UPSC 2025 Maths Optional Paper 2 Q8c-i — Step-by-Step Solution

10 marks · Section B

Hamilton's equations · Mechanics & Fluid Dynamics · asked 10× in 13 yrs · Read the full method →

Question

A particle of mass $m$ moves in a force field of potential $V(r) = -\dfrac{k\cos\theta}{r^2}$ , $k$ is constant. Find the Hamiltonian and the Hamilton’s equations in spherical polar coordinates $(r, \theta, \phi)$ .

Technique

Work in spherical polar coordinates: write the kinetic energy, form the canonical momenta $p_q = \partial L/\partial\dot q$ , invert to express $\dot q$ in momenta, build $H = T + V$ (natural, time-independent system), and write Hamilton’s equations $\dot q = \partial H/\partial p_q$ , $\dot p_q = -\partial H/\partial q$ .

Solution

Kinetic energy and Lagrangian

In spherical coordinates $(r,\theta,\phi)$ the speed-squared is

v^2 = \dot r^2 + r^2\dot\theta^2 + r^2\sin^2\theta\,\dot\phi^2.

T = \tfrac12 m\big(\dot r^2 + r^2\dot\theta^2 + r^2\sin^2\theta\,\dot\phi^2\big),\qquad L = T - V = T + \frac{k\cos\theta}{r^2}.

Canonical momenta

p_r = \frac{\partial L}{\partial\dot r} = m\dot r,\quad p_\theta = \frac{\partial L}{\partial\dot\theta} = m r^2\dot\theta,\quad p_\phi = \frac{\partial L}{\partial\dot\phi} = m r^2\sin^2\theta\,\dot\phi.

Invert:

\dot r = \frac{p_r}{m},\qquad \dot\theta = \frac{p_\theta}{m r^2},\qquad \dot\phi = \frac{p_\phi}{m r^2\sin^2\theta}.

Hamiltonian

Since the coordinate transformation is time-independent and $V$ has no velocity dependence, $H = T + V$ expressed in momenta:

\boxed{H = \frac{p_r^2}{2m} + \frac{p_\theta^2}{2m r^2} + \frac{p_\phi^2}{2m r^2\sin^2\theta} - \frac{k\cos\theta}{r^2}.}

Hamilton’s equations

Coordinate equations $\dot q = \partial H/\partial p_q$ :

\dot r = \frac{p_r}{m},\qquad \dot\theta = \frac{p_\theta}{m r^2},\qquad \dot\phi = \frac{p_\phi}{m r^2\sin^2\theta}.

Momentum equations $\dot p_q = -\partial H/\partial q$ :

\dot p_r = -\frac{\partial H}{\partial r} = \frac{p_\theta^2}{m r^3} + \frac{p_\phi^2}{m r^3\sin^2\theta} - \frac{2k\cos\theta}{r^3},

\dot p_\theta = -\frac{\partial H}{\partial\theta} = \frac{p_\phi^2\cos\theta}{m r^2\sin^3\theta} - \frac{k\sin\theta}{r^2},

\dot p_\phi = -\frac{\partial H}{\partial\phi} = 0.

The last equation shows $\phi$ is cyclic (ignorable): $p_\phi = m r^2\sin^2\theta\,\dot\phi$ is conserved (azimuthal angular momentum). The energy $H$ is conserved since $\partial H/\partial t = 0$ .

Answer

H = \frac{p_r^2}{2m} + \frac{p_\theta^2}{2mr^2} + \frac{p_\phi^2}{2mr^2\sin^2\theta} - \frac{k\cos\theta}{r^2},

with Hamilton’s equations

\dot r = \tfrac{p_r}{m},\quad \dot\theta = \tfrac{p_\theta}{mr^2},\quad \dot\phi = \tfrac{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{2k\cos\theta}{r^3},\quad \dot p_\theta = \frac{p_\phi^2\cos\theta}{mr^2\sin^3\theta} - \frac{k\sin\theta}{r^2},\quad \dot p_\phi = 0.

$\phi$ is cyclic, so $p_\phi$ is conserved.