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3D delta function

  1. Jul 25, 2009 #1
    Hi.

    How do we argue that [itex]\nabla^2\frac{1}{r}[/itex] is a three dimensional delta function? I have seen some people do it using the divergence theorem, i.e. saying that
    [tex]
    \int_V \nabla\cdot\nabla\frac{1}{r}dv=-\oint_S \nabla\frac{1}{r}\cdot ds=-4\pi
    [/tex]
    if S is a surface containing the origin, but I don't think it is legal to use the divergence theorem in this case because it requires that both the vector field and its divergence be well defined.

    Instead one could make the sequence [itex]\nabla^2\frac{1}{r+\varepsilon}[/itex] which approaches the correct function as [itex]\varepsilon\rightarrow0[/itex]. Here we can calculate the laplacian and then integrate this.. What do you think?
     
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  3. Jul 25, 2009 #2

    HallsofIvy

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    Well, of course, it not "legal" because the delta function is not a true "function". It is a "distribution" or "generalized function" and you have to use the derivative as extended to generalized functions.
     
  4. Jul 25, 2009 #3

    arildno

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    The limiting definition is just nonsense, because as epsilon goes to zero, the expression doesn't converge to anything.

    Similarly, as you have stated, the div theorem derivation is ALSO just nonsense, precisely for the reasons you state.

    The proper way to look at this is:

    1. The SURFACE integral is perfectly regular, yet has an extremely interesting property:
    Namely IRRESPECTIVE OF THE SURFACE, if the origin is strictly within the volume enclosed by that surface, then the value is -4pi, wheras if the origin lies outside we get 0!

    2. That property is, however, precisely what we would like the 3-D delta function to exhibit when considering arbitrary volumes!

    3. Thus, if we want to play, we "define" the divergence to be the delta function, and furthermore play along, pretending that this is in accordance with the divergence theorem.

    4. The only reason why this works though, is that the surface integral is the real thing here, not the delta func as a (standard) derivative, or the validity of the divergence theorem in this case.

    1+2 are the salient points, 3. is just notational garnish, or garbage if you prefer that term instead.
     
  5. Jul 26, 2009 #4
  6. Jul 27, 2009 #5
    arildno: It seems to me that what you are doing in point (3) is exactly what you said was illegal, i.e. using the divergence theorem. I don't understand point (4), could you rephrase that, please?

    jostpuur: Yes that looks good. Too bad I don't know any measure theory and only very little distribution theory. Maybe this question is too advanced for me at the moment.
     
  7. Jul 31, 2009 #6
    What I meant was this
    [tex]
    \lim_{k\rightarrow0}\int_V \nabla^2\frac{1}{r+k}dv=-\lim_{k\rightarrow0}\int_0^{2\pi}\int_0^{\pi}\int_0^{\epsilon} \frac{2k}{r(k+r)^3}r^2\sin{\theta}dr d\phi d\theta=-\lim_{k\rightarrow0}\frac{4\epsilon^2\pi}{(\epsilon +k)^2}=-4\pi
    [/tex]
    i.e. an integral on an arbitrarily small sphere which encloses the origin. Does this not prove that it is a delta function?
     
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