Simple Divergence related problem

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Homework Statement


Sketch the vector function
$$\vec{v}=\frac{\hat{r}}{r^2}$$
and compute its divergence. The answer may surprise you...can you explain it?


Homework Equations





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!
 

Answers and Replies

  • #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:
  • #3
3,812
92
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])

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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