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 $\mathrm{s^{-1}}$.
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