Solutions of Einstein’s Equations & Black Holes
1
Kostas Kokkotas
December 16, 2013
1
S.L.Shapiro & S. Teukolsky ”Black Holes, Neutron Stars and White
Dwarfs”
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Solutions of Einstein’s Equations & Black Holes
Schwarzschild Solution: Black Holes I
The light cone in a Schwarzschild spacetime
−1
2M
2M
ds2 ≡ 0 = 1 −
dt2 − 1 −
dr2
r
r
⇒
dr
2M
=± 1−
.
dt
r
r = 2M is a null surface and is called horizon or infinite redshift surface.
Kostas Kokkotas
Solutions of Einstein’s Equations & Black Holes
Schwarzschild Solution: Black Holes I
The light cone in a Schwarzschild spacetime
−1
2M
2M
ds2 ≡ 0 = 1 −
dt2 − 1 −
dr2
r
r
⇒
dr
2M
=± 1−
.
dt
r
r = 2M is a null surface and is called horizon or infinite redshift surface.
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Solutions of Einstein’s Equations & Black Holes
Schwarzschild Solution: Black Holes II
The
light
cone
for
Schwarzschild spacetime.
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Solutions of Einstein’s Equations & Black Holes
Schwarzschild Solution: Black Holes II
Collapse Diagram.
The
light
cone
for
Schwarzschild spacetime.
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Solutions of Einstein’s Equations & Black Holes
Schwarzschild Solution: Embedding diagrams
In order to get some feeling for the global geometry of the Schwarzchild
black hole we can try to represent aspects of it by embeddings in 3-space.
Let us look at the space of constant time and also supress one of the
angular coordinates, say θ = π/2. The metric becomes:
−1
2M
dl2 = 1 −
dr2 + r2 dφ2
(1)
r
The metric of the Euclidean embedding space is:
dl2 = dz 2 + dr2 + r2 dφ2
(2)
which on z = z(r) becomes
dl2 = 1 + z 02 dr2 + r2 dφ2
where z 0 = dz/dr. These are the same if
−1
p
2M
1 + z 02 = 1 −
→ z = 2 2M (r − 2M )
r
NOTE: We cannot follow the geometry further by this means for
r < 2M .
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Solutions of Einstein’s Equations & Black Holes
(3)
(4)
Schwarzschild Solution: Embedding diagrams
Embedding Diagram for
Schwarzschild spacetime. II
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An embedding diagram depicting the curvature of spacetime
around a static spherical star
Solutions of Einstein’s Equations & Black Holes
Schwarzschild Solution: Embedding diagrams II
The figure adapted from K.S. Thorne ”Black Holes and Time Warps”
(1994), Fig. 3.4 on page 132.
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Solutions of Einstein’s Equations & Black Holes
Eddington-Finkelnstein coordinates I
The Schwarzschild solution in a new set of coordinates (t0 , r0 , θ0 , φ0 )
0
r
t = t0 ± ln
− 1 , r > 2M
2M
r0
t = t0 ± ln 1 −
, r < 2M
2M
r = r 0 , θ = θ 0 , φ = φ0
can be written as
2M
4M
2M
dt02 − 1 +
dr2 ±
drdt0 − r2 dθ2 + sin2 θdφ2
ds2 = 1 −
r
r
r
While for null surfaces
2M
2M
4M
2
02
ds ≡ 0 = 1 −
dt − 1 +
dr2 ±
drdt0
r
r
r
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Solutions of Einstein’s Equations & Black Holes
Eddington-Finkelnstein coordinates II
0
(dt + dr)
2M
dt − 1 +
dr = 0
r
dr
r − 2M
and
=
dt0
r + 2M
2M
1−
r
dr
= −1
dt0
0
A manifold is said to be geodesically complete if all geodesics are of
infinite length. This means that in a complete specetime manifold, every
particle has been in the spacetime forever and remains in it forever.
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Solutions of Einstein’s Equations & Black Holes
Kruskal - Szekeres Coordinates : Maximal Extension
v
=
u =
r
1/2 r
t
4M
−1
e sinh
2M
4M
r
1/2 r
t
4M
−1
e cosh
2M
4M
—
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Solutions of Einstein’s Equations & Black Holes
Kruskal - Szekeres Coordinates : Maximal Extension
v
=
u =
r
r
1/2 r
t
−
1
er/2M
4M
−1
e sinh
2M
4M 2M
t
r
1/2 r
t
4M
4M
−1
e cosh
2M
4M
= u2 − v 2
=
tanh−1
—
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Solutions of Einstein’s Equations & Black Holes
v
u
Kruskal - Szekeres Coordinates : Maximal Extension
v
=
u =
r
r
1/2 r
t
−
1
er/2M
4M
−1
e sinh
2M
4M 2M
t
r
1/2 r
t
4M
4M
−1
e cosh
2M
4M
= u2 − v 2
=
tanh−1
—
ds2 =
32M 3 −r/2M
e
dv 2 − du2 − r2 dΩ2
r
du
= ±1
dv
In a maximal spacetime,
particles can appear or
disappear, but only at
singularities.
du2 −dv 2 = 0
⇒
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Solutions of Einstein’s Equations & Black Holes
v
u
Kruskal - Szekeres Coordinates : Maximal Extension
v
=
u =
r
r
1/2 r
t
−
1
er/2M
4M
−1
e sinh
2M
4M 2M
t
r
1/2 r
t
4M
4M
−1
e cosh
2M
4M
= u2 − v 2
=
tanh−1
—
ds2 =
32M 3 −r/2M
e
dv 2 − du2 − r2 dΩ2
r
du
= ±1
dv
In a maximal spacetime,
particles can appear or
disappear, but only at
singularities.
du2 −dv 2 = 0
⇒
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Solutions of Einstein’s Equations & Black Holes
v
u
Wormholes
A hypothetical “tunnel” connecting two different points in spacetime in
such a way that a trip through the wormhole could take much less time
than a journey between the same starting and ending points in normal
space. The ends of a wormhole could, in theory, be intra-universe (i.e.
both exist in the same universe) or inter-universe (exist in different
universes, and thus serve as a connecting passage between the two).
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Reissner-Nordström Solution (1916-18)
In contrast to the Schwarzschild (and Kerr) solutions, which are vacuum
solutions of the Einstein’s equations, the Reissner-Nordström solution is
not a vacuum solution.
The presence of the electric field gives rise to a nonzero energy
-momentum tensor throughout space.
Then we have to solve the static Einstein-Maxwell equations with a
potential
Q
, 0, 0, 0
(5)
Aµ =
r
where Q is the total charge measured by a distant observer.
1
Rµν − gµν R
2
F µν ;ν
F
µν;α
+F
να;µ
+F
αµ;ν
=
κTµν
(6)
=
0
(7)
=
0
(8)
where T µν is the energy momentum tensor of the electromagnetic field
defined in Chapter 2.
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Solutions of Einstein’s Equations & Black Holes
Reissner-Nordström Solution (1916-18)
The solution contains two constants of integration, M and Q which can
be interpreted as the mass and the electric charge2
−1
2M
Q2
Q2
2M
2
ds = − 1 −
+ 2 dt + 1 −
+ 2
dr2 + r2 dΩ2 (9)
r
r
r
r
2
The horizon is for g00 = 0
r± = M ±
p
M 2 − Q2
(10)
two horizons r− and r+ .
2
Where
G
1 G 2
M and Q2 =
Q
c2
4π0 c4
with 1/4π0 the Coulomb’s force constant
M=
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Solutions of Einstein’s Equations & Black Holes
Kerr Solution (1963) I
Kerr solution is probably the most import an solution of Einstein’s
equations relevant to astrophysics.
ds2 = −
where
2 2 ρ2 2 2 2
∆
2
2 2 sin θ
2
2
dt
−
a
sin
θdφ
(r
+
a
)dφ
−
adt
+
+ dr +ρ dθ
ρ2
ρ2
∆
3
∆ = r2 − 2M r + a2
and ρ2 = r2 + a2 cos2 θ
(11)
3
here a = J/M = GJ/M c is the angular momentum per unit mass. For
our Sun J = 1.6 × 1048 g cm2 /s which corresponds to a = 0.185.
• Obviously for a = 0 Kerr metric reduces to Schwarszchild.
• The Kerr solution is stationary but not static
• The coordinates are known as Boyer-Lindquist coordinates and they
are related to Cartesian coordinates
x
=
(r2 + a2 )1/2 sin θ cos φ
y
=
(r2 + a2 )1/2 sin θ sin φ
z
= r cos θ
3
If ∆ = r2 − 2M r + a2 + Q2 the we get the so called Kerr-Newman
solution which describes a stationary, axially symmetric and charged spacetime.
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Solutions of Einstein’s Equations & Black Holes
Kerr Solution II
Infinite redshift surface
We know that :
1 + z = νA /νB = [g00 (B)/g00 (A)]
which leads to
g00 = 1 −
2M r
=0
ρ2
→ r2 + a2 cos2 θ − 2M r = 0
R± = M ±
p
M 2 − a2 cos2 θ
which leads to
This means that there are two distinct infinite-redshift surfaces. As in the
Schwarzschild case these surfaces do not correspond to any physiccal
singularity.
The vanishing of the g00 merely tells us that a particle cannot be at rest
( with dr = dθ = dφ = 0) at theses surfaces; only a light signal emitted
in the radial direction can be at rest.
In the Kerr geometry, the infinite-redshift surface do not coincide with
event horizons.
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Solutions of Einstein’s Equations & Black Holes
Kerr Solution III
The shape of the light cone
ds2 ≡ 0 = g00 dt2 + grr dr2 + gφφ dφ2 + 2g0φ dφdt
at θ = π/2
For constant r (dr = 0) we get
2
dφ
0 = g00 + gφφ dφ
+
2g
0φ dt
dt
then at g00 = 0 we will get
and
dφ
=0
dt
(12)
gφφ
dφ
=−
dt
2g0φ
(13)
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Solutions of Einstein’s Equations & Black Holes
Kerr Black Hole
The light cone in the radial direction:
2
dr
g00
∆(ρ2 − 2M r)
2
2
0 = g00 dt + grr dr
⇒
=−
=
dt
grr
ρ4
(14)
thus the horizon(s) will be the surfaces:
∆ = r2 − 2M r + a2 = 0
⇒
r± = M ±
p
M 2 − a2
(15)
The are between r+ ≤ r ≤ R+ is called ergosphere.
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Solutions of Einstein’s Equations & Black Holes
Kerr Black Hole : Geodesics
We will consider test particle motion in the equatorial plane. If we set
θ = π/2 in the equation (11) we can get the Lagrangian
2M 2 4aM
r2 2
2M a2
2
2
2L = − 1 −
ṫ −
ṫφ̇ + ṙ + r + a +
φ̇2 (16)
r
r
∆
r
Corresponding to the ignorable coordinates t and φ we obtain two first
integrals (where ṫ = dt/dλ):
pt
=
pφ
=
∂L
= constant = −E
∂ ṫ
∂L
= constant = L
∂ φ̇
(17)
(18)
From (16) we can obtain:
ṫ
=
φ̇
=
(r3 + a2 r + 2M a2 )E − 2aM L
r∆
(r − 2M )L + 2aM E
r∆
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(19)
(20)
Solutions of Einstein’s Equations & Black Holes
Kerr Black Hole : Geodesics
A 3rd integral of motion can be derived by gµν pµ pν = −m2 which (by
using (19) & (20) is:
r3 ṙ2
= R(E, L, r)
(21)
where
R
= E 2 (r3 + a2 r + 2M a2 ) − 4aM EL − (r − 2M )L2 − m2 r∆
we can regard R as an “effective potential” for radial motion in the
equatorial plane.
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Solutions of Einstein’s Equations & Black Holes
Kerr Black Hole : Circular Orbits
Circular orbits occur where ṙ = 0. This requires:
R = 0,
∂R
=0
∂r
These equations can be solved for E and L to give
√
r2 − 2M r ± a M r
E/m = Ẽ = 1/2
√
r r2 − 3M r ± 2a M r
√
√
M r r2 ∓ 2a M r + a2
L/m = L̃ = ± 1/2 .
√
r r2 − 3M r ± 2a M r
(22)
(23)
(24)
Here the upper sign corresponds to corotating orbits and the lower sign
to counterrotating orbits.
• Kepler’s 3rd law : for circular equatorial orbits gets the form
Ω=
dφ
φ̇
M 1/2
= = ± 3/2
dt
r
± aM 1/2
ṫ
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(25)
Solutions of Einstein’s Equations & Black Holes
Kerr Black Hole : Circular Orbits
Circular orbits exist from r = ∞ all the way down to the limiting circular
photon orbit, when the denominator of Eq. (23) vanishes.
Solving for the resulting equation we find for the photon orbit
rph = 2M 1 + cos 2/3 cos−1 (∓a/M )
(26)
NOTE : For a = 0, rph = 3M , while for a = M , rph = M (direct) or
rph = 4M (retrograde)
? For r > rph , not all circular orbits are bound. An unbound circular
orbit has E/m > 1 and for an infinitesimal outward perturbation, a
particle in such an orbit will escape to infinity.
? Bound orbits exist for r > rmb , where rmb is the radius of the
marginally bound circular orbit with E/m = 1:
rmb = 2M ∓ a + 2M 1/2 (M ∓ a)1/2
(27)
rmb is the minimum periastron of all parabolic E/m = 1 orbits.
? In astrophysical problems, particle infall from infinity is very nearby
parabolic, since u∞ c. Any parabolic orbit that penetrates to r < rmb
must plunge directly into the BH. For a = 0, rmb = 4M , while for
a = M , rmb = M (direct) or rmb = 5.83M (retrograde)
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Solutions of Einstein’s Equations & Black Holes
Kerr Black Hole : Stable Orbits
Even the bound orbits are not stable.
Stability requires that
∂2R
≤0
∂r2
From eqn (22) we get
2
2M
E
≥
1−
m
3 r
The solution for rms , the “radius of marginally stable circular orbit”
h
i
1/2
rms = M 3 + Z2 ∓ [(3 − Z1 ) (3 + Z1 + 2Z2 )]
1/3 a2
a 1/3 a 1/3
Z1 = 1 + 1 − 2
1+
+ 1−
M
M
M
2
1/2
a
Z2 =
3 2 + Z12
M
(28)
(29)
is
(30)
(31)
(32)
For a = 0, rms = 6M , while for a = M , rms = M (direct) or rms = 9M
(retrograde)
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Solutions of Einstein’s Equations & Black Holes
Kerr Black Hole : Stable Orbits
• The binding energy of the marginally stable circular orbits can be taken
if we eliminate r from Eq (23) and using Eq (29) we find
√
a
4 2(1 − Ẽ 2 )1/2 − 2Ẽ
√
=∓
(33)
M
3 3(1 − Ẽ 2 )
p
p
0)
to
1/3 (a = M ) for
The quantity Ẽ decreases from p8/9 (a =
p
direct orbits and increases from 8/9 to 25/27 for retrograde orbits.
• The √
maximum binding energy 1 − Ẽ for a maximally rotating BH is
1 − 1/ 3 or 42.3% of the rest-mass energy.
• This is the amount of energy that is released by matter spiralling in
toward the black hole through a succession of almost circular equatorial
orbits. Negligible energy will be released during the final plunge from rms
into the black hole.
• Note that the Boyer-Lindquist coordinate system collapses rms , rmb ,
rph and r+ into r = M as a → M .
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Solutions of Einstein’s Equations & Black Holes
Kerr BH: Penrose Process (1969)
Particles inside the ergosphere may have negative energy
E = p0 = mg0µ uµ = m g00 u0 + g0φ uφ
(34)
Inside the ergosphere g00 < 0 while g0φ > 0 thus with an appropriate
choice of u0 and uφ it is possible to have particles with negative energy.
This can lead into energy extraction from a Kerr BH. Actually, we can
extract rotational energy till the BH reduces to a Schwarzschild one.
Starting from outside the ergosphere particle m0 enters into the
ergosphere and splits into two parts. One of them follows a trajectory of
negative energy and falls into the BH, the second escapes at infinity
carrying more energy than the initial particle!
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Solutions of Einstein’s Equations & Black Holes
Black-Hole: Mechanics
There is a lower limit on the mass that can remain after a continuous
extraction of rotational energy via Penrose process.
Christodoulou (1969) named it irreducible mass Mir
p
2
2
4Mir
= r+
+ a2 ≡ M + M 2 − a2 + a2 = 2M r+
(35)
while Hawking (‘70) suggested that the area of the BH horizon
cannot be reduced via the Penrose process. The two statements are
equivalent!
EXAMPLE : Calculate the area of the horizon
(r2 + a2 )2 sin2 θ
2
ds2 = gθθ dθ2 + gφφ dφ2 = r+
+ a2 cos2 θ dθ2 + +2
r+ + a2 cos2 θ
thus the horizon area is
Z 2π Z π
Z
√
A =
gdθdφ =
0
0
Z
2π
Z
=
0
2π
Z
0
π
√
√
gφφ gθθ dθdφ
0
π
2
(r+
2
2
2
+ a ) sin θdθdφ = 4π(r+
+ a2 ) = 8πM r+ = 16πMir
0
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Solutions of Einstein’s Equations & Black Holes
Black Hole Thermodynamics
Hawking’s area theorem opened up a new avenue in the study of BH
physics.
• The surface area of a BH is analogous to entropy in thermodynamics.
• In order to associate them one needs :
? to define precisely what is meant by the “entropy” of a BH
? to associate the concept of temperature with a BH
• Entropy is defined as the measure of the disorder of a system or of
the lack of information of its precise state. As the entropy increase
the amount of information about the state of a system decreases.
• No-hair theorem : a BH has only 3 distinguishing features : mass,
angular momentum and electric charge.
Bekenstein argued that the entropy of a BH could be described in terms
of the number of internal states which correspond to the same external
appearance. I.e. the more massive is a BH the greater the number of
possible configurations that went into its formation, and the greater is
the loss of information (which has been lost during its formation).
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Solutions of Einstein’s Equations & Black Holes
Black Hole Thermodynamics
• The area of the BH’s event horizon is proportional to M 2 .
• The entropy of the BH can be regarded as proportional to the area of
the horizon. Bekenstein has actually written
Sbh =
c
kA
4~
(36)
where A is the surface area, ~ Planck’s constant divided by 2π and k
Boltzmann’s constant.
• The introduction of BH entropy, calls for the definition of the BH
temperature!
The temperature of a body is uniform at thermodynamical
equilibrium (0th Law)
The the surface gravity of the BH (spherical or axisymmetric) plays
the role of the temperature,
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Black Hole Thermodynamics: 1st Law
• The surface gravity of the event horizon of a BH is inversely
proportional to its mass
GM
c4 1
κbh =
=
r2 r→r+
G 4M
(37)
in analogy the temperature should be inversely proportional to its mass.
• The less massive is a BH the hotter it would be !
• In more general case (Kerr-Newman) if we make a small change in the
mass M , angular momentum J and the electric charge Q we get
κ
δA + ΩδJ + ΦδQ
(38)
δM =
8πG
where the surface gravity and horizon area of a Kerr-Newman BH are:
κ =
4πµ/A
A
4π[2M (M + µ) − Q2 ]
=
µ =
(M 2 − Q2 − J 2 /M 2 )1/2
This is the 1st law of black hole mechanics
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Solutions of Einstein’s Equations & Black Holes
Black Hole Thermodynamics : 2nd & 3rd Law
Second law of black hole mechanics: the horizon can never decrease
assuming Cosmic Censorship and a positive energy condition 4
Third law of black hole mechanics: the surface gravity of the horizon
cannot be reduced to zero in a finite number of steps.
? Bekenstein-Hawking entropy formula
SH =
?
1
A
4G~
Bekenstein-Hawking temperature formula
M
~
−8
κ
≈
6.2
×
10
K
T =
4π 2 kc
M
(39)
4
Note that during the “evaporation”, M decreases and thus so does A and
S this violates Hawking’s area theorem. However one of the postulates is that
matter obeys the “strong” energy condition, which requires that the local
observer always measures positive energy densities and there are no spacelike
energy fluxes, which is not the case of pair creation i.e. there is a spacelike
energy flux.
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Solutions of Einstein’s Equations & Black Holes
Hawking Radiation
Classically: Nothing can escape from the interior of a BH.
Quantum Mechanically: Black holes shine like a blackbody with
temperature inversely proportional to their mass (Hawking 1974).
Quantum fluctuations of the zero-energy vacuum can create particle antiparticle pairs. These particles and antiparticles are considered to be
virtual in the sense that they don’t last long enough to be observed
According to Heisenberg’s principle ∆E · ∆t ≥ ~ and from Einstein’s
equation E = mc2 we get a relation which implies an uncertainty in mass
∆m · ∆t ≥ ~/c2
This means that in a very brief interval ∆t of time, we cannot be sure
how much matter there is in a particular location, even in the vacuum.
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Solutions of Einstein’s Equations & Black Holes
Hawking Radiation
Thus particle-antiparticle pairs can be created which may last for at
most, a time ~/c2 ∆m. Before, the time is up, they must find each other
and annihilate !
In the presence of a strong field the e− and the
e+ may become separated by a distance of
Compton wavelength λ = h/mc that is of the
order of the BH radius r+ .
There is a small but finite probability for one of
them to “tunnel” through the barrier of the
quantum vacuum and escape the BH horizon as
a real particle with positive energy, leaving the
negative energy inside the horizon of the BH.
• This process is called Hawking radiation
• The rate of emission is as if the BH were a hot body of temperature
proportional to the surface gravity.
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Solutions of Einstein’s Equations & Black Holes
Hawking Temperature
• Hawking calculated that the statistical effect of many such emissions
leads to the emergence of particles with a thermal spectrum:
N (E) =
8π
E2
2
c3 h3 e4π E/κh − 1
(40)
where N (E) is the number of particles of energy E per unit energy band
per unit volume.
• Notice the similarity between this formula and the Planckian
distribution for thermal radiation from a black body.
• This similarity led Hawking to conclude that a black hole radiates as a
black body at the temperature T
T =
~
~c3
κ
=
4π 2 kc
8πkGM
(41)
• Note that in the classical limit ~ → 0 the temperature vanishes and the
BH only absorbs photons, never emits them.
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Solutions of Einstein’s Equations & Black Holes
Hawking Temperature
The temperature expressed in terms of mass is
~c3
M
−8
T =
= 6.2 × 10
K
8πkGM
M
(42)
where M is the solar mass.
To estimate the power radiated, we assume that the surface of the BH
radiates according to the Stefan-Boltzmann law:
π2 k4 4
dP
= σT 4 =
T
(43)
dA
60 ~3 c2
Substituting the temperature and noting that the power radiated
corresponds to a decrease of mass M by P = (dM/dt)c2 , we obtain the
differential equation for the mass of a radiating BH as a function of t
π3
~c4
dM
=−
dt
15360 G2 M 2
By integration we get the duration of the radiation
3
M c2
G2 M 3
M
−19
t≈
≈
≈
8.6
×
10
sec
P
~c4
Kg
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(44)
(45)
Solutions of Einstein’s Equations & Black Holes
Hawking Radiation : Mini - BHs
? A BH with a mass comparable to that of the Sun would have a
temperature T ∼ 10−7 K and will emit radiation at the rate of
∼ 10−16 erg/s
? A BH with mass of about 1016 gr is expected to evaporate over a
period of about 1010 years (today).
? The critical mass M0 for a BH to evaporate after time t is
π 1
M0 =
8 101/3
~c4 t
G2 M 2
1/3
(46)
? We do not have a theory capable of explaining what happens when a
BH shrinks within the Planck radius (10−37 cm), and the answer to this
question lies in the area of speculation.
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Solutions of Einstein’s Equations & Black Holes