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Consider the Lagrangian \(L = m\dot{x}\dot{y} - m\omega_{0}^{2}xy\). If \(p_{x}\) and \(p_{y}\) denote the generalized momenta conjugate to \(x\) and \(y\), respectively, then the canonical equations of motion are

  • A.\(\dot{x} = \frac{p_{x}}{m},\ \dot{p}_{x} = -m\omega_{0}^{2}x\), \(\dot{y} = \frac{p_{y}}{m},\ \dot{p}_{y} = -m\omega_{0}^{2}y\)
  • B.\(\dot{x} = \frac{p_{x}}{m},\ \dot{p}_{x} = m\omega_{0}^{2}x\), \(\dot{y} = \frac{p_{y}}{m},\ \dot{p}_{y} = m\omega_{0}^{2}y\)
  • C.\(\dot{x} = \frac{p_{y}}{m},\ \dot{p}_{x} = -m\omega_{0}^{2}y\), \(\dot{y} = \frac{p_{x}}{m},\ \dot{p}_{y} = -m\omega_{0}^{2}x\)
  • D.\(\dot{x} = \frac{p_{y}}{m},\ \dot{p}_{x} = m\omega_{0}^{2}y\), \(\dot{y} = \frac{p_{x}}{m},\ \dot{p}_{y} = m\omega_{0}^{2}x\)

Answer: C

The canonical equations are ẋ = py/m, ṗx = -mω02y, ẏ = px/m, ṗy = -mω02x.

The Lagrangian couples the velocities crosswise, so each momentum involves the other velocity:

\[ \begin{aligned} p_{x} &= \frac{\partial L}{\partial \dot{x}} = m\dot{y}, \qquad p_{y} = \frac{\partial L}{\partial \dot{y}} = m\dot{x} \\ H &= p_{x}\dot{x} + p_{y}\dot{y} - L = \frac{p_{x}p_{y}}{m} + m\omega_{0}^{2}xy \end{aligned} \]

Hamilton's equations: \(\dot{x} = \partial H/\partial p_{x} = p_{y}/m\), \(\dot{y} = p_{x}/m\), \(\dot{p}_{x} = -\partial H/\partial x = -m\omega_{0}^{2}y\), \(\dot{p}_{y} = -m\omega_{0}^{2}x\).

The first two options pair ẋ with px, as for an ordinary kinetic term, which this Lagrangian does not have. The last option has the right pairing but the wrong sign in the force terms.

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