Why massless particles have zero chemical potential?
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$\begingroup$ 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 $ => $\mu = 0$ $\endgroup$– user346Commented Mar 11, 2011 at 14:52
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11$\begingroup$ @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... $\endgroup$– MarekCommented Mar 11, 2011 at 15:01
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2$\begingroup$ @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. $\endgroup$– MarekCommented Mar 11, 2011 at 15:20
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1$\begingroup$ @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. $\endgroup$– MarekCommented Mar 11, 2011 at 15:26
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1$\begingroup$ @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? $\endgroup$– MarekCommented Mar 11, 2011 at 17:42
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3 Answers
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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.
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3$\begingroup$ 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. $\endgroup$ Commented Mar 11, 2011 at 17:45
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1$\begingroup$ @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? $\endgroup$– user7757Commented Apr 25, 2014 at 14:02
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2$\begingroup$ 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? $\endgroup$ Commented Apr 26, 2014 at 15:12
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$\begingroup$ @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) $\endgroup$– user7757Commented Apr 26, 2014 at 15:53
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$\begingroup$ (contd)..formalism later after quantum mechanics was developed? $\endgroup$– user7757Commented Apr 26, 2014 at 15:54
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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 $$\mu=0$$.
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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.
