1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Divergence Theorem: fail?

  1. Sep 8, 2010 #1
    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

    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).

  2. jcsd
  3. Sep 8, 2010 #2


    User Avatar
    Homework Helper
    Gold Member

    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]
  4. Sep 8, 2010 #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.
  5. Sep 8, 2010 #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.
  6. Sep 8, 2010 #5


    User Avatar
    Homework Helper
    Gold Member

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Divergence Theorem: fail?
  1. Divergence theorem (Replies: 2)

  2. Divergence theorem (Replies: 5)

  3. Divergence Theorem (Replies: 5)

  4. Divergence theorem (Replies: 1)

  5. Divergence theorem (Replies: 2)