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Checking for delta function

  1. Aug 20, 2005 #1
    For a given function: [tex]r^n\hat r[/tex], find its curl.

    I formulated the divergence first. For the divergence: [tex]\nabla . (r^n\hat r) = (n+2)r^{n-1}[/tex] and the functon becomes a dirac delta at the origin in case of n = -2.

    For the curl:
    Geometrically, the curl should be zero. Likewise, the curl in spherical coordinates obviously gives zero.
    My question is how can one be certain that there is no Dirac Delta function lurking here(for the curl)? (My understanding of Dirac delta function is a bit poor, so additional explanations would help :biggrin: .)
     
  2. jcsd
  3. Aug 20, 2005 #2

    Galileo

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    You can calculate the flux of the field through a small sphere about the origin. Geometrically you can see immediately that the flux will not be zero, hence a delta function will be involved.
     
  4. Aug 20, 2005 #3

    Hurkyl

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    The main thing to realize is that there's a bit of "pretend" going on:

    Because, doing everything strictly, we have that:

    [tex]
    \iiint_B \nabla \cdot (r^{-2} \hat{r}) \, dV = 0
    [/tex]

    for any region B.

    (I put pretend in quotes because I'm pretty sure that if you really know what you're doing, you can set up the machinery so that this is no longer pretending)


    However, we would really like the divergence theorem to be true. It's fairly easy to show that

    [tex]
    \oint_S (r^{-2} \hat{r}) \cdot d\vec{A} = 4 \pi
    [/tex]

    when S is a (positively oriented) sphere centered at the origin. So, in order to save the divergence theorem in this case, we can say that [itex]\nabla \cdot (r^{-2} \hat{r}) = 4 \pi \delta^3(\vec{r})[/itex]. This presumably saves a great many other cases too, but I haven't worked out the details.


    But wait you say, if the divergence theorem is a theorem, how can it fail in the first place? The answer is because the divergence theorem assumes that the divergence is a continuous function on the domain, and that assumption fails when the region of integration contains the origin.
     
    Last edited: Aug 20, 2005
  5. Aug 21, 2005 #4
    Thank you very much Galileo and Hurkyl.

    I completely understood how the divergence theorem holds good for a function like [tex]\nabla \cdot (r^{-2} \hat{r})[/tex].

    However, I still have a few more doubts.
    For a general function like [tex]r^n\hat r[/tex], the curl gives zero. Is the function [tex]\nabla \times (r^{n} \hat{r}) = 0[/tex] a Dirac delta? Can Stokes' theorem be somehow modified to fit in this case?
     
  6. Aug 21, 2005 #5

    Hurkyl

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    Maybe someone can correct me, but I don't think it can ever make sense for the curl to involve a δ³ term... a δ² term maybe (corresponding to a one-dimensional singularity) but that can't happen here due to spherical symmetry.

    Of course, as I mentioned, I don't know all the little details about what people are doing when they use δ in this way.

    Anyways, have you tried looking at what Stoke's theorem says here?
     
  7. Aug 21, 2005 #6

    arildno

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    1. the origin is not part of your function's domain.
    2. On its domain, the divergence of your function is zero, and nothing else.
    3. You can't integrate the divergence of your function over a region including the origin, since the origin isn't part of your function divergence's domain.
    Hence, the divergence theorem can never be invoked.
    4. Because it can be proven that the flux integral over any surface bounding a region containing the origin in the interior equals one and the same constant, we may, if we like, introduce the Dirac delta formalism to create a quasi-divergence theorem result.
     
    Last edited: Aug 21, 2005
  8. Aug 26, 2005 #7
    I went through some books on mathematical physics and I did not find much reference to the curl of [itex]r^n\hat r[/itex].
    However, I found there is a corollary derived from the divergence theorem which can be applied here.
    The theorem suggests:
    [tex]\int_v(\nabla \times \vec v) d\tau = -\oint_s\vec v\times d\vec a[/tex]

    Integrating over a volume of a sphere, taking [itex]\vec v = r^n\hat r[/itex] & [itex]d\vec a = R^2\sin \theta d\theta d\phi \hat r[/itex] &[itex]\nabla \times r^n\hat r =0[/itex]
    Thus,[tex]\int_v(\nabla \times r^n\hat r) d\tau = 0[/tex]
    &[tex] \oint_s r^n\hat r\times R^2\sin \theta d\theta d\phi \hat r = 0[/tex]

    So, the theorem is verified. So I suppose there is no delta function lurking here. Is my thinking correct here?
     
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