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GATE 2024 PH – Question 38

Classical Mechanics · Hamiltonian and Hamilton's equations of motion · 2 marks · Multiple choice

For $L=m\dot x\dot y-m\omega_0^2xy$ with conjugate momenta $p_x,p_y$, the canonical equations are

  1. $\dot x=p_x/m,\dot p_x=-m\omega_0^2x,\dot y=p_y/m,\dot p_y=-m\omega_0^2y$
  2. $\dot x=p_x/m,\dot p_x=m\omega_0^2x,\dot y=p_y/m,\dot p_y=m\omega_0^2y$
  3. $\dot x=p_y/m,\dot p_x=-m\omega_0^2y,\dot y=p_x/m,\dot p_y=-m\omega_0^2x$
  4. $\dot x=p_y/m,\dot p_x=m\omega_0^2y,\dot y=p_x/m,\dot p_y=m\omega_0^2x$

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Correct answer: (C) $\dot x=p_y/m,\dot p_x=-m\omega_0^2y,\dot y=p_x/m,\dot p_y=-m\omega_0^2x$

Explanation

**Momenta** from $L=m\dot x\dot y-m\omega_0^2xy$:
$$p_x=\frac{\partial L}{\partial\dot x}=m\dot y,\qquad p_y=\frac{\partial L}{\partial\dot y}=m\dot x .$$

**Hamiltonian:** $H=p_x\dot x+p_y\dot y-L$. With $\dot x=p_y/m$ and $\dot y=p_x/m$:
$$H=\frac{p_xp_y}{m}+\frac{p_xp_y}{m}-\left(\frac{p_xp_y}{m}-m\omega_0^2xy\right)=\frac{p_xp_y}{m}+m\omega_0^2xy .$$

**Hamilton's equations:**
- $\dot x=\dfrac{\partial H}{\partial p_x}=\dfrac{p_y}{m}$
- $\dot y=\dfrac{\partial H}{\partial p_y}=\dfrac{p_x}{m}$
- $\dot p_x=-\dfrac{\partial H}{\partial x}=-m\omega_0^2y$
- $\dot p_y=-\dfrac{\partial H}{\partial y}=-m\omega_0^2x$

These match option **C**.