# Hamiltonian of charged particle

1. May 2, 2007

### Mindscrape

A charged particle of mass m is attracted by a central force with magnitude $$F = \frac{k}{r^2}$$. Find the Hamiltonian of the particle.

I'm just wondering if I did this correctly because it seemed too easy. First I used the fact that -dU/dr = F = k/r^2, so the potential (with infinite boundary) is given by
$$U(r) = \frac{-k}{r}$$

Then using plane polar coordinates the Legrangian will be
$$L = T - U = \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\theta}^2) + \frac{k}{r}$$

The general momenta will be given by
$$\frac{\partial L}{\partial \dot{r}} = p_r = m \dot{r}/[tex] and [tex] \frac{\partial L}{\partial \dot{\theta}}= p_\theta = mr^2 \dot{\theta}$$

Putting the momenta in terms of the dots of the generalized coordinates
$$\dot{r} = \frac{p_r}{m}$$
and
$$\dot{\theta} = \frac{p_\theta}{mr^2}$$

So the Hamiltonian will be
$$H(q_k, p_k) = \sum_j p_j \dot{q}_j - L(q_k, \dot{q}_k)$$
i.e.
$$H(r, \theta, \dot{r}, \dot{\theta}) = \frac{p_r^2}{m} + \frac{p_\theta^2}{mr^2} - \frac{1}{2}m(\dot{r}^2 + r^2 \dot{\theta}^2) - \frac{k}{r}$$
and with the momenta equations
$$H(r, \theta, \dot{r}, \dot{\theta}) = \frac{p_r^2}{m} + \frac{p_\theta^2}{mr^2} - \frac{1}{2}m((\frac{p_r}{m})^2 + r^2 (\frac{p_\theta}{mr^2})^2) - \frac{k}{r}$$

simplified this will give the familiar H = T + U
$$H(r, \theta, \dot{r}, \dot{\theta}) = \frac{1}{2} \frac{p_r^2}{m} + \frac{1}{2} \frac{p_\theta^2}{mr^2} - \frac{k}{r}$$

Last edited: May 2, 2007
2. May 3, 2007

### lalbatros

looks ok
if it is too easy, have look at the hamiltonian for charges and fields in the full EM theory, could be a bit more challenging and interresting

3. May 3, 2007

### Fredrik

Staff Emeritus
You got the sign of U wrong.