'decay' of photons

  1. PAllen

    PAllen 5,736
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    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?
     
  2. jcsd
  3. bcrowell

    bcrowell 5,952
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    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.
     
    Last edited: Jul 8, 2011
  4. Bill_K

    Bill_K 4,159
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    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.
     
  5. bcrowell

    bcrowell 5,952
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    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.
     
  6. tom.stoer

    tom.stoer 5,489
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    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.
     
    Last edited: Jul 9, 2011
  7. Why does it have to vanish? The diagram is just the finite light-light-scattering one, with one external leg switched from in to out.
     
  8. bcrowell

    bcrowell 5,952
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    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.
     
  9. Bill_K

    Bill_K 4,159
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    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.
     
  10. bcrowell

    bcrowell 5,952
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    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 [itex]D=\sqrt{(1-v)/(1+v)}[/itex], 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.)

    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.
     
    Last edited: Jul 9, 2011
  11. Bill_K

    Bill_K 4,159
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    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.
     
    Last edited: Jul 10, 2011
  12. PAllen

    PAllen 5,736
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    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.
     
  13. clem

    clem 1,276
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    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.
     
    Last edited by a moderator: Jul 10, 2011
  14. Bill_K

    Bill_K 4,159
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    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?
     
  15. bcrowell

    bcrowell 5,952
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    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.
     
  16. clem

    clem 1,276
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    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.
     
  17. Haelfix

    Haelfix 1,748
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    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.
     
    Last edited: Jul 10, 2011
  18. 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.
     
  19. bcrowell

    bcrowell 5,952
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    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.
     
  20. bcrowell

    bcrowell 5,952
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    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.
     
    Last edited: Jul 10, 2011
  21. PAllen

    PAllen 5,736
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    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.
     
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