Re: Phase cancellation --where does the energy go?

Neil Bates wrote:

[..cancellation of laser waves..]

You can do the same things in acoustics, where it is perhaps conceptually simpler.

Take a pipe with a wave traveling trough it. Then split it in 2. A beam splitter is simple in acoustics: It is just a T-junction! Of course, you need the width of the tube to be much smaller than a wavelength.
Now it is quite simple to arrange for a phase shift between the beams before recombining them in a second T-junction: Just make the lengths different. An wave in a pipe does not mind bends, as long as (again) the width of the tube is much smaller than a wavelength. We could arrange for a 180 degree phase difference, so that cancellation occurs: All energy is gone...


Diagram:

________
/ \
/ \
/ \
Louspeaker->>_________/ \_______>>
\______________/



So what is going on? As I see it, the paradox arises from confusing time domain and frequency domain. If you use frequency domain, you are talking about a wave which has been there for all time. So all possible reflections have "already" taken place and must be included. The pipe loop formed by the 2 paths would form a resonator as viewed from the entrance of the T-junction. A standing wave will be formed in the entrance tube and in the 2 branches. But in the exit branch, there is zero amplitude: A pressure minimum is at the second T junction, so there is no excitation of the exit branch.

If you look at the time domain, then you need to actually switch on a loudspeaker at t=0, and see what happens. A wave train will travel along the tube, split at the T junction, [The cross section of the 2 tubes need to be halved to avoid reflection!] and the 2 waves propagate further.
But at the second T-junction, where combination is to occur, the wave with the shorter path arrives first (t=t1). It is split into the "exit" tube and the other path. [ratio 2:1, because side branches must be half the area of the main branch] The part that moves into the other branch simply moves straight through the other wave. The pressures interfere constructively, the velocities destructively, and energy is conserved. Then the first wave arrives at the "entrance" T junction again, where again, it is split. Part of it moves back to the loudspeaker, as a kind of reflection. In the mean while, the other wave also arrives at the second T-junction at t=t2. In absence of the first wave, it would inject part of it into the exit branch. But because it is 180 degree out of phase with the first branch, the T junction will no longer give energy to the exit branch: The pressure is always zero from t=t2, so no work is done on the exit branch. So there will be a a short burst of sound into the exit branch lasting from t=t1 to t=t2, and then silence.

After t=t2, the waves effectively split at the first T-junction, move round without spilling into the the second T-junction, and partially reflect at the fist junction. It becomes increasingly hard to see the traveling waves, which start to look more and more like standing waves. When you have a standing wave, the work done by the loudspeaker becomes zero time-averaged, oscillating between negative and positive values.

For lasers, this will be the same. The beam splitters will become partial reflectors, and standing waves will form.

Gerard

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