GR November 19 Lecture Notes
Bobby Bond
November 19, 2014
We are going to look at the solutions to the Kerr metric.
ds2 = −(1 −
2GM r 2
GM ar sin2 θ
ρ2 2
Σ2 sin2 θ 2
2 2
)dt
−
4
dtdφ
+
dr
+
ρ
dθ
+
dφ
ρ2
ρ2
∆
ρ2
(1)
ρ2 = r2 + a2 cos θ ≥ 0
(2)
∆ = r2 − 2M r + a2 = (r − r+ )(r − r− )
(3)
Σ2 = (r2 + a2 )2 − a2 ∆ sin2 θ ≥ 0
(4)
In the last class we introduced the following null vectors
1 2
(r + a2 , ∆, 0, a)
∆
(5)
1 2
(r + a2 , −∆, 0, a)
2ρ2
(6)
lµ =
nµ =
Using the two null vectors defined in equation (5) and (6) we can define the following
Killing (0,2) Tensor
σµν = 2ρ2 l(µ nν) + r2 gµν
(7)
We can define gµν with the following equations
gµν = −2l(µ nν) + 2m(µ mν)
(8)
Where m and m are defined by the following
mµ =
1
√ (ia sin θ, 0, 1, i csc θ)
(r + ia cos θ) 2
mµ = (mµ )∗
(9)
(10)
With these defined we will now look into Geodesics of the Kerr meric
L=
m
(gµν ẋµ ẋν )
2
1
(11)
pt = m(gtt ṫ + gtφ φ̇) = −E
(12)
pφ = m(gtφ ṫ + gφφ φ̇) = Lz
(13)
If Lz = 0 then pφ = 0 we get the following
gtφ
g tφ
dφ
=−
= + tt = w(r) 6= 0
dt
gφφ
g
(14)
If equation (14) is true we call this Frame Dragging.
Lets now look at Equatorial Null Orbits
ṙ2 =
Σ2 2 4M arEL r2 − 2M rL2
Σ2
(E
−
−
)
=
(E − V+ (r))(E − V− (r))
r2
Σ2
Σ2
r2


s
!
1
φφ
g
2M ar + r2 ∆ 2
V± (r) =
L = w ± w2 − tt  L
Σ2
g
(15)
(16)
Penrose Process
1) E1 > 0 L1 > 0 carry true energy and angular momentum to infinity
2) E2 < 0 L2 < −|L1 | < 0
(M,a)→ (M − δM, J = aM − δJ)
A→ A + δA Turns out that δA > 0 to some limit.
Lets look into Non Equatorial Geodesics
K = σµν U µ U ν
(17)
We want equation (17) to be conserved to K̇ = 0
K = 2(a2 cos θ(U · l)(U · n) + r2 (U · m)(U · m))
(18)
We will define the following for U
T µ Uµ = −E
(19)
U µ ∇µ U ν = 0
(20)
U µ ∇µ (Tν U ν ) = Tν (U µ ∇µ U ν ) + U µ U ν (∇µ Tν ) = ∇(µ Tν) + ∇[µ Tν]
(21)
gµν U µ U ν = −c2 δ1
(22)
With these we can write K as the following
K=
2
ρ2
(∆ṫ−a2 ∆ sin2 θφ̇)2 − ṙ2 −δ1 a2 cos2 θ = (a sin θṫ−(r2 +a2 ) sin θρ̇)2 +ρ4 θ̇2 +δ1 a2 cos2 θ
∆
∆
(23)
2
Now lets define the following 3 variables to define τ , t, and φ
R = ((r2 + a2 )E − aLz )2 − ∆[(Lz − aE)2 + Q + δ1 r2 ]
(24)
Θ = Q − (a2 (δ1 − E 2 ) + L2z csc2 θ) cos2 θ
(25)
Q = K − (Lz − aE)2
(26)
It follows from the geodesic equations that
dθ
dτ
dr
= ±√ = ±√
ρ2
R
Θ
(27)
With equation (24)-(27) we can define τ , t, and φ with the following integrals
Z
τ=
1
t=
2
Z
1
φ=−
2
Z
r
2
r
r2 dr
√ + a2
R
1
√
Z
∂R
dr +
∆ R ∂E
1
√
θ
Z
∂R
dr −
∂L
∆ R z
3
cos2 θdθ
√
Θ
θ
Z
1 ∂Θ
√
dθ
Θ ∂E
θ
1 ∂Θ
√
dθ
Θ ∂Lz
(28)
(29)
(30)