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: F  - (dV/dx) = ma = mx (1) Hamiltonian: H = (p2)(2m)-1 + V(x) Lagrangian: L = (½)m(x)2 – V(x) 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