# Decay of massless particles

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.)

• 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).

• 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.

• 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.

• 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).

• 5) The collinearity of the decay products makes the phase-space volume vanish, but amplitudes can diverge to make up for this.

• 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.

• 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.

• 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:

1. Fiore and Modanese, "General properties of the decay amplitudes for massless particles," 1996, http://arxiv.org/abs/hep-th/9508018.
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Could you be a little more clear about what's from Fiore and what you came up with? – Dan Jul 20 '11 at 2:48
@Dan: I went back and added numbers to the paragraphs for reference. Fiore is the source for some ideas and assertions in: 1, 4, 5, 7. My own ideas occur in: 2, 3, 4, 8. P. Allen came up with 6. – Ben Crowell Jul 20 '11 at 3:55

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I don't understand the question. What do you mean with 'radioactive decay' (rather than just decay)? Do you mean a specific sort of coupling with that? Or do you just mean whether they are stable? Or if they can be composite? – WIMP Jul 20 '11 at 10:11

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I agree that the word "radioactive" is superfluous in the question's title. For clarity I think a better title would be 'Can massless particles decay?' but of course the choice is yours. – qftme Jul 20 '11 at 16:55
From a more formal point of view, the question would be why can't the pole of a propagator be located on the imaginary axis? What happens? I don't know the answer. – user1631 Jul 20 '11 at 18:08

## 3 Answers

"Is there any fundamental (and preferably simple) argument that makes decay of massless particles implausible?"

For decay into massive particles: I don't think you need any more than Special Relativity to answer this:

Massless particles necessarily travel at the speed of light. Therefore, even if they were unstable, they cannot decay in any reference frame. This is ultimately due to nature of time dilation in Special Relativity. In a sense, one could say massless particles (which must travel at $c$) do not experience time and therefore cannot decay.

In more QFT-type language:

For simplicity, consider a tree-level diagram of a hypothetical photon decay: $\gamma\rightarrow e^+e^-$

the translational-invariance of the vacuum means that momentum conservation disallows this transition; since the three-momentum of the virtual pair must be zero whilst that of the photon cannot be zero (since it travels at $c$ in all reference frames.)

Note however that in Feynman diagrams such as for: $e^+e^-\rightarrow e^+e^-$ via a virtual photon, $\gamma^*$, exchange the situation is somewhat different. A virtual photon has a finite mass, and thus a rest frame, so the three-momentum is conserved at each vertex. This however, should not be refered to as the decay of a photon and does not mean that a real photon can decay.

For decay into massless particles: ?

I realise this only addresses part of the question but if no one else contributes a more all-encompassing answer I will endeavour to expand on it once I've done a bit more research.

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4
Kinematical considerations only forbid the decay of a massless particle into massive products. The question is whether one can have processes of the from $\gamma \rightarrow \gamma +\gamma +\gamma$ or something similar in gravity. I believe one has processes like this in non-linear optics - the question would be whether Lorentz invariance necessarily removes them. – BebopButUnsteady Jul 20 '11 at 17:18
@Bebop: Good point, not sure why I only considered decay to massive particles. I'll edit the answer accordingly. – qftme Jul 20 '11 at 17:37
if they are colinear spin should not allow gamma->gamma gamma – anna v Jul 20 '11 at 18:45
@anna v: As it says in the linked paper, in actual QED all processes with a photon decaying to any number of photons are not allowed. I gather that the question is how generic is this behavior, and in particular whether it applies to gravity. – BebopButUnsteady Jul 20 '11 at 19:07
– anna v Jul 21 '11 at 3:43

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.

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Great post, thanks! Since gluons are confined, point 8 about violation of the third law of thermo seems to be resolved in the case of gluons. What about point 2? If gluons split at low energies with a lifetime proportional to energy, what fixes the constant of proportionality, which has units of mass^-2? It still seems disappointing to me that there is no overarching solution to these problems, just special cases that act different: the photon can't decay for technical reasons, the gluon doesn't violate the third law because it's confined, ... and gravitons can get away with murder? – Ben Crowell Jul 20 '11 at 23:43
@Ben Crowell gravitons are bosons too. – anna v Jul 21 '11 at 3:47
I'm not sure I understand point no. 8.  If we are talking about bosons, they should not behave like a classical gas, because they obey, well, boson statistics.  I'm also not sure how you reach your formula for the temperature, that would be interesting to see. – jdm Jul 21 '11 at 6:41
@Ben: morally, gravity is just a (lot more) complicated brother of QCD. It also has an asymptotically free range where we can treat it perturbatively (this is quantized linearized gravity on some background) while it (in a sense) confines at lower energies -> space-time itself is made of confined gravitons. Of course, take everything above with a grain of salt since we don't yet have full theory of quantum gravity (though you must've known what you're walking into with this question...). – Marek Jul 21 '11 at 7:41
@jdm: The equation for the temperature is just a rough estimate. The possible states of the particle in the box are labeled by partitions of the integer n=E/Eo into k integers, ignoring order. The entropy is the log of the number of such partitions, which is maximized when k~sqrt(n). When we have sqrt(n) particles, the average energy per degree of freedom is E/sqrt(n)=E/sqrt(E/Eo)=sqrt(EEo)=sqrt(hE/L). – Ben Crowell Jul 21 '11 at 16:18
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Spontaneous parametric down-conversion has the required properties of a decay $\gamma\to\gamma\gamma$:

... to split photons into pairs of photons that, in accordance with the law of conservation of energy, have combined energies and momenta equal to the energy and momentum of the original photon...

The process was independently discovered by two pairs of researchers in the late 1980s: Yanhua Shih and Carroll Alley, and Rupamanjari Ghosh and Leonard Mandel.
R. Ghosh and L. Mandel, "Observation of Nonclassical Effects in the Interference of Two Photons", Phys. Rev. Lett. 59, 1903 (1987)

IMO, can also be qualified as decays and I am aware that are other interpretations, the $\gamma\to e^+e^-$ with the help of a nucleus/neutron (a catalyst), and $\gamma\gamma\to e^+e^-$ in headon collisions are reported.

The gamma ray then collides with four or more laser photons to produce an electron-positron pair

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Happens in an external field, no? Then it is not the process under discussion here, as there are many way for a photon to split in the presence of a third body to take up the non-conserved quantities. The question even eludes to this. – dmckee♦ Feb 9 at 16:44
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