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Does Gauss' Law fail?

  1. Feb 9, 2010 #1
    If I have a hypothetical infinite 3 dimensional universe uniformly charged I can argue that the electric field is zero at any given point in space by symmetry. However, if I invoke Gauss' Law I find a non-zero flux of electric field through a gaussian surface encompassing any amount of charge. How can this be if the electric field is everywhere zero? I asked a friend about this and he noted that the problem is not well posed because it requires an infinite amount of energy to assemble such a system. I found this explanation lacking. After all, we use infinite lines and planes in the teaching of electrostatics all the time.
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  3. Feb 9, 2010 #2


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    When you use Coulomb's law and actually integrate, do you get zero?
  4. Feb 9, 2010 #3

    Vanadium 50

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    I can't make you feel any better, I'm afraid. Your friend is right.

    You can cast electrostatics in terms of energy and potentials instead of forces and fields if you like. If you do this, you will discover that all of your equations end up of the form infinity minus infinity, which is not defined.

    Another way of looking at it is that the infinity you describe is not the limit of a really, really big uniformly charged region. That should be a hint that it's not well defined.
  5. Feb 10, 2010 #4
    I don't think I need to calculate the Coulomb force directly. I am using the fact the distribution is infinitely largge, homogeneous, and isotropic throughout my space. By symmetry I reason that the net force on a test charge placed anywhere in this space is zero.
  6. Feb 10, 2010 #5
    However, you assume that this force exists =) (or is well-defined)
  7. Feb 10, 2010 #6
    I agree the hamiltonian is unbounded and so the constraint equations are violated. However,
    if I write down the energy density measured by a timelike obsever locally I get [tex]T_{a b}\,u^{a}u^{b}=T_{00} \sim \vec{E}^2[/tex]. If the electric field is zero locally then I obey the weak energy condition locally. Could I then proceed with a local measurement everywhere of the energy density everywhere in the space and get zero energy if E is zero.
    Last edited: Feb 10, 2010
  8. Feb 10, 2010 #7
    But [itex]\textbf{E}[/itex] is not zero! =)
    Last edited: Feb 10, 2010
  9. Feb 10, 2010 #8
    What is wrong with the symmetry argument?
  10. Feb 10, 2010 #9
    It assumes that [itex]\textbf{E}[/itex] exists.
    I guess atyy told you about the same thing: take integral not in order to find its value but to show that it is divergent.
  11. Feb 10, 2010 #10

    Physics Monkey

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    One way out of the symmetry argument arises because you can always add a solution of the homogeneous equation which is unconstrained by symmetry. For example, you might argue that in empty space the electric field must be zero by symmetry. But that isn't true. The electric field can be any constant you like. The trouble here is that this constant can be infinity depending on how you set up the problem.

    The best way to approach the problem is to consider a sensible system and take a limit. It's a quite fun game to play. For example, model the infinite space of charge by thin layers of infinite charge. Then by the usual capacitor argument the electric field is perpendicular to the layer, is independent of distance, and is proportional to the number of charged layers. Now pick any point, you have an infinite half space of charged layers above and below the point, which way does the field point? It's infinity minus infinity, so you can get whatever you. Now you may again try to argue from symmetry that the field should be zero, but if you regulate the infinities in any reasonable way (say with a finite amount of charge) then you'll find the field is generically non-zero.

    Another way to say the same thing is that the fields depend strongly on boundary conditions arbitrarily far away. This dependence arises from the long range character of the field.
  12. Feb 11, 2010 #11
    Thanks for pointing this out so clearly. Issue is resolved.
  13. Feb 11, 2010 #12
    How is it resolved?! I think this questions is pretty interesting, could you please explain the solution to me as if you were explaining it to a child? :smile:
  14. Feb 11, 2010 #13
    I think the capacitor analogy fails because thereyou have positive and negative charge so the symmetry is gone to begin with. I tend to agree with the last poster althogh I find the mathematics of physics monkeys reply satisfactory
  15. Feb 12, 2010 #14

    Physics Monkey

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    The capacitor comparison may be a bit misleading since in the setup I in mind contained only positive charge. What I meant by "the usual capacitor argument" is the statement that an infinite plane of positive charge density produces a uniform field on both sides of the plane. If the plane is [tex] z = 0 [/tex] then the field points up for [tex] z > 0 [/tex] and down for [tex] z < 0 [/tex]. An infinite number of such plates (like a half space of charge) would produce an infinite field in the region above all the plates.
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