Why is the Divergence Theorem failing for this scalar function?

In summary: In fact, the function -\frac{1}{r} is not differentiable at the origin, and so the laplacian is not defined there. However, the divergence theorem requires that the integrand be twice continuously differentiable, which is not the case for this function. So, while the result you obtained may seem contradictory, it is actually just a consequence of using a non-differentiable function in a theorem that requires differentiability.
  • #1
erogard
62
0
Hi everyone,

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

When I compute [tex]\nabla[/tex]2[tex]\Phi[/tex] is get 0.

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

[tex]\int[/tex][tex]\int[/tex][tex]\nabla[/tex][tex]\Phi[/tex][tex]\bullet[/tex]n dS

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

Yet the Divergence Theorem
6c7f5573f2059b872c1aa0a962a366bd.png


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!
 
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  • #2
erogard said:
Hi everyone,

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

When I compute [tex]\nabla[/tex]2[tex]\Phi[/tex] is get 0.

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

[tex]\int[/tex][tex]\int[/tex][tex]\nabla[/tex][tex]\Phi[/tex][tex]\bullet[/tex]n dS

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

Yet the Divergence Theorem
6c7f5573f2059b872c1aa0a962a366bd.png


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!

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,

[tex]\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][/tex]

the factor of [itex]1/r^2[/itex] 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 [itex]4\pi[/itex]

These are essentially the defining properties of the 3D Dirac Delta distribution, and it turns out that [itex]\nabla^2\left(-\frac{1}{r}\right) =4\pi\delta^3(\textbf{r})[/itex]
 
  • #3
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
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
erogard said:
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.

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

Related to Why is the Divergence Theorem failing for this scalar function?

1. What is the Divergence Theorem and why is it important?

The Divergence Theorem, also known as Gauss's Theorem, is a fundamental theorem in vector calculus that relates the flow of a vector field through a closed surface to the behavior of the vector field within the enclosed volume. It is important because it allows for the calculation of volume integrals using surface integrals, making it a useful tool in many areas of physics and engineering.

2. Why might the Divergence Theorem fail?

The Divergence Theorem may fail if the vector field is not continuous or if the surface is not smooth. Additionally, the Divergence Theorem requires that the surface be closed, so if the surface is not completely enclosed or has holes, the theorem will fail.

3. How can one recognize if the Divergence Theorem has failed?

If the Divergence Theorem fails, the results of the volume integral calculation will not match the results of the surface integral calculation. This can be recognized by comparing the two values and seeing if they are equal.

4. Are there any alternative theorems that can be used instead of the Divergence Theorem?

Yes, there are several alternative theorems that can be used in place of the Divergence Theorem, such as Green's Theorem, Stokes' Theorem, and the Divergence Theorem in higher dimensions.

5. How can one avoid failure of the Divergence Theorem?

To ensure that the Divergence Theorem does not fail, one must ensure that the vector field is continuous and that the surface is smooth and closed. Additionally, it is important to check for any boundary conditions or limitations of the vector field and surface before applying the theorem.

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