west@sdcsla.UUCP (Larry West) (11/05/84)
I have a question about two-source interference. I don't think classical physics can explain this, but I may be wrong. I'll do the example in two dimensions, but the extension to three dimensions should be obvious. All in a vacuum. Let's start with a single coherent point source of light inside a box (a circular box, but a square one is easier to type), as: -------- | | | . | <<- the "." is the source. | | -------- This point source is radiating E units of energy per second, all at a fixed wavelength W. The inside of the box will be uniformly illuminated (a circular box, but the uniformity is just for simplicity), and will receive E units of energy over its total area (some of the light will be reflected, but I don't think that matters). The source is very close to the center of the box. Were we to unroll the (circular) box and graph the intensity of the light received around the interior we would get: E Q |***************************************** etc. n | e | r | g | y | | 0 |-------------------------------------------------------------- 0 1 2 3 4 5 6 7 8 Position --> Here, "Q" is the total energy divided by the total length (in three dimensions, this would be divided by total area). Now, add a second light source, identical to the first (same energy, same wavelength, coherent and in phase with the first), except that it is placed a little away from the first source: -------- | | | .. | <<- the ".." are the two sources. | | -------- There will now be interference between the light from the two sources. At points on the box exactly equidistant from the two sources, the interfence will be constructive, because the light waves will be in phase. At those points which are at a distance X from source one and X+(W/2) from source two, there will be completely destructive interference. At other points, the interference will be partly destructive or partly constructive. Were we to unroll the (circular) box and graph the intensity of the light received around the interior (which should be roughly a sine-squared function, I think) we would get: E 2Q | *** *** n | ** ** ** ** e | ** ** ** ** r Q | * * * * g | * * * * y | ** ** ** ** etc. | ** ** ** ** ** 0 |**-----------------------***-----------------------***-------- 0 1 2 3 4 5 6 7 8 Position --> [Note that I've labeled the ordinate axis with "2Q" at the peak of the sinusoid. This may be the crux of my misunderstanding.] Now you'll note that the total energy received over the entire interior of the box (in a given unit of time) is just the integral of that function over all positions. And you will also note that this is approximately (Q*Length). However, that is the same as the total energy received (per unit time over the entire box) in the case of a single source. Perhaps a little more, but nowhere near twice the energy. What happened to the rest of the energy? There is now twice as much energy being radiated from the two sources as was being radiated by the single source, and yet the energy being received at the box is about the same. This can be restated slightly. If there was only constructive interference over the entire box (which could happen if it were shaped properly), then the total energy received at the interior surface of the circular box would be exactly doubled when the second source was added. But because there is also destructive interference, the total energy received is less than doubled by the addition of the second source. (And one could shape the box such that there was only completely destructive interference on the surface when the second source was added --> no energy received!) If anyone is confident of a cogent explanation (may include quantum physics), please reply by mail. This question has been nagging me for a few years... (If you say "it went into heat", please explain how.) Thanks for your time! -- Larry West, UC San Diego, Institute for Cognitive Science -- UUCP: {decvax!ucbvax,ihnp4}!sdcsvax!sdcsla!west -- ARPA: west@NPRDC {{ NOT: <sdcsla!west@NPRDC> }} -- -- Larry West, UC San Diego, Institute for Cognitive Science -- UUCP: {decvax!ucbvax,ihnp4}!sdcsvax!sdcsla!west -- ARPA: west@NPRDC {{ NOT: <sdcsla!west@NPRDC> }}
robison@uiucdcsb.UUCP (11/09/84)
You are right in suspecting that the ordinate axis should not be labeled 2Q.
Indeed, it should be labeled 4Q as shown:
E 4Q | *** ***
n | ** ** ** **
e | ** ** ** **
r 2Q | * * * *
g | * * * *
y | ** ** ** ** etc.
| ** ** ** ** **
0 |**-----------------------***-----------------------***--------
0 1 2 3 4 5 6 7 8
Position -->
The explanation is simple wave mechanics. Whenever we add two waves, we add
their amplitudes. The energy in a wave is proportional to the SQUARE of the
amplitude. So when you add two coherent waves constructively, the total
total amplitude is double. Therefore the total energy is quadrupled where
the two waves add constructively. By the nature of interference, however,
there will also be corresponding places where the waves interfere destructively
and the total energy is zero. If you integrate over your wall, you will find
that all the energy is conserved.
I just happen to have my prof's holography notes on hand, so I can give
the complete formula. The energy of the summations of waves with energies A
and B is given by:
A + B + 2*sqrt(A*B)*|d|*cos(a)
where d is the "normalized mutual coherence function" (wow - what mouth full!)
and a is the phase difference. If two waves have d=0, the waves are incoherent
and the resulting energy is just the sum of the energies A and B. If two waves
have d = 1, they are perfectly coherent and the total energy is dependent upon
the phase relation. For 0<d<1, the waves are partially coherent, and the
energy is partially dependent upon the phase relation.
A good analogy is random variables. The variance of a random variable
correspond to the energy of a wave. The covariance corresponds to the
mutual coherence function. Depending upon whether the variables are independent
(incoherent) or correlated (partially or perfectly coherent), the variance of
the sum can be greater than, equal to, or less than the sum of the variances.
Arch Robison @ uiucdcs