Photon Decay: Can High Energy Photons Transform?

[PAllen]

If I'm understanding properly, I think it's fundamentally pretty simple. E is the timelike component of the energy-momentum four-vector, and τ is the timelike component of the position four-vector. There is clearly no hope of relating them in a Lorentz-invariant way unless they're proportional to one another.

I had no trouble with this argument once I saw it. Looking at why this is so, what is fundamental to me is that decay is a measurement of proper time between two events. I was trying to pretend it could be something like kinetic energy, which can be measured at one event.

 

[PAllen]

I don't think they really claim that photon decay is impossible. I think they claim that it's impossible in certain field theories, not in all field theories. They have some remarks on p. 30 about field theories in which it might be possible, and those field theories seem to include field theories with a negative cosmological constant. (They were writing in 1996, so they didn't know that the cosmological constant was nonvanishing and positive.) In any case, all the material about gravity is clearly extremely speculative, since we don't have a theory of quantum gravity.

But is there some reasonably simple reason it doesn't happen according to QED (which is a theory in flat Minkowski space)?

 

[Haelfix]

Photon decay is impossible in QED and they prove it several different ways in the paper, at least perturbatively. The more difficult case apparently is graviton decay and I believe that is left open for nontrivial backgrounds.

The rigorous way they prove it is with the Ward identities section, which is difficult and I'd have to go through it more carefully. The non rigorous way is by power counting, which is pretty easy. So it suffices to show that the fermion square diagram vanishes (as the same thing will happen for all odd N > 3) which you can do by noting that the decay amplitude will always involve a positive power of the regulator and thus in the IR limit will vanish.

It is also true if you consider graviton loops instead of fermion loops. Now you could perhaps invent new physics in some way such that you generate a negative power of the regulator, but this is way beyond the scope.

 

[fzero]

The fermion square diagram (4pt function) does not vanish in general and is responsible for nonzero \gamma\gamma scattering (the 3pt function does vanish by regularization). The reason why the 4pt function vanishes for photon splitting seems to be mainly kinematical.

The collinearity constraint (together with masslessness) means that the 4-momenta p^\mu_i are proportional to some 4-vector p_0^\mu satisfying p_0^2=0. The transversality conditions p_i^\mu \epsilon_{i\mu} =0 (no sum on i) therefore actually imply that p_i^\mu \epsilon_{j\mu} =0 for all pairings i,j. We can use this to show that there are two linearly independent choices for the \epsilon^\mu_i.

One can obviously work the amplitude out explicitly, but it's easy to be convinced that any term must be proportional to a given contraction of the indices in the expression

\prod_i p^{\mu_i}_i \epsilon^{\nu_i}_i.

Since any contraction of the momenta with other momenta or polarization vectors vanishes, each possible term, and hence the whole amplitude, must vanish.

 

[bcrowell]

But is there some reasonably simple reason it doesn't happen according to QED (which is a theory in flat Minkowski space)?

I would find it extremely unsatisfying to imagine that the only answer in the context of QED was something technical about field theory. After all, QED is only one tiny piece of the standard model.

Classical kinematics says that the lifetime has to go like \tau=(\hbar/c^4m^2)E, where m is some constant with units of mass. Since we observe quasars at cosmological distances at wavelengths on the order of 10 m, m \lesssim 10^{-56} kg. For a point source, the squared wavefunction of the photon would be multiplied by an exponential factor e^{-r/c\tau}.

But this would be exactly equivalent to the result of the Proca equation for a massive, spin-1 field, which has an exponential factor of e^{-2Mcr/\hbar} for the squared field, where M is the mass of the particle.

There have been various experiments that have tested Maxwell's equations to high precision. The exponent in Coulomb's law has been constrained to be within 10^-16 of 2. The mass of the photon has been constrained to be less than about 10^{-51} kg. These experiments all use very different techniques, but suppose that you're doing such an experiment, and your technique isn't sensitive to the decay products of a \gamma\rightarrow 3\gamma decay, only to the probability of receiving the parent photon. Then you may consider yourself to be putting an upper limit on the photon mass M, but this is exactly equivalent to putting an upper limit on 2m^2/E (in units with c=1), where m is the constant with units of mass that characterizes the photon's decay rate.

Now a massive photon breaks gauge invariance. QED can accommodate it, but it would be a horrible problem for QFT: http://optica.machorro.net/Lecturas/PhotonMass_rpp5_1_RO2.pdf This seems to me like relatively nontechnical fact about QFT, too. So this suggests to me that if there is a relatively nontechnical way of prohibiting a nonzero value of M, then there might also be a relatively nontechnical way of prohibiting a nonzero m.

One interesting thing about the relationship between M and m is that if M has any nonzero value, no matter how ridiculously small, then \gamma\rightarrow 3\gamma decay is prohibited. That is, although a nonzero value of M and a nonzero value of m would act similarly in certain experiments, a nonzero M actually forces m=0.

If \gamma\rightarrow 3\gamma was possible, then the lifetimes of the products would be, on average, 1/3 the lifetime \tau_o of the parent. You would have an accelerating process of decay, and it would accelerate geometrically. Summing a geometric series, basically you expect that within a few times \tau_o, the decay would run to completion, meaning that you would produce a perfectly focused jet consisting of infinitely many photons, each with infinitesimal energy. Because of their infinitesimal energies, they would have to have infinite wavelengths. Their fields would add incoherently, so the average field would be infinitesimal, and therefore undetectable. The only way they would be detectable would be through their gravitational fields.

For the particle-in-a-box version, with the box being at a low enough temperature, the result would be that within a few times \tau_o, any photon that you initially put in the box would end up as a Bose-Einstein condensate.

 

[bcrowell]

Fzero, it looks like you've done a very nice job of extracting the relevant physics from the Fiore paper. However, the present state of my QED is pretty pathetic (from a graduate course 20 years ago), so I'm not fluent enough to follow the argument about the polarization. How about the decay of a massless scalar particle? Can the argument then be made into a form so simple that even I might have a fighting chance of understanding it?

 

[fzero]

Fzero, it looks like you've done a very nice job of extracting the relevant physics from the Fiore paper. However, the present state of my QED is pretty pathetic (from a graduate course 20 years ago), so I'm not fluent enough to follow the argument about the polarization. How about the decay of a massless scalar particle? Can the argument then be made into a form so simple that even I might have a fighting chance of understanding it?

The massless scalar decay to photons seems to vanish for the same reasons, so perhaps it's worthwhile to just fill some of the gaps in for the previous argument. The point was that all 4-momenta satisfy p^\mu_i = c_i p^\mu_0 for some numbers c_i. Therefore

0= p^\mu_i \epsilon_{i\mu} = c_i p^\mu_0 \epsilon_{i\mu}~ (\text{no sum on}~i)~\longrightarrow p^\mu_0 \epsilon_{i\mu} =0.

It follows that p_i^\mu \epsilon_{j\mu} =0 for all i,j, since each is proportional to p^\mu_0 \epsilon_{j\mu}.

As for the form of the amplitude, the factors of the polarization vectors are obviously there, while I believe that the factors of the external momenta are required by the Ward identities.

 

[kurros]

The massless scalar decay to photons seems to vanish for the same reasons, so perhaps it's worthwhile to just fill some of the gaps in for the previous argument. The point was that all 4-momenta satisfy p^\mu_i = c_i p^\mu_0 for some numbers c_i. Therefore

0= p^\mu_i \epsilon_{i\mu} = c_i p^\mu_0 \epsilon_{i\mu}~ (\text{no sum on}~i)~\longrightarrow p^\mu_0 \epsilon_{i\mu} =0.

It follows that p_i^\mu \epsilon_{j\mu} =0 for all i,j, since each is proportional to p^\mu_0 \epsilon_{j\mu}.

As for the form of the amplitude, the factors of the polarization vectors are obviously there, while I believe that the factors of the external momenta are required by the Ward identities.

Ahh, nice. In the paper if I recall (I read it a while ago) they also suggest that one considers an infinitesimal photon mass, then goes to the rest frame of the photon. Obviously in this frame the final state photons cannot be co-linear while conserving momentum (ok so you can't conserve energy either if the final photons have the same mass as the original, maybe they phrased this a bit differently. Perhaps have two sorts of photons and take some limit in which they become effectively the same thing), so one can imagine that as one takes the limit to a massless photon that the only way the co-linearity condition and momentum conservation in this pseudo-rest frame can 'converge' on being simultaneously satisfied is if the all the photon momenta go to zero, in agreement with what you say. They describe this in terms of the allowed phase space shrinking to nothing I think.

 

[bcrowell]

The massless scalar decay to photons seems to vanish for the same reasons, so perhaps it's worthwhile to just fill some of the gaps in for the previous argument.

Sorry, I was unclear. I was referring to the decay of a massless scalar into massless scalars. The point was that I was trying to avoid the mathematical complication of dealing with the polarization.

 

[bcrowell]

Let's write P for the proposition that the photon is massless, and its behavior is the same as the limit of the behavior of a massive photon as m approaches zero.

I don't think P is normally assumed in QED. From what I understand of QED (which isn't very much), the mass of the photon is never taken to be nonzero, even as a renormalization trick where it would then be allowed to approach zero.

Ahh, nice. In the paper if I recall (I read it a while ago) they also suggest that one considers an infinitesimal photon mass, then goes to the rest frame of the photon. Obviously in this frame the final state photons cannot be co-linear while conserving momentum (ok so you can't conserve energy either if the final photons have the same mass as the original, maybe they phrased this a bit differently. Perhaps have two sorts of photons and take some limit in which they become effectively the same thing), so one can imagine that as one takes the limit to a massless photon that the only way the co-linearity condition and momentum conservation in this pseudo-rest frame can 'converge' on being simultaneously satisfied is if the all the photon momenta go to zero, in agreement with what you say. They describe this in terms of the allowed phase space shrinking to nothing I think.

Here I think you're saying that if P holds, then we can't have photon decay. I think it's possible to make a much simpler argument to that effect. In the original photon's rest frame, its initial mass-energy is m. It can't decay into three photons, because then in that frame there would be a minimum mass-energy of 3m. So the behavior for all nonzero m is that it doesn't decay, and therefore the limit as m approaches 0 is that it doesn't decay. P then implies that there is no decay when m=0.

Maybe this was what you meant by the part of your post in parentheses? I don't see any reason to prefer the more complicated version of the argument.

Another simple argument based on P is this. In the original photon's rest frame, it has no observable properties that could influence its decay, other than properties that are the same for all photons. Therefore in this frame it must have some universal lifetime, in common with all other photons at rest, and in a frame where the photon isn't at rest, this lifetime is time-dilated. In the limit as m approaches zero, the photon's velocity approaches c in any frame, so its lifetime gets more and more time-dilated, approaching infinity. Therefore the limit of the behavior as m approaches 0 is that it doesn't decay.

It does seem reasonable to me to require P. Otherwise you would have a method of measurement that would be able to measure something, no matter how small it was. It seems clearly philosophically goofy if real-world measurements can distinguish between a mass of 0 kg and a mass of 10^{-10^{100}} kg. But that's what you could do if photon decay existed: by detecting photon decay you could prove that the mass of the photon was really not zero, even if it was only 10^{-10^{100}} kg.

There is also an argument in the paper by Tu et al. http://optica.machorro.net/Lecturas/PhotonMass_rpp5_1_RO2.pdf, which I don't quite understand:

It is almost certainly impossible to do any experiment that would firmly establish that the photon rest mass is exactly zero. The best one can hope to do is to place ever tighter limits on its size, since it might be so small that none of the present experimental strategies could detect it. According to the uncertainty principle, the ultimate upper limit on the photon rest mass, m_γ, can be estimated to be m_\gamma \approx \hbar/(\Delta t)c^2 , which yields a magnitude of ≈10^−66 g, using an age of the universe of about 10^10 years. Although such an infinitesimal mass would be extremely difficult to detect, there are some far-reaching implications of a nonzero value for it.

 

[kurros]

Let's write P for the proposition that the photon is massless, and its behavior is the same as the limit of the behavior of a massive photon as m approaches zero.

I don't think P is normally assumed in QED. From what I understand of QED (which isn't very much), the mass of the photon is never taken to be nonzero, even as a renormalization trick where it would then be allowed to approach zero.

I'm a little confused by your wording, but sure, in QED photons are assumed to be exactly massless and no-one usually bothers to give it a mass and then take it away again.

Here I think you're saying that if P holds, then we can't have photon decay. I think it's possible to make a much simpler argument to that effect. In the original photon's rest frame, its initial mass-energy is m. It can't decay into three photons, because then in that frame there would be a minimum mass-energy of 3m. So the behavior for all nonzero m is that it doesn't decay, and therefore the limit as m approaches 0 is that it doesn't decay. P then implies that there is no decay when m=0.

Maybe this was what you meant by the part of your post in parentheses? I don't see any reason to prefer the more complicated version of the argument.

Yes this is basically where I was going with the paranthesised bit. I was just trying to avoid the energy conservation issue because it seems to sort of discontinuously disappear when the photons become massless and you no longer have a rest frame. But perhaps you are right and it is a useful way to look at it anyway. This obviously explains why no massive particles ever spontaneously decay into slower-moving versions of themselves. Perhaps the massless limit is not really so different.

Another simple argument based on P is this. In the original photon's rest frame, it has no observable properties that could influence its decay, other than properties that are the same for all photons. Therefore in this frame it must have some universal lifetime, in common with all other photons at rest, and in a frame where the photon isn't at rest, this lifetime is time-dilated. In the limit as m approaches zero, the photon's velocity approaches c in any frame, so its lifetime gets more and more time-dilated, approaching infinity. Therefore the limit of the behavior as m approaches 0 is that it doesn't decay.

I like this line of thinking in principle, but I guess I was trying to explain why the thing is kinematically disallowed. I wanted to see the amplitude vanish more explicitly, which I think we have sort of achieved now. As a more heuristic argument however I am fond of what you are saying. Photons experience no time as they travel, so how could they possibly decay!? ;)

 

[fzero]

Sorry, I was unclear. I was referring to the decay of a massless scalar into massless scalars. The point was that I was trying to avoid the mathematical complication of dealing with the polarization.

The complication of the polarization (ultimately a consequence of gauge and Lorentz invariance) is key to the vanishing of the photon splitting amplitudes. A scalar field theory is much less restrictive and generally has terms already at tree level that would allow splitting, since there aren't any mechanisms to cause them to vanish.

Even if we were dealing with a free scalar coupled to a fermion, the most you could say is that the one-loop diagrams with an odd number of external scalar legs would vanish because they involve a trace over an odd number of gamma matrices. The ones with an even number of external lines should be nonzero in the absence of some exotic structure like supersymmetry.

 

[bcrowell]

The complication of the polarization (ultimately a consequence of gauge and Lorentz invariance) is key to the vanishing of the photon splitting amplitudes. A scalar field theory is much less restrictive and generally has terms already at tree level that would allow splitting, since there aren't any mechanisms to cause them to vanish.

Even if we were dealing with a free scalar coupled to a fermion, the most you could say is that the one-loop diagrams with an odd number of external scalar legs would vanish because they involve a trace over an odd number of gamma matrices. The ones with an even number of external lines should be nonzero in the absence of some exotic structure like supersymmetry.

Interesting! So presumably in the case of a massless spin-zero particle x coupled to a fermion of mass m, the decay x\rightarrow 3x occurs with lifetime \tau=(k\hbar/c^4m^2)E, where m is the mass of the fermion and k is some dimensionless constant of order unity? That is, it has to have the form \tau\propto E due to Lorentz invariance, and the only thing I can think of that would set the constant of proportionality would be the mass of the fermion. But this seems awfully strange to me, since it means that the lifetime can be made arbitrarily short if the mass of the fermion is large enough. So somehow low-energy experiments would be sensitive to phenomena arising from arbitrarily high energies...? Or is there something else in such a field theory that would set the constant of proportionality between the energy and the lifetime?

 

[gendou2]

A particle's decay time depends on it's velocity relative to the observer.
Fast Muons (from cosmic ray impacts) have decay times in the Earth frame that are longer than one would predict, ignoring time dilation.
For the photon, the time dilation is infinite, because it moves at c.
So, I wouldn't expect the photon to "have time" to decay.

 

[tom.stoer]

In post #6 torus said that the three photon decay channel is structurally the same as a two-photon scattering. So how is the decay related to light-light scattering? by crossing symmetry? and if so why should the matrix element vanish?

 

[bcrowell]

In post #6 torus said that the three photon decay channel is structurally the same as a two-photon scattering.

I'll expose my ignorance here. Does "two-photon scattering" refer to the reaction 2\gamma\rightarrow 2\gamma? How do you get nontrivial scattering of photons off of each other while satisfying conservation of energy and momentum? Do they just swap spins or something?

 

[tom.stoer]

What I mean is the decay γ → 3γ and the scattering 2γ → 2γ.

But thinking a little bit more about it my conclusion is that this is NOT crossing :-(

 

[Haelfix]

As Fzero explained, and as the paper notes. The QED fermion square diagram (4pt function) does NOT vanish in general, but in the case of photon splitting (I like that word better than photon decay), it is kinematically constrained to do so. You can see this by Fzero's argument, or you can see it by power counting (although you have to kinematically reduce the decay amplitude formula first). And yes, the argument does rely crucially on the assumption of Lorentz invariance and a certain amount of additional structure to the field theory in question.

To be perfectly honest, this is somewhat unsatisfying to me. I wouldn't be surprised if there was a more straightforward argument here (perhaps in eg the twistor formalism).

Incidentally, b/c the box diagram does not vanish, light on light scattering will proceed, and I believe people have looked for this signature experimentally. An even cooler process is light by light scattering proceeding via graviton exchange, where one finds the cool result that all amplitudes vanish except for the ones that proceed antiparralel (again kinematics). This is called the Tolman, Ehrenfest, Podolsky effect in classical GR.

 

[bcrowell]

What I mean is the decay γ → 3γ and the scattering 2γ → 2γ.

Oh, I see. In 1+1 dimensions, conservation of energy-momentum forbids nontrivial 2γ → 2γ scattering, but in 3+1, the plane of the incident particles doesn't have to coincide with the plane of the scattered particles.

 

[tom.stoer]

To be perfectly honest, this is somewhat unsatisfying to me. I wouldn't be surprised if there was a more straightforward argument here.

This why I started to think about crossing. I can't believe that photon decay is forbidden by such subtleties. I expected some symmetry argument.

 

[PAllen]

Ok, I will try out a non-perturbative argument that photon decay/splitting is forbidden, based on many of the ideas raised in this thread. See what you think:

1) From the SR argument Bcrowell first implied, and presented explicitly in the paper I posted, we know that Lorentz invariance required decay time to be proportional to energy.

2) However large the constant of proportionality, for any photon, there exists a frame of reference where the photon energy can be made arbitrarily small, and the required decay time arbitrarily fast.

3) Yet QED, which we take to be a consistent, Lorentz invariant theory, requires the decay to be mediated by virtual charged fermions of substantial mass. This further requires that the probability interaction is small (because the available energy can be made arbitrarily small), and the lifetime cannot be arbitrarily fast.

This is a contradiction unless the overall probability is exactly zero.

 

[bcrowell]

Here is a paper that seems relevant: Gelmini et al., "On photon splitting in theories with Lorentz invariance violation," CAP 0506 (2005) 012, http://arxiv.org/abs/hep-ph/0503130

In standard QED the photon splitting process γ → 3γ does not occur because, due to energy and momentum conservation, the three momenta must all be parallel and the amplitude vanishes in this configuration.

The paper is about using the nonobservation of γ → 3γ to put upper limits on Lorentz violation. She has a parameter with units of mass that measures the amount of Lorentz violation, by affecting the momentum-energy relation for photons. The motivation is that you might get this kind of effect at the Planck scale. The predicted half-life is of the form \tau=(\text{unitless})E^{-14}\mu^5m_e^8, where E is the energy of the photon, and my \mu is her M/\xi, the parameter that measures Lorentz violation. The result is not of the form \tau\propto E that is required by Lorentz invariance, presumably because she gets it by assuming Lorentz violation. Because her half-life gets shorter rather than longer with increasing energy, she doesn't get the decay of a photon into an infrared jet in finite time as described in my #35. She takes the tightest empirical bound on the rate to be from TeV gammas from the Crab pulsar. The result is that \mu is constrained to be at least about 10^4 times the Planck mass, whereas you'd kind of expect that if there was vacuum dispersion due to quantum gravity effects, you'd get something on the same order as the Planck mass.

For googling, a good phrase seems to be "photon splitting." You get a ton of hits that aren't vacuum effects, though, such as photons being split by interaction with matter, and photons splitting in strong magnetic fields.

 

[bcrowell]

Another paper: Jacobson et al., "Threshold effects and Planck scale Lorentz violation: combined constraints from high energy astrophysics," Phys.Rev. D67 (2003) 124011, http://arxiv.org/abs/hep-ph/0209264

p. 20:

In the Lorentz invariant case the equations of motion imply that k_a is a null vector and k_a\epsilon^a=0. Energy-momentum conservation then implies that these 4-momenta are all parallel, so being null they are orthogonal to each other and to all the polarization vectors. The rate thus vanishes for two different reasons. First, since the momenta are necessarily all parallel, the phase space has vanishing volume. Second, the rate must be a scalar formed by contracting these four field strengths using only the metric. Any such contraction vanishes since it must involve contractions of the momenta with each other or with the polarizations. Hence the amplitude vanishes. In the case of an odd number of photons, another reason for vansihing of the amplitude is Furry’s theorem, which states that the sum over loops with an odd number of electron propagators vanishes.

 

[bcrowell]

Ok, I will try out a non-perturbative argument that photon decay/splitting is forbidden, based on many of the ideas raised in this thread. See what you think:

1) From the SR argument Bcrowell first implied, and presented explicitly in the paper I posted, we know that Lorentz invariance required decay time to be proportional to energy.

2) However large the constant of proportionality, for any photon, there exists a frame of reference where the photon energy can be made arbitrarily small, and the required decay time arbitrarily fast.

3) Yet QED, which we take to be a consistent, Lorentz invariant theory, requires the decay to be mediated by virtual charged fermions of substantial mass. This further requires that the probability interaction is small (because the available energy can be made arbitrarily small), and the lifetime cannot be arbitrarily fast.

This is a contradiction unless the overall probability is exactly zero.

This sounds right to me. One thing to check about this argument is that it doesn't prove something false in cases where we know that decay of massless particles is possible. In the Fiore and Modanese paper, the only explicit example they offer of a field theory in which massless particles decay is quantum gravity with a positive cosmological constant. This is a theory with only massless particles in it (they allow other massless bosons besides the graviton), so the premise of your point #3 doesn't hold. This is as it should be, and it encourages me to believe that your argument is right.

If we look at the arguments given in the Gelmini and Jacobson papers, they both seem to say that phase space arguments, based on the collinearity of the decay products, are sufficient to prove that photon decay can't happen in standard QED. They make it sound like this argument is sufficient in and of itself, and since the argument doesn't appeal to any features of QED as opposed to quantum gravity, this would seem to contradict Fiore and Modanese's use of quantum gravity as a counterexample. I think what's happening is that in quantum gravity, there is another factor that blows up to infinity so fast that it overwhelms the vanishing phase-space factor. At least, that's my interpretation of the Fiore paper, which I don't claim to understand in detail.

One other thing that seems odd to me about all this is that in a Feynman diagram there's supposed to be a distinction between internal lines and external legs. The external legs are what you can actually observe. They represent particles that fly off and get far enough away from the others that they no longer interact, and they can be detected individually. In the case of massless-particle splitting, it's not clear to me how you make such a distinction. The three decay products fly off in the same direction at the same speed, so in some sense those three "external" legs never actually separate; you could argue that they are internal legs. In fact, splitting, unlike every other form of radioactive decay, can occur and then spontaneously reverse itself. So actually in a field theory that allows splitting, the process of formation of the infrared jet I described in #35 will not run to completion in a finite time in a vacuum, because there will be a tendency for the particles to reverse-decay just as frequently as they decay.

 

[bcrowell]

It looks like I may be the only one still interested in this thread, but anyway I thought I'd post a couple more thoughts and see if anyone has any comments.

There are some odd thermodynamic aspects to all this. Consider a one-dimensional gas of massless particles that can split and unsplit spontaneously. This is different from the usual blackbody problem, because there doesn't have to be a heat bath; the particles can change their frequencies spontaneously, without interacting with the walls of the box. So let's say the box is of length L, is perfectly insulating, and doesn't interact thermally with the particles. We're not going to get the standard blackbody results, because there's no heat bath. Let the total energy of the gas be E=nE_o, where E_o=h/2L is the ground-state energy (c=1). The entropy is S=ln p(n,k), where k is the number of particles, and p(n,k) is the number of partitions of the integer n into k nonzero terms, ignoring order. (Boltzmann's constant=1.) Playing around with Stirling's formula I convinced myself that the entropy was maximized for k\sim\sqrt{n}. So the result is that if you introduce a single short-wavelength particle of energy E into the box, it ends up in a state with a temperature of something like T\sim\sqrt{hE/L}. Letting L approach infinity, T approaches zero, so I think my original idea about the particle spontaneously decaying to an infrared jet was actually correct.

Another strange thing is that if you start off with, say, a Gaussian wavepacket, splitting causes the wavelengths to increase, so the Gaussian has to broaden over time. (The thermodynamic argument above shows that there really is a net broadening. In free space, you never reach an equilibrium between splitting and unsplitting.) That means that the corresponding classical wave equation is dispersive. But unless I'm missing something, you can't have dispersion of the vacuum for massless particles without violating Lorentz invariance. This is of course exactly what is happening in the Gelmini and Jacobson papers: they start by assuming Lorentz violation. This makes me wonder whether Fiore and Modanese's treatment, with Lorentz invariance, is even self-consistent.

For the particle in a box, assuming Lorentz invariance so that \tau\propto E, the rate of decay increases geometrically as the splitting continues, so you reach thermal equilibrium in only a few times the original particle's lifetime. This thermal equilibration time is independent of the size of the box. But this assumes that the initial particle is inserted into the box in an eigenstate of energy, so that its wavepacket fills the entire box. In free space, it's impossible to prepare a particle in an initial state such that it fills the whole infinite "box." An initial wavepacket with finite length will disperse at a rate limited by c. But even so, this seems to go against the third law of thermodynamics. As the jet's temperature approaches zero, its entropy blows up to infinity.

 

[bcrowell]

This recent paper seems relevant: Julian Heeck, How stable is the photon?, http://arxiv.org/abs/1304.2821