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Unit-4

Hamilton’s Equations
Of
Motion

Hamiltonian Formulation

Newtonian → Lagrangian → Hamiltonian



Describe same physics and produce same results
Difference is in the viewpoints
 Flexibility of coordinate transformation


Hamiltonian formalism linked to the development of





Hamilton-Jacobi theory
Classical perturbation theory
Quantum mechanics
Statistical mechanics

Lagrangian → Hamiltonian
Hamiltonian formulation  closer to Lagrangian formulation
For most systems of interest:

L = T −V

(difference of kinetic and potential energies)

Lagrange’s equations for n coordinates

d  ∂L  ∂L

 −
=0
dt  ∂q&i  ∂qi

i = 1, …
...
, qn )
(q&1, q&2 ,
...
, qn ) as a point in an n-dim
...
e
...
space

 Isn’t it more natural to consider the motion in 2n-dim space?

Phase Space



Consider coordinates and momenta as independent


State of the system is given by (q1 ,
...
, pn )



Consider it a point in the 2n-dimensional phase space



We are switching the
independent variables

(qi , q&i , t ) → (qi , pi , t )

Hamiltonian
Lagrangian describing a system where angular momentum is
conserved, does not depend on time explicitly, i
...


dL
=0
dt
we can express the dynamics in terms of the 2n + 1
variables qi, pi, and t
...
r
...
time:

dL
∂L
∂L
= ∑ q&i + ∑ q&&i
dt
i ∂qi
i ∂q&i

(1)

Lagrange’s equation:

So (1) →

∂L d  ∂L 

= 
∂qi dt  ∂q&i 
dL
d  ∂L 
∂L
 q&i + ∑ q&&i
= ∑ 
dt
i dt  ∂q&i 
i ∂q&i

(2)

Equation (2) is simply the result of differentiation of a product of two
functions,
therefore

dL
d  ∂L 

= ∑  q&i
dt
i dt  ∂q&i 
d  ∂L  dL
∑ dt  q&i ∂q&  − dt = 0

i
i

or


d   ∂L 
 ∑ q&i
 − L  = 0
dt  i  ∂q&i 


(3)

Integrating equation (3):

 ∂L 
∑ q&i
− L = H
&i 
i  ∂q
The generalised momenta pi are defined as:
So (4) →

(constant)

(4)

∂L
pi =
∂q&i

∑ {q&i pi } − L = H
i

Therefore,

H (qi , pi , t ) = ∑ q&i pi − L(qi , q&i , t )

(5)

i


Transformation of formulation of mechanics (Independent
coordinates are generalised coordinates and generalised velocities )
→ a formulation in which the independent coordinates are generalised
coordinates and generalised momenta
...
Choose the generalized coordinates qi and construct the
Lagrangian L ( qi , q&i , t )

∂L
2
...

3
...

4
...


Examples

Free particle in one-dimension
Lagrangian:

L(q, q& , t ) = 12 mq& 2

Momentum conjugate to q is:

∂L
= mq&
∂q&
p
q& =
m
p=

or

Transformation equation: H ( qi , pi , t ) = ∑ q&i pi − L( qi , q&i , t )
the corresponding Hamiltonian:

i

H (q, p ) = pq& − L
mq& 2 p 2 mp 2 p 2
= pq& −
=
−
=
2
2
2m
m 2m

So we can write:

p2
H ( q, p ) =
2m

L and H assume same values  both equal to the kinetic energy
But they are different functions  depend on different variables

Carrying out partial differentiation → the equations of motion are:

q& =

∂H p
=
∂p m

p& = −
So p is a constant of motion
...

Then Lagrange’s equations → its conjugate momentum p j is
constant
...

Let cyclic coordinate  qn
then

L = L(qi
...
q& n , t )

still, the problem to be solved is of n degrees of freedom
...
qn−1 , pi
...

The Hamiltonian:

H = q&p − L = mv 2 − L

In a conservative system:

L = T −V

so

H = 2T − T + V

or

H = T +V

Example

Motion of a particle in 2-dim
...

The total energy is then:

H = T + V (r )

where, V(r) is central potential, and K
...
is:

T = 12 mv 2 = 12 m( r& 2 + r 2θ& 2 )

Hamiltonian formulation → r,θ converted to conjugate momenta pr,pθ
So we have:

pr = mvr = mr&
pθ = mrvθ = mr 2θ&

(linear momentum)

(angular momentum)
In terms of momenta and generalised velocities:

r& =

pr
m

and

pθ
&
θ= 2
mr

therefore:

2 2
2

r
pθ 
p
r
1

T = 2 m 2 + 2 4 
m r 
m

pθ2
pr2
=
+
2m 2mr 2
2
2
p
p
θ
Hamiltonian: H = r +
+ V (r )
2
2m 2mr

θ is cyclic

pθ = constant=l

Hamilton’s equations:

∂H pr
r& =
=
∂pr m
Cyclic variable drops off by itself

pθ2 ∂V
∂H
p& r = −
= 3−
∂r mr
∂r

Derivation of Hamilton’s Equations from
Variational Principle

Variational principle  Hamilton’s principle →
useful in derivation of Lagrange’s equations
Hamilton’s principle

t2

δI = δ ∫t Ldt = 0
1


Lagrangian formulation  qi and q&i independent variables in

configuration apace

Hamiltonian formulation  q and p independent coordinates in

phase space

(12)


express Hamilton’s principle in terms of q and p
Hamiltonian

H (qi , pi , t ) = ∑ q&i pi − L(qi , q&i , t )
i

L(qi , q&i , t ) = ∑ q&i pi − H (qi , pi , t )

(13)

i


Substituting in Hamilton’s Principle:

Modified Hamilton’s
principle

δI = δ ∫tt12  ∑ q&i pi − H (qi , pi , t ) dt


i



(14)

Review  δ - variation process:

Configuration space formed by n generalised coordinates

The initial and final configurations of the system, at times t1 and t2
 each represented by a point

“variation of the integral” ⇒ variation in the value of integral
as we change the path traversed by the system between two
end points
...

therefore:

∂H
q&i =
∂pi

and

∂H
p& i = −
∂qi

Required Hamilton’s
Equations

(20)

Applications of Hamilton’s Equations
Example-1:
Use Hamiltonian method to find equations of motion of a particle
constrained to move on the surface of a cylinder defined by x2 + y2 =
R2
...
E
...


Example-2:
Use the Hamiltonian method to find the equations of motion for a
spherical pendulum of mass m and length b
...


z
x

Using spherical polar coordinates:

x = b sin θ cos φ
y = b sin θ sin φ
z = b cosθ
K
...
is given by:
T = 12 m( x& 2 + y& 2 + z& 2 )
and P
...
:

θ

b

y

φ

mg

V = − mgz

Perform simple calculations to transform the K
...
and P
...
equations
using spherical coordinates
...
pφ is actually the component
of the angular momentum along the z-axis
...
Find the Hamiltonian function and show
that the canonical equations of motion reduce to Newton’s equations
...

Solution:
For a particle moving freely in a conservative field:

T = 12 m( x& 2 + y& 2 + z& 2 )
P
...
:
V = V ( x, y , z )
The Lagrangian is:
L = T −V
= 12 m( x& 2 + y& 2 + z& 2 ) − V ( x, y, z )
K
...
:

Generalised momenta are then:

∂L
px =
= mx& ,
∂x&

∂L
py =
= my& ,
∂y&

∂L
pz =
= mz&
∂z&

H = T +V

The Hamiltonian can be written as:

H = 12 m( x& 2 + y& 2 + z& 2 ) + V ( x, y, z )
p 2y

p x2

p z2
=
+
+
+ V ( x, y , z )
2m 2m 2m
Canonical equations of motion are:

∂H
∂V
p& x = −
=−
,
∂x
∂x
however,
therefore:
so

∂H
∂V
p& y = −
=−
,
∂y
∂y

F = −∇V
∂V
Fx = −
,
∂x

p& x = Fx ,

Fy = −
p& y = Fy ,

p& z = −

∂V
,
∂y

p& z = Fz

which are simply Newton’s equations of motion
...

Solution:

x

The K
...
of the system is:
and P
...
:

V = − mgy

T = 12 mv 2

θ

l

Here the velocity v is related to angular velocity ω
and

v = lω = lθ&
y = l cosθ

y

Therefore, we write the lagrangian:

L = T − V = 12 mv 2 + mgy
= 12 ml 2θ& 2 + mgl cosθ

now

∂L
pθ =
= ml 2θ&
∂θ&

→

pθ
&
θ= 2
ml

(i)

H = pθ θ& − L

Now the Hamiltonian can be written as:

H = ml 2θ& 2 − 12 ml 2θ& 2 − mgl cosθ
= 12 ml 2θ& 2 − mgl cosθ
Substituting from (i):

pθ2
H=
− mgl cosθ
2
2ml

So the equations of motion are:

pθ
∂H
&
θ=
= 2
∂pθ ml
∂H
p& θ = −
= − mgl sin θ
∂θ
g
&
&
θ = − sin θ
or
l
For small θ : sin θ ≈ θ
so

⇒

ml 2θ&& = − mgl sin θ

g
&
&
θ =− θ
l



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