UPSC Maths 2025 Paper 2 Q8c-ii — Solution

10 marks · Section B

Question

Consider the Lagrangian $L = m\dot{x}\dot{y} - m\omega_0^2 xy$ where $m$ and $\omega_0$ are constants. Find the Hamiltonian and Hamilton’s equations of motion. Identify the system.

Technique

Take the Legendre transform of this cross-coupled Lagrangian: compute the conjugate momenta, invert them, form $H = p_x\dot x + p_y\dot y - L$ , write Hamilton’s equations, and decouple via the equations of motion to identify the system.

Solution

Canonical momenta

$p_x = \frac{\partial L}{\partial\dot x} = m\dot y,\qquad p_y = \frac{\partial L}{\partial\dot y} = m\dot x.$

Note the cross structure: $p_x$ depends on $\dot y$ and $p_y$ on $\dot x$ . Invert: $\dot x = \frac{p_y}{m},\qquad \dot y = \frac{p_x}{m}.$

Hamiltonian

$H = p_x\dot x + p_y\dot y - L = p_x\frac{p_y}{m} + p_y\frac{p_x}{m} - \Big(m\dot x\dot y - m\omega_0^2 xy\Big).$ Now $m\dot x\dot y = m\cdot\frac{p_y}{m}\cdot\frac{p_x}{m} = \frac{p_x p_y}{m}$ , so $H = \frac{2p_x p_y}{m} - \frac{p_x p_y}{m} + m\omega_0^2 xy = \frac{p_x p_y}{m} + m\omega_0^2 xy.$

$\boxed{H = \frac{p_x p_y}{m} + m\omega_0^2\,xy.}$

Hamilton’s equations

$\dot x = \frac{\partial H}{\partial p_x} = \frac{p_y}{m},\qquad \dot y = \frac{\partial H}{\partial p_y} = \frac{p_x}{m},$ $\dot p_x = -\frac{\partial H}{\partial x} = -m\omega_0^2\,y,\qquad \dot p_y = -\frac{\partial H}{\partial y} = -m\omega_0^2\,x.$

Identify the system

Differentiate $\dot x = p_y/m$ and use $\dot p_y = -m\omega_0^2 x$ : $\ddot x = \frac{\dot p_y}{m} = -\omega_0^2 x \implies \ddot x + \omega_0^2 x = 0.$ Similarly $\ddot y = \dfrac{\dot p_x}{m} = -\omega_0^2 y \implies \ddot y + \omega_0^2 y = 0.$

So $x$ and $y$ each execute independent simple harmonic motion of angular frequency $\omega_0$ . Although the Lagrangian/Hamiltonian is written in coupled (bilinear) form, the dynamics decouples: the system is two independent one-dimensional harmonic oscillators, each of angular frequency $\omega_0$ .

Answer

$H = \frac{p_x p_y}{m} + m\omega_0^2 xy,$ $\dot x = \frac{p_y}{m},\quad \dot y = \frac{p_x}{m},\quad \dot p_x = -m\omega_0^2 y,\quad \dot p_y = -m\omega_0^2 x.$ The equations of motion are $\ddot x + \omega_0^2 x = 0$ and $\ddot y + \omega_0^2 y = 0$ : the system is two independent simple harmonic oscillators of angular frequency $\omega_0$ .