Lecture 21
Dynamics of the Universe
Objectives:
• The Friedmann equations
Reading: Schutz 12; Hobson 14; Rindler 16
21.1
Friedmann’s equation
The evolution of the Universe in GR is determined as follows:
1. The FRW interval =⇒ the metric, e.g. grr = −R2 /(1 − kr2 )
2. The metric =⇒ Γα βγ , the connection.
3. The metric and connection =⇒ Rαβ , the Ricci tensor.
4. The Ricci tensor and field equations =⇒ differential equations for the
size factor R and the fluid density ρ.
Jumping straight in at step 4, consider
R̈
Rtt = 3 .
R
Use field equations in the form
Rαβ
¶
1
= k Tαβ − T gαβ .
2
µ
Assume perfect fluid:
³
p´
Tαβ = ρ + 2 Uα Uβ − pgαβ .
c
85
See handout 7
86
LECTURE 21. DYNAMICS OF THE UNIVERSE
Fluid is static in co-moving coordinates of FRW metric so U i = 0 and
gαβ U α U β = gtt U t U t = c2 ,
so since gtt = c2 , U t = 1. Hence
Ut = gtt U t = c2 ,
and
³
p´
Ttt = ρ + 2 c4 − pc2 = ρc4 ,
c
while
³
³
p´
p´
T = gαβ T αβ = ρ + 2 gαβ U α U β − pgαβ g αβ = ρ + 2 c2 − 4p = ρc2 − 3p.
c
c
Therefore
¶
µ
1 2
R̈
4
2
3 = k ρc − (ρc − 3p)c .
R
2
Putting k = −8πG/c4 we obtain
4πG
R̈ = −
3
µ
3p
ρ+ 2
c
¶
R.
(21.1)
This is the acceleration equation.
Similarly the rr, θθ and φφ components all lead to:
Ṙ2 + kc2 =
8πG 2
ρR .
3
(21.2)
This is the Friedmann equation.
Finally, taking the time derivative of the Friedmann equation and substitutProve this
ing for R̈ from the acceleration equation it is simple to show:
ρ̇ +
3Ṙ ³
p´
ρ + 2 = 0.
R
c
(21.3)
which is the fluid equation. Alternatively this comes from T αβ ;α = 0.
21.1.1
Newtonian interpretation
Each of Eqs 21.1, 21.2 and 21.3 has an approximate Newtonian interpretation. If one considers an expanding uniform density sphere then
R̈ = −
4πG
ρR.
3
LECTURE 21. DYNAMICS OF THE UNIVERSE
87
There is no Newtonian explanation for the pressure term in the acceleration
equation. Conserving energy for a particle on the edge of such a sphere gives:
1 2 4πG 2 E
ρR = .
Ṙ −
2
3
m
Newtonian equivalent for curvature term kc2 is total energy per unit mass.
Finally the fluid equation follows directly from
T dS = dU + p dV,
setting dS = 0 (reversible adiabatic, no temperature gradients) and using mass–energy equivalence. Such Newtonian interpretations are a fudge:
Eqs 21.1, 21.2 and 21.3 are relativistic.
21.2
The cosmological constant
In 1917 Einstein modified the field equations to read
1
Rαβ − Rg αβ + Λg αβ = kT αβ ,
2
where Λ is the cosmological constant. Still satisifes T αβ ;α = 0 since g αβ ;γ = 0.
Nowadays, it is usual to place the new term on the right as the stress–energy
tensor of the vacuum.
¶
µ
Λ αβ
1 αβ
αβ
αβ
.
R − Rg = k T − g
2
k
Second term in brackets on the right has the form of a perfect fluid
³
pΛ ´ α β
ρΛ + 2 U U − pΛ g αβ ,
c
if
ρΛ +
and
pΛ =
pΛ
= 0,
c2
Λ
Λc4
=−
,
k
8πG
and thus
Λc2
.
8πG
i.e. a fluid of constant density and negative pressure.
ρΛ =
This is “dark
energy”, perhaps
the most puzzling
problem in
modern physics.
LECTURE 21. DYNAMICS OF THE UNIVERSE
21.2.1
88
Einstein’s static universe
Negative pressure allows a static Universe. From
µ
¶
3p
4πG
ρ + 2 R,
R̈ = −
3
c
R̈ can be zero if
3p
= 0.
c2
Here ρ and p are the sums of contributions from all components. Considering
matter and Λ only, for matter pM ¿ ρM c2 so
ρ+
ρ+
3pΛ
3p
≈ ρM + ρΛ + 2 = ρM − 2ρΛ .
2
c
c
Thus
4πG
(ρM − 2ρΛ ) R,
3
which is zero if ρM = 2ρΛ . This is Einstein’s static universe. Unfortunately
it would not be static for long since it is unstable. Consider a perturbation
ρM = 2ρΛ + ρ′ , R = R0 + R′ . To first order
R̈ = −
R̈′ = −
4πG ′
ρ R0 .
3
If R′ > 0 we expect ρ′ < 0 since matter is diluted as the universe expands,
hence R̈′ > 0 and the perturbation will grow =⇒ instability. The universe
either contracts or expands away from R = R0 .
Λ therefore can give a static but not a stable universe. Had Einstein realised
this, he could have predicted an expanding or contracting universe. Perhaps
this was why he once referred to the cosmological constant as “my greatest
blunder” (as quoted by Gamow, 1970).