Hamiltonian applications and comparisons

1. The same system in three formulations

Hamilton's equations are most informative when the connection to the physical system remains visible. Start with the same coordinate choice and energies used in the Lagrangian calculation, then replace the velocities by conjugate momenta. This makes agreement between the formulations a check on the calculation.

The comparison assumes regular coordinates and the same physical force and constraint model. None of the formulations is universally shortest. Hamiltonian mechanics becomes particularly useful when conserved momenta can reduce the problem, as in central-force motion.

2. Applications

Example 1 · Hamilton equations for the pendulum

Take $\theta$ from the downward vertical. With $T=ma^2\dot\theta^2/2$ and $V=mga(1-\cos\theta)$, the Lagrangian gives $p_\theta=ma^2\dot\theta$. Its Hamiltonian is

Hamilton's equations are $\dot\theta=p_\theta/(ma^2)$ and $\dot p_\theta=-mga\sin\theta$. Differentiating the first gives $\ddot\theta=\dot p_\theta/(ma^2)=-(g/a)\sin\theta$, reproducing the Lagrange and tangent-projected Newton equations.

At the bottom with angular speed $\omega_0$, energy is $ma^2\omega_0^2/2$. A rod pendulum can reach the top only if this is at least $2mga$, giving $\omega_0^2\geq4g/a$. Equality describes limiting asymptotic approach rather than passage in finite time. For a string, remaining taut imposes an additional condition; the energy threshold alone is insufficient.

Example 2 · Atwood motion in phase space

With $q$ downward for mass $m_1$, let $M=m_1+m_2$ and $\Delta m=m_1-m_2$. Then $T=M\dot q^2/2$, $V=-\Delta m\,gq$, and $L=M\dot q^2/2+\Delta m\,gq$. The conjugate momentum is $p=M\dot q$, giving

Hence $\dot q=p/M$ and $\dot p=\Delta m\,g$. Integrating,

The phase curve at energy $E$ satisfies $p^2=2M(E+\Delta m\,gq)$. The momentum $p$ belongs to the motion of both masses; it is not the momentum of either mass alone. The model applies while the string is taut and neither mass reaches a stop.

Example 3 · A particle in polar coordinates

For $L=m(\dot r^2+r^2\dot\theta^2)/2-V(r)$, use $p_r=m\dot r$ and $p_\theta=mr^2\dot\theta$. Then

Take derivatives at fixed canonical variables:

Setting the conserved $p_\theta=\ell$ reduces the radial motion to $\dot p_r=\ell^2/(mr^3)-V'(r)$. Substituting $\ell=mr^2\dot\theta$ gives $m\ddot r-mr\dot\theta^2=-V'(r)$, the radial Newton equation. The term with $\ell^2$ will become the centrifugal contribution to an effective potential.

Example 4 · The spherical pendulum in canonical variables

For $T=ma^2(\dot\theta^2+\sin^2\theta\dot\phi^2)/2$ and $V=mga(1-\cos\theta)$,

Away from the polar coordinate singularities,

The azimuth is cyclic: $p_\phi=J$ is constant. Since $d(\sin^{-2}\theta)/d\theta=-2\cos\theta/\sin^3\theta$,

A nonzero $J$ produces a barrier as $\sin\theta\to0$, so the bob cannot reach the vertical axis at finite energy. For $J=0$ the barrier vanishes, though the azimuth coordinate itself is still singular at the poles. The reduced energy describes an effective one-dimensional motion in $\theta$.

3. Eliminating a cyclic coordinate carefully

Once a cyclic momentum is found, it is safe to substitute its constant value into the Hamiltonian and use the remaining canonical equations, then reconstruct the eliminated coordinate from its velocity equation. This is what we did for the polar particle and spherical pendulum.

Directly substituting a conserved velocity relation into $L$ before varying can give the wrong sign. For the central potential, substituting $\dot\theta=\ell/(mr^2)$ into $L$ gives a term $+\ell^2/(2mr^2)$, whose naive variation does not reproduce the correct radial equation. The correct reduced Lagrangian is $L-\ell\dot\theta$ after eliminating $\dot\theta$:

This subtraction is called Routh reduction; the name and general theory are optional. The practical lesson is to derive the equations first or reduce the Hamiltonian at fixed conserved momentum.

4. Transition to central forces

Central-force motion will make the advantages of all three viewpoints visible. Newtonian torque proves planarity; the cyclic angle expresses angular-momentum conservation; the Hamiltonian gives radial energy contours. The orbit equation then removes time from the description and asks for the geometric path itself.

Exercises

Hints and solutions are collected separately.

1. 15.1 Why do Hamilton’s $2n$ equations not add degrees of freedom?

2. 15.2 Construct the Hamiltonian of the Atwood machine and find its acceleration.

3. 15.3 Find all four Hamilton equations for a polar particle in $V(r)$.

4. 15.4 For the spherical pendulum, identify the cyclic coordinate and its momentum.

5. 15.5 Why does nonzero $J$ prevent the spherical pendulum reaching the vertical axis at finite energy?

6. 15.6 Find the minimum bottom speed needed for a rod pendulum to approach the top.

7. 15.7 For a string pendulum to complete a vertical circle while taut, find the bottom-speed threshold.

8. 15.8 Why can substituting $\dot\theta=\ell/(mr^2)$ directly into $L$ before variation give a wrong radial equation?

9. 15.9 Which formulation would you use to recover an unknown string tension after solving motion?

10. 15.10 For $H=p^2/(2M)-Fq$, solve with initial data $q_0,p_0$.

11. 15.11 Explain the difference between a central-force orbit in physical space and its radial phase curve.

12. 15.12 For a fixed-length pendulum, verify that Hamilton’s equations reproduce the exact angular equation.