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A question of Dirac Delta function

  1. Jan 24, 2007 #1
    A vector function

    [tex] V(\vec{r}) = \frac{ \hat r}{r^2} [/tex]

    If we calculate it's divergence directly:

    [tex] \nabla \cdot \vec{V} = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{1}{r^2} \right) = 0 [/tex]

    However, by divergence theorem, the surface integral is [tex]4\pi [/tex]. This paradox can be solved by Dirac Delta function.

    My problem is, it seems like we can't calculate V's divergence directly because V blow up at r=0, and we unwittingly divide the zero when [tex] \left( r^2 \frac{1}{r^2} \right) [/tex]

    But it seem to me that a vector function [tex] V'(\vec{r}) =\frac{ \hat r}{r} [/tex] also has the same problem, which also blows up at r=0, and we will also divide the zero when

    [tex] \nabla \cdot \vec{V'} = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{1}{r} \right) [/tex]

    But we will get the accurate answer !? Why both of the function V and V' has the same problem but we only have to use Dirac Delta function for V ??
     
    Last edited: Jan 24, 2007
  2. jcsd
  3. Jan 24, 2007 #2
    The potential is a scalar field... but the Electric field is a vector field...

    The problem is... you cannot define a divergence on a scalar field.. in other word... the
    whole thing doesn't make sense...

    However, for a scalar field, you could define a gradient... the gradient of electric potential
    is the electric vector field... but you still could not remove the singularity...
     
  4. Jan 24, 2007 #3
    My function V is vector function, not electric potential.
     
  5. Jan 24, 2007 #4

    arildno

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    Simple thing:
    The volume integral is NOT defined whenever a singularity is within it; however the surface integral about a singularity IS defined, and equals [itex]4\pi[/itex] IRRESPECTIVE of the shape of the surrounding surface.

    It is BECAUSE of this surface shape independence that it is admissible to introduce the Dirac Delta formalism in the first place.

    This has nothing to do with the divergence theorem, that has as a PRESUPPOSITION that the integrand is defined on all points in the interior.
     
  6. Jan 24, 2007 #5
    hmm..... I have one more question....

    [tex] \vec{E} = A e^{- \lambda r} (1+ \lambda r) \frac{\hat{r}}{r^2} [/tex]

    where E is electric field. If I want to find charge density, then

    [tex] \rho = \epsilon _0 ( \nabla \cdot \vec{E}) [/tex]

    In one way

    [tex] \rho = \epsilon _0 ( \nabla \cdot \vec{E}) = \epsilon _0 A e^{-\lambda r} (1+ \lambda r) ( \nabla \cdot \frac{\hat{r}}{r^2}) + \epsilon _0 \frac{\hat{r}}{r^2} \left[ \nabla(A e^{-\lambda r} (1+ \lambda r)) \right] = \epsilon_0 A \left[ 4 \pi \delta ^3 (\vec{r})- \lambda ^2 e^{- \lambda r} /r \right] [/tex]

    In another way

    [tex] \rho = \epsilon _0 ( \nabla \cdot \vec{E}) = \epsilon _0 ( \nabla \cdot \vec{E} ) = A \frac{1}{r^2} \frac{ \partial}{ \partial r} ( r^2 E_r) = A \frac{1}{r^2} \frac{ \partial}{ \partial r} ( r^2 \frac{e^{- \lambda r} (1+ \lambda r)}{r^2}) = - \epsilon_0 A \lambda ^2 e^{- \lambda r} /r [/tex]

    What is the flaw in the second way, how to avoid it??
     
    Last edited: Jan 24, 2007
  7. Jan 24, 2007 #6

    arildno

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    There is no flaw in the second method.
    Again, what is mathematical "unsound" is to perform a volume integral over a point where the function is not defined.

    What is so damn hard about understanding that trivial fact?
     
  8. Jan 24, 2007 #7

    Meir Achuz

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    What you say was the gospel 100 years ago. But work by Schwartz, Lighthill, and others has shown how to do just that using distribution theory.
    In physics applications, the basic equation turns out to be
    [tex]\nabla\cdot\left[\frac{\bf r}{r^3}\right]=4\pi\delta({\bf r})[/tex].
    You can derive all the other singular quantities from that.
     
    Last edited: Jan 24, 2007
  9. Jan 25, 2007 #8

    arildno

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    Do you know what "..." means?
     
  10. Jan 25, 2007 #9
    I don't think he actually did a volume integral at any point of that problem. He took a derivative that is now defined pretty much to leave you with the Dirac delta function.

    The problem you speak of is in Jackson's E&M book, and when I worked through it I looked at a neighborhood of radius [tex]\epsilon[/tex] around the origin, took the divergence outside it, and then inside it I Taylor expanded for small r, took a divergence of the Taylor expansion, identified the delta function hiding inside, then took the [tex]\epsilon \rightarrow 0[/tex] limit and just added things up. That seemed to satisfy the TA and professor enough.
     
  11. Jan 25, 2007 #10
    At the first place, my problem is that how should I know "what happened" at the origin?? In post 5, my second method didn't tell me any information about the origin. I think the problem is when I taking the dirivative, since the function is undifferentiable at the origin (actually it isn't defined at the origin), so the divergence I get is a function which domain is real number without origin. So I think the best way to avoid this kind of problem is to make sure that the function which I take derivative is differentiable at every point.
     
  12. Jan 26, 2007 #11

    arildno

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    If you take the ORDINARY derivative of your function, then indeed it doesn't make sense to take the derivative AT r=0!

    The issue here, however, as Meir Achuz has pointed out (and I alluded to it myself in my first post), is that it is perfectly possible to define the Dirac delta (generalized) function by distribution theory in a mathematically rigorous manner.
    However, in this rigorous manner, we have to RE-DEFINE "derivative" to what is called "weak derivative". These subtleties are generally avoided in introduction courses in, say, electro-magnetism.

    Thus, in your first manner of differentiating, you are differentiating "weakly" (getting the Dirac delta term), whereas in your second method of differentiating you are differentiating ordinarily.
     
  13. Jan 27, 2007 #12

    Meir Achuz

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    You are on the right track. The best definition of the divergence is in terms of the limit of a surface integral divided by the volume as each (V and S) goes to zero. With this definition, div(rvec/r^3)=4pi\delta(r).
    The form for div in any specific coordinate system depends on the derivative being defined, and so does not work for r-->0 in this case.
     
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