196
12 The Einstein Field Equations for Cosmology
In the presence of a pressure gradient an element of fluid will feel a force and be
accelerated, so its momentum will change. We therefore expect that the conservation
of momentum (12.7) should be modified to have a pressure gradient on the right
side. In fact the equation should become Newton’s second law for a fluid with density
ρ in the classical limit
m
a =
F → ρ
d v
dt
= −∇ p.
(12.8)
The time derivative is again the Euler derivative used in (12.7). (See Exercise 12.1).
This is the fundamental force equation of classical fluid flow. Our task in this section is
to modify the energy-momentum tensor so that it yields the dynamical equation (12.8)
replacing (12.7) for a zero-pressure fluid. This will lead to the energy-momentum
tensor to be used in the following chapters.
Our approach is thus to add to the energy-momentum tensor for dust a pressure term which will give the dynamical equation (12.8) in the classical limit. The
procedure is analogous to that which led to (12.7). The obvious quantities available to construct such a tensor are the density ρ and pressure p, which we assume
are scalars, the tensor u
α u
β , and the metric tensor g
αβ . We accordingly assume the
energy-momentum of the perfect fluid is
T
αβ
= ρu
α u
β
+ p
au
α u
β
+ bg
αβ
= (ρ + ap)u
α u
β
+ bpg
αβ
(12.9)
with a and b constants to be determined. In the flat space limit the metric tensor is
g
αβ
= η
αβ . As with the dust tensor we calculate the divergence of this tensor and
set it equal to zero as in (12.2). For α = 0 we get an equation quite analogous to the
conservation of mass (12.4).
T
0ν ,ν = T
00 ,0 + T
0i ,i
=
1
c
∂
∂t
(ρ + (a + b) p) +
∂
∂ x i
(ρ + ap)v
i
= 0.
(12.10)
It is immediately clear that for this to be consistent with the conservation of mass or
energy in (12.4) we must have both (a + b) p and ap much less than ρ.
For α = j we get an equation analogous to the momentum equation (12.5),
T
jν ,ν =
1
c 2
∂
∂t
(ρ + ap)v
j
+
∂
∂ x k
(ρ + ap)v
j
v
k
− bpc
2
δ
jk
= 0. (12.11)
The same manipulations with the product rule as used with (12.5) gives
v
j ∂
∂t
(ρ + ap) + (ρ + ap)
∂v
j
∂t
+ v
j ∂
∂ x k
(ρ + ap)v
k
+ (ρ + ap)v
k ∂v
j
∂ x k
12 The Einstein Field Equations for Cosmology
In the presence of a pressure gradient an element of fluid will feel a force and be
accelerated, so its momentum will change. We therefore expect that the conservation
of momentum (12.7) should be modified to have a pressure gradient on the right
side. In fact the equation should become Newton’s second law for a fluid with density
ρ in the classical limit
m
a =
F → ρ
d v
dt
= −∇ p.
(12.8)
The time derivative is again the Euler derivative used in (12.7). (See Exercise 12.1).
This is the fundamental force equation of classical fluid flow. Our task in this section is
to modify the energy-momentum tensor so that it yields the dynamical equation (12.8)
replacing (12.7) for a zero-pressure fluid. This will lead to the energy-momentum
tensor to be used in the following chapters.
Our approach is thus to add to the energy-momentum tensor for dust a pressure term which will give the dynamical equation (12.8) in the classical limit. The
procedure is analogous to that which led to (12.7). The obvious quantities available to construct such a tensor are the density ρ and pressure p, which we assume
are scalars, the tensor u
α u
β , and the metric tensor g
αβ . We accordingly assume the
energy-momentum of the perfect fluid is
T
αβ
= ρu
α u
β
+ p
au
α u
β
+ bg
αβ
= (ρ + ap)u
α u
β
+ bpg
αβ
(12.9)
with a and b constants to be determined. In the flat space limit the metric tensor is
g
αβ
= η
αβ . As with the dust tensor we calculate the divergence of this tensor and
set it equal to zero as in (12.2). For α = 0 we get an equation quite analogous to the
conservation of mass (12.4).
T
0ν ,ν = T
00 ,0 + T
0i ,i
=
1
c
∂
∂t
(ρ + (a + b) p) +
∂
∂ x i
(ρ + ap)v
i
= 0.
(12.10)
It is immediately clear that for this to be consistent with the conservation of mass or
energy in (12.4) we must have both (a + b) p and ap much less than ρ.
For α = j we get an equation analogous to the momentum equation (12.5),
T
jν ,ν =
1
c 2
∂
∂t
(ρ + ap)v
j
+
∂
∂ x k
(ρ + ap)v
j
v
k
− bpc
2
δ
jk
= 0. (12.11)
The same manipulations with the product rule as used with (12.5) gives
v
j ∂
∂t
(ρ + ap) + (ρ + ap)
∂v
j
∂t
+ v
j ∂
∂ x k
(ρ + ap)v
k
+ (ρ + ap)v
k ∂v
j
∂ x k