Physical Foundations of Cosmology

7.3 Hydrodynamical perturbations 307

perturbations with the help of (7.51). HereδSis constant because the entropy per
baryon is conserved.
First we need to determine the parameterτ≡(∂p/∂S)εentering (7.51). The
cold baryons do not contribute to the pressure and hence the fluctuations of the
total pressure are entirely due to the radiation:

δp=δpγ=^13 δεγ. (7.81)

In turn,δεγcan be expressed in terms of the total energy density perturbation

δε=δεγ+δεb (7.82)

and the entropy fluctuationδS.Because the energy density of the radiation,εγ,is
proportional toTγ^4 andεb∝nb,we haveS∝ε^3 γ/^4 /εband therefore

δS

S

=

3

4

δεγ

εγ

−

δεb

εb

. (7.83)

Solving (7.82) and (7.83) forδεγin terms ofδεandδS,and substituting the result
into (7.81) we obtain

δp=

1

3

(

1 +

3

4

εb

εγ

)− 1

δε+

1

3

εb

(

1 +

3

4

εb

εγ

)− 1

δS

S

. (7.84)

Comparing this expression with (7.50), we can read off the speed of soundcsand
τ:

c^2 s=

1

3

(

1 +

3

4

εb

εγ

)− 1

,τ=

c^2 sεb

S

. (7.85)

ForδS= 0 ,the general solution of (7.51) is the sum of a particular solution and
a general solution of the homogeneous equation (δS=0). To find a particular
solution, we note that

2 H′+

(

1 + 3 cs^2

)

H^2 = 8 πGa^2

(

c^2 sε−p

)

= 2 πGcs^2 εb.

Substituting this expression andτfrom (7.85) into (7.51), we immediately see that
for thelong-wavelength perturbations,for which theterm can be neglected,

= 2

δS

S

=const (7.86)

is a particular solution of this equation.
Physically, the general solution of (7.51), whenδS= 0 ,describes a mixture
of adiabatic and entropy modes. How to distinguish between them is a matter of
definition. Based on the intuitive idea that in the very early universe the entropy