1. Aug 27, 2006

### golfingboy07

Question: Show (mathematically) that a sphere charged in spherical symmetry ocillating radially will not radiate

My ideas so far:

We can use polar coordinates and choose a test point on the z axis so that the transvere components of J cancel. For radiation to occur (using the coulomb gauge) there must be a transverse component of the current density vector to drive it. Since this is not the case the sphere charged in spherical symmetry will not radiate. How do I go about proving this mathematically? Is my physical statement even correct?

Thanks

2. Aug 27, 2006

### quasar987

Just to be sure: we're talking about a sphere or a ball? The sphere is hollow, the ball is full.

3. Aug 27, 2006

### golfingboy07

The question sheet states a sphere

4. Aug 27, 2006

### quasar987

damn..

I'll think about it while I wash the dishes.

5. Aug 27, 2006

### quasar987

Would it be enough to show that at points outside the "region in which the sphere oscillates", the potentials are time independant? I was thinking of finding an expression for rho in

$$\phi(\vec{r},t)=\int\frac{\rho(\vec{r}',t-|\vec{r}-\vec{r}'|/c )}{4\pi\epsilon_0|\vec{r}-\vec{r}'|}dV'$$

6. Aug 27, 2006

### golfingboy07

Maybe. I was thinking of using the vector potential A since this is related to the current density via a similar integral as above. All we need to do is find an expression for J perhaps...not really sure though

7. Aug 27, 2006

### quasar987

I think I've found an expression for rho:

$$\rho(\vec{r}',t)=\sigma \delta^3(\vec{r}'-R_0sin(\omega t)\hat{r}')$$

where sigma is the surface charge density. If the charge Q is constant on the sphere but its radius changes, sigma is time dependant:

$$\sigma(t)=\frac{Q}{4\pi R_0^2\sin^2(\omega t)}$$

R_0 is the "radius of equilibrium" obviously and omega the frequency of oscillation.

Last edited: Aug 27, 2006
8. Aug 28, 2006

### quasar987

I meant more like

$$\rho(\vec{r}',t)=\sigma \delta^3(\vec{r}'-(R_0+\epsilon \sin(\omega t))\hat{r}')$$

$$\sigma(t)=\frac{Q}{4\pi (R_0+\epsilon \sin(\omega t))^2}$$

But is the resulting integral evluable?

9. Aug 28, 2006

### pseudovector

The electric field outside the sphere is constant because a uniformly charged sphere produces a radial, radius independent field.
The components of the magentic field perpendicular to the radius are zero due to symmetry.
Hence the Poynting vector is zero so there is no energy transmission outside the sphere.
qed.

10. Aug 28, 2006

### quasar987

Yes, but the thing is pulsating, and it takes different time for light emanating from different parts of the sphere to reach a given point, so it's a big mess and we can't use this static argument IMHO.

11. Aug 29, 2006

### pseudovector

It doesn't matter. If you freeze the sphere at any moment, you get the same electric field at a distant point regardless of the moment you chose to freeze the sphere or its radius at that time.
Moreover, since the magnetic field is zero, then according to Maxwell's equations, the electric field must be time independent.