PROBE BRANES ON FLAVORED ABJM BACKGROUND
Javier Mas
Universidad de Santiago de Compostela
Heraklion
June 2013
Niko Jokela, J. M., Alfonso V. Ramallo & Dimitrios Zoakos arXiv: 1211.0630 based on Eduardo Conde & Alfonso V. Ramallo 1105.6045
PROBE BRANES ON FLAVORED ABJM BACKGROUND Javier Mas Universidad de - - PowerPoint PPT Presentation
PROBE BRANES ON FLAVORED ABJM BACKGROUND Javier Mas Universidad de Santiago de Compostela Heraklion June 2013 Niko Jokela, J. M., Alfonso V. Ramallo & Dimitrios Zoakos arXiv: 1211.0630 based on Eduardo Conde & Alfonso V. Ramallo
Heraklion
June 2013
Niko Jokela, J. M., Alfonso V. Ramallo & Dimitrios Zoakos arXiv: 1211.0630 based on Eduardo Conde & Alfonso V. Ramallo 1105.6045
Chern-Simons-matter theories in 2+1 dimensions
bifundamentals (N, ¯ N)
S = k CS[A] − k CS[ ˆ A] − k Dµ CI † Dµ CI − Vpot(C)
Aharony, Bergman, Jafferis & Maldacena 0806.1218
N → rank of the gauge groups
k → CS level (1/k ∼ gauge coupling) ’t Hooft coupling λ ∼ N
k
amplitude relation, ...)
Fujita, Li, Ryu & Takayanagi, 0901.0924
Gaiotto&Jafferis 0903.2175
Drukker, Mariño & Putrov 1003.3837
CP3 = C4/(zi ∼ λzi)
Effective description for N
1 5 << k << N
J → Kahler form of CP3
1 2π
F2 = 2k J F4 = 3π √ 2
1
2 ΩAdS4
eφ = 2L k = 2√π 2N k5 1
4
AdS + L2ds2 CP3
Introduce quarks in the (N, 1) and (1, N) representation
Hohenegger&Kirsch 0903.1730
Gaiotto&Jafferis 0903.2175
Q1 → (N, 1)
Q2 → (1, N)
˜ Q1 → ( ¯ N, 1)
˜ Q2 → (1, ¯ N)
Q†
1 e−V Q1 + Q† 2 e− ˆ V Q2 + antiquarks
V, ˆ V vector supermultiplets for A , ˆ A
˜ Q1 AiBi Q1 , ˜ Q2 Bi Ai Q2
CI = (A1, A2, B†
1, B† 2)
plus quartic terms in Q, ˜ Q’s
ω1 = cos ˆ ψdˆ θ + sin ˆ ψ sin ˆ θd ˆ ϕ ω1 = sin ˆ ψdˆ θ − cos ˆ ψ sin ˆ θd ˆ ϕ ω3 = d ˆ ψ + cos ˆ θd ˆ ϕ where Ai = − ξ2 1 + ξ2 ωi
SU(2) instanton on S4
x1 = sin θ cos ϕ x2 = sin θ sin ϕ x3 = cos θ
Write CP3 as an S2-bundle over S4
AdS4
S4
S2
AdS + L2ds2 CP3
ds2 = L2ds2
AdS4 + L2
4 (1 + ⇠2)2 ✓ d⇠2 + ⇠2 4
+ ⇣ dxi + ✏ijkAjxk⌘2
D6-branes extended in AdS4 and wrapping RP3 ⊂ CP3
m0
θ(r)
θ(r) = π/2 ⇒ m0 = 0
AdS4
S4
S2
the idea is now to smear over positions and orientations
preserve N=1 SUSY
−flavor symmetry : U(1)Nf
Sflav = PNf
i=1
⇣ −TD6 R
M(i) d7ζ e−φ √−det ˆ
g7 + TD6 R
M(i) ˆ
C7 ⌘
m0
→ 1 κ2
10
✓ − Z d10xe3φ/4p − det g10|Ω| + Z d10x C7 ∧ Ω ◆
modified Bianchi identity
modified Bianchi identity
solution for massless flavors
modified Bianchi identity
go to vielbein basis
Ea = (E1, E2)
Si = (S1, S2, S3, S4)
along the base
along the fiber
solution for massless flavors
modified Bianchi identity
go to vielbein basis
Ea = (E1, E2)
Si = (S1, S2, S3, S4)
along the base
along the fiber
Sξ = 2 1 + ξ2 dξ S1 = ξ 1 + ξ2
S2 = ξ 1 + ξ2
S3 = ξ 1 + ξ2
E1 = dθ + ξS1 E2 = sin dϕ − ξS2
solution for massless flavors
F2 = k 2 h E1 ∧ E2 − (S4 ∧ S3 + S1 ∧ S2) i
ds2 = L2ds2
AdS + L2 4
X
i=1
(Si)2 +
2
X
a=1
(Ea)2 ! F4 = 3k 2 L2 ΩAdS4
F2 = k 2 h E1 ∧ E2 − (S4 ∧ S3 + S1 ∧ S2) i
ds2 = L2ds2
AdS + L2 4
X
i=1
(Si)2 +
2
X
a=1
(Ea)2 ! F4 = 3k 2 L2 ΩAdS4
F2 = k 2 h E1 ∧ E2 − (S4 ∧ S3 + S1 ∧ S2) i
ds2 = L2ds2
AdS + L2 4
X
i=1
(Si)2 +
2
X
a=1
(Ea)2 ! F4 = 3k 2 L2 ΩAdS4
F2 = k 2 h E1 ∧ E2 − (S4 ∧ S3 + S1 ∧ S2) i
ds2 = L2ds2
AdS + L2 4
X
i=1
(Si)2 +
2
X
a=1
(Ea)2 ! F4 = 3k 2 L2 ΩAdS4
q = 3 + 9 4 Nf k − 2 s 1 + 3 4 Nf k + ✓3 4 ◆4 ✓Nf k ◆2
1 + 3
8 Nf k − ...
q → 5
3
( Nf
k → ∞)
b = 4 + 39
16 Nf k −
r 1 + 3
4 Nf k +
⇣
9 16 Nf k
⌘2 3 + 3
2 Nf k
1 + 3
16 Nf k − ...
q → 5
4
( Nf
k → ∞)
F2 = k 2 h E1 ∧ E2 − (S4 ∧ S3 + S1 ∧ S2) i
ds2 = L2ds2
AdS + L2 4
X
i=1
(Si)2 +
2
X
a=1
(Ea)2 ! F4 = 3k 2 L2 ΩAdS4
where is related to the quark-antiquark potential screening
q = 3 + 9 4 Nf k − 2 s 1 + 3 4 Nf k + ✓3 4 ◆4 ✓Nf k ◆2
1 + 3
8 Nf k − ...
q → 5
3
( Nf
k → ∞)
b = 4 + 39
16 Nf k −
r 1 + 3
4 Nf k +
⇣
9 16 Nf k
⌘2 3 + 3
2 Nf k
1 + 3
16 Nf k − ...
q → 5
4
( Nf
k → ∞)
q Σ b 5 10 15 20 25 30 Ε
4
Nf k
q = −Q
potential screening
8 < : 1 − 3
8 Nf k − ...
→ q
k Nf
( Nf
k → ∞)
q Σ b 5 10 15 20 25 30 Ε
4
Nf k
q = −Q
potential screening dilaton shifts
eφ = 4√π ✓2N k5 ◆1/4 (2 − q)5/4 (η + q)[q(q + ηq − η)]1/4
8 < : 1 − 3
8 Nf k − ...
→ q
k Nf
( Nf
k → ∞)
q Σ b 5 10 15 20 25 30 Ε
4
Nf k
q = −Q
potential screening dilaton shifts
eφ = 4√π ✓2N k5 ◆1/4 (2 − q)5/4 (η + q)[q(q + ηq − η)]1/4
regime of validity
8 < : 1 − 3
8 Nf k − ...
→ q
k Nf
( Nf
k → ∞)
D6-branes extended in AdS4 and wrapping RP3 ⊂ CP3
new cartesian-like coordinates
AdS4
S4
S2
profile
θ(r)
S = SDBI + SW Z = TD6 ✓ − Z d7ζe−φp − det ˆ g7 + Z d7ζ ˆ C7 ◆
3 2b 1(
In our case O ∼ ¯ ψψ
2 − 3 16 Nf k + 63 512 ✓Nf k ◆2 + ... , → 7 4 ✓Nf k → ∞ ◆
In our case O ∼ ¯ ψψ
2 − 3 16 Nf k + 63 512 ✓Nf k ◆2 + ... , → 7 4 ✓Nf k → ∞ ◆
SUSY solution
Depends on the gauge for C7!!
C7 → C7 + dΛ6 generates boundary conterterms C7 = e−φK → SUSY scheme
is automatically finite
replace AdS by Schwarzschild-AdS
entropy
ds2 = L2 ✓ −r2h(r)dt2 + dr2 r2h(r) + r2(dx2
1 + dx2 2)
◆ + L2 1 b2 q
4
X
i=1
(Si)2 +
2
X
a=1
(Ea)2 !
h
blackening factor
10
replace AdS by Schwarzschild-AdS
entropy
ds2 = L2 ✓ −r2h(r)dt2 + dr2 r2h(r) + r2(dx2
1 + dx2 2)
◆ + L2 1 b2 q
4
X
i=1
(Si)2 +
2
X
a=1
(Ea)2 !
h
blackening factor
10
free energy
9 √ 2 ✓4π 3 ◆2 √ k N 3/2 ξ ✓Nf k ◆ T 3
EADM = − 1 κ2
10
p |Gtt| Z
t0,r∞
p det G8 (KT − K0)
energy density
unflavored term ∼ N
3 2
field theory match by Drukker et al. (1007.3837) !
ξ Nf k
16 q
5 2 (η + q)4
(2 − q)
1 2 (q + ηq − η) 7 2
8 > > > < > > > : = 1 + 3 4 Nf k − 9 64 ✓Nf k ◆2 + . . . Nf → 0 ∼ 1.389 r Nf k Nf → ∞
Ξ q Σ b 5 10 15 20 25 30 Ε
4 6 8
Nf k
ξ ✓Nf k ◆ = 1 16 q5/2(η + q)4 √2 − q(q + ηq − η)7/2
Comparison with 3-Sasakian
(U(Nf), N = 3 flavors)
Gaiotto&Jafferis 0903.2175
Couso-Santamaria et al. 1011.6281
Localized solution in 11d for coincident massless flavors
AdS4 × M7 with M7 a hyperkahler 3-Sasakian manifold N = 3 with U(Nf) flavor symmetry
ξ3S ✓Nf k ◆ = 1 + Nf
k
q 1 + Nf
2k
10 20 30 40 Nf k 2 4 6 8 Ξ Nf k
TriSasakian Smeared
ξ ✓Nf k ◆ = 1 16 q5/2(η + q)4 √2 − q(q + ηq − η)7/2
Comparison with 3-Sasakian
(U(Nf), N = 3 flavors)
Gaiotto&Jafferis 0903.2175
Couso-Santamaria et al. 1011.6281
Localized solution in 11d for coincident massless flavors
AdS4 × M7 with M7 a hyperkahler 3-Sasakian manifold N = 3 with U(Nf) flavor symmetry
ξ3S ✓Nf k ◆ = 1 + Nf
k
q 1 + Nf
2k
the embeddings are governed by the DBI+WZ action
rmin
h
the embeddings are governed by the DBI+WZ action
rmin
h
the embeddings are governed by the DBI+WZ action
ζ ✓Nf k ◆ = 1 32 √2 − q(η + q)4q5/2 (q + ηq − η)9/2
→ 8 < : 1 − 3
8 Nf k + . . . (Nf → 0)
q
k Nf + . . .
(Nf → ∞)
rmin
h
must be improved to get a consistent thermodynamics
represent the improvement term as follows
the angular part of must vanish at the horizon
h
Jensen 1006.3066
define (a zero point energy)
Z drL2(r) = ∆0
the total action is now
∂L ∂ ˙ θ
= 0
and satisfies that the canonical momentum vanishes at the horizon in order to fix let us compare the free energy (density) of the probe and the background
∆0
F = − S
S = N Z d3x " − 4b r3
h
Z dr r2 sin θ r 1 + r2h(r) b2 ˙ θ2 − sin θ − rh(r) b cos θ ˙ θ ! + ∆0 #
N = 2 √ 2π2 27 N 3/2√ k T 3 ζ ✓Nf k ◆
a) infinite mass limit ⇒ decoupling Consistency check:
lim
m→∞ F = N (1 − ∆0) = 0
a) infinite mass limit ⇒ decoupling b) zero mas limit ⇒ add to backreaction
Fback ✓Nf k ◆ + Fprobe(m = 0) = Fback ✓Nf + 1 k ◆ = Fback ✓Nf k ◆ + 1 k F 0
back + ...
hence Consistency check:
lim
m→∞ F = N (1 − ∆0) = 0
a) infinite mass limit ⇒ decoupling b) zero mas limit ⇒ add to backreaction
Fback ✓Nf k ◆ + Fprobe(m = 0) = Fback ✓Nf + 1 k ◆ = Fback ✓Nf k ◆ + 1 k F 0
back + ...
hence Consistency check:
lim
m→∞ F = N (1 − ∆0) = 0
1 √ 2 ✓4π 9 ◆2 N 3/2 √ k T 3 3 4ζ ✓Nf k ◆ ∆0 = 1 √ 2 ✓4π 9 ◆2 N 3/2√ k T 3 1 k ξ0 ✓Nf k ◆
a) infinite mass limit ⇒ decoupling b) zero mas limit ⇒ add to backreaction
Fback ✓Nf k ◆ + Fprobe(m = 0) = Fback ✓Nf + 1 k ◆ = Fback ✓Nf k ◆ + 1 k F 0
back + ...
hence Consistency check:
lim
m→∞ F = N (1 − ∆0) = 0
1 √ 2 ✓4π 9 ◆2 N 3/2 √ k T 3 3 4ζ ✓Nf k ◆ ∆0 = 1 √ 2 ✓4π 9 ◆2 N 3/2√ k T 3 1 k ξ0 ✓Nf k ◆
a) infinite mass limit ⇒ decoupling b) zero mas limit ⇒ add to backreaction
Fback ✓Nf k ◆ + Fprobe(m = 0) = Fback ✓Nf + 1 k ◆ = Fback ✓Nf k ◆ + 1 k F 0
back + ...
hence Consistency check:
lim
m→∞ F = N (1 − ∆0) = 0
1 √ 2 ✓4π 9 ◆2 N 3/2 √ k T 3 3 4ζ ✓Nf k ◆ ∆0 = 1 √ 2 ✓4π 9 ◆2 N 3/2√ k T 3 1 k ξ0 ✓Nf k ◆
ξ ✓Nf k ◆ = 1 16 q5/2(η + q)4 √2 − q(q + ηq − η)7/2
ζ ✓Nf k ◆ = 1 64 √2 − q(η + q)4q5/2 (q + ηq − η)9/2
the same mechanism for the entropy yields stotal = sback + s ≈ 1 3 4π 3 2 N 2 √ 2λ ξ Nf + 1 k
(m → 0) hence, massless probe entropy ≈ increase in area of the horizon Now that the action is completely fixed we may derive the correct equations of motion and the solutions, as well as the thermodynamics
S = N Z d3x " − 4b r3
h
Z dr r2 sin θ r 1 + r2h(r) b2 ˙ θ2 − sin θ − rh(r) b cos θ ˙ θ ! + 1 #
1 2 3 4 Ρ 0.0 0.5 1.0 1.5 2.0 2.5 R
Black hole embeddings
dr2 r2h(r) + dθ2 b2 = 1 u2b2
u = 2 4 ✓ r rh ◆ 3
2
+ ✓ r rh ◆3 − 1 ! 1
2 3
5
2b 3
Minkowski embeddings R(ρ) ∼ R0 + ... (ρ → 0) m + c ρ3/b−2 + ... (ρ → ∞) χ(u) = cos θ(u) ∼ χh + ... (u → uh) m u + c u3/b−1 + ... (u → ∞)
χh
R0
2 3 π
0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 m 1.5 1.0 0.5 0.0 c
Numerical c = c(m)
Mq = 21/3π 3 √ 2λ σ T m1/b
0, 0.2 , 0.4
0.5 1.0 1.5 2.0 m 0.05 0.10 0.15 ΨΨ
0.89 0.90 0.91 0.92 0.93 m Mq
1 2
Λ T 0.01 0.02 0.03 0.04 0.05 0.06 0.07 ΨΨ
√ λ Tc Mq ∼ 1 m1/b
c
σ ∼ p Nf
5 10 15 20 25 Ε
0.5 1.0 1.5 2.0 Λ Tc mq
√ λ Tc Mq
1.8 2.0 2.2 2.4 2.6 2.8 m 1.5 1.0 0.5 c
R0
R0
Mateos, Myers & Thomson hep-th/0701132
1.8 2.0 2.2 2.4 2.6 2.8 m 1.5 1.0 0.5 c
m∗
R0
R0
Mateos, Myers & Thomson hep-th/0701132
2.5 2.0 1.5 1.0 0.5 0.5 1.0 7 4 Π logR01 2 1 1 2 cc R0 132 2.5 2.0 1.5 1.0 0.5 0.5 1.0 7 4 Π logR01 2 1 1 2 mm R0 132
1.5 1.0 0.5 7 8 Π log1Χh 3 2 1 1 2 3 cc 1 Χh34
1.5 1.0 0.5 7 8 Π log1Χh 3 2 1 1 2 3 mm 1 Χh34
0.5 1.0 1.5 2.0 2.5 3.0 1 m 1.0 0.8 0.6 0.4 0.2 0.0 F
= N ⇣ G(m) − 1 ⌘
F = SE R d3x = N Z d3x " 4b r3
h
Z ∞
rmin(m)
dr r2 sin θ r 1 + r2h(r) b2 ˙ θ2 − sin θ − rh(r) b cos θ ˙ θ ! − 1 # = 2 √ 2π2 27 N √ λ T 3 ! ζ ✓Nf k ◆ ⇣ G(m) − 1 ⌘ Nf k = 0 Nf k = 10
0.3 0.4 0.5 0.6 0.7 0.8 1 m 0.2 0.1 0.0 0.1 0.2 F
0.0 0.5 1.0 1.5 2.0 2.5 3.0 1 m 0.0 0.5 1.0 1.5 2.0 2.5 3.0 T s
s = −∂F ∂T = − F N ∂N ∂T − N ∂ ∂T ✓ F N ◆ = − 3 T F N − m T (3 − 2b)c
0.0 0.5 1.0 1.5 2.0 2.5 3.0 1 m 0.0 0.5 1.0 1.5 2.0 2.5 3.0 E
E = −2F − N(3 − 2b) c m
0.0 0.5 1.0 1.5 2.0 2.5 3.0 1 m 0.0 0.5 1.0 1.5 2.0 2.5 3.0 E
E = −2F − N(3 − 2b) c m
0.0 0.2 0.4 0.6 0.8 Λ T mq 0.25 0.20 0.15 0.10 0.05 ∆vs2 Λ N
v2
s = ∂P
∂E = s cv = 1 2 + δv2
s
Nf