Thermodynamics of the BMN matrix model at strong coupling
Miguel S. Costa Faculdade de Ciências da Universidade do Porto
Crete Center for Theoretical Physics - Heraklion, May 2014
Works with L. Greenspan, J. Penedones and J. Santos
Thermodynamics of the BMN matrix model at strong coupling Miguel S. - - PowerPoint PPT Presentation
Thermodynamics of the BMN matrix model at strong coupling Miguel S. Costa Faculdade de Cincias da Universidade do Porto Works with L. Greenspan, J. Penedones and J. Santos Crete Center for Theoretical Physics - Heraklion, May 2014 Motivation
Crete Center for Theoretical Physics - Heraklion, May 2014
Works with L. Greenspan, J. Penedones and J. Santos
Motivation
Test and understand the gauge/gravity duality with observables that are not protected by SUSY or can be computed using integrability. How does gravitation phenomena, like black holes, emerge from gauge theory side?
Idea: Study thermodynamics of black holes dual to Matrix Quantum Mechanics that can be simulated on a computer using Monte-Carlo methods.
Dual CFT is renormalizable and unitary. Problem: how to decode the hologram? Unfortunately field theory very difficult in region of interest for quantum gravity (strong coupled; classical gravity , expansion loop expansion).
≡
N → ∞ 1/N
The case of D0-branes
D0-brane Open string Closed string
Closed string D0-brane geometry
[Itzhaki, Maldacena, Sonnenschein, Yankielowicz ´98]
D0-branes: field theoretic description (matrix quantum mechanics)
SD0 = N 2λ Z dt Tr (DtXi)2 + ΨαDtΨα + 1 2 ⇥ Xi, Xj⇤2 + iΨαγj
αβ[Ψβ, Xj]
Ψ ≡ SU(N) fermionic matrices (16 real components) Dt = ∂t − i[A, ] ≡ covariant derivative γi ≡ SO(9) gamma matrices
= 2δij
SO(9) global symmetry
λ = g2
Y MN =
gsN (2π)2l3
s
≡ mass3
λeff = λ E3
E → ∞ (UV ) ≡ weak coupling E → 0 (IR) ≡ strong coupling
16πGNl−8
s
= (2π)11 (λl3
s)2
N 2
Dimensionless temperature τ = T λ1/3
Low temperatures is strong coupling SD0 = N 2λ Z β dt Tr · · ·
particular strong coupled region.
Dimensionless mean energy ✏ N 2 = E N 21/3
[Catterall, Wiseman ´07,´08,´09; Anagnostopoulos et al ´07; Hanada et al ´08,´13]
D0-branes: gravitational description
ds2 = dr2 f(r) + r2dΩ2
8 +
✓R r ◆7 dz2 + f(r)dt ✓ 2dz − ⇣r0 R ⌘7 dt ◆
f(r) = 1 − ⇣r0 r ⌘7 , ✓ R `s ◆7 = 60⇡3gsN , ✓r0 `s ◆5 = 120⇡2 49 (2⇡gsN)
5 3 ⌧ 2
N − 10
21
N − 5
9
N − 5
6
IIA 11D SUGRA G-L instability Horizon at scale
lP
Horizon at scale
ls
l2
sR(r0) ⌧ 1 ) τ ⌧ 1
gseφ(r0) ⌧ 1 ) τ N − 10
21
τ = T/λ1/3
✏ N 2 = c1⌧
14 5
S = AH 4GN = d1N 2τ
9 5
d1 = 4
13 5 15 2 5
⇣π 7 ⌘ 14
5
c1 = 9 14 d1 (because dE = TdS)
in expansion
fixes both coefficient and power of correction
1/N 2
α0
1 16πGN Z d10x√−g e2φ⇣ R + · · · + α03R4 + . . . ⌘ ⇒ S N 2 = d1τ
9 5
⇣ 1 + d2τ
9 5
⌘ [Hanada et al´08]
lP
1 16πG11 Z d11x√−g ⇣ R + l6
P R4⌘
⇒ S N 2 = d1τ
9 5 + 1
N 2 d3τ − 3
5
[Hanada et al´13]
checked predicted negligible
✏ N 2 = h c1⌧
14 5 + c2⌧ 23 5 + . . .
i + 1 N 2 h c3⌧
2 5 + . . .
i + . . .
[Hanada, Hyakutake, Nishimura, Takeuchi ´08]
0.0 0.5 1.0 1.5 2.0 2.5 3.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 E/N2 T N=17, !=6 N=17, !=8 7.41T2.8 7.41T2.8-5.58T4.6
SUGRA SUGRA + corrections
Monte-Carlo simulation
✏/N 2
α0
✏ N 2 = h c1⌧
14 5 + c2⌧ 23 5 + . . .
i + 1 N 2 h c3⌧
2 5 + . . .
i + . . .
checked negligible [Hanada, Hyakutake, Ishiki, Nishimura ´13] [Hyakutake ´13]
Monte-Carlo simulation
✏/N 2
Today’s talk is not about D0-brane matrix model
suppressed and black hole is stable (positive specific heat).
Mass deformation resolves IR divergence - canonical ensemble well defined. Much richer thermodynamics with a 1st order phase transition (at large N there are two dimensionless parameters). [Berenstein, Maldacena, Nastase ´02]
in D0-brane moduli space. This is suppressed at large N (metastable state), but it is a source of tension in Monte Carlo simulations
F(T, r) N 2 ∼ Ffinite(T) + 9 N ln r
[Catterall, Wiseman ´09]
BMN matrix model
Canonical ensemble is well defined. Still “easy” to simulate on a computer. In large N ‘t Hooft limit dimensionless coupling constant
λ = g2
YMN
µ3
S = SD0 − N 2 Z dt Tr µ2 32 (Xi)2 + µ2 62 (Xa)2 + µ 4 Ψα 123
αβ Ψβ + i2µ
3 ✏ijkXiXjXk
SO(9) → SO(6) × SO(3)
a = 4, . . . , 9 i = 1, 2, 3
Many vacua
Xa = 0
Xi = µ 3 Ji
[Ji, Jj] = iijkJk
N = mn
M5−brane vacua ≡ m → ∞, n fixed Xi ∼ n × n . . . n × n . . . . . . . . . . . . . . . . . . n × n m times D2−brane vacua ≡ n → ∞, m fixed (decoupled) Focus on trivial vacuum (single M5-brane) that is SO(9) invariant
Xi = Xa = 0
Exponential growth of spectrum with energy Hagedorn transition
Tc µ = 1 12 log 3 1 + 265 34 λ − c λ2 + O(λ3)
First-order phase transition at [Hadizadeh, Ramadanovic, Semenoff, Young ’04]
N → ∞
Dimensionless temperature ≡ T µ Dimensionless coupling ≡ λ = g2
Y MN
µ3
T µ
TH µ
λ = g2
YMN
µ3
Confined phase Deconfined phase
F = O(N 2)
F = O(N 0) ?
W E A K C O U P L I N G
S T R O N G C O U P L I N G
Start here
Today: strongly coupled limit Dual geometry is SO(9) invariant non-extremal D0-brane with deformation turned on
µ → 0 , T µ fixed and large
Gravitational dual
approximately the non-extremal D0-brane solution
ds2 = dr2 f(r) + r2dΩ2
8 + R7
r7 dz2 + f(r)dt ✓ 2dz − r7 R7 dt ◆
dC = µ dt ∧ dx1 ∧ dx2 ∧ dx3
Need back-reaction to decrease temperature and study phase transition at strong coupling. In particular,
SO(9) → SO(6) × SO(3)
Non-normalizable mode responsible for massive deformation
geometries and asymptote to the M-theory plane wave solution
ds2 = dxidxi + dxadxa + 2dtdz − ✓µ2 32 xixi + µ2 62 xaxa ◆ dt2
dC = µ dt ∧ dx1 ∧ dx2 ∧ dx3
ds2 = −A (1 − y7) y7 dη2 + T4 y7 dζ + Ω (1 − y7)dη y7 2 + 1 y2 " B (dy + Fdx)2 (1 − y7)y2 + T1 4dx2 2 − x2 + T2 x2(2 − x2)dΩ2
2 + T3 (1 − x2)2dΩ2 5
# C = (M dη + L dζ) ∧ d2Ω2 Tailored to numerical implementation (domain of unknown is the unit square; everything dimensionless)
y
1
Horizon
∞
S2
S5
collapses collapses
1
| {z }
dΩ2
8
if T1=T2=T3=1 is a angular coordinate on compact 8-dimensional space with topology
x
S8
x = 1 x = 0 S2 S5 pole equator are functions of and
A, B, F, T1, T2, T3, T4, Ω, M, L
x
y y is a radial coordinate from boundary ( ) to horizon ( )
y = 0 y = 1 y = 0 y = 1 S8 × S1 × S1 S8 × S1 boundary horizon
ζ ∼ ζ + 2π M-theory circle
| {z }
dΩ2
8
if T1=T2=T3=1 ds2 = −A (1 − y7) y7 dη2 + T4 y7 dζ + Ω (1 − y7)dη y7 2 + 1 y2 " B (dy + Fdx)2 (1 − y7)y2 + T1 4dx2 2 − x2 + T2 x2(2 − x2)dΩ2
2 + T3 (1 − x2)2dΩ2 5
# C = (M dη + L dζ) ∧ d2Ω2 x = 1 x = 0 S2 S5 pole equator y = 0 y = 1 S8 × S1 × S1 S8 × S1 boundary horizon
Non-extremal D0-brane solution corresponds to
A = B = T1 = T2 = T3 = T4 = Ω = 1 , F = M = L = 0 , β = 4π 7 (Euclidean time circle)
This scaling symmetry will be important later... and need to use scaling symmetry of 11D SUGRA action with ζ ∼ ζ + 2π → ζ ∼ ζ + 2πs0 ⇒ I → s0I gµν → s2gµν , Cµνρ → s3Cµνρ ⇒ I → s9I s0 = ✓ R r0 ◆ 7
2 gs`s
r0 s = r0
invariant tensor hamonic SO(6) × SO(3)
y
1
Horizon
∞
S2
S5
collapses collapses
1
x
y = 0 At infinity ( ): A, B, T1, T2, T3, T4, Ω → 1 ,
F → 0
M → ˆ µ x3(2 − x2)
3 2
y3 , L → 3 2 ˆ µ y4x3(2 − x2)
3 2
Recall that C = (M dη + L dζ) ∧ d2Ω2
ˆ µ = 7 12π µ T
To obtain physical solution do again above scalings, then geometry has same asymptotics of non-extremal D0-brane with temperature T and mass deformation turned on. The only parameter is This important to thermodynamics, because we just learned that I = s9s0 16πGN ˆ I ⇣ µ T ⌘ = 15 28 ✓ 15 142π8 ◆ 2
5
N 2 ✓ T λ
1 3
◆ 9
5 ˆ
I ⇣ µ T ⌘ S = s9s0 4GN ˆ S ⇣ µ T ⌘ = 15π 7 ✓ 15 142π8 ◆ 2
5
N 2 ✓ T λ
1 3
◆ 9
5 ˆ
S ⇣ µ T ⌘ Regularity at the axis of symmetry: horizon ( ), pole ( ) and equator ( ). y = 1 x = 0 x = 1
S2
S5
(β = 4π/7)
Einstein-DeTurck equations
ξµ = gαβ ⇣ Γµ
αβ − ˜
Γµ
αβ
⌘
DeTurck term that makes Einstein equations elliptic Connection of reference metric
Rµν r(µξν) = 1 12 ✓ FµαβγF αβγ
µ
1 12gµνF 2 ◆
d(F) + 1 2F ∧ F = 0
F = dC
Our reference metric is the non-extremal D0-brane solution. With appropriate boundary conditions on the numerical solution so solution also solves Einstein equations. ξµ = 0 [Headrick, Kitchen, Wiseman ’09]
A Smarr formula (good to check numerics)
d ⇣ ? Kv ⌘ = 0
y1 < y < y2
0 = Z
Σ12
d ⇣ ? Kv ⌘ = Z
∂Σ12
?Kv = Z
H
?Kv − Z
y→0
?Kv
v = ∂ ∂η
area to boundary data
7 2 ˆ S = Z
y→0
?Kv
is a conserved antisymmetric tensor, i.e.
d ⇣ ? Kv ⌘ = 0
(Kv)µν = rµvν + 1 3F µναβvγCαβγ + 1 6v[µF ν]αβγCαβγ
The solution
Q1 = 1 + y4q1 , Q2 = 1 + y4q2 , Q3 = y5q3 , Q4 = 1 + y4q4 , Q5 = 1 + y4q5 , Q6 = 1 + y4q6 , Q7 = 1 + y4q7 , Q8 = 1 + y4q8 , Q9 = ˆ µ(1 + y + y2 + y3 + y4) + y4q9 , Q10 = ˆ µ + yq10
0.0 0.5 1.0 1.5 2.0 120 140 160 180 m ` S `
Horizon area
0.0 0.5 1.0 1.5 2.0 0.4 0.5 0.6 0.7 0.8 0.9 1.0 m ` R2êR5
Ratio of maximal radius of to
S2 S5
S = 15π 7 ✓ 15 142π8 ◆ 2
5
N 2 ✓ T λ
1 3
◆ 9
5 ˆ
S ⇣ µ T ⌘ After scaling symmetry to obtain physical metric: Ri = ai ✓ T λ
1 3
◆ 2
5 ˆ
Ri ⇣ µ T ⌘ Reproduces scalings predicted from strongly coupled low energy moduli estimate in [Wiseman ’13]
∆N =
AreaN AreaN+1
25 30 35 40 45 50 1034 1030 1026 1022 1018
log χ ≈ −17.5 − 0.75N
χ = max p ξνξν
25 30 35 40 45 50 1024 1022 1020 1018 1016
Numerical convergence
Discretize PDEs using a Chebyshev grid with N x N points. Derivatives are estimated using polynomial approximations that involve all points in the grid - spectral methods.
requires regularization easy to measure
Black hole thermodynamics
F(T, µ) = −c0 T
14 5 ˆ
I ⇣ µ T ⌘ , S(T, µ) = c0 14 5 T
9 5 ˆ
S ⇣ µ T ⌘
Therefore ratio of free energies and entropies
F(T, µ) F(T, 0) = ˆ I µ
T
I(0) ≡ f ⇣ µ T ⌘ , S(T, µ) S(T, 0) = ˆ S µ
T
S(0) ≡ s ⇣ µ T ⌘
✓∂F ∂T ◆
µ
= −S
F(T, 0) = −c1T
14 5
First law Free energy
✓ 1 − 5 14 ˆ µ ∂ ∂ˆ µ ◆ f(ˆ µ) = s(ˆ µ)
Analyticity
s(ˆ µ) =
∞
X
n=0
sn ˆ µn , f(ˆ µ) =
∞
X
n=0
14sn 14 − 5n ˆ µn
Critical temperature
T < Tc
[Lin, Maldacena ’05]
geometry without horizon is favoured F ∼ O
Tc µ = 7 12πˆ µc ≈ 0.091
Similar to Hawking-Page phase transition in AdS
0.0 0.5 1.0 1.5 2.0 0.0 0.2 0.4 0.6 0.8 1.0 m ` f
= −c1T
14 5 f(ˆ
µ) F(T, µ) = F(T, 0)f(ˆ µ)
c = T ✓∂S ∂T ◆
µ ⇒
c S = 9 5 − ˆ µ ∂ ∂ˆ µ log s(ˆ µ) > 0
Phase diagram at large N
T µ
λ = g2
YMN
µ3
Confined phase Deconfined phase
F = O(N 2)
F = O(N 0) ?
W E A K C O U P L I N G
S T R O N G C O U P L I N G
0.076
0.091
Very similar to SYM on a 3-sphere (µ ≡ 1/R) [Aharony, Marsano, Minwalla, Papadodimas, van Raamsdonk ’03]
To preserve depends on ratio of radii
E6
E3
Xi
Xa
θ
sin θ = R5 R2 = ✓XaXa XiXi ◆1/2 SO(6) × SO(3)
Boundary data (preliminary)
near the boundary
(y = 0)
Qi(x, y)
Qi(x, y) = X
j
yj ˜ Qj
i(x)
SO(9)
Thus we can classify the invariant perturbations according to
SO(9)
S8
SO(6) × SO(3)
˜ Qj
i(x)
µ
Ê Ê Ê ‡ ‡ ‡ ‡ ‡
1 3 5 7 9 2 4 6 1 3 5 7 9 2 4 6
ρ2
ρ4
ρ6
spin l
n(l)
vector mode
v(x, y) = X
l
⇣ ρl yn(l) + µl y˜
n(l)⌘
vl(x) + back reaction
α5 α6
β1 β2 β3
zero modes
δ7, γ7
Ê Ê Ê Ê Ê ‡ ‡ ‡ ‡ ‡
2 4 6 8 10 2 4 6 2 4 6 8 10 2 4 6
α4
spin l
n(l)
scalar mode
s(x, y) = X
l
⇣ αl yn(l) + βl y˜
n(l)⌘
sl(x) + back reaction
normalizable non-normalizable
O ∼ Tr ([Xi, Xj]XA1 . . . XAl) , l ≥ 1 odd O ∼ Tr (XA1 . . . XAl) , l ≥ 2 even
0.0 0.5 1.0 1.5 2.0
0.0 m ` a2 0.0 0.5 1.0 1.5 2.0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 m ` a5 0.0 0.5 1.0 1.5 2.0
0.00 m ` g7 0.0 0.5 1.0 1.5 2.0
0.0 m ` d7
ρ2
Numerics pass highly non-trival check! Smarr formula OK: 7 2 ˆ S = 16π5 735 (98 + 280γ7 + 63δ7 − 276α2ˆ µ)
(caveat: we really only determined upper limit on critical temperature)
What are the minimal ingredients of a quantum mechanical system such that it gives rise to classical gravity in the limit of many degrees of freedom? Future work
THANK YOU