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Hamiltonian of charged particle

  1. May 2, 2007 #1
    A charged particle of mass m is attracted by a central force with magnitude [tex] F = \frac{k}{r^2} [/tex]. 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
    [tex] U(r) = \frac{-k}{r} [/tex]

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

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

    Putting the momenta in terms of the dots of the generalized coordinates
    [tex] \dot{r} = \frac{p_r}{m}[/tex]
    [tex] \dot{\theta} = \frac{p_\theta}{mr^2}[/tex]

    So the Hamiltonian will be
    [tex]H(q_k, p_k) = \sum_j p_j \dot{q}_j - L(q_k, \dot{q}_k)[/tex]
    [tex]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}[/tex]
    and with the momenta equations
    [tex]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}[/tex]

    simplified this will give the familiar H = T + U
    [tex]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}[/tex]
    Last edited: May 2, 2007
  2. jcsd
  3. May 3, 2007 #2
    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
  4. May 3, 2007 #3


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    You got the sign of U wrong.
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