# When electromagnetic radiation does work, does its amplitude decrease?

In the beginning of his lecture on the refractive index (Chapter 31), Feynman puts a radiating source on one side of a glass plane. He then claims that the electric field on the other side of the glass would equal the sum of the radiating atoms in the glass, $\Sigma E_a$, plus the electric field of the source "if there were no [glass] material present", $E_s$.

If the electric field of the source expends energy accelerating the atoms in the glass (radiation resistance, Chapter 32), why would its amplitude/energy not be diminished on the other side of the glass?

## 2 Answers

You will see further down the page that the net field after passing through the glass has only a phase change compared to what it would have been if the glass were not there.

This means that the total energy from the field $E_s$ plus the radiating atoms $\Sigma{}E_a$ has worked out to be the same as the original energy from the source.

So it may be glossed over, but you've essentially assumed that all the energy initially absorbed by the glass is re-radiated and contributes to the energy from the radiating atoms. This does mean that the energy in the "original" field must actually be less than it would have been if the glass weren't there.

You'll also notice that he says "at the moment because we shall be limited to a calculation for a material with an index so close to 1 that very little light is reflected". This limit also means that the phase change due to passing through the glass is very small, and any energy absorbed and re-emitted by the glass is very small.

In Section 31-4 he deals with the case where the energy absorbed by the glass, but not re-emitted, is significant.

In a neutral, ideal insulating medium, that has only a real refractive index, the electromagnetic fields do not do any work.

The reason for this is that the oscillations induced by the electric field of the incoming waves are such that the velocity of the oscillating charges is $\pi/2$ out of phase with the driving force. This means that the time-averaged work done per unit volume by the electric field, given by $\vec{E}\cdot \vec{J} = \vec{E} \cdot \vec{v}\rho$, where $\rho$ is the charge density, is zero. i.e. elastic scattering.