Unit 2 Hamilton’s Equations of Motion

UNIT 2
HAMILTON’S EQUATIONS
OF MOTION
Structure

2.1 Introduction 2.4 Important Properties and

Expected Learning Outcomes Conservation Laws
2.2 Legendre Transformation 2.5 The Routhian
2.3 Hamilton’s Function 2.6 Summary
Constructing the Hamiltonian 2.7 Terminal Questions
2.8 Solutions and Answers

2.1 INTRODUCTION
In your first course on Classical Mechanics in Semester I you have studied the
Lagrangian formulation of mechanics and applied it to several problems
ranging from the simple pendulum to special relativity. As you know now, the
solution to mechanical problems with holonomic constraints in the Lagrangian
formalism hinges on the solution of the second order differential equations,
which are the Euler Lagrange (E-L) equations, for generalized coordinates
coupled with appropriate initial conditions. The central function in this
formalism is the Lagrangian, which is a function of the generalized coordinates
and their time derivatives, as well as time. Most practical problems can be
solved using this formalism.
In the first half of the nineteenth century, Sir William Rowan Hamilton (in
1835) developed his own analytic version of mechanics, the third formulation,
after the Newtonian and Lagrangian formulation. The central quantity in
Hamiltonian dynamics is the Hamiltonian of the mechanical system which
unlike the Lagrangian, is a function of the generalized coordinates, the
conjugate momenta and time. When the formalism was first introduced, it
seemed that there was nothing more to be gained from this new formalism
which was not particularly superior to the Lagrangian formalism. It was only
later that it was realized that the importance of this formalism is that is serves
as a theoretical framework for several areas of physics, like field theories,
quantum mechanics, the path-integral formulation of quantum mechanics and
others.
In Sec. 2.2 we introduce an important mathematical technique called the
Legendre transformations which is used to establish the connection between
the Hamiltonian and the Lagrangian for a mechanical system. In the next
section, that is Sec. 2.3 we derive Hamilton’s equations of motion, which can 31
Block 1 Hamiltonian Dynamics
be used to study mechanical systems. In Sec. 2.4 we discuss the important
properties of the Hamiltonian formulation and discuss how it leads to the
important conservation laws of mechanics. After working out the problems in
these sections you will see that the Hamiltonian formalism does not in general,
give us any advantage over the Lagrangian formalism. The Hamiltonian
formalism is however useful when one has cyclic coordinates. In that case one
uses a method called Routh’s procedure to solve the dynamical problem, and
you shall study about that in Sec. 2.5.
You will work out a lot of problems in this unit to understand the formalism. We
will use examples of dynamical systems for which you have already derived
the Lagrangian and the E-L equations of motion in Blocks 1 and 2 of MPH-002
and are familiar with the generalized coordinates of these systems. We will
provide the references where required.

Expected Learning Outcomes

After studying this unit, you should be able to:
 carry out a Legendre transformation;
 construct the Hamiltonian for a dynamical system;
 derive Hamilton’s equations of motion for a dynamical system;
 identify the conserved quantities; and
 construct the Routhian for particular problems.

2.2 LEGENDRE TRANSFORMATION

The Legendre transformations are a generic method for transforming a
function f  f (x ) of some variable x into a new function g  g (y ) of a new
variable y. If you were to carry out this transformation twice, you would end up
with the original function in the original variable. So:
Legendre Transforma tion Legendre Transforma tion
f ( x )          g ( y )          f ( x )
(2.1)
Note that this transformation is also for multivariable functions, so using a
Legendre transformation, we can, in general, transform from a function
f  f (u1, u2 ....., un ) which is a function of the independent variables
u1, u2 ....., un to a function g  g (v1, v 2 ,....., v n ) which is a function of the
independent variables v1, v 2 ,....., v n .

Let us now consider a known function f  f ( x, y ) of two independent variables

x and y. We can write the differential of the function f ( x, y ) as
f f
df  dx  dy (2.2)
x y y x

f
Where is the derivative of f with respect to x, holding y constant and
x y
f f
is the derivative of f with respect to y, holding x constant. Writing u 
y x x
f
and v  , we can also write this as:
y
32 df  udx  vdy (2.3)
Unit 2 Hamilton’s Equations of Motion
Now consider a new independent variable z and a function g  g ( x, y , z ) :

g  yz  f ( x, y ) (2.4)
Now, the differential of this function g  g ( x, y , z ) is:
dg  ydz  zdy  df ( x, y )
f f
 ydz  zdy  dx  dy
x y y x

 f  f
 ydz   z  dy 
 dx (2.5)
 y x xy

Suppose now that we wish to eliminate the explicit dependence on the

independent variable y in the function g  g ( x, y , z ) . This is equivalent to
saying that dg should not have a term containing dy. In that case, we must
have:
f
z (2.6)
y x

Then
f
dg   dx  ydz (2.7)
x y

The function g is only a function of the independent variables x and z,

g  g ( x, z ) , so the differential can be also written as:
g g
dg  dx  dz (2.8)
x z z x

Comparing Eqs. (2.7) and (2.8) we get the following conditions:

g g f
y ;   (2.9)
z x x z x y

Using Eq. (2.9), we can derive y  y ( x, z ) as a function of x and z . So the

function g  g ( x, z )  y ( x, z )z  f x, y ( x, z ) is a function of x and z only. We call
x the passive variable in this transformation and y the active variable.
f g
The relations involving the active variable are z  and y  and the
y x z x
relation satisfied by the passive variable x which appears in both the functions
g f
is   .
x z x y

So, a function f  f ( x, y ) of two independent variables x and y can be

transformed to a function which has the form g  g ( x, z )  yz  f ( x, y ) . g ( x, z )
has the independent variables x and z and the conditions of Eq. (2.9) would
hold. This process is completely invertible and if we carry out the Legendre
transformation of g  g ( x, z ) for the variable z, we would get back f  f ( x, y ) .

We now state the following:

Let a function f  f (u1, u2 ....., un , w1, w 2 ,....w m ) have an explicit dependence on

the n+m independent variables u1, u 2 ....., u n , , w1 , w 2 ,....w m . Let the function be 33
Block 1 Hamiltonian Dynamics
transformed to another function g  g (v1, v 2 ,....., v n ,, w1, w 2 ,....w m ) which is an
explicit function of the n+m independent variables v1, v 2 ,....., v n , w 1, w 2 ,....w m .
u1, u2 ....., un are the n active variables and w1, w 2 ,....w m are the m passive
variables. The new variables are then connected to the old variables by the
relations:
f
vi  for i  1,2,....n (2.10)
ui

And the form of the new function is:

n
g (v1, v 2 ,....., v n , w1, w 2 ,....w m )   ui v i  f (u1, u2 ,....., un , w1, w 2 ,....w m )
i 1
(2.11)
Further, the passive variables w1, w 2 ,....w m satisfy the relation:

g f
  (2.12)
w i w i

Then we can say that the variables u1, u2 ,....., un satisfy the relations:

g
ui  for i  1,2,....n (2.13)
v i

And
n
f (u1, u 2 ,....., u n , w1, w 2 ,....w m )   u i v i  g (v1, v 2 ,....., v n , w1, w 2 ,....w m )
i 1
(2.14)
Legendre transformations are used extensively in physics. In the next section
we show how it is used in classical mechanics, but it is also used in
thermodynamics to transform from the internal energy to different
thermodynamic potentials like the enthalpy and the Gibb’s free energy. You
may like to work out a few problems to understand Legendre transformations.

SAQ 1
Obtain the Legendre transforms for the functions:
a) f ( x )  x 2 to g (z )
b) f ( x )  ln x to g (z )
c) f ( x )  e x to g (z )
d) f ( x, y )  2 x 2  3 xy  y 2 to g (u, v )

2.3 HAMILTON’S FUNCTION

The Lagrangian, as you know is a function of the generalized coordinates,
generalized velocities and time, So it is an explicit function of the 2n+1
variables q1, q2 ....., qn , q1, q 2 ....., q n and t , and
L  L(q1, q 2 ....., q n , q1, q 2 ....., q n , t ) .

Now suppose that starting from the Lagrangian we wish to generate a function
34 of the generalized coordinates, the generalized momenta (the conjugate or
Unit 2 Hamilton’s Equations of Motion
canonical momenta) and time: q1, q 2 ....., qn , p1, p2 ....., pn and t . Let us pre-
emptively denote this new function by H  H (q1, q 2 ....., qn , p1, p2 ....., pn , t )
L
where pi  , as you have studied in Unit 4 of MPH-002. Let us use the
q i
method of Legendre transformations to do this:

L(q1, q 2 ....., q n , q1, q 2 ....., q n , t )

Legendre Transforma tion
        
 H (q1, q 2 ....., q n , p1, p2 ....., pn , t )

For the purpose of carrying out the Legendre transformation the generalized
coordinates q1, q2 ....., qn and t are the passive variables (since both L and H
functions contain the generalized coordinates and time) and given the relation
connecting the active variables q i to pi are (relate to Eq. 2.10 with q i as ui
and pi as v i ):
L
pi  for i  1,2,....n (2.15)
q i

We can write the function H  H (q1, q2 ....., qn , p1, p2 ....., pn , t ) using Eq. (2.11)
as:
n
H   q i pi  L (2.16)
i 1

The dual transformation (Eq. 2.13) gives us:

H
q i  for i  1,2,....n (2.17)
pi

The following relations hold for the passive variables (Eq. 2.12):
L H
 (2.18)
t t
L H
 (2.19)
qi qi

Assuming that the Euler-Lagrange (E-L) equations are valid we can write:
L d  L 
    p i (2.20)
qi dt  q i 

Therefore Eq. (2.19) reduces to:

H
p i   for i  1,2,....n (2.21)
qi

Eqs. (2.17) and (2.21) constitute Hamilton’s equation of motion for a

mechanical system for Hamilton’s function H or the Hamiltonian as it is often
n
called, given by (Eq. 2.16) H   q i pi  L .
i 1

Notice that while in the Lagrangian formulation you had one second order
differential equation for each generalized coordinate, here you have two
coupled first order differential equations for each pair of generalized
coordinates and the corresponding conjugate(or canonical) momentum. 35
Block 1 Hamiltonian Dynamics
So we have shifted from the Lagrangian description of the mechanical system
in terms of the generalized coordinates q1, q2 ....., qn and generalized velocities
q1, q 2 ....., q n to a description of the system terms of q1, q2 ....., qn and momenta
p1, p2 ....., pn using the Hamiltonian and the following set of relations 2n  1
which are called the canonical equations of Hamilton or the canonical
equations of motion:
H
q i  (2.22a)
pi

H
p i   (2.22b)
qi

L H
  (2.22c)
t t
Eqs. (2.22a) and (2.22b) give us q i and p i in terms of qi , pi and t .
Description of a mechanical system using these equations of motion is called
Hamiltonian dynamics.

Note further, that while the Hamiltonian H defined in Eq. (2.16) and the Jacobi
integral h defined in Unit 1, have the same value, they are different functions
because h is a function of qi , q i and t , H is a function of qi , pi and t .

Typically the Hamiltonian for a physical system is constructed from it’s

Lagrangian. Now we list the steps involved in this process.

2.3.1 Constructing the Hamiltonian

To obtain the Hamiltonian for any physical system we carry out the following
steps:
i) Identify a set of generalized coordinates for the system and write the
Lagrangian L  T  V  of the system, which will be a function of
qi , q i and t .
n
ii) Using Eq. (2.16) write down the Hamiltonian H   q i pi  L(q i , q i , t ) . At this
i 1
stage, H is a function of qi , q i , pi and t .

L
iii) Using the equations pi  , obtain the relation between the conjugate
q i
momenta pi and q i . Invert these relations to write the q i as a function of
qi , pi and t .

iv) Insert the q i as a function of qi , pi and t in the expression for the

Hamiltonian obtained in Step ii, to get the Hamiltonian as a function only of
qi , pi and t .

v) Once the Hamiltonian is obtained as a function of qi , pi and t , the

canonical equations can be obtained and solved.

In the following examples, we apply this method to two problems.

36 Let us now work out a familiar example using this method.

Unit 2 Hamilton’s Equations of Motion

Example 2.1
Obtain the Hamiltonian of the system and the corresponding Hamilton’s
equations for the following:
a) A particle of mass m with a single generalized coordinate x with the
following Lagrangian:
1 2 1 2
L mx  kx  ax 4  bx 2 x 2
2 2
b) An Atwood’s machine consisting of two blocks of masses m1 and m2 ,
connected by an inextensible string of length l (Fig. 2.15, Unit 2, MPH-002)
with the following Lagrangian
1
L ( m1  m2 ) y12  ( m1  m2 ) gy 1  m2 gl
2
Solution : a) Using Eq. (2.15) we can write the canonical momentum as:
L
p x  p(say )   m x  2 bx 2 x
x

We can invert this relation to write x as a function of p:

A O B
p
x (m  2bx 2 )  p or x  (i) x
m  2bx 2 a
The Hamiltonian function is (Eq. 2.16): y2

H  px  L y1
m2
1 1
 px   mx 2  kx 2  ax 4  bx 2 x 2  y
2 2 
m1
1 1
 px  x mx  2bx 2 x   kx 2  ax 4
2 2
Fig. 2.1: Atwood’s
1 1 machine
 px  x p  kx 2  ax 4
2 2
1  1 2
 px  kx  ax 4 (ii)
2 2

Using Eq. (i) for p in Eq. (ii) we get

1 p2 1
H  kx 2  ax 4 (iii)
2
2 ( m  2bx ) 2

Using Eqs. (2.22a) and (2.22b) we get the Hamilton’s equation of motion.
H p
x  
p (m  2bx 2 )

H p2
p    ( 4bx )  2kx  4ax 3
x 2( m  2bx 2 ) 2

 2bxp 2
  kx  4ax 3
2 2
(m  2bx ) 37
Block 1 Hamiltonian Dynamics
1
b) With L  ( m1  m 2 ) y 12  ( m1  m 2 ) gy 1  m 2 gl
2
The conjugate momentum for the coordinate y 1 is
L p1
p y1  p1   ( m1  m 2 ) y 1  y 1  (i)
y 1
 m1  m 2

The Hamiltonian is [using Eq. (i)]

H  p1y 1  L

1 
 p1y 1   ( m1  m 2 ) y 12  ( m1  m 2 ) gy 1  m 2 gl 
2 
p12 1 p1 2
   (m1  m 2 ) gy 1  m 2 gl
( m1  m 2 ) 2 ( m1  m 2 )

p12
  ( m1  m 2 ) gy 1  m 2 gl
2( m1  m 2 )

The equations of motion are

H p1
y 1  
p1 m1  m 2
H
p 1     ( m1  m 2 )g   (m1  m 2 ) g
y 1
O
x
 You can now work out the following SAQ.
L0

SAQ 2
m
a) A simple pendulum (Fig. 2.2 )with a bob of mass m tied to a support by
string of constant length L0 has the Lagrangian
y
1
L()  m L0 2  2  mg L0 cos  where  is the generalized coordinate
Fig. 2.2: Simple 2
pendulum. (Example 3.3a, Unit 3, MPH-002). Obtain the Hamiltonian and derive
Hamilton’s equation of motion.
1
b) The Lagrangian of a system in two dimensions is L  mx 2  mx y .
2
Determine the Hamiltonian
c) Consider a spherical pendulum of mass m and length L0 . The Lagrangian
is written in terms of the generalized coordinates  and  (TQ 3, Unit 3,
MPH-002) as:

L
1
2
 
mL20  2   2 sin 2   mg L0 sin 

Obtain the Hamiltonian and derive Hamilton’s equation of motion.

d) The Hamiltonian for a system is
ap 2q 4 b
H 
2 q2
where q and p are the generalized coordinates and momentum
respectively. Determine L(q, q ) .
38
Unit 2 Hamilton’s Equations of Motion
Recollect that in Unit 3 of MPH-002, you have derived the Lagrangian for an
electromagnetic field. In the following example you derive the Hamiltonian for
the electromagnetic field.
Example 2.2
Consider the following Lagrangian for a charged particle of charge q and mass
m moving in an electromagnetic field:
1  
L  mv 2  q  q A .v
2

where  and A are the electrodynamic potentials of the field:
A  A(r , t ) ;   (r , t ) (Example 3.4, Unit 3, MPH-002). Derive the Hamiltonian H,
and obtain the Hamiltonian equations of motion.

Solution : In Cartesian coordinates

L
1
2
 
m x 2  y 2  z 2  q  q Ax x  Ay y  Az z 

The canonical momenta are:

L
p x  L / x  mx  qAx ; p y   my  q Ay ;
y
L
pz   mz  q Az
z
Note that the canonical momentum is different from the usual form p x  mx , etc.

p  q Ax  py  q Ay  pz  q Az
x  x ;y  ;z 
m m m
H  px x  py y  pz z  L

( p x  qAx ) ( py  qAy ) ( p  qAz )

 px  py  pz z
m m m
1 1 1
 ( p x  qA x )2  ( p y  qAy )2  (p z  qA z )2  q
2m 2m 2m
( p x  qA x ) qA y ( p y  A y ) qA z ( p z  qA z )
 qA x  
m m m
2
( px  qAx )2 ( py  qAy ) ( p  qAz )2
   z  q
2m 2m 2m
 
( p  qA )2
  q
2m
Hamilton’s equations of motion are
H p  qAx py  qAy p  qAz
x   x ; y  ; z  z
px m m m

and
 H  q A
p x   q  p x  qA x  x
x x m x
39
Block 1 Hamiltonian Dynamics
 H  q Ay
p y   q  py  qAy 
y y m y
 H  q A
p z   q  pz  qAz  z
z z m z
Since H has an explicit time-dependence (, A) , it is not conserved in general. It
is conserved only for static fields.

You can also study the dynamics of a charged particle moving in a magnetic
field. Work out the following SAQ.

SAQ 3
Determine the equations of motion for a particle of mass m and charge q
moving in a uniform magnetic field B0 along the z-direction which has the
Lagrangian:
1 qB
L m ( x 2  y 2  z 2 )   0  x y  y x 

2  2c 

You have worked out the central force problem in Units 5 and 6 of MPH-002.
Let us now derive Hamilton’s equations for the central force problem.
Example 2.3
A particle moves in a central force field for which the potential function is
k
V ( r )   . Obtain the Hamiltonian, identify the cyclic coordinates and write
r
down the equations of motion. Identify the conserved momenta.

Solution : The Lagrangian can be written in spherical polar coordinates

r , ,  as:

L  T V 
1 2
2

m r  r 2  2  r 2 sin 2  2 
k
r
 (i)

The canonical momenta are

L L L
pr   mr; p    mr 2  ; p    mr 2 sin 2  (ii)
r 
 
The Hamiltonian is

1  p r2 p2 p 2  k
H  p r r  p    p    L       (iii)
2  m mr 2 mr 2 sin 2   r
 
The cyclic coordinate is  and hence the canonical momentum p is conserved.

The canonical equations of motion for the generalized momenta are obtained
using Eqs. (2.22b):

p2 p 2
p r   H    
k
(iv)
r mr 3 mr sin  r 2
3 2

H p2 cos 
p     (v)
 mr 2 sin3 

40
p   0  p   constant  a (say) (vi)
Unit 2 Hamilton’s Equations of Motion
You can check for yourself that the canonical equations for the generalized
coordinates using Eq. (2.22a) are the same as those derived in Eq. (ii):

H p H p H p a
r   r ;     ;    
p r m p  mr 2 p  mr 2 sin 2  mr 2 sin 2 
(vii)
Using Eq. (vi) we can write the Hamiltonian for the problem as:
1  p r2 p2 a2  k
H      (viii)
2  m mr 2 mr 2 sin 2   r

Hamilton’s equations can be solved to obtain the dynamics of the system. You
may like to work on another SAQ.

SAQ 4
a) A particle of mass m and coordinate q has the Lagrangian
1 a
L  mq 2  qq 2 . Determine the Hamiltonian for the system and the
2 2
canonical equations.
b) Obtain the solutions for Hamilton’s equations for the following two systems
for which the Hamiltonian functions are:
p2
i) H  xp
2
e   t p 2 e t  2 x 2
ii) H  
2 2

Let us now look at some important properties of the Hamiltonian and

Hamilton’s equations.

2.4 IMPORTANT PROPERTIES AND

CONSERVATION LAWS
1. If the Lagrangian of the system does not have an explicit dependence on
time, clearly the Hamiltonian will also not have an explicit dependence on
time. So if L  L(qi , qi ) it means that
n
H  q i pi  L(q i , q i )  H (q i , pi ) (2.23)
i 1

2. For H  H (qi , pi , t ) we can write the differential in H as:

n n
H H H
dH   q i dq i  pi dp i  t
dt (2.24)
i 1 i 1
And
n n
dH H  H  H
dt
  q i
qi   p i
pi 
t
(2.25)
i 1 i 1

Using Eqs. (2.22a) and (2.22b),

n n
dH H H
dt
  
 p i q i  q i p i 
t

t
(2.26)
41
i 1 1
Block 1 Hamiltonian Dynamics
H
If the Hamiltonian does not have an explicit dependence on t, then  0,
t
which implies:
dH
 0  H is a constant of motion .
dt
So if the Hamiltonian of a physical system does not depend explicitly on
time, then the Hamiltonian is a constant of motion sometimes also
called the Jacobi integral.
It is important to distinguish between the conditions for the Hamiltonian of
a mechanical system to be equal to its total energy and the total energy of
the mechanical system being conserved. You have already seen in Unit 4
of CM1 that if the following two conditions are satisfied, the function H
which is just the energy integral h, reduces to the total mechanical energy
E of the system:
 The equations of transformation between the Cartesian coordinates of
the system and the generalized coordinates, should not depend
explicitly of time. For such systems the kinetic energy is purely a
quadratic function of the first derivative of the generalized coordinates,
that is q . i

 The potential function should be independent of the generalized

velocity.
And then:
n
H   q i p i  L(q i , q i )  2T  (T  V )  T  V  E
i 1

For a conservative system the Hamiltonian of the system is the total

mechanical energy and is a constant of motion. You may also have
systems in which the total mechanical energy is conserved but the
Hamiltonian of the system may not be the total mechanical energy.
3. The Lagrangian of a physical system is always described by the
prescription L  T  V , which is the difference of the kinetic and potential
energies. Choosing a different set of generalized coordinates may change
the functional form of the Lagrangian, but cannot change the magnitude of
the Lagrangian. However, since the Hamiltonian is defined by
n
H   q i pi  L , a change of coordinates may change not only the
i 1
functional form but also the quantity it represents. So for one choice of
generalized coordinates the Hamiltonian function may be the total energy
of the system, but for another choice it may represent some quantity which
is not the energy.
4. You have studied that if a generalized coordinate q k is a cyclic coordinate,
that is the coordinate does not appear explicitly in the Lagrangian of the
system, then the corresponding canonical momentum (conjugate
L
momentum) pk  is a constant of motion. Now, if pk is a constant,
q k
42
Unit 2 Hamilton’s Equations of Motion
then from Hamilton’s equation, Eq. (2.22b) if
H
p k    0  q k is absent from the Hamiltonian .
q k

So a cyclic coordinate will not be present in the Hamiltonian. We can also

state that if a generalized coordinate does not appear in the Hamiltonian,
the corresponding conjugate momentum is conserved.
5. The symmetry considerations leading to the conservation laws, discussed
in connection to the Lagrangian in Unit 4 of MPH-002, hold for the
Hamiltonian formulation as well. If a physical system is to be invariant
under a rigid displacement (translationally invariant), then the generalized
coordinate representing this displacement should not appear explicitly in
the Hamiltonian. And therefore the corresponding conjugate momentum is
conserved. Similarly the condition of isotropy of space will ensure that the
Hamiltonian is invariant under a rotation about an axis, for which the
particular generalized coordinate representing the rotation would be
absent in Hamiltonian. The corresponding component of the angular
momentum about the axis would be conserved.
6. Hamilton’s equations of motion are symmetric in the generalized
coordinates and momenta, except for a change in sign. Now if we were to
introduce a new set of coordinates with the following prescription:
Qi   pi ; Pi  qi (2.27a)

We would get the following equations:

H H For an insight into the
Q i  ; Pi   (2.27b) importance of the
qi Qi
Hamiltonian formalism,
The equations, as you can see are unchanged. This also means that we you may like to read
Hamiltonian Methods and
can use either ( qi , pi ) or ( Qi , Pi ) to describe the physical system. So the
Quantum Mechanics by
generalized coordinates and the generalized momenta are dynamically P.A.M.Dirac.
equivalent sets of variables.
Now the Lagrangian of the system is a function of qi , q i and t . q i is not
dq i
functionally independent of qi because q i  but pi is dynamically
dt
independent of qi and the Hamiltonian formulation treats q i , p i as
independent variables.

SAQ 5
a) Determine the Hamiltonian and the canonical equations of motion for a a
bead of mass m sliding due to gravity on an elliptical wire (SAQ 3a, Unit 3,
MPH -002) for which the Lagrangian is:
1
L  T V  m(a 2 sin 2  b 2 cos2 ) 2  mgb sin 
2
b) Determine the conserved quantities in the Hamiltonian of Example 2.1.

There is no added advantage in using the Hamiltonian method for any

dynamical problem, in comparison to the Lagrangian formulation. The
Hamiltonian function is important when you are interested in quantizing a 43
Block 1 Hamiltonian Dynamics
dynamical system. Recall that most often in quantum mechanics you start out
by writing the Hamiltonian of the system. Then the coordinates and momenta
are replaced by their corresponding operators for setting up Schrodinger’s
equation.
However when cyclic coordinates are present, the Hamiltonian formulation
leads to fewer differential equations. In that case, it becomes convenient to
introduce a new function, the Routhian as we shall see in the next Section.

2.5 THE ROUTHIAN

First let us look at what happens to the Lagrangian and Hamiltonian methods
in the presence of a cyclic coordinate. Of the set of generalized coordinates
qi , i  1,2,...n let us say that one particular coordinate q k is a cyclic coordinate.
The Lagrangian is then a function of n-1 generalized coordinates and can be
written as L  L(q1, q 2 ,......q n 1, q1, q 2 ,......q n , t ) . Since the Lagrangian still
contains all the generalized velocities we would still have n second order
differential equations to solve.
For the Hamiltonian on the other hand, since the conjugate momentum for the
cyclic coordinate is a constant of motion, we can write
pk  a constant  a (say) and the Hamiltonian as:

H  H (q1, q2 ,......qn 1, p1, p2 ,...... pn 1, a, t ) . So the Hamiltonian reduces to a

function of n-1 generalized coordinates and their corresponding conjugate
momenta. So the cyclic coordinate in the Hamiltonian formulation is genuinely
an ignorable coordinate and reduces the complexity of the problem. Also the
time dependence of the cyclic coordinate is just:
H
q k  (2.28)
a
Routh's procedure or Routhian mechanics is a hybrid formulation of
Lagrangian mechanics and Hamiltonian mechanics developed by Edward
John Routh. He introduced a new function called the “Routhian”, in which the
transformation of coordinates from qi , q i to qi , pi in the Lagrangian function is
carried out only for those coordinates which are cyclic. For this new function,
the cyclic coordinates are governed by Hamilton’s equations of motion,
whereas the non-cyclic coordinates are still governed by Lagrange’s equations
of motion. Remember that this is not new physics, it is just a convenient way of
solving the equations of motion for a dynamical system.
Let us consider a physical system with n generalized coordinates of which r
coordinates are cyclic. So we have n  r non-cyclic generalized coordinates
(from 1 to n  r ) and r cyclic generalized coordinates (from n  r  1 to n).
The Routhian function (or the “Routhian”) is then defined as:
R(q1, q 2 ,......q n , q1, q 2 ,......q n  r , pn  r 1, pn  r  2 ,......, pn , t )
n
  q i pi  L(q1, q2 ,......qn  r , q1, q 2 ,......q n , t ) (2.29)
i  n  r 1

The first term in the RHS of Eq. (2.29) is only for the cyclic coordinates,
whereas the second term is the Lagrangian which is only a function of the
44 non-cyclic generalized coordinates, all the generalized velocities and time.
Unit 2 Hamilton’s Equations of Motion
Now, from the LHS of Eq. (2.29) we can write the differential of the Routhian
function as:
n R n  r R n R R
dR   dq i   dq i   dpi  dt (2.30)
i 1q i i 1 q i i  n  r 1 pi t


We can also write the differential using the RHS of Eq. (2.29) which is:
n  n r L n
L   L
dR    p i dq i  q i dp i     dq i   dq i 
 t
dt
 i 1 q i  i
q
i n r 1 i 1 
(2.31)
L L
Using pi  and p i  we can write Eq. (2.31) as:
qi
 qi

n  n r n  L
dR    p i dq i  q i dp i     p i dq i   p i dq i  
t
dt (2.32)
i n r 1  i 1 i 1 

In Eq. (2.32) we can see:

n n n r
 pi dq i   pi dq i    pi dq i (2.33)
i  n  r 1 i 1 i 1

And then Eq. (2.32) is just

n r n n r
L
dR    pi dq i   q i dpi   p i dqi  t dt (2.34)
i 1 i  n  r 1 i 1

We now rewrite Eq.(2.31) as

n R n  r R n  r R n R R
dR   dq i   dq i   dq i   dpi  dt
q
i  n  r 1 i i 1 q i i 1 q i p
i  n  r 1 i t

(2.35)
Comparing Eqs. (2.34) and (2.31), we therefore find that for the r cyclic
coordinates (from n  r  1 to n) we get Hamilton’s equation of motion:
R  R
p i   ; qi  i  n  r  1,....,n (2.36)
q i pi

Whereas for the remaining n  r non-cyclic generalized coordinates (from 1

to n – r) we have
R R
p i   ; pi  i  1,...., n  r (2.37)
qi q i

Using Eq. (2.27) we get back the Euler-Lagrange equations of motion for the
non-cyclic coordinates:

d  R  R
   0 i  1,...., n  r (2.38)
dt  q i  qi

Finally we can also see that:

R L
 (2.39)
t t 45
Block 1 Hamiltonian Dynamics
The total energy of the system can be written in terms of the Routhian as
follows:
n n n r n r
E  q i p i  L   q i p i   q i p i  L  R   q i pi (2.40)
i 1 i n  r 1 i 1 i 1

To summarize:
 The r cyclic generalized coordinates satisfy Hamilton’s equations of motion
with H replaced by R. The canonical momenta corresponding to the cyclic
coordinates are constants of motion
 The n – r non-cyclic generalized coordinates satisfy the Euler Lagrange
equations of motion with L replaced by R.

Example 2.4
Write down the expression for the Routhian for the central force problem of
Example 2.3 and derive the equations of motion.

Solution : The Lagrangian for the problem is:

1  2 k
L m  r  r 2  2  r 2 sin 2  2  
2   r

The cyclic coordinate for the problem is  and the conjugate momentum p is

conserved ( p  a ). The Routhian is:

1 k
R  p    L  p    m r 2  r 2  2  r 2 sin 2  2   (i)
2 r
For the cyclic coordinate  we have Hamilton’s equation (Eq. 2.36):

R
p     0  p  constant  a(say) ; (ii)


R
p   mr 2 sin 2  (iii)


Using Eq. (ii) in Eq. (iii) we get:

a
mr 2 sin2   a    (iv)
mr sin2 
2

Using the E-L equations for the non-cyclic coordinates r ,  (Eq. 2.38) is

d  R  R 2 2 2 k
 
 r   r  0  mr  mr  mr sin    2  0
 (v)
dt   r

Using the results of Eq. (iv), Eq. (v) reduces to:

a2 k
r  r 2   0 (vi)
2 3 2
m r sin  mr 2

And
d  R   R  0  mr 2   mr 2 sin  cos   2  0 (vii)
  
dt    
46
Unit 2 Hamilton’s Equations of Motion
which is
d  R  R a2
    0  mr 2   cos   0 (viii)
dt     mr 2 sin 3 

You may like to attempt an SAQ on this procedure.

SAQ 6
Identify the cyclic coordinates and obtain the Routhian for the spherical
pendulum which has the Lagrangian (SAQ 2c):

L
1
2
 
mL20  2   2 sin 2   mg L0 sin 

2.6 SUMMARY
 Legendre Transformation

A function f  f (u1, u2 ....., un , w1, w 2 ,....w m ) can be transformed to

another function g  g (v1, v 2 ,....., v n ,, w1, w 2 ,....w m ) where u1, u2 ....., un
are the n active variables and w1, w 2 ,....w m are the m passive
variables. The new variables are then connected to the old variables by
the relations:

f
vi  for i  1,2,....n
ui

The form of the new function is:

n
g (v1, v 2 ,....., v n , w 1, w 2 ,....w m )   ui v i  f (u1, u2 ,....., un , w1, w 2 ,....w m )
i 1

The variables w1, w 2 ,....w m satisfy the relation:

g f
  for i  1,2,....m
w i w i

The variables u1, u2 ,....., un satisfy the relations:

g
ui  for i  1,2,....n
v i

And also
n
f (u1, u 2 ,....., u n , w1, w 2 ,....w m )   u i v i  g (v1, v 2 ,....., v n , w1, w 2 ,....w m )
i 1

 Hamilton’s Function
For a dynamical system described by n generalized coordinates
q1, q2 ....., qn , the Lagrangian, as you know is a function of the
generalized coordinates, generalized velocities and time:
L  L(q1, q2 ....., qn , q1, q 2 ....., q n , t ) . The Hamilton’s function
H  H (q1, q2 ....., qn , p1, p2 ....., pn , t ) is a function of the generalized
coordinates, the conjugate momenta and time where: 47
Block 1 Hamiltonian Dynamics
L
pi  for i  1,2,....n
q i
n
And is defined by the relation: H   q i pi  L
i 1
L H L H
With  ; 
t t q i q i

 The canonical equations of Hamilton or the canonical equations

of motion are:
H H L H
q i  ; p i   ;  
pi qi t t

 If the Hamiltonian does not have an explicit dependence on t, then

dH
 0  H is a constant of motion .
dt
 If the following two conditions are satisfied, the function H reduces to
the total mechanical energy E of the system:
 The kinetic energy is purely a quadratic function of the first
derivative of the generalized coordinates, that is q . i
 The potential function is independent of the generalized velocity.
 Routhian
For a physical system with n generalized coordinates of which r
coordinates are cyclic the Routhian function (or the “Routhian”) is:
R(q1, q 2 ,......q n , q1, q 2 ,......q n  r , pn  r 1, pn  r  2 ,......, pn , t )
n
  q i pi  L(q1, q 2 ,......q n  r , q1, q 2 ,......q n , t )
i  n  r 1

 The r cyclic generalized coordinates satisfy Hamilton’s equations

of motion with H replaced by R. The canonical momenta
corresponding to the cyclic coordinates are constants of motion.
 The n – r non-cyclic generalized coordinates satisfy the Euler
Lagrange equations of motion with L replaced by R.

2.7 TERMINAL QUESTIONS

1. Two beads of mass m are connected by a spring of unstretched length L
1 2
x and spring constant K. The beads are free to move along a frictionless
O x x+z horizontal wire (see Fig. 2.3). The positions of the left and right beads are x
and x   x  z respectively (relative to the equilibrium position). Determine
Fig. 2.3: TQ 1.
the Hamiltonian and Hamilton’s equations of motion.
2. A bead free to slide along a frictionless circular hoop of radius which is
rotating in a vertical plane with an angular velocity  has the Lagrangian:
1  22
L m  A   A 2 sin 2  2   mg A cos  .
2  

Derive the Hamiltonian and obtain Hamilton’s equations of motion. Is the

Hamiltonian conserved?
48
Unit 2 Hamilton’s Equations of Motion
3. A block of mass m sliding down a frictionless incline of mass M (Fig. 2.4)
which in turn is sliding on a smooth horizontal plane (SAQ 3b, Unit 3,
MPH-003) has the Lagrangian:
1 1
L  M  m  X 2  ms 2  mX s cos   mg s sin   mga
2 2
y

s
.

a

X
 O x
x
b
A
LA
z yA
yB
Fig. 2.4: Block sliding down a moving incline.
xA
A
Determine the canonical momenta and show that the Hamiltonian is the
B LB
sum of the kinetic and potential energies. Is the Hamiltonian conserved?
4. A double pendulum (Fig. 2.5) is made up of two simple pendulums of
mass m and length L has the Lagrangian (TQ2, Unit 3, MPH-002 with xB
m A  mB  m and LA  LB  L ):
1  2 2 y
L m 2L  A  L2  B 2  2L2  A  B cos(  B   A )
2   Fig. 2.5: Double
pendulum.
 2 mg L cos  A  mg L cos B

Determine the Hamiltonian for the system. Is the Hamiltonian the total
energy of the system?
5. Derive Hamilton’s equations of motion for a particle of mass m constrained
to move on the surface of a right circular cylinder of radius A defined by the
2 2 2
equation x  y  A and subjected to a force derived from the potential
1
function: V (r )   kr 2 .
2
z
6. A particle of mass m moves under the influence of gravity down a helical
wire x  a cos , y  a sin  ; z  k . Obtain Hamilton’s equations of motion. 
7. Derive Hamilton’s equations of motion for a system with the following
2 L0
1  2 22  r
Lagrangian L   r  r    .
2  2

8. A particle of mass m is attached to a fixed point O by a weightless m
inextensible string of length L0 and is rotating under the gravity (Fig. 2.6).
x
as shown in the figure. The Lagrangian of the particle is:
Fig. 2.6: TQ 9.
1
L(, )  mL2  2  sin 2  2   mgL0 cos 
2
Determine the Hamiltonian and the equations of motion.
p2
9. The Hamiltonian for a particle of mass m is H   kqt where q and p
2m
are the generalized coordinate and momentum, respectively. For the initial
condition, q  0 and p  0 at t  0, q(t )  t  . Determine the value of  . 49
Block 1 Hamiltonian Dynamics
10. Consider a particle moving under the action of a central force derivable
k
from a potential function V (r )   which has the Lagrangian:
rn
1 2 k
L m r  r 2  2  
2 rn
Determine the Routhian and the equations of motion.

2.8 SOLUTIONS AND ANSWERS

Self-Assessment Questions
df z
1. a) We use z   2x, so x( z )  . The Legendre transform of f (x ) is
dx 2
2
  z z
g ( z )  zx( z )  F x ( z )  z     
2 2
z2
4
. Note that
dg ( z ) z
dz
 x
2
   
(which is what is expected) .
df 1 1
b) We have z   so x( z )  . The Legendre transform of f (x ) is
dx x z
then
1 1
g ( z )  zx ( z )  F x ( z )  z   ln   1  ln z
z z
dg ( z ) 1
Once again   x.
dz z
df
c) z   e x  x( z )  ln z
dx
gz(s )  zx  f   z ln z  z
dg
And  ln z  x
dz
d) We have:
f
u  4 x  3y (i)
x
F
v  3 x  2y (ii)
y

We can solve Eqs. (i) and (ii) to get x and y in terms of u and v:
x  2u  3v (iii)
y  3u  4v (iv)

So :
g (u, v )  xu  yv  f  u 2  3uv  2v 2
1
2. a) For L  mL0 2  2  mgL 0 cos  the canonical momentum conjugate to
2
 is

50
Unit 2 Hamilton’s Equations of Motion
L p
p   mL0 2    

 mL 0 2

So:

1 p2 
H  p   L    mL0 2 2  mgL0 cos  
2 2 
mL0
p2  p2 
    mgL0 cos  
2  2 
mL0  2mL0 

p2
  mgL0 cos 
2mL0 2

Since the potential energy is velocity-independent and the kinetic

energy is a homogenous function of the velocity  , the Hamiltonian is
also the total energy. So we can also write the Hamiltonian for the
system using the relation H  T  V :

p2
H  T V   mgL0 cos 
2mL0 2

The equations of motion are

H p H
   ; p      mgL0 sin 
p mL0 2 

b) The canonical momenta conjugate to the coordinates x and y are

L L
px   mx  m y ; p y   m x
x y

So:

py  py  px  py
x  ; p x  m    my  y 
m  m  m

H  x p x  y p y  L

2
 py   px  py  1  py  (p  py )  py
m x 
   p x    p y  m  
 m   m  2  m 2 
 m  m 

p x py p y2
 
m 2m

c) The canonical momenta corresponding to the generalized coordinates

 and  are
L p
p   m L20      ; (i)

 mL20

L p
p   mL20 sin 2     (ii)
 mL20 sin 2 
51
Block 1 Hamiltonian Dynamics
The Hamiltonian is:

H  p    p    L

 p   p  1
 p     p    mL2  2   2 sin 2    mg L0 sin 
 mL2   mL2 sin 2   2 0
 0   0 

p2 p2
   mg L0 sin  (iii)
2mL20 2mL20 sin 2 

Hamiltonian’s equations of motion are

H p H p
    ;    (iv)
p  mL2 p  mL2 sin 2 
0 0

(Note that these equations are the same as Eqs. (i) and (ii))

H 2 p  2 sin  cos  p 2 cos 

p      mg L0 cos    mg L0 cos 
 2mL20 sin 4  mL20 sin3 
H
p    0

p is conserved (you also know that because  is a cycle coordinate)

d) Using Hamilton’s equation Eq. (2.22a) we get

H q
q   a pq 4  p 
p aq 4
Substituting for p in terms of q in the equation for the Lagrangian we
get:
2
 q  aq 4  q  b
L  pq  H  q      
 aq 4  2  aq 4  q2
q 2 q 2 b
  
aq 4 2aq 4 q2
q 2 b
 
2aq 4 q2
3. The canonical momenta ( px , py , pz ) conjugate to ( x, y , z ) are
L qB0 L qB0 L
px   mx  y ; py   my  x ; pz   mz
x
 2c y
 2c z

1 qB0  1 qB0  p
 x   p x  y  ; y   p y  x  ; z  z
m 2c  m 2c  m

The Hamiltonian is
H  px x  py y  pz z  L
qB 0     qB 0  
  mx  y  x   my  x  y  mz 2
 2c   2c 
1 qB
 m x 2  y 2  z 2   0  xy  yx 
2 2c

52
Unit 2 Hamilton’s Equations of Motion


1
2

m x 2  y 2  z 2 
2 2
1  qB0  1  qB0  1 2
  px  y   py  x  pz .
2m  2c  2m  2c  2m

The canonical equations are:

H qB0  qB0 
p x     py  x
x 2mc  2c 

H qB0  qB0 
p y     px  y
y 2mc  2c 
H
pz    0  p z  constant
z
1 a
4. a) H   qp  L where L  2 mq 2  2 qq 2
L p
 p  mq  aq q  p  q ( m  aq )  q 
q
 m  aq

So:

p2 1 ( p2 ) a  p2 
H  qp  L   m  q 
(m  ap) 2 (m  aq )2 2  (m  aq )2 
 

p2 p2
  (m  aq )
( m  aq ) 2( m  aq )2

p2 p2 p2
  H 
(m  aq ) 2( m  aq ) 2(m  aq )

From Eq. (2.22 and b) we get:

H p
q  
p (m  aq )2

H  ap 2 
p     
q  2(m  aq )2 
 

b) i) The equations of motion are

H
x   px (i)
p
H
And p    p (ii)
x
Solving the second equation by the standard methods we get:
p  Ae t (iii)

Where A is a constant. Substituting for p in Eq. (i) from Eq. (iii) we

get
x  x  Ae t
53
Block 1 Hamiltonian Dynamics
Which can be solved to get: x  Ce t  Ae t / 2 (C is a constant).
ii) Hamilton’s equation of motion are
H
x   e  t p (i)
p
H
p     2 e t x (ii)
x
Differentiating Eq. (i) with respect to t, we get

x  e t p  e t p  x  2 x

Which is the standard second order differential equation

x  x   2 x  0
Which has the solution
x  Ae m1t  Be m2t
   2  42   2  42
where m1  ; m2  .
2 2

5. a) The canonical momentum is

L
p   m  a 2 sin 2   b 2  cos 2  

So we can write
1 2 2
H  p    L  p    m  a sin 2   b 2 cos 2    m g b sin 
2
1 2 2
 m  a sin 2   b 2 cos 2    m g b sin 
2

p 2
  m g b sin 
2m a 2 sin 2   b 2 cos 2 

The equations of motion are

H p
  
p  m a sin   b 2  cos 2  
2 2

H p 2 a 2  b 2  sin  cos 
p       m g b cos 
 m a 2 sin 2   b 2 cos 2  
2

b) For both examples, the Hamiltonan has no explicit dependence on time

so it is conserved
1
6. For L  mL20  2   2 sin 2    mg L0 sin  the cyclic coordinate is .
2
And we can write the Routhian as:
1
R  p    L  p    mL20  2   2 sin 2    mg L0 sin 
2
With

54
Unit 2 Hamilton’s Equations of Motion
H p H p
    ;   
p mL20 p mL20 sin2 

We get

p 2 1
R  mL20  2  mg L0 sin 
2mL20 sin 2  2

Terminal Questions
1. From Fig. 2.3 we can write the Lagrangian as:
1 2 1 1
L mx  m( x  z )2  kz 2
2 2 2
1 2 1 2 1 2 1
 mx  mx  mz  mx z  kz 2
2 2 2 2
The conjugate momenta are
L
px   2mx  mz (i)
x
L
pz   mz  mx (ii)
z
The Hamiltonian is:
H  p x x  pz z  L
1 1 1
 p x x  p z z   mx 2  m( x  z ) 2  kz 2  (iii)
2 2 2 
From Eqs. (i) and (ii) we get
p  pz
x  x (iv)
m
2p z  p x
z  (v)
m
pz
and x  z  (vi)
m
Using Eqs. (iv) and (v) in Eq. (iii) we get
2
p x  pz  2pz  p x  1 p x  pz  1 mp z2 1 2
H  px  p z   m   kz
m  m  2 m2 2 m2 2
(vii)

p x 2 p x p z 2 p z2 p x p z 1 p x2 1 p z2 p x p z 1 p z2 1 2
         kz
m m m m 2 m 2 m m 2 m 2
p x2 p z p x p z2 1 2
    kz (viii)
2m m m 2
The equations of motion are:
H 1
x   ( p x  pz )
p x m
H 1
z   (2 p z  p x )
p z m
55
Block 1 Hamiltonian Dynamics
H
p x   0
x
H
p z     kz
z
2. The canonical momentum is:
L p
p   m A 2    

 mA 2
The Hamiltonian is
H  p   L

p 1 1
 p      mA 2  2  mA 2 sin2 2  mgA cos 
2
 mA   2 2 

p 2 p 2 1
   mA 2 sin 2   2  mgA cos 
mA 2 2mA 2 2

Hamilton’s equations of motion are:

H p
  
p  mA 2
H
p     mA 2 sin  cos  2  mg A sin 

1 1
3. Given L  M  m  X 2  ms 2  mX s cos   mg s sin   mga .
2 2
The canonical momenta are:
L
pX   M  m  X  ms cos 
X
L
ps   ms  mX cos 
s
The Hamiltonian is
H  X p X  sps  L

  
 X M  m  X  ms cos   s ms  mX cos  
1 1
  M  m  X 2  ms 2  mX s cos   mg s sin   mga 
2 2 


1
2
 1

M  m X 2  ms 2  mX s cos   mg s sin   mga
2
From the solution of SAQ 3b, Unit 3, MPH-002 we know that

 the kinetic energy is: T 

1
2
 1

M  m X 2  ms 2  mX s cos 
2
 the potential energy is V  mg (a  s sin  )

So H  T  V .
For the Hamiltonian to be the total energy of the system (H  T  V )

56
Unit 2 Hamilton’s Equations of Motion
 The potential energy should be independent of the generalized
velocity which is true in this case
 The kinetic energy should be a homogenous function of degree two of
the generalized velocities. For this we must have
T
 q i q i  2T . Here

 T  X   T  s
   
 X   s 

  
 M  m  X  ms cos  X  ms  mX cos  s 
1 1
 2  M  m  X 2  ms 2  mX s cos    2T
2 2 

So as expected, the Hamiltonian of the system is also the total energy of

dH L
the system. The Hamiltonian is also conserved because   0.
dt t
4. The canonical momenta are:
L
pA  p A   2mL2  A  mL2  B cos( B   A )
 A
L
p B  p B   mL2  B  m L2  A cos(  B   A )
 B

We can solve for  A and  B in terms of p A and pB to get:

pA pB
 2 A   B cos(  B   A ) ;   A cos(  B   A )   B
mL2 mL2
p  pB cos(B   A )
 A  A (i)
mL2 1  sin2 (B   A )
 

2 pB  p A
 B  (ii)
mL2 1  sin 2 ( B   A 

The Hamiltonian for the system is

H  p A  A  pB  B  L (iii)

Using the results of Eqs. (i) and (ii) in Eq. (iii) we get
1  p 2 / 2  p 2  p p cos(   )
H  A B A B B A 
mL 1  sin (  B   A ) 
2 2  
 

 2mg L cos  A  mgL cos  B (iv)

Once again
 The potential energy is independent of the generalized velocity which
is true in this case
 The kinetic energy is a homogenous function of degree two of the
generalized velocities and for
57
Block 1 Hamiltonian Dynamics
1
T  mL2  2A  mL2  B  A  B cos(  B   A )
2  mL2 
2
T
 2m L2  A  m L2  B cos (  B   A )
 A
T
 m L2  B  mL2  A cos(  B   A )
 B

T  T
Hence A   2T

 A  B

Since both conditions are satisfied, H is the total energy of the system
(H  T  V )
5. Let us use cylindrical coordinates , , z  for this problem. The constraint
x 2  y 2  A 2 can be written as  2  A 2    A . So
1 1 1
V (r )  k ( x 2  y 2  z 2 )  k ( 2  z 2 )  k ( A 2  z 2 ) (i)
2 2 2
In cylindrical coordinates the square of the velocity is
v 2   2  2 2  z 2 (ii)

Since  is constant (=A),   0 the velocity is v 2  A 2  2  z 2 and the

kinetic energy is
1
T  m( A 2  2  z 2 ) (iii)
2
The Lagrangian is
1 1
L  T V  m( A 2  2  z 2 )  k ( A 2  z 2 ) (iv)
2 2
The generalized coordinates in this case are just  and z, and the
canonical momenta are
L p L p
p   mA 2     ; pz   m z  z  z (v)

 mA 2 z m

The Hamiltonian is:

H   p  zp z  L
p 2
p 2 1 1
  z   m( A 2  2  z 2 )  k ( A 2  z 2 )
mA 2 m 2 2 
2 2  p 2 2 
p p p 1
  z   z  k ( A 2  z 2 )
mA 2 m  2mA 2 2m 2 

p 2p 2 1
  z  k( A 2  z 2 ) (vi)
2mA 2 2m 2
As you can see, H is also the total energy expressed in terms of the
variables , p , z and p z :

 p 2 p 2 1
H  T V    z    k ( A 2  z 2 ) (vii)
 2mA 2 2m   2 
58 
Unit 2 Hamilton’s Equations of Motion
The canonical equations are:
H
p     0  p   constant (viii)

This also follows from the fact that  is a cyclic coordinate.
H
p z     kz (ix)
z
H p
   (x)
p  mA 2

H p
z   z (xi)
p z m
6. Once again, we use cylindrical coordinates , , z  for this problem. In
cylindrical coordinates the kinetic energy and the potential energy of the
particle are:
1 1
T  mv 2  m  2   2  2  z 2  (i)
2 2
V  mgz (ii)

The Lagrangian can be written as:

1
L  T V  m 2   2  2  z 2   mgz (iii)
2

Given that  2  x 2  y 2  a 2 cos 2   a 2 sin 2   a 2 ,   0 and since

z  k , z  k . The Lagrangian is then:

1
L  T V  ma 2  2  k 2  2   mgk
2

1  k 2  2
 ma 2 1    mgk (iv)
2  a2 

As you can see the Lagrangian has just one generalized coordinate which
we have taken to be . The canonical momenta conjugate to  is

L  k 2  p
p   ma 2 1      (v)
  a2   k2 
ma 2 1  
 a2 
The Hamiltonian is:

H  p    L
 k 2  2 1 2 k 2  2 
 ma 2  1     2 ma 1    mgk
2
a  2
a 
   
1  2
k  2
 ma 2 1    mgk
2  a2 

p 2
  mgk  (vi)
 k2 
2ma 2 1 
 
 a2  59
Block 1 Hamiltonian Dynamics
Hamilton’s equations of motion are:
H
p     mgk (vii)


H p
   (viii)
p   k2 
ma 2 1 
 
 a2 

Alternative Solution

Since z  k , z  k we can also choose z to be the generalized

coordinate in place of . In that case we can write the Lagrangian as
1  a2 
L m  1z 2  mgz (i)
2  k 2 

L a2  pz
pz   m  1 z  z  (ii)
z
 k 2
 a2 
m  1
2
k 
And the Hamiltonian is now written in terms of z :

p z2
H  p z z  L   mgz (iii)
a2 
2m   1
k 2
 

Hamilton’s equations of motion will be

H
p z    mg (iv)
z

H p z2
z   (v)
p z a2 
2m   1
2
k 

7. The canonical momenta are

L  L
pr  r p   r 2 
r ; 

which gives us:

p
r  pr ;   
r2

The Hamiltonian is:

1 r2 
H  p r r  p     r 2  r 2  2   
2 2 

2
1  2 p  r 2
 pr   .
2  r 2  2

60
Unit 2 Hamilton’s Equations of Motion
The equations of motion are:

H H p
r   pr ;    
p r p  r 2

2
H p  H
p r     r ; p     0.
r r 3 

p  is a constant ( is a cyclic coordinate).

8. The Hamiltonian is

1
H  p   p    L  p    p   mL0 2  2  sin 2  2   mgL0 cos 
2

The canonical momenta are:

L p L p
p     ; p   mL0 2 sin 2    
 mL0 2  mL0 2 sin 2 

The Hamiltonian is then

 2 
H
1  p 2  p    mgL cos 
 0
2mL0 2  sin 2  

9. We first write hamilton’s equations which are:

H p
q   (i)
p m

H
p    bt (ii)
q

Solving the ODE in Eq. (ii) we get

bt 2
p C (iii)
2
here C is a constant. Using the initial conditions p(0)  0 in Eq. (iii) we get
C  0 . Hence Eq. (iii) gives us:

bt 2
p (iv)
2

Substituting for p from Eq. (iv) into Eq. (i) we can write:

bt 2
q   (v)
2m

The solution for the ODE of Eq. (v) is:

bt 3
q D (vi)
6m

where D is a constant.
61
Block 1 Hamiltonian Dynamics
Using the initial conditions q(0)  0 in Eq. (vi) we get D  0 , so

bt 3
q  q t3 and   3.
6m

10. The cyclic coordinate for the problem is  and the conjugate momentum
p  is conserved ( p   a ). The Routhian is:

1 k
R  p    L  p    m r 2  r 2  2   (i)
2 rn

For the cyclic coordinate  we have Hamilton’s equation (Eq. 2.36):

R
p     0  p   constant  a(say) ; (ii)

L a
p   mr 2     (iii)

 mr 2

Using Eq. (ii) and (iii) in Eq. (i) we get:

a2 1 2 k
R ( r , r, a )   mr  (iv)
2mr 2 2 rn

Using the E-L equation for the non-cyclic coordinate r we get

d  R  R a2 nk
    0  mr   0 (v)
dt  r  r mr 3 r n 1

a2 nk
mr   (vi)
mr 3 r n 1

62