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Re: interference question



At 08:48 AM 9/12/99 +0800, romanza wrote:
I can't really follow with your argument, while qualittaively I understood
what your meant totally.
The average intensity is 2a^2, but for individual wave, the average
intensity is only (a^2)/2 (averaging cos^2 for each), and hence the total
average intensity of the two waves is a^2.
Can you pls show me step-by-step how you arrive at the conclusion that
energy is conserved.

There are two questions on the table. Let me deal with one then the other.

The first question asks how is it that when we shine two flashlight beams
on one place, we see the sum of the energy of the two beams, and don't see
any obvious interference effects -- which might give (V1 + V2)^2 or (V1 -
V2)^2 rather than the classical (V1)^2 + (V2)^2.

You requested a step-by-step mathematical explanation. Here it is. It's
straightforward. Write the voltage and energy of one beam as
V1 = v1 cos (omega t)
E1 = mean square[V1]
= .5 v1^2
and the other as
V2 = v2 cos (omega t + theta)
E2 = mean square[V2]
= .5 v2^2
using the fact that average[cos^2] is 0.5.

Now consider the wave-mechanical expression for the energy
E = mean square [v1 cos (omega t) + v2 cos (omega t + theta)]
where our eyes take the average over a time long compared to the cycle time
of the waves.

Expanding the square we get
E = .5 v1^2 + 2 v1 v2 average[cos (omega t) cos (omega t + theta)] + .5 v2^2
Then by using a trig identity this simplifies to
E = E1 + E2 + v1 v2 cos(theta)

Now we have to use a key fact from the statement of the problem, namely
that we are using classical flashlight beams (not laser beams).
Flashlights contain zillions of independent radiators, each one radiating
with its own phase. When we consider the contributions of all possible
phases theta, the cross term in the previous equations washes out, and we
have shown that the classical square-before-you-add law
E = E1 + E2
is correct because it is a consequence of the more general wave-mechanical
add-before-you-square law.

We also get a clear limitation on the validity of this law. If the
radiators are coherent, the classical square-before-you-add law is *not*
equivalent to the wave-mechanical add-before-you-square law, and only the
latter can be trusted.

High school students should be able to follow this calculation.

==========================

Somehow this question about interference of flashlight beams was at some
point mixed up with a question about conservation of energy.

The foregoing calculation is *not* about conservation of energy. It is
about interference and the lack thereof. It is about the correspondence
principle, i.e. the incoherent limit. But conservation of energy is much
more than that. Total energy is conserved even if we do not pass to the
incoherent limit.

Depending on how you phrase the question, energy of light-waves might not
even be conserved. Total energy is conserved, but in some cases it might
be convenient to divide the world into light-wave-stuff and
non-light-wave-stuff, and in such cases the energy of the part labelled
light-wave-stuff might not be separately conserved. As a particularly
simple example: consider light bouncing off a moving mirror in a
light-mill (in analogy to a wind-mill). The number of photons is
conserved, but the light changes frequency and gives energy to the
light-mill (or gains energy from it, depending on which way the mirror is
moving).

So don't look for a proof that light-wave energy is conserved. I'm not in
the business of proving things that aren't true.

OTOH if we take the full quantum-mechanical view that everything obeys a
wave equation, then the QM wave equation had darn well better conserve
total energy automatically. It does. The operator that says how
*anything* changes with time depends on the Hamiltonian. The operator that
says how much energy there is depends on the Hamiltonian. Showing that the
total energy doesn't change with time is so easy it's practically
tautological. You have to show the Hamiltonian commutes with the Hamiltonian.

But I don't expect typical high school students to follow the QM calculation.