10.3 Kerr Black Holes
10.3.1 Equatorial Circular Orbits
Time-like circular orbits in the equatorial plane of the Kerr metric are particularly important [4]. For example, they are relevant for the structure of accretion disks around astrophysical black holes [1].
Let us write the line element in the so-calledcanonical form
ds2 =gttdt2+2gtφdt dφ+grrdr2+gθθdθ2+gφφdφ2, (10.14) where the metric coefficients are independent oft andφ. For example, this is the case of the Boyer–Lindquist coordinates in (10.6). As we know, the Lagrangian for a free point-like particle is that in Eq. (3.1) and it is convenient to parametrize the trajectory with the particle proper timeτ. For simplicity, we set the rest-mass of the particlem=1.
Since the metric is independent of the coordinatestandφ, we have two constants of motion, namely the specific energy at infinity E and the axial component of the specific angular momentum at infinityLz
d dτ
∂L
∂t˙ −∂L
∂t =0⇒ pt ≡∂L
∂t˙ =gttt˙+gtφφ˙= −E, (10.15) d
dτ
∂L
∂φ˙ −∂L
∂φ =0⇒ pφ ≡∂L
∂φ˙ =gtφt˙+gφφφ˙=Lz. (10.16) The term “specific” is used to indicate that E andLz are, respectively, the energy and the angular momentum per unit rest-mass. We remind the reader that here we use the convention of a metric with signature(− + ++). For a metric with signature (+ − −−), we would have pt=E and pφ = −Lz. Equations (10.15) and (10.16) can be solved to find thetand theφcomponents of the 4-velocity of the test-particle
6A closed time-like curve is a closed time-like trajectory; if there are closed time-like curves in a spacetime, massive particles can travel backwards in time. While the Einstein equations have solutions with similar properties, they are thought to be non-physical and therefore they are usually not taken into account.
˙
t = Egφφ+Lzgtφ
gt2φ−gttgφφ , (10.17)
φ˙ = −Egtφ+Lzgtt
g2tφ−gttgφφ . (10.18)
From the conservation of the rest-mass,gμνx˙μx˙ν= −1, we can write the equation grrr˙2+g2θθθ˙2=Veff(r, θ) , (10.19) whereVeffis the effective potential
Veff = E2gφφ+2E Lzgtφ+L2zgtt
g2tφ−gttgφφ −1. (10.20) Circular orbits in the equatorial plane are located at the zeros and the turning points of the effective potential:r˙= ˙θ=0, which impliesVeff =0, andr¨= ¨θ=0, which requires, respectively,∂rVeff =0 and∂θVeff =0. From these conditions, we could obtain the specific energyE and the axial component of the specific angular momentumLzof a test-particle in equatorial circular orbits. However, it is faster to proceed in the following way. We write the geodesic equations as
d dτ
gμνx˙ν =1 2
∂μgνρ x˙νx˙ρ. (10.21)
In the case of equatorial circular orbits,r˙= ˙θ = ¨r =0, and the radial component of Eq. (10.21) reduces to
(∂rgtt)t˙2+2
∂rgtφ t˙φ˙+
∂rgφφ φ˙2 =0. (10.22) The angular velocityΩ = ˙φ/t˙is
Ω±=−∂rgtφ±
∂rgtφ 2−(∂rgtt)
∂rgφφ
∂rgφφ , (10.23)
where the upper (lower) sign refers to corotating (counterrotating) orbits, namely orbits with angular momentum parallel (antiparallel) to the spin of the central object.
Fromgμνx˙μx˙ν = −1 withr˙= ˙θ=0, we can write
˙
t = 1
−gtt−2Ωgtφ−Ω2gφφ. (10.24)
Equation (10.15) becomes
10.3 Kerr Black Holes 185
Fig. 10.1 Specific energyE of a test-particle in equatorial circular orbits in the Kerr spacetime as a function of the Boyer–Lindquist radial coordinater. Every curve corresponds to the spacetime with a particular value of the spin parametera∗. The values area∗=0, 0.5, 0.8, 0.9, 0.95, 0.99, and 0.999
0.6 0.7 0.8 0.9 1 1.1 1.2
0 1 2 3 4 5 6 7 8 9 10
E
r/M a* = 0
a* = 0.999
E = −
gtt+Ωgtφ t˙
= − gtt+Ωgtφ
−gtt−2Ωgtφ−Ω2gφφ. (10.25) Equation (10.16) becomes
Lz =
gtφ+Ωgφφ t˙
= gtφ+Ωgφφ
−gtt−2Ωgtφ−Ω2gφφ. (10.26)
If we employ the metric coefficients of the Kerr solution in Boyer–Lindquist coordinates, Eqs. (10.25) and (10.26) become, respectively, [4]
E = r3/2−2Mr1/2±a M1/2 r3/4√
r3/2−3Mr1/2±2a M1/2, (10.27) Lz = ± M1/2
r2∓2a M1/2r1/2+a2 r3/4√
r3/2−3Mr1/2±2a M1/2, (10.28) and the angular velocity of the test-particle is
Ω± = ± M1/2
r3/2±a M1/2 (10.29)
where, again, the upper sign refers to corotating orbits, the lower sign to counterro- tating ones. Figure10.1showsEas a function of the radial coordinaterfor different values of the spin parameter of the black hole. Figure10.2shows the profile ofLz.
As we can see from Eqs. (10.25) and (10.26), as well as from Eqs. (10.27) and (10.28), E andLz diverge when their denominator vanishes. This happens at the radius of thephoton orbit rγ
Fig. 10.2 As in Fig.10.1for the axial component of the specific angular momentum Lz
1 2 3 4 5
0 1 2 3 4 5 6 7 8 9 10
Lz /M
r/M a* = 0
a* = 0.999
gtt+2Ωgtφ+Ω2gφφ=0 ⇒ r=rγ. (10.30) In Boyer–Lindquist coordinates, the equation for the radius of the photon orbit is
r3/2−3Mr1/2±2a M1/2 =0. (10.31) The solution is [4]
rγ =2M
1+cos 2
3arccos ∓a
M
. (10.32)
Fora =0,rγ =3M. Fora=M,rγ =Mfor corotating orbits and 4M for counter- rotating orbits.
The radius of themarginally bound orbit rmbis defined as E = − gtt+Ωgtφ
−gtt−2Ωgtφ−Ω2gφφ =1 ⇒ r =rmb. (10.33) The orbit is marginally bound, which means that the test-particle has sufficient en- ergy to escape to infinity (the test-particle cannot reach infinity if E <1, and can reach infinity with a finite velocity ifE >1). In the Kerr metric in Boyer–Lindquist coordinates, we find [4]
rmb=2M∓a+2
M(M∓a) . (10.34)
Fora=0,rmb=4M. Fora=M,rmb=Mfor corotating orbits and rmb =
3+2√ 2
M ≈5.83M (10.35)
10.3 Kerr Black Holes 187 for counterrotating orbits.
The radius of themarginally stable orbit,rms, more often called the radius of the innermost stable circular orbit(ISCO),rISCO, is defined by
∂r2Veff =0 or ∂θ2Veff =0 ⇒ r =rISCO. (10.36) In the Kerr metric, equatorial circular orbits turn out to always be vertically stable, so the ISCO radius is determined by∂r2Veff only. In Boyer–Lindquist coordinates, we have [4]
rISCO=3M+Z2∓
(3M−Z1) (3M+Z1+2Z2) , Z1=M +
M2−a2 1/3
(M+a)1/3+(M−a)1/3 , Z2=
3a2+Z12. (10.37)
Fora=0,rISCO=6M. Fora =M,rISCO=Mfor corotating orbits andrISCO=9M for counterrotating orbits. The derivation can be found, for instance, in [5].
Note that the innermost stable circular orbit is located at the minimum of the energy E.rISCO can thus be obtained even from the equation d E/dr =0. After some manipulation, one finds
d E
dr = r2−6Mr±8a M1/2r1/2−3a2 2r7/4
r3/2−3Mr1/2±a M1/2 3/2M, (10.38) andr2−6Mr±8a M1/2r1/2−3a2=0 is the same equation as that we can obtain from Eq. (10.36), see [5]. Moreover, the minimum of the energyE and of the axial component of the angular momentumLzare located at the same radius. After some manipulation,d Lz/dris
d Lz
dr = r2−6Mr±a M1/2r1/2−3a2 2r7/4
r3/2−3Mr1/2±a M1/2 3/2
M1/2r3/2±a M , (10.39)
and we see thatd E/dr =0 andd Lz/dr=0 have the same solution.
The concept of innermost stable circular orbit has no counterpart in Newtonian mechanics. As we have already seen in Sect.8.4for the Schwarzschild metric, and we can do the same for the Kerr one, the equations of motion of a test-particle can be written as the equations of motion in Newtonian mechanics with a certain effective potential. At small radii, the effective potential is dominated by an attractive term that is absent in Newtonian mechanics and is responsible for the existence of the innermost stable circular orbit.
Figure10.3shows the radius of the event horizonr+, of the photon orbitrγ, of the marginally bound circular orbitrmb, and of the innermost stable circular orbitrISCO in the Kerr metric in Boyer–Lindquist coordinates as functions ofa∗.
0 2 4 6 8 10
0 0.2 0.4 0.6 0.8 1
r/M
a* r+
rγ rmb
rISCO
Fig. 10.3 Radial coordinates of the event horizonr+(red solid curve), of the photon orbitrγ (green dashed curve), of the marginally bound circular orbitrmb(blue dotted curve), and of the innermost stable circular orbitrISCO (cyan dashed-dotted curve) in the Kerr metric in Boyer–
Lindquist coordinates as functions of the spin parametera∗. For every radius, the upper curve refers to the counterrotating orbits, the lower curve to the corotating ones
In the end, in the Kerr metric we have the following picture forEandLz. At large radii, we recover the Newtonian limit forEandLzfor a particle in the gravitational field of a point-like massive body. As the radial coordinate decreases, E and Lz
monotonically decrease as well, up to when we reach a minimum, which is at the same radius forEandLz. This is the radius of the innermost stable circular orbit. If we move to smaller radii,EandLzincrease. First, we find the radius of the marginally bound circular orbit, defined by the conditionE =1. As the radial coordinate decreases,E andLzcontinue increasing, and they diverge at the photon orbit. There are no circular orbits with radial coordinate smaller than the photon orbit and massive particles can reach the photon orbit in the limit of infinite energy.
Note that, in the case of an extremal Kerr black hole with a=M, one finds r+=rγ =rmb =rISCO=M for corotating orbits. However, the Boyer–Lindquist coordinates are not well-defined at the event horizon and these special orbits do not coincide [4]. If we writea=M(1−ε)withε→0, we find
r+ =M
1+√
2ε+ · · ·
, rγ =M
1+2 2ε
3 + · · ·
, rmb =M
1+2√
ε+ · · · , rISCO=M
1+(4ε)1/3+ · · ·
. (10.40) We can then evaluate the proper radial distance betweenr+and the other radii [4].
Sendingεto zero, the result is
10.3 Kerr Black Holes 189 rγ
r+
r dr
√Δ → 1 2Mln 3, rmb
r+
r dr
√Δ →Mln
1+√ 2
, rISCO
r+
r dr
√Δ → 1 6Mln
27 ε
, (10.41)
which clearly shows that these orbits do not coincide even if in Boyer–Lindquist coordinates they have the same value. The energyEand the axial component of the angular momentumLzare
r=rγ E→ ∞, Lz →2E M, r=rmb E→1, Lz →2M,
r=rISCO E→ √13,Lz → √23M. (10.42)