Lecture 13
Einstein’s field equations
Objectives:
• The GR field equations
Reading: Schutz, 6; Hobson 7; Rindler 10.
13.1
Symmetries of the curvature tensor
With 4 indices, the curvature tensor has a forbidding 256 components. Luckily several symmetries reduce these substantially. These are best seen in fully
covariant form:
Rαβγδ = gαρ Rρ βγδ ,
for which symmetries such as
Rαβγδ = −Rβαγδ ,
and
Rαβγδ = −Rαβδγ .
can be proved. These relations reduce the number of independent compoHandout 6
nents to 20.
These symmetries also mean that there is only one independent contraction
Rαβ = Rρ αβρ ,
because others are either zero, e.g.
Rρ ραβ = g ρσ Rσραβ = 0,
or the same to a factor of ±1. Rαβ is called the Ricci tensor, while its
contraction
R = g αβ Rαβ ,
is called the Ricci scalar.
52
NB Signs vary
between books. I
follow Hobson et
al and Rindler.
LECTURE 13. EINSTEIN’S FIELD EQUATIONS
13.2
53
The field equations
We seek a relativistic version of the Newtonian equation
∇2 φ = 4πGρ.
The relativistic analogue of the density ρ is the stress–energy tensor T αβ .
φ is closely related to the metric, and ∇2 suggests that we look for some
tensor
the second derivatives of the metric, gαβ,γδ , which should be
à involving
!
2
a
tensor like T αβ .
0
The contravariant form of the Ricci tensor satisfies these conditions, suggesting the following:
Rαβ = kT αβ ,
where k is some constant. (NB both Rαβ and T αβ are symmetric.)
However, in SR T αβ satisfies the conservation equations T αβ ,α = 0 which in
GR become
T αβ ;α = 0,
whereas it turns out that
1
Rαβ ;α = R,α g αβ 6= 0,
2
where R is the Ricci scalar. Therefore Rαβ = kT αβ cannot be right.
Fix by defining a new tensor, the Einstein tensor
1
Gαβ = Rαβ − Rg αβ ,
2
because then
G
αβ
;α
=
µ
R
αβ
1
− Rg αβ
2
¶
;α
1
1
= Rαβ ;α − R;α g αβ − Rg αβ ;α = 0,
2
2
since ∇g = 0 and R;α = R,α . Therefore we modify the equations to
1
Rαβ − Rg αβ = kT αβ .
2
These are Einstein’s field equations.
13.3
The Newtonian limit
The equations must reduce to ∇2 φ = 4πGρ in the case of slow motion in weak
fields. To show this, it is easier to work with an alternate form: contracting
Handout 6
LECTURE 13. EINSTEIN’S FIELD EQUATIONS
54
the field equations with gαβ then
1
gαβ Rαβ − Rgαβ g αβ = kgαβ T αβ ,
2
and remembering the definition of R and defining T = gαβ T αβ ,
1
R − δαα R = −R = kT,
2
since δαα = 4. Therefore
R
αβ
µ
=k T
αβ
¶
1 αβ
.
− Tg
2
Easier still is the covariant form:
Rαβ
µ
¶
1
= k Tαβ − T gαβ .
2
The stress–energy tensor is
³
p´
Tαβ = ρ + 2 Uα Uβ − pgαβ .
c
In the Newtonian case, p/c2 ¿ ρ, and so
Tαβ ≈ ρUα Uβ .
Therefore
T = g αβ Tαβ = ρg αβ Uα Uβ = ρc2 .
Weak fields imply gαβ ≈ ηαβ , so g00 ≈ 1. For slow motion, U i ¿ U 0 ≈ c,
and so U0 = g0α U α ≈ g00 U 0 ≈ c too. Thus
T00 ≈ ρc2 ,
is the only significant component.
The 00 cpt of Rαβ is:
R00 = Γρ 0ρ,0 − Γρ 00,ρ + Γσ 0ρ Γρ σ0 − Γσ 00 Γρ σρ .
All Γ are small, so the last two terms are negligible. Then assuming timeindependence,
R00 ≈ −Γi 00,i .
But, from the lecture on geodesics (chapter 11),
Γi 00 =
φ,i
.
c2
LECTURE 13. EINSTEIN’S FIELD EQUATIONS
55
Thus
1 ∂2φ
1
1
R00 ≈ − 2 φ,ii = − 2 i i = − 2 ∇2 φ.
c
c ∂x ∂x
c
Finally, substituting in the field equations
¶
µ
1 2
1 2
2
− 2 ∇ φ = k ρc − ρc ,
c
2
or
kc4
ρ.
2
Therefore if k = −8πG/c4 , we get the Newtonian equation as required, and
the field equations become
∇2 φ = −
1
8πG
Rαβ − Rg αβ = − 4 T αβ .
2
c
Key points:
• The field equations are second order, non-linear differential equations
for the metric
• 10 independent equations replace ∇2 φ = 4πGρ
• By design they satisfy the energy-momentum conservation relations
T αβ ;α = 0
• The constant 8πG/c4 gives the correct Newtonian limit
• Although derived from strong theoretical arguments, like any physical
theory, they can only be tested by experiment.
No longer
balancing
up/down indices
since we are
referring to
spatial
components only
in nearly-flat
space-time.