Chemical potential equilibrium cosmology

[center o bass]

In Dodelson's "Introduction to Modern Cosmology" at p. 61 he introduces a non- equilibrium number density
$$n_i = g_i e^{\mu_i/T} \int \frac{d^3p}{(2\pi)^3} e^{-E_i/T}$$
and an equilibrium number density
$$n_i^{(0)} = g_i \int \frac{d^3p}{(2\pi)^3} e^{-E_i/T},$$
from which it follows that the equilibrium number density, has the same form as number density of non-equilibrium just with a chemical potential of zero.

Question: Why does the chemical potential vanish for the equilibrium case?

 

[ChrisVer]

ask yourself: What is the chemical potential? (that's a thermodynamics question).

As a fast answer: that's because in equllibrium there is not preferred direction of the reaction. So by its definition the chemical potential will vanish.

 

[center o bass]

ask yourself: What is the chemical potential? (that's a thermodynamics question).

As a fast answer: that's because in equllibrium there is not preferred direction of the reaction. So by its definition the chemical potential will vanish.

The chemical potential arises in the first law of thermodynamics as
$$dE = TdS -pdV + \mu dN$$
and hence it represents the energy added to the system when one changes the particles in the system at constant entropy and volume. I don't see why the chemical potential has to vanish in equilibrium: if two subsystems of a larger system is in equilibrium, their chemical potentials must equal, but not vanish.

 

[ChrisVer]

The chemical potential is gives you the change of the free energy as you change the number of particles:
[itex] \mu := \frac{\partial E}{\partial N}\Big|_{S,V}[/itex]
Now in equlibrium, the free energy should be minimum (otherwise work can be produced along a reaction's direction- so you get a preferable arrow).