## Amplitude & Energy in Electromagnetic Waves

Here’s a little physics riddle. As you all know, electromagnetic radiation consists of oscillating electric and magnetic fields rather like this:

(Graphic stolen from here.) The polarization state of the wave is defined by the direction of the Electric field, in this case vertically upwards.

Now the energy carried by an electromagnetic wave of a given wavelength is proportional to the square of its amplitude, denoted in the Figure by A, so the energy is of the form kA^{2} in this case with k constant. Two separate electromagnetic waves with the same amplitude and wavelength would thus carry an energy = 2kA^{2}.

But now consider what happens if you *superpose* two waves *in phase*, each having the same wavelength, polarization and amplitude to generate a single wave with amplitude 2A. The energy carried now is k(2A)^{2} = 4kA^{2}, which is twice the value obtained for two separate waves.

Where does the extra energy come from?

Answers through the Comments Box please!

Follow @telescoper
September 22, 2015 at 1:00 pm

“consider what happens if you superpose two waves in phase, each having the same wavelength, polarization and amplitude to generate a single wave with amplitude 2A. ”

In some ways you can think of this as a gauge problem. The only way you could double the amplitude, while keeping all the other geometric properties of the wave pattern the same, is to double the magnitude of the charges at the source of the wave – and power radiated goes as q^2.

I always tell my students, find out what the damned charges are doing, then you can see what’s going on.🙂

September 22, 2015 at 1:14 pm

Yes, it’s all about how the waves get generated.

September 23, 2015 at 4:23 pm

Cool! Two charges (each q) radiate twice the energy a single charge radiates, while a single double-charge (2q) radiate four times the energy of a single q charge! Unless the two separate q charges are radiating in phase…

September 23, 2015 at 9:43 am

I think the problem is that your initial statement “the energy carried…is of the form kA^2” is too imprecise. For a single monochromatic wave, both the energy density U and the energy flux density S measured at a single point vary as (cos[omega t])^2, while preserving S=c U. So energy is transported at the speed of light, but not in a completely uniform way – so your “energy carried” is implicitly a time average. Now add two modes of different frequencies and calculate U and S. In both cases you get the sum of the terms you would have got from the modes separately, plus an oscillating cross-term. This always averages to zero over a long enough time window, justifying your assumption that the kA^2 terms should add. But that derivation breaks down in the special case that the two frequencies are exactly equal.