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3D Dirac Delta Function

  1. Mar 10, 2009 #1
    Griffiths' section 1.5.3 states that the divergence of the vector function r/r^2 = 4*Pi*δ^3(r). Can someone show me how this is derived and what it means physically? Thanks in advance.
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  3. Mar 10, 2009 #2


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    First, use the divergence theorem to show that [tex]\int_{\mathcal{V}}( \vec{\nabla}\cdot\frac{\hat{r}}{r^2})d\tau=4\pi[/tex] for any surface enclosing the origin (use a spherical surface centered at the origin for simplicity).

    Then, calculate the divergence explicitly using the formula on the inside of the front cover for divergence in spherical coords. You should find that it is zero everywhere except at the origin where it blows up (because the 1/r terms correspond to dividing by zero).

    Finally, put the two together by using eq 1.98 with [tex]f(\vec{r})\delta^3(\vec{r})\equiv \vec{\nabla}\cdot\frac{\hat{r}}{r^2}[/tex]

    As for a physical interpretation, there really isn't one (although there are physical consequences as you'll see in chapter 2 and beyond); but there is a geometric interpretation....if you sketch the vector function [tex]\frac{\hat{r}}{r^2}[/tex] (figure 1.44 in Griffiths) you'll see why there must be an infinite divergence at the origin. This is all discussed in section 1.5.1
    Last edited: Mar 11, 2009
  4. Mar 12, 2009 #3
    Sorry, I'm kinda slow- why does LaTeX Code: f(\\vec{r})\\delta^3(\\vec{r})\\equiv \\vec{\\nabla}\\cdot\\frac{\\hat{r}}{r^2}?
  5. Mar 12, 2009 #4
    Can't use latex. Why does f(R)δ^3(R) = the divergence of R/r^2?
  6. Mar 12, 2009 #5


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    The fact that the divergence of [tex]\frac{\hat{r}}{r^2}[/tex] is zero everywhere, except at the origin where it is infinite, but when integrated gives a constant, means that it must be some unknown normal function [itex]f(\vec{r})[/itex] times a delta function so, you call it [itex]f(\vec{r})\delta^3(\vec{r})[/itex] and solve for [itex]f[/itex] by setting the integral (eq. 1.98) equal to the value you calculated using the diverergence theorem ([itex]4\pi[/itex])
  7. Mar 13, 2009 #6
    Ok. I think I got it. Thanks a lot!
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