Question

Why do we need to suppose the chemical potential is zero in this situation?

Answer 1

In the photon gas the chemical potential is always zero. For other cases, I'm not sure why your lecturer said it always has to be nonpositive. It istrue that the chemical potential has to be less than the energy of the lowest level (less than the lowest εε). Depending on potentials and such, that lowest energy may be positive or negative. 

Answer 2

I've been working on some statistical mechanics problems and one of them asks to compute the pressure with chemical potential zero of a boson gas whose particles do not interact and whose energies are given by ϵ=ℏckϵ=ℏck where kk is the magnitude of the wave vector.

To do so, I've used the fact that the grand thermodynamic potential ΦΦ is given by

Φ=−1βlnΞΦ=−1βln⁡Ξ

and that Φ=−PVΦ=−PV. In that case, I've computed lnΞln⁡Ξ as

lnΞ=−∑jln[1−exp(−β(ϵj−μ))]ln⁡Ξ=−∑jln⁡[1−exp⁡(−β(ϵj−μ))]

which on the thermodynamic limit may be written as an integral

lnΞ=−γV∫∞0D(ϵ)ln[1−exp(−β(ϵ−μ))]dϵ,ln⁡Ξ=−γV∫0∞D(ϵ)ln⁡[1−exp⁡(−β(ϵ−μ))]dϵ,

with D(ϵ)=Cϵ2D(ϵ)=Cϵ2. Then I've shown that on the thermodynamic limit

U=γV∫∞0ϵD(ϵ)dϵexp(β(ϵ−μ))−1U=γV∫0∞ϵD(ϵ)dϵexp⁡(β(ϵ−μ))−1

and that lnΞ=βU/3ln⁡Ξ=βU/3. Then we have Φ=−U/3Φ=−U/3 and −PV=−U/3−PV=−U/3 which gives the expected result P=13UVP=13UV.

Now, I've not used that μ=0μ=0 in any step I've presented. So why the chemical potential being zero is important? In truth this should hold just for photons, but the fact that the particles are photons is already implied by the energy spectrum. So, where μ=0μ=0 should be used?

EDIT: As asked in comment, the way I computed lnΞln⁡Ξ was the following. If we have ocupation numbers {nj}{nj} then it is easy to see that

Ξ=∑n1,n2…exp(−β(ϵ1−μ)n1)exp(−β(ϵ2−μ)n2)⋯Ξ=∑n1,n2…exp⁡(−β(ϵ1−μ)n1)exp⁡(−β(ϵ2−μ)n2)⋯

In that case we can write this as

Ξ=∑n1exp(−β(ϵ1−μ)n1)∑n2exp(−β(ϵ2−μ)n2)⋯Ξ=∑n1exp⁡(−β(ϵ1−μ)n1)∑n2exp⁡(−β(ϵ2−μ)n2)⋯

In that way we have

Ξ=∏j∑njexp(−β(ϵj−μ)nj)Ξ=∏j∑njexp⁡(−β(ϵj−μ)nj)

If then we are dealing with bosons, njnj goes from zero to infinity, so given that |exp(−β(ϵj−μ)nj)|<1|exp⁡(−β(ϵj−μ)nj)|<1 that sum reduces to

Ξ=∏j11−exp(−β(ϵj−μ))Ξ=∏j11−exp⁡(−β(ϵj−μ))

which implies the lnΞln⁡Ξ as given.