This is building off of a question I asked here.

When discussing the linked problem with some friends, the consensus seemed to be that the reason two identical semi-infinite plane waves cannot undergo perfect constructive interference is because their non-infinite spatial extent means they necessarily cannot be monochromatic, and so they cannot perfectly constructively interfere everywhere.

While this seems logical, I feel there is an assumption here that shouldn't be taken for granted: this assumes that the spatial distribution of frequencies cannot be the same for both waves. Is this the case? Or is it possible to generate two coherent semi-infinite beams whose spatial frequency distributions are perfectly in phase?

This question is, I believe, equivalent to asking if two identical semi-infinite plane waves can perfectly constructively interfere. In the linked question you can see my energy-based argument as to why I expect it is not possible. I'm looking for a more rigorous workthrough.

  • $\begingroup$ Is a semi infinite plane wave even a valid solution of Maxwell’s equations? $\endgroup$ – Dale Oct 16 '18 at 22:15
  • $\begingroup$ @Dale it is valid. It's easy to see this: just consider an initial state where a superposition of infinite plane waves forms a semi-infinite plane wave (all the wave vectors in the superposition point the same direction). Then, since vacuum is dispersionless, this sem-infinite wave will move with $c$ as a rigid structure. $\endgroup$ – Ruslan Mar 23 '20 at 9:05

I am not certain that a semi infinite plane wave is a valid solution to Maxwell’s equations. However, let us instead consider a general wave that is a solution to Maxwell’s equations. If you make another wave which has complete constructive interference with the first wave then you will add the two fields together, resulting in a double field everywhere. This also includes doubling the sources.

If you double the fields then according to Poynting’s theorem you have four times as much energy. His seems to be a violation of the conservation of energy. However, in fact it is not. The energy in the waves is indeed four times as great, but in order to produce such a wave you must have doubled the sources as well. So you doubled the E field and you doubled J. Then by Poynting’s theorem you have quadrupled the work.

So yes, the field contains four times as much energy, but it also requires four times as much energy to produce. The mistake is in believing that it only requires double the work to produce the double field.

  • $\begingroup$ You must double each of the two sources? Why? $\endgroup$ – Riley Scott Jacob Oct 18 '18 at 17:28
  • $\begingroup$ If you have complete constructive interference everywhere then by the various uniqueness theorems you must have the same sources. Therefore all sources are doubled $\endgroup$ – Dale Oct 18 '18 at 20:33
  • $\begingroup$ Two input sources. Exactly identical. Each with a field strength E corresponding to a power E^2. So the total input power is 2E^2. By superposition, output beam has field strength 2E and so a power 4E^2 - twice the input power. This is the issue. I don't find your reasoning to be rigorous or convincing $\endgroup$ – Riley Scott Jacob Oct 18 '18 at 20:37
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    $\begingroup$ This answer is rather unsatisfactory. Why do you need to double the sources? And what happens if you have complete destructive interference? What happens to the energy? $\endgroup$ – Peter Shor Oct 19 '18 at 0:25
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    $\begingroup$ You keep saying the sources need to be doubled, but you’re basically saying they need to be doubled because they need to be doubled. Can you actually provide a reason for why that’s the case, using some rigor that doesn’t just require us to trust that you’re correct? $\endgroup$ – Riley Scott Jacob Oct 19 '18 at 22:10

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