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Homework Help: Electromag - Lorenz gauge

  1. Feb 1, 2007 #1
    I've gotten out most of this question, it's really just the last part that's getting to me at this stage. I'd never seen the http://mathworld.wolfram.com/DeltaFunction.html" [Broken] before so it might be because of that. I've an idea how to do it but I just end up in a mess of partial derivatives. I'd say it's something simple I just can't see.

    1. The problem statement, all variables and given/known data
    "Demonstrate that in the Lorenz Gauge, [tex]\vec{A}(x,t)[/tex] satisfies a wave equation with the current density [tex]\vec{J}(x,t)[/tex] as source, and that for static sources this reduces to a Poisson-like equation.
    Calulate [tex]\vec{A}(x,t)[/tex] for [tex]\vec{J}(x,t)=\vec{J}_0\delta(x-x_0)[/tex]"

    2. Relevant equations
    Lorentz guage: [tex] \vec{\nabla}\cdot\vec{A}=-\mu_0\epsilon_0\frac{dV}{dt}
    delta function:[tex]\int_{I}f(x)\delta(x-x_0)dx=f(x_0)[/tex]
    (once [tex]I[/tex] includes the point [tex]x_0[/tex])
    Otherwise [tex]\delta(x-x_0)=0[/tex]
    and Maxwell's equations.

    3. The attempt at a solution

    The wave equation was relatively easy. Substituting the lorenz gauge into maxwell's equations and getting:
    [tex]-\vec{\nabla}^2\vec{A}+\mu_0\epsilon_0\frac{\partial^2 \vec{A}}{\partial t^2}=\mu_0\vec{J}[/tex]

    For static sources
    [tex] \vec{\nabla}\cdot\vec{A}=0
    So the poisson like equation that you get comes up as: [tex] \vec{\nabla}\cdot V^2=-\frac{\rho}{\epsilon_0}

    Now for the last bit :frown: , since [tex]\vec{A}=\vec{A}(x,t)[/tex] then the wave equation can be simplified down to:
    [tex]-\frac{\partial^2 \vec{A}}{\partial x^2}+\mu_0\epsilon_0\frac{\partial^2 \vec{A}}{\partial t^2}=\mu_0\vec{J_0}\delta(x-x_0)[/tex]
    (with the [tex]J_0[/tex] term substituted in)

    So how can I solve for A? My idea was to isolate [tex]d^2\vec{A}[/tex] and integrate to solve it but that gets too messy. Also over what limits would I integrate? +/- infinity? For the [tex]dx[/tex] that will give [tex]\vec{A}(x_0,t)[/tex]? What happens when I integrate the delta function in terms of [tex]dt[/tex]? I'd say there is something about the delta function that makes this fairly simple but I'm just not accustomed to it..... :grumpy:

    Thanks in advance for any hints you can give me...

    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Feb 1, 2007 #2
    maybe, you can try [tex]A(x,t)[/tex] as a plane wave solution,
    then you can get a solution
  4. Feb 2, 2007 #3


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    "Static sources' means no time dependence of the current vector, hence no time dependence of the potential vector.

    To solve the eqn, Fourier expand all terms of the equation.
  5. Feb 2, 2007 #4

    Meir Achuz

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    Each Cartesian component of A satisfies the scalar Poisson equation for a point charge, so just use the Coulomb potential for A_x, etc.
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