Holographic Thermodynamics of Accelerating Black Holes Ruth Gregory - - PowerPoint PPT Presentation

Holographic Thermodynamics of Accelerating Black Holes

Ruth Gregory

Centre for Particle Theory Andres Anabalon, Mike Appels, Finn Gray, David Kubiznak, Rob Mann, Ali Ovgun, Andy Scoins 1805.02687, 1811.04936, 1904.09660 1604:08812 & 1702:00490 [hep-th] Yamada et al. Ap J Lett 834 L3 (2016)

Black hole near SN remnant W44 in Milky Way.

Outline

• Accelerating black holes
• Thermodynamics of tension
• Acceleration and holography
• Accelerating Chemistry

Black hole is the ultimate slippery object – to accelerate we must be able to push or pull on the actual event horizon. Anything touching the event horizon must fall in, unless travelling at the speed of light. Local energy-momentum must have energy=tension. Luckily – we have a candidate..

Accelerating a Black Hole?

Cosmic String

A string produces a conical deficit, but no long range spacetime curvature (no tidal forces).

δ = 8πGµ

T µ

ν ≈ δ(2)(r) diag (µ, µ, 0, 0)

⇣ ⌘ ds2 = Ω2 " f(r)dt2 − dr2 f(r) − r2 ✓ d✓2 g(✓) + g(✓) sin2 ✓d2 K2 ◆# f = ⇣ 1 − 2m r ⌘ 1 − A2r2 + r2 `2 g = 1 + 2mA cos ✓ Ω = 1 + Ar cos ✓

The C-Metric

An accelerating black hole is described by the C-metric Where

Hong & Teo, CQG20 3629 (2003)

⇣ ⌘ ds2 = Ω2 " f(r)dt2 − dr2 f(r) − r2 ✓ d✓2 g(✓) + g(✓) sin2 ✓d2 K2 ◆# f = ⇣ 1 − 2m r ⌘ 1 − A2r2 + r2 `2 g = 1 + 2mA cos ✓ Ω = 1 + Ar cos ✓

C-Metric Deconstructed

Black hole: parameter “m”, cosmological constant, “l” f determines horizon structure – black hole / acceleration / cosmological constant

C-Metric Deconstructed

Black hole: Set A=0, mass “m”, with (negative) cosmological constant, -3/l2 Gives familiar Schwarzschild – AdS K determines conical singularity on axis / axes

ds2 = ✓ 1 − 2m r + r2 `2 ◆ dt2 − dr2

• 1 − 2m

r + r2 `2

− r2 ✓ d✓2 + sin2 ✓d2 K2 ◆

Isolating the effect of K, look on axis through black hole: K relates to tension of “cosmic string” on axis

Conical Deficits

c.f. Aryal, Ford, Vilenkin: PRD 34, 2263 (1986), Achucarro, Gregory, Kuijken: PRD 52 5729 (1995)

ds2

θ,φ ∝ dθ2 + θ2

K2 dφ2

δ = 2π ✓ 1 − 1 K ◆ = 8πµ

⇣ ⌘ ds2 = Ω2 " f(r)dt2 − dr2 f(r) − r2 ✓ d✓2 g(✓) + g(✓) sin2 ✓d2 K2 ◆# f = ⇣ 1 − 2m r ⌘ 1 − A2r2 + r2 `2 g = 1 + 2mA cos ✓ Ω = 1 + Ar cos ✓

Deconstructing A

Acceleration encoded in “A” Acceleration shows up here as a “shift” of infinity, and if m nonzero, a distortion of the spheres at a given “radius”

The Rindler Metric

With no black hole, A appears to modify the AdS length scale, but the conformal factor means the boundary has shifted and is not at infinite r

ds2 = Ω−2 " f(r)dt2 − dr2 f(r) − r2 ✓ d✓2 + sin2 ✓d2 ◆# f = 1 − A2r2 + r2 `2 Ω = 1 + Ar cos ✓

Horizon Acceleration

Slowly Accelerating Rindler

For “Al <1”, there is no horizon, and can change coords: To get (almost) global AdS! But note the factor of

1 + R2 `2 = 1 + (1 − A2`2)r2/`2 (1 − A2`2)Ω2 , R sin Θ = r sin ✓ Ω

↵ = p 1 − A2`2

ds2

AdS = −

⇣ 1 + R2 `2 ⌘ ↵2dt2 + dR2 1 + R2

`2

+ R2⇣ dΘ2 + sin2 Θd2 K2 ⌘

=/4 =0 = =/2 =3 /4 AdS Boundary = =/ = = r=

r = 0 ↔ R = A2`4 1 − A2`2

C-coordinates give AdS from an off-centre

• perspective. An observer

hovering away from centre of AdS is accelerating.

Acceleration and Black Hole

Putting together, A gives the acceleration of the black hole, which appears to lessen the effect of the cosmological constant. The acceleration also gives an imbalance between North and South axes that now have different conical deficits.

Λeff = − 3 `2

• 1 − 3A2`2
• θ → 0

ds2

θ,φ ∝ dθ2 + (1 + 2mA)2

K2 θ2dφ2

• θ → π

ds2

θ,φ ∝ dθ2 + (1 − 2mA)2

K2 (π − θ)2dφ2

Conical Deficits

Gives different tensions for N and S strings: Conventionally, we make N axis regular, with deficit on S axis

δ± = 2π ⇣ 1 − g(0) K ⌘ = 2π ⇣ 1 − 1 ± 2mA K ⌘ = “8πµ±”

δN = 0 ⇒ K = 1 + 2mA

δS = 8πmA K ⇒ µS = mA K

The Slowly Accelerating Black Hole

The slowly accelerating black hole in AdS is displaced from centre. It has a conical deficit running from the horizon to the boundary. The string tension provides the force that hold the black hole off-centre.

δ− − δ+ or δS = 8πmA K ⇒ µS = mA K

Physical Interpretation

The tension of the string provides a force to accelerate the black hole, looking like Newton’s Law:

µ = F = MA

In general, the C-metric has acceleration horizon, so thermodynamics would refer locally to black hole horizon – here we can see the nontrivial nature of the spacetime more readily. We look at slowly accelerating black holes to have only 1 horizon.

General C-metrics

Black Hole Thermodynamics

The parameters in the metric are: So expect charges and potentials associated with each. Set e=a=A=0, to understand the effect of K. Horizon defined by f=0, look at small changes in f. Changes r+, hence S Changes m, hence M Changes l, hence L

m , ` , (e , a) , A , K

f(r+ + r+) = f 0(r+)r+ + @f @mm + @f @` ` = 0

Black Hole Thermodynamics

Take a look at semi-familiar case: Schwarzschild-AdS with a deficit. Black hole horizon defined by f=0, look at small changes in f. Horizon still defined by f(r) = 0. Changes r+, hence S Changes m, hence M Changes l, hence L

f(r+ + r+) = f 0(r+)r+ + @f @mm + @f @` ` = 0

Black Hole Thermodynamics

Temperature has usual definition, but entropy depends

• n K:

So And we have to identify changes in K

T = f 0(r+) 4π S = πr2

+

K f 0(r+)δr+ = 2K r+ ✓ TδS + r2

+f 0(r+)

4 δK K2 ◆

(S): Herdeiro, Kleihaus,Kunz,Radu: 0912:3386 [gr-qc]

Changing Tension

Tension is related to K: So easily get Finally

µ = 1 4 ✓ 1 − 1 K ◆ δµ = δK 4K2 P = −Λ = 3 8⇡`2 V = 4⇡r3

+

3K

First Law with Tension

Putting together: So identify Then also get Smarr relation:

M = m K M = 2TS − 2PV

0 = 2K r+ ⇣ TδS + 2(m − r+)δµ + V δP − δ(m K ) ⌘

Thermodynamic Length

The term multiplying the variation in tension is a “thermodynamic length” Reinforces interpretation of M as enthalpy, if black hole grows, it swallows some string, but has also displaced the same amount of energy from environment.

λ = r+ − m

Kastor & Traschen: 1207:5415 [hep-th]

Does it make sense?

Consider a cosmic string interacting with a black hole: The black hole captures the string, but the string keeps moving and slides off, leaving a portion behind and the black hole heavier.

Black hole must grow:

δµ ' µ

δµ ' µ

δM + 2mδµ = TδS δM = 0 δr+ = 0

Captures string inside the horizon.

Black hole must grow:

m m0 m1 m2 1 K 1 1 − 4µ 1 S πm2 πm2

1(1 − 4µ)

πm2

2

M m0 m1(1 − 4µ) m2

m2 = m1 = m0(1 + 4µ)

Back to the acc black hole:

Full metric Kerr-Newman with the conformal factor: And importantly a possible renormalisation of t Thermodynamics has often been derived by consistency.

ds2 = 1 H2 ⇢f(r) Σ hdt α − a sin2θdϕ K i2 − Σ f(r)dr2 − Σr2 h(θ)dθ2 − h(θ) sin2θ Σr2 hadt α − (r2 + a2)dϕ K i2

f(r) = (1 − A2r2)  1 − 2m r + a2 + e2 r2

• + r2 + a2

`2 , h(✓) = 1 + 2mA cos ✓ +  A2(a2 + e2) − a2 `2

• cos2✓ ,

Σ = 1 + a2 r2 cos2✓ , H = 1 + Ar cos ✓ .

When we first derived thermodynamics of the accelerating black hole, we derived a “master formula” Where And keeping the mass as m/K, could replace dK and find a consistent first law, but without rotation.

Back to the Acc BH

δ ⇣m K ⌘ = T0δS + Ω0δJ − Φ0δQ + V0δP − λ0+δµ+ − λ0−δµ− + m δK 2K2

T0 = r2

+f 0(r+)

4⇡(r2

+ + a2)

S = ⇡ K (r2

+ + a2)

(1 − A2r2

+)

Q = e K Φ0 = − er+ r2

+ + a2

P = 3 8⇡`2 V0 = 4⇡ 3 r+(r2

+ + a2)

K(1 − A2r2

+)2

Ω0 = aK r2

+ + a2

J = ma K2

Back to the drawing board

How unique are the thermodynamic quantities, and how can we be confident, physically, that we have the right definitions? Decided to explore this in three ways: Ø Conformal computation – the Ashtekhar Das mass Ø Slow acceleration means we can compute the boundary stress tensor of the black hole and explore the dual boundary. Ø Compute the action of the black hole plus string system – is it consistent?

Return to the Metric

When computing thermodynamics of Kerr-AdS, Gibbons Perry and Pope first noticed that the standard coordinates were a rotating frame at infinity, and moreover that t was not correctly normalised.

Gibbons, Perry, Pope hep-th/0408217

ds2 = f(r) Σ hdt α − a sin2θdϕ K i2 − Σ f(r)dr2 − Σr2 h(θ)dθ2 − h(θ) sin2θ Σr2 hadt α − (r2 + a2)dϕ K i2

f(r) = 1 − 2m r + a2 r2 + r2 + a2 `2 , h(✓) = 1 − a2 `2 cos2✓ , Σ = 1 + a2 r2 cos2✓ , K = 1 − a2 `2

Return to the Metric

From their experience with the Kerr-AdS metric (and motivated by the coordinate transformation for slowly accelerating Rindler) we introduce a possible rescaling of the time coordinate This will rescale temperature, and also changes computations of the mass.

ds2 = 1 H2  f α2 dt2 − dr2 f − r2 ✓dθ2 g + g sin2 θdφ2 K2 ◆

Accelerating Thermodynamics: T

Using the usual Euclidean method, find temperature: And entropy:

S = πr2

+

K(1 − A2r2

+)

T = f 0

+

4⇡↵ = 1 2⇡r2

+↵

 m(1 − A2r2

+) +

r3

+

`2(1 − A2r2

+)

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Ashtekhar Das

Idea: do a conformal transformation to remove divergence near boundary: then integrate the conserved current formed from a Killing vector, Weyl tensor and normal to boundary:

¯ gµν = ¯ Ω2gµν

Q(⇠) = ` 8⇡ lim

¯ Ω→0

I `2 ¯ Ω N αN β ¯ Cν

αµβ⇠νd ¯

Sµ ,

M = Q(∂τ) = αm K

Fefferman-Graham

Expand the metric near the boundary: Fn and Gn determined by the requirement that

ds2 = −`2dz2 + 1 z2 ⇥ µν + z2Ψµν + z3Mµν ⇤ dxµdxν + O(z2)

1 r = −Aξ − X Fn (ξ) zn cos θ = ξ + X Gn (ξ) zn

Fefferman-Graham

For the boundary metric, get: And for the boundary fluid stress tensor: where

• 1 − A2`2g(⇠)

3 ↵2`2F 2

1 (⇠)

d⌧ 2 −

• 1 − A2`2g(⇠)
• F 2

1 (⇠)g(⇠)

d⇠2 − g(⇠)

• 1 − A2`2g(⇠)

2 K2F 2

1 (⇠)

d2

⇢E = m ↵ (1 − A2`2g)3/2(2 − 3A2`2g) Π = 3A2`2gm 2↵ (1 − A2`2g)3/2

hT µ

ν i = diag {ρE, ρE/2 + Π, ρE/2 Π}

• 1.0
• 0.5

0.0 0.5 1.0 0.5 1.0 1.5 2.0 2.5 3.0

ρE

pξ

pφ

Conical Deficit

Accelerating Thermodynamics

Integrate up the boundary stress-energy to get the mass: What is alpha? Setting m to zero, and demanding that the boundary is a round 2-sphere gives Get a consistent first law with corrections to V and TD length, and – can generalise to rotation

M = Z ρE √γ = αm K

↵ = p 1 − A2`2

General Thermo Parameters

M = m(Ξ + a2/`2)(1 A2`2Ξ) KΞ↵(1 + a2A2) T = f 0

+r2 +

4⇡↵(r2

+ + a2) ,

S = ⇡(r2

+ + a2)

K(1 A2r2

+) ,

Q = e K , Φ = Φt = er+ (r2

+ + a2)↵ ,

J = ma K2 , Ω = ΩH Ω1 , ΩH = Ka ↵(r2

+ + a2)

P = 3 8⇡`2 , V = 4⇡ 3K↵ r+(r2

+ + a2)

(1 A2r2

+) + m[a2(1 A2`2Ξ) + A2`4Ξ(Ξ + a2/`2)]

(1 + a2A2)Ξ

• ± =

r+ ↵(1 ± Ar+) m ↵ [Ξ + a2/`2 + a2

`2 (1 A2`2Ξ)]

(1 + a2A2)Ξ2 ⌥ A`2(Ξ + a2/`2) ↵(1 + a2A2)

Ξ = 1 − a2 l2 + A2(e2 + a2)

↵ = p (Ξ + a2/`2)(1 − A2`2Ξ) 1 + a2A2

Chemical Expressions

When drawing an analogy with Chemistry, it is more natural to express the chemical potentials in terms of thermodynamical charges, rather than geometrical quantities like r+. A starting point is the Christodoulou-Ruffini mass formula: Here shown for Kerr-Newman-AdS.

M 2(S, Q, J, P) = S 4π h✓ 1 + πQ2 S + 8PS 3 ◆2 + ✓ 1 + 8PS 3 ◆ 4π2J2 S2 i

The Deficits

∆ = 1 − 2(µ+ + µ−) = Ξ K C = (µ− − µ+) ∆ = mA K∆ = mA Ξ

While we originally derived the thermodynamics for each deficit separately, it is more natural to think in terms of an

• verall average deficit, and the differential deficit that

produces acceleration. We therefore encode: Then, by observing how the charges/potentials scaled with K, conjectured how the C-R formula would generalise.

✓ Ξ = 1 + e2A2 − a2 `2 (1 − A2`2) ◆

M 2 = ∆S 4π h✓ 1 + πQ2 ∆S + 8PS 3∆ ◆2 + ✓ 1 + 8PS 3∆ ◆ ✓ 4π2J2 (∆S)2 − 3C2∆ 2PS ◆i

V = 2S2 3πM ✓ 1 + πQ2 ∆S + 8PS 3∆ ◆ + 2π2J2 (∆S)2 + 9C2∆2 32P 2S2

• ,

T = ∆ 8πM " ✓ 1 + πQ2 ∆S + 8PS 3∆ ◆ ✓ 1 πQ2 ∆S + 8PS ∆ ◆ 4π2J2 (∆S)2 4C2 # , Ω = πJ SM∆ ✓ 1 + 8PS 3∆ ◆ , Φ = Q 2M ✓ 1 + πQ2 S∆ + 8PS 3∆ ◆ , λ± = S 4πM " ✓8PS 3∆ + πQ2 ∆S ◆2 + 4π2J2 (∆S)2 ✓ 1 + 16PS 3∆ ◆ (1 ⌥ 2C)2 ± 3C∆ 2PS #

RG & Scoins, 1904.09660

Black Hole Chemistry

Thermodynamics of black holes displays a rich phenomenology, and many critical phenomena. Even without charges, the Hawking Page transition occurs:

S T

T G

Accelerating Chemistry

To explore how the thermodynamics depend on the deficits, first let C=0, and consider the uncharged nonrotating black hole. It looks like D is almost irrelevant, but consider the Gibbs free energy: G = M-TS D decreases the free energy – though the HP transition remains at the same T.

G = ∆S 8πM ✓ 1 + 8PS 3∆ ◆ ✓ 1 − 8PS 3∆ ◆

0.2 0.4 0.6 0.8 1.0 T

• 0.4
• 0.3
• 0.2
• 0.1

0.0 0.1 0.2 0.3 G =1 =0.5 =0.05

Swallowtails

With charge or rotation, new critical phenomena appear. The temperature can now have two turning points with entropy, if the pressure is low enough relative to charge.

S T

0.05 0.10 0.15 0.20 0.25 T

• 0.4
• 0.2

0.0 0.2 0.4 0.6 G

Swallowtails

Gradually dropping charge causes a swallowtail to appear in the free energy.

Swallowtails

But now something interesting happens with acceleration! Let J=0, and shorten expressions by writing: Then we can factorise the expression for M: Where If q<1, these roots are real, and the enthalpy can vanish!

x = 8PS 3∆ , πQ2 ∆S = q C2 x

M 2 = ∆S 4π ✓ 1 + x − q+ C2 x ◆ ✓ 1 + x − q− C2 x ◆

q± = 2 − q ± 2 p 1 − q

Abbasvandi et al. 1812.00384

Snapped Swallowtails

q relates the magnitude of the charge to the ratio of the deficits and pressure, and as pressure drops, q drops. The swallowtail forms, then snaps at a critical Q.

0.2 0.4 0.6 0.8 T

• 0.10
• 0.05

0.00 0.05 0.10 G

RG & Scoins

Reverse Isoperimetric Inequality

The Isoperimetric Inequality in Mathematics says that the minimal boundary length enclosing a given area is a circle (or suitable higher dimensional generalisation). This is a problem if true for thermodynamic volume and black holes, since it would say that a round black hole would minimize area for a given volume – but entropy should be maximized! Cvetic et al conjectured that black hole satisfied a reverse of this mathematical inequality, and demonstrated its validity for known examples.

Reverse Isoperimetric Inequality

Focus on uncharged case: & combine the expressions

M 2 = ∆S 4π (1 + x)  1 + x − 4C2 x

• V = 2S2

3πM  1 + x + C2 x2

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4πM 2 ∆S = ✓3πMV 2S2 − 2C2 x2 ◆2 − 4 (1 + x) C2 x

New Reverse Isoperimetric Inequality

M 2 ✓3V 4π ◆2 ⇣ π S ⌘2 ≥ ✓3πMV 2S2 − 2C2 x2 ◆2 ≥ πM 2 ∆S

Now can manipulate into a new inequality appropriate for conical deficit black holes:

✓3V 4π ◆2 ≥ 1 ∆ ✓ A 4π ◆3

Recap

§ Have shown how to allow for varying tension in thermodynamics of black holes. § Conjugate variable is Thermodynamic Length § Thermodynamics of accelerating black holes is computable – non-static and non-isolated. § A key technical point is the normalisation of timelike Killing vector § Have derived extensive expressions for the TD variables and a new Reverse Isoperimetric Inequality.

Many Horizons?

We kept the slow acceleration to avoid horizons with different temperatures, but what if we have multiple horizons? In vacuo, we can have multiple black holes along an axis separated by strings or struts: Bach-Weyl or Israel-Khan solutions.

dM = X [TIdSI − λIdµI]

Tension

.. does not appear in Smarr formula but seems to have thermodynamic role….

M = 2TS − 2PV (+2ΩJ + ΦQ)