# Charged particle in EM field

1. Feb 17, 2012

### Mindscrape

If you go to the relativistic Hamiltonian, what allows us to go from
$$\dot{\vec{P}} = - \frac{\partial \mathcal{H}}{\partial \vec{x}} = e (\vec{\nabla} \vec{A}) \cdot \dot{\vec{x}} - e \vec{\nabla} \phi$$
to
$$\frac{d}{d t}\left(\frac{m \dot{\vec{x}}} {\sqrt {1 - \frac{\dot{\vec{x}}^2}{c^2}}}\right) = e \vec{E} + e \dot{\vec{x}} \times \vec{B}$$

Is there some identity with
$$\frac{\partial A^i}{\partial x^i}$$
that turns it into a curl that I'm missing?

2. Feb 18, 2012

You should try using both Hamilton equations to find the relation between P and the velocity. Then get rid of P.

For a somewhat different Hamiltonian - see the attached gif form Cordinalesi, "Classical mechanics for physics graduate students".

Last edited: Feb 18, 2012
3. Feb 18, 2012

### Mindscrape

Both Hamilton's equations are used to get to the first line. I'm looking at the RHS and wondering how to go from the RHS of eqn 1 to the RHS of eqn 2. I'm having trouble coming up with an identity that will give me a curl and divergence of the vector potential from its form in eqn 1.

4. Feb 18, 2012

The LHS is not the same, so the RHS are also different - not a surprise.

5. Feb 18, 2012

### Mindscrape

I'm missing something. Why are the two LHS not the same? P is the relativistic momentum, so dP/dt is the LHS, no?

6. Feb 18, 2012

There are two kinds of momenta: kinematic (or "kinetic") momentum and canonical momentum. When there is an electromagnetic potential present, they are not equal. You can find the difference from your Hamiltonian. P there is the canonical momentum, not the kinetic momentum.

7. Feb 19, 2012

### Mindscrape

Ah, I see. So yeah, like you said earlier, I can get a relationship between x'' and p', then go through a whole bunch of tedious algebra to get the result. :(

8. Feb 19, 2012