Destructive interference - where does the energy go?

Well there doesn't seem to be much of a problem in that case, since a standing wave has energy (constantly oscillating between the kinetic and potential form) associated with it.

Here's my guess (in an ideal case; things might not work so nicely for a physical rope with a shaker on it). I would expect the second source could either absorb or reflect the wave on the rope, or anything in between, depending on where it is located. Specifically, if the second source is located an integer multiple of the wavelength away from the first, I would expect it just produces waves propagating in both directions. The one travelling to the right combines desctructively with the original wave, while the one travelling to the left combines with it to form a standing wave. So in that case the second source would effectively reflect the wave produced by the first source, and you'd end up with a standing wave trapped between the two sources. If instead the second source is a half integer multiple of the wavelength away from the first then in that case the only possibility seems to be that the second source has to fully absorb the wave produced by the first. This is just a guess, but either way there doesn't seem to be a problem since there are at least two things a source could do with the energy of a wave (absorb or reflect it) that would satisfy energy conservation.

Another related example to think about is to take a laser beam and try to get it to interfere with itself, say using beam splitters to separate out and recombine the beam but have the two branches travel different distances so that they acquire different phases. The resolution in this case is that beam splitters always transmit and reflect with different phases (eg. the reflected beam might always be phase shifted -90 degrees relative to the transmitted beam), in such a way that if you have destructive interference on one of a beam splitter's outputs, you'll always have constructive interference on the other (this is probably the sort of thing Tach was hinting at).

As others have stated, the wave is flat but it's still moving, so all the energy is kinetic in that case.

For a more mathematical answer, if a wave is flat at t = 0, then the only possibility is for the wavefunction to be of the form

\(\psi(x,\,t) \,=\, f(x + ct) \,-\, f(x - ct)\)​

for some function \(f\). The derivative of \(\psi\) at t = 0 is \(2 c f'(x)\), which fully defines \(f(x)\) up to an (irrelevant) overall constant. So even when the wave is flat, all the information needed to reconstruct the full wave later on is encoded in the transverse velocity profile.

 

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If the pressure of the two waves travelling in the opposite direction cancel then the velocities of the two waves store the energy. If the velocities cancel then the pressure stores the energy.

 

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I only pray that some day physics will become so intuitive to me that I can make a point this elegant.

 

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I actually did some reading about this, and a book tells me that in a bounded medium like a string or a rigid rod, the total mean energy density across a section A of the medium is:

\( \frac {dE} {dt}\;= \; IA\; = \; vEA \), where the E in vEA is the energy per unit volume, and v is the propagation velocity along the string or rod.

The total energy of an oscillator is \( \frac {1} {2} m \omega^2 A^2 \); the pressure wave travels unimpeded along the string or rod (ignoring friction, heating etc), and the displacement wave is the algebraic sum of 'rotations' of the medium (vertically) perpendicular to the momentum-pressure (moving horizontally).

Energy and momentum are of course conserved, and the medium is 'rotated' by the transverse kinetic component before and after the waves pass through each other and through the same areas \( A_i \). The medium again 'reacts' to the propagating momentum-pressure by moving tranversely. And indeed, on a rotating vector diagram, the amplitude vector is perpendicular to the angular frequency vector.
Must be some kind of a rule or something. . .

 

I understand the answer when we're speaking of waves involving oscillation of physical materials which can possess kinetic and potential energy, but what about light waves? There is such thing as an "anti-laser" called a coherent perfect absorber which does exactly what you're asking about, James R, but with lasers. In a CPA the lasers are apparently dissipated as heat, eventually, but in my mind's eye I'm left wondering "where" that energy resides in the physical space of the initial laser junction? :bugeye:

 

You can't really cancel out light waves the way James R has been describing. If you send two light pulses toward each other in opposite directions in such a way that their electric fields cancel out at some moment, then at that same moment the magnetic fields will add up, and vice-versa.

 

That's interesting, so by necessity you cannot have the electric and magnetic fields both properly phased so as to be cancelled out? Why is that?

 

My first thought was that this is related to the "right-hand rule" but that made me question whether or not a left-handed EM wave is possible, maybe with negative index metamaterials? I'm basically out of my element here...

 

When the string is flat all the energy is kinetic, thus it isn't stored in the tension of the string. In the case of things like rubber bands the energy goes into the electromagnetic bonds when the two moving waves are in phase, not out of phase as occurs when the string is flat, as that's the point where there is no kinetic energy.

It's basic Lagrangian/Hamiltonian mechanics, the sum H = T+V is constant, which is the Hamiltonian. In the case of the system in question there is a point where T=0 and another where V=0 (up to arbitrary additive constant term in the Hamiltonian irrelevant to the dynamics). Given it is possible to compute analytic expressions for the string's configuration and motion and thus kinetic and potential energies you can plot T(t) and V(t) and see how they vary (though you only really not to plot one since you can work out the other from that).

If memory serves it follows from the form of Maxwell's equations in a vacuum. The variation in space of one of the fields, E or B, relates to the variation in time of the other. As such you end up with one being non-zero while the other is zero, just as the kinetic energy of the string will be zero when the potential energy is maximal and vice versa.

In fact you can see that from the explicit expressions for the Lagrangian or Hamiltonian of an EM field, where you get \(H \approx E^{2} + B^{2}\), where E is the electric field norm, not the energy. This expression is constant in time so if \(H \neq 0\) at some point in time then it means E and B can't both be zero at the same time.

 

In propagating radiation, electric and magnetic 'components' are always out of phase.
It's something like how a torus is a composition of two circles, neither of which is 'inside' the other, type of thing. I'm wingin' it here, btw.

Electronic circuits change the phases between electric and magnetic parts of the overall field by passing currents through electronic components; a current in a conductor always generates a local field, but voltages (potentials) will lead or lag currents, depending on local inductance and resistance, capacitance, and so on.

 

Alpha and Arfa, I don't think I was very clear. I was wondering out loud whether (...and if so, why it might be the case that...) one could never have two perfectly out of sync EM waves collide such that all of their fields are cancelled. This is why I mentioned the anti-laser and the right-hand rule.

 

It's not complicated really. When an electromagnetic wave propagates the electric and magnetic fields are oriented 90 degrees apart, like this:

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​

If you can imagine trying to rotate this wave 180 degrees around so it's propagating in the opposite direction, you'll see you can cancel out the electric part or the magnetic part, but never both at the same time. (In the image, whenever the electric field is pointing to the right, the magnetic field is pointing up. For a wave propagating in the opposite direction, if the electric field is pointing to the right, the magnetic field points down.)

Which way the magnetic field points relative to the electric field is a matter of convention defined by the right hand rule in the Lorentz force law, but you get the same result either way.

 

Yeah, I see that now. I think in my reply I was still somewhat following on from what Farsight said. I was mistaken in my comment anyway, as I mistakenly had this mental image of the E and B fields being out of sync by 90 degrees, which isn't the case. It's important to consider moving with a wave front, not just watching a single bit of the field go up and down. Basically ignore what I said, if you already hadn't

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I thought the electric and magnetic waves were a quarter cycle out of phase; that the magnetic field varies with the rate of change of the electric field, and vice versa, and that both rotated (unless polarised)?

 

No, for a linear wave the magnetic field is always in phase with the electric field. It sounds like you're describing circular polarised light. You get that by superposing a linear wave like the one illustrated above with another one a quarter cycle out of phase and oriented at right angles, like this:

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​

In this illustration the two linear electric field waves (green and blue) superpose to give a spiralling wave front (red). Exactly the same happens with the associated magnetic fields. The "90 degrees out of phase" is between two linear electric field waves, not between the electric and magnetic waves.

 

Thanks przyk. Maybe one day I'll learn electromagnetism properly.

 

Przyk, that answered a lot of questions I've conjured up on EM waves, thanks. Unfortunately my continuing education is on hiatus so I won't be taking any optics courses this summer as I had hoped.

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James R,

I have to agree with you! I feel a little uncomfortable with the explanation given by PRZYK also.