Why a charged sphere whose radius oscillates in and out won't radiate?

[JD_PM]

In chapter 10 (Radiation; just after example 2 from 'Radiation from an arbitrary source') of Introduction to Electrodynamics by G. Griffiths he asserts that a charged sphere with oscillating radius doesn't radiate because, by Gauss law, $E$ stays the same no matter where the charges are located (either around a inner surface enclosed by a Gaussian sphere or the center of the Gaussian sphere):

$$\vec E = \frac{kQ}{r^2}\hat {r}$$

But I don't get it. If the sphere's radius is oscillating the picture I have in my mind is the charged ball kind of bouncing back and forth, which would mean that the sphere is being accelerated and thus it should radiate...

Why is Griffiths saying it will not radiate?

I think he is considering this case as that of an electric monopole, however that one shouldn't oscillate.

Thanks.

 

[Dale]

the picture I have in my mind is the charged ball kind of bouncing back and forth

Think instead of a balloon being inflated and deflated

 

[JD_PM]

Ohh so the sphere is just expanding/contracting but its centre of mass is indeed at the same point all the time, i.e we can regard all the time Q as being at the centre.

Appreciate your answer! It was literally an enlightening bulb moment!

 

[A.T.]

Ohh so the sphere is just expanding/contracting but its centre of mass is indeed at the same point all the time, i.e we can regard all the time Q as being at the centre.

Note that EM-waves are transverse waves. In which tangential direction would the E-field vary given perfect spherical symmetry?

 

[JD_PM]

In which tangential direction would the E-field vary given perfect spherical symmetry?

$\vec E$ field of a point charge goes radially outwards/inwards (depending upon the charge; positive outwards and negative inwards by definition).

If you enclose the charge within a Gaussian surface, the $\vec E$ field will always be equal to (no matter how large the radius is):

$$\vec E = \frac{kQ}{r^2}\hat {r}$$

 

[A.T.]

$\vec E$ field of a point charge ...

That wasn't the point. In an EM-wave E oscillates perpendicularly to the propagation direction. But due to spherical symmetry in this case there is no preferred transverse direction it would oscillate about. From this one can intuitively see that no EM-waves are propagating radially outwards.

 

[vanhees71]

The reason is that the electromagnetic radiation field is a spin-1 field and thus the partial-wave expansion starts with the dipole term. The monopole term is necessarily static, i.e., the only spherically symmetric solution of Maxwell's equations is the electrostatic Coulomb field. You need at least a dipole component to get radiative solutions.

For gravity (linearized general relativity) it's even starting with the quadrupole contribution, which is because the corresponding field is a spin-2 field.