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An electron with mass $m$ and charge $-e$ moves in a magnetic field $\mathbf{B}=\nabla \times \mathbf{A}(\mathbf{r})$. It has Lagrangian $$L=\frac{1}{2} m | \dot{\mathbf{r}}|^2-e \,\dot{\mathbf{r}} \cdot \mathbf{A}(\mathbf{r}).$$ From this we can check that the Euler-Lagrange equations recover the Lorentz force law with no external electric field, and the electron satisfies $m \ddot{\mathbf{r}}=e \, \dot{\mathbf{r}} \times \mathbf{B} = e \, \dot{\mathbf{r}} \times (\nabla \times \mathbf{A})$.

I need to show that wrt cylindrical coordinates $(r,\theta,z)$, given $$\mathbf{A} = \frac{f(r)}{r} \mathbf{e}_\theta$$ that when the electron is at a distance $r_0$ from the $z-$axis, that the electron has angular velocity $$\dot{\theta} = \frac{e}{mr^2} [f(r)-f(r_0)].$$


I've tried plugging the vector potential into the equation of motion to get $$\nabla \times \mathbf{A} = \frac{f'(r)}{r} \mathbf{e}_z$$ but can't see where to actually isolate an expression for the angular velocity itself. If you work out the expression for $m \ddot{\mathbf{r}}$ it seems like you get $(e \dot{y}f'(r)/r, -e \dot{x}f'(r)/r, 0)$ but this still doesn't seem to get any closer to finding an expression for the angular velocity.

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We can figure this out just from using the Lagrangian itself, without having to resort to the Lorentz force law.

We should first write the particle velocity in cylindrical coordinates, as the vector potential is using these: $$ \dot{\mathbf{r}} = \dot{r}\hat{\mathbf{r}}+r\dot{\theta}\hat{\boldsymbol{\theta}} + \dot{z}\hat{\mathbf{z}}. $$ Note that $\dot{\theta}$ is just the angular velocity in the $xy$-plane. Expressing the velocity in these coordinates allows us to easily compute the dot product $\dot{\mathbf{r}}\cdot\mathbf{A}$ and write the Lagrangian as $$ L = \frac{1}{2}m\left(\dot{r}^2+r^2\dot{\theta}^2+\dot{z}^2\right) - ef(r)\dot{\theta}. $$ Now the Lagrangian is independent of $\theta$, so the Lagrange equation for $\theta$, $$ \frac{d}{dt}\left(\frac{\partial L}{\partial\dot{\theta}}\right) = \frac{\partial L}{\partial \theta}, $$ becomes $$ \frac{\partial L}{\partial \dot{\theta}} = mr^2\dot{\theta} - ef(r)= \textrm{constant}. $$ As this is a constant, we can set it equal to $\partial L/\partial\dot{\theta}$ evaluated at $r=r_0$. Now, assuming that $\dot{\theta}=0$ at $r=r_0$ (this may be given in your initial conditions), we obtain $$ \dot{\theta}=\frac{e}{mr^2}\left[f(r) - f(r_0)\right] $$ as required.

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