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Help with a problem

  1. Sep 20, 2006 #1
    This is from page 1 of "Photons and Atoms" by Cohen-Tannoudji et. al.

    Given a set of particles of charge [itex]q_{\alpha}[/itex] and position [itex]r_{\alpha}(t)[/itex] the charge density [itex]\rho[/itex] and current j are given by
    [tex]\rho(r,t) = \Sigma_{\alpha} q_{\alpha}\delta[r - r_{\alpha}(t)][/tex]
    [tex]j(r,t) = \Sigma_{\alpha} q_{\alpha} v_{\alpha} \delta[r - r_{\alpha}(t)][/tex]

    Show that:

    [tex]\frac{\partial}{\partial t}\rho(r,t) + \nabla \cdot \j(r,t) = 0[/tex]

    I don't know how to get started. The only thing I know of that looks like it might help is:

    [tex]x\delta'(x) = -\delta(x)[/tex]

    But I don't see how to apply it here.
    Last edited: Sep 20, 2006
  2. jcsd
  3. Sep 20, 2006 #2


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    Just use the following (product and chain) rules :

    [tex]\vec \nabla \cdot (f\vec v)=(\vec \nabla f)\vec v+f(\vec \nabla \cdot \vec v)[/tex]

    [tex]\frac{\partial}{\partial t} f(\vec r(t))=(\vec \nabla f)\cdot\frac{\partial \vec r}{\partial t}[/tex]
    And it's be a one or two line proof.

    Only thing is you get things like [tex]\vec \nabla \delta(\vec r-\vec r_\alpha(t))[/tex]. I`m not really sure myself what that would be, but just leave it like that and you're fine.
  4. Sep 20, 2006 #3
    Thanks for looking at this Galileo. I made a mistake when I posted it. The positions of the particles are a function of time [itex]r_{\alpha}(t)[/itex]. I edited the original and I also repeat the corrected formulas here:

    [tex]\rho(r,t) = \Sigma_{\alpha} q_{\alpha}\delta[r - r_{\alpha}(t)][/tex]
    [tex]j(r,t) = \Sigma_{\alpha} q_{\alpha} v_{\alpha} \delta[r - r_{\alpha}(t)][/tex]

    If I apply the partials to the formulas and use your suggestions, I get the following:

    [tex]\frac{\partial}{\partial t}\rho(r,t) = \Sigma_{\alpha} q_{\alpha} \frac{\partial}{\partial t}\delta[r - r_{\alpha}(t)] = -\Sigma_{\alpha} q_{\alpha} (v_{\alpha} \nabla \delta(r - r_{\alpha}(t)))[/tex]
    [tex]\nabla j(r,t) = \Sigma_{\alpha} q_{\alpha} (\nabla v_{\alpha} \delta(r - r_{\alpha}(t)) + v_{\alpha} \nabla \delta(r - r_{\alpha}(t)))][/tex]

    And so:
    [tex]\frac{\partial}{\partial t}\rho(r,t) + \nabla \cdot \j(r,t) = \Sigma_{\alpha} q_{\alpha} \nabla v_{\alpha} \delta(r - r_{\alpha}(t))[/tex]

    I'm not there yet, but this is a great stride forward.
    Last edited: Sep 20, 2006
  5. Sep 20, 2006 #4

    George Jones

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    Now, why is the right-hand side equal to zero?
  6. Sep 20, 2006 #5
    I figured that out on my way home from work. Thanks to Galileo and to George. I remember your help with GR earlier.
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