General Relativity 10/16/2008 Prof.Whiting’s Lecture Note
October 23, 2008
In the previous class we derived
·
¸
1 2φ d
−2Λ
G00 =
e
r(1
−
e
)
= 8πGN T00 = 8πGN (ρe2φ ),
r2
dr
1
2
G11 = − 2 e2Λ (1 − e−2Λ ) + φ0 = 8πGN T11 = 8πGN (pe2Λ ),
r
r
·
¸
φ0
Λ0
00
G22 = r2 e−2Λ φ + (φ0 )2 +
− φ0 Λ0 −
= 8πGN T22 = 8πGN (r2 p).
r
r
(1)
(2)
(3)
For static and spherically symmetric matter, we can write the metric as
ds2 = −e2φ dt2 + e2Λ dr2 + r2 dΩ2 .
(4)
At rest,
uµ = (u0 , 0, 0, 0),
where u0 = eφ
(5)
Energy momentum tensor is given by
T µν
= (ρ + p)uµ uν + pg µν
µ
¶
p
p
−2φ
−2Λ
= diag ρe , pe
, 2, 2 2
.
r r sin θ
(6)
Now let us consider
∇µ T µν = 0
(7)
∇µ T µν = ∂µ T µν + Γµαµ T αν + Γναµ T µα = 0
(8)
or
For ν = 0,
∂0 T 00 + Γµ0µ T 00 + Γ0αµ T µα = 0
1
(9)
For ν = 1,
∂1 T 11 + Γµ1µ T 11 + Γ1αµ T µα = 0
(10)
For ν = 2,
∂2 T 22 + Γµ2µ T 22 + Γ2αµ T µα = 0
(11)
For ν = 3,
∂3 T 33 + Γµ3µ T 33 + Γ3αµ T µα = 0
(12)
We need to calculate Γµ1µ =
√
√1 ∂1 g
g
and Γ1αµ
Covariant derivative for V µ is
1
√
∇µ V µ = ∂µ V µ + Γµαµ V α = ∂µ V µ + √ ∂α gV α
g
(13)
Recall the affine connection coefficients
1
Γαβγ = g αµ (gµγ,β + gµβ,γ − gβγ,µ )
2
(14)
we need to calculate the following to get the equation for ν = 1
1
Γ1βγ = g 11 (g1γ,β + g1β,γ − gβγ,1 )
2
(15)
1
1
Γ100 = g 11 (g10,0 + g10,0 − g00,1 ) = − g 11 g00,1 = +e2(φ−Λ) φ0
2
2
(16)
1
Γ111 = g 11 g11,1 = +Λ0
2
(17)
1
Γ122 = − g 11 g22,1 = −e−2Λ r
2
(18)
1
Γ133 = − g 11 g33,1 = −e−2Λ r sin2 θ
2
(19)
2
2
∂1 (pe−2Λ ) + (φ0 + Λ0 + )pe−2Λ + e−2Λ (ρφ0 + pΛ0 − p) = 0
r
r
(20)
p0 + φ0 (ρ + p) = 0
(21)
Therefore, for ν = 1,
This gives
2
Now from
we get
G00 = 8πGN T00
(22)
·
¸
d
r(1 − e−2Λ ) = 8πGρr2
dr
(23)
or
·
d r(1 − e
Now define
−2Λ
¸
) = 8πGρr2 dr
(24)
Z
ρdV ≡ m(r)
(25)
(Be careful, this does not define a mass inside r.) Then we can deduce ρr2 = dm/dr as
Z
m(r) =
Z
Z
ρr2 sin θdrdθdπ = 4π
ρdV
ρr2 dr
(26)
Using (26) and (24) we get
r(1 − e−2Λ ) = 2Gm or e2Λ =
G11 = −
1
1−
2m(r)
r
,
G=1
2m(r)
1
2
8πGN p
1
(1 − 1 +
) + φ0 =
2
2m(r)
r 1−
r
r
1 − 2m(r)
r
r
(27)
(28)
From this we get
φ0 =
r(4πGN p + m(r)/r3 )
1−
2m(r)
r
(29)
Using (21) gives
p0 =
−r(4πGN p + m(r)/r3 )
1−
2m(r)
r
(ρ + p)
(30)
Now we have 4 unknowns and 3 equations, to go further give p(ρ) (for example, p = 1/3ρ
for electromagnetic radiation) or ρ(r).
Let’s try a simplified assumption ρ(r) = ρ0 , then m(r) is
Z
m(r) = ρ0 4π
4
r2 dr = πρ0 r3
3
(31)
and
p0 =
−r(4πp + 43 πρ0 )(ρ + p)
(ρ + p)
1 − 83 πρ0 r2
3
(32)
Setting R and M as 43 πρ0 R3 = M , we get
p =
eφ =
eΛ =
ρ[(1 −
2M 2 1/2
1/2 ]
r ) − (1 − 2M
R )
R3
2M 1/2
− (1 − 2M
r2 )1/2
R )
R3
3(1 −
3
2M 1/2 1
2M
(1 −
) − (1 − 3 r2 )1/2
2
R
2
R
1
1/2
(1 − 2M
R )
and also we get the metric
¸
·
1
2M 1/2 1
2M 2 1/2 2
2
2 3
+
dr2 + r2 dΩ2
ds = −dt
(1 −
) − (1 − 3 r )
2)
2
R
2
R
r
(1 − 2M
R3
Next class, we will start from discussing “what happens to light.”
4
(33)
(34)
(35)
(36)