CHAPTER - III
HAMILTON’S PRINCIPLE AND HAMILTON’S
FORMULATION
Unit 1: Hamilton’s Principles:
•
Introduction :
In the Chapter II we have used the techniques of variational principles of
Calculus of Variation to find the stationary path between two points. Hamilton’s
principle is one of the variational principles in mechanics. All the laws of mechanics
can be derived by using the Hamilton’s principle. Hence it is one of the most
fundamental and important principles of mechanics and mathematical physics.
In this unit we define Hamilton’s principle for conservative and nonconservative systems and derive Hamilton’s canonical equations of motion. We also
derive Lagrange’s equations of motion.
•
Hamilton’s Principle (for non-conservative system) :
Hamilton’s principle for non-conservative systems states that “The motion of
a dynamical system between two points at time intervals t0 to t1 is such that the line
integral
t1
I = ∫ (T + W )dt
t0
is extremum for the actual path followed by the system” , where T is the kinetic
energy and W is the work done by the particle.
It is equivalent to say that δ variation in the actual path followed by the
system is zero. Mathematically, it means that
Classical Mechanics
Page No. 157
t1
δ I = δ ∫ (T + W )dt = 0
t0
for actual path.
•
Hamilton’s Principle (for conservative system) :
“Of all possible paths between two points along which a dynamical system
may move from one point to another within a given time interval from t0 to t1 , the
actual path followed by the system is the one which minimizes the line integral of
Lagrangian.’’
This means that the motion of a dynamical system from t0 to t1 is such that
t1
the line integral
∫ Ldt
is extremum for actual path. This implies that small δ
t0
variation in the actual path followed by the system is zero.
t1
Mathematically, we express this as δ ∫ Ldt = 0 for the actual path.
t0
Note : We will show bellow in the Theorem (2) that the Hamilton’s principle
t1
δ ∫ Ldt = 0 also holds good for the non-conservative system.
t0
•
Action in Mechanics :
Let L = L ( q j , q j , t ) be the Lagrangian for the conservative system. Then the
integral
t1
I = ∫ Ldt
t0
is called the action of the system.
Hence we can also define the Hamilton’s principle as “Out of all possible
paths of a dynamical system between the time instants t0 and t1 , the actual path
followed by the system is one for which the action has a stationary value”
Classical Mechanics
Page No. 158
t1
⇒ δ I = δ ∫ Ldt = 0
t0
for the actual path.
Theorem 1 : Derive Hamilton’s principle for non-conservative system from
D’Alembert’s principle and hence deduce from it the Hamilton’s principle for
conservative system.
Proof: We start with D’Alembert’s principle which states that
∑ ( F − p )δ r = 0 .
i
i
. . . (1)
i
i
Note that in this principle the knowledge of force whether it is conservative or nonconservative and also the requirement of holonomic or non-holonomic constraints
does not arise. We write the principle in the form
∑ F δ r = ∑ p δ r .
i
i
i
i
i
i
⇒ δ W = ∑ p iδ ri .
. . . (2)
i
where δ W = ∑ Fiδ ri is the virtual work.
i
Now consider
∑ p δ r = ∑ m r δ r ,
i
i
i i
i
i
i
=∑
i
Since we have δ ri =
d
d
( mi riδ ri ) − ∑ mi ri (δ ri ) .
dt
dt
i
d
δ ri , therefore, we write
dt
d 


1
∑ p δ r = dt ∑ m rδ r  − δ  ∑ 2 m r
i
i
i i
i
i
i i
i
i
d 
2

.


∑ p δ r = dt ∑ m rδ r  − δ T ,
i
i
Classical Mechanics
i
i i
i
i
Page No. 159
1
T = ∑ mi ri 2
i 2
where
is the kinetic energy of the system. Substituting this in equation (2) we get
δW =
d 

mi riδ ri  − δ T
∑

dt  i

⇒
δ (W + T ) =
d

mi riδ ri  .
∑

dt  i

Integrating the above equation with respect to t between t0 to t1 we get
t1


t1
∫ δ (W + T )dt = ∑ mi riδ ri  .

t0
i
 t0
Since, there is no variation in co-ordinates along any paths at the end points. i.e.,
t1
(δ ri )t
= 0 . Hence from above equation we have
0
t1
δ ∫ (W + T )dt = 0 .
. . . (3)
t0
This is known as Hamilton’s principle for non-conservative systems.
If however, the system is conservative, then the forces are derivable from potential.
In this case the expression for virtual work becomes
δ W = ∑ Fiδ ri = −∑
i
i
∂V
δ ri = −δ V .
∂ri
Hence equation (3) becomes
t1
δ ∫ (T − V )dt = 0,
t0
t1
⇒
δ ∫ Ldt = 0.
. . . (4)
t0
This is the required Hamilton’s principle for conservative system.
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Page No. 160
t1
Theorem 2 : Show that the Hamilton’s principle
δ ∫ Ldt = 0 also holds for the
t0
non- conservative system.
Proof : We know the Hamilton’s principle for non-conservative system is given by
t1
δ ∫ (T + W )dt = 0
. . . (1)
t0
for actual path. The expression for the virtual work is given by

δ W = ∑ Fiδ ri = ∑ Fi  ∑
i

i
j
∂ri 
δ q j
∂q j 

∂r 
δ W = ∑  ∑ Fi i δ q j
∂q j 
j 
 i
.
δ W = ∑ Q jδ q j ,
. . . (2)
j
Q j = ∑ Fi
where
i
∂ri
∂q j
are the components of generalized forces. In the case of non-conservative system the
potential energy is dependent on velocity called the velocity dependent potential. In
this case the generalized force is given by
Qj = −
∂U d  ∂U 
+ 
.
∂q j dt  ∂q j 
Substituting this in equation (2) and integrating it between the limits t0 to t1 we find
t1
t1
t1

∫ δ Wdt = ∫ ∑ Q jδ q j dt = ∫ 
t0
j
t0
t0

 ∂U
∑  − ∂q
j

j
+
d  ∂U   

 δ q j dt ,
dt  ∂q j   

Substituting this in equation (1) we get
t1
t1
t0
t0
∫ δ Tdt = ∫ ∑
j
Classical Mechanics
t1 
∂U

δ q j dt − ∫ 
∂q j
t0 

 d  ∂U   
∑j  dt  ∂q  δ q j dt .
  j   
Page No. 161
Integrating the second integral by parts we obtain
t
1
t1

 ∂U

∂U
δ
Tdt
=
δ
q
dt
−
δ
q
+


∑
∑
j
j
∫t
∫ ∂q j
  ∫
j  ∂q
j
t0 j
t0 t0 
0
t1
t1

 ∂U  d
 (δ q j ) dt .
 j  dt

∑  ∂q
j
t1
Since change in co-ordinates at the end point is zero. (δ q j )t = 0
0
d
(δ q j ) = δ q j ,
dt
and also
then we have
t1
 ∂U
 
∂U
δ
q
+
δ
q
∑j  ∂q j ∂q j  dt .
j
t0 
 j
 

t1

∫ δ Tdt = ∫ 
t0
Since time t is fixed along any path hence, there is no variation in time along any
path therefore change in time along any path is zero. i.e., δ t = 0
Hence we write above equation as
t0
 ∂U
 ∂U 
∂U
δ
q
+
δ
q


∑j  ∂q j ∂q j  + ∂t δ t dt .
j
t0 

 j


t1
t1
t0
t0
t1
t1

∫ δ Tdt = ∫ 
∫ δ Tdt = ∫ δ Udt .
t1
⇒
∫ δ (T − U ) dt = 0,
t0
t1
⇒
δ ∫ Ldt = 0.
t0
This proves that the Hamilton's principle holds good even for non-conservative
systems.
Theorem 3 : State Hamilton’s principle for non-conservative system and hence
derive Lagrange’s equations of motion for non-conservative holonomic systems.
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Page No. 162
Proof: Let us consider a non-conservative holonomic dynamical system whose
configuration at any instant t is specified by n generalized co-ordinates
q1 , q2 , q3 ,..., qn . Hamilton’s principle for non-conservative system states that
t1
δ ∫ (T + W )dt = 0
for actual path.
. . . (1)
t0
The virtual work done is given by

δ W = ∑ Fiδ ri = ∑ Fi  ∑
i
i

j
∂ri 
δ q j
∂q j 

∂r 
δ W = ∑  ∑ Fi i δ q j
∂q j 
j 
 i
.
δ W = ∑ Q jδ q j ,
. . . (2)
j
where
Q j = ∑ Fi
i
∂ri
∂q j
are the components of generalized forces.
t1
t1
t0
t0
∫ δ Wdt = ∫ ∑ Q jδ q j dt .
. . . (3)
j
The kinetic energy of the particle T = T ( q j , q j , t ) .
δT = ∑
j
∂T
∂T
∂T
δ q j + ∑ δ q j + δ t .
j
∂q j
∂t
j ∂q
. . . (4)
As the variation in time along any path is always zero. ⇒ δ t = 0 .
This implies that
δT = ∑
j
∂T
∂T
δ q j + ∑ δ q j
j
∂q j
j ∂q
. . . (5)
Integrating equation (5) between the limits t0 to t1 we get
t1
t1
t0
t0
∫ δ Tdt =∫ ∑
Classical Mechanics
j
t
1
∂T
∂T
δ q j dt + ∫ ∑
δ q j dt
∂q j
∂
q
j
j
t0
Page No. 163
Since we have
d
δ q j = δ q j .
dt
Therefore we write the above integral as
t1
t1
t0
t0
∫ δ Tdt =∫ ∑
j
t
1
∂T
∂T d
δ q j dt + ∫ ∑
(δ q j ) dt .
∂q j
∂q j dt
t0 j
Integrating the second integral by parts, we obtain
t1
t1
 ∂T

∂T
d  ∂T
=
+
−
δ
Tdt
δ
q
dt
δ
q


∑
∑
∑
∫
∫ ∂q j j  j ∂q j j  t∫ j dt  ∂q j
t0
t0 j



0
t1
t1
t0

δ q j dt .

Since in δ variation there is no change in the co-ordinates at the end points
⇒
(δ q )
t1
j t
0
= 0 . Hence
t1
 ∂T
t1
∫ δ Tdt =∫ ∑ 
t0
t0
j
 ∂q j
−
d  ∂T

dt  ∂q j

  δ q j dt .
 
. . . (6)
Using equations (3) and (5) in equation (1) we get
 ∂T
t1
∫ ∑  ∂q
t0
j

j
−
d  ∂T

dt  ∂q j


 + Q j  δ q j dt = 0 .


. . . (7)
If the constraints are holonomic then δ q j are independent. (Note that if the
constraints are non-holonomic, then δ q j are not all independent. In this case
vanishing of the integral (7) does not imply the coefficient vanish separately) Hence
the integral (7) vanishes if and only if the coefficient must vanish separately.
d  ∂T

dt  ∂q j
 ∂T
= Qj .
 −
 ∂q j
. . . (8)
These are the Lagrange’s equations of motion for non-conservative holonomic
system.
Classical Mechanics
Page No. 164
Theorem 4 : Deduce Hamilton’s principle for conservative system from
D’Alembert’s principle.
Proof: We start with D’Alembert’s principle which states that
∑ ( F − p )δ r = 0 .
i
i
. . . (1)
i
i
We write the principle in the form
∑ F δ r = ∑ p δ r ,
i
i
i
i
where
. . . (2)
i
i
δ ri is the virtual displacement and occurs at a particular instant. Hence
change in time δ t along any path is zero.
Now consider
∑ p δ r = ∑ m rδ r ,
i
i
i i
i
i
=∑
i
δ ri =
Since we have
i
.
d
d
m
r
δ
r
−
m
r
δ
r
( i i i) ∑ i i ( i)
dt
dt
i
d
δ ri ,
dt
therefore we write
d 


1
∑ p δ r = dt ∑ m rδ r  − δ  ∑ 2 m r
i
i
i i
i
i
i i
i
i
d 

∑ p δ r = dt ∑ m rδ r  − δ T
i
i i
i
i
2

.

. . . (3)
i
where
1
T = ∑ mi ri 2
i 2
is the kinetic energy of the system. Substituting equation (3) in equation (2) we get
d 

∑ F δ r = dt ∑ m rδ r  − δ T
i
i
Classical Mechanics
i
i i
i
. . . (4)
i
Page No. 165
Since the force is conservative ⇒ Fi = −
⇒
∂V
.
∂ri
d
∂V

mi riδ ri  = δ T − ∑
δ ri
∑

.
dt  i
i ∂ri

= δ T − δV
Integrating the above equation with respect to t between t0 to t1 we get
t1
t
1


=
m
r
δ
r
δ
∑ i i i 
∫ Ldt .
 i
 t0
t0
Since, there is no variation in co-ordinates along any paths at the end points.
t
i.e. (δ ri )t1 = 0 . Hence from above equation we have
0
t1
δ ∫ Ldt = 0 .
. . . (5)
t0
This is the required Hamilton’s principle for conservative systems.
•
Derivation of Lagrange’s equations of motion from Hamilton’s Principle :
Theorem 5 : Show that the Lagrange’s equations are necessary conditions for the
action to have a stationary value.
Proof: We know the action of a particle is defined by
t1
I = ∫ Ldt
. . . (1)
t0
where L is the Lagrangian of the system. Consider
t1
δ I = δ ∫ Ldt ,
t0
t1
 ∂L

∂L
δ qj + ∑
δ q j  dt
= ∫ ∑
j
j ∂q

t0 
 j ∂q j
As there is no variation in time along any path, hence δ t = 0 .
Classical Mechanics
Page No. 166
t1
t1
t0
t0
δ ∫ Ldt = ∫ ∑
δ
Since
dq j
=
dt
j
t
1
∂L
∂L
δ q j dt + ∫ ∑
δ q j dt.
∂q j
∂
q
j
j
t0
d
(δ q j ) ,
dt
therefore, we write
t1
t1
t
1
∂L
∂L d
δ ∫ Ldt = ∫ ∑
δ q j dt + ∫ ∑
(δ q j )dt
∂q j
∂q j dt
t0
t0 j
t0 j
. . . (2)
Integrating the second integral on the r. h. s. of equation (2) we get
t1
t1
 ∂L

d  ∂L
∂L
δ ∫ Ldt = ∫ ∑
δ q j dt +  ∑
δ qj  − ∫ ∑ 

∂q j
t0
t0 j
 j ∂q j
 t0 j dt  ∂q j
t1
t1
t0

 δ q j dt.

Since there is no variation in the co-ordinates along any path at the end points, hence
change in the co-ordinates at the end points is zero. i.e.,
(δ q )
t1
= 0.
j t
0
Thus we have
t1
 ∂L
t1
δ ∫ Ldt = ∫ ∑ 
t0
t0
j
 ∂q j
−
d  ∂L

dt  ∂q j

  δ q j dt.
 
. . . (3)
If the system is holonomic, then all the generalized co-ordinates are linearly
independent and hence we have
t1
 ∂L
t1
δ ∫ Ldt = 0 ⇔ ∫ ∑ 
j
 ∂q j
−
d  ∂L

dt  ∂q j
t0
t0
t1
∂L d  ∂L
− 
∂q j dt  ∂q j
δ ∫ Ldt = 0 ⇔
t0

 = 0.


  δ q j dt = 0
 
. . . (4)
These are the required Lagrange’s equations of motion derived from the Hamilton’s
principle. The equation (4) also shows that the Lagrange’s equations of motion for
holonomic system are necessary and sufficient conditions for action to have a
stationary value.
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Page No. 167
Worked Examples •
Example 1 : Use Hamilton’s principle to find the equations of motion of a particle of
unit mass moving on a plane in a conservative force field.
Solution: Let the force F be conservative and under the action of which the particle
of unit mass be moving on the xy plane. Let P (x, y) be the position of the particle.
We write the force
F = iFx + jFy .
Since F is conservative, we have therefore,
Fx = −
∂V
∂V
, Fy = −
.
∂x
∂y
The kinetic energy of the particle is given by
T=
1 2
x + y 2 ) .
(
2
Hence the Lagrangian of the particle becomes
L=
1 2
x + y 2 ) − V ( x, y ) .
(
2
. . . (1)
The Hamilton’s principle states that
t1
δ ∫ Ldt = 0,
. . . (2)
t0
t1
 ∂L
∂L
∂L

∂L
∫  ∂x δ x + ∂y δ y + ∂x δ x + ∂y δ y  dt = 0 ,
t0
t1
⇒

∂V
∂V

∫ ( xδ x + yδ y ) − ∂x δ x − ∂y δ y dt = 0 .
. . . (3)
t0
Consider
t1
t1
t0
t0
= ∫ x
∫ xδ xdt
Classical Mechanics
d
(δ x )dt
dt
Page No. 168
Integrating by parts we get
t1
t1
= ( xδ x )t1 − ∫ x (δ x )dt .
∫ xδ xdt
t
0
t0
t0
Since δ x = 0 at both the ends t0 and t1 along any path, therefore,
t1
t1
t0
t0
t1
t1
t0
t0
= − ∫ x (δ x )dt .
∫ xδ xdt
. . . (4)
Similarly, we have
= − ∫ y (δ y )dt .
∫ yδ ydt
. . . (5)
Substituting these values in equation (3) we get

∂V
∫t  x + ∂x
0
t1

∂V

y+
 δ x +  ∂y


 
 δ y dt = 0 .
 
Since δ x and δ y are independent and arbitrary, then we have
x+
∂V
= 0,
∂x
y+
∂V
= 0.
∂y
∂V
= Fx ,
∂x
∂V
y=−
= Fy .
∂y
x=−
. . . (6)
These are the equations of motion of a particle of unit mass moving under the action
of the conservative force field.
Example 2: Use Hamilton’s principle to find the equation of motion of a simple
pendulum.
Solution: In case of a simple pendulum, the only generalized co-ordinate is θ , and
the Lagrangian is given by (Refer Ex. 26 of Chapter I)
L=
1 2 2
ml θ − mgl (1 − cos θ ) .
2
Classical Mechanics
. . . (1)
Page No. 169
The Hamilton’s Principle states that “the path followed by the pendulum is one along
which the line integral of Lagrangian is extremum”. i.e.,
t1
δ ∫ Ldt = 0 ,
t0
t1
1
∫ δ  2 ml θ
2
2
t0

− mgl (1 − cos θ )  dt = 0 ,

t1
− mgl sin θδθ dt = 0 .
∫  ml θδθ

2
t0
δ
Since, we have
d
d
= δ.
dt dt
Therefore
t1


d
∫  ml θ dt (δθ ) − mgl sin θδθ dt = 0 .
2
t0
Integrating the first integral by parts we get
(
ml 2 θδθ
)
t1
t0
t1
− ∫ m l 2θ + gl sin θ δθ dt = 0 .
t0
t
Since (δθ )t1 = 0 , we have therefore,
0
t1
∫ m l θ + gl sin θ δθ dt = 0 .
2
t0
As δθ is arbitrary, we have
l 2θ + gl sin θ = 0
g
⇒ θ + sin θ = 0 .
l
This is the required equation of motion of the simple pendulum.
Classical Mechanics
Page No. 170
•
Spherical Pendulum :
Example 3 : A particle of mass m is moving on the surface of the sphere of radius r
in the gravitational field. Use Hamilton’s principle to show the equation of motion is
given by
θ −
pφ2 cos θ
2 4
3
m r sin θ
−
g
sin θ = 0 ,
r
where pφ is the constant of angular momentum.
Solution: Let a particle of mass m be moving on the surface of a sphere of radius r.
The particle has two degrees of freedom and hence two generalized co-ordinates
θ , φ . The Lagrangian of the motion is (Refer Ex. 28 of Chapter I) given by
L=
1 2 2
mr θ + sin 2 θφ2 − mgr cos θ .
2
(
)
. . . (1)
The Hamilton’s Principle states that “the path followed by the particle between two
time instants t0 and t1 is one along which the line integral of Lagrangian is
extremum”. i.e.,
t1
δ ∫ Ldt = 0 ,
t0
t1

∫ δ  2 mr (θ
1
2
t0
2

+ sin 2 θφ2 − mgr cos θ  dt = 0,

)
t1
+ sin
∫  mr (θδθ
2
2
+ sin θ cos θφ2δθ ) + mgr sin θδθ dt = 0.
θφδφ

t0
Since, we have
δ
dθ d
= δθ .
dt dt
Therefore,
t1

∫  mr
t0
2

d

2
2
d
2
 θ (δθ ) + sin θφ (δφ )  + mr sin θ cos θφ + mgr sin θ δθ dt = 0.
dt
 dt


(
)
Integrating the first two integrals by parts we get
Classical Mechanics
Page No. 171
(
mr θδθ
2
)
t1
t0
(
+ mr sin θ φδφ
2
2
t1
) − ∫ mr

t1
t0
t0
2
  
2 g
 θ − sin θ cos θφ − sin θ  δθ  dt −
r

 
t1
− ∫ mr 2
t0
Since
t1
(δθ )t
0
d
sin 2 θφ δφ dt = 0.
dt
(
)
t
= 0 = (δφ )t1 ,
0
we have therefore,
t1
t
1
 2
g
 
2
2 d
2
∫t  mr θ − sin θ cosθφ − r sin θ  δθ dt + t∫ mr dt sin θφ δφ dt = 0
0
0
(
)
Since θ and φ are independent and arbitrary, hence we have
g
r
θ − sin θ cos θφ2 − sin θ = 0,
mr 2
d
sin 2 θφ = 0 ⇒ mr 2 sin 2 θφ = pφ ( const.)
dt
(
)
Eliminating φ we obtain
θ −
pφ2 cos θ
2 4
3
m r sin θ
−
g
sin θ = 0 .
r
. . . (2)
as the required differential equation of motion for spherical pendulum.
Unit 2:
•
Hamiltonian Formulation :
Introduction:
We have developed Lagrangian formulation as a description of mechanics in
terms of the generalized co-ordinates and generalized velocities with time t as a
parameter in Chapter I and the equations of motion were used to solve some
problems. We now introduce another powerful formulation in which the independent
variables are the generalized co-ordinates and the generalized momenta known as
Hamilton’s formulation. This formulation is an alternative to the Lagrangian
Classical Mechanics
Page No. 172
formulation but proved to be more convenient and useful, particularly in dealing with
problems of modern physics. Hence all the examples solved in the Chapter I can also
be solved by the Hamiltonian procedure. As an illustration few of them are solved in
this Chapter by Hamilton’s procedure.
•
The Hamiltonian Function:
The
quantity
∑ p q
j
−L
j
when
expressed
in
terms
of
j
q1 , q2 ,..., qn , p1 , p2 ,... pn , t is called Hamiltonian and it is denoted by H.
H = H ( q j , p j , t ) = ∑ p j q j − L .
Thus
j
•
Hamilton’s Canonical Equations of Motion :
Theorem 6 : Define the Hamiltonian and hence derive the Hamilton’s canonical
equations of motion.
Proof : We know the Hamiltonian H is defined as
H = H ( q j , p j , t ) = ∑ p j q j − L .
. . . (1)
j
Consider
H = H (qj , p j ,t ) .
. . . (2)
We find from equation (2) that
dH = ∑
j
∂H
∂H
∂H
dp j + ∑
dq j +
dt .
∂p j
∂t
j ∂q j
. . . (3)
Now consider H = ∑ p j q j − L .
j
Similarly we find
dH = ∑ q j dp j + ∑ dq j p j − dL,
j
j
⇒ dH = ∑ q j dp j + ∑ dq j p j − ∑
j
j
j
∂L
∂L
∂L
dq j −∑
dq j − dt .
j
∂q j
∂t
j ∂q
. . . (4)
We know the generalized momentum is defined as
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Page No. 173
pj =
∂L
.
∂q j
Hence equation (4) reduces to
dH = ∑ q j dp j − ∑
j
j
∂L
∂L
dq j − dt
∂q j
∂t
. . . (5)
Now comparing the coefficients of dp j , dq j and dt in equations (3) and (5) we get
q j =
∂H
,
∂p j
∂L
∂H
∂L
∂H
=−
,
=−
.
∂q j
∂q j
∂t
∂t
. . . (6)
However, from Lagrange’s equations of motion we have
p j =
∂L
∂q j
Hence equations (6) reduce to
q j =
∂H
,
∂p j
p j = −
∂H
∂q j
.
. . . (7)
These are the required Hamilton’s canonical equations of motion. These are the set of
2n first order differential equations of motion and replace the n Lagrange’s second
order equations of motion.
•
Derivation of Hamilton’s equations of motion from Hamilton’s Principle :
Theorem 7 : Obtain Hamilton’s equations of motion from the Hamilton’s principle.
Proof: We know the action of a particle is defined by
t1
I = ∫ Ldt
. . . (1)
t0
where L is the Lagrangian of the system. If H ( p j , q j , t ) is the Hamiltonian of the
motion then we have by definition
H = ∑ p j q j − L .
. . . (2)
j
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Page No. 174
Replacing L in equation (1) by using (2) we have the action in mechanics as
t1
t1


I = ∫ Ldt = ∫  ∑ p j q j − H  dt .
t0
t0  j

. . . (3)
Now by Hamilton’s principle, we have
t1
t1
t0
t0


δ ∫ Ldt = 0 ⇒ δ ∫  ∑ p j q j − H  dt = 0 .


j
. . . (4)
This is known as the modified Hamilton’s principle. Thus we have
t1
t1
t0
t0


δ ∫ Ldt = δ ∫  ∑ p j q j − H  dt ,
t1
t1

t0
t0



j
δ ∫ Ldt = ∫ ∑ δ p j q j + ∑ p jδ q j − ∑
j
j
j
∂H
∂H
∂H 
δ qj − ∑ δ pj −
δ t dt.
∂q j
∂t 
j ∂p j
Since time is fixed along any path, hence change in time along any path is zero. i.e.,
δ t = 0 along any path. Hence above equation becomes
t1
t1

t0
t0




δ ∫ Ldt = ∫  ∑  q j −
j
∂H
∂p j


∂H
δ q j dt
δ p j + ∑ p jδ q j − ∑
j
j ∂q j


. . . (5)
Now consider the integral
t1
t1
∫ ∑ p δ q dt = ∫ ∑ p
j
t0
j
j
j
j
t0
d
(δ q j ) dt .
dt
Integrating the integral on the r. h. s. by parts we get
t1

 t1
p
δ
q
dt
=
p
δ
q
∫t ∑j j j  ∑j j j  − t∫ ∑j p jδ q j dt .
t0
0
0
t1
t1
Since (δ q j )t = 0 . We have therefore
0
t1
t1
∫ ∑ p jδ q j dt = −∫ ∑ p jδ q j dt .
t0
j
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t0
j
Page No. 175
Substituting this in equation (5) we get
t1
t1

t0
t0




δ ∫ Ldt = ∫  ∑  q j −
j
∂H
∂p j


∂H
δ p j + ∑  p j +
∂q j
j 

 
δ q j dt .
 
Now we see that
t1
t1

t0
t0




δ ∫ Ldt = 0 ⇔ ∫  ∑  q j −
j
∂H
∂p j


∂H
δ p j + ∑  p j +
∂q j
j 

 
δ q j dt = 0 .
 
For holonomic system we have q j , p j are independent, hence
t1
δ ∫ Ldt = 0 ⇔ q j −
t0
t1
⇒
δ ∫ Ldt = 0 ⇔ q j =
t0
∂H
= 0,
∂p j
∂H
,
∂p j
p j +
p j = −
∂H
= 0.
∂q j
∂H
.
∂q j
. . . (6)
These are the Hamilton’s canonical equations of motion.
Remark :
We see from equation (6) that the Hamilton’s canonical equations of motion
are the necessary and sufficient conditions for the action to have stationary value.
Example 4 : Show that addition of the total time derivative of any function of the
form f ( q j , t ) to the Lagrangian of a holonomic system, the generalized momentum
and the Jacobi integral are respectively given by
pi +
∂f
∂f
and H −
.
∂qi
∂t
Does the new Lagrangian L′ unchanged the Hamilton’s principle? Justify your
claim.
Solution: Let the new Lagrangian function after addition of the time derivative of
any function of the form f ( q j , t ) to the Lagrangian L be denoted by L′ . Thus we
have
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Page No. 176
L′ = L +
df
.
dt
. . . (1)
Thus the generalized momentum corresponding to the new Lagrangian L′ is defined
by
Thus
p′j =
∂L′
.
∂q j
p′j =
∂L
∂
+
∂q j ∂q j
⇒
⇒
p′j = p j +
p′j = p j +
. . . (2)
 df

 dt

,

∂ 
∂f
∂f
qk +
∑
∂q j  k ∂qk
∂t

,

∂f
.
∂q j
. . . (3)
This is the required generalized momentum corresponding to the new Lagrangian L′ .
Similarly, the Jacobi integral for new function L′ is given by
H ′ = ∑ p′j q j − L′,
j
H′ = ∑
j
∂L′
df

q j −  L +
∂q j
dt


.

On using equation (3) we get

∂f
H′ = ∑ pj +

∂q j
j 

df

 q j −  L +
dt


∂f
H ′ = ∑ ( p j q j − L ) − ,
∂t
j
⇒ H′ = H −

,

∂f
.
∂t
. . . (4)
This is a required Jacobi integral for the new Lagrangian L′ .
Now we show that the new Lagrangian L′ also satisfies the Hamilton’s principle.
Therefore, consider
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Page No. 177
2
2
2
1
1
1
2
2
2
1
1
1
2
2
δ ∫ L′dt =δ ∫ Ldt + δ ∫
df
dt ,
dt
δ ∫ L′dt =δ ∫ Ldt + δ ∫ df ,
δ ∫ L′dt =δ ∫ Ldt + (δ f )1 ,
2
1
1
2
 ∂f
∂f 
δ ∫ L′dt =δ ∫ Ldt +  ∑
δ qj + δt .
∂t 
1
1
 j ∂q j
1
2
2
But in δ variation time is held fixed along any path and hence δ t = 0 along any
path.
Further, co-ordinates at the end points are held fixed.
⇒
(δ q )
2
j 1
= 0.
Hence we have from the above equation that
2
2
1
1
δ ∫ L′dt = δ ∫ Ldt .
Thus the Hamilton’s Principle
2
2
1
1
δ ∫ Ldt = 0 ⇔ δ ∫ L′dt = 0 .
This shows that the new Lagrangian L′ satisfies the Hamilton’s principle.
•
Lagrangian from Hamiltonian and conversely :
Example 5: Obtain Lagrangian L from Hamiltonian H and show that it satisfies
Lagrange’s equations of motion.
Solution: The Hamiltonian H is defined by
H = ∑ p j q j − L .
. . . (1)
j
which satisfies the Hamilton’s canonical equations of motion.
q j =
Classical Mechanics
∂H
,
∂p j
p j = −
∂H
∂q j
.
. . . (2)
Page No. 178
Now from equation (1) we find the Lagrangian
L = ∑ p j q j − H .
. . . (3)
j
and show that it satisfies Lagrange’s equations of motion. Thus from equation (3) we
have
∂L
∂H
=−
, and
∂q j
∂q j
∂L
= pj .
∂q j
Now consider
∂L d  ∂L
− 
∂q j dt  ∂q j

∂H d
− ( pj ),
 = −
∂q j dt

∂L d  ∂L
− 
∂q j dt  ∂q j

 = p j − p j ,

∂L d  ∂L
− 
∂q j dt  ∂q j

 = 0.

This shows that the equation (3) gives the required Lagrangian which satisfies the
Lagrange’s equations of motion.
Example 6 : Obtain the Hamiltonian H from the Lagrangian and show that it
satisfies the Hamilton’s canonical equations of motion.
Solution: The Hamiltonian H in terms of Lagrangian L is defined as
H = ∑ p j q j − L .
. . . (1)
j
where L satisfies the Lagrange’s equations of motion viz.,
∂L d  ∂L
− 
∂q j dt  ∂q j
⇒
Classical Mechanics

 = 0 ,

. . . (2)
∂L d  ∂L 
= 
,
∂q j dt  ∂q j 
d
= ( p j ).
dt
Page No. 179
⇒
∂L
= p j .
∂q j
. . . (3)
Now from equation (1) we find
∂H
∂L
=−
.
∂q j
∂q j
. . . (4)
From equations (3) and (4) we have
∂H
= − p j .
∂q j
. . . (5)
Similarly, we find from equation (1)
∂H
= q j .
∂p j
. . . (6)
Equations (5) and (6) are the required Hamilton’s equations of motion.
••
•
Physical Meaning of the Hamiltonian :
Theorem 8 :
1.
For conservative scleronomic system the Hamiltonian H represents both a
constant of motion and total energy.
2.
For conservative rheonomic system the Hamiltonian H may represent a
constant of motion but does not represent the total energy.
Proof : The Hamiltonian H is defined by
H = ∑ p j q j − L .
. . . (1)
j
where L is the Lagrangian of the system and
pj =
∂L
∂q j
. . . (2)
is the generalized momentum. This implies from Lagrange’s equation of motion that
Classical Mechanics
Page No. 180
d  ∂L

dt  ∂q j
p j =
 ∂L
 =
 ∂q j
. . . (3)
Differentiating equation (1) w. r. t. time t, we get
dH
∂L
∂L
∂L
= ∑ p j q j + ∑ p j qj − ∑
q j − ∑
qj −
j
dt
∂t
j
j
j ∂q j
j ∂q
. . . (4)
On using equations (2) and (3) in equation (4) we readily obtain
dH
∂L
=−
dt
∂t
. . . (5)
Now if L does not contain time t explicitly, then from equation (5), we have
dH
=0
dt
This shows that H represents a constant of motion.
However, the condition L does not contain time t explicitly will be satisfied by
neither the kinetic energy nor the potential energy involves time t explicitly.
Now there are two cases that the kinetic energy T does not involve time t explicitly.
1.
For the conservative and scleronomic system :
In the case of conservative system when the constraints are scleronomic, the
kinetic energy T is independent of time t and the potential energy V is only function
of co-ordinates. Consequently, the Lagrangian L does not involve time t explicitly
and hence from equation (5) the Hamiltonian H represents a constant of motion.
Further, for scleronomic system, we know the kinetic energy is a homogeneous
quadratic function of generalized velocities.
T = ∑ a jk q j qk .
. . . (6)
j ,k
Hence by using Euler’s theorem for the homogeneous quadratic function of
generalized velocities we have
∑ q
j
Classical Mechanics
j
∂T
= 2T .
∂q j
. . . (7)
Page No. 181
For conservative system we have
pj =
∂L ∂T
=
.
∂q j ∂q j
. . . (8)
Using (7) and (8) in the Hamiltonian H we get
H = 2T − (T − V ) ,
H = T +V = E .
. . . (9)
where E is the total energy of the system. Equation (9) shows that for conservative
scleronomic system the Hamiltonian H represents the total energy of the system.
2.
For conservative and rheonomic system :
In the case of conservative rheonomic system, the transformation equations
do involve time t explicitly, though some times the kinetic energy may not involve
time t explicitly. Consequently, neither T nor V involves t, and hence L does not
involve t. Hence in such cases the Hamiltonian may represent the constant of motion.
However, in general if the system is conservative and rheonomic, the kinetic energy
is a quadratic function of generalized velocities and is given by
T = ∑ a jk q j qk + ∑ a j q j + a
j ,k
. . . (10)
j
where
∂r ∂r
1
a jk = ∑ mi i i ,
∂q j ∂qk
i 2
∂r ∂r
a j = ∑ mi i i ,
∂q j ∂t
i
. . . (11)
2
1  ∂r 
a = ∑ mi  i  .
 ∂t 
i 2
We see from equation (10) that each term is a homogeneous function of generalized
velocities of degree two, one and zero respectively. On applying Euler’s theorem for
the homogeneous function to each term on the right hand side, we readily get
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Page No. 182
∑ q
j
j
∂T
= 2T2 + T1 .
∂q j
. . . (12)
where
T2 = ∑ a jk q j qk ,
j ,k
T1 = ∑ a j q j ,
j
T0 = a
are homogeneous function of generalized velocities of degree two, one and zero
respectively. Substituting equation (12) in the Hamiltonian (1) we obtain
H = T2 − T0 + V
showing that the Hamiltonian H does not represent total energy. Thus for the
conservative rheonomic systems H may represent the constant of motion but does not
represent total energy.
•
Cyclic Co-ordinates In Hamiltonian :
Theorem 9 : Prove that a co-ordinate which is cyclic in the Lagrangian is also cyclic
in the Hamiltonian.
Solution: We know the co-ordinate which is absent in the Lagrangian is called cyclic
co-ordinate. Thus if q j is cyclic in L ⇒
∂L
=0.
∂q j
Hence the Lagrange’s equation of motion reduces to
d  ∂L

dt  ∂q j
where p j =
∂L
∂q j

 = 0 ⇒

p j = 0 ,
. . . (1)
is the generalized momentum. However, from Hamilton’s
canonical equations of motion we have
p j = −
Classical Mechanics
∂H
.
∂q j
. . . (2)
Page No. 183
Equations (1) and (2) gives
∂H
=0.
∂q j
. . . (3)
This shows that the co-ordinate q j is also absent in the Hamiltonian, and
consequently, it is also cyclic in H. Thus a co-ordinate which is cyclic in the
Lagrangian is also cyclic in the Hamiltonian.
Worked Examples •
Example 7 : Describe the motion of a particle of mass m moving near the surface of
the Earth under the Earth’s constant gravitational field by Hamilton’s procedure.
Solution: Consider a particle of mass m moving near the surface of the Earth under
the Earth’s constant gravitational field. Let (x, y, z) be the Cartesian co-ordinates of
the projectile, z being vertical. Then the Lagrangian of the projectile is given by
L=
1
m ( x 2 + y 2 + z 2 ) − mgz .
2
. . . (1)
We see that the generalized co-ordinates x and y are absent in the Lagrangian,
hence they are the cyclic co-ordinates. This implies that any change in these coordinates can not affect the Lagrangian. This implies that the corresponding
generalized momentum is conserved. In this case the generalized momentum is the
linear momentum and is conserved.
i.e.,
px = mx = const.
p y = my = const.
. . . (2)
pz = mz .
This shows that the horizontal components of momentum are conserved.
The Hamiltonian of the particle is defined by
H = ∑ p j q j − L,
j
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Page No. 184
1
H = px x + p y y + pz z − m ( x 2 + y 2 + z 2 ) + mgz .
2
. . . (3)
Eliminating x, y , z between equations (2) and (3) we get
1
px2 + p y2 + pz2 ) + mgz .
(
2m
H=
. . . (4)
The Hamilton’s equations of motion give
p x = −
and
x =
∂H
∂H
∂H
= 0, p y = −
= 0, p z = −
= − mg .
∂x
∂y
∂z
∂H px
∂H p y
∂H pz
=
=
=
, y =
, z =
.
∂px m
∂p y m
∂pz m
. . . (5)
. . . (6)
From these set of equations we obtain
x = 0, y = 0, z = −g
. . . (7)
These are the required equations of motion of the projectile near the surface of the
Earth.
Example 8 : Obtain the Hamiltonian H and the Hamilton’s equations of motion of a
simple pendulum. Prove or disprove that H represents the constant of motion and
total energy.
Solution: The Example is solved earlier by various methods. The Lagrangian of the
pendulum is given by
L=
1 2 2
ml θ − mgl (1 − cos θ ) ,
2
. . . (1)
where the generalized momentum is given by
pθ =
p
∂L
= ml 2θ ⇒ θ = θ2 .
ml
∂θ
. . . (2)
The Hamiltonian of the system is given by
H = pθθ − L,
1
⇒ H = pθθ − ml 2θ 2 + mgl (1 − cos θ ) .
2
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Page No. 185
Eliminating θ we obtain
H=
pθ2
+ mgl (1 − cos θ ) .
2ml 2
. . . (3)
Hamilton’s canonical equations of motion are
q j =
∂H
,
∂p j
p j = −
∂H
.
∂q j
These equations give
θ =
pθ
,
ml 2
p θ = − mgl sin θ .
. . . (4)
Now eliminating pθ from these equations we get
θ +
g
sin θ = 0.
l
. . . (5)
Now we claim that H represents the constant of motion.
Thus differentiating equation (3) with respect to t we get
dH pθ pθ
=
+ mgl sin θ θ ,
dt
ml 2
+ mgl sin θθ ,
= ml 2θθ
g


= ml 2θ  θ + sin θ  ,
l


dH
=0.
dt
This proves that H is a constant of motion. Now to see whether H represents total
energy or not, we consider
T +V =
Classical Mechanics
1 2 2
ml θ + mgl (1 − cos θ ) .
2
Page No. 186
Using equation (4) we eliminate θ from the above equation, we obtain
T +V =
pθ2
+ mgl (1 − cos θ ) .
2ml 2
. . . (6)
This is as same as the Hamiltonian H from equation (3). Thus Hamiltonian H
represents the total energy of the pendulum.
Example 9: The Lagrangian for a particle moving on a surface of a sphere of radius r
is given by
L=
1 2 2
mr θ + sin 2 θφ2 − mgr cos θ .
2
(
)
Find the Hamiltonian H and show that it is constant of motion. Prove or disprove that
H represents the total energy. Is the energy of the particle constant? Justify your
claim.
Solution: We are given the Lagrangian of a particle moving on the surface of a
sphere (Spherical Pendulum) in the form
L=
1 2 2
mr θ + sin 2 θφ2 − mgr cos θ .
2
(
)
. . . (1)
We see that φ is a cyclic co-ordinate. This implies the corresponding generalized
momentum is conserved. i.e.
Similarly,
pφ =
∂L
= mr 2 sin 2 θφ = const.
∂φ
. . . (2)
pθ =
∂L
= mr 2θ.
∂θ
. . . (3)
The Hamiltonian of the particle is defined as
1
H = θ pθ + φ pφ − mr 2 θ 2 + sin 2 θφ2 + mgr cos θ .
2
(
)
. . . (4)
Using equations (2) and (3) we obtain the Hamiltonian function
pφ2
1  pθ2
H=  2+ 2 2
2  mr
mr sin θ
Classical Mechanics

 + mgr cos θ .

. . . (5)
Page No. 187
The Hamilton’s canonical equations of motion give
pθ =
cos θ pφ2
mr 2 sin 3 θ
p φ = 0,
θ =
φ =
pθ
,
mr 2
pφ
mr 2 sin 2 θ
+ mgr sin θ ,
. . . (6)
.
Eliminating pθ from equation (6) we get the equation of motion of spherical
pendulum as
cos θ pφ2
mr θ − 2 3 − mgr sin θ = 0 .
mr sin θ
2
(i)
. . . (7)
Now we claim that H is a constant of motion, differentiate equation (5) with
respect to t, we get
pφ p φ
pφ2 cos θ dH pθ pθ
=
+
−
θ − mgr sin θθ,
2
2
2
2
3
dt
mr
mr sin θ mr sin θ
Putting the values of pθ , p φ from equation (6) we get
2
 p cos θ p 2
p  cos θ p
p
dH
= θ 2  2 3φ + mgr sin θ  − θ2 4 3 φ − θ 2 mgr sin θ
 m r sin θ mr
dt mr  mr sin θ

dH
= 0,
dt
showing that H is a constant of motion.
(ii)
Now consider the sum of the kinetic and potential energy of the spherical
pendulum, where
1 2 2
mr θ + sin 2 θφ2 ,
2
V = mgr cos θ
T=
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(
)
Page No. 188
Thus
T +V =
1 2 2
mr θ + sin 2 θφ2 + mgr cos θ .
2
(
)
. . . (8)
We eliminate θ, φ from equation (8) by using equation (6) to get
pφ2
1  p2
T + V =  θ2 + 2 2
2  mr
mr sin θ

 + mgr cos θ .

. . . (9)
We see from equations (5) and (9) that the total energy of the spherical pendulum is
the Hamiltonian of motion. Now to see it is constant or not, multiply equation (7) by
θ we get
−
mr 2θθ
pφ2 cos θθ
− mgr sin θθ = 0
mr 2 sin 3 θ
 d
pφ2
d  1 2 2  d 
mr
+
θ

 + ( mgr cos θ ) = 0.


2
2
dt  2
 dt  2mr sin θ  dt
Integrating we get
pφ2
 1 2 2  
mr
θ
+

 
2
2
2
  2mr sin θ

 + ( mgr cos θ ) = const.

Eliminating θ on using equation (6) we get
pφ2
1  pθ2
+

2  mr 2 mr 2 sin 2 θ

 + mgr cos θ = const. .

. . . (10)
We see from equations (5), (9) and (10) that the Hamiltonian H represents the total
energy and the energy of the particle is conserved.
Example 10: Two mass points of mass m1 and m2 are connected by a string passing
through a hole in a smooth table so that m1 rests on the table surface and m2 hangs
suspended. Assuming m2 moves only in a vertical line, write down the Hamiltonian
for the system and hence the equations of motion. Prove or disprove that
i)
Hamiltonian H represents the constant of motion.
ii)
H represents total energy of the system.
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Page No. 189
Solution: The example is solved in Chapter I. (please refer to Example 24). The
Lagrangian of the system is given by
L=
1
1
m1 r 2 + r 2θ 2 + m2 r 2 + m2 g ( l − r )
2
2
(
)
. . . (1)
We see that the co-ordinate θ is cyclic in the Lagrangian L and hence the
corresponding generalized momentum is conserved.
⇒
∂L
= m1r 2θ = const.
∂θ
. . . (2)
∂L
= ( m1 + m2 ) r = const.
∂r
. . . (3)
pθ =
Similarly, we find
pr =
Now the Hamiltonian function is defined as
H = r
H=
∂L ∂L
+θ
− L,
∂r
∂θ
1
1
( m1 + m2 ) r2 + m1r 2θ2 − m2 g ( l − r ) .
2
2
Eliminating r and θ we obtain
H=
p2
pr2
+ θ 2 − m2 g ( l − r ) .
2 ( m1 + m2 ) 2m1r
. . . (4)
The Hamilton canonical equations of motion viz.,
p j = −
∂H
∂H
, q j =
∂q j
∂p j
p r = −
p2
∂H
= θ 3 − m2 g ,
∂r m1r
give
r =
p θ = −
∂H
= 0.
∂θ
p
∂H
pr
∂H
=
, θ =
= θ2 .
∂pr ( m1 + m2 )
∂pθ m1r
. . . (5)
. . . (6)
From equations (5) and (6) we have
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Page No. 190
( m1 + m2 ) r −
pθ2
+ m2 g = 0 .
m1r 3
. . . (7)
This is the required equation of motion.
i)
To prove H represents a constant of motion, we differentiate equation (4)
with respect to t. Thus we have
p p
p 2 r
dH
pr p r
=
+ θ 2θ − θ 3 + m2 gr
dt ( m1 + m2 ) m1r
m1r
Using equations (5) and (6), we have
pr pθ2
pθ2 pr
dH
pr m2 g
m2 gpr
=
−
−
+
,
3
3
dt ( m1 + m2 ) m1r m1 + m2 ( m1 + m2 ) m1r ( m1 + m2 ) r
dH
= 0.
dt
This shows that The Hamiltonian H represents a constant of motion.
ii)
We have the kinetic and potential energies of the system are respectively
given by
1
1
m1 r 2 + r 2θ 2 + m2 r 2 ,
2
2
V = − m2 g ( l − r ) .
T=
(
)
Now consider
T +V =
1
1
m1 r 2 + r 2θ 2 + m2 r 2 − m2 g ( l − r ) .
2
2
(
)
. . . (8)
Eliminating r and θ from equation (8) on using equations (6) we obtain
T +V =
p2
pr2
+ θ 2 − m2 g ( l − r )
2 ( m1 + m2 ) 2m1r
. . . (9)
From equations (4) and (9) we see that the total energy is equal to the
Hamiltonian function. Thus Hamiltonian H represents total energy of the system. To
prove that the total energy is conserved, multiply the equation of motion (7) by r ,
we get
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Page No. 191
rr −
( m1 + m2 ) pθ2 r
+ m2 gr = 0 .
m1r 3
This we write as
d 
r 2  d  pθ2  d
m
m
+
+ 
(
)
 + ( m2 gr ) = 0 .
1
2
dt 
2  dt  2m1r 2  dt
Integrating and then eliminating r we get
p2
pr2
+ θ 2 + m2 gr = const. .
2 ( m1 + m2 ) 2m1r
. . . (10)
Equations (9) and (10) show that the total energy of the system is conserved.
Note : Equation (10) is the first integral of equation of motion. Its physical
significance is that the Hamiltonian H represents the constant of total energy.
Example 11: A particle of mass m is moving on a xy plane which is rotating about z
axis with angular velocity ω . The Lagrangian is given by
L=
1 
2
2
m ( x − ω y ) + ( y + ω x )  − V ( x, y ) .


2
Show that the Hamiltonian H is given by
1
( px2 + p 2y ) + pxω y − p yω x + V .
2m
H=
Find the equations of motion and hence prove or disprove that
i)
H represents a constant of motion and
ii)
H represents the total energy.
Solution: The Lagrangian of the particle is given by
L=
1 
2
2
m ( x − ω y ) + ( y + ω x )  − V ( x, y ) .

2 
. . . (1)
where the generalized momentum px and p y are given by
px =
p
∂L
= m ( x − ω y ) ⇒ x = x + ω y ,
m
∂x
Classical Mechanics
. . . (2)
Page No. 192
py =
∂L
= m ( y − ω x ) ⇒
∂y
The Hamiltonian H is defined by
y =
py
m
−ωx .
. . . (3)
x + yp
y −L
H = xp
1
2
2
H = px x + p y y − m ( x − ω y ) + ( y + ω x )  + V ( x, y ) .


2
Using equations (2) and (3) we eliminate x and y from the above equation to get
the Hamiltonian of the system
H=
1
( px2 + p 2y ) + ω ( px y − p y x ) + V .
2m
. . . (4)
The Hamilton’s canonical equations of motions give
p x = −
∂H
∂V
= p yω −
,
∂x
∂x
∂H px
x =
=
+ ω y,
∂px m
∂H
∂V
= − p xω −
,
∂y
∂y
∂H p y
y =
=
− ω x.
∂px m
p y = −
. . . (5)
Solving these equations we obtain the equations which describe the motion
∂V
,
∂x
∂V
m ( y + 2ω x − ω 2 y ) = −
∂y
m ( x − 2ω y − ω 2 x ) = −
Now to prove whether H is a constant of motion or not,
. . . (6)
differentiate equation (4)
w. r. t. t to get
dH 1
∂V
∂V
= ( px p x + p y p y ) + ω ( p x y + px y − p y x − p y yx ) + x
+ y
.
dt m
∂x
∂y
Using (5) we have
dH
=0.
dt
This shows that H represents the constant of motion. Now to show whether H
represents total energy or not, we have the total energy of the system
E = T +V ,
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Page No. 193
E=
1
px2 + p y2 ) + V ( x, y ) .
(
2m
. . . (7)
We see from equations (4) and (7) that the Hamiltonian H does not represent total
energy.
Example 12 : A bead slides on a wire in the shape of a cycloid described by
equations
x = a (θ − sin θ ) ,
y = a (1 + cos θ ) , 0 ≤ θ ≤ 2π .
Find the Hamiltonian H, hence the equations of motion. Also prove or disprove that
i)
H represents a constant of motion
ii)
H represents a total energy.
Solution: A particle describes a cycloid whose equations are
x = a (θ − sin θ ) ,
y = a (1 + cos θ ) , 0 ≤ θ ≤ 2π .
. . . (1)
The kinetic energy of the particle is given by
T=
1
m ( x 2 + y 2 ) ,
2
where
x = aθ (1 − cos θ ) ,
y = − a sin θθ.
Hence the kinetic energy of the particle becomes
T = ma 2θ 2 (1 − cos θ ) ,
The potential energy of the particle is given by
V = mga (1 + cos θ ) .
Hence the Lagrangian of the particle becomes
L = ma 2θ 2 (1 − cos θ ) − mga (1 + cos θ ) .
. . . (2)
The Hamiltonian H of the particle is
∂L
H = θ
−L,
∂θ
Classical Mechanics
. . . (3)
Page No. 194
where from equation (2) we have
∂L
= pθ = 2ma 2θ (1 − cos θ ) .
∂θ
. . . (4)
Using equations (2) and (4) in (3) we obtain the expression for Hamiltonian as
H = ma 2θ 2 (1 − cos θ ) + mga (1 + cos θ ) .
. . . (5)
Using equation (4) we eliminate θ from equation (5) to get the required Hamiltonian
H as
pθ2
H=
+ mga (1 + cos θ ) .
4ma 2 (1 − cos θ )
. . . (6)
The Hamilton’s canonical equations of motion are
pθ2
sin θ
∂H
pθ = −
=
+ mga sin θ ,
2
∂θ 4ma (1 − cos θ ) 2
θ =
pθ
∂H
=
.
2
∂pθ 2ma (1 − cos θ )
. . . (7)
. . . (8)
From equations (7) and (8) we obtain the equation of motion of the particle
θ(1 − cos θ ) +
pθ2
sin θ
g
− sin θ = 0
3
2 4
8m a (1 − cos θ ) 2a
. . . (9)
Eliminating pθ from equation (9) we obtain the equation which describes the motion
of the particle in the form
2ma 2 (1 − cos θ ) θ + ma 2 sin θθ 2 − mga sin θ = 0 .
. . . (10)
Now to prove
i)
H is a constant of motion, differentiate equation (6) with respect to time t we
get
2 pθ pθ
pθ2
dH
sin θθ
=
−
− mga sin θθ .
dt 4ma 2 (1 − cos θ ) 4ma 2 (1 − cos θ )2
Using equations (7) and (8) we readily get
Classical Mechanics
Page No. 195
dH
=0.
dt
This shows that the Hamiltonian H is a constant of motion.
ii)
H represents the total energy
We find from the expressions for kinetic energy and potential energy that
T + V = ma 2θ 2 (1 − cos θ ) + mga (1 + cos θ ) .
. . . (11)
Eliminating θ from equation (11) we get equation (6) that gives the required
expression for the Hamiltonian. Now multiply equation (10) by θ we get
+ ma 2 sin θθ3 − mga sin θθ = 0 .
2ma 2 (1 − cos θ ) θθ
This can be written as
d
 ma 2 (1 − cos θ ) θ 2 + mga (1 + cos θ )  = 0 .
dt
Integrating we get
H = T + V = ma 2θ 2 (1 − cos θ ) + mga (1 + cos θ ) = const.
This shows that the Hamiltonian H represents the constant of total energy.
Example 13 : Obtain the Hamilton’s equation of motion for a one dimensional
harmonic oscillator.
Solution: The one dimensional harmonic oscillator consists of
a mass attached to one end of a spring and other end of the
spring is fixed. If the spring is pressed and released then on
account of the elastic property of the spring, the spring exerts a
force F on the body in the opposite direction. This is called
restoring force. It is found that this force is proportional
F
M
x
to the displacement of the body from its equilibrium position.
F∝x
F = − kx
where k is the spring constant and negative sign indicates the force is opposite to the
displacement. Hence the potential energy of the particle is given by
Classical Mechanics
Page No. 196
V = − ∫ Fdx,
V = ∫ kxdx + c,
V=
kx 2
+ c,
2
where c is the constant of integration. By choosing the horizontal plane passing
through the position of equilibrium as the reference level, then V=0 at x=0. This
gives c=0. Hence potential energy of the particle is
V=
1 2
kx .
2
. . . (1)
The kinetic energy of the one dimensional harmonic oscillator is
T=
1 2
mx .
2
. . . (2)
Hence the Lagrangian of the system is
L=
1 2 1 2
mx − kx .
2
2
. . . (3)
The Lagrange’s equation motion gives
x + ω 2 x = 0, ω 2 =
k
.
m
. . . (4)
This is the equation of motion. ω is the frequency of oscillation.
The Hamiltonian H of the oscillator is defined as
x − L,
H = xp
1
1
x − mx 2 + kx 2 ,
H = xp
2
2
where
px =
p
∂L
= mx ⇒ x = x .
∂x
m
Substituting this in the above equation we get the Hamiltonian
H=
Classical Mechanics
px2 1 2
+ kx .
2m 2
. . . (5)
Page No. 197
Solving the Hamilton’s canonical equations of motion we readily get the equation (4)
as the equation of motion.
Example 14: For a particle the kinetic energy and potential energy is given by
respectively,
T=
1 2
1  r 2 
mr , U = 1 + 2  .
2
r c 
Find the Hamiltonian H and determine
1.
whether H = T + V
2.
whether
dH
=0.
dt
Solution: The kinetic and potential energies of a particle are given by
T=
1 2
1  r 2 
mr , U = 1 + 2 
r c 
2
respectively. The Lagrangian function is therefore given by
L=
1 2 1  r 2 
mr − 1 + 2  .
2
r c 
. . . (1)
We see that the particle has only one degree of freedom and hence it has only one
generalized co-ordinate. The generalized momentum is defined by
pr =
⇒
r =
∂L
2r
= mr − 2 ,
∂r
rc
pr r c 2
.
( mrc2 − 2 )
. . . (2)
Thus the corresponding Hamiltonian function is defined by
H = pr r − L,
1
1  r 2 
H = pr r − mr 2 + 1 + 2  .
2
r c 
. . . (3)
Eliminating r between (2) and (3) we obtain the Hamiltonian H as
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Page No. 198
H=
1.
1 pr2 r c 2
1
+ .
2
2 ( mrc − 2 ) r
. . . (4)
Now the sum of the kinetic and potential energies is given by
T +U =
1 2 1  r 2 
mr + 1 + 2  .
2
r c 
. . . (5)
Eliminating r between (2) and (5) we get
2
2
2
1 pr r c ( mrc + 2 ) 1
T +U =
+ .
2 ( mrc 2 − 2 )2
r
. . . (6)
We see from equations (4) and (6) that the Hamiltonian H does not represent the total
energy.
⇒ T +U ≠ H .
2.
Now differentiating equation (4) w. r. t. time t we get
2
dH
p p rc 2
pr2 rc
= r 2r
−
dt ( mrc − 2 ) ( mrc 2 − 2 )2
−
r
,
r2
. . . (7)
where
2  2r 2

p r = r m− 2 + 2 2 .
rc  r c

Substituting this in equation (7) and simplifying we get
dH
r3
r
= pr r + 2 2 − 2 .
dt
r c r
⇒
dH
≠0.
dt
This shows that the Hamiltonian H is not a constant of motion.
Example 15 : A particle is thrown horizontally from the top of a building of height h
with an initial velocity u. Write down the Hamiltonian of the problem. Show that H
represents both a constant of motion and the total energy.
Classical Mechanics
Page No. 199
Solution:
Let the particle be thrown horizontally from the top of a building of
y
height h with an initial velocity u. The motion
of the particle is in a plane. If P (x, y) are the
position co-ordinates of the particle at any
instant, then its kinetic energy and the potential
P(x, y)
energy are respectively given by
h
y
x
O
T=
1
m ( x 2 + y 2 ) ,
2
V = − mg ( h − y ) .
. . . (1)
. . . (2)
Hence the Lagrangian of the particle becomes
L=
1
m ( x 2 + y 2 ) + mg ( h − y ) .
2
. . . (3)
The particle has two degrees of freedom and hence two generalized co-ordinates. We
see that the generalized co-ordinate x is cyclic in L, hence the corresponding
generalized momentum is conserved.
⇒
∂L
= mx = const.
∂x
∂L
py =
= my.
∂y
px =
. . . (4)
The Hamiltonian function H is defined as
H = ∑ p j q j − L
j
1
H = px x + p y y − m ( x 2 + y 2 ) − mh ( h − y )
2
. . . (5)
Eliminating the velocities from equations (4) and (5) we obtain the Hamiltonian of
motion as
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Page No. 200
H=
1
px2 + p 2y ) − mg ( h − y ) .
(
2m
. . . (6)
The corresponding Hamilton’s canonical equations of motion are
x =
∂H px
∂H p y
=
, y =
=
,
∂px m
∂p y m
p x = −
and
∂H
= 0,
∂x
p y = −
∂H
= − mg.
∂y
Solving these equations we get the equations of motion as
x = 0, y = −g .
. . . (7)
Now differentiating equation (6) with respect to t we get
dH 1
= ( px p x + p y p y ) + mgy
dt m
⇒
dH
= 0,
dt
This proves that H is a constant of motion.
Now to see whether H represents total energy or not, we consider
T +V =
1
m ( x 2 + y 2 ) − mg ( h − y )
2
Putting the values of x and y we obtain the expression for the Hamiltonian as
T +V =
1
px2 + p y2 ) − mg ( h − y ) .
(
2m
. . . (8)
This represents the Hamiltonian H, proving that H represents the total energy of the
particle.
Example 16: A particle is constrained to move on the arc of a parabola x 2 = 4ay
under the action of gravity. Show that the Hamiltonian of the system is
2a 2 px2
mg 2
H=
+
x .
2
2
m ( 4a + x ) 4 a
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Page No. 201
Is the Hamiltonian of the particle representing total energy? Is it a constant of
motion?
Solution: Given that a particle is constrained to move on the arc of the parabola
x 2 = 4ay
. . . (1)
where y is vertical axis, under the action of gravity. The kinetic energy of the particle
is given by
T=
1
m ( x 2 + y 2 )
2
. . . (2)
and the potential energy is given by
V = mgy .
. . . (3)
However, x and y are not the generalized co-ordinates, because they are related by
the constraint equation (1). Eliminating y from equations (2) and (3) on using (1) we
obtain
T=
1 2
x2 
x2
mx 1 + 2  , V =
mg .
2
4a
 4a 
Hence the Lagrangian function is
L=
1 2
x2  x2
mx 1 + 2  − mg .
2
 4a  4a
. . . (4)
Now we see that the system has one degree of freedom and only one generalized coordinate x.
⇒
px =
⇒ x =

∂L
x2 
= mx 1 + 2  ,
∂x
 4a 
4a 2 px
.
m ( 4a 2 + x 2 )
. . . (5)
Now the Hamiltonian H becomes
x −L
H = xp
Classical Mechanics
Page No. 202
x H = xp
1 2
x2  x2
mx  1 + 2  +
mg .
2
 4a  4a
. . . (6)
On using (5) we write equation (6) as
2a 2 px2
mgx 2
+
.
H=
4a
m ( 4a 2 + x 2 )
. . . (7)
This is the required Hamiltonian function. Now to see whether this H represents total
energy or not, we consider
1 2
x2  x2
T+V = mx 1 + 2  +
mg .
2
 4 a  4a
. . . (8)
Using equation (5) we obtain
2a 2 px2
mgx 2
T +V =
+
.
4a
m ( 4a 2 + x 2 )
. . . (9)
Which is the Hamiltonian of the motion, showing that it represent the total energy of
the particle. Now to show that the Hamiltonian H represents constant of motion, we
first find the equation of motion. From equation (4) we have
∂L
m
mgx
= 2 xx 2 −
,
∂x 4a
2a
∂L 
x2 
= 1 + 2  mx.
∂x  4a 
Hence the equation of motion becomes
d  ∂L  ∂L
=0 ⇒
 −
dt  ∂x  ∂x
⇒
( 4a
2
d 
x2   m
mgx
2
= 0,
1 + 2  mx  − 2 xx +
dt  4a   4a
2a
+ x 2 ) x + xx 2 + 2agx = 0 .
. . . (10)
Now differentiating equation (7) with respect to t we get


x2
xxp
dH 4a 2  px p x
 + mg xx .
=
−
2
2
2
dt
m  ( 4 a + x ) ( 4a 2 + x 2 )  2 a


Classical Mechanics
Page No. 203
Eliminating px , p x we obtain
dH
m
= 2 ( 4a 2 + x 2 ) x + xx 2 + 2agx  x .
dt 4a
This implies from equation (10) that
dH
=0.
dt
This shows that the Hamiltonian H is a constant of motion.
Example 17 : Set up the Hamiltonian for the Lagrangian
L ( q, q , t ) =
m 2 2
 q sin ωt + qq sin 2ωt + q 2ω 2  .
2
Derive the Hamilton’s equations of motion. Reduce the equations in to a single
second order differential equation.
Solution: The Lagrangian of the system is given by
L ( q, q , t ) =
m 2 2
 q sin ωt + qq sin 2ωt + q 2ω 2 
2
. . . (1)
The system has only one degree of freedom and hence only one generalized coordinate q. The generalized momentum is given by
p=
⇒ q =
∂L m
= ( 2q sin 2 ωt + qω sin 2ωt )
∂q 2
1 p q

− ω sin 2ωt  .
2

sin ωt  m 2

. . . (2)
. . . (3)
Now the Hamiltonian function H is defined as
H = pq −
m 2 2
q sin ωt + qqω sin 2ωt + q 2ω 2 ) .
(
2
. . . (4)
Substituting the value of q from equation (3) in (4) and simplifying we get
H=
p2
q 2 mω 2
m
−
pq
ω
cot
ω
t
+
cos 2 ωt − q 2ω 2 .
2
2m sin ωt
2
2
. . . (5)
This is the Hamiltonian of the system. The Hamilton’s canonical equations of motion
give
Classical Mechanics
Page No. 204
q =
and
∂H
p
=
− qω cot ωt .
∂p m sin 2 ωt
. . . (6)
p = pω cot ωt − qω 2 m cos 2 ωt + mqω 2 .
. . . (7)
From equation (6) we find
p=
m
 2q sin 2 ωt + qω sin 2ωt 
2
. . . (8)
Differentiating equation (8) w. r. t. t we get
p =
m
 2q sin 2 ωt + 4qω sin ωt cos ωt + qω sin 2ωt + 2qω 2 cos 2ωt  . . . (9)
2
Equating equations (7) and (9) we get
q + 2ω q cot ωt − 2qω 2 = 0 .
. . . (10)
This equation determines the motion of the particle.
Example 18 : A Lagrangian of a system is given by
L ( x, y, x, y ) =
where
m
k
+ cy 2 ) − ( ax 2 + 2bxy + cy 2 ) ,
ax 2 + 2bxy
(
2
2
a, b, c, k , m are constants and b 2 − ac ≠ 0 . Find the Hamiltonian and
equations of motion. Examine the particular cases a = 0, c = 0 and b = 0, c = − a .
Solution: Given that
L ( x, y, x, y ) =
where
m
k
+ cy 2 ) − ( ax 2 + 2bxy + cy 2 ) ,
ax 2 + 2bxy
(
2
2
. . . (1)
a, b, c, k , m are constants and b 2 − ac ≠ 0 . We see that the system has two
generalized co-ordinates x and y. Hence the corresponding generalized momenta are
and
px =
∂L
= m ( ax + by ) ,
∂x
. . . (2)
py =
∂L
= m ( bx + cy ) .
∂y
. . . (3)
Solving these equations for x and y we get
Classical Mechanics
Page No. 205
x = −
cpx − bp y
m ( b − ac )
2
,
y =
bpx − ap y
m ( b 2 − ac )
.
. . . (4)
The Hamiltonian H is defined by
H = ∑ p j q j − L,
j
H = px x + p y y −
m
k
+ cy 2 ) + ( ax 2 + 2bxy + cy 2 ) . ...(5)
ax 2 + 2bxy
(
2
2
Using equations (4) in (5) we obtain after simplifying
H=
1
a
c  k

bpx p y − p y2 − px2  + ( ax + 2bxy + cy 2 ) .. . . (6)

2
2  2
m ( b − ac ) 
2
This is the required Hamiltonian of the system. The Hamilton’s equations of motion
corresponding to two generalized co-ordinates x, y are
∂H
= − k ( ax + by ) ,
∂x
∂H
p y = −
= − k ( bx + cy ) .
∂y
p x = −
. . . (7)
and
x =
1
∂H
=
( bp y − cpx )
2
∂px m ( b − ac )
∂H
1
=
y =
( bpx − cp y )
2
∂p y m ( b − ac )
. . . (8)
From equations (2), (3) and (7) we have
m ( ax + by) + k ( ax + by ) = 0,
m ( bx + cy) + k ( bx + cy ) = 0.
. . . (9)
These are the required equations of motion. Solving these equations for x and y we
obtain respectively
Classical Mechanics
mx + kx = 0.
. . . (10)
my + ky = 0.
. . . (11)
Page No. 206
The solutions of these equations are
and
k
k
x = c1 cos   t + c2 sin   t ,
m
m
. . . (12)
k
k
y = d1 cos   t + d 2 sin   t.
m
m
. . . (13)
Now the cases a = 0, c = 0 and b = 0, c = − a yield from equations (9) the same set
of equations (10) and (11).
Example 19 : The Lagrangian for a system can be written as
L = ax 2 + b
y
+ gy − k x 2 + y 2 ,
+ fy 2 xz
+ czy
x
where a, b, c, f, g and k are constants. What is Hamiltonian? What quantities are
conserved ?
Solution: The Lagrangian of the system is
L = ax 2 + b
y
+ gy − k x 2 + y 2 ,
+ fy 2 xz
+ czy
x
. . . (1)
where a, b, c, f, g and k are constants. The system has three degrees of freedom and
has three generalized co-ordinates (x, y, z), of which z is cyclic. This implies the
corresponding generalized momentum pz is conserved.
⇒
pz =
∂L
= cy + fy 2 x = const.
∂z
. . . (2)
Similarly, we find
and
px =
∂L
= 2ax + fy 2 z ,
∂x
. . . (3)
py =
∂L b
= + cz + g .
∂y x
. . . (4)
Solving these equations for x, y , z we get
Classical Mechanics
Page No. 207
x =
1 
fy 2 
b

p
−
x
 py − − g  ,

2a 
c 
x

1
fy 2 
fy 2 
b
 
y =  pz −
p
−
 x
 py − − g   ,
c
2a 
c 
x
 
1
b

z =  p y − − g  .
c
x

. . . (5)
The Hamiltonian of the system is defined as
H = px x + p y y + pz z − L
p y y+p
z z − ax 2 − b
H = px x+
y
− fy 2 xz
− gy + k x 2 + y 2 .
− cxy
x
. . . (6)
The required Hamiltonian is obtained by eliminating x, y , z from equation (6).
••
Unit 3: Routh’s Procedure :
Introduction:
The presence of cyclic co-ordinates in the Lagrangian L is not much
profitable because even if the co-ordinate q j does not appear in L, the corresponding
generalized momentum q j generally does, so that one has to deal the problem with all
variables and the system has n degrees of freedom. However, if q j is cyclic in the
Hamiltonian then p j is constant and then one has to deal with the problem involving
only 2n-2 variables, i.e., only n-1 degrees of freedom. Hence Hamiltonian procedure
is especially adapted to the problems involving cyclic co-ordinates. The advantage of
Hamiltonian formulation in handling with cyclic co-ordinates is utilized by Routh
and devised a method by combining with the Lagrangian procedure and the method
is known as Routh’s Procedure. The Method is described in the following theorem.
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Page No. 208
Theorem 10: Describe the Routh’s procedure to solve the problem involving cyclic
and non-cyclic co-ordinates.
Proof: Consider a system of particles involving both cyclic and non-cyclic co-
ordinates. Let q1 , q2 , q3 ,..., qs of q1 , q2 , q3 ,..., qn are cyclic co-ordinates, then a new
function R, known as the Routhian is defined as
s
R ( q1 , q2 ,..., qn ; p1 , p2 ,..., ps ; qs +1 , qs + 2 ,..., qn , t ) = ∑ p j q j − L ( q j , q j , t ) . . . (1)
j =1
The Routhian R is obtained by modifying the Lagrangian L so that it is no longer a
function of the generalized velocities corresponding to the cyclic co-ordinates, but
instead involves only its conjugate momentum. The advantage in doing so is that p j
can then be considered one of the constants of integration and the remaining
integrations involve only the non-cyclic co-ordinates.
Now we take R = R ( q1 , q2 ,..., qn ; p1 , p2 ,..., ps ; qs +1 , qs + 2 ,..., qn , t ) , and find the total
differential dR as
n
s
∂R
∂R
dq j + ∑
dp j +
j =1 ∂q j
j =1 ∂p j
n
dR = ∑
∂R
∑ ∂q dq
j = s +1
j
j
+
∂R
dt .
∂t
. . . (2)
Now we consider
s
R = ∑ p j q j − L ( q j , q j , t )
j =1
and find the total differential as
s
s
j =1
j =1
s
s
j =1
j =1
dR = ∑ p j dq j + ∑ q j dp j − dL,
n
n
∂L
∂L
∂L
dq j − ∑
dq j − dt.
j
∂t
j =1 ∂q j
j =1 ∂q
dR = ∑ p j dq j + ∑ q j dp j − ∑
Classical Mechanics
Page No. 209
s
s
 s ∂L
dR = ∑ p j dq j + ∑ q j dp j −  ∑
dq j +
 j =1 ∂q
j =1
j =1
j

 s ∂L
− ∑
dq j +
 j =1 ∂q
j

s
n
∂L
∑ ∂q dq
j = s +1
n
j
j
∂L
∑ ∂q dq
j = s +1
j
j

 −

 ∂L
dt.
 −
∂
t

s
n
∂L
∂L
dq j − ∑
dq j −
j =1 ∂q j
j = s +1 ∂q j
dR = ∑ q j dp j − ∑
j =1
n
−
∂L
∑ ∂q dq
j = s +1
j
j
−
∂L
dt.
∂t
. . . (3)
Now equating the corresponding coefficients on both the sides of equations (2) and
(3) we obtain
∂R
= q j ,
∂p j
and
j = 1, 2,..., s
. . . (4)
∂R
∂L
=−
= − p j ,
∂q j
∂q j
j = 1, 2,..., s
. . . (5)
∂R
∂L
=−
= − p j ,
∂q j
∂q j
j = s + 1, s + 2,..., n
. . . (6)
∂R
∂L
=−
= − pj,
∂q j
∂q j
j = s + 1, s + 2,..., n
. . . (7)
We see that for cyclic co-ordinates q1 , q2 ,..., qs equations (4) and (5) represent
Hamilton’s equations of motion with R as the Hamiltonian, while equations (6) and
(7) for the non-cyclic co-ordinates q j
( j = s + 1, s + 2,..., n )
represent Lagrange’s
equations of motion with R as the Lagrangian function. i.e., from equations (6) and
(7) we obtain
d  ∂R

dt  ∂q j
 ∂R
= 0,
 −
 ∂q j
j = s + 1, s + 2,..., n
. . . (8)
Thus by Routhian procedure a problem involving cyclic and non-cyclic co-ordinates
can be solved by solving Lagrange’s equations for non-cyclic co-ordinates with
Classical Mechanics
Page No. 210
Routhian R as the Lagrangian function and solving Hamiltonian equations for the
given cyclic co-ordinates with R as the Hamiltonian function. In this way The
Routhian has a dual character Hamiltonian H and the Lagrangian L.
Worked Examples •
Example 20 : Find Lagrangian L, Hamiltonian H and the Routhian R in spherical
polar co-ordinates for a particle moving in space under the action of conservative
force.
Solution: Let a particle be moving in a space. If (x, y, z) are the Cartesian co-
ordinates and ( r ,θ , φ ) are the spherical co-ordinates of the particle, then we have the
relation between them as
x = r sin θ cos φ ,
y = r sin θ sin φ ,
. . . (1)
z = r cos θ .
The kinetic energy T =
1
m ( x 2 + y 2 + z 2 )
2
of the particle, in spherical polar co-
ordinates becomes
T=
1
m r 2 + r 2θ 2 + r 2 sin 2 θφ2 .
2
(
)
. . . (2)
Since the force is conservative, hence the potential energy of the particle is the
function of position only.
⇒
V = V ( r ,θ , φ ) .
. . . (3)
Hence the Lagrangian function of the particle becomes
L=
1
m r 2 + r 2θ 2 + r 2 sin 2 θφ2 − V ( r , θ , φ ) .
2
(
)
. . . (4)
We see that φ is cyclic in L, hence the corresponding generalized momentum is
conserved. i.e.,
Classical Mechanics
Page No. 211
pφ =
∂L
= mr 2 sin 2 θ φ = const.
∂φ
. . . (5)
Similarly we find
∂L
= mr,
∂r
∂L
pθ =
= mr 2θ.
∂θ
pr =
. . . (6)
Now the Hamiltonian function is defined as
H = ∑ p j q j − L,
j
1
H = pr r + pθ θ + pφ φ − m r 2 + r 2θ 2 + r 2 sin 2 θφ2 + V .
2
(
)
. . . (7)
Eliminating the generalized velocities r,θ, φ between equations (5), (6) and (7) we
get
H=
1  2 1 2
1
2
 pr + 2 pθ + 2 2 pφ  + V .
2m 
r
r sin θ

. . . (8)
Now the Routhian R is defined by
R = pφ φ − L,
. . . (9)
This becomes after eliminating r,θ, φ between (5), (6) and (9) we get
(
)
R r , θ , φ , r, θ, t =
pφ2
1
− m r 2 + r 2θ 2 + V .
2mr sin θ 2
2
2
(
)
. . . (10)
Example 21 : A planet moves under the inverse square law of attractive force, Find
Lagrangian L, Hamiltonian H, and the Routhian R for the planet.
Solution: A motion of a planet is a motion in the plane. If ( r ,θ ) are the generalized
co-ordinates of the planet then it’s kinetic and potential energies are respectively
given by
T=
Classical Mechanics
1
K
m r 2 + r 2θ 2 , V = − .
2
r
(
)
Page No. 212
Hence the Lagrangian function is defined by
L=
1
K
m r 2 + r 2θ 2 + .
2
r
(
)
. . . (1)
We see that θ is the cyclic co-ordinate in L. This implies that the corresponding
angular momentum of the planet is conserved.
Also
pθ =
p
∂L
= mr 2θ = const. ⇒ θ = θ 2 .
∂θ
mr
. . . (2)
pr =
∂L
p
= mr ⇒ r = r
∂r
m
. . . (3)
Now the Hamiltonian function is defined as
H = ∑ p j q j − L,
j
1
K
H = pr r + pθ θ − m r 2 + r 2θ 2 −
2
r
(
)
On using equations (2) and (3) we obtain
H=
1  2 pθ2  K
 pr + 2  − .
r  r
2m 
. . . (4)
This is the required Hamiltonian.
Now the Routhian is defined as
R ( r , θ , pθ , r, t ) = pθθ − L
1
K
R ( r , θ , pθ , r, t ) = pθθ − m r 2 + r 2θ 2 − .
2
r
(
)
Eliminating θ we get
R ( r ,θ , pθ , r, t ) =
pθ2
1
K
− mr 2 − .
2
2mr
2
r
. . . (5)
pθ2
pr2 K
−
− .
2mr 2 2m r
. . . (6)
This can also be written as
R ( r ,θ , pθ , r, t ) =
Classical Mechanics
Page No. 213
•
Principle of Least Action :
Action in Mechanics :
In Mechanics the time integral of twice the kinetic energy is called the action.
Thus
t1
A = ∫ 2Tdt
t0
is called the action.
t1
A = ∫ ∑ p j q j dt
i.e.
j
t0
is called action in Mechanics.
Principle of Least Action :
There is another variational principle associated with the Hamiltonian
formulation and is known as the principle of least action. It involves a new type of
variation which we call the ∆ - variation.
In ∆ - variation the co-ordinates of the end points remain fixed while the time is
allowed to vary. The varied paths may terminate at different points, but still position
co-ordinates are held fixed.
Mathematically, we have
δI =
∂I
dα ,
∂α
∆I =
dI
dα .
dα
Thus for the family of paths represented by the equation
q j = q j (α , t ) , t = t (α )
We have
∆q j =
Classical Mechanics
 ∂q
dt 
dα =  j + q j
 dα .
dα
dα 
 ∂α
dq j
Page No. 214
 ∂q j

dt
∆q j = 
dα + q j
dα 
dα
 ∂α
.
∆q j = δ q j + q j ∆t
This shows that the total variation is the sum of two variations.
Worked Examples –•
Example 22 : If f = f ( q j , q j , t ) then show that
∆f = δ f + ∆t ⋅
df
.
dt
Solution: Consider a system of particles moving from one point to another. Let the
family of paths between these two points be given by
q j = q j ( t ,α ) .
. . . (1)
In ∆ variation time is not held fixed, it depends on the path. This implies that
Since
t = t (α )
. . . (2)
f = f ( q j , q j , t )
. . . (3)
then we find ∆ variation in f as
However, we have
 ∂f
 ∂f
∂f
∆f = ∑ 
∆q j +
∆q j  + ∆t .

 ∂t
∂q j
j  ∂q j

. . . (4)
∆q j = δ q j + q j ∆t
. . . (5)
Similarly we find
∆q j = δ q j + qj ∆t ,
. . . (6)
Using equations (5) and (6) in equation (4) we get
 ∂f
 ∂f
∂f
∆f = ∑ 
δ q j + q j ∆t ) +
δ q j + qj ∆t )  + ∆t .
(
(

 ∂t
∂q j
j  ∂q j

Classical Mechanics
Page No. 215
 ∂f
 ∂f
∂f
∂f 
∂f
∂f 
∆f = ∑ 
δ qj +
δ q j + δ t  + ∑ 
q j +
qj + ∆t

∂q j
∂t  j  ∂q j
∂q j
∂t 
j  ∂q j
Note here that the term δ t added because it is zero, since in δ variation time t is
held fixed and consequently change in time t is zero. This can be written as
∆f = δ f + ∆t ⋅
df
.
dt
. . . (7)
Since f is arbitrary, we can write it as
∆ = δ + ∆t ⋅
d
.
dt
. . . (8)
••
Theorem 11 : For a conservative system for which the Hamiltonian H is conserved,
the principle of least action states that
t1
∆ ∫ ∑ p j q j dt = 0 .
j
t0
Proof: Consider a conservative system for which the Hamiltonian H is conserved.
Let AB be the actual path and CD be the varied path. In ∆ - variation the end points
of the two paths are not terminated at the same point. The end points A and B after
∆t take the positions C and D such that the position co-ordinates of A, C and B, D
are held fixed. Now we know the action is given by
t1
y
A = ∫ ∑ p j q j dt
B
t0
D
j
t1
A = ∫ ( L + H )dt
t0
t1
A = ∫ Ldt + H ( t )t1 ,
t
0
A
O
t0
C
t0 t0 + ∆ t0
Classical Mechanics
t1 t1 + ∆ t1
x
Page No. 216
t1
A = ∫ Ldt + H ( t1 − t0 ) .
. . . (1)
t0
Thus
t1
∆A = ∆ ∫ Ldt + H ( ∆t )t1 .
t
. . . (2)
0
t0
Since time limits are also subject to change in ∆ -variation, therefore ∆ can’t be
taken inside the integral. Let
t1
∫ Ldt = I
⇒ I = L .
t0
Therefore
∆I = δ I + I∆t .
Thus we have
t1
t1
t0
t0
∆ ∫ Ldt = δ ∫ Ldt + L ( ∆t )t1 .
t
0
t1
t1
  ∂L
 ∂L 
∂L
t
δ qj +
δ q j  + δ t  dt + L ( ∆t )t1 .
∆ ∫ Ldt = ∫  ∑ 
0


∂q j
t0
t0 
 ∂t 
 j  ∂q j
Since in δ variation, time is held fixed along any path, hence there is no variation in
time, therefore change in time is zero. Thus we have
 ∂L

∂L
t
∆ ∫ Ldt = ∫ ∑ 
δ qj +
δ q j dt + L ( ∆t )t1 .


0
∂q j
∂q j
t0
t0 j 

t1
t1
Using Lagrange’s equations of motion we write this equation as
t1
t1
t0
t0
∆ ∫ Ldt = ∫ ∑ ( p jδ q j + p jδ q j )dt + L ( ∆t )t1 .
t
0
j
Since
δ
Classical Mechanics
dq j
dt
=
d
δ qj .
dt
Page No. 217
Hence we have
t
t
t1
t1
t0
t0
1
1
d
t


∆ ∫ Ldt = ∫ ∑  p jδ q j + p j δ q j dt + L ( ∆t )t1 .
0
dt

t0
t0 j 
⇒
∆ ∫ Ldt = ∫

d 
t1
 ∑ ( p jδ q j )  dt + L ( ∆t )t0 .
dt  j

Since
∆ = δ + ∆t
d
dt
Hence above integral becomes
t1
t1

d  
t

∆ ∫ Ldt = ∫ d  ∑ p j  ∆ − ∆t q j  dt + L ( ∆t )t1 .
0
dt  

 j
t0
t0
t1
t1

 

t
∆ ∫ Ldt =  ∑ p j ∆q j  −  ∑ p j q j ∆t  + L ( ∆t )t1
0
 j
 t0  j
 t0
t0
t1
Since in ∆ variation, position co-ordinates at the end points are fixed.
⇒
( ∆q )
t1
j t
0
= 0.
Consequently above equation reduces to
t1


∆ ∫ Ldt = − ∑ ( p j q j − L ) ∆t 
t0
 j
 t0
t1
t1
∆ ∫ Ldt = − ( H ∆t )t1
t
0
t0
Substituting this in equation (2) we get
∆A = 0,
t1
i.e., ∆ ∫ ∑ p j q j dt = 0
t0
.
j
Thus the system moves in space such that ∆ -variation of the line integral of twice
the kinetic energy is zero. This proves the principle of least action.
Classical Mechanics
Page No. 218
Example 23 : A system of two degrees of freedom is described by the Hamiltonian
H = q1 p1 − q2 p2 − aq12 + bq22 , a, b are const.
Show that i)
p1 − aq1
,
q2
ii )
p2 − bq2
,
q1
iii ) q1q2
iv) H are constant of motion.
Solution: The Hamiltonian of a dynamical system is given by
H = q1 p1 − q2 p2 − aq12 + bq22 , a, b are const.
. . . (1)
where we see that q1 , q2 are the generalized co-ordinates. The Hamilton’s canonical
equations of motion are
p j = −
∂H
⇒
∂q j
p1 = 2aq1 − p1 ,
. . . (2)
p 2 = p2 − 2bq2 ,
and
q j =
∂H
⇒ q1 = q1
∂p j
. . . (3)
q2 = − q2 .
Now to show
1)
p1 − aq1
is a constant of motion, consider
q2
d  p1 − aq1  q2 ( p1 − aq1 ) − ( p1 − aq1 ) q2
.

=
dt  q2 
q22
Using equations (2) and (3) we obtain
d  p1 − aq1 

=0⇒
dt  q2 
p1 − aq1
= const.
q2
Similarly we prove that
p1 − aq1
= const.,
q2
Classical Mechanics
p2 − bq2
= const.,
q1
q1q2 = const.
Page No. 219
Now to prove the Hamiltonian H is also constant, we differentiate equation (1) with
respect to t to get
dH
= q1 p1 + q1 p1 − q2 p2 − q2 p 2 − 2aq1q1 + 2bq2 q2 .
dt
Using equations (2) and (3) we see that
dH
= 0 ⇒ H = const. .
dt
This shows that H is a constant of motion.
Example 24 : A Lagrangian for a particle of charge q moving in the electromagnetic
field of force is given by
L=
1 2
mv + q ( v ⋅ A ) − qφ .
2
Find the Hamiltonian H, the generalized momenta.
Solution: The Lagrangian of a particle moving in the electromagnetic field is given
by
L=
1 2
mv + q ( v ⋅ A ) − qφ .
2
. . . (1)
We write this expression as
L=
1
x + yA
y + zA
z ) − qφ .
m ( x 2 + y 2 + z 2 ) + q ( xA
2
. . . (2)
where φ is a scalar potential function of co-ordinates only. We see that x, y, z are
the generalized co-ordinates. Hence the corresponding generalized momenta become
pj =
∂L
∂q j
⇒
px = mx + qAx ,
p y = my + qAy ,
pz = mz + qAz .
Solving these equations for velocity components we get
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Page No. 220
1
( px − qAx ) ,
m
1
y = ( p y − qAy ) ,
m
1
z = ( pz − qAz ) .
m
x =
. . . (3)
The Hamiltonian of the particle is given by
H = ∑ p j q j − L,
j
1
x + yp
y + zp
x + yA
y + zA
z − m ( x 2 + y 2 + z 2 ) − q ( xA
z ) + qφ . . . (4)
H = xp
2
Eliminating x, y , z from equation (4) by using equation (3) we get
H=
1
q
1 2 2
px2 + p y2 + pz2 ) − ( px Ax + p y Ay + pz Az ) +
q ( Ax + Ay2 + Az2 ) + qφ . . (5)
(
2m
m
2m
This can be written in vector notions as
H=
2
1
p − qA ) + qφ .
(
2m
. . . (6)
This is the required Hamiltonian of the particle moving in the electromagnetic field.
The Hamilton’s equation of motion q j =
while the equation p j = −
p x = −
∂H
gives the same set of equations (3),
∂p j
∂H
gives
∂q j
∂H q ∂
q2 ∂
∂φ
=
p
A
+
p
A
+
p
A
−
Ax2 + Ay2 + Az2 ) − q
.
(
(
x x
y y
z z)
∂x m ∂x
2m ∂x
∂x
This can be written as
p x = q
∂
∂φ
v ⋅ A) − q
.
(
∂x
∂x
Similarly, other two components are given by
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Page No. 221
∂
∂φ
v ⋅ A) − q ,
(
∂y
∂y
∂
∂φ
p z = q ( v ⋅ A ) − q .
∂z
∂z
p y = q
All these three equations can be put in to the single equation as
p = − q∇φ + q∇ ( v ⋅ A ) .
. . . (7)
Exercise:
1.
The Lagrangian of an anharmonic oscillator of unit mass is
1 2 1 2 2
x − ω x − α x3 + β xx , α , β are constants.
2
2
L=
Find the Hamiltonian and the equation of motion. Show also that
(i)
H is a constant of motion
(ii)
H ≠ T +V .
Ans : H =
and
1
1
2
( px − β x ) + ω 2 x 2 + α x 3 .
2
2
Equation of motion x + ω 2 x + 3α x 2 = 0 .
2.
Find the Hamiltonian and the equations of motion for a particle constrained to
move on the surface obtained by revolving the line x = z about z axis. Does
it represent the constant of motion and the constant of total energy?
Hint: Surface of revolution is a cone x 2 + y 2 = z 2
pφ2
pr2
Ans.: H =
+
+ mgr .
4m 2mr 2
r−
pφ2
2 3
2m r
Classical Mechanics
+
g
= 0,
2
pφ = mr 2φ − a const. of motion .
Page No. 222
3.
Let a particle be moving in a field of force given by
1  r 2 − 2rr 
F = 2 1 −
.
r 
c2 
Find the Hamiltonian H and show that it represents the constant of motion
and also total energy.
Ans. : Refer Example (25) of Chapter I; the potential energy of the particle is
given by
1  r 2 
V = 1 + 2  .
r c 
The Hamiltonian becomes H =
4.
pr2
2 

2 m − 2 
rc 

+
1
.
r
A sphere of radius ‘a’ and mass m rests on the top of a fixed trough sphere of
radius ‘b’. The first sphere is slightly displaced so that it rolls without
slipping. Obtain the Hamiltonian of the system and hence the equation of
motion. Also prove that H represents a constant of motion and also total
energy.
Ans. : H =
5.
7
2
m ( a + b ) φ2 + mg ( a + b ) cos φ .
10
A particle is constrained to move on the plane curve xy = c , c is a constant,
under gravity. Obtain the Hamiltonian H and the equations of motion. Prove
that the Hamiltonian H represents the constant of motion and total energy.
Ans. : Refer Example (20) of Chapter I for the Lagrangian L and is given by
L=
1 2  c 2  mgc
mx 1 + 4  −
.
2
x
 x 
The Hamiltonian H becomes H =
Classical Mechanics
px2
 c 
2m  1 + 4 
 x 
2
+
mgc
.
x
Page No. 223
6.
A body of mass m is thrown up an inclined plane which is moving
horizontally with constant velocity v. Use Hamilton’s procedure to find the
equations of motion. Prove that the Hamiltonian H represents the constant of
motion but does not represent the total energy.
Ans. : For the Lagrangian function, refer Example (28) of Chapter I. The
Hamiltonian of motion is
H=
7.
pr2
1

− pr v cos θ + mgr sin θ −  mv 2 sin 2 θ  .
2m
2

A particle moves on the surface characterized by
x = r cos φ ,
y = r sin φ , z = r cot θ .
Find the Hamiltonian H and prove that it represents the constant of motion
and also the constant of total energy.
2
p
p 2 sin 2 θ
+ φ 2 + mgr cot θ .
Ans. : H = r
2m
2mr
The equation of motion is r − r sin 2 θφ2 + g cos θ sin θ = 0 .
8.
Find the Hamiltonian and the Hamilton’s canonical equations of motion for
the Lagrangian given by
1
1
2
L r , r, θ , θ = m r 2 + r 2θ 2 + mgr cos θ − k ( r − r0 ) ,
2
2
(
)
(
)
where k , m, g , r0 are constants.
Ans : H =
p2
pr2
1
2
+ θ 2 − mgr cos θ + k ( r − r0 )
2m 2mr
2
Equations of motion:
mr − mrθ 2 − mg cos θ + k ( r − r0 ) = 0,
2
r
θ + r θ +
g
sin θ = 0.
r
••
Classical Mechanics
Page No. 224