# Divergence Theorem: fail?

1. Sep 8, 2010

### erogard

Hi everyone,

so let me introduce the scalar function $$\Phi$$ = -(x2+y2+z2)(-1/2) which some of you may recognize as minus one over the radial distance from the origin.

When I compute $$\nabla$$2$$\Phi$$ is get 0.

Now if I do the following integral on the surface S of the unit sphere x2+y2+z2= 1 :

$$\int$$$$\int$$$$\nabla$$$$\Phi$$$$\bullet$$n dS

I obtain 4$$\pi$$ which is just the surface area of the unit sphere since the laplacian of $$\Phi$$ dotted with the unit radial vector turns out to be one.

Yet the Divergence Theorem

tells me precisely that those two quantities should be equal, but integrating the laplacian of phi would give me zero since the latter quantity is already 0.

Any idea on what I'm doing wrong?
(please only suggestion but do not give me the answer straight away if you think you know it, I'd like to figure it out by myself for my hw).

Thanks!

2. Sep 8, 2010

### gabbagabbahey

Your problem is that the Laplacian of that function is undefined at the origin. You can see that by looking at the formula for the laplacian in either Cartesian or Spherical coordinates. In spherical coordinates,

$$\nabla^2\left(-\frac{1}{r}\right)= \frac{1}{r^2}\frac{\partial}{\partial r}\left[r^2\frac{\partial}{\partial r}\left(-\frac{1}{r}\right) \right]$$

the factor of $1/r^2$ is what causes the problem.

So, what can you conclude? You know 3 things:

(1) The Laplacian is zero everywhere except at the origin
(2) The Laplacian is undefined at the origin
(3) If you integrate the laplacian over any region that includes the origin you will get $4\pi$

These are essentially the defining properties of the 3D Dirac Delta distribution, and it turns out that $\nabla^2\left(-\frac{1}{r}\right) =4\pi\delta^3(\textbf{r})$

3. Sep 8, 2010

### erogard

gabbagabbahey: thank you for pointing that out. I realize that the theorem holds for continuously differentiable function on such a volume T, yet that is not the case for the gradient phi.

4. Sep 8, 2010

### LawlQuals

Further, if you are encountering this in a course on electrodynamics (Griffiths, or Jackson), the motivation for the definition of the dirac delta function is precisely this argument, and they both spend a few pages discussing this. If you would like further discussion on this that is more detailed, I recommend you to their texts.

Why does this apparent contradiction appear to happen? There is a caveat in the divergence theorem regarding how such functions must behave.

5. Sep 8, 2010

### gabbagabbahey

The problem isn't with the divergence theorem, but rather with your calculation of the laplacian (divergence of $\mathbf{\nabla}(-\frac{1}{r})$). You weren't careful enough wih your calculation near the origin.