25 Jun 2025
Hamiltonian Dynamics and Mechanical Examples
Canonical momenta, cyclic coordinates, conservation laws, Hamilton equations, and representative systems.
From the Lagrangian to Hamilton’s equations
For a regular Lagrangian, define the generalized momenta and Hamiltonian by
\[p_i=\frac{\partial L}{\partial\dot q_i}, \qquad H(q,p,t)=\sum_i p_i\dot q_i-L.\]Regularity means that the velocity Hessian is non-singular, so the equations $p_i=L_{\dot q_i}$ can be inverted to express $\dot q_i$ in terms of $(q,p,t)$. Starting with
\[dL =\sum_i\frac{\partial L}{\partial q_i}\,dq_i +\sum_i p_i\,d\dot q_i +\frac{\partial L}{\partial t}\,dt,\]the differential of $H$ is
\[dH =\sum_i\dot q_i\,dp_i -\sum_i\frac{\partial L}{\partial q_i}\,dq_i -\frac{\partial L}{\partial t}\,dt.\]Using $\dot p_i=\partial L/\partial q_i$ and comparing coefficients gives Hamilton’s equations,
\[\boxed{\dot q_i=\frac{\partial H}{\partial p_i}}, \qquad \boxed{\dot p_i=-\frac{\partial H}{\partial q_i}}, \qquad \boxed{\frac{\partial H}{\partial t}=-\frac{\partial L}{\partial t}}.\]Cyclic coordinates and conservation theorems
If $q_k$ does not occur explicitly in $L$, it is cyclic and
\[\dot p_k=\frac{\partial L}{\partial q_k}=0.\]Thus its conjugate momentum is conserved. Along a Hamiltonian trajectory,
\[\frac{dH}{dt} =\sum_i\left( \frac{\partial H}{\partial q_i}\dot q_i +\frac{\partial H}{\partial p_i}\dot p_i \right) +\frac{\partial H}{\partial t} =\frac{\partial H}{\partial t}.\]Hence $H$ is conserved when it has no explicit time dependence. For a natural Lagrangian $L=T-V$ with $T$ quadratic in the velocities and with time-independent constraints, $H=T+V$, the mechanical energy. Translational invariance makes the corresponding Cartesian momentum constant, while rotational invariance makes the corresponding angular momentum constant.
Hamilton’s equations in different coordinate systems
In Cartesian coordinates,
\[H=\frac{p_x^2+p_y^2+p_z^2}{2m}+V(x,y,z),\]so $\dot x_i=p_i/m$ and $\dot p_i=-\partial V/\partial x_i$.
In cylindrical coordinates $(\rho,\phi,z)$,
\[T=\frac m2(\dot\rho^2+\rho^2\dot\phi^2+\dot z^2),\] \[p_\rho=m\dot\rho, \qquad p_\phi=m\rho^2\dot\phi, \qquad p_z=m\dot z,\]and
\[\boxed{ H=\frac{p_\rho^2}{2m} +\frac{p_\phi^2}{2m\rho^2} +\frac{p_z^2}{2m}+V(\rho,\phi,z)}.\]Hamilton’s equations are
\[\dot\rho=\frac{p_\rho}{m}, \qquad \dot\phi=\frac{p_\phi}{m\rho^2}, \qquad \dot z=\frac{p_z}{m},\] \[\dot p_\rho=\frac{p_\phi^2}{m\rho^3}-\frac{\partial V}{\partial\rho}, \qquad \dot p_\phi=-\frac{\partial V}{\partial\phi}, \qquad \dot p_z=-\frac{\partial V}{\partial z}.\]In spherical coordinates $(r,\theta,\phi)$,
\[T=\frac m2\left(\dot r^2+r^2\dot\theta^2+r^2\sin^2\theta\dot\phi^2\right),\] \[p_r=m\dot r, \qquad p_\theta=mr^2\dot\theta, \qquad p_\phi=mr^2\sin^2\theta\dot\phi,\]and
\[\boxed{ H=\frac{p_r^2}{2m} +\frac{p_\theta^2}{2mr^2} +\frac{p_\phi^2}{2mr^2\sin^2\theta} +V(r,\theta,\phi)}.\]Thus
\[\dot r=\frac{p_r}{m}, \qquad \dot\theta=\frac{p_\theta}{mr^2}, \qquad \dot\phi=\frac{p_\phi}{mr^2\sin^2\theta},\] \[\dot p_r =\frac{p_\theta^2}{mr^3} +\frac{p_\phi^2}{mr^3\sin^2\theta} -\frac{\partial V}{\partial r},\] \[\dot p_\theta =\frac{p_\phi^2\cos\theta}{mr^2\sin^3\theta} -\frac{\partial V}{\partial\theta}, \qquad \dot p_\phi=-\frac{\partial V}{\partial\phi}.\]The extra momentum terms are consequences of the coordinate-dependent scale factors, not additional forces.
One- and two-dimensional harmonic oscillators
For one dimension,
\[H=\frac{p^2}{2m}+\frac12m\omega^2q^2.\]Hamilton’s equations give $\dot q=p/m$ and $\dot p=-m\omega^2q$, hence $\ddot q+\omega^2q=0$. For two independent directions,
\[H=\frac{p_x^2+p_y^2}{2m} +\frac m2(\omega_x^2x^2+\omega_y^2y^2).\]The pairs $(x,p_x)$ and $(y,p_y)$ obey independent Hamilton equations. When $\omega_x=\omega_y$, the oscillator is isotropic.
Central-force motion
For $V=V(r)$, angular momentum is conserved, so the motion lies in a plane. In plane polar coordinates,
\[H=\frac{p_r^2}{2m}+\frac{p_\phi^2}{2mr^2}+V(r), \qquad p_\phi=mr^2\dot\phi=\ell.\]The radial equations are
\[\dot r=\frac{p_r}{m}, \qquad \dot p_r=\frac{\ell^2}{mr^3}-\frac{dV}{dr}.\]Equivalently, the radial motion has the effective potential
\[V_{\mathrm{eff}}(r)=V(r)+\frac{\ell^2}{2mr^2}.\]Charged particle in an electromagnetic field
For
\[L=\frac12m\dot{\mathbf r}^{\,2} +q\dot{\mathbf r}\cdot\mathbf A-q\phi,\]the canonical momentum is
\[\mathbf p=m\dot{\mathbf r}+q\mathbf A.\]It differs from the mechanical momentum $m\dot{\mathbf r}$; both $\mathbf p$ and $q\mathbf A$ have SI units $\mathrm{kg\,m\,s^{-1}}$. Solving for the velocity and taking the Legendre transform gives
\[\boxed{ H=\frac{[\mathbf p-q\mathbf A(\mathbf r,t)]^2}{2m} +q\phi(\mathbf r,t)}.\]The first Hamilton equation gives $\dot{\mathbf r}=(\mathbf p-q\mathbf A)/m$. The second gives
\[\dot p_i=q\dot x_j\frac{\partial A_j}{\partial x_i} -q\frac{\partial\phi}{\partial x_i}.\]Differentiating $m\dot x_i=p_i-qA_i$ then recovers
\[m\ddot{\mathbf r}=q(\mathbf E+\dot{\mathbf r}\times\mathbf B).\]Compound pendulum
Let a rigid body of mass $M$ have moment of inertia $I$ about its pivot, with its centre of mass a distance $a$ from the pivot. Taking the lowest position as zero potential,
\[L=\frac12I\dot\theta^2-Mga(1-\cos\theta), \qquad p_\theta=I\dot\theta.\]Therefore
\[\boxed{ H=\frac{p_\theta^2}{2I}+Mga(1-\cos\theta)}.\]Hamilton’s equations yield
\[\dot\theta=\frac{p_\theta}{I}, \qquad \dot p_\theta=-Mga\sin\theta,\]and hence $I\ddot\theta+Mga\sin\theta=0$. For $\lvert\theta\rvert\ll1$ rad, $\sin\theta\simeq\theta$, so
\[\boxed{\omega_0=\sqrt{\frac{Mga}{I}}},\]with $\omega_0$ measured in $\mathrm{s^{-1}}$.
Solved Problems
1. Circular orbit in an attractive inverse-square field
For the central potential $V(r)=-\kappa/r$, find the radius and energy of a circular orbit with non-zero angular momentum $\ell$, and test its radial stability.
The planar Hamiltonian is
\[H=\frac{p_r^2}{2m}+V_{\mathrm{eff}}(r), \qquad V_{\mathrm{eff}}(r)=\frac{\ell^2}{2mr^2}-\frac{\kappa}{r}.\]A circular orbit has $p_r=0$ and constant $r=r_0$, so $\dot p_r=-dV_{\mathrm{eff}}/dr=0$. Therefore
\[-\frac{\ell^2}{mr_0^3}+\frac{\kappa}{r_0^2}=0,\]which gives
\[\boxed{r_0=\frac{\ell^2}{m\kappa}}.\]Substitution into the Hamiltonian gives
\[E_0=\frac{\ell^2}{2mr_0^2}-\frac{\kappa}{r_0} =\frac{\kappa}{2r_0}-\frac{\kappa}{r_0} =\boxed{-\frac{\kappa}{2r_0} =-\frac{m\kappa^2}{2\ell^2}}.\]The second derivative is
\[\frac{d^2V_{\mathrm{eff}}}{dr^2} =\frac{3\ell^2}{mr^4}-\frac{2\kappa}{r^3}.\]Using $\ell^2/m=\kappa r_0$,
\[\left.\frac{d^2V_{\mathrm{eff}}}{dr^2}\right\rvert_{r_0} =\frac{3\kappa}{r_0^3}-\frac{2\kappa}{r_0^3} =\frac{\kappa}{r_0^3}>0.\]The circular orbit is therefore a stable minimum of $V_{\mathrm{eff}}$. The small radial angular frequency is $\sqrt{\kappa/(mr_0^3)}$, which for $\kappa=GMm$ reduces to $\sqrt{GM/r_0^3}$.
2. Compound pendulum released from a finite angle
A compound pendulum has $M=2.0\ \mathrm{kg}$, $a=0.25\ \mathrm m$, and $I=0.18\ \mathrm{kg\,m^2}$. It is released from rest at $\theta_0=60^\circ$. Find its angular speed at the bottom and its small-oscillation period.
Because the Hamiltonian has no explicit time dependence,
\[H=\frac{p_\theta^2}{2I}+Mga(1-\cos\theta)\]is conserved. At release, $p_\theta=0$, so
\[H_0=Mga(1-\cos\theta_0).\]At the bottom, $\theta=0$ and $p_\theta=I\dot\theta$, hence
\[\frac12 I\dot\theta_{\mathrm b}^{\,2} =Mga(1-\cos\theta_0).\]Thus
\[\dot\theta_{\mathrm b} =\sqrt{\frac{2Mga(1-\cos\theta_0)}{I}} =\sqrt{\frac{2(2.0)(9.81)(0.25)(1-0.5)}{0.18}} =\boxed{5.220\ \mathrm{s^{-1}}}.\]For $\lvert\theta\rvert\ll1$, the period is
\[T_0=2\pi\sqrt{\frac{I}{Mga}} =2\pi\sqrt{\frac{0.18}{(2.0)(9.81)(0.25)}} =\boxed{1.204\ \mathrm s}.\]The finite-angle release determines the bottom speed, whereas the displayed period is the small-angle limit; using it at large amplitude would neglect the nonlinearity of $\sin\theta$.
Descriptive Questions
- Derive Hamilton’s equations as a Legendre transform of a regular Lagrangian, stating what fails when the velocity Hessian is singular.
- Explain how cyclic coordinates encode translational or rotational symmetries and derive the corresponding conservation law.
- Construct the Hamiltonian in spherical coordinates from the kinetic energy, carefully identifying all scale-factor terms.
- Distinguish canonical and mechanical momentum for a charged particle and explain why the Hamiltonian is gauge-dependent while the Lorentz force is not.
Numerical Problems
- A one-dimensional oscillator has $m=0.50\ \mathrm{kg}$ and $\omega=4.0\ \mathrm{s^{-1}}$. At one instant $q=0.10\ \mathrm m$ and $p=0.60\ \mathrm{kg\,m\,s^{-1}}$. Find its energy and amplitude. Answer: $0.4000\ \mathrm J$, $0.3162\ \mathrm m$.
- A two-dimensional oscillator has $\omega_x=2.0\ \mathrm{s^{-1}}$ and $\omega_y=3.0\ \mathrm{s^{-1}}$. Find the two periods and the least positive time at which both phase-space pairs return simultaneously. Answer: $T_x=3.142\ \mathrm s$, $T_y=2.094\ \mathrm s$, $T_{\mathrm{return}}=6.283\ \mathrm s$.
- For $m=0.020\ \mathrm{kg}$, $q=0.010\ \mathrm C$, $\mathbf A=0.40\hat{\mathbf y}\ \mathrm{T\,m}$, $\phi=3.0\ \mathrm V$, and $\mathbf p=(0.10\hat{\mathbf x}+0.080\hat{\mathbf y})\ \mathrm{kg\,m\,s^{-1}}$, find $\dot x$, $\dot y$, and $H$. Answer: $5.000\ \mathrm{m\,s^{-1}}$, $3.800\ \mathrm{m\,s^{-1}}$, $0.4244\ \mathrm J$.
- A free particle of mass $2.0\ \mathrm{kg}$ has cylindrical phase-space data $\rho=0.50\ \mathrm m$, $p_\rho=0.40\ \mathrm{kg\,m\,s^{-1}}$, $p_\phi=0.30\ \mathrm{kg\,m^2\,s^{-1}}$, and $p_z=0$. Find $\dot\rho$, $\dot\phi$, and $\dot p_\rho$. Answer: $0.200\ \mathrm{m\,s^{-1}}$, $0.600\ \mathrm{s^{-1}}$, $0.360\ \mathrm N$.
Maxima verification: central-orbit residuals and all solved/numerical values.
References
- Hamiltonian mechanics — Wikipedia
- H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Chapters 3 and 8, Pearson (2002).
- L. D. Landau and E. M. Lifshitz, Mechanics, 3rd ed., Sections 40–45, Butterworth-Heinemann (1976).
- J. R. Taylor, Classical Mechanics, Chapter 13, University Science Books (2005).
Discussion