# Why the chemical potential of massless boson is zero? [duplicate]

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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–  joshphysics Apr 9 '13 at 2:10

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

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.