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I have a problem with energy conservation in case of interfering waves.

Imagine two harmonic waves with amplitudes $A$. They both carry energy that is proportional to $A^2$, so the total energy is proportional to $2A^2$. When they interfere, the amplitude raises to $2A$, so energy is now proportional to $4A^2$ and bigger than before.

The equivalent question is what happens to the energy with the superposition of two waves that interfere destructively.

Also, if someone could comment on the statement about this problem in my physics book (Bykow, Butikow, Kondratiew): the sources of the waves work with increased power during the interference because they feel the wave from the other source.

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4 Answers

It is guaranteed that finite wave packets always create places where the interference is constructive as well as places where it is destructive: the energy simply flows from the maxima to the minima.

In your convention, it's guaranteed that the total energy at the end is always "in between" the energy from the constructive interference and the energy from the destructive one, which is simply the average $$ (4 A^2 + 0 ) / 2 = 2A^2,$$ exactly as the original energy. Otherwise, the energy conservation can be proved even locally - as a continuity equation - directly from Maxwell's equations so it always holds. This is particularly easy to prove for vacuum Maxwell's equation - enough for propagation and interference of light,

http://en.wikipedia.org/wiki/Electromagnetic_stress-energy_tensor

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What do you mean by saying that it is desctructive in some places? Lets for instance take two waves propagating in circles, like the waves on the water, that have the same centre. It there is no phase difference, then the displacement at any point is twice the displacement due to the single wave. –  malina Mar 28 '11 at 21:27
    
And how does this explain the case of two waves in a counterphase that interfere only destructively? –  malina Mar 28 '11 at 23:21
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@malina: And how do you generate two waves at exactly the same point? How is this different from generating a wave with twice the same amplitude or no amplitude at all? See there is no violation of conservation of energy. –  Raskolnikov Mar 30 '11 at 10:33
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The energy does not double. This is prevent in part by the Born rule. Consider the state vector $|\psi\rangle~=~\sum_n c_n|n\rangle$, where it is easy to see $c_n~=~\langle\psi|n\rangle$. The energy eigenvalues are computed as $H|n\rangle~=~E_n|n\rangle$. For the single wave function the amplitudes are found by the modulus square $\langle\psi|\psi\rangle~=~1$ of the wave, where $$ \langle\psi|\psi\rangle~=~\sum_nc_n^*c_n~=~\sum_nP_n~=~1 $$ Now consider the expectation of the Hamiltonian $$ \langle\psi|H|\psi\rangle~=~\sum_nP_nE_n. $$

To address this question directly we then consider the sum of two waves in a superposition with the state vectors $|\psi_1\rangle$ and $|\psi_2\rangle$, $$ |\psi_1\rangle~=~\sum_n c_{n,1}|n\rangle,~|\psi_2\rangle~=~\sum_n c_{n,2}|n\rangle, $$ where the two waves are expanded in their own coefficients which are normalized for the sum of the two waves. Consider the modulus square of the sum $|\psi_t\rangle~=~|\psi_1\rangle~+~|\psi_2\rangle$ of the two waves $$ \langle\psi_t|\psi_t\rangle~=~\sum_n(c_{n,1}^*c_{n,1}~+~ c_{n,2}^*c_{n,2}~+~ c_{n,1}^*c_{n,2}~+~ c_{n,2}^*c_{n,1}) $$ At this point you can now see the answer to the question, for the amplitude coefficients $c_{n,1}$ and $c_{n,2}$ are normalized to counter the growth by four of the modulus square of the amplitudes. This is particularly easy to see if $c_{n,1}~=~c_{n,2}$, where now the amplitudes for the two waves are normalized to $1/\sqrt{2}$ their independent values. One might have $c_{n,1}~=~exp(ikx_1)c_n$ and $c_{n,2}~=~exp(ikx_2)c_n$, where the interference term $$ c_{n,1}^*c_{n,2}~+~ c_{n,2}^*c_{n,1}~=~c_n(e^{ik(x_1~-~x_2)}~+~ e^{ik(x_2~-~x_1)}) $$ which gives a oscillating term with respect to the difference $ x_1~-~x_2$. Then further the expectation of the Hamiltonian $\langle\psi_t|H|\psi_t\rangle$ is similarly expanded as $$ \langle\psi_t|\psi_t\rangle~=~\sum_n(c_{n,1}^*c_{n,1}~+~ c_{n,2}^*c_{n,2}~+~ c_{n,1}^*c_{n,2}~+~ c_{n,2}^*c_{n,1})E_n $$ where there is clearly no increase in the energy.

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I think that you are dealing with classical electrodynamics and I will answer within this domain.

In Kostya answer in PSE about interference look at the equations of parallel polarization that I copy here:

Total field: $\vec{E} = > \vec{i}E_0\left(\cos\omega > t+\cos(\omega t+\Delta)\right)$.
Intensity: $I_\parallel\sim > E_0^2\langle\cos^2\omega t+2\cos\omega > t\cos(\omega t+\Delta)+\cos^2(\omega > t+\Delta)\rangle=E_0^2(1+\cos\Delta)$, which nicely depends on the phase shift between the waves.

When $\Delta= \pi$ we got $\vec{E} = \vec{0}$ and $I = 0$. If the antennas are headed on the Poyinting vector also cancels.
Two waves and zero net field. And this means trouble as I will point out later and justifies the naive words of Bykov. This image from sbu.edu constructive and destructive interference of light waves we see in the image at left a constructive pattern and at right a destructive pattern. This game to 'shape the field' in space is played with antenna arrays always. What means the obvious diference in the intensities? Physically one radiator is moved apart from the other half wavelength, and the feed (current intensity, frequency,phase) of both radiators was kept invariant. It appears that when the field cancels one (strange property) must say to both radiators: Stop radiating! Or as Bykov 'they feel the wave from the other source'. But this is complete nonsence.
Think of this: Does the electrons in the radiators have sensor to all space to say : radiate in this direction and distance; stop radiating because the other radiator is changing position; then add N radiators,... We can use light originated in the stars and do they play this game?
There is no theory to cover Bykov words.

The problem is, as you say, 'the conservation of energy'. When I studied EM field (is light), radiation, antennas, I never used the concept of 'photons' as particles, but only the 'EM field'. We can cancel the field, as equations show, but can not cancel the 'particles' and lost the energy. A recent experiment on 'anti-laser' motivated me to ask a question that is similar to your's. The experiment is a head-on photon-photon cancelation consistent with the above EM equations.

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Later I will have to substitute the image because it has a copyright notice that I saw just now.There are several available in the net. Sorry for the inconvenience. –  Helder Velez Mar 29 '11 at 19:31
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When sound waves interfere, a they "cancel" each other out. Does this means sound energy + sound energy = zero energy? We know that we just can't "cancel" energy. What actually happens is that a "sound wave" is actually the pressure or potential energy part of an acoustic wave. There is also a velocity component, or kinetic energy part that people keep forgetting about. When two waves interfere the potential energy density part of the two waves does go to zero but the kinetic energy density parts of the two waves double. Energy is conserved. I prefer to think of the collision as a way to change or transform the form of the energy density but not the fact of the energy density. The problem of "where did the energy go" is always that only one half of the total energy of the wave is being tracked. If you track both forms, you will see that the energy density stays constant, only its form changes. Art Noxon, Acoustical Engineer

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