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PHY390, The and Black Holes

James Lattimer

Department of Physics & Astronomy 449 ESS Bldg. Stony Brook University

April 1, 2021

Black Holes, Neutron and Gravitational Radiation [email protected]

James Lattimer PHY390, The Kerr Metric and Black Holes What Exactly is a ?

Standard definition: A region of space from which nothing, not even light, can escape.

I Where does the escape velocity equal the ? r 2GMBH vesc = = c RSch

This defines the RSch to be

2GMBH M RSch = 2 ' 3 km c M

I The horizon marks the point of no return for any object. I A black hole is black because it absorbs everything incident on the and reflects nothing. I Black holes are hypothesized to form in three ways:

I of a

I A high energy collision

I Density fluctuations in the early universe I In , the black hole’s is concentrated at the center in a singularity of infinite density.

James Lattimer PHY390, The Kerr Metric and Black Holes John Michell and Black Holes

The first reference is by the Anglican priest, John Michell (1724-1793), in a letter written to Henry Cavendish, of the Royal Society, in 1783. He reasoned, from observations of radiation pressure, that light, like mass, has inertia. If affects light like its mass equivalent, light would be weakened. He argued that a with 500 times its radius and the same density would be so massive that it’s escape velocity would exceed light speed. He proposed using a prism to measure the gravitational weakening of starlight due to the surface gravity of the star. He predicted one day they could be observed if some were in binary stars, by their gravitational effect on their companions. Michell, around 1783, designed the experiment now attributed to Cavendish which first accurately measured the force of gravity between . This resulted in the first accurate values for G and M⊕. He invented the torsion balance for the experiment but didn’t live to put it to use. His device was willed to his friend Henry Cavendish, who performed the experiment in 1797-8.

James Lattimer PHY390, The Kerr Metric and Black Holes Michell’s Apparatus – The Torsion Balance κθ = LF = LGmM/r 2 r r I mL2 T = 2π = 2π κ 2κ 2π2Lr 2 G = θ MT 2

Harvard Lecture Demonstration

Cavendish 1798, Phil. Trans. Roy. Soc. Lon., 469

2 M⊕ = gR⊕/G

James Lattimer PHY390, The Kerr Metric and Black Holes More About John Michell He discovered that double stars are more common than statistically probable, which must be a product of their mutual gravitation. Michell tried to measure the radiation pressure of light, but when he focused sunlight onto a compass needle, it melted. He first proposed the accepted explanation for the twinkling of starlight. He proposed that parallax, although too small at present to be observed, could in the future determine a star’s distance, which together with its apparent brightness, could be used to measure it’s true luminosity. He sold a 30” telescope to William Herschel, and introduced him to astronomy. In geology, he proposed that earthquakes were experienced as seismic waves of elastic compression travelling through the Earth, and determined the epicenter of the 1755 Lisbon earthquake. He also first suggested that tsunamis were caused by earthquakes. He laid the foundations of English stratigraphy, especially concerning the Mesozoic strata, without any knowledge of fossils. He first explained how to manufacture articial magnets much stronger than natural ones, and discovered the inverse square law of magnetism. James Lattimer PHY390, The Kerr Metric and Black Holes Other Early Work on Black Holes

I The mathematician Pierre-Simon Laplace made similar arguments in 1796 in the first two editions of his book Exposition du syst´eme du Monde, although the discussion was removed from later editions. I Einstein’s paper describing the advance of the perihelion of Mercury was published on Nov. 25th, 1915. It had an approximate metric, in Cartesian coordinates, valid in the weak-field limit. I within two weeks found the exact analytical solution for the spherical symmetry metric, valid for arbitary field strength, despite the fact that we was at the Russian front in WWI, and was ill with an autoimmune disease. He mailed it to Einstein, who read it on Dec. 15th. This solution was published in Jan. 1916. I A few months later, Johannes Droste, Lorentz’s student, independently derived the solution to the metric. I The metric was followed by its uniform density solution in Feb. 1916. I Schwarzschild died in May 1916 at age 41. I The metric had a bad behavior (a singularity) at the so-called Schwarzschild radius, but it was only later (1939) that Oppenheimer, Tolman and Volkov interpreted this radius as the boundary of a bubble where time stopped. This led to the idea of “frozen stars”. James Lattimer PHY390, The Kerr Metric and Black Holes The Trouble With Singularities

I In 1958, David Finkelstein identified the Schwarzschild radius with the event horizon, a perfect unidirectional membrane: causal influence can cross it in only one direction.

I He and Martin Kruskal extended the Schwarzshild solution into the interior of the event horizon (so it could be applied to infalling observers) by means of a coordinate transformation, thereby showing it was not a true singularity.

I The origin, however, contains a true singularity.

I Although black holes seem exotic, a massive one has the interesting property that, within the event horizon, it has a low average density

3M 3c6 109M 2 ρ¯ = = ' g cm−3. 4πR3 32πG 3M2 M

I But back to that singularity.

James Lattimer PHY390, The Kerr Metric and Black Holes The Trouble With Singularities Coordinate singularities are certainly not confined to general relativity. Consider spherical coordinates at the poles. The north pole has θ = 0, but 0 ≤ φ ≤ 2π. So that point has infinitely multi-valued coordinates. Points separated by zero distance must be the same point, even if the coordinate distance between them does not vanish. This is a symptom of a bad choice of coordinates. In relativity, the situation is more subtle because of certain curves, called null curves, that have zero invariant distance between them. It wasn’t until 1960 that the true nature of the Schwarz- schild event horizon became apparent.

James Lattimer PHY390, The Kerr Metric and Black Holes Infalling Particles and

Let a particle fall inwards to r = 2M from R > r. How much proper time elapses (the time measured on a clock attached to the particle)? For radial infall,  dr 2 2M dr = E˜2 − 1 + , dτ = − dτ r q E˜2 − 1 + 2M/r

where E˜ = −p0/m = −U0 is the energy per unit mass of the particle. We choose the minus sign to represent infall. ˜2 R 2M If E > 1 (unbound particle), R > 0. If E˜ = 1 (particle at rest at ∞), the integral is 4M  r 3/2R τ(R) − τ(2M) = > 0. 3 2M 2M 2 2 If E˜ < 1, the particle can’t get closer than 1 − E˜ = 2M/ri or 2 ri = 2M/(1 − E˜ ) > 2M. Any particle reaching 2M gets there in a finite proper time. In fact, why stop at 2M? The proper time stays finite to any r.

James Lattimer PHY390, The Kerr Metric and Black Holes and Infalling Particles

Now let’s ask the same question about coordinate time. Using the fact that the time component of the four-velocity is U0 = dt/dτ,

dt p  2M −1 U0 = = g 00U = g 00 0 = −g 00E˜ = 1 − E˜. dτ 0 m r

Ed˜ τ (1 − 2M/r)−1Edr˜ dt = = − . 1 − 2M/r (E˜2 − 1 + 2M/r)1/2 Define ε = r − 2M, so that

(ε + 2M)3/2Ed˜ ε 2Mdε dt = − −→ − . ((2M + ε)E˜2 − ε)1/2ε ε→0 ε

This diverges logarithimically for all E˜, so an infinite coordinate time elapses. It is behaving badly. Nothing happens to the particle as it falls through the event horizon. Tidal forces don’t suddenly become infinite there.

James Lattimer PHY390, The Kerr Metric and Black Holes Inside 2M

Now define ε = 2M − r, so that the metric becomes ε 2M − ε ds2 = dt2 − dε2 + (2M − ε)2dΩ2. 2M − ε ε

On a trajectory for which t, θ and φ are constant, ds2 < 0, which means it is timelike. Thus ε, and r, is timelike, while t is now spacelike. The infalling particle must follow a timelike , so r perpetually decreases. Eventually, r = 0 is reached, a point at which there is finally a true singularity. A particle sending an outwardly directed photon does so futiley, because the photon must also go forward in time as seen locally; it also has to follow a timelike path, which means to decreasing r. Nothing inside r = 2M gets out and is doomed to encounter the singularity at r = 0. Nothing inside the event horizon can be seen outside of it; hence the name horizon.

James Lattimer PHY390, The Kerr Metric and Black Holes The Coordinate Singularity Consider the light cones of various events occuring outside the event horizon. These are cones showing radially ingoing and outgoing null lines emanating from the events. They are computed by solving ds2 = 0 for dθ = dφ = 0 (radial), dt  2M −1 = ± 1 − . dr r Far from the star, dt/dr = ±1, 45◦. For events close to r = 2M, dt/dr → ±∞; the cones close up. Particles are confined to move within light cones, limited in velocity by c. As r → 2M, they can’t get to r = 2M until t → ∞. In fact, no particle can reach r = 2M for any finite t. Perhaps the line r = 2M is not a line but a point in : a single event has been expanded into the whole line r = 2M. This is analogous to spherical coordinates at the pole, where a whole line represents but a single point. But in reality, light cones don’t close up near the horizon because particles can get there in finite proper time and aren’t destroyed automatically.

James Lattimer PHY390, The Kerr Metric and Black Holes Kruskal-Szekeres Coordinates u = (r/2M − 1)1/2er/4M cosh(t/4M)  r > 2M v = (r/2M − 1)1/2er/4M sinh(t/4M) u = (1 − r/2M)1/2er/4M cosh(t/4M)  r < 2M v = (1 − r/2M)1/2er/4M sinh(t/4M) 32M3 ds2 = − e−r/2M (dv 2 − du2) + r 2dΩ2, r

u2 − v 2 = (r/2M − 1)er/2M gives r(u, v) implicitly. The transformation is singular at r = 2M, but removes the coordinate singularity there. A singularity exists at r = 0. A radial null line (dθ = dφ = ds = 0) is a line dv = ±du, i.e., always 45◦, and never closed.

James Lattimer PHY390, The Kerr Metric and Black Holes Black Holes in Kruskal Coordinates

I θ and φ are suppressed, each point represents a two sphere of events. 2 2 I u(r, t) and v(r, t), but u − v = f (r). Fixed r is a u-v hyperbolae. I These hyperbolae are vertical for r > 2M, but horizontal for r < 2M. I For r < 2M, timelike lines confined within a light cone cannot remain at fixed r; for r > 2M, timelike lines cannot remain at fixed t.

I The ’point’ r = 0 is actually an entire hyperbola in u-v space, and is the end of space, being a singularity.

I Lines of t = constant are orthogonal to those of fixed r, and are straight lines radiating outwards from the origin. ◦ I t = 0 is horizonal; t → ∞ is 45 . I All t = constant lines pass through (u, v) = (0, 0); this origin can be expanded into a whole line in a (t, r) diagram. The dashed world line, after passing through the t = ∞ line, can’t go back to finite t. This is the horizon, r = 2M. But both u and v remain finite.

James Lattimer PHY390, The Kerr Metric and Black Holes Notes

I A distant observer sees that τ = t. An infinite proper time elapses to get the information that a particle passed through the horizon. I If an infalling object sends out regular pulses, only a finite number will be emitted before reaching the horizon. But these pulses appear to the observer as stetched out over longer and longer times, i.e., they are redshifted. I The horizon r = 2M is itself a null line, because it is the boundary between a null ray that can get out and one that can’t. I Spacetime is divided into four quadrants: I I is the exterior r > 2M.

I II is the interior r < 2M.

I III and IV have no applicability to us. I The dashed worldline could represent the surface of a spherically-symmetric collapsing star, since everything exterior to it is described by the Schwarzschild geometry and, thus, this diagram. I Everything interor to the dashed line, the stellar interior, is described by a different metric and its geometry possibly has no relation to this diagram. This includes all of III and IV. I Far from the star u and v are inconvenient, but r and t are fine. James Lattimer PHY390, The Kerr Metric and Black Holes Spherically Symmetric Time-Dependent

We’d like to understand further the black hole formation process resulting from gravitational collapse. To do so, we have to generalize the spacetime to include time dependence. It is possible to show that the most general form would be

ds2 = −e2Φ(r,t)dt2 + e2Λ(r,t)dr 2 − r 2d 2Ω. (1)

A variant of this is to use co-moving coordinates, in which the time coordinate is the proper time to a local freely-falling observer:

ds2 = −dt2 + U(r, t)2dr 2 + V (r, t)2dΩ2. (2)

ν The stress-energy tensor for a perfect fluid is Tµ = diag(−ε, p, p, p). We will utilize the Einstein relations

ν ν µ Gµ = 8πTµ , ∇µTν = 0.

James Lattimer PHY390, The Kerr Metric and Black Holes Einstein, metric and stress-energy tensors: 1 8πG G = R − g = T . µν µν 2 µν c4 µν Ricci tensor and Ricci scalar: α µ µα Rµν = Rµνα, R = Rµ = g Rµα. : β β β α β α β Rνρσ = Γνσ,ρ − Γνρ,σ + ΓνσΓαρ − ΓναΓασ. Christoffel symbols: 1 Γµ = g µλ(g + g − g ). νσ 2 λν,σ λσ,ν νσ,λ β β β α β α β Rµν = Rµνβ = Γµβ,ν − Γµν,β + ΓµβΓαν − ΓµνΓαβ. Commas indicate derivatives: ∂Γβ Γβ = νσ = ∂ Γβ . νσ,ρ ∂x ρ ρ νσ James Lattimer PHY390, The Kerr Metric and Black Holes Gravitational Collapse of Dust

We are going to consider a homogenous sphere of dust (p = 0, ε = ε(t)). µ Energy conservation is ∇µTr = 0, which is automatically satisfied. conservation is ∂(εV 2U) ∇ T µ = 0 = = 0. µ t ∂t The needed components of the Einstein tensor can be shown to be ! 1  F 0  1 2R0 F˙ 2 G 0 = 2VH˙ − , G 1 = − H + , G 1 = H, 0 VV 0 V 1 V V˙ U2 V 0 V ! U˙  V 02  ∂V ∂V H = V˙ 0 − V 0 , F = V 1 − + V˙ 2 , V˙ = , V 0 = . U U2 ∂t ∂r

1 0 1) G0 = 0 gives H = 0, or ∂(V /U)/∂t = 0. Thus, we can set U = R(t)f (r) and V = R(t)g(r) by a rescaling of f and g. 1 ˙ −2 02 2 ¨ ˙ 2 2) G1 = 0 gives F = 0, leading to g (1 − g /f ) = −2RR − R . 0 0 2 0 2 0 3 3) G0 = −8πε gives F = 8πεV V = 8πg g εR . James Lattimer PHY390, The Kerr Metric and Black Holes Dust Collapse Continued

2) and 3) therefore give that 1  g 02  F 0 1 − = κ = −2RR¨ − R˙ 2, = α = εR3. g 2 f 2 8πg 2g 0 Both κ and α have to be positive constants, since both LHSs are functions of r and both RHSs are functions of t. We still have freedom to redefine the radial coordinate, so we do so by defining g(r) = r. The relations above then give

−1/2 F (r) 8π f (r) = 1 − κr 2 , = α = R[κ + R˙ 2]. r 3 3 The last equation describes the collapse evolution. The metric becomes  dr 2  ds2 = −dt2 + R2(t) + dΩ2 . 1 − κr 2 This is just the Robertson-Walker metric used in cosmology. 3 We set R(t = 0) = R0 and ε(t = 0) = ε0. Then α = ε0R0 , and  R 3 4π ε(t) = ε(0) 0 , M = ε R3. R(t) 3 0 0 James Lattimer PHY390, The Kerr Metric and Black Holes Black Hole Formation

Consider our evolution equation:

2M  1 1  R˙ 2 = − κ = 2M − R R R0

assuming the initial condition R˙ (t = 0) = 0. Elapsed proper time since start of collapse:

r 1 Z R  1 1 −1/2 t = − − dR 2M R0 R R0 r " r # R3 r R R = 0 1 − + atan 0 − 1 . 2M R0 R

Note each mass shell reaches the singularity r = 0 at exactly the same time t(R = 0) = p 3 τcol = π R0 /(8M), which does not depend 3 R0 but on ε0. This is about 3541 s if M = 4πε0R0 /3 = 1M and R0 = R . In comparison, R /c ' 2.3 s.

James Lattimer PHY390, The Kerr Metric and Black Holes More About Black Hole Formation

The event horizon is the boundary in spacetime dividing events that can communicate with distant observers and events than cannot. It is a 3-dimensional surface that separates trapped from untrapped events. The test of trapping is whether one can send light rays, i.e., null rays, to infinity. As a boundary, the horizon is itself composed of (marginally trapped) null world lines. The event horizon grows from zero radius. As matter falls in, trajectories of photons beginning from the star’s center are more and more affected. The horizon is marked by the null ray that just gets trapped and remains on the horizon. Anything emitted later is trapped, so the marginal null ray is in fact marking the horizon at all times.

James Lattimer PHY390, The Kerr Metric and Black Holes Some Theorems

What happens to the horizon, if at a later time more mass falls in? The new horizon must enclose the old surface at an earlier time. The old horizon was only an apparent horizon. It’s not possible to determine the location of a horizon at a particular time; we must look at its entire evolution, including the future, to determine the boundary between trapped and untrapped regions. Eventually, any hoizon should become stationary, determined by just M and J as seen at infinity, not by any interior integrals. The exterior metric is affected by any mass outside the horizon (accretion disc, for example). Hawking’s area theorem says in any dynamical process involving black holes, the total area of all the horizons cannot decrease in time. This analogy to entropy leads to black hole thermodynamics. Quantum mechanics violates the area theorem because it allows negative energies. Inside horizons there are curvature singularities, where tidal forces become infinite. Does quantum mechanics prevent this? Cosmic Censorship Hypothesis says there can be no singularites outside horizons (naked singularities). These would not allow predictive science. However, the Big Bang is a ! James Lattimer PHY390, The Kerr Metric and Black Holes The Kerr Metric

Roy Kerr found the metric valid outside a rotating object in 1963. It depends only on M and the spin parameter a = J/M. In Boyer-Lindquist coordinates: ds2 = −dt2 +Σ(∆−1dr 2 +dθ2)+(r 2 +a2) sin2 θdφ2 +2MrΣ−1(a sin2 θdφ−dt)2, Σ = r 2 + a2 cos2 θ, ∆ = r 2 − 2Mr + a2.

I It does not depend on t or φ, so it’s stationary and axisymmetric.

I It’s not invariant upon time reversal, but is invariant on the inversion of both t and φ, corresponding to rotation in the opposite direction.

I For r → ∞ it reduces to the Minkowski metric, so it’s asymptotically flat.

I For a → 0 it reduces to the .

I In the limit M → 0 with a 6= 0 it reduces to the metric of flat space in spheroidal coordinates p p x = r 2 + a2 sin θ cos φ, y = r 2 + a2 sin θ sin φ, z = r cos θ.

I The metric is singular for ∆ = 0 and for Σ = 0, but ∆ = 0 is only a coordinate, not a curvature, singularity. In the Schwarzschild metric, r = 0 is a curvature singularity and r = 2M is a coordinate singularity.

James Lattimer PHY390, The Kerr Metric and Black Holes Frame Dragging

Define the Killing fields kµ = (1, 0, 0, 0) and mµ = (0, 0, 0, 1). Consider an µ φ infalling observer with zero , L = u mµ = uφ =x ˙ = 0. For r → ∞, the metric is flat, so uφ = 0. However, this does not vanish for finite r, φ φt u = g ut , and the observer obtains a finite angular velocity dφ (dφ/dτ) ω =x ˙ φ = = 6= 0. dt (dt/dτ) φ t uφ = gφφu + gφt u , angular velocity of a zero-angular momentum particle is uφ g 2Mar ω(r, θ) = = − φt = t 2 2 2 2 2 u gφφ (r + a ) − a ∆ sin θ which satisfies ω/(Ma) > 0. The observer is dragged around in the same direction as the source is rotating. This phenomenon occurs only because 3 gφt 6= 0. It weakens roughly as 1/r , and is a way to measure a in principle. This effect has a close analogue to magnetism. In electromagnetism, a spinning charge creates additional effects we call magnetism. The Lense-Thirring effect is the small precession experienced by a gyroscope placed in orbit around a rotating star; it is proportional to the star’s angular momentum, just as a spinning electron precesses passing through a magnetic field. It has been experimentally confirmed for the Earth (). The effect is observed in the double pulsar system PSR J0737-3039, and might be observable in X-ray emission near black holes.

James Lattimer PHY390, The Kerr Metric and Black Holes The Horizon

Consider the case when ∆ = r 2 + a2 − 2Mr = 0, which has no real solution if a2 > M2. Then there is no horizon, and the singularity where Σ = 0 is ”uncovered” or ”naked”, which would bring all sorts of paradoxes. Thus |a| ≤ M. This is called the Cosmic Censorship Hypothesis. We can write

2 2 p 2 2 ∆ = r − 2Mr + a = (r − r+)(r − r−), r± = M ± M − a .

r+ (r−), where ∆ = 0 and grr = ∞, is the outer (inner) horizon. The outer horizon is the true event horizon, as no null rays can escape from inside r+. This horizon is a surface of constant r and t, and has an intrinsic metric found from ds2 using dt = dr = 0: (r 2 + a2)2 − a2∆ d`2 = sin2 θdφ2 + Σ2dθ2. Σ2 The proper area of this surface is the integral of the square root of the determinant of this metric over θ and φ: Z 2π Z π A(r) = p(r 2 + a2)2 − a2∆ sin θdθdφ = 4πp(r 2 + a2)2 − a2∆, 0 0

but the horizon r = r+ has ∆ = 0, so 2 2 p 2 2 Ahorizon = 4π(r+ + a ) = 8πM(M + M − a ). 2 For a Schwarzschild hole, Ahorizon = 16πM . James Lattimer PHY390, The Kerr Metric and Black Holes The Ergoregion

When ∆ = 0, one finds at r = r+ φ u 2Mar+ a a ω(r+) = t = 2 2 = = 2 2 , u (r+ + a ) 2Mr+ r+ + a which can be considered to be the angular velocity of the black hole itself. Since 2Mr r 2 − 2Mr + a2 cos2 θ (r − r )(r − r ) g = −1 + = − = − E+ E− = 0 tt Σ Σ Σ √ 2 2 2 the radii rE± = M ± M − a cos θ are where gtt changes sign and are

infinite surfaces. rE+ (rE− ) defines the outer (inner) edge of the . The ergosphere lies outside the event horizon, except at the poles where it is tangent to it. Consider photons moving tangentially to r in the ± direction. From ds2 = 0, s  2 2 2 dφ gtφ gtφ gtt 0 = gtt dt + 2ttφdtdφ + gφφdφ , = − ± − . dt gφφ gφφ gφφ

When gtt = 0, dφ/dt = 0 or −2gtφ/gφφ, i.e., photons moving backwards are stationary! Any massive particle is dragged around with the hole.

James Lattimer PHY390, The Kerr Metric and Black Holes Kerr Black Hole

James Lattimer PHY390, The Kerr Metric and Black Holes Orbits in the Kerr Geometry

Take sin θ = 1 and solve the Euler-Lagrange equations for t and φ, 2M  2   r 2 t˙2 = t˙ − aφ˙ + r 2 + a2 φ˙2 + r˙2 − κ, r ∆  2MaE  2M  φ˙ = ∆−1 − L 1 − r r  2Ma  t˙ = ∆−1 E(r 2 + a2) + (Ea − L) r As before, L and E are the conserved angular momentum and energy per unit mass of a (or the angular momentum and energy of a photon), κ = −1 for a massive particle, and κ = 0 for a photon. Eliminate φ˙ and t˙: L2 − E 2a2 + a2 2M(Ea − L)2 κ∆ r˙2 + − − =r ˙2 + Φ(r) = E 2, r 2 r 3 r 2 This can be written as " 2 22 2 # r + a − a ∆ κ∆ r˙2 = (E − V )(E − V ) + , r 4 + − r 2 √ where 2MaL ± r|L| ∆ V± = r , V+ ≥ V−. (r 2 + a2)2 − a2∆ James Lattimer PHY390, The Kerr Metric and Black Holes Photon Orbits in the Kerr Geometry

Thus Φ can be written as

" 2 22 2 # r + a − a ∆ κ∆ Φ = E 2 − (E − V )(E − V ) − . r 4 + − r 2

One finds that V+ + V− ∝ a which → 0 in the Schwarzschild limit, but 2 V+V− → (L/r) (1 − 2M/r) in that limit, so we have

 2M   L2  r˙2 = E 2 − 1 − Φ(r), Φ(r) = 1 − − κ − 1. r r 2

In the Kerr case, √ 2 2 2 2 2 4M a − r ∆ 2|L|r ∆ V+V− = (Lr) 2 , V+ − V− = 2 . (r 2 + a2)2 − a2∆ (r 2 + a2) − a2∆

In the case of photons, which follow null with κ = 0, we have 2 r˙ ∝ (E − V+)(E − V−) so either E < V− or E > V+, which forbids occupancy in the interval V− < E < V+.

James Lattimer PHY390, The Kerr Metric and Black Holes Behavior of the Kerr Potentials

V− < E < V+ is forbidden.

When ∆ = 0, i.e.,√ when 2 2 r = r+ = M + M − a , 2Mr+La V+ = V− = = Lω. 2 2 2 (r+ + a ) For r → ±∞, V± → 0. When L > 0 (co-rotating orbit), V > 0 + a → M =⇒ r = M for all r, but V vanishes when ph √ − a → 0 =⇒ rph = 3M r ∆ = 2Ma, or when r = 2M, the a → −M =⇒ rph = 4M location of the ergosphere rE+ in the equatorial plane. When L < 0 (counter-rotating orbit), V− < 0 for all r, and V+ vanishes at the ergosphere r = 2M = rE+. Both potentials have zero derivatives only at a single point, rph: 2 2 rph(rph − 3M) = 4Ma , http://www.roma1.infn.it/teongrav/leonardo/bh/bhcap4.pdf the location of unstable circular orbits.

James Lattimer PHY390, The Kerr Metric and Black Holes The Bizarre Case When E < 0

Normally, we consider E to be the energy per particle at infinity, which cannot

be negative. Nevertheless, it does appear from the potentials that for r < RE+ negative values of E seem to be allowed. µ µ Define the energy measured by an observer u to be E = −u pµ. A static µ observer at infinity has u = (1, 0, 0, 0) therefore has E = −p0 which must be positive or else the particle is moving backwards in time. And the existence of negative energy states would imply it is energetically favorable to create more and more such particles. Now consider a massless particle moving within the ergosphere, where static observers can’t exist. Consider the zero angular momentum observer (ZAMO): 2Mar uµ = C(1, 0, 0, Ω), ω = , ZAMO (r 2 + a2)2 − a2∆ µ where C > 0 is determined by gµν u uν = −1. The ZAMO sees the particle ZAMO µ ZAMO having energy E = −pµuZAMO = C(E − µL), so E > 0 means E > Lω, or V− < LΩ < V+.

Geodesics with E > V+ are allowed and those with E < V− are forbidden. In the case that La < 0 (counter-rotating particles), V+ < 0 so that it appears E < 0 is allowed. This is not a contradiction, since it is only at infinity that E represents the physical energy, and these geodesics don’t reach infinity.

James Lattimer PHY390, The Kerr Metric and Black Holes The

µ µ We define here E˜ = −E uµ and L˜ = m uµ, so that E˜ and L˜ represent the total (i.e., not per unit mass) energy and angular momentum at infinity. This discussion thus applies to both massive and massless particles. Assume a > 0. Shoot a particle at infinity towards the black hole in the equatorial plane, so µ µ Pµ = (−E˜, Pr , 0, L˜). PµE = −E˜ and pµm = L remain constant, although pr changes. Now arrange to have the particle decay when it gets into the ergosphere. The µ µ daughters i = 1, 2 have piµE = −E˜i and piµm = L˜i . Conservation of four-momentum is pµ = p1µ + p2µ. Contract pµ with -Eµ to find E˜ = E˜1 + E˜2 and L˜ = L˜1 + L˜2, as expected. Further arrange that particle 1 falls into the black hole but particle 2 reaches infinity. If particle 1 never goes outside the ergosphere, we can arrange V−(r+) < E˜1 < 0 if L˜1 < 0. Thus

E˜2 = E˜ − E˜1 > E˜, L˜2 = L˜ − L˜1 > L˜.

Rotational energy has been extracted from the black hole, which spins down, and the escaping particle has more energy than the original particle initially had.

James Lattimer PHY390, The Kerr Metric and Black Holes Massive Particle Equatorial Orbits

In this case, κ = −1, and A ∆ r˙2 = (E − V )(E − V ) − , A = r 3 + a2r + 2a2M, r 3 + − r 2 2aML ± X r 3r˙2 = A(E − V 0 )(E − V 0 ), V 0 = , X 2 = r∆(L2r + A). + − ± A This can also be written as 3 2 2 h 2  i r r˙ = (r − rcirc ) E − 1 (r + 2rcirc ) + 2M . For circular orbits, E = V 0 and dV 0 /dr = d(X /A)/dr = 0, which give + √ + √ 2 2 2 2 2 r − 2Mrcirc + a Mrcirc M r + a − 2a Mrcirc E 2 = circ √ , L2 = circ √ . circ 2 2  circ 2  rcirc rcirc − 3Mrcirc + 2a Mrcirc rcirc rcirc − 3Mrcirc + 2a Mrcirc

Because E > 0, rcirc reaches an extremum when dE/drcirc = 0, or when

2 √ 2 rISCO − 6MrISCO + 8a MrISCO − 3a = 0.

When a = 0, we find rISCO = 6M, the ISCO for the Schwarzschild solution. When a = ±M, one finds 2 3/2 1/2 2 rISCO − 6MrISCO ± 8M rISCO − 3M = 0. 4 3 2 2 3 4 rISCO − 12MrISCO + 30M rISCO − 28M rISCO + 9M = 0. 3 This is (rISCO− − 9M)(rISCO+ − M) = 0, so rISCO− = 9M and rISCO+ = M. James Lattimer PHY390, The Kerr Metric and Black Holes Accretion in Kerr

From the above expression for E 2, we can determine the maximum energy of accretion in the Kerr geometry. One cannot simply insert a = M and rcirc = M into this equation, however, because both numerator and denominator vanish.

However, from the equation for rISCO(a), one can determine that √ drISCO 4 MrISCO − 3a = p → ∞ da rISCO − 3M + 2a M/rISCO

in the limit that a → M, showing that a → M faster than rISCO → M. Therefore setting a = M in the original expression, we have

3/2 2 2 2 (1 − 2M/rISCO + (M/rISCO ) ) x /4 1 E = 3/2 → 2 → 1 − 3M/rISCO + 2(M/rISCO ) 3x /4 3

where we used rISCO ' M(1 + x) and used the Taylor expansion (1 + x)n ' 1 + nx + n(n − 1)x 2/2 + ... . Therefore, the maximum energy release is 1 − p1/3 ' 42.3%. For a counter-rotating particle, E 2 = 121/135, and the released energy is 5.3%.

James Lattimer PHY390, The Kerr Metric and Black Holes Orbit Summary (1972)

The marginally bound orbits are found by setting E 2 = 1 in the above. One finds

2 2 rmb(rmb − 4M) =

2 2 2 a (8Mrmb−2rmb−a ). For a = 0, rmb = 4M.

For a = −M, √ rmb = M(3 + 8). For a = M, rmb = M.

Note rms = rISCO .

James Lattimer PHY390, The Kerr Metric and Black Holes Kepler’s 3rd Law in Kerr

A circular satisfiesr ˙ =r ¨ = 0, which becomes ∂g ∂g ∂g tt t˙2 + 2 tφ t˙φ˙ + φφ φ˙2 = 0. ∂r ∂r ∂r The angular velocity in a circular orbit is constant, ω = φ/˙ t˙. Using the notation gtt,r = ∂gtt /∂r, one obtains

1 2gtφ,r gφφ,r 2 + + = 0 ω ωgtt,r gtt,r Using

2M Ma  Ma2  g = − , g = , g = 2 r − , tt,r r 2 tφ,r r 2 φφ,r r

we find r 1 r 3 ±M1/2 = a ± , ω = . ω M r 3/2 ± aM1/2 Therefore, a co-rotating particle has a smaller orbital frequency and longer orbital period than a counter-rotating particle. For a → 1, ω of a co-rotating particle is (1 + (M/r)3/2)−1 times the Newtonian or Schwarzschild value.

James Lattimer PHY390, The Kerr Metric and Black Holes General Motion in Kerr

One property of the Kerr metric is the existence of a mysterious third constant of the motion when motion outside the equatorial plane is considered. We need to use the metric coefficients and their inverses:  2Mr  4aMr Σ g = − 1 − , g = − sin2 θ, g = , tt Σ tφ Σ rr ∆ r 2 + a22 − a2∆ sin2 θ g = sin2 θ, g = Σ; φφ Σ θθ 2 22 2 2 r + a − a ∆ sin θ 2aMr ∆ − a2 sin2 θ g tt = − , g tφ = − , g φφ = , ∆Σ ∆Σ ∆Σ sin2 θ ∆ 1 g rr = , g θθ = . Σ Σ The energy per unit mass E = −ut and angular momentum per unit mass µν L = uφ are conserved, and we must have g uµuν = −1, giving

2 h 2 2 2 22i 2  2 2 E a ∆ sin θ − r + a + 4aMrEL + L ∆ csc θ − a ∆(u )2 + (u )2 −1 = + r θ . ∆Σ Σ In the Kerr case, we can’t just rotate the coordinate frame to put a particle in the equatorial plane because of Kerr’s preferred axis.

To find the general particle motion, we need to find ur and uθ.

James Lattimer PHY390, The Kerr Metric and Black Holes The

µ µ µ Evaluate duθ/dτ, usingx ˙ = dx /dτ = u : du d(Σuθ) 1 1 1 θ = = g uµuν = g g αµg βν u u = − g αβ u u . dτ dτ 2 µν,θ 2 µν,θ α β 2 ,θ α β Since Σ appears in the denominator of g αβ , it’s easier to write du 1 h i Σ θ = − (Σg αβ ) u u + Σ , dτ 2 ,θ α β ,θ αβ 2 using g uαuβ = −1. Since Σ,θ = −2a sin θ cos θ and 2 Σduθ/dτ = (1/2)d(uθ) /dθ, we have d(u )2 θ = −(Σg αβ ) u u + 2a2 sin θ cos θ. dθ ,θ α β Now only Σg tt and Σg φφ depend on θ, and the derivatives don’t depend on r. d(u )2 d sin2 θ d csc2 θ θ = −a2E 2 − L2 + 2a2 sin θ cos θ. dθ dθ dθ Integrate this: 2 2 2 2 2 2 2 2 (uθ) = −a E sin θ − L csc θ + a sin θ + constant = a2(E 2 − 1) cos2 θ − L2 cot2 θ + C, where C is known as the Carter constant, an accidental conserved quantity.

James Lattimer PHY390, The Kerr Metric and Black Holes Periodicity of Motion out of the Equatorial Plane √ For bound orbits, E < 1, C > 0; uθ has a maximum equal to C at θ = π/2 and a minimum of −∞ for θ = 0, π.

uθ = 0 when θ = π/2 ± i, where i is an inclination angle that bounds the θ motion of the particle. Note that C = a2(1 − E 2) sin2 i + L2 tan2 i. √ In the Newtonian limit, where E ∼ 1 and L >> 1, we have tan i = C/L. For retrograde orbits, L < 0 and tan i < 0. Particles reach the poles only if L = 0. θ We can’t integrate dθ/dτ = u = Σuθ because Σ depends on both r and θ. However, if we define dλ = dτ/Σ, we find

Z θ dθ = λ − cθ p 2 2 2 2 2 π/2−i a (E − 1) cos θ − L cot θ + C

with cθ a constant.

In the Newtonian or Schwarzschild limit, cos θ ' sin i cos[L(λ − cθ)].

θ is a periodic function of λ with some period Pθ, which is 2π/L in the Schwarzschild limit. James Lattimer PHY390, The Kerr Metric and Black Holes Periodicity in the Radial Motion

µν Substituting uθ into the normalization condition g uµuν = −1,

2 h 2 2 2 22i 2 2 E a ∆ sin θ − r + a + 4aMrEL + L ∆ − a + ∆C ∆(u )2 −1 = + r , ∆Σ Σ or  2  2 2  2 2 2 2  2 2 2 0 = ∆R + E r + a + a ∆ + 4aMrEL + L ∆ − a + ∆C + ∆ (ur ) .

2 2 4 2 3 h 2 2 2 i 2 −∆ (ur ) = r (1 − E) − 2Mr + a (1 − E ) + L + C r h i −2M (aE − L)2 + C r + a2C ≡ V (r).

Particles are trapped in regions where V < 0. V (r+) ≤ 0 and V >> 0 for r → ∞. There are either 1 or 3 real roots of V . If 1 root, particles fall into the hole and there are no stable orbits; if 3, there is a bounded trapped region where V is negative. r √ Trajectories are found by noting that dr/dλ = Σu = ∆ur = −V , Z r dr p = λ(r) − cr . rmin −V (r)

The function λ(r) is periodic, with period Pr = 2[λ(rmax ) − cr ].

James Lattimer PHY390, The Kerr Metric and Black Holes Longitudinal and Temporal Motions

The remaining motions are

" 2 22 # dt r + a 2aMrL = Σut = − a2 sin2 θ E − , dλ ∆ ∆ dφ  a2  2aMrE = Σut = csc2 θ − L + . dλ ∆ ∆

These functions have average (over λ) values bt =< dt/dλ >λ and bφ =< dφ/dλ >λ, so that

t = ct + bt λ + Dtr (λ) + Dtθ(λ), φ = cφ + bφλ + Dφr (λ) + Dφθ(λ),

where Dµr are periodic with period Pr and Dµθ are periodic with period Pθ.

James Lattimer PHY390, The Kerr Metric and Black Holes Precession

Quasi-circular orbits at a given value of r with inclination i will precess at a rate equal to the rate of motion in the longitudinal direction minus the rate of bobbing up and down in latitude:   −1 2π Ωprec = Ωφ − Ωθ = bt bφ − . Pθ √ In the case of particles at large r, E ∼ 1 and L = Mr cos i. Then

2 22 r + a 2aMr b = E − L − a2E < sin2 θ > → r 2. t ∆ ∆ λ 2aMr a2L b = E − + < csc2θ > . φ ∆ ∆ λ The second term can be ignored for large r. Subtracting 2π/Pθ,

 Z Pθ  −1 2 Pθ L csc θdλ − 2π ' 0 0 to the order we seek (lots of algebra omitted here). Thus, 2aM 2GJ Ω ' = prec r 3 c2r 3 where J is the angular momentum of the rotating object; for a black hole J = aM, for a J = I Ωp where Ωp = 2π/P and P is the spin period.

James Lattimer PHY390, The Kerr Metric and Black Holes Limits to Energy Extraction from a Kerr Hole

In the Penrose process, a particle entering the ergosphere decays; daughter 1 falls into the hole and daughter 2 goes to infinity. If the angular momentum of particle 1 is fixed, the energy that can be extracted is

2Mar+L1 a δM = E1 ≥ V+(r) ≥ V+(r+) = = L1ΩH = δJ 2 2 2 2 2 (r + a ) r+ + a

The equality holds if particle 1 is produced at r = r+ with infinitesimal radial 2 2 velocity. ΩH = a/(r+ + a ) = a/(2Mr+) is the angular velocity of the horizon. Using a = M/J in the last equality, this inequality becomes JδJ  J2  MδM ≥ =⇒ δ M2 − ≥ 0, r 2 + a2 r 2 + a2 which can be proved since+ +  2   2    2 J 4J r+ JδJ δ M − 2 2 = 2 + 2 MδM − 2 2 ≥ 0. r+ + a (r+ − M)(2Mr+) r= + a 2 2 2 2 2 2 But M − J /(r+ + a ) = (r+ + a )/4, so it appears the irreducible mass s pr 2 + a2 r 2 + Mr+ 2 J Mirr = = = M − 2 2 2 4Mirr

satisfies δMirr ≥ 0. James Lattimer PHY390, The Kerr Metric and Black Holes Black Hole Area Theorem

Only reversible processes leave Mirr fixed; otherwise it increases. To extract the maximum possible energy per unit black hole mass , perform a sequence of reversible Penrose processes until J = 0, so that r 2 + a2 Mr M2 = M2 = + = + . fin irr 4 2 Then ∆M M − M r r = fin = 1 − + . M M 2M √ So if a = M, r+ = M and ∆M/M = 1 − 1/ 2 = 0.29; if a = 0, ∆M/M = 0.

Insight into Mirr stems from a comparison to black hole’s horizon area, which is the 2-surface with t = constant and r = r+. The metric is 2 22 2 2 2 2 2 2 r+ + a sin θ 2 dσ = (r+ + a cos θ)dθ + 2 2 2 dφ . r+ + a cos θ The area is 2 q 2 d A = gθθgφφ − gθφdθdφ, Z 2 2 2 2 2 A = dθdφ(r+ + a ) sin θ = 4π(r+ + a ) = 16πMirr . During a merger involving two black holes, the Area Theorem must apply.

James Lattimer PHY390, The Kerr Metric and Black Holes Black Hole Thermodynamics

Consider the relation δM ≥ ΩH δJ. To make this an equality, we need to consider the change in Mirr . Begin with 2 2 2 2 J A 4πJ M = Mirr + 2 = + . 4Mirr 16π A This gives ∂M ∂M κ dM = dJ + dA = ΩH δJ + δA ∂J ∂√A 8π 2 2 2 2 if we define the surface gravity κ = M − a /(r+ + a ), which is the acceleration of a ZAMO at the horizon. κ = 0 for a = M, and κ = 1/(4M) for a = 0.

dM = ΩH dJ + κdA/(8π) resembles the first law of thermodynamics, with entropy SBH = αkB A and temperature TBH = κ/(8πkB α).

δSBH = αδA/kB ≥ 0 resembles the second law of thermodynamics. Bekenstein argued that if entropy was lost when matter fell into a black hole, this would violate the 2nd law of thermodynamics. Thus black holes must gain entropy from these processes. Hawking showed that α = /(4~) and that a black hole radiates a boson spectrum with

~κ −6 M TBH = ' 10 K. 2πkB M James Lattimer PHY390, The Kerr Metric and Black Holes Black Hole Temperature

There is no easy way to derive the proportionality constant α between 2 SBH and A. However, SBH is dimensionless, area has dimension length , so α must have geometrized dimension length−2. Since black hole evaporation is a quantum mechanical process it stands to reason that α −2 p 3 should scale like λP , where λP = G~/c is the Planck length. The 2 3 relevant quantal area is λP = G~/c . There is a final numerical factor of order unity, which turns out to be 4. An escaping photon will carry energy E rather than the effective blackbody temperature TBH . Indeed, s √ −2 2 F4(0) 1 − 2 < E > ∼ −4 T ' 3.2T . F2(0) 1 − 2

where Fi are ordinary Fermi integrals. Hawking derived c3 α = , 4~G so, for a Schwarzschild black hole, GM2 κ ~c3 SBH = αA = 4π , TBH = = . ~c 8πkB α 8πkB GM James Lattimer PHY390, The Kerr Metric and Black Holes Fate of Black Holes

How long does it take a black hole to evaporate? With T ∝ M−1 and A ∝ M2, we expect the luminosity L ∝ AT 4 ∝ M−2. Using 2 4 3 2 σB = π kB /(60~ c ), 1 c6 dM L = Aσ T 4 = ~ = −c2 . B 256 · 60 G 2M2 dt Integrating this from M to 0, one sees that the evaporation time is

2 3  3 5120G M 66 M tevaporation = 4 ' 6.7 · 10 yr. ~c M Therefore stellar-mass black holes are essentially permanent. On the age 10 −19 of the Universe, 1.4 · 10 yr, a black hole of ∼ 10 M would just now be completely evaporated. However, black holes are bathed in the cosmic background radiation of 2.7 K. The temperature of a black hole is c3 M T = ~ ' 6.2 · 10−4 K. 8πkB GM M

James Lattimer PHY390, The Kerr Metric and Black Holes Black Hole Cross Section

−4 Thus, black holes of mass greater than 2.3 · 10 M will have a temperature smaller than the cosmic background and will absorb more radiation than it evaporates and therefore ultimately grow. We need to know the effective cross section for capturing photons. The black hole bends light, so this cross section is larger than the geometrical cross-section, which for a Schwarzschild hole is σ = 4πM2. Recall that the effective Schwarzschild potential for photons: L2  M  V 2 = 1 − . eff r 2 r 2 2 2 It’s maximum is at r = 3M where Vmax = L /(27M ). When E > Vmax , an incoming photon enters r = 2M and is captured. When E < Vmax , an incoming photon is scattered√ back to infinity and is not captured. Capture requires L/E < 27M.

Define bcap to be the impact parameter of a photon that is just captured, 2 so the cross-section will be πbcap. b = L/p, where p = E. Thus 27 σ = πb2 = 27πM2 = σ. cap cap 4

James Lattimer PHY390, The Kerr Metric and Black Holes Feeding on the CMB The black hole mass when it’s temperature equals that of the CMB, T = 2.725 K, is CMB 3 ~c −8 MCMB = = 6.2 · 10 M . 8πGkB TCMB Smaller black holes are hotter, and evaporate faster than they absorb. Larger black holes are cooler, absorb the background radiation faster than they evaporate, and cool further. The equilibrium point is unstable. 4 A black hole absorbs the CMB radiation at the rate AσB TCMB . It doesn’t 4 absorb at the rate σcapσB TCMB because it absorbs photons from all directions. Those that are bent into the hole at one location are not available to be absorbed at another location. The mass gain is  4  2 dM σB 4 −34 TCMB M −1 = A 2 TCMB = 5.4 · 10 M yr . dt c 2.725 K M −1 However, TCMB ∝ t due to the cosmic expansion and subsequent redshift. One can show a black hole grows with time according to   −1 3 aM0 1 1 M0t0 M(t) = M0 1 + 3 − 3 → 3 . 3 t t0 t0 − aM0/3 We used (dM/dt)0 = a; M0 and t0 are current values. But the mass −24 increases by only a fraction of 5(M/M )10 even after an infinite time!

James Lattimer PHY390, The Kerr Metric and Black Holes Apparent Size of the Event Horizon

It follows from the derivation of the capture√ cross section that a photon from infinity with impact parameter b = 27M will enter a circular orbit at the photon radius, r = 3M for a Schwarzschild black hole.

Conversely, a light ray coming from√ the unstable circular orbit at r = 3M will have an impact parameter b = 27M. This will be the apparent radius to an distant observer of a non-. For a Kerr black hole, the photon radius is given by the solution of  2 −|a| r (r − 3M)2 = 4Ma2, r = 2M 1 + cos cos−1 , ph ph ph 3 M

At the photon radius, we have E = V+, from which we find L (r 2 + a)2 − a2∆ b = = ph √ . E 2 2Marph + rph ∆ ∗ 6 The black hole at the ’s center, Sagitarius A , has M ' 4 · 10 M and D = 8 kpc distant.√ If Schwarzschild, its apparent angular diameter is 27GM θ = = 1.24 · 10−10 radian = 25.6µ00 Dc2 7 times larger than the apparent angular diameter of its event horizon.

James Lattimer PHY390, The Kerr Metric and Black Holes Tidal Effects

One important way a black hole interacts with its environment is through tidal effects on nearby objects. Tides originate from the fact that geodesics may converge or diverge. For example, in gravitational collapse, a matter parcel not only falls to the center, it also falls towards its neighbors. Even a neutron star can be disrupted by a black hole. Balancing the star’s gravity Gm/R2 with the expected tidal force 3 1/3 2 GMRf /rt , where f ∼ 3, we expect rt ∼ R(fM/m) . If rt > 6GM/c , then m/M > (6β)3/2/f 1/2 is necessary. For a Kerr hole, the 6 is reduced. One can define a separation vector between two adjacent geodesics in terms of a parameter ν that distinguishes geodesics: ξα = dx α/dν. From this one can deduce (after algebra we won’t worry about) the equation of geodesic deviation D D dx α ∇ ∇ ξα = Rα V βV µξν , ∇ = V α = , V α = . V V βµν V Dx α Dτ dτ ~ ~ ~ R is the Riemann tensor. A property of V is that ∇V V = 0 because V is α the tangent vector to ~x. Also Df /Dx is the covariant derivative f;α.

James Lattimer PHY390, The Kerr Metric and Black Holes Tidal Effects in the Schwarzschild Geometry

Because this is a description of the vacuum outside a spherically symmetric object, Rµν = 0. Define a spatial separation vector µ µ ν η = hν ξ where h is a spatial projection tensor orthogonal to g nµν = gµν + umuuν , h · g = 0. Then the surviving expressions from the geodesic deviation equation for ~η are

D2ηr 2M D2ηθ M D2ηφ M = ηr , = − ηθ, = − ηφ. Dτ 2 r 3 Dτ 2 r 3 Dτ 2 r 3 Positive means stretching and negative means compression. Note these deformations are finite when r = 2M so a body can survive into the interior of the event horizon. The deformations are infinite at r = 0. The tidal fields are larger at the horizon for smaller black holes since M/r 3 ∝ 1/M2 there.

James Lattimer PHY390, The Kerr Metric and Black Holes An Inertial Frame in Orbit Around a Black Hole

In the Kerr geometry, the effective tidal potential is −GM 3∆ − P  1 3∆ − 2P   Φ = x 2 − x 2 − x 2 . t r 3 2P 1 2 2 2P 3

Here xi are the local coordinates relative to the center of a frame attached to an object orbiting a black hole√ of mass M at distance r. ∆ = r 2 − 2Mr + a2, P = r 2 − 3Mr + 2a Mr. The centrifugal potential in this frame turns out to be GM Φ = − (x 2 + x 2), c 2r 3 1 2 which implies the frame is rotating with an angular velocity Ω2 = GM/r 3, which is independent of a and the same as the Newtonian result despite the effects of frame dragging! The sum of these potentials is GM 3∆ 3∆ − 2P   Φ = − x 2 − x 2 ; t+c 2r 3 P 1 P 3

note the tidal and centrifugal effects cancel in the x2 direction. James Lattimer PHY390, The Kerr Metric and Black Holes Hydrostatic Equilibrium

For simplicity, treat the gravitational potential in the orbiting inertial frame as a Newtonian point mass m: Gm Φ = − . g p 2 2 2 x1 + x2 + x3 The equation of hydrostatic equilibrium is then

∇p/ρ = ∇µ = −∇(Φg + Φt+c ), which we can integrate to form the Bernoulli Integral

H = µ(R) + Φg + Φt+c = C = constant.

Evaluate this at the stellar surface at three points (x1, x2, x3) = (a1, 0, 0), (0, a2, 0) or (0, 0, a3), and further assume that m = 4πρ¯a1a2a3/3: 3∆ GM 4π (x , x , x ) = (a , 0, 0) : C = − a2 − Gρ¯a a , 1 2 3 1 2P r 3 1 3 2 3 4π (x , x , x ) = (0, a , 0) : C = − Gρ¯a a , 1 2 3 2 3 1 3 3∆ − 2P GM 4π (x , x , x ) = (0, 0, a ): C = a2 − Gρ¯a a . 1 2 3 3 2P r 3 3 3 1 2

James Lattimer PHY390, The Kerr Metric and Black Holes The Roche Limit

We now use the shorthand f = 3∆/(3∆ − 2P), α2 = a2/a1 and α3 = a3/a1, and eliminate C: 2 f + α3 α2(1 − α3) M 4 1 − α3 2 = , 3 ≡ η = (f − 1)α3 3 . α3 α2 − α3 πρ¯r 3 f + α3 We note the facts that in the Newtonian case, f = 3 and, in the GR case, f = 2 corresponds to the ISCO for all a. Irrespective of f , α2 ' α3, and the quantity η → 0 for both α3 → 0 and α3 → 1. Therefore η has a maximum, ηmax ' 0.158, near α3 = 0.4. This configuration represents the smallest density that can remain in hydrostatic equilibrium; stars with lower density tidally disrupt. The Roche limit is  M 1/3  4M 1/3 rRoche = = R . πρη¯ max 3mηmax 2 For a = 0, we want rRoche > 6GM/c , or m/M > 0.34(6β)3/2. For β = Gm/(Rc2) ' 0.2, we need m/M ∼> 0.4.

James Lattimer PHY390, The Kerr Metric and Black Holes More Realistic Roche Limits This estimate for the limits for tidal disruption is unduly pessimistic. One problem is the unrealistic estimate for the gravitational potential. A better study using an incompressible fluid finds ηmax = 0.0901(0.0644) for the Newtonian (GR ISCO) case, about 1.75 (2.45) times smaller. The disruption condition can be written m r c2 3/2 r3η > ISCO β max , M ∼ GM 4 which requires m/M ∼> 0.34 (but only 0.02 for extreme Kerr); as a increases, rISCO shrinks. 2 Usinga ¯ = a/M andr ¯ = rISCOc /(GM), √ r¯2 − 6¯r − 3¯a2 ± 8¯a r¯ = 0. p r¯ = 3 + Z2 ∓ (3 − Z1)(3 + Z1 + 2Z2), q 2 2 Z2 = 3¯a + Z1 ,

21/3 h 1/3 1/3i Z1 = 1 + 1 − a¯ (1 +a ¯) + (1 − a¯) .

James Lattimer PHY390, The Kerr Metric and Black Holes Towards More Realistic Roche Limits

The unrealistic assumption of a point-like gravitational potential can be rectified by assuming the tidally-disrupted body has an ellipsoidal shape. Assuming it to be incompressible means that its volume 4πa1a2a3/3 = m/ρ¯ is constant. The gravitational potential is

" 3 # q X 2 2 2 2 Φg = −πGρ¯ I − Ai xi , D = (a1 + u)(a2 + u)(a3 + u) i Z ∞ du Z ∞ du I = a1a2a3 , Ai = 2 . 0 D 0 (ai + u)D The integrals are forms of (not surprisingly) so-called elliptical integrals. They obey

2 2 2 A1 + A2 + A3 = 2, I = A1a1 + A2a2 + A3a3. The potential satisfies Poisson’s Equation

2 ∇ Φg = 4πGρ.¯

James Lattimer PHY390, The Kerr Metric and Black Holes Special Cases

Note that these integrals can be generally put into a more convenient 2 2 2 form for computation with the substitutions y = a1/(u + a1), α2 = a2/a1 and α3 = a3/a1. For example, Z ∞ Z 1 2 du 2a1α2α3dy I = a1a2a3 = . D p 2 2 p 2 2 0 0 1 − y (1 − α2) 1 − y (1 − α3)

The integrals become analytic in the case of a sphere, a1 = a2 = a3, for 2 which I = 2a1 and Ai = 2/3, or in the two cases when two axes are equal:

I The oblate case a1 = a2 > a3, suitable for rotating configurations, arcsin p1 − α2 I /a2 − 2α2 I = 2a2α 3 , A = A = 1 3 ; 1 3 p 2 1 2 2(1 − α2) 1 − α3 3

I The prolate case a1 > a2 = a3, suitable for tidally deformed configurations, " # a2α2 1 + p1 − α2 I /a2 − 2 I = 1 3 ln 3 , A = 1 . p 2 p 2 1 1 − α2 1 − α3 1 − 1 − α3 3

James Lattimer PHY390, The Kerr Metric and Black Holes Remnant Mass From a Black Hole-Neutron Star Merger

When tidal disruption occurs outside RISCO , simulations show that not all the neutron star mass will fall into the black hole, at least immediately. A schematic estimate for the fraction of mass remaining in a disc outside the black hole after disruption is

M  m 1/3  2Gm  r t = α0 1 − − β0 ISCO . m M Rc2 R We add a GR correction to the self-gravity of the neutron star. The disruption threshold we previously obtained implies

α0  4 1/3 0 ' = 2.75. β 3ηmax Remarkably, simulations agree with this prediction. Disc masses of simulations are well fit by α0 ' 0.427 and β0 = 0.171.

James Lattimer PHY390, The Kerr Metric and Black Holes Binaries

For the case of two neutron stars, the stars will always tidally disrupt before coalescence. We first consider the Newtonian case. For a binary with component masses M1 and M2, we define the reduced mass µ = M1M2/M where M is the total mass. The mean distances of the stars from the center of mass are a1/a2 = M2/M1, and the mean separation of the stars is a = a1 + a2. The orbit is generally elliptical with eccentricity e, so at periastron the separation is a(1 − e) and at apastron it is a(1 + e). The total energy and orbital angular momentum are 1 GM1M2 p E = − , J = µpGaM(1 − e2) = µΩa2 1 − e2. 2 a Kepler’s Law is 2π 2 GM Ω2 = = . P a3 The projected orbital velocity of star 1 along the line of sight is v1 = Ωa1 sin i. We define the mass function (M sin i)3 v 3 f (M , M , i) = 2 = 1 , 1 1 2 M2 GΩ which depends only on observables v1 and P. It is also possible in some systems to determine f2. Then, M2/M1 = v1/v2 independent of i. In an eclipsing binary, the angle i can be estimated. James Lattimer PHY390, The Kerr Metric and Black Holes Roche Lobes

The total potential of a binary is Φg + Φc , GM GM 1 −Φ = 1 + 2 + d 2Ω2, r1 r2 2

where d is the distance to the rotation axis, r1 and r2 are the distances to the stars. In the orbital plane, with the origin at the center of mass, GM GM 1 GM −Φ(x, y) = 1 + 2 + (x 2 + y 2) . p 2 2 p 2 2 3 (x − a1) + y (x + a2) + y 2 a

We convert to dimensionless coordinatesx ¯ = x/a,y ¯ = y/a, m1 = M1/M and m2 = M2/M: " # GM m m x¯2 +y ¯2 −Φ(¯x, y¯) = 1 + 2 + . p 2 2 p 2 2 a (¯x − m2) +y ¯ (¯x + m1) +y ¯ 2

The next figure shows contours of constant Φ, showing deep minima at the stellar centers, and extrema at 5 so-called Lagrangian points.

James Lattimer PHY390, The Kerr Metric and Black Holes The equipotential surface passing through L1 is the Roche Lobe; its size depends on the mass ratio and the separation. A fit by Kopal gives 1/3 1/3 RR /a = 0.46(M1/M) . Compare coefficient to (3βmax /4) ' 0.48. A better fit is by Eggleton: 2/3 1/3 −1 RR /a = 0.49[0.6 + (M2/M1) ln(1 + (m1/M2) )] .

James Lattimer PHY390, The Kerr Metric and Black Holes Neutron Star Mergers

Neutron star mergers will involve mass ratios of order unity. For q = m2/m1 = 1, Eggleton’s formula gives RR = 0.379a so that the Roche limit is always reached before coalescence. Note that the shapes of the stars, which fill the Roche Lobes, are not as distorted as in the case of neutron stars in a black hole-neutron star merger. Simulations show that mass is both tidal- and shock-ejected, and further mass ejection occurs from a remnant disc via a neutrino-driven wind. The combined masses of the neutron stars likely exceed the neutron star maximum mass, even accounting for mass loss, so the final remnant will ultimately be a black hole. However, a black hole should not form immediately, because there is too much angular momentum in the remnant, and angular momentum supports additional mass. A uniformly rotating body at the Keplerian (mass-shedding) spin rate can support about 1.18 times the non-rotating maximum mass, Mmax .

Differential rotation can support up to 1.5Mmax or even more.

Since it is unlikely that Mtot , the total inspiralling mass, will exceed > 1.5Mmax , where Mmax ∼ 2M from pulsar observations, a black hole probably never forms immediately following a merger.

James Lattimer PHY390, The Kerr Metric and Black Holes The Initial Remnant

Bauswein and Stergioulas fit from simulations the initial angular momentum in the remnant:

2 J ' [4.04(Mtot /M ) − 4.66] M ,

using a kind of geometrized units. Neutron stars with maximal rotation are found to have 2 < amax /M = cJmax /(GM ) ∼ 0.7. For Mtot ' 2.5M , one finds 2 a/M = cJ/(GM ) ' 0.87 > amax /M so the initial remnant will be differentially rotating and mass is likely shed into a disc.

Indeed, as long as Mtot < 4.18M , likely true for any binary, uniform rotation is not possible and a differentiallly rotating remnant is formed. Within hundreds of milliseconds, the excess angular momentum will be dissipated and either a long-lived uniformly-rotating neutron star forms, or if the maximum mass limit is breached, a collapse to a black hole ensues.

Thus, if Mtot > 1.18Mmax , a black hole will form after a short delay of less than a second. Otherwise, collapse to a black hole, if it occurs at all, will happen on much longer timescales of seconds to millions of years.

James Lattimer PHY390, The Kerr Metric and Black Holes