Gravitation:Rotating and/or Charged Black Holeslaguna.gatech.edu/Gravitation/notes/Chapter06.pdfPablo Laguna Gravitation:Rotating and/or Charged Black Holes is calledsurface gravitybecause

Gravitation: Rotating and/or Charged BlackHoles

An Introduction to General Relativity

Pablo Laguna

Center for Relativistic AstrophysicsSchool of Physics

Georgia Institute of Technology

Notes based on textbook: Spacetime and Geometry by S.M. CarrollSpring 2013

Pablo Laguna Gravitation: Rotating and/or Charged Black Holes

More about Black Holes

No-hair Theorem

Event Horizons

Killing Horizons

Mass, Charge and Spin

Charged Black Holes

Rotating Black Holes

Penrose Process

Black Hole Thermodynamics


No-hair Theorem

No-hair Theorem

Stationary, asymptotically flat black hole solutions to general relativity coupled to electromagnetism that arenon-singular outside the event horizon are fully characterized by the parameters mass, electric and magnetic chargeand angular momentum


Information Loss Paradox

Classical picture

If the final state of a system is a black hole, the information regarding the complicated collection of matter thatproduced this black hole is hidden behind the event horizon.

Quantum picture:

When quantum effects are taken into account, black holes radiate (Hawking) and thus evaporate. Then, eventuallyblack holes disappear. Where is the information trapped by the event horizon? (Information Loss Paradox).


Event Horizons

Event Horizon

A hypersurface separating those events that are connected to infinity by a timelike path from those that are not.

Figure : Kruskal DiagramFigure : Penrose Diagram


I -

I +

i+i+

I +

I -

i 0

i - i -

i 0

r = const

t = const

r = 2G

M

r = 2GM

r = 0

r = 0

i+ = future timelike infinityi0 = spatial infinity

i− = past timelike infinityI+ = future null infinityI− = past null infinity

The points i± are the limits of spacelike surfaces whose normals are timelike.

The point i0 is the limit of timelike surfaces whose normals are spacelike.

Radial null geodesics are at±45

The event horizon is a null hypersurface

The event horizon is a global concept

Timelike geodesics begin at i− and end at i+ or at the singularity.

Null geodesics begin at I− and end at I+ or at the singularity.

There are plenty of timelike paths that do not hit the singularity.


Penrose Diagram of Minkowski Spacetime

Null Geodesics



Timelike Geodesics



Spacelike Geodesics



In Minkoski spacetime

ds2 = −dt2 + dr2 + r2dΩ

where−∞ < t <∞ and 0 ≤ r <∞

Introduce the null coordinatesu = t − r and v = t + r

where−∞ < u, v <∞ and u ≤ v

thus

ds2 = −du dv +1

4(v − u)2dΩ2



Bring infinity to a finite coordinate value with

U = tan−1(u) and V = tan−1(v)

such that−π

2< U, V <

π

2and U ≤ V

thus

ds2 =1

4 cos2 U cos2 V

[−4 dU dV + sin2(V − U)dΩ2

]Reintroduce timelike and radial coordinates

T = U + V and R = V − U

such that|T | + R < π and 0 ≤ R < π

Thusds2 = ω

−2(T ,R)(−dT 2 + dR2 + sin2 RdΩ2

)where

ω = 2 cos U cos V = cos T + cos R


That is, the Minkowski metric is conformal to a metric with line element

ds2 = ω2 d2 = −dT 2 + dR2 + sin2 RdΩ2

i+ = future timelike infinity(T = π,R = 0)

i0 = spatial infinity(T = 0,R = π)

i− = past timelike infinity(T = −π,R = 0)

I+ = future null infinity(T = π − R, 0 < R < π)

I− = past null infinity(T = −π + R, 0 < R < π)


Penrose Diagram of Collapsing Matter Spacetime


Penrose Diagram of Extended Swarzschild Spacetime


Singularity Theorems

The ubiquity of singularities is guaranteed by the Singularity Theoremsby Hawking and Penrose

The theorems demonstrate that once gravitational collapse reachescertain point, evolution to a singularity is inevitable.

The presence of a singularity is detected via geodesic incompleteness.

An incomplete geodesic is one that cannot be extended within themanifold but ends at a finite value of the affine parameter.

The appearance of a trapped surface signals reaching the point of noreturn during collapse.

Trapped Surface: a compact, spacelike, 2-dimensional submanifold suchthat future directed light rays converge in both directions.

Trapped Surfaces


Singularity Theorem

Let M be a manifold with a generic metric gµν , satisfying the Einstein’s equations with a strong energy conditionimposed. If there is a trapped surface in M, there must be either a closed timelike curve or a singularity (asmanifested by an incomplete timelike or null geodesic)

Cosmic Censorship Conjecture

Naked singularities cannot form in gravitational collapse from generic, initially non-singular states in anasymptotically flat spacetime obeying the dominant energy condition

Hawking’s Area Theorem

Assuming the weak energy condition and cosmic censorship, the area of a future event horizon in an asymptoticallyflat spacetime is non-decreasing

Energy Conditions:

Weak Energy Condition: Tµν tµtν ≥ 0

Null Energy Condition: Tµνkµkν ≥ 0

Dominant Energy Condition: Tµν tµtν ≥ 0 and (Tµν tµ)(Tν α tα) ≤ 0

Null Dominant Energy Condition: Tµνkµkν ≥ 0 and (Tµνkµ)(Tν αkα) ≤ 0

Strong Energy Condition: Tµν tµtν ≥ 12 Tλ λtσ tσ

where tµ and kµ are arbitrary time-like and null vectors, respectively.


Killing Horizons

Recall that a Killing vector satisfies∇(µKν) = 0

In the Schwarzschild spacetime, the Killing vector K = ∂t is a Killing vector; that is, ∂t gµν = 0:

ds2 = −(

1−2M

r

)dt2 +

(1−

2M

r

)−1dr2 + r2dΩ2

For r > 2 M, K = ∂t is timelike.

In the hypersurface Σ defined by r = 2 M, K = ∂t is null.

For r < 2 M, K = ∂t is spacelike.

Killing horizon: A null surface Σ with null Kiliing vector field χµ.

Killing horizons are not necessarily the same as event horizons. In spacetimes with time translation, the twohorizons are related.

Every event horizon Σ in a stationary (i.e. timelike Killing vector at infinity), asymptotically flat spacetime is aKilling horizon for some Killing vector field χµ.

If the spacetime is static (i.e. timelike Killing vector orthogonal to t = constant hypersurfaces), χ = K = ∂t

If the spacetime is stationary but not static, it will be axisymmetric with a rotational Killing vector R = ∂φand χ = K + ΩH R for some constant ΩH .


Not every Killing horizon is an event horizon.

Example:

Consider Minkowski, ds2 = −dt2 + dx2 + dy2 + dz2. There are no event horizons.

Consider the Killing vector that generates boosts in the x − direction, namely χ = x∂t + t∂x

It is easy to check that χ satisfies the Killing equation∇(µχν) = 0, and its norm is χµχµ = −x2 + t2.

Notice that χµχµ = 0 at surfaces Σ such that x = ±t . Therefore Σ are Killing horizons.


Surface Gravity

In Newtonian Physics: surface gravity is the acceleration experienced by a test body in the surface of an object.

In General Relativity: the surface gravity κ of a Killing horizon Σ with Killing vector field χµ is defined from

χµ∇µχν = −κχν

from which one can show that

κ2 = −

1

2(∇µχν )(∇µχν )

Notice that the equation looks like the geodesic equation for an arbitrary parametrization.


Since Σ is a null hypersurface, then χµ are integral curves of null geodesics in Σ, called the generators of Σ.

PROOF:

Since χµ is null, it is both normal and tangent to Σ.

Let xµ(α) be the integral curves of χµ; thus χµ = dxµ/dα

If xµ(α) are geodesics, we need to show that they satisfy χµ∇µχν = η(α)χν

The r.h.s. of this equation does not necessarily vanish because α in general is not an affine parameter (i.e.α 6= aτ + b).

Since χµ is normal to Σ, then χµ = ∇µf and

χµ∇µχν = χ

µ∇µ∇ν f = χµ∇ν∇µf

= χµ∇νχµ =

1

2∇ν (χµχµ)

On Σ, one has that χµχµ = 0, but not necessarily∇ν (χµχµ) = 0

Thus, we use χµχµ = 0 as the quantity that defines Σ. Then

∇ν (χµχµ) = g∇ν f = g χν

and

χµ∇µχν =

1

2g χν = −κχν

with κ = −g/2


κ is called surface gravity because in a static, asymptotically flat spacetime, κ is the acceleration of a staticobserver near the horizon, as measured by a static observer at infinity.

PROOF:

For a static of observer with 4-velocity Uµ, one has that Kµ = V (x) Uµ for Kµ the time-translationalKilling vector and V (x) =

√−KµKµ.

Notice that V = 1 at infinity and V = 0 at the horizon.

V is also called the redshift factor since ω = −pµUµ = −pµKµ/V = E/V

The 4-acceleration aµ = Uν∇νUµ takes the form aµ = ∇µ ln V

∇µV 2 = −∇µ(KνKν ) = −2 Kν∇µKν = 2 Kν∇νKµ

∇µ ln V = V−2 Kν∇νKµ = Uν∇ν (V−1 Kµ)− Uν Kµ∇νV−1

∇µ ln V = Uν∇νUµ + Uν Uµ∇ν ln V

(δνµ − Uν Uµ)∇ν ln V = Uν∇νUµ

then

Uµ(δνµ − Uν Uµ)∇ν ln V = 2 Uν∇ν ln V = Uµ Uν∇νUµ = 0

thus

∇µ ln V = Uν∇νUµ = aµ


Then its magnitude a =√

aµaµ =√∇µ ln V∇µ ln V = V−1√∇µV∇µV

a→∞ at the Killing horizon.

An observer at infinity, however, well detect the acceleration redshited; thus, κ = a V =√∇µV∇µV

Which can be shown to yield κ2 = − 12 (∇µχν )(∇µχν )


Schwarzschild Case:

ds2 = −(

1−2M

r

)dt2 +

(1−

2M

r

)−1dr2 + r2 dΩ2

The Killing vector and 4-velocity are

Kµ = (1, 0, 0, 0) Uµ =

[(1−

2 M

r

)−1/2, 0, 0, 0

]

thus the redshift factor is

V =

(1−

2 M

r

)1/2

The accelerations is then

aµ =M

r2

(1−

2 M

r

)−1∇µr

and

a =M

r2

(1−

2 M

r

)−1/2

so the surface gravity is

κ = V a =M

r2=

1

4 M


Charge

In Electromagnetism, we have that one of the Maxwell’s equations reads∇µFµν = Jµ, thus

Q = −∫

Σd3x√γ nµJµ = −

∫Σ

d3x√γ nµ∇νFµν

using Stoke’s theorem we get

Q = −∫∂Σ

d2x√γ(2) nµ σν Fµν


Mass - Energy

In general relativity, we have seen in Chapter 3 that a current Jµ = Kν Tµν with Kµ a Killing vector is conserved,i.e.

∇µJµ = (∇µKν )Tµν + Kν (∇µTµν ) = 0

Thus, if Kµ is time-like, we can define an energy as

E =

∫Σ

d3x√γ nµ Jµ =

∫Σ

d3x√γ nµ Kν Tµν

but this is not well defined for vacuum spacetimes.

Consider instead Jµ = Kν Rµν . Is this current conserved?

∇µJµ = ∇µ(Kν Rµν )

= Rµν∇µKν + Kν ∇µRµν

= Rµν∇(µKν) + Kν1

2∇νR

=1

2Kν ∇νR = 0

In the last step, we used that the directional derivative of R along a Killing vector vanishes (see Eq. 3.178).


We are now in the position to define

E =1

4π

∫Σ

d3x√γ nµ Jµ =

1

4π

∫Σ

d3√γ nµ∇ν (∇µKν )

where we have used that for Killing vectors∇µ∇νKµ = KµRµν .

Using Stoke’s theorem, we arrive to

Komar Integral

E =1

4π

∫∂Σ

d2x√γ(2) nµ σν∇µKν

In the Schwarzschild case, n0 = −(1− 2 M/r)1/2 and σ1 = (1− 2 M/r)−1/2, one can show thatnµσν∇µKν = −∇0K 1 = M/r2.

Since√γ(2) = r2 sin θ, then

E =1

4π

∫∂Σ

d2x√γ(2) nµ σν∇µKν =

1

4π

∫dθ dφr2 sin θ

(M

r2

)= M


ADM Energy

Arnowitt, Deser and Misner energy

EADM =1

16π

∫∂Σ

d2x√γ(2) σ

i(∂j h

ji − ∂i h

jj

)where hµν is a small perturbation of flat spacetime (i.e. gµν = ηµν + hµν ).

Positive Energy Theorem

The ADM energy of a non-singular, asymptotically flat spacetime obeying Eintein’s equation and the dominantenergy condition is non-negative. Minkowski is the only spacetime with vanishing ADM energy.


Angular Momentum

Consider a rotational Killing vector R = ∂φ, then we can define a conserved current Jµ = RνRµν . Then, the totalangular momentum is defined as

J = −1

8π

∫∂Σ

d2x√γ(2) nµ σν∇µRν


Charged Black Holes

We will consider only the spherical symmetry

ds2 = −e2α(r,t)dt2 + e2β(r,t)dr2 + r2dΩ2

The energy-momentum tensor for electromagnetism is given by

Tµν =1

4π(FµρFν

ρ −1

4gµνFρσFρσ) ,

where Fµν is the electromagnetic field strength tensor, with F0i = E i and F ij = εijk Bk

Because of spherical symmetry, only radial components of the electric and magnetic field are allowed, thus

Ftr = f (r, t)Fθφ = g(r, t) sin θ

where f (r, t) and g(r, t) are functions to be determined by the field equations.

Ftr corresponds to a radial electric field, while Fθφ corresponds to a radial magnetic field.


Maxwell’s equations read:

gµν∇µFνσ = 0∇[µFνρ] = 0

The solution to these equations coupled to the Einstein equations is the Reissner-Nordstrom metric

ds2 = −∆dt2 + ∆−1dr2 + r2dΩ2

where

∆ = 1−2M

r+

(p2 + q2)

r2.

Above, M is the mass of the hole; q is the total electric charge, and p is the total magnetic charge, if they exist!

The electromagnetic fields associated with this solution are given by

Er = Ftr = −q

r2

Br =Fθφ

r2 sin θ=

p

r2


There is a true curvature singularity at r = 0

There is a horizon at ∆ = 0 which occurs at

r± = M ±√

M2 − (p2 + q2)

Case: M2 < p2 + q2

∆ > 0

The metric is completely regular in the (t, r, θ, φ) coordinates all the way down to r = 0.

The coordinate t is always timelike, and r is always spacelike.

There is the singularity at r = 0, which is now a timelike line.

There is no event horizon, there is thus naked singularity.

The singularity is repulsive

As r →∞ the solution approaches flat spacetime.

To form these type of black holes via gravitational collapse requires negative mass.


Case: M2 > p2 + q2

The energy in the electromagnetic field is less than the total energy.

∆(r) is positive at large r and small r , and negative inside the two vanishing points

r± = M ±√

M2 − (p2 + q2).

The metric has coordinate singularities at both r+ and r−.

The surfaces r = r± are both null; they are event horizons

The singularity at r = 0 is a timelike line (not a spacelike surface as in Schwarzschild).

Case: M2 = p2 + q2

This case is known as the extreme Reissner-Nordstrøm solution

The extremal black holes have ∆(r) = 0 at a single radius, r = M.

r = M is not an event horizon; the r coordinate is never timelike; it becomes null at r = M, but is spacelikeon either side.

The singularity at r = 0 is a timelike line. So the singularity can be avoided.


Rotating Black Holes

The metric solution to the Einstein equations representing a rotating black hole was found by Kerr in 1963. The Kerrmetric reads:

ds2 = −dt2 +ρ2

∆dr2 + ρ

2dθ2 + (r2 + a2) sin2θ dφ2 +

2Mr

ρ2(a sin2

θ dφ− dt)2

where ∆(r) = r2 − 2Mr + a2 and ρ2(r, θ) = r2 + a2 cos2 θ

The parameter a measures the rotation of the hole.

a = J/M with J the angular momentum of the black hole.

M is the mass.

To include electric and magnetic charges q and p replace 2Mr with 2Mr − (q2 + p2); the result is theKerr-Newman metric.

The coordinates (t, r, θ, φ) are known as Boyer-Lindquist coordinates.

As a→ 0, one recovers Schwarzschild.

ds2 = −(

1−2M

r

)dt2 +

(1−

2M

r

)−1dr2 + r2dΩ2

As M → 0 with a fixed, one recovers flat spacetime.


Ellipsoidal Coordinates

The spatial coordinate are ellipsoidal coordinates.

r = 0 is a two-dimensional disk.

r = 0, θ = π/2 is a ring.

There are two Killing vectors ζµ = ∂t and ηµ = ∂φ.

ζµ is not orthogonal to t = constant hypersurfaces, hence this metric is stationary, but not static. (It’s notchanging with time, but it is spinning.)

r =

r = 0

= const const

a


ds2 = −dt2 +ρ2

∆dr2 + ρ

2dθ2 + (r2 + a2) sin2θ dφ2 +

2Mr

ρ2(a sin2

θ dφ− dt)2

Singularities seem to appear at both ∆ = 0 and ρ = 0

For ∆ = r2 − 2Mr + a2 = 0 with M2 < a2, there is a naked singularity.

For ∆ = r2 − 2Mr + a2 = 0 with M2 = a2, the black hole is extremal but unstable.

For ∆ = r2 − 2Mr + a2 = 0 with M2 > a2, there are two radii at which ∆ vanishes, given by

r± = M ±√

M2 − a2

Both radii r± are null surfaces and also event horizons.

The Killing vector ζµ = ∂t is not null at the outer event horizon; in fact, since

ζµζµ = −

1

ρ2(∆− a2 sin2

θ)

one has that at r = r+

ζµζµ =

a2

ρ2sin2

θ ≥ 0 .

So the Killing vector is already spacelike at the outer horizon, except at the north and south poles (θ = 0)where it is null.


The Killing horizon is found where

ζµζµ = −

1

ρ2(∆− a2 sin2

θ) = 0

The outer event horizon is found at (r+ − M)2 = M2 − a2

The inner event horizon is found at (r− − M)2 = M2 − a2

The region in between the Killing and event horizons is known as the ergosphere.

Inside the ergosphere, you cannot move against the φ-direction of the rotation of the black hole (framedragging).

You can, however, escape the ergosphere.

The true curvature singularity occurs at ρ = 0.

Since ρ2 = r2 + a2 cos2 θ, the singularity is located at r = 0, θ = π/2; that is, the shape of the singularityis a ring.

-

r

r=0

r

outer event

ergosphere

+

Killinghorizon

inner event horizon

horizon


Dragging of Inertial Frames

Consider a particle with 4-velocity Uµ with zero angular momentum

Recall, ∂σ∗gµν = 0⇒ dpσ∗/dτ = 0; thus, a particle with vanishing angular momentum hasm Uφ = −L = 0.

The particle’s trajectory is then such that

dφ

dt=

Uφ

U t=

gφt Ut + gφφUφgtt Ut + gtφUφ

=gφt

gtt≡ ω(r, θ)

Where ω is the coordinate angular velocity of a zero angular momentum particle.

The particle is dragged by the influence of gravity an acquires angular velocity.

Notice that this effect is only present if gφt 6= 0



Consider a photon is emitted in the φ direction at some radius r in the equatorial plane (θ = π/2); therefore

ds2 = 0 = gtt dt2 + 2 gtφdt dφ + gφφdφ2

which implies

dφ

dt= −

gtφ

gφφ±

√√√√( gtφ

gφφ

)2

−gtt

gφφ.

If gtt 6= 0, then dφ/dt 6= 0.

Evaluate this quantity when gtt = 0 (stationarity limit),

dφ

dt=

a

2 M2 + a2and

dφ

dt= 0

Thus the photons are dragged and forced to have coordinate angular velocity dφ/dt ≥ 0.



Any surface for which gtt = 0 is then called the an stationary limit surface.

For a Kerr black hole this surface coincides with the Killing horizon and takes place when r = 2 M

At the event horizon,dφ

dt≡ ΩH =

a

r2+ + a2

This is the minimum angular velocity of a particle at the horizon.


Penrose Process

Consider conserved quantities associated with the Killing vectors ζµ = ∂t and ηµ = ∂φ, restricting theattention to massive particles with four-momentum pµ = m dxµ/dτ

The two conserved quantities are the energy and angular momentum of the particle,

E = −ζµpµ = m(

1−2Mr

ρ2

) dt

dτ+

2mMar

ρ2sin2

θdφ

dτ

and

L = ηµpµ = −2mMar

ρ2sin2

θdt

dτ+

m(r2 + a2)2 − m∆a2 sin2 θ

ρ2sin2

θdφ

dτ.

Inside the ergosphere, ζµ becomes spacelike; one can then have particles for which E = −ζµpµ < 0 .


Let p(0)µ be the initial 4-momentum of a massive particle outside the ergosphere; thus,E(0) = −ζµp(0)µ > 0.

Inside the ergosphere, the particle splits into two particles m1 and m2; thus, p(0)µ = p(1)µ + p(2)µ andequivalently E(0) = E(1) + E(2)

Arrange E(2) < 0 in such a way that particle m2 falls into the black hole and m1 leaves the ergosphere.

From conservation of energy, E(1) > E(0) ; that is, m1 emerges with more energy than the one entered bym0

The energy gained by particle m1 is at the expense of decreasing the angular momentum of the black hole.

p

p

p

Killing horizonergosphere

(top view)

(2)

(0) µ

µ

(1)µ


Penrose Process: Extracting energy from a rotating black hole by decreasing its angular momentum.

Proof:

Define the Killing vector χµ = ζµ + Ω ηµ with ζµ = ∂t and ηµ = ∂φ

On the outer horizon χµ is null and tangent to the horizon.

Consider a co-rotating (i.e. r and θ constant) observer inside the ergosphere with 4-velocity Uµ = U t χµ

that observes particle (2) as it passes by.

This observer will measure an energy

E = −p(2)µ Uµ = −U t (p(2)

t + Ω p(2)φ

) = E(2) − Ω L(2)> 0

which yields L(2) < E(2)/Ω

Since E(2) < 0 for the Penrose process, thus L(2) < 0. The upper limit occurs at the horizon whenΩ = ΩH

Let

δM = E(2)

δJ = L(2).

the contributions to the mass and angular momentum of the black hole by particle m2. Thus,

δJ <δM

ΩH


Black Hole Horizon Area

Recall: In flat spacetime in spherical coordinates, the area of a sphere with radius R is found from the inducedmetric in the sphere surface:

γij dx i dx j = ds2(dt = 0, dr = 0, r = R) = R2(dθ2 + sin2θ dφ2)

The area is found from integrating the ”volume” element; that is

A =

∫√γdθ dφ =

∫R2 sin θ dθ dφ = 4π R2

In Kerr, and taking R = r+

γij dx i dx j = (r2+ + a2 cos2

θ)dθ2 +

[(r2

+ + a2)2 sin2 θ

r2+ + a2 cos2 θ

]dφ2

and√γ = (r2

+ + a2) sin θ

so the horizon area is given by

A =

∫√γdθ dφ =

∫(r2

+ + a2) sin θ dθ dφ = 4π (r2+ + a2)


Black Hole Thermodynamics

The area of the event horizon A = 4π(r2+ + a2) can never decrease

Proof: Consider the irreducible mass of the black hole, defined by

M2irr =

A

16π

=1

4(r2

+ + a2)

=1

2

(M2 +

√M4 − (M a)2

)=

1

2

(M2 +

√M4 − J2

).

ThenδMirr =

a

4 MirrΩH√

M2 − a2(δM − ΩH δJ) .

whereΩH =

a

r2+ + a2

Then, form the Penrose process

δJ <δM

ΩH

we have thatδMirr > 0


The irreducibility of Mirr implies that the area A can never decrease

δA = 32πMirr δMirr = 8πa

ΩH√

M2 − a2(δM − ΩHδJ )

where we usedδMirr =

a

4 MirrΩH√

M2 − a2(δM − ΩH δJ) .

ThusδM =

κ

8πδA + ΩHδJ

where

κ =

√M2 − a2

2M(M +√

M2 − a2).

with κ the surface gravity of the black hole. Recall the surface gravity κ of a Killing horizon with Killing vector field

χµ is defined fromχµ∇µχν = −κχν

from which one can show that

κ2 = −

1

2(∇µχν )(∇µχν )

Use χµ = ζµ + ΩH ηµ to derive κ.


Comapre

δM =κ

8πδA + ΩHδJ

with the first law of thermodynamics,

dU = T dS − p dV

with T the temperature, S the entropy, p the pressure and V the volume.

There is then natural correspondence

U ↔ MS ↔ A/4T ↔ κ/2π

−p dV ↔ ΩHδJ

where the term ΩHδJ is viewed as the work.


Zero Law Thermodynamics

In thermal equilibrium, T = contant

Zero Law of BH Thermodynamics

For stationary black holes, κ = constant

First Law Thermodynamics

dU = T dS − p dV

First Law BH Thermodynamics

δM =κ

8πδA + ΩHδJ

Second Law Thermodynamics

δS ≥ 0

Generalized second Law BH Thermodynamics (Bekenstein)

δ

(S +

A

4

)≥ 0


Documents

Gravitation:Rotating and/or Charged Black Holeslaguna.gatech.edu/Gravitation/notes/Chapter06.pdfPablo Laguna Gravitation:Rotating and/or Charged Black Holes is calledsurface gravitybecause