# The vector potential and the Hamiltonian?

1. Oct 28, 2009

### jeebs

Hi,
I have a problem involving the Hamiltonian of a particle of mass m, charge q, position r, momentum p, in an external field defined by vector potential A and scalar potential X. Here's the Hamiltonian:

H(r,p) = (1/2m)[p - qA(r,t)]2 + qX(r,t) = (1/2m)(pjpj - 2qpjAj + q2AjAj) + qX

The problem I have to solve is that i have to show that

$$\dot{A}$$ = $$\partial _{t} A + (v \bullet \nabla)A$$

My attempted solution:

When I put this into Hamilton's equations of motion, (which are the following:)

$$\frac{dr_{j}}{dt} = \frac{\partial H}{\partial p_{j}}$$ and $$\frac{dp_{j}}{dt} = - \frac{\partial H}{\partial r_{j}}$$

I get : $$\ \frac{\partial H}{\partial p_{j}} = (1/m)(p_{j} - qA_{j}) = \frac{dr_{j}}{dt} = v_{j}$$

and : $$\ \frac{\partial H}{\partial r_{j}} = (\frac{q^{2}A_{j}}{m} - \frac{qp_{j}}{m})\frac{\partial A_{j}}{\partial r_{j}} + q\frac{\partial X}{\partial r_{j}}= - \frac{dp_{j}}{dt} = F_{j}$$

I am 99% certain these are correct.

Using what I got for the first differential equation, i find that A_{j}=(1/q)(p_{j} - mv_{j})

hence $$\ \dot{A} = \frac{1}{q}(\frac{dp_{j}}{dt} - m\frac{dv_{j}}{dt})$$

$$\ = \frac{1}{q}((-(\frac{q^2A_{j}}{m} - \frac{qp_{j}}{m})\frac{dA_{j}}{dr_{j}} + q\frac{\partial X}{\partial r_{j}}) - m\frac{dv_{j}}{t})$$

$$\ = (\frac{p_{j}}{m} - \frac{qA_{j}}{m})\frac{\partial A_{j}}{\partial r_{j}} - \frac{\partial X}{\partial r_{j}} - m\frac{dv_{j}}{dt})$$

$$\ = (v_{j}.\nabla)A_{j} - \frac{\partial X}{\partial r_{j}} - m\frac{dv_{j}}{dt}$$

and this is as far as I get really.

First off, am I correct in the way i got to the (v.\nabla) part? (ie. using the first Hamilton equation for velocity?)

Assuming that is correct, I haven't been able to find a way to deal with the other terms in the equation... the advice given on the question sheet is to use the chain rule in index notation, but I don't see where this comes in handy for this question.

Can anyone offer any advice on how to tackle this one?

Thanks.

2. Oct 28, 2009

### gabbagabbahey

You're making things much too complicated...$A$ is a function of $r_i$ and $t$, so the chain rule tells you

$$\dot{A}=\frac{dA}{dt}=\partial_t A+\frac{dr_i}{dt}\frac{\partial A}{\partial r_i}$$

Last edited: Oct 28, 2009