chapter 20 geodesics of the kerr metric · chapter 20. geodesics of the kerr metric 309 motion in a...

Chapter 20

Geodesics of the Kerr metric

Let us consider a geodesic with affine parameter λ and tangent vector

uµ =dxµ

dλ≡ xµ (20.1)

in Boyer-Lindquist coordinates (in this section we use the overdot with the meaning ofderivation with respect to the affine parameter λ). The tangent vector uµ is solution of thegeodesic equation

uµuν;µ = 0 . (20.2)

As we have seen in Chapter 11, the geodesic equation (20.2) is equivalent to the Euler-Lagrange equations

d

dλ

∂L∂xα

=∂L∂xα

(20.3)

associated to the Lagrangian

L (xµ, xµ) =1

2gµν x

µxν . (20.4)

One can define the conjugate momentum pµ to the coordinate xµ as

pµ ≡ ∂L∂xµ

= gµν xν . (20.5)

In terms of the conjugate momenta, the Euler-Lagrange equations can be written as

d

dλpµ =

∂L∂xµ

. (20.6)

Notice that, if the metric does not depend on a given coordinate xµ, the correspondingconjugate momentum is a constant of motion. The existence of such a constant can also beseen in a simpler way. We know that, if the metric is independent by a coordinate, there is aKilling vector ξµ corresponding to this symmetry, and ξµx

µ is a constant of motion; actually,this constant of motion coincides with the one arising from Euler-Lagrange equations (as wewill see, for instance, in the case of t and φ for the Kerr metric).

Let us come back to the geodesic equation for Kerr spacetime. Equation (20.2) (or,equivalently, (20.3)) in Kerr spacetime is very complicate to solve directly. To study geodesic

308

CHAPTER 20. GEODESICS OF THE KERR METRIC 309

motion in a simple way, we would need to find a way to express the geodesic uµ in terms ofconserved quantities, as we have done in Chapter 11 in the case of Schwarzschild spacetime.For this, we need four algebraic relations involving uµ.

As we have seen in Section 19.2, Kerr spacetime has two Killing vectors: a timelike Killingvector kµ = (1, 0, 0, 0) and a spacelike killing vector mµ = (0, 0, 0, 1). This corresponds tothe fact that the metric (in Boyer-Lindquist coordinates) does not depend explicitly by tand φ: the spacetime is static and axially symmetric.

Therefore, in geodesic motion there are two conserved quantities:

E ≡ −kµuµ = −gtµu

µ = −pt constant along geodesics

(20.7)

L ≡ mµuµ = gφµu

µ = pφ constant along geodesics .

(20.8)

As explained in Section 19.2, in the case of massive particle E is the energy at infinity permass unit while L is the angular momentum per mass unit; in the case of massless particles,E is the energy at infinity and L is the angular momentum.

Furthermore, we have the relation

gµνuµuν = κ (20.9)

where

κ = −1 for timelike geodesics

κ = 1 for spacelike geodesics

κ = 0 for null geodesics . (20.10)

The relations (20.7), (20.8), (20.9) give us three algebraic relations involving uµ, but theyare not sufficient to to determine the four unknowns uµ. This is different from the caseof Schwarzschild spacetime, where we have the further condition of planarity of the orbit(uθ = 0 if θ(λ = 0) = π/2) arising from the Euler-Lagrange equations, which have a simpleform in that case.

Without this further condition, we cannot study geodesic motion using only (20.7), (20.8),(20.9). Actually, there is a further conserved quantity, the Carter constant, which allows tofind the tangent vector uµ with algebraic relation.

Anyway, it is possible to study geodesic motion with the only help of (20.7), (20.8), (20.9)in the particular case of equatorial motion, i.e. with θ ≡ π/2. Therefore, we will first studyequatorial geodesics, and then, in Section 20.2, we will consider the case of general motionand derive the Carter constant.

20.1 Equatorial geodesics

In this section we study equatorial geodesics, i.e. geodesics with

θ ≡ π

2. (20.11)


First of all, let us prove that such geodesics exist, i.e. that equatorial geodesics are solutionsof the geodesic equation (20.2), or, equivalently, of the Euler-Lagrange equations (20.3). Theθ component of (20.3) is

d

dλ(gθµx

µ) =d

dλ(Σθ) = Σθ + Σ,µx

µθ =1

2gµν,θx

µxν (20.12)

where we have used the fact that the only non-vanishing gθµ component in (19.1) is gθθ = Σ.The right-hand side is

1

2gµν,θx

µxν =1

2

Σ,θ

(

(r)2

∆+ (θ)2

)

+ 2 sin θ cos θ(r2 + a2)(φ)2

− 2Mr

Σ2Σ,θ

(

a sin2 θφ− t)2

+4Mr

Σ

(

a sin2 θφ− t)

2a sin θ cos θφ

(20.13)

where Σ,θ = −2a2 sin θ cos θ and Σ,r = 2r. It is easy to check that, when θ = π/2, equation(20.12) reduces to

θ = −2

rrθ . (20.14)

Therefore, if θ = 0 and θ = π/2 at t = 0, then for t > 0 θ ≡ 0 and θ ≡ π/2. Thus, a geodesicwhich starts in the equatorial plane, remains in the equatorial plane at later times.

This also happens in the Schwarzschild metric. But, while in that case it is possibleto generalize the result to any orbit, thanks to the spherical symmetry, and prove that allSchwarzschild geodesics are planar, such a generalization is not possible for the Kerr metricwhich is axially symmetric. We only can say that equatorial geodesics are planar.

On the equatorial plane, Σ = r2, therefore

gtt = −(

1 − 2M

r

)

gtφ = −2Ma

r

grr =r2

∆

gφφ = r2 + a2 +2Ma2

r(20.15)

and

E = −gtµuµ =

(

1 − 2M

r

)

t +2Ma

rφ (20.16)

L = gφµuµ = −2Ma

rt+

(

r2 + a2 +2Ma2

r

)

φ . (20.17)

To solve (20.16), (20.17) for t, φ we define

A ≡ 1 − 2M

r


B ≡ 2Ma

r

C ≡ r2 + a2 +2Ma2

r(20.18)

so that (20.16), (20.17) can be written as

E = At +Bφ (20.19)

L = −Bt + Cφ . (20.20)

We also have

AC +B2 =(

1 − 2M

r

)

(

r2 + a2 +2Ma2

r

)

+4M2a2

r2= r2 − 2Mr + a2 = ∆ . (20.21)

Therefore,

CE − BL = [AC +B2]t = ∆t

AL +BE = [AC +B2]φ = ∆φ (20.22)

i.e.

t =1

∆

[(

r2 + a2 +2Ma2

r

)

E − 2Ma

rL

]

φ =1

∆

[(

1 − 2M

r

)

L+2Ma

rE]

. (20.23)

Equation (20.9) can be written in terms of A,B,C:

gµνuµuν = κ

= −At2 − 2Btφ+ Cφ2 +r2

∆r2

= −[At +Bφ]t + [−Bt + Cφ]φ+r2

∆r2

= −Et + Lφ+r2

∆r2 (20.24)

where we have used (20.19), (20.20). Therefore,

r2 =∆

r2(Et− Lφ + κ)

=1

r2

[

CE2 − 2BLE − AL2]

+κ∆

r2

=C

r2(E − V+)(E − V−) +

κ∆

r2(20.25)

where V±(r) are the solutions of the equation in E

CE2 − 2BLE − AL2 = 0 , (20.26)


i.e.

V± =BL±

√B2L2 + ACL2

C=L

C[B ±

√∆] . (20.27)

The quantity C can be rewritten as follows:

(r2 + a2)2 − a2∆

r2=

1

r2[(r2 + a2)(r2 + a2) − a2(r2 + a2 − 2Mr)]

=1

r2[(r2 + a2)r2 + 2Mra2] = r2 + a2 +

2Ma2

r= C (20.28)

therefore (20.25) and (20.27) can be written as

r2 =(r2 + a2)2 − a2∆

r4(E − V+)(E − V−) +

κ∆

r2(20.29)

V± =2Mar ± r2

√∆

(r2 + a2)2 − a2∆L . (20.30)

Notice that since (r2 + a2)2 − a2∆ > 0, C > 0.In the Schwarzschild limit a→ 0, we have

V+ + V− ∝ a→ 0 , V+V− → L2∆

r4(20.31)

therefore, if we define V ≡ −V+V−, Eqns. (20.29), (20.30) reduce to the well known form

r2 = E2 − V (r)

V (r) = −κ∆r2

+L2∆

r4

=(

1 − 2M

r

)

(

−κ+L2

r2

)

(20.32)

where we recall that κ = −1 for timelike geodesics, κ = 0 for null geodesics, κ = 1 forspacelike geodesics.

20.1.1 Null geodesics

In the case of null geodesics (20.29) becomes

r2 =(r2 + a2)2 − a2∆

r4(E − V+)(E − V−) (20.33)

with

V± =2MLar ± Lr2

√∆

(r2 + a2)2 − a2∆. (20.34)

In principle we would have four possibilities, corresponding to L positive and negative anda positive and negative. In practice, the physically different possibilities are two: La > 0


and La < 0, i.e. the photon and the black hole corotating and counterrotating. If we changesimultaneously the signs of L and a, the potentials V± remain the same, they only exchangetheir names: V+ becomes V− and viceversa. To avoid this exchange, it is better to redefinethe names of the potentials (20.34), by exchanging them when L is negative:

V± =2MLar ± |L|r2

√∆

(r2 + a2)2 − a2∆. (20.35)

With this definition, we always have

V+ ≥ V− . (20.36)

In the following we will study (20.33), (20.35) considering the two cases of geodesic and holecorotating (La > 0) and counterrotating (La < 0).

Since r2 must be positive, from eq. (20.33) we see that, and being (r2 + a2)2 − a2∆ > 0,null geodesics are possible for massless particle whose constant of motion E satisfies thefollowing inequalities

E < V− or E > V+ . (20.37)

Thus, the region V− < E < V+ is forbidden.In general, we find that:

• The two curves coincide for ∆ = 0, i.e. at

r = r+ = M +√M2 − a2 (20.38)

while for r > r+, ∆ > 0 and then V+ > V−. Furthermore,

V+(r+) = V−(r+) =2Mr+La

(r2+ + a2)2

, (20.39)

which is positive if La > 0, negative if La < 0.

• In the limit r → ∞, V± → 0.

• If La > 0 (corotating orbits), the potential V+ is definite positive; V− (which is positiveat r+) vanishes when

r√

∆ = 2Ma ⇒ r2(r2 − 2Mr + a2) = 4M2a2 (20.40)

which gives

r4 − 2Mr3 + a2r2 − 4M2a2 = (r − 2M)(r3 + a2r + 2Ma2) = 0; (20.41)

thus V− vanishes at r = 2M , which is the location of the ergosphere in the equatorialplane.

• If La < 0 (counterrotating orbits), the potential V− is definite negative and V− (whichis positive at r+) vanishes at r = 2M .


Figure 20.1: The potentials V+(r) and V−(r), for corotating (aL > 0) and counterrotating(aL < 0) orbits. The shadowed region is not accessible to the motion of photons or othermassless particles.


• The study of the derivatives of V±, which is too long to be reported here, tells us thatboth potentials, V+ and V−, have only one stationary point.

In summary, V+(r) and V−(r) have the shapes shown in Figure 20.1 where in the two panelswe show the case La > 0 (up) and La < 0 (down), respectively. Once we have drawn thecurves V+(r), V−(r) for assigned values of a, M and L, we can make a qualitative study ofthe orbits. To this purpose it is useful to compute the radial acceleration. By differentiatingeq. (20.33)

(r)2 =C

r2(E − V+)(E − V−) (20.42)

with respect to the geodesic parameter λ, we find

2rr =

[

(

C

r2

)′(E − V+)(E − V−) − C

r2V ′

+(E − V−) − C

r2V ′−(E − V+)

]

r (20.43)

i.e.

r =1

2

(

C

r2

)′(E − V+)(E − V−) − C

2r2

[

V ′+(E − V−) + V ′

−(E − V+)]

. (20.44)

Let us consider the radial acceleration in a point where the radial velocity r is zero, i.e. whenE = V+ or E = V−:

r = = − C

2r2V ′

+(V+ − V−) if E = V+

r = = − C

2r2V ′−(V− − V+) if E = V− . (20.45)

Since

V+ − V− =2|L|r2

√∆

(r2 + a2)2 − a2∆=

2|L|√

∆

C, (20.46)

we find

r = −|L|√

∆

r2V ′± if E = V± . (20.47)

This result shows that if, for example, E = V+(rmax) where rmax is a stationary point forV+ (i.e. V ′

+(rmax) = 0), the radial acceleration vanishes; since when E = V+(rmax) the radialvelocity also vanishes, a massless particle with that value of E can be captured on a circularorbit, but the orbit is unstable, as it is the orbit at r = 3M for the Schwarzschild metric.It is possible to show that rmax is the solution of the equation

r(r − 3M)2 − 4Ma2 = 0 . (20.48)

Note that this equation does not depend on L, thus the value of rmax is independent of L.The solution of (20.48) is a decreasing function of a, and, in particular,

rmax = 3M for a = 0

rmax = M for a = M

rmax = 4M for a = −M . (20.49)


Therefore, while for a Schwarzschild black hole the unstable circular orbit for a photon islocated at r = 3M , for a Kerr black hole it can be located much closer to the black hole, inparticular for large values of a; in the case of an extremal (a = M) black hole, rmax = M ,which is the position of the horizon for such a black hole (see also Figure 20.1.4).

A photon coming from infinity with constant of motion E > V+(rmax), falls inside thehorizon, whereas if 0 > E > V+(rmax), it approaches the black hole reaching a turning pointwhere E = V+(r) and r = 0, then reverts its motion (because r > 0 there, see eq. (20.47)),and escapes free at infinity. In all this cases the constant of motion E associated to thetimelike Killing vector is positive.

It remains to consider the case when E < V−, and in particular to see whether negativevalues of E which are in principle admitted by eq. (20.37) have a physical meaning.

20.1.2 How do we measure the energy of a particle

In special relativity, the energy E of a particle with four-momentum P µ is E = P 0 (rememberwe are using geometric units). More generally, we define the energy measured by an observeruµ, as

E (u) = −uµPµ . (20.50)

For instance, the energy measured by the static observer uµst = (1, 0, 0, 0) is E (ust) = −P0.

From this is clear that a particle cannot have a negative energy, because it would be a particlemoving backwards in time; the energy measured by a different observer will be different, butit will still need to be positive.

In general relativity, we still define the energy of a particle measured by an observer uµ

as in eq. (20.50). Indeed, eq. (20.50) is a tensor equation which holds in a locally inertialframe, and consequently in any other frame by the principle of general covariance. For thesame reason, it must be

E (u) > 0 . (20.51)

Having said this, let us now consider an observer in the Kerr spacetime, but located atradial infinity, where a static observer with uµ

st = (1, 0, 0, 0) can exist. Thus, according tothe definition E (u) = −uµPµ, the enrgy measured by the static observer will be E (ust) = −P0

and, according to eq. (19.20)1

E (ust) = −P0 = E (20.52)

We conclude that for a particle coming from radial infinity, the constant of motion E canbe identified with the energy as measured by a static observer at infinity, and consequentlyorbits with negative values of E are not allowed.

Referring to figure 20.1, let us now consider a massless particle, say a photon, that movesin the egoregion, i.e. between r+ and r0. The problem here is that a static observer cannotexist in this region, therefore we need to consider a different observer, for instance the ZAMO,

uµZAMO = C(1, 0, 0,Ω) (20.53)

1Note that eq. (20.52) holds for a massless particle; in the case of a massive particle, where E is theenergy per mass unit, we have to multiply E with the particle mass.


where (see eq. (19.28) )

Ω =2Mar

(r2 + a2)2 − a2∆ sin2 θ(20.54)

and C is found by imposing gµνuµZAMOu

νZAMO = −1. We have C > 0, otherwise the ZAMO

would move backwards in time.The energy of the photon with respect to the ZAMO is

EZAMO = −PµuµZAMO = C(E − ΩL) . (20.55)

Thus, the requirement EZAMO > 0 is equivalent to

E > ΩL . (20.56)

Notice that, by comparing (20.54) with (20.35) we find that

V− < ΩL < V+ . (20.57)

Therefore, the geodesics with E > V+ satisfy (20.56), have positive energies with respectto the ZAMO, and are then allowed. The geodesics with E < V−, instead, do not satisfy(20.56), have negative energies with respect to the ZAMO, and are then forbidden.

In the case La < 0 (counterrotating particles), V+ is negative in the ergoregion; thus, therequirement

E > V+ (20.58)

(necessary and sufficient to have a particle with positive energy E) allows negative values ofthe constant of motion E. This is not a contradiction, because it is only at infinity that Erepresents the physical energy of the particle; the geodesics we are considering never reachinfinity.

We can conclude that the constant E (which is the energy at infinity) must always belarger than V+

E > V+(r) (20.59)

and this guarantees that the energy of the particle, as measured by a local observer, ispositive; in the counterrotating case La > 0, E can be negative still preserving (20.59),because inside the ergosphere V+ < 0.

20.1.3 The Penrose process

**************************A short comment about notation: in this section we will use a slightly different notation

for the constants of motion E, L, which have been defined in Section 19.2 to be the energyat infinity and angular momentum per unit mass in the case of massive particles, and to bethe energy at infinity and angular momentum in the case of massless particles, namely,

E = −kµuµ , L = mµuµ . (20.60)

Here we define E,L to be the energy ay infinity and angular momentum, both for massiveand massless particles, i.e.

E = −kµPµ , L = mµPµ . (20.61)


This means that we have multiplied E and L, in the case of massive particles, for the massm.

**************************

An interesting consequence af the possibility of having particles moving with negative Eis that we can imagine a process which allows to extract rotational energy from a Kerr blackhole, through as mechanism called Penrose process.

Let us assume that a > 0. The same reasoning can be repeated in the case of a < 0,going to the same conclusions.

Let us assume we shoot massive particle with energy E and angular momentum L frominfinity radially towards the black hole in the equatorial plane. Its four-momentum, incovariant components, is

Pµ = (−E, 0, 0, L) . (20.62)

When the particle is in the ergoregion, it decays in two photons, with momenta

P1 µ = (−E1, P1 r, 0, L1) P2 µ = (−E2, P2 r, 0, L2) . (20.63)

The conservation of four-momentum requires that Pµ = P1 µ + P2 µ, thus

E1 + E2 = L1 + L2 = 0 . (20.64)

We also assume that r1 < 0 and r2 > 0, i.e. the photon 1 falls in the black hole, while thephoton 2 comes back to infinity.

The photon 1 never goes outside the ergosphere, thus we can require E1 < 0, if L1 < 0.It must be

V−(r+)(= V+(r+)) < E1 < 0 for the particle 1

E2 > V max+ for the particle 2 (20.65)

and the photon 2 reaches infinity with

E2 = E − E1 > E

L2 = L− L1 > L . (20.66)

In this way, we end at infinity with a particle more energetic than the initial one. Wesay we have extracted rotational energy from the black hole because the photon 1, withE1 < 0, L1 < 0, reduces the mass-energy M of the black hole and its angular momentumJ = Ma to the values Mfin, Jfin respectively:

Mfin = M + E1 < M (20.67)

Jfin = J + L1 < J . (20.68)

The expressions (20.67), (20.68) could appear strange, after we have stressed so much thatthe constant E is not really the energy of the particle, but only its energy at infinity. Toprove (20.67), we note that, as shown in Chapter 17, the total mass-energy of the system is

P 0tot =

∫

Vd3x(−g)(T 00 + t00) (20.69)


where V is the volume of a t = const. three-surface. If we neglect the gravitational fieldgenerated by the particle, t00 is only due to the black hole, thus

P 0tot =

∫

Vd3x(−g)T 00

particle +M . (20.70)

If we compute the integral at a time when the particle is far away from the black hole, thenthe spacetime is flat where T 00

particle 6= 0; the stress-energy tensor for a point particle withenergy E, in Minkowskian coordinates, has

T 00particle = Eδ3(x − x(t)) . (20.71)

Integrating in V we getP 0

tot = E +M (20.72)

when the initial particle is still far away from the black hole.On the other hand, this quantity is conserved, because P µ

tot ,µ = 0 and we neglect theoutgoing gravitational flux generated by the particle. Therefore, repeating the computationwhen the photon 2 has reached infinity, when the mass of the black hole is Mfin,

P 0tot = E2 +Mfin . (20.73)

This proves the relation (20.67). A similar proof holds for the angular momentum.

20.1.4 The innermost stable circular orbit for timelike geodesics

The study of timelike geodesics is much more complicate, because equation (20.25), whichbecomes when κ = −1

r2 =C

r2(E − V+)(E − V−) − ∆

r2, (20.74)

does not allow a simple qualitative study as in the case of null geodesics. Therefore, here weonly report on some results one finds by a detailed study of these equations.

A very relevant quantity (with astrophysical interest) is the location of the innermoststable circular orbit (ISCO), which, in the Schwarzschild case, is at r = 6M . In Kerrspacetime, the expression for rISCO is quite complicate, but its qualitative behaviour issimple: there are two solutions

r±ISCO(a) , (20.75)

one corresponding to corotating orbits, one to counterrotating orbits. For a = 0, obviouslythe two solutions coincide to 6M ; by increasing |a|, the ISCO moves closer to the black holefor corotating orbits, and far away from the black hole for counterrotating orbits. Whena = ±M , the ISCO, in the corotating case, coincides with the horizon, at r = M . Thisbehaviour is very similar to that we have already seen in the case of unstable circular orbitsfor null geodesics.

In Figure 20.1.4 we show (in the case a ≥ 0) the locations of the last stable and unstablecircular orbits for timelike geodesics, and of the unstable circular orbit for null geodesics.This figure is taken from the article where these orbits have been studied (J. Bardeen, W.H. Press, S. A. Teukolsky, Astrophys. J. 178, 347, 1972).


20.1.5 The 3rd Kepler law

Let us consider a circular timelike geodesic in the equatorial plane. We remind that theLagrangian (20.4) is

L =1

2gµν x

µxν (20.76)

and the r Euler-Lagrange equation is

d

dλ

∂L∂r

=∂L∂r

. (20.77)

Being grµ = 0 if µ 6= r, we have

d

dλ(grrr) =

1

2gµν,rx

µxν . (20.78)

In the case of a circular geodesic, r = r = 0, and this equation reduces to

gtt,r t2 + 2gtφ,r tφ+ gφφ,rφ

2 = 0 . (20.79)

The angular velocity is ω = φ/t, thus

gφφ,rω2 + 2gtφ,rω + gtt,r = 0 . (20.80)

We remind that

gtt = −(

1 − 2M

r

)

gtφ = −2Ma

r

gφφ = r2 + a2 +2Ma2

r, (20.81)

then

2

(

r − Ma2

r2

)

ω2 +4Ma

r2ω − 2M

r2= 0 . (20.82)

The equation(r3 −Ma2)ω2 + 2Maω −M = 0 (20.83)

has discriminantM2a2 +M(r3 −Ma2) = Mr3 (20.84)

and solutions

ω± =−Ma ±

√Mr3

r3 −Ma2= ±

√M

r3/2 ∓ a√M

r3 −Ma2

= ±√M

r3/2 ∓ a√M

(r3/2 + a√M)(r3/2 − a

√M)

= ±√M

r3/2 ± a√M

. (20.85)


This is the relation between angular velocity and radius of circular orbits, and reduces, inSchwarzschild limit a = 0, to

ω± = ±√

M

r3, (20.86)

which is Kepler’s 3rd law.

20.2 General geodesic motion: the Carter constant

To study generic geodesics in Kerr spacetime, we need to apply the Hamilton-Jacobi ap-proach, which allows to indentify a further constant of motion (not related to any isometryof the metric).

To apply the Hamilton-Jacoby approach, we first have to express our system with anHamiltonian description. Given the Lagrangian of the system

L(xµ, xµ) =1

2gµνx

µxν (20.87)

we define the conjugate momenta2

pµ =∂L∂xµ

= gµν xν . (20.88)

By inverting (20.88), we can express xµ in terms of the conjugate momenta:

xµ = gµνpν . (20.89)

The Hamiltonian is a functional of the coordinate functions xµ(λ) and of their conjugatemomenta pµ(λ), defined as

H(xµ, pν) = pµ xµ(pν) − L (xµ, xµ(pν)) . (20.90)

In our case, we have

H =1

2gµνpµpν . (20.91)

The geodesics equation is equivalent to the Euler-Lagrange equations for the Lagrangianfunctional, which are equivalent to the Hamilton equations for the Hamiltonian functional:

xµ =∂H

∂pµ

pµ = − ∂H

∂xµ. (20.92)

To solve equations (20.92) is as difficult as to solve the Euler-Lagrange equations. Anyway,it is possible to apply the Hamilton-Jacobi approach, which here we briefly recall.

2Not to be confused with the four-momentum of the particle, which we denote with P µ.


In the Hamilton-Jacobi approach, we look for a function of the coordinates and of thecurve parameter λ,

S = S(xµ, λ) (20.93)

which is solution of the Hamilton-Jacobi equation

H

(

xµ,∂S

∂xµ

)

+∂S

∂λ= 0 . (20.94)

In general such a solution will depend on four integration constants.It can be shown that, if S is a solution of the Hamilton-Jacobi equation, then

∂S

∂xµ= pµ . (20.95)

Therefore, once we have solved (20.94), we have the expressions of the conjugate momenta(and then of xµ) in terms of four constants, which allows to write the solutions of geodesicmotion in a closed form, through integrals.

In general, it is more difficult to solve (20.94), which is a partial differential equation,than to solve the geodesic equation, but in this case we can find such a solution.

First of all, we can use what we already know, i.e.

H =1

2gµνpµpν =

1

2κ

pt = −E constant

pφ = L constant . (20.96)

These conditions require that

S = −1

2κλ− Et + Lφ+ S(rθ)(r, θ) (20.97)

where S(rθ) is a function of r and θ to be determined.Furthermore, we look for a separable solution, by making the ansatz

S = −1

2κλ− Et + Lφ+ S(r)(r) + S(θ)(θ) . (20.98)

Substituting (20.98) into the Hamilton-Jacobi equation (20.94), and using the expression(19.15) for the inverse metric, we find

−κ +∆

Σ

(

dS(r)

dr

)2

+1

Σ

(

dS(θ)

dθ

)2

− 1

∆

[

r2 + a2 +2Mra2

Σsin2 θ

]

E2 +4Mra

Σ∆EL +

∆ − a2 sin2 θ

Σ∆ sin2 θL2 = 0 .

(20.99)

Using the relation (19.27)

(r2 + a2) +2Mra2

Σsin2 θ =

1

Σ

[

(r2 + a2)2 − a2 sin2 θ∆]

(20.100)


and multiplying by Σ = r2 + a2 cos2 θ, we get

−κ(r2 + a2 cos2 θ) + ∆

(

dS(r)

dr

)2

+

(

dS(θ)

dθ

)2

−[

(r2 + a2)2

∆− a2 sin2 θ

]

E2 +4Mra

∆EL+

(

1

sin2 θ− a2

∆

)

L2 = 0

(20.101)

i.e.

∆

(

dS(r)

dr

)2

− κr2 − (r2 + a2)2

∆E2 +

4Mra

∆EL− a2

∆L2

= −(

dS(θ)

dθ

)2

+ κa2 cos2 θ − a2 sin2 θE2 − 1

sin2 θL2 .

(20.102)

We rearrange equation (20.102) by adding to both sides the constant quantity a2E2 + L2:

∆

(

dS(r)

dr

)2

− κr2 − (r2 + a2)2

∆E2 +

4Mra

∆EL− a2

∆L2 + a2E2 + L2

= −(

dS(θ)

dθ

)2

+ κa2 cos2 θ + a2 cos2 θE2 − cos2 θ

sin2 θL2 .

(20.103)

In equation (20.103), the left-hand side does not depend on θ, and is equal to the right-handside which does not depend on r; therefore, this quantity must be a constant C:

(

dS(θ)

dθ

)2

− cos2 θ[

(κ+ E2)a2 − 1

sin2 θL2]

= C

∆

(

dS(r)

dr

)2

− κr2 − (r2 + a2)2

∆E2 +

4Mra

∆EL− a2

∆L2 + E2a2 + L2

= ∆

(

dS(r)

dr

)2

− κr2 + (L− aE)2 − 1

∆

[

E(r2 + a2) − La]2

= −C .

(20.104)

Notice that in rearranging the terms between the last two lines, we have used, for the LEterms, the relation

−2aLE + 2aLEr2 + a2

∆= −4aMr

∆LE . (20.105)

If we define the functions R(r) and Θ(θ) as

Θ(θ) ≡ C + cos2 θ[

(κ+ E2)a2 − 1

sin2 θL2]

R(r) ≡ ∆[

−C + κr2 − (L− aE)2]

+[

E(r2 + a2) − La]2,

(20.106)


then

(

dS(θ)

dθ

)2

= Θ

(

dS(r)

dr

)2

=R

∆2(20.107)

and the solution of the Hamilton-Jacobi equation has the form

S = −1

2κλ− Et + Lφ+

∫

√R

∆dr +

∫ √Θdθ . (20.108)

Once we have the solution of the Hamilton-Jacobi equations, depending on four constants(κ,E, L, C), it is possible to solve the geodesic motion. Indeed, from (20.95) we know theexpressions of the conjugate momenta

p2θ = (Σθ)2 = Θ(θ)

p2r =

(

Σ

∆r)2

=R(r)

∆2(20.109)

therefore

θ =1

Σ

√Θ

r =1

Σ

√R (20.110)

which can be solved by simple integration. The constant C is called Carter constant, andcharacterize, together with E and L, the geodetic motion. Notice that differently from Eand L, the Carter constant is not related to any isometry of the spacetime.

chapter 20 geodesics of the kerr metric · chapter 20. geodesics of the kerr metric 309 motion in a...

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