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We know for EM waves undergoing destructive interference, there's constructive interference somewhere else that compensates for the loss in energy during destructive interference. However, I tried constructing a case (below) where such arguments seemingly don't apply.

Statement: Laser A, placed at the origin, produces a monochromatic EM wave with E field $\vec{E}_a(x,t)=\vec{E}_0e^{i(k_x x-\omega t)}$. Laser B, placed at $x=-\frac{\pi}{k_x}$, produces a monochromatic wave $\vec{E}_b(x,t)=\vec{E}_0e^{i(k_xx-\omega t+\pi)}$. At $t=0$, both lasers are fired for one cycle. Let's say it takes an energy $U_0$ for each laser to produce this wave (since they produce the same wave, they obviously take the same amount of energy). Notice that after a long time, the two waves propagate through space with the EM waves canceling for half a cycle in the region of overlap -- we effectively have one cycle of the same wave, transporting $U_0$, not $2U_0$ (as one would expect). Where did the energy go?

Three things to note with the construction of the problem:

  1. The resulting wave is still a monochromatic plane wave (locally), so the usual calculations (mechanical properties) apply: energy density is still uniform along the waves
  2. I've avoided unrealistic waves that extend out to infinity by restricting the waves to one period. (if you find this troubling, the same problem arises with any plane wave)
  3. The $(\vec{k},\vec{E},\vec{B})$ triad are all oriented in the same direction, so the EM waves do indeed cancel in both electric and magnetic components in the region of overlap (ie. there are no fields in the region of interference)

It seems like a problem that can be solved in classical EM, but I just can't figure out where the energy is going... With waves in media, you can always argue that the energy goes into the medium through which the wave propagates (ex.potential energy in the particles in a string) but this doesn't apply to EM waves in vacuum. Any help is appreciated.

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  • $\begingroup$ See physics.stackexchange.com/q/434024/291677 and physics.stackexchange.com/q/196650/291677. The former says you can't perfectly arrange these two beams to overlap the way you are describing, while the latter says you need to bit a bit more careful to define energy flux when you've got multiple beams superposed $\endgroup$ Apr 29, 2023 at 20:10
  • $\begingroup$ @QuantumMechanic thank you for your input. However, I kind of don't see why these two posts are relevant: 1. I'm not combining two sources: I have two separate sources that are individually producing waves (our notion of superposition suggests there's no field in the region of overlap) 2. the second post is precisely the reason this issue arises: I'm not adding the energies separately but rather looking at the energy of the wave after interference. In the region the waves do not overlap, we have half a cycle of the original wave $\endgroup$
    – Huchaney
    Apr 29, 2023 at 20:19
  • $\begingroup$ wait actually, maybe it's the fact that laser B goes through laser A at some point. The wave source of A somehow affects the EM wave generated by laser B $\endgroup$
    – Huchaney
    Apr 29, 2023 at 20:26

1 Answer 1

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Where did the energy go?

The energy went into source A.

Let's say it takes an energy U0 for each laser to produce this wave (since they produce the same wave, they obviously take the same amount of energy)

This actually is not the case. Although they are producing the same wave, their EM environments are different. They are not symmetrical.

Source B has no external field while generating the pulse. Source A is generating half of its pulse in the presence of an external field that is out of phase with A. This results in energy going into A. So the process is as follows:

Both A and B produce a half-pulse with energy $U_0/2$. A’s half pulse continues on, while B’s is absorbed by A and B produces another half-pulse with energy $U_0/2$.

A output a total of $U_0/2$. B output a total of $U_0$, the first half of which was absorbed by A and the second half continued on after A stopped absorbing it.

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  • $\begingroup$ Yeah it makes sense now haha. If we don't have absorbtion (so we just have both lasers turned on at all times when the waves are being generated) laser B inevitably changes as it passes through laser A's source, am I correct? Is there a way to make calculations with that? $\endgroup$
    – Huchaney
    Apr 29, 2023 at 20:59
  • $\begingroup$ Like an oscillating dipole as the source $\endgroup$
    – Huchaney
    Apr 29, 2023 at 20:59
  • $\begingroup$ @Huchaney yes. It is much easier to use a pair of dipole antennas. Then the energy transfer into or out of the sources is easier to calculate directly since you know the current density at each point $\endgroup$
    – Dale
    Apr 29, 2023 at 21:17
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    $\begingroup$ Right. Thank you so much for the answer! $\endgroup$
    – Huchaney
    Apr 29, 2023 at 21:30

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