# Poisson brackets for a particle in a magnetic field

1. Jun 2, 2012

### joriarty

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 $$\dot{r}_a = p_a - \frac{e}{c}A_a (r)$$ with $$A_a = \frac{1}{2}\epsilon _{abc}B_br_c$$

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. Jun 2, 2012

### Dickfore

Notice that:
$$\left\lbrace p_a, f(\mathbf{r}) \right\rbrace = \sum_{b}{\left(\frac{\partial p_a}{\partial x_b} \, \frac{\partial f(\mathbf{r})}{\partial p_b} - \frac{\partial p_a}{\partial p_b} \, \frac{\partial f(\mathbf{r})}{\partial x_b}\right)} = -\frac{\partial f(\mathbf{r})}{\partial x_a}$$

Therefore:
\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*}

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.

3. Jun 2, 2012

### joriarty

$x_{a}$ stands for the a-th component of the position vector, just a $p_{a}$ is the a-th component of the momentum vector.