Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Divergence-less Fields

  1. May 14, 2009 #1
    So I am reading through Griffith's E&M and am on page 54. (This isn't a homework problem). He has a "Theorem 2" where he says if and only if you have a divergence-less field can you have these following conditions.

    So if you have a field Div F = 0, then F = Curl A. That's easy.
    Here is the confusing part: Then he says integral over the surface of F.da is independent of surface and integral over the surface of F.da =0 for a closed surface.
    I am not sure how to prove these things. I get 0 as a result no matter what because of the levi-cevita symmetry. Please could somebody let me know what I am doing wrong.
    Thank you so much.
  2. jcsd
  3. May 14, 2009 #2


    User Avatar
    Homework Helper
    Gold Member

    It follows from Gauss' theorem. The flux of a vector field F out of a closed surface is equal to the integral of the divergence of F over the volume bounded by that surface.
  4. May 14, 2009 #3
    Okay, so I don't need to introduce Curl A?
    I just say Integral over S of F.da = Integral over V of Div F?
    How then do I show it's independent of surface?
  5. May 14, 2009 #4


    User Avatar
    Homework Helper
    Gold Member

    It's only independent of the surface if div F = 0.

    If div F = 0, then clearly ∫ div F = 0 no matter volume you do the integration over.
  6. May 14, 2009 #5
    I see. It's because Integral over 0 is Cdxdy = Cdydz = Cdxdz and the only way that can work is if C = 0.
    How do you then show that integrating over an open surface results in something independent of surface?
  7. May 14, 2009 #6


    User Avatar
    Homework Helper
    Gold Member

    Integrating what over a surface? A zero vector field?
  8. May 14, 2009 #7
    Yes. Integral over an open S of F.da, when Div F = 0. How is it indep of surface? Is it that since S1 + S2 = S (where S is a closed surface), integrating F over S1 would result in the opposite of integrating over S2? Since the results are equal but opposite, the integral only depends on how the surfaces are split?
    How do you show this symbolically?
  9. May 14, 2009 #8


    User Avatar
    Homework Helper
    Gold Member

    Yes, that's correct. We write that symbolically like this:

    [tex]\int_{S_1} F \cdot dA + \int_{S_2} F \cdot dA = \int_{V} \nabla \cdot F = 0 [/tex]


    [tex] \int_{S_1} F \cdot dA = - \int_{S_2} F \cdot dA [/tex]
    Last edited: May 14, 2009
  10. May 14, 2009 #9
    Okay, so then you use Stoke's Theorem to get a path integral. Thank you so much dx :)
  11. May 14, 2009 #10
    One way to get the same result, w/o introducing line integrals, would be as follows.

    You have a divergence-less vector field F, so you know:

    [tex]\nabla \cdot F = 0[/tex]


    [tex]F = \nabla \times A[/tex]

    for some vector field A. From Gauss we have that:

    [tex]\large{ \int_V (\nabla \cdot F) dV = \int_{\partial V} F \cdot dA }[/tex]

    where V is any arbitrary volume, and [tex]\partial V[/tex] is the closed surface that defines its boundary. Since [tex]\nabla \cdot V = 0[/tex], we conclude that the surface integral of F over any closed surface must be zero:

    [tex]\int_{\partial V} F \cdot dA = 0[/tex]

    Furthermore, since F must be the curl of some vector field A, we can further conclude that:

    [tex]\int_{\partial V} (\nabla \times A) \cdot dA = 0[/tex]

    or in other words, that the surface integral of the curl of the vector function A over a closed surface must be zero.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook