:Pkqk - L(q, q, t). According to Hamilton's principle, 8I = 8 J Ldt =
o.
Thus,
1I
k t2
51 = 8 j[LPkqk - H(q,p,t)]dt tl
=0
k
t2
:::::} 51 = 8 j[LPk d:; - H(q,p, t)]dt tl
=0
k
t2
t2
:::::} 51'= 8 j LPkdqk - j H(q,p, t)dt = 0 tl
k
tl
which is called modified Hamiton's principle for any holonomic bilateral system. t2
The function 1=
J L(q, q, t)dt is called Hamilton's principal function.
This principle
tl
leads to Hamilton's canonical equation of motions. Deduction 5.4.1. Here we are to derive the Lagrange's equation of motion for systems involving forces not derivable from potential functions. Let 8W be the work done by the force on the system during the virtual displacement from actual to the
136
CHAPTER 5. HAMILTONIAN MECHANICS
rest conceived or rather varied paths. Then ~ ---t ---t T l
c5W = 15 ~ F.
=
~ ---t
~ F
---t
.15 T
i
Let q and q stand for the sets of n generalised co-ordinates and n generalised velocities respectively, then we can construct an n dimensional space and varied paths and drawn in it. Let the possible paths be referred to as qk(t, a). The transformation equations can be written as
~ a7 i
Ti= Tlqkt,a,t =*c5Ti=~8c5Qj·
---t
---t [ (
)]
---t
q]
j
Lr:t the extended Hamilton's principle with fixed end points be assumed of the form
J+ J +J t2
c5I = 15
(T
W)dt = O.
tl
t2
=* 15
t2
L Qj c5qj dt = 0
Tdt
1I
tl
J
~
JL[Z~ JL[Z~
=* 15
qJ
J
tl
~
c5qj
~
=*
tl
"q]
c5qj]dt
]
t2
J
tl
" J
L Q j c5qj dt
tl
J
J
= 0;
-
as T
= T(qj,qj)
J"
J:t [Z~lc5qjdt + J +J ~
dd (aa~ )]c5qj dt ,t qJ
aT d aT L[aq" - dt(aq)
J
J
+ L "qJ Z~ c5qjl~~
j' L["uqJ ~T -
.
tl
=*
qJ
J
~
=*
+ :.. c5qj]dt +
~
qJ
tl
tl
L Qj c5qj dt = 0 " J
~
tl
L Qjc5qjdt = 0 as c5qj = 0 " J
+ Qj]c5qj dt =
O.
Since, the constraints are assumed to be holonomic, c5qj are independent of each other, and the above integral can vanish if and only if the coefficients of c5qj separately vanish, i.e.,
aT d aT aqj - dt ( aqj ) + Qj = 0 which are Lagrange's equation of motion for non-conservative system. Similarly, we can derive the Lagrange's equation of motion for any bilateral holonomic system
5.4. HAMILTON'S PRINCIPLE
137
having no non-potential forces of n degrees of freedom. As we have observed, La~ grange's equations and the Hamilton's principle are equivalent from an information content point of view for a physical system.
Deduction 5.4.2. Here we are to derive the Hamilton's canonical equations of motion from Hamilton variational principle. Using the relation between the Lagrangian L and Hamiltonian H, i.e., H = ~Pkqk - L, we get k
L(q, q, t) = LPkqk - H(q,p, t) k
=> oL = LPkOqk + L OPkqk k
k
L[~H Oqk + ~H 0Pk] k
qk
aqk
" d . 8H . = 'L.)dt (PkOqk) - (Pk + )oqk + (qk -
k
Pk
a8HPk )OPk].
Thus the variations of L, i.e., oL can be expressed in terms of Oqk and OPk through the Hamiltonian function. Both can be arbitrary, provided we consider the variation of the paths in a 2n dimensional phase space and ne longer in the n dimensional configuration space. Allowing OPk = -9t(OPk) and Oqk = -9t(<5qk), we have, using Hamilton's principle, t2
t2
{, j L(q, q, t)dt = j {,L(q, q, t)dt = 0 tl
tl
=>
t2
[pkoqkn~ -
j(Pk
t2
+ ~~)Oqk + j(qk - ~~)OPk =
tl t2
0
tl t2
=> - j(Pk +
~~)Oqk + j(qk - ~~)OPk =
tl
0 at tl,t2.
tl
Since the system is holonomic and now described in the phase space, qk and Pk'S are all independent and Oqk's and 0Pk'S are arbitrary at all points of the path. Thus the above integral vanish, only if .
8H
0
Pk+- = ; 8 qk which are the Hamilton's canonical equations of motion. Ex 5.4.1. Using Hamilton's principle write down the equations of motion in spherical polar co-ordinates.
CHAPTER 5. HAMILTONIAN MECHANICS
138
SOLUTION: In spherical polar co-ordinates the KE T can be written as T = !m[i'2 + + 'r 2 sin2 O¢2l. Let the potential energy V = V(T, 0,
T2iJ2
tt
tl
J
J[~m(i'2 +
to
to
8
[T - Vldt = 8
r 2iP
+ r2 sin2 O¢2) -
Vldt = 0
::} J h
[{i'6i' + riP8r + r 2(J8iJ + r sin2 O¢28r + r2¢2 sinO cos 080
to 2 2 •. av av av +r sin O
tl
[Now,
J
i'8rdt = -
J :} J
r2 sin 2 O¢
J J J:t
:t
J:t tl
r 2iJ8iJdt = -
r8rdtj
to
to
tl
tl
tl
to
(r 2iJ)80dt
to
tl
(8
to
2
(r2 sin O¢)8¢dt]
to
tl
[(r - riJ2 - r sin2 O¢2
+ ~~ )8r + {:t (r 2iJ) -
r2 sin 0 cos O¢2
to
Since 8r,80 and 8
rsm6 84>
Deduction 5.4.3. Here we are to deduce the Newton's second law of motion from Hamilton's principle. Let us suppose that, a particle of mass M at the position -;;7 ---+ is moving under the action of force F. Let -;;7 be its position vector at any instant. The kinetic energy of the particle is given by .
T
=
1 1~12 -m r
2
=
1 (~~) -m r. r
2
1·· .. ::} 8T = 2[-;;7.8-;;7 + 8-;;7.-;;7]
..
= m-;;7.8-;;7.
Also, small variation in potential energy may be expressed as 8V Now, a,ccording to Hamilton's variational principle,
J -:t t2
[m .8-:t +
tl
F.8-;;7]dt =
0
=-
---+
F .8-;;7
= 8W.
139
5.5. ACTION PRINCIPLES
::}
Im~.c5~I~~
Jm7·.c5~dt+ F.c5~dt = t2
-
0
tl t2
::::} -
r .c5 r dt + -F.c5 r dt = 0 as c5r =0 j m ...;tl t2
::} j[F - m7l.c5~dt = o. tl
- -
Again, as c5~ vanishes at end points, the above equation is true for every virtual .. = 0, which displacement c5~ and hence the integrand must vanish, i.e., F - m~ is Newton's second law of motion.
5.5
Action Principles
P. M. L. Maupertuis (1740) asserted that all activities of nature are performed with a least possible expenditure of 'action', which is defined as the product of mass t2
velocity and distance. In mechanics, the quantity A =
J L,Pkqkdt
is known as k action. A variation of the path in configuration space is always considered to be infinitesimal and the variation in the generalised co-ordinates qk is always taken at the same instant. There are certain physical situations in which moving boundaries are relevant. This is in contrast to the case of Hamilton's principle where the varied path shares with the actual path the same and points which are fixed. tl
5.5.1
Action as function of co-ordinates
We have defined the action as A = J Ldt; L = Lagrangian, so that ~1 = L, where the integral is taken along a path between ql and q2 at t = tl and t = t2 respectively, so ql = q(tl), q2 = q(t2). Of the many paths that can be found between these points only one path is the actual one for which A is minimum. For the actual path also, the Lagrange's equation of motion (~t) - ~~ = 0 is satisfied. Let us now consider another aspect of A, when we regard A as a quantity characterizing the motion along the actual path, and compare the values of A for paths having a common beginning at ql, but passing through different points at t = t2, i.e., end point variation at t = t2 different from 0 but c5q = 0 at t = tl. Thus,
fit
140
CHAPTER 5. HAMILTONIAN MECHANICS = LPk8qk;
taking 8q for 8q(t2)
k
8A
8L
=> 8qk = Pk; Pk = 8qk' Now, L ,
= =
dA dt
=
8A at
+L
8A dqk k 8qk 8t
8A ' " 8A qk 8A ' " . at + L.J a . = at + L.JPkqk k qk k
= LPkqk -
H(q,p, t)
k
=> dA = LPkdqk - H(q,p, t)dt k
= LPk(t2)dqk(t2) -
H2 dt 2 - LPk(tl)dqk(tl) + H1dtl
k
k
if the end variations at t = tl and t = t2 are not zero. Thus equation shows whatever the external forces be present in the system, the RHS is a perfect differential.
5.5.2
Principle of Least Action
The principle of least action is of fundamental importance in classical mechanics. An important variational principle associated with Hamiltonian formulation is the principle of least action. In 8-variation process (i) the varied path in configuration space always terminated at endpoints representing the system configuration at the time tl and t2 as the correct path, for the derivation of Hamilton's principle. (ii) the varied path return to the same endpoints in configuration space, to derive the Lagrangian equation of motion.
In this general type of variation, known as 6-variation of the path of a system, we allow to vary the position co-ordinates as well as time. The principle of least action t2
for conservative system is then expressed as 6
J E Pkqkdt = 0 on the actual path tl
k
as compared with some neighboring paths, where 6 is the variation in A with time. PROOF: Let a configuration of dynamical conservative system be represented by n generalised co-ordinates ql, q2,.'" qn. Let us consider, a family of possible varied paths defined by the functions
where ex is an infinitesimal parameter and for the correct path it is zero. The continuously differentiable function 1}k do not necessarily have to vanish at the endpoints,
5.5. ACTION PRINCIPLES
141
either the original or the varied. We are to evaluate the 6- variation of the action integral as ~
J
6
Ldt =
tl
~+6~
J
~
L(o:)dt -
tl +6tl
J
L(O)dt
tl
where L(o:) means the integral is evaluated along the varied path and L(O), to the actual path of the motion. Since the variation includes the change in time for every path, i.e., t is a function of 0: and therefore, each point qk will be function of 0: and t, i.e., qk = qk(O:, t), so that
as change in t occurs as a result of 6-variation, so, j!do: = 6t. From the above expression, we see that, the change in generalised velocities is not same as the time derivative of the change in the generalised co-ordinates. Thus a point P on the actual path now goes over to pIon the path where qk -+ q~ = qk + 6qk and 6qk = 0 at the end points of the path. Now the 6-variation in the Lagrangian L = L(qk, qk, t) is given by
Let 6-variation consists of two paths : (i) The first arises from the change in the limits of the integrand. For first order approximation, this part is simply the integrand on the actual path times the difference in the limits in time. ~ ..
(ii) The second is caused by the change in the integrand on the varied path, but now between the same time limits as the original integral. Thus the 6-variation of the action integral can be written as t2
6
J
Ldt = 6I(t2) - 6I(tl);
tl
142
CHAPTER 5. HAMILTDNIAN MECHANICS
= [8I(t2) + j(t2)6t2J - [8I(tl) + j(tl)6tlJ = [8In~ + L6t2 - L6tl; as j = L t2
=6
J
Ldt + L6tl~~·
t1
The 'variation of the first integral can be carried out through a parameterization of the varied path, as like Hamilton's principle except, the variation qk does not vanish at the end points. Thus
t2
=> 6
J
Ldt
=
t1
[L6t -
L:Pkqk6tJ~~. k
In this equation 6qk refers to the variation in qk at the original endpoint times tl and t2. To obtain the principle of lest action we restrict our further considerations by three important qualifications : (i) Only systems are considered for which L, and so H, are not explicit functions of time, and consequently H is conserved.
(ii) The variation is such that H is conserved on the varied path as well as on the actual path: (iii) The varied paths are constrained requiring that 6qk vanish at the endpoints but not 6t.
5.5. ACTION PRINCIPLES
143 t2
Under the above prescriptions, taking the action A
= J L,Pktlkdt,
we get,
tl k
t2
t2
A = j LPkizkdt = j[L + Hjdt; as H is conserved tl
tl
k
t2
= j Ldt + H(t2 - tl) tl t2
=> 6A = 6 j Ldt + H 6(t2 - tI) = [L6t - LPktlk6t + H 6tn~ 4
k
= [(H
+ L - LPkizk6t)6tj~~ = 0 asH = LPktlk k
L
k
t2
=> 6A = 6 j LPkizkdt = 0 tl
k
which is the principle of least action. In recent times it is more customary to refer to the integral in Hamilton's principle as the action. Thus the physical trajectory is the one for which the action is stationary. The end points are fixed in both the variational principles. In Hamilton's principle, the time (t2 - tl) is prescribed ( fixed as the body moves from one point to another ) in the configuration, while in the principle of the least action, there is no such restriction on the time (t2 - tl), but the total energy between the end points is prescribed. This is the basic difference between Hamilton's principle and principle of least action.
5.5.3
Second Form of Least Action
Let the co-ordinate transformation equations of a dynamical system do not contain the time explicitly, then L = T - V and H = T + V, so that,
L Pktlk = H
+L =
T
+ V + (T -
V) = 2T.
k
In this case the principle of least action takes another form t2
6 j 2Tdt = 0 => 6A = 0 tl
144 where A
CHAPTER 5. HAMILTONIAN MECHANICS
=
t2
I2Tdt. Now we are to derive the parametric form of the principle. Now, tl t2
A
t2
J = .l =
V2TV2Tdt
J
=
y2(H -
tl
V)VL m(~:)2dt
tl
t2
V)VL m(:~ )2d)';
).
J
dql dq2
dqn
where 1(ql, q2, ... ,qn; qL ... ,q~.); function. Thus,
q~
J2(H -
= arbitrary parameter
h P2
=
1(ql' Q2,· .. , qn; d)' , d)' , ... , d)' )d)'
PI
81
d
=
~; k
= 1,2, ... n independent of t, is a
81
8qk - d)' (8q~) = 0; k = 1,2, ... ,n
which is the the required parametric form. If the system does not involve any external force, then T is conserved, and then t2
6.
J
dt = 0
tl
'* 6.(t2 -
tl) = O.
5.5. ACTION PRINCIPLES
145
This conditions leads to Fermat's principle in geometrical optics, "a light ray travels between two points along such a path that time taken is the least." Ex 5.5.1. Brachistochrone Problem:
Brachistochrone Problem is concerned with finding a plane curve that joins two points such that the time required for a particle falling under gravity along it, from an upper point to the lower, is a minimum. Let x axis be horizontal and z axis vertically upwards. P and Q are initial and final positions of the particle respectively. If the velocity and height at P is (v, z) and the same at Q is (Vq, Zq), then the total energies at P and Q are respectively given by,
Since it is a conservative force field, by conservation of energy, we have Tp i.e., 1 2 2mv
Now, the velocity v
=
=
TQ,
1 2 + mgz = 2mVq + mgZq 2 =} v = V~ + 2g(Zq - z).
~: for an element of arc ds of PQ in time dt. Hence
p
t
= f[v; + 2g(zq - z)r 1/ 2ds Q Zo
=
viI + x'2 ,dx ---;======d x . fZ Jv~ + 2g(zq - z) Z; = dz'
as ds 2 = dx 2 + dz'2
Zo
= f L(x, z, x')dz;
, dx x =-. dz
Z
The equation has the form of the action with x playing the role of 'generalised coordinate', Z that of 'time' and the integrand that of the 'Lagrangian L', so L = vi ~ Z ). Since L(x, z, x') does not contain x explicitly, so x is ignorable. Also v q +2g
Z'J-
as L does not contain time t explicitly, Brachistochrone is the path of minimum time obtainable from (/::::,. is replaced by J as t ignorable), Zo
' d aL aL J f L(x, z, x )dz = 0 =} dz (ax') - ax = 0 z =}
aL ax' = constant
= c( say)
146
CHAPTER 5. HAMILTONIAN MECHANICS X, ~
VI + Xl\/V~ + 2g(zq l+x'2 X'2
~
x'2
=
1
2 2
2
1
2+2gZq
2
1!q
29
dx=j
j
~ x = a + b[0 2
[v~
x'2
+ 2g(zq - z)] c2[v~ + 2g(zq - z)]
where a = -c Vig~2gC Zq and b = integration since we have,
1
= 1 + - = - - : : - - - - - - - 0 - - : -2
c2[v~ 1-
=c
z)
sin 0]
•
+ 2g(zq -
_ b- Z - a+Z
_ -
z)]c
[dX]2 dz
Setting, Z = _[a;b]_ [at b] cosO facilitates
l+cosO[a+bsinO]dO I-cosO 2
+A =
4 1 2 [0 - sin 0] gc
+ A.
Thus we get the curve for minimum time, the Brachistochrone, in parametric form with x and z appearing as functions of 0 and this represents the equation of the cycloid.
5.5.4
Maupertui's Principle
Let we are interested only in the motion but not the trajectory, we can write a restricted form of variational principle. Let us take the situation in which the initial and final positions are fixed and a given initial time, however we allow the variation in final time. Hence
If further we are interested in real motion in which energy is conserved, then H = E and so b"A + EJi = O. Also, if the Lagrangian L does not contain time t explicitly, then, Pk = = Thus writing the action for real motion, we have for A,
:t gr·
t
A = j[LPkQ.k - H]dt· to
k
tl
= j LPkQ.kdt - E(t - to); to
as H = E
k tl
= Ao - E(t - to); Ao = j LPkQ.kdt to
b"A = b"Ao - Eb"t b"Ao = b"A + EJi = 0
~
~
k
147
5.5. ACTION PRINCIPLES
The quantity Ao is called Hamilton's characteristic action or observed action. This equation satisfies least action principle for all paths for which energy is conserved. Let the co-ordinate transformation equations do not contain the time explicitly, then the KE of the dynamical system is a homogeneous function of velocities. Thus
2T = 2[E - V(q)] = LLajkq/lk J
'* dt =
k
2[E - V(q)]
=/
[LLajkdqjdqk]2[E - V(q)]. k
j
Thus the principle of least action takes the form c5Ao
[L L ajkdqjdqk]2[E - V(q)] =
= c5 /
o.
k
j
For a single particle, the principle of least action takes the form t2
~/
2Tdt
= ~/
m
d(3
= O.
1I
Further, for such a system, the Hamiltonian H = T + V(q), so that the special form (known as Jacobi's form )of the least action principle becomes
~
J
J2[H - V(q)] d(3
= O.
For one particle system, the principle of least action can be expressed in terms of arc length ~ f J2m[H - V]ds = O. Ex 5.5.2. Kepler's Problem: For the planetary problem, the KE T and PE V are respectively given by T = ~m[r2 + r2(j2] and V = -~. Since the system is conservative, the total energy in Kepler's problem is given by H
'* H
= T + V = !m[i·2 + r2(j2] - ~ = E 2
- V
r
= !m[r2 + r2(j2] = E +~. 2
r
148
CHAPTER 5. HAMILTONIAN MECHANICS
The Jacobi's form of principle of least action gives ~ J y'2[H - V] df3 = 0, so that
~
J
2m(E +~) y'm 2(r 2d0 2 + dr 2)
or, ~
J
k 2m(E + -;:-)[1"2
or, ~
dr
+ (dO)2]dO = 0
J(
I) 1 dr f r, r dO = OJ r = dO·
+ ~)[r2 + r /2 ] do not contain the co-ordinate e explicitly, using same pattern as Lagrangian [2: g~ qk - L = H], we get, k qk
Now, f(r, r') so,
=0
=
J2m(E
af 1 - f ( r,r ') = lj l = constant ar,r r-----2m(E + klr) 12 2 m (E + -k)[ r 2 + r 12] -- l --';:;---;:-....:...r r2 + r /2 r 12 2mr 2 2 l2 =} r = -l2-[Er + kr - 2m]
or,
o
=}
J
J
o
~
r
dO = _l_ J2m
dr r[Er 2 + kr -
2
_I P/2 2m
We shall discuss it in the next chapter. Ex 5.5.3. A particle of unit mass is projected so that its total energy is h in a field of force whose potential energy is ¢( r) at a distance r from the origin. Deduce the differential equation of the path zs c2[r2 + (~~)2] = r 4 [h - ¢(r)], where c is a constant. SOLUTION: Given that, the force field have the potential energy V = ¢(r), r is the distance from the origin and the total energy is h, so by the energy equation, •
H
Now, the
~,ction
= T +V
=}
T = h - ¢( r).
A is given by,
J =] =J t1
A
= h
=
2Tdt
to
J Sl
=
y'2[h - ¢(r)]ds
So
SO
Sl
y'2[h - ¢(r)][l + r 20/2]dr
so
J Sl
=
f(r, O')dr( say),
So
149
5.6. SYMMETRY, CONSERVATION LAWS
where ()' = ~~ and f(r, ()I) = )2[h -
d df dr ( dOl) df
=> d()'
=
constant
r2()1
df
= dO = 0 In
= y 2C( say)
=> V2[h -
hc
r2()1
=> 2[h -
2
=> C 2[r2 + (dr )2] = 1.4[h -
which is the required differential equation of the path. Ex 5.5.4. A uniform heavy flexible string is fixed at two ends. Find the equation of the curve describe the form of the string. SOLUTION: Assuming Jj- axis is positive in the downward direction. Let the two points within which the string is fixed be given by A(Xl' Yl) and B(X2' Y2). The condition for the stable configuration of the string is the condition of the minimum potential energy, then 6 J mgyds = 0, where y is vertical coordinate of the element of mass m. The variational equation becomes X2
6/ VI +y/2 ydx = O. Xl
YVI
= + y/2 as Lagrangian, then y plays the role of q and y' as x as t. Since the Lagrangian L does not contain x, so the corresponding Hamiltonian must be constant, i.e.,
If we consider L
q and
8L 8y'
H = _y' -
=> =>
yy/2 _ y(I
+ y/2)
VI + y/2 V1 y+ y/2
=> J
5.6
YV I + y /2 = =
dy Vy2 - c2
=
constant constant
constant
= c( say)
=! Jdx => y = ccosh(~). C
C
Symmetry, Conservation Laws
Here we shall discuss how the conservation laws follow in Hamiltonian mechanics. The conservation laws i.e., law of conservation of linear momentum, law of conservation of angular momentum and law of conservation of energy are derivable from more
CHAPTER 5. HAMILTONIAN MECHANICS
150
general principles, i.e., homogeneity of space, isotropy of space and homogeneity of flow of time respectively. These are called principles of space-time symmetry. As applications of Lagrange's and Hamilton's equations we shall now derive the conservation laws from the space-time symmetry principles. The motivation for discussing them in Hamiltonian theory is that this gives a direct passage for its discussion in quantum theory.
5.6.1
Conservation of linear momentum
The properties of space are same at every point, i.e., the space is homogeneous. Thus the Lagrangian of a system will be unaffected by the translation of the system in space. Let us suppose that, the position vector of a particle changes from 7 to 7 + fJ7 due to an infinitesimal translation fJ7. The change fJH in Hamiltonian H due to infinitesimal translation is given by fJH
=
aL L a7 1
--t
• 1
fJr
=
aL L[c 'a7 --t
1
i
]
=
aL c. L a7
--t
1
i
where E' = fJ7 is a constant infinitesimal small in magnitude. Further, as the particle has undergone only an infinitesimal displacement, therefore E' are not explicit of implicit functions of time. Thus
[/-;;.+ = fJ[d7] = dt
~[157] = O. dt
Since the space is homogeneous, the change in Hamiltonian of the particle due to infinitesimal translation must be zero, i.e., 8L .fJ --t fJH = '"' L...J 87 r i = 0 1
1
We require that fJH = 0, for any arbitrary translation, which implies, for the whole system, acqJ; = O. Using Lagrange's equation of motion, for all particles, we have,
: : } L Pi = K =
constant.
Therefore, the total linear momentum of any closed system is conserved due to homogeneity of space. It is the law of conservation of linear momentum. Thus, the homogeneity of space leads to the law of conservation of linear momentum. For example, consider the motion of infinite cylinder in xy- plane with a finite base, but infinitely long and homogeneous - it is translationally invariant along z axis, so pz is conserved.
5.6. SYMMETRY, CONSERVATION LAWS
5.6.2
151
Conservation of angular momentum
The space is isotropic, i.e., all the directions in the space are equivalent. If an event occurs in space in a direction in some manner, it will occur exactly in the same manner in other directions. It implies that the mechanical properties of a system are unaffected by the orientation of the system, i.e., the Lagrangian of the system will not change, if the single particle. Suppose that when the particle is rotated through an infinitesimal angle 15 0 about a certain axis, its position vector changes from -:;? to -:;? + r5-:;? so that, r5-:;? = 15 0 x -:;? Due to infinitesimal rotation, the velocity vector of the particle will change by ~
~
~~~~~~
r5v =r5r =150 x r =150 x v. The change in Hamiltonian H due to infinitesimal rotation is given by ~
,,[oH
r5H = ~ ~~ .. r5 r i t
ur
t
DH + UV ~~
~J
.15 v i
t
Since space is isotropic, the change in Hamiltonian H of t.he part.icle due t.o infinit.esimal rotation must be zero, i.e., d,,~
~
~
d(~ ri x pd= 0
t
~
where L
=
~
'r X
.
t
~ ~ '* ,,~ ~ ri x Pt= L(say) . t
~
p, the angular momentum of the particle. Thus we have, ~
dL
-
dt
~
= 0
~
'* L
=
a constant vector.
It is the law of conservation of angular momentum. Thus, isotropy of space leads to the law of conservation of angular momentum.
5.6.3
Conservation of energy
Here we shall discuss on the time-translation invariance on Hamiltonian. The time flows uniformly, i.e., forces acting at a point does not depend upon time explicitly. In other words, any event occurring under some conditions but different instants of
CHAPTER 5. HAMILTONIAN MECHANICS
152
time, must occur exactly in the same way. It is called principle of homogeneity of time. From the Hamiltonian function H = H(q,p, t), we have
As a consequence of homogeneity of time, Hamiltonian of the system does not depend upon time explicitly, i.e., aa~ = O. Therefore dH
dt = 0 => H = constant
=> E
= T +V =H =
constant .
It is the law of conservation of energy. Thus homogeneity of flow of time leads to the law of conservation of energy.
Ex 5.6.1. Spherical Pendulum:
A spherical pendulum is essentially a simple pendulum which is free to move through the entire space about the point of suspension. The bob of the pendulum can swing on the surface of the sphere in any direction whose radius is equal to the length of the pendulum. It is obvious that spherical polar co-ordinates are most suitable to locate the position of the bob of the pendulum. Moreover, since 1 is constant. the generalised co-ordinates are 0 and ¢. As the mass Tn traces out a sphere of constant length l. the KE of the pendulum is given by
T
= !m[r2 + r 2iP + r2 sin2 O¢2] 2
= !ml 2[(P + sin2 0¢2]. 2
Potential energy V of the bob due to gravity is V = -mgl cos 0; relative to the horizontal plane z = O. Hence, the Lagrangian function L for the system is thus given by
Then, the Lagrange's equations corresponding to
e and ¢ are given by
.!!:.-(ml2iJ) - ml 2 sinOcosO¢2 -mglsinO dt ., '2 g => 0 = sin 0 cos O¢ + I sin 0 and
:t
(mZ2 sin 2 O¢)
= O.
=0
5.6. SYMMETRY, CONSERVATION LAWS
153
The second equation implies that the angular momentum in the vertical direction, z, i.e., direction is a constant of motion. Hence the corresponding momentum is conserved. Thus
= ml 2 sin2 O¢ = constant = . A ~ = ml 2 sin2 () . PcP
A(say)
Substituting the value of in terms of PcP, the one dimensional equation of motion in 0 turns out to be 2"
ml 0 = mgl sin 0 + '2
A2
2
P~ cos 0 2' 3
ml sm
2g
e
~ 0 = - m2l2cot () + TCOS(}
+B
where B is a constant. We can find the total energy of the pendulum. Since ~f the total energy E is given by
·8L E = "~ qjpj - L = ()-.
. .1
8(}
= 0,
·8L - L + -.
8
= ~m12e2 + ~ml2 sin 2 (}¢2 + mgl cos () 2 1
= '2ml
1
2 '2
e + '2 ml
= '2ml2e2 ~
2 1 1
2. 2
sm
A2
em 2l4 sin4 () + mgl cos ()
A2
+ '2 ml2 cosec2() + mgl cos () = T + Ve
dE . 2'" A2 2 . = ml (}() - --2 cosec (}cot() - mgl sin 00 dt ml
-
= o.
Hence it is a constant of motion. Further Ve = ~ ~ cosec20 + mgl cos 0 is known as the effective potential energy and it depends only on O. When the pendulum is restricted to move only in one plane, say = 0 plane, then the equation of motion becomes .. g () = I sinO which is the familiar equation occurring in the case of simple pendulum with the difference that () here is measured with respect to the vertically upward direction.
5.6.4
Noether's Theorem
In classical mechanics, Noether's theorem occupies an important position. This theorem states that " if a symmetry is found to exist in a dynamical problem then there is a corresponding constant of motion." Discussion : We see that symmetry properties of the Lagrangian or Hamiltonian
CHAPTER 5. HAMILTONIAN MECHANICS
154
imply the existence of conserved quantities. Thu~, if the Lagrangian L does not explicitly a particular co-ordinate of displacement, then the corresponding canonical momentum is conserved. The absence of explicit dependence on the co-ordinate means the Lagrangian L is unaffected by the transformation that alters the value of the co-ordinate; it is said to be invariant, or symmetric under the given transformation. Similarly, invariance of Lagrangian under time displacement implies conservation of energy. The formal description of the connection between invariance or symmetry properties and conserved quantities is contained in Noether's theorem. This theorem enables us to obtain first integrals of systems of Euler's equations for furictionals admitting of some invariance properties.
5.7
Conclusion
The Hamiltonian formulation has several advantages over that of Lagrangian, as (i) Hamiltonian formulation provides deeper insight into the behavior of a dynamical system. (ii) Hamilton's formulation proves superior in the indirect integration ofthe equations of motion by means of suitable transformation.
(iii) The transformation used in Hamiltonian mechanics are known as canonical transformation, which is characterized by a single function, namely, the generating function. If a generating function can be found, any dynamical problem can be reduced to one of the differentiations and elimination.
5.8
Exercise
Ex 5.8.1. For the Lagrangian L
= !q2 - qq + q2,
find the Hamiltonian.
Ex 5.8.2. Show that the constant of motion associated with the infinitesimal transformation for the spatial displacement leads to the constant of motion of the linear momentum. Ex 5.8.3. Set up Hamilton's equations of motion for the Lagrangian L(q, q, t) = mlq2 sin 2wt + qqw sin 2wt + q2w2]/2. Ex 5.8.4. For the Lagrangian L = qt+q~+qlq2-qt-q~, find the conjugate momenta and obtain the Hamiltonian. Write down the Hamiltonian's canonical equations and solve them. Ex 5.8.5. Write down Hamilton's equations in spherical polar co-ordinates for the Lagrangian' L
. 2
= !mT7 + 7;
J.L constant.
5.8. EXERCISE
155
Ex 5.8.6. Consider the Lagrangian L = ~qlog((?) - )..q + qf(q); where f(q) an arbitrary function and)" > O. Write down the Lagrange's equation of motion. calculate the Hamiltonian of the system. Is total energy of a constant of motion. Does the Lagrangian transform covariantly under the transformation q -+ aq for a a real constant. GATE - 2001 Ex 5.8.7. For the Lagrangian L
=L
fk(qk)q~ - LVk(qk), obtain the Lagrangian
k
k
and Hamiltonian equation of motion. Solve the Lagrange's equation. Ex 5.8.8. Define phase space. Starting with a Lagrangian L = L(q, q, t), obtain the Hamilton's canonical equations of motion aaH = qk and aaH = -'Pk. Pk qk' Ex 5.8.9. State Hamilton's principle. Deduce it from D'Alemberts principle. Ex 5.8.10. Solve the simple pendulum problem by using Hamilton's canonical equations of motion. Ex 5.8.11. Write Hamilton's canonical equations of motion for the compound pendulum. Ex 5.8.12. Using Hamilton's canonical equations find the equations of motion of a p;rojectile in space. Ex 5.8.13. A particle moves under the influence of gravity on the frictionless inner surface of a cone x 2 + y2 = c2 z2. Obtain the equations of motion. Ex 5.8.14. A heavy bead of mass rn is freely movable on a smooth circular wire pf radius a which is made to rotate about a vertical diameter with a spin w. Prove that, the action will be 62
A
=
rna 2
J[ rna2~ + 2ga cosO +
w 2 sin 2
Ol~dO
61
where H is the Hamiltonian and with downward vertical.
e is the angle made by the radius through the bead
Ex 5.8.15. Show that the shortest distance between two points on the surface of the sphere is a great circle. Ex 5.8.16. Show that the area of the surface of revolution of a curve y = y(x) is X2
21l"
J yy'l + yl2dx.
Hence show that for this to be minimum, the curve must be a
Xl
catenary. Ex 5.8.17. A particle of unit mass moves along x axis under a constant force .f starting from the rest at the origin at time t = O. If T and V are KE and P E of to
the particle, respectively. Calculate J(T - V)dt also calculate for the varied path o in which the position of the particle is given by x = !ft2 + c.ft(t - to), where c is a constant. Show that the result is in agreement with Hamilton's principle.
"This page is Intentionally Left Blank"
Chapter 6
Transformation Theory We have seen that the integration of a dynamical system which is solvable by quadratures can generally be effected by transforming it into another dynamical system with fewer degrees of freedom. The way of canonical transformation to obtain solution of a dynamical problem is to transform old set of co-ordinates into new set of co-ordinates that are cyclic. In this way, the new equations of motion can be integrated much easily to give a solution. In this cha~ter, we shall discuss several bracket notations and canonical transformation which underlies this procedure, and, indeed, underlies the solution of all dynamical systems. In the next chapter, we will consider Hamilton-Jacibi theory of equation of motion which is another aspect of canonical transformation theory. At first, we introduce some Bracket not.at.ions to deal with the transformation theory.
6.1
Poisson Brackets
Given any two dynamical variables u(q,p, t) and v(q,p, t), of any dynamical system, the Poisson brackets of two functions u, v with respect to the canonical canonical variables (q, p) is defined as
(6.1) Here (q,p) refers to the set (qk,Pk, t) of independent variables pertaining to a holonomic system with the number of degrees of freedom 11, with respect to which poisson bracket is evaluated. The poisson brackets (PB) are found to be very useful tool in quantum mechanics and field theory. In this notation the equations of motion can be written in a symmetrical form. PBs can be a useful tool for finding out some of the hidden constants of motion. Ex 6.1.1. If ql and q2 are generalised co-ordinates and Pl and P2 are corresponding generalised moments. Find [X, YJ where X = q~+q~ andY = 2Pl+P2.GATE -1991
157
CHAPTER 6. TRANSFORMATION THEORY
158
SOLUTION: Let X bracket, we get [X,
=
q~
+ q~
and Y
Yl = (ax DY
_
=
2Pl
Using the definition of Poisson
ax DY ) + (DX
Dql DPI BPI Dql X 2 - 0) + (2q2
= (2ql 6.1.1
+ P2.
X
DY _ DX DY ) Dq2 DP2 DP2 Dq2 1 - 0) = 4ql + 2q2.
Properties of PB
In this section, we shall derive some of the properties of poisson brackets. Let 11.(q,p,t),v(q,p,t) and w(q,p,t) are assumed to be any three dynamical variables pertaining to a holonomic system with degrees of freedom n, whose generalised coordinates and momenta are denoted by (q, p). The algebraic properties of PB are therefore of considerable interest. Property 6.1.1. The PB of any two dynamical variables is non-commutative, i.e., [u, vlq,p
=
-[v, ulq,p,
PROOF: Using the definition of PB, we have [u, vlq,p
= L( DDu k
=-
qk
DDv - DDu DDv) Pk Pk qk
"" uu Dv Du Dv L:(DPk Dqk - Dqk DPk)
= -[v,ulq,p'
Hence the PBs are anti-commutative. In particular, when, v = u, then [u, ulq,p = -[u, ul'l,p = 0, i.e., the PB of a function with itself is identically zero. Property 6.1.2. If c is a constant independent of (q,p, t) then [cu, vlq,p
= c[u, vlq,p =
[u, cVlq,p'
PROOF: Using the definition of PB, we have
Similarly, [u, CVlq,p
= c[u, vlq,p'
This is called homogeneous property of PBs.
Property 6.1.3. The PBs also satisfy the distributive property with respect to addition, i.e., [u + v, wlq,p = [u, wlq,p + [v, wlq,p'
6.1. POISSON BRACKETS
159
PROOF: Using the definition of PB,
[u -I- v, wlq,p
a -I- v)."!l ow a ow = "" L) "!l(u - "!l(u -I- v)."!l} k
uqk
UPk
UPk
·uqk
=L{ou.oW _ ou.ow}-I-L{ov.OW _ ov.ow} k oqk OPk OPk oqk k oqk OPk OPk oqk = [u, wlq,p -I- [v, wlq,p. Also,[u-l-v-l-w, xlq,p = [u, xlq,p-l-[v, xlq,p-l-[w, xlq,p. In general, when a, b are constants then [au -I- bv, wlq,p = a[u, wlq,p -I- b[v, wlq,p, which is the linear property of PBs. Ex 6.1.2. If ql and q2 are generalised co-ordinates and PI and P2 are corresponding generalised moments. Find [X, Yl where X = qr -I- q~ and Y = 2PI -I- P2.
SOLUTION: Using the definition of Poisson bracket, we get
[X, Yl = [q~ -I- q~, 2PI -I- P2l = [q~,2PIl -I- [q~,2pIl -I- [q~,p2l -I- [q~,p2l = 2qI -I- 2qI -I- q2 -I- q2 = 4qI -I- 2q2.
Property 6.1.4. [uv, wlq,p = [u, wlq,pv -I- u[v, wlq,p, for three dynamical variables u,v and w. PROOF: For dynamical variables, u, v, w, we have
This is the elementary property of differentiation of PB. Also,
[uvw, xlq,p = [yw, xlq,p; y = uv = y[w, xjq,p -I- [y, xjq,pW = uv[w, xlq,p -I- [uv, xlq,pw.
Property 6.1.5. If c is a constant, then poisson bracket of c with any dynamical variable u is given by [e, ul = O. PROOF: Since e is a constant, we have aac qk dynamical variable is given by
[e, ulq,p
"" oc au
= 0 = aacPk .
= L....,,["!l"!l k uqk UPk
oc au
"!l"!ll
UPk uqk
Now, the PB of e with any
= o.
160
CHAPTER 6. TRANSFORMATION THEORY
Property 6.1.6. Let u, v, w be three dynamical quantities, having continuous second order derivatives, then the double PB relation holds
[u, [v, wjq,pjq,p + lv, [w, ujq,pjq,p + [w, [u, vjq,pjq,p = 0 i.e., the sum of the cyclic permutation of the double PB of the three quantities u, v, w is zero. PROOF: There is no simple algebra to prove the identity of PBs. We shall prove this by introducing some new variables. The definition of PB can be written as ~ au a au a [u, vjq,p = 6 (aq ap. - ap aqk)v k=l k k k 2n av
=Du v =
where
~i
Laia~i
i=l
represents a set of (qk,Pk) variables and Du is an operator defined by
2n
Du
= L: 0i a~.· .=1
2n
Similarly, we can define Dv
= ?: f3j J=l
at. Now
where we have used variables qi and Pi and the coefficients of as constants Aj., Bj respectively. First let w = Pj, then
g, ~ are expressed
[u[v, pjll - [v, [u, pjll = Aj av au => Aj = [u, -a j - [v, -a j qj
qj
av au a =[u'-a l+[-a ,v1=-a [u,v1· qj
Similarly, taking w
= qj
[u, [v, W1q,p1q,p
we have B j
qj
qj
= -k[u, v1.
+ [v, [w, u1q,p1q,p =
Hence
~awa
awa 6( -a. -a. [u, v1- -a. a-:-[u, v]) i
P. q]
[u, vll => [u, [v, W1q,p1q,p + [v, [w, u1q,p1q,p + [w, [u, v1q,p1q,p =
q] p]
= -[w,
o.
6.1. POISSON BRACKETS
161
This is an important identity known as Jacobi '8 identity satisfied by Poisson brackets. This identity says that the bracket product is non associative. This property of PBs is very important in defining the nature of the PBs. Let VI, V2, .. . , Vn be a set of dynamical quantities (all functions of q, p, t) and let f(vl, V2, ... , vn ) be differentiable functions of VI, V2, ... , Vn . Then
of [u, f(vl, V2,···, v1£ )] = ~[u, VI] UVI
of
of
+ ~[u, V2] + ... + n-[u, v n ]. UV2 UV
n Since the PB have the properties anti-commutative, Linearity and Jacobi's identity, the PB defines a particular type of non-associative algebra, called Lie algebra. Property 6.1.7. If [u, v]q,p be the poisson bracket of u and derivative of any PB relation is 9t[u, v]q,p = [~~]q,p + [u, ~~]q,p. PROOF: Using the the definition of PB, we have,
a ot [u, v]q,p
V
then the partial
~oa,u
=~
ov ou ov ot (oqk OPk - OPk Oqk)
= I)~(ou) ov + ou ~(OV) _ ~(ou) ov _ OU ~(01J)] k oqk at OPk oqk OPk ot OPk ot oqk OPk oqk ot =
L[~(ou) ov _ ~(ou)~] k oqk ot OPk
6.1.2
OPk
at oqk
+ I) ou ~(OV) _ ~(ou)~ k oqk OPk ot
OPk ot oqk
Basic PBs
The PBs constructed out of the canonical co-ordinates as [qi, qj] and momenta [Pi, pj] themselves are called elementary PBs. These are called fundamental PBs as with the help of these, one can calculate all PBs of daily use. Now by definition of PB, we have
_ _] _ ~(Oqi oqj _ oqi Oqj) [q~, q] q,p - ~ 0 0 0 0 . k qk Pk Pk qk Since q~( or qj) is not a function of Pk, we have
Oqi UPk
~
oqj UPk
= ~ = 0 =}
[qi, qj]q,p
= 0 = [Pi,Pj]q,p'
~(Oqi Opj
Also, [qi,Pj ]q,p = ~ k
oqi OPj) 8 8- 8 8 qk Pk Pk qk
= L
Oqi OPj; as oqi = 0 = oq] k oqk OPk OPk OPk
=
L dikdjk = dij. k
Thus for the elementary basic or fundamental PBs we have,
[qi, qj]q,p
= 0 = [Pi,Pj]q,p
and [qi,Pj]q,p
= dij = -[Pj, qi]q,p.
(6.2)
162
CHAPTER 6. TRANSFORMATION THEORY
The invariance of the fundamental PBs is always equivalent to the symplectic condition for the canonical transformation. Result 6.1.1. As qi = qi(t) and Pi = Pi(t) are explicit functions of the time t, it is possible to invert these relations. Thus we can write the dynamical variable t as t = t(CJi, Pi) and
6.2
Lagrange Bracket
For any two independent dynamical variables u(q,p, t) and v(q,p, t) pertaining to a dynamical system with the number of degrees of freedom = n, the Lagrange's bracket of (u, v) with respect to the basis (q, p) is defined in the analogous way of Poisson bracket as
OPk _ OPk oqk} (u, v )Q,l1 = ~{Oqk ~ OU ov ou ov .
(6.3)
k=l Here (q, p) refers to the set (qk, Pk, t) of independent variables pertaining to a holonomic system with the number of degrees of freedom n with respect to which Lagrange bracket is evaluated. Property 6.2.1. Lagrange bracket is antisymmetric, i.e., (u, v)q,p = -(v, u)q,p. PROOF: Using the definition of Lagrange bracket we get
(u, v)
q,p
= ~ {a qk apk _ apk aqk } ~
k=l
au av
au av
= _ I)aPk aqk _ aqk aPk} k=l au av
au ov
= -(v,u)q,p" Property 6.2.2. The fundamental brackets have the same value in both the bracket systems. PROOF: Here we are to show that (qi, qj) = 0; (Pi,Pj) = 0; (qi,Pj) = &ij. Since the co-ordinates (q, p) are independent in nature, the relations hold from the general expression of Lagrange bracket. Since q's and P's are independent, we have
Using the definition of Lagrange bracket, we get
'(qi,qj)= t{a qk apk _ aqk aPk }. k=l aq.i aqj aqj aqi
163
6.2. LAGRANGE BRACKET Similarly, (pi, pj)
= O. Further (% pj) =
t
=
t
{Oqk OPk _ OPk Oqk} k=l Oqi OPj OPj Oqi
k=l
~qk ~Pk = qt PJ
t
8ki8kj
= 8ij .
k=l
Property 6.2.3. Lagrange and PBs stand in some kind of inverse relationship to each other, but the precise form of the relation is complicated. When the coordinates are not canonical, then the receiprocal character of the two brackets manifests itself the relation 2n (6.4) L[U/,Ui](U/,Uj) = lSij, 1=1 where [Ul, Ui] and (u/, Uj) are the PB and LB respectively. PROOF: Let us suppose that there are 2n independent functions U1, U2, ... , 'U2n of the variables (qi,Pi); i = 1,2, ... ,n, so that ql,q2, ... ,qn;Pl,P2'''',Pn may be regarded as functions of Ul, U2, ... , U2n. From definitions of PB and LB we have
I)t(
I)Ul,Ui](U/,Uj) = OUI OUi _ OUI OUi) f)Oqk OPk _ oqk OPk 1=1 l=l m=l oqm 0Pm 0Pm oqm k=l OUI OUj OUj OUI
=
-_
n
f
{ t (OUI OUi _ OUl OUi) t(Oqk OPk _ oqk OPk)} /=1 m=l oqm 0Pm 0Pm oqm k=l OUI OUj OUj OUl ~
~(OUi 0Pm ~
1=1 0Pm OUj
OUi Oqm) _ + oqm OUj
OUi ( ) _ OUi _ ~ .. qm, Pm - UtJ' OUj OUj
The properties of the LB and PB parallel to each other in many aspects but LBs do not obey Jacobi's identity. So LBs do not obey the 'product' operation in the Lie algebra.
6.2.1
Angular Momentum and PB relations
The canonically conjugate momentum corresponding to any angle variable represents a component of angular momentum. Nevertheless, the total angular momentum of a system is best defined in known of cartesian co-ordinates and momenta of the individual particles. The angular momentum L can be expressed in terms of linear momentum P and co-ordinate q as ~
L
= q x P;
i.e., Li = f.ijk qj Pk
where the alternating tensor f.ijk is given by f.ijk
=
+1 when i,j,k are in even permutation ofl,2,3 0 when any two of the indices i,j, k are equal { -1 when 'i, j, k are in odd permutation ofl, 2, 3
(6.5)
CHAPTER 6. TRANSFORMATION THEORY
164
Property 6.2.4. Let us consider a one-particle system with the ith cartesian component of its total angular momentum vector is Li = €ijkXjPk. Now the PB of Li with x J is [Li' Xj]
= [€ikIXkP/, x J ] = €ikl[XkPI, Xj] = -€ikI X k c5jl - €ikjXk = €ijkXk
Property 6.2.5. For the angular momentum [L'i, Lj]
Li
and
Lj,
we have,
= [€inkXnPk, €jlmXIPm] = €inkEjlm[XnPk, XIPm] = EnkiEjlm{XnPk[X n , Pm]
+ XnPm[Pk, xd}
= EnkiEjlm {c5nm X lPk - c5kl X n Pm}
= EnkiEjlnXlPk - EnkiEjknXnPm = EnmiEjlnXlPk - ElkiEjkmXlPm = (EnmiE.1ln - EkiIEkm.i)XlPm = (c5mj c5il - c5ml c5ij - c5im c5l j + c5ij c5lm )XIPm
= (c5jm c5il Similarly, [Lj, nents, we get
Lkl
=
Li;
[Lk' Lil
[L~:, Lyl
= Lz;
c5im c5jl )XlPm
=
Lj.
= EkijEklmXlPm = EijkLk.
When expressed in terms of x, y, z compo-
= Ly = [Lx, Lxl = [Ly, Ly] = [Lz, L z ] = O. [Ly, Lzl
= Lx;
[Lz, Lxl
Thus if we have any two components of L as constants of motion, then the third component and hence L as a whole must be constant in motion. So, if any two components of the angular momentum are constant, the total angular momentum is conserved. If a and b be any two vectors then we have the following PB relation [a.L, b.Ll
= (a x b).L.
Using this result we can generalised the above results. Thus if a and b represent any two unit vectors along which the components of L are separately conserved, the component of L perpendicular to the plane formed by a and b is also conserved. Similarly, we get the relation [L2, Lil = 0, for i = 1,2,3. Property 6.2.6. We know, [Pi,P.il = 0, but [Li' L j ] = EijkLk =F o. Hence if one component of the angular momentum along a fixed direction is taken as the canonical momentum, the two perpendicular components cannot be simultaneously the canonical momenta. But [Li' L j ] =F 0 shows that the scalar magnitude of the angular momentum and anyone of its components can be simultaneously canonical.
6.3. POINT TRANSFORMATION
165
Property 6.2.7. We know, the Hamilton's canonical equations can be written as [p, Hjq,p = P and [q, Hjq,p = q. Now the PB of Li with H is
[Li' Hlq,p = [EtJkqjPk, Hl = Etjk[qjPk, Hl = Eijkqj{ qj[Pk, Hl + Pk[H, qJl}
= Eijk{ qjPk - pk<.ij} d
= dt [Etjk{ qjPk Hence if L is conserved then [Li' Hlq,p =
Pktij}l
=
dL·t dt .
o.
Property 6.2.8. For angular momentum Li, we have
[[Lt, Ljlq,p, Lklq,p
+ [[Lj, Lkjq,p, Lilq,p + [[Lk, Lilq,p, LJlq,p = {EtJIElkm + E)klEltm + EkilEljm} Lm = {6ik Jjm - 8tm 8Jk + 8j!8km - 8jm 8kt + 8kj 8im -
6krn8ij }Lm
=0.
Hence the Jacobi identity for angular momentum Li is satisfied.
6.3
Point Transformation
If a co-ordinate transformation from one set of old co-ordinates qj to a new set of co-ordinates Qj can be expressed as
(6.6) then such transformations are called point transformation. We know that configuration space is adequate only providing the information about the position co-ordinates q) and not about the velocities ti). So transformations of the type Qj = Q)(qj, t) can be referred to as the transformations of the configuration space or in other words point transformations are the transformations of configuration space.
6.4
Legendre Transformations
The basic idea behind Legendre transformation is to construct the Hamiltonian function H (q1 , Q2, ... , qn; t) from the Lagrangian L (q1 , q2, ... , qn; til, (12, ... , qn; t), by eliminating tir's (1' = 1,2, ... , n) from L in favour of Pr's through the equation Pr = 88!-; r = 1,2, ... , n. For simplicity, let us consider a function of two variables, q,. f(x. y), so that df
=
Df
ax dx
+
Df
ay dy = udx
+ vdy
CHAPTER 6. TRANSFORMATION THEORY
166
U.
where u = ~ and v = Let us consider another function g(u,y) as a function of u and y defined by the equation
g(u,y) =} dg 8g Also, dg = 8y dy
= f(x,y) - ux = df - udx - xdu = vdy - xdu. 8g
+ au du =}
8g 8y
=v=
8f 8g 8y j 8u
= -x.
Thus if 7/. = ~, then the relation g(u, y) = f(x, y) - 11,x would be appropriate to effect a change from the basis x, y.to 11" y. Now we construct a function g( u, v) from f(x, y) as
9 (11" v) =}
= ux + vy - f (x, y)
8f 8f dg = udx + xd11, + vdy + ydv - 8x dx - 8y dy
= 11,d11,+ ydv which shows that 9 is a function of 11, and v only. Let us consider L P = ~J. Identify L with f and q - t X, q - t y, then
8L P = aq
= L(q, q)
and
af
= ax = 11, =} g(11" y) = g(q,p) = 11,:» - f = pq - L(q, q). Let g(p, q) = Hamiltonian function H, as H(q, p) = pq- L(q, q). Let L = L(q!, q2, ... j (h, Q2, ... ; t) and PI =
gt
j
P2
=
g:~, etc., then
H(ql, Q2,··· jPl,P2,··· j t)
= L:Pkqk - L(ql, q2,··· j ql, (h,··· j t) k
which is the definition of the Hamiltonian, used to obtain Hamilton's equation of motion. The quantities qk and Pk form a set of independent variables in Hamiltonian formalism, whereas qk and qk form the independent set in Lagrangian Formulation.
6.5
Canonical Transformation
Here we will be concerned with the application of Hamilton's ideas, to get a general case of a dynamical system with any number of degrees of freedom. We shall first define a contact-transformation in n dimensional space. The principal utility of canonical transformation is to transform the Hamiltonian H(q,p, t) such that the corresponding equations of motion become easier to solve and also display all possible cyclic co-ordinates and conserved quantities. Let (qi,Pi) be a set of 2n variables and (Qi, Pd be 2n other variables ofpo~ition and momentum in old and new co-ordinates respectively and they are defined in terms of 2n equations by (6.7)
6.5. CANONICAL TRANSFORMATION
167
=
If in the new co-ordinate :3 a new Hamiltonian K ordinates Qk, Pk such that
K (Qi, Pi, t) in the new co-
(6.8) then the transformation is called canonical transformation and the co-ordinates (Qi, Pd are known as canonical co-ordinates. These are also called called contact transformation, as the two curves in both phase spaces have at least one point( configuration) of contact. Note that, any arbitrary transformation will not be canonical, there must be some constraint on the choice of such transformation. Therefore, canonical transformations are the transformations of phase space. They are characterized by the property that they leave the form of Hamilton's equations of motion invariant. The Hamiltonian in old and new co-ordinates are H(q,p) = LPkqk - L(q, qk, t); K(Q, P) = L k
PkQk - L'(Q, Qk, t)
(6.9)
k
Thus to obtain solution of a mechanical problem in the way of canonical transformation is to transform old set of co-ordinates into new set of co-ordinates that are all cyclic and should satisfy the Hamilton's equation of motion. If all the co-ordinates are cyclic then we could solve very easily the Hamilton's equations. Ex 6.5.1. Prove that the transformation P SOLUTION:
=
~(p2 +q2), Q
= tan-l(~)
is canonical.
The given transformation equations are p
8P
= ~(p2 + q2), 8P
Q
= tan-l(~)
8Q
=> op = p, oq = q; op = - p2
q
+ q2'
8Q oq
P
=
p2
+ q2 .
Differentiating the transformation equations with respect to P and Q, we get
Since the generating function does not involve time, so it is obvious that H(q,p) J((Q, P). Therefore, dH oK oQ oK OP p oJ( oK Dq = DQ Dq + DP Dq = p2 + q2 DQ + q DP 8H DK OQ 8K OP q DK DK =-+- = - p2 + q2 +p-. 8p 8Q 8p 8P 8p 8P
aq
=
168
CHAPTER 6. TRANSFORMATION THEORY
As the old set co-ordinate (q,p) are canonical, and q = q(Q, P), we have uH . up· up· uH . uq· uq· oq =-P=-oPP- oQQ; oq =q= oPP+ oQQ
and
p
or, p2
+ q2
oj( oQ
+ q oP = - p2 + q2 P + qQ
oj(
p..
- p2
+ q2
q
oj( DQ
+ P0P =
oj(
=?
q ? + Q' p2 + q2 P oj( . oK . P = - oQ and Q = oP'
This shows that the new variables P and Q are canonical and consequently the transformation is canonical.
6.6
Generating Functions
Canonical transformations are the transformations of phase space. We know every natural path in the phase space spanned by the co-ordinates q and the momenta P for a system. In Hamiltonian formulation, we admit the existence of one more independent variable called momentum and, therefore, above general form must be widened to accommodate of the new variable. Consequently, the simultaneous transformation of the independent co-ordinates and momenta (qj, pj) to a new set (Qj, PJ ) can be represented in the form
(6.10) For (QJ , PJ ), the new set of co-ordinates to be canonical as is demanded in Hamiltonian formulation, they should be able to be expressed in Hamiltonian form of equations of motion,
where K = K(Q, P, t) and is substitute for the Hamiltonian H of old set in new set of co-ordinates. Since Q k and Pk are canonkal co-ordinates, they must satisfy the modified Hamilton's principle of"the form t2
&J[ iI
LPkQk-K(Q,P,t) jdt=O.
(6.11)
k
Also, the old co-ordinates qk and Pk are canonical, so t2
&
J[
tl
LPk
= o.
(6.12)
6.6. GENERATING FUNCTIONS
169
The simultaneous validity of these equations does not mean that the integrands in both expressions are equal. Therefore t2
0= 8
J[ LPkqk - H(q,p, t) tl
jdt
k
t2
= 8
J[ LPk(lk - K(Q, P, t) tl
k
t2
= 8
J
.
~
[L..J PkQk - K(Q, P, t)
tl
jdt
dF
+ dt
jdtj
k
where F is any function of the phase space co-ordinates with continuous second order derivatives and A is a constant scale factor, known as valence of canonical transformation, independent of the canonical co-ordinates and the time. When A =1= 1, then the transformations of canonical co-ordinates is called an extended canonical transformation. When A = 1, the transformation is called univalent of canonical transformation. Thus for A = 1, the condition becomes ~
~
k
k
.
L..JPkqk - H(q,p,t) = L..JPkQk - K(Q,P,t)
dF + dt
(6.13)
From this equation we see that F is a function of (4n + 1) mixture phase space co-ordinates q,p, Q, P and the time t. But the two sets of canonical co-ordinates are connected by the transformation equations Qj = Qj(q,p, t)j Pj = Pj(q,p, t) so there are only 2n independent variables. The function F is useful for specifying the exact form of the canonical transformation only when out of 2n independent variables n of the variables are form the old set and n are form the new. So the possible principal form of Fare H(q, Q, t), F2(Q, P, t), F3 (p, Q, t), F4(p, P, t). The exact form of the generalised function depends upon the circumstances. As the function F connects between old and new set of co-ordinates, it can affect on the sets (6.10) and thus termed as generating function of the transformation. Now we are to show that how the generating function specifies the equation of transformations.
170
6.6.1
CHAPTER 6. TRANSFORMATION THEORY
First Form
Let F be given as a function of the old and new co-ordinates of the form F Fl (q, Q, t), then the total time derivative of H is given by
=
(6.14) Since the old and new co-ordinates qk and Qk are independent variables equation (6.14) hold only if the coefficient of qk and Qk vanish separately, (6.15) Integrating the first two equations we get a desired transformation equations Q k = Qk(qbPb t); Pk = Pk(qk,Pk, t) and the equation gives the relation between the old and new Hamiltonian. Thus when the generating function Fl(q, Q, t) is given, one can uniquely construct the new Hamiltonian K in the following ways : (i) Construct the n equations of the old momenta Pk using the relations Pk (q, Q, t) = &..t k = 1,2, ... n. From these relations, we try to solve Qk'S in the form a'F; qk Qk = Qk(q,P, t) by algebraic manupulation, as the transformation of configuration space is invertible. Hence we obtain the new co-ordinates as the function of old co-ordinates, old momenta and time, i.e., half of the required equations for the canonical transformation are constructed. (ii) Construct the next n equation of the new momenta Pk by the relation Pk(q, Q, t) - gQ}'!; k = 1,2, ... ,n and then substitute all the Qk'S. Hence we obtain the k new momenta Pk = Pk(q,p, t) as the function of old co-ordinates, momenta and time. (iii) Find the inverse transformations Pk
= Pk(Q, P, t);
qk
=
= qk(Q, P, t).
(iv) Construct the new Hamiltonian K(Q, P, t) using the relation K = H(q,p, t) + ~ by substituting all P's and q's as Pk = Pk(Q, P, t); qk = qk(Q, P, t). Conversely, for the given transformation equations F1(q, Q, t) is not given. We are to derive an appropriate generating function F1(q, Q, t) in the following procedure: (i) First the given transformation equation in canonical space Q~ = Qk(qk, Pk, t); Pk Pk(qk,Pb t) are inverted in the order relationpk = Pk(Q, P, t); qk = qk(Q, P, t).
=
·6.6. GENERATING FUNCTIONS
171
(ii) From the transformation equations in configuration space we eliminate all p's so that after rearrangement we get Pk = Pk(q, Q, t). Similarly, eliminate all Pk'S so that after some rearrangement we get Pk = Pk(q, Q, t). (iii) Now for the newly obtained expressions, constitute a coupled set of partial differential equations as Pk(q, Q, t) =~; Pk = -g~~ and integrate both of them separately as
J =- J
H(q, Q, t) = Fl(q, Q, t)
Pk(q, Q, t)dqk
+ '1/J1(Q, t)
Pk(q, Q, t)dQk + 'l/J2(q, t).
To find the correct solution for H(q, Q, t), we shall combine these two expressions suitably. (iv) Once H(q,Q,t) is obtained, we are to find the new Hamiltonian K from K = H + ~ by substituting for p's and q's from the inverse transformations. This equation is also connected with Hamilton-Jacobi equation, discuss later.
Ex 6.6.1. Solve one dimensional harmonic oscillator using canonical transformations. SOLUTION: The Hamiltonian H for a simple, one dimensional harmonic oscillator in usual co-ordinates (q,p) is given by
where m = mass and k is the force constant. Since the Hamiltonian H is the sum of two square terms, we can construct a transformation in which H is cyclic in new co-ordinate. The problem is given canonical co-ordinates q and p. We shall transform this into new set of canonical co-ordinates Q and P in which Q is cyclic and the corresponding momenta is constant. Here we are to show that how to solve a dynamical problem by using canonical transformations. Let us take a canonical transformation of the form P = f(P) cos Q;
q = f(P) sinQ
mw
such that the Hamiltonian K in new co-ordinate is given by
Since the new Hamiltonian K does not contain the co-ordinate Q explicitly, so Q is cyclic. Now we are to find the unspecified function f(P) such that the transformation
CHAPTER 6. TRANSFORMATION THEORY
172 is canonical. Now,
p = m w q cotQ = mwq2
aFl{q, Q) aq
../km
2
=> F1{q, Q) = -2- cotQ = -2- q cotQ => Q = tan-1{../km q ). p
Since Fl does not contain time t explicitly, the Hamiltonian is unaffected by the transformation. Hence, the new momenta is given by
P
aFt
mwq2
2
= - aQ = -2-cosec Q =
1
rm
2
p2
2V k{kq + m)
~ ../2mwP.
=> q = J2P sinQ => f{P) mw
Therefore, the Hamiltonian H in terms of the new variables (Q, P) becomes
H
1 = _[p2 + m 2w2q2]
2m 1 2P = -[mwq2.mwcot2Q+-sin 2 Q] 2m mw
= P[wcos 2 Q +
+
mw
sin2 Q]
= wP =
~p,
V~
which is quite simple form. Since the Hamiltonian H is cyclic in Q, hence, the conjugate momentum P is a constant, i.e., P
H E = - = - = a{
constant,) say w w where E is the constant energy. The equation of motion for co-ordinate Q reduces to the simple form
.
Q
aH = -aP = w => Q = wt + 8
where <5 is the integration constant, determined by the initial conditions. Thus, the transformation has changed the problem of the oscillator in such a way that the new co-ordinate P is constant of motion and the new co-ordinate Q increases linearly with time t. The equation for Q gives translational motion of the oscillator and the transformation is equivalent to changing the oscillatory motion into translation one. Hence the solution for q is given by,
q=
P. -mw [!f
smQ
=
{gE. --2
mw
sm{wt + <5)
.
= Asm{wt + 8)
which is the equation for SHM and is the desired solution of the one dimensional harmonic oscillator problem, where A
=
J
2E2 and <5 is a constant.
mw
6.6. GENERATING FUNCTIONS
173
Ex 6.6.2. For the simple harmonic oscillator, let the transformation (q,p) be Q = log[~ sinpji P = qcotp. Find F1(q, Q, t) and K(Q, P, t).
--+
(Q, P)
The Hamiltonian H for a simple, one dimensional harmonic oscillator in usual co-ordinates (q,p) is given by SOLUTION:
where m = mass and k is the force constant. Here the given transformation is Q
. = log[-q1 smp]i -
P
= qcotp
P = qcotp '* eQ = -1. smpi q '* Pe Q = cosp; p2 = q2cot 2p '* P = cos-1(PeQ)i q = Ptanp = [e- 2Q Also, using the transformation equations Q P
= log[~ sinp]i P = qcotp,
= qcotp = q c~sp = smp
and sinp = qe Q
J =J
we get
[e- 2Q _ p 2]1/2
'* p = sin-l(qeQ).
FI:om the partial differential equation Pk(q, Q, t) H(q, Q, t) =
P2]1/2.
Pk(q, Q, t)dqk
= £El 8qk
we get,
+ 'l/Jl(Q, t)
sin-l(qeQ)dq + 'l/Jl(Q, t)
= e-
Q
J
OcoSOdO+'l/Jl(Q,t)i sin-l(qe Q) = 0
= qsin-l(qe Q) + [e- 2Q - q2j1/2
+ 'l/Jl(Q, t).
From the partial differential equation Pk = - gg~ we get,
Therefore, the generating function Fl (q, Q) is given by
174
CHAPTER 6. TRANSFORMATION THEORY
and the new 1!amiltonian K (Q, P, t) is K(Q, P, t)
8Ft
= H(q,p) + at = 1 Q 2m [cos- l (Pe )]2
=
p2 2m
1
+ "2 kq
2
k
+ 2" [e- 2Q -
p 2].
Ex 6.6.3. Find the canonical transformation whose generating function is Fl(q, Q) = qcos- l y'1 - q2e2Q + y'e- 2Q _ q2. SOLUTION: The canonical equations are equation, we have p
*
= p and ~ =
qe Q y'1 - q2e2Q 2Q sin2 p e =-q2
8F = _ 1 = cos- l y'1 _ q2e2Q _ 8q /
1 - q2e2Q = cos p =}
=}
V
=}
Q = log(sin p ).
-P. From the first Q
qe + --;:.==~~ 2Q
y'1 - q2e
q
From the second equation, we have
p
8Ft
= -- =
8Q
-
6.6.2
-q2 eQ y'1 - q2e2Q
1
y'1- q2e2Q
2sinp [-q q
e- Q
+ -;::=:::::::=;:::;;: 2 2Q y'1 - q e
+ -.q-] = qcotp. smp
Second Form
Let the canonical transformation be suitable described by the functions of q, P and t rather than q, Q, t. In this case, the generating function of the form F2 = F2(q, P, t) is more fruitful than Fl' Since the new variables Pk = - g~~ are independent of Q's take the Legendre dual transformation of Ft we get F = F2(q, P, t) - L
PkQk, 't.e., F2(q, P, t)
= Ft(q, Q, t) +
L PkQk k
k
Substituting F in (6.13), we get LPkqk - H = k
=
L PkQk -
K k ,,",' L...J PkQk - K k
+ :t {F2(q, P, t) -
L PkQk}
k ""' aF2 ""' 8F2 · + L...J aC[k + L...J ap, Pk k qk k k
+ 8F2 8t
,,",' ,,",' - L...J PkQk - L...J PkQ, k k
6.6. GENERATING FUNCTIONS
175
Since qk and Pk are independent variables, this equation is satisfied when
_ 8F2. Q _ 8F2. K-H 8F2(q, P, t) Pk - aqk' k - aPk' + at
(6.16)
Integrating the first two equations we get a desired transformation equations Qk = Qk(qk,Pk, t); Pk = Pk(qk,Pk, t) and the equation gives the relation between the old and new Hamiltonian. Ex 6.6.4. Find the canonical transformation in which the generating function F2 is given by F2 = q cot- 1 + P.
f
SOLUTION: The canonical equations are ~
= P and
VI = Q. Thus,
8F2 -1 P qP P = - 8q = cot q + p2 + q2 . Also, from the second equation we get,
6.6.3
Third Form
In the same manner, to construct the third generating function F3(Q,P, t) from Fl by replacing q's by p's given by Pk = m, Legendre'S dual transformation is 8qk
F3 (q, P, t) = F1 (q, Q, t) - LPkqk k
Substituting F in (6.13), we get
CHAPTER 6. TRANSFORMATION THEORY
176
Since Qk and Pk are independent variables, this equation is satisfied when (6.17) Integrating the first two equations we get a desired transformation equations qk = qk(qk,Pk, t); Pk = Pk(qk,Pk, t) and the equation gives the relation between the old and new Hamiltonian.
6.6.4
Fourth Form
Similarly, to construct the fourth generating function F4(p, P, t) from F l Legendre's dual transformation gives
,
double
L:: QkPk + F4(p, P, t) F4(p, P, t) = Fl (q, Q, t) - L::Pkqk + L:: PkQk F
= L::Pkqk k
k
k
k
= F3 (Q,p, t) + L::PkQk = F2 (q, P, t) - L::Pkqk' k
k
Substituting F in (6.13), we get
Since qk· and Pk are independent variables, this equation is satisfied when
Qk
=
__ 8F4 . K _ H aF4 8Pk; qk 8Pk ' -
+
8F4 at
(6.18)
Integrating the first two equations we get a desired transformation equations Qk = Qk(qk,Pk, t); Pk = Pk(qk,Pk, t) and the equation gives the relation between the old and new Hamiltonian. From the above discussions, we say that appropriate form of the generating function depends upon the circumstances. Also we note that th~ value of the transformed Hamiltonian K is related to the value with the old Hamiltonian H by similar relation, but the value difi'erers by the partial time derivative of anyone of the generating functions F. If there is an explicit time dependence in the transformations, the
177
6.6. GENERATING FUNCTIONS
generating functions F must also leave the same as Q(t) and P(t) imply moving frame of reference and hence Hamiltonian cannot be same in the two frame even though with respect to either frame they may be a constant in motion. Ex 6.6.5. A canonical transformation is given by F3 FI, F2 and F4 .
=
-Q[1
+ log(p2/4Q)].
Find
SOLUTION: Here the transformation is given by
F3
= F3(p, Q, t) = -Q -
Q[2logp -log(4Q)].
The transformation equations give,
P
oF 2Q - -3 = -
=
1 -pq.
= - aQ = log(p /4Q)
~ P
q=
op
p
o~
~
Q
2
2
P
= log(2q)·
Since the transformation equations does not involve the time t explicitly, so K Also, we know that,
pdq - PdQ + (K -qdp - PdQ + (K pdq + QdP + (K -qdp + QdP + (K As K = H, so F3 F4 are given by
-
H)dt = dFI H)dt = d(FI - pq) = dF3 H)dt = d(F1 + PQ) = dF2 H)dt = d(Fl - pq + PQ) = dF4.
= FI - pq,F2 = FI + PQ,F4 = H - pq+ PQ.
Thus, F1 ,F2 and
H = F3 +pq = -Q - Qlog(p2/4Q)
= -Q F2
Qlog{
(2Q/q? Q 4Q } = -Q - QlOg(q2)
= FI + PQ = -Q - Q log(p2 /4Q) + PQ = - pq - pq log( pq ) + Ppq 2
2
2q2
= H.
2
= H(q, Q, t)
178
6.6.5
CHAPTER 6. TRANSFORMATION THEORY
Cyclic relations
Here we are to obtain the cyclic relationships. If the generating function is of the first form F 1 (q, Q, t), then, we may have,
Again, if the generating function is of the second form F2 (q, P, t), then, we may have, Bpk BQk BPk = Bqk; k = 1,2, ... , n.
Similarly, if the generating function is of the third form F 3 (p, Q, t), then, we may have, BPk --; k=1,2, ... ,n. BPk
Finally, if the generating function is of the fourth form F4 (p, P, t), then, we may have,
These four relations, referred to as cyclic relations, are very much similar to well known Maxwell equations in Electromagnetism. These relations are useful in establishing certain properties of canonical transformations. These symmetrical form of these relations is their spaciality, i.e., they are independent of the Hamiltonian or generating functions.
6.6.6
Examples of CT
In this section, the simple but well-known examples are illustrated the nature of CTs given the generator and the role played by the generating functions. Ex 6.6.6. Now we consider a generating function F2 of the second type with particular form
which shows that qk ---... Qk and Pk ---... Pk, i.e., the old and new co-ordinates are same. Therefore F2 generates the identity transformation. If we take F2 = -qkPk, we get the reflection transformations as qk ---... -Qk and Pk ---... -Pk. Since Pk = £El. is a Bqk function of q, Q, t but not in P on which the identity transformation depends only,
•
179
6.6. GENERATING FUNCTIONS
so the generating function Fl cannot be used to generate the identity transformations. One can overcome this difficulty by defining an Fl function thro~h Legendre transformation Ft(q,Q,t) = F2(q,P,t) - PkQk. Since
Fl(q, Q, t)
= F2(q, p, t) - PkQk = F2(Q, P, t) - PkQk = QkPk - PkQk = 0, II- ~..
Similar argument shows that, constructing F2 we can gen~ate exchange transformations. Similarly, it can be shown that the generating function F3 can generate the identity transformation and the exchange transformation can be derived from F4 functions. Ex 6.6.7. Consider, a generating function of the first type with particular form
We are to show that it generates the exchange transformation in which position co-ordinates and the momenta can be interchanged. Hence the corresponding transformation equations becomes
which shows that, the transformation interchanges the momenta and co-ordinates except for a sign. This exchange transformation is canonical follows directly from Hamilton's equation. Also, Hamilton's equations of motion are symmetric in (q,p) if qk is replaced by -Pk. Hence generalised (q,p) are completely equivalent in the description of the phase trijectories of a system obeying Hamilton's equations of motion. Since in the exchange transformation Pk = ~ is not a function of P at all but depends only on new co-ordinate Qk, so the generating function F2 is not suitable for such an exchange transformations. Ex 6.6.8. Let us consider a mote general type transformation chosen by the generating function
where fk may be any desired set of independent functions. ordinates are given by
Hence the new co-
In this generating function, the new co-ordinates Qk depend only on the old co. ordinate qk and the time t, but independent of old momenta p. This type of transformation is an example of the class of point transformations. Also, to define a p~int transformation, h must be independent and invertible, so that qk can be expressed in terms of the new co-ordinate Qk. Since fk are completely arbitrary,
CHAPTER 6. TRANSFORMATION THEORY
180
so all point transformations are canonical. The new Hamiltonian in terms of the old and of the time derivatives of the fk functions were given by 6.16. Note that F2 = Ik (ql, ... , qn; t) is not only generating function leading to the point transformation specified by Ik. Clearly, the same point transformation is implicit in the more general form
where g(q, t) is any differentiable function of the old co-ordinates and time. Ex 6.6.9. Sometimes, we consider mixed type generating function. For example, in a system of two degrees of freedom, the transformation QI = qI, PI = PI; Q2 = P2, P 2 = -q2 is generated by the function F - qIPI + q2Q2 which is a mixture of the FI and F2 type. This type of transformation is also canonical. Ex 6.6.10. Here we are to derive the perfect differential as a condition for a tr"lnsformation to be canonical. Let us consider the generating function F2 as F2
= LLaijq)P, )
where at) is an orthogonal matrix. Therefore Pk
=
;F2 qk
=L
atjPt . t
Since akZ is orthogonal, we have
of
Pz = LPkakZ and Qz = __2 = L aijqj· k BPz j
Ex 6.6.11. Determine the canonical transformation defined by the generating function FI(q, Q, t) = ~mw[q - ~~~j2cotQ. What happens to the Hamiltonian of the simple harmonic oscillator when transformed (q, p) -+ (Q, P). SOLUTION: Using the transformation equations for FI, Pk H + iHl at ' we have
= f/j;; P k = -
BFI F(t) P = ! : l = mw[q - - - 2 JcotQ uq mw F(t) mwq F(t) = mwqcotQ - --cotQ =? tanQ = - - - - - . w
P
= -BFI -= BQ
2
=?
and K
=
WP
1 F(t) 2 2 -mw[q - --J coec Q 2 2 mw
2Psin Q = [q _ mw BFI H + -B = H t
P
F(t)J2 =? q = F(t) + J2P sinQ rnw 2 m,w 2 mw q . F(t)F(t) -F(t)cotQ + 3 cotQ. w mw
g~~; K =
6.6. GENERATING FUNCTIONS
181
Hence the old and new co-ordinates are given by
When a simple harmonic oscillator is acted upon by a force F(t) then the old Hamiltonian is H = ~ + !mw 2q2 - qF(t). Therefore the new Hamiltonian is given by
K
p2 1 2 2 q. F(t)F(t) + -mw q - -F(t)cotQ + 3 cotQ 2m 2 w mw 1 F(t) 1 F(t) /!fP. = -[mwqcotQ - --cotQf + -mw 2[--2 + -smQf 2m w 2 mw mw q . F(t)F(t) - -F(t)mtQ + 3 cotQ w mw 2 2 1 F F 2P 2 - = wP(cos Q + sin Q) - - - -cosQ 2 2mw w mw
=-
= wP -
1 2 --2F (t) 2mw
{gPo
'/!f
-3F(t)cosQ mw
Ex 6.6.12. Find the canonical transformation for the free particle Hamiltonian. SOLUTION: Let an electric charged particle of mass m, having charge e be moving in an uniform electric field E, where E is pointing along the one dimensional positive q axis. The Hamiltonian H for such motion is given by
p2 H(q,p) = - eEq. 2m Here the transformed Hamitonian K of a non-free system may look like that of a free particle. The Hamilton's canonical equations of motion gives
. qk
DH = -Dpk
an d'Pk
Pan d 'P = = -DH - => q. = -m Dqk
eE
=> q = -P t + qo and P = eEt + Po; qo = q( t = 0), Po = p( t = 0) m t
=> q = -[eEt + pol + qo and P = eEt + Po m
eE
2
Po
=> q = - t + - t + qo and p = eEt + Po. m
m
For the free-particle Hamiltonian for mass m, the transformed Hamiltonian K(Q, P) assumes the form K(Q, P) = [>2. The new co-ordinate and momenta (Q, P) must m,
CHAPTER 6. TRANSFORMATION THEORY
182
be satisfied the Hamilton's canonical equations of motion
. Qk .
8K
= 8Pk
. and Pk
oK
= - 8Q k
=?
P . Q = - and P = 0
=?
p Q = - t + Qo and P = Po·
'm
m.
Eliminating t, between the equations, we get cErn
q=~[p(Q-Qo)]
or, q = eEm[
Q - Qo
P
2
]
2
[IT"
po ,m
+ mlp(Q-Qo)]+qo;p=eE p(Q-Qo)]+Po
+ Q + (qO
- Qo)];p
m
= eEl P (Q - Qo)] + Po·
Thus the required canonical transformation is given by (qO Q2
q = eEm p2
6.7
= Qo = 0; Po = Po = P)
Q
+ Q;p = meE P'
Condition for Canonicality
Let us take time as independent parameter. The conditions of canonicality of a given phase space transformation are seem to be evaluated at constant t, that indicates the purely geometrical nature of the canonical transformations that take place in terms of the phase space co-ordinates alone. We shall now obtain the explicit analytical expression for conditions for which a transformation to be canonical. The condition for canonicality must be such that as to preserve the form of Hamilton's canonical equations of motion.
6.7.1
Exact Differential Form
Let we are to show that perfect differential as a condition for a transformation to be canonical. Let the generating function F which transforms variables (q, p) -+ (Q, P), when the time is held fixed, then
For a transformation to be canonical, the equation (6.13) must be satisfied. Thus """ ~Pkqk - H(q, P, t) k
k
"". =? """ ~Pkqk - " ~PkQk k
. - K(Q, P, t) + dt dF = """ ~ PkQk dF as K = H = dt
k
=? I:[PkdqJ,. - PkdQk] = dF k
6.7. CONDITION FOR CANONICALITY
183
where dqk and dQk are the differential changes in the old and new co-ordinates in real time dt. Thus if the transformation (qk, Pk) - t (Qk, Pk) be a contact transformation, n
then the expression
L: [PrdQT -
Prdqr 1 would be an exact differential and this result
1'=1
can be obtained by using any generating function. Since the transformations in phase space between two sets qk's and Qk'S are essentially geometrical, we can look upon dqk and dQh: as the elements of the static geometrical curves rather than evolving paths, so far as the testing of the canonical condition is concerned. Ex 6.7.1. Calculate pdq - PdQ for the transformation Q = log(1 + .;qcosp); P = 2(1 + .;q cos p).;q sin p. Show that zt is canonical and find the generating function. SOLUTION: The given transformation Q = log(I+.;qcosp); P = 2(1+.;qcosp).;qsinp is canonical if the expression pdq - PdQ is an exact differential. Now, pdq - PdQ
cosp - 2qsinpdp
.
= pdp - 2(1 + y'qcosp)y'qsmp x 2 yq in(1 in ) + yqcosp = pdq + 2q sin 2 pdp - cos p sin pdq
= d(pq)
1 - d[2"(sin2p)q] = a perfect differential.
Since the expression pdq - PdQ is an exact differential, the given transformation is canonical. Now the generating function is given by Q sin 2p -1 e - 1 F1(q, Q) = pq - -2-. q = qcos /q-
F:.,,(p, (.2)
-
(e
Q
./
- l)y q - (e Q -1)2
sin2p Fl - pq = --2-· Q
=
= - sinpcosp
(e Q - 1)2 2 = -tanp(eQ _1)2. cos p
But neither q nor p can be expressed explicitly in P, as a result other two forms of generating function may not be written. However, as
~~ a at> a = IJI = ap
1, the
8q 8p
transformation are invertible, and 4(e Q q
=
-
1)2(e Q + 1)2 + p2 4(e Q + 1)2 ; cosp
Ex 6.7.2. Show that the transformation Q ical. Find generating function.
=
=
2(e Q - 1)(eQ + 1) Jp2 + 4(e Q - 1)2(e Q + 1)2
y'2(je t cos p; P
=
y'2(je- t sin p is canon-
SOLUTION: The given transformation Q = y'2(jc t cosp; P = y'2(jc- t sinp is canonical if the expression PdQ - pdq is an exact differential. Now, PdQ - pdq
= y2qc- t sinp[(2q)-1/2c t cos pdq = sin p[cos pdq - 2qsinpdp] - pdq =
81.
~[-qsm2p - pq]dq
uq 2
y2qc t sin pdp] - pdq
81 + ~[-qsin2p up 2
pq]dp
= dF.
CHAPTER 6. TRANSFORMATION THEORY
184
Since it is an exact differential, the given transformation is canonical and the generating function is F ( q, p) = ~ q sin 2p - pq. From the transformation equations Q = J2qe i cos p; P = y'2ije- t sinp, we have Q . cos P = rrr;:. t => S111 p y2qe
=
Rf,,2 1-
--2t·
2qe
-
So, P(q, Q. t) = qsinpcosp - pq - Q - ei
~J
V2"
1
2t
Q2 _ -1 .j2qe 2qe2i qtan Q
-
Q'
.
Ex 6.7.3. Show that the transformation Q = logei~p)j P = qcotp is canonical. Find generating function. SOLUTION: The given transformation Q = 10gCi~P)j P expression pdq - PdQ is an exact differential. Now, pdq - PdQ
= pdq -
8
qcotp is canonical if the
oQ oQ P[ oq dq + op dp]
dq = pdq - qcotp[ - q = 8q [qp
=
+ cotpdpj 8
+ qcotpjdq + 8p [qp + qcotpjdp = dF.
Since it is an exact differential, the given transformation is canonical and the generating function is F(q,p) = qp + qc;otp. From the transformation equations, we have
6.7.2
Poisson Bracket Form
PBs can be used to. test whether a given transformation is canonical. Now, we shall give another form to the conditions that a transformation (q,p) --+ (Q, P) may be a contact transformation. The fundamental PB as [Qi,n]q,p = ,",(8Q 2 8P) _ 8Qi BPj 8qk OPk 8Pk 8qk
7
= 2:(8 Q i k iJQi = &Qj Similarly, [Qi, Qjjq,p
=0=
8qk 8qk 8Qj
+ 8Qi
= 6i,j =
[QI,' PjjQ,p.
)
8 Pk ) 8Pk 8Qj
[Qi, QjjQ,P and [Pl , Pj1q,p
=0=
[Pi, PjjQ,p.
6.7. CONDITION FOR CANONICALITY
185
The PBs such as [Qi, P j ] = Jij , [P,;, P j ] = [Q~, Ql] = 0 are elementary useful for testing the canonically of any given phase space transformation. Ex 6.7.4. Let (q,p) be canonical variables and let Q = Q(q,p), P = P(q,p) be transformation equation from (q,p) to (Q, P). Show that the transformation will be canonical if [Q, P]q,p = 1. Hence show that Q = q cos fJ - !u;; sin fJ; P = mwq sin fJ + p cos fJ is canonical. SOLUTION: Let the transformation Q = Q(q, p), P = P(q,p) be canonical. Then the expression pdq - Pdq is an exact differential. Also,
pdq - Pdq = pdq -
P[~Q dq + 8 Q dp] 8p
oq
= [p _ p8 Q ]dq _ p8Q dp.
8q
8p
The condition for perfect differential is
~[p _ p8 Q ] = ~[_p8Q]
8p 8q 8q 8p =? 1- 8P 8Q _ p~(8Q) = _ 8P 8Q _ p~(8Q) 8p 8q 8p 8q 8q 8p 8q 8p 8P8Q 8P8Q =? --8 -8 + -8 -8 = 1 =? [Q,P]qp = 1 p q q p , which is the required condition for the transformation to be canonical. Conversely, let [Q, P]q,p = 1 i.e., Poisson bracket of Q and P is given as 1. But the Jacobean of transformation .J from (q,p) to (Q, P), where Q = Q(q,p), P = P(q,p) is defined as J
=
-8(Q P)
'
8(q,p)
£9. £9.
= jJp fjp 8q
ap
= 8P8Q _ 8P8Q
8q 8p
Dp 8q
= [Q,P]
q,p
= 1.
Thus the given transformation is canonical. Now
[Q, P]q-p
8P 8Q
8P 8Q
sinO.
= -8 -8 - -8 -8 = cosf) + -.mwsmf) = 1. qp pq mw
Hence the given transformation is canonical. Ex 6.7.5. Under what condition will the transformation P would be canonical, where a, b, c, d are constants?
= ap + bq, Q = cp + dq
SOLUTION: It would be canonical if [Q, P] = 1, when it is given that [q,p] :::;: 1. To see this, we calculate the PB of [Q, Pl. Then,
[Q, P]
= [cp + dq; ap + bq] = cb[p, q] + da[q, p] = ad -
be.
For [Q, P] to be unity we must have ad - bc = 1.
186
6.7.3
CHAPTER 6. TRANSFORMATION THEORY
Lagrange Bracket Form
Here we shall give another form to the conditions that the transformation (q, p) -+ (Q, P) may be contact transformation. Let the transformation Qi = Qi(q,P, t); Pi = p/,(q,p, t); i = 1,2, ... , n be canonical. Then Pic5qi - Pi 8Q, = an perfect differential.
Since the old variables (qi,Pi) may be regarded as functions of new variables Qi, P~; the necessary and sufficient condition for a transformation to be canonical is that
for a fixed value of time. The conditions for perfect differential gives
However, Qi and Pi are independent variables, so that these conditions reduces to
6.7.4
Bilinear Covariant Form
Let (qI, q2, ... ,qn) be any set of n variables and consider a differential form ( known as Pfaff's expression) in the variables (qI, q2, ... ,qn) as
where Q1, Q2, . .. , Qn are the functions of (qI, q2, ... ,qn)' Let us write
where 8 is the symbol of an independent set of increments. Therefore 6e 1 - (If}2
= 6[Ql d ql + Q2 dq2 + ... + Qndqnl
- d[Q1 8q1 + Q2 c5q2 + ... + Qnc5qnl = c5Ql d ql + ... + c5Qn d qn + Ql c5(dql) + ... + Qn c5(dqll) - dQl c5ql - ... - dQn c5qn - Ql d(c5qn) - ... - Qnd(c5qn).
6.7. CONDITION FOR CANONICALITY
187
S"ince the variations d and <5 are independent we can use the result 6 (dqr) = d( 6qr ). Also, using the results dQr = ~~; dq1 + ... + ~~: dqni 6Qr = ~~; 6q1 + ... + ~~: 6qn, we have n
n
= LLaiJdq~8qji
6(h -dfh
/.=1 )=1
Let (PI, P2, ... ,Pit) be a new set of variables derived from (q1, q2, ... ,q,.) by some transformation and the differential form when in terms of the variables be
and let the quantity ~P, - aap,] be denoted by biJ·. Then since the expression Mh - dfh has obviously the same value whatever be the variables in terms of which it is 8xpressed, we have P
'-Ip)
Il
L ~=1
n
The expression
n
n
'It
L a~jdqi8qJ = L L biJdp~8pj. j=l z=l j=l
n
2: 2:
n
(L
ij dq z 8qJ
is called the bilinear covariant of the form
2: Qidqi.
'=1J=1 i=l It is clear that, the bilinear covariant of a differential form is not affected by the addition of an exact differential to the form, as it (iepends only on the quanti-
J ties ~ - u)Q which are all zero when the form is an exact differential Now, let uqJ (q, (Q1, Q2, ... , Qn; PI, P2, . .. ,Pn ) be variables connected with (q1, Q2, ... ,QniP1, P2, ... ,Pn by a canonical transformation, so that
n
n
r=l
r=l
We know the bilinear covariant of a form is transformed by any transformation into the bilinear covariant of the transformed form. It follows that the bilinear covariant n n of the forms 2: PrdQ.,. and 2: Prdq.r are equal, i.e., ,'=1 r=l n
n
Thus, if the transformation from (q, p)
-t
(Q, P) is a canonical transformation, the
n
expression
2: (8p r dqr -
r=l
dq r 8Pr) is invariant under the transformation.
Ex 6.7.6. Show that the transformation Q = ical transformation.
I¥ cos
p;
P = J2qk sin p is a canon-
CHAPTER 6. TRANSFORMATION THEORY
188 SOLUTION:
If the transformation from (q1, q2,· .. , qni PI, P2, ... , Pn) to (Q1, Q2, ... , Qni n
PI, P2, ... , Pn ) is a contact-transformation, then the expression is invariant under the transformation, i.e., n
I: (JP, dQT 1'=1
.sp
# ~ f£ ~
1'=1
n
=
I¥
COSPi
P = J2qksinp, we have,
J
oQ
~ 2~k sin poq
sinp.sq + .,j2qkcosplip; dQ
~ J2~k cospdp
sin pdq +
dqrJp1')
dPI'JQ,,) = I:(6p.,dqr - dq.rJPT). 1'=1
From the transformation equations Q
dP
L: (Jprdq1' -
.,j2qk cos pdp;
-/¥ + /¥
sin p1ip sin pdp
sin pdq + yi2qkcosPdP][V lk sinpJq - (2qk sinpJp] V(k 2q 2q ~ Vk fJ
dPJQ = [
= 8 Pdq
~
2. sin p cos pdqJq 2q
[f£
sin poq +
.,j2qk cos 1'01'11
= 2. sin p cos pdpJq -
=?
2q dPJQ - JPdQ
sin2 pdqJp + cos2 pdpJq - 2q sin p cos pdpJq.
J2~k
cos pdp +
/¥
sin pdp]
sin2 pdpJq + cos 2 pdpJp - 2q sin p cos pdpJp.
= -sin2p[dqJp - 6qdp] + cos2 p[dpJq - Jpdq]
= dpJq - Jpdq.
Hence the given transformation Q
=
I¥ cos
Pi
P=
V2rik sin p is canonical.
Ex 6.7.7. Prove that the transformation x = X cos A + ~ sin A, y = Y cos A + ~ sin A" Px = -Tnw Y sin A + Px cos A, py = -mwX sin A + Px cos A is canonical. Determine the New Hamiltonian, if the old is H(q,p) = 2!n (p; + p~) + ~(x2 + y2).
The given transformation x = X cos A+ ~ sin A, y = Y cos A+ ~ sin Ai = -mw Y sin >'+ Px cos A, py = -mwX sin A+ Px cos A is canonical if the expression
SOLUTION:
Px
6.8. PROPERTIES OF CT
189
~Pkdqk - ~ PkdQk is an exact differential. Now, k
k
Pxdx
+ pydy =
. sin A [-mwY sm A + Px cos A][COS AdX + --dPv] mw
. ~nA + [-mwXsm>. + PyCOSA][COSAdY + --dPxl ntw P;rdx + ]Jydy - PxdX - PydY = -mw sin ACOS A(YdX + XdY)
+ sin A cos A (PxdPy + PydPx)
r 2 +XdPx ) + cos A(PxdX + PydY)
.
= -mw sm A cos Ad( XY)
mw
. 2 - sm Ad[Y dPy
- sin 2 A(YdPy' - PxdX - PydY
+ X dPx] + sin A cos A d( Px Py )
mw sin >.cos >. = d[-mwsmAcos>,(XY) - sm A(YdPy + XdPx) + (PXPy). mw .
. 2
Hence the given transformation is canonical. Now, using the relations, x = X cos A+ sin >., y = Y cos A + ,,~"w sin >.; Px = -mw Y sin>. + P x cos A, Py = -mwX sin>. + ~Qx cos >., k = mw 2 , we get
!f!
K(Px, Py, X, Y) =
2~ [( -mwY sin A + PTcos A)2 + (-mwX sin>. + P.T cos A)2]
+ ~2" [(X cos A + 2
= ntw (X2
2
6.8
y P sin A)2
+ (y,cos A +
'InW
x P sin >.)2] nLW
p2) + y2) + _1_(p2 2m x + y.
Properties of CT
In this section we shall discuss some properties of canonical transformations. Property 6.8.1. From the transformation equations, in generating functions, diffcnmtiatillg Ollce more the first two relatiolls of each of sets of t.he t.ransformation equations, we get
where 'i; j = 1,2, ... , n. These symmetrical formed equations are to be satisfied by all univalent canonical transformations, which are totally independent of the generating functions or the Hamiltonian H. These sets of equations together are known as the direct conditions for a canonical transformations. Thus given a set of canonical transformations either in the form of point transformations or in the form of transformation of configuration space, they must satisfy all the above sets of equations. Property 6.8.2. Any finite volume in the phase space before and after the transformation remains canonically invariant. This is known as Liouville's Theorem.
190
CHAPTER 6. TRANSFORMATION THEORY
Let ql, q2,···, qn and PI,P2,'" ,Pn be a set of generalised co-ordinates and their corresponding conjugate momenta respectively to characterize a given state of a physical system. Then the corresponding phase-space (q, p) is 2n-dimensional space with ql, q2, ... ,qn and PI, P2, ... ,Pn denoting the 2n-number axes system. The elementary volume elements dVI , dV2 in this phase space (q,p) and (Q, P) are respectively given by PROOF:
dVI = dql.dq2 ... dqn.dPI.dp2 ... dpn dV2 = dQI·dQ2· .. dQn.dPl·dP2···dPn. We are to show that, under canonical transformation this elementary volume is an invariant. Let us consider an univalent canonical transformations carries the set (q,p) ~ (Q,P) as Qk = Qk(q,P,t); Pk = Pdq,p,t);k = 1,2, ... ,n. Then the Jacobian of transformation is defined as
.7= 8(Q,P) = 8(q,p)
Hence the Jacobian determinant of the univalent canonical transformations is unity. Now the change in the volume element in two different bases (q,p), (Q, P) is related with the Jacobian as
dql.dq2 .. ·dqn. dPI.dP2 .. ·dpn
=
IJldQI.dQ2 ... dQn.dPI.dP2 ... dPn
= dQI.dQ2 ... dQn.dPl.dP2 ... dPn
which shows that the phase space volume is invariant through canonical transformations. This is generally valid when the phase space is a constructed in cartesian fashion. Ex 6.8.1. Show that the cartesian to polar co-ordinate transformation in phase space given by (Xl, X2, X3, PI, P2, P3) ~ (1', e, , PI, P2, P3) is not canonical. SOLUTION:
The transformation equations from cartesian (Xl, X2, X3) to polar (T, e,
6.8. PROPERTIES OF CT
191
is given by Xl
= r sin () cos ¢;
X2
= r sin () sin ¢;
X3
= r cos ()
~~~
or 00 04> or 00 o ~~~ or 00 o sin () cos ¢ r cos () cos ¢ -r sin () sin ¢ sin () sin ¢ r cos () sin ¢ r sin () cos ¢ cos () -r sin () 0 ~~~
= r2 sin () i=
1.
Since the Jacobian of the transformation is not unity so the given transformation is not canonical. Now, let us take a transformation (x!, X2, X3,P!'P2,P3) (r, (), ¢, Pr, Po, P
+ P2r sin ()sin¢ + P3r cos ()]
. ().A. • () • .A. () => Pr = - 8F3 8r = PI sm cos 'I' + P2 sm sm'l' + P3 cos Po
=-
3 8:e
= PI r cos () cos ¢ + P2r cos () sin ¢ -
P
=-
8F3 8¢
. () . .A. • () .A. = -PI r sm sm'l' + P2 r sm cos '1"
P3r sin ()
Using these values of Pr,pO and P' we have
p~ + r12 p~ + r 2 sm ~ 2 () P~ = [PI sin () cos ¢ + P2 sin () sin ¢ + P3 cos ()]2 + [PI l' cos () cos ¢ + P2r cos () sin ¢ - P3r sin ()]2 + [-PI r sin () sin ¢ + P21' sin () cos ¢]2 = pi + p~ + p~ = p2 ( say ). Also, for this transformation, J
= 1, so that
the transformation is canonical.
Property 6.8.3. The volume of any arbitrary region in the phase space in canonically invariant. This is known as Poincare '8 Theorem. PROOF: Here we are to show that, the volume JI = L: dqtdpi' taken over an ar-
JJ B
i
bitrary two dimensional surface S on the 2n dimensional phase space is canonically invariant, i.e., (6.19) Let the position of a point on any two dimensional phase space is specified completely by two parameters, u, v as qk = qk(U, v); Pk = Pk(U, v)
=> dqidpi = I~{~:~)) Idudv =
jJldudv;
1 0Ji
~I
J= ~~
av av
192
CHAPTER 6. TRANSFORMATION THEORY
where J
= DJ?~:~))
is the Jacobian of transformation. From equation (6.19), we get dudv = ;.;. L B(Qt, Pi) Jdudv ; .;. L B(%Pi) B( u, v) .. B( v)
..
S
'll"
S
t
t
provided Qi and Pt are derived from qi and Pi through a canonical transformation. The theorem can be proved if we are to prove that the sum of the Jacobian is invariant which would itself imply the invariance of h. In order to prove it, let us consider the transformation (q, p) ~ (Q, P) through the generating function P2(q, P, t). With this form of generating function, we have, DP2 ap2 Pk = aqk and Qk = aPk
"* Bpk = ~(ap2) = L au
and BPk
au aqk
= ~(ap2) = L
ov
"* L k
ov oqk
L
!!Ek !2P..t I au au _ 0A!"!Es 1 av
2P a 2 aql I aqlaqk Bu
av
k
I
2 a P2 Bql aqlaqk av
+L
2 a H aPI BPIBqk au
I
+L
2P a 2 aPI aPlaqk av
I
+ ~ ~ aPI ~ aPlaqk au I ~ ~ t&2 0n. ~ ~{!!1 av 7 ql qk av + 7 aP1aqk av ~ a 2F2 iJ!l1. au ~ aqlaqk Bit I
0z.!E.
B2P2 1$h: ~ 1 a2P2 1~ $l1.1 L aP1aqk #;!}i + L L aq1aqk #; ~ p. &. ~ = L L aa a ~ ;First is antisymmetric k ql qk
= LI
av
k
2
2
I
1 aqk
a p.
I
k
iJv
t
gp 1 ¥Pv
av
2
a p.
$L 1
~ ~ aqla;k fv- ~ + ~ ~ aPIB~k 1!!Ek
iJv
1
av
2
=
av
=
L k
ga. ?!p ¥J1
{!gJ. 1
8v
1!ms.
1
av
Which shows that the integral J1 is invariant under canonical transformation. Extending it to I n , we arrive at .In
=
J... J
dq1 ... dqndp1 ... dpn·
6.8. PROPERTIES OF CT
193
The volume integral 111, is the final member of a sequence of canonical invariants, known as the integral invariants of Poincare, comprising integrals over subspaces of phase space of different dimensions. Property 6.8.4. All univalent transformations possess the group property. PROOF: Let us consider, an canonical transformation. We are to show that this satisfies the four axioms to satisfy the group property.
(i) Let us consider the following two canonical transformation equations
Q
= Q(Q,P, t)
Q = Q(q,p, t)
P=P(Q,P,t) P = P(q, p, t),
in which F, F are generating functions. Now for a fixed time, the condition of canonicality gives,
Pk8Qk Pk8qk -
'*
Pk8Qk = dF(Q, Q);Pk8qk -
Pk8Qk
= dF(q, Q)
Pk8Qk = d(F + F).
This shows that, F + F is generating function of the canonical transformation (q,p) ---+ (Q,p), i.e. the result of performing two successive canonical transformations in sequence is to obtain a change of single variable which is itself a contact transformations. ( Closure Property). (ii) Consider the following canonical transformations
c:
(qCD) , pCD))
B:
(qCl) ,pCl))
A:
(qC2) , p(2))
---+
F(O)
(qCl), pCl))
(qC2) , p(2)) ---+
F(2)
( q C3), p C3)).
Using the procedure as in (i) we see that the generating function for both A(BC) and (AB)C are FCD) + F(1) + F(2) shows that the compositions of canonical transformations is associative. (iii) Consider a generating function F2(q, P, t) with the particular form F2(q, P, t) = qkPk. Then from the transformation equation we have
Hence the old and new generalised co-ordinates are same. This shows that if the transformation (q, p) ---+ (Q, P) is a contact transformation, then the transformation (Q, P) ---+ (q, p) is also contact transformation, i.e., the inverse of a contact transformation is a contact transformation.
194
CHAPTER 6. TRANSFORMATION THEORY
(iv) Let us assume that the derivatives of a univalent canonical transformation Q = Q(q,p, t); P = P(q,p, t)
(6.20)
be continuous. Then the Jacobian of transformation is given by
J = 8(Q,P) 8(q,p)
= 1(# 0).
Since the Jacobian determinant of (6.20) is non-zero at every point in the domain of definition, by inverse mapping theorem of differential calculus, the inverse transformation
q = q(Q, P, t); P = p(Q, P, t)
(6.21)
exists. Now for a fixed time, the condition for canonicality gives
=}
Pk6qk - Pk6Qk = df(q,p) Pk6Qk - Pk6qk = d[- j(Q, P)];
using (6.21).
Hence the inverse of a canonical transformation exists and is itself canonical transformation. (Existence of inverse) Thus the canonical transformations have the four properties that characterize a group. In other words, the algebra of the group are uniformly applicable to all possible examples of canonical transformations.
Theorem 6.8.1. Poisson brackets are invariant under canonical transformations. Let us consider canonical transformation Qk = Qk(ql, ... , qn;PI,·· .Pn; t); Pk = PA: (ql, ... , qn; PI, ... Pn; t) which transforms two dynamical variables u (q, P, t) and v(q,p, t) to ij,(Q, P, t) and v(Q, P, t). PROOF:
6.8. PROPERTIES OF CT
Similarly [u',Prl
= aau. qr
195
Hence from (6.22), we get
[7J., vlQ,p =
. '""
au av
r
Pr qr
L,.. {--a -a
au av
+ -a -a } = [7J., vlq,p' qr Pr
Result 6.8.1. We know Lorentz and Galilean transformations are all canonical transformations. Any poisson bracket defined for a pair of dynamical variables must remain unchanged for all the above general transformations. So any vector or scalar quantity that is expressible as a poisson bracket must remain invariant under rotation, translation, Galilean transformations, Lorentz transformations and so on. Result 6.8.2. We have shown that a canonical transformation can be generated from functions F1 (qk, Qk, t), F2(qk, Pk, t), F3 (Pk , Qk, t) and F4 (pk, Pk , t).
(i) When H is an generating function, we have
(ii) When F2 is an generating function, we have
*
(iii) Similarly, when F3 is an generating function, we have ~ is an generating functi~m, we have
=~
and when F4
= - ~.
Theorem 6.8.2. Lagrange bracket is invariant under canonical transformations. PROOF:
We know that, the volume J 1 =
J J 2:= dqtdpi is a canonical invariant.
Let
S i
the position of a point (q, p) in phase space is specified completely by two parameters a, fJ as qk = qk(o., {J); Pk = Pk(o., {J) so that a(qk,Pk) dqkdpk = [ a(o., {J) [do.d{J = [J[do.d{J
CHAPTER 6. TRANSFORMATION THEORY
196 where J =
!!.!lk. !!.!lk.
!k~ ~ 00<
is the Jacobean. Let (qk,Pk)
-+
(Qk, Pk) be the canonical
a(J
transformation. Using integral invariants of Poincare, we get
where (ct, (3) is the Lagrange Bracket. Thus the Lagrange bracket is invariant under canonical transformations.
6. 9
Equation of Motion
Theorem 6.9.1. The total time rate of evaluation of any dynamical variable u(q, p, t) is ~~ = [u, H] + ~~. This is known as Poisson's first theorem.
PROOF: Here we are to write the canonical equations in terms of the notation of PBs. Now, the PB of qk with the Hamiltonian H is given by [qk, H]
=
aaH PI.;
= qk; [pk, H] = -
aa
H qk
= Pk·
For the dynamical variable u = u(q,p, t), the total time derivative is given by du dt
au.
au.
au
= aqk qk + apk Pk + at
=
au aH _ au aH aqk apk apk aqk
+ au = at
[u, H]
+ au. at
This equation may be looked as the generalised equation of motion for the arbitrary function 1J, in the PB formulation and it is valid in whatever set of canonical variables q, P is used to express the function 1J, or to evaluate the PB. As a result of the canonical invariance of the PBs, the relations so obtained will also be invariant in form under a canonical transformations. In particular, if u = H to a particular set of canonical variables, then ~~ = ~~ as [H, H] = O. Also, if 1J, is a constant
6.9. EQUATION OF MOTION
197
of motion so that ~~ = 0, we have [u, H] + ~~ = O. Furthermore, if u does not contain time explicitly, we have ~~ = 0, so that [u, H] = 0 which is the required condition for u to be a constant in motion. Conversely, the PB of a constant of motion with Hamiltonian H will be zero. Thses properties can be used for finding out the constant of motion.
Ex 6.9.1. Show by use of PEs that for a one dimensional harmonic oscillator there
is a constant of motion u defined as u(q, p, t) = log(p+ imwq) - iwt, w = 28 the physical szgmficance of this constant of motion?
4.
What
SOLUTION: The total time derivative of u(q,p, t) is given by
du
It
=
au aH _ oq op op oq at 2 2 p au 2m w q au au =? m aq 2m ap + at = 0 dp mdq dt du =? =--=-=2 -mw q p 1 0 mdq p2 + m 2w 2l = u = C2; - - = dt =?
=? =?
J
=?
=?
log[
J
c
q
1
p
u
= C2
idq = iwJdt; u = C2; Jc 2 + i 2q2
iq log[ - +
imwq + J c2 =?
cr;
mdq - Jdt. JCT - m 2w 2q2 ,
=?
au
+ at = 0 au oH + au = 0
[u,H]q,p
-
g2
1 - 2" ] = iwt + C3; c
m 2w2q2
] =
log[p + imwq]- iwt
iwt + c3;
'U
= C2
U
= C2
= c; = u(q, p, t).
Theorem 6.9.2. If the dynamical variables 'U and 'V are any two constants of motion of any given holonomic, dynamical system, then PE is also constant of motion.
PROOF: The total time derivative of a dynamical variable F(q,p, t) can be written as~' ~)~ + [F, H]. Since u, 'V are constants of motion, we have ~~ = 0 = ~~.
CHAPTER 6. TRANSFORMATION THEORY
198 Taking, F
= [u, v]q 1"
we get
d D dt [11, v] = at ['U, v] =
au
[at' U]
+ [Il,
av
Dt]- [[v, H], uJ
-
1
[[H, u], VJ
O'U
. (1) at + rlil, H ],v] + lu, at + [v. H ]'J ,d-u dv 1- v] + [u -1 = o. 'dt' , dt·
= [ =
+ [[11, v], H]
Thus the PI3 of two constants of motion is also a constant of motion. This is known as Jacobi-Poisson theorem on the PB relation. This theorem leads to the possibility of constructing new constants of motion from known ones. Deduction 6.9.1. Equations of motion in Poisson bracket form: The total time derivative of a dynamical variable F(q,p, t) can be written as rlF
of
.
dI = at + IF,HJ. If F does not contain the time explicitly, then The total time derivative of a dynamical variable F(q, p, t) can be written as ~f = [F, H], giving that if the poisson bracket of F with H vanishes then dynamical variable F will be a constant of motion. This requirement does not, however, require that H should be a constant of motion. Let such dynamical variables be qJ and }J), then
dpi = [p H] dt
(6.23)
.1'
which are identical with Hamilton's canonical equations of motion, since
lq), H] =
L(aqJ 8H k
_ 8qj 8H)
8qk 8Pk
DH
8H
apk
Dpj
= - JkJ = -
8Pk aqk
Dqj
as -
8Pk
=
o.
Equations (6.23) can thus be referred to as the equation of motion in poisson bracket form. Also, when the poisson bracket [p), H] vanishes, then Pj =constant, i.e., linear momentum is conserved, which implies that corresponding co-ordinate is cyclic.
Ex 6.9.2. Show that if the Hamiltonian and the quantity F are constants of the motion, then the nth partial derivative of F with respect to t must also be constant of the motion. SOLUTION: Given that, the Hamiltonian H and the quantity F are two constants of motion, so
dH = 8H dt dF
di
=
8t 8F
[H H] = 8H = 0
+,
at + [F, H] = O.
8t
6.9. EQUATION OF MOTION We first prove that,
199
c:: is a constant of the motion. Now,
Since Hand F are two constants of motion, the Poisson bracket [H, constant of motion and so d
FJ
is also a
of
d t
-d [H,FJ = O,i.e., -d (-0 ) = O. t
t
¥t-
This shows that is a constant of motion. Now since, Hand !!Jf are two constant of motion, their Poisson bracket [9ft, H] is a constant of motion; which proves that ~l is a constant of the motion. Hence by principle of'mathematicJl induction, ~e conclude that ~:!' is a constant of motion if Hand F are so.
*.
Ex 6.9.3. Consider the uniform motion of a free particle of mass m. The Hamil~ tonian is certainly conserved and there exists a constant of the motion F ,= x S{I,ow by direct computation that the partial derivative of F with t, which is a constant of motion, agrees with [H, FJ.
SOLUTION: For the uniform motion of a free particle of mass m, we have, T=
=> L
~mx2. 2
=T
V=O
'
- V
= ~m:i:2
oL
2
.
.
p
=>P=-o' =mx=>x=-. x m Thus, the Hamiltonian H of the system is given by, p2 H=px-L=-. 2m
Therefore,
Thus, H is a constant of the motion. Also, given thp.t, ,f. =: x - ~, so, ,
dF of dt = at + [F,H] = of + (oF.oH
ot p
ox op
P
t
_ of.oH) op ox
= -+ 1.+ -.0 = O. m m m
•
CHAPTER 6. TRANSFORMATION THEORY
200
Thus, F is a constant of the motion. Thus [H, F] is a constant of the motion. Now,
uH uF aH uF [H,F]=-a '-a --a '-a x P P x
= o.(-~) - ~.1 = -~. m
m
Since Hand F are constant of motion, so is
m
0:.
Now,
of = _~ = [H,F].
at
Tn
Hence the result is verified. Ex 6.9.4. Write Hamilton's equation in matrix form and obtain the sympletic condition for a canonical transformation. SOLUTION: Hamilton's canonical equations are prescribed in a phase space. Let "7 be the column matrix of order 271, x 1, defined by "7 = (ql, q2, ... , qn;PI,P2,'" ,Pn)T. Let J be the antisymmetric matrix of order 271, x 271, given by,
=
J
(~~)
where Q is the null matrix of order 71, and I is the identity matrix. Hamilton's canonical equation can be written as
.
"7=J
aH
ary ·
Therefore,
(6.24)
Let us consider the canonical transformation, where time does not appear and is given by Pi = Pi(qbPk), Qi = Qi(qk,Pk), i.e., a transformation from 2n variables (ql,q2, ... ,qn;PI,P2, ... ,Pn) and (QI,Q2, ... ,Qn;PI ,P2, ... ,Pn) = ((say). For the restricted canonical transformation, ( = (("7) and clearly,
So, in matrix notation, we have,
.
aH
(=M,,=MJ-
ary
where NI is a 271, x 271, matrix, whose (i, j)th element is given by ~. This matrix 111 is called Jacobean matrix of transformation. Now, by the inverse transformation H can be considered as a function of ( and in fact
6.10. INFINITESIMAL CONTACT TRANSFORMATION
201
Now, for restricted canonical transformation, the Hamiltonian H is same as the new Hamiltonian K and therefore K = H. Since the Hamilton's canonical equations of motion for the ( system is
(= J~~. For a restricted canonical transformation H MJM
T
=K
and hence
= J.
(6.25)
This is known as sympletic condition for the canonical transformation and the matrix AI satisfying the condition is called sympletic matrix.
6.10
Infinitesimal Contact 'fransformation
If a canonical transformation is introduced very near the identity transformation, then it is known as infinitesimal canonical transformation or infinitesimal contact transformation. As in case of infinitesimal rotations such a transformation is one in which the new co-ordinates differ from the old by i.lfinitesimal amounts. Let the transformation equation, when we retain term correct to first order terms, can be written as (6.26) where 6qk and 6Pk are the infinitesimal changes in the co-ordinates and momenta. Thus to find a transformation mean the identity transformation, the generating function F2 (q, P) will also differ only by an infinitesimal amount and is given by
F2(q, P) = qkPk
+ EG(q,p, t)
where E is the some infinitesimal parameter of the transformation, indcpcndent of qk, Pk and G is arbitrary differentiable function of its (2n + 1) arguments. The transformation equations become are
Pk
aG
aG
= Pk + E a - ; Qk = qk + Eap qk
=> 6Pk = Pk - Pk =
k
aG -E aqk ;
6qk
aG
= Qk - qk = E aPk .
These equations gives the infinitesimal changes in the conjugate variables which are generated by arbitrary function G. Since the terms under consideration is of the first order in E, so Pk - Pk is infinitesimal. Thus, we can replace Pk by Pk and G(q, P) ~ G(q,p), which gives (6.27) The diffen~nce involved ill this replacemcnt is of second order of smallncss. The new generating function G(q,p) is called infinitesimal generator of the infinitesimal canonical transformation.
CHAPTER 6. TRANSFORMATION THEORY
202
..
Deduction 6.10.1. Here we are to derive if G is the infinitesimal generator of the canonical tnt~sformation, then the change of Hamiltonian is given by the Poisson bra,cket o(H with G. The change in the value of the Hamiltonian H = H(q,p) as a result of substitution of Q for q and P for p at instant t and t + dt, in one dimension nndc£ the infil¥tesimRI transformation it; 8H oH_ fJH = -fJp + -oq 8p oq 8H 8G 8H 8e = --;:;}€!=} + -8 €-8 = €[H, Gl q p
fJX
= X(q,p, t),
= €[X, Glq,p'
which is the relation between infinitesimal transformations and Poisson Brackets. In particular, if € = dt, then fJX = [X, Glq,pdt. Let X = qk, then
fJqk
= dt[qk, Hl = qkdt = dqki
and fJpk
= dt[Pk, Hl = Pk dt = dpk·
Hence the Hamiltonian H is the generator of system motion. Thus for an infinitesimal canonical transformation, if we take, the arl;>itrary function G is the Hamiltonian H(q,p) and € is an infinitesimal interval of time dt(i.c.,€ = dt), then the correspondiug iufinitesimal changes in the co-ordinates and momenta when the system moves for a time dt are
Oqk fJpk
=
= dt8~H = dtqk = dqk Pk
oH -dtaqk
= dtPk = dpk
which occurs when the system of its motion. Thus the infinite transformation changes the co-ordinates and momenta at time t to the values they have at the time t + dt. Hence the motion of a system during an time interval dt can be described by an infinitesimal canonical transformation generated by the Hamiltonian. The motion of the system may be regarded as continuous evolution of a canonical transformation corresponds to the motion of a mechanical system. Hence, Hamiltonian is the generator of the system motion.
Deduction 6.10.2. Let G(q,p) be a constant of motion, i.e., does not contain the time explicitly. Then
6H
= [H, Gl = o:::} H
=
Constant in motion.
It generates an infinitesimal canonical transformation that does not change the Hamiltonian. Hence, if G does not contain the time explicitly, then the infinitesimal generator G of au infinitesimal canonieal transformation is a constant in motion. Equivalently " the constants of motion are the generating functions of those infinitesimal canonical transformation that leave the Hamiltonian invariant" .
(j.l0. INFINITESIl\IAL CONTACT TRANSFORlvIATION Deduction 6.10.3. Under the infinitesimal time translation t tOll remains invariant. l.e.,
dH
203 ~
t + 8t, the Hamil-
DH
8H=O=> - = =0. dt at => Energy is constant of motion.
Thus, Energy is the generator of the infinitesimal time-translation, i.e., Hamiltonian II is itself the generator of the time translation.
Deduction 6.10.4. Let us consider translation q displacement of co-ordinate q, i.e., 8q = E. Hence JqE =
DC
BC
=> -
f,-
op
op
~
q+
E,
so
E
is the infinitesimal
= 1.
Under the space translation p does not change, so, 8q
UC
DC
oq
Dq
= o.
= 0 => E-.- = 0 => -
=> C = p = momentum of the system Thus for an infinitesimal space translation q - t q + E, the generator of the transformation is the momentum p. If the Hamiltonian H is invariant under the infinitesimal space translation, then the generator i~ constant of motion. This is conservation of linear momentum. Thus if a co-ordinate is cyclic, its conjugate momentum is a constant of motion.
Deduction 6.10.5. Here we are to show that the angular momentum is the generator of the infinitesimal rotational motion of the system. Let us consider the infinitesimal contact transformation of dynamical variables that produces a rotation 8 c/J of a system so that the co-ordinates r changes to r +8r where, 8r = J c/J x 7. Let the axes be so oriented that the infinitesimal rotation is along the z axis, i.e., ~
---4
r
.........
=
(dx, dy, dz);
c/J = (0,0, J¢).
Hence the infinitesimal changes in the co-ordinates are
15x = -yJrp.
Jy = xJcp,
Jz = 0
and similarly the equations involving the changes in the components of momenta conjugate to particle co-ordinates are
Jpx = -pyJrp, Again from the relation, !5q =
f.
Jpv
= Px r5¢, Jpz = O.
~;, we get
DC
J.r =
E-.-
8y =
E-;--)
opx
= -yJ¢
fJG
= - yE => -.-
opx
BC. ( Py
= x15cp =
XI'
= -y
HC => -D = :c. Py
CHAPTER 6. TRANSFORMATION THEORY
204
Hence the infinitesimal generating function G becomes,
G
= XPy -
Y]Jx
=
Lz
= (r x p) z
so that fix infinitesimal three-dimensional rotation the infiuitesimal generator is the angular momentum with de in the present case as the infinitesimal parameter L Since the z axis is chosen arbitrary, generally the generating function corresponding to an infinitesimal rotation about an axis -t
G= L.'n where n is the unit vector along the direction of the infinitesimal rotation vector. Thus, angular momentum is the generator of the infinitesimal rotational motion of the system. Hence if Hamiltonian is invariant under infinitesimal rotation, the angular momentum is conserved.
6.11
Finite Transformations
The fillit(~ transformation could be looked on as a sum of an infinite succession of infinitesimal canonical transformations, each corresponding to an infinitesimal displacement along the continuous trajectory in phase space. By integrating the expression for the infinitesimal displacements, it is possible to obtain the finite transformation. Note that, each point on the trajectory in the phase space corresponds to a particular value of the parameter a, say, starting from the initial system configuration say a = o. Let 1L = 1L(a) be some continuous function of system configuration along the trajectory where 1L0 = 1L(0). The infinitesimal change of 1L on the trajectory is given by
61L = [1L, G]q,pda
=?
du dO'
= [u, G].
By integrating this differential equation, we can get u( a) and the effect of finite canonical transformation. Also, by the repeated application of the above equation, taking [u, G], flu, G], G], ... , etc as a function of the configuration.
d2 u -d 2 0:
=
d3 u da 3
= do([[u, G], G])
d d1L
-Z (-Z ) (0:
d
(a
=
d
-z [u,G] = [[u,G],G] (a = [[[u, G], G], G], ... etc.
Thus the formal solution can be obtained by expanding u(a) in a Taylor's series as
6.12.
LIOUVILLE'S THEOREM
205
where the zero subscript meaning that the value of the Poisson bracket taken at the initial point a = o. Let us consider a particular case, G = II and the parameter is the time t. Then the equation of motion for u is ~~ = [u, II] and the formal solution for u(t) is given by
u(t)
t3
t2
= uo + t[u, II]o + 2! flu, II], II]o + 3! [flu, II], II], H]o + ...
where the zero subscript refers the initial conditions at t
= o.
Ex 6.11.1. Obtain the motion in time of the linear harmonic oscillator by means
of the formal solution for the Poisson bracket version of the equation of motion. Assume that, at time t = 0, the initial values are xo and PO. SOLUTION: The Hamiltonian H for a one dimensional harmonic oscillator is H(q, p) = + ~x2 where x = the position co-ordinate, m = mass and k is the force constant. From the definition of Poisson Bracket, we have
Fm
[x H] = ax aH _ ax aH , ax ap ap ax [[x,H],H]
= 1 E- _ 0 x (kx) = E.m
m
= [E-,II] = ~[p,H] = ~{apaH _ apaH} m
TIL
TIL
1
p
k
m
m
m
Dx Dp
Dp ax
= -{O- - l(kx)} = --x. [[[x, H],H], H]
k
k
m
'Tn
k ax aH m Dx Dp
ax oH Dp ax
= [--x,H] = --[x,H] = - - { - - - - - } =
k p - - { l - - O(kx)} m
m
=
kp --2
m
=
constant.
Since [[ [x, H] , H], H] is constant, the higher order bracket vanishes identically. The formal solution for the Poisson Bracket version of the equation of motion is given by
x(t) = Xo
t2
t3
+ t[x, H]o + 2! flu, H], H]o + 3! [flu, H], H], H]o + ...
Po kxo t 2 kpo t 3 =·-z:o+-t- - - - - 2- . m m 2! m 3! This is the familiar elementary solution to the problem.
6.12
Liouville's Theorem
This is fundamental theorem of statistical mechanics. We know, at some instant of time, every possible dynamical state of a system in classical mechanics can be uniquely represented by a point in the phase space. This point in the phase space
206
CHAPTER 6. TRANSFORMATION THEORY
is known as the image point of the system. Also, this point will not be stationary bur will move along a definite trijectory determined from the equation of motion q] = PJ and pj = - q) where H is the Hamiltonian of the system. Now, if we consider many mechanical systems with slightly different q's and p's, all of them can be represented by a set of distinct points in the phase space, occupying an infinitesimal volume in it. In course of time, the system points move along different paths in the phase space due to different initial conditions. As a result, the density of systems in the neighborhood of some given systems in phase space changes with time. Liouville's theorem states that "the density of systems in the neighborhood of some given dynamical system in phase space remains constant in time." PROOF: Case 1:0ne degree of freedom: We shall first prove the theorem with one degree of freedom. Let p be the density of a dynamical system with one degree of freedom, in the neighborhood of a given system of point in two-dimensional phase
gH
gil,
p
C '~ A Ii B D
L -_ _ _ _ _ _ _ _ _ _ _
q
Figure 6 1..\rea element in phase space
space described by q and p co-ordinates. When we are to evaluate the density p at a fixed point in the phase space, it may still vary with time. Since p = p( q, p, t), the total time derivative is given by dp dt
as
(j)' .
~H, PJ
up).
-
=
ap at
ap .
ap .
+ aq q + apP =
ap at
~H, where the Poisson bracket
uq)
.
+ [p, Hj; [p, H] arises from the implicit
dependence and !fit from the explicit dependence. Let us consider the area element dqdp of two dimensional phase space. At time t = 0, the boundary of the area is formed by some surface of neighboring system points. As the time progresses, the different shapes of infinitesimal areas, defined by system points will move in phase space. Let the speed with which the representative points enter the element ABeD through AD be q. Then, the number of representative points entering the element through AD per unit time will be pqdp. Similarly, the number of points leaving the element through BC will be pqdp + gq (pq)dqdp. Thus the number of representative points that remain in the element ABC D are found to be pqdp - pqdp - :q (pq)dqdp = - :q (pq)dqdp.
Similarly, considering the number of representative points extering through AB and leaving through BC, we find that the number of points thnt remains in the element
6.12.
LIOUVILLE'S THEOREM
207
ABC D are found to be - gp (pp )dqdp. Hence the increase in the representative points per unit time in the typical element as -[gq (pq)
+ ~ (pp)]dqdp.
But the rate of increase of points per unit time in the above typical element id ?!j[dqdp. Therefore, '::: =
dp or, -d t
-[:q aq
(pq)
+ ~ (pp)]
ap
+ p( -aq + -ap ) = O.
(6.28)
Now, we know the motion of a system point in time is simply the evolution of a canonical transformation generated by Hamiltonian. Therefore, the co-ordinates (q, p) at time t2 are related to the variables at time tl by a particular canonical transformation. The change in the infinitesimal area element about the system point over the time interval is given by same canonical transformation. Recalling t 's canOl1lca . 1 equat'IOns qj . = op]' 8H· H amI' Ion Pj = - 8H' oqj' I.e., ~ dq + £i!. up = 0 , tlIe equat'Ion (6.28) becomes,
ap =}
at
+ [p, H]
= 0
dp =}
dt = 0
which is the Liouville '8 theorem. Case 2:n degrees of freedom: Let p be the density of a dynamical system with n degrees of freedom, in the neighborhood of a given system of p,oint in phase space. Through the phase space, there will be an implicit an explicit dependence of time because the 2n co-ordinates (qk,Pk) ofthe system vary with time. Since p is function of (q,p, t), the total time derivative is given by
dp
dt
=
ap
at
~ ap .
ap.
+ L.Ja-qk + a k=l
qk
Pk
Pk ] =
8p
at
+ [p, H]
where the Poisson bracket [p, H] arises from the implicit dependence and !Jff from the explicit dependence. Consider an infinitesimal volume of a typical element as
surrounded a given system point in phase space. At time t = 0, the boundary of the volume is formed by some surface of neighboring system points. As the time progresses, the different shapes of iufinitesimal volumes, defined by system points will move in phase space. The transformation in the infinitesimal volume is a canonical transformation which is a composition of an infinite sequence of infinitesimal canonical transformations. Now Poincare '8 Theorem states that the volume of any
208
CHAPTER 6. TRANSFORMATION THEORY
arbitrary region in the phase space in canonically invariant, by the virtue of the fact that its Jacobian is unity. Thus, the size of the volume element about the system point cannot vary with time. Thus,
So both the number of systems in infinitesimal region dM and the volume dV are constants, and so the density p = ~t! must also be constant in time. This shows that the numerical density satisfies the equation = 0, which is the Liouville's theorem. From this theorem, we have the following discussions :
*
(i) The above theorem can be written alternatively by the partial differential equation + [p, H] = o. In statistical arguments it is useful to employ, instant of the numerical density p, a probability density (J obtain by dividing p by the total number of dynamical system considered.
!fit
(ii) When the number of systems dM in a given state is constant in time, then the systems is in statical equilibrium. Hence the density p of system points at an instant in phase space does not change in time. Thus the statistical equilibrium condition can be written as [p, H] = o. The characteristic of the essence will be determined by the choice of the function p. Conversely, we can insure equilibrium by choosing the phase density p to be function of the constants of motion of the system, for then its PB with H would vanish. (iii) For such a conservative system, p is a function of the energy E, i.e., stationary and the statistical equilibrium condition is automatically satisfied. So a region of a phase space containing the particles evolves as a whole but the density of the pha.'le particle remain unchanged. (iv) Liouville's theorem can not be applied to the dissipative system for which the constant energy surface containing the trajectories in the phase space keeps on shrinking in volume due to dissipation.
6.13
Exercise
Ex 6.13.1. Calculate the commutators [x, L z ] and [x 2 , L z ].
GATE -1996
Ex 6.13.2. A particle of mass m is subjected to a potential V(x) = kx; k > O. If H is the Hamiltonian of the particle, calculate [H, V(x)]. GATE - 2001 Ex 6.13.3. Find the relation between rotation and angular momentum through the PB relations.
6.13. EXERCISE
209
Ex 6.13.4. For what value of a and j3, the transformation Q qCt sin j3p is canonical?
=
qCt cos j3p and P
=
Ex 6.13.5. The transformation Q = y1qcos2p,P = y1qsin2p is canonical. Show that the generating function is Fl = ¥cos -1 ( ~) - ~ q - Q2. Find also F2 , F3, F4.
J
Ex 6.13.6. Find out the canonical transformation in which the generating function F2 is given by F2(q, P) = qcot- 1 ~ + P. Ex 6.13.7. Using fundamental PB, find the values of a and j3, for which Q = P = qf3 sin ap represents a canonical transformation. Find the generating function F3 (Q, P) for this case. qf3 cos ap,
Ex 6.13.8. Prove that, the form of the Lagrange's equations is invariant under a point transformations. Ex 6.13.9. Prove that, the transformation Q canonical, where a is any constant.
= tan":'I(~n, P = ~aq2[1 + Ct~:2] is
Ex 6.13.10. Show that the Hamiltonian for the simple harmonic oscillator, namely H = ~ + !mw 2 x 2 , can be written in the form H = waa*, where a = vfTI¥(x+~) and a* its conjugate. Evaluate the PBs, [a, a*], [a, H], [a*, H]. Ex 6.13.11. A particle of mass m is acted upon by a constant force F. Using the techniques of PB show that x and p are in the forms x = Xo + ~} + !!fit 2 and p = Po + Ft. Ex 6.13.12. Find the position x and momentum p at time t for the motion of a particle given by the Hamiltonian H = ~ + !mw 2x 2. Ans: x = Xo cos wt + ..E.!L sin wt; p = -mwxo sin wt + Po cos wt. m,w
"This page is Intentionally Left Blank"
Chapter 7
Hamilton-Jacobi Theory We have seen that if the Hamiltonian H is conserved, we can solve the problem by the transforming to new canoilical co-ordinates which would all cyclic. HamiltonJacobi equation provides a general method to find the trajectories of the particles with Lagrangian and Hamilton's canonical equation. Mechanical problems are easily solved by the general procedure named canonical transformations. There are two ways of effecting such transformations. (i) If the Hamiltonian H is conserved, we can solve the mechanical problem by transforming old set of co-ordinates into new set of co-ordinates which would be ,all cyclic, as the equation in the new form are easily integrable. Consequently all momenta are constants. (ii) Alternative way to obtain the solution the problem is to seek canonical transformation from co-ordinates and momenta (q, p) at time t, to a new set of constant quantities, which may be the known 2n initial values (qO, po) at t = O. Let the relation between the old (initial values) and new canonical variables for a mechanical problem be
q = q(qo,po, t); p
= p(qO,PO, t)
(7.1)
which is actually the desired solutions of the problems as the function of initial values and time. This is most general procedure and is applicable at least in principle, to a case when the Hamiltonian involves time.
7 .1
Hamilton-Jacobi Equation
The equations of motion in terms of the transformed Hamiltonian K and the new K(Q, P) and the old Hamiltonian H(q,p) are
. Qi
=
oK·
oK
BPi;
BQi ; K
P.l = -
211
of = H + at
CHAPTER 7. HAMILTON-JACOBI THEORY
212
where F is the generating function of the transformation. As the new variables Qi and Pi are constant co-ordinates so, . Qi
.
= 0 = Pt
BK BPi
=? -
BK OQi
= 0 = -,-.
If the transformed Hamiltonian function K to be zero, then the system will be said to be transformed into the equilibrium problem. Thus if the transformed Hamiltonian K will be zero, then the generating function F satisfies the equation
H(q,p, t)
+
of
Bt = O.
(7.2)
Now the generating function F is a function of old co-ordinates q1., the new constant momenta Pi and the time t, so we take the generating function of the form F2 = F 2 (q, P, t). For such a generating function, the equation of transformation is Pi = HF2~'l,TJ,t) , so that the equation (7.2) assumes the form q,
BF2
oF2
H(q1, . .. qn; ~ •... ,-;-; t) uQ1 uqn
BF2
+~ = O. uqi
(7.3)
This first order partial differential equation in (n + 1) variables q1, q2, ... , qn, t of the desired unknown function F2, is known as the time dependent Hamilton-Jacobi equation for a holonomic system having n dJ. Like the Lagrangian and Hamilton's canonical equation, Hamilton-Jacobi equation also provides a general method to find the trajectories of the particles.
7.2
Hamilton principal function
We know, every first order partial differential equation like (7.3), has a solution, depending on an arbitrary function; such a solution is called general integral of the equation. The general solution of equation (7.3) is called Hamilton principal function and is denoted by S, which will merely provide dependence on old-coordinates and time as 5 = S(Q1,Q2, ... ,qn,t). Every first order PDE has a solution, termed as general integral, depending on an arbitrary function. But general integral of HJ equation is not very useful for mechanical problems, so we need to know the complete integral of such equations which contains many arbitrary constants as there are independent variables. Since the PDE (7.3) in (n + 1) independent variables qI, £]2,···, qn, t, its complete solution must involve (n + 1) independent constants of integration aI, 02, ... a n +1. However, S does not appear directly in PDE (7.3), only partial derivatives with respect to Q and t appear in (7.3). Further S is a solution of (7.3), (S+k) must also be a solution, where k is some constant. Hence out of (n+1) constants of integration mentioned above. one is an additive constant attached to S. But as this additive constant will have no effect in a generating function, we can take S to be involving only n constants aI, a2, .. . , an, and the complete solution of (7.3)is, therefore, of the form
(7.4)
7.2.
HAMILTON PRINCIPAL FUNCTION
213
in which none of the n constants aI, a2,. " ,an (may depend on time) is additive and the constant A appears as an additive constant. Thus S is a function of n co-ordinates qi, n constants at and time t.
7.2.1
Connection with canonical transformation
Let a solution containing n arbitrary constants in addition to the additive constant, be known. We now, investigate how the solution, given by (7.4), is related to the solution of the equation of motion variables qk(t); k = 1,2, ... n. To find this relation we employ canonical transformations. Consider the canonical transformation (q,p) ~ (Q, P) via the generating function F 2 (ql, q2,.'" qn, PI"'" PrI , t), where we can take liberally the n constants aI, a2, ... , an of integration to be the new momenta at = Pi; (i = 1,2, ... , n). Since qk's are old co-ordinates and ak's are new moInenta, so S(ql, .. . , qn, 0'1, Ct2,· .. , CY n , t) corresponds to F 2(ql, ... , 'In, PI, . .. , Pn , t), i.e., F 2(r./1, ... ,f]n,Pl, ... ,Pn ,t) = S('ll, ... ,f]n,O'l, 0'2, ... ,an ,t). In order that the transformation be canonical we must have (i) First, let us write the transformation equation as
which are n complete equations and at the time t relating a's in terms of initial values q and P, as
=
0 these give n values
It does not indeed satisfy the definition of momenta in terms of Hamilton's principal functions.
(ii) Further, the second transformation equation which provide the new constants co-ordinates appear as (7.6) which are n-equations and at time t = 0 give the n values of (3i in terms of the known initial values of qt as qi = qi(fJi, at, t), and so Pi = Pt(fJi, ai, t). This set of equations defines a set of new co-ordinates Qi in terms of qt, ai, t. (iii) The relation between the new Hamiltonian K(Q, P, t)
aF2
= H + at
J(
= H
and old H as
as + at =
0
(7.7)
The equations (7.5) and (7.7) clearly identify ('i)S to be Hamilton's principal function (ii) the desired complete integral of the HJ equations of motion. As the transformed
214
CHAPTER 7. HAMILTON-JACOBI THEORY
= 0, the corresponding Hamilton's equations of motion . oK . oK Qt = o~. = 0; Pi = - OQi = 0 =* Qi = canst. = f3i and P = canst. = ai
Hamiltonian K(Q, P, t)
t
where D.! are the constants of integration in the HJ equation. Also Qi = f3, = aas a, is the new results of the identification of S with the generating function F2, i.e., F2 = S. fJt are called the second integrands of motion. Thus the set (f3i, D t ) is canonically related to the set (lji, Pi) through a generating function the same as Hamilton's principal function. Thus the 71, co-ordinates qk's can be expressed in terms of time t and 2n constants ak, fJk. This gives the solutions of the equation of motion for the q's. Also, since S satisfies Hamilton's equation, the Hamiltonian function of the new system is zero, and consequently the equations of the system are da t dt
df3t
"
= 0; ill = 0; z =
1,2, ... ,71,
so that (aI, a2,···, an, (31, f32, .. . , f3,,) are constant throughout the motion. It follows that, if S denotes a complete integral of Hamilton's partial differential equation, containing 71, arbitrary constants (1:1, 0'2, ... , an, the the equations {3,
=
aS
-~; U(l:i
Pi
=
aS
!:l';i uljt
=
1,2, ... ,71,
constitute the solution of the dynamical problem, since they express the variables (fJl, .... ljn, P1, P2, ... ,]In) in terms of t and 271, arbitrary constants (Dl, D2, ... ,O:n, (31, /h, .. . , (3n)· In this way the solution of any dynamical system with n degrees of freedom is made to depend on the solution of a single partial differential equation of the first order in (n + 1) independent variables.
7.2.2
Significance of S
Hamilton's principle function S is the generator of the canonical transformation to constant canonical co-ordinates and momenta. The choice of the new momenta (l:i, to a certain extend, is arbitrary, so one can choose any n quantities "Ii which are independent functions of the constants of integration O:t, i.e., "It = "It(a1,··· an). According to this choice, the Hamilton's principal function is S(qi, "Ii, t) and the rest derivations remains unaltered. In order to understand the physical significance of Hamilton's principal function S, let us write the total time derivative of S as
d~ dt
=
L. t
aS q.t + oqi
= LPtqt =* S =
J
H
L. t
aS C¥i oat
=L
Ldt + constant
+ aS at
7.2.
HAMILTON PRINCIPAL FUNCTION
215
where g;i, = Pi ai = 0, as a, are constants and H + ~~ = 0. Thus the Hamilton's principal function S differers from the indefinite time integral of the Lagrangian by a constant term. If qt and p, are known functions of time t, then the integration is performed and the problem is solved but practically it has no use. Thus the same integral when indefinite form shapes the Hamilton's principle which results in the Holution of mechanical problems, while the indefinite form it shapes the Hamilton's principal function that provides an alternative way of solving the problem.
Theorem 7.2.1. Let S(q,p, t) be any complete integral of the Hamilton-Jacobi equation ~~ + H (q, ~~, t) = 0, then each motion with associated momenta given by ~~ = P = - (3 is a natural motion satisfying the canonical equations. and
g;
PROOF:
The Hamilton-Jacobi equation for a dynamical system is
as at
+ H(q,p =
oS
oq' t)
= 0.
(7.8)
Let S = S(q, a, t) + A be the complete integral of the HJ equation (7.8), which contains (2n + 2) independent quantities ql, ... , qn, al, ... , an, t, A. Differentiating equation (7.8) partially with respect to ak and qk respectively, we get
a2 s oakat a 2s and - aqk ot
11.
a2 s
oH
+ ~ OPr oakaqr = aH
+-
oqk
°
02S 2: -aH = 0. apr aqkaqr 11.
T
(7.9) (7.10)
r=l
g;
Further, we are given = -(3, which represents n equations. Now, differentiating this equation with respect to t, we get
a2s
11.
a2s
ataak
r=l
aqroak
- - + 2:
qr =0.
(7.11)
This equation represents n linear equations and can be solved to get ql, q2, ... , tin. It may be noted that, equations (7.9) and (7.11) are precisely of the same form ulI , ... , aoH . Thus we have the first and so equation (7.9) can be solved to get auH ,a PI P2 Pn Hamilton's canonical equation qk = aaH. Also we note that, aasq = p. Differentiating Pk this equation partially with respect to t, we obtain 02S ata qk
11.
o'1.s
+ 2: r=l aqr a qk tir = Pk·
(7.12)
Comparing equations (7.10) and (7.12), we have at once, the second canonical equation Pk = - aaH . Thus the two equations constitute HamiJton's canonical equations qk and hence each motion described by the characteristic function S is a natural motion. This is known as Jacobi's theorem, follows from Hamilton's equation of motion in terms of the transformed quantities, with the fact that K = Constant = E and it will help us to determine the natural motion for the given dynamical system.
216
CHAPTER 7. HAMILTON-JACOBI THEORY
Ex 7.2.1. Simple Harmonic Oscillator The form of the Hamiltonian H of the one dimensional harmonic oscillator as 2 = + ~ kq2, which is independent of time, k = force constant. Put p = ~~, where S = Hamilton principal function, in the equation of H, we have I-I (q, p)
tm
H(q,p)
= 0,
Now the new Hamiltonian K
1 as 2 1 2 = -(-) + -kq
aq
2m
2
for that requirement, the HJ equation becomes
as
K=H+-=O
1 as 2 =:;. - ( - ) 2m
aq +
at
1 2 -kq 2
+ -as at = O.
Since the only term that involves an explicit dependence of S on it is the last term, so the solution can be written in the form of separated variable as
S(q, 0', t) = W(q, 0')
-
at
where 0' is the constant of integration. With this choice of solution the time can be eliminated and so
where c is an additive constant, which will not affect the transformation. The constant 0' is thus to be identified with the total energy E. We, however, do not need the value of S, but g~. Now, the solution for the co-ordinates q arises out of the transformation equation
1m] J
as - aO' -
Vk
fi
sin- 1
(3 -
=:;. (3
+t =
=:;. q
=
I¥-
dq _ t 2f: _ q2
q{f
sin{ {f(f3 +
tn =
I¥-
sinw(f3 +
tn
7.2.
HAMILTON PRINCIPAL FUNCTION
217
jF;,.
where w = This is a usual solution for a one-dimensional harmonic oscillator problem. Finally the solution for the momenta p arises out of the transformation equation p
=
as aq
=
aw aq
=
V2ma -
mkq2
= V2macosw(f3 + t).
Now we are to evaluate the constants (a, 13) in terms of initial values (qO, po) at time t = 0. Let the particle is at rest at t = 0, i.e., Po = 0, but it is displaced from its equilibrium position by an amount qo. Then
which is the initial total energy of the system. Thus qo
=
V% k
=?
q = qo sin w (13
+ t).
Since at t = 0, q = qO, we see that 13 = 0, and thus for particular initial condition, the solution becomes q = qo sin wt. Thus Hamilton's principal function S is the generator of the contact transformation transferring harmonic oscillator with a canonical momentum a = H, the total energy and a co-ordinate 13 which vanishes initially at t = 0. Thus Hamilton's principle function can be written as S(q, a, t) = Vmk j[2; -
q2]~dq 1
=Vmk
j
[q5-q 2J2dq-
= kq5 j[cos 2 w(f3 + t) -
at
kq2 20t
~]dt;
q
=
qo sinw(f3 + t).
Also the Lagrangian becomes
which shows that S is the time integral of the Lagrangian. Ex 7.2.2. Show that the function S = ~w [q2 + a 2 ] cot wt - mwqa cosecwt is a solution of the HJ equation for H principal function for the Harmonic Oscillator with H = 2;n [p2 + m 2w 2q2]. Prove that this function generates a correct solution to the motion of the Harmonic Oscillator in time.
CHAPTER 7. HAMILTON-JACOBI THEORY
218 SOLUTION:
Hamilton-Jacobi equation satisfied by Hamilton's principal function is 8S H (q, 8q)
1 8S 2 or, 2m [( 8q) For the given S
=
~ [q2
+
8S 8t
+m
+ a 2]cot wt -
=0
2 2 2]
w q
8S
+ at = o.
mwqa cosecwt, we have
2 2 2] - 1 [(8S)2 +mwq 2m 8q
8S +8t
1 ·mw 2q2 mw 2 x m 2w2(qcotwt - acosecwt)2 + - __ [q2(1 2m 2 2 + a 2cosec2wt - 2qacosecwtcotwt] 1 1 1 = 2mw2(qcotwt - acosecwt)2 + 2mw2q2 - 2mw2q2
=-
+ cot2wt)
1 - 2mw2(qcotwt - acosecwt)2 = O.
Thus given S satisfies Hamilton-Jacobi equation. The transformation equations give 8S
Q = f3 = 8a = mwacotwt - mwqcosecwt =}
q=
mwacotwt - f3 mwcosecwt
= acoswt -
f3 . smwt, 'mw
which is familiar solution for a harmonic oscillator. Also, 8S
p = 8q = mw(qcotwt - acosecwt)
= mwcotwt[ar.oswt cos 2 wt
= mwa[. smwt = -mwasinwt -
L sinwt]- mwacosecwt m,w
1 - . - ] - f3 cos wt smwt f3coswt = mq.
Thus the given S generates correct solution to the motion of one dimensional harmonic oscillator.
7.3
Hamilton's characteristic function
We see that, if the Hamiltonian H does not contain the time t explicitly, the Hamiltonian principle function S should be separated into two parts - one involving coordinate q only and the other involving time f, only. Hence in this case also the integration of the HJ equation is possible and HJ equation for S can be written as (7.13)
7.3. HA.MILTON'S CHARACTERISTIC FUNCTION
219
The first term of (7.13) is concerned only with the dependence of S on the qk's and the second term involves only the dependence of t. Hence the solution for S can be written in the form
This substitution gives the reduced time independent differential equation as
as
H(qk; - ) = aqk
a1 =
constant.
One of the constants a1 of the integration appearing in S, so this equal to the constant value of H. Although, when H is constant, the time independent function W appears here as a part of principal function S, it can be shown that W separately generates its own canonical transformation that has properties very much different from that generated by principal function S. Consider a canonical transformation in which the new momenta are all constants of motion ak and where a1 in particular is the constant of motion H. Let the generating function for this transformation be W(q, P), then
Since W is independent of time, so the old K and new Hamiltonian H are equal, i.e., J( = O:t. The function W = W(qk, O!k) is known as Hamilton's characteristic function, which generates a canonical transformation in which all the new co-ordinates are cyclic. The canonical equations for Pk and Qk are
Ther~fore,
Q1 is the only co-ordinate which is not a constant in motion. Q1 is related with time and we have an instance of the conjugate relationship between the time t as a co-ordinate and the Hamiltonian as its conjugate momentum.
CHAPTER 7. HAMILTON-JACOBI THEORY
220
7.3.1
Significance of W
The total time derivative of the Hamilton's characteristic function W = W(qk, Uk) is given by,
=
action of the system.
Ex 7.3.1. A particle of mass m projected with a velocity u at an angle () .with xn,xis. Write down the HJ equation for the motion. Find the complete integral and the equation of the trijectory. SOLUTION: Let P(x, y) be the position of a particle of mass m at any time t. The KE (T) and PE (V) are given by
T = .!m(i;2 + ii)j V = mgy 2 The Lagrangian and hence the conjugate momenta Px and Py are given by 1
L = T - V = 2m(x2 Px
=
8L ai;
.
= mXj
Py
+ ii) =
aL ay
mgy.
.
= my.
We use the HJ technique for solving such mechanical system. The Hamiltonian H is given by H (Px,Py,y )
1 + (2 1 m (x.2 +y.2) +mgy= 2m =2 Px+P2) y +mgy.
Since H does not contain X explicitly, so Px [2
H(py, y)
=
2m
=
Constant
= 1 (say).
p2
+ 2:'n + mgy.
Let S be the action and put Py = ~;, then the HJ equation is
as
H(ay'Y)
as + at
= 0
l2 1 as 2 as ::::} -2m + -(-) + mgy+ - = 0 2m ay at
Hence
7.3. HAMILTON'S CHARACTERISTIC FUNCTION
221
Since H is constant of motion, so the solution can be written in the form S(y, a, t) = W(y) - at, where a is a constant of integration. Hence ~~ = -a and so the HJ equation becomes H - a = 0, i.e., H = a. Thus,
l2 2m
1 oW + _(_)2 +mgy = 2m oy
oW oy
= J2ma - 2m 2gy - l2
=> => -
=> W(y, a) =
a
J
J2ma - 2m2gy -[2dy + c
2V2
2ma -l2 ~ - Y]2 + C 2 3 2m g 2V2 2ma _l2 3 --3-mj9[ 2m 2g - y]"2 - at + c
= --mj9[ => S(y, a, t) =
where c is the additive constant and this gives the complete integral for S. For the general solution of the equation of motion, we put {3=
oS
oa
=> (3 + t = m
oW _ t
=
J
oa
•
dy J2ma - 2m 2gy - [2
1 2 2 1 = --[2ma -l - 2m gy]"2
mg a l2 g 2 => y = mg - 2m 2g - 2({3+t) .
Let the particle be initially projected from 0 with velocity u at an angle () with the x-axis. Then Velocity at t = 0 along x axis = u cos () Velocity at t = 0 along y axis = u sin (). and put Px = 1 given 1 = mu cos (). Also the constant a is evaluated from H at t = 0 as H = !mu 2 = a. Therefore {3
= -t +
J
dy Ju 2 sin 2 ()
-
2gy
= -t - ~ 9
viu 2 sin
2
() -
2gy.
When t = O,y = 0, then {3 = -~usin(),x = ucos().t. Hence the equation becomes t -
~ucos () = _~vlu2 sin2 () 9
2gy
9
. ()]2 or, [-X- - -Usm ucos () 9
= -1
g2
[u 2 sm . 2 () - 2 gy ]
which is the equation of the trijectory and this represents a parabola.
CHAPTER 7. HAMILTON-JACOBI THEORY
222
7.4
Separation of Variables
Under suitable conditions, the variables in the HJ equation can be separated. Consider systems in which the Hamiltonian is one of the constants of motion, which may not necessarily be the total energy. In such a case, we need consider only the contact transformation generated by the Hamilton's characteristic function W and its corresponding HJ equation. A co-ordinate qi occurring in the HJ equation is said to be separable, if the Hamilton's principal function can be written in the form
W
= L Wi(qi, 0:1,···, O:n, t) i
which will split the HJ equation in n equations
Hi(qi,
aW -a qi t
j 0:1, ... , O:n)
= O:i·
The constants O:i are referred to as the separation constants. Each of the equations involves only one of the co-ordinates qi and first order partial derivative of Wi with respect to qt. These equations are solved for a first order differential equation ~ and the integrand with respect to qi to get W. Now if the Hamiltonian H does not explicitly contain time, one can linearly decouple time from the rest of 8 and write
Under this assumption, the HJ equation with this trial solution becomes
Since the RHS and the LHS are now functions of totally different variables, they can be equal for all values of these variables only if the LHS and the RHS are separately equal to constant independent of all the above variables. Let us write
a82 -8 = t
-0:1,
8W then H(qi, -8 ) qi
= 0:1
where 0:1 the constant value of Hamiltonian, is the constant of separation. The first gives the solution 82 = -O:lt and the second can be viewed as the time independent HJ equation with W to be identified as Hamilton characteristic function. In any case, the solution to the time dependent HJ equation for conservative dynamical systems, thus has the form, apart from a trial additive constant term
Now, let all the co-ordinates except q1 are cyclic. Since q1 in non-cyclic coordinate, the conjugate momentum PI is constant, and the solution for W be of the form n
W
= LWi(qi,Pi ). i=l
7.4. SEPARATION OF VARIABLES
223
Since all the coordinates except q1 are cyclic, we have
aWi -a = qi
p
i
= O:i =}
UT
Hi
. ....L 1 = O:iqij~;.
Thus the HJ equation becomes
H(ql,
aWl -a ,0:2,"" q1
O:n)
= 0:1
which is the first order ordinary differential equation for WI in the independent variable q1 and hence can be solved easily. Thus for a single cyclic co-ordinate the completely specify function W can be written in the separated form as n
W
= W 1(q1, 0:) +
L
O:iqi·
i=2
Ex 7.4.1. Hydrogen Atom Problem Here we consider the well known Hydrogen atom problem or Kepler problem. This problem, in general, consists in finding the trajectory of a charged particle under the action of central force in an inverse square field. Let a particle of mass m with charge e is moving about the stationary nucleus with positive charge ,e, where , = atomic number. Thus the field of central potential is given by
,2e V(r) = - - = r
k
k = ,2e > O.
--j
r
In three-dimension spherical polar co-ordinate (r, (), ¢), the Hamiltonian His,
Since H does not contain ¢ explicitly, so ¢ is cyclic co-ordinate and so, PcP = mr2 sin 2 ()¢ = Lz = Constant.
This shows that the motion is confined to a two dimensional xy-plane and so the Hamiltonian H in two dimensional polar variables is
1 .2 H = -m[r 2
+ r 2 ()'2 ] -
k r
- =
1
2
-[p 2m r
+ P~ -J --kr r2
where Pr = mr and Po = mr 2 iJ are the conjugate momentum along the rand () directions respectively. Further, the last equation shows that H does not contain () explicitly, so () is cyclic and so Po = constant = I (say), which also characterized by for a central force motion the angular momentum is conserved. Therefore
1 H = -[p~ 2m 1
[2
k
+ -J --. r2 r
CllAPTER
224
r
HAMILTON-JACOBI THEORY
As the system is conservative, the Hamiltonian H represents the total constant energy E of the system and hence 1 _[p2
2m
r
Z2 + _] __k = E
r2
r
Z2
=> Pr = [2m(E - 2" r
k
1
+ -)]2. r
Now H does not involve the time T explicitly, so we can separate the variables. Hence our Hamilton's principal function S has the spatial part lV which is to be taken as the generating function for the transformation. Hence the Hamilton's characteristic function can be written as
W(r, 0) = Wr(r)
+ Wo(O)
fr
= 8;:r j Po = ~~ = 8':0°. Also Po = 8':0° = Z(constant) so that Wo = l() + q(constant). Hence the HJ equation can be written as
where Pr =
aW Pr = -r = [2m (E. - -l2
ar
=> Wr (-r) = "hm
z2
J
(E - 2" r
r2 k
+ -kr )] !.2. 1
+ -r )2" d-r + c
Therefore the generating function for the transformation is
S = W(r, (}) = Wr(r)
+ Wo((})
=l()+v'2m
J
Z2 r2
k r
1
(E--+-)2"dr+c
where c = q + C2 is the constant of integration. Using this expression for Hamilton's characteristic function the transformation equations become t + i31 = ~ = ~~ and
which is the equation of the conic with eccentricity
7.4. SEPARATION OF VARIABLES
225
The path is ellipse, parabola, hyperbola according as e < 1, e = 1, e > 1, i.e., according as E < 0, E = 0, E > 0. In Hydrogen-atom problem, we are interested in elliptic path. If the length of major axis is 2a then, 2 _ 2Z 2 /k _
1- e2
a -
-
-
k E'
When the trijectory is ellipse, the energy can be expressed as k 2a
,2£
E=--=-where 2a
=
2a
length of the major axis of the ellipse.
Ex 7.4.2. A particle of mass m is constrained to move on a smooth parabolic wire, x 2 = by under the action of gravity. Write down the Hamilton-Jacobi equation for the system and find the general solution of the equation of motion. Let P(x, y) be the position of the particle of mass m at any time t. The KE (T) and PE (V) of the particle moving on the parabola are SOLUTION:
2
1 [.2 .2] 1 4x . 2 mg 2 T= 2 mx +y = 2m[I+t;2]x; V=mgy= bX'
4X 2
1
'2
mg
2
=> L = T - V = 2 m [1 + t;2]X - b X . Hence Px
aL
4x2 .
= ax = m[1 + t;2]x.
Therefore, the Hamiltonian (H) is given by
H = Px x. - L 1
1
p2
= _
[1
4x2]
= 2m + b2
x
2m 1 + 4x 2
·2
x
mg X 2
+b
+mgx2
IT
Let S be action and Px = ~~. Since H is constant in motion, put S and ~7 = -a. Hence the HJ equation becomes
as H ( ax ' x)
as
= W(x, a) -
aw
+ at = 0 => H ( ax ,x) = a 1 1 oW 2 2 => -2 4 2 (-a ) + mgx = ml+ x x
a
V
aw ax
=> W(x, a)
=
2
4x = [2m(1 + -)(a - mgx 2 )Jfrac12 2
b 4x2 1 [2m(I+t;2)(a-mgx2)]"2dx
J
=> S(x, a, t) = W(x, 0') -
at
=
J
[2m(1
4x2
+ b2 )(0' -
1
mgx 2)]"2dx - at.
at
CHAPTER 7. HAMILTON-JACOBI THEORY
226
For the general solution, we put f3
=
g~
=
~~
-
t so that,
Ex 7.4.3. A particle of unit mass moves in a field given by the potential V = ~ Fz, where k and F are constants. Prove that HJ equation is completely separable in parabolic co-ordinates (C rJ, ¢) given in terms of cylindrical polar co-ordinates (p, ¢, z) by the following relations z = ~(e2 - rJ2), P = erJ, ¢ = 'P. SOLUTION: Here we use Pilrabolic cylindrical co-ordinates, where the transformation equations are,
x = pcos¢ = erJcos¢; y = psin¢ = erJsin¢; z = r
~(e - rJ2)
= vx2 + y2 + z2 = JerJ 2 + 4 ~(e2 - rJ2)22 = ~(e +· rJ2)
In this co-ordinate system, the KE in terms of e, rJ and 1 variables is given by, - -1 [x·2 T -2
+ y·2 + z.2J
= ; [(irJ cos ¢ + ~r,cos ¢ - erJ sin ¢¢)2 + (irJsin¢ + er,sin¢
+ ~rJcos¢¢)2 + (~~ - rJ1])2J = ~' [(e 2 + rJ2)(e + r,2) + erJ2¢2]. In terms of
~,17
and ¢ variables, V
k
= -:; - Fz =
e 2k+ rJ2 2
F
2
2
"2(e - rJ ).
Therefore, the Lagrangian and so the momenta are given by,
The Hamiltonian for the system is
7.4. SEPARATION OF VARIABLES
227
Since H is free from explicit time dependence, the system is conservative, so the HJ equation is H = constant = E, E being total energy of the system. Hamilton-Jacobi equation satisfied by Hamilton's characteristic function W can be written as
Since cP is cyclic co-ordinate so, P
=
~~
= constant =
a
We separate the equation in ~ and 'f/ variable by writing W = W~(~)+W1]('f/)+W
The expression within first third bracket is ~ dependent and the expression within second third bracket is 'f/ dependent. We put,
Therefore, the ( equation becomes,
am
a2
~)2 + -1!. _ Fm(4 - 2mEc 2 + 4mk + a = 0 ( __ a~ ~2 ~ 1]'
Hence, W is completely separable in W
=
= +
W~(~)
~,'f/,
cP and we obtain the complete integral as,
+ W1]('f/) + W
J Fm~6 + 2mE~4 J + J
J2mE'f/4 - Fm'f/6
Et
- (4mk
+ a1](2
a1]'f/2 -
-
a~ ~(
a~ ~'f/ + a
Et.
228
7.5
CHAPTER 7. HAMILTON-JACOBI THEORY
Action-Angle Variables
We know that, in classical mechanics the action or the phase integral is defined as
Solution of HJ equations by separation of variables plays an important role in the physical system in which the motion is periodic. Periodic system are those which repeat their state of motion after a constant interval of time known as period. For example, in the hydrogen atom, the electron moves about the nucleus, the proton, like a planet about the sun. We are interested in knowing the frequency of the periodicity rather than the orbit itself. In the phase space all such systems need not reference their paths. Those which retrace their paths in phase space are called librating systems. For the rotating systems some momenta are periodic functions of the conjugate co-ordinates, but the latter do not return to their original value and increase or decrease monotonically with time. Also, the orbit in the (q,p) phase is a closed one. In the procedure, we do not choose fl'i as a new momenta but, instead, we use suitable defined constants Ji which form a set of n independent function of the ai's and are known as action variables. Action variables Ji which form a set of n independent function are useful parameters for systems which are periodic, conservative and orthogonally decomposable. Let Hamilton's characteristic function be completely separable in all its variables. Complete separability means that, the equations of the canonical transformation have the form,
which furnishes each Pi is a function of the corresponding qi and the n number of integration constants aI, a2, ... , an, i.e.,
This equation gives p as a function of q, i.e., in the phase space, it represents the orbit equation of the projection of the system point on the (qi,Pi) plane. The motion of the system is said to be periodic, if the projection of the system point orbit on each (qi, pd plane is simply periodic in the sense as discussed in the case of the system with one degree of freedom. Since the variables are separable, these projection points are independent of each other. So, it is not necessary that all (qi,Pi) sets will have same frequency. In a three dimensional harmonic oscillator the force constants along the -three axes may all be different but the system may still be called periodic. We can define action-angle variables for the system, when the orbit equations, for all pairs (qt,Pi) describe either closed paths or periodic functions of q~. It is not necessary that all (qi,Pi) sets exhibit same frequencies of the periodic motion. Indeed, if the separate frequencies are not rational fraction of each other, the particles will not
7.5. ACTION-ANGLE VARIABLES
229
transverse a closed path in space, but will describe an open' figure. Such a motion is called conditionally or multiply periodic. It is the advantage of the action-angle variables that, they lead to an evaluation of all frequencies involved in conditionally periodic motion without requiring a complete solution of the motion. The action variables Ji corresponding to a pair of separation variables (qt, Pt) are defined as a phase integral, given by, (7.14) where t~e integration has to be preformed over a complete periodic variation or .rotation of qi, as the case may be. In case of qt is a cyclic co-ordinate, its conjugate 27r
momenta Pi is constant and then Ji
=
J dqt = 2npi for all cyclic co-ordinates.
o
Since
J t has dimensions of angular momentum, co-ordinates conjugate of it should be an angle and so its name is action variables. Let the complete integral of the time independent HJ equation H (ql, ... , qn; aas , ... , -2§...) = E, be defined by the action ql aqn SasS = j"pdq, then n
S(ql, ... , qn; al,···, an)
=L
Si(qi; al,···, an),
i=l where one of the a's is the energy E. Further, since, each pair (qi,Pt) in the sepa'ration of variables for the action S the Ji'S are independent set of function of the constant ai's and so they are suitable for the use as a set of new constant momenta. As one the {ai} is the energy E, so, Ji
=
f
f
= =
Ji(al, ... , an); i
f
t dS dqi qi dqi a function independent of qi but dependent on al, ... 1 an Pidqi
=
-aas dqi =
= 1,2, ... n.
Inverting these n relations, we get the ai's as a function of the Ji's as
By means of these relations the complete integral for the action, can be written as
where Ji's are here n constants of motion while Hamiltonian appears as function of Ji's only H = al = H(Jt, ... I n ). From the generator f(qi, Ji), we can find the conjugate Wi corresponding to J i as
CHAPTER 7. HAMILTON-JACOBI THEORY
230
These variables Wi are called angle variables. It is to be treated like co-ordinate Qi and as angular frequency. As Ji'S are new momenta, f is the generating function, Wi is the new co-ordinate, Ji has the dimension of angular momenta, Wi is an angle. The constants of motion Jt can be regarded as new momenta Pi for a canonical transformation with a new set of co-ordinates Qi = Wt , conjugate to these momenta Pi; i = 1,2, ... , n. This canonical transformation may be generated by a generating function of the type F2, same as the complete integral of the time independent HJ equation or the Hamilton's characteristic function f(ql, ... , qn; Jl, ... , I n ). The conditions are
with the Hamiltonian K(Q, P) as
K(Q, P)
= H(q,p) = E(J1 , ... I n )
which is independent on {Qi}. The canonical equations of motion corresponding to the new Hamiltonian K are
.
aK
Pi = - - =0 and aQt
.
Qi
. aK aE =W= - =- = aPi
aJi
Constant
= 'Yi(Jl, h,··., I n ).
where "Yi'S are another set of constant functions of the action variables Ji. The transformation equations to be (w, J) set of variables implies that, each qj (and pj) is a function of the constants J i and the variable Wi. The solution of the equations of motion for the angle variables are given by
where (3i are the constants of integrations and are known as phase constants. Because, the 'Yi'S are constants, functions of the action variables only, the angle variables Wi are linear functions in time. In general, the separate Wi'S increase in time at different rates. Physical Significance of 'Y's : Here we are to show that 'Yi'S look like frequencies of oscillation of the co-ordinate qi which gives rise to k We know Wi = 8!CQ1,Q2'''·'li'Jih ,h, ... ,Jn). We consider the change in Wi when qi makes one complete revolution. Let 5Wi be an infinitesimal change when qi --t qi + 5qi. Thus,
The total change 6.Si is the ith modulus of periodicity offunction S, i.e., the change suffered by S when qk runs through a complete cycle of its values. Thus the total
7.5. ACTION-ANGLE VARIABLES change /:::,wi in
Wi
231
for one complete revolution, is, therefore,
shows that only when qi goes through a complete period, Wi changes by unity. The differential operator a~ comes outside the integral sign, because throughout the cyclic motion of qi all the J~'s are of course constant. Also, .
As we know just stated /:::,wi = 1, when qi goes through a complete period, i.e., when 6t = Ti, where T~ is the period associated with qi. Therefore
Thus "Ii is the frequency associated with the one complete revolution of the periodic motion of qi and f3i are the phase constants. Complete Solution: If the relation E = E(JI, ... I n ) is known, the frequencies of the oscillation "Ii's for all the co-ordinates given by "I. = can be obtained. If two or IlIDre frequencies of oscillation are identical, then such a periodic system is called a degenerate periodic system. If further, all the frequencies {"Ii} are found to be identical, the system is known as a completely degenerate periodic system . . Now, the new set of variables {"Ii, Jd should completely specify the dynamical state of any conservative periodic system and Wi = "lit + f3i can be produce the general solution
gf
Hence "Ii is identified as the frequency associated with the periodic motion of qi and hence we calculate the frequency of periodic motion without finding a complete solution for the motion of the system.
Ex 7.5.1. Planetary Motion Let the Hamiltonian in spherical polar co-ordinates be
If the variables in the spherical polar co-ordinates corresponding to the HJ equation are separable, then the characteristic function must have the form
CHAPTER 7. HAMILTON-JACOBI THEORY
232
As H does not contain (
+ ~[(aWO)2 + .a~2 ar
r2
sm ()
]_ 2mk r
= at/> =
= 2mE.
Now, the quantity in the square brackets must be constant, i.e., (
+ a~2 = a~
aW O)2 ar
sin
()
for such condition, the HJ equation for Wr only becomes (aWr)2 ar
+ a~ = 2m[E + ~]. r2
r
Thus all the variables in HJ equation are completely separable. Now.let the motion in each of the co-ordinates be periodic- liberation in r, () and rotation in are defined as k 0'.2 2m(E+ -) - ~dr. r r2
f =f
Jo =
pod()
Jt/>
pt/>d
f aa~ = f = f ~~ = f
=
2
a~ - Si:~ ede.
de
d
CYt/>d
For the complete revolution 0 ::;
Jt/>
=
f
at/>d
= 21fat/>.
o To evaluate Jo we take the circular path so that ~
JJa~ f =
~
2
Jo
=4
-
a~cosec2ed() = 40'.0
o
J 2
Jk2 - cos 2 ecosec()d();
0
~
2
2
4k 0'.0
.
o
2
cos , 2 . 2 d,; cose 1- k sm ,
. = ksm,
7.5. ACTION-ANGLE VARIABLES
233
Next we consider the first integral, which is to be performed over a closed orbit. We see that for the finite motion E < 0 and further the integrand is Pr where Pr = mr. Hence the limits of integration are obtained from the equation r = 0, i.e., k a2 2m[E+ -]- -.-!t l' 1'2
= 0 =} l' =
-
B ±
v' B2 + AC = 1'1,1'2 A
where A = 2mE, B = mk, C = a~. For applying the angle variable method we require one complete cycle of co-ordinate l' which consists of the motion 1'1 to 1'2 and back of 1'1, where 1'1 is the inner bound. Now 2m(E
Jv' =J v'
a2
k
+ -) l'
-.-!td1' 1'2
J Jv'
r2
= 2
r2
A1'2
+ 2B1' -
A1'2 + 2B1' - C d1' V7'(A1'2 + 2B1' - C)
C d1' = 2
l'
rl
rl
~
~
2A1' + 2B d1' A1'2 + 2B1' - C
+ 2B
rl
rl
= [VA l' 2 + 2 B l' _ r2
+ 2C
J v' •
c]r2 rl
+ 2B1' + [2B VA Iogl A1' + B + v' A1'2 A
+ 2Br -
dr
C)
C]r2 rl
1
+ 2Bu -
Cu 2
dUj
U
-1 = -2BIog (A1'2+B) + 2 V [;:::,C[' l./ sm
=-
r
C-B1'2 . -1 - sm v'AC + B2
+ B' = 2;; log( -1) + 2VC[sin- 1 ( -1) _ sin- 1 (1)] =
VA
d1'
V1'(A1'2
rl
d1' A
J ~
d1' d1' - C A1'2 + 2B1' - C
A1'1
vA
C-B1'1 ] v' AC + B2
2mk1r
- 27r0'0.
v'-2mE
From the above three relations we get Jr+Jo+J = 7rk~. This equation supplies the functional dependence of H upon the action variables, and 2m7r 2 k 2 H = E = - ...,-------:--:::-
(Jr
If we calculate J r + .10 T
=
7rkJ
-;nE3'
+ J¢
+ Jo + J)2'
in terms of energy function, the period of the orbit is
If 'Yr, 'Yo, 'Y¢ are the frequencies then
'Yr
=
8H 8Jr
=
= (Jr + J() + J¢)2
8H 4m7r 2 k 2 = 8Jo (Jr + Jo + J = (Jr + Jo + J
'Y()= 'Y
4m7r 2 k 2
CHAPTER 7. HAMILTON-JACOBi THEORY
234
As all the frequencies are equal, so the motion of the given system is degenerate. Thus we see that planetary motions are completely degenerate in all the three coordinates r,8 and ¢, i.e., all bounded orbits of any two body system moving under any inverse square law of an attractive central force are closed.
Ex 7.5.2. Obtain Jacobi's complete integral for a simple harmonic oscillator of mass' m and frequency w. Plot the phase space trajectory. Show that the action angle are ( ~, ¢), where E is the energy and ¢ is the phase angle. The form of the Hamiltonian H of the one-dimensional harmonic os2 cillator as H(q,p) = ~ + !kq2, which is independent of time and leads to the HJ equation in the form SOLUTION:
~ (aw ? + ~kq2 + aw = 0
aq
2m
Let W(q, t) = WI (q)
=}
dW2 dt W2 (t) Wl(q)
+ W2(t),
(7.15)
at
2
then from (7.15) we get,
= _~(aW2)2
_ ~kq2 =-E 2m 2 Et + constant and
aq
=1
[kq2
2E.
r-;-
~
= 2 vmk [T sm- 1 y 2E + qy T - q2] + constant
Hence the complete integral is W(q, t) = W 1 (q, E) -Et+C. Apart from an additive constant, a new constant E has appeared in the solution and since this is a one variablf~ problem, we expect only one constant in the class of ai's. Here al = E, hence
rm sm. Y[kq2 2E
aw aE = i31 = -t + Yk
-1
where (Jl is a constant, whose value can be obtained by q(t = to) = q(O) so that
to = -t+ =}
fi
sin - 1
J~~
q(E, t)
=
{k aw YfiE. T sm[y -:;;;,(t + to)] and p = aq = v'm(2E -
Now J
=
f
pdq
f ~! = f J =
=
dq
v'm(2E - kq2)dq
27r
= 2Efi
2
cos 8d8
o =}E=
~ {k. 21r
V-:;;;,
2fiE1r
kq2).
7.5. ACTION-ANGLE VARIABLES
235
Since the system is conservative H is constant say H Thus the frequency of the Harmonic oscillator is
~
27r')'
=W =
Vfk ;;, =
=
a, then H
=
a
=
irr jF;..
.
angular velocIty
Ex 7.5.3. A particle of mass m moves in two dimensions (x, y) in a non-isotropic simple harmonic oscillator V(x, y) = !mw;x 2 + !mw~y2, where Wx :I wy . Obtain Jacobi's complete integral in terms of the action variables. SOLUTION: The KE of the two dimensions non-isotropic simple harmonic oscillator is T = !m[i;2 + :!P]. Thus the Lagrangian L and the corresponding momenta are given by,
Thus, the Hamiltonian H is given by, H(x, Y,P;r,Py)
= Pxx + pyy - L 1 2 2 1 2 2 = -p; + -P~ - -m(PX)2 - - -m(PY)2 - + -mw x + -mw y m m 2m 2m 2 x 2 Y =
1 [p2 2m x
1 2 2 1 2 2 + Py2] + "2mwxx + "2mwyy .
H-J equation satisfied by Hamilton's characteristic function W can be written as awaw H(x,y, ax' ay)
= al
1 aw 2 aw 2 1 2 2 2 2] ~ 2m[( ax) + ( ay) ] + "2 m [wx X +Wyy
= al·
We separate the equation in x and y variables by writing W = Wx(x) where Px = ~,py = aa~' Thus, Hamilton-Jacobi equation becomes, m 2 2] - - +-w.x [- 1 (aWX)2 2m ax 2 x
Let, 2!n(aruV )2
+ ~w~y2 =
m 2 y 2] =al. + [1 -2m (aWY)2 -ay- +-w 2 y
a2, then the above equation becomes,
1 awx 2 m 2 2 -(-_\ + -w x = al 2m ax I 2 x
a2
= a3,
say.
+ Wy(y),
CHAPTER 7. HAMILTON-JACOBI THEORY
236 Thus, the action variables are given by,
Hence, W is completely separable in x, y and we obtain the complete integral as,
W = Wx =
+ Wy
=
f ..j
2mn 3
fV
-
mJ;wx - m 2w'ix 2dx
m 2w'ix 2dx
+
+
f
f vm~wy
V2mn2 -
-
m2w~y2dy
m2w~y2dy.
!
Ex 7.5.4. A particle of mass m moves in a one dimensional potential V = [kq2+A J, where k, A, q > O. Find the energy in terms of action J. Show that the period is independent of amplitude. SOLUTION: The Lagrangian and so the corresponding momenta are given by 1 .2 1 2 A L=-mq --[kq +-] 2 2 q2 8L . . p =}p=-. =mq=}q=-. 8q m
The Hamiltonian H of the system is H
.
p2
1
A
2
= pq - L = -2m + -2 [kq + -]. q2
Since H is free from explicit time dependence, the system is conservative, so the HJ equation is H = constant = E, E being total energy of the system. Thus,
p2
H = 2m =}
P=
1
).
+ _[kq2 + -] = 2 q2 1
E ).
± 2m[E - _(kq2 + -)] 2 q2
vmk 2 2 1 = ±-[(,B - q )(q - a)]2, q
where a, ,B are the roots of the equation,
2;; x -
x2
-
i
say,
= 0, i.e., say a = E-~
7.5. ACTION-ANGLE VARIABLES and f3
=
237
E+q=u, where f3 > a > O. The action variable J is given by,
7r /2
J J
= v'mk[
7r /2
{a(l
+ cos 20) + f3(l -
cos 20) }dO
+ 2( a + f3)
o
7r/2
2
+ af3-f3
o
a
J
dO
0 2
sec 0dO /f3 201; put, ~
+tan
2· 2
= a cos 0 + f3sm ' . 0
Now the period of oscillation is given by
so that it is independent of amplitude. Also the angle vfl,riable is, w = vt + c, .where c is a constant. Hamilton-Jacobi equation satisfied by Hamilton's characteristic function W can be written as H = constant = E, E being total energy of the system. Thus,
CHAPTER 7. HAMILTON-JACOBI THEORY
238
Now, w = a~~j,J) gives q in terms of wand J. Thus the co-ordinate q is given by a function of t as,
E
= [k +
q
VE2 - >'k . k sm(//t + c)]1/2.
Ex 7.5.5. A particle of mass m is constrained to move under the action of gravity in the vertical xz- plane on a smooth cycloid curve given perimetrically by, x = 2
a( e + sin e), z = a( e - cos e). Show that a suitable Hamiltonian is H = 4ma2(i~cos e) + mga(l - cos e). Use action-angle variables to show that the frequency of oscillation of the particle is independent of its amplitude with lei < 1r. SOLUTION: The equation of the given cycloid is x The KE T of the moving particle is given by
= a(e +
sine), z
= a(e -
cos e).
1 T = _m(±2 + z2) 2
= ~mZ2 [(1 +
cos e)2 + sin2 e](j2 = mZ2(1 + cos e)(j2.
= mgz = mgl(l - cos e). Thus the Lagrangian is given by L = mZ2(1 + cosO)(j2 - mgZ(l - cosO).
The PE V is V
=> Po =
8L
80 = 2ml
2
.. Pe (1 + cose)e => 0 = 2m12(1 + cosO)'
Now, the Hamiltonian of the system is,
Pe2
.
= poO - L = 4m12(1 + cosO)
H
+ mgl(l- cos e).
Since the system is conservative, so the energy equation gives H = E, where E is the total energy of the system. Thus, H = E => Pe = ± 4m -=Z2-;'"(l-+-co-s-=-e:-:-)[E=----m-g--=Z-;-(1---cl)-s-=-O:-:')1
Jr:-
= ±4mZJ9icos ~Ja2 -
sin
2
~j
2 a = 2!9Z > O.
Thus the action variable J is defined by,
J
=
f
=
8va 2 2 x 8mZJ9i[ 2·
PedO
=
2 x 4mZ J9i
J ~J cos
a2 - sin
2
~ j where sin ~ = a
-0
~2 => E =
X
4~
The frequency is
4mly'9ia'"
IfJ 1/
=
82
a2 8 (J +"2 sin-l(~)l~aj sin 2 =
~ 2 x 4mly'9i"
= H.
WJ = 4~ If
X
2!91
8
~ 4"E~
Chapter 8
Reference Frames The solution of many physical problems in Physics, Applied Mathematics can be readily obtained by introducing the moving frames of reference, i.e., non-inertial frames of reference, for example, while discussing the motion of the particles near earth's surface, separation of liquids of different density in a centrifuge, the analysis of the motion of the rigid body, the experiences of an astronaut in an orbiting satellite, etc. Usually for such problems, we are to consider two frames - the inertial frame and the moving frame. The frame which is at absolute rest or in uniform motion, is called an inertial frame. Relative to a fixed frame of reference, a moving frame can posses a translation or rotational velocity or both. If the frame of reference moving with constant velocity relative to a fixed frame, then it is called an inertial frame of reference. If a frame of reference has some acceleration or retardation relative to a fixed frame, then it is called non-inertial frame of reference. Basically, earth 1S itself a non-inertial frame and any frame associated with earth in noninertial. Introduction of non-inertial systems also enables us to explore some of the conceptual difficulties of classical mechanics, such as properties of space, validity of Newton's law and meaning of inertia.
8.1
Translation motions
Let us consider two co-ordinate systems 0' (x' y' z') fixed in space and O{ xy z) fixed to a certain body whose orientation in space are observed to remain fixed. O{xyz) is moving with a translational velocity with respect to the frame of reference O'{x'y' z') so that the axes are parallel. The physical principle involved are not affected by such a particular simple choice of translation. At time t = 0, the origins 0 and 0' of the two co-ordinate systems will coincide. At a certain time t, let R be the instantaneous position vector of the point 0 with respect to the point 0', i.e., 00' = R. A single observer, who sees the two co-ordinate systems, is able to describe the motion of a particle with respect to the either of the reference frames. Let the position vectors of the same point P with respect to the base points 0 and 0' be -:;;+ and -:;;+, in the ~
~
239
~
CHAPTER 8. REFERENCE FRAMES
240
p
~-------------y
)C-------------y' x'
Figure 8.1: Translation motion
two frames respectively, then ---+
r" = R + r' ~,
~
~
::::} 11' =
~,~
+ rand r = 11' + V; and 0;' =
::::} r = R
~
+r 0; + a: R
(8.1)
where 11", 0;1' mean the velocity.,and .. acceleration of a with respect to a'. Thus the velocity and acceleration (r' ,-:t)' of a particle at point P measured in the ~~
fixed system is the vector SUlll of velocity and acceleration (R, R) of the moving frame and the velocity and acceleration (-t, ~) of the particle at the point P in the ---+ moving frame. Let F be the total external force acting on the particle P. Then the equations of motion of the particle at the point P in the fixed frame and moving frame of reference are .. ,
---+
~
---+
mr' = ..
.. ,
and m -:t = m r'
-
F ~---+
m R = F - m R = F ef f
(8.2)
---+
where F eff is the effective force acting on the particle at the point P. The equation {8.2)has the same form as Newton's equation of motion in an inertial frame, where ---+
---+
~
~
F is replaced by F - m R. The quantity -m R is called pseudo or fictitious force, which arises out of the acceleration of the reference frame and is not due to particle interactions. This pseudo force sharply acts in direction opposite to the acceleration of the frame. Hence, an acceleration frame is non-inertial. From equation (8.2), we have the following discussions : ~
(i) If a moving frame of reference has an acceleration R, the effective force acting ~
on the particle at point P is smaller than the actual force by an amount m R , ~
due to a result of the acceleration m R of the moving frame. ~
(ii) If 0 moves with constant velocity relative to a', then R = 0 and the Newton's equation of motion will hold in both the systems. Thus all co-ordinate system
241
8.2. ROTATING FRAMES OF REFERENCE
which move with uniform velocity relative to one another will be inertial if any one of them is inertial. This is known as principle of Newtonian relativity. (iii) There is in fact no dynamical way to distinguish one inertial system from another.
Ex 8.1.1. A pendulum in a Lift Let us consider a pendulum of length l and bob mass m within a lift, which is moving The pendulum is oscillatory in a vertical plane with upward with an acceleration -t small angular amplitude (). Let T be the tension of the string with respect to an observer in the lift, the total force acting on the bob is
a.
-t
T
+m
-t
9 -
-t
m
a
-t
A
A
= T - mgk - mak -t = T -m(g+a)k A
where k is a unit vector in the vertically upward direction. This equation at once suggests that the pendulum oscillates in a gravitational field which is effectively increased from -mgk to -m(g + a)k. Now the radial and tangential components of the gravitational force is meg + a) cos () and m(g + a) sin () respectively. The radial components of the forces supply the necessary centripetal acceleration to keep the bob moving on the circular arc of radius l. Thus, the equation of motion is, d2 s
m dt 2
d 2 ()
= -m(g + a) sin() (g
+a
.
=> dt 2 = - - l - sm();
as s
= l()
2
=> ddt () _ (g l +-a ()., when () is small (:-:;- 40). -2 This shows that the motion is simple harmonic, and the period of oscillation is Tl = 27f 9 +1 a. • Similarly, when the lift descends with a downward acceleration a( < g),
V
V
then the period of the simple pendulum becomes T2 = 27f g~a' When the begins to fall freely, then a = 9 and T2 - t 00. This shows that the pendulum stop oscillating.
8.2
Rotating frames of reference
The transformation from an inertial system to a rotating system is fundamentally different from that to a translating system. The system translating uniformly with respect to an inertial system is also inertial) whereas a uniformly rotating system is intrinsically non-inertial and involves several fictitious forces including one which is velocity dependent. Let us consider a reference frame 0' (x' y' Zl) fixed in space and another frame O(xyz) fixed in t.he body. Their origin and base vectors of the two systems coincide at time t = 0, but at any time t, they may not necessarily coincide. When the body rotates with angular velocity c;J about some instantaneous axis
242
CHAPTER 8. REFERENCE FRAMES z' z
p
y
~~---------------y'
x' x
Figure 8.2: Rotdional motion passing through 0, frame O(xyz) will rotate with respect to the frame O(x'y'z') with angular velocity W. The frame O(x'Y' z') is the rotating frame of reference. Let the unit vectors i',]', /';' as the fixed unit vectors forming a rectangular cartesian triad in the primed system O(x'y'z') and the unit vectors i,l,/.; in the unprimed system O(xyz) which are changing their directions along with the rotating axes. The position vector of a particle at P can be written as ~
= x'i' + y']' + z' /';'; = xi + Yl + z/.;;
space set of axes body set of axes.
Now the transformation equations from unprimed system to primed system are ,
X
=
~
~,
r.2
~ ~,
~
~,
~,
= n.2 + YJ.2 + zk.2 A
, A. A., A Y = T.J = X2.J + YJ.J + zk.J , A, A, A, A A, Z = T.k = xz.k + YJ.k + zk.k ~~,
~
~
Le.,
~,
~
(~:) = (!f. z'
~,
~
nH) (~) .
i.k' j .k' k.k,'
(8.3)
z
where the dot products i.i',l.i', .. .etc. on the right hand side of the above equation are simply the cosines of the angle between the corresponding axes. Similarly we get the equations for inverse transformations.
8.2.1
Rate of change of arbitrary vector
Here we are to find the transformation rule relating the time derivatives of an arbitrary vector G = G (t) in the inertial and rotating systems. Physically, the vector G is the same in both the frames, i.e., G f = G r , the suffices f and r are used for fixed and rotating frames. The components of G will be different in the two systems because the unit vectors in both the frames are pointing differently. Thus, ~
~
~
~
~
~
~
Gf
,..
....
....
= G~i' + G~j' + G~k'
A A A =G1 i+G2j+G3 k = Gr· ~
8.2. ROTATING FRAMES OF REFERENCE
243
---t
The time derivative of G will differ in the above two co-ordinate systems: Since the ---t rate of variation of G is measured by an observer in the fixed frame so the time ---?
derivative of G f is denoted by constants, so ~:
d,--40
Z.
f.
In the primed syste~ as the unit vectors are
= ~ = ~' = 0 and so the time derivative in the primed system is ---t
d' G f = dG~ i' dG; "./ dG~ k' dt dt + dt J + dt . ---t
,~
is the rate of change of G with respect to the rotating frame because in this £, £., dk' d'G ---t case d~t = Tt = dt = o. Also is called rate of growth of G. Similarly, the time ---t derivative of G r with respect to an observer in the rotating frame in which vectors i,3, k are treated as constant unit vectors is ' d :;; f
Tt
---t
dG r _ dG 1i dG 2 ". dG 3 k dt - dt + dt J + dt . In the rotating system, however, the unit vectors are changing in directions. Thus ---t the time derivative of G with respect to 0' frame in unprimed components is
*,
The remaining terms ~, ~~ arise as a result of the rotation of the system. Since the unit vectors i, k are moving (changing directions) with time with respect to 0', they can not be treated as constant vectors in 0' frame. is the velocity of
3,
that point on OX' whose position vector is primed components.
8.2.2
i.
*
Similarly, we can also find
d~r in
Velocity and Acceleration
Let us assume that, the origin of the fixed and rotating frames are the same, but their axes may not be parallel. Let -:t be the position vector of a point with respect to the base point O. The body is rotating with angular velocity C;; about ~he axis passing through the origin. When a vector changes, then in both the frames ---t
---t
---t
dr=rsin(Jdn=}dr =dn x r ---t
---t
dr dn ---t ---t ---t ---t =}v=-=-xr=wxr. dt dt
(8.4)
The quantity C;; x -:t is called rate of transport. This expression is the linear velocity of a particle having position vector -:t and rotating with angular velocity C;;. In
CHAPTER 8. REFERENCE FRAMES
244
dX
this expression, 7 is any position vector rotating in a body and is the time derivative of 7. So the expression can be used for particular cases to unit vectors i, 3, k. Applying this to the unit vectors i, 3, k in a system rotating with angular velocity of rotation TJ, we have,
di dt
~ d3
-t
= w x z, dt =
~ dk x J, dt
-t
W
=
-t
W
A
x k.
Hence the rate of change of vector 7, when the unprimed system is rotating with an angular velocity TJ with respect to primed system is given by
d7 dt
d
A
A
A
= dt(xi+yj+zk) dx ~ dy ~ = dt Z + dt J + dx dy = dt i + dt j +
dz dt k dz dt k
dx ~ dy = dt Z + dt J
dz
A
A
A.
di
A
A
d]
dk
+ x dt + y dt + z dt -t
A
+ x( w
A
x i)
-t
-t
+ y( w
-t
A
x j)
+ z( w
A
x k)
-t
+ dt k + w
x r
d7, d7 r -t -t =?--=--+wxr dt dt
(8.5)
dJ;
where r represents the velocity of the rotating system and the term TJ x 7 arises as the result of rotation of the system. The relation can be used to obtain expressions for velocity of the particle situated at the point P. In operator symbol this equation can be written as d'
d
-dt = -+ dt
-t
(8.6)
wX
it
where is the time derivative in the rotating co-ordinate system. This relation may be used whenever we wish to discuss the motion of a particles, relative to a rotating co-ordinate system. In particular, if we operate it in TJ, we get, d-t d-t d,-t W W -t-t W -t-t-t (8.7) dt = dt + W x W = dt as W x W = O. This shows that the angular acceleration rotating frames. Now the acceleration is
d,27 d' d'7 d dt2 = dt (dt) = (dt d27 dt d2-t r
-t
+W
-2-
-t
x r
+ -Wt x
d7 dt
will be the same in both fixed and
[d7 x) dt
-t
+W
-tJ
x r
d7 - t -t -t + W x (w x r) dt d-t d-t -t r -t -t -t) W -t = -d 2 + 2w x -d + W x (W x r + x r. t t ~ -t -t dr -t, -t -t -t -t -t) dw -t a =a+2wx dt +wx{wxr + dt xr.
=
+ -dTJ dt
iJ
+ -Wt x
(8.8)
8.2. ROTATING FRAMES OF REFERENCE
245
dJ!
The term W x (w x ~) + X ~ is called acceleration of the transport. The expression is the acceleration of the particle situated at the point P, relative to the rotating frame where, (i)
(ii)
d;X = rt is the acceleration as measured in the unprimed system. dX = velocity relative to the rotating axes. 2w x dX is called Coriolis acceleration. This acceleration is present only when the particle is moving in the unprimed system with velocity 7. It arises from the motion of the particle in the non-inertial frame.
(iii)
w
(w
x ~ = velocity due to rotation of the axes. Wx x~) is called Centripetal acceleration. This term reduces to w 2 r if w.~ = O. This acceleration of the particle situated at point P is directed towards the axis of rotation and is perpendicular to it. Its magnitude 2
Iw x (w x ~)I = w 2 r sin () = ~() rsm where v = wr sin () is the speed of the particle when it rotates in a circle of radius r sin (). (iv)
dJ!
x ~ = angular acceleration of the particle due to the acceleration of the rotating axes. It is known as Euler acceleration. When W is constant, it vanishes.
The equation (8.8) is sometimes referred to as Coriolis theorem. Thus we see that, in a rotating non-inertial frame, two pseudo forces will appear, one of which is called coriolis force and other is centrifugal force. The action of the first one is conditional and is not effective for the rest particle with respect to rotating frame, where as the second force always acts in rotating frame. Deduction 8.2.1. Velocity and acceleration in plane polar co-ordinates: Let P(r, ()) be the position of the particle at any time t, i = unit vector in the radial direction (r increasing), ] = unit vector in the cross-radial direction (() increasing) and let i x 3 = k. Now with the rotation of particle P, i,3, k form a rotating set at -+ .A ---+-+A 0, rotating with the angular velocity w = ()k. Now, OP = r i. So, -+ V
d
-+':
~':
-+
= dt ( r ~) = r ~ + w ~"
= r i
-+':
x ( r ~)
." --+..... -:..,." --+.,.. + ()k( r i) = r i + ( r ())j.
Hence the velocity components in radial and cross-radial directions are respectively ~ -+' 1° and r (). Also, -+
a
d1J
d ~A
-+' A
= dt = dt [ r i + ( r ())j]
= 7i + (70 + ~(j)] + 70] - ~02i = [7 - ~02]i + [270 + ~(j]).
CHAPTER 8. REFERENCE FRAMES
246
Hence the acceleration components in radial and cross-radial directions are respectively -:P - -;;7iP and 270 + -;;70.
8.2.3
Equation of motion
In a inertial frames of reference, the Newton's second law of motion is valid. Let us assume that Newton's law of motion holds in the primed system. If we are to consider the motion of a particle of mass m, then its equations of motion from the point of view of an observer who is at rest with respect to primed system as ---+/
ma
= ---+ F
d2 -;;7 ---+ d-;;7 ...... ---+ ---+ dc;J ---+ ---+ or,m-2-+2mw x-+mw x(w x r)+m- x r =F dt dt dt d2 -;;7 ---+ ---+ d-;;7 ---+ ---+ ---+ dc;J---+ ::::} m dt 2 = F - 2mw x dt - mw x (w x r) - m dt x r
(8.9)
---+
where F is the force act on mass m. This relation shows that Newton's equation ---+ of motion retains its form in a rotating frame if to the applied force (F) we add a number of fictitious forces. Thus, the fictitious forces acting on the particle in the rotating frame are ---+
(i) The real force F
= m ddt->f , it 2
is as usual term as Newton's second law.
(ii) The force -m c;J x (c;J x -;;7) arising as a result of the rotation of co-ordinate axes. The term -mc;J x (c;J x -;;7) is called the centrifugal acceleration, in the general sense, for nonuniform in rotation which always points radially outwards. This
N ~------:,.:P,----
o
cox(coxr)
rXm
Figure 8.3: Centrifugal force acceleration will be forced to be zero at the pO,les where c;J is parallel to -;;7, . and has the maximum value at the equator. It's magnitude is mw2 p, where p = sin e is the distance of the point, from the axis of rotation. (iii) The coriolis force -2mc;J x d:;; is an velocity dependent force arising as a result of the motion of the particle in the rotating system. Like the dissipative forces it depends on the velocity. It is obviously perpendicular to the plane containing c;J and When a particle is fixed with respect to the rotating .... -> ---+ co-ordinate system, dd; = 0 and the coriolis force is zero.
d:;;.
8.3. LAGRANGIAN AND HAMILTONIAN
247
(iv) The force mTJ x (TJ x 7). For a uniformly rotating system, TJ is constant and hence TJ x (TJ x 7 is zero. Thus the acceleration (11 referred to a uniformly rotating system is ---t ---t 1
a
=
-F
m
---t
-
---t
(---t
---t)
w x w x r
- 2w Xd -r . ---t
dt
The centrifugal and coriolis force are arising because of use of non-inertial frame. Thus non-inertial forces are often helpful in dealing with the problems concerned ---t with rotational motion. Now the effective force F ef f is given by ---t
F eff
= ---t F -
d7 - mw ---t (---t ---t dTJ---t x w x r) - m - x r. dt dt
---t
2mw x -
---t
=F +
non-inertial forces.
The distinction between a true force and a fictitious force is that while the former is due to interaction between the particles, the later is due to acceleration of the reference frame. Thus, for a satellite moving around the earth in the fixed system, the only real force acting on it is the gravitational force that produces the centripetal acceleration.
8.3
Lagrangian and Hamiltonian
Since the Lagrangian L is a scalar quantity, we still use the same form as L = T - V in this frame also. The Lagrangian of the system has to be evaluated in the inertial frame only and then should be expressed in terms of quantities defined in the rotating frame. For a single particle, the PE is V = V(171) and the KE T is given by, T =
1
~2
2m r
=
1
---t
---t
---t
2
2m[ v + w x r 1
= ~ml"if12 + ~mltt x TJI2 + m"if.(TJ x tt) where "if is the velocity of rotating frames. Therefore, the Lagrangian L is
L
1 ---t 2 1 ---t ---t 2 ---t ---t ---t' () = T - V = 2ml v I + 2ml r x w I + m v .( w x r) -'- V r 8L '* 8"if = m v + m( w ---t
---t---t
x r) = Pr
where Pr is the conjugate momenta corresponding to the co-ordinate 7. Thus the true momentum of the particle is not just m 1/ but m[1/ + TJ x ttl which is the momentum observed in the fixed frame. Also,
8L
---+
---t
87 = m v x w
---t) ---+ + m( ---t w x r x w -
---+
V V.
CHAPTER 8. REFERENCE FRAMES
248
The Lagrange's equation of motion gives,
~(8~)_ 8~=O
--=-t ='?mv=
dt 811 8r 2mvxw (--+ --+) +mu.,xr (--+ --+) xW-v. --+ MV
The first term 2m(11 x w) of the RHS of the above expression is called Coriolis force and the second term m(w x ~) x w is known as centrifugal force. The Hamiltonian H of the system is H = --+ P r' --=-t r - I
8.4
J
2 2 1 1--+1 1 1--+ = 2m v - 2m r x --+1 w + V(--+) r.
Motion of rotating earth
The earth rotates about its north-south axis from west to east with an angular velocity 0.7292 x 1O-4 ra dj sec. When we ordinarily think of co-ordinate system, we generally take the laboratory system, i.e., one fixed to the earth, in which our necessary equipments are at rest. In fact, this system is rotating with the earth, so that it becomes a non-inertial system and so the fictitious forces appear. To find the effect of rotation, we need a fixed frame. We may imagine one such with its origin 0 at the centre of earth, its z axis north-ward along the earth's rotation axis and x, y axes at right angle to this. The origin 0' of the other co-ordinate system is fixed to the earth at the place of observation. It's z' axis is the upward vertical at 0', x' axis is eastward along the line of latitude and the y' axis is northward along the line of longitude. Let a particle of mass m be situated at P near or on the surface --+ of the earth and suppose some applied force F acts on it in addition to its weight Let ~, be the position vector of the particle at P with respect to 0 and with --+ respect to 0' is ~. Let 00' = R, R = radius of the earth. Then,
mg.
--+ ~'=~+ R
='?
--+ d~' d~ --+ --+ dR ='?--=-+w x r +dt dt dt d2--+ d--+ d 2~ r r --+ (--+ --+) --+ r ---;]l2 = dt2 + w x w x r + 2 w x dt
+
d--+ w--+ dt x r --+
+
d2--+R dt 2
where the rotating system has a translation as well. Also, R is a vector of constant magnitude and rotates with 0' frame with an angular velocity W about 0 Z axis.
iii
--+
--+
Thus, dt = w x Rand 2 --+ --+ --+ --+ d R = ~(dR) = dw x x dR dt 2 dt dt dt + dt --+ (--+ --+) iasWIS --+. fi xe. d =wx wxr
11 w
8.4. MOTION OF ROTATING EARTH
249
(;J x (w x 7) is directed towards the rotation axis of the earth and perpendicular to it. In the inertial frame, the equation of motion of the particle will be
d2 7'
m---;]i2 =
-+-+
m
9
+F
d 2-+
T -+ -+ -+ (-+ => m-= m 9 + F - mw x w
X
-+ -+ (-+ = m -+ 9 + F - mw x w
X
~
d-+
-+
-+ T) - 2mw x -
-+
-+
'r
&
2-+
d R - m-~
d7
T) - 2m w x dt'
This is the equation of motion on the surface of the earth rotating with constant angular velocity It shows that if the form of Newton's law (valid in inertial frame) is to be preserved in the rotating laboratory frame (no~-inertial frame).
w.
Deduction 8.4.1. Effect of g: Let earth be a sphere of radius T and centre O. ~hoose the co-ordinate frame OXYZ, such that Z-axis coincides with the axis of rotation. Then, X~ -plane represents the equitorial plane of the earth. The earth
z OJ N ___- r - _
c
Fe
~'-------:r--Y
s Figure 8.4: Effect of g
w
rotates about Z axis with uniform angular velocity from west to east, so that = wk. Consider a particle of mass m placed at a point P on the surface of earth, where the geographical latitude is A. The position vector of the particle is given by 7 = r cos A} + r sin Ak. If r is the unit vector along OP, then f = cos A} + sin Ak. The particle at point P is acted upon by the following forces:
w
(i) The real force equal to the weight mg of the body along PO, i.e.,
mg = mg(-r) = -mg[coSA}
+ sin Ak].
CHAPTER 8. REFERENCE FRAMES
250 ~
(ii) The fictitious force Fe is given by ~
~A
A
A
Fe = -m w k x [wk x (r cos Aj
= -mwrk:
+ r sin Ak)] A
+ sin AD] = mw 2r cos A].
x [cosA(-i)
~
The central force Fe acts along CP, i.e., radially outwards along the circular path traversed by the particle on the surface of the earth due to rotational motion. If m 9' be the effective weight of the body, then, ~,
~
m9 = m9
*
~
+ Fe =
-mg( cos Aj + sin Ak) A
2
~,
A
A
+ mw
2
A
r cos Aj
A
9 =-[gCOS)..-w rcosAJj-gsinAk.
Therefore, the magnitude of the effective acceleration due to gravity ,is given by, g' = J(g cos A - w2rcosA)2 + (gsinA)2 2w 2r w4r2 = g[l - - - cos2 A + - - cos 2 AJ1 / 2. 9 g2 Now,'r
= 6400km = 6.4 w2r
9 2
x 1Q6m ;w 21T
= 24x~~x60rad8-1 2
= [24 x 60 x 60 J x
6.4
and 9
= 9.8ms- 2, so that
6 X 10 9.8 = 0.003436«
Since w r « 1, the third factor containing ~ 9 9 fore, the above equation reduces to
=
(w2r? 9
1.
can be neglected. There-
When the particle lies at equator, i.e., A = 00 , geq = 9 - w2r and when the particle lies at poles, i.e., A = 900 , gpo = g. From these results, it follows that due to the rotation of earth, thee effect on the acceleration due to gravity is zero at poles and it goes on increasing as one moves from the poles to the equator, and becomes maximum at the equator. Ex 8.4.1. How fast than its present speed of rotation, earth should start rotating, so that the effective acceleration due to gravity at the equator will become zero? SOLUTION: Due to rotation of the earth, the effective acceleration due to gravity at latitude 'A is 9' = 9 - w 2 r cos2 A, where w is present speed of rotation of earth and r is its radius. At equator, A = 00 , we have, geq = 9 - w2r. Suppose that Wo is the speed of rotation at which, if the earth rotates, the effective acceleration due to gravity at equator will become zero. Therefore, if w = wo, geq = O. Thus
8.4. MOTION OF ROTATING EARTH
251 24X~~X60rads-l and
Now, radius .of the earth, r = 6400km = 6.4 x 10 6 m; w g = 9.8ms- 2 , so that w 2r
9
=
27r 2 6.4 x 10 6 [24 x 60 x 60] x 9.8
=
1 291
~
w2
J = 291 ~
Wo
~
17w.
Thus, if the earth starts rotating with angular speed of about 17 times its present value, then the acceleration due to gravity at equator will become zero. Deduction 8.4.2. Effect of coriolis force on a particle moving on the surface of earth : Here we are now interested about the effect of coriolis force on a
T y
IS Figure 8.5: Local co-ordinate system
particle moving on the surface of earth surface. Let earth of radius r and centre o rotates about its axis of rotation from west to east with uniform velocity TJ. Consider a mass m lying at point P in the northern hemisphere where latitude is ).. Consider that a co-ordinate frame P XY Z is attached to the surface of earth at point P, such that its Z axis, i.e., PZ coincides with the vertical line OP, Y axis, i.e., PY is directed towards north of the earth, so that X axis, i.e., P X is directed towards east. Let the particle moves with velocity 11 in the XY plane, then 11 = vxi + vyJ. The angular velocity TJ lies in Y Z plane makes an angle ). with the Y axis, so TJ = w cos).) + w sin ),k. Due to the earth rotation, the coriolis force acting on the particle is given by -+
-+-+
i
j
k
Fe = -2mw x v = -2m 0 wcos). wsin). Vx Vy 0 = 2mwsin).[vyi - Vx)J
+ 2mwcos).(vxk).
•
CHAPTER 8. REFERENCE FRAMES
252
The acceleration due to the coriolis force is given by ---+
a
1 ---+ m
~
~
~
= -Fe = 2wsin>,[vyi - vxjl + 2wcos>.(vxk)
a:
a:
w here H = 2w sin>. [v yi - vxYl and v = 2w cos>. (Vx k) are the horizontal and the vertical components of the acceleration due to the coriolis force.
a:
(i) When the particle is moving along X axis, then Vy = 0 and H = -2w sin >'vxY, i.e., the acceleration of the particle will be along negative Y axis. Therefore, a particle moving in northern hemisphere along X axis will be deflected towards negative Y axis. Similarly, it can be shown that in case the particle is moving in southern hemisphere along X axis it will be deflected towards left.
a:
(ii) When the particle is moving along Y axis, i.e., Vx = 0, then H = 2w sin >.vyi, i.e., the acceleration of the particle will be along positive X axis. Therefore, a particle moving northern hemisphere along Y axis will be deflected towards positive X axis. Similarly, it can be shown that in case the particle is moving in southern hemisphere along Y axis, it will deflected towards left. Thus the maximum magnitude of the coriolis acceleration is at the north pole, or, the south pole and is ac
= 2w sin ~v = 2 x
7.29
X
1O-5 v
where v is the velocity in the horizontal plane. Thus we conclude that although the magnitude of coriolis acceleration is small, it plays an important role in many phenomenon in the earth. It is take into consideration the effects of the coriolis acceleration in the flight of missiles, the velocity and the time of flight which are considerably large. Hence the velocity 11 of the moving particle in the horizontal plane on earth surface, we can immediately find out the angle of deflection for particle motion along the duration t as, distance travelled in time t in the deflected direction a = distance travelled in time t in the direction of projection
_ ~(2wsin>'v)t2 -"'------- = w sin >.t. vt Thqs, on the north pole (>. = 900 ), a = wt, which is maximum and for t = 1808, then a = 7 x 1O-4 rad = 0.040 , which is found to be quite small, but it assumes considerably important in guided missiles.
Ex 8.4.2. Deflection of a freely falling body
253
8.4. MOTION OF ROTATING EARTH
Let us assume that the angular velocity of the earth's rotation is small so that the variation in 9 can be neglected. In this situation, we are to find the horizontal deflection from the vertical of a freely falling particle (neglecting effects of wind. viscosity) caused by the coriolis force near the earth's surface, falling in the earth's gravitational field. Let P be the position of the particle (Fig.8.5) of mass m, which is falling freely from rest under gravity towards the earth from a height h( P A = h) in the northern hemisphere. Although the coriolis force will give rise to small velocity components in the horizontal plane, we can neglect them in comparison with V z • Thus the velocity 11 at P is given by
11 = gt = -gtk = (0,0, -gt). OPZ is along z-axis which is in vertically upward direction and makes an angle () = 900 - ). with ON. The particle is falling along t.he z-axis freely under gravity. o is the centre of the earth and P is the origin of the xyz system. OB is the polar axis of rotation of the earth with angular velocity (;J. P B is parallel to ON, then LNOZ = (J and the angular velocity is along P B. Since the particle velocity is almost vertical, the angular velocity vector (;J lies in north-south vertical plane, i.e., YZ plane and makes angle.(J with y-axis, so (;J = (O,wcos(J,wsin(J). The component of (;J along the vertical direction is w cos (J and dhected upward in the northern hemisphere and downward in the southern hemisphere. Hence, a body falling freely in the Northern hemisphere will be deflected to the east. PT is the tangent at P and is along the north direction denoting y-axis. The line P R 1S along x-axis towards eastern direction. The particle is acted upon by the following forces: (i) The real force equal to weight m 9 of the body along PC, i.e., m 9
=
mg(-k) = -mgk. (ii) The fictitious force F curialis = -2m (;J x 11. To find the coriolis force on the particle, let us evaluate the deflecting force 2m d;J: x (;J is in the east-west direction as
--
i
j
vxw=x
iJ
k
Z ow cos (J w sin (J
i 0
3
k
0 -gt = wgt cos (Ji. ow cos (J w sin (J
-
Therefore the coriolis force on the particle is F curialis = 2mwgt COS (Ji~ and so the effective force on the particle is given by
F
Effective
- -
= F Real + F Coriolis = -mgk~ + 2mwgt cos (Ji~
Hence the path of the particle will be deflected towards the right in the northern hemisphere and left in the southern hemisphere due to coriolis acceleration. The
CHAPTER 8. REFERENCE FRAMES
254
equation of motion is given by ~
= 2m ~ v
m r =}
= my =
mx
~
"
x w - mgk
i
=2 x
°
J
y
z
-mgk
wcosOwsinO 2m(ywsinO - zw cos O)j East ward -2mxw sin OJ
North ward
mz = 2mxw cos 0 - mg. Vertically upward dy .. 2 . 0 dz 0 =} x = dt w sm - dt w cos =}
d .. dt x
=
d2 y . d2 z 2 dt 2 w sm 0 - dt 2 W cos 0
= 2w sin O[-2xw sin 0] = -4w 2 [x - :w cosO]. Let p =
2w cos O[2xwcos 0 - g]
x - fw cos 0, then this equation becomes, 2
ddt p 2 =}
x
+ 4w 2P =
°
=}
P = A cos 2wt + B' sm 2wt
= JL cosO + A cos 2wt + B sin 2wt.
2w The body falls under the action of gravity, so the velocity will be almost opposite to z-axis. The small correction on deflection of V z . produced by the coriolis force is quite small and hence the value of the force have negligible components along x and y direction. As 9 is directed along z direction the components of acceleration of the particle at time t = are
°
= 0, y = 0, Z =
x
-g
=}
x
= 0, y = 0, Z = -gt
as the particle is falling along z- axis. Using these initial conditions, we get x = :w sinO(1- cos2wt).
The x- component equation changes. It is evident from this result that the particle falls following a spiral path about the vertical. Again integrating the above equation, with respect to t, we get the coriolis displacement Xc toward east, as Xc
J
dx =
2~ cos 0
o =} Xc
t
J
(1 - cos 2wt)dt
0
9 sin2wt = -2w cos O[t - 2w ] 9 1 (2wt)3 = 2w cosO[t - 2w {2wt - -3-!-
1
= '3Wgt3 cos e.
+ ... }]
8.4. MOTION OF ROTATING EARTH
255
Hence if h is the height from which the the particie falling from rest, the time. of falling T is T = ~, follows from the equation z = -g. Hence the deflection of a particle towards the east when it is dropped from rest at a height h, at latitude () in the Northern hemisphere will be deflected eastward by a distance Xl
1 2h 2V2 = -w cos ()g{_)3/2 = --w cos () 3 9 3
fJ3 -
9
where we neglect the resistance of air. This is the eastward effect on a freely falling body, due to the effect of coriolis force. For example, if the particle be dropped from a height of 100 m from rest at latitude e = 45°, it will be deflected by about 1.55 x 1O- 2 m toward east in northern hemisphere. The coriolis deflection in a vertical fall is greater at the equator e = 0, where 1/ is perpendicular to (;J and it is absent at the poles e = I' where 1/ is parallel to (;J.
Ex 8.4.3. A particle is projected vertically upward with a velocity V from a point 0 on the earth's surface. Prove that when it returns to the horizontal plane through 0 it will have a westerly deviation 4V3~80SA, ), being the latitude of the place 0 and w the earth's angular velocity of rotation. SOLUTION: Let P{x, y, z) be the position of the particle at any time t, then the equation of motion of the particle relative to the rotating earth,
x - 2wysin), = 0; ii + 2w[xsin), + zcos),] = 0; z - 2wycos), = -g where), is the latitude of the point P{x, y, z). The given initial conditions are x(o) = y(o) = z(O) = 0; x(O) = y(O) = 0; z{O) = V. Integrating the first equation and using initial condition we get
x - 2wy sin), = constant
= 0 => x = 2wy sin ),.
Integrating the third equation and using the initial condition, we get
z - 2wy cos),
= -gt +
Substituting the values of x and
constant => z - 2wy cos), = -gt + V.
z in the second equation and neglecting w2 , we get
ii + 2w[sin )'(2w sin),) + cos )'(2wy cos), + V => ii + 2w{V - gt) cos), = 0 => y + 2w{Vt - ~gt2) cos), = constant = 0 1
- gt)]
=0
1
=> y + 2w{ '2 V t 2 - '6 gt 3 ) cos), = constant = O. Substituting for
y in the expression for x and neglecting w2 we obtain, x
= 2wsin)'[2wcoS)'{~Vt2 - ~gt3)] = 0 =>
X
= constant = O.
256
CHAPTER 8. REFERENCE FRAMES
Substituting for
y in the expression for z and neglecting w 2 we obtain,
z= V
- gt
=}
z - Vt+
~gt2 = 2
constant
= O.
Let at· time t = T the particle returns to the horizontal plane, then
1 2V VT - 29T2 = 0 =} T =
g'
At time T =
2:,
the deviation is given by
y
= -2wcOSA[~ 4V3
_
2 g2
f!. 8V3] = _ 4V3wCOSA 3g2
6 g3
Since this is negative, the particle suffers a westerly deviation of amount 4V3~l)s>,.
Ex 8.4.4. A particle is fired at latitude .X with velocity V directed towards the west and at an angle 0 with the horizontal. Prove that the time taken to reach the maximum is ~ sin 0 - 2~f cos Asin 0 cos O.
SOLUTION: Let P(x, y, z) be the position of the particle at any time t, then the equation of motion of the particle relative to the rotating earth,
x-
2wy sin A = OJ ii + 2w[x sin A + Z cos A]
= OJ Z -
2wycosA
= -g
where A is the latitude of the point P(x, y, z). The given initial conditions are x(O) = y(O) = z(O) = 0 x(O) = OJ y(O) = -V cos OJ z(O) = V sin O. ~ntegrating the first equation and using initial condition we get x - 2wy sin A = constant
= 0 =} x = 2wy sin A.
Integrating the third equation and using the initial condition, we get Z - 2wy cos A = - gt + constant =}
z - 2wycos A =
Let at time t = T the particle reach the
-gt + V sinO.
m~imum
hight and then Z = 0,
2wy cos A - gT + V sin 0 = 0 V. 2wy =} T = - sm 0 + -- cos A.
9
9
Substituting the values of x and Z in the second equation and neglecting w2 , we get
ii + 2w cos A[gt + V =}
sinO] = 0
Y+ 2w [V sin Ot + ~ gt 2] cos A = constant
=}.y
+ 2w[~ V sinOt 2 + ~gt3] cos A =
= - V cos 0
-V cosOt.
8.4. MOTION OF ROTATING EARTH
257
Neglecting the term containing w2 , we get the time as T
V.
2wy
= -smO + --COSA 9
9
= V sin 0 + 2w cos A[ - 2w( ~ V sin Ot2 + ~ gt3 ) cos A - V cos Ot] 9
2
9
6
= V sinO- 2w cos AV cos O.t 9 9 V . 0 2wV2 = - SIn - - - 29
COS
\. sm 0 COS 0
A
9
Ex 8.4.5. Foucault's pendulum
If a freely suspended simple pendulum with a heavy bob and a long suspension, is
~--~~-------y
x
m
Figure 8.6: Foucault's pendulum set swinging at the north pole in the given plane in space, then its linear momentum perpendicular to the plane is zero, and it will continue to swing for a long time in the invariable plane while the earth rotates beneath it. J.L. Foucault [1851] established that, the plane of oscillation of a pendulum is caused by the coriolis force to precess about the vertical axis through the point of suspension. To analyze the motion of Foucault's pendulum, the following assumptions are made (i) The displacement of the bob from the equilibrium position is very small, since the suspension is very large, the motion of the bob will be practically confined in a horizontal plane. As such, z becomes constant and negligible and so z, Z becomes negligible, i.e., Z « :i; or iJ and Z ~ o.
(ii) When the displacement of the bob is very small and suspension is very long, the tension of the suspension becomes practically vertically upward and nearly equal to the weight of the bob. In this pendulum, the size of the bob is of no consequence. Here we discuss the effect of coriolis force, on the oscillation of the simple pendulum. The pendulum is under the action of the following forces :
CHAPTER 8. REFERENCE FRAMES
258
(i) The true force on the pendulum is the downward force of gravity. Thus the real force, i.e., the weight m 9 of the bob is acting vertically downwards and
mg
= mg(-k) = -mgk. ~
(ii) The real force equal to the tension T in the string. When the displacement of the bob is very small, in case of a pendulum of very large suspension, the tension T in the suspension will almost vertically upwards, i.e., T is nearly equal to mg. ~
~
~
- - -
(iii) The fictitious force F Coriolis = - 2m W x 11, where v = xi + iJj + ik is the velocity of the bob at any instant. The angular velocity vector W lies in yz-plane and makes angle () with the y-axis, so W = (O,wcos(),wsin()). Therefore, the coriolis force on the bob is given by ~
F Coriolts
~
= 2m v
~
x w
i
=
j
k
. z o w cos () w sin ()
. 2m x
iJ
= 2mw[( - i cos () + iJ sin ())i -
xsin ()] + xcos ()k].
Now, as coriolis force is always at right angles to W the effective component in the horizontal plane will be due to w sin ()k, where () is the geographical latitude of the place. Hence the x and y component of coriolis components are 2mw sin ()iJ and -2mw sin ()x, which acts perpendicular to the plane of oscillation of the pendulum and is solely responsible for rotation of it. ~
(iv) The fictitious force F Centrifugal = -m Wx (w x 7), where 7 is position vector of the bob with respect to the origin of the co-ordinate frame. The angular velocity of the earth is very small. Since the magnitude of the centrifugal force is proportional to w 2 , it can be neglected.
.
~
(v) The fictitious force F Transverse = -m W X 7. Since the angular velocity of the earth is constant, F Transverse = O. Hence the effective force on the bob of Foucault's pendulum is given by ~
~
~
~
~
FT=FR+Fc m -it = (m 9 + T)
+ 2mw[( -i cos () + iJ sin())i - x sin ()] + x cos Ok].
Assume that the displacement of the bob from the equilibrium position is very small. Since the suspension is very large, the motion of the bob will be practically horizontal, so that z will be constant negligible and so i, Z will be zero. Now the motion of the pendulum is confined to the horizontal xy-plane, so we shall consider only the horizontal component of coriolis force. The equation of motion for the pendulum is ~ mr
= 2m
d7~~ dt x w - V'V
~~~ x w - V'V
= 2m v
8.4. MOTION OF ROTATING EARTH
259
where V = !k2m(x2 + y2) and k 2 ~ sr-; l =the effective length of swing. Hence the equations of motion in terms of cartesian components becomes,
+ k2x = ij + k 2 y =
x
2w(iJ sin () - i cos ()) ~ 2iJw sin () -2xw sin ().
From this, we see that the equations carry the contributions from the coriolis acceleration in addition to the usual ones of the harmonic oscillator. In order to solve the equations, multiply the second equation by i = P and then add with the first equation we get
(x + iij) - 2w sin ()(iJ - ix) + k 2 (x + iy) = 0 or, ~ + i20e + k 2 = 0; 0 = wsin()
e
=> e=
Ae-i[n+~ ]t
+ Be-i[n-~
=> x + iy = [Ae-i~ t + Bei~ = e-iflt[(A + B) cos Jk2
]t
t]e- iflt
+ 0 2 t + i(A -
B) sin Jk2
+ 0 2 t]
= e-iflt(X + iY) where X = (A + B) cos "/P + 0 2 t and Y = (A - B) sin ../k2 + 0 2 t are solutions when coriolis force is absent and (A!~)2 + (A~~)2 = 1, which is the equation of the ellipse. Thus, due to the rotation of the earth, the plane of oscillation of the bob traces an elliptical path. The factor e- iflt tells that the plane of the oscillation of the bob is rotating with an angular velocity 0 = w sin () in the direction opposite to the direction of rotation of the earth. The continuous change in the plane of oscillation is known as precession. Now we have the following discussions (i) The period of complete rotati.on of the plane of oscillation of the Foucault's pendulum is given by
T _ 211" _
211"
_ 24h sin()
- n - wsin() -
where () is the geographical latitude of the place. At the equator, () = 00 and hence
T=
w
211" =00 sin 00
which shows that, at the equator, the plane of vibration of the Foucault's pendulum does not rotate at all.
(ii) At latitude (), the period of rotation of the plane of oscillation of a Foucault's pendulum is T = wsm 2?," o' Now, 211" w = 24h = 1 day, the period of the earth, so
CHAPTER 8. REFERENCE FRAMES
260
that T = si!oday. If n be the angular velocity of the plane of vibration of Foucault's pendulum, then 21f
21f
.
n = -T = 1/.sm a = 21fsm(} rad/day. At the pole, A = 90 0 , so that
npo1e =
21f sin 90 0 = 21f rad/ day
which is universally true. At pole, the plane of vibration of a Foucault's pendulum rotates through 21f radian in a day. Thus, once an ideal pendulum is set swinging in a given vertical place, it will observed to suffer deflection a precession, in the course of 24hrs, the plane of oscillation will be turned through an angle 21f sin (). (iii) At the north and south pole,
e = 90
0
and 270 0 respectively and
n = ±w.
Thus Foucault's pendulum freely swings in an invariable plane while the earth rotates beneath it. The precessional motion of a Foucault's pendulurr. is then not perceptible to an ~bserver on the north and south pole, but an observer at any other place on the earth, if he were able to watch the pendulum at the pole, could see the plane of oscillation precessing and completing the rotation once a day. Thus Foucault's pendulum is used to demonstrate the rotation of the earth.
Ex 8.4.6. Oceanographic Phenomena The coriolis force -2mTJ x U arising from the earth's rotation, plays a significant role in oceanographic phenomena involving displacements of masses of matter over long distances, such as the river flow on the surface of the earth. Basically, rivers flow approximately in a horizontal plane for which a slight downward slope in the direction of flow is required to maintain the speed of the flow. When the river flows, there is a small component of gravitational force in the direction of flow (forced direction). In sllch a situation, the coriolis force acts across the river flow, which tends to change the direction of flow in the direction of the force. Let the direction of the flow to be the x axis, the transverse horizontal axis to the left of the flow direction as the y axis and the local vertical as the z axis. Let the direction of the flow at a place having geographical latitude A, make an angle ¢ with respect to the geographical north direction. Then the angular velocity TJ of the earth can be expressed as
TJ
= w[sin Ak + cos A cos ¢i
The velocity U of the river flow is U (/ c
= 2u
x TJ
=
= vi
-
cos>. sin ¢)].
and so, the coriolis acceleration is
-2wv[sin >.) + cos>. cos ¢k].
Hence the transverse component is - 2vw sin A, which is independent on the direction (¢) of the river flow but depends on the latitude (A) of the place. Thus the river /
8.4. MOTION OF ROTATING EARTH
261
deviation occurs to the right of the river flow in the northern hemisphere and to the left of flow in the southern hemisphere. This other effect of the coriolis force also tends to slightly raise the right banks of the rivers in the northern hemisphere than the left banks. The opposite is the case for southern hemisphere. Ex 8.4.7. Change of Meteorological Phenomena (a) Cyclones: When a low pressure zone in atmosphere is generated ay any particular place of the earth, pressure gradient will set up and the pressure gradient
High pressure
Low !Jt'eSRure
Figure 8.7: Cyclones
flows occur towards the low pressure zone for equilibrium of pressm·e. Due to the horizontal pressure gradient the long range motions of air masses take place, which we perceive as winds. In the absence of coriolis force, the direction of the wind velocity occurs perpendicular to the isobars (the equi-pressure lines). On the northern hemisphere, coriolis force acts on the wind to deviate its direction towards its rig aIHI this process is continuous so that the wind flows spirally towards the centre of the low pressure in an anticlockwise direction. The deflection to the right continues, until the wind motion becomes parallel to the isobars. The wind, then continues parallel to the isobars, circulating in anticlockwise direction about the centre of low pressure. Such circulating winds give rise to whirlwinds and cyClones. This cyclone activity is limited to a smaller area on the earth. Since the coriolis force acts perpendicular to the trajectory, it will provide the centripetal force. Let R be the radius of the wind trajectory, the centripetal acceleration is v2
v
R
2w
-=2wv~R=-.
III the southern hemisphere, the coriolis force will deflect the wind motion towards left and so in a whirlwind or a cyclone set up in the southern hemisphere, the wind whirls in clockwise direction. At equator, >. = 0°, so the horizontal component of
CHAPTER 8. REFERENCE FRAMES
262
acceleration due to coriolis force is zero and so no cyclones are found at the equator. (b) Trade Winds and Tropical Winds : If the moderate pressure gradients form over large distances, the wind flow is still from high pressure zones to a low pressure zones, but due to deflection by coriolis force ultimately becomes parallel to the low pressure isobar and the resulting coriolis force just balances the force due to
Low pressure
High 'pressure
Figure 8.8: Trade winds pressure gradient. The wind flow the continuous along isobar around the centre of the pressure. If such a flow occurs in tropical zones, these winds are called tropical winds. The same effect is responsible for the direction of trade winds. The heating of the earths surface near the equator causes the air to rise, and to fill up the vacuum cooler air from north rushes to these areas. Because of the effect of coriolis force, the wind does not flow directly north to south, but deviates towards the west in the northern hemisphere, this giving rise to north-eastly trade winds. In the southern hemisphere, the same effect causes the south-east trade winds.
Ex 8.4.8. Larmor Precession: Larmor frequency has a great importance in atomic and nuclear physics. According to classical electromagnetic theory, electrons revolve around the nucleus in discrete orbits under the influence of an central attractive electrostatic forces exerted on then by the nucleus. Let in a magnetic field, a particle of charge q is moving in an orbit round a fixed point of charge q'. The magnetic force of the moving charged particle is F = q 7 x B, where B is the magnetic density. This frame has the same form as the coriolis force. The magnetic moment AI and the angular momentum L of a system of moving changes, relative to a particular origin are given by
- - -
- -
-
-q---M=2c[r x v];
-
L=m[r x v].
Here both M and L have similar form. Including electric force and magnetic force B, for such a bound system, the equation of motion of the particle of mass" m can
to#f,
8.4. MOTION OF ROTATING EARTH
263
be written as ~ qq' ~ q ~ ~ 'mr =--er +-['I' x BJ '1'2 e d11 qq' ~ q ~ ~ or m - = --e + - [v x BJ 'dt '1'2 r e
where Er is the unit vector along T7, 11 is the velocity, i.e., absolute time-derivative of the position vector T7. Let the co-ordinate system is rotated about the origin with angular velocity TJ. Then using the expression, 11 = -:t + TJ x T7, we have
d11~ dt = 'I' + 2[ ~~J~~~J w x 'I' + w X [w x 'I' • Thus the equation of motion of the charge in the rotating system is ~ ~J ~ [~ ~J '1'+ 2[~ wxr+wxwxr
qq' ~ q ~ ~ ~ =---2er+-[r + w x '1'] mer me
=-
qq' 2 Er mer
~ X
B
+ (-2 q )2(iJ x (iJ x T7)J; me
~
q~
w =---B. 2me ~
Since the magnetic force that we have eliminated in linear in B, whereas the magnetic term (~)2 [B x (B x T7) ] is quadratic in B, it is possible in the limit of weak magnetic fields to neglect this term. Let us assume that the magnetic field B is weak, such that ~
~
~
~
i.e., it is equivalent to requiring that TJ be much less than any frequency associat.ed with the motion in absence of the magnetic field. Then with this weak field approximation, the equation of motion becomes ~ 'I'
qq' ~ = - mer2 - - er
which is the equation of motion of a particle in an inverse square-law field. Therefore, in a rotating frame, the orbit is precessing ellipse, precessing with angular velocity TJ . Since w2 < < w5, the axis of the ellipse will precess only through a small angle in each revolution. This effect is known as Larmor effect and for the classical gyromaganetic ratio r = ~, the precession angular velocity TJ = ~ B is known as the Larmor frequency. For electrons-2 is negative, and the Larmor precession is anticlockwise around the direction of B. Hence we conclude that the effect of magnetic field on the electron in its orbit round the nuclear remains essentially the same except for a slow precession of its orbit around the field direction. The effect results in an ~
CHAPTER 8. REFERENCE FRAMES
264
observable shift in the ~ectrallines of atom in presence of a magnetic field, known as Zeeman effect. Let 0 be the angular momentum, then ~
dO q~ [~ ~J - = - r X vxB. dt c ~
~
Let us choose B k B and the orbit is to be circular. Thus in the weak field is rotating around B with Larmor approximation, dd~ = - 29,~Jk X "!::1], i.e:,
n
~
frequency with the orbital plane precessing around B. Larmor frequency is found as the rate at which the magnetic moment vector precesses about the magnetic field vector. Larmor's theorem can be expanded to include a collection of charged particles, all with the same charge-to-mass ratio, acted on by mutual electrostatic forces and an external central force directed toward a common centre. This generalization is possible because these forces depend only on the inner particle distances, and the distances of the partiCles from the attracting centre of the external force which must coincide with the origin.
8.5
Exercises
Ex 8.5.1. Let an S' frame be rotating with respect to a fixed frame S having the same origin with the angular velocity V = 2ti - t 2) + (2t + 4)/::, where t is the time. The position vector of a typical particle at an instant t as observed in S' frame is T' = (t 2 + l)i - 6t) + 4t 3 k. Find the acceleration of transport and the coriolis acceleration at time t = 1. Ex 8.5.2. If a particle is dropped from rest at height h, show that any time t the position of the particle is x = 0, y = lwg cos ).,t3, z = h- ~gt2. Find also the equation of the trajectory. Ex 8.5.3. An object is thrown downward with initial speed Vo. Prove that after time t, it is deflected east of the vertical by an amount wvot2 cos () + ~wgt3 cos ().
Chapter 9
Rigid Body Motion By a rigid body we mean, a set of dynamical discrete mass points subject to the holonomic constraints where the distances between all pair~ of constituent points remains invariance throughout the motion and whose orientation does not change under any external force. However, in the strict sense, no material body of the world is perfectly rigid in nature. Before proceeding to discuss the problems in the dynamics of rigid· bodies (which can be solved by quadratures), it is convenient to introduce a number of constants, which can be assigned to a rigid body, depending on its constitution.
9.1
Moments, Products of Inertia
The moment of inertia determine the dynamical behavior of a body. Let a~y rigid body be considered, and let the particles of which it is constituted by typified by a particle of mass mk situated at a point, whose position vector is r+ k = Xki+YkJ + zkk. The scalar quantity
(9.1)
3
where n = either i or or k, is known as the moment of inertia about an fixed axis. Let us introduce the nine symbols lxx, l xy , l xz , l yx , l yy , l yz , l zx , l zy , lzz as n
lxx
= L mk(Y~ + z~) = A k=l
n
lyy
n
= L mk(Y~ + z~) = B;
lzz
= L mk(x~ + Y~) = C
k=l
k=l
n
lxy
=L
n
mkxkYk
= lyx = H;
lyz
k=l
k=l
n
lzx
=L
= L mkYkZk = jzy = F
mkxkzk
= lxz = G
k=l
265
266
CHAPTER 9. RIGID BODY MOTION
where the summation extended over all particles of the system. The three scalar physical quantities lxx, I yy , I zz are called moment of inertia about x, y, z axes respectively and the remaining six scalar physical quantities I xy , I xz , lyx, I yz , I zx , lzy, are called products of inertia of the body associated with corresponding planes. Each scalar component has the dimension [Ml L2]. If in a body, the matter is continuously distributed and the density function is p = p(?7), then the moments and products of inertia can be written as lxx
=
J
p(?7)(r2 - x 2)dT; lxy = -
J
p(?7)xydT
where the integration is performed over the entire volume of the body.
9.1.1
Axes theorem for moment of inertia
The moment of inertia for a rigid body about any axis parallel or perpendicular to the axis passing through centre of gravity, can be obtained by two axes theorem, which are (i) Perpendicular axis theorem. (ii) Parallel axis theorem. (i) Perpendicular axis theorem: For a two dimensional laminated body, if Ix, ly, lz be the respective moment of inertia for rotation about x, y and z axis respectively, then perpendicular axis theorem states that,
+ ly; 1:1; = ly + lz; ly = Ix + lz; lz = Ix
when the body exists in xy plane when the body exists in yz plane when the body exists in xz plane.
This means that the sum of the moments of inertia of a lamina about two mutually perpendicular axes in its plane is equal to its moment of inertia about a perpendicular axis passing through the point of intersection. For a three dimensional body, if Ix, ly, lz be the respective principal moment of inertia for rotation about x, y and z axis respectively, then perpendicular axis theorem states that, Ix
+ ly + Iz = 2 L
m(Op)2
when P is imaginary point at which total mass is supposed to be concentrated, and o is arbitrary chosen point. This theorem is applicable to laminar bodies, i.e., when the particles may be considered as confined to a plane. (ii) Parallel axis theorem: This theorem states that if AB and CD be two parallel axes such that the former axis passes through the centre of gravity of the body then for moment of inertia lAB and ICD about AB and CD axis respectively, ICD
= lAB + Ma 2
where M = mass of the body, a is the normal separation of two parallel axis lAB and ICD. Thus the moment of inertia (1) of a body about any axis is the sum of its
9.1. MOMENTS, PRODUCTS OF INERTIA
267
moment of inertia about parallel axis through the centre of mass and the product of mass of the body by the square of the distance between the two axes. This is known as parallel axes theorem.
9.1.2
Radius of gyration
The dimension of moment of inertia are [M L 2 J. SO the moment of inertia of a body can be equated to product of mass M = E mi of the body and square of some i
appropriate length K. This means that, we can always write
This quantity K in this equation is called the radius of gyration of the given body about the given axis. If we had placed a particle of mass equal to that of the body at a distance K from the given axis, the moment of inertia of the particle would be equal to that of the body. So far as rotation about the given axis is concerned, the body behaves like particle so placed. The concept of radius of gyration helps us to replace a body with extension by a particle.
9.1.3
MI of some symmetrical bodies
1. Uniform rod of length 2a and mass M : (i) About a line through the middle point and perpendicular to the rod = iMa 2 . (ii) About a line through one end and perpendicular to the rod = 1Ma2 . 2. Rectangular plate of sides 2a, 2b and mass M: (i) About a line through the centre and parallel to the side 2a is = 1 M b2 . (ii) About a line through the centre and perpendicular to the plate= iM(a 2 +b2 ). 3. Rectangular parallelopiped of sides 2a, 2b and 2c: About a line through the centre and parallel to the side 2a is = iM(b2 + c2 ). 4. Circular ring of radius a and mass M: (i) About its diameter is = ~ M a 2 • (ii) About a line through the centre and perpendicular to its plane = Ma 2 . 5. Circular plate of radius a and mass M: (i) About its diameter is = ~Ma2. ' (ii) About a line through the centre and perpendicular to its plane = ~Ma2. 6. Hollow sphere of radius a and mass M: About its diameter = ~Ma2. 7. Solid sphere of radius a and mass M: About its diameter = ~M0,2. 8. Elliptic disc of axes 2a and 2b and mass M: (i) About the axis 2a = IMb 2 • (ii) About the axis 2b = ) M 0,2. (ii) About a line through the centre and perpendicular to its plane = ~M(a2+b2).
268
9.2
CHAPTER 9. RIGID BODY MOTION
Independent Co-ordinates
The position of a rigid body is fixed as soon as three non-collinear points in the body are fixed. Any arbitrary point in the body can be localized with reference to these three basic co-ordinates in some convenient co-ordinate system. Thus a rigid body with n particles can almost have 3n co-ordinates. But, for a rigid body, as r 1.j = Cij, where r 1.) is the distance between ith and yth particle and ct ) are constants, the actual number of degrees of freedom can not be obtained by the 3n - ~n(n -1) relations. Thus to find a point in the rigid body, it is necessary to specify its distances to all other points in the body, one needs only the distances at any three ot.her non-colli~lear points. Hence to fix 3-points AI, A 2 , A3 in a body we need nine co-ordinates. But the three reference points are themselves not independent, there are in fact three equations of rigid constraints imposed on them
which reduce the number of independent degrees of freedom to six. Three independent co-ordinates out of six will be required to locate a point on the instantaneous axis of rotation in the rigid body, which is underlying a translation motion relative to some fixed or inertial frame of reference. Two independent co-ordinates are required to locate the axis of rotation passing through the above mentioned point. Lastly, the orientation of the body can be specified in terms of an angle. Therefore, to discuss the motion of a rigid body, we shall use two co-ordinate system (i) Inertial frame, fixed in space and (ii) Non-inertial or Body co-ordinate system, fixed in body. Thus to specify the configuration of a rigid body in space, we need six independent generalised co-ordinates : 3 degrees of freedom represent the translational motion (each particle of the rigid body is the same direction and of equal magnitude) and another 3 degrees of freedom represent the rotational motion (the body rotates about some point itself) of the body.
9.2.1
Generalised Co-ordinates
Let a rigid body be free to rotate about a fixed point. To specify the orientation of the rigid body the Lagrangian formulation needs three independent parameters. When these generalised co-ordinates are known, one can write a Lagrangian for the system and obtain Lagrange's equation of motion. The most common and useful set of parameters are Eulerian angles. In the Eulerian angles, we have a set of three co-ordinates, defined rather unsymmetrically it is true, yet suitable for use as a generalised co-ordinates describing the orientation of the rigid body.
9.2.2
Eulerian angles
So far, we referred the motion of a rigid body to a body frame of axes of which are coincident with the principal axes of the body. These axes rotate along with
269
9.2. INDEPENDENT CO-ORDINATES z
y
x
Figure 9.1
the body. The position of a rigid body, one of whose points is fixed, is completely determined by three co-ordinates, namely, the three Eulerian angles (),
In the second stage, the intermediate axes ~"'( are rotated anticlockwise about the (-axis by an angle () to produce another set of axes e",'('. The e-axis is at the
(,'
Figure 9.2
intersection of the xy and
e",' -planes and is known as line of nodes. Since we are
CHAPTER 9. RIGID BODY MOTION
270
to rotate the axes
~'r/(
(' =~;
about the
O~axis
'r/' = 'r/cos()
t
=
through an angle (), we have
+ (sin();
(' = -'r/sin() + (cos()
[~:] [~0 -co~sin()() cos Si~ ()]() [~]( B~. (' =
=
Finally the axes ('77'( are rotated anticlockwise by an angle 'lj; about (' -axis to produce the desired x' y' z' system of axes (fixed in the body). Since we are to rotate
c;
z'
E,
Figure 9.3
the axes ( 'r/' (' about the 0(' axis through an angle 'lj;, we have x' = (' cos't/J +.,/ sin'lj;; y' = -(' sin 'lj; +.,/ cos'lj;; z' = ('
X' =
[~;] [~~~:'lj; ~~~~~] [~:] Ce. =
z'
=
0
0
1
('
The angles ¢, f), '1/); 0 ~ ¢ ~ 27r, 0 ~ f} ~ 7r, 0 ~ 1/) ~ 27r, acting as the three needed generalised co-ordinate, are known as Eulerian angles and they will specify completely the motion of a rigid body about a point fixed in the body. These tools will now be applied to obtain the dynamical equations of motion of the rigid body in their most convenient form. The relation between old and new set of co-ordinates is given by
x' = C(' = [
CB~ = CBAX
COS 'lj; sin 'lj; 0] [ 1 0 0] - sin 'lj; cos 'lj; 0 0 cos () sin () o 0 1 0 - sin () cos () COS
'l/J cos ¢ - cos () sin ¢ sin 'lj;
[
COS ¢ sin ¢ 0] - sin ¢ cos ¢ 0 X o 0 1
cos 'lj; sin ¢ + cos f} cos cb sin 'lj; sin 'lj; sin f} ]
- sin 'l/J cos ¢ - cos () sin ¢ cos 'l/J - sin 'lj; sin ¢ + cos () cos ¢ cos 'lj; cos'l/J sin () X.
= [
sin () sin ¢
- sin () cos ¢
cos ()
Note that, z, ( and z' axes are in one plane. Also, the resultant transformation P = CBA is an orthogonal transformation.
9.3. ANGULAR VELOCITY
9.3
271
Angular Velocity
Consider a rigid body composed of n particles having masses mki k = 1,2, ... n and rotating with instantaneous angular velocity c;J. Let one of the points be fixed in the body, so that translational motion is absent. The distance between any two points of the rigid body remain fixed. this means that the position vector of any point P relative to the fixed point O(origin) of the body set of axes is constant in magnitude and so, it changes in direction only when the body is in motion(rotation). Let 11k be the linear velocity of the kth particle of mass mk situated at P, whose position vector is 7 k with respect to the fixed point, then
N ow we are to express the angular velocity vector in terms of the Eulerian angles and their derivatives. Suppose a rigid body rotates about a point 0 fixed in both body and space. Let OX, OY, OZ be three mutually perpendicular axes through o and fixed in space. At any time t, suppose the body has an angular velocity c;J referred to these axes. Then this angular velocity c;J can be considered as consisting of three successive rotations with angular velocities
where w¢ is along the z-axis, Wo is along the line of nodes (Le., along e-axis) and wl/J is along the body z'-axis. Hence the components of the angular velocity vector (O,O,w¢) on to (x',y',z') is given by CBA [oOw¢f. Since w¢ is parallel to the space z-axis, its components along the body axes x', y', z', are given by
°
Likewise, components of (we, 0, 0) vector on (x', y', z') is CB [we of. The lines of nodes, which is the direction of We, coincides with ~'-axis, so the components of We with respect to the body axes are
(W.8)x'
= ocos 'l/J;
(wo)y'
= -Osin'l/Ji
(we)z'
= 0.
°
The components of (O,O,w'I/J) on (x',y',z') is C [0 w'l/Jf. The components of w'I/J lies along the z-axis. Now adding the components of the separate angular velocities, we obtain the components of w along the body set axes
[~~ 1~CBA UJ
+CB
[Wi
[
¢cose + 'l/J
+c
1 'l/J .
¢ sin esin 'l/J + 0cos 'l/J
= ¢ sin ~ cos 'l/J - ~ sin
1 UJ
CHAPTER 9. RIGID BODY MOTION
272
Rotating axes xyz by an angle ¢ counterclockwise about the z-axis we get the resultant co-ordinate system as the axes ~TJ(. Again rotate ~TJ( about ~-axis anticlockwise by an angle e to produce eTJ'('. The e-axis is at the intersection of the .1:]j and TJ' planes and is known as the line of nodes. Finally, TJ' (' axes are rotated anticlockwise by an angle 'ljJ about the (' -axis to produce the x' y' z' system of axes, fixed in the body.
e
9.3.1
e
Precession and Nutation
We know, the angular velocities ;p, 0, -J; are directed along the axes of z, ~ and z' respectively. If the body is spinning about axis ;p, -J;, then -J; represents the spin angular velocity. When 0 = 0, angular velocity ;p represents the angular velocity with which axis z, is rotating about the z-axis. Thus ;p represents the precessional velocity. Axis z' moves on the surface of a cone which has semi-vertical angle e and oz as its axis. Angular velocity iJ indicates that axis z' may not remain on the surface of the cone. This motion is called nutation. Thus, in general, a rotating body can perform three types of motion - a spin motion, a precessional motion and nutational motion.
9.4
Angular Momentum
Here we are to derive the angular momentum of a rigid body due to rotation about any line, at any instant of its motion. Let us consider a rigid body, consisting of n particles having masses mk(k = 1,2, ... , n) and rotating with instantaneous angular velocity W. Let one of the points in the body be fixed, so that the translation motion is absent. Then the angular momentum I k of the kth particle of mass mk with respect to inertial frame is
Since the rigid body is taken as a rigid collection of particles, the total angular momentum L is the sum of angular momenta 1 k of the individual particles and is given by ~
~
n
=
L mk[(wxi + WyJ + wzk)r~ k=l
(~k.w)(xi + YJ
+ zk)]
9.5. KINETIC ENERGY
273
n
=> Lx = Wx
n
L mk(rk -
L mkXkYk -
xk) - Wy
k=l
Lz
=
=
L
-W,l;
Wz
L mkYkZk
k=l
n
Ly
n
k=l
n
mkXkYk
n
+ Wy L
mk(rk - Yk) - Wz
k=l
k=l
n
n
n
k=l
k=l
k=l
L mkYkZk k=l
L mkXkzk - Wy L mkYkZk + Wz L mk(r~ -
-Wx
z~)
which contains the linear function of all components of (J'. Let us introduce the symbols lxx, lxy, lxz. lyx, lyy, lyz, lzx, lzy, lzz for the coefficients of w x , w y , Wz as n
lxx =
L mk(r~ -
n
x~)
k=l
=L
n
mk(Yk
+ Zk);
lzz
k=l
= L mk(x~ + Yk) k=l
n
lxy
=-
+ z~)
k=l
n
lyy
mk(Y~
= L
L
n
mkxkYk
= lyx;
k=l
lyz
=-
L
mkYkzk
= lzy
k=l
n
lzx = - L'mkxkzk = lxz· k=l
The quantities lxx, lyy, lzz are called moment of inertia about x, y, Z axes respectively and lxy, l~·z, lyx, lyz, lzx, lzy, are called products of inertia associated with corresponding planes. Thus, the angular momenta can be written as
The symmetric second order tensor I which operates on angular velocity (J' is known as moment of inertia tensor, with three diagonal elements, principal moment of inertia where as, the off diagonal elements are product of inertia. In tensor notation we can write, lij
=L
m[ r20ij - rirj]
where Oij is the Kronecker delta. The coefficients lxx, ... can always be calculated if the distribution of the particles about the axis is known.
9.5
Kinetic Energy
The kinetic energy of a rigid body which is in motion, can be calculated in the same way as angular momentum. The kinetic energy of motion of the rigid body is a
CHAPTER 9. RIGID BODY MOTION
274
scalar quantity and is defined by 1~ 2 1~ ~ 12 (9.2) T=2"~mkVk=2"~mklvk k=l k=l where the summation is taken to all the particles of the body and Vk is the linear velocity of the kth particle whose position vector is 7 k with respect to the fixed point o. The expressions for the inertial velocity 11k must come from the general equation 11k = 11 + 01 x 7 ki 11k = velocity of mk relative to the inertial frame, 11 is the velocity of translation of the body relative to the inertial frame and so, 2T
n
n
k=l
k=l
= I: mkv~ = I: mk 11k. 11 k n
= Lmk(11 + 01 x 7 k ).(11 + 01 x 7 k ) k=l n
11
= L mkl 11 12 k=l
+ 11. I: ffik(W
n
x
k=l
n
= Mu 2
+ I: m}.; 7
n
k (11 x
01) +
k=l = Mu = Mu
2
2
+ (11
+ (11
I: mk(w x 7 k).(W x 7 k)
7 k) +
k=l
I: mk(w x 7 k ).(w x 7 k ) k=l
x 01).
x 01).
n
n
k=l
k=l
I: mk 7k + I: mk(w x 7 k ).(w x 7 k ) n
n
k=l
k=l
I: mk 7 k + w. I: mk[(7 k.7 k)W n
(01.7 k)7 k]
n
1 M u 2 + 2" 1 (u ~x~ 2 (~~ )2] ~ T = 2" w). '"' ~ mk ~ T k + "21 '"' ~ mk [2 W Tk W. T k k=l k=l = T(t) + T(m) + T(r) where, T(t) = translational kinetic energy of the whole body at the given instant. T(m) = is the term having mixtures of rotational and translational velocities of individual particles and can therefore be termed as the mixed term. It simply vanishes if either the centre of mass is chosen to be origin of reference frame or the whole body translates in a direction parallel to the axis of rotation of the body AND T( r) is the rotational kinetic energy of the body. Also the rotational KE can be written as n
2TCr)
= ~.[7 k
x (~ x
7 k)] = ~. L mk[7 k k=l
x (~ x
7 k)]
9.5. KINETIC E)reRGY
275
~.,
, '-.. where we can easily put the expression for KE in simpler form if we assume in much that the centre of ljass of the rigid body is at origin of the body. Using the principal axes transformdHoll, we write Iij = Ii8ij and in this case,
.'
which involve the principal moments of inertia Ix, I y , Iz along x, y, z directions. Thus the products of inertia are all zero and the moments of inertia Ix, I y , Iz are the non zero elements. This can be achieved, as Iij is symmetric, by choosing the axes in the body such that the products of inertia are zero. These axes are called the principal axes of inertia. In many practical problems, we are handling the rigid bodies with regular shapes, which helps us to locate the principal axes, (i) For each point of a rigid body, there exists a set of principal axes and the corresponding moment of inertia and the principal inertia. If two of the three principal components Ix, I y , Iz of the moment of inertia about the principal axes are equal, i.e., Ix = Iy -1= Iz then the body is called symmetrical top. In general, bodies for which Ix = Iy -1= Iz behave like homogeneous bodies of revolution about an axis of symmetry. For example, let us consider a right circular cylinder, the axis of the cylinder is taken as z axis. The symmetry of the cylinder shows that Ix = I y. (ii) When all the components of moment of inertia about principal axes are equal,
i.e., Ix
= Iy = Iz
then the body is called spherical top. Such equality may also be observed for bodies of an absolutely arbitrary shape when their mass is properly distributed. For a body with central symmetry such as homogeneous sphere or cube etc. are the examples of spherical top. (iii) When all the components of MI about principal axes are not equal, i.e.,
for all points of the body, then the body is called asymmetric top. (iv) When for a body Ix = I y , but I z = 0, then the body is called rotator ,Le., it has two degrees of rotational symmetry. When we are to e~press the KE T in terms of Eulerian co-ordinates, then T will depend upon angulat velocities ~, 8, -¢ and co-ordinates (},,,p. Since the KE term does not contain 4> explicitly, it will be ignorable co-ordinate, if it is absent in the PE also.
276
9.6
CHAPTER 9. RI6ID BODY MOTION
'"
Euler's Theorem
)
To describe the motion of 8, rigid body completely, we must specify two independent motions of rigid body: (i) a translation motion (ii) a rotational motion. Euler's theorem is one of the basic theorem to confirm the description of the motion of a rigid body. Theorem 9.6.1. A general displacement of a rigid body with one point of which is fixed, is a rotation about some axts through the fixed point. Deduction 9.6.1. The most general displacement of a rigid body is a translation with a rotation about some axis. This is known as Chasle's theorem.
9.7
Euler's Equation of Motion
Let the motion of rigid body with one point 0 of which is fixed,(chosen as our base ---t point) will take place under the action of torque N acting on it. We take principal axes at the body at 0 as a set of moving axes about which the components of angular velocity of the body are W x , w y , W z • Newton's approach leads to a set of equations known as Euler's equation of motion for such a rigid body. Thus the rotational motion of the rigid body in the inerti,al frame of reference attached to 0 is
(9.3) The time derivative in a fixed frame of reference is related to the time derivative of a vector in a rotating frame of reference rotates with an instantaneous angular velocity (;] by the following relation
(9.4) ---t
Here the angular momentum L about origin and the instantaneous angular velocity (;] are quantities that refer to a fixed inertial frame, known as 'body frame of reference'. Hence the equation (9.3) can be written in terms of body axes is
(9.5) which is the appropriate form of the Newtonian equation of motion relative to body axes. Let W x , w y , W z be the components of angular velocity (;] along the principal axes, then for such choice of principal axes, ---t
~
~
~
~
~
~
L = hwxi + 12wyj + 13wzk
---t
=?
dL dt
= hwxi + 12wy .i + Iswzk
(9.6)
9.7. EULER'S EQUATION OF MOTION
277
where the body base vectors i,], k are constants in time with respect to the body co-ordinate system. Now, components of moments of inertia are, however, constant with respect to the body frame of reference, and hence, --+
--+
dL) ( dt T
[dL] dt f
=
[d --+] dt (I w ) f
=
....:...
(9.7)
= I w.
Also, -;;; is same in the fixed and body frames of reference. Hence equation (9.5) can be written as ....:...
--+
--+
--+
(9.8)
Iw+wxL=N
from which the angular momentum will be easily calculated, if we orient the axes of the body frame of reference such that they coincide with the principal axes of the body. Thus the x component of (9.8) is obtained as Nl
=
--+ ";
N.2
•
--+--+
= hwx - (w
x L)x
= hwx - wyL z - wzLy = hwx - (I2 - h)wywz .
Similarly for others. Thus the equations of motion of the rigid body becomes
h~x - (h - h)wyw z :: Nl } 12wy - (h - h)wzwx - N2 hw z - (II - 12 )wx wy = N3
(9.9)
which are Euler's equations for the motion of the rigid body moving with one point of it is fixed. The components of torque and of angular velocity in equations (9.9) are taken along the axes of the body frame of reference, which coincide with the principal axes at the origin. The solution of equations (9.9) will enable us to understand how angular velocity of a rigid body changes with time with respect to the principal axes under the cation of the known torque. Deduction 9.7.1. Rate of work done by the torque: If the torque acting on a rigid body is zero, according to the law of conservation of angular momentum, the angular momentum of the body is constant of motion. Thus, --+
--+
--+
--+
--+
--+
wxL=O::::}wx(Iw)=O. --+
--+
Hence the angular momentum L is parallel to the angular velocity TJ, i.e., L = ITJ, where I is the moment of inertia taken about one of the principal axis. As the torquefree motion of the rigid body ITJ is parallel to TJ, the angular velocity TJ is directed along the principal axis of the body. Hence --+ --+
w .N =
--+
dTJ
w.I dt;
--+ --+
= ~~(TJ.ITJ) = dT 2 dt
--+
w .w x L = 0
dt
278
CHAPTER 9. RIGID BODY MOTION
where the differential takes place either in the fixed or in the body frame of reference and T = ~ (w .JLJ) is the kinetic energy. This is the relation between the rate at which work is done by the torque and the rate of change of kinetic energy with respect to time.
Deduction 9.7.2. Rotation of a free rigid body: The total KE is just the rotational KE if the rigid body is rotating about a fixed point. Let a rigid body --t --t rotating freely, i.e., in the absence of any external torque (N = 0). Hence from equation (9.5) we get, --t
dL --t --t --t Tt+wxL=O --t
dL L. dt
--t
=}
=}
d
1
=}
L2
+
--t
--t
=0 --t
dt(2L2) =0;
--t
--t
L.( w xL)
L=ILI;
= ILI2 = constant.
--t
--t--t
Hence, when w is not parallel to L, and N = 0, then --t
dL dt
--t
+w
--t
x L
#
--t
2
--t
0 and L = 1 L
--t
--t
2 1
= constant.
--t
This shows that, L is fixed in space (as N = 0), but with respect to the body --t frame, only the direction of L appears to change without changing its magnitude. Also, --t
dL w.Tt
--t
--t + --t w.(w x
--t)
L =0
--t
dL dT w '-d = 0 =} -d = 0 =} T = constant. ,t ,t
--t
=}
dli
Thus for freely rotating rigid bodies LJ and are perpendicular to each other and the rotational kinetic energy T is a constant in motion. Hence when a rigid body is rotating about a fixed point, then L2 and T are the two constants of motion.
9.8
Force/Torque-free motion
Let us assume that a rigid body is rotating about a fixed point freely, i.e., the motion of the body about a fixed point not subject to any external forces or torques acting --t --t on it. In the absence of any torques we have (N = 0) i.e., Nl = N2 = N3 = 0 and hence the Euler's equations becomes
9.8. FORCE/TORQUE-FREE MOTION
279
The solution of this equation (9.10) will give the required rotational motion. But two integrals of great importance can immediately be obtained. MUltiplying equations of (9.10) by wx,wy,w z respectively and adding, we get,
=}
hwxWy h w; + hw~
+ hW2W2 + hW3W3 = 0 + hw; = constant = 2T;
(9.10)
Integrating
where T is the KE of the body. Multiplying equations of (9.10) by hwx , hwy, hwz respectively and adding and integrating, we get,
But we have already seen that the components of W, i.e., the angular 'momentum of the body about the principal axes through the base point 0 are hwx , I 2w y , hwz . Therefore, (9.11) Thus there are two constant of motion H2 and T for rigid body motion, rotating freely about a fixed point, in absences of any external forces. Since there are no external force or torque acting of the system, the KE and the total angular momentum vector must be constant in time, and so they are two integrals or constant of motion. Again, if l, m, n be the direction cosines of the resultant of W = (wx, wy , wz ), then
J[2 + m 2 + n 2 V(hw x )2 + (I2Wy)2 + (hw z )2
1. H'
This gives the direction of the invariable lines. From the equations (9.10) and (9.11), we might be able to eliminate only two of the three unknowns wx,wy,wz ' But the solution of the equations in the remaining unknown can only in general be expressed in terms of elliptic functions. Using the identity, w; + w; + w; = 1, we have trom Cramer's rule,
1
h h 2T
h 12 h
Ii Ij H2
Ii Ii Ij
1 1 1
111 2 Al
2T(I2
=
+ h) -
H2
hl3
A~ = 2T(I3 + h) - H2 hh \2 _ "3 -
2T(h
..
+ 12 ) h/2
H
2
280
CHAPTER 9. RIGID BODY MOTION
Differentiating the identity w;
+ w; + w; = 1, we get
where A = w 2 and c is the constant of integration. Thus using initial conditions it is possible to integrate the (9.9) completely in terms of elliptic functions. But this -PTOcess is very laborious.
Deduction 9.8.1. Integration of equation (9.9) under certain conditions : When there are no external forces the Euler's dynamical equation is given by the equation (9.9). We have seen that, the general solution of Euler's dynamical equation involves Jacobean of three fundamental relative function. However there are two cases in which the equation can be integrated without using any relative function. The first one when two of the principal moments are equal and the second, if the motion so started that H2 = 2hT. Case I: Let us take a rigid body symmetrical about z principal axis so that the two principal moments of inertia are equal, i.e., II = I 2 . Hence the equations take the form (9.12) The third equation shows that, the component of the angular velocity W z along the z axis is constant, i.e., W z = constant = n(say). Differentiating, the first two equations of (9.12), with respect to t, we get
hw x = (h - I3)W y W z ~ I{wx = -(h - h)2w y w; .. (h h- h)2 2 ~ Wx = Wz Wx = -0 Wx
i/
where 0 2 = It 3 W z = constant. The differential equation represents the equation of SHM. The solution is Wx = acos(Ot + b), where a and b are constants. Now,
ThiR shows that, the angular velocity components Wx and Wy change in such a manner that the vector c;J p = wxi + WyJ has a constant magnitude a and rotate uniformly about the z axis of the body with angular frequency O. The sense of 0 is the same as that of W z if h > h and opposite to that of W z if h < I3. For finding the position
9.S. FORCE/TORQUE-FREE MOTION of the body in space at any time t, when It equation,
= 12wy = hw z =
11 Wx
281
= 12 ,
we get from Euler's dynamical
= H sin () sin 'l/J O2 = H sin () cos 'l/J 0 3 = H cos ().
resolve part of c;J about 0 1 resolve part of c;J about resolve part of c;J about
Dividing the first two equations, we get,
tan'l/J
Itw x = Wx = -1- = -cot(Dt + b) lWy
~ 'l/J
Also for Ix
= Iy , iJ
=
1r
"2
Wy
+ (Dt + b).
from Euler's first kinematical equation of motion gives
= Wx cos'l/J - Wy sin 'l/J 1r 1r = acos(Dt + b) cos["2 + (Dt + b)] + asin(Dt + b) sin["2 + (Dt + b)] = 0 ~ () = constant.
Also, () = cos
-1
hw z
as hwz = H cos ().
(H);
Also from Euler's second kinematical equation of motion gives
;p = Wx sin 'l/J + Wy cos 'l/J =
H sin () sin 'l/J . nl. H sin () cos 'l/J nl. sm'f/ + cos 'f/
It
It
= Hsin(} ~
¢ = Ht.
It
It
Case II: Let H2 = 2I2T and It > 12 > 13 . If H2 (9.10) and (9.11), we get
= 2I2T,
then using equations
(9.13) Now, by Euler's second equation, we have
(It - h)(I2 - h) [2T _ I 2] It/3 2Wy
hwy = =f ~ Wy
= =fJL(n 2 - w;)
where JL2 = (II-I~~~~2-13) > 0 as It > this equation is Wy
h > hand n =
= ±ntanh(nJLt).
I¥i = i;·
The integral of
282
CHAPTER 9. RIGID BODY MOTION
Thus, when t (9.13), we get
Wx
-+ 00,
= ±n
tanh(nJ-lt)
-+
1, so that, JWyJ
h(h - h) h (h - 13 sech(nJ-lt);
Wz
= ±n
-+
n
=
%as t
-+ 00.
Now from
12(h - h) 1 ( sech(nJ-lt). 3h-h
As t -+ 00, w x , W z -+ 00. Consequently, the axis ofrotation ultimately coincides with the mean axis of 12 axis or the mean axis of the momental ellipsoid. The angular velocity about this axis, tends to the value but the instantaneous axis never actually coincides with it. The equation of the double sign can be effected by noting the initial values of w x , w y , W z .
%,
9.8.1
Poinsot's geometrical representation
Let a rigid body is rotating freely through a fixed point 0, about an instantaneous axis 01 so that the KE T is constant in motion. However, it is possible to describe the geometrical representation of motion of rigid body, known as Poinsot's construction. based on the integrals of motion without solving the problem completely. For a proper choice of the axes in the body, the products of inertia may be made vanish, for this the quadratic surface is (9.14) Momental ellipsoid: axis at 0 be
Let the equation of the momental ellipsoid to the principal
(9.15) where !vI is the total mass of the body and c is sublinear factor referred to the -.. principal axis of the body at O. Thus, if we consider any radius vector OP through z
~~-------+~--y
x
Figure 9.4: Momental Ellipsoid
o meeting the ellipsoid P. Momentum inertia of the body about OP is inversely proportional to the square of 0 P. Let the instantaneous axes of the rotation 01
9.B. FORCE/TORQUE-FREE MOTION
283
meets the ellipsoid at P(xo, Yo, zo), where OP be l, m, n, then
= r.
Let the direction cosines of OP
Xo = I = Wx ,. Yo = m = Wy., Zo = n = r w r W r rwx rwy 1"Wz *Xo=-; Yo=-; ZO=-· W
W
,
wz. W
W
Thus, from equation (9.15), we have, T 2 1lW x2 + .l2Wy
2
+ 12 3W z =
4
2
4w M c'
2
2r
{IT
* (2T).r =Mc.w *w=rv~
which shows that, the resultant angular velocity is proportional through the radius vector. The equation of the tangent plane to the momental ellipsoid is (9.16) Let A, fL"
be the direction cosines of the normal to the tangent plane, then
This shows that, this normal is parallel to the invariable line and hence tangent is parallel to the invariable plane. Hence the equation of the tangent plane can be written as AX + fLY
+
,Z = p.
(9.17)
Now since equations (9.16) and (9.17) represent the same plane, we have
Since p is constant, the tangent plane to the ellipsoid at Q is fixed in direction and at a fixed distance from 0 and hence the tangent plane is fixed in space. Hence it follows that" The motion of the body is such that the momental ellipsoid rolls on the invariable plane with its centre fixed at 0 with angular velocity which is proportional to the radius vector of the ellipsoid in the direction of the instantaneous axis. The radius vector to the point of contact is the instantaneous axis of rotation." The momental ellipsoid stands for the space of the instantaneous angular velocity. Higher the amount of rotation KE the bigger·is the size of this inertial ellipsoid. Ellipsoid of inertia: When a rigid body rotates about some fixed point 0, the
CHAPTER 9. RIGID BODY MOTION
284
instantaneous axis of rotation always passes 'through 0 but its direction changes with time. The MI of the body about these different axes through 0 will, in general, be different. These different values can be represented geometrically in a very elegant way if we define a vector X in the form -+
X
-+
w
= wVi =
-+
w m;
-+
w= Iwl·
Using this notation, equation (9.14) becomes
F(X) =
XT1X =
1
which belongs the angular velocity space in which every points corresponds to a set of values of w x , w y , W z . Also, the expression F(X) is always finite and never zero but
, Figure 9.5: Inertia ellipsoid positive and so it is a closed surface. This equation of quadratic surface is elliptic in nature in w-space and is known as inertia ellipsoid relative to the point 0, which is the centre of the ellipsoid. The axes of the inertia ellipsoid lie along the principal axes of the body which are fixed in the body. This equation looks like the equation of the ellipsoid of inertia provided we identify with X. Now
k
Vx F = 21-+X =
(2 L. m = Vf
2I(;]
Thus (;] will always move such that the corresponding normal to the inertia ellipsoid is in the direction of the angular momentum, where the distance from the centre of the inertial ellipsoid to any point on the surface is ~. Let us draw a tangent plane to the surface of the inertial ellipsoid at the point P. The perpendicular distance of this plane from the origin of the inertial ellipsoid is given by
ON
=
-+
-+
0 P cos (); () = angle between wand L -+
_H
rnrr.
-+-+
w.~ =
m 1(;]11 L I
-+2T
IL 1m
= v~T;
IL I
T=KE
9.8. FORCE/TORQUE-FREE MOTION
285
---+
FDr all time, the KE T and L are constants of motion, and the tangent plane is so always a fixed distance ON from the origin 0, of the ellipsoid. A line through the ---+ point. in the fixed directioll of L is called invariable line. Also, the normal to the ---+ plane has the fixed direction, i.e., along the direction of L, so the tangent plane is called invariable or fixed plane. Hence we conclude that, the inertia ellipsoid of a rigid body under free rotation always touches the invariable plane, the perpendicular to which drawn from the origin points always toward the fixed direction of the angular momentum of the body.
9.8.2
Polhode and Herpolhode
The motion of a rigid body about a fixed point under no forces, may be represented by the rolling of an Poinsot's ellipsoid fixed in the body upon an invariable plane. The invariable plane is fixed in space. The point of contact of the ellipsoid and the invariable plane lies on the instantaneous axis of rotation and so there can be no slipping at the point. Hence the ellipsoid rolls on the invariable plane. Now we have the following facts ---+
(i) Both the invariable plane and the angular momentum L are fixed in space. (ii) The invariable plane has to meet tangentially the inertial ellipsoid at some or other point and this point of constant determines the direction of the instan- . taneous angular velocity c;J of the body. (iii) As c;J changes with time, this point of contact on the surface of the inertial ellipsoid ought to change with time. We know, for all time, the invariable plane has to remain fixed in space and so the point of contact can change only if the inertial ellipsoid itself changes its orientation with time. Also as the direction of c;J changes continuously with time, the force free motion of the rigid body as being such that the inertia ellipse rolls over the invariable plane, without slipping. As the ellipsoid of inertia rolls without slipping over the invariable plane the point of contact between the ellipsoid of inertia and the invariable plane indicating the direction of instantaneous c;J with respect to the body as well as the fixed plane traces out simultaneously two curves. The curve traced out on the inertia ellipsoid by the point of contact P between the inertia ellipsoid and the invariable plane is called polhode and similar curve on the invariable plane is called herpolhode. The equations of any polhode may be found from the fact that it is the locus of a point on the momental ellipsoid such that the tangent plane at the point is at a constant distance from the centre. If p be the distance from the centre of (9.15), to the tangent plane at (x, y, z), then (9.18)
286
CHAPTER 9. RIGID BODY MOTION
The equations, (9.15) and (9.18) represent, for different values of Hand T, the polhode on the ellipsoid (9.15). Eliminating Mc4 from these two equations, we get,
which describes a cone in the body, called polhode cone. This cone will be real if, and only if, p lies between the greatest and the least semi-axis of the ellipsoid (9.15), or if ~; lies between the greatest and the least of the principal moments h, h, h. The projections of the polhode on the co-ordinate planes, found from (9.15) and (9.18), are
If h > h > h, then the first and the last equations represent ellipses and the middle represents hyperbola, so that the polhodes are closed curves surrounding the axes of greatest and least moment. Now we consider the polhode cone for different values H2 of 2T' (i) If ~; = h, the cone is the line y = 0, Z = 0; i.e., the axis of greatest moment is the instantaneous axis, and the distance of the plane from the centre does not permit of contact at any point but the en<:l..Tof the axis.
°
(ii) If h > ~; > 12 > h, then the cone cuts the plane x = in imaginary lines and the planes y = 0, Z = in real lines, so that in this case the cone surrounds the axis of greatest moment, and therefore so does the polhode.
°
(iii) If ~; = h, the mean moment, the cone is two planes passing through the axis of mean moment, cutting the ellipsoid in two ellipsoid in two ellipses which separate the polhodes round the axis of moment h from the polhodes round the axis of moment h.
°
°
(iv) If 12 > ~~ > C, the cone cuts the planes x = 0, y = in real lines and Z = in imaginary lines, so that the cone surrounds the axis of least moment and so does the polhode. (v) If ~; = h, the cone is the line x = 0, y = 0; i.e., the axis of least moment is the instantaneous axis and only possible for it.
Though the invariable line is fixed in space, as the body moves the line traces a cone in the body and is called the invariable cone.
9.8. FORCE/TORQUE-FREE MOTION
9.8.3
287
Body and Space cone
Let us consider a torque free motion of symmetric body having its principal moments of inertia It = 12 . described by Euler's equation of motion, such that, Poinsot ellipsoid is an ellipsoid of revolution. The components of (J, i.e., WI, W2, W3 are along the principal axes of the ellipsoid. The KE T = ~ I: 1iw; and the magnitude of the ---'> angular momentum 1L 12 = L2 = I: l]w; arc constants of motion. Let us define k = [31/1; h > It, then the torque free Euler's equation becomes
+ kW2W3 = 0; ~ WI + k2WIW~ = 0; WI
W2 - kWIW3 W2
= 0;
+ k2W2W~ =
= O. W = O.
W3
0;
The solution of the third equation gives W3 =constant = n, say, which indicates the component of the angular velocity in the direction of the inertial symmetry axis of
z
L
OJ
z Body cone
o
o
Figure 9.G: Body and Space cone
the ellipsoid does not change with time. The first two equations give
WI
+ k 2 n 2 wI
+ iW2 = ~ w~ + w~ =
~ WI
2
~w =
= 0; W2
Ae(kntHO)i;
A,
+ iW2)(WI w~ + w~ + w~ = (WI
+ k 2 n 2w2 =
eo
0 constant andi =
iW2) = A2 2 A2 n =
+
V-I
constant.
This shows that the angular velocity vector (J rotates about the body z axis describes a cone with the vertex at origin. Let ab be the angle between (J and the symmetry axis, then W3
cos a b = 1---'>1] ~ tanab W
A = -. W3
Both A and W3 being constants, ab is fixed in magnitude. As ab remains constant with respect to an observer fixed to the body, the instantaneous axis (J rotates about symmetry axis (also called body axis) in the form of a cone about the inertial
288
CHAPTER 9. RIGID BODY MOTION
symmetry axis of the body of the constant semi angle O:b with the inertial symmetry axis of the ellipsoid. This cone is called body cone. Also, A = Iwl sin O:b and W3 = Iwl cos O:b. Thus the polhode on the ellipsoid is clearly a circle about the symmetry axis, and the herpohode on the invariable plane is likewise a circle. With respect to the body co-ordinate system, the angular velocity vector W is also constant in magnitude and it traces out the body cone with respect to the rigid body, whose intersection with the inertia ellipsoid is the polhode. Accordingly, an observer fixed in the space axes sees W move on the surface of a space cone whose intersection with the invariable plane is a hepolhode. Now let us see how an observer fixed in the spase would view this motion of W. In a fixed inertial frame as N = 0, so L has a fixed direction (also magnitude) in space. Therefore, it is easier to locate W with respect to L. Also, ~
~
~
~
~~
2T = w. L
~~
=
constant =*
w.L
IwllLI
= cos O:s = ~
constant ~
where o:~ is the constant angle between the two vectors wand L. Thus the vector --> W rotates about the vector L in a cone with semi-vertical angle O:s given by, cosO: s =
This means that, the instantaneous axis W makes a constant angle O:s with the. axis of L, i.e., moves on the surface of a cone with the direction of L as axis and apex at the centre of mass. This is called space cone. The instantaneous axis of rotation is the line of contact between the space cone and the body cone at any instant. This axis in the body being instantaneously at rest, the body cone rolls without slipping around the space cone. Now ~
~
Since h = 12, so this' shows that the space cone lies inside the body cone when n > 0, i.e., 13 > h and outside when n < 0, i.e., h < h.
9.8.4
Binet Ellipsoid
The Poinsot construction shows how the angular velocity W moves, but gives no information as to how the momentum L vector appears to move in the body system of axis. Here we describe geometrically, the path of L as seen by an observer in the ~
~
9.8. FORCE/TORQUE-FREE MOTION
289 -+
principal axes system. With respect to the set of axes each of the components of L would involve only the corresponding components of W as
Ll
= hWI; L2 = hW2; L3 = I3w3.
1
2
2
2
Li
L~
L~
=> T = KE = 2[It Wl + hW2 + h W3J = 2It + 2h + 2h
(9.19)
As T is constant, equation (9.19) represents an ellipsoid fixed in the body axes, -+ known an Binet ellipsoid. Now the angular momentum vector L must lie on the sphere
(9.20) follows from the conservation law of angular momentum. For a given initial condition, KE and angular momentum - the path of L is a body given by the intersection of the ellipsoid (9.19) with the sphere (9.20). Also equation (9.19) can be written as
Li L~ L~ 1 -2+ - 2+ - = a a a2 123
where ai = .j2TIi ; i (9.20) it follows that
= 1,2,3 are the semimajor axis of the ellipsoid.
From equation
a; < L2; aj = smallest semi major axis
> L2; aJ = largest semi major axis Let the principal moments of inertia are labelled as h < h < h, then
J2Th < L < J2Th. Hence the momentum sphere of radius L always intersects the KE ellipsoid. Ex 9.8.1. A body moves under no force about a point 0, the principal moments of inertia at 0 being 3A,5A and 6A. Initially, the angular velocity of the body has components WI = n, W2 = 0, W3 = n about the principal axes. Show that at any later time, W2 = ~ tan( ~) and ultimately the body rotates about the mean axis. SOLUTION: Here given that It = 3A, h = 5A and h = 6A. Euler's equation of motion of the rigid body under no forces, with one point of it is fixed is given by,
hWl - (12 - I 3 )w2 w3 = 0; hW2 - (13 - It)W3Wl hW3 - (It - h)WIW2
= 0; =0
=> 3Wl = -W2W3; 5W2 = 3W3Wl; 3w = -WI W2. From the last two equations, we get
9WIWI
+ 5W2W2 = 0 => 9wr + 5w~
= a(say).
290
CHAPTER 9. RIGID BODY MOTION
Initially, WI = n, W2 = 0, so a = 9n 2. Using the first and last equations, we have
Initially, when WI = n, W3 = n, we get b = 0, so that w~ = w~. From the equation 5W2 = 3W3WI, of the Euler's equation, we have,
Using the relation, 9w~
+ 5w~
= 9n 2, we get,
Thus at any time t, the angular velocity components are obtained, which satisfy the given initial conditions. Further, when t -- oo,Le., after very long time period of motion, tanh[fsl -- 1, so W2 -- ~. Using the relation 9w~ + 5w~ = 9n 2 , we get WI = and W3 = 0. Ultimately, body keeps rotating about the mean axis with a limiting angular velocity ~. Now we can find the direction ratios of the invariable
°
line. The direction ratios of the line are
hlP, 12J12 , 13113 , where H
is given by,
Thus, the direction ratios are ]gsech(fs)' tanh(fs)' Jssech(fs)' At t = 0, these are ]g,0, As t -- 00, these become 0,1,0, Le., invariable line, Le., axis of rotation coincides with mean principal axis.
Js.
9.9
Motion of Symmetrical Top
Let us consider the motion of a symmetrical top (like child's top and gyroscope) about the axis of symmetry as an application of the methods of rigid dynamics. The top is fixed at its lower tip 0 and is acted upon by only the gravitational force mg acting at G downward, where OG = l. G is located on the symmetric axis z'. The symmetry axis is one of the principal axes and will be chosen as z' -axis of the co-ordinate system fixed in the body, so h = 12 . As one point is stationary so
9.9. MOTION OF SYMMETRICAL TOP
291
z
z'
~~---------------y
x' Line of nodes
Figure 9.7: Symmetrical top
the description of the motion of symmetric top is completely specified by the three convenient set of generalised co-ordinates as Euler angles : () : it gives the inclination of the z axis from the vertical. cp : it measures the azimuth of the top from the vertical. 'ljJ : The rotation angle of the top about its own z axis. To obtain the solution of the motion of the top, we use Lagrangian procedure. The angular velocities along individual axes we have the components of W as
Wx' = ¢ sin () sin 'ljJ + iJ cos '1jJ Wyl = ¢ sin () cos 'ljJ - iJ sin 'ljJ WZI
= ¢ cos () +
;po
Let h, h, h be the moment of inertia about ox', oy', oz' respectively, where h = I 2 . The KE of the symmetrical top can be written in the form as
T. =
1 2 "2h[Wxl
1
2
+ Wy/] + "2hwz1
=
~h
=
~h[¢2sin2()+iJ2] + ~h(¢cos()+;P)2
[(¢sin () sin 'ljJ + iJ cos 'ljJ)2 + (¢sin () cos'ljJ - iJ sin 'ljJ)2] +
~h(¢cos () +
;P)2
which is the expression for T in terms of Eulerian co-ordinates and it depends upon only one generalised co-ordinate (). Let us take the horizontal plane through 0 is as the zero level, then the potential energy of the top is V = M gl cos (), in terms of Euler's angles, where 9 is the constant vector for the acceleration of gravity. The Harpiltonian H of the symmetric top is 1 '2 2 '2 1· . 2 H = T + V = "2 h [cp sin () + () ] + "2h(cpcos() + 'ljJ) + Mglcos(). Since H does not contain time t explicitly, H = E =Energy and hence it is conserved, Now the Lagrangian function L of the conservative system is given by 1 '2 2 '2 1, , 2 L = T - V = "2 h [cp sin () + () ] + "2h(cpcos() + 'ljJ) - Mglcos(),
(9,21)
292
CHAPTER 9. RIGID BODY MOTION
Taking, e, ¢, 'ljJ as generalised co-ordinates, where ¢ and 'ljJ are ignorable co-ordinates, the Lagrange's equation of motions are (i)dd
(8~)
t 8¢
dd [h¢sin2 e+hcose(¢cose+¢)] =0
=
o=}
=}
h("j; + ¢cos e) = constant = ha(say).
t 2 2 =} h ¢ sin e + hJ; cos e + h¢ cos e = constant = h b( say) d 8L d·· (ii)-d (-.) = 0 =} -d [h(lj, + ¢cose)] = 0 t 8'1/) t
The motion due to change in e is called nutation and that due to change in ¢ is called precession. The general motion of the top about the fixed point 0 is the combination of nutation and precession.
9.9.1
Integrals of motion
Now from (9.21), we see that the Lagrangian function L does not contain ¢ and 'IjJ explicitly, hence they are cyclic co-ordinates. Hence the total energy E and two canonical conjugate momentum p¢ and P'I/J are constants of motion. Only three additional quadratures are needed to solve the problem, and they can easily be obtained from these three first integrals without directly using Lagrange's equation. Hence the two integrals of motion are
8L . 2 . . 2 8¢ 8L .. P'I/J = - . = h['ljJ + ¢cose] = I3Wz = ha; a = constant
P¢ = - . = h¢sin e + h'ljJ cos e + h¢cos e = hb; b = constant
8'ljJ
.
=}
13'1/) . =} ¢
.
= 11 0, =
.
2
.
I3¢COS () and h¢sin e + I3'IjJ cos f} b - a cos e ni. h - b - a cos e cos e . 'f/ = -a 2 sin ( ) ' h sin 2 e
.
+ h¢cos
2
()
= hb (9.22)
from which rb and 'ljJ are known if e is known. Here the two constants of motion P¢ and Pw are expressed in terms of new constants a and b. Another integral is in the above E = T + V = H. Now, W z is constant in time and the term
t;a
{ta
2 equation is also constant. Hence E is also another constant of the motion and so the energy equation can be written as
E
,
=E
If
2
1
'2
2
'2
1·
.
2
If
2
- - a = -h[¢ sin e + e ] + -I3(¢cose + 'IjJ) + MglcosO - - a 2h 2 2 2h 1 b - a cos e 2 . 2 '2 1 h 2 If 2 =-h[( .2 ) sm O+O]+-h[-a] - - a +MglcosO 2 sm e 2 h 2h 1 '2 1 (b-acosO)2 = -2hO + -2h sm . 2 + Mgl cosO 0 1 '2 1 (b - acose)2 = -2he + V'(O); V'(e) = -2 h . 2 + Mgl cos e (9.23) . sm
e
9.9. MOTION OF SYMMETRICAL TOP
293
which becomes an one-dimensional problem in () co-ordinate, and V' (()) is the effective potential as a function of () alone. The problem of motion of the symmetric top is solved in principle, when we know the dependence of () on time. Hence from (9.23) we get
iJ = dO = dt
~[E'(()) h
! JX = VI;! = -VI;! = _! =-!
'* t(()) =
V'(O)]
d()
[E'(()) - V'(())] sin ()d()
y'2E' sin 2 0 - h (b - a cos ())2 - 2Mgl sin2 () cos 0 du ; y'2E'(1 - u 2) - (b - au)2 - 2Mglu(1 - u 2) du
. a
y'a(l - u 2) - (b - au)2 - (3u(l - u 2)'
=
2E' (3
h '
u
= cos()
=
2M gl
h
du 2 )(1 - u )(a - (3u) - (b - au)2·
The method of solution involves elliptic integrals, but the general nature of the motion can be discussed without performing the integration. Knowing ()(t) ¢ and 'I/J can be evaluated from (9.22) by quadratures. Now, we consider the variation of effective potential energy function V'(()) with the co-ordinate O. The effective potential energy function V' (()) assumes infinite values in 0 < 0 < 7r and V' (()) ~ 00 for () = 0, 7r. Now, 8V'(()) _ M 1 . () J [b-acos()] -(b-acosO)cos() + asin 2 () !)() - 9 sm + 1 . 0' x . 2 () u sm sm . () J (b-acosO)(a-bcosO) = - M 9 1sm + 1 . 3 sm 0 8V'(()) 0 8() > ; for () ~ 7r and a =1= b
< 0;
for ()
~
0 and a =1= b.
Hence a~Jf}) must vanish in 0 ::; () ::; For extreme value of V' (()), we have
7r,
where the minimum value occurs for V'(()).
8V'(()) _ -M 1 . () A.J a - bcos() _ . 8() 9 sm + 'f' 1 sin 0 - 0,
'* -l'vIgl sin() + sm.¢() [h(~ + ¢cos()) -
as
A. _ b - acos() sin2 ()
'f' -
(h¢sin 2 () + h~cos() + h¢cos 2 ()) cosO]
=0
'* - Mgl sin2 0 + ¢[h( ~ + ¢ cos 0) - h ¢ sin2 () cos () - h cos2 ()(~) + ¢ cos ())] = 0 '* -1vIgl sin2 () + ¢[h( ~ + ¢ cos ()) sin2 () - h ¢ sin2 () cos ()] = (j '* ¢2 [h - h] cos () - ~h¢ + M gl = 0 (9.24)
CHAPTER 9. RIGID BODY MOTION
294
If the top has a particular value of energy E~ = E, () motion is bounded motion, confined between the two values ()l and ()2, which are the roots of the equation ,
1
El = M gl cos 0 + - h [ 2
b-acos() 2 . () J. sm
These two points represent turning points and they coincide if the symmetric top has the energy E' = E~, which corresponds to the minimum value of V'(O) for fixed () = 00, then the symmetric top has the constant angular velocity given by ¢ = b-~ cos 2 eo about the vertical axis. 8m 00
9.9.2
Precession without Nutation
The variation in the angle () is referred to as the nutation of the symmetry axis of the symmetric top and is an up and down motion of the symmetry axis. We have seen that, if the top has energy E' = E~, then the motion of the top is restricted between two values of 0, say ()I, 02, known as turning points, where () is the root of the equation E~ = MglcosO + ~h[b-s~~~SeJ2. Thus the axis of symmetry performs nutational motion confined between values ()l and ()2. If the axis of symmetry z' -axis I
I I I I
V(8)
I I I
I I --------- ----------------------------------- - --------r----I
I
:
:
: E'l
I I I I I I I I I I I I
I I I I I I I I I I I I
I I I I I I I I I I I I
!
: E'z
---------+---------------
i
I
---------------+---------.. ----
I I I I I I I
I I I I I I I
O~----~I----------~--------~I
I I I I I I I
_____'~_____.
°2
1t
Figure 9.8: Precession with nutation is inclined at an angle ()o, the top possesses uniformly about the vertical z-axis, with the precessional velocity ¢o = b-~1o:eo. From equation (9.24), it follows that, for a sm 0 given
eo, there are two angular frequencies of precision ¢* given by . ¢*
[hJ ± -'-V______________ ft#,-J;2 - 4(h - h)coseMgl _ J = ___ 2(h - h) cos 0
=
h
2(h - h) cos 00
[1±
9.9. MOTION OF SYMMETRICAL TOP
=?
=?
295
b - acos()o 13 sin2 eo = 2(h - Is) cos eo [1 ± b - a cos ()o
sin2 ()o
Is
= 2(11-3COSO I ) () [1 ±
1 - 4Mgl(
h - Is cos eo [2
) - .-
3
'ljJ2
.
]'IjJ (9.25)
The +ve sign of radial precession will be fast while that with negative sign is called a slow precession. Also, ;p* is real if 1 - 4M 1(h - h ) co~ ()o > 0
Ij
9
'2
[2 =?
cos ()o ::; I
1 -
'ljJ2-
3I
'IjJ
.
4Mgl'
3
1["
for ()o <
"2
and for ()o > ~, the discriminant is always positive. Thus for a given value of ()o, there are two values of precessional velocity ;p given by equation (9.25). Using the relation WZI = ;Pcos() + ¢ we get from (9.24), . 2
..
cos ()o(h - h)¢* - Is'IjJ¢* + M gl = 0 . 2
.
or, cos ()o(h - 13)¢* -
[WZI -
.
¢cos()]13¢* + Mgl
J
=0
. * 13 wz ± Ijw;, - 4h cos ()oM gl or, ¢ = ______2-~-------------2h cos eo 1
l* - -1 13 [Wzi ± or 'I-' , - 2 h cos()o
Hence for
h cos()o W;, - 4Mgl([2)
].
3
eo < ~, physical motion is possible and hence the condition becomes
W;, - 4Mgl( ;;) cos ()o 2:: 0 3
2
=? [Wz/]min
2:: Is J M glh cos ()o.
This is the condition that must be satisfied by the value of WZI if there is to be a 'precession without nutation'. The equality sign gives the minimum spin angular velocity at which the top will be just able to perform precession without nutation. If the spin angular velocity is below this value, the top will not be able to perform uniform precessional motion, at an angle ()o. When WZI > [Wz/]min, quantity under radical will not be zero. In this case, the two values of ;Po can be represented as (i) For the case of fast top the ;Po is l (f
'1-'0
"" Is- WZI as t) -_ b - acos()o "" --. 2 sin ()o
h cos eo
(ii) For the 'slow' precession, b - a cos eo M gl ~ -2 sin eo 13 w z ' which is ordinarily observed with a rapidly spinning top. l (l
'1-'0
)
s ow =
296
9.9.3
CHAPTER 9. RIGID BODY MOTION
Nutational Motion
For a given energy E', the axis of the top oscillates between two turning angles (h and (h at e = 0o, thc cff'cctive potcutinl function V'(e) has the minimum value. The angles are given by
E
,
,
1
= V (0) = Mgl cos e + - It [ 2
=> j('/1,) = (1- '/1,2) (a - (3'/1,) - (b -
b - a cos () 2
.
S111
e 1 =0
a'l1i = 0;
'/1, =
cose.
(9.26)
This is a cubic equation in '/1, = cos e, having three roots, where the cubic polynomial
--~~~--~~------------~----x
Figure 9.9: Turning angles furnish the angles at which '/1, = cos e changes sign. Also, j (u) -+ {3'/1,3, for large '/1" where (3 = 2~~gl > O. Hence j(oo) > O,j(-l) = -(b+a)2 < O,j(l) = -(b-a)2 < 0 and j ((0) < O. When '/1, = ± 1, the roots correspond to a vertical top which is unusual case. Hence at least one of the roots of the cubic must satisfy '/1, > 1, a region that does not correspond to real angles. Physical motion is possible when f(u.) is positive in 1J, E (-1,1), i.e., e E (0,11"). The curve of the effective potential energy indicates that, there are two real roots '/1,1 = cos el, '/1,2 = cos e2 , lies between -1 and +1 and the third root on consideration of equation (9.26), 'U3 = cos e3 > 1 and is not physical interest, Initially, when the top is spinning fast, nutation motion is absent and cos el = cos e2 = cos eo. Also, when iJ = 0, initially, the initial value cos e1 of cos e satisfies equation (9.26), During nutation, the precessional velocity is J; = b~~1(~o, from which it is clear that ¢ changes with e. If Ibl < lal, i.e., Ip¢1 < Ip'l,bl, let us define an angle e* as
cose* With this value of
e > e*,
=!!.; Ibl < lal. a
e, the precessional velocity is given by . ¢
For any
= P¢ P'I,b
=
a -'-2-(cOS e* S111
e
we have, cos e* > cos e and
-
cos e).
¢ has
the same sign of a and that of
9.9. MOTION OF SYMMETRICAL TOP·
297
z
Figure 9.!O: Turning angle
If however, () < ()*, ;p has a sign opposite to W z'. Also, dV:JB) IB=Bo < O. Hence ()* corresponds to the angle satisfying ()* < ()o. Thus for the two angles ()1 and ()2, if fh > 0*, then ;p has the same sign as W z' throughout the nutational motion, whereas if ()1 < ()* < ()2, ;p has opposit.e directions as shown in t.he figures, which show the loci of the axis of the top on a unit sphere. Now we consider the case of symmetric top start.ed with a spin angular velocity W z ' = at· the given angle () = ()1. It is clear that initially iJ = 0 and ;p = o. The constants of motion are
W z'.
;p
In this case, where
()*
= ()1 the motion of the symmetric top is in the following figure z
Figure 9.11: Ocillation of spin axis· nutation
298
CHAPTER 9. RIGID BODY MOTION
In this case, the effective potential energy function is given by V'((J)
= Mgl cos (J + ~h[hwzl COS;l ~ ~WZI cos (J]2 Ism
2
_ ~1' 1 (J lVJ 9 cos
-
+ ~1 2 1 X
_ l§w;1 [2M g l 2h l§w;1 _ l§w;1 [ (J
- - - --cos -
2
I
1
a cos
+
(J
1§ 2 [cos (J1 - cos (J]2 12 WZI . (J 1 sm cos (J)2] sin (J
+ (COS(J1 -
(cos (h - cos (J )2]. _ 2M glh . (J , a [2 2 sm 3Wz1
The angle 01 is one of the roots of the equation V'
l§w;1 [
given by
0 + (cos Olsm!7 .- cos 0 )2] = M 91cos 01 = al§w;1 21 cos 1 1 + a cos 0 sin2 0 = a cos 01 sin2 0 cos 0)2 + 0' sin2 O( cos 0 - cos Ol) = 0 0
-1- a cos 2 1 or, (cos Ol - cos 0)2
or,( cos Ol -
= E',
II
or,(cos Ol - cos 0) [cos Ol - cos 0 - a sin2 OJ
=0
or,( cos Ol - cos O)[a cos 0 - cos 0 + (cos Ol - a)] = 0 1 or, cosO = COSOl; 2a[1 ± }1- 4a(cosOl - a)] 2
or, cos 0 = cos Ol, cos O~, cos O~ where cos Of = 2~ [1 + VI - 40' cos (h + 40'2] > 1, which has no physical value and cos O~ = cos (J2. Thus, when a spinning top initially at rest is released at an angle (h, it falls slightly under the action of gravity and gains in the precessional and the nutational motion. The axis of the top reaches the limiting angle 02 where it has maximum precessional velocity and zero nutational velocity. It describes successive cusps with the turning points on the surface 0 = Ol. The precessional velocity at o = (h has the value ).. _ b 'f'-
a, cos 0 2
sin 0
_
_ 2M gl
-wz1a---. hWzl
Thus the axis of the top undergoes precession very slowly ( as ¢ = 0) and nutation at very large velocity at the smaller angle 0 = Ol. The situation is exactly opposite at the larger angle 0 = (}2.
9.9.4
Steady motion
The motion of the symmetrical top is said to be steady, if 0 = constant = a, say. We know, one of the constants of motion is
..
'I/J
h
+ ¢cos () = h a =
constant
= k(say).
9.9. MOTION OF SYMMETRICAL TOP
299
This means that the total angular velocity k about the axis of symmetry remains constant, i.e., -J; = constant = n(say). From Lagrange's 0 equation, we have
he - h¢2 sinO cos 0 + I 3(;P + ¢cos O)¢sin 0 - Mgl sinO = 0 ::::} -hn 2 sin a cos a + hknsina - Mglsina = 0 ::::} h n 2 cos (Y - hkn + M gl = 0; a i= 0 ::::} n
=
hk ± JIjk 2 - 4h cosaMgl
2h
= nI, n2·
This shows that there is a pair of real and distinct zeros of n, provided
I§k 2 - 4h cosaMgl > O. Thus, it)!' a sufficiently great spin k about the axis of symmetry the top can execute a steady motion 0 = a about the vertical under two possible precessional angular velocities nI, n2·
.9.9.5
Stability of steady motion
The motion of the symmetrical top is said to be steady, if 0 = constant = a, say. Here we are to discuss the stability of steady motion of the top when the axis is not vertical. From first integrals of motion, we have,
., + ¢cosO
'IjJ
II
= ha=
h¢sin2 0 + I3n cos 0
= hb =
= n(say)
constant constant
= k(say).
Since 0 = constant = a, we have from the second equation, ¢ The Lagrange's 0 equation of motion gives,
=
constant
= w(say).
he - h¢2 sin 0 cos () + h[;P + ¢cos (;I]¢sin () - Mgl sinO = 0 he - h¢2 sin 0 cos 0 + I3n¢sinO - Mgl sin (;I = 0 he + F(O) = 0 where, F( (;I) = - h ¢2 sin (;I cos (;I + 13n¢ sin (;I - M gl sin (;I
() I [k - I3ncosO] . (;I M l . 0 . 0 = - I 1 [k - I3ncosO]2 . 2 sm cos + 3 n sm 9 sm . 2 hsm 0
-- -
cosO [k - I 3ncos 0]2 h sin 3 ()
hsm 0
+ -I3n[k-I3ncosO] . h sm 0
Ml' 9 sm 0.
Now, we calculate P'(()), which is given by,
= [k - hn cos Oj2 + I§n 2 sin2 () - 4Mglh sin2 Ocos() F(a) = -h¢2 sin a cos a + I3n¢sina - Mgl sin a = -Ie = O·, as 0 = constant = a
h sin2 ()P'(O) Also,
l\1gl ::::} J,n = hwcosa + --. w
3h sin()cosOP(O).
CHAPTER 9. RIGID BODY MOTION
300
Suppose that the axis of the top is disturbed slightly from its original position, then e = a + e, where e is very small quantity. Thus,
= P(a + e) = P(a) + ep'(a) = ep'(a); as F(a) = 0 =? hi + f.P'(a) = 0 p(e)
=?
a? I§n2 4M 1 ]C 0 I 1<"e+ [(k - I3n. cos 2 + -19 cos a <" = hsm a 1 .. '2 2 I§n 2 he + [h4> sin a + T - 4Mgl cosa]e = 0
=?
he + [hw sin a
=?
..
2
2
Mgl 2 + -I1 (hw cos a + --) w
1
=?
hi + [hw 2 + hw 2 cos 2 a
=?
.. he + [hw 2 - 2Mgl cos a
=?
.. Mgl f. + [(w - -1- cos a)2 w
1
4Mgl cos a]f. = 0
+
M2g2l2 21 - 2Mgl casale = 0
+
M2g2l2 2I]f. = 0
+
w
1
w
1
M2g2l2 2 212 sin alf. = 0 W
1
.. M 1 M2g2l2 2 =?e+p2e=0; p2=(w--.fLcosa)2+ 22 sin a>0 wh w 11 which is the equation of the SHM. This shows that, the axis of the top, if disturbed slightly from its position in the steady motion, will tend to come back to the original position. This steady motion of the top, with axis inclined at an angle a to the vertical, is stable. Ex 9.9.1. A top consists of thin uniform spherical shell of radius a and centre C is free to move about a fixed point 0 of its centre. The radius OC makes an angle e with upward drawn vertical OZ and 4> is the angle between the plane ZOC and a
fixed vertica.l plane Z 0 X. Initially,
e = ~,¢ = ~n,
where n
=
I¥i.
Show that at
time t, cos e = tanh 2[tJi']. SOLUTION: From the equation of the top, we have,
+ ¢2 sin20] + 2mgh cos 0 = and A¢2 sin2 e + C*n cos e =
A[(j2
P(say)
Q( say)
where P and Q are constants to be determined by using the initial conditions 0 = ~, iJ = 0, ¢ = ~n. Thus, P = A 4~2 , Q = A Also, the moment of inertia of a hollow sphere about-a diameter is C* = 2Ma;. Now, A is the moment of inertia about OA. In other words, moment of inertia of hollow sphere about a line perpendicular to OC is
2;.
9.9. MOTION OF SYMMETRICAL TOP and h
=a=
301
distance of CG from O. Hence the first equation of motion gives, ·2
+ ¢·2 sin2 fJ] + 2mghcosfJ =
A[fJ
=>
~.
5M "3¢ sin •
2
~
+ 2M "3 n cos fJ = 5
fJ
+ 2ncosfJ = 2n => ¢ =
2
=> 5¢sin
~
()
5M
.
4n
2
25 A
~
"3 2n
5(1 + cosfJ)
Putting the value of A and using this value of (p, form second equation we get, ·2
2
=>
·2 5a fJ
=>
5afJ
A¢ sin fJ
where we use n
=
·2
JTfi.
As,
5a02
2n + C*n cos fJ = 5A
+
4n2a(1 - cos fJ) 5(1 + cosfJ)
+
6
fJ
gcos
4n 2 a
1 - cos fJ
+ 6g cos fJ = --[1fJ] 5 1 + cos 8n 2a cos fJ
12g cos fJ
5(1 + cos fJ)
1 + cos fJ'
0 is in the direction of the fJ decreasing sense, we have,
= 6g cos fJ[
2
-1]
=
l+cosfJ
=>
0= _
f69
V~
Vf69 ~ o
6gcosfJ(1 - cosfJ) l+cosfJ
cos fJ(l - cos fJ) l+cosfJ
J =- J t
=>
4n 2a
+ -5-
t
1 + cos () dO; cos fJ(l - cos fJ)
dt
Put x 2 = cos fJ
211" cos- 1 (x 2 )
=2
J
1 ~ x 2 dx
= 2 tanh -1 x
1
=>
6 g t = 2 tanh-l [Vcos fJ] => cosfJ = tanh 2 [tJ 3g J. f69 lOa
V~ 5
Ex 9.9.2. The principal moments of inertia of a body at the centre of mass are A, 3A, 6A. The body is so rotated that its angular velocities about the axes are 3n, 2n, n respectively. If in the subsequent motion under no force WI, W2, W3 denote the angular velocities about principal axes at that time t, show that WI = 3W3 = ~sechu and W2 = 3ntanhu, where u = 3nt + ~ log~ 5. Hints: Here the given principal moments are A, 3A and 6A, so the Euler's equation of motion for no force becomes
CHAPTER 9. RIGID BODY MOTION
302
with the initial condition WI = W2 = 3n, W3 = n at t = O. These results gives, 1 3n
t = -tanh
-1
W2 lr,:: (-) - -10gv5. 3n 3n
Now using the relations tanh-I(~) = ~ log ~~~ and u = 3nt required result.
+ ~ loge 5,
we get the
Ex 9.9.3. A solid cube is in motion about an angular point which is fixed. If there are no external forces and WI, W2, W3 are the angular velocities about the edges through the .fi:uod point, prove tha.t WI + w2 + W3 = constant and also WI + + wj = constant.
wi
Hints: Let 0 be the fixed angular point and G be the CG of the cube. The line OG and two lines through 0 perpendicular to OG form a set of principal axis, for which the DCs are (~, ~, ~), (~, - ~, 0), (~, ~, - ~). Obviously the angular velocities about OG and the other principal axes respectively are Wx = WI +~+W3 ,Wy = 2w3 Also I = 2ma 2 l = llm(J2 I = llma 2 Then uSl'ng Euler's wI- W 2 W V2 ' z = WI +W2,j6' , 1 3 ,-2 3' 3 3' equation with no forces and 12 = h, gives the result. Ex 9.9.4. If a rectangular parallelopiped with its edge!; 2a, 2a, 2b rotates about its centre of gravity under no forces. Prove that, its angular velocity about one principal axis ~s constant and about the other axis it is periodic, the period about the first axis is (a 2 + b2) : (b 2 - a2) . Hints: As the body moves under no forces, we use the Euler's equations of motion. Let m be the mass of the parallelopiped, then 11 = W(a 2 +b2) = 12,13 = ~(a2+a2). From the last equation, we get W3 = 0 => W3 = constant = n say.
Thus, W3 = n implies that the angular velocity about one principal axis is constant and is given by TI = ~. Using the other two equations we get,
which shows that WI is periodic with period T2 = ~(~;i!~). Find
R.
Ex 9.9.5. A rectangular plate spins with constant velocity W about a diagonal. Find the couple which must act on the plate in order to maintain this motion. Hints: Let 0 be the centre of the lamina. The unit vectors along the principal axes at 0 are i, k, where k is normal to the plane of the plate and they are fixed in body. Also, It = k mb2 ,12 = kma2,h = W(a 2 +b2) and WI =wcosa,W2 =wsina and W3 = 0, where tan a = ~. Using Euler's equation, we get NI = 0, N2 = 0, N3 = 1 2 a 2 -b 2 "3 mw ab a2 + F ·
3,
9.9. MOTION OF SYMMETRICAL TOP
303
Ex 9.9.6. A rigid body which is symmetric about an axis has one point fixed on this axis. Discuss the rotational motion of the body, assuming that these are no forces acting on the body, other than the reaction force at the fixed point. Hints: Let us assume that, the axis of symmetry coincides with one of the principal axis, say the direction of k and A = B. Using Euler's equation with no force, we get .. W2
°
+ £"'12 W2 = ; H
£"'12
H
C- A 2 = -A--W3'
This equation can be solved by using the initial condition t = 0, W2 = 0, so W2 = bsin(Ot),wl = bsin(Ot). Hence w2 = constant, which shows that the angular velocity is constant in magnitude and the axis of W precesses round the k axis with frequency 2~' Ex 9.9.7. A circular disc of radius a and mass m supported on a point at its centre. It is set spinning with angular velocity Wo about a line making an angle a with the normal to the disc. Find the angular velocity of the disc at any subsequent time. Hints: We choose unit vector k along normal to the plane of the disc at the centre o of the disc. i and 3 vectors on the plane of the disc are chosen suitably to form a right-handed system. We assume that, the initial angular velocity vector w lies on the k - 3 plane. Let at subsequent time w = WI i + W2) + W3k and initially, WI = 0, W2 = Wo sin a, W3 = wo cos a. For the circular disc, A = B = tma2 and C = !ma 2 . Use Euler's equation of no force to obtain WI = -Wo sin'a sin(wo cos a.t), W2 = Wo sin a. cos(wo cos a.t) and W3 = Wo cos a. Ex 9.9.8. A uniform circular disc of mass m be initially at rest but is free to move in any manner, is suddenly set in motion by an impulse applied to a point A of its circumference. The initial velocity of A has components U along the radius of the circle through A, V along the tangent to the circle at A and W along a third perpendicular. Prove that T, the initial KE of the disc is given by 2T = m[U 2 + iV2 + ~W21· Hints: Let C be the centre of the disc. With C as origin, we consider a set of rectangular axes CXYZ. These are principal axes of the disc and A = B = C = Let w= wli + W2) + W3k be the angular velocity of the disc at A(a, 0, 0), then the velocity components f) = VI i + 1)2) + V3k of C will be given by,
mt,
m;2.
i ) k ("11)c = (l1)A
+Wx 7
= (U, V, W)
+
WI
a
W2 W3
°0
then, the velocity of the eM of the body VI = U, V2 = V + aW3, 713 = W - aW2. The angular momentum about C = (AWl, BW2, CW3), so that the momentum about A is (AWl)i + (BW2 - amv3)) + (CW3 + amv2)k. Since the disc was rest before impulse, using conservation law of momentum about A,
CHAPTER 9. RIGID BODY MOTION
304
Using these equations, find
9.10
aW2, aW3
and then use the formula
Exercise
Ex 9.10.1. Find the moment of inertia tensor of a uniform square plate of side a and mass m, taking the origin at one corner and the x and y axes along the edges. What are the values of the principal moments of inertia? CATE - 1994 Ex 9.10.2. Three particles of mass m each are situated at the corners of an equilateral triangle of side a. Consider the inertia tensor about point 0, the midpoint of base BC. Using symmetry arguments, suggest a set of principal axes. Evaluate the principal moments. CATE -1999 Ex 9.10.3. A body moves under no force about a point 0, the principal moments of inertia at 0 being 6A, 3A and A. Initially, the angular velocity of the body has components WI = n, W2 = 0, W3 = 3n about the principal axes. Show that at any later time, W2 = -vgntanh( vgnt) and ultimately the body rotates about the mean axis. Ex 9.10.4. An elliptic disc ~ + ~ = 1 at rest, is put into motion by making the points (a, 0, 0) and (0, b, 0) move with velocity (0,0, U) and (0,0, V) respectively. v. Find the resulting motion and prove the velocity of the centre is
Ut
Ex 9.10.5. A rigid body is rotating about a fixed point at which A, A, C are principal moments of inertia, under a couple ->'w x , ->'w y , ->'w z , where w x , w y , W z are the angular velocity components under the principal axis. Prove that, at any time t,
where
(J'
=
n(CiA); E,
n, a are arbitrary constants.
Ex 9.10.6. A top an axes of symmetry OC, where G is the CM and it spins with the end 0 on a rough horizontal table. The mass of the top is M and its M I about OC and any axis through 0 perpendicular to OG are C and A respectively. Initially, OC is vertical and the top is set spinning with spin n about its axis. It is then slightly displaced. If the subsequent motion e is the angle OG makes with the vertical and ¢; is the angular velocity about the vertical, show that, A¢ sin2 e = Cn(l - cos e) and A( ¢2 sin 28 + (;i2) = 2M gh(l - cos 8), where OC = h. Ex 9.10.7. A body rotates under no forces about its one point, the principal moments at this point are 4k, 3k and k. Initially angular velocity of the body has components WI = n, W2 = 0 and W3 = nv'2 about the principal axes. Find the angular velocity components at any time t and direction ratios of the invariable line.
Chapter 10
Central Force Field In classical mechanics, 'central force' plays an important role, as the knowledge about central force is important for dealing with several inverse square force, specially, in case of planetary motion. Newton's gravitational force, Ven'dar Waal's force between atoms and molecules in atomic structure, are examples of central forces. This plays also an important role in spacecraft dynamics. Here we shall discuss the characteristics and several applications of central forces.
10.1
Central force
If a force (attractive or repulsive) acting on a particle is always directed from or towards a fixed centre along the line joining the origin to the particle and magnitude of which is a function only the distance from the fixed point, then it is called central force. The fixed point is called centre of force. Both the gravitational force between two masses and the electrostatic force between the two stationary point charges are central' forces. Mathematically, it is written as -+
F
= F(r)r
(10.1)
where r is unit vector in the direction of the force along the line joining the centre of force and the particle. The central forces have the following properties, which are independent of the form of the force law : (i) The force may be repulsive or attractive, which is directed towards or away from the centre of force. Gravitational forces between two mass particles, electrostatic force between two unlike charges, etc. are the examples of attractive forces.
(ii) Central forces, which do not depend an time explicitly, are conservative forces, both have a natural occurance in classical mechanics. (iii) Since central forces are conservative forces, they can be expressed as the neg-+ ative gradient of corresponding potential energy V(r), i.e., F = -~~ r. 305
306
CHAPTER 10. CENTRAL FORCE FIELD
(iv) When a body moves under the action of a central force, its energy and angular momentum are conserved. (v) When a body moves under the action of a central force, its motion is confined to a plane, known as orbital plane. In this orbital plane, the motion lies and it is the normal to the angular momentum. (vi) Only radial dependence in F(r) implies that the isotropy of spacve is preserved by F(r) about the origin. The idea about central force is, in general, applicable in discussion on tides, dynamical manipulations of the orbits of spaceships in the space age, the geometry of the orbits of planets, etc. The motion of a particle under a central force field can be classified as 1. Bounded motion: In such type of motion, the distance between two bodies never exceeds a finite limit. For example, the motion of the planet's around the sun. 2. Unbounded motion: In such type of motion, the distance between two bodies is infinite, initially and finally. For example, scattering of alpha particles by nuclei of a gold foil as the Rutherford experiments. Apply the Lagrange's equation of motion, we shall discuss the problem of two bodies moving under the influence of a mutual central force. Also, under mutual interaction, the two bodies move in such a way that their centre of mass remain fixed in space. Here we take this centre of mass remains fixed in space. Here we take this centre of mass as the centre of force. This is an important part of theoretical physics as this type of motion performed by the planets around the sun, by satellites around the earth, by two charged particles with respect to each other.
10.1.1
Angular momentum
Let us consider, the motion of a particle in a plane under central force. So, the torque acting on it is ---t
T
= ---t r
---t
x F
= F(r)( ---t r
---t
x r)
= ---t O.
---t
Since we have for effective angular momentum L, ---t
---t
T
=
dL
---t
dt = 0 =} L = constant.
So far particle motion in a plane under conservative central force, its angular momentum will remain conserved.
10.2. CENTRE OF MASS
10.2
307
Centre of Mass
Let us consider a-system of n-particles having masses ml, m2,"" m n . Suppose that the particles constituting the system are situated at points in space, whose position vectors are r\, r' 2, ... , r' n respectively. Then the position vector of the centre of mass of the system is given by --t
R =
where M =
L
Li
mi T'i --==----
Lmi
= --M
(10.2)
mi is the total mass of the system. The centre of mass of a system
i
have the following properties (i) The total linear momentum of the system is equal to the product of the total mass of the system and the velocity of the center of mass. Form the equation (10.2), we get --t
MR
~
--t
= L..tmiTi i
(10.3)
--t
~
--t
~
where V = R and v i = T i are the velocities of the centre of mass and of the ith particle of the system respectively. Hence the result fo1l9wS from (10.3). --t
(ii) If there is no force act on the system, i.e., F = 0, then,
L mi 11 --t
:::::} V
2
= a constant vector = a constant vector.
Thus, if no external force acts on the system, then the centre of mass moves with constant velocity and the total linear momentum of the system must be conserved. (iii) The product of the total mass of the system and the acceleration of the centre of mall is equal to the sum of the external forces acting on the individual particles of the system. For this, from (10.3), we get
MV
= Lmitfi i
Ma: = Lmiit = Lmia:i = LFi i
(10.4)
308
CHAPTER 10. CENTRAL FORCE FIELD ~
~
~
~
where a = V and a i = v i are the accelerations of the centre of mass and of the ith particle respectively. Thus, when a system of particles mov~s under the action of external forces acting on the particles of the system, the centre of mass moves as through it were a particle of mass equal to that of the whole system with all the external forces acting directly on it. (iv) Centre of mass of a system of the two particles divides the distance between the two particles in the inverse ratio of their masses.
10.3
Two-Body Problem
Here we shall consider the motion of the two particles which are free to move in space under the influence of mutual forces of attraction or repulsion, acting in the line joining the particles and dependent on their distance from each other. Here we shall assume that, the masses involved are spherically symmetric, homogeneous and Newton's laws hold. Let us consider the interaction of two particles under a mutually attractive force characterised by a potential energy V of the system, which depends only on the position of particles relative to each other, between the two
o Figure 10.1. Two-body'problem particles or their relative velocity. Let --;;71 and --;;72 be the position vectors of the monogenic system two particles of masses m1 and m2 respectively. The KE T and PE V can be written as: 1 ~2 1 ~2 T = 2m1 T 1 + 2m2 T 2j V = V (1--;;7 1 - --;;721) = V (T) where the masses m1 and m2 separated by distance for centre of gravity, then ~
R
~
=
~
1
7n1 ~
~
T.
Let R be the position vector
~
+ 'fn2 (mIT 1 + m2 T 2)
=:';>T1=R+
7n1
m1 +m2
~
~
~
T;T2=R-
(10.5) m1 ~ r. m1 +m2
It is always possible to reduce the problem of the motion of two bodies so that of an equivalent one body problem in a central force field of mutual interaction, by
10.3. TWO-BODY PROBLEM
309
introducing the concept of reduced mass. Such a reduction of two body problem to an equivalent one body problem is found to be very convenient. Now the Lagrangian for such system is given by
Let ]7~ and ]7~ be the position vectors of the masses mI, m2 with respect to the centre of mass, where ]7 = ]71 - ]72 = ]7~ - ]7~, then ~I
'"""
L.J mk r k
~I m2 ~I =~ 0 =} r 1 = - - T 2
m1 m2 ~ ~I m1 ~ =}r1= rir2=r. m1 +m2 m1 +m2 k
~I
Thus the kinetic energy is given by ~
.
T( R, J7)
= KE
eM + KE of motion about the eM
of motion of the
1 ~2 = 2(m1 +m2)R +T1 1 ~2 1 . 12 1 . 12 = 2(m1 + m2) R + 2m1]71 + 2m2r2 1 ~2 1 m1 m 2 .~2 =-2(m1 +m2)R +-2 r m1+ m 2
1
~2
1
~2
= -M R + -J-L r 2 2
where M = m1 + m2 = total mass of the system and J-L =
m 1+m 2
ml
m,2
= (_1_ + ..1..-)-1 m,l m,2 ~
is known as reduced mass of the system. In terms of two variables Rand Lagrangian L becomes ~
~
T,
the
.
L = T(R, J7) - VCr)
=
1 ~2 -M R
2
1 ~2
+ -J-L T 2
-
VCr). ~2
Since V is a function of relative co-ordinates, the first term ~ M R
represents the
~
free motion of the centre of mass. As the three components of R does not appear in L, so they are cyclic and consequently the centre of mass is either at rest or moving uniformly. Lagrange's equations of motion in two variables Rand r will be d(8L) 8L 0 ( z.) dt fiR - 8R = => tt(MR) = 0 =} R = constant which shows that the velocity with which the eM moves in a straight line with uniform velocity. .. ) d (8L) 8L 0 ( Z'/, dlN-8r= => dtd ( J-Lr.) + 8V ar = 0 =} J-Lr.. = - 8V ar = f() r
CHAPTER 10. CENTRAL FORCE FIELD
310
which represents the equation of motion for the system under consideration at a distance T. This equation shew that, the motion of m2 relative to ml is the same as if mi were fixed and m2 were attracted to ml with the force derived from the potential energy mlml +m2 V. Thus, if the two free particles move in space under any law of mutual attraction, the tangents to their paths meet an arbitrary fixed plane in two points, the line joining which passes through a fixed point. Hence the problem of two-body motion has an exact solution. Dropping the first term, the Lagrangian becomes L
=
1 2
~2
-/1 T - V(T)
(10.6)
which is the Lagrangian, if we had a fixed center of force with a single particle at a distance 7 from it having mass /1 = (2+ _1 )-1. Thus under the central ml m2 force the motion of two bodies about their center of mass can always be reduced to an equivalent one-body problem. Now we have the following discussions about the Lagrangian L in equation 10.6. (i) In this Lagrangian, 7 is the position vector of the mass m2 with respect to the mass mI. Hence it represents the Lagrangian of mass /1 is situated at the position of mass m2 and its position vector with respect to mass mi is 7. (ii) In order to appreciate the importance of the reduction of a two body problem to a one body problem, consider that mass ml is very very large as compared to mass m2, i.e., ml > m2. The reduced mass of the system is given as 1
1
1
1
/1
ml
m2
m2
-=-+-~-.
This shows that the reduced mass of the system will be practically the same as the mass of the heavier body. This is used in Bohr's theory of the Hydrogen atom, Satellite motion around the earth, Planetary motion around the sun etc. Further let us suppose that the centre of mass are equal, m2 mlTI = m2T2 ::::} TI = - T 2 mi ::::} Tl ~
0 as
ml » m 2 ·
Thus the centre of mass of the system of two masses m 1 and m2 coincides with the heavier mass mI. For example, for studing the motion of artificial satellites, ballistic missiles or space problems orbiting the earth or the motion of a planet about the sun, the mass of the orbiting body is very much less than that of the central body. Such a reduction of a two body problem to an one-body problem is very convenient and is very often used in classical as well as quantum mechanics. The motion of an electron around the nucleus, before the advent of quantum mechanics, was studied as two body problem in classical mechanics.
10.3. TWO-BODY PROBLEM
311
Deduction 10.3.1. From Newton's laws of motion, we have,
where (! 1 and (! 2 are the constant vectors. If M = pair, then the centre of mass of the system is
ml
+ m2, the total mass of the
which shows that, the centre of mass of the system moves in a straight line in space. Ex 10.3.1. Calculate the changes in the values of energy and angular momentum, when a two body system interacting through gravitational force is reduced to an equivalent one body case. SOLUTION: Let us consider a system of two bodies of masses ml and m2. Let at any instant, -:t 1 and -:t 2 be their position vectors and 111 and 112 are their respective --t --t velocities. If R 1 and V 2 be the position vector and the velocity of the centre of mass respectively, then --t
--t
--t--t
Ii = ml r 1 + m2 r 2 j V = ml v 1 + m2 v 2 ml
+m2
ml
+m2
Now, the reduced mass J..L of the system, is J..L = :::'~2 and let the reduced mass J..L is located at the position of the body of mass m2 and -:t is the position vector of the reduced mass w.r.t. mass mI. Then, --t
--t
--t
...!.t
...!.t
...!.t
r=r2- r l=?r=r2- r l --t --t --t =?V=V2- V I,
where 11 = 7 is velocity of the mass m2 or the reduced mass w.r.t. mass mI. Let be the distance between the two bodies. Then the total energy E and E' of the system of two masses and reduced mass are given by, r
Therefore, the change E - E' ·in the energy, is given by,
312
CHAPTER 10. CENTRAL FORCE FIELD
which is the KE of the centre of the mass system. Therefore, on reducing a two body system to a one body case, the total energy of the system decreases by an amount equal to the KE of the centre of mass. Let us now calculate the corresponding change in the angular momentum. The total angular momentum of the system of two bodies is given by, ~L
~
= r 1
~
~
~
x ml v 1 + r 2 x m2 v 2
and the angular momentum of the reduced mass is given by
The change in the angular momentum is given by ~
~,
~
~
~
~]
ml v 1 + r 2 X m2 v 2 mlm2 [~ ~ ---+ ~ ~ ~ ~ ~ 1 rlXV1+r2XV2-rlXV2-r2XVl ml+m2 1 [~ ~ ~ ~ ~ ~ ~ ~l - - - ml r 1 x ml v 1 + m2 r 2 x m2 v 2 + ml r 1 x m2 v 2 + m2 r 2 x ml v 1 ml + '!n2 m 1 T 1 + m2 T 2] [m 1 v 1 + m2 v 2] = (rnl + m2 ) [ x ml +m2 ml +m2
L - L = [r
1 X
~
= (ml
~
~
+ m2) R x V
~
~
=R
~
~
~
x (ml + m2) V ,
which is the angular momentum of the centre of mass about point O. Therefore, on reducing two body system to a one body case, the angular momentum of the system decreaSes by an amount equal to the angular momentum of the centre of mass.
10.4. LAGRANGE'S EQUATION OF MOTION
10.4
313
Lagrange's equation of motion
Let the origin be centre of force and a particle of reduced mass J.t at a distance r, the separation distance between the two particles. Let us describe the position of the hypothetical particle in spherical polar co-ordinates (r, e, ¢), then x = Tsinecos¢; y = l' sine sin ¢; Z = TCOSe X = rsin ecos ¢ + r cos ecos r sin esin ¢¢
¢a -
y = rsin esin ¢ + l' cos esin ¢O + l' sin ecos ¢¢
z = rcos e- l' sin eo KE=
~/j,[±2 + y2 + z2]
=
~J.t[r2 + r 2iP + 1'2 sin
The potential energy V is a function of the distance of the hypothetical particle from the centre of force only and does not depend upon the orientation. Hence the Lagrangian L will be
Since L does not contain ¢ explicitly, so it is cyclic co-ordinate, the conjugate momentum to ¢ must be constant of motion, i.e., Pcp
8L 8¢
2 • 2
.
= -. = Ji,T sm e¢ = constant = lz
~
where 1 (lx, lv, lz) is the angular momentum vector. Since lz is constant, we choose that motion in xy-plane, where the Lagrangian is
Since e is cyclic co-ordinate, the conjugate momentum to motion, i.e., Po
e must
be constant of
8L = -, = J.tT 2'e = constant = ly
8e
Lagrange's equations of motion of the particle in terms of two variables
T,
e will be
(10.7)
From the second equation, we get d
2'
dt (J1,T 8) = 0 =}
".
J.tT~8
= Constant = l(say)
CHAPTER 10. CENTRAL FORCE FIELD
314
where l is the constant magnitude of the angular momentum. This is the integral corresponding to the ignorable co-ordinate () and can be physically interpreted as the integral of angular momentum of the particle about the centre of force. Since angular momentum of the deduced mass about the origin, l = J.Lr 2 it follows that angular momentum of the reduced mass moving under the effect of the central force is conserved. Also, the second equation can be written as
e,
d 1 2, 1 2' dt ("2 r ()) = 0 => "2 r () = h
=
Constant; J.L
i= 0
which interprets an interesting geometrical significance. The factor ~ is inserted, because ~r2e is the areal velocity. Let, in the figure, The differential area dA is the area swept out by the radius vector in time dt, then
dA =
~r(rd()) => dA 2
dt
=
~r2d() 2
dt
=
~h 2
dA
=> dt = Constant. Thus the conservation of the angular momentum is equivalent to saying the areal velocity of the particle in a central force field is constant. Hence the rate at which t+d/
o ----'-------11--Figure 10.2: Areal velocity the area is swept out by the radius vector is constant, which is Kepler's second law of planetary motion. Hence the conservation of areal velocity is not only restricted to an inverse square law of force but is a general property of central force.
10.4.1
Expression for total energy
From the Lagrange first equation of motion 10.7, we have the second order differential equation in r as
.
l J.Lr2
()= -
315
lOA. LAGRANGE'S EQUATION OF MOTION
This first order differential equation in r is known as energy integral and it solve for the motion. Here! f..Lf2 represents the KE, 2~:2 the centrifugal potential energy (energy due to rotational motion) while V(r) represents the gravitational potential energy of the system. Since the forces are conservative, we get another first integral of motion, which is actually the total energy E of the system. Thus
-
E
12
1
= 2f..Lf2 + 2f..Lr2 + V(r)
(10.8)
= ~f..L[f2 + r 202] + V(r)
= T
+V
= Constant
which shows that the conservation of energy E can be derived directly from the equation of motion.
10.4.2
Potential energy
By orbit, we mean a scheduled path of any object moving under a central force. The equation of the orbit in the differential form is found to be a great use. The one dimensional equation of motion of the particle moving under a central force is
f..LT
.
dV dr
= f..LrrP - -
=-
[2
f..Lr 3
+ F(r) = Fe
(10.9)
where the additional term ~ is known as centrifugal force. This equation of motion in r, with iJ expressed in terms of [, involves only r and its derivatives. An equivalent statement can be obtained from the conservation theorem of energy. Equation (10.9) can be looked upon as a one-dimensional equation of motion in r with the effective potential energy Ve. As the centrifugal force does not arise out of interaction between two particles, so it is not strictly a force and is known as fictitious force or pseudo or false force. It arises due to a accelerated motion of the particle in the orbit. The effective force Fe and the the effective or fictitious potential energy Ve are given by
Fe
[2
= F(r) + -f..Lr3 ; Ve(r) = -
J
Fe dr
[2
= V(r) + -f..Lr 22'
When the particle situated at 00, Le' r --+ 00, then the attractive force is asymp2 totically vanishes. Also, the terms 2!r 2 in Ve(r) is the potential energy associated with centrifugal force. Now the total energy for such a system is
E
= ~f..Lf2 + Ve(r).
t
This equation for energy suggests that the radial KE f..Lf2 and the effective PE for the radial motion in Ve(r). Again, Ve(r) consists of two parts, namely V(r), which is the actual PE and ~ is the centrifugal PE for the radial motion. If the motion is bounded in r, then f vanishes at the extreme values of r say 7' = rmin, r = rmax. Both of these bounds must exist for a bounded motion, Hence
!
E
= Ve(r)
at r
= rmin, rmax
and E - Ve(r)
= ~f..Lf2 > 0, "ir.
316
10.4.3
CHAPTER 10. CENTRAL FORCE FIELD
Integrals of motion
To solve the equations of motion, a total four integrations are needed, as there are two variables rand o. The first two integrals are J.Lr 2iJ = l, and the equation (10.8), the remaining two integration can be accomplished in a variety of ways. For simplicity, from the equation (10.8) we get,
1
[2
2
"2.'1.r + 2J.Lr2 + V(r) 2
r=
=}
= E [2
1
[-(E - V(r) - -)J2 J.L 2J.Lr2
J r
t
=}
=
dr
(10.10)
1
[~(E - V(r) - ~12 )J2
ro
~
~r
where r(t = 0) = TO. From this we calculate r = r(t), the constants of integration are E, l, roo Using this value of t, we can calculate 0 as
[J 0= t
. de e = -d
t
=
[ -2 J.Lr
=}
J.L
o
dt r t
~()
+ eo
where e(t = 0) = 00. Also the equation of the orbit in the integral form as,
de
iJ
= ...,.dr = t
dr
l -2
2
1
J.Lr [~(E - V(r) - 2~r2)]2
+ -[J r
=}
e
= eo
J.L ro
dr 2 1 r2[~(E - V(r) - 6)J2 ~ 2~r
(10.11) .
which is. the equation of the orbit in integral form. When V(r) is given these integrations are easily performed consisting of four constants E, [, ro, eo. The above two expressions for t and 0 are the two remaining integrations. But practically, the integrals are usually quite unmanageable and in any specific problem it is often more convenient to perform the integration in some other fashion.
Ex 10.4.1. Kepler's problem: For all central force problems, the inverse square law plays an important role. The differential equation of the orbit of a particle subjected to an attractive force F(r) = -f?z, Le., potential force V(r) = -~, is
J.LT = F(r)
[2
k
[2
+ -J.Lr 3 = -r2 -+- 3 J.Lr
where k is a constant signifying the strength of the attractive force be,tween sun and planet, J.L denotes the reduced mass of sun and the planet. Putting 'It = ~, the
10.4. LAGRANGE'S EQUATION OF MOTION
317
differential equation of the orbit is d2u J1k . d2y d0 2 + u = ~.e., => d0 2
=> y = u 1
r' + y = 0; rJ1k = Acos(O - (
y=u-
J1k
r
0)
J1k
=> -r = -l2 + Acos(O -
(0)
l2 j J1k
l2 A (10.12) cos(O - (0) l' kJ1 where 00 and A are constants of integration. The value of the cQnstant 00 can be obtained as a value of 0 when y has the maximum value. 0 is called the true anomaly or the angle between the perihelon. The words aphelion and perihelion are generally used instead of apocentre and pericentre when the centre of force is supposed to be the sun. Equation (10.12) gives the orbit of the body moving under the effect of inverse square force field. Comparing this with the standard equation of the general conic section with the origin as focus as = 1 + ecos(O - a), we say that the orbit is a conic section with semi latus rectum L = and eccentricity e = l~:. L j e is the distance of the focus from the directrix. The parameters e and L determine the shape of the conic. In order to know which of the conic sections can possibly represent a planarity orbit, it is necessary to obtain a relation between the orbital eccentricity (e) and the energy (E) of the planet. To calculate eccentricity (e) in terms of the energy (E) and angular momentum l, we have
=> - - = 1 + -
f
0-00
J =- J v'
ljr2dr J2J1(E _ V -
=
2!:2) = -
du ; A A + Bu + Cu 2
=> l2 j J1k = 1 + r
k:
J
J
= 2J1E
rl
+ ~u _
B
= 2J1k, C = -1
2
z2 '
·du
2El2 1 + J1k 2 cos(O -
u2
l2
(0);
where the quantity 00 is a constant of integration determined by the initial conditions, but not necessarily the same as the initial angle 00 can now be identified as one of the turning angles of the orbit. Also, three of the four constants of the integration appear in the equation of orbit. In effect, the fourth constant locates the initial position of the particle on the orbit. Note that, at the ends of the transverse axis of the conic, i.e., apses r = O. The turning points are given by an apsidal quadratic in l' whose roots are say 'T' = 1'1, 'T'), for which cos a = ±1 and they are given by 1
J1k
-1'1 = -l2 +A;
(10.13)
318
CHAPTER 10. CENTRAL FORCE FIELD
Now the turning points are given by 1
ILk
/l,k
2El2
=+ l2TI l2
1 + ILk 2
.
The value of r, thus obtained correspond to those at the ends of the transverse axis of the conic. Comparing these with (10.13), we get
As E, l, IL and k are known the values of A and e are known. As the eccentricity is a function of E, for possible values of E, we get different types of orbit as : Values of E E> 0 E=O vemin < E < 0
1 k2
,
E=~mm=-2~
Values of e Nature of motion Nature of Orbit Non-parabolic Hyperbola e> 1 Bound and Non-parabolic Parabola e=l Ellipse Parabolic O<e
Thus the orbital equation for motion in a central inverse square force law can thus be solved in a straightforward manner, with simple closed relations.
Deduction 10.4.1. Circular orbit: For the circular orbit, T and V are constants in time, and from the virial theorem
E
V
V
k
= T + V = -2" + V = 2" = - 2R
Also, - -
k
l2
[2
= --R2 mR3
=::}
E
=-
ILk
=::}
R=ILk
2
2[2
which is the condition for circular orbit.
Deduction 10.4.2. Elliptic orbit: In case of elliptic orbit the major axis depends solely on the energy. Let a be the length of the semi major axis of the ellipse. If r = 1'1 and l' = 1'2 be the turning points, then apsidal distance is 1'1
where
1'1
and
1'2
+ 1'2 =
2a
are the roots of the equation [2 k E---+-=O 2 2m1'
k
=::}
1'2
+ -1' E
l'
l2 -2mr2
= O.
10.4. LAGRANGE'S EQUATION OF MOTION
319
The line of the major axis is also called the apsidalline or apsis. Thus the relation between energy E and the semi major axis of the ellipse is given by k
1
a="2(r 1 +r2 )=-2E which shows that all ellipses with the same major axis have the same energy. The eccentricity in terms of semi major axis of the ellipse can be written as, e
2
= 1+
2El2
l2
k
2
- - => - = --(1 - e ) . J.tk 2 J.tE 2E
= a(1 -
2
e ).
Using these results, the polar equation of the elliptical orbit becomes, r
=
a(1 - e2 ) 1 + ecos(e - eo)
--~--:--'---:-
from which it follows that the two apsidal distances are a(1 - e) and a(1 + e). For the elliptic motion, if we defined an auxiliary variable (3, by r = a( 1 - e cos (3), then {3 is known as eccentric anomaly. From the equation of the orbit, it is clear that, as e goes through a complete revolution, {3, covers [0,27r] and the perihelion occurs at {3 = e = 0, and the aphelion at {3 = e = 7r. Hence, the polar angle 8 is given by cos {3 - e 1 - e2 e => cose = 1 a 1 + e cos - e cos I--' l-cose l-ecos{3-cos{3+e => = -------'--1 + cos e 1 - e cos {3 + cos {3 - e l-e l-e => tan 2 e = --tan 2 {3 => tane = --tan{3 l+e l+e which gives the relation between e and (3. For a attractive inverse square law, the requirement that the orbit be bounded corresponds to -~ < E < 0 => 0 < e < 1. Thus, the4 planetary orbits must be ellipsoid. When, E < 0, the system is bound as the earth is bound to sun. Hence" each planet moves round the sun in an elliptic orbit, with sun situated one of the foci"- which is one of the laws of Kepler. To describe the motion of particle in time as it traverses the elliptic orbit, we have 1 - ecos{3 =
t-
-
~J J _ -
r
2k
ro
r
rdr
r2 _
2a
a(1-e2 ) 2
where, the convention as the starting time, ro is the perihelion distance. Thus, using the relation r = a(1 - e cos (3); e < 1, we have, {3
t-
-
2k o J ~J VJ.t;3 J(1 - e cos (3)d{3.
a(1- e cos (3)ae sin {3 (1 (3) - a(l-e 2cos (3)2 a - ecos
-
{3
=
o
d{3 a(1-e 2 ) 2
CHAPTER 10. CENTRAL FORCE FIELD
320
This equation provides an expression for the period T, of the elliptic motion, if the integral is carried over the full range in f3 of 27T, i.e.,
= 27TV J.L;3 = 27Ta3/2!f
T
Now, let us introduce the frequency of revolution was, w =
t=
~
2; =
a,
then,
f3
J
a(l - ecos(3)df3
o :::::} wt
= f3 -
(10.14)
esinf3
which is the well known Kepler's equation. The solution of the transcendental equation gives the value of f3 corresponding to a given time (we can use one of the numerical methods- say, Newton Raphson method; see Author's Numerical Book). For a complete orbital revolution, wt varies through the range 0 to 27T along with (), f3 and is denoted as mean anomaly.
Deduction 10.4.3. Hyperbolic orbit : Let 2a be the distance between the two vertices of a hyperbola. For such an orbit, E > 0 and e > 1. The polar equation of the orbit is a(e 2 - 1) r=±1 + ecos(O - 00)'. To describe the motion of particle in time as it traverses the parabolic orbit, we have
Now, the polar equation of the parabola, with focus as pole, is, ~a = 1 + cos 0, where, 4a is the latus rectum. Also, the polar angle 0 and the time are counted through 2 the conservation of angular momentum as dt = TdO. Hence the time of describing the arc is,
J+ 8
l3
t = J.lk2
[1
80
dO cos(O - 00)]2
from the radius vector at the point of closest approach, known as perihelion. For simplicity, setting 00 = 0, the time measured from the moment of perihelion passage is given by,
J 8
[3
t = 4J.Lk 2
4 ()
sec ("2)dO
l3
0
1
3 ()
= 2J.lk 2 [tan("2) + "3 tan ("2)]'
o This is the relation for t as a function of O. When t is given then we are to solve a cubic equation for tan(~), to find O.
321
10.5. GENERAL EQUATION OF THE ORBIT
10.5
General equation of the orbit
The most important case of motion under central forces is that in which the magnitude of the force depends only on the distance r. Denoting the force by F(r), the differential equation of the orbit is,
2 d u F(r) du 2 2 d(}2 +u= h 2u 2 ~'[d(}] =c- h 2
J l'
2
F(r)dr-u; integrating
TO T
T
~ () = j[c - ~2 j F(r)dr _ ~]_1/2dr r2
h
r2
TO
1'0
which is the equation of the orbit in polar co-ordinates, where c is a constant. Using energy equations, we have T
j
() - () + i -
0
J.l
TO
r 2 [1(E
JL
dr - V - 6)]-1/2 2JLT
U
j V/2JLl _du2JLV _ u2; u = ;:. 1
= (}o -
,
Uo
yr
I
Since the expression can not be expressed in terms of well-known function, so the integration is not always a practicable solution. We shall now discuss the cases in which the quadrature can be effected in terms of known functions, the central force being supposed to vary as some power law (podtive or negative) of the distance r, say, F(r)aT~' i.e., V(r) = arn+1. Under the law of force, the equation ofthe orbit becomes, 1J.
() = (}o -
J
Uo
du J2'/2E I
•
-n-1 _ u2 pU 2JLQ:
If the quantity in the radical is of no higher power in u than quadratic, for determination of (), the denominator is of the form (au 2 + bu + c) 1/2, a, b, c constants, so that the integration can be effected in terms of circular functions, excluding n = -1, i.e., when a logarithm replaces the term in u- n - 1 . If the problem is to be solve in terms of circular functions, the polynomial under the radical in the integrand must be at most of the second degree. Hence the restriction is equivalent to
-n - 1 = 0,1 or 2, ; i.e., n
= -2, -3, excluding n = -1
which corresponds to inverse square or inverse cube force laws. The case n = 1 is to be added, since in this case, the irrationality becomes quadratic, when u 2 is taken as a new variable. Thus, a solution in terms of simple functions is obtained
322
CHAPTER. 10. CENTRAL FORCE FIELD
for the exponents n = 1, -2, -3. Next, let us find the cases in which the integration can be effected by the' aid of elliptic functions. For this, it is necessary that, the irrationality to be integrated should be of the third or fourth degree in the variable with respect to which the integration is taken. Thus, the condition for which the integral can be evaluated in terms of elliptic functions is n
n
= 0, -4, or, = 3,5, or, -
- 5; when u is taken as the independent variable 7; when u 2 is taken as the independent variable.
Hence it follows that, the problem of motion under a central force which varies as the nth power of the distance is soluble by circular or elliptic functions in the cases n
= 5,3,1,0, -3, -4, -5, -7.
The cases in which motion under a central force varying as a power of the distance is soluble by means of circular functions are of special interest. Deduction 10.5.1. Let us consider n = 1, i.e., the attractive force is f(r) = kr. In this case, the equation of the orbit is
() = =>
2()
/[c - _k_ _ u 2]-1/2du h 2u 2
2
= _~ /[(2 _.!5..-) _ (t _ 2
2:)2rl/2dt 2
= cos- 1
constant
2
=> 2(() - (3)
= _~ /[ct _ .!5..-2 _ t 21- 1/ 2dt·
4
h
t - c/2 ; f3 = . / c2
h
·
,
k
VT-h'!
=>
1 1,2
c
~
= '2 + V'4 - h2 cos 2( () - (3)
°
which represents the equation of the ellipse when k > or hyperbola, when k < 0 referred to its centre. The orbits are therefore conics whose centre is at the centre of force. Deduction 10.5.2. Let us consider n = -3, in this case, the attractive force is F{-r) = -fir. So the equations of the orbit becomes,
() = - / [c +
(:2 -
1)u2rl/2du.
k =>u=Acos(p()+c); p2=1_ h 2 andk
= .!5..-2 h
1 and k > h 2
= h2
where A and 10 are constants of integration. The first two curves are sometimes known as Cotes' spirals and the last is the reciprocal spiral.
10.5. GENERAL EQUATION OF THE ORBIT
10.5.1
323
Equation of orbit
The relation between rand 0 gives the equation of the orbit. In order to obtain the differential equation of the orbit, we define a new variable u as r = ~, then, . r
Substituting rand
r,
=
dr dt
=
1 du .. -;, dO; r
=
d 2r dt2
=-
12 2 d 2u J.L2 u d02'
the equation of motion becomes 12 2 d 2u z2u 3 J.L( - J.L2 U d( 2 ) - -;;: 12
=}
=F
d 2u
-_[u2 _ 2 +u3 ] = F J.L d0
which is the differential equation of the orbit of a body moving under central force. From this equation, for a given law of force F we are to find the orbit, and conversely if the orbit is given, we are to find the law of force.
Ex 10.5.1. Find the law of force, if a particle which is on the circumference of a unit circle g,iven by 7" = 2 cos 0 directed towards the origin, is under central force. SOLUTION: The equation of the unit circle is given by, r = 2cosO
=}
1 u = -2 0°
cos
Differentiating twice, with respect to 0, we get, 2 du dO
=
2
sinO and 2d u cos2 0 d0 2
= 1 + sin
2
O.
cos3 0
The differential equation of the orbit is gives,
Hence, the law of force is an inverse fifth power.
10.5.2
Stability of orbits
An orbit is said stable if, keeping either the energy or the angular momentum unchanged, when a slight perturbation is given to the initial radial co-ordinates the orbit will perturbated slightly. The condition of stability in radial motion is given by the existence of a local minimum in the effective potential function Ve(r), Le., d2~(r)
dr2
.
> 0 at r = ro gIven by
d~(r)
dr
= O.
CHAPTER 10. CENTRAL FORCE FIELD
324
Let the equation of the orbit under any central force f{r) is given by
d2 u J-l d(}2 +u=-[2u2F{u) where u
uo,
2
d u dJj'I
=
~ and f{~) 0 . Hence
=
= F(u).
Now for a circle of radius r
= ro,
i.e., u
=
r~
=
Hence the equation reduces to
For a small displacement along a radial direction, we take u = u + 0: where 0: is a small perturbation to u, i.e., 0: « Uo. Then the differential equation of the orbit becomes
d2u d(}2
u5 F{uo+o:) F{uo)· (uo + 0:)2 Uo 0:. 2 = --F() .F{uo + 0:)[1 + - j Uo Uo Uo , 20: = --p() [F{uo) + o:F (uo) + ... ][1- Uo Uo
+ Uo + a = -
+ ...l.
Neglecting the squares and higher powers of Uo and keeping the angular momentum constant we get,
d2u, d(}2
+ Uo + 0: = d2u d(}2
F'{uo) -Uo + 0:[2 - F(uo) uo]. F'(uo)
= -0:[3 - F(uo) uo] = -m
2 ll'.
The general solution of the differential equation is 0:
= C1 cos m(} + C2 sin m(}j m > 0 = qcoshm(} + c2sinhm(}j m 2 < 0 = q(} + C2j m = O.
,
The first solution is interesting but the other solution 0: increase indefinitely with continuously increase of (). The approximation to the path is
u
= un + Cl cosm(} + C2 sin m(} =
Uo
+ Acos(m(} + B).
10.5. GENERAL EQUATION OF THE ORBIT
325
Thus for such small deviations from the circularity conditions, the particle execute simple harmonic motion in u about uo, where A is the amplitude, that depends on the deviations of the energy from the circular orbit and m 2 = [3 - ~(~~? uo]. Hence the orbit will be elliptic close to the circular orbit. Hence a circular orbit of radius r = TO = is stable if m > 0, i.e.,
;0
F'(uo) F( uo )
3 uo
< - = 3ro
d
=> [du log F(u)]u=uo < 3ro
J3 -
with the apsidal angle 7r/ ~(~~?uo. Under this condition, the maximum and minimum values of u are a + A and a - A. Hence the new path will be ellipse with semi-axes given by the reciprocal of a - A and a + A. Ex 10.5.2. A system is in motion under no force and its K E is ~ (A0 2+ B J>2), where A, B are functions of () alone. Show that a steady motion is possible in which () = a, and J> = w, a constant, provided that in the position () = a, B is a maximum or a minimum and further that this state of motion is stable if B is a maximum and unstable if B is a minimum.
SOLUTION: The KE and PE in motion of the particle are T = ~(Ae2 V = 0 and so the Lagrangian L is given by
L
+ BJ>2)
and
= T - V = ~(Ae2 + BJ>2).
As A, B are functions of () alone, so 4> is ignorable. Therefore,
aL
..
{3
aJ> = constant = (3(say) => B4> = (3 => 4> = B(())
The Routhian function R is given by, R
=L
. - (34>
1'2
'2
= "2(A() + B4>
. ) - (34>
1
'2
1· {32
= "2 A () -"2 B(())"
The equation of motion gives,
d aR dt ( ae)
aR
- a() = 0 => ..
d· {32 , d() (A()) - 2[B(())]2 B (())
1"2
(32
=0
,
=> A(())() + "2A (())() - 2[B(())]2 B (()) = o. Steady motion is possible at () = a is possible, if jj = O. From the above equation, we see that, B'(()) = 0 for () = a is possible if B(()) is either maximum or minimum. In that position 4>
(3
= B(a) =
constant
= w.
326
CHAPTER 10. CENTRAL FORCE FIELD
Now, to examine, the stability of the motion, take, () compared to unity. Thus, ..
1
1'2
..
1
I
fJ2
I
A(())e + '2A (())e - 2[B(())J2 B (()) '2
A(a + O~ + '2A (a + ~)~
=
a
+ €,
where ~ is small
=0
fJ2
- 2[B(a + ~)J2 B
I
(a +~) = O.
Expanding in Taylors series and neglecting the square and higher order terms, we get,
Since A(a) > 0, it follows that the motion is stable if B"(a) < 0 and unstable, if B"(O') > O.
10.6
Motion under arbitrary potential field
Here solution for a specific force field, we use only the equations of motion and conservation laws, without performing the integration. To discuss the motion of a particle in an arbitrary potential field, consider a system of known energy and angular momentum, the magnitude and direction of the velocity of the particle can be written as
r=
J~[EVel = J![E- V 11 m
_l2_l. 2 2mr
The conservation of angular momentum gives the angular velocity iJ and this together
l'
Figure 10.3: Curve of arbitrary potential field
10.6. MOTION UNDER ARBITRARY POTENTIAL FIELD
327
with r gives both the magnitude and direction of r. In the figure, we draw an arbitrary potential energy continuous curve. At the points A, B we have r = 0 and they are correspond to the intersection of Ve = E = constant, and on the potential energy curve. Such points where the radial velocity r = 0 are defined as turning points. At the intermediate points between A and B, we have
which is the KE of the particle at that point. Now we discuss the motion of the particle at the different range of the value of r. Case 1: Let r < rl. In that region the PE Ye > E, the total energy, so that r will be imaginary, as KE is '-'ve. Hence the region is forbidden for the particle. Even if E < Ve , the particle always has some probability of penetrating the potential barrier in quantum mechanical description. This is known as turning effect. Case 2 : Let l' E h,1"2], in this region, E > Ye. The points l' = 1'1 and r = 1·2 are the turning points. The motion of the particle is, therefore, oscillatory in the potential wall and the particle WIll be confined to the region. The particle does
Figure 10.4: Bounded motion not possess enough KE to cross over the potential barriers on either side. The orbit of the particle may not be closed and might be bounded in the region between two circles of radii rl and r2. rl rv r2 is called the apsidal distances. In quantum mechanical description, the particle does not possess any arbitrary energy E in the potential well, but the energy will have discrete values. Case 3: Let r E (r2, r3). Here also Ye > E and like case 1, the region is forbidden for the particle. Case 4: Let r 2 r3. In that region, r = r3 is the only turning point. A particle initially at an infinite distance from the center of force is able to approach up to r = 1'3 where it gets rebound and proceeds unhindered to infinity ag tin. This phenomenon is known as scattering. Thus two regions [1'1,1'2] and r 2 '13 are respectively the bounded and unbounded motions. With the help of the arbitrary potential energy curve, we are to understand the nature of the orbits.
CHAPTER 10. CENTRAL FORCE FIELD
328
~~----------~-------.x
Figure 10.5: Scattering
10.7
Kepler's Law
Kepler's law of planetary motion were (i) Each planet moves round the sun in an elliptic orbit, with sun situated one of the foci. (ii) The radius vector drawn from the sun to the planets sweeps out equal ares in equal time. (iii) The square of the time period of the planet around the sun is directly proportional to the cube of the semi major axis of the elliptical orbit of the planet.
Deduction 10.7.1. Kepler's Law from Newton's law: For the verification of the third law, let a and b be the major axes of the orbit. Then the area of the orbit is nab and the rate of describing area ~h. Hence the periodic time T, i.e., the time of one revolution along the orbit, is
nab 2nub - h/2 - .Jiil
T------
=}
T2
4n 2
a
p,
3" = -
-
2nab 2n 3 --a2 p,b 2 /a - VIi
J
= a constant
=} T
2
aa 3
Thm; we have shown that, all the three laws of Kepler can be verified from Newton's law of gravitation.
Deduction 10.7.2. Newton's law from Kepler's Law: Here we are to deduce the Newton's law of gravitation from Kepler's law. Second law of Kepler states that It, is constant, hence T2iJ is constant. But we know that the expression for cross-radial acceleration is ~fh,C,.2iJ) and hence it vanishes. There is no cross radial force, i.e., the force is only radial. Hence the force is directed towards the sun. Again, from the first law, we know that the path is an ellipse, whose equation may be written as 1
l
;;: = 1 + e cos e =} u = y (1 + e cos e) d2 u =}
e
de 2 = -y cos e.
329
10.8. UNBOUNDED MOTION Hence, for the central orbit, the force F per unit mass is given by,
F
= h2
2 2[d u U d()2
Again, for an elliptic orbit, we have = constant. Therefore, we have a
+U T
=
1= h
2 2 u =
l
h
l
2
~
r2'
2~ab, and from the law of Kepler, we get
5-
constant
h2 1
=
=> - =
(27rab)2 47r 2 b2 47r 2 ,2 3 = -,-2- = h2 1 1 a 1 a 2 47r = constant for all points constant
= J1( say)
J1
=> F= 2"' r Thus, for each planet the force per unit mass (i.e., acceleration) is towards the sun and varies inversely as the square of the distance from the sun. Hence Newton's law of gravitation " every particle in this universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely as the square of the distance between them", follows from Kepler's law. .
10.8
Unbounded Motion
The scattering or collisions experiments constitute an important source of information regarding the nature of the interaction between the particles. We assume that, the velocities of the particles are very small as compared to the velocity of light, i.e., the treatment is non-relativistic. The discussion of collision between two bodies is found to be very simple and symmetric, if we describe the effects of collision in two co-ordinate systems : (i) Laboratory system: In general, the collision between two particles is oblique collision where both moving particles approaches to each other in a non-linear manner with respect to their line of impact. After collision, they move away from each other in the same fashion. So, in general case, collision takes place in a plane and obviously, the direct or linear collision is a special case of oblique collision in which particles moves along their line of impact before and after collisions. Experiments are often carried out in which one of the particles is rest in the laboratory and the other approaches it and collision takes place. Such a set up in which a particle collides with another particle is at rest is called laboratory frame of reference. (ii) Centre of mass system: The discussion of collision between two particles becomes very simple and symmetric for motions of particles before and after collision which are taken such that, the centre of mass of the system be rest throughout the collision. Such set up is thus called centre of mass (eM) frame, and, obviously, in this system, the collision between two particles is treated as they have equal and opposite momenta initially.
330
10.9
CHAPTER 10. CENTRAL FORCE FIELD
Scattering
Below, we are discussing three types of collision, in which the law of conservation of linear momentum and angular momentum holds.
10.9.1
Elastic Scattering
The collision between two particles is said to be elastic in nature, if the linear
-> X' ______ , _ _ II,,,-!_
... ______ , _ _ ___
In!
Before collision
Figure 10.6: Elastic Scattering momentum and kinetic energy of the system of the two particles is conserved and the masses of the scattered and recoil target particles are same as those of the incident and target particles respectively. Let, two particles A and B have masses ml and m2 respectively. A is moving with velocity ttl (called incident particle) collide against the particle B moving with velocity tt 2 (called the target). After collision, the particle A moves with velocity 1/1 (called the scattered particle) in the direction making an angle a with the incident direction and the particle B moves with velocity 1/2 (called the recoil target) in the direction making an angle j3 with the incident direction. Using the law of conservation of linear momentum, we have
and the conservation of kinetic energy of the system,
In such a process, the internal nature, or the structure of the particles does not change and without taking into consideration, the internal energies at all, we can apply the law of conservation of energy.
Elastic Scattering in Laboratory SystE'm Let us consider :the incident particle of mass ml moving with velocity ttl along x-direction collides against the target particle of mass m2 at rest tt 2 = 'o. Suppose, after the collision, the scattered particle moves with a velocity 1/1 making an angle
10.9. SCATTERING
331 y
1."
Figure 10.7: Elastic Scattering in L~boratory System
a with the direction of the incident particle, while the target particle recoils with velocity 7J 2 in the direction making an angle (3 with the incident direction. According to the law of conservation of linear momentum, we have ---t
---t
---t
---t
ml U I + m2 0 = ml v I + m2 v 2 =? ml UI i = ml (VI cos ai + vI sin a]) = (ml VI cos a =?
mIVI cos a
+ m2( V2 cos (3i - V2 sin (3]) + m2v2 cos (3)i + (ml VI sin a - m2v2 sin (3)3
+ m2v2 cos (3 = mlUlj
mIVI
sin a -
m2v2
sin (3
= O.
These equations describe the conservation of linear momentum of the system along x and y direction respectively. Now, using the law of conservation of KE of the system, we have, 12121212 "2mIUI
+ "2m20 = "2mIVI + "2 m2v2
=?
1 2 1 2 1 2 "2mIUI = "2mIVI + "2 m2v2 '
The above mentioned three equations contain four unknowns VI, V2, a and (3, but the values of these four unknowns can not be found from the above three equations, either the angle a or (3 is measured experimentally. Deduction 10.9.1. Let ml VI
cos a
=? (UI -
in this case the above equations become
+ V2 cos (3 = Ulj VI sina - V2 sin (3 = 0 and u~ = v~ + v~ VI cos a)2 + (VI sina)2 = (V2 cos (3)2 + (V2 sin (3)2 =? U~ + V~ - 2UI VI cos a = V~ = U~ - V~ =?
=? (UI
= m2 = m,
VI
= UI cosaj
cos a) sin a =
(UI
V2
= UI sina
cos a) sin (3
=?
a
+ (3 = ~.
Thts shows that, if in the lab system, incident particle collides with the target particle of equal masses at rest, the two particles move at right angles to each other after the elastic collision.
CHAPTER 10. CENTRAL FORCE FIELD
332 Elastic Scattering in
eM
System
Let us consider the incident particle of mass ml moving with velocity 1/ 1 with respect to the observer ('If the lab system against the target particle of mass m2 at
--> --> =1 1 - V 0
x' ---It;- - _ "
nn_no
_
V
~ n
G
In!
_ _ _ _- -
tn2
Figure 10.8: Elastic Scattering in eM System -->
rest u
-->
2
-->
= o. Let V be the velocity of the eM of the system of the particles, then -->
ml u
1
+ m2 --> 0 =
(ml
--> + m2) --> V =} V =
m1 ml+ m 2
-->
u
1·
(10.15)
From the equation (10.15), it follows that the eM moves in the direction of the incident particle. Let 1/~ and 1/~ be the velocities of the two particles in the eM system. Then according to Galilean transformations, we have (10.16) (10.17) From these equations (10.16) and (10.17), it follows that the two particles move towards each other in the eM system. Figure 1 represents the velocities of the incident particle, target particle and the centre of mass in the eM system. Let 111 ->
v'!
--"-- --------- X
Figure 10.9 and 112 be the velocities of the two particles with respect to the observer of the lab system. If 11~ and 11~ be the respective velocities of the two particles in eM system, then according to Galilean transformations, --> I
v
1 =
-->
v
1 -
-->
--> I
V and v
2 =
-->
V 2 -
-->
V.
10.9. SCATTERING
333
Using the law of conservation of linear momentum in CM system, we have ml
ml m 2 ::::}
~/
ml +m2
~/
+ m2 u 2 =
~ U 1
ml m 2 - . ml +m2
u
1
~/
~/
+ m2 --+/ v2 ~ ~/ --+/ u 2 = ml v 1 + m2 v 2
ml
v
1
ml~/
::::} V 2 = - - V l ' m2
This shows that, after collision, the two particles in CM system move in opposite directions. Further, the equations suggest as to how the velocity vectors v~ and v~ of the particles after collisions are oriented in the xy-plane. If we draw the velocity --+ vectors VI, 112 after collision in LAB system and V, then the velocity vectors v~ and v~ was shown in figure 10.8. As in CM system the particles move towards each other before collision, and move away from each other after the collision, the collision between two particles in CM system may be depicted as shown in figure 10.9.
10.9.2
In-elastic scattering
The collision between two particles is said to be inelastic in nature, if the linear momentum of the system is conserved (KE is not conserved) and the masses of the scattered and the recoil target particles are same as those of the incident and target particles respectively. An inelastic collision, the KE of the system decreases, i.e., 12121212
rm1'U1
+ 2m2u2 >
2m1v1
+ 2 m2v2 '
The decreases in KE of the particles appears in some other form of energy such as heat, sound, etc. This loss of KE must be taken into account while using the law of conservation of energy. When changed particles interact, they may, due to acceleration acquired as a result of the force of interaction, radiate energy in the form of electro-magnetic waves. Most of the collisions that occur in nature are Inelastic.
10.9.3
Reaction
In some interactions, the masses of the particles may change or new particles may be created or some particles may coalesce into one. Charges are known to take place in the internal structure of the particles, or the total number of particles in the final state may be different from that in the initial state. The collision between two particles is said to be reaction, if the linear momentum of the system is conserved, but the masses of the scattered and the recoil target particles are entirely different from those of the incident and target particles. In case of reaction between the two particles, the scattered and the recoil target particles are better known as outgoing particle and the residual particle respectively. In a reaction, the total KE of the system may be less or greater than the final KE of the system, i.e., 1212
2m1u1
1212
+ 2m2u2 >, < 2m1v1 + 2 m2v2 '
CHAPTER 10. CENTRAL FORCE FIELD
334
10.10
Direct /linear collisions
Two smooth spheres of masses ml and m2 moving with respective velocities UI and Let e be the coefficient of restitution between them and VI, V2 be the velocities of the two spheres just after impact in the direction of UI and U2 respectively. Then the velocity of approach is ttl -11'2 and the velocity of separation is V2 - VI. By, Newton's experimental law of impact, we have, U2 in the same direction impinge directly.
(10.18) Again, since the impulsive action and reaction is along the line of centres, by the principle of conservation of momentum, we have, (10.19) From these equations, we get the velocities impact, as
10.10.1
VI
and
V2
of the two spheres just after
Loss of KE
The loss of KE t:.T, due to impact is given by
t:.T
1
= 2[(ml'u~ + m2'u~) 2(
-
1 ) [(ml + ml+m2
(mlv~
+ m2v~)1
m2)(mlu~ + m2u~) -
(ml +
m2)(mlv~ + m2v~)1
1 2 2
) [(mIUI + m2 U2) + mlm2(UI - U2) 1 ml+m2 - 2( mlm2 ) ( 1 - e 2)( UI - U2 )2 ; 0::; e ::; 1. ml+m2 = 2(
Now we have the following conclusions : (i) For perfectly inelastic collision of bodies, e = 0, and so and mlm2 ( )2 t:.T = 2( ) UI - U2 ml+m2
VI
= ml~!t:~U2 = V2
and this two particles will move together after collisions. (ii) For perfectly inelastic collision of bodies, e = 0, and so m2) + 2m2tt2 u2(m2 - ml) + 2mlUI ; V2 = -'"-----'-'----ml +m2 ml +m2 and so t:.T = 0, i.e., for elastic collision, particles will move separately along the same line of impact after collision without any loss in KE. and this two particles will move together after collisions. VI
= uI(ml -
10.10. DIRECT/LINEAR COLLISIONS
10.10.2
335
Characteristic of direct collision
We see that, for direct collision, between two bodies moving in the same direction before and after collision along their line of impact. The velocities after collision are
°
and the loss of KE for such collision is 6.T = 2f;~~2) (1 - e 2 )(-ul - U2)2 j ~ e ~ l. With these results, we can have the following c aracteristics for direct impact, (i) For ml = m2, VI = U2, V2 = UI, for e = 1, i.e., for elastic collision between two bodies having same mass, the velocity exchange takes place between them. (ii) For elastic collision, e = 1,6.T collision between two particles. (iii) For elastic collision, e initially, then,
= OJ, i.e., no loss of KE occurs for elastic
= 1, if U2 = 0, i.e., if the second particle is at rest
For, ml = m2, VI = 0, V2 = UI, i.e., the first body is stopped and the second one takes off with the velocity which the first one originally had. Both the momentum and KE of the first are completely transferred to second. For m2 » ml,VI ~ -UI and V2 = 0, i.e., when a light body collides with a much heavier body at rest, the velocity of the light body is approximately reversed and the heavier body remains approximately at rest. For m2 « ml, VI ~ UI and U2 ~ 2ul, i.e., when heavy body collide elastically which remains practically unchanged, but the light body rebounds approximately with twice the velocity of the heavy body. Deduction 10.10.1. Maximum energy transfer due to head on elastic collision : Consider a particle of mass ml which collides with another mass m2, initially at rest. For such head of collision, if the initial velocity UI of mass ml changes to the velocity VI, then for U2 = 0, we have VI
=
ml-m2 UI· ml+m2
The fractional loss of the KE of mass 6.T T
ml
for such collision is
CHAPTER 10. CENTRAL FORCE FIELD
336
where x
=
~. ml
The transfer in energy will be maximum if 1
2
'2m2V1
=0*
4x (1 + x)2
=1*
.
:1:
= 1; z.e., m1 = m2
which is the condition maximum energy transfer for head on collision.
10.11
Oblique collisions
Two smooth spheres of masses m1 and m2 moving with respective velocities U1 and at angles a1 and a2 with their line of centres impinge directly. Let e be the coefficient of restitution between them and VI, V2 be the velocities of the two spheres just after impact in the direction making angles (h and (h with the line of impact. Since the spheres are smooth, the impulsive action and reaction will be along the line of centres and, therefore, there will be no change in velocity components in the direction perpendicular to the line of impact. Resolving the velocities perpendicular to the line of impact, we get, U2
VI
sin 81 = U1 sin a1; V2 sin 82
= U2 sin a2·
Again, the velocity of approach is U1 cos a1 - U2 cos a2 and the velocity of separation is V2 cos a2 - VI cos a1. By, Newton's experimental law of impact, we have, (10.20) Again, since the impulsive action and reaction is along the line of impact, by the principle of conservation of momentum, we have, (10.21) From these equations, we get the velocities VI and V2 of the two spheres just after impact, as VI
cos 81 =
V2cOS 82=
10.11.1
U1 'U2
cos a1 (m1 - em2) + m2 u 2 cos a2(1 + e) cosa2(m2 -
m1+ m 2 em1) + m1 u 1 cosa1(1
+ e)
ml+m2
; .
Loss of KE
The loss of KE 6T, due to oblique impact is given by 6T
= ~[(mlu~ + m2u~)
-
(mlv~ + m2v~)]
= 'I[ 2 mlul2( cos 2 al +
. 2al) + m2 u 22( cos 2 a2 + sm . 2a2) sm 2 2 - {m1v~(cos2 81 + sin 81) + m2v~(cos2 (h + sin 82)}]
= '12 [m1 Ul2 cos 2 al + (
1
2 ml +m2
) [(ml
2 cos2 a2 m2 u 2
- {ml VI2 cos2 81 + m2 v 22 cos 2 82}]
+m2)(mlu~cos2a1 +m2u~cos2a2)
10.12. EXERCISES
337
- (ml +m2)(mlV~Cos2(h +m2v~cos202)]
2(
1 Tnl +m2
) [(mlUICOS a l +m2U2COSa2) 2 +mlm2(UICOSal-U2COSa2) 2
- {(ml vI cos fh ml m 2
(
2 ml + m2
10.12
+ m2v2 cos O2)2 + mlm2(V2 cos (h - VI cos 01 )2] 2
2
)(1-e )(Ulcosal-U2cosa2); O~e~1.
Exercises
Ex 10.12.1. In elliptic motion under the inverse square law, the time occupied in describing any arc depends only on the major axis, the sum of the distances from the centre of force to the initial and final points, and the length of the chord joining these points, so that if these three elements are given, the time is determinate, whatever be the form of the ellipse-prove this. This is known as Lambert's theorem. Ex 10.12.2. Describe thR nature of orbits for a particle in motion in an inverse square law force field. Ex 10.12.3. Describe the nature of orbits for a isotropic harmonic oscillator. Ex 10.12.4. The total energy expression of a body moving under the effect of a central force is given by E = ~mr2 + 2;~2 + U. Discuss the centrifugal potential energy of the body. Ex 10.12.5. A particle moving under the action a central force describes a spiral path r = e bO . Find the law of force.
-:!&
Ex 10.12.6. A particle of mass m moves under the action of a central force directed towards a fixed point. Show that the velocity of the particle at any point in its orbit is V = [b(e 2 - 1) + !~P/2, where e is the eccentricity of the orbit and L is the angular momentum of the particle.
"This page is Intentionally Left Blank"
Chapter 11
Theory of Oscillations In everyday life we come across various things that move. The motion of physical systems can be classified into two categories - translation and vibration motion. If the position of a body varies linearly with time its motion is translatory. A motion that repeats itself in equal intervals of time is called periodic motion. If a body in periodic motion moves back and forth over the same path, its motion is called vibratory or oscillatory. Oscillations are the most general form of motion of dynamic systems about the equilibrium position. For small derivations from the equilibrium position, the oscillations are harmonic and acquire a special significance. In this chapter, we shall study three types of problems : (i) First, we shall study the simplest and smoothest type of free oscillatory motion of undamped physical systems. (ii) Secondly, we consider the forced and damped oscillation. (iii) Thirdly, we shall discuss, the theory of free oscillations of small amplitude about the position of stable equilibrium that only the fundamental frequencies are excited.
11.1
Equilibrium
While dealing with oscillations executed by various systems, we shall frequently come across the condition of eqUilibrium. For any periodic motion, there exists a position of a particle in which the net force acting on the particle is zero. This is the equilibrium position.
11.1.1
Stable equilibrium
If a slight displacement of the system from its position of equilibrium results only. in small bounded motion about the point of equilibrium, then the system is said to be stable equilibrium. At the extreme point xo at which the potential energy = function V is minimum, if the particle is displaced, then the force F( x) = -
%:
339
CHAPTER 11. THEORY OF OSCILLATIONS /
340
slope of the potential energy curve will tend to return it and it will oscillate about eqUilibrium point, performing bound motion. These points are actually points of stable equilibrium. For example, the bob of a simple pendulum in its equilibrium state satisfies the stability criteria.
11.1.2
Unstable equilibrium
If a slight displacement of the system from its equilibrium results in bounded motion, then the system is said to be in an unstable equilibrium. If the particle is displaced sightly from the point of equilibrium Xo then it will be acted upon by a force F(x) = ~~ which will tend to push the particle away from equilibrium point, when released. For this reason such points are called the points of unstable equilibrium. The egg standing on end is an example of unstable eqUilibrium. Hence the equilibrium is unstable if the extreme value of the PE V is a maximum.
Ex 11.1.1. Simple pendulum: Taking the zero of the potential energy at the bottom of the swing, the potential energy of the bob of mass m of the pendulum at a position of angular deflection o is V(O) = mgl(l - cosO), where I is the length of the pendulum and 9 is the acceleration due to gravity. The equilibrium position of the pendulum is given by,
:~ = mgl sin 0 = 0 => 0 = 0, 7r. Now, ~:¥ = mgl cos 0 is positive at 0 = 0 and negative at 0 = 7r. Therefore, 0 = 0 is the position of stable equilibrium at which the b01tcan hang for an indefinite period, which is not at all possible at the position of uri"'Lable equilibrium at 0 = 7r. The pendulum can oscillate only about the position of stable equilibrium.
11.2 11.2.1
Oscillatory problems Simple pendulum
Simple pendulum is an example of a dynamical system. It consists of a heavy particle of mass m called the bob attached to one end of a weightless string of length l, which oscillates back and forth in a vertical plane, while the other end of the string is attached to a fixed point O. Taking the fixed point 0 as the origin of the co-ordinate system, let the typical position of the particle at time t be given by (x, y) or (r,O) in polar co-ordinates. Here, the particle moves under the action of two forces, namely the gravIty and the tension T of the string. The tangential and normal equations of motion are d2 s m d0 2
or,
=
v2 -mgsinO; mT
zii = -g sin 0;
v2 mT
=T
=T
- mg cos 0
- mgcosO, as s
= Ie
11.2. OSCILLATORY PROBLEMS If () is small, then sin ()
~
341
() and the first equation reduces to .. l()
..
g.
= -g() => () = -1 SIll()
which is a simple harmonic motion of period
T, = 27r/fr = 27r/f If () = a(very small) when iJ = 0, then () at any instant is given by () To solve the general solution for the motion exactly, we have
iJ.
=
a cos
j¥t ..
()
j d(();) = -y j sin()d() o
a '2
2g
=> () = T[cos() - cosal· Since () decreases with t, we have
.
. /2g
() = -Y T[cOS () =>
f29T -
YT
cosal
o
-
-
j ·Jcos() - cos a d()
~ v'2
-
a
Q
j 0
d()
·!sin2 g
V
_
2
sin2!!.
2
where T is the time to reach A. To integrate the right hand side, let us put sin ~ sin ~ sin
Jv a
1
v'2 o
=
J ".
2
1 2 sin ~ cos
g _
".
2
=
v'2j o
VI -
d
This is an elliptic integral and since sin ~ and sin
342
CHAPTER 11. THEORY OF OSCILLATIONS
Hence the time of complete oscillation is
4T = 27f
If
12 2 a 13 a -[1 + (-) sin - + (--) 2 sin4 - + .] 9 2 2 24 2' . 1 2 . 2a ,c, 2
13 2 . 4 a 2
=Tl[l+(~) sm -+(-2-4) sm -+ ...].
11.2.2
Damped harmonic oscillation
A particle moves in a straight line with an acceleration k 2 x towards a fixed origin o on the line when at a distance x from it and under a small resistance equal to J.L x velocity. Let P be the position of the particle at any time t and a the fixed origin on the line a X. The equation of motion of the particle is, .. = X
- 2J.Lx' -
k2x
or,x + 2J-Lx + k 2 x = 0
(11.1)
This equation is a linear homogeneous differential equation of the second order, with constant coefficients. Let x = eQt be a trial solution of (11.1). Then the auxiliflry equation is a 2 + 2J.La + k :=;. a
where >-.
=
=
2
=0
-J.L ± J J.L2 - k 2 = J.L
± >-.
JJ.L2 - k 2. The general solution of the equation (11.1) can be written as
(11.2) This general solution represents different physical situations depending upon relative values of J.L and k. Following are the cases:
Deduction 11.2.1. Over damped motion: Let J-L > k, i.e., p. > >-. as well. Thus, both the terms on the right hand side of equation (11.2) decay exponentially, the first one decreasing faster than the other. Let us take the initial condition x(t = 0) = Xo, x(t = 0) = vo, then from (11.2), we have
A = _ Vo + (p. - >-.)xo. B = Vo + (p. + >-.)xo . 2>-.'
2>-'
When Xo, Vo are positive, then B is positive and A is negative. Moreover IBI > IAI. Hence the displacement x, will be positive at all instants. If, on the other hand, Vo is negative such that Vo < -(p. + >-')xo, B is negative and A is positive, IAI > IBI. Since the term containing A decays more rapidly than the term containing B, the term containing B will be predominant after some time when the term containing A becomes insignificant. Thus, the positive displacement will become negative crossing the equilibrium position and once again will tend monotonically to the equilibrium position.
11.2. OSCILLATORY PROBLEMS
343
J
Deduction 11.2.2. Under damped motion: Let JL < k, then JL2 - k 2 is an imaginary quantity, say k'. Then the solution of (11.2) can be written as, x
= Ce- JLt sin(k't + c:)
(11.3)
where C and c: are constants - both real numbers. Equation (11.3) shows that the system performs sinusoidal oscillations having the period T' = ~'J = h2a n d 2 k
-Jl
amplitude Ce- JLt • Thus, we observed that: (i) period T --
T'
in the presence of a damping force is different from the natural period
27r k
(ii) the amplitude of oscillations, Le., Ce- JLt decreases exponentially with time. The ratio of the amplitude at two successive points separated by time T' is Ce~~.~t~7"') eW
'.
Quantity JLT'
=
n
k 2 _JL2
is called the logarithmic decrement of the motion.
Deduction 11.2.3. Critically damped motion:Let JL ~ k, we can treat this as a limiting case when JL - t k. In this case, the solution can be written as x
= [G + HtJe-Jlt.
It will observe that for the same initial conditions, the system will return to the equilibrium position most quickly if it is critically damped. Hte- p·t decays rapidly than the term Ge- pt . Hence, depending upon the relative ~agnitudes and signs of G and H, we get cases similar to those obtained in the case of over damped motion.
11.2.3
Forced harmonic oscillation
A particle moves in a straight line with an acceleration k 2 x towards a fixed origin o on the line when at a distance x from it and is simultaneously acted on by a periodic force F cos pt per unit mass, F being a constant. Let P be the position of the particle at any time t and 0 the fixed origin on the line OX. The equation of motion of the particle is, (11.4) where the driving force F(t) = Fo sinwt is sinusoidal and 1/ = 2~ is the frequency of the applied force. To solve the linear inhomogeneous differential equation (11.4), let us assume the solution is of the form x = A sin(wt - e), where we are to determine the values of the constants A, known as amplitude and e, known as phase. Thus from the equation (11.4), we get
- Aw 2 sin(wt - e)
* *
+ k 2A sin(wt - e) = Fo sin wt = Fo sin[(wt - 0) + eJ = Fo sin(wt - e) cos e + Fo cos(wt - e) sin e. 2 - Aw + k 2A = Fo sin e; 2JLwA = Fo sin e tane
+ 2JLAw cos(wt -
2JLw
e)
Fo
= 2 2; A = . k - w J(k 2 _ w2)2 + 4p. 2w2
CHAPTER 11. THEORY OF OSCILLATIONS
344
With the values of the constants, the solution of equation (11.4) can be written as,
This solution of the damped forced oscillator shows that the particle performs forced oscillations having amplitude and phase dependent on damping and forcing frequency w. Deduction 11.2.4. Amplitude resonance: Amplitude A of the forced oscillations varies with frequency of the applied force. There exists some frequency called resonant frequency at which the amplitude becomes maximum. At this stage maximum transfer of energy occurs from the driving force to the driven system. This is called resonance. The frequency at the time of resonance, i.e., resonant frequency will obviously correspond to the maximum value of the amplitude A. Thus the maximum value of A is given by,
The resonant frequency is
Vr
=
~;
and the maximum value of the amplitude is
This shows that the value of Amax depends upon the damping factor /1-. For undamped oscillations (/1- = 0) when forcing frequency w = k, the natural frequency, the amplitude of the oscillations becomes very large, theoretically infinite. However, due to damping, which is always present in any naturally oscillating system, the amplitude becomes large but finite. Deduction 11.2.5. Velocity resonance: The velocity of the particle performing forced oscillations is obtained by differentiating the solution w.r.t time. Thus, x
=
Faw cos(wt - 8) J(k 2 - w 2 )2 + 4j1,2W 2
---r~===i<~=='i<=;;:
= Aw cos(wt -
8); where Aw =
J(
Fa 2
2
k :W )2
+ 4/1-2
.
The velocity amplitude will be maximum, when the denominator is minimum. The only quantity in the denominator that can be changed is (k2:w2)2, as /1- =constant and w can be changed by changing frequency of the force. But, it is squared quantity and hence its minimum value must be zero. This gives w = k, i.e., v = Va. Thus velocity resonance occurs when the inducing frequency equals to the natural frequency of the oscillator. It should be noted that this frequency and the frequency of amplitude resonance given by Wr = Jk 2 - 2/1- 2 are not the same in general. The
11.3. SMALL OSCILLATION
345
two frequencies are equal for free oscillations. Further, the maximum value of Aw will be given by
Fo
[Aw]max = 2JL'
Thus [Ai.cJ]max also depends upon the damping factor and becomes infinite for free oscillations (JL = 0).
11.3
Small oscillation
Here we shall discuss the theory of small amplitudes oscillatory motion developed in a conservative system by applying the Lagrange equations of motion. When it is slightly displaced from its stable equilibrium position, although it can be applied to small oscillations about the stable motion as well. The theory of small oscillation can be physically applied in variety of fields such as molecular spectra, acoustics, vibrations of mechanics and coupled electrical circuits.
11.3.1
Generalised co-ordinates
We shall suppose that the system is defined by its KE T and its PE V, and that the position of the system is specified by the co-ordinates (qI, q2, ... , qn) independently of the time, so that T does not involve time t explicitly. We shall also suppose that no co-ordinates have been ignored. There is evidently no loss of generality, in assuming that the equilibrium configuration corresponds to zero values of the coordinates qI, q2,.··, qn; so qI, q2,···, qn; qI, q2, ... , qn are very small throughout the motion considered.
11.3.2
Kinetic and Potential energy
Here we are to derive the expressions for the kinetic and potential energy of the system with many holonomic degrees of freedom, when the system is displaced around its stable equilibrium position. We shall consider only conservative system, so the forces are derivable from potential function, which is independent of time but functions of position only. Let qI, q2, .. . qn be the set of generalised co-ordinates which represent the system after removing the time independent holonomic constraints. Let Qj be the generalised forces acting on each particle, then Qj
=
-aaqj V(qI,q2, ... qn);
j
= 1,2, ... n.
(11.5)
As the system of particles is in equilibrium, the forces acting on the system is zero. Hence in a conservative force field, generalised forces acting on each particle must vanish, i.e., (11.6)
346
CHAPTER 11. THEORY OF OSCILLATIONS
which shows that PE V has an stationary value in the equilibrium configuration of the system. Thus the equilibrium solution is given by q] = q.iO; j = 1,2, ... , n, which is the solution of (11.6). Also, if the equilibrium is stable, the potential energy is minimum at qjo. Let 1]](j = 1,2, ... , n) be the small displacements of the generalised co-ordinates to the motion, given to the system around the equilibrium points q] = q]o, then
Since qjO are fixed, we can treat 1]j as generalised co-ordinates for the system. Let us assume that during the motion of the system, the departures from the configurations of the stable equilibrium are negligibly small. Hence every dynamical quantity may be appropriately expanded in a Taylor's series about the equilibrium, retaining only first order non-vanishing terms. Here expanding the potential energy function V, we get
The term V(qjo) in the series represents the potential energy in the equilibrium position, which may be arbitrarily be chosen as reference level, with the potential energy vanish there. Using equation (11.6), the equilibrium conditions (g~ )qji=qjO = 0; j = 1,2, ... , n, we have,
where the second derivative of V have been designated by the constants Vij depending only on the equilibrium values of qj's. Here the potential energy function V as measured from the minimum value can be written with the quadratic terms as the first approximation to V. Here
V =
1
"2
LL j
VjkT/j1]k
(11. 7)
k
where the coefficients Vjk = (~:~Jqj=qJO,qk=qkO are constants (depends upon the equilibrium values qjo), characteristic of equilibrium configuration and are symmetric (Vjk = Vkj ), with respect to indices. Under a variety of circumstances, the coefficients Vjk can vanish, so that it is independent of a particular co-ordinate. Hence the equilibrium occurs at any arbitrary value of that co-ordinate. Again, it may also happen that the potential behaves like a quadratic at a point, again causing
347
11.3. SMALL OSCILLATION
one or more of the 1tjk'S vanish. In matrix notation, equation (11.7) can be written as V = ~1]TV 1], where
1]1) 1]2 .. ;
1]=.
( .
(vn V=
1]n
Using
V12 ... V1n) V21 V22 .. , V2n ....
..
.. ..
..
Vn1 Vn2 ... Vnn
1]) 's as n generalised co-ordinates, the generalised force on the lh particle is
Let us consider a time-independent holonomic system with generalised co-ordinates q1, q2, ... , qn' Now the KE expression of the system will be homogeneous quadratic function of generalised velocities. Here the KE T is generalised co-ordinates can be written in the form
(11.8) where in general, the coefficients D:ik are functions of the co-ordinates qk. We may suppose that, these functions are expanded in powers of co-ordinates, we only require the dynamical equations to be correct to the first order of small quantities, which they contain. Hence at the points qj = qjO + 1]j; (j = 1,2, ... ,n), the expression for the total KE becomes T
=
~L
L i
=~L
+ 1]i·)i/ir,k
L[D:ik(qiO)
i
=
D:ik(qiQ
k
~L
k
L i
+ L(7aD:ik )qioql + " I
D:ik (qlO, Q20,' .. qnO)r,ir,k
k
1~~'T'
..
l.T. T1]
= 2 ~ ~ .L~k1]i1]k = 21] i
.]'i]d'k
ql
(11.9)
k
where the constants are symmetric, i.e., 1':Jk = O!ik(qlO, q20,··· qnO) = Tkj, as the individual terms are uneffected by the interchange of indices. Thus the problem of small oscillation about a configuration of equilibrium depends on the solution of Lagrangian equations of motion in which the KE and PE are homogeneous quadratic forms in the velocities and co-ordinates respectively, with constant coefficients.
348
11.4
CHAPTER 11. THEORY OF OSCILLATIONS
Lagrange's Equation of Motion
The problem of oscillatory motions about a configuration of equilibrium depends on the solution of Lagrange's equations of motion in which the KE and PE are homogeneous quadratic forms in the velocities and co-ordinates respectively, with constant coefficients. Here we are to derive the Lagrange's equation of motion of a conservative system having many degrees near its equilibrium configuration. Near equilibrium configuration of the system, the Lagrangian L can be written as
L=T- V =
~L i
L[1ik7J{r7k - Vik17i17k] k
where 17k'S are taken as generalised co-ordinates. The Lagrange's equation of Illotion
!i( 8~) _ 8L = 0 =} ~ L {!i(Tik7Jk) dt 817i 8rJi 2 k dt =}
L {Tlkijk
+ Vik17k} = 0;
i
+ Vik17d = 0
= 1,2, ... , n
(11.10)
k
where explicit use has been made of the symmetry property of the coefficients Tik and Vik. These n equations represents the equations of motion near equilibrium and the solution describes the motion of the system near the equilibrium. In matrix notation, the Lagrange's equation of motion can be written as
rij + V1] = O.
(11.11)
These Lagrange's equations for small oscillations is the set of n simultaneous linear differential equation with constant coefficients that must be solved completely, containing 2n constants, to obtain the motion of the system near the equilibrium.
11.4.1
The Eigenfunctions and Eigenfrequencies
The equations of motion (11.11) are linear differential equation with constant coefficients. Since we are dealing with an oscillatory motion, the solution of (11.10) can be put in the form (11.12) where C is taken as a scale factor, common to all co-ordinates and Cak are the complex amplitudes of the oscillation corresponding to the co-ordinates 17k, w is real. The real part of RHS of equation (11.12) correspond to the actual motion and the physical motion. Frequency w is determined by the equation of motion. If w is the real quantity, then only equation (11.12) represents the oscillatory motion. Substituting the trial solution (11.12) into the equation of motion (11.10), we get
Va - w 2,ra = L(Vik - w 2 1ik)ak = 0; i = 1,2, ... ,n. k
(11.13)
11.4. LAGRANGE'S EQUATION OF MOTION
349
Equation (11.13) represents a set of n linear homogeneous, algebraic, equations in n unknowns ak. The condition that, ak has non-trivial solution is IVik - W2Tiki = 0 2 Vu - w Tu V 12 - w2T12 ... V ln - w2Tln
or,
(11.14)
=0.
This n degree equation in frequency w 2 represented by the determinant is called characteristic or secular equation of the system. Equation (11.14) may be called the period or frequency equation as it determines the periods ~:, ~: ' ... ,~: of the oscillations. Let the n roots of the equation (11.14) be wr,w~, ... ,w;, which are called eigenfrequencies of the system. Also from equation (11.14), we have w 2 = k V,kX,X > O. Since Vik and Ttk are symmetric, the eigenfrequencies are real and T ,kX,X" positive. If two or more of Wl' are real, the phenomenon is known as degeneracy. We shall confine our attention to the ca..<;e in which these roots are posItive and distinct. Taking anyone root Wl' and values of the ratios of aI, a2, . .. , an and they give al h(w
) T
=
a2 h(w
) T
= ... =
an In(w
) T
=0:1'; r=1,2, ... ,n
where h, h, ... , In denote the minors of any row or column in equation (11.14). Each root of the characteristic equation can substituted in equation (11.14) and the (n - 1) values of ak for each value of Wk can be determined or the ratios of all amplitudes ak in terms of the remaining value of ak can be found out. Since :J n values of w~, we can form n sets of the values of ak. Each of the n vales of ak corresponding to one value of w~ can be considered to define the components of n dimensional vector ak. The vector ak is called eigenvector associated with eigenfrequency Wk of the system. Property 11.4.1. Orthogonal property of Eigenvalues: Here we are to show that the eigenvectors aT are orthogonal. Let akl represents the kth component of the zth eigenvector al. Hence from equation (11.14), we get n
n
n
k=l
k=l
k=l
L Vikakl = wr L 1ik a kl = >"1 L Ttkakl
(11.15)
where the summation is carried over k without affecting the equation as Tik and Vik are symmetric. Similarly, for mth root w~, the complex conjugate of (11.15) can be written as n
n
n
k=l
1=1
i=l
L Vikaim = w~ L Tikaim = >..;;' L Tikaim.
(11.16)
350
CHAPTER 11. THEORY OF OSCILLATIONS
Multiplying (11.15) by aim and (11.16) by akl and then subtracting we get, n
(AI - A:n) L
n
n
n
~kaklaim = L L[Vikaklaim - Vikaimakz]
L
k=li=l n n
k=li=l
=> (AI - gn) L L ~kaklaim = O.
(11.17)
k=li=l
of.
If Al
A::n, then (AI - A::n) is non zero in general, and hence n
n
L
L
~kaklaim = OJ
of. m.
l
k=li=l
Now, we consider Al - A~
=
n
0, i.e., Al or n
terminate. We are to show that
n
wl are real and 2: 2: ~kaklaim is indek=ll=l
n
L: 2: ~kaklaim of. 0, for l = m.
For this purpose
k=ll=l
separate akl into real and imaginary components as
n
With this separated value ok akl the expression
n
L: 2: ~kaklarm can be written as k=ll=l
L k
LTjkakla;m = L
LTjk(Qkl
k
j
=L
+ i{3kl)(Qjm -
i{3jm)
j
LTjk [QklQjl
k
+ {3kl{3jz] + i L k
j
= LLTjk[QkIQJI + {3kl{3jz]j k
as Tjk
LTjk [{3kI Qjl - Qkl{3jz] j
= T kj .
j
2: 2: TJkaklajm > 0 and this sum vanish only if the system is not k j Using this result we also conclude that the KE T = 2: 2: TjkT]kT]j is also
This shows that, in motion.
k
j
positive and
Hence the eigenfrequencies are all real and the motion for a given w 2 will be completely o~cillatory about the position of stable equilibrium. Also, if wl of. w~, then L k
LTjkaklaJm J
= OJ
l
of. m
(11.18)
11.4. LAGRANGE'S EQUATION OF MOTION
351
wl
which is called the orthogonal property. Also when the values of are substituted in equation (11.14), we shall determine the ratios of akl. This indeterminacy can be removed by requiring the additional condition L
LTJkaklajm
k
= 1;
l
=m
(11.19)
J
to be satisfied by ak's, i.e., akl are normalized. The equation (11.18) and (11.19) can be written as the orthognormality conditions LLTjkaklajm = 81m , j
k
i.e., aFTa m = 81m ; in matrix notation, where 8kl is the Kronecker delta and al is the vector corresponding to frequency WI. The eigenvectors a r so defined are said to form an orthonormal set. They are orthogonal in the sense of the condition of equation (11.18) and they are normalised in accordance with the condition of equat,ion (11.19).
11.4.2
Normal co-ordinates, Frequencies of Vibration
We see that, the equations of motion will be satisfied by an oscillatory solution of the form (11.12) not merely for one frequency but in general for a set of n frequencies Wk. So a complete solution of the equations of motion therefore involves a superposition of oscillations with all the allowed frequencies. Thus if the system is displaced slightly from equilibrium and released the system performs small oscillations about the equilibrium with the frequencies WI, W2, ... , W n . The solutions of the characteristic equation (11.14) are therefore often designed as the frequencies of the vibration or resonant frequencies of the system. The normal co-ordinates of a system are the parameters in terms of which the equations of motion of the system can be written each equation contains only one dependent variable. If a solution of equations of motion involve only one single frequency, the co-ordinate appearing in it will be called the normal co-ordinate. Thus, the generalised co-ordinates, each one of them executing oscillations of one single frequency are called normal frequency. The general solution of the equations of motion for T/j may now be written as n
T/j
=L
Ckajke-iwkt; i
= V-I
k=1
where for each resonant frequency Wk, there being a complex scale factor Ck' The solution for each co-ordinate is, in general, a sum of simple harmonic oscillations in all the frequencies Wk satisfying the characteristic equation. Unless it happens that all of the frequencies Wk are commensurable, i.e., rational fractions of each other, T/J" never repeats its initial value and is therefore not itself a periodic function of time. From algebra, "when two homogeneous quadratic functions of arbitrary number of variables, one of which is positive for all values of the variables, then by a real
352
CHAPTER 11. THEORY OF OSCILLATIONS
transformation we may clear both expressions of the terms containing products of the variables and at the same time change the coefficients of the square terms in one of the expressions so that each is equal to unity." However it is always possible to find a linear transform of the co-ordinates 1]j to a new set of generalised co-ordinates, that are all simple periodic functions of time, i.e., the KE and PE are expressed only the new generalised co-ordinates in simple way. Let us now define the new set of co-.g.rdinates (J related to the original co-ordinates TIJ by linear relations as
where alj's are constants. From this equation, it is observed that (k = Cke-iwkt are the quantities that would undergo oscillations with only use of frequency Wk. Then the KE reduces to the simpler form in the new co-ordinates as
The PE reduces to the simpler form in the new co-ordinates as V =
1
2L j
=
1
2L r
=
L ~k1]j17k
1
=
2L
k
J
L V)k [L ajr(r] [L aks(s] k
s
r
L[L L V)ka.7raks](r(s s
k
j
~ LLw;[LLTjkajraks](r(s = ~ LW;(;' r
s
k
J
r
It should be remembered that the off-diagonal elements of T and V rise to the coupling between the particles. Thus, when the normal co-ordinates are used, the KE and PE are become simultaneously diagonal and hence this will uncouple the co-ordinates and the problem will become completely separable into the independent motions of the normal co-ordinates. In terms of the new variables, the Lagrangian L can be written as
The Lagrange's equation of motion gives
L[e + W~(k] = 0 k
which are a set of n independent equations. The solution of the single equation is (k = Cke-iwkt. Thus, each of the new co-ordinates is thus a simply periodic function
11.4. LAGRANGE'S EQUATION OF MOTION
353
involving only one of the resonant frequencies. These quantity ('s are called the normal co-ordinates or principal co-ordinates of the vibrating system. Each normal co-ordinate corresponds to a vibration of the system with only one frequency, and these component oscillations are spoken of as the normal modes of vibration, Le., the system is vibrating in a normal mode. Further physical properties of all normal modes that when the system is vibrating in a normal mode is simply periodic, passing twice through the equilibrium position in every complete oscillation, and the velocity of every particle of the system vanishes at the same instant twice in every complete oscillation. All of the particles in each mode vibrate with the same frequency and with the same phase, the relative amplitudes being determined by the matrix elements ajk. The complete motion is then build up out of the sum of the normal modes weighted with amplitude and phase factors contained in the Ck'S. Ex 11.4.1. Two Coupled Pendulums Here we shall discuss the motion of two oscillating systems in which the motion of one system can influence that of the other. The two systems are then said to be
o 0' --~~--------------------~------~y
"--____...+-____--'l B X2
mg
Figure 11.1: Two coupled pendulums coupled to each other and the resulting oscillations are called coupled oscillations. The system consists of two identical simple pendulums, each having length 1, a bob of mass m coupled to each other by means of a massless horizontal spring of force constant /-t. The separation between the two bobs is such that the spring is relaxed in the equilibrium position. The system is disturbed slightly from the equilibrium position as shown in the figure and relaxed. At any time t, let Xl, X2 represent the horizontal displacements and YI, Y2 be the vertical displacements of the two bobs from their equilibrium position, such that they move in the same vertical plane. We shall consider, the small horizontal oscillation about the position of equilibrium. The spring is stretched or compressed depending on whether Xl > X2 or Xl < X2. Let us choose (h and e2 as generalised co-ordinates, we get from figure, Xl
YI
= 1sin el; X2 = 1sin e2 = 1(1 - cos el ); Y2 = l(1
- cos e2 ).
CHAPTER 11. THEORY OF OSCILLATIONS
354
The kinetic energy T of the two bobs can be written as
. = e. T Te· ,
T
(ml2 0 ) 0 ml 2
=
((h)
·e = (h' ' X2-Xl
I
The potential energy due to extension of spring is
J-lxdx
=
~(X2 - Xl)2 and so
o the PE V of the two bobs measured from their equilibrium position is V
1
= mg(Yl + Y2) + "2J-l(x2 = mgl[(l - COSOl)
- Xl)
+ (1 -
2
cos ( 2)]
+ ~J-ll2(sin02 -
sinOl)2
oi ( 8~ 1 1 2( . . 2 2 -···)+ 1-1+ 2 -",) +"2J-ll sm82-sm8l) 1 2 2 1 2 2 ~ "2 mgl (8 l + ( 2 ) + "2J-ll (8 2 - ( 1 ) [
=mg l(1-1+
=
~(mgl + J-ll2) (0; + O~) -
= ~(Ol' (2) (mgl + 2J-ll2 -J-ll
2
2 J-ll 0l02
(81)
2 -J-ll 2 ) mgl + J-ll
=
82
eTve.
Now the characteristic equation for determination of normal frequency of vibration IV - w 2 TI = 0, which gives
w 2 is
1
=> mgl + /1,[2 => (mg[ WI
=
- w
- J-ll2
+ J-l[2 -
2 2 ml mgl
w2m[2)2
1-
- J-ll2 0 - w 2ml2 -
+ J-ll2
= J-l2[4
If, JT+ ~ Jwi + ~, W2
=
=
are the normal frequencies of vibration. Let the normal co-ordinates 1]1,7}2 corresponding to WI, W2 respectively, then the equations of motion in terms of normal co-ordinates as
WI, W2
The angular frequency WI of the oscillation is the same as that of either pendulum oscillating in isolation. This case arise as both pendulums are always in phase and the spring is at its natural length throughout the motion. N ow we shall proceed
11.4. LAGRANGE'S EQUATION OF MOTION
355
to express 1]1,1]2 in terms of generalised co-ordinates{h, rh. Obviously, we are to calculate the eigenvectors aI, a2 corresponding to the normal frequencies WI, w2 of vibration. (i) For WI = let al = (an, a21f, then
j¥,
J
=?
(!D2 _:z;2) (:~~) = (~) =?
an = a21 = a.
Using the normalization condition, we get
JEt + ~
=
(ii) For the angular frequency W2 (aI2' a22f, then
=
(0)
-a22
= (3.
-JLl~ -JLl~) (a I2 ) ( -JLl -JLl a22 =?
U~ing the normalization, we get (3
=
a12
=
=
vwt
+~ >
WI, let a2
0
lvkn. Hence, the normal co-ordinates
1]l, 1]2
corresponds to WI, W2 respectively are given by
~*r ) ((hfh) = (l1m v'2m v'2m 1
=? 1]1
1
(1]1) 1]2
= lJ2mUh + O2); 1]2 = lJ2m(01
- 02).
The two pendulums are oscillating harmonically at the same frequency W2 and are always out of phase. The spring is either extended or compressed during the motion and the coupling is effective. Also, it is clear that the two vectors are orthogonal to each other. The co-ordinates 771 and 7]2 are the desired co-ordinates and they are referred to as the normal co-ordinates of the system because equations of motion in them contain a single frequency of normal modes of oscillation. Now, we shall interpret modes of vibration. (i) To discuss the mode associated with the normal co-ordinate 1]1 and hence with WI, we put 7]2 equal to zero. Thus, 01 = O2 , which means two pendulum oscillate in phase, the frequency of motion is WI = Obviously, the spring constant JL is not occurring in the expression of WI, the two pendulum can vibrate as if they are independent. Hence there is no scratching or compressing of spring
j¥.
CHAPTER 11. THEORY OF OSCILLATIONS
356
at any time during the motion justifying the inphase condition. The normal mode of oscillation is called a symmetric mode and the particles are oscillating always in phase. (ii) To discuss the normal mode associated with the normal co-ordinate TJ2 and hence with W2, we put TJI equal to zero. Now, fh = -(h, so they are out of phase which is not hard to understand, because the frequency associated with
V
this mode W2 = 7+ ~ involves spring constant /1. The spring is compressed and elongated, and exerts a restoring force. Hence restoring force due to spring will come into action. These oscillations are called mode of oscillations since the particles always move out of phase with frequency W2. We see that, the normal co-ordinates T}1 and TJ2 are independent of each other, as the system can oscillate with one frequency only while the other is suppressed.
Ex 11.4.2. Linear Triatomic Molecule: Here we obtain the resonant frequencies and normal co-ordinates of vibration of a linear symmetrical triatomic molecule, and in particular C02-molecule. The configuration of such molecule is given in the following figure: All three atoms are
m M m o~----~~--~Or--~~~----o
Figure 11.2: Linear Triatomic Molecule on one straight line. The oxygen atoms are co.nnected to the carbon atom by springs. For simplicity we shall consider only vibration along the line of the molecule, and the actual complicated interatomic potential will be approximated by two springs of spring constant /1 joining the three atoms. Let TJl, TJ2, TJ3 be the displacements of Oxygen, Carbon, Oxygen atoms from their equilibrium positions. Let m be the mass of Oxygen and A1 be the mass of Carbon atoms. Let the atoms interact harmonically, then the potential energy as a function of co-ordinates TJI, TJ2, TJ3 is V =? 2V
=
1
2,/1(TJ2 - TJI)
2
1
+ 2,/1(TJ3 -
= /1[TJ~ + 2rJ~ + TJ5 = (T}1, TJ2, TJ3)
TJ2)
2
- 2TJ1TJ2 - 2TJ2TJ3J
(~/1 ;: _0/1) (~~) = TJTVTJ· ° -/1 /1 TJ3
Now the KE of such linear triatomic molecule is T =?
2T
1 [.2 TJI
= 2,
+ TJ3.2] + 2,1 M TJ3.2
= (1h, il2, il3)
(~ ~ ~ )
°°
m
(ill) .
~~
·T
= TJ
.
TTJ·
11.4. LAGRANGE'S EQUATION OF MOTION
357
Now the characteristic equation for determination of frequency w2 is i.e.,
IV -
w 2 rl
= 0,
J.L - w2m J.L 0 J.L 2J.L-Mw 2 J.L =0 2 o J.L J.L - mw 2 2 w [J.L - mw ][J.L{ M + 2m) - M mw 2] = 0 WI WI
= 0,W2 =
(Ii
/ J.L
2m
V-;;;"w3 = VmY + M]'
= 0 refers to translatory motion but not oscillatory motion of the atoms and the
rest two to the oscillatory motion. Now we are to find the eigenvectors aI, a2, a3 corresponding to WI, W2, W3. (i) For WI = 0, let al = (an, a21. a31)T, then the components aij are determined for WI = 0 by the equations,
j
=}
~J.L) (:~~) (~)
(!J.L ; : = o -J.L /1, a31 0 =} an = a21 = a31 = a(say).
This of course, is exactly what was expected from the translational nature of the motion. In this situation, the displacements of all the atoms are in the same direction
and in equal amount as shown in the figure. Using the normalization condition to fixes the values of aij, we get, 2
m{an
2 2 + a31) + M a21 =
1 =} a
=
1 J2m+M
~:===::;=;;
(ii) For W2 = ~, let a2 = (aI2' a22, a32)T, then L(\tij - w~Tij)aj
= Q; i = 1,2,3
j
=}
(~J.L 2J.L __J.L!:f!: ~J.L ) ( :~~) = ( ~ ) o
=}
-J.L
0
a22 = 0; al2
a32
0
= -a32 = (3(say).
In this situation, the central atom does not take part in motion and the end atoms oscillate with actual amplitude but opposite in phase as shown in the figure. Using
CHAPTER 11. THEORY OF OSCILLATIONS
358
the normalization, we get the numerical value of the quantities as 2
m(a12
2 2 + a32) + Ma22 = 1 =? f3 =
1 y2m
!7'L:'
In this case, the centre of the atom is at rest, while the two outer ones vibrate exactly out of phase, because they must in order to converse linear momentum. (iii) For W3
= ...j~[1 + ~],
let a3
L)~'J
= (a13,a23,a33)T,
-
w~1ij)aj
then
= Qi i = 1,2,3
J
=?
(-2o -
-~1
';lJ.L J.L - J.L m -J.L
=? a13
-
0
fJ
)
-2!!Jjf
= a33 = ')'i
(a13) a23 a33
a23
rrL
= -2 M
(0) 0 0
')'(say).
This shows that, the end terms vibrate in phase with equal amplitude whereas the ~
,,-
Figure 11.5:
central atom vibrates in opposite phase with a different amplitude as shown in the figure. Any general longitudinal vibration of the molecule that does not involve a rigid transformation will be some linear combination of the normal modes W2 and W3. The amplitudes of the normal modes and their phases relative to each other will be determined by the initial conditions. Using the normalization, we get
Since in the third case, the molecule is asymmetrically stretched, oscillating dipole moment associated with the motion and corresponding absorption based will appear in the infarct region. Hence the normal co-ordinates (1, (2, (3 associated with normal frequencies WI, W2, W3 respectively as 1]1) 1]2 ( 1]3
=
M ]1/2) 1 1 [2m(M+2m) v'2m+M v'2m 1 0 [ M ]1/2 v'2m+M - 2m(M+2m) ( 1 1 [ M ]1/2 v'2m+M - v'2m 2m(M+2m)
Hence the relations complete the discussions of the linear triatomic molecule. Ex 11.4.3. Oscillation of a Particle on a String
11.4. LAGRANGE'S EQUATION OF MOTION
359
Figure 11.6: Oscillation of a Particle on a String
Here we consider a light string of length L = (n + 1)1 with n equal masses m placed along it at regular equal spacing interval 1. The string is stretched to a force F. We shall consider small transverse or longitudinal oscillations of the particles. Let the transverse displacements be Xi from the equilibrium positions of the ith particle, where they are taken as generalised co-ordinates. Thus the KE of the system of particles is
The equilibrium length betweeni h and U+1)th particles is l. But when the particles are displaced, the changed length between lh and U+ l)th particles is 01 and hence
Now, F is the tension, hence in the change ol = (xJ~Xj? the work dOlle = F.ol. Since the displacement Xj are transverse, the PE comes from the tension of the string. Hence the PE is V
= F.ol =
F
2l [x~
+ (X2
- Xl)2
+ ... + (xn -
F n+l
= 2l
L:)Xj-l - Xj)2;
XO
= Xn+1 = o.
j=l
Thus, the Lagrangian L of the system of particle is
X n _l)2
+ x~l
360
CHAPTER 11. THEORY OF OSCILLATIONS
and the Lagrange's equation of motion for Xj are
d aL dt ( ax) J
=>
aL
- axJ_=
mx-J -
=> Xj
0
F(x_ l J- 1
2 = Wo(Xj-l
-
2x-J
- 2Xj
+ x-+!) = J
0
+ Xj+!) = OJ
2
Wo
F = -. ml
Thus the equation of motion for the particle clearly shows that it depends on displacements Xj-l and Xj+! of its neighboring particles. We seek solution
Xj
= ajeiwtj
i
= F1
then we get the following set of secular equations (2W5 - w2)al - W5a2 = 0
-W5al -W5an-l
+ (2w5 + (2w5 -
w2)a2 - W5a3 w2)an = O.
The condition that an's have solution for w2 is 2W5 - w2 -w5 o· .. -Wo2 2Wo2 - W2 -W 2 '"
=0
o o
O
o o =0
o
O· .. 2w5 - w2 -w5 0 -w5 2w5 - w2
0 0
o
where Dn is the determinant of order n. Expanding the determinant with respect to the first row, we get Dn - (2w5 - w2)Dn_l + w6Dn-2 = 0 2
=> D~ - (2 - W2)D~_1 + D~-2 = OJ Dn = w5nD~.
Wo To solve the linear homogeneous difference equation, put
D~
w2 ,sin{n + 1)0 - (2 - '2) sin nO:+ sin(n - 1)0
Wo
=>
;
= sin(n + 1)0, then
=0
w2 Wo
2 - '2 = 2cosO
=> w2 = w5(2 - 2 cos 0)
= 4w5 sin2 ~
=> Dn = W5 n sin(n + 1)0. Now the zeros of Dn are given by sin{n + 1)0
= 0 => 0 = ~j n+l
k
= 1,2, ... , n.
Hence the eigenvalues, i.e., the normal modes are given by w~ = 4w5 sin2 2(!~1); k = 1,_2, ... ,no
11.4. LAGRANGE'S EQUATION OF MOTION
11.4.3
361
Oscillation under constraint
Here we shall consider the effect produced on the periods of normal vibration of a dynamical system about a configuration of stable equilibrium when the number of degrees of freedom of the system is diminished by the i~troduction of an additional constraint. Suppose that, the original system is specified in terms of its normal co-ordinates (1]1, 1]2,. " ,1]71.), so that the KE and PE of the constraint system have the form _
T -
[.2.2 .2]. 21 1]1 + 1]2 + ... + 1]71.'
_ 1[ 2 2
V -
2 2 2 2J. 2 A11]1 + A21]2 + ... + An1]n ,
and the additional constraint be expressed by the equation
Since 1]1, 1]2, ... ,1]71. are small, we can expand the function 1 in ascending powers of 1]1, 1]2,··.,1]71. and retain only the first terms of the expansion. As the relation 1(1]1, 1]2, .. . ,1]71.) = 0 is to be satisfied in the position of equilibrium in which the co-ordinates vanish, this constraint may be expanded and can be express by the equation, A 1'1]1
+ A 21]2 + ... + An1]n =
0
where AI, A 2, ... , An are constants. As the equilibrium configuration is supposed to be compatible with the constraint, there will be no constant term. By means of this equation, we can eliminate qn, we get T
1 [1]1' 2 2 + ... + 1]n-1 ·2 1 (A' ' )2] =2 +' 1]2 + A2 11]1 + A' 21]2 + ... + A n-11]n-1 71.
V
=
~[Ai1]~ + A~1]~ + '"
+ A~-l1]~-l + ~~ (A11]1 + A21]2 + '" + An-11]n-1)2J. 71.
The Lagrange's equation of the constrained system are therefore the (n-l) equations .
!!.. ( aT ) + 8V + p 81 dt 81]7'
.. or, 1]r
81]r
\2 1 (A 11]1 .. + ... + A ) + .l\r1]r + A r [ A2 n-1 7" J71.-1
71.
+ or, i]r where, p
= 0
81]r
A2 A~ (A11]1 n
+ A21]2 + ... + An- 11]n-dJ =
+ A;1]r + Ji,Ar = 0; =
~271. (A1ih + '"
iin A~1]n =-An-A ' n
r
= 1,2, ... , n -
0
1
+ An- 1i]n-1) + ~~ (A11]1 + '" + An- 11]n-1) n
CHAPTER 11. THEORY OF OSCILLATIONS
362
Hence the equations of motion of the constraint system can be written in the form of the n equations as ilr
+ )..~TJr + J.LAr = 0;
T
= 1,2, ... , n
where J.L is undetermined. Now, consider a normal mode of vibration of the modified system, defined by the equations TJr = {r cos )..t, for T = 1, 2, ... ,n; and J.L = v cos )..t. Substituting in the equations of motion, we have
Substituting the values of Qr'S given by these equations in the equation Al Ql ... AnQn = 0, we get an equation in )..2 as
This is of degree n - 1 in
)..2,
+
and by writing it in the form
we see that if, )..1 < )..2 < '" < )..n, F()..I) is positive, F()..2) is negative and so on. So this equation in )..2 has (n - 1) roots, which from the form of the equation are evidently interspaced between the quantities )..i, )..~, ... , )..;, i.e., the periods under constraint separate the free periods. The quantities corresponding to these roots are the periods of the normal modes of vibration of the constrained system, and it therefore follows that the (n - 1) periods of normal vibration of the constraint system are spaced between the n periods of the original system.
2;
11.4.4
Stationary property of the normal modes
Let frictionless constraints be imposed until the system retains only one degree of freedom. Here we shall consider the effect of adding constraints to a dynamical system to such an extend that only one degree of freedom is left to the system. Let (TJl, TJ2, ... , TJn) be the normal co-ordinates of the original system. Since the constraints may be represented by linear equations between the normal co-ordinates, so they may put in the form
where Tl, T2, . .. , Tn are constants and ( is the new variable which may be taken as defining the configuration of the constrained system at time t. So the KE and PE of the original system have the form
11.4. LAGRANGE'S EQUATION OF MOTION
363
27r are I·t speno . d s 0 f normal·b A1' 27r A2 ' ... , An VI rat·Ion. Then t h e KE an d PE 0 f t h e so 27r constraint system have the form
T
=
1[ 2 2" rl
Hence the equation
+ ... + r n2] <:·2 ;
V
ft(%f) + ~~ =
= 2"1 [>'.2I rl2 + A22r 22 + ... + A2nr n2],...2 ~ .
0 has a solution <:
period of a vibration of the constraint system is A2 =
\ 2, 2 AIrl
=
Asin(At
+ c).
Thus the
2;, where A is given by the equation
+ A2 \ 2 r 22 + .... + Anrn \2 2 17I + ... + 'Ij,~
This gives the square of the frequency of the constrained vibrations and it clearly lies between the greatest and least of the quantities Ai, A~, . .. , A~, which are proportional to the squares of the frequencies in normal modes. If the constraints are varied, this expression has a stationary value when (n - 1) of the quantities J-LI, J-L2,···, J-Ln are zero and the stationary value is one of the quantities Ai, A~, ... , A~. Thus, when constraints are put on the system so as to reduce its number of degrees of freedom to unity, the period of the constrained system has a stationary value for those constraints which make the vibration to be a normal vibration of the unconstrained system.
Ex 11.4.4. Consider the system of three degrees of freedom for which 2T =
±2 +
y2 + i 2 , 2V = p2(x 2 + y2) + 2az(x + y) + q2 z 2; a being small in comparison with p, q, and find the principal co-ordinates and periods. Prove that if such a system be let go from rest with y = 0, z = 0 initially, the vibration in x will have temporarily ceased after a time p7r(p~2a2) and there will then be vibration of the same amplitude in y as the original one was in x.
SOLUTION: Since the KE is in the normalized form but the potential energy is not the normalized form, so, let us take a transformation x + y = 2~ and x - 11 = 21], for which the given KE and PE becomes T
= ~[(~ + 'Ij)2 + (~_ 'Ij)2 + i 2] = ~2 + iP + ~i2 2
1
V = 2"[P2{(~ + 1])2
+ (~ -
1])2}
+ 4az~ + q2 z 2]
1
= p2(e + r?) + 2az~ + _q2 z2. 2
The Lagrange's
1]
equation is given by, !!.-(8T) _ aT = _ 8V dt 8'1j 81] 81] =? ij = _p21].
CHAPTER 11. THEORY OF OSCILLATIONS
364
2;.
Therefore, the variable 'fJ is a normal co-ordinate and the time period is To reduce the remaining terms in the kinetic and potential energies to sum of squares, we write, 2a
a
z = (- q2 -p2
.
a·
2
1·
.2
2a·
= [
2a 2 ·2 (q2 _ p2)2l
.2
1
2
2a·2
+ 'fJ + 2[1 + (q2 _ p2)2K
and the term of PE becomes,
where we are to neglect the last term as a is small as compared with p and q. The La.grange's
Therefore, the variable
11.4. LAGRANGE'S EQUATION OF MOTION
365
Therefore, the variable, is a normal co-ordinate and the time period is W2 211". Thus the variables ¢, 'T},' are therefore the normal co-ordinates. Let us suppose that initially, we hav~
x
= k; y = 0; z = 0, x = 0; iJ = 0; z = 0,
and suppose that k is so small that its product with other small quantities can be neglected. Then to this degree of approximation, we have initially 'T} = ~k, ¢ = ~k, ( = O. The vibrations of the normal co-ordinates 'T} is therefore given by the equation 'T} = ~k cos pt and the vibrations of the normal co-ordinates ¢ is
¢
=
1 2kcos[
2
p -
a 2 (2q2_4p2) (qLp2)2
20:2] t
1 + (qLp2)2 a 2t
1
= 2kcosptcos p(q2 _
P q
P
a 2t
1
p2)
02
1
= 2kcos[pt{1 - 2( 2 _ 2) }],
+ 2ksinptsin p(q2 _
p2)'
The motion can therefore be approximately represented initially by 1 1 .
'T}
= 2kcospt; ¢ = 2kcospt;2.e.,x = kcospt,y =
O.
After an interval of time P7r(P~20:2) , the motion is approximately represented by 'T}
1 1 .
= -kcospt; ¢ = --kCospt,2.e.,x = O,y = -kcospt 2
2
.
which establishes the result stated.
Ex 11.4.5. A ring slides on a smooth circular hoop of equal mass and of radius a which can turn in a veritcal plane about a fixed point in its circumference. Prove that the periods of small oscillations are 27r
Iii
and 27r ~.
SOLUTION: Let 0, (j> be the inclinations to the vertical of the radius through the fixed point and the radius through the ring. Then the KE and PE are given by
2T
= ma2[3()2 + 20Jy + Jy2]; 2V = mga[20 2 + ¢2].
Here we use the stationary property of the normal modes. Suppose that, we impose a constraint represented by the relation ¢ = rO. Then the KE and PE becomes
2T = ma2[3 + 21' + r2]()2; 2V = mga[2 + r2]02. The Lagrange's equation of motion for 0 gives d 8T dt ( => ma 2[3
av
ao )+ BO = 0
+ 21' + 1'2]8 + mga(1 + r2)O = .. 9 2 + r2 =>0+0=0 a 3 + 21' + ,. 2
0
CHAPTER 11. THEORY OF OSCILLATIONS
366
which represents oscillations with period value of the function p2 are given by dp2 = 0
dr
2
=}
PI
Thus the periods of oscillations are
2;, where p2 = ~ 3';t~r2'
=} 7'
= -2 1
2g
2
'
= -; P2 = a 27r Pl
The stationary
9
-2 . a
= 27f VG. a and 2 2g
27r P2
= 27f V!2Q. 9
Ex 11.4.6. A light string stretched to tension P has its ends fixed at a distance 91 apart. Masses 2m, m are attached to the string at distances l,51 respectively from one of its fixed ends. Find the normal modes of transverse vibration, and show that
the periods are
47f~, 47f1fj.
Let ql, q2 denote the small displacements of the particles of masses 2m, m at time t, gravity is neglected. We may neglect vibrations in the tension due to the slight extension of the string as they would only introduce terms of higher order in ql, q2 than those which we retain. Thus the KE of the system is given by SOLUTION:
m
2m
Figure 11.7
2T
·2 = 2mql·2 + mq2'
Let QI: Q2 denote forces which would keep the particles at rest in the displaced position, then
The infinitesimal increment in PE, negative of which will be equal to work done for infinitesimal displacements in the masses, can be written as
11.4. LAGRANGE'S EQUATION OF MOTION
367
Also for such conservative system the PE can be obtained as
The Lagrange's equation are then
These equations might have been written down at once as the equations of motion of the two particles. To solve them, we put, ql = Ml cos(pt + c), q2 = M2 cos(pt + c) and we get p
_M:p2
+ Blm (5Ml
- M2)
and - M2p2
+ B,m P (4M2
- 2Ml)
==>
_p2 + 5A 2A
2 =
==> p
z
A _p2
+ 4A;
=0 =0 P
A=
SZ;;;
2 9A ± VBIA - 4.1BA2 = 6A 3A = 3P 3P 2 ' 4 l m ' Blm'
Thus the periods of vibrations are Tl =
;7 = 41r/fji and T2 =
;~
= 41rJf/s.
Ex 11.4.7. A light string of length 3l is stretched horizontally between two fixed points, gravity is neglected. Masses 15m, 7m are attached to the points of trisection. The tension in equilibrium is Aml. The particle of mass 15m is drawn aside a distance a, the other remaining undisplaced, and both are simultaneously released. Prove that in the subsequent motion the displacement of the particle of mass 7m is 15 [!3i. 26 a cos 35 t -
V
cos
Vr;"3 t 1.
Let ql, q2 denote the small displacements of the particles of masses 15m,7m at time t respectively. The KE of the system is given by
SOLUTION:
2T = 15mq~
+ 7mq~.
Let Ql: Q2 denote forces which would keep the particles at rest in the displaced
7m 15 m
Figure 11.8
368
CHAPTER 11. THEORY OF OSCILLATIONS
position, then ql Ql = >..ml [T
+ ql -l
q2 - ql Q2 = >..ml [ l
q2]
=
q2] +T =
(
>..m 2ql - q2
)
) >..m{2q2 - ql .
For such conservative system the PE is given by
The Lagrange's equation are then
These equations might have been written down at once as the equations of motion of the two particles. To solve them, we put, ql = MICOS{pt+e),Q2 = M2COS{pt+e) and we get
=:}
{2>.. - 15p2)Ml - 7M2 = 0; ->..M1 + {2>.. -7p2)M2 = 0 (2X - 15p2)(2).. - 7p2) - >..2 = 0 3>" >..
=:}
p~
= -35
and p~
= -. 3
Substituting the two values of p2 in succession gives two sets of corresponding values for !vII, 1\12 which may be write, Mu M21 M12 M22 -7- = -5- = 01; -1- = -3 = 02·
Thus the complete solution of the differential equations in
Ql,
q2 can be written as
ql = 701 COS(plt + e1) + 02 COS(P2t + e2) q2 = 501 COS{p1t + e1) - 302 COS{p2t + e2) where aI, 0'2, e1, e2 are arbitrary constants, to be determined by using initial conditions. The initial conditions are q1 = a, q2 = 0, til = 0, ti2 = o. The vanishing of the velocities shows that e1 = 0 = C2, then we have a
= 701 =:}
+ 02; 0 = 501 - 302 3 5 01 = 26 a ; 02 = 26 a .
Hence the displacement of the second particle of mass 7m is
11.4. LAGRANGE'S EQUATION OF MOTION
369
Ex 11.4.8. A uniform bar of length 2a is hung from a fixed point by a string of length b fastened to one end of the bar. Show that, when the system makes small normal oscillations in a vertical plane, the length l of the equivalent simple pendulum is the root of the quadratic [2 - (1a + b)l + ~ab = O. Let AB be the string of length b and Be be the rod of length 2a, where centre of gravity at G(x, y). The string AB makes an angle 0 with the vertical and the rod Be makes an angle ¢ with the vertical. Then x = bcosO + acos¢, y = bsin 0 + a sin q;. The KE of the rod Be is given by SOLUTION:
T=
~m[x2 + 1?1 + ~Iw2 2 2 1
.
.
2
•
• 2
1
(2a?'2
+ asin¢¢) + (bcosOO + acos¢¢) 1+ 2'm 12 ¢
= 2'm[(bsinOO
= ~m[b2e2 + ~a2¢2 + 2abe¢cos(0 - ¢)l 1
4
2 '2
.. + 2abO1;cos(0 -
= 2'm[b 0 + 3a The PE of the rod is V equation gives
2'2
d aT aT av dt ( ae ) - ao = - ao
'*
'* be + a¢ =
-gsinO be + a¢ + gO = 0; sin 0 ~ 0, for small oscillation.
the Lagrange's ¢ equation gives d aT aT av dt ( a¢) - a¢ = - a 4.. .. 3a¢ + bO + g¢ = O.
'*
To solve them, we put, 0 = Ml cos(pt + c:), ¢ = M2 cos(pt + c:) and we get 4 (_bp2 + g)Ml - ap2 M2 = 0; and (-3ap2 + g)M2 - bp2 Ml = 0
+9 '* _bp2 -bp2 -
'* (g -
4
ap2 - ap2 + 9
----,.--.0..._ _
1
bp2)(g - 3 ap2 ) = -abp4.
The two pendulums are equivalent if they have the same time periods. The time period of simple pendulum is 27r above equation becomes,
If, so that 27rIf = 2;, i.e., p2 = 7' Hence the
bg 4 ag g2 (g - -)(g - --) = -abI 3 l 241
'* [ - (3a + b)[ + 3ab =
I2
O.
CHAPTER 11. THEORY OF OSCILLATIONS
370
Ex 11.4.9. Two 'C-qual uniform rods AB, Be, each of length I, smoothly jointed together at B, are suspended from A and collide in a vertical plane through A. Show that the periods of normal oscillations are where n 2 = (3 ±
2:,
J.r)ff.
SOLUTION: Let G(x, y) and GI(XI, YI) be the centre of gravities of the two rods AB and Be respectively, which makes an angle and
e
. e. x = -I cos e Y = -I SIn 2 ' 2 ' Xl
= I cos () + ~ cos
The KE of the rods AB and Be are
1 2 TI = 2"[x
ml 2 • 1= 12,wI = ()
1
+ I?] + 2"Iw?;
= ~mI2lj2. 6
1'-
2
1 [.2
= 2"
X
1I
.2]
+ Y + 2"
I
2 w2;
ml
2
= 12' W2 = 'f'1.
= ~mI2[02 + ~c,&2 + Oc,&] 2
where we use, cos((} -
~
3
1, for small oscillation. Thus the total KE of the system
T=
~mI2[i02 + ~c,&2 + Oc,&l. 2
3
3
The PE of the system is given by
3
1
V = -mgx - mgxl = -2"mglcos(} - 2"mgl cos
..
3g
....
3g
"3(} +
= Ml cos(nt + e:),
-in2 + c 3n2 :::} n
11.5
2
n2 -2n2 +c 6 c [3 ± y'71"3 -~-;
=
3g
c=-
I
6
9
= (3 ± y'7)Z'
Exercises
Ex 11.5.1. Explain a method of determining the periods of small oscillations of a dynamical systems of n d.f. when the geometrical equations do not contain time explicitly.
11.5. EXERCISES
371
Ex 11.5.2. Discuss the small oscillations about the state of steady motion of a symmetrical top. Ex 11.5.3. Two uniform rods of the same mass and length 2a, freely jointed at a common extremity rest upon two smooth pegs, which are in the same horizontal plane so that each rod is inclined at the same angle 0: to the vertical. Show that, the time of oscillations, when the joint moves in a vertical straight line through the
centre of the line joining the pegs is 27r!ii
1+~o~o~2 ex
Ex 11.5.4. A ring slides on a smooth circular hoop of equal mass and a radius a where can turn in a vertical plane about a fixed point 0 in its circumference. If () and (jJ be the inclinations to the vertical of the radius through 0 and of the radius through the ring, show 2() + (jJ and ¢ - () are normal co-ordinates and the periods of normal oscillations are 27r and 27r ~.
jig
Ex 11.5.5. A light string of length 3l is stretched under the tension P between two fixed points, gravity is neglected. Masses 5m,8m are attached to the points of trisection. The whole is at rest until a small transverse velocity u is suddenly given to the particle of mass 5m. Prove that in the subsequent motion the displacement of
the particle of the other particle is
1~~ [ yfii- sin
Vio:t - yI2 sin ~~ l, where
0:
2
= ::'Z.
Ex 11.5.6. One point of a uniform circular hoop of mass M and radius a fixed, and the hoop is free to move in a vertical plane through the fixed point. A bead of mass m slides on the hoop, and there is no friction. Prove that, the position of stable
equilibrium are 27r ~, 27r of vibrations.
Jl\f'!g ~. Find the normal co-ordinates and normal mode
Ex 11.5.7. A smooth circular wire, of mass 8m and radius a, swings in a vertical plane, being suspended by an inextensible string of length a attached to one point of it. A particle of mass m can slide on the wire. Show that, the angle between the radius to the bead and the diameter through the point of suspension is a normal co-ordinate. Prove that, the periods of the normal oscillations are
27r ~, 27r
/ii, 27r /fi and find the other normal co-ordinates.
Ex 11.5.8. A light string of length 3l is suspended from one end and has particles of masses 6m, 2m, m attached to it at distances l, 2l, 3l from this end. Find the normal mode of small oscillations of the system in a vertical plane. Ex 11.5.9. A uniform thin hollow cylinder of mass 2m and radius 2a rolls on a horizontal plane. Inside it rolls a second uniform thin hollow cylinder of mass m and radius a loaded at one point with a particle of mass 2m. All the surfaces are perfectly rough. Find the normal co-ordinates.
"This page is Intentionally Left Blank"
Chapter 12
Relativistic Mechanics The motion of body, involves in position as a function of time. According to Einstein, all motions are relative. Therefore, one can specify the position of one body only relative to another. Accordingly, it is necessary to fix a frame of the reference with respect to which the measurements should be made. When a body occupies the same position, with respect to a frame of reference, it is said to be in motion. In this chapter, we shall discuss of brief discussion of the theory of relativity. In applied Mathematics, Physics various physical quantities are measured. The relativity theory reveals that the measurements depend upon the state of motion of the observer as well as upon the quantities that are being measured. The idea of relativity incorporated into mechanics gives rise to relativistic mechanics - in which we come across some peculiar phenomena, particularly when the particles forming a system are moving with high velocity comparable to that of light. The theory of relativity enables us to understand the high energy phenomena in the microscopic as well as macroscopic world.
12.1
Inertial frame of reference
The absolute space is an imaginary frame work, in which bodies move, but it itself is immovable and is same everywhere. A frame of reference, which is at absolute inertial frame of reference. An inertial frame of reference is one, in which the law of inertia i.e., Newton's laws of motion holds. Newton assumed that 'absolute motion is the translation of a body from one absolute space to another absolute space. Also a translatory motion can be detected only in the form of a motion relative to another material bodies.' Since the motion involves the passage of time, so a single time scale would be valid everywhere.
12.2
Newtonian relativity
According to Newton, the motion of frame of reference moving with a uniform velocity along a straight line cannot be detected by performing some physical experiment
373
CHAPTER 12. RELATIVISTIC MECHANICS
374
in the frame itself. He formulated it as the principle, known as principle of relativity, which states that "If a frame of reference is moving with a uniform velocity along a straight line with respect to some other frame, then by performing physical experiments entirely in that inertial frame, we can not determines it motion with respect to the other inertial frame". Thus (i) The uniform motion of a frame of reference cannot be detected by an observer situated within the same frame. (ii) The motion of a frame of reference can be detected and measured by an observer of some other frame but only relativity and not absolutely. The kind of relativity mentioned in the above is called the Newtonian Relativity.
12.3
Galilean transformation
Galilean transformation is the transformation, which relates the two inertial frames. Let us consider two inertial frames of reference 8 and 8' having their respective axes y
y'
s·
v
s
p
~
o
o·
x'
z· ________________________ _
x
z Figure 12.1 parallel. For simplicity, let the inertial frame 8' be moving with a relative constant velocity v along the x-direction relative to the frame 8. Let us suppose from the instant (t = t' = 0), the origin 0 and 0' of the frames 8 and 8' coincide with each other. Suppose that as measured by the observer of the frame 8, a particle which is moving in space, is at a point P at any time t, whose co-ordinates are (x, y, z) with respect to the origin of the frame 8. Let the observer of frames 8' measures the co-ordinates of P as (x', y', z') at any time t. The reference frame 8 is conditionally called stationary, while the other 8' is called moving. We introduce cartesian coordinate system in each reference. According to the Newton's concept of time the time flows uniformly without any reference to anything external, i.e., there is no effect of the motion of the frame 8' on the flow of time. Therefore
t' = t.
12.4.
GALILEAN INVARIANCE
375
Since frame 8' is moving with a constant velocity v along x-axis, the displacement of the origin of frame 8' with respect to 8,
00'
= vt
and x'
= x - vt.
Since there is no motion of the frame 8' along y and z axes, we have y'
=y
and z'
= z.
Thus the co-ordinates are related to each other by the transformation equations
x' = x - vt; and y' = y; z' or , -:;7' = -:;7 - 1/ t·, t' = t.
= z; t' = t
(12.1)
These set of equations (12.1) are called Galilean Transformation from stationary frame 8 to moving from 8'. The transformations assume that at each instant of time, the same relation exists between the co-ordinates and the time of the systems (x, y, z) and (x', y' , z'), which would exist between them, if systems were at rest relative to each other. The more general Galilean transformation between two frames respectively uniform translatory motion relative to each other is given by,
x' = x - 'Vxt; y' = Y - Vyt; z' = z - 'Vzt; t' = t
(12.2)
where V x , v y , V z are the components of velocity of the primed frame relative to the unprimed frame. Now, as 1/ is a constant we have, ~,
T
=
~
T
-> -
v and
~,~
T
=
(12.3)
T.
From the equations (12.3) we see that the velocity of a particle is different in the two different systems but the acceleration will have the same form in these two frames. Hence, the second law of motion is said to be Galilean invariant. To the observer of frame 8', the frame 8 appears to move with the same speed, but along negative x-axis i.e., with velocity -v. Thus we have the following set of equations x'
= x + vt;
and y'
= y; z' =
z; t'
=t
(12.4)
These set of equations (12.4) are called inverse Galilean Transformation.
12.4
Galilean invariance
The numerical values of the various geometrical and physical quantities generally change as a result of co-ordinate transformations. If the measurement of a physical quantity does not undergo a change under Galilean Transformation or its measurement is independent of relative motion of the observer, it is called a Galilean Invariance. They are of prime importance to physics.
376
12.4.1
CHAPTER 12. RELATIVISTIC MECHANICS
Invariance of space
Let us consider, two inertial frames of reference 8 and 8', where the frame 8' is moving with a constant velocity v along x-direction. Let a rod AB be placed in the frame S' parallel to the direction of x-axis. For the observer of the frame 5', the rod AB is at rest. Let the distances of the two ends A and B of the rod from the origin of frame 5' be x~ and x~ respectively. Then, length of the rod as measured by the observer of the frame 5' is
where the suffix '0' tells that the observer of the frame 5' measures the length of the rod at rest. For the observer of frame 5, the rod is in motion. Hence to measure the length correctly, the observer of the frame 5 measure the distance Xl and X2 of the ends A and B of the rod from the origin simultaneously. From Galilean transformations, we have,
=}
.T~ =
.T1 -
Lo =
(X2 -
x~ =
vt;
vt) -
X2 -
(Xl -
vt
vt) =
X2 -
Xl
=L
as
where L is the length of the rod measured by the observer of the frame 5. Therefore, when the length is measured in two inertial frames moving with uniform velocity relative to each other, the length interval remains unchanged.
12.4.2
Invariance of Time interval
Suppose that an event, say a flash of light, occurs in frame 5' at time ti and then again at time t~. Then, as observed by an observer of the frame 5', the time interval between the t.wo flashes is given by, 'To
= t'2 - t ,l ·
Here, the suffix zero tells that the event of flashing of light occurs at rest with respect to the observer of the frame 8'. Suppose that the observer of frame 5 records the time of occurrence of the two flashes as tl and t2' According to Galilean transformations, t~ = tl and t~ = t2' Thus,
where 1" = t2 - tl is the time interval between the flashes as recorded by the observer of the frame S. Thus, TO = T. Hence, the time interval does not change, when it is measured in two inertial frames moving with uniform velocity relative to each other.
12.4.3
Invariance of Velocity
Here we are to discuss how the velocity of a particle in one inertial frame of reference are measured by the observer of another frame moving with a uniform velocity
12.4.
377
GALILEAN INVARIANCE
relative to each other. The Galilean Transformation from frame 8 to 8' are given by
x' = x - vt, y' = y, z' = z, t' = t dx' dx dy' dy dz' dz => dt = dt - v; dt' = dt; dt' = dt ,
I
=> U x = U x -- v; u y => 11' = 11 - V'
,
= uy; U z =
(12.5)
Uz
where 11 = (u x , u y , u z ) is the velocity of the particle with respect to the observer of the frame 8 gnd 11' = (u~, u~, u~) is of the frame 8'. The set of transformation equations (12.5) is the classical velocity addition formula in Newtonian Mechanics. Also from equation (12.5), we conclude that velocity of a particle is not a Galilean invariant. Also, the inverse Galilean velocity transformations are
12.4.4
Invariance of acceleration
Differentiating the transformation set of equations (12.5), w.r.t time t, we get du~
dt =>
dux du~
=
du~
dt; dt dux du~ dt ' dt'
=
du y d11.~ du z dt ; dt = dt du y du~ dt ' dt'
, du z ast=t dt .
-=-'- =-'-= -
dt'
Since the components of the acceleration of the particle measured in the two inertial frames are independent of the uniform relative velocity of the two frames, acceleration of the particle is invariant under Galilean transformation.
12.4.5
Invariance of Newton's Law
In fact, Newton's first law of motion, whether a given frame of reference is an inertial frame of reference or not. Since Newton's first law of motion always holds good in an inertial frame of reference, it will also hold in two inertial frames of reference moving with uniform velocity relative to each other. We know, that the measurement of acceleration of a particle is independent of the uniform relative motion of the two inertial frames, i.e., 71' = 71. Let m be the mass of the particle. In Newtonian mechanics, mass is a constant quantity and is independent of the motion of the observer. Multiplying both sides of the above equation with m, we have, ~,
ma ~
~
~
~
= ma =>
~,
F
=
~
F,
~
where F' = m a' and F = m a. Thus, the observers of the two inertial frames of reference moving with uniform velocity with respect to each other measure the force on the particle to be the same. Therefore, equations of motion (F = m 71) of a ~
CHAPTER 12. RELATIVISTIC MECHANICS
378
particle is invariant under Galilean transformation. In other words, Newton's second law of motion is Galilean invariant. Since measurement of force is not affected by the uniform relative motion between the two inertial frames of reference, Newton's third law of motion F 12 = - F 21 must also hold in the two inertial frames moving with uniform velocity relative to each other. Hence Newton's laws of motion are invariant under Galilean transformations. ~
12.4.6
~
Conservation law of linear momentum
Consider two particles of masses m1 and m2 moving with velocities 111 and 112 respectively in an inertial frame 8 at rest. Suppose that, the two particles collide and after the collision, they move with velocities 1/ 1 and 1/2 respectively. According to the law of conservation of linear momentum in frame 8, we have,
Let us now view this collision between the two particles from another inertial frame of reference 8' moving with a uniform velocity 1/ with respect to the frame 8. Suppose that the observer of frame 8', the velocities of the two particles before and after collision appear as 11~, 11~ and 1/~, 1/~ respectively. According to inverse Galilean transformations for velocities,
Thus the law of conservation becomes, m1(U~
+ 11) + m2(11~ + 1/) =? 'rn 1 'U
= ml(1/~
+ 1/) + m2(11~ + 1/)
1 + m2 u 2 = ml v 1 + m2 v 2·
~,
~,
~,
~,
It is the law of conservation of linear momentum in frame 8'. It has the same form in
the frames 8 and 8' moving with a uniform velocity relative to each other. Hence, the law of conservation of linear momentum is invariant under Galilean transformation.
12.4.7
Conservation law of Energy
Consider two particles of masses m1 and m2 moving with velocities 111 and 112 respectively in an inertial frame 8 at rest. Suppose that, the two particles collide and after the collision, they move with velocities 1/1 and 1/2 respectively. If F is the amount of energy that appears as heat or in some other form, then according to the law of conservation of energy in frame 8,
1 ~2 1 ~2 1 ~2 1 ~2 2m1 u 1 + 2m2 u 2 = 2m1 v 1 + 2m2 v 2 + F. Let us now view this collision between the two particles from another inertial frame of reference 8' moving with a uniform velocity 1/ with respect to the frame 8. Suppose that the observer of frame 8', the velocities of the two particles before
12.4.
GALILEAN INVARIANCE
379
and after collision appear as 1ti,.1t~ and 1ii, 1i~ respectively. According to inverse Galilean transformations for velocities, ........ ........, ................ ........ , ................ ........ , ................ ........, ........ Ul=ul+v; U2=u2+ v ; Vl=vl+v; V2=v2+ v . According to the law of conservation of linear momentum in frame 8, we have, ........
, + m2 ........,2 = ml ........v ,1 + m2 ........v ,2
ml U 1
=? [mll1i
U
+ m21t~].11 =
[mll1i
+ m211~].11
Also, the law of conservation of energy becomes,
1
2m1
(........ , U1
) ( ........ , ........ ) 1 ( ........ , ........ ) ( ........ , ........ ) + ........ V . U 1 + V + 2m2 '!L 2 + V . U 2 + v
= 1 =? [2m1
........ '2 U1
1
2ml
( ........ , ........ ) ( ........ , v 1 + v . V1
) 1 ( ........ , ........ ) ( ........ , ........ ) F + ........ V + 2m2 v 2 + V . V 2 + V +
1 ........ '2] [ . . . . , ........ '] . . . . + 2m2 U 2 + ml U 1 + m2 U 2 . V 1 ........ '2 1 ........ '2] [ ........ , . . . . ']........ F = [2m1 v 1 + 2m2 v 2 + ml v 1 + m2 v 2 . V +
1
=? 2m1
........ '2 U1
1 ........12 + 2m2 U2 =
1
2m1
........ '2 v1
1 ........ '2 + 2m2 v 2 + F.
It is the law of conservation of energy in frame 8'. It has the same form in the frame 8 and 8' moving with a uniform velocity relative to each other. Hence, the law of conservation of energy is invariant under Galilean transformation. 82~
82~
a2~
1 a2~
Ex 12.4.1. Show that, the electromagnetic wave equation ~ + 8? + Fz"2" - ? ift"T is not invariant under Galilean transformation, where 1> is scalar potential. SOLUTION:
=0
The Galilean transformation from unprimed to primed frame to refer-
ence are
x' = x - vt; and y' = y; z' = z; t' = t 8x' 8y' 8z' 8t' 8y' 8x' 8z' 8t' = 1 and = - = - =0· - = 1 and = - = - =0
fu
~
Bz =
fu
fu
fu
'~
~
~
~
~
1 and 8z = 8z = 8z = 0;
~
~
8t = -v;
~
~
at = at = 0;
~
~
8t = 1.
Here the electromagnetic wave equation in frame S is 2
2
2
2
8 1> 8 1> 8 1> -2- -c12-88t-1>2 0 2 8y2+ 8z 8x+ - . In frame 8', the scalar potential 1> is function of x', y', z' and t', i.e., 1> = 1>( x', y', Thus,
z', t').
380
CHAPTER 12. RELATIVISTIC MECHANICS
a2 dJ For this, Now, we calculate iJi'T'
Thus, the electromagnetic wave equation in frame 8' becomes,
a2¢ a2¢ v 2 a2¢ 1 a2¢ 2v a2¢ ax,2 ay,2 3 z'2 c ax'2 c at'2 c ax' at' v 2 a2¢ a2¢ a2¢ 1· a2¢ 2v a2¢ =} [1- - j - + - + - - - - + - - - =0 c2 ax'2 ay,2 az'2 c2 at'2 c2 ax' at' . fJ2¢
-+-+---- - -2+ - -2 - = 0 2
Thus, the two equations do not have the same form in the two inertial frames of reference moving with uniform velocity v relative to each other. Hence, the electromagnetic wave equation is not invariant under Galilean transformations.
12.5
Special Theory of Relativity
In 1905, the special theory of relativity which deals with the problems involving the inertial frames of reference is developed by Einstein. He realized that a relativity principle existed both for mechanics and electromagnetism. This theory is the outcome of the analysis of the physical consequences that would result if no uniform frame of reference is in existence. The special theory of relativity is based on the following two postulates: (i) All the basic laws of Physics which can be expressed in the form of equations have the same form in all the inertial frames of reference moving with an uniform velocity with respect to one another. (ii) The measured speed of light in free space has always the same value c for all the observers irrespective of the relative motion of the source of light and the observer. The first postulate bears the fact that, there is no existence of an universal frame. If we reformulate the different laws of Physics for observers in a relative constant motion with respect to one another, we obtain other set of transformations in place of Galilean Transformation, under which the laws would be invariant. If there is no difference, then this difference enable us to determine which objects are 'stationary' in space and which objects are 'moving'. Practically, an universal frame of reference is not available and we are not able such differences in physical laws. The second postulate follows directly from the negative results of MichelsonMorley experiment, i.e., the speed of light in free space is constant irrespective of the motion of the source or the observer. Since the speed of the light does not depend upon the motion of the earth, there is no path difference and hence no
12.6. LORENTZ TRANSFORMATION
381
fringe shift is expected. Further, the velocity of light is not invariant under Galilean Transformations, they need to be discarded in view of the second postulates also. These postulates of the special theory of relativity do not appear so radical at first sight, but lead to very revolutionary ideas. The concepts of the invariance of mass, length interval and time interval and their mutual independence are neither invariant nor independent of one another in special theory of relativity.
12.6
Lorentz Transformation
In 1904, H.A. Lorentz derived a set of transformations for space and time coordinates of an event in two inertial frames moving with a uniform velocity relative to each other. However, Einstein obtained these transformations quite independently on the basis on the two postulates of the special theory of relativity. These equations are known as Lorentz-Einstein transformations or simply Lorentz transformations. This tra'1sformation should confirm to the following general requirements: (i) The transformation must be linear, i.e., any single event in one inertial frame must transform to a single event in another frame, with a single set of coordinates. (ii) The postulate of the equivalence of all interval frames requires that the direct and inverse transformations should be symmetrical with respect to each other. (iii) In case, there are no relative motion between the two frames of reference, Lorentz transformations should be reduce to identity transformations. (iv) In case, the velocity of moving frame is very small as compared to the velocity of light, Lorentz transformation should reduce to Galilean transformations. (v) The relativistic law of addition of velocities as derived from the Lorentz transformations should leave the velocity of light as invariant. Let us consider two inertial frames of reference Sand S' having their respective axes parallel to each other. The frame S' is moving with a relative constant velocity v along x-direction. According to Galilean transformation, the velocity transformations are
which violate the postulates of the special theory of relativity. The first postulates of the special theory of relativity implies that a uniform linear motion is one frame of reference must appear as the same in the other inertial frame also and likewise a linear relationship among x, y, z and t must go over a linear relationship among x' , y', Zl and t'. Let us assume that x and x' may be connected by the relation
x'
= k(x -
vt)
(12.6)
382
CHAPTER 12. RELATIVISTIC MECHANICS
where k is a constant of proportionality, which does not depend upon x, t but may • be a function of v. When k = 1 and v «< c, then x' = x - vt which is known to be correct in Newtonian Mechanics. Further as the motion of the frame S' is only along x-direction, we have y'
=y
and z'
= z.
(12.7)
In the special theory of relativity, space and time are independent and they form a space time continumn. Therefore time co-ordinates t and t' may be assumed to be connected by the relation t'
= pt + qx
(12.8)
where p and q are some constants. To determine the constants k, p, q we use the second postulate of special theory of relativity. Let at t = t' = 0, the origins 0 and 0' of the frames coincide with each other and at that instant, a light signal is emitted in the space from the origin 0 of the frame S. With the passage of time, this light signal will grow into a.spherical wavefront of ever increasing radius for the observers of the two frames, with the origin of that frame as the centre of the spherical wavefront. Let the velocity of light in frame S is c and that in S' is c' = c - v. If according to the observer of the frame S, the spherical wavefront passes through the point P(x,]}, z) in space at time t, then (12.9) Let the point P in the frame S be observed as P'(x', y', z') at a time t' in frame S'. Since according to second postulate of the special theory of relativity, the speed of the light is same in all the inertial frames, i.e., x,2 + y,2 + z,2 = (ct')2 2 2 =} k (x - vt)2 + y2 + z2 = c (pt + qx)2 2 2 2 2 =} (k _ c q2)x + y2 + z2 _ 2(vk2 + c pq)xt
= (C 2p2 -
k 2v 2 )t2.
Since (12.1O)is identical with (12.9), we get k 2 _ c2q2 = 1; vk 2 + c2pq = 0; c2p2 _ k 2v 2 = c2 2 2 2 =} v[l + c q J + c pq = 0; _v 2[c 2q2 + 1J + c2p2 = c2 1- p2 =} vpq + p2 = 1 =} q = - vp 1p2 2 2 =} vk + c p. _ _ = 0 vp 2 2 c c2 vp 2 =} P = c2 _ v 2 and so k = c2 _ v 2 ; q = - c2 v/c2 q = - 731=_=(;=v'if'2/;=c2irr)
-V
(12.10)
12.6. LORENTZ TRANSFORMATION
383
Hence, putting these values in (12.6) to (12.8), we obtain,
x' =
x - vt . y' VI - (v 2Jc2) '
y' z'
=,
z' t'
=,
2
=
t - (vx) Jc (v 2Jc 2)
Vl-
(12.11)
These set of equations are called Lorentz transformations. From this set of equations, one can obtain x, y, z and t in terms of x', y', z' and t' as
x
=
x' + vt' , -y'r=I=_=(~v:;<=2/~c2""");' Y = y;
,
z
= z;
t
=
t'
+ (vx')/c 2
VI _ (v2 / c2)
(12.12)
which is called inverse Lorentz transformations. Thus, the measurements of position and time are found to depend upon the frame of reference of the observer. It is possible to get (12.11) from (12.12) without using relativity principle. For this purpose, we must consider (12.11) as a system of equations in unprimed quantities and then solve the system. As a result, we obtain the expression (12.12). Following are some important discussions of the Lorentz transformations. (i) The direct and inverse Lorentz transformations are symmetrical with respect to each other.
(ii) When v ---+ 0, the Lorentz transformation relations (12.11) reduces to x' = x, y' = y, z' = z and t' = t, which are termed as identity transformations. Thus, in the limit, when v ---+ 0, the Lorentz transformation reduces to identity transformations. 2
•
(iii ) When v < < c, the factor ~ can be neglected as compared to 1 and ~ III comparison to t. So, when v < < c, the, the Lorentz transformation (12.11) reduces to x' = x - vt, ; y' = y, z' = z and t' = t, which is the Galilean transformations. Thus the Lorentz transformation equations reduce to the Galilean transformations when when relative velocity v is very small in comparison with velocity c of light.
(iv) The velocity of light is same in all the inertial frames, regardless of the relative motion between the frames. (v) Note that Lorentz transformation put an upper limit on the velocity, a moving frame of reference (or an object) can possess. It follows that, in case v = c, or v > c, the Lorentz transformation gives unphysical results. Hence the relative velocity of an inertial frame with respect to another must always be less than the velocity of light. (vi) Lorentz transformations are merely the orthogonal transformations of four dimensional space, which is known as world space or Minkowski's space. Hence the peculiarities of the special theory of relativity will be observable only when velocity v is practically comparable with the velocity of light. Now, we shall discuss some of the most important consequences of the Lorent~ transformations.
CHAPTER 12. RELATIVISTIC MECHANICS
384
Ex 12.6.1. An event occurs in frame 8 at x = 50km, y = lOkm and z = 2km at t = 1 X 1O- 6 s. If frame 8' moves with velocity 0.94c along the common xx' axis, the origins coinciding at t = t' = 0, find co-ordinates x', y', z' and t' of this event in 8'. SOLUTION:
The transformation equation, from 8 to 8' is given by the Lorentz
formula as, x'
=
x - vt
VI - (v
2/
, c2); Y
For the given data, we have,x c = 3 x 108 m / s. Thus,
=
= y;
VI -
12.6.1
=
= lOkm; z' t - (vx)/c 2
VI -
(v 2 /
=
VI _ (v2 / c2) . X
10-6 s and
VI - (0.94)2
50 - 0.94 x 3 x 10(0.94)2
t'
t
50km, y = 10km, z = 2km at t = 1
VI - (v 2 / c2)
y' = Y
z = z;
50 - 0.94c x 1 x 10- 6
x - vt
x'
, t - (vx) / c2
,
c2)
=
=
2
=
146.47km.
= 2km. 10- 6 - 0.94
Z
c
X
50
VI - (0.94)2
= 0.2885
x 10- 4 sec.
Relativity of simultaneity
The problem concerning the relativity and simultaneity and the physical meaning of Lorentz's transformations was solved by Einstein. We known that space and time are relative in character. One of the most important consequences of this character is that simultaneity is relative. Here we are to find out whether the two events occurring at two different points in some inertial frame of reference at the same time are simultaneous or not to the observer of another inertial frame of reference moving relative to the first. Let us suppose that two events occur at two distinct locations (Xl, YI, Zl, tl) and (X2' Y2, Z2, t2) in the frame of reference 8. Now, the frame 8', whose respective axes parallel to each other, is moving with a constant velocity v along x-direction with respect to the frame 8. Let those two events occur at the locations (x~, y~, z~, t~) and (x~, y~, z~, t~) in the frame of reference 8'. Using Lorentz transformation (12.11),
If the two events are simultaneous in the moving frame 8', we have, ti
= t~
and so
(12.13)
12.6. LORENTZ TRANSFORMATION
385
Since the two events occur simultaneously in moving frame S' at two different points, i.e., xl =f. x~, it follows that tl =f. t2. Thus two distinct events observed as simultaneous in frame 8' will not appear as simultaneous in frame 8. The concept of simultaneity is therefore not absolute in the sense that it depends on the co-ordinate system. Now we have the following discussions: (i) If x; > xl, it follows that t2 > tI, i.e., the first event precedes the second in the :.lnprimed frame. It implies that the event, which occurs at a distance point in moving frame will appear to occur later in the frame 8. (ii) If the two events occur in the frame 8' at the same point, i.e., xl = x~, then tl = t2. It implies that the two events will be simultaneous in frame 8 also, if they occur simultaneously at the same point in the frame 8'. On the other hand, if it is less than zero, then the events are observed in reverse order in the two frames. This will occur if 2
x; - xl > :"'-(t2 'V
tl) < C(t2 - tl).
This corresponds to the case of the two events happening at such a distance apart that a ray of light leaving the first event could not have reached to the place of second event in time to cause that event. Since the simultaneity of events is relative, physical theories that involve simultaneity in events at different positions must be modified. As no signal can travel faster than light, so no signal can bridge the interval between the two events separated by the relation (12.13). Consequently, one event can not have knowledge of the prior occurrence of the other.
12.6.2
Length contraction
Suppose that a rod AB is placed along the axis of x' in the moving frame of reference 8', whose respective axes are parallel to the reference frame 8 and which is moving with a constant velocity 11 along X-direction relative to frame 8. A frame in which the object AB is at rest is known as proper frame and the length of the rod in such a frame is called the proper length. Let xl and x; be the distances of the ends A and B of the rod from the origin 0' of the moving reference frame 8', then if Lo be the proper length,
To an observer of frame 8, the rod AB is moving with a constant velocity v and its length L (say) as measured by the observer offrame 8 is called non-proper or moving length of the rod. The process of measuring the length of moving object requires that the distances of the two ends from the origin must be measured simultaneously. Thus, if at any instant t, the distances of the ends A and B of the rod AB from the origin 0 of the frame 8 are Xl and X2 respectively, then L
=
X2 -
Xl·
386
CHAPTER 12. RELATIVISTIC MECHANICS
Now, let us find length L of the rod in system 8 relative to which the rod is in motion with velocity v. The direct Lorentz Transformation equations give
Here t2 and tl are the times at which the end co-ordinates X2 and Xl of the rod are measured. Since the measured should be simultaneous in frame 8 for determining the length of the rod, we have,
If follows that L < La, i.e., the length of the rod in motion with respect to an observer appears to the observer to be shorter than when it is at rest with respect to him/her, and the length appears to contract by a factor v 2 / c2 . Hence, it is clear that, the length of an object is a maximum in the frame of reference in which it is stationary. This phenomenon is known as Lorentz-Fitzgerald contraction. Now we have the following discussions ;
VI -
(i) The length contraction effect appears to take place only, when the relative velocity of the frame 8' with respect to the frame 8 is comparable to the velocity of light. In case v < < C, we have
Also, the effect of length contraction become significant only when the velocity of the objects approaches the velocity of light. Thus if, v ~ 0.ge, then L = 0.4359Lo and the effect occurs only in the direction of the relative motion. (ii) The length contraction effect is reciprocal in nature. It is because, if the rod is placed along the x- axis of the stationary frame 8, then to the observer of the frame 8', it will appear to be moving with velocity -v. Now,
so L remains unchanged when v is replaced by -v. So, to the observer of frame 8' also, the rod will appear to be contracted by the .same factor. (iii) The contraction effect does not appear in the dimensions of a moving object, which are transverse to the direction of motion, as y' = y and z' = z. Therefore, if a square lamina is placed in XY on ZX plane, it would appear
12.6. LORENTZ TRANSFORMATION
3S7
rectangular. On the other hand, a cube placed with its edges parallel to the coordinate axes, would appear as a rectangular parallelepiped. Generally, to an observer, if an object having volume V is moving with a uniform velocity v with respect to the observer, then its ,apparent volume will be V' = V v2 /c 2 .
VI -
(iv) In special theory of relativity, the length of an object has no absolute meaning. We may talk of absolute length of an object only in approximation, where the velocity of the moving object is negligibly small as compared to the velocity of light. (v) The length contraction effect is only an apparent effect. No real contraction takes place in the dimensions of the object due the motion of the object. Ex 12.6.2. A rod has a length of SOcm. When it is moving with a speed of 0.75c along tlu: diTectwn of its length, find the length of the 1'Od with 1'espect to an observe',. at Test.
SOLUTION: Here we use Lorentz-Fitzgerald contraction formula, L = LOV1 - v2 /c 2 , where c is the velocity of light = 3 x 108 m/ s. From the given data, we have, L = SOcm, v = 0.75c and so,
so = Lo[l -
(0.75)2]1/2
=}
Lo
= SO
x [1 - (0.75)2r1/2
= 52.915cm.
To the observer moving with the rod, the rod is at rest. So in this frame the rod will appear to be of length SOcm. Ex 12.6.3. The length of the space vehicle is 100m on the ground. When it is in flight, it:; length as obseTved on the g1'Ound is 99m. Calculate its speed.
SOLUTION: Here we use Lorentz-Fitzgerald contraction formula, L = LOV1 - v2 /c 2 , where c is the velocity of light 3 x 108 m/ s. From the given data, we have, L = 99cm, Lo = 100cm and so, _ _ v 2 1/2 99 - 100[1 c2 ] =}
=}
(3
_ v2 _ 99 2 _ ~ 108)2 - 1 (100) - (100)2
X
J199 v = -- x 3 100
X
108 = 42.32 x 106 m/s.
Ex 12.6.4. A rod of length 5m lies in xy- plane of a stationary frame 8 and makes an angle 60 0 with x axis. Find its length and inclination with x direction in other frame 8' moving with a relative velocity of 0, .Sc along x direction.
SOLUTION: Consider the two frames 8 and 8', where the frame 8' is moving with a constant velocity v = O.Sc along x direction. Let a rod of length L = 5m is placed in xy plane of the frame 8, such that it makes an angle 600 with x direction. If Lx and Ly are components of the length of the rod along x and y axis respectively, then Lx
= 5 cos 600 = 2.5m;
Ly
= 5 sin 600 = 4.33m.
388
CHAPTER 12. RELATIVISTIC MECHANICS
Suppose that, to the observer of the frame S', the x component of the length of the rod appears as L~ and the y component of the length of the rod appears as L~. Since motion of the frame is along x direction only, we have,
L~ = Lx[l -
2 V2l1 /
c
2 = 2.5J1 - (0.8)21.5m; L~ = Ly = 4.33m.
Hence, the length of the rod L' as it appears to the observer of frame S' is given by,
L' =
L'}
+ L~2 =
J(1.5)2 + (4.33)2 = 4.58m.
If the rod approaches to make an angle () with the x direction, then
L~ 4.33 tan () = -L' = = 2.8867 x 1.50
0
=} ()
= 70.893 .
Ex 12.6.5. A cube of rest volume L~ is moving with a speed v parallel to one of its edges. When observed relativistically from the laboratory frame, show that it has a volume L8 [1 - ~ ]1/2.
SOLUTION: according to length contraction, the edge of the cube placed along the direction of motion will contract from its original length Lo to Lo [1 - ~ ]1/2. the other two edges of the cube, which are at right angles to the direction of motion, will not undergo any change. Thus,the observed volume of the cube is 2
= Lo[l - ~ll/2 2 c
12.6.3
2
X
Lo x Lo = L8[1 - ~F/2. 2 c
Time dilation
Let us suppose that a light signal is emitted in space from P(x', 0, 0) in frame S' (which is moving with a constant velocity v along x-direction with respect to the frame S) ;;tt time t~ and again at time t~ from the same point. Thus the proper time TO is the time interval between the two light signals and so
Suppose that tl and t2 are two instants recorded by the observer in the frame S, then the non-proper time T is the time interval between the two signals and is,
The observer in frame S, however measures these instants given by the Lorentz transformation equations give
12.6. LORENTZ TRANSFORMATION
389
Since }1 - v 2 / c2 is a fraction, it follows that 7 > 70, i.e., the time interval between two light signals as recorded by the clock of the frame 8' will be smaller than the interval in the laboratory frame. Since the event of emitting of light signal takes place in frame 8' and it appears to the observer of frame 8 in motion, the time interval between the two light signals is recorded by the clock of frame 8 in motion. In other words, to an observer in motion relative to the clock, the time intervals appears to the lengthened. The phenomenon of kinematical effect of relativity of time is called time dilation. Now we observe the following phenomena: (i) The time dilation effect is expected only, when the relative velocity of frame S' is comparable to the velocity of light. Therefore, in case v < < c, i.e., ~ < < 1, then 7 = 70. (ii) The opinion of the observer will also reciprocal in nature, like length contraction effect, 70
Hence to the observer of 8', the clock of frame 8 will appear to run slower. (iii) Time does not run backward for any observer. The sequence of the events in a series of events is never altered for any observer. Since the velocity available for communication is always less than or equal to c, the intervals of the time between any two events may be different. (iv) No observer can see an event before it takes place, i.e., there is no way of probing into the future.
Ex 12.6.6. The mean life of time of a j.l-meson, when it is at rest is 2.2 x 10- 6 . Calculate the average distance, it will travel in vacuum before decay, if its velocity is 0.9c. SOLUTION: From the given data, we have, 70 = 2.2 x 10-6 , v = 0.9c. Thus, the mean life of a j.l-meson in the laboratory frame is given by TO
7
=
-Vri==_=v:;;=2/=:=c~2
2.2 x 10-
6
---r=7=~
}1 - (0.9)2
=
5
.05 x
10-6
s.
Therefore, average distance travelled by the j.l-meson before decay is given by, L=
VT
= 0.9
= 0.9c x 5.05 x
10-6
x 3 x 108 x 5.05 x 10- 6
= 1.36km.
Ex 12.6.7. A clock keeps correct time. With what speed should it be moved relative to an observer, so that it may seem to lose Imin. in 24 hrs.?
CHAPTER 12. RELATIVISTIC MECHANICS
390
SOLUTION: From the given data, we have, TO = 24hrs. = 24 x 60min. = 1440min. Let v be the speed of the clock relative to the observer, when it will lose 1min. in 24 hrs., i.e., T = (1440 + 1) = 1441min. Now, using the Lorentz transformation · equat lOn, T = . / Tg 2 2' we get , v I-v Ie 2
1440 =? 1 _ v = (1440)2 = 0.99861 1)2/c2 c2 1441
1441 =
J1 -
=? (~)2 = 1- 0.99861 = 0.00139 =? v = 0.03728c. c
Ex 12.6.8. With what velocity should a rocket move, so that every year spent on it corresponds to 4 years on earth? SOLUTION: From the given data, we have, TO = 1year; T Lorentz transformation equation, T = JI":-~2/e2' we get,
4= 11
=? ~ =
= 4years. Now, using the
J1- v2 /c 2
1)2 1 15 = ?2- = 1 - - = c 16 16
JI5 4 =? v =
0.9682c.
1
Ex 12.6.9. A clock in a spaceship emits signals at intervals of 3s as observed by an astronaut in the spaceship. If the spaceship travels with a speed 1.8 x 108 m/ s, find the interval between successive signals as received by an observer on the ground. From the given data, we have, v = 1.8 x 108 m / S; TO = 3s. If T be the time interval between the two signals as received by an observer on the ground, then, using the Lorentz transformation equation, T = J Tg 2 2' we get, I-v Ie SOLUTION:
T
=
3 ,====::== = 1- (1.8xIOB)2 3x108
3 - = 3.75s. 0.8
Ex 12.6.10. Two events separated by a distance 15m and time interval 3 x 1O-8 s occur in frame 8. Find the distance and the time separation between the same two events as observed from 8' moving with speed ~ along the direction of 15m line. SOLUTION: Let us suppose that, the two events occur in frame 8 at points, whose locations are Xl and X2 and at times tl and t2 respectively. Then X2 - Xl = 15m and t2 - tl = 3 X 10-8 8. Further, suppose that, the same two events appear to occur at points x~ and x~ and at times t~ and t~ in frame 8', which is moving in a velocity v = ~. Using Lorentz transformations, we have
12.7. ADDITION OF VELOCITIES
391
Also, the time dilation equation gives,
12.7
Addition of velocities
Let us consider a body that moves with respect to both inertial frames of reference Sand S' having their respective axes parallel and the frame S' is moving with a constant velocity v along x- direction relative to frame S. Let P(x, y, z) be the position of the particle at time t with respect to the observer of the stationary co-ordinate frame S and it possesses velocity 11 = (u x , u y , u z ), then dx
Ux
dy
=-;
uy
dt
uz
= dt;
= dz. dt'
u
=
2 vux2 + u 2y + u z·
At the instant, let P(x', y', z') be the position of the particle in moving co-ordinate system at time t'. An observer of the frame of reference S' makes the following measurements of velocity u' = (u~, u~, u~) of the body measured by him/her as ~
, u'x
.' = dy'dt" .
dx'
= dt';
dz' =' dt"
u'
z
.J,y
u
=
J,.-----u'2 + u,2 + u,2 . x
y
z
The velocity components of the particle in the two frames can be related to each other by using the Lorentz transformation as
, x
=
x-v t
VI _ v2 / c2 ;
,
,
= y;
y
z
= Z;
t'=
t
vx
-C2"
VI _ v2 / c2
We must establish the connection between the projection between the projections of the velocities of the point mass in the moving and stationary co-ordinate system. Differentiating both sides, we can write,
U' Y U
,
-
Z -
, _
dx' _
Ux -
dt' -
!!:JL _ dt' d ' dt' -
2- -
d",
Jff -v _ 1- v d", -
C2"
dt
1L-V
l-";;u", C
~-=-....".
1tJ1-v2/c 2 _ uyJ1-v 2/c 2 1-
v d",
C1dt
-
~y'1-v2/c2 dt 1- v dx -
C2"
dt
l-;;u",
r
(12.14)
C ::"""",;,...,...".
2 2 .u z y'1-v /c -"-,_~..--.!...1-4u", c ..
These equations, which transform the velocity components of a particle in unprimed frame S to the velocity components in primed frame S' moving with a constant velocity 'V with respect to the reference frame S, constitute the relativistic transformation law equations of addition of velocities. In accordance with the relativity
CHAPTER 12. RELATIVISTIC MECHANICS
392
principle, the formulas for the inverse transformations are obtained, as usual, by interchanging the primed and un primed quantities and by replacing v by (-v) as U
u~
+v
. x- 1 + v l ' 2~
'lJ,
-
y-
'lJ,~J1 - v2 jc2 • 1+vl
U
-
,z-
~~
u~J1 - v 2 jc2
(12.15)
-"--"-----'---
1
VI
-2~
which are the inverse transformation laws. Following are some points about the relativistic law of addition of velocities: (i) In case, the reference frame S' moves with a velocity very small as compared to the velocity of light, then both ~ and ?!- can be neglected as compared to 1. Hence the direct velocity transformations reduces to
i.e., the relativistic law of addition of velocities reduces to Galilean law of addition of velocities. (ii) In case the motion of the particle is confined to x direction, i.e., U z = 0, then the direct velocity transformation reduces to
Ux
= U, u y =
(iii) Let U x = c, u y = U z = 0, i.e., the ray of light signal is emitted with velocity c along x direction in S. Applying the direct velocity transformation, we get,
Thus we find that the two inertial frames moving relative to each other, the velocity of light remains the same, which is consistent with the second postulate of the theory of relativity. Also, the composition of velocities never leads to veloeities exceeding the velocity of light. Thus we conclude that c is the highest limit to the velocity that can be acquired by material bodies. (iv) Suppose that a ray of light propagates along y' axis in the S' co-ordinate system, i.e., u~ = 0 = 'lJ,~ = u~. In a stationary co-ordinate system, we obtain 'U x
=
'V,
'uy
=
c-J1 -
~, U z = O.
Consequently, the ray of light forms an angle
(J with the y axis in stationary co-ordinate systems, given by
tan (J
Ux
v
uy
c-J1-~'
1
= - = - ----,===
For v < < c, this equation becomes identical with the classical theory formula tan (J = ~. However, this relation now has a different meaning. In the classical theory, it was necessary to distinguish between the situations involving
TRANSFORMATION OF ACCELERATION
12.8.
393
a moving source and a stationary observer from those involving a moving observer and a stationary source. In the theory of relativity, there is just one case involving the relative motion of the source and the observer. This fact is known as aberration. Ex 12.7.1. Two particle come toward each other, each with speed 0.9c with respect to the laboratory. What is their relative speed? SOLUTION: Here we refer to relativistic law of addition of velocity. From the given data, we have,. Ul = 0.9c, which is the speed of the particle 1 with respect to the laboratory, U2 = -0.9c, which is the speed of the particle 2 with respect to the laboratory. Let U be the speed of the second particle w.r.to the first particle. Then from addition law of velocities, U2
=?
=
-1.8c =
Ul
+U
UlU2
u[l - - -2 ] 1+~ c c (0.9c)( -0.ge)J 1.8 U [1 2 =? U = ---lc = -0.9945c. c 1.8 =? U2 - Ul =
Ex 12.7.2. A rocket is chasing an enemy's spaceship. An observer on the Earth observes the speed of the rocket as 2.5 x 108 m/sand that of the spaceship as 2 x 108 m/ s. Calculate the velocity of the enemy's spaceship seen by the rocket. SOLUTION: Let us consider the rocket to be in S' frame and the earth in S frame. Suppose u' be the velocity of the enemy's spaceship as seen by the rocket; Ul, the velocity of the rocket relative to earth; U2, the velocity of the spaceship relative to earth. The addition law of velocities gives U2 = Ul +u',. The given data are l+~
2.5 x 108 m/ s, U2 velocities give til
= 2.0 x 108 m/ s, c = 3 x
=
Ul
=? U =? U
,
[1 -
2.5
X
8
+ U'
- 1 + u~t
=? U2
,
=
U2 -
[1 -
ulU2 -2-J C
+
U2 -
Ul U2U'
c
2
108 m/s.
The addition law of
= U 1 + u'
Ul
8
10 x 2.0 X 10 J ) 8' 8 / (3 x 108)2 = (2.0 - 2.5 x 10 =? U = -1.125 x 10 m s.
The negative sign indicates that the enemy's spaceship is approaching the rocket with velocity 1.125 x 108 m/ s.
12.8
Transformation of acceleration
Suppose that, a particle undergoes in a primed co-ordinate system an acceleration, whose projections are given by a~, a~, a~, while the particle is instantaneously at rest in the proper frame 8', i.e., U'x
= u'y = u'z
= 0
CHAPTER 12. RELATIVISTIC MECHANICS
394
at that instant of time. But the instantaneous acceleration is not necessarily zero in the proper frame. Now, the acceleration in the two frames Sand S' are respectively, ax =
dux
ill;
a
y
du du z az = dt' dt
=-Y.
, du~ , du~ , du~ a x = - ' a y = - ' a z = dt' . Jt' ' dt' '
The quantities dt, dux, du y and du z are determined from (12.12) and (12.15), while the quantities u~, u~, u~ can be set equal to zero after the differentials have been evaluated. From Lorentz transformation, we have, t
2
= (t' - ~x')(1 - ~2 )-1/2 c
c
2
= (1 - ~ )-1/2 [dt' ~
=?
dt
=
2 d ' (1 - ~ )-1/2[dt' - ~~l 2 c c dt
2
- ~u' 1= (1 - ~ )-1/2dt'; c x ~
as u~
= o.
Taking the differentials of the velocity transformation equations (12.15), we get, the expression for dux as
where, in accordance to our assumption, set u~ = O. The differentials duy is calculated in a similar manner as d ' 2 , 2 _ u Y .( _ ~)1/2 _ uy ~(_ ~)1/2~ y du - 1 + v u' 1 c2 (1 + v u' ) c2 1 c2 c2 dux ?' x ?' x 2
-- -- (1 - 2v )1/2d11.' ' as 11.,y -- 0 -Y C
=?
a1 = J
d y 11. dt
=
,
11. x
2 2 d' (1- ~)1/2(1_ ~)1/2 u y 2 2 c c dt'
=
2
(1- ~)a'. c2 11
The differentials du z is calculated in a similar manner. Thus we obtain the following transformation formulas for the acceleration as 2 2 2 '. a y -_ ( 1 _~) '. _ ( 1 _~) , ax -_ ( 1 _ ~)3/2 2 ax' 2 ay, a z 2 az · c c c
(12.16)
The particle moves at a velocity v in the S frame. Hence equation (12.16) has the following meaning. A moving particle can be assigned in inertial co-ordinate system in which it is at rest at a given instant of time. Such a co-ordinate system is called the body axes system. If the particle moves with an acceleration in this
12.9.
VARIATION OF MASS WITH VELOCITY
395
system, it will also move with an acceleration in any other system. However, this acceleration will be different but always smaller. The projection of the acceleration on the direction of velocity is reduced in proportion to the factor (1 - ~ )3/2, where to v is the velocity of the body axes co-ordinate system. The transverse component of acceleration perpendicular to the velocity of the particle varies to a smaller extent, its decreasing being proportional to the factor (1 _ ~ )3/2.
12.9
Variation of mass with velocity
. The quantity of matter contained in the body is defined as the mass of the body. Consider two bodies A and B each of mass m moving in frame S' with velocities u' and -u' respectively along x' axis. Let these bodies collide inelastically and coalesce into one body of mass 2m, which will be at rest in frame S'. If the collision of the two bodies is viewed from the frame of reference S, the x component velocities Ul and U2 of the two bodies A and B as observed from S is given by the inverse velocity transformation formula as 1J, 1
=
+V 1 + .:zzu' u'
and
U2
-1J,' + V
= ---::-:---,-
1-
.:zzu'·
When viewed in frame S, let ml and m2 be the masses of the two bodies A and B. Before collision, let the two bodies A and B appear to be moving with velocities Ul and U2 along x direction in frame S. After collision, by conservation law os mass, the combined mass is (ml + m2) and is moving with velocity v along x direction with respect to S (i.e., of the frame S'). In the reference frame S, by the law of conservation of linear momentum, we have,
(12.17) Now, we simplify the RHS of (12.17) in terms of the velocities bodies in the frame S. For this,
Ul
and
U2
of the two
(12.18)
396
CHAPTER 12. RELATIVISTIC MECHANICS
which is the relation between the values of masses m1 and m2. If the velocity of the second body as observed with respect to the reference frame S is zero (i,e., U2 = 0), then its mass m2 can be denoted by mo. The symbol mo gives the mass of a body when it is at rest with respect to the frame of reference being used. In general, if a body possesses mass mo, when at rest and mass m (m1 = m) when moving with velocity V(U1 = v), then equation (12.18) becomes
v'I"=O
m mo
-=
J1- v
2
jc
mo
2
(12.19)
~m=-,==;;:::;:::::;;:
J1- v2 jc2 '
It follows that m > mo, i.e., mass of a body in motion is greater than its mass, when at n~st. The variation of mass with velocity has the experimentally verified for high speed electrons and the {3 particles emitted by some radio active substances. Now we have the following discussions drawn from the relativistic mass variation formula:
(i) If the two effects length contraction and mass variation are combined, measured from the earth, the length of a rocket (which is in motion) is shorter while its mass is greater. Since the speeds acquired by the rockets are considerably small as compared to the velocity of light, these effects are quite small. (ii) When '11« c, then ~ « 1 and equation (12.19) gives m = mo. It indicates that the increase in mass of a body is observable only, when a body moves with a velocity comparable to the velocity of light. For this reason equation (12.19) is called relativistic mass variation formula with relativistic mass m.
Ex 12.9.1. The rest mass of the electron is 9.1 x 1O- 31 kg. What will be its mass, if it were moving with a velocity of 0.8c? SOLUTION: From the given data, we have, mo relativistic mass of the electron is given by,
m=
mo
VI - ~
=
9.1 x 10- 31
= 9.1 =
J1 - (0.8)2
x 1O- 31 kg and v
=
0.8c. The
15.17 x 10- 31 kg.
Ex 12.9.2. With what velocity should a particle move, so that the increase in its mass may be 25 percent of its rest mass? SOLUTION: Suppose that, mo is the rest mass of the particle and v is the velocity, at which its mass increase by 25 percent, i.e., it becomes
m
= mo +
mo x 25 100
= 1.25mo·
Using the given data, the relativistic mass of the particle is given by, mo mo m = 1 v 2 ~ 1. 25mo = 1 v 2 -"22 -"22
A
'()2
~ 1- c2
A
1
2
16
= (1.25) = 25
~v
'
= 0.6c.
12.10. MASS-ENERGY RELATION
12.10
397
Mass-energy relation
Let us consider a body of rest mass mo moving with velocity v. Then according to relativistic mass variation formula, the relativistic mass m is given by m =
mo
}1 - v2 jc2 => mvdv
=> m 2 (c2 _ v 2 )
+ v 2 dm =
= m6 c2
c2 dm; differentiating.
Let a force F be acting on the body which displaced the body through a small distance dx in the direction of the force. Now the KE T of the moving body is the work done in giving it that state of motion starting from rest. The increase of KE of the body caused by the work done by the frame is dT = Fdx. Since the force acting on the body is defined by the time rate of change of its linear momentum, so, dT
=
d dt (mv)dx
dv
dm
= [m dt + vYtl dx
dx dx =m-dv+v-dm dt dt 2 = mvdv + v dm = c2 dm.
Let us assume that when the velocity of the body increases from 0, i.e., the mass of the body increases from mo to m, its KE increases from 0 to T. Therefore, T
J
dT
o
rna
2
=c
J
drn => T
=
(m - mo)c
2
.
(12.20)
rn
Thus, the KE of a moving body is equal to the product of the increase in the mass with square of the speed of light. Since the body possesses mass mo even when at rest, it may, therefore, be assumed that the rest mass of the body is due to an internal store of energy Eo = moc2 . The quantity Eo is called the rest mass energy of the body. Now equation (12.20) can be written as
mc2 = T + moc2 = T + Eo. Now, E = Total energy of the body
= T + moc2 = mc2 . This famous relation is known as Einstein's mass-energy relation and it shows the equivalence of mass and energy. The expression Eo = moc 2 shows that mass is yet another form of energy. Since mass and energy are related to each other, we have to consider a principle of conservation of mass and energy. Mass can be created or destroyed, provided that an equivalent amount of energy vanishes or is being created and vice versa. Now, when v « c (non-relativistic approximation), using
398 m --
CHAPTER 12. RELATIVISTIC MECHANICS m.p ha:ve, J I-v 2 2' we Ie
which is the expression for KE of the particle in Newtonian Mechanics. Thus we ha:ve seen that, at low speeds, all formulas of relati:vity reduce to the corresponding formula of classical mechanics. So the relati:vistic formulation of mechanics is more accurate approach while the classical formulation in just an approximation to the former. When v « c, i.e., ~ ---7 0, the transition from relati:vistic to classical mechanics is obtained. Following important conclusions can be drawn from the mass energy equi:valence relation: (i) In classical mechanics, the laws of conser:vation of mass and energy are separate principles independent of each other. The relation E = mc2 leads to the unification of the two laws into one law, which is the law of conser:vation of relati:vistic energy. (ii) From classical mechanics, we know mass is considered something fundamental to matter, while energy is a property of the matter acquired by :virtue of its configuration; The relation E = mc2 puts an end to such a distinction between mass and energy. (iii) The special theory of relati:vity ascribes energies to all masses and masses to all energies, follows from the relation
where 6.m is the increase in mass corresponding to the increase in energy 6.E. (i:v) When v ---7 c, then T ---7 00, i.e., in order to accelerate a body so as to make it move with the speed of light, an infinite amount of work would need to be done. Hence the mass energy equivalence relation can be put as a limit on the velocity, a material body can be accelerated to.
Ex 12.10.1. Calculate the effective mass of a photon of wavelength 5890AO. SOLUTION: The given data are h = 6.62 X 10-34 J s;). = 5890AO = 5890 x lO- lO m and c = 3 x 108 m/ s. The energy of a photon of wavelength), is given by, he
E
=T =
6.62 x 10- 34 x 3 X 108 5890 X 10- 10
= 3.372 x
-19
10
J.
12.10. MASS-ENERGY RELATION
399
If m is the effective mass of the photon, then from the relation E = mc2 , we have,
m - E _ 3.372 X 10- c2 (3 x 108 )2
19
= 3.7467 X
10-36.
Ex 12.10.2. Calculate the incr"ease in mass of a proton, when accelerated to 500M eV. SOLUTION:
Here, the KE of the proton is given by, T
= 500M eV = 500 x
1.6
10- 13 J
X
=8X
10- 11 J.
Let mo be the mass of proton and m its relativistic mass, then using the formula,
T = (m - mo)c2, we get
m - mo
T
= c2 =
8.0 X 10- 11 (3 X 108 )2
= 0.889 x
10
-27
Kg.
Ex 12.10.3. Calculate the decrease in mass of 19 of water at OoC, when it turns into ice ofOoC. SOLUTION: Let L be the latent heat of the fusion of ice, then the amount of heat energy required to convert 19 of water into 19 ice,
6E
= mL = 1 x 80cal = 80 x 4.2J = 336J.
This energy is supplied to the water by itself at the expanse of the mass. If 6m is the decrease in the mass of water, then, 6E
= 6mc2 = 6(3 x
16
108 )2
336
10 16 6mJ. -12 3.73 x 10 g.
=9X
=> 9 x 10 6m = 336 => 6m = 9 x 1016 kg =
Ex 12.10.4. Calculate the increase of mass of500g of metal of specific heat 0.15cal/g/oc when it heated through 900° . SOLUTION: From the given data, we have,m = 500g; s = 0.15cal/g/oC; 6T = 900°C. Thus the heat absorbed by the metal, is given by,
6E
= ms6T = 500 x 0.15
x 900cal
= 500 x 0.15 x 900 x 4.2J = 2.835 x 105 J. If the heat energy absorbed by the metal appears as increase in mass 6m, then,
6E
= 6mc2 = 6m x (3 x 2.835 x 105
108)2J
=> 6m = (3 x 108 )2 = 3.15 x 10
-12
kg = 3.15 x 10
-9
g.
Ex 12.10.5. A proton has a total relativistic energy 9000MeV. Find its KE and its speed, assume that the rest mass of proton is 1.6 x 1O- 27 kg.
CHAPTER 12. RELATIVISTIC MECHANICS
400
From the given data, we have, mo = 1.6 x 1O- 27 kg; E 9000 x 1.6 X 10- 13 J. Thus the KE T of the proton is given by,
=
SOLUTION:
T
=E-
moc2
= [9000 x = 12.96 X
1.6 x 10- 13
-
1.6
X
10- 27 x (3
mo
. 11
--;====;~::;;: =}
V
\11 - v2 /c 2
Jl _ v2/c2 =}
12.11
v2 1- 2 c
108 )2]J
10- 10 J.
If m be the relativistic mass of the proton, then using m
=}
X
9000MeV =
- V
2/ 2 _ C
= 1.6 X 10-
moc E
= J I-v m0 2' 2 Ie
2
---
27
x (3 X 10 9000 x 1.6 X 10- 13
=
0.01
=} V
we get,
8
)2 =
0.1
= 0.995c.
Momentum and energy
Consider a body of rest mass mo moving with a velocity 11. Then its relativistic mass is m = J m02 2' The relativistic linear momentum of the body is, given by, I-v Ie -+
-+
p=mu=
mO-+
Jl- v2 /c 2
u
(12.21)
which depends upon the velocity 11 of the body. If 11 = (u x , U y , u z ) and p = (px, Py, Pz), then we can find the components of relativistic linear momentum of the body. Now, according to Einstein's mass-energy relation, the body of mass m is equivalent to energy E, given by
(12.22) This is called relativistic energy relation. From equation (12.22), it follows that the energy of the body may be positive or negative. The physical meaning to the negative energy by predicting the existence of antiparticles was described by P.A.Dirac. Deduction 12.11.1. We know, the relativistic energy of a particle of rest mass mo
12.12. TRANSFORMATION OF ENERGY AND MOMENTUM and momentum pis E
=
401
J p2c2 + m6c4. Also,
E = moc2 + T
=}
2 Vp 2c2 + m6c4 = moc + T p2c2 = T2 + 2moc2T
=}
p=
=}
J~: + 2moT.
This is the relation between the momentum and KE.
Ex 12.11.1. Calculate the linear momentum of a 25GeV proton, assuming the rest mass energy of the proton is IGeV. SOLUTION: The given quantities are moc2 energy of the proton is given by,
E Using the relation E
pc = =}
12.12
P=
IGeV; T
=
25GeV. The relativistic
= moc2 + T = (1 + 25)Gev = 26GeV.
= ±Jp2c2 + m5c4,
V
E2 - m6c4
=
the relativistic energy of the proton is,
J26 2 - 12
25.9S x 1.6 x lO lD 8 kgm/s 3 x 10 .
= 25.9SGeV =
= 1.3S6 x
10
-17
25.9S x 1.6 x 1010 J kgm/ s.
Transformation of energy and momentum
Consider two inertial frames of reference 8 and S' having their respective axes parallel and the frame 8' is moving with a constant velocity v with respect to frame 8 along X - direction. Consider that a particle of rest mass mo is moving with a velocity u with respect to the observer of frame 8. If U x , u y and U z be the components of velocity of the particle, then
(12.23) To the observer of the frame 8, the mass of the particle will appear as m
= J l-u rn P • 2 /c 2
Suppose that to the observer of the frame 8', the particle appears to move with velocity u'. If u~, u;, and u~ be the components of velocity of the particle, then
(12.24) To the observer of the frame 8', the mass of the particle will appear as m' rnp According to velocity transformation law, we have J1-'u /2 /c 2 • ,
Ux
=
'U x
1
'V
v
- CIux
, -
;
uy -
u Y Jl 1
- v 2 /c 2 . v
- CIux
,
2
U'
z
u z ' __ /1 - v / c = _--'v ---'_ 1-.3i.-u c~ x
2
=
CHAPTER 12, RELATIVISTIC MECHANICS
402
The X - component of the relativistic momentum of the particle in frame 8' is
",
= rnu x =
Px
rna
ux-v
' VI - u,2/ c2'1 v - ~ux
To express the X - component of momentum in frame 8' in terms of the X - component of momentum in frame 8, let us first evaluate the factor u,2 / c2 , From ' th e re1a t lon, 11,'2 = U '2 U z ,we h ave, x + U '2 y +'2
VI -
=?
11,,2
1- -
c2
(u x - v)2 + (u 2 - u;)(1 - v 2/c2) = 1 - ..:....-:=------.:....-::-:.....:.....-~~_:::__-!.-.:... 2 2
c (1 - vu x /c )2 2 (c - v ) - (u; + u 2 - u;)(1 - v 2/c2) c2 (1 - vu x /c 2)2 2 2 (1 - u /c )(1 - v 2/c 2) = c2(1 - vux /c2)2 2
Thus, the X - component of the relativistic momentum of the particle in frame 8' is given by
It is how the X - component of momentum of the particle in frame 8 transforms to X-component of the momentum in frame 8', Now, the Y - component of the momentum of the particle in frame 8 f is given by,
12.12. TRANSFORMATION OF ENERGY AND MOMENTUM
403
Similarly, p~ = pz. Therefore, the transformation equations for momentum components in unprimed frame 8 to primed frame 8' are ,
Px
=
PT -
./1
V
vE/c2 2/2;
-v c
,
,
Py=Py;
Pz=pz·
(12.25)
It follows that just as in case of Lorentz space-time transformations, the time t joins the space co-ordinates x, y, z as the fourth co-ordinate; in momentum transformations, the energy E combines with momentum components Px,Py and pz as the fourth co-ordinate. Let us now obtain the transformation equation for energy from unprimed to primed frame. If E' be the energy of the particle in frame 8', then 2 E' = m'c2 = mo c2 = mo(l- vu x /c ) c2 Jl - u'2/c2 Jl - u 2/c 2 Jl _ v 2/c 2 ' _ mc2 - muxv _ E - vPx . as Px = mu x , E = mc2. (12.26) - Jl- v2/c 2 - Jl- v2/c 2 ' The equations (12.25) and (12.26) are the momentum- energy transformations from unprimed frame 8 to primed frame 8' moving with constant velocity v relative to 8. The inverse momentum- energy transformations can be obtained by changing v to -v, primed quantities to unprimed and vice-versa, i.e., _
p~+vE/c2 .
,_
,_
Px - Jl-v 2 /c 2 ' Py - Py,PZ E _ E'+'Vp~
-
pz
} .
(12.27)
~h-v2/c2
-
Ex 12.12.1. 8how that, the quantity [P2 - ~] is invariant under Lorentz transformations. SOLUTION: According to momentum energy transformations, we have , Px - v E / c2 " , E - vP~ x P = Jl-v 2/c 2 ; PY=PlI,pz=pzandE = Jl-v 2/c 2 Using these relations, the quantity [p'2 E'2 p'2 _ _ c2
Ff)-] in 8' frame becomes,
E'2 = p'2x + p'2y + p'2z---;}2 2
]2 2 2 [E - vp~ ]2 = rPx - vE/C +p +p . }1- v 2/c 2 y z }1- v 2/c 2 1 2] 2 2 E 2 v 2") - 2" (E - vpx) + Py + Pz C c 1 2 E2 1)2 2 E2 2 2 1 -v 2/ c2l(px - 2c - "2(Px 2] + Py + pz c C
1
1
2/ C2[(Px V
2 E2 2 2 =Px - 2 +Py +pz c E'2 E2 :::} p'2 _ _ = p2 c2 -~.
404
CHAPTER 12. RELATIVISTIC MECHANICS
Thus, the quantity [P2 - ~] is invariant under Lorentz transformations.
12.13
Relativistic force
---> ---> The momentum p and the force F are connected by the Newton's second law of motion. In the non-relativistic case, we have, ---> d Ii --> ---> F = Tt; p =mv ---> =d - (---> 'rn'u)=rna;
(J = constant. dt In fact, for the relativity theory, we have to use the relations (12.25) for momentum, and so we have, 2
--->
d ---> d11 ---> d(3 v 1/2 F = dt(m(3v) =m(3 dt +mv dt; (3=(1- c2 )- .
ft
The second term m11 is in the direction of the velocity 11. Also, the first term contains d]{, which may be in any direction. So we resolve it into a component perpendicular to the velocity and a component parallel to the velocity. Writing the perpendicular component as the perpendicular component of the acceleration, i.e., d1i (lIth -_ ---> a1., we have,
---> d11 ---> F 1. = m(3( dt h = m(3 a1..
(12.28)
Now, we calculate the component of force parallel to the velocity. Now, 2 2 d(3 = ~(1 _ v )-1/2 = _1_(1 _ v )-3/2~(v2) dt dt c2 2c 2 c2 dt where v 2 = 11112. If () be the angle between the vectors 11 and 0;, we have
~( 2) dt v
= ~(1--->12) = 2---> d-V = 2--->.--> dt v v . dt v a = 2111110;1 cos () = 2vl0;1 cos ().
Thus, 10;1 cos () is the component of the acceleration in the direction of the velocity, i.e., 10;1 cos() = alI-. so that ft(v 2 ) = 2van· Thus, d11 ---> d(3 v v 2 3/2 Fll = m(3- + m v - = m(3an + mv-(l -)an dt dt c2 c2 2 2 2 2 = m[(l - ~ )-1/2 + ~(1 - ~2 )-3/2]an = m(l - ~ )-3/2 an . (12.29) 2 2 2 c c c c Equations (12.28) and (12.29) give respectively the components of force, perpendicular and parallel to the direction of motion. The mass m(l - ~) -3/2 is called
congitudinal mass" where as m(l - ~ )-1/2 is called the tr~nsverse mass. The difference between these two masses implies that at high speed (Le., in the relativistic region) the acceleration vector is not parallel to the force vector.
12.14. EQUATION OF MOTION
12.14
405
Equation of motion
-
Suppose that, a particle moves along a trajectory. We denote the unit vectors normal and tangents to the trajectory ny nand f. The total force F acting on the particle can be decomposed into normal and tangential components as
Each component of the force produces an acceleration in the appropriate direction, determined by the inertia of the body in the direction. Since the normal acceleration is '~ and R is the radius of curvature of the trajectory, and v is the particle's velocity and the tangential acceleration is ~~. Thus the force can be written as -F
=
mo v2 mo dv n+ I 'T 2 2 2 2 Jl - v /c R (1- v /c )3 2 dt mo df mo dv Jl - v 2/(;2 v dt + (1 - v 2/c 2)3/2 dt'T A
A
A
=
d movf dt [Jl _ v2/c21
which is the relativistic equation of motion of the particle and it is a generalisation of Newton's equation of motion. It can be written in a more convenient form as dp dt
= Fj
p
= mV,m =
mo J l - v 2 /c2
(12.30)
The quantity m is called the relativistic mass, or simply mass, mo is the rest mass.
12.14.1
Relativistic rocket
In the relativistic case, a mass M must be considered relativistic, i.e. M
= J 1-v M02 2' Ie
where Mo is the variable rest mass of a rocket. The rest mass of the rocket decreases during its motion. Taking this into account, we can write, (12.31) If necessary, we can easily take into account the external forces acting on the rocket. As a result, we arrive at the relation
(12.32)
The difference 'U - v is not the velocity of ejection of gases relative to the rocket because we must use use formula for the velocity addition in the relativistic case.
406
12.15
CHAPTER 12. RELATIVISTIC MECHANICS
Inelastic collision
Let us consider two inertial frames of reference Sand S' having their respective axes parallel and the frame S' moving with a constant velocity v along x-direction w.r.t. frame S. Suppose that two identical bodies A and B, each of mass m, are moving in frame S' with velocities u' and -u' respectively along x'- axis. Further, that the two bodies collide inelasti'cally, so that after collision they stick together to form a single body. In the frame S', the combined mass of the two bodies will be at rest and let it be equal to Mo. When viewed in the frame S, suppose that the two bodies A and B appear to be of masses ml and mao Further, suppose that in the frame S, the two bodies appear to move with velocities Ul and U2 along x- axis before collision. According to the velocity transformations, we have, u' +v . - 1 + vu'lc2 '
Ul -
and
U2
-U' +v = -----.,,-2 1 + v( -u')lc
-U'+v 1-vu'lc2 '
Let us assume that the frame S' is moving with velocity u' w.r.t. frame S. Then, setting v = u' in expressions for Ul and U2, we have, Ul
=
+ u' 2u' = . 2 1 + u' u'1c 1 + u'2 1c 2 ' U'
and
U2 =
-U' + u' 1 -u'21 c2 =
o.
As such, the body B of mass m appear at rest and the combined mass of the two bodies will appear to be moving with velocity v = u' in frame S. Suppose that in the frame S, the combined mass of the two bodies appears as M, then according to mass variation formula, M = J M02 2' If mo is the rest mass of the either of the l-u'
Ie
two bodies A and B, then we have, m1 =
rno
rno
"-t==;;r:;=::;:;:
2
J1- [1+~%le2]2/c2 mo[l + u'2/c2] mo[l + u'2/c2] J[l + u'2/c2J2 - 4u'2/c2 1 - u'2/c2 J1- uVc
From the law of conservation of linear momentum in frame S, we have,
Since the factor J1 - u r2 1c2 < 1, from the above equation, it follows that, Mo > 2mo, i.e., during an inelastic collision, the mass of the system increases. The increase
12.16.
407
MINKOWSKI SPACE
in mass is given by,
6m
= Mo - 2mo = = 2mo[
Vl-
2mo u,2/ c2
Vl-
- 2mo
1 -1] . u,2/ c2
The energy equivalent of the increase in mass during the inelastic collision,
It may again be pointed out that the increase in mass in an inelastic collision is due to materialization of the kinetic energy possessed by the two colliding bodies before collision. The KE possessed by the two bodies before collision,
Now, the mass of the either body w.r.t. the observer of the frame 8', is m rnQ and so Jl-u /2 /c 2 ' T
= 2[
mo 2 - mo]c u,2/ c2
VI -
= 2mo[
VI -
1 2 - l]c , u,2/ c2
which is same as the expression for 6E. In other words, energy equivalent of the increase in mass of the system is equal to the total KE of the two bodies before collision.
12.16
MiIikowski space
In special theory of relativity, the concept of space and time are independent of each other. The length of a radius vector in three dimensional geometry, i.e., x 2+y2+z2 is not invariant under Lorentz transformation. However, the quantity x 2+y2+z2_c2t 2 is invariant under Lorentz transformations, i.e.,
The above equation may be written as, (12.33)
where x', y', z' and ict' are the four coordinates in the frame 8', while x, 1], z and ict are those in frame 8. Just as the radius vector in the three dimensional space has three mutually perpendicular components x, y and z, the space components x, y, z and time components ct may be treated as the components of the radius vector in the four dimensional space, called Minkowski space or space time continuum. Since
CHAPTER 12. RELATIVISTIC MECHANICS
408
it is not possible to draw four mutually perpendicular axes through a point in space, the fourth component ct has been multiplied with i so as to remind that this fourth dimension is imaginary. Thus, in Minkowski space, the fourth-radius vector having the co-ordinates (x, y, z, ict) may be represented as --+
...
r = xi
..
,..
+ yj + zk + ctw,
(12.34)
where w is the unit vector along the fourth imaginary axis. The four coordinates x, y, z and ict define a vector in the four-space. Such a vector is often termed a four vector. The representation of space-time can be extended to give a geometrical interpretation of the relativity of simultaneity, length contraction, time dilation etc. The fact that L2 is invariant suggests that a Lorentz transformation can be merely regarded as a rotation of coordinate axes x, y, z and ict in space time. Introduction of four-vectors in the four space has another advantage. When physical laws are expressed in terms of four-vectors then the law takes the same form in any frame moving with a uniform velocity with another frame. Thus, the invariance of the physical law is automatically proved when expressed in four vectors. The Lorentz transformation (12.11) for x and t co-ordinates are x'
=
t - (vx)/c 2
x - vt I \1"1 _ (v2 / c2)' t
= -vr=I=_=(O=v~2/:;::;c2~)
which can be written in the form, I ct - (v/c)x ct = -Vr=I=-=(~v:;;=2/~c2~)
I x - (v/c).ct or, x = -vr=I=_=(-Fv2~/:=;c2~)
I
or. x
=
X -
{3w
\1"1- {32;
I
w
=
W -
{3x
v
Jl- (32;{3 = ~,w =
ct.
(12.35)
Note that, the Lorentz transformations for x and t co-ordinates are not symmetrical. But introducing ct as the fourth dimension, the Lorentz transformations assume the form of the equations (12.35), which become just symmetrical to each other. Minkowski referred to space-time as the world and a point in space-time as the \ -world po-into The motion of a particle in space-time can be represented by a curve called wprld line, which gives the loci of space-time points corresponding to the motion. The world line of a particle must lie in its light cones. Now we are to apply this concept to understand some implications of the theory of relativity. We draw only x and waxes of the frame of reference and ignore the y and z axes, because y' = y and Zl = Z. The tangent to the world line at any point is always inclined at an angle 0« 45°). It is because, the slope of the tangent to the world line at any point, m
dx
Idx
'v
= tan () = -dw = -edt - = -. c ,
Since v <,< c,Le., velocity of the particle is always less than that of light, we have tan () < 1. ~ () < 45°. Obviously, the world line of light wave (v = c) is a straight line making an angle 45° with the x and waxes.
12.16.
409
MINKOWSKI SPACE
12.16.1
Proper time ~
A four-vector X which is used to represent the position vector of a point in a four space or a four position vector is written as ~
X
(x, y, z, ict) = (Xl, X2, X3, X4).
=
I-'
~
~
The different.ial of X along the world line will involve a change in the position XI-" denoted by the four vector dX J.L' given by ~
dX J.L
= (dx!, dX2, dX3, dX4 = icdt) =
(d3!, ict).
The length element in the world space can be written as 4
(ds)2 =
L
3
IdX J.L12
=
(dXI)2
+ (dX2)2 + (dX3)2 -
c2(dt)2 = L(dxj)2 - c2(dt)2
1-'=1
j=l
which is invariant under Lorentz transformation. Now, we define a quantity dr as, (12.36)
which is also invariant under Lorentz transformation as it is ~ds. The quantity dr is called the element of proper time or world time in the Minkowski four space. As one of the four components of the four radius vector is imaginary, the square of its mag'nitude may be positive, negative or zero, i.e., x2 + y2 + z2 - c2t 2 may be positive, negative or zero. The four radius vector is called space like interval, if (dT)2 < O. Also, if (dT)2 > 0, then it is called time like interval. If (dr)2 = 0, then the corresponding interval is called a light-like interval.
12.16.2
Four velocity ~
Having defined a four-dimensional length vector XI-" we can now define a fourvelocity. The rate of change of the position vector dX I' of the particle with respect ~
to Its invariant proper time dT, i.e., the four vector quantity
d~"
is called the four
~
velocity and is denoted by V Ii' Tnus,
Thus the four-velocity is the particle proper time rate of change of its four position vector. It should be noted that the three dimensional velocity 11 = (~, ~, ~) of
410
CHAPTER 12. RELATIVISTIC MECHANICS
the particle has magnitude v which is the same as that occurring in factor . ~, 2 V 1-8
since we are finding the proper time rate of change positIOn vector of the particle. For the velocity four vector, we find d Xi)2 1 _ (32 ~ dt 3
V2
J.l
= _1_[,",(
_
c
2J =
2
2
v - c 1 _ (32
=_ 2 0 C
< .
t=l
Since squares of the world velocity is less than zero, it IS tIme like and has a constant magnitude. Thus, V~ is invariant under Lorentz transformation. The four acceleration can be defined as the particle's proper time rate of change of the four velocity. Thus, the four acceleration A J.l is given by, ~
~
2~
-; dV J.l d XJ.l d [ 1 -; J AJ.l=-d-=-d 2 =-d ~(v,zc). T T T VI - (32 Thus, the first three components A J ; j = 1,2,3 of the four acceleration are non zero when 7J is not constant and the fourth component A4 is zero.
12.16.3
Four momentum
In Newtonian mechanics, the momentum of a particle is defined as the product of the mass and the velocity of the particle. We are to extend the same definition in relativistic mechanics, on the basis of the following assumptions: (i) Since time t is not Lorentz invariant, it should be replaced by proper time
T.
(ii) The mass of the particle is truly the characteristic of the particle and is independent of the velocity. For the purpose, we take mass of the particle measured in a frame in which it is at rest. Such a frame is called rest frame or the proper frame of the particle and the mass so measured is called rest frame or the proper frame of the particle and the mass so measured is called rest mass or the proper mass of the particle and is denoted by rna. Then the four momentum is defined by, ~
P = rna dX J.l = ( L
I
dT
~
rna v , irnac );"V' Jl _ (32 Jl - (32
= (dX1 , dX2 . dX3). dt
dt
dt
-;
The first three components of four momentum P dimensional momentum. Thus,
and the fourth component of momentum is P4
=
IL
are the components of the three
~
~
~. Ab:" P J.l" P J.l 2
V 1-13
=
_m,2r2
< O.
12.16.
12.16.4
MINKOWSKI SPACE
411
Four force
In. Newtonian mechanics, the generalised equation of motion is given by,
which is not invariant under Lorentz transformation. Its relativistic generalization should be a four-vector equation, in special part of which reduce to the above equation in the limit as fJ - t O. Now we have the following assumptions: (i) Since time t is not Lorentz invariant, it should be replaced by proper time r. (ii) The mass m can be taken as an invariant property of the particle. (iii) In place of the velocity
Vi,
world velocity vJ1. should be substituted.
(iv) Force Fi should he replaced by some four vector KJ1.' called the Minkowski force. Thus taking into account all these changes, the equation of motion is given by
-t
The first three components of four force K J1. are the components of the three dimensional force. Thus,
-t
The time like part K4 of the four vector K J1. can be obtained as ~
--->
K 11.' V p.
d
-t
--->
= V Jl.' dr (m V I') d
mc2
= - dr (-2-) = 0; V·
=> K
J
J
yh -
(J2
+ K4
d 'm - t
= dr ( 2" V
as
mc2
2
-t
J1.'
V J1.)
is constant
zc
JI - fJ2 = 0 'l.moc
VI - f32
CHAPTER 12. RELATIVISTIC MECHANICS
412
12.16.5
Kinetic energy
In the classical case, the KE T is given by, ~~
dT i dT F. v = - =? K4 = ---;===.:::-dt c {32dt
VI -
Also, from the fourth component of Minkowski force, we write dId ( - dr ' {32 dt dT 1 d mic f3 2dt = f32 dt ( (32) K4
i
=?
-
~ VI -
- ( mv4)
VI -
VI -
dT
=?
dt =
mic ) {32
VI -
VI -
d ( "me? ) T dt f32 =?
VI -
=
mc2
VI _ f32 + To
(12.37)
where To is the constant of integration. Also, equation (12.37) can be written as T
= (1- (32)-1/2mc2 + To = (1
1 v2
+ "2 c2 )mc2 + To;
For v
< < c.
Hence T will be classical KE if To = -me? Generally, we take, To = 0, so that K4 = t(i~). Using this relation, the fourth component of momentum is P4 = ~T. To express KE in terms of four momentum, we use the Lorentz invariant of four momentum. Now,
12.17
The relativistic Lagrangian
Our prior discussion of Lagrangian and Hamiltonian dynamics is correct only in the non relativistic limit. The extension to a relativistically correct formulation is straight forward, however. For a single particle moving in a velocity-independent potential, we defined the generalised momentum as Pk = ~t. We shall adopt this definition and require that relativistic Lagrangian, when differentiated with respect to qk, must yield the components of the relativistic expression for the momentum . {)L mqk, = -{)'; qk
,= Vl- f32.
12.18. EXERCISE
413
Since this imposed requirement involves only the velocity-dependent part of the Lagrangian, we would expect only this part (the KE) to differ from the non relativistic Lagrangian. If we write the relativistic Lagrangian as
then, we have, mc2 - - - , Integrating ,
,
where the constant of integration can be ignored. Thus, the relativistic Lagrangian becomes,
Note that, this expression is not equal to T - V, as in the non relativistic case. The equations of motion are found by operating the above Lagrangian with the Euler-Lagrange equations in the usual way. The Hamiltonian is given by, H= Li]kPk-L= L k
= mc2,
k
m i]2
r:;--;;? . k +mc2 yl-{32+V }1 - {32
+ V = T + Eo + V = E.
Therefore, the relativistic Hamiltonian is equal to the total energy, including the rest energy.
12.18
Exercise
Ex 12.18.1. Show that, the expression dx 2 + dy2 Lorentz transformations.
+ dz 2 -
c2dt 2 is invariant under
Ex 12.18.2. Prove that, the four dimensional volume element dxdydzdt is invariant under Lorentz transformation. Ex 12.18.3. Show that, under the Lorentz transformation, the wave equation for the propagation of the electromagnetic potential ,,2<jJ - ~~ = 0, where 6,2 = ~~ is the D'Alembert operator remains invariant.
,,2 _
Ex 12.18.4. Find the relativistic expression for KE of a particle, whose rest mass is mo, moving with velocity 11. Obtain the relation E2 = p 2c2 + m5c4. Show also that, in the limit of low velocities the usual expression for KE can be obtained from the relativistic expression of it. Ex 12.18.5. Two electrons move towards each other. The speed of each being O.gc in a Galilean frame of reference. What is their speed relative to each other?
414
CHAPTER 12. RELATIVISTIC MECHANICS
Ex 12.18.6. What should be the speed of a particle of rest mass mo in order that its relativistic momentum is mac. Find its total relativistic energy. v'2moc2. Ex 12.18.7. A rod 1m long is moving along its length with a velocity of O.6c. Calculate its length as it appears to (i) an observer on the earth (ii) moving with the rod itself. Ans : O.8m, 1m Ex 12.18.8. The earth revolves around the sun with an orbital speed of 30Km/s. Due to this motion, the diameter of the earth will appear shortened to an observer sitting on the sun. Find the percentage contraction in the diameter of the earth. Ans : 0.5 x 10- 8 . Ex 12.18.9. A rod of proper length L rests in the xy plane inclined at an angle e to the x axis. What does the observer moving with speed v along the x' axis. Find its length and the angle of inclination. Ex 12.18.10. A rod of length l, always parallel to the x-axis of frame 8, moves with a speed 'lJ, in the y direction with respect to s-frame. The ends of the rod are at (0, nt) and (I, ut) in the xy-plane of 8. Another frame 8' is moving with respect to 8 along their common xx' axes with speed v so that the usual Lorentz transformations are valid. Find the tangent of the angle which the rod makes with the x' - axis as observed in the frame 8'. GATE - 1996
