where does the light energy "go"?

Question

Here is a physics thought experiment examining conservation of energy and EM propagation. In answering, please address how the experiment is resolved from a purely classical interpretation (i.e., how might Maxwell have answered this question?); and, how a QM interpretation might result in a different explanation? Here is the experiment. Let us arrange a coherent monochromatic light beam, e.g. a laser beam, to be split and directed along two paths. With repect to the light's wavelength, the paths are suitably arranged such that a half-cycle net time delay exists between them. These beams are then recombined downstream, so as to destructively interfere there. Question: where does the light energy "go"? Or, worded another way, what physical part of this experiment "gets hot"?

Thanks deeptimer

Answer 1

A question I have wondered myself.

The best I can manage is this...

I know that it is a thought experiment but it is actually impossible for the waves to be perfectly recombined as you suggest.

For example, if one of them passes through a partially reflective surface with the other undergoing its phase shift and then being sent back to reflect from the surface into the path then some of this energy will be transmitted through the partial reflector while some of the original ray will reflect off it. These will then interfere constructively (there will be an additional phase shift).

I think that there will always be this kind of loop-hole, no matter how you do it.

If you only do it approximately, so that the waves are very close to converging (with a phase shift) then you get an interference pattern that is pretty much zero along the central path (ie the first order solution is destructive to zero as expected) but with constructive interference elsewhere so that energy is conserved overall.

I hope this is clear!

Answer 2

I asked a similar question to this once - if you place a torch (flashlight) upside down on a perfect mirror so that the light is reflected back from the mirror and trapped between it and the reflector in the torch, where does all the light energy go when you turn the torch on, assuming none can leak out from where the torch and mirror meet ?

Most of the answers were that it is transformed into heat in the space between the torch and the mirror, but I tried doing this as an experiment at home with an almost perfect mirror, and didn't measure any temperature rise over a period of 4 hours. Needless to say there wasn't an excess of light energy when I lifted the torch off the mirror again ! So where did all the light go ?

Maybe all the photons eventually cancel each other out, but you'd expect there to be a residue somewhere wouldn't you ?

Answer 3

Simply put, the first piece and the last piece of equipments are heated -- the first and the last piece can be the same piece like a quarter-wavelength thick glass, depending on your setup. To Maxwell, the two waves cancelled each other. In quantum mechanics, photons have zero probability appearing in the last segment.

Answer 4

Wow this looks like a midterm question!

I cant speak for maxwell but the as for the energy in the light:
It will drop wavelengths to release engergy in the form of heat and continue to be light. (just of a different color, and likely a new direction)

Answer 5

Your answer lies in the Quantum Field world of propagation of potential, We see how photons can be emitted spontanously from the collisions or quarks, leptons and protons. We see virtual particles that seem to exist but not really, so it is we see existing photons evidently anihlate one another or return back to particle/energy potential.

Answer 6

Stop trying to be clever, you have set yourself up as a pratt.
and will now be ignored. :o)

And I along with others `aint gonner tell yer.