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The need for the Dirac delta function

  1. Aug 8, 2014 #1
    So part of the idea presented in my book is that:
    div(r/r3)=0 everywhere, but looking at this vector field it should not be expected. We would expect some divergence at the origin and zero divergence everywhere else.

    However I don't understand why we would expect it to be zero everywhere but the centre, because if you draw it, the arrows get smaller as we move out radially. If you consider placing a little cube somewhere in the field not at the centre, the arrows entering that cube would be larger than those leaving. Surely that would give a negative divergence at these points. I obviously understand why it should be large at the centre.

    This leads me to my other point. Consider the r/r2 field - it is similar to the above field, just falling off less rapidly. This field has 1/r2 divergence however. My first issue is, why does it not give negative divergences by my above argument of arrows into a little cube, and because of the similarities for the above field, why does it not give zero divergence apart from at the origin? Secondly, why does the maths behave so much more nicely for such a similar field?

    Thanks for any help :)
     
  2. jcsd
  3. Aug 8, 2014 #2

    Ray Vickson

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    You say "However I don't understand why we would expect it to be zero..."; well, have you actually sat down and computed ##\text{div}\: (\vec{r}/r^3)## at ##\vec{r} \neq \vec{0}##? If you keep track of things during the calculation you will see how and why some cancellations occur, leaving you with 0.
     
  4. Aug 8, 2014 #3

    vanhees71

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    This is an utmost important exercise. So you should do it very carefully. The correct equation to prove is
    [tex]\vec{\nabla} \cdot \frac{\vec{r}}{r}=4 \pi \delta^{(3)}(\vec{r}).[/tex]
    The proof for [itex]\vec{r}[/itex] is almost trivial, just brute-force derivatives will do.

    To check that the [itex]\delta[/itex]-distribution is correct, use Gauss's Law for an arbitrary volume containing the origin in its interior with a ball of infinitesimal radius around the origin taken out to integrate the expression on the left-hand side together with an arbitrary test function.
     
  5. Aug 9, 2014 #4

    vela

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    Because of the spherical symmetry of this scenario, a cube isn't the best choice to use to analyze the situation. Try using the infinitesimal volume element for the spherical coordinate system instead.

    For an infinitesimal cube, I expect any differences in flux correspond to second-order terms and can therefore be neglected as the volume of the cube tends to 0. For a finite cube, I suspect the difference in angle at which the field lines intersect the surfaces is just enough to cancel out the effect of the changing field strength. It's left to the reader (you) to verify this is indeed the case. :wink:


    In this case, the field doesn't fall off fast enough with increasing ##r## so that there's a net positive flux.
     
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