```A Simple Example
• Another simple example: An illustration of my point
that the Hamiltonian formalism doesn’t help much in
solving mechanics problems. In fact, one is usually
led directly back to the SAME 2nd order differential
equations of motion you’d get from Lagrange’s
formalism (or even Newton’s formalism!)
• A particle in 1 dimension (x) subject to conservative
force.  A potential V(x) exists.
Newton:
F  - (dV/dx) = ma = mx
Hamiltonian:
H = (p2)(2m)-1 + V(x)
Lagrangian:
L = (½)m(x)2 – V(x)
Newton:
Hamiltonian:
Lagrangian:
F  - (dV/dx) = ma = mx
H = (p2)(2m)-1 + V(x)
L = (½)m(x)2 – V(x)
(1)
• The Lagrangian formalism first:
(d/dt)[(∂L/∂x)] - (∂L/∂x) = 0  (d/dt)[mx] + (dV/dx) = 0
Or: mx = - (dV/dx)
 The SAME as (1) !
• The Hamiltonian formalism: x = (H/p), p = - (H/x)
x = (H/p) = (p/m)  p = mx
(a)
(The definition of p in terms of x! No new information!)
p = - (H/x) = - (dV/dx)  F
(b)
(Newtons 2nd Law!)
Take the time derivative of (a) & equate to (b).
 mx = -(dV/dx) The same as (1)! Note the extra steps required!
Marion’s Example 7.11
• See the figure: Use the Hamiltonian method to find the
equations of motion of a particle, mass m, constrained to move
on the surface of a cylinder defined by x2 + y2 = R2 & subject
to a force directed towards the origin & proportional to the
distance of the particle from the origin: F = - kr
Potential: V = (½)kr2
= (½)k(x2 + y2 + z2) = (½)k(R2 + z2)
KE: T = (½)mv2
= (½)m(R2θ2 + z2) (R = const)
Lagrangian:
L =T-V
= (½)m(R2θ2 + z2) - (½)k(R2 + z2)
Conjugate Momenta to θ & z:
pθ = (L/θ) = mR2θ
pz = (L/z) = mz
θ is cyclic so pθ = mR2θ = constant
Hamiltonian: H(z,pθ,pz)
H = T + V = (pθ)2(2mR2)-1 + (pz2)(2m)-1 + (½)k(R2 + z2)
Hamilton’s Equations of Motion:
θ = (H/pθ) = (pθ)(mR2)-1 (1)
pθ = - (H/θ) = 0
(2)
z = (H/pz) = (pz)(m)-1
(3)
pz = - (H/z) = - kz
(4)
(1) & (2) along with previous result:
 pθ = mR2θ = constant
(angular momentum around
the z axis is conserved)
(3): the definition of pz = mz
(no new information!)
Taking the time derivative of (3) & equating the result to (4)
gives (same as Lagrange Eqtns!): z + (ω0)2z = 0; (ω0)2  (k/m)
The motion in the z direction is simple harmonic motion!
Marion’s Example 7.12
• Using the Hamiltonian method, find the equations of motion
for a spherical pendulum, mass m, length b. (Figure). (PE  U)
 U= 0
```