Gravitation:Schwarzschild Black Holes - Pablo...

Gravitation: Schwarzschild Black HolesAn 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: Schwarzschild Black Holes

Schwarzschild Black Holes

Schwarzschild Metric

Birkhoff’s Theorem

Singularities

Geodesics

Experimental Tests

Stars and Black Holes


One of the most important solutions to the Einstein’s equations is that of spherically symmetric vacuumspacetimes.

The solution was discovered by Karl Schwarzschild 1915.

It represents the solution outside a spherical, static body.

Schwarzschild metric

ds2 = −(

1−2GM

r

)dt2 +

(1−

2GM

r

)−1dr2 + r2dΩ2

where dΩ2 = dθ2 + sin2 θ dφ2 and the parameter M is interpreted as the mass of the gravitating object.


Derivation

Starting point, Einstein’s equation in vacuum, i.e. Rµν = 0

We require the spacetime to be static and spherically symmetric.

Static: i) The metric is time-independent and ii) the metric does not have time-space terms dt dx i

Example, of a static and spherically symmetric spacetime is Minkowski ds2 = −dt2 + dr2 + r2dΩ2

We will begin with the following ansatz:

ds2 = −e2α(r)dt2 + e2β(r)dr2 + e2γ(r)r2dΩ2

Introduce the following change of coordinates: r = eγ(r)r

Then

dr =

(1 + r

dγ

dr

)eγdr

so

ds2 = −e2αdt2 +

(1 + r

dγ

dr

)−2e2β−2γdr2 + r2dΩ2

and redefine

e2β =

(1 + r

dγ

dr

)−2e2β−2γ

to get

ds2 = −e2αdt2 + e2βdr2 + r2dΩ2

r is an areal coordinate.


Derivation

Compute the Christoffel symbols (drop bars)

Γttr = ∂rα Γr

tt = e2(α−β)∂rα Γrrr = ∂rβ

Γθrθ = 1r Γr

θθ = −re−2β Γφrφ = 1

r

Γrφφ = −re−2β sin2 θ Γθφφ = − sin θ cos θ Γ

φθφ

= cos θsin θ .

Compute Riemann tensor

Rtrtr = ∂rα∂rβ − ∂2

r α− (∂rα)2

Rtθtθ = −re−2β

∂rα

Rtφtφ = −re−2β sin2

θ ∂rα

Rrθrθ = re−2β

∂rβ

Rrφrφ = re−2β sin2

θ ∂rβ

Rθφθφ = (1− e−2β ) sin2θ

Compute Ricci tensor

Rtt = e2(α−β)[∂2r α + (∂rα)2 − ∂rα∂rβ +

2

r∂rα]

Rrr = −[∂2r α + (∂rα)2 − ∂rα∂rβ −

2

r∂rβ]

Rθθ = e−2β [r(∂rβ − ∂rα)− 1] + 1Rφφ = Rθθ sin2

θ


Derivation

Notice that

e2(β−α)Rtt + Rrr =2

r(∂rα + ∂rβ)

Since each component of Rµν vanishes independently, then α = −β.

Consider now Rθθ = 0,1 = e2α (2 r ∂rα + 1) = ∂r (r e2α)

which has the following solution

e2α = 1−RS

r

with RS a constant of integration.

Thus

ds2 = −(

1−RS

r

)dt2 +

(1−

RS

r

)−1dr2 + r2dΩ2

It is easy to check that e2α = e−2β = 1− RS/r satisfies Rtt = Rrr = 0.

Recall: in the weak field limit we saw that

gtt = −1 + htt = −(1 + 2 Φ) = −(

1−2 G M

r

)

thus if we take the weak field limit of our solution

gtt = −(

1−RS

r

)⇒ RS = 2 G M Schwarzschild Radius


Birkhoff’s Theorem

Theorem: The Schwarzschild metric is the unique vacuum solution with spherical symmetry.Proof consist of showing that:

A spherically symmetric spacetime can be foliated by two spheres.

The spatial metric can always be recast as ds2 = dτ2(a, b) + r2(a, b)dΩ2 with (a, b) coordinatestransverse to the sphere.

the solution to Einstein’s equation is the Schwarzschild metric.

STEP 1

Our spacetime manifold M has the same symmetries as a S2 sphere, i.e. the ordinary rotations in 3-dimEuclidean space (special orthogonal group SO(3).

The Killing vectors associated with these symmetries are:

R = ∂φS = cosφ∂θ − cot θ sin θ∂φT = − sinφ∂θ − cot θ cos θ∂φ

The algebra of these Killing vectors is

[R, S] = T[S, T ] = R[T ,R] = S


Then, M admits a foliation of S2 spheres if every point x ∈ M is on exactly one of these spheres.

An example of this type of foliation is the space R× S2 (a wormhole)

STEP 2:

Our submanifolds are 2-spheres with coordinates (θ, φ) and metric dΩ2 = dθ2 + sin2 θ dφ2

Our spacetime metric is then in general

ds2 = gaa(a, b)da2 + 2 gab(a, b) da db + gbb(a, b)db2 + r2(a, b)dΩ2

Without loss of generality, we can pick r = b, so

ds2 = gaa(a, r)da2 + 2 gar (a, r) da dr + grr (a, r)dr2 + r2dΩ2


Next, find a function t(a, r) such that we eliminate the cross term dt dr in the metric and haveds2 = m dt2 + n dr2 + r2dΩ2.

Given t(a, r),

dt =∂t

∂ada +

∂t

∂rdr

so

dt2 =

(∂t

∂a

)2da2 + 2

(∂t

∂a

)(∂t

∂r

)da dr +

(∂t

∂r

)2dr2

Then,

ds2 = m dt2 + n dr2 + r2dΩ2

= m

[(∂t

∂a

)2da2 + 2

(∂t

∂a

)(∂t

∂r

)da dr +

(∂t

∂r

)2dr2]

+ n dr2 + r2dΩ2

= m(∂t

∂a

)2da2 + 2 m

(∂t

∂a

)(∂t

∂r

)da dr +

[m(∂t

∂r

)2+ n

]dr2 + r2dΩ2

Comparison withds2 = gaa(a, r)da2 + 2 gar (a, r) da dr + grr (a, r)dr2 + r2dΩ2

yields the following three equations for the three unknowns (t,m, n)

m(∂t

∂a

)2= gaa

n + m(∂t

∂r

)2= grr

m(∂t

∂a

)(∂t

∂r

)= gar


After solving these equations our metric will have the form

ds2 = m(t, r)dt2 + n(t, r)dr2 + r2dΩ2

Since we are dealing with a Lorentzian metric, either m or n will have to be negative. We select t to be thetimelike coordinate and write the metric as

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

STEP 3: Solving the Einstein equations.

Calculate Christoffel symbols, using labels (0, 1, 2, 3) for (t, r, θ, φ).

Γ000 = ∂0α Γ0

01 = ∂1α Γ011 = e2(β−α)∂0β

Γ100 = e2(α−β)∂1α Γ1

01 = ∂0β Γ111 = ∂1β

Γ212 = 1

r Γ122 = −re−2β Γ3

13 = 1r

Γ133 = −re−2β sin2 θ Γ2

33 = − sin θ cos θ Γ323 = cos θ

sin θ .

Calculate the Reimann tensor

R0101 = e2(β−α)[∂2

0β + (∂0β)2 − ∂0α∂0β] + [∂1α∂1β − ∂21α− (∂1α)2]

R0202 = −re−2β

∂1α

R0303 = −re−2β sin2

θ ∂1α

R0212 = −re−2α

∂0β

R0313 = −re−2α sin2

θ ∂0β

R1212 = re−2β

∂1β

R1313 = re−2β sin2

θ ∂1β

R2323 = (1− e−2β ) sin2

θ .


Calculate the Ricci tensor

R00 = [∂20β + (∂0β)2 − ∂0α∂0β] + e2(α−β)[∂2

1α + (∂1α)2 − ∂1α∂1β +2

r∂1α]

R11 = −[∂21α + (∂1α)2 − ∂1α∂1β −

2

r∂1β] + e2(β−α)[∂2

0β + (∂0β)2 − ∂0α∂0β]

R01 =2

r∂0β

R22 = e−2β [r(∂1β − ∂1α)− 1] + 1R33 = R22 sin2

θ .

From R01 = 0 we get ∂0β = 0

Taking the time derivative of R22 = 0 and using ∂0β = 0, we get ∂0∂1α = 0

Therefore, β = β(r) and α = f (r) + g(t)

The metric becomesds2 = −e2(f (r)+g(t))dt2 + e2β(r)dr2 + r2dΩ2


Redefine the time coordinate by replacing dt → e−g(t)dt to finally get

ds2 = −e2α(r)dt2 + e2β(r)dr2 + r2dΩ2

That is, any spherically symmetric vacuum metric can be written in coordinates such that all its componentsare independent of the time coordinate.

Therefore, any spherically symmetric metric possesses a time-like Killing vector.

Recall, we have already shown that the metric above can be brought to have the form

Schwarzschild Metric

ds2 = −(

1−2GM

r

)dt2 +

(1−

2GM

r

)−1dr2 + r2dΩ2


Stationary and Static

Stationary metric: one that possesses a time-like Killing vector near infinity.

For a stationary metric, one can choose coordinates (t,~x) such that the Killing vector is ∂t and the metrictakes the form

ds2 = g00(~x) dt2 + 2 g0i (~x) dt dx i + gij (~x) dx i dx j

Static metric: one that possesses a time-like Killing vector orthogonal to a family of hypersurfaces.

In the coordinates for which the metric is independent of the time coordinate, the Killing vector is orthogonalto the t = constant hypersurfaces.

In these coordinates, the metric reads

ds2 = g00(~x) dt2 + gij (~x) dx i dx j


Singularities

ds2 = −(

1−2GM

r

)dt2 +

(1−

2GM

r

)−1dr2 + r2dΩ2

The Schwarzschild metric becomes singular at r = 0 and r = 2GM

Are these real singularities? After all the form of the metric depends on the coordinate system used.

Could they be coordinate singularities? Example: ds2 = dr2 + r2dθ2 becomes degenerate at r = 0 andthe component gθθ = r−2 of the inverse metric blows up.

We need a coordinate independent quantity to identify singularities.

Examples: R = gµνRµν , RµνRµν , RµνρσRµνρσ , RµνρσRρσλτRλτµν

In the case of the Schwarzschild metric,

RµνρσRµνρσ =12G2M2

r6.

thus r = 0 represents an honest singularity.

Notice that at r = 2 G M this quantity is not singular. We will show later that we can transform to a newcoordinate system for which the metric is regular at this point.

Form most realistic spherical object, their radius R is such that R RS = 2 G M. For example, the Sun:R = 106 G M


Geodesics

Christoffel symbols for Schwarzschild:

Γ100 = GM

r3 (r − 2GM) Γ111 = −GM

r(r−2GM)Γ0

01 = GMr(r−2GM)

Γ212 = 1

r Γ122 = −(r − 2GM) Γ3

13 = 1r

Γ133 = −(r − 2GM) sin2 θ Γ2

33 = − sin θ cos θ Γ323 = cos θ

sin θ

Geodesic equations Uν∇νUµ = 0 with Uµ = dxµ/dλ

d2t

dλ2+

2GM

r(r − 2GM)

dr

dλ

dt

dλ= 0

d2r

dλ2+

GM

r3(r − 2GM)

( dt

dλ

)2−

GM

r(r − 2GM)

( dr

dλ

)2− (r − 2GM)

[( dθ

dλ

)2+ sin2

θ

( dφ

dλ

)2]

= 0

d2θ

dλ2+

2

r

dθ

dλ

dr

dλ− sin θ cos θ

( dφ

dλ

)2= 0

d2φ

dλ2+

2

r

dφ

dλ

dr

dλ+ 2

cos θ

sin θ

dθ

dλ

dφ

dλ= 0

with λ an affine parameter.


Recall that for any Killing vector Kµ,

∇(µKν) = 0 ⇒ pµ∇µ(Kνpν ) = 0

with p = m Uµ the 4-momentum of a particle.

Since Schwarzschild space-time have 4 Killing vectors (3 rotations and time translation), there are then 4constants of motion:

Kµdxµ

dλ= constant

In addition, from the geodesic equation Uν∇νUµ = 0 and metric compatibility∇αgµν = 0

ε = −gµνdxµ

dλ

dxν

dλ= constant

For time-like curves we typically pick λ = τ thus ε = 1. For null geodesics, ε = 0.

Invariance under time translation implies conservation of energy

Invariance under spatial rotation implies conservation of angular momentum

Conservation of angular momentum consists of magnitude and direction

Conservation of the angular momentum direction implies particles move in a plane thus we can, without lossof generality, choose θ = π/2


With θ = π/2, the two Killing vectors corresponding to conservation of energy and magnitude of theangular momentum are given, respectively, by K = ∂t and L = ∂φ or equivalently byKµ = (∂t )µ = (1, 0, 0, 0) and Lµ = (∂φ)µ = (0, 0, 0, 1).

Also,

Kµ =

(−(

1−2GM

r

), 0, 0, 0

)Lµ =

(0, 0, 0, r2 sin2

θ).

Therefore, the conserved quantities (energy and angular momentum) Kµ dxµdλ = constant are

(1−

2GM

r

) dt

dλ= E

r2 dφ

dλ= L


From

ε = −gµνdxµ

dλ

dxν

dλ= constant

we have that

−(

1−2GM

r

)( dt

dλ

)2+

(1−

2GM

r

)−1 ( dr

dλ

)2+ r2

( dφ

dλ

)2= −ε

Using E and L

−E2 +

( dr

dλ

)2+

(1−

2GM

r

)( L2

r2+ ε

)= 0

Rewrite as1

2

( dr

dλ

)2+ V (r) = E

where

V (r) =1

2ε− ε

GM

r+

L2

2r2−

GML2

r3

E =1

2E2

Notice: The geodesic equation has the same structure as that for a classical particle of unit mass, energy E moving

in a 1-dimentional potential V


We are interested on the solutions r(λ), t(λ), φ(λ)

There are different curves V (r) for different values of L

The type of orbit can be obtained by comparing the 12 E2 to V (r) and in particular looking for the turning

points where V (r) = 12 E2.

Circular orbits with radius rc = const happen if the potential is flat, i.e. dV/dr = 0. That is

εGMr2c − L2rc + 3GML2

γ = 0

where γ = 0 in Newtonian gravity and γ = 1 in general relativity.

Circular orbits will be stable if they correspond to a minimum of the potential, and unstable if theycorrespond to a maximum.

Bound orbits which are not circular will oscillate around the radius of the stable circular orbit.


Orbits in Newtonian gravity

0 10 20 300

0.2

0.4

0.6

0.8

r

1

23

4L=5

Newtonian Gravity

massless particles

0 10 20 300

0.2

0.4

0.6

0.8

r

L=1

2

3

4

5

Newtonian Gravity

massive particles

Circular orbits in Newtonian gravity appear at

rc =L2

εGM.

For massless particles ε = 0. Thus, there are no circular orbits (no bound orbits in left Figure).

Massless particles move in a straight line since the Newtonian gravitational force on massless particles iszero.

For massive particles, we have the standard situation of bound orbits for E < 1 are either circles or ellipses,while unbound for E ≥ 1 ones are either parabolas or hyperbolas.


Orbits in General Relativity

0 10 20 300

0.2

0.4

0.6

0.8

r

1

2

3

4L=5

General Relativity

massless particles

In general relativity the situation is different only for r → 0 when thepotential goes to−∞.

At r = 2GM the potential is always zero.

For massless particles there is always a barrier.

At the top of the barrier there are unstable circular orbits.

For ε = 0, γ = 1, the radius of this orbit is at rc = 3GM.


Orbits in General Relativity

0 10 20 300

0.2

0.4

0.6

0.8

r

L=1

2

3

4

5

General Relativity

massive particles

For massive particles, circular orbits are at

rc =L2 ±

√L4 − 12G2M2L2

2GM.

For large L there will be two circular orbits, one stable and one unstable.For L→∞ their radii are

rc =L2 ± L2(1− 6G2M2/L2)

2GM=

(L2

GM, 3GM

).

As we decrease L the two circular orbits come closer together; theycoincide when

L =√

12GM ,

for whichrc = 6GM ,

and disappear entirely for smaller L.

6GM is the smallest possible radius of a stable circular orbit.

In summary, Schwarzschild spacetimes possesses stable circular orbitsfor r > 6GM and unstable circular orbits for 3GM < r < 6GM.


Experimental Tests of General Relativity

Most the experimental tests involve motion of particles or photons (i.e. geodesics)Precession of periheliaGravitational redshiftDeflection of lightGravitational time or Shapiro delay

Early tests were confined to the Solar System

The deflection of light and the Shapiro delay arise in the weak-field limit; thus, they do not constitute astrong test of GR.

The ultimate test of GR is the detection of gravitational waves.


Precession of the Perihelia

In GR, elliptical orbits are not closed, they are ellipses that precess.

The precession of Mercury was the first test of GR.

The observed precession of Mercury’s major axis is δφ ≈ 5601 arcsecs /100 yrs.

Most of the precession, δφ ≈ 5025 arcsecs /100 yrs, is due to the precession of equinoxes in ourgeocentric coordinate system.

The gravitational perturbations of the other planets contribute an additional δφ ≈ 532 arcsecs /100 yrs.

The remaining δφ ≈ 43 arcsecs /100 yrs are explained by GR.


Precession of the PeriheliaDERIVATION:Recall from the geodesic equation:

( dr

dτ

)2= E2 −

(1−

2M

r

)(1 +

L2

r2

)dφ

dτ=

L

r2

Thus ( dr

dφ

)2=

E2 − (1− 2M/r)(

1 + L2/r2)

L2/r4r

Introduce u ≡ 1/r . ( du

dφ

)2=

E2

L2− (1− 2Mu)

( 1

L2+ u2

)In the Newtonian limit this equation reads

( du

dφ

)2=

E2

L2− (1− 2Mu)

1

L2− u2

A circular orbit in Newtonian theory has u = M/L2. Define y = u − M/L2 to represent the deviation from circularorbit. We then have for the GR case neglectingO(y3),

( dy

dφ

)2=

E2 + M2/L2 − 1

L2+

2M4

L6+

6M3

L2y −

(1−

6M2

L2

)y2


The solution to this equation has the form

y = y + A cos (kφ + B)

where

k =

(1−

6M2

L2

)1/2

y =3M3

k2L2

A =1

k

[E2 + M2/L2 − 1

L2+

2M4

L6− y

]1/2

with B an arbitrary constant. For comparison, in the Newtonian case

y =

[E2 + M2/L2 − 1

L2

]cos (φ + B)

Notice that in this case after φ advances 2π, the orbit returns to the same radius (i.e. closed orbits). In the GR case,the situation is different because of the constant k . The orbit returns to the same radius after kφ goes through 2π.That is

∆φ =2π

k= 2π

(1−

6M2

L2

)−1/2


For nearly Newtonian orbits

∆φ = 2π

(1−

6M2

L2

)−1/2

≈ 2π

(1 +

3M2

L2

)

Therefore, the perihelium advance is given by

δφ =6πM2

L2

For orbits about a non-relativistic star L2 ≈ Mr . Thus,

δφ =6πM

r

Consider the case of Mercury, r = 5.55× 107 km and M = 1 M = 1.47 km. Then

δφ = 4.99× 10−7 radians per orbit

Or equivalentlyδφ = 0.43′′/yr = 43′′/century


Gravitational Redshift

The gravitational redshift is another effect which is present in the weak field limit.

It will be predicted by any theory of gravity which obeys the Principle of Equivalence.

Over large distances, the amount of redshift will depend on the metric, and thus on the theory underquestion.

DERIVATION:

Consider the Schwarzschild geometry.

Consider two observers who are not moving on geodesics at fixed spatial coordinate values (r1, θ1, φ1)and (r2, θ2, φ2).

From the line element with dθ = dφ = 0,

dτi

dt=

(1−

2M

ri

)1/2

Suppose that the observerO1 emits a light pulse which travels to the observerO2, such thatO1 measuresthe time between two successive crests of the light wave to be ∆τ1.

Each crest follows the same path toO2, except that they are separated by a coordinate time

∆t =

(1−

2M

r1

)−1/2

∆τ1 .


The second observer measures a time between successive crests given by

∆τ2 =

(1−

2M

r2

)1/2

∆t =

(1− 2M/r21− 2M/r1

)1/2

∆τ1

∆τi measure the proper time between two crests of an electromagnetic wave, the observed frequenciesωi ∝ 1/∆τi are then related by

ω2

ω1=

∆τ1

∆τ2=

(1− 2M/r11− 2M/r2

)1/2

In the weak-field limit r >> 2M and

ω2

ω1= 1−

M

r1+

M

r2= 1 + Φ1 − Φ2

Consider an emitter at the location r = r1 near the object M and an observer at a location r2 >> r , then

λe

λ= 1−

M

ror equivalently

λ

λe= 1 +

M

r

Define the redshift z as

z ≡λ − λe

λe=

M

r


Deflection of Light

Recall that for photons the orbits are given by

dt

dλ= E

(1−

2 G M

r

)−1

dφ

dλ=

L

r2( dr

dλ

)2= E2 − V (r)

where

V (r) =L2

r2

(1−

2 G M

r

)Therefore

dφ

dr= ±

L

r2

(E2 − V (r)

)−1/2

= ±1

r2

[E2

L2−

1

r2

(1−

2 G M

r

)]−1/2

dφ

dt=

1

r2

L

E

(1−

2 G M

r

)


For r →∞

φ ≈b

rdr

dt≈ −1

dφ

dt=

dφ

dr≈

b

r2

butdφ

dt=

1

r2

L

E

(1−

2 G M

r

)≈

1

r2

L

E

thus the impact parameter b is given by

b =L

E

thereforedφ

dr= ±

1

r2

[ 1

b2−

1

r2

(1−

2 G M

r

)]−1/2

which yields a deflection angle

∆φ = 2∫ ∞

r1

dr

r2

[ 1

b2−

1

r2

(1−

2 G M

r

)]−1/2

with r1 the turning point radius, the radius where the square bracket vanishes.


Define r = b/w . Thus

∆φ = 2∫ ∞

r1

dr

r2

[ 1

b2−

1

r2

(1−

2 G M

r

)]−1/2

becomes

∆φ = 2∫ w1

0dw[

1− w2(

1−2 G M w

b

)]−1/2

expanding in powers of 2 G M/b one gets

∆φ = 2∫ w1

0dw(

1 +G M w

b

)[1− w2 +

2 G M w

b

]−1/2

which yields after integration

∆φ ≈ π +4 G M

b

The deflection angle is then

δφ = ∆φ− π =4 G M

b

which for the Sun is 1.7”


Time of Light

From

dφ

dr= ±

1

r2

[ 1

b2−

1

r2

(1−

2 G M

r

)]−1/2

dφ

dt=

1

r2

1

b

(1−

2 G M

r

)

we have thatdt

dr= ±

1

b

(1−

2 G M

r

)−1 [ 1

b2−

1

r2

(1−

2 G M

r

)]−1/2

Consider a pulse of light originating at the Earth r⊕ that grazes the Sun and is reflected back to Earth from a pointrR . Assume origin at the Sun.

The total travel time is(∆t)tot = 2 t(r⊕, r1) + 2 t(rR , r1)

where r1 is the distance of closest approach to the Sun and t is the travel time computed from

t(r, r1) =

∫ r

r1dr

1

b

(1−

2 G M

r

)−1 [ 1

b2−

1

r2

(1−

2 G M

r

)]−1/2


To first order in M, the integral

t(r, r1) =

∫ r

r1dr

1

b

(1−

2 M

r

)−1 [ 1

b2−

1

r2

(1−

2 M

r

)]−1/2

becomes

t(r, r1) =√

r2 − r21 + 2 M log

r +√

r2 − r21

r1

+ M

(r − r1r + r1

)1/2

thus(∆t)excess ≡ (∆t)tot − 2

√r2⊕ − r2

1 − 2√

r2R − r2

1

For r1/rR << 1 and r1/r⊕ << 1

(∆t)excess ≈4 G M

c3

[log

(4 rR r⊕

r21

)+ 1

]


Schwarzschild SpacetimeTo understanding the geometry of the Schwarzschild spacetime, we explore its causal structure as defined by thelight cones. Because of its spherical symmetric symmetry, let’s consider only radial null curves:

ds2 = 0 = −(

1−2GM

r

)dt2 +

(1−

2GM

r

)−1dr2

,

thusdt

dr= ±

(1−

2M

r

)−1.

Notice

dt

dr= ±1 for r →∞

dt

dr= ±∞ for r → 2M

The light that approaches 2m neverseems to get there!

We need to investigate if this is becauseof the choice of coordinates.

r

t

2GM


QUESTION: Does a free-falling observer ever reaches or even crosses r = 2M in a finite amount of proper time?

Consider the following coordinate transformation:

t = ±r∗ + constant

where the tortoise coordinate r∗ is defined by

r∗ = r + 2M ln( r

2M− 1)

The metric in these coordinates becomes

ds2 =

(1−

2M

r

)(−dt2 + dr∗2

)+ r2dΩ2

with r = r(r∗)

Notice that the light-cones do not close up, i.e. dt/dr∗ = ±1 everywhere.


ingoing Eddington-Finkelstein coordinate

Introduce the following ingoing coordinate u = t + r∗

The metric becomes

ds2 = −(

1−2M

r

)du2 + 2 du dr + r2dΩ2

At r = 2M, although guu = 0, there is no degeneracy since g = −r4 sin2 θ

For radial null curvesdu

dr=

0 , (infalling)

2(

1− 2Mr

)−1(outgoing)

Integration yiels

u = constant

u − 2(

r + 2M ln∣∣∣∣ r

2M− 1∣∣∣∣) = constant


At r = 2M

Light-cones tilt inwards. It is impossible to seethe inside.

The metric is perfectly regular

The surface is null

This surface is defines as the event horizon

The event horizon is a global object, requiresthe knowledge of the entire spacetime todefine it.

u

r = 2GM

u =

r = 0

const

~

~

r


Maximally Extended Schwarzschild SolutionRecall the metric in the form

ds2 =

(1−

2M

r

)(−dt2 + dr∗2

)+ r2dΩ2

Introduce both an ingoing coordinate u = t + r∗ and an outgoing coordinate v = t − r∗

The metric becomes

ds2 = −(

1−2GM

r

)du dv + r2dΩ2

with r = r(u, v) such that1

2(u − v) = r + 2M ln

( r

2M− 1)

Notice that in these coordinates r = 2M is infinitely far away (at either u = −∞ or v = +∞).

Introduce the following null coordinates: u′ = eu/4M and v′ = e−v/4M

The metric becomes

ds2 = −32M3

re−r/2GM du′ dv′ + r2dΩ2

Notice that none of the metric coefficients behave in any special way at the event horizon r = 2GM.

The transformation to the original coordinates reads:

u′ =

( r

2M− 1)1/2

e(r+t)/4M

v′ =

( r

2M− 1)1/2

e(r−t)/4M.


Kruskal-Szekeres coordinates

Introduce time-like and space-like coordinates

T =1

2(u′ − v′) =

( r

2M− 1)1/2

er/4M cosh(t/4M)

R =1

2(u′ + v′) =

( r

2M− 1)1/2

er/4M sinh(t/4M)

The metric becomes

ds2 =32G3M3

re−r/2GM (−dT 2 + dR2) + r2dΩ2

,

where

T 2 − R2 =

( r

2M− 1)

er/2M

The coordinates (T ,R, θ, φ) are known as the Kruskal-Szekres coordinates.


The event horizon (r = 2M) is defined byT = ±R

Surfaces of constant r = satisfy hyperbolas

T 2 − R2 = constant

Surfaces of constant t satisfy straight lines

T

R= tanh(t/4M)

Notice that t → ±∞ are the same surfacesas the horizon r = 2GM.


Schwarzschild vs Kruskal-Szekeres coordinates


Coordinates (T ,R) range over every value withouthitting the real singularity at r = 2GM; that is,−∞ ≤ u ≤ ∞ and T 2 < R2 + 1.

The T − R plane is known as the Kruskal diagramand has 4 regions.

Region I: The original region covered by theSchwarzschild coordinates.

Region II:

What we think of as the black hole.

Anything inside of this region can neverescape.

Every future-directed path in region II ends uphitting the singularity at r = 0.

The boundary of region II is called the futureevent horizon.

II

IV

III

I


Region III:

The time-reverse of region II

Things can escape to us, while we can neverget there.

It is often called a white hole

There is a singularity in the past, out of whichthe universe appears to spring.

The boundary of region III is called the pastevent horizon

Region IV:

Cannot be reached from region I.

It is another asymptotically flat region ofspacetime, a mirror image of region I.

It is connected to region I via a wormhole

II

IV

III

I


Wormholes

A

B

C

D

E

A B C D E

r = 2GM

v


Stars and Black Holes

The spacetime outside a spherically symmetric star is that of Schwarzschild (Birkhoff theorem).

Stars eventually collapse under their own gravitational pull once hydrostatic equilibrium cannot be longermaintained.

For stars with mass M < 8 M, gravitational collapse is stopped by electron degeneracy pressure (i.e.Pauli exclusion principle: no two fermions can be in the same state). Hydrostatic equilibrium is once againachieved and the resulting object is a white dwarf. (Mwd ≈ 1.4 M and Rwd ≈ R⊕).

For stars with mass 8 M < M < 18 M , neutron degeneracy pressure stops the gravitational collapseand a neutron star is formed (1.2 M < Mns ≈ 2 M and Rns ≈ 10 km).

For stars with mass 18 M < M, the stars will shrink down to below r = 2GM and further into a singularity,resulting in a black hole.

r = 2GMr = 0

vacuum(Schwarzschild)

interiorof star 0.5

1.0

1.5

log R

white dwarfs

neutron stars

1 2 3 4

D

B

CA

10

M/M

(km)


Stellar Models

Consider the general, static, spherically symmetric metric:

ds2 = −e2α(r)dt2 + e2 β(r)dr2 + r2 dΩ2

We need solutions to the Einstein equations Gµν = 8π Tµν such that

Tµν = (ρ + p)UµUν + p gµν

which ρ the energy density, p the pressure and Uµ the 4-velocity which for static solutions is given byUµ = (eα, 0, 0, 0).

The components of Tµν and Tµ ν are

Tµν = diag(e2αρ, e2βp, r2 p, r2 sin2

θ p)

Tµν = diag(−ρ, p, p, p)


The components of Gµν and Gµ ν are

Gtt =1

r2e2(α−β)

(2 r ∂rβ − 1 + e2β

)Grr =

1

r2

(2 r ∂rα + 1− e2β

)Gθθ = r2 e−2β

[∂

2r α + (∂rα)2 − ∂rα∂rβ +

1

r(∂rα− ∂rβ)

]Gφφ = sin2

θ Gθθ

and

Gtt = −

1

r2e−2β

(2 r ∂rβ − 1 + e2β

)Gr

r =1

r2e−2β

(2 r ∂rα + 1− e2β

)Gθ

θ = e−2β[∂

2r α + (∂rα)2 − ∂rα∂rβ +

1

r(∂rα− ∂rβ)

]Gφ

φ = Gθθ

respectively


The above yields the following three independent equations:

1

r2e−2β

(2 r ∂rβ − 1 + e2β

)= 8π ρ

1

r2e−2β

(2 r ∂rα + 1− e2β

)= 8π p

e−2β[∂

2r α + (∂rα)2 − ∂rα∂rβ +

1

r(∂rα− ∂rβ)

]= 8π p

Introduce the following new variable

m(r) =r

2(1− e−2β )

Φ = α

The metric takes the form

ds2 = −e2Φdt t +

(1−

2 m

r

)−1dr2 + r2 dΩ2

The first equation above becomesdm

dr= 4π r2

ρ


Integrate this equation to get

m(r) = 4π∫ r

0ρ(r) r2 dr

If the star has a radius R, then let M = m(r = R).

Question: Is M the mass of the star?

Recall: The proper spatial volume element is

√γ d3x = eβ r2 sin θ dr dθ dφ

whereγij = diag(e2β

, r2, r2 sin2

θ)

In the integral for m(r), the spatial volume element used was

√η d3x = r2 sin θ dr dθ dφ

whereηij = diag(1, r2

, r2 sin2θ)


It seems then that the “true” mass from the integrated energy density should be

M = 4π∫ R

0ρ(r) r2 eβdr

= 4π∫ R

0

ρ(r) r2[1− 2 m

r

]1/2dr

The difference between M and M is the binding energy; that is

Eb = M − M > 0

The binding energy is the energy needed to disperse the matter in the star to infinity.


Consider now the rr component equation

1

r2e−2β

(2 r ∂rα + 1− e2β

)= 8π p

which in terms of m and Φ readsdΦ

dr=

m + 4π r3 p

r [r − 2m]

From the energy-momentum conservation equation∇µTµν = 0, one gets that

(ρ + p)dΦ

dr= −

dp

dr

thus one arrives to

Tolman-Oppenheimer-Volkoff equations

dp

dr= −

m ρ

r2

(1 +

p

ρ

)(1 +

4π r3 p

m

)(1−

2m

r

)−1

dm

dr= 4π r2

ρ


Tolman-Oppenheimer-Volkoff equations

dp

dr= −

m ρ

r2

(1 +

p

ρ

)(1 +

4π r3 p

m

)(1−

2m

r

)−1

dm

dr= 4π r2

ρ

To close the above system, one need to provide an equation of state that relates pressure in terms ofenergy density and specific entropy, that is p = p(ρ, S)

In many situation the entropy is very small, so p = p(ρ)

Astrophysical systems can be often modeled with a polytropic equation of state: p = K ρΓo with K and Γ

constants and ρo the rest mass density.

The total energy density is then given by ρ = ρo(1 + ε) with ε the specific internal energy.

With the Gamma-law p = (Γ− 1)ρo ε one can show that

ρ =

( p

K

)1/Γ+

p

Γ− 1


EXAMPLE: Constant density sphere

ρ(r) =

ρ∗ r < R0 r > R

Integration of the dm/dr equation yields

m(r) =

43π r3ρ∗ r < R

43π R3ρ∗ = M r > R

Integration of the dp/dr equation yields

p(r) = ρ∗

√

1− 2 MR −

√1− 2 M

Rr2

R2√1− 2 M

Rr2

R2 − 3√

1− 2 MR

Notice that p(r = R) = 0 and

p(r = 0) = ρ∗

√

1− 2 MR − 1

1− 3√

1− 2 MR

Notice also that the pressure becomes infinity if the mass of the star exceeds Mmax = 4 R/9.


Chandrasekhar limit

WHITE DWARFS:

As stars run out of burnable fuel, they collapses.

A star burned-out star with mass M < 8 M , consists of a gas of electrons and ions.

When the electron separation becomes comparable to the de Broglie wavelength, λ = h/p, the starbecomes supported by electron degeneracy pressure because of the Pauli exclusion principle

The resulting object is a white dwarf with masses Mwd ≤ 1.4 M and radius Rwd ≈ R⊕ .

The maximum mass of a white dwarfs for which hydrostatic equilibrium can be maintained isMwd = 1.4 M . This limit is known as the Chandrasekhar limit


Oppenheimer-Volkoff limit

NEUTRON STARS:

For stars with mass 8 M < M < 18 M , electron degeneracy pressure is not capable to stop thegravitational collapse.

Electrons combine with protons to make neutrons and neutrinos.

The results is an object supported by neutron degeneracy pressure called neutron star with masses1.2 M < Mns ≈ 2 M and radius Rns ≈ 10 km).

The maximum mass of a neutron star for which hydrostatic equilibrium can be maintained isMns = 3− 4,M . This limit is known as the Oppenheimer-Volkoff limit

Highly magnetized, rotating neutron stars that emit a beam of electromagnetic radiation are called pulsars.


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