3.6 The Kerr Black Hole and the Laws of Thermodynamics |
65 |
Fig. 3.5 An example of a closed orbit described by the exact analytic solution (3.5.61) of the geodesic equations for the Schwarzschild metric. The values of the roots chosen in this example are:
{e1, e2, e3} = {2.7, 11.7, 25.6} corresponding to
{L2, m, E 2} = {19.98, 1, 0.95}. As we see, in this case E 2 < 1 and for this reason the orbit is closed
Fig. 3.6 An example of an open orbit described by the exact analytic solution (3.5.61) of the geodesic equations for the Schwarzschild metric. The values of the roots chosen in
this example are:
{e1, e2, e3} = {−17.1, 2.1, 10.9} corresponding to
{L2, m, E 2} = {100, 1, 1.5}. As we see, in this case E 2 > 1 and for this reason the orbit is open
3.5.5 About Explicit Kerr Geodesics
In the Schwarzschild case we demonstrated the use of the complete integration formulae. The classification of all time-like and null-like geodesics encoded in the final integration formulae is still very laborious for the general Kerr case because of the implicit form of the solution. Indeed there are very many different type of geodesics spherical, and non-spherical, open and closed, retrograding and advancing and so on. We stop our discussion at this level and we turn to the most intriguing analogy with thermodynamics.
3.6 The Kerr Black Hole and the Laws of Thermodynamics
Let us now focus on the case of a neutral rotating black-hole by setting q = 0 and let us reconsider the results we obtained for the horizon area AH of a pure Kerr
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3 Rotating Black Holes and Thermodynamics |
solution and for its angular velocity ΩH . In terms of the black-hole mass m and of its angular momentum J , (3.4.13) and (3.4.11) can be rewritten as follows:
AH (m, J ) |
= |
8π m m |
/m2 |
J 2 |
(3.6.1) |
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+ |
− m2 |
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ΩH (m, J ) |
= |
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J |
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(3.6.2) |
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2m2(m + m2 − mJ 22 ) |
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Let us now introduce an additional function, whose interpretation we will later retrieve:
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2 |
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κ(m, J ) |
= |
m2 − mJ 2 |
(3.6.3) |
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2m(m + m2 − mJ 22 ) |
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Calculating the variation of AH (m, J ) in the standard way:
δAH = ∂mAH δm + ∂J AH δJ |
(3.6.4) |
we can verify the following variational identity:
1
δm = κ δAH + ΩH δJ (3.6.5) 8π
What is it special about this identity? The answer is striking: it is formally identical to the first law of thermodynamics if we introduce the following interpretations:
1 |
m = U |
internal energy |
(3.6.6) |
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AH = S |
entropy |
(3.6.7) |
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8π |
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1 |
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κ = |
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inverse temperature |
(3.6.8) |
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T |
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ΩH = −p |
pressione |
(3.6.9) |
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J = V |
volume |
(3.6.10) |
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At first sight this might seem just an arbitrary, meaningless, formal exercise yet a little bit of further consideration starts revealing the profound significance of the analogy. First of all if (3.6.5) is the first law of thermodynamics then the second law should also apply in the form:
δAH ≥ 0 in all physical processes |
(3.6.11) |
thirdly if κ is the inverse temperature, it should be an intensive quantity, namely constant over the body which in our analogy is the event horizon. Clearly the function κ(m, J ) introduced in (3.6.3) as such a property yet the interesting point is that we
3.6 The Kerr Black Hole and the Laws of Thermodynamics |
67 |
can identify this expression with a quantity defined in terms of the black-hole geometry that is constant over the horizon and has a well defined physical interpretation. Let us postpone this identification for a moment and consider the last implication of the thermodynamical interpretation of (3.6.5). Indeed if all the rest is as we claimed the term
δW = ΩH δJ |
(3.6.12) |
should be interpreted as some work extracted from a thermodynamical process involving the black-hole. The whole point is precisely this. Do such processes exist by means of which we can extract energy from a rotating black-hole and do they satisfy the second law of thermodynamics (3.6.11)? The answer is yes and involves in a crucial way the near horizon region that we named ergosphere in previous pages. The gedanken experiment showing the mechanism of energy extraction was found by Penrose in 1969.
3.6.1 The Penrose Mechanism
The Killing vector field k defined in (3.3.1) which becomes the standard time translation in the asymptotic flat space-time far from the hole is instead space-like inside the ergosphere as we already noted. Thus for a massive test particle of four momentum pμ = μuμ the energy:
E ≡ pμkμ |
(3.6.13) |
is not necessarily positive inside the ergosphere. Therefore, by making a black hole absorb a particle with negative total energy we can actually extract energy from the black hole! Let us see how we can do this. Suppose that from our laboratory, located far from the hole and at rest with respect to the reference frame of the fixed stars, we throw a rocket towards the black-hole. Let us denote p0μ the momentum of our missile that will navigate along a time-like geodesic. Its energy:
E0 ≡ (p0, k) |
(3.6.14) |
stays constant along the trajectory since it is the scalar product of a Killing vector with the tangent vector to a geodesic. Suppose that when it enters the ergosphere the rocket splits into two fragments as illustrated in Fig. 3.7. Conceptually this can be arranged for instance by means of an explosive connected to a suitable clock. By local conservation of the energy-momentum we have:
p0μ = p1μ + p2μ |
(3.6.15) |
where p1μ,2 are the four-momenta of the two fragments. Contracting |
equation |
(3.6.15) with the Killing vector kμ we obtain: |
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E0 = E1 + E2 |
(3.6.16) |
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3 Rotating Black Holes and Thermodynamics |
Fig. 3.7 Schematic view of the Penrose gedanken experiment
However, inside the ergosphere, we can arrange the breakup of the rocket in such a way that one of his fragments has negative total energy:
E1 < 0 |
(3.6.17) |
Therefore, if the other fragment will make return to asymptotically flat infinity following its own geodesic it will have an energy E2 which is greater than the initial energy of our projectile. In other words we have extracted energy from the black hole which has made some work for us! What has it happened? It is easily understood. The fragment with negative energy from the ergosphere has crossed the event horizon and it has fallen inside the whole. The latter having absorbed a negative energy particle has now a slightly smaller mass: m = m − |E1|. Let us now consider the angular momentum of the infalling negative energy particle. By definition we have:
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1 |
= − ˜ 1 |
) |
(3.6.18) |
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(k, p |
where ˜ is the rotational Killing vector defined in (3.3.1). On the other hand since k
the Killing vector χ (ΩH ) is null-like and future-directed on the horizon it follows that for any physical particle of momentum pμ crossing the horizon we must have:
p, χ (ΩH ) ≡ E − ΩH > 0 |
(3.6.19) |
This applies to all particles also to our negative energy rocket-fragment. It follows that, not only the energy, but also the angular momentum of this latter is negative and we have:
1 < |
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E1 |
(3.6.20) |
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ΩH |
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At the end of the process our black hole has swallowed an object of energy E1 < 0 and of angular momentum 1 < 0. As a result both its mass and its angular momentum have been decreased since:
m = m − |E1|
(3.6.21)
J = J − | 1|