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Simple Divergence related problem

  1. Oct 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Sketch the vector function
    $$\vec{v}=\frac{\hat{r}}{r^2}$$
    and compute its divergence. The answer may surprise you...can you explain it?


    2. Relevant equations



    3. The attempt at a solution
    I have recently started with Introduction to Electrodynamics by David J Griffiths and according to the book notation,
    $$\hat{r}=\frac{\vec{r}}{r}=\frac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{x^2+y^2+z^2}}$$

    I have calculated the divergence to be zero. The problem is how do I draw the vector function? I mean the book shows a few drawings of vector functions but they all are in 2-D. I have three variables, x,y and z. How do I sketch the function in this case?

    And what is special about the result?

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Oct 13, 2013 #2

    arildno

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    Remember that the divergence is zero, wherever it is DEFINED in a normal manner.
    That is everywhere, execept at r=0.

    To interpret it for regions about r=0, it means that for every annulus about r=0, the divergence is 0.
    ------------------------
    The really interesting feature is, of course, when you make a surface integral around a volume (containing no holes) containing r=0.

    Your result will be [itex]4\pi[/itex]

    This motivates to represent the divergence as [tex]div(\vec{v})=4\pi\delta(\vec{r})[/tex], where we use the Dirac Delta function.

    By using this, we have in a clever manner formally achieved to regard this special case as ALSO being in accordance with Gauss' theorem:
    [tex]\int_{V}(div(\vec{v})dV=\int_{S}\vec{v}\cdot{d\vec{S}}[/tex]

    (Note that Gauss' theorem is typically derived for functions DEFINED on the whole volume (otherwise, how can you integrate it over the volume?), so there is no a priori contradiction between Gauss' theorem in that the divergence is zero, yet a surface integral around r=0 equals 4*pi. For example, for every region on which the vector is defined, say an annulus about r=0, Gauss' theorem is trivially valid, since the surface integral for such regions reduces to [itex]4\pi-4\pi=0[/itex])
     
    Last edited: Oct 13, 2013
  4. Oct 13, 2013 #3
    Thanks arildno for the effort you have put here. I will return to this thread when the book introduces the surface integrals and the dirac delta function. The problem I posted is an intext question just after the divergence section in the book and much of what you wrote is introduced later in the chapter.

    Thanks! :)
     
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