Photon Decay: Can High Energy Photons Transform?

[PAllen]

I was recently wondering about this. A very high energy photon cannot transform into any collection of particles with mass without interacting with another photon or particle, else it is trivial to show energy/momentum cannot be conserved. Interacting with another photon allows particle/antiparticle production, for example.

However, I could not think of any conservation rule that would prevent, say a gamma ray from 'decaying' into a plethora of low energy photons (following the same path). Energy and momentum would be conserved, no quantum number rules would be violated; and since photons are bosons, there wouldn't seem to be any difficulty with all the 'decay photons' occupying the same path.

Yet... I've never heard of such a thing. What prevents this?

 

[bcrowell]

Cool idea.

Answer #1: Say a photon with wavelength λ becomes two photons, each with wavelength 2λ. This violates Maxwell's equations, and I don't think you get a free pass to violate Maxwell's equations just because your wave is quantized. Maxwell's equations play the same role for photons as Schrodinger's equation plays for massive particles.

Answer #2: Even without appealing to Maxwell's equations, suppose your original wavetrain consists of one half-cycle of a sinusoidal wave. It can't double its wavelength without instantaneously stretching out to encompass more space, which I think would violate causality. If a single half-cycle wavetrain can't do it, then it seems implausible that a full-cycle wavetrain can, since the latter can be seen as a superposition of two of the former.

Answer #3: What would the half-life be, and what frame would it be measured in? I think this is actually the most ironclad argument against it.

 

[Bill_K]

There is one conservation law that restricts such a process - charge parity. The C-parity of a photon is -1, and the C-parity of an n-photon state is (-1)ⁿ. Therefore if a photon could decay it would have to decay into an odd number of photons.

 

[bcrowell]

It's interesting to try to apply the same arguments to gravitons. I guess a graviton has C-parity +1, so there is no constraint to an odd number of gravitons in the decay. Because the Einstein field equations are nonlinear, you can't necessarily argue that it violates them. And because propagation at c is only an approximation that applies in the weak-field limit, the argument that there is no rest frame in which to express the half-life is not necessarily valid.

I know that the 1/r² form of Coulomb's law has been verified to absurd precision. I don't know how tight the corresponding results are for gravity.

 

[tom.stoer]

One could try to find a Feynman diagram describing the process gamma => 3 gamma

- the in-photon could turn into a electron-positron pair
- the wo particles in this loop could emit two "Bremsstrahlung" photons
- eventually the electron-positron pair recombines into one photon

I bet the matrix element vanishes due to symmetry or parity reasons.

 

[torus]

Why does it have to vanish? The diagram is just the finite light-light-scattering one, with one external leg switched from into out.

 

[bcrowell]

I bet the matrix element vanishes due to symmetry or parity reasons.

I suppose any time the probability for a process vanishes, there is probably some kind of symmetry involved, but I don't think it has anything to do with parity in this case. There is certainly a symmetry-based reason that it has to vanish, which is Lorentz invariance: there is no rest frame in which to express its half-life. I don't know how that translates into the language of Feynman diagrams. There is also the fact that all three photons are constrained by conservation of energy-momentum to come off in the same direction as the incoming one, but again I don't know how to express that in Feynman-diagram language.

 

[Bill_K]

What if it turned out that the half-life of a photon was inversely proportional to its energy. Not unreasonable, since it would mean that only very high energy photons would have a short enough half-life to have been observed decaying.

Then wouldn't this satisfy Lorentz invariance? In a moving frame where the half-life was measured to be smaller, the energy of the photon would be measured to be greater by the same amount.

A way to express it invariantly would be to say that the propagation 4-vector of the photon has a small imaginary part, so the wavefunction falls off exponentially in both space and time.

 

[bcrowell]

What if it turned out that the half-life of a photon was inversely proportional to its energy. Not unreasonable, since it would mean that only very high energy photons would have a short enough half-life to have been observed decaying.

Then wouldn't this satisfy Lorentz invariance? In a moving frame where the half-life was measured to be smaller, the energy of the photon would be measured to be greater by the same amount.

If you don't introduce a new scale, then the only way the lifetime could be determined would be, as you suggest, through an inverse proportionality to the energy. You could write it as τ=bħ/E, where b is a unitless constant. (This τ would be a half-life that was different from every other half-life, because it would be defined not in the particle's rest frame but in some other frame. It would also violate the correspondence principle, because both E and τ are measurable classically, so you don't recover the classical limit by letting ħ approach zero. Or you could use some other constant with units of J.s instead of ħ, but then you'd be introducing a new scale.)

But then I think this violates Lorentz invariance. Suppose you do a boost by v in the direction of propagation. The energy transforms by the Doppler shift factor D=\sqrt{(1-v)/(1+v)}, so the half-life goes up. But when you chase after a relativistic particle, its half-life decreases based on the Lorentz transformation. E.g., cosmic-ray muons have a shorter half-life if you chase after them than they do in the frame of the earth. Therefore the relation τ=bħ/E can't have the same form in all frames.

(I could have made a mistake in the above argument. If so, call me on it.)

A way to express it invariantly would be to say that the propagation 4-vector of the photon has a small imaginary part, so the wavefunction falls off exponentially in both space and time.

But then you have to specify how the imaginary part is determined. It's directly related to τ, so I don't think rephrasing it in terms of this imaginary part changes anything about the Lorentz invariance argument.

 

[Bill_K]

Ben, If you're claiming that Lorentz invariance is what prevents a photon from decaying, then I do disagree with you.

From a Feynman diagram you obtain an 'invariant amplitude' M: a Lorentz invariant whose square |M|² represents the transition probability per unit space-time volume. To calculate from this the transition rate in a particular frame, you must insert a phase space factor for each ingoing line. Each massive particle requires a factor m/2E. So for example, the total transition rate Γ for a collision process is (mm'/4EE')|M|². For a decay it is just Γ = (m/2E)|M|². Stating the obvious, the rate is different in every frame and goes to zero as 1/γ for a frame in which the particle is relativistic. Again stating the obvious, the rate can be measured in any frame, it does not have to be measured in the particle's rest frame.

An ingoing massless particle works the same way, except that the phase space factor is (1/2ω). If a photon does decay (and this leaves open the question of whether it actually does or not) the total transition rate per unit time will be Γ = (1/2ω)|M|². Again the rate is different in every frame. But it can be nonzero without violating Lorentz invariance. True, there is no rest frame in which to say τ₀ = 1/Γ and call it "the" lifetime, but there is still a nonzero invariant |M|² associated with the transition rate.

 

[PAllen]

I think Ben's argument does show that half life of a photon (if it existed) would not follow the same law as for a massive particle. Part of Bill_k's response is 'so what?' Having thought about this more, I am inclined to favor Bill_k's point of view, as follows:

For a massive particle we happen to have law for half life of form:

gamma(o,p) t0

where o is observer 4-velocity, p is particle 4-velocity, t0 is 'rest half life'. Expressing gamma in strictly coordinate independent language is a bit of a pain, but it is scalar invariant function of two unit 4-vectors.

For light, a hypothesis is:

k/E, E= o dot P, where P is light 4 momentum, k is some constant.

While this is a different law, it is again a manifestly scalar invariant function of the two relevant vectors.

So, I'm still looking for a sharp reason this doesn't happen. Bill_k notes that if k is large enough, observations don't exclude this (e.g. if half life is a million years for a 10^20 ev gamma). I've done quite a bit of searching to see if this is addressed in any paper available on line, but haven't found anything. Very strange.

 

[Meir Achuz]

A decay rate is proportional to a phase space factor \int d^3p for each final particle. Since the three final state photons must be in the forward direction, this integral vanishes.

 

[Bill_K]

clem, interesting idea. But the constraint is a standard one that always holds: momentum conservation and particles on the mass shell. So maybe the integrals are over a delta function?

 

[bcrowell]

I think Ben's argument does show that half life of a photon (if it existed) would not follow the same law as for a massive particle.

My argument is not that it doesn't follow the same law as for a massive particle. My argument is that the only law that can be constructed that doesn't set a new scale is a law that violates Lorentz invariance. My argument may be right or wrong, but as far as I can tell, Bill_K has not addressed it.

 

[Meir Achuz]

clem, interesting idea. But the constraint is a standard one that always holds: momentum conservation and particles on the mass shell. So maybe the integrals are over a delta function?

Four photons are only connected via an internal electron loop.
The interaction can't have a delta function.
Zero phase space is zero phase space.

 

[Haelfix]

Maybe I am missing something here. Suppose a photon decays into two photons x and y (a vertex which doesn't exist in QED in general, but suppose it did).

Conservation of energy-momentum implies that photons x and y are collinear.
However in the case where you have two collinear photons, the decay is prevented by conservation of total angular momentum.

I can't do it in my head, but I think this generalizes to N decay products and more or less follows what Clem is saying.

 

[meopemuk]

Check out this paper. It discusses the impossibility of the photon decay in QED.

Fiore, G.; Modanese, G., General properties of the decay amplitudes for
massless particles, Nucl. Phys. B 477, 623 (1996). hep-th/9508018

Eugene.

 

[bcrowell]

Conservation of energy-momentum implies that photons x and y are collinear.
However in the case where you have two collinear photons, the decay is prevented by conservation of total angular momentum.

Why is it prevented by conservation of angular momentum? You can couple two spin-1's to make a spin-1.

As Bill_K has pointed out, decay to two photons is prevented by C-parity, so you need at least three photons as the output of the decay. But three spin-1's can be coupled to make a spin-1.

 

[bcrowell]

An ingoing massless particle works the same way, except that the phase space factor is (1/2ω). If a photon does decay (and this leaves open the question of whether it actually does or not) the total transition rate per unit time will be Γ = (1/2ω)|M|². Again the rate is different in every frame. But it can be nonzero without violating Lorentz invariance. True, there is no rest frame in which to say τ₀ = 1/Γ and call it "the" lifetime, but there is still a nonzero invariant |M|² associated with the transition rate.

I don't see how this addresses the argument of the second paragraph of my #9. Note that my argument is purely classical, so if it's wrong, it should be possible to show it's wrong on purely classical grounds, without resorting to QFT.

 

[PAllen]

But then I think this violates Lorentz invariance. Suppose you do a boost by v in the direction of propagation. The energy transforms by the Doppler shift factor D=\sqrt{(1-v)/(1+v)}, so the half-life goes up. But when you chase after a relativistic particle, its half-life decreases based on the Lorentz transformation. E.g., cosmic-ray muons have a shorter half-life if you chase after them than they do in the frame of the earth. Therefore the relation τ=bħ/E can't have the same form in all frames.

I don't see how this addresses the argument of the second paragraph of my #9.

I thought I addressed this as follows:

For light, a hypothesis is:

k/E, E= o dot P, where P is light 4 momentum, k is some constant.

While this is a different law, it is again a manifestly scalar invariant function of the two relevant vectors.

How is this law not Lorentz invariant? It takes the same form in all frames (in fact is expressed coordinate independent). It is just different from the law for a massive particle.

 

[Haelfix]

You are right. Conservation of angular momentum when all particles are collinear prevents the process for N even but not for N odd (thus reproducing the CP requirement).

 

[bcrowell]

Check out this paper. It discusses the impossibility of the photon decay in QED.

Fiore, G.; Modanese, G., General properties of the decay amplitudes for
massless particles, Nucl. Phys. B 477, 623 (1996). hep-th/9508018

Excellent! Now we don't have to reinvent the wheel :-)

Property 6 on p. 5 is a purely classical fact about the decay, and it says that the lifetime of a massless particle has to be related to the energy of the particle by a relation of the form \tau=\xi E. This seems to be consistent with my argument in #9 that, for classical reasons, Lorentz invariance forbids the decay of a massless particle unless we introduce a new scale. The constant \xi constitutes a new scale. Of course, just because the result of my argument is confirmed by their (presumably correct) argument (they say it's "elementary" and give a reference to another paper), that doesn't prove that my argument is correct. But at least it gives me a little more confidence :-)

 

[PAllen]

Check out this paper. It discusses the impossibility of the photon decay in QED.

Fiore, G.; Modanese, G., General properties of the decay amplitudes for
massless particles, Nucl. Phys. B 477, 623 (1996). hep-th/9508018

Eugene.

Wow, great to see the question was analyzed. I couldn't believe this hadn't come up before. They discuss most of the ideas thrown out in this thread. In particular, they state a 'property 6' proved in a referenced paper I haven't located, that the proposed 1/E half life is impossible; it must be proportional to E.

It seems to me (I could easily be misunderstanding) that they don't take the momentum phase space integral argument to be conclusive. They present that integral, discuss implications for co-linear decay products (odd number, as discussed here), and it seems they need other arguments which are beyond my expertise to clinch the conclusion that the decay is impossible.

Can someone give a succinct summary, from this paper, of what makes the photon decay impossible?

 

[bcrowell]

Since CMB photons don't decay in 10 billion years, we can estimate that \xi c^4/\hbar \gtrsim 10^{108} kg^-2. This corresponds to a mass scale of \lesssim 10^{-54} kg. This is many orders of magnitude less than the Planck mass, and also many orders of magnitude less than the smallest mass occurring in the standard model.

 

[PAllen]

Here is the paper where they prove 'property 6', that the decay of massless particle must be proportional to its energy:

http://arxiv.org/abs/hep-th/9501123

 

[Haelfix]

Yea the paper is wonderfully unlovely, and I am somewhat surprised that it is so difficult. But the power counting argument that it gives is simple enough and applicable here.

 

[bcrowell]

I thought I addressed this as follows:

For light, a hypothesis is:

k/E, E= o dot P, where P is light 4 momentum, k is some constant.

While this is a different law, it is again a manifestly scalar invariant function of the two relevant vectors.

How is this law not Lorentz invariant? It takes the same form in all frames (in fact is expressed coordinate independent). It is just different from the law for a massive particle.

I didn't read your #11 carefully because its first sentence was a misunderstanding of my #9. Given my later attempt to clarify my #9, do you think #11 is still relevant? #11 would seem to me to contradict Fiore's property 6. I don't understand what you mean by "While this is a different law, it is again a manifestly scalar invariant function of the two relevant vectors." You seem to be proposing τ=k/E, but clearly k/E can't be a Lorentz scalar, since τ is not a Lorentz scalar...??

 

[bcrowell]

Here is the paper where they prove 'property 6', that the decay of massless particle must be proportional to its energy:

http://arxiv.org/PS_cache/hep-th/pdf/9501/9501123v1.pdf

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.

 

[PAllen]

I didn't read your #11 carefully because its first sentence was a misunderstanding of my #9. Given my later attempt to clarify my #9, do you think #11 is still relevant? #11 would seem to me to contradict Fiore's property 6. I don't understand what you mean by "While this is a different law, it is again a manifestly scalar invariant function of the two relevant vectors." You seem to be proposing τ=k/E, but clearly k/E can't be a Lorentz scalar, since τ is not a Lorentz scalar...??

My argument is clearly inconsistent with the paper's property 6 (whose proof I found in the other paper; it is based on Lorentz transform of the pair of events: emission, and decay).

My argument proposes (impossibly, I now see) to define a half life as a strictly local measurement (like measurement of KE of a particular particle by a particular observer). My definition is a scalar function of a contraction to two vectors. This *is* a manifestly Lorentz invariant. Not only a Lorentz boost, but a completely arbitrary coordinate transform leaves it unchanged (for a given observer, and a given photon).

 

[bcrowell]

Can someone give a succinct summary, from this paper, of what makes the photon decay impossible?

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.

 

[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.