MIT OpenCourseWare http ocw mit edu 8 821 String Theory Fall 2008 For information about citing these materials or our Terms of Use visit http ocw mit edu terms 8 821 F2008 Lecture 14 Wave equation in AdS Green s function Lecturer McGreevy November 11 2008 Topics for this lecture Find 0 x by Green functions in x space e cient Compute OO counter terms Redo in p space general References Witten hep th 9802150 GKP hep th 9802109 Solving Wave Equation I Witten s method Let s study the wave equation in AdS in some detail This rst method uses a trick by Witten which is e cient but slightly obscure If we know bulk to boundary Green s function K regular in the bulk such that m2 Kp z x 0 Kp z x D x p z where p is some point on the boundary then the eld in the bulk 0 Ren z x dD x Ren 0 x 0 x Kx z x z solves 1 1 1 2 Euclidean AdS Recall the metric on AdS with curvature scale L in the upper half plane coordinates ds2 L2 dz 2 dx2 z2 Now here comes some fancy tricks thanks to Ed Trick 1 Pick p point at This implies that the Green s function K z x is x independent The wave equation at k 0 0 z D 1 z z D 1 t m2 K z can easily be solved The solution is power law recall that in the general k wave equation it was the terms proportional to k 2 that ruined the power law behavior away from the boundary K z c z c z We can eliminate one of the constants c 0 whose justi cation will come with the result Trick 2 Use AdS isometries to map p to nite x Let xA x z take xA x A xA xB xB The inversion of this mapping is x z 2x x2 I z z z 2 x 2 Claim I A is an isometry of AdS also Minkowski version see pset 4 B is not connected to 1 in SO D 2 C maps p to x 0 i e I K z x K z x K0 z x c z z 2 x2 Some notes i That this solves the wave equation 1 as neccessary can be checked directly ii The Green s function is Kx z x c z K z x x z 2 x x 2 iii The limit of the Green s function as z 0 i e the boundary is cz 0 if x x K z x x cz if x x 2 recall that 0 for any D m More speci cally the Green s function approaches a delta function K z x x const D x x Clearly it has support only near x x but to check this claim we need to show that it has nite measure c dD x K0 x dD x 2 x2 c D 2 D 1 dD x 2 1 x 2 D 2 c D 2 We will choose the constant c to set this last expression equal to one Hence 0 z x dD x Kx z x Ren 0 x x dD x c 2 Ren x z x x 2 0 this solves 1 and approaches Ren 0 x as z The action is related to expectation values of operators on the boundary R S 0 ln e 0 O n 2 AdS D zz d x g g z x z z x 2 z Ren dD x1 dD x2 Ren 0 x1 0 x2 F x1 x2 2 where the ux factor is F x1 x2 dD x K z x x1 z z K z x x2 zD z The boundary behavior of K is D 2 K z x x x x O z c O 2 x x 2 D the rst terms sets c 1 2 D 2 the second term is subleading in z 1 z z K z x x x x cz x x 2 z Ok now for the 2 point correlation function on the boundary G2 x1 x2 O x1 O x2 c S 0 F x1 x2 0 x1 0 x2 3 We must be careful when evaluating the cases x1 x2 and x1 x2 which we do in turn Firstly if x1 x2 G2 x1 x2 D cz D D 2 2 d xz z x x1 O z ignore by x1 x2 O z 2 z x1 x2 2 1 c D O 2 2 x1 x2 2 c 2 x1 x2 2 Good This is the correct form for a two point function of a conformal primary of dimension in a CFT this is a check on the prescription Secondly if x1 x2 G2 x1 x2 2 D D x1 x2 c c2 2 D x1 x2 2 1 d x 2 x x1 x x2 2 D As 0 the rst term is divergent the second term is nite and the third term vanishes The rst term is called a divergent contact term It is scheme dependent and useless Remedy Holographic Renormalization Add to Sgeometry the contact term 2 S Sc t dD x 2 D Ren x 0 2 bdy 2 z x 2 AdS z Note that this doesn t a ect the equations of motion Nor does it a ect G2 x1 x2 Solving Wave Equation II k space Since the previous approach isn t always available for example if there is a black hole in the spacetime let s redo the calculation in k space Return to wave equation 0 z D 1 z z D 1 z m2 L2 z 2 k 2 fk z with k 2 2 k2 0 The solution is D D fk z AK z 2 K kz AI z 2 I kz with D 2 2 m2 L2 D 2 Assume k R real time issues later As z K e kz and I ekz The latter is not okay so AI 0 4 At boundary K n n 2 4 a0 a1 n a2 n n b0 b1 n2 b2 n4 R n ln n b0 b1 n2 b2 n4 R Hence D fk z AK z D 2 K kz z 2 z as z 0 5