PHY5210 W08

Chapter 13: Hamiltonian Mechanics

Reading:

In Taylor, read sections 13.3 to 13.4 for today, and 13.5 to 13.6 for Friday.

Recall

H = T + U

for natural generalized coordinates. Hamilton's equations in one dimension are

pdot = -∂H /∂ q        qdot = ∂H /∂ p

Hamilton's Equations for Several Dimensions

The derivation of Hamilton's equations for multiple dimensions is outlined in the text. The result is the straightforward generalization of the case for one dimension.

pidot = -∂H /∂ qi        qidot = ∂H /∂ pi

For n generalized coordinates, Hamilton's equations yield 2n first order differential equations, in contrast to the n second order differential Lagrange equations. Another result that can be generalized to multiple dimensions is the fact that for natural generalized coordinates

∑i pi qidot = 2T

so that H = T + U. And it is interesting to look at the total time derivative of the Hamiltonian. The total time derivative is

dH/dt = ∑i[ (∂H/∂q) qdot + (∂H/∂p) pdot] + ∂H/∂t .

The Hamiltonian can vary due to explicit time dependence, and possibly implicitly through the time dependence of the coordinates and momenta. But, using Hamilton's equations, the terms in square brackets cancel pairwise, so that

dH/dt = ∂H/∂t .

That is, the Hamiltonian is time dependent only if it explicitly depends on time.

Hamilton's Equations for a Particle in a Central Force Field

A particle of mass m is subject to a central force with potential U(r). Using spherical coordinates (r, φ, θ) write the Hamiltonian and Hamilton's equations.

We know from previous studies that the motion is confined to a plane. Make this the θ = π/2 plane and concern ourselves with the r and φ coordinates. From Taylor Eq. (8.20) we have

L = ½ m (rdot² + r²φdot²) - U(r)

The two generalized momenta are

pr = ∂L/∂rdot = mrdot

and

pφ = ∂L/∂φdot = mr²φdot

The coordinates are natural, so the Hamiltonian is

H = T + U = pr²/2m + pφ²/2mr² + U(r)

Hamilton's equations read

rdot = pr/m    and    prdot = pφ²/mr³ - dU/dr

and

φdot = pφ/mr²    and    pφdot = 0

The first of each pair represents an expression of momentum. The second radial equation is the equivalent to Eq. (8.24), relating ma to the sum of the central force and the centrifugal force. The second angular equation is conservation of angular momentum.

The procedure for the Hamiltonian technique can be broken down into 6 steps.

  1. Choose suitable generalized coordinates q1, ..., qn.
  2. Write down T and U in terms of the q's and qdot's.
  3. Find the generalized momenta from pi = ∂T/∂qi. (Assumes U is independent of qdot's, otherwise need to use ∂L/∂qi.)
  4. Solve for qdot's in terms of q's and p's.
  5. Write down H in terms of q's and p's using either T + U (natural coordinates) or the formal definition.
  6. Write down Hamilton's equations.

In familiar circumstances, one can shortcut some of the steps, for instance directly writing down the kinetic energy in terms of linear or angular momentum.

Hamilton's Equations for a Particle on a Cone

A particle of mass m is constrained to move without friction on the survace of a circular cone, pointing up along the z--axis with vertex at the origin. The radius of the cone is related to the z coordinate by ρ = cz in polar coordinates. For uniform gravity in the z direction, write down the Hamiltonian, and find Hamilton's equations of motion. Show that for any solution the motion has a maximum and minimum height (z), and that for any z > 0, there is a solution where the path is a circle of constant z.

In this case, the generalized coordinates are specified to be z and φ. The second step is to write down the kinetic and potential energies

T = ½ m [ ρdot² + (ρφdot)² + zdot² ] = ½ m [ (c² + 1)zdot² + (czφdot)² ]
U = mgz

Find the generalized momenta for the third step. The potential has no velocity dependence so we can use the partial derivative with respect to the kinetic enery

pz = ∂T/∂zdot = m (c² + 1)zdot
pφ = ∂T/∂φdot = m c² z² φdot

The fourth step is to find zdot and φdot in terms of the generalized momenta and coordinates. It is obvious from the expressions of the generalized momenta and need not be explicitly written here. The fifth step is to write down the Hamiltonian. The coordinates z and φ have no explicit time dependence, so they are natural and we can use H = T + U. Just remember to replace the velocities by the results of the fourth step.

H = pz²/2m(c² + 1) + pφ²/2mc²z² + mgz

The sixth step is to calculate the derivatives for Hamilton's equations. Since we have 2 generalized

zdot = ∂H/∂pz = pz/m(c² + 1)    and    pzdot = -∂H/∂z = -mg
φdot = ∂H/∂pφ = pφ/mc²z²    and    pφdot = -∂H/∂φ = 0

The last equation tells us that the angular momentum around the z axis is conserved.

We can see that there is a maximum and minimum height by noting that the Hamiltonian is the total energy and must be a constant, and looking at the behavior as z gets large or small. There are terms that get very large in either case, so z must be bounded on both sides so that this doesn't happen.

The existence of a circular orbit implies that z is constant, so pz = 0 and pzdot = 0. Looking at Hamilton's equation for pzdot reveals that this happens when pφ = ±(m²c²gz³)½

Ignorable Coordinates

When a coordinate does not appear in the Hamiltonian, then it is obvious that its conjugate momentum is conserved since

pidot = -∂H/∂qi = 0

This is the same result that appears for Lagrangians, but since the Hamiltonian is written in terms of the generalized momenta, it has the additional advantage of reducing the degree of the problem. That is, a problem has n generalized coordinates and n geneeralized momenta, but one of the coordinates is missing in the Hamiltonian, then the conjugate momentum is a constant, so that the Hamiltonian is now a function of (n-1) coordinates and momenta, plus the constant.

© 2015 Robert Harr