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The Kerr Metric for Rotating,

Electrically Neutral Black Holes:

The Most Common Case of BlackHole Geometry

Ben Criger and Chad Daley

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Assumptions

Non-zero angular momentum

Insignificant charge

Axial symmetry

No-Hair Theorem

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Derivation (Abridged)

Null Tetrad: Any set of four vectors (one timelikeand three spacelike such as m) for which the nullcondition (defined below) is met.

0

a

a

a

a

mmmm

Frame Metric:b

j

a

iabij eegg

Where is our metric of choice, and is one

of the null vectors in our tetrad.abg

b

je

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But Whats the Point?

To represent any metric in null tetrad / framemetric form.

Meaningful Example: Schwarzschild Metric

aal

1 aaa

r

mn

10

21

2

1

aaa

i

r

m32

sin2

1

aaa

i

rm

32

sin2

1

abbaabbaab mmmmnlnlg

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We perform an ingenious substitution of co-ordinates and obtain the following null vectors as

valid for a new metric:

aal1

' aaaar

mrn

12220cos'

'21

2

1'

aaaaa

iia

iarm

3210sin

sincos'2

1'

aaaaa

iia

iarm

3210

sinsin

cos'2

1'

We use the process detailed in the previous slides tofind the metric for these null vectors. Drumroll please . . . .

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The Kerr Metric in Boyer-LindquistCo-ordinates

We present the Kerr Metric in Boyer-Lindquist co-ordinates (first, wepresent BL co-ordinates here in comparison with spherical co-ordinates):

Where and

222

22

222

2

2

2

2

2

2sin

2sin421

d

arMrddrdtd

aMrdt

Mrds

2222 cosar 22 2 aMrr

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However. . .

We dont have any physical intuition at this point about the metric!

We will have to prove (or convince ourselves) that ais an angularmomentum parameter, etc.

We start by setting this parameter a = 0and seeing what happens.

222

22

222

2

2

2

2

2

2sin

2sin421

d

arMrddrdtd

aMrdt

Mrds

22222222sin

21

121 drdrdr

r

Mdt

r

Mds

dtdt dtdt dd dada

Now, we can say with confidence that arepresents angularmomentum (and has dimensions of radius).

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Nothing Succeeds Like Success

Now, we try removing mfrom the equation, and leaving a fixed a.

222

22

222

2

2

2

2

2

2sin

2sin421

d

arMrddrdtd

aMrdt

Mrds

222222222

22

222

22sincos

cos

dardardrar

ar

dtds

This metric may look deceptively complex, but this is simply theexpression of flat space in Boyer-Lindquist Co-ordinates.

Here, we have confirmed thatm

anda

are what they appear to be,and that our metric (chosen through a convenient, if unintuitivemethod) is a valid solution for rotating black holes.

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Just one more thing. . .

We need to prove that the metric is flat at infinity.

22222

22

2222222

22

222

222

2

2

222

2sin

cos

22cos

2

cos

cos

sin4

cos

21

d

ar

arMraMrrdardr

aMrr

ardtd

ar

aMrdt

ar

Mrds

22222222 sin drdrdrdtds

We recover flat space in spherical co-ordinates. We have effectively

argued that this metric is valid, and we can apply this method to theReissner-Nordstrom metric to obtain. . .

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The Kerr-Newman Metric

Represents a rotating, charged, black hole

Can devolve to any of the Schwarzschild, Kerr, or Reissner-Nordstrom metrics.

We use the following definitions:

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Singularities and Horizons

2 categories, essential and coordinate

Schwarzschild Solution

Essential singularity at r =0 Event horizon at Schwarzschild radius, r= 2m

Reissner-Nordstrm Solution Retain essential singularity at r =0

0 - 2 coordinate singularities at 22 qmmr

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Horizons and Singularities of theKerr Metric

Looking at our metric we find an essential singularity for;

Remembering the definitions of our co-ordinates we find;

This corresponds to a ring of radius a

0cos

0cos

22

2222

r

ar

022 zayx

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Horizons and Singularities fora2 < M2

A surface of infinite gravitational red shift can bedetermined by;

Setting a= 0, or = /2these reduce to;

21222

222

200

cos

0cos2

02

1

aMMr

aMrr

Mrg

S

0

2

S

S

r

Mr

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Horizons and Singularities Cont

We can also recover two event horizons settingthe radial coefficient to zero

In the case of a= 0, these surfaces reduce to:

2122

2202

aMMr

aMrr

0

2

r

Mr

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Summary of Kerr Geometry(a2 < m2)

Essential ring singularity at:

Two surfaces of infinite

red shift at :

Two event horizons

at:

022 zayx

21222 cos aMMrS

21

22 aMMrS

Particle

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Possible Energy Source?

Equip yourself with a large mass

Within the ergophere throw the mass against therotation

Upon exiting the ergosphere you will havegained energy

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A word about cases of a2 m2

For a2 > m2 we find only the essentialsingularity at r= 0

This naked singularity violates Penroses

cosmic censorship hypothesis

The solution for a2 = m2 is unstable

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Observed Black Holes

Cygnus X-1 widely accepted as the firstobserved black hole candidate

Jets observed companioning black holes termedquasars

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