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In Bose-Einstein condensation, the chemical potential is less than the ground state energy of the system($\mu<\epsilon_g$). But why does the massless boson such as photon have zero chemichal potential($\mu=0$)?

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marked as duplicate by Chris White, Waffle's Crazy Peanut, Qmechanic Apr 9 '13 at 22:23

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The chemical potential is a complementary variable to $N$, the number of particles (of a certain kind), and they get combined in the same sense as $-\beta,H$ and similar pairs. The chemical potential "punishes" too high or too low number of particles in grand canonical and similar distributions such as $$\exp(-\beta(H-\mu N))$$ In the derivation of similar terms in the exponential in the distribution, it's important that all the extensive quantities such as $H, N$ are conserved. You may view $\beta,\beta\mu$ and similar coefficients as Lagrange multipliers that impose the conservation of $H,N$ etc.

The distribution is maximizing the number of microscopic rearrangements given the fixed specified values of the conserved quantities such as $H,N$ etc.

However, for massless bosons, there doesn't exist any sense or approximation in which the number $N$ of these particles would be conserved. So the states with higher or lower numbers $N$ can't be punished by any $\exp(\beta\mu N)$ factor. It always takes "zero work" to change the number of these massless bosons by one. For example, it's trivial to create a photon; in fact, an accelerating charge is emitting an infinite number of photons (a source of infrared divergences in quantum field theories). The number $N$ of particles like photons isn't even finite so it's clear that the coefficient multiplying it has to be zero for the product to be well-defined.

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Thanks for your answer. Can you please explain more about"The chemical potential "punishes" too high or too low number of particles in grand canonical "? What do you mean by "punishes"? Does the system not allow too high or too low number of particles? –  Jeremy Apr 12 '13 at 2:12

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