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Dirac Delta

  1. May 5, 2009 #1
    Hi guys.

    I play now a bit with EM fields and I have encountered some problems connected with Dirac delta. By coincidence I visited this forum and I thought I could find some help in here.

    The problem is that in order to get a potential in some point from a single charge you need to just solve such thing:

    [tex]
    \square \phi = - \frac{ \rho}{\epsilon}
    [/tex]

    and there by the way you need to use such equation:
    [tex]

    \nabla^2 \frac{ 1}{ | \vec{r} - \vec{r_0} |} = - 4 \pi \delta ( \vec{r} - \vec{r_0})
    [/tex]

    I would appreciate if someone could show me where it comes from :)

    PS. Sorry for my mistakes or improper names for some mathematical or physical stuff but I am from Poland :P
     
  2. jcsd
  3. May 5, 2009 #2

    dx

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    The second equation just means that the charge distribution δ(r - r0) has the electrostatic potential 1/4πε0|r - r0| because it satisfies the Poisson equation.
     
  4. May 5, 2009 #3
    Hmm right! I have just realized that thanks to you but still something is not clear for me.

    [tex]\rho = q \delta (\vec{r} - \vec{r_0} )[/tex]
    right?
    So where [tex]q [/tex] and [tex]\epsilon [/tex] are missing in this equation?
     
  5. May 5, 2009 #4

    dx

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    They cancel out because they're on both sides of the equation:

    [tex] \nabla^2 \frac{q}{4\pi\epsilon_{0}|r - r_0|} = -\frac{q\delta(r - r_0)}{\epsilon_0} [/tex]
     
  6. May 5, 2009 #5
    Oh yes! Thank you very much!

    One more thing, my professor looking on that equation said something about sham contraction conected with the fact that divergence is zero all around that charge. That was surely connected somehow with Dirac delta, but I do not get it somehow. Do you know something about it?
     
  7. May 5, 2009 #6

    dx

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    Sham contraction? Hmm.. never heard of it.
     
  8. May 5, 2009 #7
    I am a little bit embarassed as it can be coused by a lingul problem. I meant ofcourse apparent discrepancy connected with the fact that divergence of E field around this single charge is 0, where it is not true. But if just you have not not heard about it as you wrote thanks for all your help :)
     
  9. May 5, 2009 #8

    dx

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    The divergence of E is always zero in places where charge density is zero. That's one of Maxwell's equations.
     
  10. May 5, 2009 #9
    Oh yes and thats why there is this paradox I think. Thanks again.
     
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