Black Holes Galore in D 4 - UV•D=3has no black holes –GM is dimensionless can't construct a...
Transcript of Black Holes Galore in D 4 - UV•D=3has no black holes –GM is dimensionless can't construct a...
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Black Holes Galore
in D>4
Roberto Emparan
ICREA & U. Barcelona
Collaborations w/ R.Myers, H.Reall, H.Elvang, P.Figueras,
A.Virmani, T.Harmark, N.Obers, V.Niarchos, M.J.Rodríguez
(2000-2007)
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Higher-dimensional gravity
Motivations
As applications:
• String / M theory
• Large Extra Dimensions & TeV gravity
• AdS/CFT
• Mathematics: Lorentzian geometry
But, not least, also of intrinsic interest:
• D as a tunable parameter for gravity and black holes
What properties of black holes are
• 'intrinsic' ?
• D-dependent?
What can a black hole (i.e. spacetime) do?
Laws of bh mechanics…
Uniqueness, topology, shape, stability…
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• Why is D>4 richer?
– More degrees of freedom
– Rotation:
• more rotation planes
• gravitational attraction , centrifugal repulsion
– 9999 extended black objects: black p-branes
• Why is D>4 harder?
– More degrees of freedom
– Axial symmetries: U(1)'s at asymptotic infinity appear only every
2 more dimensions -- not enough to reduce to 2D σ-model if D>5
FAQ's
x2 x4 x6
x1 x3 x5
…
¡ GMrD¡3
+J2
M2r2
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Phases of black holes
• Find all stationary solutions that
– are non-singular on and outside event horizons
– satisfy Einstein's equations Rµν = 0
– with specified boundary conditions Asymp Flat
• What does the phase diagram look like?
• Which solutions maximize the total horizon area
(ie entropy)?
• N.B.: I (almost) won’t discuss stability today
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Phases of 4D black holes• Static: Schwarzschild
• Stationary: Kerr
Horizon: ∆=0 ) ) ) ) M r a : Upper bound on J for given M
J b GM 2
ds2 = ¡dt2 + ¹r
§
¡dt + a sin2 µdÁ
¢2+§
¢dr2 + §dµ2 + (r2 + a2) sin2 µdÁ2
ds2 = ¡³1¡ ¹
r
´dt2 +
dr2
1¡ ¹r
+ r2¡dµ2 + sin2 µdÁ2
¢
§ = r2 + a2 cos2 µ ; ¢ = r2 ¡ ¹r + a2 ; a =J
M
¹ = 2GM
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Phase plots• To compare solutions we need to fix a common scale
• Classical GR doesn't have any intrinsic scale
We'll fix the mass Mequivalently factor it out to get dimensionless quantities
• Uniqueness theorem: End of the story!multi-bhs are unlikely
j
extremal Kerr J b GM 2
non-singular, finite area
aH Kerr black holes aH ´ AH(GM)2
j ´ J
GM2
1
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Black holes in D>4
• Schwarzschild is easy:
• Unique static black hole Gibbons+Ida+Shiromizu
• Dynamically linearly stable Ishibashi+Kodama
Tangherlini 1963
ds2 = ¡³1¡ ¹
rD¡3
´dt2 +
dr2
1¡ ¹=rD¡3 + r2d(D¡2)
¹ /M
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• Kerr is harder. But Myers+Perry (1986) found
rotating black hole solutions with angular
momentum in an arbitrary number of planes
• They all have spherical topology
x2 x4 x6
SD¡2
x1 x3 x5
…
J1 J2 J3
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• Consider a single spin:
gravitationalcentrifugal
¢
r2¡ 1 = ¡ ¹
rD¡3+a2
r2
ds2 = ¡dt2 + ¹
rD¡5§
¡dt+ a sin2 µ dÁ
¢2+§
¢dr2 +§dµ2 + (r2 + a2) sin2 µ dÁ2
+r2 cos2 µ d2(D¡4)
§ = r2 + a2 cos2 µ ; ¢ = r2 + a2 ¡ ¹
rD¡5; ¹ /M
a / J
M
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• Consider a single spin:
D=5:
) ) ) ) a 2 b µ ) upper bound on J for givenM
• similar to 4D
• but extremal limit a 2 = µ ) rh=0
this is singular, zero-area
Horizon: ∆=0 ¢ = r2 + a2 ¡ ¹
rD¡5
r2h + a2 ¡ ¹ = 0 ) rh =
p¹¡ a2
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∆
r
~r 2
Dr6:
For fixed µ there is an outer event
horizon for any value of a
))))No upper bound on J for given M
))))9 ultra-spinning black holes
must have a zero 8a
~ -r -(D - 5)
Horizon: ∆=0 ¢ = r2 + a2 ¡ ¹
rD¡5
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JD = 5upper (extremal) limit
on J is singular
JD r 6 no upper limit on J
for fixed M
j
D = 5
extremal, singular
• Single spin MP black holes:
aH aH
j
D r 6
aH ! 0 ; j !1
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Several spins turned on:
D=5
aH
j2
j1
D=6
j1
j2
• If all j1~j2~…~jb(D-1)/2c ) ) ) ) ~ Kerr
• 9 ultra-spinning regimes if
one (two) ji are much smaller
than the rest
D r 6
j1
j2
j1~j2 ) ) ) ) ~ Kerr
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Is this all there is in D>4?
Not at all
Combine black branes & rotation:
)))) Black Rings + other blackfolds in Dr5
)))) Pinched black holes in Dr6
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D=D=55
End may be in sightEnd may be in sight
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The forging of the ring
bend into a circlespin it up
black string:Schw4D x R
Horizon topology S 1x S 2
Exact solution available -- and fairly simpleRE+Reall 2001
S2
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5D: one-black hole phases
thin black ring
1
fat black ring
Myers-Perry black hole
(naked singularity)
(slowest ring)
q2732
3 different black holes with the same value of M,J
A
J2
(fix scale: M=1)
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Multi-black holes
• Black Saturn:
• Exact solutions available
• Co- & counter-rotating, rotational dragging…
Elvang+Figueras
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Filling the phase diagram
• Black Saturns cover a semi-infinite strip
A
J
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Multi-rings are also possible
• Di-rings explicitly constructed
• Systematic method available (but messy)
Iguchi+Mishima
Evslin+Krishnan
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Thermodynamical equilibrium
is not in thermo-equil
• Beware: bh thermodynamics makes sense only with Hawking radiation
• Radiation is in equilibrium only if
Ti=Tj , Ωi=Ωj
)))) continuous degeneracies removed
• Multi-rings unlikely
Tr À Th ; r 6= h
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5D phases in thermal equilibrium
A
J
is there anything else?
Tbh=Tbr , Ωbh=Ωbr
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Towards a complete classification of
5D black holes
• Topology: S 3, S 1x S 2
Galloway+Schoen
• Rigidity: stationarity one axial U(1), but not (yet?) necessarily twoHollands et al
• If Rt xU(1)φxU(1)ψ then
– complete integrability Pomeransky
– ''uniqueness'' Hollands+Yazadjiev
• Essentially all bh solutions may have been found:
MP, black rings, multi-bhs (saturns & multi-rings)
(including also with two spins: Pomeransky+Sen'kov, Elvang+Rodríguez)
(bubbly black holes?)
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Black Holes Galore
in D>4 (II)
Roberto Emparan
ICREA & U. Barcelona
Collaborations w/ R.Myers, H.Reall, H.Elvang, P.Figueras,
A.Virmani, T.Harmark, N.Obers, V.Niarchos, M.J.Rodríguez
(2000-2007)
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D=D=55
End may be in sightEnd may be in sight
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DDrr66
Terra incognitaTerra incognitaTerra incognitaTerra incognitaTerra incognitaTerra incognitaTerra incognitaTerra incognita
Here be dragons!Here be dragons!Here be dragons!Here be dragons!Here be dragons!Here be dragons!Here be dragons!Here be dragons!
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Difficulties:
Explicit construction techniques
• Newman-Penrose formalism: unwieldy in D>4
• In D=4, 5: R xU(1), R xU(1) xU(1)
)))) 2D σ-model
But in Dr6: R xU(1)b(D-1)/2c not enough:
1+b(D -1)/2c < D -2
• Kerr-Schild class:
MP black holes are K-S, but black rings are not
g¹º = ´¹º + 2Hk¹kº
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Difficulties:Horizon topology
• Hawking's 4D theorem relies on Gauss-Bonnet thm:
• Galloway+Schoen: +ve Yamabe R(D-2) > 0 S 3, S 1x S 2
• D=6: Helfgott et al: S 4, S 2x S 2, S 1
x S 3, (H2/Γ)g x S2
So far: S 4(exact soln), S 1
x S 3(approximate soln)
• D>6: essentially unknownS D-2
(exact soln), S 1x S D-3
(approximate soln),
Tp x S D-p-2
(in progress)
RHR(2) > 0)H = S2
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So, very limited success in extending 4D
approaches
)))) Need new ideas
• more qualitative & less rigorous methods
(physics-guided)
• may guide later numerical attacks
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Pinched (lumpy) black holes in Dr6
Ultraspinning regime in Dr6
Black strings and branes exhibit – Gregory-Laflamme instability
– GL zero-mode branching into non-uniform horizons
J
)))) black membrane along rotation plane
¸GL ' r0L<r0
r0L GL zero-mode
Gubser
Wiseman
L ' r0 branch of static lumpy black strings
Gubser, Wiseman
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J
Pinched (lumpy) black holes in Dr6RE+Myers
L
L
a
Gregory-Laflamme-like
axisymmetric zero-mode
Ultra-spinning = membrane-like
Gregory-Laflamme zero-mode
L
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L
Replicas: multiple pinches
fit one GL zero-mode wavelength
fit two GL zero-mode wavelengths
etc(fixed mass)
Black membrane in T2
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Multiply pinched black holes from axisymmetric zero-modes:
• Not yet found --- presumably numerically or approximately
• They're necessary to complete the phase diagram) ) ) ) connect to black rings
• Pinched plasma balls recently found by Lahiri+Minwalla:
dual to (large) pinched black holes in AdS
…more axisymmetric pinches
J
GL-like
axisymmetric zero-mode
central pinch
circular pinch
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Thin black rings in D>5
• Heuristic: seems plausible
• Thin black rings > circular boosted black strings
• Equilibrium can be analyzed w/in linearized
gravity:– balance: tension ,,,, centrifugal repulsion
– gravitational self-attraction is subdominant
• Approximate construction via matched
asymptotic expansion T. Harmark's talk
SD¡3
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Matched asymptotic expansion
2- perturbations of a boosted black string
1- linearized soln around flat space
r0r¿ 1
r
R¿ 1
r0
Match in overlap zone
r0 ¿ r¿ R
R
equivalent delta-source
need bdry conditions to fix integration constants
very thin ring
r0
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Dr6 phase diagram
A(M,J) from approx solution
A
J
(fix M)
Black rings dominate the entropy at large J
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Black membrane ,,,, Rot Black Hole
(iv)
r0
r0
L
L
a
(i)
(ii)
(iii)
(fixed mass)
A
L
(i)
(ii)
(iv)(iii)
Black membrane in T2
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A
J
A
L
Dr6 phase diagram: a proposal
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Dr6 phase diagram: a proposal
A
J
caveat: branchings must be tangential, but direction is unknown
A
L
arguments for this phase structure can be made in detail less tight arguments, but still plausible
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Possible extensions
• Curving thin black branes:
– dynamical constraints on possible topologies
• Branes in gravitational potentials (AdS, dS, black
saturns, etc)
• Charged and susy black holes
Law of brane dynamics (Carter)
extrinsic curvature
K¹º½T¹º = F ½
eg, for tori
K¹º½T¹º = 0
T11R1
+T22R2
+ ¢ ¢ ¢ = 0easily satisfied!
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Conclusions: More is differentVacuum gravity Rµν = 0 in
• D=3 has no black holes
– GM is dimensionless can't construct a length scale (Λ, or h, provide length scale)
• D=4 has one black hole
– but no 3D bh no 4D black strings no 4D black rings
• D=5 has three black holes (two topologies); black strings black
rings, infinitely many multi-bhs…
• Dr6 seem to have infinitely many black holes (many topologies,
lumpy horizons…); black branes rings, toroids…, infinitely many multi-bhs…
we've just begun
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The End
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Curving a black string
Boosted black string:
z
r0
R
equivalent delta-source
SD-3
Thin boosted black string along a circle
What fixes σ ? or: how is R fixed in terms of M,J ?
)))) A(M,J,R)
ADM stress tensor:
Ttt / rD¡40 ((D ¡ 4) cosh2 ¾ + 1)Ttz / rD¡40 (D ¡ 4) cosh¾ sinh¾Tzz / rD¡40 ((D ¡ 4) sinh2 ¾ ¡ 1)
M = 2¼R
Z
SD¡3
T(±)tt
J = 2¼R2Z
SD¡3
T(±)tz
A = 2¼R
Z
SD¡3
pghor
T (±)¹º = TADM¹º ±(r)
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R
Compute 1/R corrections to linearized black string
Solution is singular unless Tzz=0
Overlap zone analysis: r0 ¿ r¿ R (still thin ring)
equilibrium conditions
) sinh2 ¾ =1
D¡ 4
MR =D ¡ 2pD ¡ 3
J
Curving a black string: equilibrium condition
equivalently from
r¹T¹º = 0