Non-radiation charge density

Question

For all $l,m$ but $l=0,m=0$, can we find $r_0,w_0$ such that the following charge distributions can represent a charge field that does not radiate:

$$ \rho(r,\theta,\phi) = \Re(c_{l,m} Y_{l,m}(\theta,\phi)\exp(iw_0t)\delta_{r=r_0}) $$

with $\Re(\cdot)$ the real part, $c_{l,m}$ constants, $Y_{l,m}$ spherical harmonics, $\delta_{r=r_0}$ the delta measure of a spherical shell of radius $r_0$. The values are real because we take the real part and the charge is conserved because the integral of $Y_{l,m}=0$ for all $l,m$ but $l=0,m=0$. Also we need to consider the current sources so that they exists and the continuity equation is satisfied. To construct a current field that satisfy the continuity equation take $$ j = \Re(a_{l,m} \nabla_S Y_{l,m} \exp(i w_0 t) \delta_{r=r_0}), $$ with $a_{l,m}$ constants and $\nabla_S$ the spherical gradient at $r=r_0$ and $\nabla_S \cdot (\cdot)$ the spherical divergence at $r=r_0$. $\Re$ is linear and commute with the differential operators so we can skip that from now and consider values in C in stead. Then use the property of the spherical harmonics (eigenfunction of the spherical part of the laplace operator), $$ \nabla_S \cdot j = \nabla_S \cdot \nabla_S a_{l,m} Y_{l,m} \exp(i w_0 t) \delta_{r=r_0}= a_{l,m} \lambda_{l,m}Y_{l,m}\exp(i w_0 t) \delta_{r=r_0}, $$ with $\lambda_{l,m}$ the eigenvalues associated with the spherical harmonics. So $$ \frac{\partial}{\partial t} \rho = i c_{l,m} w_0 Y_{l,m}\exp(i w_0 t)\delta_{r=r_0}. $$ To satisfy the continuation equation we need this to be equal to (note the charge is on the sphere so we restrict the gradiant to e.g. $\nabla_S$), $$ \nabla_S \cdot j, $$ which according to the above is the same with the right choice of the constants $a_{l,m}$.

Note that multiplying this with the conjugate will lead to a non time varying charge distribution that of cause will not radiate. But to keep it interesting and according to a claim in a paper from the 92 I let it be time varying and hence less obvious that this is non radiating. It's possible to keep the charge distribution of the same sign by adding a constant to it. I also seen a proof of this fact that is overly complex spanning many pages and in the end, to my knowledge, is wrong. One way to prove it is to take the fourier transform in time and space of the charge distribution (Goedecke) and show that the transform is zero for all parameters that is associated with waves at the speed of light.

For a reference see: Force Free Time Harmonic Plasmoids, Jack Nachamkin, University Of Dayton Research Institute 1992. Especially the statement on p. 11. The question I formulate is not directly stated in the paper, but I derive it. You can also find these fields in Randell Mills GUTCP, which is a work of dispute. But there is two proofs submitted there for this fact, I could follow the first of them and I think it contains an error. Finally consider this paper for non radiation conditions Goedecke, G. H. (1964). "Classically Radiationless Motions and Possible Implications for Quantum Theory". Physical Review. 135: B281–B288. Bibcode:1964PhRv..135..281G. doi:10.1103/PhysRev.135.B281.

Answer 1

If we can agree to that according to Goedecke it is enough to prove that the fourier transform in time and space is zero for all waves associated with the speed of light we can use the following route. Now the real part function can be excluded from the analysis because it commutes with the fourier transform and we need to calculate $$ F(\rho) = \int \exp(itu + i \hat k \cdot \hat r) c_{l,m} Y_{l,m}\exp(i w_0 t)\delta_{r = r_0}\,drdt $$

Now a we can expand the plane wave in spherical harmonics according to $$ \exp(i \hat k \cdot \hat r) = \sum_{l,m} d_{l,k} j_l(|\hat k||\hat r|)Y_{l,m}(\hat k) Y_{l,m}^*(\hat r), $$ with $j_l$ the spherical bessel function.

Use that the time part becomes a delta measure on $w_0$ and insert the expansion above and sketchfully exchange integral and summation and you have $$ F(\rho) = \sum_{l',m'} j_{l'}(|k|r_0) d_{l',m'}c_{l,m} \int_{r=r_0} Y_{l,m}(\hat r) Y_{l',m'}(\hat k) Y_{l'm'}^*(\hat r) dS_{r=r_0} \delta_{u=w_0} $$ So using the orthogonality e.g. $l=l', m=m'$ we have a constants $e_{l,m}$ so that, $$ F(\rho) = e_{l,m} j_l(|k|r_0)Y_{l,m}(\hat k)\delta_{u=w_0} $$

Now for all $l,m$, $u=w_0$ and for waves at the speed of light we have $|k|c = |w_0|$ so for all $r_0,w_0$ so that $|w_0| r_0 / c$ is a zero to the bessel function $j_l$ we have that the fourier transform is zero for all values $k,u$ comensurate with the speed of light hence it looks like indeed the question is answered affirmatively.