Solutions of Einstein s Equations & Black Holes 2

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1 Solutions of Einstein s Equations & Black Holes 2 Kostas Kokkotas December 19, S.L.Shapiro & S. Teukolsky Black Holes, Neutron Stars and White Dwarfs Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 3

2 Schwarzschild Solution: Black Holes I The light cone in a Schwarzschild spacetime ds 2 0 = 1 2M ) dt 2 1 2M r r ) 1 dr 2 dr dt = ± 1 2M r r = 2M is a null surface and is called horizon or infinite redshift surface. ). Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 4

3 Schwarzschild Solution: Black Holes I The light cone in a Schwarzschild spacetime ds 2 0 = 1 2M ) dt 2 1 2M r r ) 1 dr 2 dr dt = ± 1 2M r r = 2M is a null surface and is called horizon or infinite redshift surface. ). 4 S.L.Shapiro & S. Teukolsky Black Holes, Neutron Stars and White Dwarfs Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 5

4 Schwarzschild Solution: Black Holes II The light cone for Schwarzschild spacetime. Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 6

5 Schwarzschild Solution: Black Holes II The light cone for Schwarzschild spacetime. Collapse Diagram. 6 S.L.Shapiro & S. Teukolsky Black Holes, Neutron Stars and White Dwarfs Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 7

6 Schwarzschild Solution: Embedding diagrams In order to get some feeling for the global geometry of the Schwarzchild black hole we can try to represent aspects of it by embeddings in 3-space. Let us look at the space of constant time and also supress one of the angular coordinates, say θ = π/2. The metric becomes: dl 2 = 1 2M ) 1 dr 2 + r 2 dφ 2 1) r The metric of the Euclidean embedding space is: which on z = zr) becomes dl 2 = dz 2 + dr 2 + r 2 dφ 2 2) dl 2 = 1 + z 2) dr 2 + r 2 dφ 2 3) where z = dz/dr. These are the same if 1 + z 2 = 1 2M ) 1 z = 2 2Mr 2M) 4) r NOTE: We cannot follow the geometry further by this means for r < 2M. Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 8

7 Schwarzschild Solution: Embedding diagrams Gravitational redshift Embedding Diagram for Schwarzschild spacetime. Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 9

8 Eddington-Finkelnstein coordinates I The Schwarzschild solution in a new set of coordinates t, r, θ, φ ) ) r t = t ± ln 2M 1, r > 2M ) t = t ± ln 1 r, r < 2M 2M r = r, θ = θ, φ = φ can be written as ds 2 = 1 2M r ) dt M r ) dr 2 ± 4M r drdt r 2 dθ 2 + sin 2 θdφ 2) While for null surfaces ds 2 0 = 1 2M r ) dt M r ) dr 2 ± 4M r drdt Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 10

9 Eddington-Finkelnstein coordinates II [ dt + dr) 1 2M r ) dt 1 + 2M r ) ] dr = 0 dr dt = 1 and dr dt = r 2M r + 2M A manifold is said to be geodesically complete if all geodesics are of infinite length. This means that in a complete specetime manifold, every particle has been in the spacetime forever and remains in it forever. Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 11

10 Kruskal - Szekeres Coordinates : Maximal Extension v = u = r ) ) 1/2 2M 1 r t e 4M sinh 4M r ) ) 1/2 2M 1 r t e 4M cosh 4M Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 12

11 Kruskal - Szekeres Coordinates : Maximal Extension v = u = r ) ) 1/2 2M 1 r t r ) e 4M sinh 4M 2M 1 r ) ) 1/2 2M 1 r t e 4M cosh 4M e r/2m = u 2 v 2 t 4M = v ) tanh 1 u 12 S.L.Shapiro & S. Teukolsky Black Holes, Neutron Stars and White Dwarfs Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 12

12 Kruskal - Szekeres Coordinates : Maximal Extension v = u = r ) ) 1/2 2M 1 r t r ) e 4M sinh 4M 2M 1 r ) ) 1/2 2M 1 r t e 4M cosh 4M ds 2 = 32M 3 e r/2m dv 2 du 2) r 2 dω 2 r du 2 dv 2 = 0 du dv = ±1 In a maximal spacetime, particles can appear or disappear, but only at singularities. e r/2m = u 2 v 2 t 4M = v ) tanh 1 u 12 S.L.Shapiro & S. Teukolsky Black Holes, Neutron Stars and White Dwarfs Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 12

13 Kruskal - Szekeres Coordinates : Maximal Extension v = u = r ) ) 1/2 2M 1 r t r ) e 4M sinh 4M 2M 1 r ) ) 1/2 2M 1 r t e 4M cosh 4M ds 2 = 32M 3 e r/2m dv 2 du 2) r 2 dω 2 r du 2 dv 2 = 0 du dv = ±1 In a maximal spacetime, particles can appear or disappear, but only at singularities. e r/2m = u 2 v 2 t 4M = v ) tanh 1 u 12 S.L.Shapiro & S. Teukolsky Black Holes, Neutron Stars and White Dwarfs Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 13

The ends of a wormhole could, in theory, be intra-universe i.e. both exist in the same universe) or inter-universe exist in

14 Wormholes A hypothetical tunnel connecting two different points in spacetime in such a way that a trip through the wormhole could take much less time than a journey between the same starting and ending points in normal space. The ends of a wormhole could, in theory, be intra-universe i.e. both exist in the same universe) or inter-universe exist in different universes, and thus serve as a connecting passage between the two). Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 14

15 Reissner-Nordström Solution ) Solution of static Einstein-Maxwell equations with a potential ) Q A µ =, 0, 0, 0 r 5) where Q is the total charge measured by a distant observer ds 2 = 1 2M r The horizon is for g 00 = 0 two horizons r and r +. ) + Q2 r 2 dt M r ) 1 + Q2 r 2 dr 2 + r 2 dω 2 6) r ± = M ± M 2 Q 2 7) Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 15

16 Kerr Solution 1963) I ds 2 = ρ 2 [ dt a sin 2 θdφ 2] 2 + sin 2 θ ρ 2 [ r 2 + a 2 )dφ adt ] 2 + ρ 2 dr2 +ρ 2 dθ 2 where 16 = r 2 2Mr + a 2 and ρ 2 = r 2 + a 2 cos 2 θ 8) here a = J/M = GJ/Mc 3 is the angular momentum per unit mass. For our Sun J = g cm 2 /s which corresponds to a = Obviously for a = 0 Kerr metric reduces to Schwarszchild. The coordinates are known as Boyer-Lindquist coordinates and they are related to Cartesian coordinates x = r 2 + a 2 ) 1/2 sin θ cos φ y = r 2 + a 2 ) 1/2 sin θ sin φ z = r cos θ 16 If = r 2 2Mr + a 2 + Q 2 the we get the so called Kerr-Newman solution which describes a stationary, axially symmetric and charged spacetime. Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 17

17 Kerr Solution II Infinite redshift surface We know that : 1 + z = ν A /ν B = [g 00 B)/g 00 A)] g 00 = 1 2Mr ρ = 0 2 i.e. r 2 + a 2 cos 2 θ 2Mr = 0 which leads to R ± = M ± M 2 a 2 cos 2 θ ds 2 0 = g 00 dt 2 + g rr dr 2 + g φφ dφ 2 + 2g 0φ dφdt at θ = π/2 ) 2 ) for constant r i.e. dr = 0 we get 0 = g 00 + g dφ φφ dt + dφ 2g0φ dt then at g 00 = 0 we will get dφ dt = 0 and dφ dt = g φφ 9) 2g 0φ Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 18

18 Kerr Black Hole The light cone in the radial direction: ) 2 dr 0 = g 00 dt 2 + g rr dr 2 = g 00 = ρ2 2Mr) dt g rr ρ 4 10) thus the horizons) will be the surfaces: = r 2 2Mr + a 2 = 0 r ± = M ± M 2 a 2 11) The are between r + r R + is called ergosphere. Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 19

19 Kerr Black Hole : Geodesics We will consider test particle motion in the equatorial plane. If we set θ = π/2 in the equation 8) we can get the Lagrangian 2L = 1 2M ) ṫ 2 4aM ) ṫ r r φ + r2 ṙ2 + r 2 + a 2 + 2Ma2 φ 2 12) r Corresponding to the ignorable coordinates t and φ we obtain two first integrals where ṫ = dt/dλ): From 12) we can obtain: p t = L ṫ = constant = E 13) p φ = L = constant = L 14) φ ṫ = r3 + a 2 r + 2Ma 2 )E 2aML r r 2M)L + 2aME φ = r 15) 16) Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 20

20 Kerr Black Hole : Geodesics A 3rd integral of motion can be derived by g µν p µ p ν = m 2 which by using 15) & 16) is: where r 3 ṙ 2 = RE, L, r) 17) R = E 2 r 3 + a 2 r + 2Ma 2 ) 4aMEL r 2M)L 2 m 2 r we can regard R as an effective potential for radial motion in the equatorial plane. Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 21

21 Kerr Black Hole : Circular Orbits Circular orbits occur where ṙ = 0. This requires: R = 0, These equations can be solved for E and L to give R r = 0 18) E/m = Ẽ = r 2 2Mr ± a Mr r r 2 3Mr ± 2a ) 1/2 19) Mr Mr r 2 2a ) Mr + a 2 L/m = L = ± r r 2 3Mr ± 2a ) 1/2. 20) Mr Here the upper sign corresponds to corotating orbits and the lower sign to counterrotating orbits. Kepler s 3rd law : for circular equatorial orbits gets the form Ω = dφ dt = φ ṫ = ± M 1/2 r 3/2 ± am 1/2 21) Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 22

22 Kerr Black Hole : Circular Orbits Circular orbits exist from r = all the way down to the limiting circular photon orbit, when the denominator of Eq. 19) vanishes. Solving for the resulting equation we find for the photon orbit r ph = 2M { 1 + cos [ 2/3 cos 1 a/m) ]} 22) NOTE : For a = 0, r ph = 3M, while for a = M, r ph = M direct) or r ph = 4M retrograde) For r > r ph, not all circular orbits are bound. An unbound circular orbit has E/m > 1 and for an infinitesimal outward perturbation, a particle in such an orbit will escape to infinity. Bound orbits exist for r > r mb, where r mb is the radius of the marginally bound circular orbit with E/m = 1: r mb = 2M a + 2M 1/2 M a) 1/2 23) r mb is the minimum periastron of all parabolic E/m = 1 orbits. In astrophysical problems, particle infall from infinity is very nearby parabolic, since u c. Any parabolic orbit that penetrates to r < r mb must plunge directly into the BH. For a = 0, r mb = 4M, while for a = M, r mb = M direct) or r mb = 5.83M retrograde) Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 23

23 Kerr Black Hole : Stable Orbits Even the bound orbits are not stable. Stability requires that 2 R r ) From eqn 18) we get 1 ) 2 E 2 M m 3 r 25) The solution for r ms, the radius of marginally stable circular orbit is [ r ms = M 3 + Z 2 [3 Z 1 ) 3 + Z 1 + 2Z 2 )] 1/2] 26) ) 1/3 [ Z 1 = a2 M a ) 1/3 + 1 a ) ] 1/3 27) M M Z 2 = 3 a2 M 2 + Z2 1 ) 1/2 28) For a = 0, r ms = 6M, while for a = M, r ms = M direct) or r ms = 9M retrograde) Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 24

24 Kerr Black Hole : Stable Orbits The binding energy of the marginally stable circular orbits can be taken if we eliminate r from Eq 19) and using Eq 25) we find a 21 M = 4 Ẽ 2 ) 1/2 2Ẽ 3 31 Ẽ2 ) 29) The quantity Ẽ decreases from 8/9 a = 0) to 1/3 a = M) for direct orbits and increases from 8/9 to 25/27 for retrograde orbits. The maximum binding energy 1 Ẽ for a maximally rotating BH is 1 1/ 3 or 42.3% of the rest-mass energy. This is the amount of energy that is released by matter spiralling in toward the black hole through a succession of almost circular equatorial orbits. Negligible energy will be released during the final plunge from r ms into the black hole. Note that the Boyer-Lindquist coordinate system collapses r ms, r mb, r ph and r + into r = M as a M. Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 25

Kerr BH: Penrose Process 1969) Particles inside the ergosphere may have negative energy E = p 0 = mg 0µ u µ = m g 00 u 0 + g 0φ u φ) 30) Inside the ergosphere g 00 < 0 while g 0φ > 0 thus with an

25 Kerr BH: Penrose Process 1969) Particles inside the ergosphere may have negative energy E = p 0 = mg 0µ u µ = m g 00 u 0 + g 0φ u φ) 30) Inside the ergosphere g 00 < 0 while g 0φ > 0 thus with an appropriate choice of u 0 and u φ it is possible to have particles with negative energy. This can lead into energy extraction from a Kerr BH. Actually, we can extract rotational energy till the BH reduces to a Schwarzschild one. Starting from outside the ergosphere particle m 0 enters into the ergosphere and splits into two parts. One of them follows a trajectory of negative energy and falls into the BH, the second escapes at infinity carrying more energy than the initial particle! Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 26

26 Black-Hole: Mechanics There is a lower limit on the mass that can remain after a continuous extraction of rotational energy via Penrose process. Christodoulou 1969) named it irreducible mass M ir 4Mir 2 = r+ 2 + a 2 M + ) M 2 a 2 + a 2 = 2Mr + 31) while Hawking 70) suggested that the area of the BH horizon cannot be reduced via the Penrose process. The two statements are equivalent! EXAMPLE : Calculate the area of the horizon ds 2 = g θθ dθ 2 + g φφ dφ 2 = r a 2 cos 2 θ ) dθ 2 + r2 + + a 2 ) 2 sin 2 θ r a 2 cos 2 θ thus the horizon area is A = = 2π π 0 0 2π π 0 0 gdθdφ = 2π 0 π 0 gφφ gθθ dθdφ r a 2 ) sin θdθdφ = 4πr a 2 ) = 8πMr + = 16πM 2 ir Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 27

27 Black Hole Thermodynamics Hawking s area theorem opened up a new avenue in the study of BH physics. The surface area of a BH is analogous to entropy in thermodynamics. In order to associate them one needs : to define precisely what is meant by the entropy of a BH to associate the concept of temperature with a BH Entropy is defined as the measure of the disorder of a system or of the lack of information of its precise state. As the entropy increase the amount of information about the state of a system decreases. No-hair theorem : a BH has only 3 distinguishing features : mass, angular momentum and electric charge. Bekenstein argued that the entropy of a BH could be described in terms of the number of internal states which correspond to the same external appearance. I.e. the more massive is a BH the greater the number of possible configurations that went into its formation, and the greater is the loss of information which has been lost during its formation). Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 28

28 Black Hole Thermodynamics The area of the BH s event horizon is proportional to M 2. The entropy of the BH can be regarded as proportional to the area of the horizon. Bekenstein has actually written S bh = c ka 32) 4 where A is the surface area, Planck s constant divided by 2π and k Boltzmann s constant. The introduction of BH entropy, calls for the definition of the BH temperature! The temperature of a body is uniform at thermodynamical equilibrium 0th Law) The the surface gravity of the BH spherical or axisymmetric) plays the role of the temperature, Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 29

29 Black Hole Thermodynamics: 1st Law The surface gravity of the event horizon of a BH is inversely proportional to its mass GM κ bh = r )r r+ 2 = c4 1 G 4M 33) in analogy the temperature should be inversely proportional to its mass. The less massive is a BH the hotter it would be! In more general case Kerr-Newman) if we make a small change in the mass M, angular momentum J and the electric charge Q we get δm = κ δa + ΩδJ + ΦδQ 34) 8πG where the surface gravity and horizon area of a Kerr-Newman BH are: κ = 4πµ/A A = 4π[2MM + µ) Q 2 ] µ = M 2 Q 2 J 2 /M 2 ) 1/2 This is the 1st law of black hole mechanics Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 30

30 Black Hole Thermodynamics : 2nd & 3rd Law Second law of black hole mechanics: the horizon can never decrease assuming Cosmic Censorship and a positive energy condition 31 Third law of black hole mechanics: the surface gravity of the horizon cannot be reduced to zero in a finite number of steps. Bekenstein-Hawking entropy formula S H = 1 4G A Bekenstein-Hawking temperature formula T = 4π 2 κ kc M M ) K 35) 31 Note that during the evaporation, M decreases and thus so does A and S this violates Hawking s area theorem. However one of the postulates is that matter obeys the strong energy condition, which requires that the local observer always measures positive energy densities and there are no spacelike energy fluxes, which is not the case of pair creation i.e. there is a spacelike energy flux. Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 32

31 Hawking Radiation Classically: Nothing can escape from the interior of a BH. Quantum Mechanically: Black holes shine like a blackbody with temperature inversely proportional to their mass Hawking 1974). Quantum fluctuations of the zero-energy vacuum can create particle - antiparticle pairs. These particles and antiparticles are considered to be virtual in the sense that they don t last long enough to be observed According to Heisenberg s principle E t and from Einstein s equation E = mc 2 we get a relation which implies an uncertainty in mass m t /c 2 This means that in a very brief interval t of time, we cannot be sure how much matter there is in a particular location, even in the vacuum. Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 33

Hawking Radiation Thus particle-antiparticle pairs can be created which may last for at most, a time /c 2 m. Before, the time is up, they must find each other and annihilate!

32 Hawking Radiation Thus particle-antiparticle pairs can be created which may last for at most, a time /c 2 m. Before, the time is up, they must find each other and annihilate! In the presence of a strong field the e and the e + may become separted by a distance of Compton wavelength λ = h/mc that is of the order of the BH radius r +. There is a small but finite probability for one of them to tunnel through the barrier of the quantum vacuum and escape the BH horizon as a real particle with positive energy, leaving the negative energy inside the horizon of the BH. This process is called Hawking radiation The rate of emission is as if the BH were a hot body of temperature proportional to the surface gravity. Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 34

33 Hawking Temperature Hawking calculated that the statistical effect of many such emissions leads to the emergence of particles with a thermal spectrum: NE) = 8π E 2 c 3 h 3 e 4π2 E/κh 1 36) where NE) is the number of particles of energy E per unit energy band per unit volume. Notice the similarity between this formula and the Planckian distribution for thermal radiation from a black body. This similarity led Hawking to conclude that a black hole radiates as a black body at the temperature T T = 4π 2 kc κ = c3 8πkGM 37) Note that in the classical limit 0 the temperature vanishes and the BH only absorbs photons, never emit them. Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 35

34 Hawking Temperature The temperature expressed in terms of mass is T = c3 8πkGM = M M ) K 38) where M is the solar mass. To estimate the power radiated, we assume that the surface of the BH radiates according to the Stefan-Boltzmann law: dp da = σt 4 = π2 k c 2 T 4 39) Substituting the temperature and noting that the power radiated corresponds to a decrease of mass M by P = dm/dt)c 2, we obtain the differential equation for the mass of a radiating BH as a function of t dm = π3 c 4 dt G 2 M 2 40) By integration we get the duration of the radiation t Mc2 P G2 M 3 ) 3 M c sec 41) Kg Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 36

35 Hawking Radiation : Mini - BHs A BH with a mass comparable to that of the Sun would have a temperature T 10 7 K and will emit radiation at the rate of erg/s A BH with mass of about gr is expected to evaporate over a period of about years today). The critical mass M 0 for a BH to evaporate after time t is M 0 = π 8 1 c 4 ) 1/3 t 10 1/3 G 2 M 2 42) We do not have a theory capable of explaining what happens when a BH shrinks within the Planck radius cm), and the answer to this question lies in the area of speculation. Kostas Kokkotas Solutions of Einstein s Equations & Black Holes 37

Solutions of Einstein s Equations & Black Holes 1

Solutions of Einstein s Equations & Black Holes 1 1 March 8, 2018 1 S.L.Shapiro & S. Teukolsky Black Holes, Neutron Stars and White Dwarfs Schwarzschild Solution: Black Holes I The light cone in a Schwarzschild spacetime ds 2 0 = 1 2M ) dt 2 1 2M r r