discretized minimal surface and gluon scattering ... · introduction ads/cft correspondence and...

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IntroductionAdS/CFT correspondence and Gluon amplitudes

Numerial Solutions of minimal surface in AdSConclusions and Outlook

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Discretized Minimal Surface and Gluon ScatteringAmplitudes in N=4 SYM at Strong Coupling

Katsushi Ito

Tokyo Institute of Technology

June 29– July 2, 2009@Integrability in Gauge and String TheoryS. Dobashi, K.I. and K. Iwasaki, arXiv:0805.3594, JHEP 07 (2008)088

S. Dobashi and K.I., arXiv:0901.3046, Nucl. Phys. B819 (2009) 18

Katsushi Ito gluon amplitudes

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.Gluon Scattering Amplitudes in N = 4 SYM

Planar L-loop, n-point amplitude

A(L)n (k1, · · · , kn) = A(0)

n (k1, · · · , kn)M(L)n (ε)

the BDS conjecture Bern-Dixon-Smirnov, Anastasiou-Bern-Dixon-Kosower

lnMn(ε) =A2

ε2+

A1

ε

− 1

16f(λ)

nX

i=1

“

ln

„

µ2

−si,i+1

«

”2

− g(λ)

4

nX

i=1

ln

„

µ2

−si,i+1

«

+f(λ)

4F (BDS)

n (0) + C

For n = 4

F BDS4 =

1

2log2

“s

t

”

+2π2

3

For n ≥ 5, F BDSn (0) = 1

2

Pni=1 gn,i ( Mandelstam variables:

t[r]i ≡ (ki + ... + ki+r−1)

2)

gn,i = −[n/2]−1X

r=2

ln

−t[r]i

−t[r+1]i

!

ln

−t[r]i+1

−t[r+1]i

!

+ Dn,i + Ln,i +3

2ζ2


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.Test of the BDS conjecture (Weak Coupling)

explicit loop calculation

4-point up to 3-loops [BDS]5-point up to 2-loops [Cachazo et. al. , Bern et. al.]n(≥ 6)-point 1-loop [Bern et. al.]

Discrepancy in 6-point 2-loop amplitude[Bern et. al. 0803.1465]Gluon amplitudes=Wilson loop [Drummond et. al. 0803.1466]

ln MMHV6 = ln W (C6) + const., F WL

6 = F BDS6 + R6(u), R6 6= 0

Non-trivial dependence comes from the function of the cross-ratio

uij,kl =x2

ijx2kl

x2ikx2

jl

, (x2ij = t

[j−i]i )

For n = 6, u13,46, u24,15, u35,26 are independent cross ratios.


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Gluon amplitudes = Wilson loop with light-like boundaries

ends at r = R2

zIR→ 0 (zIR → ∞)

yµ: surrounded by the light-likesegments

Mn ∼ exp(−SNG)

k1

k2

r=0

y

SNG: the value of the Nambu-Goto action for the surface surrounded light-likesegments = Area of the minimal surfacestatic gauge: surface y0(y1, y2), r(y1, y2) parametrized by (y1, y2)

SNG =R2

2π

Z

dy1dy2

p

1 + (∂ir)2 − (∂iy0)2 − (∂1r∂2y0 − ∂2r∂1y0)2

r2

Euler-Lagrange equations

∂i

„

∂L

∂(∂iy0)

«

= 0, ∂i

„

∂L

∂(∂ir)

«

− ∂L

∂r= 0,

non-linear differential equations, difficult to solve

Test of the BDS conjecture at Strong Coupling


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.Alday-Maldacena’s Solution

4-point amplitude (s = t): s = −(k1 + k2)2, t = −(k1 + k4)

2

boundary condition:r(±1, y2) = r(y1,±1) = 0,y0(±1, y2) = ±y2,y0(y1,±1) = ±y1

solution: y0 = y1y2,

r =p

(1 − y21)(1 − y2

2)

-1-0.5

00.5

1y1

-1

-0.500.5

1y2

-1

-0.5

0

0.5

1

y0

-1-0.5

00.5

1y1

-1

-0.500.5

1y2

-1-0.5

00.5

1y1

-1

-0.5

0

0.5

1

y20

0.250.5

0.751

r

-1-0.5

00.5

1y1

general (s, t)-solution (SO(2, 4) transformation)conformal boost (b)+scale transformation (a)

r′ =ar

1 + by0, y′

0 =a√

1 + b2y0

1 + by0, y′

1 =ay1

1 + by0, y′

2 =ay2

1 + by0

−s(2π)2 = 8a2

(1−b)2, −t(2π)2 = 8a2

(1+b)2

The area agrees with the BDS formula. Higher-point amplitudes?


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.Numerical Solutions of Minimal Surfaces in AdS

discretization

square lattice with spacing h = 2M

(i, j) (i, j = 0, · · · , M)y0[i, j] = y0(−1 + hi,−1 + hj)r[i, j] = r(−1 + hi,−1 + hj).

h

h

(i,j)

-1 +1-1

1

y_1

y_2

E-L Equations→ 2(M − 1)2 nonlinear simultaneous equations for y0[i, j] and r[i, j]

Use the same momentum configurations as in Astefanesei-Dobashi-Ito-NastaseSolve these equations numerically. (Newton’s method)Evaluate the action S =

P

L[i, j]h2 using the radial cut-off regularization

S[rc] =R

r(y1,y2)≥rcdy1dy2L

Sdis[rc] =P

r[i,j]≥rcL[i, j]h2

r

y

y

1

2

r=rc

quantitative check of the gluon amplitude/Wilson loop duality and theBDS conjecture at strong coupling

some hints to obtain exact solutions


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.Minimal surface:4-point amplitude

-1 -0.5 0 0.5 1 -1-0.5

0 0.5

1-1

-0.5 0

0.5 1

y0M=50

exact sol.

y1y2

y0

-1 -0.5 0 0.5 1 -1-0.5

0 0.5

1 0

0.2 0.4 0.6 0.8

1r M=50

exact sol

y1y2

r

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

-1 -0.5 0 0.5 1

y 0

rho

M=300y0=rho2/2

y1 = y2 section (M = 300)

S4[rc, b] =

Z

S

dy1dy2L, L =1

(1 − y21)(1 − y2

2)

S: region surrounded by the cut-off curve C

r2c = (1 − y2

1)(1 − y22)

1

(1 + by1y2)2

S4[rc, b] =1

4log

2

r2c

−8π2s

!

+1

4log

2

r2c

−8π2t

!

−1

4log

2(

s

t)−3.289... + O(r

2c log r

2c).

FBDS4 = −

1

4log

2(

s

t) −

π2

3= −

1

4log

2(

s

t) − 3.28987...


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.Numerical check of the BDS formula: 4-pt amplitude

M = 520

5

10

15

20

25

30

35

0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

S 4[r c

,b]

rc

b=0.0b=0.0(S4)

b=0.4b=0.4(S4

b=0.8b=0.8(S4)

S[rc, b] vs Sdis[rc, b]

-0.07

-0.06

-0.05

-0.04

-0.03

-0.02

-0.01

0

0.01

0.02

0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

(Sdi

s 4-SBD

S 4)/S

dis 4

rc

b=0.0b=0.0(exact)

b=0.4b=0.4(exact)

b=0.8b=0.8(exact)

(Sdis4 − SBDS

4 )/Sdis4

Finite rc correction ≤ 6%

numerical error becomes large rc ≤ 0.2 (and large b)


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.n-point amplitude (conjecture)

S̃n[rc] =18

n∑i=1

(log

r2c

−8π2si,i+1

)2

+Fn(p1, · · · , pn)+O(r2c log2 rc),

Fn = −12FBDS

n + Rn(uij,kl)

remainder function: Rn

It is difficult to distinguish finite rc correction and remainderfunction, numerically.

Gdisn [rc, b] = Sdis

n [rc, b] − Sdisn [rc, 0] : smaller rc correction


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.Difference of areas with different boost parameters

4-point amplitude:

0

1

2

3

4

5

6

7

8

0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

G4[r

c, b]

rc

b=0.2b=0.2 (BDS)

b=0.4b=0.4 (BDS)

b=0.6b=0.6 (BDS)

b=0.8b=0.8 (BDS)

Gdis[rc, b] vs GBDS [rc, b]

-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.2 0.25 0.3 0.35 0.4 0.45 0.5

(Gdi

s 4-G

BDS 4)

/Gdi

s 4

rc

b=0.4b=0.4(exact)

b=0.6b=0.6(exact)

b=0.8b=0.8(exact)

(Gdis − GBDS)/Gdis

small rc: large fluctuation, large rc: large rc correction(∼10%)

difference 5% at rc = 0.3If we find deviation larger than finite rc correction, thissuggests the existence of the remainder function Rn.


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.Highre-point amplitudes

6-point solution1

-1 -0.5 0 0.5 1 -1-0.5

0 0.5

1-1

-0.5 0

0.5 1

y0M=50

eq. (2.7)

y1y2

y0

-1 -0.5 0 0.5 1 -1-0.5

0 0.5

1 0

0.2 0.4 0.6 0.8

1 1.2

r M=50eq. (2.7)

y1y2

r

6-point solution2

-1 -0.5 0 0.5 1 -1-0.5

0 0.5

1-1

-0.5 0

0.5 1

y0M=50y1 |y2|

y1y2

y0

-1 -0.5 0 0.5 1 -1-0.5

0 0.5

1 0

0.2 0.4 0.6 0.8

1r M=50

eq. (2.9)

y1y2

r

8-point

-1 -0.5 0 0.5 1 -1-0.5

0 0.5

1 0

0.2 0.4 0.6 0.8

1y0

M=50|y1 y2|

y1y2

y0

-1 -0.5 0 0.5 1 -1-0.5

0 0.5

1 0

0.2 0.4 0.6 0.8

1 1.2

r M=50eq. (2.11)

y1y2

r

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

y 2

y1

b=0.0b=0.4b=0.8

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

y 2

y1

b=0.0b=0.4b=0.8

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

y 2

y1

b=0.0b=0.4b=0.8


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.

.6-point amplitude solution 1

Sdis[rc, b] vs SBDS

5

10

15

20

25

30

35

0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

S 6(1

) [r c,b

]

rc

b=0.0b=0.0(BDS)

b=0.4b=0.4(BDS)

b=0.8b=0.8(BDS)


-2.4

-2.2

-2

-1.8

-1.6

-1.4

-1.2

-1

-0.8

-0.6

-0.4

0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

G6(1

) [r c, b

]

rc

b=0.2b=0.2 (BDS)

b=0.4b=0.4 (BDS)

cross ratios are independent of b: u1 = u2 = u3 = 1

SBDS(1)6 [rc, b] =

1

8

n

2 log2(

r2c(1 − b)

8) + 2 log

2(

r2c(1 + b)2

8) + log

2 r2c

4+ log

2(

r2c(1 + b)2

16)o

−1

2

n

log 2 log(1 − b) − 2 log 2 log(1 + b) − 2 log(1 − b) log(1 + b)

+1

2log

2(1 − b) + 3 log

2(1 + b)

o

−3π2

16

=⇒ Rdis6 does not depend on b and is non-zero constant.


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.

.6-point amplitude solution 2

Sdis[rc, b] vs SBDS [rc, b]

5

10

15

20

25

30

35

40

45

0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

S 6(2

) [r c,b

]

rc

b=0.0b=0.0(BDS)

b=0.4b=0.4(BDS)

b=0.8b=0.8(BDS)


0

1

2

3

4

5

6

7

8

9

0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

G6(2

) [r c, b

]

rc

b=0.2b=0.2 (BDS)

b=0.4b=0.4 (BDS)

b=0.6b=0.6 (BDS)

b=0.8b=0.8 (BDS)

cross ratios are independent of b: u1 = u2 = u3 = 1

SBDS(2)6 [rc, b] =

1

8

n

log2(

r2c(1 − b)2

8) + log

2(

r2c(1 + b)2

8) + 2 log

2(

r2c(1 + b)

8) + 2 log

2(

r2c(1 − b)

8)o

−1

2

n 3

2log

2(1 − b) +

3

2log

2(1 + b) − 2 log(1 − b) log(1 + b)

o

−3π2

16


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.

.8-point amplitude

Sdis[rc, b] SBDS [rc, b]

0

5

10

15

20

25

30

35

40

45

0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

S 6(2

) [r c,b

]

rc

b=0.0b=0.0(BDS)

b=0.4b=0.4(BDS)

b=0.8b=0.8(BDS)


-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

F 8[r c

, b]

rc

b=0.2b=0.2 (BDS)

b=0.4b=0.4 (BDS)

b=0.6b=0.6 (BDS)

b=0.8b=0.8 (BDS)

cross ratios are independent of b: uijkl = 1

SBDS8 [rc, b] =

1

8

n

4 log2(

r2c(1 + b)2

8) + 4 log

2(

r2c

4)o

−1

2

n

4 log2(1 + b) − 4 log 2 log(1 + b) −

π2

6

o

−π2

2


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.Conclusions and Outlook

For the 4-point amplitude, M = 520 data is numerically consistent withthe BDS formula.

For the 6 and 8-point amplitudes, the present numerical solutions suggestnon-zero constant remainder functions Rn.

Non-trivial momentum configurations =⇒Rn(u).Compare the deviation with the results from weak coupling analysisAnastasiou et al., 0902.2245

Improve numerical solutions (larger M)

AdS3 constraints: r2 − y20 + y2

1 + y22 = 1

Jevicki-Jin-Kalousios 0712.1193, Alday-Maldacena 0903.4701; 0904.0663

Newton method→contragradient method

Hints to obtain exact solution ( without AdS3 constraints)Numerical check of underlying integrable structure at strong coupling(dual conformal symmetry, fermionic T-duality, Yangian)

non-AdS geometry (finite temperature) Ito-Nastase-Iwasaki


discretized minimal surface and gluon scattering ... · introduction ads/cft correspondence and...

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