0
$\begingroup$

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 $\phi$.

$\endgroup$
0
$\begingroup$

I've finally come to the conclusion that $\vec p(t)$ doesn't necessarily vanish because $\rho$ does not necessarily vanish although it's constant thru time. So that there will be no radiation due to the electric dipolar term, but I can't conclude that $\vec p(t)=0$ with the information given.

$\endgroup$

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.