Special and General Relativity Lecture Notes:
Day 19 (11/18/08)
Shawn Mitryk and Steve Hochman
11/18/08
Contents
1 Kerr Metric
1.1 Where does g tt go to ∞? . . . . . . . . . . . . . . . . . . . . . .
1.2 Killing Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
2
2
2 Next Class
2.1 Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
4
1
1
Kerr Metric
1.1
Where does g tt go to ∞?
Starting with the Kerr Metric:
ds2
=
∆
=
2GM r 2 4GM arsin2 θ
ρ2 2
]dt
−
(dtdφ)
+
dr
ρ2
ρ2
∆
sin2 (θ) 2
[(r + a2 )2 − a2 ∆ sin2 (θ)]dφ2
+ρ2 dθ2 +
ρ2
r2 − 2GM r + a2
−[1 −
2
(2)
2
=
r + a cos θ
(3)
J
=
aM
(4)
ρ
2
(1)
2
Inverting the Matrix we can find g tt :
g tt
−((r2 + a2 )2 − a2 ∆ sin2 θ)
ρ2 ∆
=
(5)
This goes to ∞ at ∆ = 0 thus we can solve:
r2 − 2GM r + a2 = 0
p
= M ± M 2 − a2
∆
=
r±
(6)
(7)
We can also solve for:
g tφ
=
g φφ
=
2GM r
2a(∆ − (r2 + a2 ))
=
ρ2 ∆
ρ2 ∆
2
2
∆ − a sin θ
ρ∆ sin2 θ
2−
(8)
(9)
Despite that all of these go to ∞ at ∆ = 0 there exists a component that
remains finite (On the r± horizon)
Expanding out g µν ∂x∂ µ ∂x∂ ν :
g µν
∂ ∂
∂xµ ∂xν
=
−
∂
∂ 2
((r2 + a2 ) ∂t
− a ∂φ
)
ρ2 ∆
+[
∂
a sin2 θ ∂t
−
ρ2 sin2 θ
∂
∂φ 2
]
(10)
Note as a → 0 this becomes the Schwarzschild Solution.
1.2
∂
∂φ
Killing Vectors
∂
and ∂t
are killing vectors.
Writing out the components explicitly:
∂
∂t
∂
∂φ
=
T µ = (1, 0, 0, 0)
= Tµ = (−(1 −
2M r
−4GM ra sin2 θ
),
0,
0,
)
ρ2
ρ2
2
(11)
(12)
Calculating the scalar T µ Tµ :
T µ Tµ
2M r
)<0
ρ2
= −(1 −
(13)
for (∆ − a2 sin2 θ) > 0
∂
∂
Note that any α ∂t
+ β ∂φ
is a killing vector for α and β constant.
∂
∂
2
Consider the numerator (r + a2 ) ∂t
− a ∂φ
in the diverging component of Eq.
(10).
At a fixed r, this points in the direction of a killing vector but it is not a
killing vector unless the coefficients are constant.
Evaluating this at r+ :
∂
∂
2
(r+
+ a2 ) ∂t
− a ∂φ
2 + a2 )
(r+
2
r+
a
+ a2
=
∂
a
∂
+ 2
∂t r+
+ a2 ∂φ
(14)
=
ΩH
(15)
This ΩH can be interpreted as the angular momentum of the horizon.
If we throw something into a black hole (radially from ∞):
M
= M + δM
A
=
(16)
16π(M + δM )2
(17)
Thus δM > 0 and δAH > 0.
Considering Kerr Black Holes:
Given
AH
AH
=
2
4π(r+
+ a2 )
=
2
16πG
(18)
2
Mirreducable
whereδMirr
>0
(19)
Plugging in r± gives:
r
AH
=
=
ΩH
√
2
8πG [M +
Thus as we increase M, AH
Finding δAH :
δAH
J2
]
(20)
G2
increases and as we reduce J, AH increases.
2
M4 −
8πGa
8π
=
(δM − ΩH δJ) > 0
2
2
2
κ
G M −a
(21)
Penrose considered throwing a mass into the black hole with an angular
momentum opposite to that of the spinning black holes. This decreases the
angular momentum but it increases the Area.
Gravitational Radiation (spin 2) carries away both spin and mass.
Next we consider whether we can throw radiation (w/ entropy) into a black
hole thus reducing the systems entropy and violate thermodynamics.
A
Consider δ(S + 4Gh̄
) > 0 which combines the entropy derived from QFT
with the classical Area and Gravitational constant but still holds true.
This implies the temperature TBH ∝ κ = (8πM )−1 .
3
2
Next Class
2.1
Reading
• Chapter 7 in Spacetime and Geometry (Carroll)
4