Built on Facts

Gauss’ Law PROVED WRONG!

Just though I’d try writing a post title in the style of a crank. Kinda fun!

Gauss’ law, of course, is not wrong. But I got a question from a reader that deceptively simple and an interesting example of a theorem not quite working the way you’d expect. I’ve gone over Gauss’ law before, so as a quick refresher I’ll just say that it relates the electric charge at a location to the way the electric field lines diverge at that location. Symbolically (and in, appropriately, Gaussian units):

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

E is the electric field, ρ is the charge density. Draw a closed surface around the charge in question and integrate over the enclosed volume:

i-88516ca505d84f969b7ab15a965c3fb3-2.png

The integral of the charge density within the surface is of course just the total charge enclosed within the surface:

i-d18063683dcc0d42b9be45451a84d1e3-3.png

Quoting Wikipedia’s article on the divergence theorem, it’s true that the volume integral of the divergence is equal to the surface integral of the vector field itself:

i-8d7070ce28f0af28047c7a5fb8246d53-4.png

Wikipedia’s notation for the integrals is a tiny bit different, but you get the gist. Making this substitution in Gauss’ law gives:

i-bb5c2c6b0452df43a61e3974bd9b473f-5.png

Where n is the normal vector – the direction locally outward perpendicular to the surface. This is useful because if the symmetry of the problem can guarantee that E is constant and parallel to n, E.n is just the scalar E and comes outside the integral. Which is nice, and allows such neat things as a proof of Coulomb’s law for a point charge, the shell theorem, and so forth.

Ok, so here’s the reader question: consider a constant ρ(x,y,z) = &rho0;. In other words, a uniform charge that fills all of space. By symmetry, it’s clear that there’s no preferred direction and thus the electric field is zero everywhere. But pick a point as an origin and draw a sphere around it – inside the sphere is some positive amount of charge Q, but the symmetry requires the total flux be zero. There seems to be a contradiction in a fairly simple-seeming example of Gauss’ law.

I have to issue my standard disclaimer that I’m not a mathematician (or even a theorist!), but I think I’ve sussed out the problem. The problem is not with Gauss’ law, but with the divergence theorem itself. As you’d expect, a theorem is a mathematically proven statement about abstract objects that satisfy certain conditions. Of course in this case the vector field F has to be a well-behaved function in some sense, which basically all functions in physics are. But less well known is that F must be compact supported – ie, that it’s only nonzero over some finite region. This ρ isn’t. The divergence theorem fails to hold, and it’s no longer possible to draw conclusions about the electric field in the usual way. You have to try something else.

Exercise: The usual undergrad method of Gaussian surfaces works just fine with lines of charge and sheets of charge, even though they don’t vanish at infinity. Why? (Seriously, why? I don’t know for sure – I suspect it’s because the charge distributions are zero at infinity except for a set of measure zero, but I couldn’t say for sure.)

Comments

  1. #1 Adam Jermyn
    May 17, 2011

    It’s not actually an issue with either Gauss’s law or with the Divergence Theorem, it’s more an issue with how you define an infinite space with uniform charge density. Specifically, it’s an issue with how you approach infinity. To see this, consider a universe with a thin plane of charge (extending infinitely in two dimensions) with spatial charge density p. It produces a uniform electric field away from itself which does not diminish with distance. If we wanted to calculate using Coulomb’s law the electric field at a point inside a uniform charge density filling the entire volume of space, we could start by integrating the contributions from an infinite number of these sheets. The issue with this is that it is not an absolutely convergent integral. That is, if you approximate it by an infinite sum, its value depends on the order in which you add the terms. You could add the contribution for one sheet on the left, two on the right, one on the left, two on the right, and so on, and find that a uniform charge distribution filling all space actually has a net electric field in one direction! On the other hand, you could add two on the left, one on the right, and so on, and find a field in the opposite direction. This is a specific example of a general problem, which is that the integral is really an improper integral defined as a limit which does not exist (the limit varies based on the direction you approach it from). In calculating this sort of situation physically we usually make an implicit assumption; we normally assume that the uniform distribution is no “bigger” in one direction than in another, so we approach the limit symmetrically and get an answer of 0. The reason that this works is that we never deal with truly infinite space in real calculations, and so we usually implicitly assume that the contributions from things further away will diminish, allowing us to assume that the integral converges no matter how you approach it (Cauchy Principle Value) and to assume that infinite just means really big. This assumption, however, falls apart for uniform charge distributions filling all of space, and so no matter how far out we take the integral, it will never properly converge.

  2. #2 Daan van Vugt
    May 17, 2011

    The problem occurs even before using the divergence theorem, in the differential version of gauss’ law.

    ∇ · E = 0 by symmetry.
    Therefore ρ = 0 and no such configuration is allowed.

  3. #3 Eric Lund
    May 17, 2011

    suspect it’s because the charge distributions are zero at infinity except for a set of measure zero, but I couldn’t say for sure.

    At least in the line charge case, you don’t have to assume infinitesimal thickness. Just consider a cylindrically symmetric charge distribution over a length L and let L -> infinity. Each additional slice along the z axis gives a contribution proportional to dz/z^2 (in the limit of large z), so the integral converges. I suspect this is also true of a sheet distribution of finite thickness, but it is less obvious that this is true.

    Another way of looking at it: There is a version of the divergence theorem in arbitrary dimensional space, as long as there actually is a space. An infinite line charge gives you a 2-D subspace (the directions perpendicular to the line charge) and an infinite plane charge gives a 1-D subspace, and there are divergence theorems in both subspace. But there is no divergence theorem in a 0-D subspace (AKA a point), which is where you would be trying to go by extending the analogy to a space-filling charge distribution. So Gauss’ Law fails in this case, quite aside from self-energy problems.

  4. #4 Torbjörn Larsson, OM
    May 17, 2011

    we usually implicitly assume that the contributions from things further away will diminish, allowing us to assume that the integral converges no matter how you approach it (Cauchy Principle Value)

    To cut the difference you could use distributions, in which case support and explicit requirement of convergence goes hand in hand. Then Matt is “more generally correct” on the solution, and you “more generally correct” on the problem. Potato, potaeto.

  5. #5 Ahcuah
    May 17, 2011

    I don’t think so.

    The ρ inside produces an electric field just as advertised by Gauss’ Law. And the ρ outside produces the exact opposite field.

    And by superposition, it all adds to zero, just as it should.

  6. #6 Andy Wood
    May 17, 2011

    By symmetry, it’s clear that there’s no preferred direction and thus the electric field is zero everywhere.

    This is incorrect.

    Consider the vector field with components (x-x0,y-y0,z-z0), ie it points radially outwards from the arbitrarily chosen origin. It’s divergence is 1 everywhere, and so, up to a constant of proportionality, will satisfy Gauss’s Law for an infinite, homogeneous charge distribution.

    Notice that the solution is not unique and I think that is really where the problem lies. Gauss’s Law does not define the electric field uniquely. You need additional constraints, such as the field vanishing at infinity, or Dirichlet boundary conditions on some enclosing surface. It’s not clear to me that it’s even possible to define boundary conditions in this problem.

    I believe that Newton was aware of this problem when considering the gravitational field of an infinite universe. It wasn’t until Einstein formulated GR that an infinite universe could be studied in self-consistent manner.

    I’m in a bit of hurry writing this comment, so forgive any errors I’ve made.

  7. #7 Knightly
    May 17, 2011

    You nearly broke my railguns! D:

  8. #8 rob
    May 17, 2011

    What #5 Ahcuah said.

    To find the electric field everywhere you have to use the superposition principle.

  9. #9 Matt Springer
    May 17, 2011

    I still have to disagree with #5 (and #8). What you’re outlining is essentially the proof of the divergence theorem, and that proof method (as noted in the technical link) doesn’t work in this scenario.

    I do like the argument in #6. (My E = 0 argument is physical, not mathematical.) It really says the same sort of thing as the compact support argument, but in a more intuitive way.

  10. #10 In Hell's Kitchen (NYC)
    May 17, 2011

    the divergence theorem is fine, it’s the reader’s question that is nonsensical.

    Try solving -\del^2\Phi=\rho_0 without boundary conditions. What do you get ? Then put boundary conditions. What do you get now ?

  11. #11 kevin
    May 17, 2011

    The partial differential equation div E = 4 pi rho is missing boundary conditions. As Andy Wood said, the solution is not unique; replace E by E + curl F for any vector field F and you have another solution. In physics the boundary conditions are often somewhat obvious from physical considerations and you use them without realizing it. With appropriate boundary conditions (in the case of EM, fields decay to zero at infinity, suitably interpreted in the case of infinite wires and such) you have a unique solution which does what you expect. Not paying attention to the assumed boundary conditions can get you in trouble, as in the “it’s clear” in the question as posed – it’s clearly not clear.

    To put it another way: physics says “div E = 4 pi rho and E goes to zero at infinity”. There is no solution to this equation with constant non-zero rho.

  12. #12 In Hell's Kitchen (NYC)
    May 17, 2011

    One more thing…the following is wrong: “But less well known is that F must be compact supported – ie, that it’s only nonzero over some finite region.”

  13. #13 Joshua Zelinsky
    May 17, 2011

    I haven’t thought about this in a while, but compact support seems like a very strong restriction. I suspect some sort of statement involving the resulting integral being Lebesgue measurable is probably enough.

  14. #14 Andy Wood
    May 18, 2011

    Some follow ups to my previous comment.

    Consider the vector field with components (x-x0,y-y0,z-z0), ie it points radially outwards from the arbitrarily chosen origin. It’s divergence is 1 everywhere,…

    That should read “It’s divergence is 3 everywhere,…”.

    If you wish to invoke symmetry arguments to find a solution to a PDE, it’s not enough to note the symmetries of the source term. The boundary conditions must possess the same symmetries. As noted, the boundary conditions for this problem were unstated.

    It’s not clear to me that it’s even possible to define boundary conditions in this problem.

    I’ve changed my mind. It should be possible to specify asymptotic behaviour as r -> infinity.

    I think it should be obvious that you don’t have complete freedom of choice in specifying boundary conditions. Some boundary conditions will be inconsistent with the PDE.

  15. #15 Paul Murray
    May 19, 2011

    Couldn’t you duplicate this effect without resorting to infinity by having an isolated charge and a hollow charged sphere (at the same voltage) around around it? We know that the electric field is zero inside the hollow charged sphere, but nevertheless the isolated charge inside the sphere has a charge on it. What’s the deal there?

  16. #16 Andy Wood
    May 19, 2011

    #15 Paul,

    The boundary conditions in problem you’ve stated are over-specified.

    In effect you’ve said:

    phi(a) = phi(b) = some_arbitrary_reference_voltage

    and

    phi'(a) = -q/(4 pi eps0 r)

    where ‘phi’ is the potential, ‘a’ and ‘b’ the inner and outer radii, q the charge and r the distance from the origin.

    That amounts to 3 boundary conditions on 2 boundaries. To solve Laplace’s equation for the potential, you only need one condition per boundary.

  17. #17 Andy Wood
    May 19, 2011

    Correction:

    phi'(a) = -q/(4 pi eps0 r)

    should be

    phi'(a) = -q/(4 pi eps0 a^2)

  18. #18 Eric Lund
    May 20, 2011

    We know that the electric field is zero inside the hollow charged sphere, but nevertheless the isolated charge inside the sphere has a charge on it.

    The electric field due to the hollow charged sphere is zero inside the hollow charged sphere. The isolated charge inside the sphere still contributes an electric field.

  19. #19 #raghuvir#
    May 20, 2011

    The paradox arises due to symmetry at every point. So any point can be taken as the centre of a spherical symmetry.So every point can be taken as origin. Gauss law also says that, at the centre of a spherical symmetry, the field vanishes. So according to Gauss law also, field becomes zero everywhere.

  20. #20 Tevong
    May 21, 2011

    I have to agree with Ahcuah. It’s very simple, the divergence theorem still holds for a uniform charge distribution even though the electric field is zero. It’s just that when you set a boundary that isolates a region with a certain amount of charge, the outgoing electric field flux lines are cancelled by the adjacent region that shares a boundary and contains the same amount of charge. (For the layman: imagine dividing up all of the uniformly charged space by cubes, each of which obeys gauss’s law with flux lines emerging from each cube, and it should be clear why they all cancel so the total electric field is zero). As for infinite charged plane sheets the charge is only uniformly distributed in one plane so obviously there will be emanating flux lines in the direction where no charge exists to cancel it out.

    Yes it’s essentially the proof of the divergence theorem but that’s why it works. Apply it to an infinite uniform charge distribution and it’s just what you physically expect, there’s no paradox or problem here: It’s *because* of the divergence theorem that the electric field is zero.

  21. #21 pallabbasu
    May 22, 2011

    Gauss law is kind of violated in QED due to vacuum polarization.

  22. #22 Andy Wood
    May 23, 2011

    #19 and #20 are both wrong.

    If the electric field vanishes everywhere, then its divergence vanishes everywhere and so the charge density must vanish everywhere.

    A non-zero charge density cannot, under any circumstances, give rise to a an electric field that vanishes everywhere.

    Like I said in comment #14, if symmetry arguments are to be applied to this problem, then the boundary conditions must possess those symmetries. And, as has been remarked in several comments, the boundary conditions were left unstated.

    For example, you could say that as r -> infinity, the electric field tends asymptotically to a field that is isotropic about the origin. In this case, you can conclude that the electric field vanishes at the origin. However, this does not imply that the electric field is also isotropic about any other point and it is incorrect to conclude that the electric field vanishes everywhere.

  23. #23 Anonymous Coward
    May 23, 2011

    What #1 and #6 said.

  24. #24 BlackGriffen
    May 25, 2011

    Agreed with the “you need boundary conditions” responses, and the 0D one, too.

    There is, of course, another way to look at this. Laplace’s equation for the electric potential is the E&M equivalent of F=ma. In looking at the div(E) = rho version of the equation, you’ve effectively said, “a=0 so I’m looking for static solutions.” The problem you’re having is that you’ve got one force set up, rho ~ force, but you don’t have any countervailing forces that come from imposing boundary conditions or other things. For instance, you could consider the potential in a massive field: – Laplacian( phi ) – m^2 phi + rho = 0 . The solution in this case is clear: phi = rho / m^2 (constant). Take the limit as the mass of the particle goes to zero and the potential diverges uniformly. That is, the massless case without boundary conditions permits no static solutions in the face of infinite charge. It’s just like trying to look for a static solution for applying a force to a mass on a spring when there’s no spring.

  25. #25 dmabs
    May 26, 2011

    sh*thead pz

    WRONG AGAIN

    clubconspiracy.com/forum/f29/judgment-day-may-21-2011-a-13663.html

  26. #26 dmabs
    May 29, 2011

    Sunday Sacrilege pz’s blaspheming head

    debunkingskeptics.com/forum/viewtopic.php?f=7&t=1756

  27. #27 Gael
    June 2, 2011

    I have a counter-example for you where you can have a field for an infinitely defined region.

    Say you have a sphere with charge density rho and radius r=L. Defining the center of the sphere as the center of our coordinate system, our answer for the electric field is E=(4/3)Pi*rho*r when rinfinity. Thus it is NOT zero due to symmetry.

    Basically the lesson to be learned is you need to be very careful with the assumptions you make and the boundary conditions you define.

  28. #28 Shouvik Datta
    June 5, 2011

    I feel the notion of a compact support still holds good here since we are integrating over a finite region (which is the sphere about the origin.) The field configuration still forms a compact set within the region of integration.

    The problem why we get a non zero value is slightly different. One single sphere is not the end of the story. In fact its ugly to think of spheres in this problem since it won’t fill up the space completely. So think of small cubes filling up the entire space and by symmetry arguments and Gauss’ law the field on the boundary of the cube is constant and points perpendicularly away from the surface. Okay, now its the same story for all the cubes. So, the fields will cancel when you consider the neighbouring cubes. Reduce the size of the cubes and you get a zero field everywhere!

  29. #29 John Peacock
    June 6, 2011

    Nice problem. As has been commented, the issues are the same in the gravitational field of an infinite uniform mass distribution. Treating this Newtonianly, you would pick an origin, and conclude from Gauss that points on the surface of a sphere or radius R are accelerated towards the origin by the mass within R. You would then appeal to the Newtonian shell theorem to argue that the gravitational effect of more distant shells of matter can be neglected. This yields the correct dynamical equation for this universe, the Friedmann equation; thus, we conclude that the mass distribution cannot be static and must either expand or contract uniformly (the second time derivative of R is proportional to R). If it is expanding, this “mass inside the sphere” argument correctly gives the evolution of the rate of expansion. The argument that the gravitational field must be zero by symmetry is false: the gravitational field is observable only via the relative acceleration of pairs of particles, and you get the same answer for this independent of where you place the origin (which is the only constraint imposed by symmetry).

    You can combine gravity and electrostatics in this approach and use Gauss to predict the acceleration of the expansion caused by a nett charge density. If the universe had a slight imbalance between protons and electrons, the resulting electric field would cause plasma to expand at a different rate from dark matter. Since we don’t see any such effect, the universe must be neutral to high precision.

    Of course, this is a cheat to an extent: space-time curvature means we can’t use Newton’s shell theorem to argue that the shells at large distances still cause no acceleration. This requires a fuller relativistic approach (Birkhoff’s theorem) to prove that Newton’s result still applies. Still, I think the basic point stands: you can use Gauss to calculate fields within a domain that is small by comparison with the scale of any spacetime curvature and you get the right answer for the relative acceleration of pairs of particles, in gravity and electrostatics.

  30. #30 Neil Bates
    June 13, 2011

    This should be a cautionary tale about how glibly people should accept various framings of issues. Matt, are you the first one to notice this loose end? If so, you deserve much credit. Consider writing it up for American Journal of Physics, it’s their kind of thing.

  31. #31 Laura
    July 4, 2011

    I wonder what happens to Gauss’ law when there’s a compact space with uniform curvature, like the surface of a sphere in 4 dimensions.
    I agree with comment #6: an electric field with constant divergence satisfies the problem, and the assumption that “symmetry implies the electric field is zero everywhere” is false. This assumption wasn’t given any justification.

  32. #32 J
    October 10, 2011

    Hey there, I do agree that Gauss’ Law only works if there is a high degree of symmetry. For in this case, we could use the integral form since it works for cases like this. However, I have a bit of a challenge, if one were to consider a parallel plate capacitor with a distance d between them connected to a battery and a vacuum space between them. Will the integral form of Gauss’ Law to any closed surface between the plates that does not cross either plate hold? If not, what in Gauss’ Law would prevent this from working?

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