Nicole Crisosto
PHZ 6607
GR Class Notes
11/24/2014
Kerr Geometry has the metric
ds2 = −(1 −
2M r 2 4GM ar sin2 θ
ρ2 2
dr + ρ2 dθ2 + sin2 θ[Σ2 ]dφ2
)dt
−
dtdφ
+
2
2
ρ
ρ
∆
where
ρ2 = r2 + a2 cos2 θ ≥ 0
∆2 = r2 − 2M r + a2
Σ2 = (r2 + a2 ) − a2 ∆ sin2 θ ≥ 0
and Killing tensor
K = σµν U mu U ν
ρ4 ṙ2
1
(r2 + a2 )E − aLz +
− δ1 r2
K=
∆
∆
2
Lz
K = aE sin θ −
+ ρ4 θ̇2 + δ1 a2 cos2 θ.
sin θ
Here δ1 = 1 when spacelike, δ1 = 0 when null, and δ1 = −1 when timelike.
dτ
dr
dθ
=√ =√
2
ρ
R
Θ
Null Geodesics
ds2 = 0 = −gtt dt2 + 2gtφ dtdφ + gφφ dφ2
s
2
2
−gtt gφφ + gtφ
−gtφ
dφ
gtφ
gtt
−Det
=
±
−
=
= 2
2
dt
gφφ
gφφ
gφφ
gφφ
gφφ
√
4M ar ± ∆ sin θ
=
Σ2
At horizon, ∆ = 0,
a
ΩH = 2
> 0, (a > 0)
r+ + a2
The Killing vector is timelike outside the ergosphere, null on ergosphere, and spacelike inside.
Kµ =
∂
∂t
Killing vector is null just outside horizon, future pointing,
X µ = K µ + ΩH R µ
1
Outside:
−K µ pµ > 0 =⇒ going forward in time.
Inside:
−X µ pµ > 0
Penrose Process:
E (0) > 0
E (0) = E (1) + E (2)
E (2) < 0, E (1) > 0
=⇒
L2z <
E (2)
<0
ΩH
throw with negative angular momentum
δJ <
δM
=⇒ δM − ΩH δJ > 0
ΩH
Schwarzchild:
δM > 0 =⇒ δA ≥ 0
Kerr
δA ≥ 0
on horizon
2
2
A = 4π(r + a ) =
2
4π(r+
2
+ a ) = 4π (M +
√
M2
−
a2 )2
+a
√
1 2
M + M M 2 − a2
2
Mirr M
=√
(δM − ΩH δJ)
M 2 − a2
2
Mirr
=
δMirr
1
(M − Mirr ) = M 1 − √
≈ .29M
2
which comes out of rotation of black hole.
Rewritting δMirr in terms of A,
δA =
8πG
(δM − ΩH δJ)
K
δM =
kδA
+ ΩH δJ
8πG
2
2
2
= ... = 16πMirr
If
δA > 0.
In relation to thermodynamics, generalized 2nd law,
S+
Ac3
>0
4G~
√
M 2 − a2
T =
2
8πG~Mirr
Spinning up the black hole, a → M , TextremeKerrBH → 0
Adding charge,
KδA
δM =
+ ΩδJ + ΦδQ + V dΛ + µδB
8πG
can throw entropy into a black hole
1974: Hawking - blackholes can evaporate,
lifetime ∼ M 3
solar mass ∼ 1067 yrs
mp ∼ 10−5 g =⇒ planck time
mp ∼ 1015 g ∼ Mt. Everest mass =⇒ age of universe from
Semitted << SBH
Can toss in QM pure state, comes out thermally, not pure anymore. Is information lost?
How can the final state be described as a pure state?
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