# Violation of conservation of energy?

Imagine a perfectly spherical mirror with 100% reflection. Imagine a point source of light in the center. The point source keeps radiating light. Will the light undergo destructive interference completely? If so, how does this not violate conservation of mass-energy. Normally in interference, constructive interference makes up for the destructive, protecting conservation of mass-energy. I doubt that this is a violation, but why doesnt the sphere lead to perfect interference?

• Related: physics.stackexchange.com/q/21301/2451 and links therein. – Qmechanic May 23 '15 at 9:28
• The mirror needs to be very massive to reflect light backwards with almost the same wavelength. So firstly, that reduces conservation of energy, the energy of the massive mirror increases when its velocity changes by a very small amount. And since the wavelength reflected is tiny a bit longer, there never was perfect destructive interference anyway. If you ignored the mass of the mirror you'd have violation of conservation of momentum. – Timaeus May 23 '15 at 17:29

If you setup a perfect cavity where no modes of light are possible, then the light will not be emitted in the first place (0 probability).

You run into problems if you consider the emittor as a classical light-source and then combine it with a quantum-mechanical reasoning regarding interference in cases such as this.

• I believe this is the right answer, although I don't agree with the second paragraph fully. You calculate the emission amplitudes and the Purcell-effect-wavelength shifting through an analysis of the interference of a lone photon with itself, which is exactly the same as the Maxwellian description, as I describe here physics.stackexchange.com/a/165020/26076. Classical optics and one-photon QED are almost the same thing – Selene Routley May 24 '15 at 5:10
• @WetSavannaAnimalakaRodVance Thanks, the lesson is you run into problems when you try to handwave in answers ;) I guess the core of what I meant was that you have to consider the source emission in the analysis and not just that "there is a photon to start with, now what happens". And as you note in your interesting writeup of maxwell<->one-photon QED states, there are many analysis methods possible to reach the conclusion. – BjornW May 25 '15 at 7:44

The point source keeps radiating light. Will the light undergo destructive interference completely?

Point particle as a source of light is OK, but it would need to move with acceleration to produce EM radiation. Static source of light of zero size seems to be reduction ad absurdum, at least from the standpoint of common theory of light based on EM theory.

For radiation to be present, the charged particles constituting the source have to move with acceleration. Thus if the radiation from the particles is present in the cavity, they cannot not be at rest in the center of the cavity.

If a charged particle stays at the center at all times, the EM field in the cavity is static electric field, thus no light is there.

If the particle oscillated a little around the geometrical center of the cavity, radiation would be present and would add geometrically with the radiation generated by the reflective surface. But total cancellation of the fields leading to zero total field inside seems impossible.

• This answer is interesting if you do try to model the source as a classical emitter! I think maybe it ends up in conflict with the premise or spirit of the question as a philosophical question of perfect destructive interference though. – BjornW May 23 '15 at 12:48

You are describing a resonant cavity with a discrete set of allowed modes and your thought experiment is a good illustration of the Purcell Effect (see Wiki page here), which is the influence on quantum mechanical photon emission probability amplitudes by an emitter's surroundings. The resonator simply shifts the emission frequency to one where the interference is imperfectly destructive: the destructive interference then lengthens the emission lifetime greatly. If you put an exited atom into a cavity where the fundamental mode were of much higher frequency than the usual emission frequency for that atom, the emission could only happen near that much higher frequency and therefore the emission lifetime becomes greatly lengthened. This is another way of describing BjornW's answer. However, I don't fully agree with Bjorn's comments about photon / classical optics because one photon quantum optics is described precisely by Maxwell's equations, as described in my answer to the question "Relation between Wave equation of light and photon wave function?" here. You calculate the Purcell-altered probability amplitudes through a Maxwellian analysis. I describe exactly this problem in:

Rod W.C. Vance and François Ladouceur, "One-photon electrodynamics in optical fiber–fluorophore systems: III. One-polariton propagation in fluorophore-driven fibers", JOSA-B, 24, #6, p1369

To fully define the problem, you need to describe your source boundary conditions fully. One instructive way to think of this is at microwave frequencies, where your source is an antenna, say a dipole, and it will be fed by a waveguide. The returning waves, out of phase with those emitted, simply pass through the incoming waves linearly, are received by the antenna (which, reciprocally, is also a receiver) and become reflected waves in the feeding waveguide.

Another situation of interest is to imagine an energized conducting ring, radiating into the spherical cavity or an excited atom in the center of the sphere. Without the spherical mirror, the excited atom is coupled to all modes of freespace equally, as in the simple quantum model I talk about in my answer to the Physics SE question "Does an electron move from one excitation state to another, or jump?" here. However, even so, as it radiates its phase oscillates with center frequency $\omega_0$ so as the probability amplitude to find a photon in a mode with frequency $\omega_1$ far from $\omega_0$ grows with time, its phase oscillating at that frequency gets out of step with the radiator and destructive interference keeps the probability of finding the photon at $\omega_1$, whereas the photon has maximum amplitude to be found in the mode at frequency $\omega_0$ in resonance with the emission. However, we see that there is still a nonzero amplitude to find the photon in other modes: this qualitatively is the reason why emission spectra have a basic Lorentzian lineshape even in the absence of Doppler broadening. The linewidth is proportional to the square magnitude of the of the coupling amplitude per unit frequency and the decay lifetime inversely proportional thereto.

Now we imagine putting the sphere in place. Let us assume at first that there is a small transmission co-efficient across the mirror, so that the emitted photon has a small amplitude to cross the mirror, whereas most of its amplitude is reflected. If the sphere is small compared with the linewidth above, most of that amplitude is reflected back to the emitter before there is a significant amplitude to find the photon has been emitted. If it interferes destructively at a particular frequency, it simply lowers the probability amplitude that the photon will be emitted in its normal frequency. Instead, the photon's greatest emission amplitude is shifted by the cavity to another frequency where the destructive interference is imperfect. Moreover, the emission lifetime will be greatly lengthened by the resonator.