Suppose 2 sinusoidal waves undergo totally destructive
interference. This results in wave with zero amplitude (and hence zero
intensity). What happens to the energy associated with wave (1) and wave (2)?
Asked by: Achilles
Oh! I like questions like this one, which lead
one to arrive at an absurdity using apparently
plausible, but impossible in detail arguments.
So, lets analyze how this one works:
The main argument is, any wave has an energy
associated with it, which is proportional to its
amplitude squared, therefore necessarily positive.
Then, you take two such waves, let them interfere
with phase difference Pi (1.57 will *almost* work)
and destroy the two waves. No wave, means no
energy, so what happened to the energy?
On a side note, you don't really need total
destructive interference -- even partial
interference will seemingly violate conservation
of energy. But total destructive interference
makes the argument a little more drastic.
To start with, when we say a sinusoidal wave, we
are talking about waves of the sort
A sin (kx - wt)
The first thing to note is that, such a wave
extends indefinitely in time and space, and
exists everywhere and every time. With such waves
you either have them, or you don't. So you can't
cause any sort of interference. Then we must
restrict ourselves to waves which are sinusoidal
only in part of space, or in part of time.
Now, to have interference, the waves will have
to have the same k and same w. Otherwise they
will not interfere in terms of energy -- this
can be shown, but is probably better left to
The best way to attack the question is with an
example. But, the idea can be generalized to any
Lets take an electrical circuit, with two AC
sources of the same frequency, acting on a purely
reactive circuit (one that is made of capacitors
and/or inductances only) so that no energy is
drained by the circuit itself. Let the two sources
be exactly in opposite phase -- a phase difference
Now, turn on one source. You get a sinusoidal wave
on the circuit. It has an energy. Where did the
energy of the wave come from? Of course, from the
source. Turn it off, and turn on the other source.
Again, same result. Now, turn the second one on,
too. Result? No wave on the circuit. (Of course,
one may need to adjust things rather carefully
to get this in practice -- so don't try this at
home plugging a capacitor into the AC outlet).
Now comes the big question: What happened to
the energy of the wave source #1 had set up
before we turned on source #2??? The capacitors
had charges oscillating, inductors had currents
oscillating, all at the same frequency... Where
did all that go? The answer is simple. It was
absorbed by the second source you just turned
on. (Those who know how to analyze circuits, can
figure it with two current sources, one capacitor,
and letting the second source come on slowly.
Then think of the limit of turning it on in an
instant). What would happen if I turned on the
two sources at exactly the same time? There would
not be a wave to start with!
So, the essence of the answer is:
* All waves have sources (and drains).
* One must include those in the analysis.
* The source that causes the destruction of
the wave absorbs the energy.
After all, there is not much point in talking
about waves which don't have sources and drains,
since they don't interact with anything, and
there is not much point in talking about their
Answered by: Yasar Safkan, Ph.D. M.I.T., Software Engineer, Istanbul, Turkey
Consider the case of two radio transmitters in space, transmitting to earth
on the same frequency. Then there will be some places on earth where the two
waves cancel out and the received amplitude is zero. But there will be
other places where the two sources reinforce each other. So energy is not
lost, just moved around. The exact geometry governs the patterns of peaks
Answered by: Phil Freedenberg, E.E.D., Exec. VP, Federal Engineering, Inc., Fairfax, VA
'Arrows of hate have been shot at me too, but they have never hit me, because somehow they belonged to another world with which I have no connection whatsoever.'