blackhole notes

8/14/2019 Blackhole Notes

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Geometric Aspects of Black HolesinGeometric, Topological and Algebraic Methods for Quantum Field Theory

Bruno Carneiro da CunhaDepartamento de Fsica Universidade Federal de Pernambuco

Villa de Leyva July 15-27 2013

BCdC (DF-UFPE) Geometry of Black Holes... VdL 2013 1 / 35
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Black Holes

Newtons Gravity: Fermi electronic pressure can support a body against its own

gravity if (c= 1):

M

G

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m2B

M (1)

In GR, metricds2 = e2(r)dt2 +h(r)dr2 +r2d2.Perfect fluid (static)Tab=uaub+P(uaub+gab). u

a =et. EinsteinsEquations:

Gtt: h(r) = 1 2m(r)

r

1

, m(r) = 4 r

0

(r)r2 dr

Grr: d

dr =

m(r) + 4r3P

r(r 2m(r))



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Remaining components give the Tolman-Oppenheimer-Volkoff equation:

dP

dr =

(P+)

m(r) + 4r3P

r(r 2m(r)) (2)

A greater(r)needed for equilibrium in GR.

ForR 1/3, orR 9M/4theres no equilibrium for constant density.Assuming a reasonable equation of state (P=P()), the existence of asuperior limit forMcan be proved.

Taking0 nuclear densities and a plausible behavior beyond0, one arrives at

Mmax (2 3)M

Extremelyhard to show the existence of collapsing solutions. Cf. D.

Christodoulou, The Formation of Black Holes in General Relativity.



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Endpoint of collapse: Birkhoffs theorem: Schwarzschilds metric:

ds2 = 1

2M

r

dt2 + dr2

1 2Mr

+r2d2

The problem atr= 2Mmay be either because of physical singularities, or just thefailure of the coordinate system chosen.



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Endpoint of collapse: Birkhoffs theorem: Schwarzschilds metric:

ds2 = 1

2M

r

dt2 + dr2

1 2Mr

+r2d2

The problem atr= 2Mmay be either because of physical singularities, or just thefailure of the coordinate system chosen.

Preamble: Rindlers metric:

ds2 = x2dt2 +dx2

problem atx= 0may be solved by changing coordinates:

T= xsinh t, X= xcosh t = ds2 = dT2 +dX2

Q: how to find a change a coordinates which solve these singular points?



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In the Schwarschild case, null trajectories (geodesics) satisfy:

dt= dr

1 2Mr

x=t

r+ 2Mlog

r 2M

2M



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Writing the metric in terms ofx:

ds

2

=

1

2M

r

dx+dx+r

2

d

2

, 1

2M

r =

er/2M

r e

(x+x)/4M

The exponential terms can be absorbed by a redefinition of coordinates:

X+=exp(x+/4M)eX= exp(x/4M):

ds2 = 16M2

er/2M

r dX+dX +r2d2 = 32M

2

er/2M

r (dT2+dX2)+r2d2

which shows no problem atr= 2M(orT=X): metric can be continued down tor= 0(RabcdR

abcd = 48M2/r6).

T=

1

2M

r

1/2er/4Msinh

t

4M

, X=

1

2M

r

1/2er/4Mcosh

t

4M



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The horizon in the place where the time translation symmetryta =tbecomesnull: gtt= (1 2M/r) = 0. Working withX, the metric is given by

ds2 = 32M3

r

er/2MdX+dX+r2d2 (3)

In terms of the Kruskal coordinates:

dr=4M2

r er/2M(X+dX+XdX+), dt= 2M

dX+

X+

dX

X

.

At the horizonX+= 0orX = 0, we have null norm and

atb=8M3

r3 er/2M

1

2M

r

(dT)[a(dX)b] =

16M3

r3 er/2MdX+ dX.

The quantity, computed atr= 2M:

2 = 1

2(atb)(atb) =

M2

r4

r=2M

= 1

4M,

gived the acceleration of the horizon. Numerically, it is equal to a force (per unit mass)

exerted at infinity, to keep a particle at the pointr= 2M.BCdC (DF-UFPE) Geometry of Black Holes... VdL 2013 7 / 35


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From the expression of the time translation field:

ta

=

1

4M

X+

X+ X

X, ta=

4M2

r er/2M

(XdX+X+dX), (4)

We define the acceleration of the time translation by:

b =ta

atb=4M3

r3 er/2M(X

+dX+XdX

+), (5)

withbb= 2M3/r5er/2MXX+. The surface gravity can be defined by

2 = bb/tctc, so=M/r

2. On the horizon,

a= 12eXdX+= ta= 14M2Me

r/2MXdX+. (6)

So, on the horizon, the Killing vector field is also geodesic.



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Kruskals Extension

From the coordinate transformation:

X2 T2 =

r2M

1

er/2M, t4M

=arc tanh

TX

r>2Monly coversX>T,X>0. What are the other regions?



11/25

Kruskals Extension

From the coordinate transformation:

X2 T2 =

r2M

1

er/2M, t4M

=arc tanh

TX

r>2Monly coversX>T,X>0. What are the other regions?The trajectory of null geodesics is unchanged by a conformal transformation of the

metric:

gab gab=e2gab, k

aak

b = 0 kaakb = (2kcc)k

b

So the causal structure is the same. Through the scale transformation, we can study

the infinity structure.

ds2 = dt2 +dr2 =dpdq+ 14sin

2(p q)d2

cos p cos q



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Causal Diagram for Kruskal Coordinates



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Reissner-Nordstrm

Black hole with massMand electric chargee.

ds2 = 1 2M

r + e

2

r2

dt2 + dr

2

1 2Mr

+ e2

r2

+r2d2

using the same methods as before, we can introduce null coordinatesx=t r

where

r =

dr

1 2Mr

+ e2

r2

=r+r2+

r+ rlog(rr+)

r2r+ r

log(rr), e2


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We arrive at the form of the metric

ds2 =4r4+

(r+ r)2(r r)

r2/r

2+e

r+r

r2+

r

r2 dX+dX+r

2d2

Withrdefined implicitly:

XX+= exp

r+ rr2+

r

in terms of the new variables, X can be continued beyondr=r+ (X+= 0or

X= 0. Metric can be continued beyondr=r by another set of transformations.



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And, again, the covariant derivative oft

atb =

2

r3 (Mr e2

)(dt)[a(dr)b]

results a constantat the horizonr=r+:

2 = (Mr e2)2

r6r=r+

= (M2 e2)1/2

2M[M+ (M2 e2)1/2] e2

Once more, the singularity at r= 0is physical:

RabcdRabcd =

8(6M2r2 12Mre2 + 7e4)

r8

Problem: Verify thataab =b at the horizon and find.



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Causal Diagram for Reissner-Nordstrm



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Near the Horizon

Makingr= r++y,y1, ther tpart of the metric changes to:

ds2 r+ r

r2+ydt2 r

2+

r+ r

dy2

y = r

+ r

4r2+x2dt2 + r

2+

r+ rdx2

Rindler! Part of Minkowski space!



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Near the Horizon

Makingr= r++y,y1, ther tpart of the metric changes to:

ds2 r+ r

r2+ydt2 r

2+

r+ rdy2

y = r

+ r

4r2+x2dt2 + r

2+

r+ rdx2

Rindler! Part of Minkowski space!

Whenr+ =r, however, have a different behavior:

ds2 x2

r2+dt2 +

r2+

x2dx2

AdS2! This is a space of constant negative curvature! Its isometries include a

dilation:

a = t

t+x

x


A S


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The causal structure ofAdS2 can be obtained through a series of coordinate changes:

X0= x

2

1

x2 t2 + 1

, X1=xt, X2=

x

2

1

x2 t2 1

And one arrives at the isometric embedding of AdS2 in R1,2:

ds2 = dX20+dX21 dX

22

WithX2

0

X2

1

+X2

2

= 1. Now we make:

X0=cosh rcosT, X1=sinh r, X2=cosh rsinT

And arrive at the global metric of anti-de Sitter (cosh r= 1/ sinX):

ds2 = cosh2 r dT2 +dr2 = dT2 +dX2

sin2 X

This space can be now continued to allT. Causal structure is curious: light rays arrive

atX= with finite affine parameter!


Th l h h i


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The geometry close to the horizon:Rindler AdS2


K N


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Kerr-Newman

Solution for a black hole with charge and angular momentum.

ds2 =

a2 sin2

dt2

2asin2 (r2 +a2 )

dtd+

+(r2 +a2)2 a2 sin2

sin

2

d2

+

dr2

+ d2

Aa= er

[(dt)a asin2 (d)a], and

=r2 +a2 cos2 , =r2 +a2 +e2 2Mr

a= J/Mis a measure of the angular momentum of the Black Hole.



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Horizon atr=r=M (M2 a2 e2)1/2. Gravity at the horizon:

=

(M2 a2 e2)1/2

2M[M+ (M2 a2 e2)1/2] e2

constant at the horizont, as well as the angular velocity:

H= a

r2

++a2

Complicated causal structure. Physical (curvature) singularity at

=r2 +a2 cos2 = 0

with a ring topologyS1.

Problem: Work out the near horizon geometry for charglesse= 0Kerr-Newman inthe extremal casea= M.


Diagram for Kerr


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Diagram for Kerr


What do the Black Hole solutions teach us?


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What do the Black Hole solutions teach us?

Curvature singularitys hidden behind event horizons.

Horizon atr=r+ are stable by small metric pertubations. There is nobifurcation.

Horizon is defined by the region where an isometry generator becomes null.

andHare constant at the horizon.

Near horizon geometry Rindler (except in the extremal case).


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