In the standard undergraduate treatment of E&M, Gauss's Law is loosely stated as "the electric flux through a closed surface is proportional to the enclosed charge". Equivalently, in differential form, and in terms of the potential (in the static case):

$$\nabla^2 \phi = -\frac{\rho}{\epsilon_0}$$

Now, when using the integral form, one typically uses the symmetries of a known charge distribution to deduce related symmetries in the electric field, allowing the magnitude of the field to be factored out of the integral. To do so, one usually relies on intuitive, heuristic arguments about how the field in question "ought to" behave$^1$.

I'm wondering how one goes about formalizing this notion in precise mathematical terms. In particular, it seems that there ought to be an equivalent statement for Gauss's law in differential form, along the lines of "symmetries in $\rho$ induce related symmetries in $\phi$". Is there a way to formally state this claim? In particular:

  1. How would one formulate a proof of the conditions (necessary and sufficient) under which a symmetry of $\rho$ induces a symmetry in $\phi$?
  2. When it exists, how does one explicitly state the induced symmetry in terms of the known symmetry?
  3. Can such a result be generalized for arbitrary linear PDEs whose source terms exhibit some symmetry?

It seems to me like there must exist a concise, elegant, and general way to state and prove the above, but I can't quite seem to connect all the dots right now.

$^1$ See, for example, in Griffiths, Example 2.3, p. 72: "Suppose, say, that it points due east, at the 'equator.' But the orientation of the equator is perfectly arbitrary—nothing is spinning here, so there is no natural "north-south" axis—any argument purporting to show that $\mathbf{E}$ points east could just as well be used to show it points west, or north, or any other direction. The only unique direction on a sphere is radial."

  • 1
    $\begingroup$ I have a follow-on question... Broken symmetry, where the symmetry of a solution is less than the symmetry of the equation, is common in quantum field theory. Why doesn’t it happen in, say, classical electrostatics? $\endgroup$ – G. Smith Nov 15 '18 at 1:36
  • $\begingroup$ If you really want to say something simple in a more formal way, maybe think about symmetries as operators acting on some space of functions. If the operator maps a function to itself, it is symmetric. If they operator commutes with the Laplacian (or whatever your diff eq happens to be) then your equation is symmetric. Then it's easy to see that if a function on one side of a symmetric equation is symmetric, the other must be. $\endgroup$ – octonion Nov 15 '18 at 2:01

Let $\mathcal{D}$ be the operator corresponding to your equation ($-\epsilon_0 \nabla^2$ in this case). Let $U$ be some operator corresponding to the symmetry. It might be a rotation or parity transformation, etc.

If $U f = f$, we say the function $f$ is symmetric.

If $U\mathcal D=\mathcal D U$ as operators, then we say your equation is symmetric.

Let, $\mathcal {D} f= g$. If $\mathcal{D}$ is symmetric and $g$ is symmetric, then we can easily show $\mathcal{D}f = \mathcal{D}\mathcal U f$. If we can take an inverse of $\mathcal{D}$ we've proven $f$ is symmetric.

Taking an inverse of $\mathcal{D}$ is the the same thing as being able to solve the equation uniquely. In your particular case we can solve the equation uniquely if we restrict our function space to have some boundary conditions, say vanishing at infinity.

So this is the example you had in mind and this a formalization of the argument that $\phi$ must be symmetric.

Now if we can't solve the equation uniquely then there may be a loophole in the argument. A particular case I have in mind is a magnetic monopole which is rotationally symmetric, but the vector potential solution has a Dirac string and is not. But any two solutions $f$ and $Uf$ in this case are connected by a gauge transformation.

  • $\begingroup$ That loophole is (I think, if I understand correctly) the point that I'm stuck on. It's perfectly clear to me that, if your equation has some symmetry, then acting on some particular solution $f$ with the relevant symmetry operator produces another particular solution $Uf$, but the stronger requirement that these two solutions are, in fact, equal doesn't seem to follow in a straightforward way, if at all. I guess what I'm looking for is a more detailed treatment of how/when former implies the latter? $\endgroup$ – TheMac Nov 15 '18 at 3:31
  • $\begingroup$ I don't think you were touching on the loophole in your question. In your particular example the argument that $\phi$ is symmetric is valid. I've edited my answer to be clearer (this is why I prefer to answer in the comments btw) $\endgroup$ – octonion Nov 15 '18 at 4:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.