A particle of mass \(m\) and charge \(q\) is moving around a fixed point charge \(-q^{\prime}\left(q q^{\prime}>0\right)\), and in a uniform magnetic field \(\boldsymbol{B}\). The motion is confined to the plane perpendicular to \(\boldsymbol{B}\). Write down the Lagrangian function in polar co-ordinates rotating with the Larmor angular velocity \(\omega_{\mathrm{L}}=-q B / 2 m\) (see \(\left.\S 5.5\right) .\) Hence find the Hamiltonian function. Show that \(\varphi\) is ignorable, and interpret the conservation law. (Note that \(J_{z}\) is not a constant of the motion.)

First, we need to express the position and velocity of the particle in rotating polar coordinates. Let \(\boldsymbol{r}\) and \(\boldsymbol{v}\) be the position and velocity of the particle in the lab frame. In polar coordinates, they are given by: \begin{align*} \boldsymbol{r} &= r\hat{r}\\ \boldsymbol{v} &= \dot{r}\hat{r} + r\dot{\theta}\hat{\theta} \end{align*} In the rotating frame with angular velocity \(\omega_{L}\), the position remains the same but the velocity vector is modified by the Coriolis effect. The velocity of a point in the rotating frame is given by: \(\boldsymbol{v'} = \boldsymbol{v} - \boldsymbol{\omega}_{L} \times \boldsymbol{r}\) So, we have: \(\boldsymbol{v'} = \dot{r}\hat{r} + r\dot{\theta}\hat{\theta} + \frac{qB}{2m}r\hat{\theta}\) Now, the kinetic energy of the particle is given by: \(T = \frac{1}{2}m\boldsymbol{v'}^2 = \frac{1}{2}m(\dot{r}^2 + r^2(\dot{\theta} + \frac{qB}{2m})^2)\) The potential energy is given by the electrostatic interaction between the charges: \(U = \frac{qq'}{4\pi\epsilon_{0}r}\) The Lagrangian function is then: \(L = T - U = \frac{1}{2}m(\dot{r}^2 + r^2(\dot{\theta} + \frac{qB}{2m})^2) - \frac{qq'}{4\pi\epsilon_{0}r}\)

We will use the following formula to find the Hamiltonian function from the Lagrangian: \(H = p_{r}\dot{r} + p_{\theta}\dot{\theta} - L\) In order to find the conjugate momenta, we take the derivatives of the Lagrangian with respect to \(\dot{r}\) and \(\dot{\theta}\): \begin{align*} p_{r} &= \frac{\partial L}{\partial\dot{r}} = m\dot{r}\\ p_{\theta} &= \frac{\partial L}{\partial\dot{\theta}} = mr^2(\dot{\theta} + \frac{qB}{2m}) \end{align*} Next, we solve for \(\dot{r}\) and \(\dot{\theta}\): \begin{align*} \dot{r} &= \frac{p_r}{m}\\ \dot{\theta} &= \frac{p_{\theta}}{mr^2} - \frac{qB}{2m} \end{align*} Substitute these expressions back into the formula for the Hamiltonian: \(H = p_{r}\frac{p_{r}}{m} + p_{\theta}\left(\frac{p_{\theta}}{mr^2} - \frac{qB}{2m}\right) - L\) After simplification we get: \(H = \frac{p_{r}^2}{2m} + \frac{p_{\theta}^2}{2mr^2} - \frac{qBp_{\theta}}{2m}- \frac{1}{2}m r^{2}\left(\frac{qB}{2m}\right)^{2} +\frac{qq'}{4\pi\epsilon_{0} r}\)

Notice that the Lagrangian does not depend on \(\varphi\), so we can say that \(\varphi\) is ignorable. Consequently, the conjugate momentum associated with \(\varphi\), which is \(p_\theta\), is conserved. This conservation law means that the angular momentum of the particle projected onto the \(z\)-axis is conserved during the motion. It is important to note that the total angular momentum \(J_{z}\) is not conserved because the magnetic field \(\boldsymbol{B}\) applies a torque on the particle. However, the projection of the angular momentum onto the \(z\)-axis, given by \(p_{\theta}\), is conserved due to the ignorable \(\varphi\) coordinate.