# Thread: Destructive interference - where does the energy go?

1. Originally Posted by James R
So, to spell it out, what is your answer for the counterpropagating case you describe?
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.

And then, what about the co-propagating case?
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).

Originally Posted by James R
I'm not interested in what happens to the energy eventually due to dissipative processes. I'm interested in what happens to it while the string is flat and there's no visible wave on it. And, as I said before, how does the string "know" how to reconstruct the two pulses from the flat string, and where does the energy come from to do that?
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.

2. Originally Posted by James R
Consider two counter-propagating waves on a string with equal frequencies but 180 degrees out of phase so that there is complete destructive interference.

Each wave carries energy - one to the left and the other to the right, let's say. But when they interfere we get a stationary string and, apparently, no energy.

Where did the energy go?

I'd like to nail down the answer to this question. If you look on the internet, there are all kinds of different explanations. Let's see if we can do better at sciforums.
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.

3. Originally Posted by przyk
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
$\psi(t) \,=\, f(x + ct) \,-\, f(x - ct)$
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.
I only pray that some day physics will become so intuitive to me that I can make a point this elegant.

4. 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. . .

5. 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?

6. Ok. Here's my understanding from what people have said so far.

First, for the two pulses propagating in opposite directions along the string: when the string is flat and the pulses are overlapped, the energy of the wave is still there as kinetic energy of the moving string, since even when the string is flat it is still moving. It's about to be non-flat in the next instant.

In general for waves on a string, the energy is shared between potential energy of the stretched string (akin to a spring force) and kinetic energy of the parts of the string as they move.

In the extended-wave counter-propagating case, a standing wave is formed on the string. There is no energy at the nodes, which remain stationary at all times. But the "missing" energy is distributed to the antinodes, meaning that overall energy is conserved.

For the extended-wave co-propagating case, we need to investigate the sources. If the second shaker is to the right of the first, it will generate waves that propagate both left and right. If, to the right of the second shaker, the string is stationary, then either the energy supplied by the second shaker is distributed to the left, or else the second shaker itself is actually absorbing energy from the first shaker. I'm not 100% sure which of these is correct.

7. Originally Posted by RJBeery
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?
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.

8. Originally Posted by przyk
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?

9. 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...

10. Originally Posted by Farsight
Then I'm back to electromagnetism, and telling James that the energy is in his flat string is still there in the electromagnetic field between the atoms, like in the compressed then stretched spring in the harmonic oscillator.
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).

Originally Posted by RJBeery
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?
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.

11. Originally Posted by RJBeery
you cannot have the electric and magnetic fields both properly phased so as to be cancelled out? Why is that?
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.

12. 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.

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

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.

14. Originally Posted by RJBeery
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.
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

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

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)?

16. 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:

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.

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

18. 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.

19. Originally Posted by przyk
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 doesn't sound right to me. The electric and magnetic fields in a linearly-polarised n EM wave are always in phase with one another.

Also, if you could cancel out the electric portion and while doubling the magnetic portion, for example, then you'd have at that instant of time just a magnetic field in space, which could not regenerate the electric field component necessary for the propagating wave. Again, the question becomes in this case - where is the information stored to enable the reconstruction of the electric part of the wave? (Oh wait, I think I might know the answer to that one, but my other questions still remain.)

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

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.
I see the argument, but I'm not sure if I'm convinced by it yet. Maybe I need to think some more about this.

20. James R,

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