78 The hot universe
μ≡λ 2 T(kB=1). The distribution function (3.19) then takes the form
n=1
exp((−μ)/T)− 1. (3.20)
This spectrum describes bose particles in a state of maximal possible entropy and is
known as the Bose–Einstein distribution. A similar derivation can be carried out for
fermi particles, the only difference being that we have to take into account the Pauli
exclusion principle, which forbids two fermions from simultaneously occupying
the same microstate.
Problem 3.4Derive the following expression for the entropy of fermi particles:
S({n})=∑
[(n− 1 )ln( 1 −n)−nlnn]g, (3.21)and show that it takes its maximal value for
n=1
exp((−μ)/T)+ 1. (3.22)
Problem 3.5According to (3.20) and (3.22), the energy of a single particlecan, in
principle, be larger than the total energy of the whole systemE,which contradicts
our assumptions. Where does the above derivation fail forcomparable to or larger
thanE?
In quantum field theory particles can be created and annihilated, so their total
number is generally not conserved. In this case the number of particles in equilib-
rium is determined solely by the requirement of maximal entropy for a given total
energy. This removes the need to satisfy the second constraint (3.18). If there are
no other constraints enforced by conservation laws, then the chemical potentialμ
is zero and there remains only one free parameter,λ 1 , to fix the total energy. For
example, the total number and the temperature of photons are entirely determined
by their total energyE.
Because of the conservation of electric charge, electrons and positrons can be
produced only in pairs. Therefore, the difference between the numbers of electrons
and positronsNe−−Ne+does not change. With this extra constraint, the Lagrange
variational principle takes the form
δ[
S
({
ne
−
})
+S
({
ne
+
})
+λ 1 (Ee−+Ee+)+λ 2 (Ne−−Ne+)]
= 0 , (3.23)
where we vary separately with respect tone
−
andn
e+
.It is not hard to show that
the variation vanishes only if the electrons and positrons both satisfy the Fermi dis-
tribution (3.22) withT=− 1 /λ 1 andμe−=−μe+=Tλ 2 .Thus, the chemical po-
tentials of the electrons and positrons are equal in magnitude and have the opposite