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Laplacian of 1/r in Darwin term

  1. Apr 26, 2010 #1
    The http://en.wikipedia.org/wiki/Fine_structure#Darwin_term" contains a (3D-)delta function as a result of taking the Laplacian of the Coulomb potential. I'm trying to find out why. I've been searching, and I've so far come across different views of the Laplacian of 1/r at the origin. Either it's considered zero, or

    [tex]
    \nabla^2\frac 1 r = -\,\frac{\delta(r)}{r^2},
    [/tex]

    but I can't find any reference that says it's [itex]\delta^3(r)[/itex]. It has the right units, but that's about all I can say about it. Could someone clarify this?
     
    Last edited by a moderator: Apr 25, 2017
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  3. Apr 26, 2010 #2

    samalkhaiat

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    Science Advisor

    The relation

    [tex]\nabla^{2}|\frac{1}{\vec{r}}| = -4\pi \delta^{3}(\vec{r})[/tex]

    is proved in post #10 of;

    www.physicsforums.com/showthread.php?t=200580

    In spherical coordinates, the delta function can be written as

    [tex]\delta^{3}(\vec{r}) = \frac{1}{r^{2}} \delta (r) \delta (\phi) \delta (\cos \theta )[/tex]

    regards

    sam
     
  4. Apr 26, 2010 #3
    Ah, I see. I assumed [itex]\delta(r)[/itex] and [itex]\delta(\vec{r})[/itex] were pretty much the same thing, but I guess not. Thanks. Is there actually any point in including those delta functions in [itex]\phi[/itex] and [itex]\theta[/itex]? The delta function in [itex]r[/itex] narrows things down to a single point.
     
  5. Apr 27, 2010 #4

    samalkhaiat

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    Yes, thse are the same thing! However, [itex]\delta(r)[/itex] is different from [itex]\delta^{3}(r)[/itex].

    Well, the theta and the phi are there! you can't just leave them out. For certain potential you can integrate or average over the angular dependence. But, in general, potentials do depend on theta and phi.

    sam
     
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