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Divergence of inverse square field and Dirac delta

  1. Mar 14, 2013 #1
    [tex]\nabla \cdot \frac{\mathbf{r}}{|r|^3}=4 \pi \delta ^3(\mathbf{r})[/tex]
    What's the proof for this, and what's wrong with the following analysis?

    The vector field
    [itex]\frac{\mathbf{r}}{|r|^3}=\frac{1}{r^2}\hat{r}[/itex] can also be written [tex]\mathbf{F}=\frac{x}{\sqrt{x^2+y^2+z^2}^3}\hat{x}+ \frac{y}{\sqrt{x^2+y^2+z^2}^3}\hat{y}+ \frac{z}{\sqrt{x^2+y^2+z^2}^3}\hat{z}[/tex]

    Take the z-part of the divergence of F:
    [tex]\frac{\partial}{\partial z}\mathbf{F_z}=\frac{\partial}{\partial z}\frac{z}{\sqrt{x^2+y^2+z^2}^3}=\frac{1}{\sqrt{x^2+y^2+z^2}^3}+\frac{-3}{2}\frac{z}{\sqrt{x^2+y^2+z^2}^5}2z[/tex]
    Evoking symmetry to find the other parts of the divergence is easy:
    Why am I incorrect at [itex]x,y,z=0[/itex]?
  2. jcsd
  3. Mar 14, 2013 #2


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    0 / 0 is undefined.
  4. Mar 14, 2013 #3
    Fair enough, but why is this particular 'undefined' equal to [tex]4 \pi \delta ^3(r)[/tex]?
  5. Mar 14, 2013 #4
    It comes from the Divergence Theorem and relating the expression [itex]\frac{\hat{r}}{r^2}[/itex] over the volume/surface of a sphere. In particular, you find that the surface integral of the divergence of that quantity over a sphere of any radius is equal to [itex]4\pi[/itex]. Normally the effect of increasing radius should have some effect on the value of the volume integral, but it doesn't here. This implies that its divergence must be a delta function to "sift out" that value of [itex]4\pi[/itex]. (As Griffiths would say, it is a strange integral to have all of its value concentrated at one point.)

    This is easiest shown in spherical coordinates. I can show you the math if it isn't clear after trying to work it out on your own.
    Last edited: Mar 14, 2013
  6. Mar 15, 2013 #5
    Thanks, that factor of 4 pi makes sense now.

    Unfortunately, I'm now having trouble grokking why [tex]\frac{0}{|\mathbf{r}|^5}=\delta ^3( \mathbf{r})[/tex]. Can you take any function where the limit approaches infinity at only one point (and is 0 elsewhere), and call it the dirac delta function? Surely all these functions cannot be the same, and there must be something more than 'it is undefined when r=0, 0 elsewhere, so it's the dirac delta'. However, I can see no other way from my above argument to the dirac delta function.

    But, as I said, once that's established, 'normalising' the dirac delta function is something I can bellyfeel.
  7. Mar 15, 2013 #6
    It is not true that [tex]\frac{0}{|\mathbf{r}|^5}=\delta ^3( \mathbf{r})[/tex] (See below.)

    Regarding the definition of the Dirac delta, you can create a delta from basic functions as long as it's clear what you are limiting. The limit of a Dirac delta as you approach the active point is always zero--not infinity. It's the point itself when evaluated that gives you infinity. (It also doesn't really make sense to talk about the value of a Dirac delta outside of an integral.)

    Here's how you can create a Dirac delta from basic functions. First, you need to find a bounded function with a well-defined area and central peak (e.g. a Gaussian). Then you "squeeze" the function's peak into zero width while keeping the area constant. The only way you can keep the area constant is to have the height of the peak approach infinity in relation to the shrinkage. The area is preserved, and that's why the integral has a well-defined value (area) even though the height is infinite and the width is zero. I admit it is difficult to apply this reasoning to the original problem you asked about, but it is indeed happening.

    Also, not all infinite discontinuities are Dirac deltas. The divergence of [itex]\frac{\hat{\bf r}}{r^n}[/itex] for integer [itex]n > 2[/itex] is not even defined at the origin. [itex]n = 2[/itex] is a special case because its divergence exactly cancels out the dependence on radius for a sphere's surface area.

    Edit: Check this link for a picture of what it looks like to create a Dirac delta from a Gaussian: http://en.wikipedia.org/wiki/Dirac_delta . From the picture, you should notice that the area in each limiting step is the same, but the width is decreasing and the height is increasing. When the width reaches zero, you have a Dirac delta.
    Last edited: Mar 15, 2013
  8. Mar 15, 2013 #7


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    Keep in mind that a Dirac delta makes sense only inside an integral. Here's an even simpler definition than using a Gaussian: ##\delta(x-a) =## the limit as ##\epsilon \rightarrow 0## of ##1/(2\epsilon)## for ##a-\epsilon < x < a+\epsilon##, and 0 everywhere else. This is a very thin, tall rectangle with area 1, centered at x = a. Here's an example of how to apply this definition:

    $$\int {x^2 \delta(x-a) dx} = \lim_{\epsilon \rightarrow 0} \int^{a+\epsilon}_{a-\epsilon} {x^2 \left( \frac{1}{2\epsilon} \right) dx}$$

    Evaluate the integral, then take the limit, and you should get ##a^2##.
  9. Mar 16, 2013 #8
    Thanks for the help, both of you. So it seems that the divergence is 0 everywhere but the origin, and the dirac delta function is used so the integral over a sphere centred at the origin has the correct value.
  10. Mar 16, 2013 #9


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    The Divergence Theorem states that for any well-behaved vector-valued function [itex]\vec{F}[/itex],

    [itex]\int (\vec{\nabla} \cdot \vec{F}) dV =\int \vec{F} \cdot d\vec{S}[/itex]

    where the first integral is a volume integral over some volume, and the second integral is a surface integral over the boundary of that volume.

    In the particular case of [itex]\vec{F} = \dfrac{1}{r^3} \vec{r}[/itex], with the volume being a sphere of radius [itex]R[/itex] centered on the origin, the right-hand side becomes:
    [itex]\int \dfrac{1}{r^3} \vec{r} \cdot d\vec{S} = 4 \pi[/itex]

    On the other hand, if you calculate [itex]\int \vec{\nabla} \cdot \vec{F}[/itex] for this case, you get zero, unless [itex]r = 0[/itex], in which case, the divergence is undefined. What this really means is that this particular [itex]\vec{F}[/itex] is not one of the "well-behaved" functions that were under consideration in the derivation of the divergence theorem. However, we can, in a physicist-style handwavy way, define a 3-D delta function [itex]\delta^3(\vec{r})[/itex] to be a "function" such that:

    1. [itex]\delta^3(\vec{r}) = 0[/itex] whenever [itex]\vec{r} \neq 0[/itex]
    2. [itex]\delta^3(\vec{r})[/itex] is undefined at [itex]\vec{r} = 0[/itex]
    3. [itex]\int \delta^3(\vec{r}) dV = 1[/itex] for any volume integral enclosing the point [itex]\vec{r} = 0[/itex]

    So given these rules for [itex]\delta^3(\vec{r})[/itex], we can conclude that

    [itex] \vec{\nabla} \cdot (\dfrac{1}{r^3}\vec{r}) = \delta^3(\vec{r})[/itex]
    Last edited: Mar 16, 2013
  11. Mar 17, 2013 #10
    That's what I said...

    Thanks, anyway.
    Shouldn't the last equality really have the delta function multiplied by a constant for the integral to check as 4∏?
  12. Mar 17, 2013 #11


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