'decay' of photons
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? 
Re: 'decay' of photons
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 halfcycle 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 halfcycle wavetrain can't do it, then it seems implausible that a fullcycle wavetrain can, since the latter can be seen as a superposition of two of the former. Answer #3: What would the halflife be, and what frame would it be measured in? I think this is actually the most ironclad argument against it. 
Re: 'decay' of photons
There is one conservation law that restricts such a process  charge parity. The Cparity of a photon is 1, and the Cparity of an nphoton state is (1)^{n}. Therefore if a photon could decay it would have to decay into an odd number of photons.

Re: 'decay' of photons
It's interesting to try to apply the same arguments to gravitons. I guess a graviton has Cparity +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 weakfield limit, the argument that there is no rest frame in which to express the halflife is not necessarily valid.
I know that the 1/r^{2} form of Coulomb's law has been verified to absurd precision. I don't know how tight the corresponding results are for gravity. 
Re: 'decay' of photons
One could try to find a Feynman diagram describing the process gamma => 3 gamma
 the inphoton could turn into a electronpositron pair  the wo particles in this loop could emit two "Bremsstrahlung" photons  eventually the electronpositron pair recombines into one photon I bet the matrix element vanishes due to symmetry or parity reasons. 
Re: 'decay' of photons
Why does it have to vanish? The diagram is just the finite lightlightscattering one, with one external leg switched from in to out.

Re: 'decay' of photons
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Re: 'decay' of photons
What if it turned out that the halflife 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 halflife to have been observed decaying.
Then wouldn't this satisfy Lorentz invariance? In a moving frame where the halflife 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 4vector of the photon has a small imaginary part, so the wavefunction falls off exponentially in both space and time. 
Re: 'decay' of photons
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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 [itex]D=\sqrt{(1v)/(1+v)}[/itex], so the halflife goes up. But when you chase after a relativistic particle, its halflife decreases based on the Lorentz transformation. E.g., cosmicray muons have a shorter halflife 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.) Quote:

Re: 'decay' of photons
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 spacetime 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 τ_{0} = 1/Γ and call it "the" lifetime, but there is still a nonzero invariant M^{2} associated with the transition rate. 
Re: 'decay' of photons
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 4velocity, p is particle 4velocity, 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 4vectors. 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. 
Re: 'decay' of photons
A decay rate is proportional to a phase space factor [itex]\int d^3p[/itex] for each final particle. Since the three final state photons must be in the forward direction, this integral vanishes.

Re: 'decay' of photons
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?

Re: 'decay' of photons
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The interaction can't have a delta function. Zero phase space is zero phase space. 
Re: 'decay' of photons
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 energymomentum 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. 
Re: 'decay' of photons
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). hepth/9508018 Eugene. 
Re: 'decay' of photons
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As Bill_K has pointed out, decay to two photons is prevented by Cparity, so you need at least three photons as the output of the decay. But three spin1's can be coupled to make a spin1. 
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