If all the multipole expansions apart from the monopole is zero, then is the charge density spherically symmetric? Say I have a charge density $\rho(\vec{x})$ in some finite volume $V$, such that all of the multipole expansions apart from the monopole moment (meaning dipole, quadrupole and so forth) are zero. Does this mean that $\rho(\vec{x})$ is spherically symmetric? I know the opposite is true - meaning, if $\rho(\vec{x})$ is spherically symmetric, then all of the multipole expansions apart from the monopole moment are zero, since we can treat $\rho(\vec{x})$ as a point charge density. But it seems the original statement doesn't have to be true, I just can't find a counterexample.
Thanks!
 A: The charge density does not have to be spherically symmetric. Imagine that you start with a spherically symmetric charge distribution, which as you say will have zero multipole moments (except possibly for a monopole moment). Now add to this original distribution a new distribution, which is spherically symmetric about a different point, and which has zero net charge. For example, you could imaging taking an "off-center" spherical chunk inside the original spherically symmetric distribution and moving the charge around radially (with respect to the center of the off-center chunk) without adding or removing any charge.
The second distribution is spherically symmetric and has zero net charge, so by Newton's shell theorem its contribution to the electric potential vanish identically outside of itself, and so all of its multipole moments vanish (including the monopole moment). So adding it to the original distribution changes the charge distribution without affecting any of its multipole moments, and you end up with a non-spherically symmetric charge distribution whose electric potential only has a monopole moment.
The lesson here is that the multipole expansion does not uniquely determine a source charge distribution; it only determines the asymptotic form of the faraway electric field.
A: If all the higher order multipole expansions are zero, then you would effectively just have a monopole, i.e., you could reduce the problem to that of a point charge. Then the charge density is spherically symmetric. See here. A more strict mathemtical approach would involve spherical harmonics.
