3.3 Rudiments of thermodynamics 79signs as a consequence of electric charge conservation. Only if the total electric
charge of the electron–positron plasma is equal to zero do they vanish.
Problem 3.6Assume particles of typesA,B,C,Dare in equilibrium with each
other due to the reaction
A+BC+D.It is easy to see that the following combinations are conserved:NA+NC,NA+ND,
NB+NC,NB+ND.Using this fact, show that the chemical potentials satisfy the
relation
μA+μB=μC+μD. (3.24)Note that, if electrons and positrons are in equilibrium with each other and with
radiation due to the interactione−+e+γ+γ,then from (3.24) we recover the
result,μe−=−μe+,since the chemical potential of radiation is equal to zero.
The above consideration can be directly applied to matter in ahomogeneous and
isotropicexpanding universe. If the interaction rate is much larger than the rate of
expansion, the entropy of matter reaches its maximal value very quickly. In a ho-
mogeneous universe there are no external sources of entropy, and therefore the total
entropy of matter within a given comoving volume is conserved. If the interactions
of some particles become inefficient, they decouple and evolve independently and
their entropy is conserved separately. For example, after recombination photons
propagate freely and they are not in thermal equilibrium. Nevertheless, they still
have maximal possible entropy and hence satisfy the Bose–Einstein distribution
as if they were in equilibrium. A similar situation occurs for neutrinos when they
decouple from matter.
The simple arguments above are not valid when the universe becomes highly
inhomogeneous as a result of gravitational instability. For this reason the initial
state of the universe, which looks like a state of “ thermal death” where nothing
could happen, can evolve to a state where very complicated structures, such as
biological systems, occur. Nonequilibrium processes and gravitational instability
will be considered later in detail and here we concentrate on the local equilibrium
state. It is rather remarkable that in this state general arguments involving only the
entropy and conservation laws are sufficient to describe the system completely and
we do not need to use a kinetic theory or go into the details of quantum field theory.
3.3.2 Energy density, pressure and the equation of state
To calculate the energy density and pressure for a given distribution functionn,
we have to determineg,the total number of possible microstates for asingle