Physical Foundations of Cosmology

86 The hot universe

To estimate the number density of ultra-relativistic bosons we setm=μb=0in
(3.27) and then obtain

nb

ζ( 3 )

π^2

gT^3 , (3.50)

whereζ( 3 )≈ 1. 202 .From (3.49) and (3.50) one might be tempted to conclude
that at high temperatures the excess of particles over antiparticles is always small
compared to the number density of the particles themselves. This conclusion is
wrong, however. The expression in (3.49) is applicable only ifμb<m.Asμb→m,
new particles added to the system fill the minimal energy state=m,which is
not taken into account in (3.49). These particles form a Bose condensate which can
have an arbitrarily large particle excess.

Problem 3.10Given a particle excess per unit volumen,find the temperatureTB
below which the Bose condensate forms. Assume thatTBmand determine when
this condition is actually satisfied. How much does a Bose condensate contribute
to the total energy density, pressure and entropy?

If no Bose condensate is formed, the excess of bosons over antibosons is small
compared to the number density. In this case the energy densities of particles and
antiparticles are nearly equal and it follows from (3.35) and (3.45) that

εb

εb+εb ̄

2



π^2

30

gT^4. (3.51)

The pressure and the entropy density are

pb

εb

3

, sb

4

3

εb

T

=

2 π^4

45 ζ( 3 )

nb, (3.52)

respectively. For massless bosons the chemical potential should be equal to zero.
In this case (for example, for photons) all equations above are exact.

FermionsThe chemical potential for fermions can be arbitrarily large and can
exceed the mass. We first derive theexactformulae for anarbitraryμfin the limit
of vanishing mass.Takingα→0 in (3.44) and (3.45) and substituting the result
into (3.35), we obtain

εf+εf ̄=

7 π^2

120

gT^4

[

1 +

30 β^2

7 π^2

+

15 β^4

7 π^4

]

, (3.53)

whereβ=μf/T.The pressure is equal to one third of the energy density as
expected for massless particles. It follows from (3.37) that the excess of fermions