Madhya Pradesh Bhoj (Open) University, Bhopal
M.Sc. (Previous) Physics
PAPER –II
CLASSICAL AND STATISTICAL MECHANICS
BLOCK-I
Lagrangian and Hamiltonian Mechanics
1
Syllabus
UNIT-I
Lagrangian Mechanics
Constraints, Generalised coordinates D' Alembert Principle and derivation of
Lagrangian equation velocity dependent potentials and Rayleigh's dissipination
function.
Variational Principle. Euler – Lagrange equation, Derivation of Lagrange's
equation from Hamilton’s principle.
UNIT-II
Kepler's Problem and Hamiltonian Mechanics
Two-body central force problem, Kepler's problem, inverse square law of force.
Scattering in a central force field.
Derivation of Hamilton's equation from variational principle of least action.
Equations of canonical transformation. Lagrangian and Poisson brackets.
Angular momentum and Poisson bracket relation. Equation of Motion in Poisson
bracket notation.
2
UNIT 1
LAGRAGIAN MECHANICS
Structure
1.0
Introduction
1.1
Objectives
1.2
Constraints
1.2.1
1.3
Classification of Constraints
Generalised coordinates
1.3.1
Generalised Notation
1.3.2
Advantages of the Generalised Notation
1.4
DAlembert Principle
1.5
Derivation of Lagrangian equation from
1.5.1
Velocity dependent potentials
1.5.2
Rayleigh dissipation function
1.6
Variational principle
1.7
Euler –Lagrange Equation
1.8
Lagrange equation from Hamilton’s Principle
1.9
Let Us Sum Up
1.10
Check Your Progress: The Key
3
Lagrangian and Hamiltonian Mechanics
1. 0
INTRODUCTION
Mechanics is the study of the motion of physical bodies .The possible and actual
motions of physical objects, whether large or small, fall under the domain of mechanics.
In the present century the term “Classical mechanics” has come in to wide to denote this
branch of physics in the contradiction to the newer theories especially quantum
mechanics. “Classical mechanics has been customarily used to denote that part of the
mechanics which deals with the description and explanation of the motion of the objects,
neither too big so there exists a close agreement between theory and experiment nor too
small interacting objects, more precisely like the systems on molecular or subatomic
scale.” We shall follow this usage, interpreting theories the name to include the type of
mechanics. Classical mechanics may be classified in to three subsections (i) Kinematics
(ii) Dynamics (iii) Statics.
In this unit we deals with the structure and law of mechanics with the
applications, starting from basic fundamental concepts .Having established the essential
pre-requisites, the Lagrangian formulation known for its mathematical elegance.
1.1
OBJECTIVES
After completing this unit we will able to,

Define constraints, its types and Generalised coordinates.

State the DAlembert Principle.

Derive the Lagrangian equation from(i) Velocity dependent potentials (ii) Rayleigh dissipation function.
1.2

State and define the Variational principle.

Derive the Euler –Lagrange Equation.
CONSTRAINTS
Constraints are the geometrical or kinematical restrictions on the motion of the particle
or system of the particles. Systems with such constraints of motion are called as
4
Lagrangian Mechanics
Constrained systems and their motion is known as constrained or restricted motion.
Some examples of restricted motions are The motion of the rigid body is restricted to the condition that the distance
between any two particles remains unchanged.
 The motion of the gas molecules with in the container is restricted by the walls of
the vessels.
 A particle placed on the surface of a solid sphere is restricted so that it can only
move either on the surface or outside the surface.
1.2.1 Classification of Constraints
The constraints can be classified in to the following categories:
(i) Holonomic and non-holomonic constraints (ii) Scleronomic and rhenomic constraints
Holonomic constraints:-Constraints are said to be holomonic if the conditions of all the
constraints can be expressed as equations connecting the coordinates of the particles and
possible time in the form
  

f ( r1,r2,r3……..,rn,t) =0
(1.1)
 

Where r1, r2, r3……..,r
n represent the position vectors of the particles of a system and t
the time. In Cartesian coordinates equation (1.1) can be written as,
f (x1, y1, z1; x2, y2, z2,……… xn, yn, zn,t) =0
(1.2)
Examples of holonomic constraints:1. The constraints involved in the motion of rigid bodies. In rigid bodies, the
distance between any two particles is always constant and the condition of
constraints are expressed as

ri - rj2 - Cij2 =0
(1.3)
2. Constraints involved in the motion of the point mass of a simple pendulum.
3. The constraints involved when a particle is restricted to move along any curve
(circle or ellipse) or in a given surface.
Non-holonomic constraints: - If the conditions of the constraints can not be expressed
as equations connecting the coordinates of particles as in case of holomonic, they are
called as non-holomonic constraints. The conditions of these constraints are expressed in
the form of inequalities. The motion of the particle placed on the surface of sphere under
the
5
Lagrangian and Hamiltonian Mechanics
action of the gravitational force is bound by non-holonomic constraints, for it can be
expressed as an inequality, r2 - a2  0.
Examples of non-holonomic constraints
1. Constraints involved in the motion of a particle placed on the surface of a solid
sphere
2. An object rolling on the rough surface without slipping.
3. Constraints involved in the motion of gas molecules in a container.
(ii) Scleronomic and Rhenomic Constraints: - The constraints which are independent
of time are called Scleronomic constraints and the constraints which contain time
explicitly, called rhenomic constraints
Examples: - A bead sliding on a rigid curved wire fixed in space is obviously subjected
to Scleronomic constraints and if the wire is moving is prescribed fashion the constraints
become Rhenomic.
1.3
GENERALISED COORDINATES
Generalised co-ordinates:- These are the coordinates which are used to eliminate the
dependent coordinates and can be expressed in another way by the introduction of (3N-p)
independent coordinates of variables called the Generalised coordinates, where N
represent the number of particles of a system and p represent the holonomic constraints.
Thus any ‘q’ quantities which completely define the configuration of the system having
‘f’ degree of freedom are called Generalised co-ordinates of the system and are denoted
by q1, q2, q3,…… qf , or just qi ( i=1,2,3,4…f )
Principles for the choosing a suitable set of Generalised co-ordinates - For this three
principles are used –
1. They should specify the configuration of the system.
2. They may be varied arbitrarily and independently of each other, with out violating
the constraints on the system.
3. There is no uniqueness in the choice of the generalised coordinates
6
Lagrangian Mechanics
It may be noted that generalised co-ordinates need not to have the dimensions of length
or angles. Generalised co-ordinates need not to be Cartesian co-ordinates of the particles
and the condition of the problem may render some other choice of co-ordinates which
may be more convenient.
1.3.1 Generalised Notations
(i)
Generalised Displacement – A small displacement of an N particle system is


defined by changes ri in position co-ordinates ri ( i =1,2,3….,N) with time ‘t’ held fixed.


An arbitrary virtual displacement ri, remembering that ri ’s are function of generalised


co-ordinates i.e. ri = ri (q1, q2,….. q3N,t), can be written by using Euler’s theorem as,

3N
ri =
ri
qj
qj
(1.5)
j =1
qj is called the generalised displacement or virtual displacement. If qj is an angle coordinate, qj is an angular displacement.
(ii)
Generalised velocity – The time derivative of the generalised qk ,is called
generalised velocity associated with particular co-ordinates qk for an unconstrained
system,

ri
Then,

= ri (q1, q2,….. q3N,t),


ri 
qj  ri
qj
t

ri 
qj  ri
qj
t

ri =
3N



(1.6)
j =1
If N-particle system contains k-constraints, the number of generalised co-ordinates are
3N-k=f and,

f

ri =


(1.7)
j =1
(iii) Generalised Acceleration- components of generalised acceleration are obtained by
differentiating equation (1.6) or (1.7) w.r.t. time and finally we obtain the expression

 3N
ri =
j =1

3N 3N

ri 
qj
qj
+
j =1 k =1

2ri q q
j k +2
qj qk
3N
2ri
qj t
2

qj +  2ri
t
j =1
(1.8)
7
Lagrangian and Hamiltonian Mechanics
From the above equation it is clear that the cartesian components are not linear functions
 qj alone, but depend quadratically and linearly
of components of generalised acceleration

on generalised velocity component qj as well.

(iv) Generalised Force – Let us consider the amount of work done W by the force Fi
during an arbitrary small displacement

 
Fi .ri

Qj .qj
N
W =
 
N
=
3N

Fi .
i=1
i
j=1

ri of
i
i
the system

ri
qj =
qj
 
N
i=1
3N
j=1

Fi .

ri
qj
qj
3N
=
i=1
Where,
(1.9)

N
Qj
=
j=1

Fi

.
ri
qj
(1.10)
Here we note that Qj depends on the force acting on the particles and on the co-ordinate qj
and possibly on time t. Therefore, Qj is called the generalised force.
1.3.2 Advantages of Generalised co-ordinates
The main advantage in the formulating laws of mechanics in terms of generalised coordinates and the associated mechanical quantities is that the equation of motion looks
simpler and can be solved independently of each other since generalised co-ordinates are
all independent and constraints have no effect on them. The equations of motion are then
called Lagrange’s equation of motion.
Check Your Progress 1
Note: a)
Write your answers in the space given below.
b)
Compare your answers with the ones given at the end of the units.
(i)
Define constraints and Write down its type with examples?
(ii)
What are the generalised co-ordinates? Write the expression for
generalised force?
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
8
…
Lagrangian Mechanics
D’ALEBERT’S PRINCIPLE
1.4
This method is based on the principle of virtual work. The system is subjected to an
infinitesimal displacement consistent with the forces and constraints imposed on the
system at a given time t. This change in the configuration of the system is not associated
with a change in time i.e., there is no actual displacement during which forces and
constraints may change and hence the displacement is termed virtual displacement.
From the principle of virtual work

N



Fi a . ri
=0
(1.11)
i

Here Fia represent the applied force and ri denote the virtual displacement.
To interpret the equilibrium of the systems, D’Alembert adopted an idea of reverse force.
He conceived that a system will remain in equilibrium under the action of a force equal to
.

the actual force Fi plus reversed effective force pi.Thus
or,
.

Fi
+ (- pi) = 0

Fi
– pi = 0
(1.12)

.

Thus the principle of virtual work takes the form,
.
  
(Fi - pi) .ri = 0
i



Again writing Fi = Fia + fi
.  
 
a
(Fi - pi).ri + fi.ri = 0
i
Dealing with the systems for which the virtual work of the forces of constraints is zero,


we write

.

 
a
(F i - pi).ri =
0
i
Since force of constraints are no more in picture, it is better to drop the superscript ‘a’.
.
Thus
  
(Fi - pi) .ri = 0
(1.13)
i
The equation (1.13) is called D’Alembert
principle. To satisfy the above equation, we can


not equalate the coefficient of ri to zero since ri are not independent of each other and

9
Lagrangian and Hamiltonian Mechanics

hence it is necessary to transform ri in to generalised co-ordinates , qj which are
independent of each other .The coefficient of qj will then equated to zero.
DERIVATION OF LARANGE’S EQUATION
1.5
The Lagrange’s equations can be obtained from Hamilton’s variational principle, velocity
dependent potentials and also by Rayleigh’s dissipation function. In the present article we
shall discuss the derivation of Lagrange’s equations from velocity dependent potential
and by Rayleigh’s dissipation function.
1.5.1 Lagrange’s Equations from velocity dependent potential
The co-ordinate transformation equations are

ri

= ri ( q1,q2……,qn,t)
So that,


dri
dri dq1
=
dt
q1 dt
So that

vi



+
dri dq2
q2 dt
+………...+
dri
t
dt
dt

ri
=
j qj
. r
qj + i
t
(1.14)

Further infinitesimal displacement ri can be connected with qi



ri =  ri qj + ri t
j qj
t
But the last term is zero since in virtual displacement only co-ordinate displacement is
considered and not that of time. Therefore,


ri
ri = 
qj
j qj
Now we write equation (1.13) as,
.

 
r
(Fi - pi)  i qj = 0,
j qj
i



Fi
i,j

. ri qj
qj


Fi
-

i,j
.

pi

. ri qj
qj
(1.15)

. ri qj = Qj as the component of generalised force. So the above
qj
equation becomes

.

(1.16)
Qj qj pi . ri qj = 0
qj
i,j
j
We define


10
Lagrangian Mechanics
The evaluation of second term in equation (1.16) gives the expansion as

.

pi
i,j

. ri qj =
qj

j
d 
2
. (  (½) mivi )
dt qj i
 ( ( ½) m v 2 )
i i
qj
qj
(1.17)
With this substitution equation (1.16) becomes

Qj qj
-

j
j
d
dt
T.
qj
T
qj
qj = 0
Where (1/2) mivi2 = T, is written since it represents the total kinetic energy of the
system, further the above equation may be

j
d
dt
T.
qj
T - Q
j
qj
qj = 0
Since the constraints are holonomic, qj are independent of each other and hence to satisfy
above equation the coefficient of each qj should necessary vanish, i.e.
d
dt
T.
qj
T = Q
j
qj
(1.18)
As j ranges 1 to n, there will be ‘n’ such second order equations.
If potential are velocity dependent, called generalised potentials, then through the system
is not conservative, yet the above form Lagrange’s equations can be obtained provided
.
.
Qj, the components of the generalised force, are obtained from a function U(qj,qj) such
that
U + d U
Qj =
.
qj dt qj
Qj
Hence the from equation (1.18) and equation (1.19) ,we have
(1.19)
d  (T-U)
 (T-U)
=0
.
dt
qj
qj
If we take L = T-U, the Lagrangian function, where U is generalised potential, then above
equation becomes
(1.20)
d
L
L
=0
.
dt
qj
qj
Which are the Lagrangian equations for holonomic constraints systems.
11
Lagrangian and Hamiltonian Mechanics
1.5.2 Lagrange’s equations from Rayleigh’s dissipation function
It can be shown that if a system involves frictional forces or dissipative forces, then in
suitable circumstance, such
a system can also be described in terms of extended
Lagrangian formulation. Frictional forces are found to be proportional to the velocity of
the particle so that in cartesian co-ordinates components are,
.
(1.21)
Fjd = - kixj ,
Where kj are constants. Such frictional forces are defined in terms of a new quantity
called Rayleigh dissipation function given as,
.
 =(1/2)kix2j
Which yields

.
Fjd =  xj
(1.22)
Writing equation (1.18) in cartesian co-ordinates, assuming that this still holds for such a
system,
L.
qj
d
dt
L = Q
j
qj
Where L contains the potential of conservative forces as described earlier; Qj represents
the forces which do not arise from a potential, i.e.
Qjd = Fjd = - .
(1.23)
xj
Thus equation (1.18) can be written as,
d
dt
L.
xj
L

+
. = 0
xj
xj
The above equation may be expressed as in terms of generalised co-ordinates qj
d L.
L

+
. = 0
dt qj
qj
qj
(1.24)
Thus for such a system, to obtain equations of motion, two scalar L and  are to be
specified.
Check Your Progress 2
Note: a)
Write your answers in the space given below.
b)
Compare your answers with the ones given at the end of the units.
(i)
Write the principle and expression for the D’Alembert principle?
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
12
Lagrangian Mechanics
1.6
VARIATIONAL PRINCIPLE
t2
This principle state that the integral ( T-V )dt shall have a stationary value or extremum
t1
value, where T, kinetic energy of the mechanical system, is a function of co-ordinates and

their derivatives and V is the potential energy of the mechanical system, is a function of
co-ordinate only. Such a system for which V is purely a function of co-ordinates is called
conservative system.
Statement: The variational principle for the conservative system is stated as follows
“The motion of the system from time t1 to time t2 is such that the line integral
t2
t2
I = ( T-V )dt = L dt, is extemum for the path of motion” .Here L=T-V is
t1
t1
the Lagrangian function .


EULER –LAGRANGE EQUATION
1.7
The integral I, representing a path between the two points 1 and 2 will be written as
t2
.
.
I=
f [y1(x) y2(x),……. ……..y1(x)y2(x)……..….,x]dx
t1
(1.25)
Now to account for all possible curves between the two points1,2,we assign different

values of a parameter  to these curves, so that yj will also be a function of , i.e. curves
being represented by yj (x, ).The family of the curves may be represented as
y1(x,) = y1(x,0) + 1(x)
y2(x,) = y2(x,0) + 2(x)
…………………………...
Where 1 and 2 etc. are completely arbitrary functions of x,which vanishes at end points
and the curves y1(x,0), y2(x,0) etc. for =0 are paths for which the integral I is extemum
The integral I will be the function of  and hence its variation can be represented as
t2
 I()
 f yj
 f yj d dx
d = 
d + .
j
 ()
y

()
yj  ()
j
t1

Integrating by parts the second term of the integrand we get,
t2
t2
.
2
 I()
 f yj
 f yj
d  f  yj d dx
d = 
d dx +  .
d

.
j
 ()
j yj  ()
t1j yj  ()
t1 dx  yj   (1.26)
1


13
Lagrangian and Hamiltonian Mechanics
Since at end points, which are held fixed, all paths meet,so
equation (1.26) becomes
t2
 I()
 f yj
d = 
d dx
 ()
t1j yj  ()

t2
=
Let us put
t
1

j
f
yj
t2
t

1
d f
.
dx  yj
 yj
I
d = yj
d = I &


So that
t2
d f
f
.

I=
dx  yj
yj
j
t1

For the integral to be extremum
t2
d f
f
.

I=
dx

y
j
y
j
t1 j

j
yj

2
= 0 . Therefore
1
d  f  yj d dx
.
dx  yj  
 yj d
dx

yj dx
yj dx =0
Since yj are independent of each other, coefficient of yj should separately vanish if
above equation is to be satisfied. Thus.
f
d f
.
= 0, j=1,2,3,…n
yj
(1.27)
dx  yj
The set of differential equations represented by equation (1.27)are known as EulerLagrange differential equations. Thus solutions of Euler-Lagrange equation represent
those curves for which the integral I=

2
f (yj, yj, x)dx assumes an extremum value.
1
Check Your Progress 3
Note: a)
Write your answers in the space given below.
b)
Compare your answers with the ones given at the end of the units.
(i)
Write the statement of the Variational principle?
(ii)
Write the Lagrange’s equations from Rayleigh’s dissipation function
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
…
14
Lagrangian Mechanics
1.8
DERIVATION OF LAGRANGE’S EQUATION FROM HAMILTON’S
PRINCIPLE
According to Hamiltonian’s variational principle, motion of a conservative system from
time t1 to time t2 is such that the variation of the line integral
t2
.
, is zero
I=
L [qj(t), qj(t), t]dt
i.e.

I =  
t1
t2

(1.28)
L [qj(t), qj(t), t]dt =0
t1
Now we shall show that Lagrange’s equations of motion follow directly from Hamilton’s
principle. If we account for all possible paths of motion of the system in configuration
space and label each with a value of a parameter ,then since paths are being represented
by qj(t,),I also becomes a function of  so that we can writ,
I () =
So that,
I()
=
 ()

t2
(1.29)
.
L [qj(t, ), qj(t, ), t]dt
t1

.
L qj
L qj
L t

+
.
+ t  dt
j
q

q

j
j
t1
t2
Since in  variation, there is no time variation along any path and also at end points and
hence (I/) is zero along all paths. Therefore, on multiplying by d, above equation is
. .
t2
t2 L qj
I()
L qj d dt
d dt +
.
d = 

(1.30)
j q 
j
 ()
qj 
j

t1

t1
Integrating second term of L.H.S. by parts
 j
t2
=
t1
L qj
d dt +
qj 
j
L qj
d
.
qj 
t1
 j
t2
t2
t1
d L
.
dt qj
qj d dt

The middle term is zero since  variation involves fixed end points.
t2
t2
d  L  qj d dt
So,  I() d =   L qj d dt

.
 ()
q

()
j
j
j
t1
t1 dt  qj  t
t2
d  L  qj
L
.
(1.31)

=
qj
dt  qj  t dt
j
t1
Since qj are independent of each other, the variations qj will be independent. Hence
 I()=0 if and only if the coefficients of qj separately vanish, i.e.
d L
L
(1.32)
.
=
0
qj
dt  qj
Which are Lagrange equations of motions for a conservative system. It is obvious that
these equations follow directly from Hamilton’s principle.



15
Lagrangian and Hamiltonian Mechanics
Check Your Progress 4
Note: a)
Write your answers in the space given below.
b)
Compare your answers with the ones given at the end of the units.
(i)
Write the Euler-Lagrange differential equation? For what its solutions
represents for?
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
………………………………………………………………………………………….
1.9
LET US SUM UP
After going through this unit, you would have achieved the objectives stated earlier in the
unit. Let us recall what we have discussed so far.

Constraints are the geometrical or kinematical restrictions on the motion of the
particle or system of the particles. Systems with such constraints of motion are
called as constrained systems and their motion is known as constrained or
restricted motion.

Generalised co-ordinates:- These are the coordinates which are used to eliminate
the dependent coordinates and can be expressed in another way by the
introduction of (3N-p) independent coordinates of variables called the
Generalised coordinates, where N represent the number of particles of a system
and p represent the holonomic constraints.

Advantages of Generalised co-ordinates - The main advantage in the
formulating laws of mechanics in terms of generalised co-ordinates and the
associated mechanical quantities is that the equation of motion looks simpler and
can be solved independently of each other since generalised co-ordinates are all
independent and constraints have no effect on them. The equations of motion are
then called Lagrange’s equation of motion.
16

Lagrangian Mechanics
D’Alembert principle- The equation below is called D’Alembert principle
equation .
 
(Fi - pi) .ri = 0
i


To satisfy the above equation, we can not equalate the coefficient of ri to zero

since ri are not independent of each other .

The variational principle for the conservative system is stated as follows
“The motion of the system from time t1 to time t2 is such that the line integral
t2
t2
I = ( T-V )dt = L dt, is extemum for the path of motion” .Here L=T-V is
t1
t1
the Lagrangian function .


1.10
CHECK YOUR PROGRESS: THE KEY
1. ( i) Constraints are the geometrical or kinematical restrictions on the motion of
the particle or system of the particles. The constraints can be classified in to
following categories: (i) Holonomic and non-holomonic constraints (ii)
Scleronomic and rhenomic constraints.
(ii) A st of generalised co-ordinates is any set of co-ordinates which describe the
configuration. The expression for the generalised force.

N
Qj
=

Fi

.
j=1
ri
qj
Here we note that Qj depends on the force acting on the particles and on the coordinate qj and possibly on time t. Therefore, Qj is called the generalised force.
2
(i)
This method is based on the principle of virtual work.
D’Alembert principle- The equation below is called D’Alembert principle
equation
.



(Fi - pi) .ri = 0
i
3
(i)
The variational principle for the conservative system is stated as follows
“The motion of the system from time t1 to time t2 is such that the line
integral
t2
t2
I = ( T-V )dt = L dt, is extemum for the path of motion”
t1
t1
Here L=T-V is the Lagrangian function.


17
Lagrangian and Hamiltonian Mechanics
(ii)
The Lagrange’s equations from Rayleigh’s dissipation function equation
may be expressed as in terms of generalised co-ordinates qj
d L.
L

. = 0
+
dt qj
qj
qj
Thus for such a system, to obtain equations of motion, two scalar L and
 are to be specified.
4. (i)
Thus the Euler-Lagrange differential equations are
f
yj
d f
.
dx  yj
= 0, j=1,2,3,…n
The set of differential equations represented by equation (1.27)are known
as Euler-Lagrange differential equations. Thus solutions of EulerLagrange equation represent those curves for which the integral
I=

2
f (yj, yj, x)dx, assumes an extremum value.
1
18
UNIT-II
KEPLER’S PROBLEM AND HAMILTONIAN MECHANICS
Structure
2.0
Introduction
2.1
Objectives
2.2
Two body central force problems
2.3
Inverse square law of force: Kepler Problem
2.4
2.3.1
Case of elliptic orbits
2.3.2
Kepler’s law
Scattering in a central force field
2.4.1
Angle of scattering
2.4.2
Differential Cross section
2.4.3
Nucleon Scattering
2.5
Derivation of Hamilton’s equation from variational principle of least action
2.6
Principle of Least action
2.7
Equations of canonical transformation
2.8
2.7.1
Definition of transformation
2.7.2
Canonical Transformation
2.7.3
Generating function for the Canonical Transformation
2.7.4
Advantages of the canonical transformation
2.7.5
Conditions of a transformation to be canonical
Lagrangian and Poisson Brackets
2.8.1
Poisson’s Bracket
2.8.2
Poisson Bracket in Quantum Mechanics
2.8.3
Lagrange’s Bracket
2.8.4
Relation between Poisson and Lagrange’s Brackets.
2.9
Angular momentum and Poisson bracket relation
2.10
Equation of Motion in Poisson Bracket notation
2.11
Let Us Sum Up
2.12
Check Your Progress :The Key
19
Lagrangian and Hamiltonian Mechanics
2.0
INTRODUCTION
In this unit we shall study the problem of two bodies moving under the influence of a
mutual central force as an application of the Lagrangian formulation. The problem of
finding the motion of a particle under a central force is one of the most important
problems in Physics because it closely related to the mechanics of nature e.g., motion of
the planets with respect to sun, that of satellite about earth and the motion of two charged
particles with respect to each other. Central force is that force which is always directed
away or towards a fixed centre and magnitude of which is a function only of the distance
from that fixed centre. The force between two interacting particles is primarily a central
force. In this unit we shall derive the equation of motion by using the concepts of central
force, inverse square law and Hamilton’s, Lagrangian formulation with the Poisson
Brackets notations.
2.1
OBJECTIVES
The main objective of this Unit is the study of two body problems .After going through
this Unit you should able to

Solve two body problems in one body problem.

Know about the inverse square law applicable in two body problem.

Solve Scattering problems of particle in the central force field.

Derive the Hamilton’s Equation from variation principle and derive the canonical
equations.

Lagrangian and Poisson’s Bracket with their relations with angular momentum
and Equation of Motion.
2.2
TWO BODY PROBLEM EQUIVALENT TO ONE BODY PROBLEM
Let us consider a conservative system of two mass points m1 and m2 .Such a system has
six degrees of freedom because the motion of the system consisting of m1 and m2 can be
thought as made up of motion of centre of mass and motion about the centre of mass. Let

us choose them to be the three components of R, the position vector of the centre of mass
and three components of difference vector.


r = (
r1-r
2 )
20
Kepler’s problem and Hamiltonian Mechanics
The Lagrangian for such a system can be written as
. .
L = T(R,r ) -V(r)
. .
.
Where T (R,r) = ½ (m1+m2)R2 +T,
(2.1)
Which expresses the kinetic energy as the sum of kinetic
m1

energy of motion of the centre of mass, plus the kinetic
of motion about the centre of mass T.It is assumed that

r1

V is purely function of r, the separation between two
T = ½
R

r2
Particles. T is given by
.

(m1r1 2 )+ (½


r1
1C.M.

r2
1m2
Fig.2.1
. .
)m2
r2. 2

.


[Relation among
the various co-ordinate
Where 
r1 and 
r2 are the position vectors with respect to the centre of mass and
difference vector is





r = r1 –r2 = r1 –r2
(2.2)
We know that a body balances about the centre of mass and hence mi ri =0.Therefore


m1r1 + m2r2 =0 or,

r1
=-
m2 
r2
m1
(2.3)
Putting equation (3) in equation (2), we get



r = r1 + (m1/m2) r1 = [(m1+m2)/m2] 
r1
Or,

r1=

[ m2/(m1+m2)] r
also, r2 = - [m1/ (m1+m2)] 
r
Kinetic energy is
.

.
(2.4)
.
2
T =[ ½ (m1+m2)R2 + ½ (m1r12 ) + ½ (m
2r2 )]
Putting the values from equation (2.4) in equation (2.5) we obtain
.

(2.5)
.
T = ½ MR2 + ½ r2
(2.6)
m1m2
Where M =m1+m2 is total mass of the system, and  = (m +m )
1
2
.

.

The Lagrangian can thus be expressed as, L= ½ MR2 + ½  r2 – V(r)
(2.7)

From which we note that the three components of R do not appear in the Lagrangian L
and consequently are cyclic. In this case none of the equation of motion 
r will contain
21
Lagrangian and Hamiltonian Mechanics
terms involving R and R. Consequently ,we can ignore the first term in equation (2.7) and
.

we write, L= ½  r2 – V(r)
(2.8)
which is effective in describing the motion of the component of r .But the Lagrangian
given by equation (2.8) is the same as would be in the case of single particle of mass,
moving at a distance r from a fixed centre of force which gives rise tom the potential
energy V(r) and this type of two body problem can always be reduced to the equivalent
on body problem.
INVERSE SQUARE FORCE: KEPLER’S PROBLEM
2.3
The important central force is that which varies inversely as the square of the distance
,i.e.
f ( r) = -
k
r2
Where k is a constant .The corresponding potential energy will be
V( r) = -
k
r
Equation of the orbit is
d2u
+u=d 2
m
lu
2 2
f
(2.9)
1
u
Putting f (r) = f (1/u) = -( k /r2 ) = - ku2, we get
d2u
mk
+u= 2
d 2
l
Let us put y = u - m2 k
l
Then equation becomes,
2
2
so that d u2 = d y2
d
d
(2.10)
2
dy
+y =0
d 2
The general solution of this equation is
y = u cos ( - )
or, u =
mk
+ u cos ( - )
l2
1 mk
u = r = 2 + u cos ( - )
l
22
Kepler’s problem and Hamiltonian Mechanics
Where u and  are constants. For simplicity, if we consider the co-ordinate system so
1 mk
that  =0, then u = = 2 + u cos ( )
r
l
Or,
r=
l2 / mk
1+ (ul2/mk)cos
(2.11)
 which we have made zero can be recognized as one of the angles at which the orbit
turns because = or  = + results in a maximum and minimum value for ‘r’. Thus 
can be called a turning point. The equation (2.11) represents a conic section with the
centre of force at one of the foci. We define the conic section as a curve for which the
distance .i.e.
r
= constant = 
d
d
r

Where  is the eccentricity.
focus
From figure 2.2, we see that
P = d+ r cos  = (r/+ + r cos )
Let P =  p, so that
P
r
+ r cos 
=
Or,


p
directrix
Fig.2.2 A conic section
p
r=
1+  cos 
(2.12)
which represents a conic section .Equation (2) and equation (3), the orbit under an inverse
square force is always a conic section. To determine the shape of the orbit can be
obtained by the resulting conic equation (2.12). The total energy of the system is
.
E = ½ mr2 + (l2/2mr2) + V(r)
Here V ( r ) = - (k/r)
So that E = ½ mr. 2 + (l2/2mr2) – (k/r)
(2.13)
Now because E is constant, we can evaluate it at any convient point on the orbit. Let we
.
take the turning point at which r is minimum, say rmin .Thus at the turning point r will be
zero. Equation (2.13) then becomes
.
E = 0 + (l2/2mrmin2) – (k/rmin)
From equation (2.12),
23
Lagrangian and Hamiltonian Mechanics
rmin = [( p /1+  )]
Putting the value of p = (l2 /mk), then above equation becomes, rmin =[ ( l2 /mk) / (1+ )]
Therefore,
E = (l2/2mrmin2) – (k/rmin)
2
Which on substituting for rmin , becomes E = mk2 (2-1)
2l
giving,
2
 = 1+ 2El 2
mk
(2.14)
1/ 2
(2.15)
It is obvious that  is now known because constants E, l, m and k are known .Now putting
the value of  and p = (l2/mk) in equation (2.12), we get
1
r=
1/ 2
2
2El
cos 
1+ 1+
mk2
(2.16)
If
E >0 giving  >1 –conic is hyperbola non periodic motion
E=0 giving  =1- conic is parabola
E <1 giving  <1 – conic is ellipse
Periodic motion
2.3.1 Case of elliptic orbits
(a) Relation between energy and semi major axis of ellipse
We have shown that in the case of elliptic orbits, semi major axis is given by
p
(1- 2 )
a=
Putting, p= ( l2 / mk), then a =
Or,
( l2 / mk),
( 1- 2 )
(2.17)
l2
(1- 2) =
mka
Substituting the value of (1-2 )in equation (2.14), we get
2
E = - mk2
2l
l2 = - (k/2a)
mka
(2.18)
Which show that all ellipses with the same major axis have the same energy.
24
Kepler’s problem and Hamiltonian Mechanics
(b) Period of elliptic motion 
It is defined as the ratio of the total area of the ellipse to the rate at which the area is
swept out. Suppose dA is the area swept out by radius vector in time dt, then the rate of
sweeping the area will be dA/dt, hence
Area
=
dA/dt
Area of the ellipse is ab, and
dA r d
r
dt = 2
dt
.
= ½ r2 = l/2m
Thus,
2mab
 = ab = l
(l/2m)
2
Putting b = a ((1- ) then the above equation becomes
2ma2 ((1-2 )
ab
=
= l
(l/2m)
2 2 4
2 = 4 m2 a (1-2)
l
Using equation (2.17),
2
2 = 4 m
k
(a3)
Or,
2  (a3)
(2.19)
Equation (2.19) is known as Kepler’s third law which states that the square of the period
of elliptical motion is proportional to the cube of the semi major axis and is independent
of the minor axis.
2.3.2 Kepler’s law
In seventeenth century, Kepler announced the following three laws
1. The planet move in elliptical orbits with sun as one of foci.
2. Area swept out by the radius vector from the sun to a planet in equal times are
equal.
3. The square of the period of revolution is proportional to the cube of the semi
major axis.
25
Lagrangian and Hamiltonian Mechanics
2.4 SCATTERING IN A CENTRAL FORCE FIELD
The knowledge of orbits, i.e. the path adopted by a particle during its motion under the
action of a central force, provides us the information about the dependence of potential
energy on r, the radial distance, and consequently gives the information about a force
law.
The total energy in the orbit is
.
E = ½ mr2 + (l2/2mr2) + V(r)
.2
- mr
2
l2
2mr2
Putting u = (1/r), equation becomes
So that V(r) = E -
1
V
u
-
=E -
l2 2
u
2m
l2
2m
du
d
2
(2.20)
Which is the equation of orbit, for it gives u=u().It is thus obvious that from the
knowledge of the orbit u=u() , we can find the dependence of potential energy on r, the
radial distance. But while dealing with atomic particles, it is not always possible to
observe the orbits and consequently, the method of finding the dependence of potential
energy on r with the help of orbit observation fails. For such cases another method, called
the method of scattering, is employed to find this dependence. In this method of
scattering, a particle is fired at another particle (assumed to be fixed) and the deflection
first particle is measured. This particle will follow an unbound orbit; since initially being
at infinite distance from the centre of force it will start journey at a straight line trajectory
as it approaches centre of force.
At some later instant it attains the closest
Orbit
approach to the centre of force and then

subsequent motion carries the particle
away form the centre of force and finally
it picks up straight line trajectory again.
In general, final direction of motion is
not the same as the incident direction
and the particle is said to be scattered.
s
Centre of
force
0
0

Particle
Fig.2.3 Hyperbolic orbit in
repulsive inverse square field
26
Kepler’s problem and Hamiltonian Mechanics
The angle between these two directions is called the angle of scattering denoted by. It is
shown in figure, which is drawn in case of repelling centre of force. It is obvious that
 =  - 20
(2.21)
where 0 is the angle which initial and final radius vectors from the centre of force make
with the symmetry axis.
2.4.1 Angle of Scattering
Let the particle have an initial velocity v0 and let it be traveling in such a direction that it
undeflected; it would pass at a distance s from the centre of force. This distance defined
as the perpendicular distance between the centre of force and the incident velocity is
called impact parameter. Here we can compute the energy and angular momentum in
terms of speed and impact parameter.
E = ½ mv02 and l =mv0s
From these two relations we can obtain
l = s  (2mE)
From the fig.2.3, tan ( / 2) = cot 0
But cot 0 = (2 -1)- 1/2 = (mk2 / 2E l2)1/2 , putting the value of  from previous relations.
Putting the value of l,
tan( / 2) = ( k2/ 4s2E2)1/ 2 = (k / 2sE)
(2.22)
Thus once E and s are fixed, the angle of scattering  is then determined uniquely.
2.4.2 Differential Scattering Cross section
When we consider a uniform beam of particles all of the same mass and energy, incident
up on the centre of force. The incident beam is charactrised by specifying its incident I,
defined as the number of particles crossing unit area normal to the beam in unit time. We
consider the distribution of scattered particles per unit solid angle per unit time.such an
information is contained with in the quantity  (,) called differential scattering cross
section and is defined as,
Number of particles scattered per unit solid angle per unit in the direction
specified by spherical angle  and 
 (,) =
Incident intensity I
27
Lagrangian and Hamiltonian Mechanics
The expression for differential cross section is given by
s
ds
(2.23)
sin  d
The minus sign is introduced here because  usually decreases as s increases or in other
 ( ) = -
words the larger the impact parameter, smaller is the angle through which the particles
will be scattered. The differential cross –section is a measure of the probability that a
particle will be scattered through a solid angle d along direction (, ).It can be stated as
an effective area posed by the scatterer to the incident particles.
2.4.3 Nucleon scattering
Rutherford scattering is the specific example of scattering of charged particles by
coulomb field. We can take a simple case of proton –proton scattering .Suppose charge at
the fixed centre of force is Ze which repeal the incident particles having a charge Ze. The
force between them will be
ZZe2
r2
Which is repulsive square law of force. Comparing with f = -(k/r2 ), we write the value of
f =
k =- ZZe2
(a) Shape of the orbit: - From equation (2.16) r=
1
2
1+ 1+ 2El 2
mk
1 mk
= 2
r
l
1/ 2
cos 
2El2
1+ 1+
mk2
1/ 2
cos 
Which represents a conic section for f = - k/r2 .This equation takes the form for this case,
after putting the value of k, as
1
ZZe2 m
=
r
l2
1/ 2
2El2
cos 
1+ 1+
(2.24)
m ( ZZe2
)2
since eccentricity  in equation (2.24) is greater than one, conic section is a hyperbola .
28
Kepler’s problem and Hamiltonian Mechanics
as equation (2.24) is for the conic, i.e. for the conic, i.e. hyperbola, involves a negative
sign, values of  will be restricted to angles such that
cos  < -(1/)
(b) Cross section :- To calculate scattering cross section, from equation (2.22) we have
tan 2 ( / 2) = (k / 2sE)2 = [ (ZZe2 )2 / 4s2E2 ]
or,
ZZe2
cot (/2)
2E
From which
ds
ZZe2 1

Cosec2
=
d
2E
2
2
s=
(2.25)
(2.26)
Putting this value in (2.23)
 () =
( ZZe2 / 2E).cot  / 2 x
sin
ZZe2
Cosec2 
4E
2
(2.27)
ZZe2 2 1
2E
sin4/2
This result gives the famous Rutherford scattering cross section. This also agrees with the
1
 () = 4
quantum mechanical result with in non-relativistic limit.
Check Your Progress 1
Note: a)
Write your answers in the space given below.
b)
Compare your answers with the ones given at the end of the units.
(i)
Write the equation for the shape of a conic section and also write the
condition of orbit shape?.
(ii)
Define the angle of scattering and differential cross section?
(iii) Find the expression of nucleon scattering?
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
………………………………………………………………………………………….
29
Lagrangian and Hamiltonian Mechanics
DERIVATION OF HAMILTON’S EQUATION FROM VARIATION
PRINCIPLE
2. 5
Lagrange’s equations have been shown to be the consequence of a variational principle,
namely, the Hamilton’s principle. Indeed the variational method has often proved to be
the preferable method of deriving equations, for it is applicable to types of systems not
usually comprised with in the scope of mechanics. It would be similarly advantageous if
a variational principle could be found that leads directly to the Hamilton’s equation of
motion.
Hamilton’s principle is stated as
t2
L dt = 0
t1

I= 
(2.28)
Expressing L in terms of Hamiltonian by the expression by the expression
.
H=  piqi – L,
i
We find,
t
I=
t

t2
t2
1
.
 pi dqi
i
dt
t
t2
 pi dqi -  H (qi, pi, t)dt =0
i
1
- H (qi, pi, t) dt
(2.29)
1
Equation (2.29) is some times is referred as the modified Hamilton’s principle. Although
it will be used most frequently in connection with transformation theory ,the main interest
is to show that the principle leads to the Hamilton’s canonical equations of motions.
The modified Hamilton’s principle is exactly of the form of the variational problems in a
space of 2n dimensions as
t
t2
(2.30)
.
I =  f (q, q, p, p, t) dt =0
.
1
For which the 2n Euler-Lagrange equations are
d
dt
f
q. j
f
qj
d
dt
f
p. j
f
pj
(2.31)
J=1,2,3….n
(2.32)
J=1,2,3….n
30
Kepler’s problem and Hamiltonian Mechanics
.
.
The integrand f as given as (2.29) contains qj only through the piqi term, qj only in H.
Hence equation (2.30) leads to
H
.
pj +
=0
qj
(2.33)
On the other hand there is no explicit dependence of the integrand in equation (2.30) on
.
pj. Equation (2.29) therefore reduce simply to
H
.
qj =0
pj
(2.34)
Equation (2.33) and (2.34) are exactly Hamilton’s equations of motion .The Euler –
Lagrange equations of the modified Hamilton’s principle are thus the desired canonical
equations of motion .From the above derivation of Hamilton’s equations we can consider
that Hamiltonian and Lagrangian formulation and therefore their respective variational
principles, have the same physical content.
2.6
PRINCIPLE OF LEAST ACTION
The important variational principle associated with Hamiltonian formulation is the
principle of least action. The principle of least action for the conservative system is
expressed as,

t
t2
.
 pi qi dt = 0
(2.35)
i
1
Where  is the variation
Features of the  variation: In  variation process, we shall restrict the comparison to
all paths involving no violation of the conservation of energy but relax the condition that
all paths take the same length of time. In brief this variation can be described as

Ends point’s time may be different for every path, i.e. time of travel along
different paths, may be different, and will, in fact, happen if all the paths are real.

End point’s position co-ordinates are held fixed, which is also possible in real
paths.

H is conserved along every path.
31
Lagrangian and Hamiltonian Mechanics
Other forms of least action principle
(A) If the co-ordinate transformation equations do not involve time explicity, then
.
 pi qi =2T ,so that the principle of least action assumes the form
i
t2

t 2 T dt = 0
t2
Or, 
1
t T dt = 0
(2.36)
1
Further, if the system is not involving any external force, T is conserved, giving
t2

t dt = 0
or  ( t2 – t1) =0
(2.37)
1
This condition leads to a very important principle called Fermat’s principle in geometric
optics. This principle predicts that out of all possible paths, consistent with the
conservation energy, the system moves along that particular path for which the transit
time is the least or more strictly as extremum.
(B)
When the transformation equations do not involve time, kinetic energy cab always
be expressed as a homogeneous quadratic function of the velocities then the principle of
least action can be expressed as,

  2[H-V(q)] d = 0
(2.38)
(C ) If the system consists of only one particle then the principle of least action becomes

  2[H-V ] ds = 0
(2.39)
This expression is quite similar to equation (2.38).The principle of least action is here
expressed in terms of the arc length of the particle trajectory.
Check Your Progress 2
Note: a)
Write your answers in the space given below.
b)
Compare your answers with the ones given at the end of the units.
(i)
What
is
the
modified
Hamilton’s
principle?
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………32
…………………………………………………………………………………………
………………………………………………………………………………………….
Kepler’s problem and Hamiltonian Mechanics
2.7
EQUATIONS OF CANONICAL TRANSFORMATION
2.7.1 Definition of Transformation
A given system can be described by more than one set of generalised co-ordinates. We
can choose a set which is more convenient for the solution of the problem under
consideration. For example, to discuss the motion of a particle in a plane ,we may use as
generalised, the cartesian co-ordinates
q1 = x, q2 = y,
or, in plane polar coordinates q1 = r, q2 = ,
Thus here we want to discuss a specific procedure for transforming one set of variable in
to some other which may be more convenient .If a problem has been formulated in the
form of Hamilton’s canonical equations, the canonical transformation can be aimed to
put these equations in to more easily soluble form, i.e. to make integration of the equation
of motion more simpler. Suppose we transform from cartesian to plane polar co-ordinate,
then transformation equations are
r = (x2 + y2 ) = r (x, y) and  = tan-1 (y/x) =(x, y)
This is an example of co-ordinate transformation
2.7.2 Canonical Transformation
The canonical transformation is the transformation of phase space. They are charactrised
by the property that they leave the form of Hamilton’s equations of motion invariant. We
know that while deducing Lagrangian equations, no stress was given to any particular
choice of co-ordinates system .In fact, Lagrangian equations of motion are invariant in
form with respect to the choice of the set of any generalised co-ordinates. Therefore, in
new set Qi, Lagrange’s equations will be
d L.
dt Qj
L
=0
Qj
i.e. Lagrange’s equations are covariant with respect to point transformation and iif we
.
L
define Pi as, Pi = . ( Qi, Qi)
Qi
33
Lagrangian and Hamiltonian Mechanics
The Hamilton’s canonical equations will also be covariant, i.e.
.
L
Qi =
( Qi, Pi)
Pi
.
L
Pi =
(Q, P)
Qi i i
(2.40)
(2.41)
Therefore this transformation is extended to Hamiltonian formulation. The simultaneous
transformation of the independent co-ordinates and mementa qi and pi to a new set Qi,Pi
can be represented in the form
Qi = Qi (q, p, t)
Pi = Pi (q, p, t)
For Qi and Pi the new set of coordinates to be canonical, as it demanded in Hamiltonian
formulation, they should be able to be expressed in Hamiltonian form of equations of
motion i.e.
Qi =
.
Pi =
K
 Pi
(2,42)
K
 Qi
(2.43)
Where k is a function of (Q,P,t) i.e. K(Q,P,t) and is substitute for the Hamiltonian H of
old set of co-ordinates. Moreover, if Qi, Pi are to be canonical co-ordinates, they must
also satisfy the modified Hamilton’s principle of the form
t2
 [  PiQi – k (Q, P,t)]dt = 0
i
t1
The old co-ordiantes pi, qi are already canonical; therefore


t2
[  pi qi – k (q, p, t)]dt = 0
i
t1

(2.44)
(2.44)
The simultaneous validity of above relations (2.44) and (2.45), as their right hand side is
zero, does not mean that the integrands of the two integrals are equal. We can, therefore,
write

t2
t
.
( pi qi – H (q, p, t) - ( Pi Qi – K ) dt = 0
i
i
(2.45)
1
In which the integrand to a certain extent in unknown
34
Kepler’s problem and Hamiltonian Mechanics
Equation (2.45) will not be affected if we add to or subtract from it a total time derivative
of a function F =F(q, p ,t) because
t2
t2
 dF dt =  [F (q, p, t)] = 0
t1
t1 dt
Since at end points the variation in qi and pi vanishes. Therefore, we can write equation

(2.45)

t2
t
.
( pi qi – H (q, p, t) - ( Pi Qi – K )
i
i
dF
dt
dt = 0
1
Thus, it follows that
.
dF
( pi qi – H (q, p, t)) - ( Pi Qi – K ) =
dt
i
i
(2.46)
2.7.3 Generating function of canonical transformation
The first bracket of equation (2.46) is regarded as a function of qi, pi and t ,the second as
a function of Qi, Pi and t .F is thus, in general a function of (4n+1) variables qi, pi, Qi ,Pi
and t. Now F is a function of both old and new co-ordinates and therefore out of 2n
variables, n should be taken from new and n from old set, i.e. one variable should be out
of pi and qi and other should be from Qi and Pi . Thus following four forms of function F
are possible
F1(q, Q, t), F2(q, p, t), F3(p, Q, t) and F4(p, P,t)
As f is a function of new and old co-ordinates, it can affect the transformation from old
set to new set, i.e. transformation relations can be derived by the knowledge of the
function F. It is thus termed as the generating function. Out of these above four forms
choice of one particular will depend upon the problem.
2.7.4 Advantage of Canonical Transformation
The way of canonical transformation is to obtain solution of a mechanical problem is to
transform old set of co-ordinates in to new set of co-ordinates that are all cyclic. In this
way the new equations of motion can be integrated much easily to give a solution.
Transformation equations in this case can be written as
Qi = Qi(q,p,t)
Pi = Pi (q,p,t)
35
Lagrangian and Hamiltonian Mechanics
Where p, q, t belong to old set and Pi, Qi, t to new set of co-ordinates.
2.7.5 Condition for a transformation to be canonical
We shall arrive at the following conditions for a transformation to be canonical
(A)
An exact different condition
If the expression
 (Pi dQi – pi dqi)
or  (pi dqi - Pi dQi )
(2.47)
be an exact differential then transformation from (qi, pi) set to (Qi, Pi) set is canonical.
(B)
Bilinear Invariant Condition
The condition is that “if a transformation from (qi, pi) set to (Qi, Pi) set is canonical then
the bilinear form
 ( pi dqi - qi dpi ) remain invariant”.
This statement means
 ( pi dqi - qi dpi ) =  ( Pi dQi - Qi dPi )
(2.48)
Check Your Progress 3
Note: a)
Write your answers in the space given below.
b)
Compare your answers with the ones given at the end of the units.
(i)
Write the Hamilton’s canonical equations?
(ii)
What you mean by the principle of least action ?Write the important
forms of the principle of least action.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
………………………………………………………………………………………….
36
Kepler’s problem and Hamiltonian Mechanics
2.8
POISSON AND LAGRANGE’S BRACKETS
2.8.1
Poisson Brackets: Definition
Let F be any dynamical variable of a system. Suppose F is the function of conjugate
variables qi and pi and t ; then
dF dF
F .
F .
F
qi + 
pi +
(q
=
i, pi, t) = 
dt
dt
t
i qi
i pi
=  F H - F H + F
i qi pi
pi qi
t
On using Hamiltonian’s canonical equations of motion. The first bracketed term is called
Poisson Bracket of F and H. In general, if X and Y are two dynamical variables then their
Poisson bracket is defined as,
[ X,Y]q , p =  F H - F H
i qi pi
pi qi
(2.49)
From which it is quite easy to see that
[X,Y]=-[ Y,X]
(2.50)
[X,X] =0
[X,Y+Z] =[X,Y]+[X,Z]
[X,YZ]=Y[X,Z]+[X,Y] Z
Also,
[qi,qj ]q , p = 0 = [pi,pj ]q , p
[qi,pj ]q , p = ij = 0 ,if i  j
(2.51)
=1 , if i = j
The quantities in equation (2.51) are known as the fundamental or basic Poisson bracket.
Poisson bracket are invariant under canonical transformation. We can express this
mathematically as
[X,Y]q, p = [X,Y]Q, P
(2.52)
37
Lagrangian and Hamiltonian Mechanics
2.8.2 Poisson Bracket in Quantum Mechanics
In quantum mechanics, dynamical variables are described by operators which do not
obey the computational rules of ordinary algebra. These operators are represented by
matrices. Quantum Poisson brackets of two operators X and Y is defined as
[X,Y]  -
i
[XY-YX]
h
(2.53)
If the matrices X,Y commute,[X,Y] will be zero; this means that operators X and Y then
behave like dynamical variables of classical mechanics .We can also interpret that if the
Poisson bracket of two variables in classical mechanics is zero, the operators which
represent these variables in quantum theory should commute. According to quantum
mechanics, it is physically interpreted as that the variables may be simultaneously
observed. Moreover, invariance of any classical form of Poisson brackets in quantum
mechanics, when definition (2.53) is adopted, also leads to the inclusion of classical
mechanics as a valid approximation for all but not in atomic phenomenon.
2.8.3 Lagrange’s Brackets
Lagrange’s bracket of (u, v) with respect to the basis (qi ,pi) is defined as
( u,v)q , p =  qi pi - pi qi
i u v
u v
(2.54)
For Lagrange’s bracket we note that
(a)
Lagrange Bracket is invariant under canonical transformation and it is worthless
to designate any basis to the bracket i.e. we should, henceforth, drop the
subscripts q, p or Q,P.
(b)
( c)
i.e. {u, v}q, p = {u, v}Q,P
Lagrange brackets do not obey the commutative law.
i.e.
{u, v} = - { v ,u }
For the Lagrange’s brackets
{qi, qj} =0 and {pi, pj} =0
{qi, pj} =ij
38
Kepler’s problem and Hamiltonian Mechanics
2.8.4 Relation between Lagrange and Poisson Brackets
The relation between Lagrange and Poisson bracket show that
(2.55)
2n
 {ul, ui}[ui, uj] = ij
i
Where {ul, ui} is Lagrange bracket and [ui, uj] is the Poisson bracket
From the definition of the Lagrange’s bracket and Poisson brackets, we at once arrive at
2n
2n
i
l =1
n
 {ul, ui}[ui, uj] = 

K =1
qk pk
p q
- k k
ul uj
ul uj
n

m =1
ul uj
uj ul
qm pm qm pm
(2.56)
The first of the four terms on the right hand side that shall be obtained on multiplication
are
n

k,m=1
n
=
k,m=1
pk uj
ui pm
n
qk ul
pk uj qk

=
ul qm
k,m=1 ui pm qm
2n

i=1
pk uj

ui pm k m
But  k m is also expanded as
 km =
pm
pk
So that
n

k,m=1
n
=
k,m=1
pk uj
ui pm
2n

i=1
n
qk ul
pk uj qk

=
ul qm
k,m=1 ui pm qm
n p
pk uj pm
k uj

=
ui pm pk
ui pk
k
(2.57)
The last of the four terms will be
n

k,m=1
n
=
k,m=1
qk uj
ui qm
2n

i=1
n
pk ul
qk uj pk

=
ul pm
k,m=1 ui qm pm
n q
pk uj qm
uj
k

=
ui pm qk
ui qk
k
(2.58)
The rest terms will be zero so the right hand side of equation (2.56) is simply becomes
with the help of equation (2.57) and equation (2.58),
39
Lagrangian and Hamiltonian Mechanics
pk ui
 u p
i
k
k
+
uj
(qk, pk ) =
ui

k
uj
ui
qk ui
ui qk
=
k
ui pk
pk ui
+
uj qk
qk ui
(2.59)
= ij
Therefore
2n
 {ul, ui}[ui, uj] = ij
(2.60)
i
Which sets a relation between Lagrange and Poisson brackets. This relation between the
two types of brackets holds even if the co-ordinates are not canonical and it is true for
any arbitrary transformation from qi, pi to qi, pi.
If we use the matrices concepts to find out the relation between the Lagrange and Poisson
brackets we find the relation with the sign changed i.e. if we denote L and P the matrices
of the Lagrange and Poisson brackets respectively, then we have
L P = -1
2.9
ANGULAR MOMENTUM AND POISSON BRACKET RELATION
The identification of the canonical angular momentum as the generator of a rigid rotation
of the system leads to a number of interesting and important Poisson bracket relations.
Equation for the change in the value of a function under infinitesimal canonical
transformation is given by the relation,
u =  [u, G]
Where  is some infinitesimal parameter of the transformation and G is any
(differentiable) function of its 2n+1 argument. Thus, if F is a vector function of the
system configuration, then
 Fi =[ Fi, G]
It is important to note that the direction along which the component is taken must be
fixed i.e., not affected by the canonical transformation .If the direction itself is
determined in terms of the system variables, then the transformation changes not only the
value of the function but the nature of the function, just as with the Hamiltonian. With
40
Kepler’s problem and Hamiltonian Mechanics
this understanding the change in a vector F under a rotation of a system about a fixed
axis n , generated by L. n, can be written in vector notation
F = d[F, L. n ]
(2.61)
For a system vector F, the change induced under and I.C.T. generated by L.n can
therefore be written as
F = d[F, L. n ]= n d x F
(2.62)
Equation (2.62) implies an important Poisson bracket identity obeyed by all system
vectors;
[F, L.n ] = nx F
(2.63)
Here it should be noted that in equation (2.63) there is no longer any reference to a
canonical transformation or even to a spatial rotation. It is simply a statement about the
value of certain Poisson brackets for a specific class of vectors and, as such, can be
verified by direct evaluation in any given case. Suppose, for example, we had s system of
an unconstrained particle and used the Cartesian co-ordinates as the canonical space coordinates. Then the Cartesian vector p is certainly a suitable system vector .If n is taken
as a unit vector in the z-direction, they by direct evaluation we have
[ px, x py – y px] = - py
[ py, x py – y px] = px
(2.64)
[ pz, x py – y px] = 0
The right –hand sides of these identities is clearly the same as the components of k x p. If
any two components of the angular momentum are constant, the total angular momentum
vector is conserved .As a further instance, let us assume that in addition to Lx and Ly
being conserved there is a Cartesian vector of canonical momentum p with pz a constant
of motion .Not only then is Lz conserved but we have two further constants of the motion,
[pz, Lx] = px
and
[pz,Ly] = -px
(2.65)
i.e., both L and P are conserved. We have here as instance in which Poisson’s theorem
does not yield new constants of the motion .Then their Poisson brackets are
41
Lagrangian and Hamiltonian Mechanics
[px,py] =0,
[px,Lz]= py
and
(2.66)
[py,Lz] =px
Here no new constants can be obtained from the Poisson theorem. It will be remembered
from the fundamental Poisson brackets, that the Poisson brackets of any two canonical
momenta must always be zero.
Check Your Progress 4
Note: a)
Write your answers in the space given below.
b)
Compare your answers with the ones given at the end of the units.
(i)
Define Poisson and Lagrange Brackets with suitable expressions?
(ii)
Write the relation between the Poisson and Lagrange’s brackets?
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
………………………………………………………………………………………….
2.10
EQUATION OF MOTION IN POISSON BRACKET NOTATION
The total time derivative of a dynamical variable F( qi, pi, t) can be expressed as
.
F
F= [F,H] +
t
If F does not involve time t explicity then ,
(2.67)
.
F =[F,H]
Giving that if the Poisson bracket of F with H vanishes then dynamical variable F will be
constant of motion. This requirement does not, however, require that H should be a
constant of motion. Suppose such dynamical variables are qi and pi, then
42
Kepler’s problem and Hamiltonian Mechanics
.
qi = [qi, H]
.
and p. i = [pi, H]
(2.68)
Which are identical with the Hamilton’s canonical equations of motion because
[qi, H ] = 
j
And since ,
qi H
qj pj
qi H
pj qi
qi
pj = 0
We find that
[qi,H] =
H
ij
pj
Therefore
.
H
qi =
= [qi, H]
pi
.
H
pi =
= [pi,H]
qi
Equation (2.68) thus be referred to as the equation of motion in Poisson bracket form.
From this equation it is also concluded that if Poisson bracket [pi,H] vanishes,
Then
.
pi = 0 and pi = constant
that is, linear momentum is conserved, which implies that corresponding co-ordinates are
cyclic . With the help of Poisson brackets we have a general test for seeking and
identifying these constants of motion since all functions whose Poisson bracket with
Hamiltonian vanish will be constants of motion and conversely Poisson brackets of all
constants of motion with H must be zero.
Check Your Progress 5
Note: a)
Write your answers in the space given below.
b)
Compare your answers with the ones given at the end of the units.
(i)
Write the relation for angular momentum with Poisson Bracket.
(ii)
Write the equation of motion in Poisson Bracket notation and its
significance.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
………………………………………………………………………………………… 43
…………………………………………………………………………………………
…………………………………………………………………………………………
Lagrangian and Hamiltonian Mechanics
2.11 LET US SUM UP
After going through this unit, you would have achieved the objectives stated earlier in the
unit. Let us recall what we have discussed so far.

Lagrangian of Two body Problem equivalent to one body problem
.
L= ½  r2 – V(r)

Equation for the shape of a conic section
r=
1
2
1+ 1+ 2El 2
mk
1/ 2
cos 
If
E >0 giving  >1 –conic is hyperbola non periodic motion
E=0 giving  =1- conic is parabola
E <1 giving  <1 – conic is ellipse
Periodic motion

Kepler’s law
In seventeenth century, Kepler announced the following three laws

1. The planet move in elliptical orbits with sun as one of foci.
2. Area swept out by the radius vector from the sun to a planet in equal times are
equal.
3. The square of the period of revolution is proportional to the cube of the semi
major axis.
The knowledge of orbits, i.e. the path adopted by a particle during its motion
under the action of a central force, provides us the information about the
dependence of potential energy on r, the radial distance, and consequently gives
the information about a force law.

If E and s are fixed, the angle of scattering  is then determined uniquely.
tan( / 2) = ( k2/ 4s2E2)1/ 2 = (k / 2sE)

The quantity  (,) called differential scattering cross section and is defined as,
Number of particles scattered per unit solid angle per unit in the direction
specified by spherical angle  and 
 (,) =
Incident intensity I
44
Kepler’s problem and Hamiltonian Mechanics

This result gives the famous Rutherford scattering cross section. This also agrees
with the quantum mechanical result with in non-relativistic limit.
1
 () = 4

ZZe2 2 1
2E
sin4/2
The important variational principle associated with Hamiltonian formulation is
the principle of least action. The principle of least action for the conservative
system is expressed as,

t
t2
.
 pi qi dt = 0
1
Where  is the variation

The Hamilton’s canonical equations are

.
L
Qi =
( Qi , P i )
Pi
.
L
Pi =
(Q, P)
Qi i i
Poisson bracket is defined as,
[ X,Y]q , p =  F H - F H
i qi pi
pi qi

Lagrange’s bracket of (u, v) with respect to the basis (qi ,pi) is defined as
( u,v)q , p =  qi pi - pi qi
i u v
u v

The relation between Lagrange and Poisson bracket show that
2n
 {ul, ui}[ui, uj] = ij
i
Where {ul, ui} is Lagrange bracket and [ui, uj] is the Poisson bracket

The equation below represent the equation of motion in Poisson Brackets
.
H
qi =
= [qi, H]
pi
.
H
pi =
= [pi,H]
qi
45
Lagrangian and Hamiltonian Mechanics
2.12
1.
CHECK YOUR PROGRESS: THE KEY
(i) Equation for the shape of a conic section
r=
1
1/ 2
2
1+ 1+ 2El 2
mk
cos 
If
E >0 giving  >1 –conic is hyperbola non periodic motion
E=0 giving  =1- conic is parabola
E <1 giving  <1 – conic is ellipse
Periodic motion
(ii) If E and s are fixed, the angle of scattering  is then determined uniquely.
tan( / 2) = ( k2/ 4s2E2)1/ 2 = (k / 2sE)
The quantity  (,) called differential scattering cross section and is defined as,
Number of particles scattered per unit solid angle per unit in the direction
specified by spherical angle  and 
 (,) =
Incident intensity I
(iii)
The result of nucleon scattering gives the famous Rutherford scattering
cross
section. This also agrees with the quantum mechanical
result with in non1
4
 () =
2.
(i)
ZZe
2E
2
relativistic limit.
1
sin4/2
Equation below is some times is referred as the modified Hamilton’s
principle.
t

3.
2
t2
t
 pi dqi -  H (qi, pi, t)dt =0
i
1
t2
1
H
.
, qj - p = 0
j
(i)
H
.
pj +
=0
qj
(ii)
The principle of least action for the conservative system is expressed as,

t
,are the Hamilton’s canonical Equations
t2
.
 pi qi dt = 0
i
1
Where  is the variation
46
Kepler’s problem and Hamiltonian Mechanics
Various form of principle of least action
(a) If the co-ordinate transformation equations do not involve time explicity, then
.
 pi qi =2T so that the principle of least action assumes the form
i
t2

t 2 T dt = 0
t2
Or, 
1
t T dt = 0
1
Further, if the system is not involving any external force, T is conserved, giving
t2

t dt = 0
or  ( t2 – t1) =0
1
This condition leads to a very important principle called Fermat’s principle
(b)
When the transformation equations do not involve time, kinetic energy cab
always be expressed as a homogeneous quadratic function of the velocities then
the principle of least action can be expressed as,

  2[H-V(q)] d = 0
(c ) If the system consists of only one particle then the principle of least action
becomes
  2[H-V ] ds = 0

The principle of least action is here expressed in terms of the arc length of the
particle trajectory.
4.
(i)
Poisson bracket is defined as,
[ X,Y]q , p =  F H - F H
i qi pi
pi qi
Lagrange’s bracket of (u, v) with respect to the basis (qi ,pi) is defined as
(ii)
( u,v)q , p =  qi pi - pi qi
i u v
u v
The relation between Lagrange and Poisson bracket show that
2n
 {ul, ui}[ui, uj] = ij
i
47
Lagrangian and Hamiltonian Mechanics
5.
(i)
[pz, Lx] = px
and
[pz,Ly] = -px
i.e., both L and P are conserved. We have here as instance in which Poisson’s
theorem does not yield new constants of the motion .Then their Poisson
brackets are
[px,py] =0,
[px,Lz]= py
[py,Lz] =px
(ii)
Equations of motion in Poisson Bracket form
.
H
qi =
= [qi, H]
pi
.
H
pi =
= [pi,H]
qi
From this equation it is also concluded that if Poisson bracket [pi,H] vanishes,
Then
.
pi = 0 and pi = constant
that is, linear momentum is conserved, which implies that corresponding coordinates are cyclic . With the help of Poisson brackets we have a general test for
seeking and identifying these constants of motion since all functions whose
Poisson bracket with Hamiltonian vanish will be constants of motion and
conversely Poisson brackets of all constants of motion with H must be zero.
REFERENCES AND SUGGESTED TEXT BOOKS
1.
2.
3.
4.
5.
6.
7.
Classical Mechanics by Herbert Goldstein (Second Ediition).
Calculus of Variation by G.A.Bliss.
Variational Principles of Mechanics by C. Lanczos.
Mechanics by D. Landau and E.M. Lifshitz
Generalised Co-ordinates by W. B . Byerly
Lagrangian Dynamics by D.W. Wells.
Classical Mechanics by Dr.S.L.Gupta, Dr.V.Kumar and Dr.H.V.Sharma.
48
49