We don't normally consider the possibility that massless particles could undergo radioactive decay. There are elementary arguments that make it sound implausible. (A bunch of the following is summarized from Fiore 1996. Most of the rest, except as noted, is my ideas, many of which are probably wrong.)
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1) Normally we state the lifetime of a particle in its rest frame, but a massless particle doesn't have a rest frame. However, it is possible for the lifetime $\tau$ to be proportional to energy $E$ while preserving Lorentz invariance (basically because time and mass-energy are both timelike components of four-vectors).
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2) The constant of proportionality between $\tau$ and $E$ has units of mass-2. It's strange to have such a dimensionful constant popping up out of nowhere, but it's not impossible.
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3) We would typically like the observables of a theory to be continuous functions of its input parameters. If $X$ is a particle of mass $m$, then a decay like $X\rightarrow 3X$ is forbidden by conservation of mass-energy for m>0, but not for m=0. This discontinuity is ugly, but QFT has other cases where such a discontinuity occurs. E.g., historically, massive bosons were not trivial to incorporate into QFT.
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4) In a decay like $X\rightarrow3X$, the products all have to be collinear. This is a little odd, since it doesn't allow the clear distinction one normally assumes in a Feynman diagram between interior and exterior lines. It also means that subsequent "un-decay" can occur. Strange but not impossible.
So what about less elementary arguments? My background in QFT is pretty weak (the standard graduate course, over 20 years ago, barely remembered).
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5) The collinearity of the decay products makes the phase-space volume vanish, but amplitudes can diverge to make up for this.
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6) If $X$ is coupled to some fermion $Y$, then one would expect that decay would correspond to a Feynman diagram with a box made out of $Y$'s and four legs made out of $X$'s. If $Y$ is a massive particle like an electron, $P$. Allen on physicsforums argues that when the energy of the initial $X$ approaches zero, the $X$ shouldn't be able to "see" the high-energy field $Y$, so the probability of decay should go to zero, and the lifetime $\tau$ should go to infinity, which contradicts the requirement of $\tau\propto E$ from Lorentz invariance. This seems to rule out the case where $Y$ is massive, but not the case where it's massless.
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7) If $X$ is a photon, then decay is forbidden by arguments that to me seem technical. But this doesn't forbid decays when $X$ is any massless particle whatsoever.
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8) There are some strange thermodynamic things going on. Consider a one-dimensional particle in a box of length $L$. If one $X$ is initially introduced into the box with energy $E=nE_o$, where $E_o$ is the ground-state energy, then it undergoes decays and "undecays," and if I've got my back-of-the-envelope estimate with Stirling's formula right, I think it ends up maximizing its entropy by decaying into about $\sqrt{n}$ daughters at a temperature $\sim \sqrt{hE/L}$. If you then let it out of the box so that it undergoes free expansion, it acts differently from a normal gas. Its temperature approaches zero rather than staying constant, and its entropy approaches infinity. I may be missing something technical about thermo, but this seems to violate the third law.
So my question is this: Is there any fundamental (and preferably simple) argument that makes decay of massless particles implausible? I don't think it can be proved completely impossible, because Fiore offers field theories that are counterexamples, such as quantum gravity with a positive cosmological constant.
References:
- Fiore and Modanese, "General properties of the decay amplitudes for massless particles," 1996, http://arxiv.org/abs/hep-th/9508018.
Best Answer
I agree with qftme's answer for the case of massive decay products. By energy conservation alone, $\gamma \rightarrow e^+e^-$ should be allowed, but momentum conservation forbids it (as well as the opposite case, $e^+e^-$ annihilation). It is only allowed if you have some other particle involved to take care of the photon momentum.
In case of a massless particle decaying into other massless particles, this only works if your particles have a self-interaction. As far as I know, you can only have this for Bosons. QED (photons) has no self interaction, and the weak bosons (W, Z) are massive, but this is well known to occur in QCD in form of $g \rightarrow gg$. In fact, gluons are more likely to interact at lower energies, than at highest energies where QCD is asymptotically free - that means the interaction is well-behaved and weak, and we can use perturbation theory. At low energies, gluons just keep splitting and producing even lower energy quarks or gluons, until there are in a sense "infinitely many" of "infinitely low" energy. This is what is called infrared divergence. It may sound strange, but in practice all observable quantities remain finite, so there is nothing to worry about.
Your point number 5 is also applies here. The decay products are not only very low energetic, but tend to be very collinear (collinear divergence). This is the source of the famous "jets" that appear in high energy collider experiments.