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):
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:
The integral of the charge density within the surface is of course just the total charge enclosed within the surface:
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:
Wikipedia’s notation for the integrals is a tiny bit different, but you get the gist. Making this substitution in Gauss’ law gives:
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.)