Hello to all: Suppose we have two photons that are 180 degrees out of phase. We can assign one unit of energy to each photon. When we superpose the two photons, the combined electric and magnetic field goes to zero as does the Poynting vector and the energy density. So where does the energy go? I am investigating this because some people claim that the energy goes into what are called "scalar waves." Thank you for your help. I am an experimentalist who has been out of graduate school now for about 28 years. So, if I ever did know the answer to this question, I have forgotten it. RoKo
Can you really isolate photons like that, saying "oh this photon has this field"? I didn't think that was allowed. And shouldn't this question be on the quantum forum?
This problem arises not only with two photons but also in the classical electrodynamics where two classical EM waves can be "180 degrees out of phase". They cannot be "180 degrees out of phase" in the whole space because each wave is created by its own source. So there are space intervals where the two waves cancel each other (destructive interference), for example, out of both sources, and there are space intervals where these waves add (constructive interference), for example, between the sources. As the energy of any wave (including that of photon) is the volume integral over the whole space, the energy does not disappear but redistribute in space. It is a typical interference picture: there are minima and maxima of the field intensity and of the energy density in space, and they can even move. Bob.
I know this is nitpicking a little, but technically they can cancel in the whole space if they originate in the same place exactly. Also, while the energy in the wave certainly is redistributed when cancellation happens, it does not necessarily need to stay the same in the integral over all space. What can happen is that the waves canceling in some ways can change the effective impedance seen by each source, so that less total energy can be present in the wave, but then each source is actually doing less work, and therefore putting out less total energy, so it still all works out. I unsuccessfully tried to explain this to a couple people on an acoustics thread a while back. This is a common source of confusion about waves in general, and I'd like to figure out a clearer way to explain it to people.
Dear Bob: Thank you for your reply. I used to be a professor of physics, so I suppose I should know all of this spot on. But that was a long, long time ago. Let' see. I believe you are somewhat right that this is a problem in classical electrodynamics. But only if there are no gauge degrees of freedom for the photons. I will explain what I mean later. Okay, I will be honest: I don't see what you mean when you say that two photons cannot be 180 degrees out of phase "because each wave is created by its own source. "Are you saying, for instance, that atoms in a system cannot be made to coherently cancel, sort of like a MASER or LASER system in reverse? Could I not electromagnetically pump a metal box or a transmission line in such a way as to induce standing photon waves that are 180 degrees out of phase? Why would that be impossible to do? And if you say it is impossible, are you really sure that you are right? Can a particle be at once indivisible and divisible, a quantum and a wave? There are many strange things in this universe. So maybe we should be careful about what we say is impossible. Is it possible that a mother can be younger than her children? Special relativity says it is possible. I also believe you are invoking the idea that electromagnetic energy is conserved. If so, then the E and B fields of photons in a system must always provide us with a basis to account for total energy. But isn't it true that electromagnetic energy is not conserved when a system can make gauge transformations and thus generate gauge waves? Do not gauge waves carry energy? I recall that it used to be believed that these kinds of gauge transformations had no relevance to the real world. But is that still an idea that is believed today? My impression is that those old ideas were wives tales. I think they were. Jackson, in his famous text on Classical Electrodynamics appears to have been one person who told us these things. Magnetic Field Cancellation With Two Parallel Wires Wound Into a Coil Consider two parallel wires which are soldered together at one end. Let's imagine then that a current is made to flow through the wire. The current passes through one wire and then comes back in the opposite direction in the second wire. We would expect with this system that there will be quite a bit of magnetic field cancellation in the regions around the wires. Winding this system into a large coil might then lead to a very large region of very high magnetic field cancellation. What can we say about the cancellation region? It does not have to be the case that nothing happens because we know from the Aharonov–Bohm effect that magnetic field cancellation can indeed affect the behavior of particles that move in the cancellation zones. What then goes on with canceling fields? Might there not be something of interest in this? Well, thanks again for your replies. I will look for an answer. Take care. But please forgive me if I am late in answering any replies that might be made. I had a fire in my home yesterday and there was quite a bit of damage. The fire was in the kitchen and was just about to really take off when I was able to put it out. One or two more minutes more and the house would probably have been lost. The curtains in the kitchen were on fire, and the fire was starting to get in between the walls. Close, so close. Take care. RoKo
Okay, but your original question was where the energy went. In this case, my previous reply gives you a complete answer, I believe. The fields cancel, so no energy gets transferred out of the coil, or, to put it in a more superpositioney way, the piece of the coil in one direction transfers its energy to the piece of the coil in the other direction, and vice versa. The net effect is that applying an alternating voltage to the coil requires no power. The impedance of the coil is zero. As far as the Aharonov-Bohm effect, I don't know much about that stuff, but asking about the quantum mechanical effect of the wave on external particles isn't exactly the same as asking where the energy goes in the case of cancellation, is it? You are setting up a different case. Having that other particle there (even though it isn't classically supposed to see the field) changes the system being analyzed. Now the field doesn't (perfectly) cancel anymore, because some little tail of a particle's wave function is passing right through the wires and possibly grabbing energy from the infinitesimal area between the windings in the two directions, where there actually is some field. Right? Sounds extremely interesting. However, I still say the energy doesn't "go anywhere" when fields actually cancel. :)
Exactly that: if there are two excited atoms, they most likely will radiate a couple of coherent photons that add, not cancel each other (both atoms will recoil in the opposite to photons direction). Total canceling means E*sin(omega*t-k*x) - E*sin(omega*t-k*x) = 0 everywhere and every time. A non radiating source (E=0) creates exactly this kind of zero and no work is done to radiate. It is easy to understand: a charge to radiate such counter-phased waves should move in opposite directions simultaneously, in other words, to stay still. Two waves of the same amplitude but propagating in the opposite directions can create a standing wave. Its energy density is not equal to zero because there is no cancellation: E*sin(omega*t-k*x) + E*sin(omega*t+k*x) =/= 0. Yes, it is conserved. The temporal and the scalar gauge degrees of freedom do not interact with charges. They cannot receive/give up energy-momentum. The quantum theory of electromagnetism (QED) can be formulated in terms of the field E and B tensions, without “gauge” degrees of freedom. It is done just by the variable change. And the conservation laws are based on the experimental data, of course. Apart from magnetic field cancellation, there is a simpler example: two equal but opposite in sign charges put close to each other. They create a dipole electric field proportional to the distance between them. If you put them together at one point, the total electric field is equal to zero. This is cancellation in the whole space and in time that happens in reality when a proton captures an electron and emits the excess of energy as a neutral neutrino (K-capture). This cancellation happens also while radiative recombination of ionized atoms: an electron + ion= neutral atom + photon(s). Anyway, the energy is conserved. In your case of wires with the magnetic field there may be cancellation at long distanced but with extreme field tension in some local places: the wires attract each other that creates their mechanical displacement and mechanical wire tension at wire touching points (potential energy increase), for example. Bob.
To the OP: I think you're mistaking a mathematical concept (vectors) for physical reality. Two EM waves that exactly cancel each other means that there is nothing there, not that two EM waves are somehow hiding underneath the covers. Think of a plane wave coming from the left that's incident on a thin conducting sheet. Mathematically, EM theory says that the sheet emits two plane waves, one to the left and one to the right. The right emitted em wave exactly cancels the incident wave. The one emitted to the left is the usual "reflected" wave. Of course, in reality, nothing gets past the conducting sheet -- the two self-cancelling waves on the right are a mathematical accounting mirage. This is actually a very slippery subject in EM theory, IMHO. What exactly happens physically when two em waves are coincident but don't cancel each other out? Does that have any implications for "static" E and B fields? For example, take a long, lossless transmission line and connect sources to each end. Have each source send a 1V 1ns pulse at the same time. At the middle of the transmission line, the two pulses will perfectly overlap. The voltage on that one foot section of line will jump to 2V and the current will drop to zero. Then the 1V pulses will continue on their merry way as if nothing happened. The thing is, during that overlap, the electric field in the middle of the one foot overlap will look exactly like a static E field yet we know that it's composed of two oppositely moving em waves.
Thanks for your excellent posting. I agree that energy would be conserved, but what about electromagnetic energy? Would not gauge waves carry energy? I believe it to be the case that classical electrodynamics allows for non-conservation of the electromagnetic energy because of gauge transformations. Thanks again. RoKo
The electromagnetic energy is also conserved. Gauge waves (or degrees of freedom), if present in the theoretical formalism, are decoupled from the charges. They live with their own life. There are many theoretical formalisms and amongst them there is a Coulomb gauge where there is no longitudinal and temporal degrees of freedom (only transversal electromagnetic field). All formalisms are obtained from each other with the variable change, so if there is no sink/source of energy due to gauge degrees of freedom in the Coulomb gauge, it is absent in the other formalisms. The gauge degrees of freedom are introduced to cast the equations in formally more symmetric form but not for real (physical) interaction with them. Bob. Bob.
Hello. Thanks for your posting. Gauge waves "live with their own life?" I don't know what that means. Do you imagine that there are real waves in the real world that don't carry energy and don't have any physical implications?" But the gauge waves are linked to the original charges, for the original charges determine the nature of the gauge waves. And this takes us back to my original question: "If we superpose two identical photons that are 180 degrees out of phase, then what happens to the energy of the photons? Where does it go?" Is not the energy density zero? Is not the Poynting vector zero? I have asked two very good, senior professors of physics about this and they said they don't know the answer. One of those professors is an expert in electrodynamics, whereas my Ph.D. is in experimental nuclear physics. Also, while what you say obeys conventional ideas that I was first taught 30 years ago, I have come to the conclusion that those ideas are wrong. I believe that gauge waves do have physical meaning. In a gauge transformation, the E and B fields do not change, but that does not mean that there is nothing physical that takes place. And by definition, a gauge transformation involves a shift in the electromagnetic potential energy. I believe you are a very bright fellow, and you write well, but I ask that you think about this for awhile and examine the conventional wisdom. You might find that there is at least one potentially wrong idea in the conventional analysis. One more thing: I am cheating a little because I essentially know from experimental results that gauge waves do involve something physical. I also know that they carry energy -- or can. Part of this involves data that is a little sensitive; it relates to weapons of rather large energy output. Don't forget: We physicists have often made mistakes in our theories. Please think about the matter for a day or two, will you? It could be very important. Actually, it is very important. RoKo
The shift in potential energy U in Classical Mechanics is the brightest and the simplest example what the gauge degrees of freedom represent in the theory: it is a liberty to add a constant to U to detract it from the total energy E. Moreover, you may just add a constant to the potential energy without detracting it from E - the trajectories of particles will stay the same (the force is a derivative of U). In Electrodynamics the liberty is even larger - you can add some functions to the four-vector potential A without changing the forces. Still you obtain the same trajectories, etc. These functions can look as waves but they are anyway kind of "shifts" on the paper, not in the reality, not in course of the real interactions. So they do not exchange the energy with the real fields and particles. Bob.