Lecture course on theoretical mechanics: educational manual 9786010423350

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АL-FARABI KAZAKH NATIONAL UNIVERSITY

N. Beissen H. Quevedo

LECTURE COURSE ON THEORETICAL MECHANICS Educational manual

Almaty «Qazaq university» 2017 1

UDC 53 / (075) LBC 22.2 я 73 B 40 Recommended for publication by the decision of the Academic Council of the Faculty of Physics-Technical Faculty and Publishing Council of the Kazakh National University named after Al-Farabi (Protocol № 3 dated 17 March 2017) Educational-methodical association on groups of specialties «Natural sciences», «Humanities», «Social sciences, economics and business», «Technical sciences and technology» and «Art» of Republican educational-methodical council of higher and postgraduate education of the Ministry of Education and Science of the Republic of Kazakhstan (Protocol № 1 dated 26 January 2016) Reviewers: doctor of Physics and Mathematics sciences, Professor K.N. Jumagulova doctor of Physics and Mathematics sciences, Professor L.M. Chechin doctor of Physics and Mathematics sciences, Professor A.A. Bekov

Beissen N. B 40 Lecture course on theoretical mechanics: educational manual / N. Beissen, H. Quevedo – Almaty: Qazaq university, 2017. – 156 p. ISBN 978-601-04-2335-0 This book is intended to review several standard topics of modern theoretical mechanics, which have been selected and briefly explained from a theoretical point of view. In addition, the definitions and theoretical explanations have been complemented with particular examples in which most of the intermediate calculations are presented. This is the first tutorial of theoretical mechanics written in English language for undergraduate students of Kazakhstan universities and institutes. It is based on lectures and seminars given by the authors for specialties «Physics» and «Nuclear Physics» at the al-Farabi Kazakh National University. The examples included in this tutorial are intended to help the students to solve the learning problems that have been detected during the lectures and seminars. Publishing in authorial release. Предлагаемая книга является изложением курса теоретической механики, в которой рассмотрены и объяснены ряд стандартных тем этой дисциплины с теоретической точки зрения. Кроме того, определения и теоретические объяснения дополнены конкретными примерами, в которых представлены большинство промежуточных расчетов. Это первый учебник по теоретической механике, написанный на английском языке для студентов высших учебных заведений казахстанских университетов и институтов. Он основан на лекциях и семинарах, прочитанных авторами в КазНУ им. АльФараби для специальностей «Физика» и «Ядерная физика». Примеры, включенные в этот учебник, призваны помочь студентам решить проблемы с обучением, обнаруженные во время лекций и семинаров. Издается в авторской редакции.

UDC 53 / (075) LBC 22.2 я 73 ISBN 978-601-04-2335-0

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© Beissen N., Quevedo H., 2017 © Al-Farabi KazNU, 2017

CONTENTS

PREFACE.................................................................................................. 5 THE SUBJECT OF THEORETICAL MECHANICS ............................... 6 I. EQUATIONS OF MOTION ................................................................. 7 1. Generalized coordinates. Coordinate planes and lines. Lame coefficients. Orthogonal system of coordinates ............................... 7 2. Curvilinear Motion. Curvilinear motion in polar coordinates. Example problem of curvilinear motion .................................................... 11 3. Newton's laws of motion. Newton's Universal law of gravitation ......... 18 4. Galilean transformation. Galilean Principle of Relativity ...................... 25 5. First and second integrals of motion ...................................................... 27 6. Lagrangian for a free particle. Lagrangian for a system of particles. The principle of least action or Hamilton's principle. Lagrange equations of the second kind or Euler-Lagrange equations ........ 30 7. Hamilton function. Derivation of the canonical equations from Hamilton's principle of least action. Using Hamilton's equations .............. 35 II. CONSERVATION LAWS ................................................................... 39 8. The law of conservation of energy ......................................................... 39 9. Momentum............................................................................................. 41 10. Center of mass ..................................................................................... 44 11. Angular momentum ............................................................................. 45 III. INTEGRATION OF THE EQUATIONS OF MOTION ..................... 51 12. Motion in one dimension ..................................................................... 51 13. Motion in a central field. Sectorial velocity, Kepler’s second law ....... 53 14. Kepler’s problem ................................................................................. 59 IV COLLISIONS BETWEEN PARTICLES ............................................. 65 15. Disintegration of particles ................................................................... 65 16. Elastic collisions .................................................................................. 72 17. Scattering. Rutherford’s formula. Effective cross-section .................. 80 V. SMALL OSCILLATIONS ................................................................... 87 18. Free oscillations in one dimension ....................................................... 87 19. Forced oscillations ............................................................................... 90 20. Damped oscillations ............................................................................. 94 21. Forced oscillations under friction......................................................... 98 22. Anharmonic oscillations ...................................................................... 100 3

VI. MOTION OF A RIGID BODY ........................................................... 105 23. Motion of a rigid body. Degrees of freedom and coordinates for a rigid body. Angular velocity. Instantaneous axis of rotation ............. 105 24. Inertia tensor. Moments of inertia. Principal axes of inertia and principal moments of inertia. Symmetric top and spherical top. Rotator .......................................................................... 106 25. Angular momentum of a rigid body The regular precession of a top .................................................................. 111 26. The equations of the motion of a rigid body ........................................ 114 27. Eulerian angles..................................................................................... 119 28. Euler's equations of motion for an absolute solid body with one fixed point ................................................................................... 123 29. Motion in a non-inertial frame of reference ......................................... 128 VII. THE CANONICAL EQUATIONS .................................................... 134 30. Poisson brackets. Properties of the Poisson brackets. Jacobi's identity and Poisson’s theorem ..................................................... 134 31. Abbreviated action. The variational principle determining the trajectory of the system or Maupertuis’ principle ................................ 137 32. The method of Hamilton-Jacobi. The mathematical structure of the equations of Hamilton-Jacobi. Their full integral ............................ 144 33. The method of separation of the variables ........................................... 148

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PREFACE

This tutorial is intended to serve as an introduction to the field of theoretical mechanics. It is a short book, and is not intended to explain in detail and to review all the topics of modern theoretical mechanics. Instead, several standard topics have been selected and briefly explained from a theoretical point of view. In addition, the definitions and theoretical explanations have been complemented with particular examples in which most of the intermediate calculations are presented. With this method we intend to help novice students to develop their own skills in solving problems independently. This is the first tutorial of theoretical mechanics written in English language for undergraduate students of Kazakh universities and institutes. It is based on lectures and seminars given by the authors at the al-Farabi Kazakh National University. The examples included in this tutorial are intended to help the students to solve the learning problems that have been detected during the lectures and seminars. We offer this book to our students with the aim to help them to understand a little bit the wonderful world of theoretical physics.

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THE SUBJECT OF THEORETICAL MECHANICS Mechanics is the branch of physics that studies the behavior of physical bodies when subjected to forces or displacements, and the subsequent effects of the bodies on their environment. This discipline has its origins in Ancient Greece with the studies performed by Aristotle and Archimedes. Later on, scientists such as Galileo, Kepler, and especially Newton, formulated the basis of what is now known as classical mechanics. Later developments made it necessary to split classical mechanics into two correlated branches which are usually known as Newtonian and theoretical mechanics. Newtonian mechanics uses vector quantities such as velocities, accelerations and forces to describe the motion of physical bodies. The basis of Newtonian mechanics consists of a number of empirical physical laws which are the result of observations and experiments. Today, we know that the validity of such laws is restricted to the case of velocities much less than the speed of light. The laws of Newtonian mechanics relate the underlying vector quantities in such a way that it is possible to predict the future displacement of a physical body, once its position and velocity are known at a given moment of time. This property is known as determinism. Theoretical mechanics implements more advanced mathematical tools such as variational principles and differential equations with the same goal of describing the free motion of physical bodies and their behavior under the action of external forces. The fundamentals of theoretical mechanics were developed by many physicists and mathematicians during the XVIII century in order to represent in a different manner and to generalize Newtonian mechanics. Instead of assuming the validity of the Newtonian laws, theoretical mechanics derive them by assuming the validity of a variational principle. In this sense, theoretical mechanics is a more profound approach that allows its generalization to include more complicated physical systems like fields.

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I EQUATIONS OF MOTION

1. Generalized coordinates. Coordinate planes and lines. Lame coefficients. Orthogonal system of coordinates A collection of Cartesian coordinates or curvilinear coordinates which represent the position of a mechanical system with respect to a reference frame is called a set of generalized coordinates, qi (i=1, 2, 3… n). The derivative of qi with respect to the time t defines the generalized velocity and is denoted by q i . Consider, for instance, a particular system which can be described by means of Cartesian coordinates x, y and z. Generalized coordinates qi (i=1,2,3) can also be used if there exists a set of transformations q1  q1  x, y, z   (1) q 2  q 2  x , y , z  , q  q  x, y , z  3  3

such that the Jacobian is different from zero, i.e,

 q J  i  x k

   0. 

(2)

On the contrary, if the Jacobian is equal to zero, it is not possible to introduce generalized coordinates. In fact, the non-vanishing of the Jacobian implies that it is possible to solve the equations (1) in terms of the generalized coordinates:  x  xq1 , q 2 , q 3    y  y q1 , q 2 , q 3  .  z  z q , q , q  1 2 3 

(3)

7

Let us consider the case of spherical coordinates which are related to Cartesian coordinates by means of the transformation

x  r cos  sin  y  r sin  sin  z  r cos  .

(4)

Fig. 1

Then, the Jacobian is calculated as  x   r x J     x   

 cos  sin   J    r sin  sin   r cos  cos  

y r y  y 

z   x  r   r z   y     r z   z     r

sin  sin  r cos  sin  r sin  cos 

x  y  z 

x     y    z   

cos    0   r 2 sin   0. (6)  r sin  

Consequently, we can compute the inverse transformation 8

(5)

  2 2 2 r   x  y  z y  J  0  x, y, z  r ,  ,  ,   arctg x  z  .   arccos 2  x  y2  z2 

(7)

Coordinate planes and lines        Consider the radius vector r  xi  yj  zk , where i , j , k are unit orthonormal vectors along the axes of a Cartesian coordinate system. Equivalently, the radius vector can be represented as   r  r x, y, z  . In the same manner, we can introduce the radius   vector r  r q1 , q 2 , q3  for a system of generalized coordinates. If we fix one of the coordinates, the remaining two coordinates form a coordinate plane. In the case of Cartesian coordinates, the coordinate planes are represented by flat surfaces. The intersection of any two coordinate planes constitute coordinate lines which are called q1 , q 2 , q3 -lines. For instance, in the case of Cartesian coordinate systems the coordinate lines are straight lines and, therefore, they are called straight line coordinate systems. For other coordinate systems, the coordinate lines are curved and hence they are called curvilinear coordinate systems. Lame coefficients and orthogonal system of coordinates    Consider the unit vectors e1 , e2 , e3 along the generalized coordinates q1 , q 2 , q3 . Then, we define the line element

  3 3  r   r ds  dr  dr       m 1n 1 q  m q n 2

   dq m  dq n . 

It is convenient to introduce the notation

(8)

  r  hm  e m q m 9

3 3   ds 2    hm hn ет еn   dq m  dq n .

(9)

m 1n 1

If the coordinate lines are orthogonal the corresponding system is called orthogonal and satisfy the condition

em en    mn ,

(10)

ds 2   hm2 dq m2  h12 dq12  h22 dq 22  h32 dq 32 .

(11)

then 3

m 1

The quantity

dV  h1  h2  h3  dq1  dq2  dq3 .

(12)

determines an infinitesimal volume. Moreover, the quantities  r  hm are known as the Lame coefficients. The Cartesian, q m cylindrical and spherical coordinate systems are examples of orthogonal coordinate systems. The Lame coefficients are given by   x r hm    q m  q m

2

  y      q m

2

  z      q m

2

  . 

(13)

The line element can be written in terms of the metric tensor g  as ds 2  g  dx  dx where the convention of summation over repeated indices is used. For instance, in the case of four generalized coordinates, we obtain ds2  gdx dx  g00dx0dx0  g01dx0dx1  g02dx0dx2  g03dx0dx3  g10dx1dx0  g11dx1dx1  g12dx1dx2  g13dx1dx3  g20dx2dx0  g21dx2dx1  g22dx2dx2  g23dx2dx3  (14)  g30dx3dx0  g31dx3dx1  g32dx3dx2  g33dx3dx3  g00dx0dx0  g11dx1dx1   g22dx2dx2  g33dx3dx3  c2dt2  h12dq12  h22dq22  h32dq32 ,

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where we have introduced the coordinates 0 1 2 3 x  ct , x  q1 , x  q2 , x  q3 , and have used the orthogonal   condition em en   g   diag (1,h12 ,h22 ,h32 ). For spherical coordinates

x  r cos  sin  y  r sin  sin  z  r cos  , we found that the Jacobian J  r 2 sin   0 so that the Lame coefficients are h1  1, h2  r , h3  r sin  and the volume can be calculated as

V   dx  dy  dz   h1  h2  h3  dr  d  d   r

 0

 h1  h2  h3  J  2  0

2

2  r sin dr  d  d 

0

4 3 r . 3

(15)

In addition, the line element reads 3

ds 2   hm2 dq m2  h12 dq12  h22 dq22  h32 dq32  dr 2  r 2 d 2  r 2 sin 2 d 2 . (16) m 1

2. Curvilinear Motion

Curvilinear motion is defined as the motion that occurs when a particle travels along a curved path. The curved path can be in two dimensions (a plane) or in three dimensions. This type of motion is more complex than the rectilinear (straight-line) motion. Threedimensional curvilinear motion describes the most general case of motion for a particle. To find the velocity and acceleration of a particle experiencing curvilinear motion, one only needs to know the position of the particle as a function of time. Let us suppose we are given the position of a particle P in three-dimensional Cartesian (x,y,z) coordinates, with respect to time, where 11

x p  x p t ,

y p  y p t ,

(1)

z p  z p t .

Fig.2

The velocity and acceleration of the particle is given by dx p vx  , dt dy p vy  , dt dz p vz  . dt

(2)

and ax  ay  az  12

d 2xp dt 2 d 2 yp dt 2 d 2zp dt 2

, , .

(3)

It follows that if we know the position of a particle as a function of time, it is a fairly simple exercise to find the velocity and acceleration. One simply calculates the first derivative to find the velocity and the second derivative to find the acceleration. The magnitude of the velocity of the particle P is given by v p  v x2  v y2  v z2 where as the magnitude of the acceleration is defined as a p  a x2  a y2  a z2 . Note that the direction of the velocity of the particle is always tangent to the curve (i.e. the path traveled, denoted by a curve in the figure above). Nevertheless, the direction of the acceleration is in general not tangent to the curve. However, the acceleration component tangent to the curve is equal to the time derivative of the magnitude of the particle velocity (along the curve). In other words, if vt is the magnitude of the particle velocity (tangent to the curve), the acceleration component dv of the particle tangent to the curve (at) is simply at  t . In dt addition, the acceleration component normal to the curve (an) is v2 given by a n  t where R is the radius of curvature of the curve at R a given point on the curve x p , y p , z p . The figure below illustrates the acceleration components at and an at a given point on the curve xp , yp , zp .

For the specific case where the trajectory of the particle is given by y  f x  (two-dimensional motion), the radius of curvature R reads   dy  2  1       dx    R  d2y dx 2

3

2

.

(4)

13

Fig. 3

Where x means the «absolute value» of x. For example, |-2.5| = 2.5, and |3.1| = 3.1. However, it is usually not necessary to know the radius of curvature R along a curve. But nonetheless, it is informative to understand it on the basis of its relationship to the normal acceleration an . Curvilinear motion in polar coordinates It is sometimes convenient to express the planar (twodimensional) motion of a particle in terms of polar coordinates R, , so that we can explicitly determine the velocity and acceleration of the particle in the radial ( R -direction) and circumferential (  -direction) directions. For this type of motion, a particle is only allowed to move along the radial R -direction for a given angle  . For a particle P determined in polar coordinates (as shown below), we can derive a general equation for its radial velocity ( vr ), radial

acceleration ( a r ), circumferential velocity ( vc ), and circumferential acceleration ( ac ). Note that the circumferential direction is perpendicular to the radial direction. 14

Fig. 4

The position of the particle P is given in terms of time, i.e., x t   R cos  , y t   R sin  . To find the velocity, we need to take the first derivative of xt  and y t  with respect to time: dx dR  cos   R sin  dt dt dy dR  sin   R cos  dt dt

d , dt d . dt

(5)

To find the acceleration, we compute the second derivative of

xt  and y t  with respect to time:

2

d 2x d 2R dR d d 2  d  R R     cos 2 sin cos  sin  ,        dt 2 dt 2 dt dt dt 2  dt  2

d 2 y d 2R dR d d 2  d  R R      sin  2 cos   sin   cos  .   dt 2 dt 2 dt dt dt 2  dt 

(6)

Without loss of generality we can evaluate the velocities and accelerations at the angle   0 , which implies that at this angle the radial velocity and radial acceleration are in the x -direction, whereas the circumferential velocity and circumferential acceleration are in the y -direction. Setting   0 we have: 15

vr 

dx , dt

d 2x , dt 2

ar 

vc 

ac 

dy , dt

d2y , dt 2

vr 

dR ; dt

(7) 2

ar 

d 2R  d   R  ; 2 dt  dt 

(8)

d ; dt

(9)

vc  R

ac  2

d 2 dR d  R 2 . dt dt dt

(10)

Equations (7), (8), (9), and (10) fully describe the curvilinear d motion of a particle P in polar coordinates. The term is called dt angular velocity. It has units of rad . One rad (radian) = 57.296 s d 2 is called angular acceleration. It has units of degrees. The term dt 2 rad 2 . Since v and v are perpendicular to each other, the r c s magnitude of the velocity of the particle P is given by v p  vr2  vc2 . Moreover, since ar and ac are perpendicular to each other, the magnitude of the acceleration of the particle P is given by a p  a r2  ac2 . Example problem of curvilinear motion A slotted link is rotating about a fixed pivot O with a counterclockwise angular velocity of 3 rad , and a clockwise s rad angular acceleration of 2 . The movement of the link is s2 causing the rod to slide along the curved channel, as shown in the figure. The radius of the channel as a function of  is given by, 16

R  0.7 (with R in meters and  in radians). Determine the velocity and acceleration components of the rod at   45

Fig. 5

Solution: The angle   45 0 is equal to 

radians. In the 4 equations, counterclockwise angular velocity is positive, and clockwise angular acceleration is negative (since it acts to «slow down» the rotational speed of the link). The radial velocity of the rod is given by equation vr 

dR d  0,7  0,73  2,1 m . s dt dt

(11)

The radial velocity is in the direction of increasing R and the circumferential velocity of the rod is given by the equation vc  R

d    0,7   3  1,65 m . s dt 4

(12)

The circumferential velocity is in the direction of increasing θ and the radial acceleration of the rod is given by the equation 17

2

ar 

d 2R d 2  d      2 2  R   0,7 2  0,7   3  0,7 2  - 0,7   3  -6/35 m s 2 dt dt  dt  4 4

. (13)

The radial acceleration is in the direction of decreasing R and the circumferential acceleration of the rod is given by the equation 2

ac  2

dR d d   d     d    R  0,7  2     2   0,7  dt dt dt  dt  dt   4

   2  0,733  0,7    2   13,7 m 2 . s 4

(14)

3. Newton's laws of motion. Newton's Universal law of gravitation

They consist of three physical laws that form the basis for classical mechanics. They describe the relationship between the forces acting on a body and its motion due to those forces. They have been expressed in several different ways over nearly three centuries, and can be summarized as follows: 1. First law: If an object experiences no net force, then its velocity is constant: the object is either at rest (if its velocity is zero), or it moves in a straight line with constant speed (if its velocity is nonzero).  2. Second law: The acceleration a of a body is parallel and  directly proportional to the net force F acting on the body, is in the direction of the net force, and is inversely proportional to the mass m   of the body, i.e., F  ma .  3. Third law: When a first body exerts a force F1 on a second   body, the second body simultaneously exerts a force F1   F2 on the   first body. This means that F1 and F2 are equal in magnitude and opposite in direction. The three laws of motion were first compiled by Sir Isaac Newton in his work Philosophiæ Naturalis Principia Mathematica, first published in 1687. Newton used them to explain and investigate 18

the motion of many physical objects and systems. For example, in the third volume of the text, Newton showed that these laws of motion, combined with his universal law of gravitation, explained Kepler's laws of planetary motion. Newton's laws are applied to bodies (objects) which are considered or idealized as a particle, in the sense that the extent of the body is neglected in the evaluation of its motion, i.e., the object is small compared to the distances involved in the analysis, or the deformation and rotation of the body is of no importance in the analysis. Therefore, a planet can be idealized as a particle for analysis of its orbital motion around a star. In their original form, Newton's laws of motion are not adequate to characterize the motion of rigid bodies and deformable bodies. Leonard Euler in 1750 introduced a generalization of Newton's laws of motion for rigid bodies called the Euler's laws of motion, later applied as well for deformable bodies assumed as a continuum. If a body is represented as an assemblage of discrete particles, each governed by Newton’s laws of motion, then Euler’s laws can be derived from Newton’s laws. Euler’s laws can, however, be taken as axioms describing the laws of motion for extended bodies, independently of any particle structure. Newton's laws hold only with respect to a certain set of frames of reference called Newtonian or inertial reference frames. Some authors interpret the first law as defining what an inertial reference frame is; from this point of view, the second law only holds when the observation is made from an inertial reference frame, and therefore the first law cannot be proved as a special case of the second. Other authors do treat the first law as a corollary of the second. The explicit concept of an inertial frame of reference was not developed until long after Newton's death. In the given interpretation mass, acceleration, momentum, and (most importantly) force are assumed to be externally defined quantities. This is the most common, but not the only interpretation of the way one can consider the laws to be a definition of these quantities. Newtonian mechanics has been superseded by special relativity, but it is still useful as an approximation when the speeds involved are much slower than the speed of light. 19

Explanation of Newton's first law. The first law states that if the net force (the vector sum of all forces acting on an object) is zero, then the velocity of the object is constant. Velocity is a vector quantity which expresses both the object's speed and the direction of its motion; therefore, the statement that the object's velocity is constant is a statement that both its speed and the direction of its motion are constant. The first law can be stated mathematically as   dv 0. F  0 dt

(1)

Consequently, – An object that is at rest will stay at rest unless an unbalanced force acts upon it. – An object that is in motion will not change its velocity unless an unbalanced force acts upon it. This is known as uniform motion. An object continues to do whatever it happens to be doing unless a force is exerted upon it. If it is at rest, it continues in a state of rest (demonstrated when a tablecloth is skillfully whipped from under dishes on a tabletop and the dishes remain in their initial state of rest). If an object is moving, it continues to move without turning or changing its speed. This is evident in space probes that continually move in outer space. Changes in motion must be imposed against the tendency of an object to retain its state of motion. In the absence of net forces, a moving object tends to move along a straight line path indefinitely. Newton placed the first law of motion to establish frames of reference for which the other laws are applicable. The first law of motion postulates the existence of at least one frame of reference called a Newtonian or inertial reference frame, relative to which the motion of a particle not subject to forces is a straight line at a constant speed. Newton's first law is often referred to as the law of inertia. Thus, a condition necessary for the uniform motion of a particle relative to an inertial reference frame is that the total net force acting on it is zero. In this sense, the first law can be restated 20

as: In every material universe, the motion of a particle in a preferential reference frame Φ is determined by the action of forces whose total vanishes for all times when and only when the velocity of the particle is constant in Φ. That is, a particle initially at rest or in uniform motion in the preferential frame Φ continues in that state unless compelled by forces to change it. Newton's laws are valid only in inertial reference frames. Any reference frame that is in uniform motion with respect to an inertial frame is also an inertial frame, i.e. Galilean invariance or the principle of Newtonian relativity. Explanation of Newton's second law. The second law states that the net force on an object is equal to the  rate of change (that is, the derivative) of its linear momentum p in an inertial reference frame:   dp d mv  . (2)  F dt dt The second law can also be stated in terms of an object's acceleration. Since the law is valid only for constant-mass systems, the mass can be taken outside the differentiation operator by the constant factor rule in differentiation. Thus,    dv F m  ma , dt

(3)

  where F is the net force applied, m is the mass of the body, and a is the body's acceleration. Thus, the net force applied to a body produces a proportional acceleration. In other words, if a body is accelerating, then there is a force on it. Consistent with the first law, the time derivative of the momentum is non-zero when the momentum changes direction, even if there is no change in its magnitude; such is the case with uniform circular motion. The relationship also implies the conservation of momentum: when the net force on the body is zero, the momentum of the body is constant. Any net force is equal to the rate of change of the momentum. 21

Any mass that is gained or lost by the system will cause a change in momentum that is not the result of an external force. A different equation is necessary for variable-mass systems (see below). Newton's second law requires modification if the effects of special relativity are to be taken into account, because at high speeds the approximation that momentum is the product of rest mass and velocity is not accurate.   Impulse. An impulse p occurs when a force F acts over an interval of time t and it is given by p   Fdt . t

(4)

Since force is the time derivative of momentum, it follows that    p  p  mv .

(5)

This relation between impulse and momentum is closer to Newton's wording of the second law. Impulse is a concept frequently used in the analysis of collisions and impacts. Variable-mass systems. Variable-mass systems, like a rocket burning fuel and ejecting spent gases, are not closed and cannot be directly treated by making the mass a function of time in the second law. The falsehood of this formula can be seen by noting that it does  not respect Galilean invariance: a variable-mass object with F  0 in  one frame will be seen to have F  0 in another frame. The correct equation of motion for a body whose mass m varies with time by either ejecting or accreting mass is obtained by applying the second law to the entire, constant-mass system consisting of the body and its ejected/accreted mass; the result is    dm dv . F u m dt dt 22

(6)

 where u is the relative velocity of the escaping or incoming mass as seen by the body. From this equation one can derive the Tsiolkovsky rocket equation.  dm Under some conventions, the quantity u on the left-hand dt side, known as the thrust, is defined as a force (the force exerted on the body by the changing mass, such as rocket exhaust) and is  included in the quantity F . Then, by substituting the definition of   acceleration, the equation becomes F  ma . Explanation of Newton's third law. The third law states that all  forces exist in pairs: if one object A exerts a force FA on a second  object B , then B simultaneously exerts a force FB on A , and the   two forces are equal and opposite: FA   FB . The third law means that all forces are interactions between different bodies, and thus that there is no such thing as a unidirectional force or a force that acts on only one body. This law is sometimes referred to as the action-reaction law,   with FA called the «action» and FB the «reaction». The action and the reaction are simultaneous, and it does not matter which is called the action and which is called reaction; both forces are part of a single interaction, and neither force exists without the other. The two forces in Newton's third law are of the same type (e.g., if the road exerts a forward frictional force on an accelerating car's tires, then it is also a frictional force that Newton's third law predicts for the tires pushing backward on the road). From a conceptual standpoint, Newton's third law is seen when a person walks: they push against the floor, and the floor pushes against the person. Similarly, the tires of a car push against the road while the road pushes back on the tires and road simultaneously push against each other. In swimming, a person interacts with the water, pushing the water backward, while the water simultaneously pushes the person forward both the person and the water push against each other. The reaction forces account for the motion in these examples. These forces depend on friction; a person or car on ice, for example, may be unable to exert the action force to produce the needed reaction force. 23

Newton's universal law of gravitation Every point mass attracts every single other point mass by a force pointing along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them:

 mm  F G 13 2 r , r

(7)

Fig. 6

 where: F is the force between the masses, G is the gravitational  constant, m1 is the first mass, m2 is the second mass, and r is the distance between the centers of the masses. Newton's Universal law of gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Moreover, large spherically symmetrical masses attract and are attracted as if all their mass were concentrated at their centers. This is a general physical law derived from empirical observations by what Newton called induction. It is a part of classical mechanics and was formulated in Newton's work Philosophiæ Naturalis Principia Mathematica («the Principia»), first published on 5 July 1687. (When Newton's book was presented in 1686 to the Royal Society, Robert Hooke made a claim that Newton had plagiarized the inverse square law from him – see History section below). Newton's law of gravitation resembles Coulomb's law of electrical forces, which is used to calculate the magnitude of 24

electrical force between two charged bodies. Both are inverse-square laws, in which force is inversely proportional to the square of the distance between the bodies. Coulomb's law has the product of two charges in place of the product of the masses, and the electrostatic constant in place of the gravitational constant. Newton's law has since been superseded by Einstein's theory of general relativity, but it continues to be used as an excellent approximation of the effects of gravity. Relativity is required only when there is a need for extreme precision, or when dealing with gravitation for extremely massive and dense objects. 4. Galilean transformation. Galilean Principle of Relativity

The Galilean transformation is used to transform between the coordinates of two reference frames which differ only by constant relative motion within the constructs of Newtonian physics. Galileo formulated these concepts in his description of uniform motion. In essence, the Galilean transformations embody the intuitive notion of addition and subtraction of velocities. The assumption that time can be treated as absolute is at the heart of the Galilean transformations, so they presume absolute time and space. The notation below describes the relationship under the Galilean transformation between the coordinates  x, y , z , t  and x , y , z , t  of a single arbitrary event, as measured in two coordinate systems K and K  in uniform relative motion (velocity v ) in their common x and x  directions, with their spatial origins coinciding at time t  t '  0 :

Fig. 7

25

    r  i x  jy  kz .

   r   r  v0 t .  x '  x  v0 t ,  '  y  y,  z '  z. 

(8) (9)

(10)

Standard configuration of coordinate systems for Galilean transformations. Galilean Principle of Relativity. This is the principle of the physical equality of inertial systems in classical mechanics, which is manifested in the fact that the laws of mechanics are identical in all such systems. Hence, it follows that it is impossible to determine by any mechanical experiment conducted in any inertial system whether the given system is at rest or moving uniformly and rectilinearly. The principle was first established by Galileo in 1636. Galileo demonstrated the identical nature of the laws of mechanics for inertial systems by using as an example the phenomena occurring below the deck of a ship that is at rest or moving uniformly and rectilinearly (with respect to the earth, which may to a sufficient degree of accuracy be considered as an inertial system): «Have the ship proceed at any speed you like so long as the motion is uniform and not fluctuating this way and that. Mathematically, the Galilean principle of relativity expresses the invariance of the equations of mechanics with respect to transformations of the coordinates of moving points (and time) upon a transition from one inertial system to another (Galilean transformations). In classical mechanics the motion of a particle is defined by Newton’s second law: f x'  m

26

d 2 x' d2 d2x d       m v 0   f x . m x v t m 0 2 2 2 dt dt dt dt

(11)

 Where m is the mass of a point and F is the resultant of all forces applied to it. Here the forces (and masses) are invariants in classical mechanics that is, they are quantities that do not change on transition from one frame of reference to another. Therefore, during Galilean transformations, equation (3) does not change. This is the mathematical expression of the Galilean principle of relativity.

5. First and second integrals of the motion

The main problem of dynamics is to find the solution of the   equations of motion r  r t  which represents the trajectory equation. If it is not possible to specify the effective force, it is necessary to consider an initial point t 0 where the position and velocity must be given:

  r t t  r0 ;

  v t t  v0 .

0

0

(1)

In a Cartesian coordinate system, this is equivalent to:

x t t  x0 ; x t  t  x 0 . 0

0

y t t  y0 ; y t  t  y 0 . 0

0

(2)

z t t  z 0 ; z t  t  z0 . 0

0

The effective force can be obtained from Newton’s second law





   mr  f r , r , t ,

(3)

which in a Cartesian coordinate system reads:

mx  f x x, y, z , x. y , z, t , my  f y x, y, z, x. y , z, t ,

(4)

mz  f z x, y, z, x. y , z, t .

27

  If the effective force is a single-valued function of its arguments r , v , t , then there will be one solution of the equations of motion (3) satisfying the condition (2). The mechanical state of a material point is determined by its  position r and its velocity v . This can be used to explain the single value solution of the motion equation: the mechanical state of a material point at time t is determined uniquely by its initial mechanical state and by the motion equation. We will consider a mechanical state as determining a point of the right-hand side of (4) expressed in terms of the resultant force  f . This concept represents the principle of the mechanical effect or the determinism of mechanics. Then, the mathematical function   f r , r , t is the initial mechanical state of a material point and the initial condition of motion. For t  t 0 the mechanical state changes in accordance with the equation of motion. One can say that the initial conditions are the reason and the motion is the effect. If a second-degree equation satisfies the conditions





d1 x, y, z, x, y , z, t   0  1 x, y, z, x, y , z, t   C1 , dt d 2 x, y, z, x, y , z, t   0   2 x, y, z, x, y , z, t   C2 , dt d 3 x, y, z, x, y , z, t   0   3 x, y, z, x, y , z, t   C3 . dt

(5)

Here Ci are known as constants of integration and (5) the first integrals of motion. To solve the equation, we calculate d1 x, y, z, t , C1 , C2 , C3   0 , dt d 2 x, y, z, t , C1 , C2 , C3   0 , dt d 3 x, y, z, t , C1 , C2 , C3   0 . dt 28

(6)

Then

1 x, y, z , t , C1 , C2 , C3   C4 ,  2 x, y, z , t , C1 , C2 , C3   C5 ,  3  x, y, z, t , C1 , C2 , C3   C6 .

(7)

The expressions give in Eq.(7) represent second integrals of motion. If we solve the system (7) for the coordinates, we obtain

x  xt , C1 , C2 , C3 , C4 , C5 , C6  , y  y t , C1 , C 2 , C3 , C4 , C5 , C6  , z  z t , C1 , C2 , C3 , C4 , C5 , C6  .

(8)

These functions are the solution of the differential equation of motion. They have six integral constants, and are the general solution of the equation of motion. To find the general solution of this equation is equivalent to solving the main problem of dynamics. If we substitute the initial conditions at time moment t 0 in C1 , C2 , C3 , we will find the constants of the first integral:

Ci   i t 0 , x0 , y0 , z 0 , x0 , y 0 , z0  .

(9)

Then, if we replace these constants in the equations (7) at t  t0 , we find the constants of the second integral.

C4  1 x0 , y0 , z 0 , t 0 , C1 , C2 , C3  , C5   2  x0 , y0 , z 0 , t 0 , C1 , C2 , C3  , C6   3  x0 , y0 , z 0 , t 0 , C1 , C2 , C3  .

(10)

So, depending on the value of qi and qi some functions will keep these values as a result of the initial conditions. These functions are integrals of motion. 29

Control questions: 1. 2. 3. 4. 5.

Main task of dynamics. Determinism of mechanics. First integral of motion. Second integral of motion. How to give initial conditions?

6. Lagrangian for a free particle. Lagrangian for a system of particles. The principle of least action or Hamilton's principle. Lagrange equations of the second kind or Euler-Lagrange equations

The core element of Lagrangian mechanics is the Lagrangian function, which summarizes the dynamics of the entire system in a very simple expression. The physics analyzing a system is reduced to choosing the most convenient set of generalized coordinates, determining the kinetic and potential energies of the constituents of the system, then writing down the equation for the Lagrangian to use in Lagrange's equations. It is defined by

L

mv 2 U T U , 2

(1)

Where T is the total kinetic energy and U is the total potential energy of the system. In general, the Lagrangian function depends on the generalized coordinates and velocities:

L  Lqi , qi   Lq1 , q2 ,...qi , q1 , q 2 ,...qi  . The next fundamental element is the action time integral of the Lagrangian: t2

S   Ldt.

(2)

S , defined as the

(3)

t1

The action contains the dynamics of the system, and has deep theoretical implications. Technically, the action is a functional, rather 30

than a function: its value depends on the full Lagrangian function for all times between

t1 and t2 .

Hamilton's principle of stationary action

Let q0 and q1 be the coordinates at the initial and final time

t0

and t1 . Using the calculus of variations, it can be shown that Lagrange's equations are a consequence of Hamilton's principle: The

trajectory of the system between t 0 and t1 has a stationary action S. By stationary, we mean that to first order the action does not vary under the application of infinitesimal deformations of the trajectory, with the end-points ( q0 , t 0 ) and ( q1 , t1 ) fixed. Hamilton's principle can be written as:

S  0 .

(4)

Hamilton's principle is sometimes referred to as the principle of least action, however the action functional need only be stationary, not necessarily a maximum or a minimum value. Any variation of the functional gives an increase in the functional integral of the action. Lagrange equations of the second kind. Euler-Lagrange equations Below, we sketch out the derivation of the Lagrange equations of the second kind. Then writing down the variation L :   L L qi  q i . q i   qi

L    i

(5)

Then t

 L

L



S     qi  qi dt . t0 i qi   qi

(6)

31

The integral of the second term t L  dq  L q i dt     i dt , i  i  dt  t0 q t0 q t



(7)

can be calculated by using the commutativity condition dqi d  qi dt dt



(8)

and takes the form t L d L  dqi  q dt .  dt   i  dt  i dt i t0 q t0 q t



(9)

Furthermore, using the method of integration by parts, we obtain t

t L d qi dt  L qi   d L qi dt   i dt i t0 dt q t0 q qi t t

0

t



   L  L L qi   qi    qi   0  qi  t0  t  qi  qi t0 d L q dt. i i t0 dt q t

 

(10)

The condition of fixed end-points qt0   qt   0 implies that

 L d L  qi dt  0 .  i 1 t0 q  i dt q i  n t

S    

(11)

Since the variations qi are arbitrary, the vanishing of the terms in brackets is equivalent to the condition S  0 . For any system with s degrees of freedom, the Lagrange equations include s generalized coordinates and s generalized 32

velocities. The equations of motion in Lagrangian mechanics are the Lagrange equations of the second kind, also known as the Euler– Lagrange equations:

L d L   0, i  1,2,...s . qi dt q i

(12)

where i  1,2,...s represents the i th degree of freedom, qi are the generalized coordinates, and qi are the generalized velocities. To obtain Newton’s equations, we write the Lagrangian in Cartesian coordinates as:

q1  x, q2  y,

(13)

q 3  z.

L

m 2 x  y 2  z 2   U x, y, z . 2

(14)

Using Lagrange’s equations: Li L U   ; qi x x

d Li d  L  d     mx  mx . dt q i dt  x  dt

(15)

In other words, Newton’s equations are: U , x U my   , y U mz   . z mx  

(16)

Consider now the case of polar coordinate with: 33

q1  r ,

(17)

q2   . Then, the kinetic energy becomes

T

m 2 r  r 2 2  2

(18)

and for the potential energy we choose the particular case

U 

 r

.

(19)

Consequently, Lagrange’s function is

L T U 

m 2 r  r 2 2    , 2 r

(20)

which leads to the following Lagrange’s equations

L d L  d    2  mr 2  mr   0, r dt ri r dt L d L d    mr  0,   dt  i dt d mr  mr dt

(21)

(22)

(23)

or equivalently,

  2 mr   2  mr , r  mr  const.

34

(24)

This shows that once the Lagrange function is given, the calculation of the corresponding motion equations is straightforward. 7. Hamilton function. Derivation of the system of canonical equations from Hamilton's principle of least action. Technique of using Hamilton's equations

The Hamiltonian H represents the energy of the system (provided that there are no external forces and that no energy is added to the system), which is the sum of the kinetic and potential energy, traditionally denoted T ( pi ) and U ( qi ) , respectively. Here qi is the coordinate and pi is the momentum, mqi . Then

T

H  T U,

(1)

p2 , 2m

(2)

U  U(q i ).

Note that T is a function of pi alone, while U is a function of qi alone. In Lagrangian mechanics, the equations of motion are based on generalized coordinates. Hamiltonian mechanics aims to replace the generalized velocity variables with generalized momentum variables, also known as conjugate moments. For each generalized velocity, there is one corresponding conjugate momentum, defined as: pi 

L . q i

In Cartesian coordinates, the generalized moment is precisely the physical linear momenta. In circular polar coordinates, the generalized momentum corresponding to the angular velocity is the physical angular momentum. For an arbitrary choice of generalized coordinates, it may not be possible to obtain an intuitive interpretation of the conjugate moment. The Hamiltonian is the Legendre transform of the Lagrangian: 35

H (qi , pi , t )   pi q i  L(qi , q i , t ).

(3)

i

If the transformation equations defining the generalized coordinates are independent of t , and the Lagrangian is a sum of products of functions (in the generalized coordinates) which are homogeneous of order 0, 1 or 2, then it can be shown that H is equal to the total energy E  T U .

(4)

Deriving Hamilton's equations. Hamilton's equations can be derived by looking at how the total differential of the Lagrangian depends on time, generalized positions qi and generalized velocities qi . In terms of the Hamiltonian, the action can be written as t S     Pi q i  H dt.  t0  i

(5)

Then, the variation of the action is given as t t S      Pi q i  H dt      Pi qi  H dt . t0





i

t0





i

(6)

On the other hand, the variation of the Hamiltonian can be expressed as

H   i

H H qi  Pi . qi Pi

Consequently, h side in the definition of differential: t



H

H

(7)

H 

produces a

S     Pi q i  q i Pi  qi  Pi dt , t0 i Pi q i   36

(8)

t

t

t0

t0

t dqi d dt   Pi qi dt , t0 dt dt

 Pi q i dt   Pi 

t

 Pi

t0

d qi dt  Pi qi dt

t



t0 i



t

t dP dPi qi dt    i qi dt , t 0 dt t 0 dt t

 t0

   Piq i  q i Pi 

t  dP H      i  t0 i   dt qi

(9)

(10)

 H H qi  Pi dt  qi Pi 

  dq H   qi   i  Pi dt 0.   dt Pi  

(11)

Since the above equation must be satisfied for any values of the variations qi and Pi the terms inside the brackets must be zero, i. e.,



dPi H  0, dt qi

dqi H  0 dt Pi

(12)

or

dPi H dqi H  , .  dt qi dt Pi

(13)

These the Hamilton equations of motion which consist of 2n first-order differential equations. Instead, Lagrange's equations consist of n second-order equations. However, Hamilton's equations usually do not reduce the difficulty of finding explicit solutions. They still offer some advantages, since important theoretical results can be derived because coordinates and momenta are independent variables with nearly symmetric roles. Hamilton's equations have another advantage over Lagrange's equations: if a system has a symmetry, such that a coordinate does not occur in the Hamiltonian, the corresponding momentum is conserved, and that coordinate can be ignored in the other equations of the set. Effectively, this reduces the problem from n coordinates to 37

(n-1) coordinates. In the Lagrangian framework, of course the result that the corresponding momentum is conserved still follows immediately, but all the generalized velocities still occur in the Lagrangian – we still have to solve a system of equations in n coordinates. The Lagrangian and Hamiltonian approaches provide the groundwork for deeper results in the theory of classical mechanics, and for formulations of quantum mechanics. Control questions: 1. 2. 3. 4. 5.

38

Number of degrees of freedom. Generalized coordinates, generalized velocities. Principle of least action or Hamilton’s principle. Lagrangian and Hamiltonian. Definition of the action.

II CONSERVATION LAWS

8. The law of conservation of energy

As mentioned above, during the motion of a mechanical system, the 2s quantities qi sand qi (i=1,2,…,s) which specify the state of the system vary with time. There exist, however, functions of these quantities whose values remain constant during the motion, and depend only on the initial conditions. Such functions are called integrals of motion. Not all integrals of motion, however, are of equal importance in mechanics. There are some whose constancy is of profound significance, deriving from the fundamental homogeneity and isotropy of space and time. The quantities represented by such integrals of the motion are said to be conserved, and have an important common property of being additive: their values for a system composed of several parts whose interaction is negligible are equal to the sums of their values for the individual parts. It is due to this additivity that the quantities concerned owe their especial importance in mechanics. Let us suppose, for example, that two bodies interact during a certain interval of time. Since each of the additive integrals of the whole system is, both before and after the interaction, equal to the sum of its values for the two bodies separately, the conservation laws for these quantities immediately make possible various conclusions regarding the state of the bodies after the interaction, if their states before the interaction are known. Let us consider first the conservation law resulting from homogeneity of time. By virtue of this homogeneity, the Lagrangian of a closed system does not depend explicitly on time. The total time derivative of the Lagrangian can therefore be written dL L dqi L dqi   i i dt qi dt qi dt

(1)

39

L would have to be t L added on the right-hand side of the above equation. Replacing qi in accordance with Lagrange’s equations,

If L would explicitly on time, a term

L d L  qi dt q i

(2)

we obtain dL d L d L d  L  L dq i L dq i  (3) q i   q i        q i i q i q  i dt i dt q i  i dt i dt  q i  dt i dt q i

or

 L d   q i  L   0. dt  i q i 

(4)

Hence we see that the quantity  qi i

L LE qi

(5)

remains constant during the motion of a closed system, i.e. it is an integral of the motion; it is called the energy of the system. The additivity of the energy follows immediately from that of the Lagrangian, since (5) shows that it is a linear function of the latter. The law of conservation of energy is valid not only for closed systems, but also for those in a constant external field (i.e. one independent of time): the only property of the Lagrangian used in the above derivation, namely that it does not involve the time explicitly, is still valid. Mechanical systems whose energy is conserved are sometimes called conservative systems. As we have seen, the Lagrangian of a closed system (or one in a constant field) is of the form 40

L  T q, q   U q ,

(6)

where T is a quadratic function of the velocities. Using Euler’s theorem on homogeneous functions, we have L T   q i  2T . i q i q i

 q i i

(7)

Substituting this in (5) gives E  T q, q   U q ,

(8)

in Cartesian coordinates, 1 E   mva2  U r1 , r2 ,.... a 2

(9)

Thus the energy of the system can be written as the sum of two quite different terms: the kinetic energy, which depends on the velocities, and the potential energy, which depends only on the coordinates of the particles. 9. Momentum

A second conservation law follows from the homogeneity of space. By virtue of this homogeneity, the mechanical properties of a closed system are unchanged by any parallel displacement of the entire system in space. Let us therefore consider an infinitesimal  displacement  , and obtain the condition for the Lagrangian to remain unchanged. A parallel displacement is a transformation in which every particle in the system is moved by the same amount, the radius    vector r becoming r   . The change in L resulting from an infinitesimal change in the coordinates the velocities of the particles remaining fixed, is L  ra



L ra

L    ra     , a

a

(1) 41

 where the summation is over the particles in the system. Since  is arbitrary, the condition L  0 is equivalent to  a

L  0. ra

(2)

From Lagrange’s equation we therefore have L L d L d     0. a a qa dt va dt va

(3)

Thus we conclude that, in a closed mechanical system, the vector  L , P a v a

(4)

remains constant during the motion; it is called the momentum of the system. Differentiating the Lagrangian, we find that the momentum is given in terms of the velocities of the particles by

  P   ma v a . a

(5)

The additivity of the momentum is evident. Moreover, unlike the energy, the momentum of the system is equal to the sum of its values   pa  ma va for the individual particles, whether or not the interaction between them can be neglected. The three components of the momentum vector are all conserved only in the absence of an external field. The individual components may be conserved even in the presence of a field, however, if the potential energy in the field does not depend on all the Cartesian coordinates. The mechanical properties of the system are evidently unchanged by a displacement along the axis of a coordinate, which does not appear in the potential energy, and so the corresponding component of the momentum is conserved. For example, in a 42

uniform field in the z -direction, x and y components of momentum are conserved. Equation (11) has a simple physical meaning. The derivative L U       Fa ra ra

(6)

is the force acting on the a -th particle. Thus, equation (11) signifies that the sum of the forces on all the particles in a closed system is zero:  (7)  Fa  0. a

  In particular, for a system of only 2 particles, F1  F2  0 : the force exerted by the first particle on the second is equal in magnitude, and opposite in direction, to that exerted by the second particle on the first. This is the equality of action and reaction (Newton’s first law). If the motion is described by generalized coordinates qi , the derivatives of the Lagrangian with respect to the generalized velocities

pi 

L qi

(8)

are called generalized momenta, and its derivatives with respect to the generalized coordinates

Fi 

L qi

(9)

are called generalized forces. In this notation, Lagrange’s equation are

p i  Fi .

(10) 43

In Cartesian coordinates the generalized momenta are the  components of the vectors pa . In general, however the pi are linear homogeneous functions of the generalized velocities qi and do not reduce to products of mass and velocity. 10. Center of mass

The momentum of a closed mechanical system has different values in different (inertia) frames of reference. If a frame K  moves   with velocity V relative to another frame K , then the velocities  'a  and  a of the particles relative to the two frames are such that     a   'a V . The momentum P and P ' in the two frames are therefore related by       P   m a v a   m a v a   m aV  P   V . a

a

a

(1)

In particular, there is always a frame of reference K ' in which  the total momentum is zero. Putting P '  0 in (1), we find the velocity of this frame:    ma v a  P a  . (2) V  ma  ma a

a

Differentiating:   dR d a ma ra  . dt dt  ma

(3)

a

If the total momentum of a mechanical system in a given frame of reference is zero, it is said to be at rest relative to that frame. This is a natural generalization of the term as applied to a particle.  Similarly, the velocity V given by (2) is the velocity of the «motion as a whole» of a mechanical system whose momentum is not zero. 44

Thus we see that the law of conservation of momentum makes possible a natural definition of rest and velocity, as applied to a mechanical system as a whole.  Formula (2) shows that the relation between the momentum P  and the velocity V of the system is the same as that between the momentum and velocity of a single particle of mass    ma , the sum of the masses of the particles in the system. This result can be regarded as expressing the additivity of mass. The right-hand side of formula (2) can be written as the total time derivative of the expression:

  a ma ra . R  ma

(3)

a

We can say that the velocity of the system as a whole is the rate of motion in space of the point whose radius vector is (3). This point is called the center of mass of the system. 11. Angular momentum

Let us now derive the conservation law which follows from the isotropy of space. This isotropy means that the mechanical properties of a closed system do not vary when it is rotated as a whole in any manner in space. Let us therefore consider an infinitesimal rotation of the system, and obtain the condition for the Lagrangian to remain unchanged. We shall use the vector  of the infinitesimal rotation, whose magnitude is the angle of rotation  , and whose direction is that of the axis of rotation (the direction of rotation being that of a righthanded screw driven along  ). Let us find, first of all, the resulting increment in the radius vector from an origin on the axis to any particle in the system undergoing rotation. The linear displacement of the end of the radius vector is related to the angle by

45







r  r sin  .

(1)

  The direction of r is perpendicular to the plane of r and  . Hence it is clear that

 



r    r  .

(2)

When the system is rotated, not only the radius vectors but also the velocities of the particles change the direction, and all vectors are transformed in the same manner. The velocity increment relative to a fixed system of coordinates is



 

v      .

(3)

Fig. 8

If these expressions are substituted in the condition that the Lagrangian is unchanged by the rotation:

 L 

L

 

L     ra    a   0 a  a  ra 

46

(4)

and the derivative

  L  L    pa replaced by pa and   p a by p a the  a ra

result is





       p a   ra   pa   a   0 a

(5)

or permuting the factor and taking  outside the sum, 













a





d    ra  p a  0 . dt a

  ra  p a   a  pa   

Since  is arbitrary, it follows that (

(6)

  d ) ra  pa  0 and we dt

conclude that the vector

   M   ra  pa ,

(7)

a

called the angular momentum or moment of momentum of the system, is conserved in the motion of the closed system. Like the linear momentum, it is additive, whether or not the particles in the system interact. There are no other additive integrals of motion. Thus every closed system has seven such integrals: energy, three components of momentum, and three components of angular momentum. Since the definition of angular momentum involves the radius vector of particles, its value depends general on the choice of origin.   The radius vectors ra and ra of a given point relative to origins at a distance a apart are related by

   ra  ra  a .

(8)

Hence

 

         M   ra  pa    ra  pa   a   pa   M   a P .   a a  a 

(9) 47

It is seen from this formula that the angular momentum depends on the choice of origin except when the system is at rest as a whole  (i.e. P  0 ). This indeterminacy, of course, does not affect the law of conservation of angular momentum, since momentum is also conserved in a closed system. We may also derive a relation between the angular momenta in two inertial frames of reference K and K  , of which the latter  moves with velocity V relative to the former. We shall suppose that the origins in the frames K and K  coincide at a given instant. Then the radius vectors of the particles are the same in the two frames, while their velocities are related by

   va  va  V .

(10)

Hence we have

 

    M   ma ra va    ma ra va    ma raV . a

a

a

(11)

The first sum on the right-hand side is the angular momentum M  in the frame K  ; using in the second sum the radius vector of the center of mass (11), we obtain

 

   M  M    RV .

(12)

This formula gives the law of transformation of angular momentum from one frame to another, corresponding to formula (11) for momentum and for energy. If the frame K  is that in which the system considered is at rest   as a whole, then V is the velocity of its center of mass, V its total  momentum P relative to K , and

 

   M  M   RP .

(13)

 In other words, the angular momentum M of a mechanical system consists of its «intrinsic angular momentum» in a frame in 48

 

 which it is at rest, and the angular momentum RP due to its motion as a whole. Although the law of conservation of all three components of angular momentum (relative to an arbitrary origin) is valid only for a closed system, the law of conservation may hold in a more restricted form even for a system in an external field. It is evident from the above derivation that the component of angular momentum along an axis about which the field is symmetrical is always conserved, for the mechanical properties of the system are unaltered by any rotation about that axis. Here the angular momentum must, of course, be defined relative to an observer lying on the axis. The most important such case is that of a centrally symmetric field or central field, i.e. one in which the potential energy depends only on the distance from some particular point (the center). It is evident that the component of angular momentum along any axis passing through the center is conserved. In other words, the angular  momentum M is conserved provided that it is defined with respect to the center of the field. Another example is that of a homogeneous field in the z – direction; in such a field, the component M z of the angular momentum is conserved, whichever point is taken as the origin. The component of angular momentum along any axis (say the z – axis) can be found by differentiation of the Lagrangian:

Mz   a

L , a

(14)

where the coordinate  is the angle of rotation about the z – axis. This is evident from the above proof of the law of conservation of angular momentum, but can also be proved directly. In cylindrical coordinates r ,  , z we have (substituting):

xa  ra cos a ,

y a  ra sin  a ,

(15)

49

M z   ma xa y a  y a x a    ma ra2 a , a

a

(16)

The Lagrangian is, in terms of these coordinates, L   ma ra2  ra2 a2  za2   U , a

and substitution of this in (16) gives (17). Control questions: 1. 2. 3. 4. 5.

50

Homogeneity of time. Conservative systems. Homogeneity of space. Angular momentum. Centrally symmetric field or central field.

(17)

III INTEGRATION OF THE EQUATIONS OF MOTION

12. Motion in one dimension

The motion of a system having one degree of freedom is said to take place in one dimension. The most general form of the Lagrangian of such a system in fixed external conditions is

L

1 aq q 2  U q  , 2

(1)

where a q  – is some function of the generalized coordinate q . In particular, if q is a Cartesian coordinate ( q  x ) then

L

mx 2  U x  . 2

(2)

The equations of motion corresponding to these Lagrangians can be integrated in a general form. It is not even necessary to write down the equation of motion; we can start from the first integral of this equation, which gives the law of conservation of energy. For the Lagrangian (2) (e.g.) we have

mx 2  U x   E . 2

(3)

This is a first-order differential equation, and can be integrated immediately. Since dx  dt

2 E  U x  , m

(4) 51

it follows that

t

m dx  const .  2 E  U x 

(5)

The two arbitrary constants in the solution of the equations of motion are here represented by the total energy E and the constant of integration. Since the kinetic energy is essentially positive, the total energy always exceeds the potential energy, i.e. the motion can take place only in those regions of space where U x   E . For example, let the function U x  be of the form shown in Fig.1. If we draw in the figure a horizontal line corresponding to a given value of the total energy, we immediately find the possible regions of motion. In the example of Fig.9, the motion can occur only in the range AB or in the range to the right of C . The points at which the potential energy equals the total energy,

U x  E ,

(6)

give the limits of the motion. They are turning points, since the velocity there is zero. If the region of the motion is bounded by two such points, then the motion takes place in a finite region of space, and is said to be finite. If the region of the motion is limited on only one side, or on neither, then the motion is infinite and the particle goes to infinity. A finite motion in one dimension is oscillatory, the particle moving repeatedly back and forth between two points (in Fig.1, in the potential well AB between the points x1 and x2 ). The period T of the oscillations, i.e. the time during which the particle passes from x1 to x2 and back, is twice the time from x1 to x2 (because of the reversibility property) or, by (5), x2  E 

T E   2m 

x1  E 

52

dx , E  U x 

(7)

Where x1 and x2 are roots of equation (6) for the given value of E . This formula gives the period of the motion as a function of the total energy of the particle.

Fig. 9

13. Motion in a central field. Sectorial velocity, Kepler’s second law

On reducing the two-body problem to one of the motion of a single body, we arrive at the problem of determining the motion of a single particle in an external field such that its potential energy depends only on the distance r from some fixed point. This is called a central field. The force acting on the particles is  U r   dU F      r  dr

 r  ; r

(1)

 its magnitude is likewise a function of r only, and its direction is everywhere that of the radius vector. As has already been shown above the angular momentum of a single particle is

 

  M  rP .

(2) 53

   Since M is perpendicular to r , the constancy of M shows that, throughout the motion, the radius vector of the particle lies in  the plane perpendicular to M . Thus the path of a particle in a central field lies in one plane. Using polar coordinates r ,  in that plane, we can write the Lagrangian as m (3) L  r 2  r 2 2   U r  . 2 This function does not involve the coordinate  explicitly. Any generalized coordinate qi which does not appear explicitly in the Lagrangian is said to be cyclic. For such a coordinate we have, by Lagrange’s equation, d L L  0, dt qi qi

(4)

L is an qi integral of the motion. This leads to a considerable simplification of the problem of integrating the equations of motion when there are cyclic coordinates. In the present case, the generalized momentum

so that the corresponding generalized momentum Pi 

P  mr 2  M ,

(5)

is the same as the angular momentum M z  M and we return to the known law of conservation of angular momentum: M  mr 2  const .

(6)

This law has a simple geometrical interpretation in the plane motion of a single particle in a central field. The express on 1  dS  r  r d is the area of the sector bounded by two neighboring 2 54

radius vectors and an element of the path. Calling this area df , we can write the angular momentum of the particle as

M  2mf ,

(7)

where the derivative f is called the sectorial velocity. Hence the conservation of angular momentum implies the constancy of the sectorial velocity: in equal times the radius vector of the particle sweeps out equal areas (Kepler’s second law).

Fig. 10

The complete solution of the problem of the motion of a particle in a central field is most simply obtained by starting from the laws of conservation of energy and angular momentum, without writing out the equations of motion themselves. Expressing  in terms of M from (6) and substituting in the express on for the energy, we obtain

Hence

r 

Е

m 2 r  r 2 2   U r  , 2

(8)

Е

mr 2 M2   U r . 2 2mr 2

(9)

dr 2E  U r  M 2  2 2  . m m r dt

(10)

or integrating, 55

dr

t

2 2 E  U r   M2 2 m m r

 const.

(11)

Writing (6) as M  mr 2 , d 

Mdt , mr 2

(12)

substituting dt from (10) and integrating, we find



M dr r2 M2 2mE  U r   2 r

 const .

(13)

Formulae (11) and (13) give the general solution of the problem. The latter formula gives the relation between r and  , i.e. the equation of the path. Formula (11) gives the distance r from the center as an implicit function on of time. The angle  , it should be noted, always varies monotonically with time, since (6) shows that  can never change sign. The expression (9) shows that the radial part of the motion can be regarded as taking place in one dimension in a field where the «effective potential energy» is U eff  U r  

The quantity  of r for which

(14)

M2 is called the centrifugal energy. The values 2mr 2

U r   56

M2 . 2mr 2

M2 E 2mr 2

(15)

determine the limits of the motion as regards distance from the center. When equation (15) is satisfied, the radial velocity r is zero. This does not mean that the particle comes to rest as in true onedimensional motion, since the angular velocity  is not zero. The value r  0 indicates a turning point of the path, where r t  begins to decrease instead of increasing, or vice versa. If the range in which r may vary is limited only by the condition r  rmin , the motion is infinite: the particle comes from, and returns to, infinity. If the range of r has two limits rmin and rmax , the motion is finite and the path lies entirely with in the annulus bounded by the circles r  rmax and r  rmin . This does not mean, however, that the path must be a closed curve. During the time in which r varies from rmax to rmin and back, the radius vector turns through an angle  which according to (13), is given by

rmax

  2 

rmin

M dr r2 M2 2mE  U r   2 r

.

(16)

The condition for the part to be closed is that this angle should m be a rational fraction of 2 , i.e. that   2 , where m and n n are integers. In that case, after n periods, the radius vector of the particle will have made m complete revolutions and will occupy its original position, so that the path is closed. At the turning point the square root in (10), and therefore the integrands in (11) and (13), change sign. If the angle  is measured from the direction of the radius vector to the turning point, the parts of the path on each side of that point differ only in the sign of  for each value of r , i.e. the path is symmetrical about the line   0 . Starting, say, from a point with r  rmax , the particle traverses a segment of the path as far as a point with r  rmin then follows a 57

symmetrically placed segment to the next point where r  rmax , and so on. Thus the entire path is obtained by repeating identical segments forwards and backwards. This applies also to an infinite path, which consists of two symmetrical branches extending from the turning point r  rmin  to infinity.

Fig. 11.

 The presence of the centrifugal energy when M  0 , which 1 becomes infinite as 2 when r  0 , generally renders it impossible r for the particle to reach the center of the field, even if the field is an attractive one. A «fall» of the particle to the center is possible only if  the potential energy tends sufficiently rapidly to   as r  0 . From the inequality M2 m 2    0, r E U r    2mr 2 r2

or r 2U r   zero only if 58

(17)

M2  Er 2 , it follows that r can take values tending to 2m

r U r  2

r 0



M2 , 2m

i.e. U r  must tend to   either as  proportionally to  

(18)

 r2

with  

M2 or 2m

1 with n  2 . rn

14. Kepler’s problem

An important class of central fields is formed by those in which the potential is inversely proportional to r , and the force accordingly 2 inversely proportional to r . They include the fields of Newtonian gravitational attraction and of Coulomb electrostatic interaction; the latter may be either attractive or repulsive. Let us first consider an attractive field, where U 



(1)

r

with  a positive constant. The «effective» potential energy U eff  

 r



M2 2mr 2

(2)

is of the form shown in Fig. 12. As r  0 , U eff tends to   , and as r   it tends to zero from negative values; for r 

M2 it has a m

minimum value U eff ,min  

m 2 . 2M 2

(3)

59

Fig. 12

It is seen at once from Fig.1 that the motion is finite for E  0 and infinite for E  0 . The shape of the path is obtained from the general formula



M

dr r2

M2 2mE  U r   2 r

Substituting there U  

 r

 const .

(4)

and effecting the elementary

integration, we have



M dr r2 M2 2mE  U r   2 r



M dr r2 M2 2mE  2m  2 r r

1 U; r   dr dU   2 r 60







MdU

MdU



 2 m 2 2  m    MU  2mE   M2 M   M m  r M (5)  arccos  const . m 2 2 2mE  M2

2mE  2mU  M 2U 2

Taking the origin of  such that the constant is zero, and putting

P

e  1

M2 , m 2 EM 2 , m 2

(6)

(7)

we can write the equation of the path as

r

P . 1  e cos

(8)

This is the equation of a conic section with one focus at the origin 2 p is called the Latus rectum of the orbit and e the eccentricity. Our choice of the origin of  is seen from (8) to be such that the point where   0 is the point called the perihelion. In the equivalent problem of two particles interacting according to the law (1), the orbit of each particle is a conic section, with one focus at the center of mass of the particles. It is seen from (6) and (7) that, if E  0 then the eccentricity e  1, i.e. the orbit is an ellipse (Fig. 13) and the motion is finite, in accordance with what has been said earlier in this section. According to the formulae of analytical geometry, the major and minor semiaxes of the ellipse are 61

a

P  P and b    1  e2 1 e2 2 E

M 2m E

.

(9)

Fig.13

The least possible value of the energy is (3), and then e  0 , i.e. the ellipse becomes a circle. It may be noted that the major axis of the ellipse depends only on the energy of the particle, and not on its angular momentum. The least and greatest distances from the center of the field (the focus of the ellipse) are rmin 

P P  a1  e  and rmax   a1  e  . 1 e 1 e

(10)

These expressions, with r and  given by (9) and (6), (7), can, of course, also be obtained directly as the roots of the equation U eff r   E .

(11)

The period of revolution in an elliptical orbit is conveniently found by using the law of conservation of angular momentum in the form of the area integral M  2m 62

df . dt

(12)

Integrating this equation with respect to time from zero to dt we 1  have 2mf  TM , where dS  r  r d is the area of the orbit. For 2 an ellipse f    a  b , and by using the formulae (9) we find T  a

3

2

m



 

m 2E

3

.

(13)

The proportionality between the square of the period and cube of the linear dimension of the orbit has already been demonstrated in §12. It may also be noted that the period depends only on the energy of the particle. For E  0 the motion is infinite. If E  0 , the eccentricity e  1 , i.e. the path is a hyperbola with the origin as internal focus (fig.3). the distance of the perihelion from the focus is rmin  where a 

P  ae  1 , e 1

(14)

P   is the «semi-axis» of the hyperbola. e 2  1 2E

Fig. 14 63

If E  0 , eccentricity e  1 , and the particle moves in a parabola P with perihelion distance rmin  . This case occurs if the particle 2 stars from rest at infinity. Control questions: 1. 2. 3. 4. 5.

64

Central field. Sectorial velocity. Turning points. Eccentricity. Latus rectum.

IV COLLISIONS BETWEEN PARTICLES

15. Disintegration of particles

In many cases the laws of conservation of momentum and energy alone can be used to obtain important results concerning the properties of various mechanical processes. It should be noted that these properties are independent of the particular type of interaction between the particles involved. Let us consider a «spontaneous» disintegration (that is, one not due to external forces) of a particle into two «constituent parts», i.e. into two other particles which move independently after the disintegration. This process is most simply described in a frame of reference in which the particle is at rest before the disintegration. The law of conservation of momentum shows that the sum of the momenta of the two particles formed in the    disintegration is then zero; p  p10  p20 that is, the particles move apart with equal and opposite momenta.

Fig. 15

If the particle is initially at rest, then

 p 0,

(1)

  p10  p20  0   p10   p 20 ,

(2)

This is to say that

65

or that their magnitudes are

p 20  p10  p0 .

(3)

The magnitude p 0 of each momentum is given by the law of conservation of energy:

E1  E2  Edis  0 ,

(4)

 m1v102  E  1 2  2  E  m2 v 20  2 2

(5)

or equivalently for the internal energy of the particles  p102  E  1 2m1   2  E  p 20 . 2  2m 2 

(6)

Using p10  p 20  p 0 , we obtain E1 

p02 , 2m1

E2 

p02 . 2m2

(7)

Here m1 and m2 are the masses of the particles. If we denote their internal energies as E1i and E2i , and as Ei the internal energy of the original particle, then the "disintegration energy" is determined by the equation

E dis  Ei  Ei1  Ei 2  , 66

(8)

which must obviously be positive, then E dis  E1  E 2 

p 02 p2 p2 m1 m2  0  m  0 , 2m1 2m2 2m m1  m2

(9)

which determines p0 ; here m is the reduced mass of the two p p particles. The velocities are v10  0 , v 20  0 . m1 m2 Let us now change to a frame of reference in which the primary particle moves with velocity V before the break-up. This frame is usually called the laboratory system, or L system, in contradistinction to the center-of-mass system, or C system, in which the total momentum is zero. Let us consider one of the   resulting particles, and let v0 and v be its velocities in the L and the C system-respectively. Evidently

   v  V  v0

or

   v  V  v0

(10)

and so v02  v 2  V 2  2vV cos1 .

(11)

Fig. 16

where  is the angle at which this particle moves relative to the direction of the velocity V . This equation gives the velocity of the 67

particle as a function of its direction of motion in the L system. In Fig.16 the velocity V is represented by a vector drawn to any point  on a circle of radius v0 from a point A at a distance V from the center. The cases V  v0 and V  v0 are shown in Figs. 17a,b, respectively. 1) The case V  v0

Fig. 17a

v sin   a v0 sin  0  a.

(12)

From Fig. 17a v cos  V  v0 cos 0 v sin   v0 sin  0 ,

(13)

v sin  v0 sin  0  , v cos v cos

tg 

v0 sin  0 . V  v0 cos 0

(14)

2) In the V  v0 former case  can have any value, but in the latter case the particle can move only forwards, at an angle  which does not exceed  max given by

68

Fig. 17b

V cos max  v , V sin  max  v0 , sin  max 

v0 , V

(15)

this is the direction of the tangent from the point A to the circle. The relation between the angles  and  0 in the L and C systems is evidently (Fig.17)

tg V  v0 cos 0   v0 sin  0

(16)

if this equation is solved for cos  0 , we obtain cos 0  

V V2 sin 2   cos 1  2 sin 2  , v0 v0

(17)

tg 2 V  v0 cos 0   v02 sin 2  0  v02 1  cos 2  0  . 2

For v0  V the relation between  and  0 is one-to-one (Fig. 17a). The plus sign must be taken in (18), so that  0  0 when 69

  0 . If v0  V , however, the relation is not one-to-one: for each value of  there are two values of  0 , which correspond to vectors

v0

drawn from the center of the circle to the points B and C (Fig.17b), and are given by two signs in (17). In physical applications we are usually concerned with the disintegration of not one but many similar particles, and this raises the problem of the distribution of the resulting particles in direction, energy, etc. We shall assume that the primary particles are randomly oriented in space, i.e. isotropically on average. In the C system, this problem is very easily solved: every resulting particle (of a given kind) has the same energy, and their directions of motion are isotropically distributed. The latter fact depends on the assumption that the primary particles are randomly oriented, and can be expressed by saying that the fraction of particles entering a solid angle element dO0 is proportional to dO0 , ie. equal dO0 . The distribution with respect to the angle  0 is obtained by to 4 putting

dO0  2 sin  0 d 0

(18)

i.e. the corresponding fraction is dN 0 dO 1   sin  0 d 0 N 4 2

(19)

The corresponding distributions in the L system are obtained by an appropriate transformation. For example, let us calculate the kinetic energy distribution in the L system. Squaring the equation    v  v0  V : v 2  v02  V 2  2v0V cos 0

we have d v 2   d v02   d V 2   2d v0V cos 0   2v0Vd cos 0  70

whence d cos 0 

d v 2  . 2v0V

(20)

mv2 Using the kinetic energy T  , where m is m1 or m2 2 depending on which kind of particle is under consideration, and substituting in (19), we find the required distribution: d cos 0 

dT . 2mv0V

(21)

The kinetic energy can take values between

m v0  V 2 2 m 2  v 0  V  . 2

Tmin  Tmax

(22)

The particles are, according to (21), distributed uniformly over this range. When a particle disintegrates into more than two parts, the laws of conservation of energy and momentum naturally allow considerably more freedom as regards the velocities and directions of motion of the resulting particles. In particular, the energies of these particles in the C system do not have determinate values. There is, however, an upper limit to the kinetic energy of any one of the resulting particles. To determine the limit, we consider the system formed by all these particles except the one concerned (whose mass is m1 , say), and denote the «internal energy» of that system by E'i . Then the kinetic energy of the particle m1 is, by (8) and (9), T10 

P02 M  m1 Ei  Ei1  Ei  ,  M 2m1

(23)

71

where M is the mass of the primary particle. It is evident that T10 has its greatest possible value when E'i is least. For this to be so, all the resulting particles except m1 must be moving with the same velocity. Then E'i is simply the sum of their internal energies, and the difference Ei  E1i  E 'i is the disintegration energy  . Thus

T10 max  M  m1 Eы . M

(24)

16. Elastic collisions

A collision between two particles is said to be elastic if it involves no change in their internal state. Accordingly, when the law of conservation of energy is applied to such a collision, the internal energy of the particles may be neglected.

Ei  Ei .

(1)

The collision is most simply described in a frame of reference in which the center of mass of the two particles is at rest (the C system). As in §16, we distinguish by the suffix 0 the values of quantities in that system.

  p10  p 20  0   ,   0 p10  p 20   p10   p 20    .   p10   p 20

(2) (3)

The velocities of the particles before the collision are related to   their velocities  1 and  2 in the laboratory system by:

   v1  v01  V   ,   v2  v02  V  72

(4)

   v1  v01  V   ,   v2  v02  V 

(5)

  P m  m1  m2 m1v1  m2 v2 V      , m P  P1  P2 m1  m2

(6)

  m1v1  m2 v 2   v1  v01  m  m  1 2     m v m   2 v2 v  v  1 1 , 02  2 m1  m2

(7)

 v01  ?  v02  ?        m v  m2 v 2 m1  m2 v1  m1v1  m2 v 2 v01  v1  1 1   m1  m2 m1  m2       m2 v1  v 2  v  v1  v 2 m2 v    , 9 m1  m2 m1  m2  m2 v  m1  m2   . m1v   v02   m1  m2 

 v01 

(8)

Because of the law of conservation of momentum, the momenta of the two particles remain equal and opposite after the collision, and are also unchanged in magnitude, by the law of conservation of energy. Thus, in the C system the collision simply rotates the velocities, which remain opposite in direction and unchanged in  magnitude. If we denote by n0 a unit vector in the direction of the 73

velocity of the particle m1 after the collision, then the velocities v10 , v20 of the two particles after the collision (distinguished by primes) are

Fig. 18

 m2 vn0   v10  v10 n0  m  m  1 2   m v n   v   v n   1 0 . 20 0  20 m1  m2

(9)

In order to return to the L system, we must add to these  expressions the velocity V of the center of mass. The velocities in the L system after the collision are therefore

   m2 vn0 m1v1  m2 v 2   v    1 m m m1  m2  1 2     v    m1vn0  m1v1  m2 v 2 .  2 m1  m2 m1  m2

(10)

No further information about the collision can be obtained from the laws of conservation of momentum and energy. The direction of  the vector n0 depends on the law of interaction of the particles and on their relative position during the collision. 74

The results obtained above may be interpreted geometrically. Here it is convenient to use momenta instead of velocities. Multiplying equations (10) by m1 and m2 respectively, we obtain





  m1 m2  m1   P1  m  m vn0  m  m P1  P2  1 2 1 2    m m m  2  P    1 2 vn  P1  P2 , 2 0  m1  m2 m1  m2   m1    P1  mvn0  m  m P1  P2  1 2     P    mvn  m2 P1  P2 , 2 0  m1  m2







(11)







(12)

m1 m2 is the reduced mass. We draw a circle of radius m1  m2  mv and use the construction shown in Fig. 19.

where m 

Fig. 19

 If the unit vector n0 is along OC , the vectors AC and CB give the momenta p1 and p 2 respectively. When p1 and p2 are given, the radius of the circle and the points A and B are fixed, but the point C may be anywhere on the circle. 75

     OC  mv  mvn0     m1 P1  P2  AO  m1  m2      m2 P1  P2 . OВ   m1  m2









(13)

Let us consider in more detail the case where one of the particles ( m2 , say) is at rest before the collision. In that case the distance   m1 OB  P1  mv1 is equal to the radius, i.e. B lies on the m1  m2  circle. The vector AB is equal to the momentum p1 of the particle m1 before the collision. The point A lies inside or outside the circle, according as m1  m2 or m1 > m2 . The corresponding diagrams are shown in Figs. 20a, b.

a

b Fig. 20

    АС  P1  АО  ОС     СВ  P2  ОВ  ОС. The angles 1 and  2 in these diagrams are the angles between the directions of motion after the collision and the direction of 76

 impact (i.e. of p1 ). The angle at the center, denoted by  , which gives the direction of

 n0 , is the angle through which the direction of

motion of m1 is turned in the C system. It is evident from the figure that 1 and  2 can be expressed in terms of  by

Fig. 21

P1sin 1  OC sin  0 P1cos1  AO  OC cos 0 tg1 

OC sin  0 m2 sin  0  , AO  OC cos 0 m1  m2 cos 0

(14)

then  2 find from OCB – is isosceles triangle, next C  B C  B  O  180   O  2B     0

2 2   0    0 2  . 2

(15)

We may give also the formulae for the magnitudes of the velocities of the two particles after the collision, likewise expressed in terms of  : 77

v1 

m

2 1

 m 22  2m1 m 2 cos  m1  m 2

v,

v 2 

2m1 v 1 sin  . (16) 2 m1  m 2

The sum 1   2 is the angle between the directions of motion 1 of the particle after collisions. Evidently 1   2   if m1  m2 2 1 and 1   2   if m1  m2 . 2 When the two particles are moving afterwards in the same or in opposite directions (head-on collision), we have    , i.e. the point  C lies on the diameter through A , and is on OA (Fig. 21b; p1 and    p2 in the same direction) or on OA produced (Fig. 16a; p1 and p 2 in opposite directions). In this case the velocities after the collision are v1 

m1  m2 v, m1  m2

v2 

2m1v . m1  m2

(17)



This value of v2 has the greatest possible magnitude, and the maximum energy which can be acquired in the collision by a particle originally at rest is therefore E 2 max 

4m1 m2 1 2 m2 v 2 max  E, 2 m1  m2 2 1

(18)

1 2 m1v1 is the initial energy of the incident particle. If 2 m1  m2 the velocity of m1 after the collision can have any direction. If m1  m2 , however, this particle can be deflected only through an angle not exceeding  max from its original direction; this maximum value of 1 corresponds to the position of C for which AC is a tangent to the circle (Fig. 21b). Evidently where E1 

78

sin  max  OC/OA 

m2 . m1

(19)

The collision of two particles of equal mass, of which one is initially at rest, is especially simple. In this case both B and A lie on the circle (Fig. 22).

Fig. 22

Then 1 2

1   ,

1  v1  v cos  , 2

2 

1    , 2

(20)

1  v2  v sin  . 2

(21)

After the collision the particles move at right angle to each other. 17. Scattering. Rutherford’s formula. The effective cross-section of the scattering

As already mentioned in §16, a complete calculation of the result of a collision between two particles (i.e. the determination of 79

the angle  ) requires the solution of the equations of motion for the particular law of interaction involved. We shall first consider the equivalent problem of the deflection of a single particle of mass m moving in a field U r  whose center is at rest (and is at the center of mass of the two particles in the original problem). As has been shown in §15, the path of a particle in a central field is symmetrical about a line from the center to the nearest point in the orbit ( OA in Fig. 23). Hence the two asymptotes to the orbit make equal angles (  0 , say) with this line. The angle  through which the particle is deflected as it passes the center is seen from Fig. 23 to be

    2 0 .

(1)

Fig. 23

The angle  0 itself is given, according to by

0  

80

M

dr r2

 M2 2mE  U r   2  r  

 constant ,

(2)

taken between the nearest approach to the center and infinity. It should be recalled that rmin is a zero of the radicand. For an infinite motion, such as that considered here, it is convenient to use instead of the constants E and M the velocity v of the particle at infinity and the impact parameter  . The latter is the length of the perpendicular from the center O to the direction of v , i.e. the distance at which the particle would pass the center if there were no field of force (Fig. 23). The energy and the angular momentum are given in terms of these quantities by

m 2 , 2       M  mr  v   mvr sin v  r eM E

       M  mv rsin r eM  mv eM v 

M  m  ,

  r sin r    

(3)

and formula (2) becomes



0  

rmin



 

rmin

m dr r2   m2  m 2  2 2  U r   2m r2  2 

m dr r2 2m 22 m 2  22  2mU r   2 r2



81



 

rmin

m 

m   dr dr  2 r2 r   . rmin  2 2U 2mU r  m 2  2 2 1 2  1  2 2 2 r m2 m 2 2 m  r

(4)

Together with (1), this gives  as a function of  . In physical applications we are usually concerned not with the deflection of a single particle but with the scattering of a beam of identical particles incident with uniform velocity v on the scattering center. The different particles in the beam have different impact parameters and are therefore scattered through different angles  . Let dN be the number of particles scattered per unit time through angles between  and   d . This number itself is not suitable for describing the scattering process, since it is proportional to the density of the incident beam. We therefore use the ratio d 

dN , n

(5)

where n is the number of particles passing in unit time through unit area of the beam cross-section. This ratio has the dimensions of area and is called the effective scattering cross-section. It is entirely determined by the form of the scattering field and is the most important characteristic of the scattering process. We shall suppose that the relation between  and  is one-toone; this is so if the angle of scattering is a monotonically decreasing function of the impact parameter. In that case, only those particles whose impact parameters lie between    and     d   are scattered at angles between  and   d . The number of such particles is equal to the product of n and the area between two circles of radii  and   d , i.e. dN  2d  n . The effective cross-section is therefore d  2d . 82

(6)

In order to find the dependence of d scattering, we need only to rewrite (6) as d  2  

on the angle of

d    d . d

(7)

d , since the d derivative may be (and usually is) negative. Often d is referred to the solid angle element dO0 instead of the plane angle element d . The solid angle between cones with vertical angles  and   d is dO0  2 sin d . Hence we have from (7)

Here we use the modulus of the derivative

d 

   d  dO 0 . sin  d 

(8)

Returning now to the problem of the scattering of a beam of particles, not by a fixed center of force, but by other particles initially at rest, we can say that (7) gives the effective cross-section as a function of the angle of scattering in the center-of-mass system. To find the corresponding expression as a function of the scattering angle  in the laboratory system, we must express  in (7) in terms of  by means of formula tg1 

m2 sin  , m1  m2 cos 

2 

1     . 2

(9)

This gives expressions for both the scattering cross-section for the incident beam of particles (  in terms of 1 ) and that for the particles initially at rest (  in terms of  2 ).

83

Rutherford’s formula One of the most important applications of the formulae derived above is to the scattering of charged particles in a Coulomb field.

Putting in (1) U 

 r

and the

 

0  

rmin

r2

dr (10)

2

2U 1 2  r m 2

effecting the elementary integration, we obtain M m  0  arccos r M m2 2 2mE 2 M

 arccos



m M   r  arccos  m2 2 2mE 2 M

0

  r  r

min

rmin



m m m 2 m 2 2 2m   2 2 m 2  2

2 m

 arccos

    1   m 2   

2

,

(11)

 cos  0 

cos  0

84

m 2   1   2   m 

  1   2   m 

2

  

2

,

    m 2 

(12)

   cos  0 1      m 2  2

   

2

2     m 2  2 4 

2

   2   2 2 4 cos  0  cos  0  2   m  m   2

2

2

2

       1  cos 2  0     sin 2 0 cos  0   2  2  m m           2

   2  m 

2

   ctg 2 0 

whence

2 

or putting  0 

  2

2 tg 2 0 m 2 4

(13)

,

2 

2     tg 2  , 2 2 m   2 

2 

2  ctg 2 . 2 4 m  2

Differentiating this expression with respect to substituting in

d 

(15)



and

d    d d

(16)

   d dO0 sin  d

(17)

d  2  

or

(14)

gives 85

  1   ctg     2  2 sin 2 2

d 



2  m sin  2 2 





2 , m 2 d 1  1 ,  2  d 2 m  sin 2 2



ctg

ctg

dO0 m 2 sin 2

 2 cos

 2

4m 2 2 cos



 sin   sin2

2

(19)

 2

 2 cos

 2

sin

 2



2

dO0



(18)

sin 4





2

  d   2  2 m 

 2 dO0 4m 2 4 sin 4



,

(20)

2

2

 dO0  .  sin 4  2

(21)

This is Rutherford’s formula. It may be noted that the effective cross-section is independent of the sign of  , so that the result is equally valid for repulsive and attractive Coulomb fields. Control questions: 1. 2. 3. 4. 5.

86

Elastic collisions. Reduced mass. Effective scattering cross-section. Impact parameter. Rutherford’s formula.

V SMALL OSCILLATIONS

18. Free oscillations in one dimension

A very common form of motion of mechanical systems is what are called small oscillations of a system about a position of stable equilibrium. We shall consider first of all the simplest case, that of a system with only one degree of freedom. Stable equilibrium corresponds to a position of the system in which its potential energy U q  is a minimum. A movement away dU from this position results in the setting up of a force – which dq tends to return the system to equilibrium. Let the equilibrium value of the generalized coordinate q be q0 . For small deviations from the equilibrium position, it is sufficient to retain the first non-vanishing term in the expansion of the difference U q   U q 0  in powers of q  q0 U q   U q0  

dU dq

q  q0

 q  q0  

1 d 2U 2 dq 2

 q  q0      . 2

q  q0

(1)

In

general, in this the second order term: 1 2 U q   U q0   k q  q0  , where k is a positive coefficient, the 2 value of the second derivative U q  for q  q0 . We shall measure the potential energy from its minimum value, i.e. we put U q0   0 , and use the symbol x  q  q0 for the deviation of the coordinate from its equilibrium value U q  

1 d 2U 2 dq 2

 q  q 0       2

q  q0

1 kx 2 2 . k q  q 0       2 2

(2)

87

Thus U x  

kx 2 . 2

(3)

Fig. 24

The kinetic energy of a system with one degree of freedom is in general of the form T

1 1 a q q 2  a q x 2 . 2 2

(4)

In the same approximation, it is sufficient to replace the function a q  by its value at q  q 0 . Putting for brevity a q   m , we have the following expression for the Lagrangian of a system executing small oscillations in one dimension: L

mx 2 kx 2  . 2 2

(5)

The corresponding equation of motion is or 88

mx  kx  0

(6)

where

x   2 x  0 ,

(7)

k  2 . m

(8)

The solution of this equation we should finding as:

x  cet .

(9)

(9) characterizesequation 2   2  0 . From that   i , if

x  ce it  c1eit  c2eit .

(10)

e i t  cos  t  i sin  t – using Euler's formula,two independent solutions of the linear differential equation (7) are cos t and sin t , and its general solution is there fore

x  c1 cos t  c2 sin t .

(11)

This expression can also be written

c1  a cos ; c2  a sin  , in the simple form

x  a cos cos t  a sin  sin t  a cost    x  a cost   .

(12)

Since cos cos t  sin  sin t  cost    , a comparison with (11) shows that the arbitrary constans a and  are related to c1 and c 2 by a  c12  c22 ,

(13)

c2  tg . c1

(14)



89

Thus, near a position of stable equilibrium, a system executes harmonic oscillations. The coefficient a of the periodic factor in (12) is called the amplitude of the oscillations, and the argument of the cosine is their phase;  is the initial value of the phase, and evidently depends on the choice of the origin of time. The quantity  is called the angular frequency of the oscillations; in theoretical physics, however, it is usually called simply the frequency, and we shall use this name henceforward. The frequency is a fundamental characteristic of the oscillations, and is independent of the initial conditions of the motion. According to formula (8) it is entirely determined by the properties of the mechanical system itself. It should be emphasized, however, that this property of the frequency depends on the assumption that the oscillations are small, and ceases to hold in higher approximations. Mathematically, it depends on the fact that the potential energy is a quadratic function of the coordinate. The energy of a system executing small oscillations is E T U 

m 2 k 2 m 2 m 2 a 2 . x  x  x   2 x 2   2 2 2 2

(15)

It is proportional to the square of the amplitude. 19. Forced oscillations

Let us now consider oscillations of a system on which variable external force acts. These are called forced oscillations, where as those discussed in §19 are free oscillations. Since the oscillations are again supposed small, it is implied that the external field is weak, because otherwise it could cause the displacement x to take too large values. kx 2 , The system now has, besides the potential energy U x   2 the additional potential energy U c x, t  resulting from the external field. Expanding this additional term as a series of powers of the small quantity x , we have 90

U c x, t   U c 0, t  

dU x, t  dx

x 0

 x      U c 0, t   F t   x .

(1)

The first term is a function of time only, and may therefore be omitted from the Lagrangian, as being the total time derivative of another function of time. In the second term dU x, t  dx

x 0

   F t 

(2)

is the external «force» acting on the system in the equilibrium position, and is a given function of time, which we denote by F (t ) . Thus the potential energy involve a further term  xF (t ) , and the Lagrangian of the system L

mx 2 kx 2   x  F t      . 2 2

(3)

The corresponding equation of motion is x   2 x 

F t  , m

(4)

where we have again introduced the frequency  of the free oscillations. The general solution of this inhomogeneous linear differential equation with constant coefficients is x  x0  x p .

(5)

x 0 is the general solution of the corresponding homogeneous equation and x p is a particular integral of the inhomogeneous equation. In the present case x 0 represents the free oscillations discussed in §19. 91

The equation of motion (4) can integrated in a general form for an arbitrary external force F (t ) . This is easily done by rewriting the equation as F t  m d x  ix   i x  ix   F t  , dt m x  ix  ix   2 x 

(6)

where y  x  ix

(7)

and F t  m

(8)

y0  Ae it .

(9)

y  iy  is a complex quantity

A  At  is necessary to find complete solution. For this we use the method of variation, that is, we take a constants as a function of time. y  At e it , y  At ie it  A t e it . Substituting

so

F t  it A t   e , m F t  it e dt  c , m  F t  it  y  e it At   e it  e dt  c  ,  m  At   

92

(10) (11) (12)

(13) (14)

y  x  ix  G t  give as

x p  e it Bt   e it  G t e  it dt  c.

(15)

Let us consider a case of especial interest, where the external force is itself a simple periodic function of time, of some frequency



x   2 x 

F0 cost    . m

(16)

The solution x  x0  x p , ал x0  A cost    , x p  B cost    (17) B

F0 m 2   2 

x  A cost    

F0 cost    . m 2   2 

(18)

(19)

Thus a system under the action of a periodic force executes a motion which is a combination of two oscillations, one with the intrinsic frequency  of the system and one with the frequency  of the force. The solution (19) is not valid when resonance occurs, when    . To find the general solution of the equation of motion in this case, we rewrite (19) as x   2 x 

F0 cost    m

the particular solution  x p  Bt sin t   

(20) 93

rewriting (16) B

F0 . 2m

(21)

So xp 

F0 t sin t    . 2m

(22)

The total solution

x  A cos  t    

F0 t sin  t    2m

(23)

In the case of resonance, the oscillation amplitude increases to infinity. The energy of the system is E

m 2 x   2 x 2   m  2 , 2 2

(24)

where

    2

E

2 1  it   F t e dt ,  m 2 

2 1  it  F t e dt . 2 m 

(25)

(26)

If the external acting for small time, e  it  1 , and: E

2 1     F t dt  .  2m   

(27)

20. Damped oscillations

So far we have implied that all motion takes place in vacuum, or else that the effect of the surrounding medium on the motion may be neglected. In reality, when a body moves in a medium, the latter exerts a resistance which tends to retard the motion. The energy of the moving body is finally dissipated by being converted into heat. 94

Motion under these conditions is no longer a purely mechanical process, and allowance must be made for the motion of the medium itself and for the internal thermal state of both the medium and the body. In particular, we cannot in general assert that the acceleration of a moving body is a function only of its coordinates and velocity at the instant considered; that is, there are not equations of motion in the mechanical sense. Thus the problem of the motion of a body in a medium is not one of mechanics. There exists, however, a class of cases where motion in a medium can be approximately described by including certain additional terms in the mechanical equations of motion. Such cases include oscillations with frequencies small compared with those of the dissipative processes in the medium. When this condition is fulfilled we may regard the body as being acted on by a force of friction which depends only on its velocity. If, in addition, this velocity is sufficiently small, then the frictional force can be expanded in powers of the velocity. The zeroorder term in the expansion is zero, since no friction acts on a body at rest, and so the first non-vanishing term is proportional to the velocity. Thus the generalized frictional force f fr acting on a system executing small oscillations in one dimensions (coordinate x ) may be written f fr  x , (28) where  is a positive coefficient and the minus sign indicates that the force acts in the direction opposite to that of the velocity. Adding this force on the right-hand side of the equation of motion, we obtain mx  kx  x .

(29)

We divide this by m and put

k 2  m   0     2 .  m

(30)

95

0 is the frequency of free oscillations of the system in the absence of friction, and  is called the damping coefficient or damping decrement. Thus equation is

x   02 x  2x  0 .

(31)

We again seek a solution x  e rt

(32)

and obtain r for the characteristic equation r 2  2r   02  0 ,

(33)

whence

r1, 2  

2  42  4 02

   2  02

2 r1    2   02

r2    2   02 .

(34)

The general solution of equation (34) is

x  c1er1t  c2er2t .

(35)

Two cases must be distinguished. If   0 , we have two complex conjugate values of r . The general solution of the equation of motion can then be written as x  e  t cost    x  c1e  t cost  c2 e  t sin t  e  t c1 cost  c2 sin t   ae  t cost    . (36) 96

a  c12  c22 ,

(37)

x  ae  t cost    ,

(38)

with    02  2 , a and a being real constants. The motion described by these formulae consists of damped oscillations. It may be regarded as being harmonic oscillations of exponentially decreasing amplitude. The rate of decrease of the amplitude is given by the exponent 1 and the «frequency»  is less than that of free oscillations in the absence of friction. For   0 , the difference

between  and 0 is of the second order of smallness. The decrease in frequency as a result of fraction is to be expected, since friction retards motion. If   0 , the amplitude of the damped oscillation as almost unchanged during the period

2



. It is then meaningful to consider

the mean values of the squared coordinates and velocities, neglecting the change in e  t when taking the mean. These mean squares are evidently proportional to e 2 t . Hence the mean energy of the system decreases as E  E 0 e 2  t ,

(39)

where E0 is the initial value of the energy.

Next, let   0 . Then the values of r are both real and negative. The general form of the solution is x  c1e

     2  02  t  

 c2e  t  e

     2   02  t  

.

(40)

We see that in this case, which occurs when the friction is sufficiently strong, the motion consists of a decrease in x , i.e. an 97

asymptotic approach (as t   ) to be equilibrium position. This type of motion is called aperiodic damping. In the special case where   0 , the characteristic equation has the double root r   . The general solution of the differential equation is then x  c1  c2 t e  t .

(41)

This is special case of aperiodic damping. 21. Forced oscillations under friction

The theory of forced oscillations under friction is entirely analogous to that given for oscillations without friction. Here we shall consider in detail the case of a periodic external force, which is of considerable interest. Adding to the right-hand side of the equation an external force f 0 cos t and dividing by m , we obtain the equation of motion: x  2x   02 x 

f0 cos t . m

(42)

The solution of this equation is more conveniently found in complex form, and so we replace cos t on the right by eit : x  2x  02 x 

f 0 it e . m

(43)

We seek a particular integral in the form x  c1e r1t  c2 e r2t ,

(44)

x p  1 cos t   2 sin t ,

(45)

x p   1 sin t   2 cos t 98

xp   1 2 cos t   2 2 sin t   1 2 cos t   2  2 sin t  2   1 sin t   2  cos t     02  1 cos t   2 sin t  

f0 cos t m

cos t  1 2  2 2   02 1  sin t    2 2  2 1   02  2 

1 02   2   2 2 

f0 m

 2  02   2   21  0 , 2 

2 

f 21  0 , 2 2   m  0

m 02   2   42 2 2



f 0  02   2  2 02   2  m 02   2 2  42 2

2 





2f 0

m    2   42 2 2 0

(47)



f 0 02   2 



(46)

21 02   2

1  02   2   2

1 

f0 cos t m

2



.

(48)

Substituting in the solution x p  1 cos t   2 sin t  b cost   

(49) 99

and b  12   22 

f0 m

tg 



1 2 0





2 2

 42

2

2 .    02 2

,

(50)

(51)

The full solution is x  c1e r1t  c2 e r2t  b cost    .

(52)

The expression (50) for the amplitude b of the forced oscillation increases as  approaches 0 , but does not become infinite as it does in resonance without friction. 22. Anharmonic oscillations

The theory of small oscillations discussed above is based upon the expansion of the force which tends to return the system to equilibrium f  kx  x 2  x 3    

(1)

Although such an expansion is entirely legitimate when the amplitude of the oscillations is sufficiently small, in higher approximations (called anharmonic or non-linear oscillations) some minor but qualitatively different properties of the motion appear. The corresponding equation of motion is

x   02 x  x 2 .

(2)

We shall seek the solution as a series of successive approximations:

x  x1  x2 , 100

(3)

when (3) is substituted in (2)

x12   02 x1  0

(4)

the solution of (4) turns out to be

x22  02 x2  x12 .

(5)

x1  c cos0 t    .

(6)

The initial phase in x1 can always be made zero by a suitable choice of the origin of time

x t 0  a; x t 0  0

(7)

x1  a cos  0 t

(8)

then

and substituting (8) to (5), we obtain x 2   02 x2  a 2 cos 2  0 t . 1 1  cos 2  2 a 2 x2  02 x2  1  cos 20t  2

Furthermore, using cos 2  

(9)

the general solution can be chosen as x2  c1  c2 cos 20 t

(10)

and the particular solution as x2 

a 2 a 2  cos 2 0 t 2 02 6 02

(11) 101

so that (9) becomes

x  A cos 0 t  B sin 0 t  xд x  A cos 0 t  B sin 0 t 

 a 2 a 2  cos 2 0 t . 202 602

(12)

Solving the inhomogeneous linear equation in the usual way, we have  1 a  a 2  cos   x  a1  t 0 2  202  3 0 

  1 1  3 cos 2 0 t  .  

Control questions: 1. 2. 3. 4. 5.

102

What are anharmonic oscillations? Solution as a series of successive approximations. Small oscillations. What is the damping coefficient or damping decrement. Forced oscillations under friction.

(13)

VI MOTION OF A RIGID BODY

23. Motion of a rigid body. Degrees of freedom and coordinates of a rigid body. Angular velocity of a body. The instantaneous axis of rotation

A rigid body may be defined in mechanics as a system of particles such that the distances between the particles do not vary. This condition can, of course, be satisfied only approximately by systems which actually exist in nature. The majority of solid bodies, however, change so little in shape and size under ordinary conditions that these changes may be entirely neglected in considering the laws of motion of the body as a whole. In what follows, we shall often simplify the derivations by regarding a rigid body as a discrete set of particles, but this in no way invalidates the assertion that solid bodies may usually be regarded in mechanics as continuous, and their internal structure disregarded m   mi . The passage from the formulae which involve a i

summation over discrete particles to those for a continuous body is effected by simply replacing the mass of each particle by the mass m   dV contained in a volume element dV (    r, t  – being V

the density) and the summation by an integration over the volume of the body. To describe the motion of a rigid body, we use two systems of coordinates: a "fixed" (i.e. inertial) system XYZ , and a moving system x  x , y   y , z   z , which is supposed to be rigidly fixed in the body and to participate in its motion. The origin of the moving system may conveniently be taken to coincide with the center of mass of the body. The position of the body with respect to the fixed system of coordinates is completely determined if the position of the moving system is specified. Let the origin O of the moving system have the 103

 radius vector R (Fig. 25). The orientation of the axes of that system relative to the fixed system is given by three independent angles,  which together with the three components of the vector R make six coordinates. Thus a rigid body is a mechanical system with six degrees of freedom. Let us consider an arbitrary infinitesimal displacement of a rigid body. It can be represented as the sum of two parts. One of these is an infinitesimal translation of the body, whereby the center of mass moves to its final position, but the orientation of the axes of the moving system of coordinates is unchanged. The other is an infinitesimal rotation about the center of mass, whereby the remainder of the body moves to its final position.

Fig. 25

 Let r be the radius vector  of an arbitrary point P in a rigid body in the moving system, and a the radius vector of the same point in  the fixed system (Fig. 25). Then the infinitesimal displacement da  of P consists of displacement dR , equal to that of the center of mass,   and a displacement d  r  , relative to the center of mass resulting  from a rotation through an infinitesimal angle d :     da  dR  d  r  . 104

(1)

Dividing this equation by the time dt during which the displacement occurs, and putting

   da  dR  d   v,  V, dt dt dt

(2)

we obtain the relation

 

   v  V  r .

(3)

 The vector V is the velocity of the center of mass of the body,  and is also the translational velocity of the body. The vector  is called the angular velocity of the rotation of the body; its direction,   like that of d , is along the axis of rotation. Thus the velocity v of any point in the body relative to the fixed system of coordinates can be expressed in terms of the translational velocity of the body and its angular velocity of rotation. It should be emphasized that, in deriving formula (3), no use has been made of the fact that the origin is located at the center of mass. The advantages of this choice of origin will become evident when we come to calculate the energy of the moving body. Let us now assume that the system of coordinates fixed in the body is such that its origin is not at the center of mass O , but at some point O at a distance a from O . Let the velocity of O be V  , and the angular velocity of the new system of coordinates be  . We again consider some point P in the body, and denote by r  its radius vector with respect to O . Then    r  rb

(4)

and substitution in (3) gives

  

    v  V  b  r  .

(5)

105

 

    The definition of V  and  shows that v  V   r  . Hence it follows that

 

     

 

           V   v  r  ; V  b  r   r   V  b or

  / =  .

(6)

(7)

The second of these equations is very important. We see that the angular velocity of the rotation, at any instant, of a system of coordinates fixed in the body is independent of the particular system  chosen. All such systems rotate with angular velocities  which are equal in magnitude and parallel in direction. This enables us to call

 

the angular velocity of the body. The velocity of the translational motion, however, does not have this "absolute" property.   It is seen from the first formula (7) that, if V and  are, at any given instant, perpendicular for some choice of the origin O , then

  V  and  are perpendicular for any other origin O . Formula (3)

shows that in this case the velocities v of all points in the body are  perpendicular to  . It is then always possible to choose an origin  O whose velocity V  is zero, so that the motion of the body at the instant considered is a pure rotation about an axis through O . This axis is called the instantaneous axis of rotation. In what follows we shall always suppose that the origin of the moving system is taken to be at the center of mass of the body, and so the axis of rotation passes through the center of mass. In general,  both the magnitude and the direction of  vary during the motion. 24. Inertia tensor. Moments of inertia. Principal axes of inertia and principal moments of inertia. Symmetrical top and spherical top. Rotator

To calculate the kinetic energy of a rigid body, we may consider it as a discrete system of particles and put 106

T 

mv 2 , 2

(1)

where the summation is taken over all the particles in the body. Here, and in what follows, we simplify the notation by omitting the suffix which numerates the particles. Substitution of

 

   v  V  r

gives

 

2 1 T    r dV 2

 r

2

where

(2) (3)

 

   2 r 2 sin 2   2 r 2 1  cos 2    2 r 2  r

2

(4)

  is the angle between the vectors  and r ,



 r    x  

 2   2x   2y   2z , r 2  x 2  y 2  z 2 ,

2

x

y   z z    2x x 2   2y y 2   2z z 2  2 x  y xy  2 x  z xz  2 y  z yz 2

y

then 1 2 1 1  x    y 2  z 2 dV   2y   x 2  z 2 dV   2z    y 2  x 2 dV  (5) 2 2 2   x  y  xydV   x  z  xzdV   z  y  zydV .

T

  The velocities V and  are the same for every point in the body. Since we take the origin of the moving system to be at the  center of mass, this term is zero, because  mr  0 . Finally I xx     y 2  z 2 dV , I xy  -  xydV , I yy    x 2  z 2 dV , I xz  -  xzdV ,

I zz     y 2  x 2 dV , I yz  -  yzdV .

(6)

107

The components I xx , I yy , I zz are called the moments of inertia about the corresponding axes. So, the kinetic energy of a rigid body becomes 1 1 T  I xx2x  I yy2y  I zz2z  2I xyxy  2I xzxz  2I yzyz   Iikik . (7) 2 2

And the total kinetic energy of a rigid body T

V 2 2



1 I ik  i  k . 2

(8)

The Lagrangian for a rigid body is obtained from (8) by subtracting the potential energy: L

1 1 V 2  I ik  i  k  U . 2 2

(9)

The potential energy is in general a function of the six variables which define the position of the rigid body, e.g. the three coordinates X, Y, Z of the center of mass and the three angles which specify the relative orientation of the moving and fixed coordinate axes. The tensor I ik is called the inertia tensor of the body. It is symmetrical, i.e. I ik  I ki

(10)

as is evident from

I ik   mxl2 ik  xi xk  . Like any symmetrical tensor of rank two, the inertia tensor be reduced to a diagonal form by an appropriate choice of directions of the axes x , y , z  .These directions are called principal axes of inertia, and the corresponding values of 108

(11) can the the the

diagonal components of the tensor are called the principal moments of inertia; we shall denote them by I1 , I 2 , I 3 .When the axes x , y , z  are so chosen, the kinetic energy of rotation takes the very simple form

T

1 I112  I 2 22  I 332  . 2

(12)

None of the principal moments of inertia can exceed the sum of the other two. For instance, I1  I 2    x 2  y  2 dV  I 3 .

(13)

A body whose three principal moments of inertia are all different is called an asymmetrical top. If two are equal I1  I 2  I 3 , we have a symmetrical top. If all three principal moments of inertia are equal, the body is called spherical top, and the axes of inertia may be chosen arbitrary as any three mutually perpendicular axes. The determination of the perpendicular axes of inertia is much simplified if the body is symmetrical, for it is clear that the position of the center of mass and the directions of the principal axes must have the same symmetry as the body. For example, if the body has a plane of symmetry, the center of mass must lie in that plane, which also contains two of the principal axes of inertia, while the third is the perpendicular to the plane. An obvious case of this kind is a coplanar system of particles. Here there is a simple relation between the three principal moments of inertia. If the plane of the system is taken as the x , y  – plane, then z   0 for every particle, and so I 1   y  2 dV , I 2   x 2 dV , I 3    x  2  y  2 dV , whence I1  I 2  I 3 .

(14)

If a body has an axis of symmetry of any order, the center of mass must lie on that axis, which is also one of the principal axes of inertia, while the other two are perpendicular to it. If the axis is of order higher than the second, the body is a symmetrical top. For any 109

principal axis perpendicular to the axis of symmetry can be turned through an angle different from 180° about the latter. A particular case here is a collinear system of particles. If the line of the system is taken as the z  -axis, then x  y   0 for every particle and so two of the principal moments of inertia are equal and the third is zero: I 1  I 2   z  2 dV ,

I 3  0.

(15)

Such a system is called a rotator. The characteristic property which distinguishes a rotator from other bodies is that it has only two, not three, rotational degrees of freedom, corresponding to rotations about the x and y  axes: it is clearly meaningless to speak of the rotation of the speak line about itself. Finally, we may note one further result concerning the calculation of the inertia tensor. Although this tensor has been defined with respect to the system of coordinate whose origin is at the center of mass, it may sometimes be more conveniently found by first calculating a similar tensor

I ik/   mxl/ 2 ik  xi/ xk/  ,

(16)

defined with respect to some origin O . If the distance OO is represented by a vector a , then

   r  r  a ,

(17)

or

xi/  x/  ai . Since by the definition of O ,

  mr   0,  mr  0 , R  m we have 110

(18)

I ik/  I ik   a 2 ik  ai a k 

(19)

   r  r  a I   mx  ik  x x   / ik

/2 l

/ i

/ k

xi/  xi  ai x k/  x k  a k   mr  0



  mxl/ 2 ik  ai/ 2 ik  2ai xi  ik  xi  ai  x k  a k     mxi/ 2 ik  ai/ 2 ik  xi x k  xi a k  x k ai  ai a k  

(20)

 I ik   a  ik  ai a k . 2 i

Using this formula, we /2 I ik   mxl  ik  xi xk , with  m   .

can

easily

calculate

25. Angular momentum of a rigid body. The regular precession of the top

The value of the angular momentum of a system depends, as we know, on the point with respect to which it is defined. In the case of a rigid body, the most appropriate point to choose for this purpose is the origin of this moving system of coordinates, i.e. the center of mass of the body and in what follows we shall denote by

  M   mr v 

(1)

the angular momentum so defined. According the formula

 

   M  M   RP ,

(2)

111

when the origin is taken at the center of mass of the body, the  angular momentum M is equal to the «intrinsic» angular momentum resulting from the motion relative to the center of mass. In the     definition (1), we therefore replace v  V  r by r :

   

  



 

      M   m r r   m r 2   r r  ,

(3)

or in tensor notation, M i   mxl2 i  xi xk  k   i   k  ik   k  mxl2 ik  xi xk . (4)

Finally, using the definition of the inertia tensor,





2  m xl  ik  xi xk  I ik ,

(5)

we have M i  I ik  k .

(6)

If the axes x, y , z  are the same as the principal axes of inertia, Eq.(6) gives M 1  I 11 , M 2  I 2  2 ,

M 3  I 3 3 .

(7)

In particular, for a spherical top, where all three principal moments of inertia are equal, we have simply   M  I ,

(8)

i.e. the angular momentum vector is proportional to, and in the same direction as, the angular velocity vector. For an arbitrary body,   however, the vector M is not in general in the same direction as  ; this happens only when the body is rotating about one of its principal axes of inertia. Let us consider a rigid body moving freely; i.e. not subject to any external forces. We suppose that any uniform translation motion, 112

which is of no interest, is removed, leaving a free rotation of the body. As in any closed system, the angular momentum of the freely rotating body is constant. For a spherical top the condition   M  const gives   const ; that is, the most general free rotation of a spherical top is the uniform rotation about an axis fixed in space.   The case of a rotator is equally simple. Here also M  I  , and  the vector  is perpendicular to the axis of the rotator. Hence, the free rotation of a rotator is a uniform rotation in one plane about an axis perpendicular to that plane.

Fig. 26

The law of conservation of angular momentum can also be used to determine the free rotation of a symmetrical top. Using the fact that the principal axes of inertia x, y  (perpendicular to the axis of symmetry z  of the top) may be chosen arbitrarily, we take the y  –axis perpendicular to the plane containing the constant vector   M and the instantaneous position of the z  –axis. Then M 2  0 and  formulae (7) show that  2  0 . This means that the direction of M ,   and the axis of the top are at every instant in one plane (Fig. 26). 113

 

  Hence, in turn, it follows that the velocity v  r of every point on the axis of the top is at every instant perpendicular to that plane. That is, the axis of the top rotates uniformly (see below) about the  directions of M describing a circular cone. This is called regular precession of the top. At the same time the top rotates uniformly about its own axis. The angular velocities of these two rotations can easily be  expressed in terms of the given angular momentum M and the angle   between the axis of the top and the direction of M . The angular velocity of the top about its own axis is just the component  3 of  the vector  along the axis:

3 

M3 M  cos  . I3 I3

(9)

 To determine the rate of precession  pr the vector  must be  split into components along z  and along M . The first of these components gives no displacement of the axis of the top, and the second component is therefore the required angular velocity of precession. Fig.26 show that

sin    pr  1 ,

and since

1 

(10)

M 1 M sin   , I1 I1

(11)

M . I1

(13)

we have

 pr 

26. The equations of the motion of a rigid body

Since the rigid body has, in general, six degrees of freedom, the general equation of motion must be six in number. They can be put 114

in a form which gives the time derivatives of two vectors, momentum and angular momentum of the body. The first equation is obtained by simply summing the equations    p  f for each particle in the body, p being the momentum of the  particle and f the force acting on it. In terms of the total momentum      of the body P   p  V and total force acting on it F   f , we have  dP   F. (1) dt   Although F has been defined as the sum of all the forces f acting on the various particles, including the forces due to other  particles, F actually includes only external forces: the forces of interaction between the particles composing the body, must cancel out, since if there are no external forces the momentum of the body, like that of any closed system, must be conserved, i.e. we must have  F  0. If U is the potential energy of a rigid body in an external force  F is obtained by differentiating U with respect to the coordinates of the center of mass of the body:

 U F  . R

(2)

For, when the body undergoes a translation through a distance    R , the radius vector r of every point in the body changes R , and so the change in the potential energy is U  r

 U r





 

U    r  R    R  f i   FR .

(3)

It may be noted that equation (1) can also be obtained as the Euler-Lagrange equation for the coordinates of the center of mass

115

d L L  , dt V R

(4)

with the Lagrangian L

1 1 V 2  I ik  i  k  U R  , 2 2

(5)

for which

  L   V  P , V

(6)

L U     F, R R

(7)

d   PF, dt

(8)

  PF.

(9)

Let us now derive the second equation of motion, which gives  the time derivative of the angular momentum M . To simplify the derivation, it is convenient to choose the «fixed» (inertial) frame of reference in such a way that the center of mass is at rest in that frame at the instant considered. We have

   

 d     M   r Pi   rPi   r Pi  .   dt

(10)

 Our choice of the frame of reference (with V  0 ) means that the    value of v at the instant considered is the same as v  r . Since the    vectors v and P  mv are parallel,

116

  rPi  0 ,

 

(11)

  Pi  f i .

(12)

  Replacing p by the force f , we have finally

  dM  K, dt where

 

  K   rfi .

(13) (14)

 Since M has been defined as the angular momentum about the center of mass, it remains unchanged when we go from one inertial frame to another. We can therefore deduce that the equation of motion (13), though derived for a particular frame of reference, is valid in any other inertial frame, by Galileo’s relativity principle.   The vector r f i is called the moment of the force, f i and so  K is the total torque, i.e. the sum of the moments of all the forces  acting on the body. Like the total force F , the sum (14) includes only the external forces; by the law of conservation of angular momentum, the sum of moments of the internal forces in a closed system must be zero. The moment of a force, like the angular momentum, in general depends on the choice of the origin about which it is defined. In (13) and (14) the moments are defined with respect to the center of mass of the body.  When the origin is moved a distance a , the new radius vector r  of each point in the body is equal to

 

   r  r a .

(15)

Hence

     

    K   r f i   r f i   af i .

(16) 117

Hence we see, in particular, that the value of the torque is  independent of the choice of origin if the total force F  0 . In this case the body is said to be acted on by a couple. Equation (13) may be regarded as Lagrange’s equation d L L    dt  

(17)

for the «rotational coordinates». Differentiating the Lagrangian L

1 1 V 2  I ik  i  k  U R  , 2 2

(18)

 with respect to the components of the vector  we obtain

L  I ik   k  M i .  i

(19)

The change in the potential energy resulting from an  infinitesimal rotation  of the body is 





 



  

 

U    f i  r    f i    r      r f i   K . (20) So that

L U     K .  

(21)

  Let us assume that the vectors F and K are perpendicular.   Then a vector a can always be found such that K   0 , according to Eq. (16), and

 

  K  aF . 118

(22)

  The choice of a is not unique, since the addition to a of any  vector parallel to F does not affect equation (22). The condition  K   0 thus gives a straight line, not a point, in the moving system of   coordinates. When K is perpendicular to F , the effect of all the  applied forces can therefore be reduced to that of a single force F acting along this line. Such a case is that of a uniform field of force, in which the force    on a particle is f  eE , with E a constant vector characterizing the field and e characterizing the properties of the particle with respect to the field. Then   F  E e ,





   K   er  E .

(23) (24)

 Assuming that  e  0 , we define a radius vector r0 such that

   er r0  . e

(25)

Then the total torque is simply

 

   K  r0 F .

(26)

Thus, when a rigid body moves in a uniform field, the effect of  the field reduces to the action of a single force F applied at the point whose radius vector is (25). The position of this point is entirely determined by the properties of the body itself. In a gravitational field, for example, it is the center of mass. 27. Eulerian angles

As has already been mentioned, the motion of a rigid body can be described by means of the three coordinates of its center of mass 119

and any three angles which determine the orientation of the axes x, y , z  in the moving system of coordinates relative to the fixed system X , Y , Z . These angles may often be conveniently taken as the so-called Eulerian angles.

Fig. 27

Since we are here interested only in the angles between the coordinate axes, we may take the origins of the two systems to coincide (Fig.27). The moving xy  -plane intersects the fixed X , Y – plane in some line ON, called the line of nodes. This line is evidently perpendicular to both the Z – axis and the z  – axis; we take its  positive direction as that of the vector product zz  (where z and  z  are unit vectors along the Z and z  axes). We take, as the quantities defining the position of the axes x, y , z  relative to the axes X , Y , Z the angle  between the Z and z  axes, the angle  between the X -axis and ON, and the angle  between the x -axis and ON. The angles  and  are measured around the Z and z  axes, respectively, in the direction given by the corkscrew rule. The angles are defined within the intervals 0     , 0    2 and 0    2 . 120

Let us now express the components of the angular velocity 

vector  along the moving axes x, y , z  in terms of the Eulerian angles and their derivatives. To do this, we must find the components along those axes of the angular velocities , , . The angular velocity  is along the line of nodes ON, and its components are .

1   cos ,

.

.

2   sin , 3  0 .

(1)

Furthermore,  represents the angular velocity along the Z – axis; its component along the z  – axis is  3   cos  , and in the x, y  – plane  sin  . Resolving the latter along the x and y  axes, we have

1   sin  sin ,  2   sin  cos .

(2)

Finally,  is the angular velocity along the z  – axis. Collecting the components along each axis, we have  1   sin  sin    cos     2   sin  cos    sin   .   3   cos    

(3)

If the axes x, y , z  are taken to be the principal axes of inertia of the body, the rotational kinetic energy in terms of the Eulerian angles is obtained by substituting (3) in T





1 2 2 2 I11  I 2  2  I 3 3 . 2

(4)

For a symmetrical top I1  I 2  I 3 , a simple reduction gives

T





I I1 2 2 2  sin    2  3  cos    . 2 2

(5) 121

This expression can also be more simply obtained by using the fact that the choice of directions of the principal axes x, y  is arbitrary for a symmetrical top. If the x axis is taken along the line of nodes ON, i.e.   0 , the components of the angular velocity are simply 1   ,  2   sin  ,  3   cos   .

(6)

As a simple example of the use of the Eulerian angles, we will determine the free motion of a symmetrical top, already found before. We take the Z – axis of the fixed system of coordinates in the 

direction of the constant angular momentum M of the top. The z  –axis of the moving system is along the axis of the top; let the x – axis coincide with the line of nodes at the instant under consideration. Then, according to Eq.(6), the components of the 

vector M are M 1 I11  I11 , M 2 I 2  2  I1 sin  , M 3 I 3 3  I 3  cos    .

(7)

Since the x  – axis is perpendicular to the Z – axis, we have

M 1 0 ,

M 2 M sin  , M 3 M cos .

(8)

A comparison gives

  0 , M  I 1 , M cos   I 3  cos     .

(9)

The first of these equations gives   const , i.e. the angle  between the axis of the top and the direction of M is constant. The M second equation gives the angular velocity of precession   , in I1 agreement with 122

M pr  . I1

(10)

Finally, the third equation gives the angular velocity of the top with respect to its own axis: 3 

M cos  . I3

(11)

28. Euler's equations of motion for an absolute solid body with one fixed point

The equations of motion in the fixed system of coordinates are

 dP   F, dt

  dM K, dt

(1)

 where the derivatives are the rates of change of the vectors P and  M with respect to that system. The simplest relation between the  components of the rotational angular momentum M of a rigid body and the components of the angular velocity occurs, however, in the moving system of coordinates whose axes are the principal axes of inertia. In order to use this relation, we must first transform the equations of motion to the moving coordinates x,y,z.   dA Let be the rate of change of any vector A with respect to the dt  fixed system of coordinates. If the vector A does not change in the moving system, its rate of change in the fixed system is due only to the rotation, so that  dA    A , (2) dt because

 



 

r    r ,



 

v  v  r  .

(3) 123

In the general case, the right-hand side includes also the rate of  change of the vector A with respect to the moving system. Denoting  d A this rate of change by , we obtain dt

  dA d A     A . dt dt

 

(4)

Using this general formula, we can immediately write equations (1) and in the form

 d P     P  F , dt

 

(5)

    d M  M  K . dt

(6)

 

Since the differentiation with respect to time is here performed in the moving system of coordinates, we can take the components of equations (5) and (6) along the axes of that system as

 d P  dP1 ,  ,     dt 1 dt

(7)

dM 1  d M  ,...,    dt  dt  2

(8)

where the suffixes 1, 2, 3 denote the components along the axes   x,y,z. In the first equation we replace P  V , obtaining  dV1    2V3   3V2   F1 , dt  



124

 dV2    3V1  1V3   F2 , dt  



 dV3   1V2   2V1   F3 .  dt 



(9)

If the axes x,y,z are the principal axes of inertia, we can put M 1  I 11 , M 2  I 2  2 , M 3  I 3  3 , etc., in the equations (5) and (6), obtaining d1 I1  I 3  I 2  2  3  K1 , dt I2

d 2  I 1  I 3  3 1  K 2 , dt

I3

d 3  I 2  I 1 1 2  K 3 . dt

(10)

These are Euler's equations. In the case of free rotation, K  0 , Euler's equations become d 1  I 3  I 2    23  0 , dt I1 d 2  I 1  I 3    3 1  0 , dt I2

d 3  I 2  I 1   1  2  0 . dt I3

(11)

As an example, let us apply these equations to the free rotation of a symmetrical top, which has already been discussed. Putting   0    const . I1  I 2 , we find from the third equation  3 3 125

I 3  I 1   I1

3



(12)

where  is a constant. We then write the first two equations as

   ,     .  1 2 2 1 Multiplying the second equation by equation, we have

i

(13)

and adding the first

d 1  i 2   i 1   2  , dt

(14)

1  i 2  Aeit ,

(15)

so that

where A is a constant, which may be made real by a suitable choice of the origin of time. Thus 1  A cos t ,  2  A sin t .

(16)

This result shows that the component of the angular velocity perpendicular to the axis of the top rotates with an angular velocity  , satisfying the relation 12   22  A . Since the component  3  const along the axis of the top is also constant, we conclude  that the vector  rotates uniformly with angular velocity  about the axis of the top, remaining unchanged in magnitude. Taking into account the relations M 1  I 11 , M 2  I 2  2 , M 3  I 3  3 between   the components of  and M , we conclude that the angular  momentum vector M evidently executes a similar motion with respect to the axis of the top. This description is naturally only a different view of the motion already discussed before, which was referred to the fixed system of  coordinates. In particular, the angular velocity of the vector M (the 126

Z-axis in Fig. 28) about the z  -axis is, in terms of Eulerian angles, is the same as the angular velocity  Using equations

  0 , M  I 1 , M cos   I 3  cos     .

(17)

we have

Fig. 28

 

 I I  M cos    cos   M cos     I3  I 3 I1 

(18)

or     3

I 3  I1 , I1

(19)

in agreement with

I 3  I 1   I1

3

 .

(20)

127

29. Motion in a non-inertial frame of reference

Up to this point we have always used inertial frames of reference when discussing the motion of mechanical systems. For example, the Lagrangian L0 

mv02 U 2

(1)

and the corresponding equation of motion mv02 U   2 r

(2)

for a single particle in an external field are valid only in an inertial frame. Let us now find out how the equations of motion will look like in a non-inertial frame of reference. The basis of the solution of this problem is again the principle of least action, whose validity does not depend on the choice of the frame of reference. Lagrange's equations

d L L   dt v r

(3)

are likewise valid, but the Lagrangian is no longer of the form (1), and to derive it we must carry out the necessary transformation of the function L0 . This transformation is done in two steps. Let us first consider a frame of reference K  which moves with a translational velocity    V t  relative to the inertial frame K 0 . The velocities v0 and v  of a particle in the frames K 0 and K  respectively are related by

   v0  v   V t  . Substitution of this in (1) gives the Lagrangian in K  : 128

(4)

    mV 2 mv  2 L   m vV  U . 2 2

 

(5)

Now V 2 t  is a given function of time, and can be written as the total derivative with respect to t of some other function; the third term in

L can therefore be omitted. Next,   dr  v  , dt

 where r  is the radius vector of the particle in the frame K  . Hence

    dr  d   dV mV t v   mV mVr   mr   . dt dt dt





(6)

Substituting in the Lagrangian and again omitting the total time derivative, we have finally L 

   mv  2  mW t r   U , 2

(7)

  dV where W  is the translational acceleration of the frame K  . dt The Lagrange equation derived from (7) is

 d  U mv      mW t  , dt r 

(8)

  dv  U     mW t  . dt r 

(9)

m

Thus, as regards the equations of motion of a particle, the accelerated translational motion of a frame of reference is equivalent to the application of a uniform field of force equal to the mass of the 129

 particle multiplied by the acceleration W , in the direction opposite to this acceleration. Let us now consider a further frame of reference K , whose origin coincides with that of K  , but which rotates relative to K   with angular velocity t  . Thus, K executes both a translational and a rotational motion relative to the inertial frame K 0 .



The velocity v  of the particle relative to K  is composed of its   velocity v relative to K and the velocity r of its rotation with K:    (10) v   v  r .

 

 

  The radius vectors r  and r in the frames K and K  coincide. Substituting this in the Lagrangian (7), we obtain

L

 

 

   m 2 mv 2  mv r  r  mWr   U . 2 2

(11)

This is the general form of the Lagrangian of a particle in an arbitrary, not necessarily inertial, frame of reference. The rotation of the frame leads to the appearance in the Lagrangian of a term linear in the velocity of the particle. To calculate the derivatives appearing in Lagrange's equation, we write the total differential

 

    

          dL  mv dv  mdv r  mv dr  m r dr 

 

 

  U         mWdr   dr  mv dv  mdv r  mdr v   r

  

     U   m r  dr  mWdr   dr . r   The terms with dv and dr give

130

(12)

 

  L   mv  m r , v

(13)

    

   U  L   m v   m r   mW   . r r

(14)

Substitution of these expressions in (2) gives the required equation of motion:

       dv U m     mW  m r   2m v   m  r  .   dt r

    

(15)

We see that the "inertia forces" due to the rotation of the frame   consist of three terms. The force m  r   is due to the non  uniformity of the rotation, but the other two terms appear even if the  rotation is uniform. The force 2m v  is called the Coriolis force; unlike any other (non-dissipative) force hitherto considered, it   depends on the velocity of the particle. The force m  r  is called

 

   



the centrifugal force. It lies in the plane through r and  is  perpendicular to the axis of rotation (i.e. to  ), and is directed away from the axis. The magnitude of this force is m 2 , where  is the distance of the particle from the axis of rotation. Let us now consider the particular case of a uniformly rotating frame with no translational acceleration. Putting in (11) and (15)     const and W  0 , we obtain the Lagrangian L

 

 

  m 2 mv 2  mv r  r  U 2 2

(16)

and the equation of motion

    dv U m     2m v   m  r  . dt r

    

(17) 131

The energy of the particle in this case is obtained by substituting

 

 L   P    mv  m r v

(18)

in E  pv  L , which gives

E

 

mv 2 m   2  r  U . 2 2

(19)

It should be noticed that the energy contains no term linear in the velocity. The rotation of the frame simply adds to the energy a term which depends only on the coordinates of the particle and is proportional to the square of the angular velocity. This additional m 2 term  r is called the centrifugal potential energy. 2  The velocity v of the particle relative to the uniformly rotating frame of reference is related to its velocity v0 relative to the inertial frame K 0 by    v0  v  r . (20)

 

 

 The momentum p (18) of the particle in the frame K is   therefore the same as its momentum p0  mv0 in the frame K 0 . The angular momenta

    M 0  r p0  and M  r p 

(21)

are likewise equal. The energies of the particle in the two frames are  not the same, however. Substituting v from (20) in (19), we obtain

E

 

 mv02 mv 2    mv0 r  U  0  U  mrv0  . 2 2

(22)

The first two terms are the energy E0 in the frame K 0 . Using the angular momentum M , we have 132

 E  E0  M .

(23)

This formula gives the law of transformation of energy when we change to a uniformly rotating frame. Although it has been derived for a single particle, the derivation can evidently be generalized immediately to any system of particles, obtaining again Eq. (23). Control questions: 1. 2. 3. 4. 5.

Degrees of freedom of a rigid body. Coriolis force. Principal axes of inertia and principal moments of inertia. Centrifugal potential energy. Principal moments of inertia.

133

VII THE CANONICAL EQUATIONS

30. Poisson brackets. Properties of the Poisson brackets. Jacobi's identity and Poissson’s theorem

Let f  p, q, t  be some function of coordinates, momenta and time. Its total time derivative is

 f  f df f     q k  p k . pk dt t k  qk 

(1)

Substitution of the values of Hamilton's equations

p k  

H H ; q k  , qk p k

(2)

leads to the expression

df f   H , f , dt t

(3)

where



f H f  .   pk qk qk pk 

H , f     H k

(4)

This expression is called the Poisson bracket of the quantities H and f . Those functions of the dynamical variables which remain constant during the motion of the system are, as we know, called integrals of motion. We see from (3) that the condition for the df quantity f to be an integral of motion  0 can be written as dt 134

f  H , f   0 . t

(5)

If the integral of motion does not depend explicitly on time, then

H , f   0 ,

(6)

i.e. the Poisson bracket of the integral and the Hamiltonian must be zero. For any two quantities f and g , the Poisson bracket is defined analogously to (4):

 f g f g  .   pk qk qk pk 

 fg    k

(7)

The Poisson bracket has the following properties, which are easily derived from its definition. If the two functions are interchanged, the bracket changes sign; if one of the functions is a constant c , the bracket is zero:

Also

 f1 

 fg  gf ,

(8)

 fc  0 .

(9)

f 2 , g   f1 g   f 2 g,

(10)

 f1 f 2 , g  f1  f 2 g  f 2  f1 g.

(11)

or

Taking the partial derivative of (7) with respect to time, we obtain

  fg   f , g    f , g . t  t   t 

(12)

135

If one of the functions f and g is one of the momenta or coordinates, the Poisson bracket reduces to a partial derivative:

 f , qk  

f , pk

(13)

or

 f , pk   

f . qk

(14)

Formula (13), for example, may be obtained by putting g  qk qk in (7); the sum reduces to a single term, since   kl and ql qk  0 . Putting in (13) and (14) the function f equal to qi and pi pl we have, in particular,

qi , qk   0 , pi , pk   0 , pi , pk    ik .

(15)

 f , g , h  g , h, f   h,  f , g  0 .

(16)

The relation

is known as Jacobi's identity. An important property of the Poisson bracket is that, if f and g are two integrals of the motion, their Poisson bracket is likewise an integral of the motion:

 f , g   const .

(17)

This is Poisson's theorem. The proof is very simple if f and g do not depend explicitly on the time. Putting h  H in Jacob i's identity, we obtain

H  f , g    f g , H   g H , f   0 . 136

(18)

Hence, if H , g   0 and H , f   0 , then H ,  f , g   0 , which is the required result. If the integrals f and g of the motion are explicitly time-dependent, we put, from (3), d  f , g     f , g   H ,  f , g . t dt

(19)

Using formula (19) and expressing the bracket H ,  f , g  in terms of the two others by means of Jacobi's identity, we find d  f , g    f , g    f , g    f , g , H   g , H , f   dt  t   t   f   g   df   dg     H , f , g    f ,  H , g    , g    f , , t t        dt   dt 

(20)

which evidently proves Poisson's theorem. 31. Abbreviated action. The variational principle determining the trajectory of the system or the Maupertuis principle

In formulating the principle of least action, we have considered the integral t2

S   Ldt

(1)

t1

taken along a path between two given positions q1 and q2 which the system occupies at given instants t1 and t2 . In varying the action, we compared the values of this integral for neighboring paths with the same values of q(t1 ) and q(t 2 ) . Only one of these paths corresponds to the actual motion, namely the path for which the integral S has its minimum value. Let us now consider another aspect of the concept of action, regarding S as a quantity characterizing the motion along the actual path, and compare the values of S for paths having a common 137

beginning at q(t1 )  q1 , but passing through different points at time t2 . In other words, we consider the action integral for the true path as a function of the coordinates at the upper limit of integration. The change in the action from one path to a neighboring path is given (if there is one degree of freedom) by the expression t2

t 2  L  L  d L  qdt. S   q       q  t1 t1  q dt q 

(2)

Since the paths of the actual motion satisfy Lagrange's equations, the integral in S  0 is zero. In the first term we put q(t1 )  0 , and denote the value of q(t 2 ) by q simply. Replacing L  p , we have finally S  pq or, in the general case of any q number of degrees of freedom,

S   piqi . i

(3)

From this relation it follows that the partial derivatives of the action with respect to the coordinates are equal to the corresponding momenta: S  pi . (4) qi The action may similarly be regarded as an explicit function of time, by considering paths starting at a given instant t1 and at a given point q1 and ending at a given point q2 at various times t2  t . The S thus obtained may be found by an appropriate partial derivative t variation of the integral. It is simpler, however, to use formula (3), proceeding as follows. From the definition of the action, its total time derivative along the path is 138

dS  L. dt

(5)

Next, regarding S as a function of coordinates and time, in the sense described above, and using formula (3), we have dS S S S   qi    pi qi . dt t t qi t t

(6)

A comparison gives S  L   pi q i t t

(7)

or S  H . t

(8)

Formulae (3) and (8) may be represented by the expression dS   pi d qi  Hdt t

(9)

for the total differential of the action as a function of coordinates and time at the upper limit of integration in (1). Let us now suppose that the coordinates (and time) at the beginning of the motion, as well as at the end, are variable. It is evident that the corresponding change in S will be given by the difference of the expressions (9) for the beginning and end of the path. It is of interest to note that Hamilton's equations can be formally derived from the condition of minimum action in the form S     pi d qi  Hdt , t 

(10)

which follows from (9), if the coordinates and momenta are varied independently. Again assuming for simplicity that there is 139

only one coordinate and momentum, we write the variation of the action as 

 H  q

S   pdq  pdq   

  H qdt     p

  pdt .  

(11)

An integration by parts in the second term gives 

 H  p

S   p dq   

    H dt    pq    q dp      q 

  dt .  

(12)

At the limits of integration, we must put q  0 , so that the integrated term is zero. The remaining expression can be zero only if the two integrands vanish separately, since the variations p and q are independent and arbitrary:  H dq    p

 dt , 

 H dp    q

 dt , 

(13)

which, after division by dt , are Hamilton's equations. Maupertuis' principle The motion of a mechanical system is entirely determined by the principle of least action. By solving the equations of motion which follow from that principle, we can find both the form of the path and the position on the path as a function of time. If the problem is the more restricted one of determining only the path, without reference to time, a simplified form of the principle of least action may be used. We assume that the Lagrangian and therefore the Hamiltonian, do not involve the time explicitly, so that the energy of the system is conserved: H ( p, q )  E  const . According to the principle of least action, the variation of the action, for given initial and final coordinates and times ( t0 and t say), is zero. If, however, we allow a variation of the 140

final time have

t,

the initial and final coordinates remaining fixed, we

S   Ht.

(14)

We now compare, not all virtual motions of the system, but only those which satisfy the law of conservation of energy. For such paths we can replace H in (14) by a constant E , which gives

S  Et  0.

(15)

Writing the action in the form (10) and again replacing H by E , we have S    pi d qi  E (t  t 0 ) . (16) t

The first term in this expression, S 0    p i d qi t

(17)

is sometimes called the abbreviated action. Substituting (16) in (15), we find that

S 0  0.

(18)

Thus the abbreviated action has a minimum with respect to all paths which satisfy the law of conservation of energy and pass through the final point at any instant. In order to use such a variational principle, the momenta (and so the whole integrand in (17)) must be expressed in terms of the coordinates q and their differentials dq . To do this, we use the definition of momentum: pi 

 dq L( q, ) q i dt

(19)

and the law of conservation of energy: 141

E ( q,

dq )  E. dt

(20)

Expressing the differential dt in terms of the coordinates q and their differentials dq by means of (20) and substituting in (19), we have the momenta in terms of q and dq , with the energy E as a parameter. The variational principle so obtained determines the path of the system, and is usually called Maupertuis' principle, although its precise formulation is due to Euler and Lagrange. The above calculations may be carried out explicitly when the Lagrangian takes its usual form as the difference of the kinetic and potential energies: L

1  aik q q i q k  U q  . 2 i ,k

(21)

The momenta are pi 

L   aik q q k q i k

(22)

and the energy is E

1  aik q q i q k  U q  . 2 i ,k

(23)

The last equation gives   a dq dq  dt   ik i k   2 E  U  

(24)

substituting this in  pi dqi   aik i

142

i ,k

dqk dqi , dt

(25)

we find the abbreviated action: S 0   2( E  U )  aik dqi dq k  .   i ,k

(26)

In particular, for a single particle the kinetic energy is 2 1  dl  T  m  , where m is the mass of the particle and dl an 2  dt  element of its path; the variational principle which determines the path is



2m( E  U )dl  0,

(27)

where the integral is taken between two given points in space. This form is due to Jacobi. In the case of a free particle, U  0 , Eq.(27) gives the trivial result   dl  0, i.e. the particle moves along the shortest path between the two given points, i.e. in a straight line. Let us return now to the expression (16) for the action and vary it with respect to the parameter E . We have

S 

S 0 E  (t  t 0 )E  Et E

(28)

substituting in (15), we obtain S 0  (t  t 0 ) . E

(29)

When the abbreviated action has the form (26), this gives   aik dqi dqk  i ,k   t  t0 ,    2( E  U )   

(30)

143

which is just the integral of equation (24). Together with the equation of the path, it entirely determines the motion. 32. The method of Hamilton-Jacobi. The mathematical structure of the equations of Hamilton-Jacobi. Their full integral

The principle of least action states that t2

S    Lq, q , t dt  0 ,

(1)

t1

where the action is a function of time and coordinates: t2

S   Lq, q , t dt .

(2)

t1

The variation of this action with respect to the trajectory q(t ) can be expressed as t2

t 2  L d L  L qdt . S  q     q t1 t1  q dt q 

(3)

Since the Euler-Lagrange equation is satisfied along the trajectory, the integrand in Eq. (3) is equal to zero. Without loss of generality, we can assume that the lower limit of the path is L qt1   0 and qt 2   q . On the other hand, since  p , we have q that S  pq , an expression which for an arbitrary number of degrees of freedom becomes

S   piqi . i

(4)

As we can see from this expression, the partial derivative of the action with respect to the generalized coordinates is equal to the impulse, i.e., 144

S  pi . qi

(5)

Furthermore, according to the definition of the action, its total time derivative is: dS  L. dt

(6)

On the other hand, since the action can be considered as a function of the coordinates and time, we can calculate the total derivative of S as dS S S S   q i    pi q i , i i t dt t qi

(7)

where we have used Eq. (5). The last equation can be rewritten as the expression S  L   pi qi , i t

(8)

S  H . t

(9)

or equivalently,

Remembering that H  H  pi , qi  and equation satisfied by the function S q, t  :

S  pi , we obtain the qi

 S S S   H  q1 ,..., q s ; ,..., ;t   0 . t q1 q s  

(10)

145

This is a first-order partial differential equation which is known as the Hamilton-Jacobi equation. It represents one of the methods for integrating the equations of motion. Let us consider as a particular example the Hamiltonian of a particle for which the motion is confined by the potential U, i.e., H

p2  U r  . 2m

(11)

Then, the Hamilton-Jacobi equation S p2   U r  t 2m

(12)

under the consideration of the definition relation  S p  r

(13)

becomes 2

S 1  S       U r  . t 2m  r 

(14)

In a Cartesian coordinate system, we have that S 1   S   S   S         t 2m   x   y   z  2

2

2

   U r   

(15)

and in a polar coordinate system: 2

2

1  S  1  S  S    U r  .      2mr 2    t 2m  r 

(16)

Now let us consider the method of Hamilton-Jacobi in detail. In general, the solution of a first-order partial differential equation 146

depends on an arbitrary function. Such solution is called a generalized solution or general integral. In mechanics, however, the so-called complete integral plays a more important role because it contains as many independent arbitrary constants as there are independent variables. The independent variables in the HamiltonJacobi equation are the time and the coordinates. Therefore, the complete integral of the equation with s-degrees of freedom contains s + 1 arbitrary constants. The total integral of the Hamilton-Jacobi equation is therefore

S  f t , q1 , ..., qs ;  1 , ...,  s   A .

(17)

Now let us consider the relation between the total integral of the Hamilton-Jacobi equation and the equation of motion. This procedure can be interpreted as a canonical transformation in which we pass from the set of coordinates q and p to new set of coordinates α and β, according to the relationships pi 

f , qi

i 

f f , H  H  .  i t

(18)

Where as the function f satisfies the Hamilton-Jacobi equation, we can see that the new Hamiltonian becomes zero: H  H 

S f 0. H  t t

(19)

Therefore the canonical equations in terms of the new variables become  i  0 , i  0 (20) and the solutions are

 i  const ,  i  const .

(21)

On the other hand, the equation 147

f  i  i

(22)

allows us to express the new s coordinates in terms of the time,  and  . To obtain the main solution to the problem of motion of a mechanical system by using Hamilton-Jacobi method, one can use Eq.(19) and the relations S  i , (23)  i which determine the new coordinates as functions of time. The time dependence of the impulse can be obtained by using the equation pi 

S . qi

The Hamilton-Jacobi equation simplifies drastically when the function H is independent of time and therefore the system is conservative. The time dependence of the action becomes in this case S  S 0 q   Et ,

(24)

where E is the total energy of the system. In this case, from Eq. (10) it follows that the Hamilton-Jacobi equation can be written as  S S  H  q1 , ..., qs ; 0 , ..., 0   E . q1 qs  

(25)

33. The method of separation of the variables

In most cases, to find the complete integral of the HamiltonJacobi equation one can use the method of separation of the variables. We know that energy is conserved for the closed mechanical systems, i.e., 148



Then,

S  H  E  const . t

(1)

S   Et  S q   S 0 .

(2)

Let us consider the particular case of a mechanical system with two degrees of freedom in polar coordinates, i.e., S r , , t    Et  S r ,   S 0 .

(3)

It is always possible to choose S 0  0 so that the only unknown is the function S r ,  . For concreteness, we focus on the case of a particle in a Kepler potential so that 2

2

1  S  1  S   S        2mr 2    r t 2m  r  Hamilton-Jacobi equation we obtain 

and

for

the

stationary

2

2

1  S  1  S      E    2m  r  2mr 2    r

(4)

or 2

2

 1  S  1  S       E    0 .    2  2m  r  2mr     r

(5)

The method of separation of variables consists in assuming that

S r ,   Rr     .

(6)

Then, from Eq.s (5) we obtain 2

2

 1  R  1         E    0    2  2m  r  2mr     r

(7)

or equivalently, 149

2

2

       R   . r    2mr 2  E     r   r     2

(8)

Since the left-hand side depends only on the coordinate r and the right-hand side depends only  , then the equality holds only if both expressions are equal to the same constant. This means that the following two equations must be satisfied  , 

(9)

   R      2m E    2  0 ,  r r r   

(10)

2

where β is an arbitrary constant. The solution of Eq. (9) is     const ,

(11)

where as the solution of Eq.(10) is

  dR   2m E    2 . dr r r 

(12)

Finally, according to Eq. (3), the solution can be expressed as: S r , , t    Et    

   2m E    2 dr  S 0 . r r 

(13)

We see that the solution to the Kepler problem is given in terms of two independent functions which depend on the coordinates r and  . In addition, it contains the two multiplicative constants E and β and the additive constant S 0 .

150

We will now consider as a second example the case of a particle falling along the axis y under the action of the gravitational acceleration g. The corresponding Hamilton-Jacobi equation reads   S S S   H  ; ; x; y  t   x y

(14)

with the Hamiltonian H

2 p x2 p y   mgy . 2m 2m

(15)

The total energy of the system is 2 2 1  S   S   E        mgy . 2m  x   y    

Considering 

(16)

S  E , the action can be expressed as the t

function S   Et  S x  S y ,

(17)

p x x  S x ; S x  p x x , 2

p2 1  S     mgy , E x  2m 2m  y 

(18)

2

 mgy  E 

p x2 1  S    ,  2m 2m  y 

2mE  p x2  2m 2 gy 

S , y

(19)

(20)

151

S y   2mE  p x2  2m 2 gy dy .

(21)

If we introduce the notation a  2 mE  p x2 ; b  2m 2 g ,

then 1 b

1

1

 a  b  2 dy    a  by 2 d a  by   

Sy

2mE  p 

S   Et  p x x 

3 2 a  by 2 , 3b

(22)

 2m 2 gy 2 , 3m 2 g

(23)

3 1 2mE  p x2  2m 2 gy 2 . 2 3m g

(24)

3

2 x

Moreover, the following constants can be introduced S  c1 , E S  c2 , p x

c1  t 

(25) (26)

p y2  2m 2 gy mg

p y2  m 2 g 2 c1  t 

,

(27)

.

(28)

2

y

2m 2 g

Assuming the initial conditions:

t  0 ; y  0 ; c1 

152

py mg

,

(29)

where p y  mv0 y , we obtain y  mv0 y t 

gt 2 , 2

(30)

and p y2  2m 2 gy  p x

c2  x 

m2 g

.

(31)

Moreover, by using the initial conditions

t  0 ; x  0 ; from this c2 

p y px m2 g

,

(32)

we get

x  mvox t .

(33)

Finally, we consider the case of a particle with total energy E

p2 p2 mv 2 and H  x  y  const . 2 2m 2m

(34)

Then 

S mv 2  t 2

(35)

and the Hamilton-Jacobi equation for the particle reads 2 2 S 1  S   S           . t 2m  x   y    

(36)

153

Furthermore, S   Et  p x x  S y ,

and from 

(37)

S  E we obtain t p2 1  S y  E x  2m 2m  y

2

  , 

(38)

dy 2mE  p x2  dS ,

(39)

S   Et  p x x  2mE  p x2 y .

(40)

We now introduce the constants S  c1 , E S  c2 , p x

(41) (42)

to obtain c1  t 

y

c1  t 

2mE  p x2 m



my 2mE  p x2

,

c1  t  p y c1  t mvoy m



m

(43)

 c1  t voy . (44)

Using the initial conditions

t  0 ; y  0 ; c1  0 ,

(45)

it follows that y  voy t . 154

(46)

Moreover, c2  x 

yp x 2mE  p

2 x

x

mvoy tmvox yp x x  x  mvox t . (47) py mvoy

Finally, with the initial conditions

t  0 ; x  0 ; c2  0 ,

(48)

we get x  vox t .

(49)

Control questions: 1. 2. 3. 4. 5.

Poisson brackets. Jacobi identity. General and total integral. The Hamilton-Jacobi equation. Separation of variables.

155

Educational issue Beissen Nurzada Quevedo Hernando LECTURE COURSE ON THEORETICAL MECHANICS Educational manual Computer page makeup and cover designer N. Bazarbaeva

IS No.10823 Signed for publishing 27.04.17. Format 60x84 1/16. Offset paper. Digital printing. Volume 9,75 printer’s sheet. Edition 160. Order No.2005 Publishing house «Qazaq university» Al-Farabi Kazakh National University, 71 Al-Farabi, 050040, Almaty Printed in the printing office of the «Qazaq university» publishing house

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