Nearing Extremal Intersecting Giants and New Decoupled Sectors in N = 4 SYM M.M. Sheikh-Jabbari
IPM, Tehran-Iran Based on [arXiv:0801.4457], JHEP0808:070. in collaboration with
- C. N. Gowdigere, R. Fareghbal and A.E. Mosaffa.
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Nearing Extremal Intersecting Giants and New Decoupled Sectors in N - - PowerPoint PPT Presentation
Nearing Extremal Intersecting Giants and New Decoupled Sectors in N = 4 SYM M.M. Sheikh-Jabbari IPM, Tehran-Iran Based on [arXiv:0801.4457], JHEP0808:070. in collaboration with C. N. Gowdigere, R. Fareghbal and A.E. Mosaffa. p. 1/119
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J
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10 =
5 +
5
5 = −
3
5 = 3
i + µ2 i [dφi + ai dt]2
5d BH = (H1H2H3)1/3ds2 5
3
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5 is the metric for a deformed S5 and
1
2
3
3
i
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N
N is the five-dimensional Newton constant and is
N = G(10) N
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N
N
sgs).
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N factor is equal to the sum
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1 3r2,
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1 = 1 case
i − ǫ1/2ˆ
i ≤ π/2, i = 2, 3
i , xi, L fixed.
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S
BTZ + dΩ2 3
S
C4
BTZ = −(ρ2 − γ2)dτ 2 +
1
S =
for
S =
1
for
1
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C4 have different
C4 =
i + x2 i dψ2 i )
1 = 1 case
C4 =
i + (µ0 i )2dψ2 i )
2 = sin θ0 1 cos θ0 2, µ0 3 = sin θ0 1 sin θ0 2,
1 cos θ0 2dˆ
1 sin θ0 2dˆ
2 sin θ0 1dˆ
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BTZ = −(ρ2 − γ2)dτ 2 +
1
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1
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10 = µ1 (R2 AdS3 ds2 3 + R2 S dΩ2 3 ) + 1
M4
3 = −(ρ2 − ρ2 0)dτ 2 +
3 is the metric for a three-sphere of unit radius and
M4 = L2
S
2 + µ2 2 dψ2 2) + q3 (dµ2 3 + µ2 3 dψ2 3)
S ≡ √q2q3 =
AdS3 = R2 S
0 = M
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3 ∼ R2 AdS3ǫ2ρ2(−dt2 + dφ2 1) ,
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2 dψ2 + ˜
3 dψ3
2 = q2 2(1 + q3
3 = q2 3(1 + q2
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S
rot.BTZ + dΩ2 3
S
C4
S = ˆ
rot.BTZ is the metric for a
c
1 = 1 and µ1 ≃ 1 cases. As
1, R4 S = ˆ
1)2.
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+ − a2 −
+ + a2 −)r2 + (a2 + − a2 −)2.
+ = −M + J
− = −M − J
+ ≤ a2 −, i.e. J ≥ 0 and J ∈ Z.
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+, a2 − > 0,
− is a rational number
− ≤ 1/4. In terms of M, J that is
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+ < 0, a2 − > 0, corresponding to −J < M < J. The
+, a2 − ≤ 0,
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+ = a2 − > 0, keep half of the maximal
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AdS ds2
S dΩ2 3
4
S = q2q3, R2 AdS = R2 S/f0
BTZ
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N
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µν)2 + e−2Φ(F 2 µν)2.
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s |Near BPS = πǫL2 · T3 =
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N = G(10) N
2µ0 3 ǫ2
1
N
sl8 s,
s
N = π2
2µ0 3
1, we have
1 so that,
S = √ˆ
1 = 1, and µ0 1 = 1 cases .
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N
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10 =µ1 g(6) µν dxµdxν+ 1
M4
M4 = L2
S
2 + µ2 2dψ2 2) + e−Φ(dµ2 3 + µ2 3dψ2 3)
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2 ∧ dχ2 ∧ dxµ ∧ dxν ∧ dxρ
3 ∧ dχ3 ∧ dxµ ∧ dxν ∧ dxρ,
N = G(10) π2 2 L4 = π2L4
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s
N
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h
N
h
h(H1H2H3)1/2|r=rh .
N
sl8 s,
N = G(10) N
s ,
h
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h = µ − µc
BH
BH
h =
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s
s ∼ ǫ → 0 limit together with scaling
s fixed.
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h = µ − µc
BH
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2 .
s ∼ N −1 ∼ ǫ → 0.
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BH = A(3)
N
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N =
N
S
S
2µ0 3
2 and µ0 3 is hence
BH = 8π ˆ
2µ0 3,
BH over values of µ0 2, µ0 3, yielding
BH = π ˆ
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N =
N
S
S
BH = πRAdSR3 S
S/L2.
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s to infinity.
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i Ji
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Y MN ∼ N → ∞
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i Ji = N2N3 4 ˆ µ ˆ µc.
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i=2,3 Ji ∼ J1 ∼ N 2ǫ2 ∼ N → ∞.
i=2,3 Ji it is more
3
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∂ψi = −i ∂ ∂φi = Ji and
i
i = L2 R2
S qi.
i /qi.
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s M)
S
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s M, indicating that it can be
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S) = R3 S
S.
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N
N =
S
2µ0 3
2+µ2 3≤1
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24a2 +, ¯
24a2 −.
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i=2,3 Ji, J1, we
i=1,2,3 Ji ≥ 0
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i=1 Ji = 0 and N ˆ µc L2 respectively
3
3
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i=1 Ji ≥ c 12;
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N
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0 for the AdS3 × S3
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(Qe,Qm) = T (6) s
e gs + N2 mg−1 s
s gs.
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