# Chemical potential equilibrium cosmology

1. Jun 6, 2015

### 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?

2. Jun 6, 2015

### 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.

Last edited: Jun 6, 2015
3. Jun 7, 2015

### center o bass

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 dont 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.

4. Jun 7, 2015

### ChrisVer

The chemical potential is gives you the change of the free energy as you change the number of particles:
$\mu := \frac{\partial E}{\partial N}\Big|_{S,V}$
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).