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Why massless particles have zero chemical potential?

Andy Bale
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  • E = mc^2 = 0. So adding or removing a massless particle from a system does not change the internal energy of the system $dE = \mu dN = 0 $ => $\mu = 0$ –  Mar 11 '11 at 14:52
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    @Deepak: nonsensical argument. Even massless particles do carry energy (and the $E=mc^2$ formula is completely invalid in this case) and adding them changes internal energy. Not to mention that internal energy can be also changed by interactions between those particles... – Marek Mar 11 '11 at 15:01
  • @Marek do you have a better explanation? –  Mar 11 '11 at 15:02
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    @Deepak: well, in my opinion it has nothing to do with energy or being massless but instead with whether the number of particles (of the total system) is conserved. When it's not then $N$ loses its thermodynamic meaning and so does $\mu$. One typically encounters this non-conservation with massless particles (photons in box, phonons in lattice) but massive particles could also decay. – Marek Mar 11 '11 at 15:20
  • @Marek you say * ... it has nothing to do with energy or being massless but instead with whether the number of particles (of the total system) is conserved* . Well if the mass is zero then that is exactly what happens. Also if a massive particle with chemical potential $\mu_0$ decays into other massive particles with chemical potentials $\mu_i$, then $\mu_0 = \sum g_i \mu_i$, where $g_i$ is the multiplicity of the $i$th particle type in the decay result. One cannot make any such statement if an excitation is massless to begin with. –  Mar 11 '11 at 15:25
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    @Deepak: although with massive particles you probably wouldn't get equilibrium unless everything decays to lightest particles (respecting other conservation laws) and then a) the number of massive particles stabilizes; or b) everything decays to massless particles. So in the end one probably indeed gets $\mu \neq 0$ for massive particles and $\mu = 0$ for massless particles. – Marek Mar 11 '11 at 15:26
  • @Deepak, "that is exactly what happens" -> what do you mean? This happens all the time also for massive particles. There is no god given conservation of number of particles (if they don't have other charges). Therefore the discussion of chemical potential is completely orthogonal to your discussion of energy. – Marek Mar 11 '11 at 15:29
  • @Marek I mean that when the particle is massless, particle number is not conserved. –  Mar 11 '11 at 15:38
  • @Deepak: not true. It's easy to consider model where you have ideal gas (i.e. absolutely no interactions) with massless particles. And it's not just a toy model. Consider a perfectly reflective box (as an idealization) with radiation inside. The number of photons would be approximately conserved. Once again, energy and mass have nothing to do with conservation, it's totally orthogonal issue. – Marek Mar 11 '11 at 15:42
  • @Marek if the reflective box is allowed to come to equilibrium, the photons within will have $\mu = 0$ (see @dbrane's answer below). In fact one could assert that unless the energies of photons exceed 1 Mev (the amount required for $e^-,e^+$ pair-creation), the chemical potential will be zero even for a gas away from equilibrium. Once pair creation becomes a possibility one might have to assign a non-zero $\mu$ to the gas. Look, I agree that there can be situations when photons could have $\mu \ne 0$. But it seems we're splitting hairs, here. –  Mar 11 '11 at 17:07
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    @Deepak: forget about photons for a second. Do you agree that at least theoretically (disregarding physical significance) there is no problem with having ideal gas of massless relativistic particles with non-zero chemical potential? If you do, we're finished because this shows that chemical potential is unrelated to being massless (and in particular proves your answer invalid). If not, what is your problem with this model? – Marek Mar 11 '11 at 17:42
  • @Marek there is no end to arguing with you. We have both made our viewpoints quite clear. I think the readers can make their own informed decisions. –  Mar 12 '11 at 05:56
  • @Deepak: well, I am not really arguing. I am constructively pointing holes in your argument (supported by up-votes of my comments) and as of now you haven't addressed anything but as usual you rather prefer talking about how I should not argue with you. Whatever suits you the best... :) – Marek Mar 12 '11 at 07:25
  • As I said - there really is no end ... –  Mar 12 '11 at 08:31

3 Answers3

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Massless particles don't always have zero chemical potential. Suppose that you have a box full of photons and other particles, and it's possible for the photons to exchange energy with other particles, but the number of photons cannot change. Then the system will reach a thermal equilibrium in which the photons are described by a Bose-Einstein distribution with (in general) a nonzero chemical potential.

The reason this doesn't usually happen with photons is that the number of photons is often not conserved in situations like this. If there are photon-number-changing processes, then the equilibrium state for the photons will have zero chemical potential (since otherwise entropy could go up by creating or destroying a photon).

In summary, the rules are that the chemical potential must be zero if particle-number-changing interactions are possible, but not otherwise. That distinction often coincides with the massless or massive nature of the particles, but not always.

By the way, there was a period of time in the early Universe when we were in precisely this situation: photons could thermalize via Compton scattering with electrons, but at the temperature and density at the time, photon-number-changing processes essentially did not occur. That means that the cosmic microwave background radiation today could have a nonzero chemical potential. People have tried to measure the chemical potential, but as it turns out it's consistent with zero to quite good precision. This makes sense, as long as the photons and electrons came into thermal equilibrium at an earlier epoch (when photon-number-changing processes did occur), and nothing happened during the later epoch to mess up that equilibrium. Various theories in which particle decays inject energy into the Universe during the constant-photon-number epoch are ruled out by this observation.

Ted Bunn
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    Obviously, this is the correct answer. Zero chemical potential $\mu$ means that its dual variable, $N$, is not conserved, so the ensemble is not allowed to punish states with different values of $N$. All the exponentials, including $\exp(-E/kT)$, in the distribution always have an exponent proportional to the conserved quantity. That's how they're derived by maximizing the number of microstates while keeping conserved quantities fixed. – Luboš Motl Mar 11 '11 at 17:45
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    @LubošMotl: If I consider just a photon gas, doesn't the non conservation of photon number contradict conservation of energy, as each photon has Energy$\hbar \omega$? Please could you explain why photons don't obey conservation of number? –  Apr 25 '14 at 14:02
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    Hi @ramanujan_dirac, the energy of the photon is going somewhere else, or from somewhere else, like it excites atoms, makes it decay and speed up the products, and so on, and so on. Free photons are not the only carriers of energy, are they? – Luboš Motl Apr 26 '14 at 15:12
  • @LubošMotl: Thanks for the reply. Now I understand it better. In one of your other answer you have remarked that how Boltzmann while working in classical statistical mechanics, was unknown to him using intuition from quantum statistical mechanics. Could you elaborate on how this is so? AFAIK, this is true because you can see things like indistinguishability and particle statistics from classical statistical mechanics. Is there any other such example? Also, was the semi-classical quantization (breaking the phase space into discrete parts)donecby Boltzmann,or was it added in the formalism(contd) –  Apr 26 '14 at 15:53
  • (contd)..formalism later after quantum mechanics was developed? –  Apr 26 '14 at 15:54
  • What I meant was that he pioneered the framework to answer questions probabilistically - and he even realized that physics should be not about "how things are" but "what is the probability that a proposition is correct" - not necessarily assuming that a full description exists. He would also realize that the volume of the phase space would ultimately behave as a dimensionless number (in QM, an integer). But he lived before the era of quantum mechanics so of course he couldn't have found none of the actual properties of quantum mechanics in its own language. – Luboš Motl Apr 26 '14 at 15:58
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Photons require matter in order to come into thermal equilibrium (if you ignore the negligible contribution from photon-photon scattering). This means that the particle number is not conserved. This further implies that in order for the free energy to be minimum (for a given temperature and volume), you also need $$\partial F/\partial N=0$$ because now $N$ can vary too. Since $$(\partial F/\partial N)_{T,V}=\mu$$ we have $$\mu=0$$.

dbrane
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The short answer: if the system is in thermal equilibrium the photons are absorbed/emitted and do not change the internal energy of the system. The chemical potential is zero. In all other occasions the absorption/emission of a photon (say single photon e/a) changes the internal energy and the photon has chemical potential.

Alexei
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