M07M2

M07M2 is a short name for the second problem in the Classical Mechanics section of the May 2007 Princeton University Prelims. The problem statement can be found in the problems list. Here is the solution.

(a)

Since there is no gravity, the Lagrangian is simply:

$$\mathcal{L}=\frac{1}{2}m\dot{\vec{x}}^2$$

$$\dot{\vec{x}}=\left(acos\theta cos\phi \dot{\theta}-a\sin\theta sin\phi\dot{\phi}\, ,\, acos\theta\dot{\theta}sin\phi+asin\theta cos\phi\dot{\phi}\, ,\, -bsin\theta\dot{\theta}\right)$$

$$\dot{\vec{x}}^2=a^2cos^2\theta \dot{\theta}^2+a^2\sin^2\theta \dot{\phi}^2+b^2 sin^2\theta \dot{\theta}^2$$

$$\mathcal{L}=\frac{1}{2}m\left(a^2cos^2\theta \dot{\theta}^2+a^2\sin^2\theta \dot{\phi}^2+b^2 sin^2\theta \dot{\theta}^2\right)$$

$$\frac{d}{dt}\frac{\partial{\mathcal{L}}}{\partial{\dot{\theta}}}=\frac{\partial{\mathcal{L}}}{\partial\theta}$$

$$\frac{d}{dt}\frac{\partial{\mathcal{L}}}{\partial{\dot{\phi}}}=\frac{\partial{\mathcal{L}}}{\partial\phi}$$

$$-a^2\dot{\theta}^2+a^2cot\theta\ddot{\theta}+b^2\dot{\theta}^2+b^2tan\theta\ddot{\theta}=a^2\dot{\phi}^2$$

$$\label{conserved}\frac{dA}{dt}=0 \quad A=\dot{\phi}sin^2\theta$$

(b)

Notice that since there is no potential energy, the total energy is just the kinetic energy, and it equals the lagrangian:

$$E=\mathcal{L}=\frac{1}{2}m\left[a^2\left(cos^2\theta\dot{\theta}^2+\frac{A^2}{sin^2\theta}\right)+b^2sin^2\theta\dot{\theta}^2\right]$$

The endpoints of the motion will be those for which the particle is instantaneously at rest, i.e., $$\dot{\theta}=0$$. Those are found at:

$$E=\frac{1}{2}m\left[a^2\left(\frac{A^2}{sin^2\theta}\right)\right]$$

$$2Esin^2\theta=ma^2A^2$$

Then we can solve the energy equation for $$\dot{\theta}$$:

$$\dot{\theta}^2=\frac{2Esin^2\theta -ma^2A^2}{msin^2\theta(a^2cos^2\theta +b^2sin^2\theta)}\equiv -V_{E,A}$$

The period of motion is:

$$T=\int_0^T dt=\oint \frac{dt}{d\theta} d\theta$$

We can visualize the motion as being from one endpoint to the other and then back, or equivalently twice between endpoints (since the energy is conserved, the integral is independent of path):

$$T=2\int_{\theta_-}^{\theta_+} \frac{d\theta}{\sqrt{-V_{E,A}}}$$

This completes the proof, after reminding that $$\theta_-$$ and $$\theta_+$$ are the roots of the equation that defines the endpoints of the motion:

$$\theta_{\pm}=arcsin\left(\pm\sqrt{\frac{ma^2A^2}{2E}}\right)$$

Note that these are also the roots of $$V_{A,E}=0$$, so the proof is complete.