Microstates of Neutral Black Holes

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Microstates of Neutral Black Holes

• Gary Horowitz
• UCSB

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The entropy takes the following form This is
the exact entropy of any black hole with these
charges - even 5D Schwarzschild x T5. S and T
duality can permute the charges and the formula
is clearly invariant under this. It has never
been reproduced by a microscopic counting of
states at weak coupling.
Mysterious duality invariant formula
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Four dimensional analog
(Lowe, Maldacena, G.H., 1996)
D4 black holes need four charges to have SUSY
extremal limits. For later convenience, we
consider charges associated with wrapped
D3-branes as follows Compactify on T6 T2 x T2
x T2. Each 3-brane wraps one of the fundamental
cycles on each T2.
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• Consider the four 3-branes wrapping an even
  number of vertical cycles. Generically they
  dont intersect. But we can move the branes so
  that
• - Any two 3-branes intersect over a line and
• All four intersect at a point
• To preserve supersymmetry, one can choose the
  orientation of three of the branes arbitrarily.
  (The orientation of the last brane is then fixed.)

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The general 4D black hole with these charges is
Where, as usual
In the 10D solution, the dilaton and volume of
the T6 are both constant so the 4D dilaton is
constant and the above metric is both the
Einstein and string metric.
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Assuming each T2 is square with radii R1, R3, R5,
the volume of each 3-brane is
V3 (2?)3 R1R3R5 The integer normalized
charges, black hole energy and entropy are
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In addition to the 4 charges and energy, the 10D
solution has 3 pressures coming from the T6.
Once again we can exchange these 8 parameters
for the numbers of noninteracting branes and
antibranes wrapping each of the four cycles So
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The entropy becomes Mysterious duality
invariant formula in D4 In the extremal limit,
only one term in each factor remains e.g. if
there are no antibranes This has been
reproduced by counting states of strings at weak
coupling.
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At weak coupling this solution describes four
stacks of D3-branes. By duality, one can map this
to other systems where the counting is easier
e.g. three D4-branes and D0, or three M5-branes
and momentum. Changing the sign of the charge
doesnt change the extremal entropy counting.
Since the entropy is independent of the size and
shape of the torus, it seems clear that the
states are associated with the common
intersection point of the branes.
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If you accept branes and anti-branes can coexist,
there is a simple argument for the general
formula (M. Roberts, G.H.)
We have four stacks of 3-branes each containing
both branes and antibranes. There are 16
different ways of choosing one type of brane in
each stack. For each such choice, there are eS
states at the common intersection. The total
entropy is thus the sum of these 16 terms. This
reproduces the general formula.
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Precise Counting of Microstates of Kaluza-Klein
Black Holes

• Part 1 J0
• R. Emparan and G. H., 2006
• M. Roberts and G.H., to appear

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Outline

• Review basic properties of these black holes
• Embed in M theory and give precise counting of
  extremal entropy in a special case
• Show extremal entropy agrees for more general
  cases
• Discuss entropy of nonextremal solutions

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Kaluza-Klein Black Holes
These are five-dimensional vacuum solutions. In
terms of their four-dimensional reduction, they
are characterized by three parameters M, Q, P
satisfying
If Q P, these black holes are
Reissner-Nordstrom. Otherwise, the scalar field
is nontrivial. These black holes are never
supersymmetric.
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Integer normalized charges
Q is momentum in fifth direction y, and P
describes twisting of y circle over S2. If
y y 2?R, the integer normalized
charges are Q60 Horizon is S2 x S1 and
solution describes a boosted (neutral) black
string Q61 Extremal limit with Q0 0 is the KK
monopole. Otherwise, have S3 horizon.
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Taub-NUT
The Kaluza-Klein monopole is the product of time
and the euclidean Taub-NUT metric.
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black hole
Taub-NUT
One can think of these solutions as a 5D rotating
Myers-Perry black hole sitting at the tip of
Taub-NUT. Q now describes 5D rotation If Q ?0
(and J 0), then J1 J2 (Gaiotto et al. 2005).
In the limit R??, one recovers the
asymptotically flat rotating 5D black hole.
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The entropy is simple in the extremal limit
SBH ? Q0 Q6 Independent
of the circle radius R (due to attractor
mechanism). This reproduces the entropy of the
Myers-Perry black hole in the limit R ? ?. Since
the extremal entropy is such a simple function of
the integer charges, a microscopic counting of
states should be possible.
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Embed in M Theory
Take product of KK black hole with T6 and view KK
circle as the M theory circle, so R g ls. Q0
is the number of D0-branes. Q6 is the number of
D6-branes. The mass of the extremal black hole
is where
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Recall Entropy Counting for SUSY 4D Black Holes
Need four charges in four dimensions. Can take
four stacks of D3-branes. Any two stacks
intersect over a line, all four intersect over a
point.
With N branes in each stack, the solution is the
product of T6 and extreme Reissner-Nordstrom.
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Assuming a square torus with V6 (V3)2 These
represent the mass and entropy of both the black
hole and the configuration of branes. Since
entropy of the branes is independent of the size
(and shape) of the torus, it is natural to assume
that it is associated with the common
intersection point of the branes.
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D0-D6 bound states
D0 and D6 branes repel each other, but Taylor
(1997) showed that there are still perturbatively
stable bound states. He found a gauge
configuration on four D6-branes describing four
D0-branes bound to the 6-branes. That is
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Under T-duality on x2, x4, x6, Taylor showed this
is equivalent to four D3-branes wrapping the
diagonals of the torus. Explicitly Call x2 ?
x1 the and - cycle, and similarly for the other
two T2s (oriented so the even coordinate always
increases). The 3-branes are ( ), (
- -), (- -), (- - )
x2
This looks just like a rotated version of the
supersymmetric configuration.
x1
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Since there are four D3-branes, and they wrap
diagonal cycles, one would expect many nonzero
charges associated with branes wrapping various
three-cycles. But most of these vanish. It is
nonzero only over (135) and (246), each with
charge four. This is exactly what should arise
after three T-dualities of a D0-D6 system.
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With N D3-branes in each of the four stacks, the
mass of the configuration is The mass is larger
than the SUSY case simply because the branes are
wrapping cycles of larger volume.
This agrees with the mass of the KK black hole.
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R1
R3
R5
If each T2 is square with radii R1, R3, R5,
after the three T-dualities, V6 ls6 ? M0
M6. The string coupling is also rescaled g ? g
ls3/V3. Since Q0 Q6 4N
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What about the entropy? Assume that the entropy
of the intersecting D3-branes is unaffected by
the rotation. The black hole entropy is
SBH ? Q0Q6 16 ? N2 Which does not seem to
agree with the intersecting brane entropy Sbranes
2 ? N2
x2
But the branes have two intersection points on
each T2 !
x1
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With two intersection points on each T2, there
are eight intersection points on T6. Each
intersection contributes an entropy equal to the
SUSY case so Sbranes 8 (2 ? N2)
16 ? N2 SBH
This is a precise counting of the entropy of a
neutral black hole.
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Generalization to Q0 ? Q6
In terms of gauge fields on the 6-brane, this
corresponds to just rescaling F (Dhar and
Mandal,1998 Larsen, 2000). The T-dual
description consists of D3-branes wrapping cycles
more general than the diagonal. If Q04Nk3 and
Q64Nn3 , then the branes wrap the cycles x2 ? ?
kx1/n on each T2. The mass of the branes is now
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This again agrees with the mass of the black
hole! With Q04Nk3 and Q64Nn3
x (previous result)
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The entropy also agrees since there are now 2kn
intersection points on each T2 With a total
of (2kn)3 intersection points, the entropy is
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Comparison of Kaluza-Klein BH with earlier four
charge solution
The 4D metric for the Kaluza-Klein black holes is
with
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The coefficient of the 1/r2 vanishes unless both
charges Q, P are nonzero. This would agree with
four charge solution if we could factor with
the four constants ci, di all real. But this
turns out to be possible only if Q P, when the
solution reduces to Reissner-Nordstrom.
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Stability
Taylor showed his D0-D6 bound states were stable
to quadratic fluctuations of the gauge
field. Non-SUSY intersecting 3-branes are stable
since strings only know about two of the branes.
Any two are consistent with supersymmetry. These
extremal black holes have TH 0. But they can
still decay by emitting individual 0-branes or
6-branes. For a large charge black hole, decay
rate e-kQ. This is a nonperturbative
instability.
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Comments

• We dont understand why a simple system of
  D-branes can describe a vacuum black hole.
• It is intriguing that one needs four sets of
  branes even though (from IIA standpoint) there
  are only two charges. Reminiscent of mysterious
  duality invariant formula.