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Homework Help: The vector potential and the Hamiltonian?

  1. Oct 28, 2009 #1
    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

    [tex]\dot{A}[/tex] = [tex]\partial _{t} A + (v \bullet \nabla)A[/tex]

    My attempted solution:

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

    [tex]\frac{dr_{j}}{dt} = \frac{\partial H}{\partial p_{j}}[/tex] and [tex]\frac{dp_{j}}{dt} = - \frac{\partial H}{\partial r_{j}}[/tex]

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

    and : [tex]\ \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} [/tex]

    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 [tex]\ \dot{A} = \frac{1}{q}(\frac{dp_{j}}{dt} - m\frac{dv_{j}}{dt}) [/tex]

    [tex]\ = \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})[/tex]

    [tex]\ = (\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}) [/tex]

    [tex]\ = (v_{j}.\nabla)A_{j} - \frac{\partial X}{\partial r_{j}} - m\frac{dv_{j}}{dt} [/tex]

    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?

  2. jcsd
  3. Oct 28, 2009 #2


    User Avatar
    Homework Helper
    Gold Member

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

    [tex]\dot{A}=\frac{dA}{dt}=\partial_t A+\frac{dr_i}{dt}\frac{\partial A}{\partial r_i}[/tex]
    Last edited: Oct 28, 2009
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