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.