I'm interested to show that given a ring with radius a where there's a current $I_0 \cos \omega t$ ($I_0$ is a constant) there is no radiation due to the electric dipole term (appearing in the multipole expansion).
Now the electric dipole term is worth $\vec p (t)=\int \vec r \rho (\vec r) dV$ where rho is the charge density. The radiation due to this dipole term is proportional to $|\ddot {\vec p} (t)|^2$.
My first attempt was to simply say that since $\rho=0$, $\vec p (t)=0$ and so there's no radiation but this did not convince my professor. He told me that I could have used the continuity equation $\nabla \cdot \vec J + \frac{\partial \rho}{\partial t}=0$ to deduce that there would be no radiation due to the electric dipole term.
So I tried to use his tip. Using spherical coordinates $(r, \theta , \phi)$, $\vec J=I_0 \cos (\omega t) \frac{1}{r}\delta (r-a) \delta(\cos \theta)\hat \phi$. I then took the divergence in spherical coordinate, of $\vec J$, which is worth 0 if I didn't make any mistake. So that $\frac{\partial \rho}{\partial t}=0$. Now looking back at the expression for the radiation and at the expression for the electric dipole moment, it's obvious that $\ddot {\vec p} (t)=\vec 0$ and so there is no radiation due to the electric dipole.
Does this argument sound convincing to you? Do you have a simpler way to prove or show what I tried to achieve?
Here are the details for the math related to the divergence: $\nabla \cdot \vec J=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2J_r)+\frac{1}{r\sin \theta}\frac{\partial}{\partial \theta}(J_\theta \sin \theta ) + \frac{1}{r\sin \theta} \left ( \frac{\partial J_\phi}{\partial \phi} \right )=0$ because the first two terms are simply 0 because $\vec J$ has only a $\hat \phi$ component; in other words $\vec J = J_\phi \hat \phi$. The last term is 0 because $\vec J$ does not depend on $\phi$.
