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Poisson brackets for a particle in a magnetic field

  1. Jun 2, 2012 #1
    I'm struggling to understand Poisson brackets a little here... excerpt from some notes:

    I am, however, stumped on how this Poisson bracket has been computed. I presume ra and Aa(r) are my canonical coordinates, and I have [tex] \dot{r}_a = p_a - \frac{e}{c}A_a (r) [/tex] with [tex] A_a = \frac{1}{2}\epsilon _{abc}B_br_c [/tex]

    Unfortunately, my calculations on paper aren't getting anywhere! Could someone please shed some light here? I suspect something is wrong with the canonical coordinates I'm trying to use to do the derivatives for the Poisson brackets, or maybe I'm getting my indices muddled.

    Thanks :)
     
  2. jcsd
  3. Jun 2, 2012 #2
    Notice that:
    [tex]
    \left\lbrace p_a, f(\mathbf{r}) \right\rbrace = \sum_{b}{\left(\frac{\partial p_a}{\partial x_b} \, \frac{\partial f(\mathbf{r})}{\partial p_b} - \frac{\partial p_a}{\partial p_b} \, \frac{\partial f(\mathbf{r})}{\partial x_b}\right)} = -\frac{\partial f(\mathbf{r})}{\partial x_a}
    [/tex]

    Therefore:
    [tex]
    \begin{align*}
    \left\lbrace m \, \dot{r}_a, m \, \dot{r}_b \right\rbrace & = \left\lbrace p_a - \frac{e}{c} \, A_a(\mathbf{r}), p_b - \frac{e}{c} \, A_b(\mathbf{r}) \right\rbrace \\
    & = \left\lbrace p_a, p_b \right\rbrace - \frac{e}{c} \, \left\lbrace p_a, A_b(\mathbf{r})\right\rbrace + \frac{e}{c} \, \left\lbrace p_b, A_a(\mathbf{r}) \right\rbrace + \left(\frac{e}{c} \right)^2 \, \left\lbrace A_a(\mathbf{r}), A_b(\mathbf{r}) \right\rbrace
    \end{align*}
    [/tex]

    Two of these are identically zero, and for two of them you can use the hint I gave in the beginning. Then, use the properties of the Levi-Civita symbol and the definition of curl of a vector potential.
     
  4. Jun 2, 2012 #3
    Thanks for your help Dickfore.

    I'm not sure where your first expression comes from. I suppose Aa(r) is f(r), but what is xa? Thus, I'm having trouble seeing how to apply that to where you've broken down the Poisson brackets by linearity (I see that the first and last terms there are zero, though).

    Classical mechanics usually makes sense to me, but the whole Hamiltonian formalism just isn't clicking very well...
     
  5. Jun 2, 2012 #4
    [itex]x_{a}[/itex] stands for the a-th component of the position vector, just a [itex]p_{a}[/itex] is the a-th component of the momentum vector.
     
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