TFI Theories of the Fundamental Interactions. Andrea Marini Università di Perugia & INFN Perugia

Transcription

1 TF 2015 Theories of the Fundamental nteractions Università di Napoli Federico, November 18-20, 2015 Born-nfeld/Gravity Duality Andrea Marini Università di Perugia & NFN Perugia Based on: G. Grignani, T. Harmark, AM and M. Orselli arxiv:1512.xxxx [hep-th]

2 Outline 1 Overview 2 Born-nfeld side 3 Gravity side 4 Conclusions

3 Overview New duality between 4d Born-nfeld (B) theory and 5d classical gravity manifestation of the open/closed string duality Born-nfeld side low energy effective theory for open strings ending on a D3-brane in a slowly varying background Kalb-Ramond field gravity side gravitational (closed string) description of D3-branes in the same background Kalb-Ramond field the duality is a correspondence between effective theories in a similar sense as in the fluid/gravity correspondence it can be used to derive higher derivative correction to the B action

4 Setup N coincident D3-branes in 10d Minkowski background Mink 10 x 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 D3 flat embedding i = 4, 5,..., 9 x a (σ) = σ a, for a = 0, 1, 2, 3, and x i (σ) = 0 for B SUGRA background has φ = 0 and H c = 0 turn on B D3 world-volume (while B ij = 0) F = B gauge invariant field strength on the D3 B SUGRA EOMs db (2) = 0 [a F bc] = 0

5 Setup Assume F slowly varying over the D3 R l s (g s N) 1/4 and R l s 1 R = minimal length scale of variation of F when g s N 1 open string description Born-nfeld action when g s N 1 closed string description type B SUGRA

6 Outline 1 Overview 2 Born-nfeld side 3 Gravity side 4 Conclusions

7 Born-nfeld theory B action: low energy effective action for open strings ending on a single D3 valid when g s 1 (perturbative string theory) S B = T D3 d 4 σ det (η + F ) F = B + 2πl 2 sf and T D3 = ( g s l 4 s(2π) ) 1 [a F bc] = 0 df = 0 useful notation: F a b = F ac η cb B action G a b = δ b a F a c F c b S B = T D3 m 4 = det(g a b ) d 4 σ m 2

8 B EOMs EOMs by varying the action b (mg 1 F) ba = 0 Born-nfeld energy-momentum tensor τ = T D3 m(g 1 ) energy-momentum tensor conservation b τ ba = 0 b (m(g 1 ) ba ) = 0 these eq.s along with df = 0 the D3 constraints on how F can vary along

9 Outline 1 Overview 2 Born-nfeld side 3 Gravity side 4 Conclusions

10 Gravity setup SUGRA description of N D3-branes in the background of a slowly varying Kalb-Ramond potential Type B Supergravity Lagrangian (bosonic part) L = ( g R (10) 1 2 µφ µ φ 1 12 e φ H µνρh µνρ 1 2 e2φ µχ µ χ 1 12 eφ F µνρf µνρ 1 ) 4 5! F µνρλσf µνρλσ ! ɛµ 1...µ 10 A µ1...µ 4 µ5 B µ6 µ 7 µ8 A µ9 µ 10 lim B = 0 r 0 boundary conditions lim g µν = η µν (Mink 10 ) r lim B = F with df = 0 r lim φ, χ, r A(2) = 0 r = radial coordinate in the transverse space of the D3 (D3 are at r = 0)

11 Gravity setup using symmetries and gauge freedom ten-dimensional metric can be written as g µν dx µ dx ν = e 1 2 η ( h mn dx m dx n + e 2ρ dω 2 ) 5 m, n = 0,..., 4 with ρ = log r and h mn and η only depends on x m impose that we have N D3-branes set the RR 5-form field strength to F (5) = 4r 4 c(ω 5 + ω 5 ) r 4 c = N 2π 2 T D3 r c sets the length scale of the thickness of the D3 only the metric and F (5) involve the S 5 directions 5d reduction

12 5d EOMs m ( h e 2η+5ρ m φ ) e 5ρ h m ( h e 2φ+2η+5ρ m χ ) e 5ρ h = 1 6 eφ+η F mnk H mnk = 1 12 eη φ H mnk H mnk eη+φ F mnk F mnk + e 2η+2φ mχ m χ ) m ( h e η+5ρ (e φ H mnk χe φ F mnk ) = 4r4 c 6 ɛnklpq (F lpq + χh lpq ) m ( h e η+5ρ e φ F mnk ) = 4r4 c 6 ɛnklpq H lpq 4e 2ρ 5 m ρ mρ 1 2 m η mη 13 4 m η mρ D m D mρ 1 4 Dm D mη = 1 48 e φ η H mnk H mnk 1 48 eφ η F mnk F mnk + 4r 8 c e 2η 10ρ (R (5) ) mn = 5 mρ nρ+5d md nρ 1 2 mη nη+2dmdnη+ 1 2 hmn k η k η+ 1 4 hmndk D k η hmn k η k ρ mφ nφ e2φ mχ nχ e φ η (3H m kl H nkl 1 4 hmnhklp H klp ) eφ η (3F m kl F nkl 1 4 hmnf klp F klp ) 4r 8 c e 2η 10ρ h mn

13 D3 with constant F Known solutions D3 solution F = 0 F1-D3 solution E 0, B = 0 D1-D3 solution B 0, E = 0 0 E 1 E 2 E 3 E 1 0 B 3 B 2 F = E 2 B 3 0 B 1 E 3 B 2 B 1 0 (F1 D1)-D3 solution E B = 0 E = (E 1, E 2, E 3 ) B = (B 1, B 2, B 3 ) (F1 D1)-D3 solution E B = 0 Grignani, Harmark, AM, Orselli arxiv:

14 D3 with constant F New (general) solution generalization of the previous solutions h mn dx m dx n = h dx a dx b + dr 2 (a, b = 0,..., 3) ρ = log r h = {( + mrcr 4 4 G 1 ) 1 } c a η cb h = e 2η ( ) 2 e 2φ = e 2η 1 + m r4 c r 4 χ = Tr(F F) 4 e 2η = 1 + Tr(G) rc 4 2m r 4 + r8 c r 8 r 4 c r m r4 c r 4 B = F a c h cb A = r4 c r 4 F a c h cb low energy description of D3 dynamics r c l s g s N 1 low energy effective description of the strongly coupled Born-nfeld theory

15 Derivative expansion consider the D3 solution with general F let F be a slowly varying function of the world-volume coordinates x a analogy with fluid/gravity correspondence set up a perturbative procedure to build a corrected solution world-volume derivative expansion zeroth order D3 with constant F

16 First order neglect in the EOMs all the terms with more than one world-volume derivative equations with one world-volume derivative correction are b ( e η φ H bar e η+φ χf bar) = 4r4 c 6r 5 ɛcd (F bcd + χh bcd ) (1) b ( e η+φ F bar) = 4r4 c 6r 5 ɛcd H bcd (2) 1 2 b(h bc r h ca ) 1 2 Γc h bd r h dc = 1 2 aη r η aφ r φ e2φ a χ r χ e φ η H a bc H bcr eφ η F a bc F bcr (3)

17 First order Consider the ove equations to leading order at large r Eq. (1) b H bar = 0 b (mg 1 F) ba = 0 Eq. (2) b F bar = 0 b ( F) ba = 0 df = 0 Eq. (3) b r h ba = 0 b (mg 1 ) ba = 0 these are the constraints on the variation of F Exactly the same eqs we got from the B!

18 Higher-derivative correction starting from second order corrections to the zeroth order solution general corrected solution Y (x a, r) = Y (0) (x a, r) + Y (2) (x a, r) + Y (4) (x a, r) + O( 6 ) Y = (χ, φ, A, B, η, h ) no corrections with an odd number of world-volume derivatives covariant expression in terms of F no

19 Second order Eq.s that determine Y (2) H m, (0) Y (2) +H m, (1) involve two world-volume derivatives (2) ry +H m, (2) 2 r Y (2) = K m,, a b Y (0) +L m,,j ay (0) b Y (0) solution in a large r expansion has the form to leading order at large r χ (2) = B (2) η (2) = r4 c 16r 2 a a (F F) + O(r 6 ) Y (2) = n=0 = r4 c 4r 2 3 c [c (mg 1 F) ] + O(r 6 ) A (2) ( r4 c TrG 16r 2 c c m C (2),n r 2+4n φ (2) = r4 c 4r 2 a a ( m TrG 4m = r4 c 4r 2 3 c [c F ] + O(r 6 ) ) + O(r 6 ) ) + O(r 6 ) h (2) = r4 c 4r 2 c c(mg 1 ) + O(r 6 )

20 Third order Third order eq.s to leading order at large r Eq. (1) b H bar = 0 b c [c (mg 1 F) ] = 0 Eq. (2) b F bar = 0 b c [c F ] = 0 Eq. (3) b r h ba = 0 b c c (mg 1 ) ba = 0 these are trivially satisfied assuming the first order constraints

21 Fourth order fourth order corrections at large r Y (4) = C (4),0 ( ) C(4),4 1 log r + r 4 + O r 8 one can easily compute C (4),0 ( C (4) χ,0 = r4 c (F F) C (4) φ,0 = r4 c m TrG ) 4m C (4) B,0 = r4 c c [c (mg 1 F) ] C (4) A,0 = r4 c c [c F ] ( C (4) η,0 = r4 c TrG m where 2 = a a the next-to-leading coefficient C (4),4 cannot be fixed ) C (4) h,0 = r4 c (mg 1 )

22 Comparison with the B the constraints we got at first and third order are consistent with the B ones what out the corrections to the fields at second and fourth order? corrections to the metric corrections to EM tensor compare this corrected EM tensor with B EM tensor derivative corrections B higher

23 Reading off the energy-momentum tensor EM tensor can be read off from the metric in the asymptotic region in our approximation the asymptotic region is r c r R linearize the metric h = η + h η = η with h 1, η 1 make the gauge choice η h = 4 η and b hba = 0 imagine an infinitely thin 3-brane sitting at r = 0 with EM tensor τ (x c ) on the brane linearized Einstein equations in the region r c r R [ 2 r + 5 r r + c c ] h = 16πG δ 6 (r)τ

24 Linearized Einstein eq.s assume τ varies slowly along the world-volume expand τ in higher-derivative corrections τ = τ (0) + τ (2) + τ (4) + O( 6 ) expand h in higher-derivative contributions h = h (0) linearized Einstein eq.s become [ r r r [ r ] r r [ r r r + h (2) + h (4) + O( 6 ) ] h (0) = 16πG δ6 (r)τ (0) h (2) + c c h (0) = 16πG δ6 (r)τ (2) ] h (4) + c c h (2) = 16πG δ6 (r)τ (4)

25 EM tensor Solution of the linearized Einstein eq.s h (0) h (2) h (4) = r4 c NT D3 = r4 c r 4 τ (0) τ (2) r 4 + r4 c c c τ (0) NT D3 4r 2 = r4 c NT D3 τ (4) r 4 + r4 c c c τ (2) NT D3 4r 2 NT D3 16 log r c c d d τ (0) r4 c NT D3

26 Zeroth order EM tensor solution of linearized Einstein h(0) = r4 c NT D3 r 4 τ (0) leading correction to flat space metric at zeroth order plug h (0) h (0) = r4 c r 4 m(g 1 ) into ove solution of linearized Einstein eq. τ (0) = NT D3m(G 1 ) this is the leading order EM-tensor for N D3-branes = N Born-nfeld EM-tensor

27 Second order EM tensor solution of linearized Einstein h(2) τ (2) r 4 + r4 c c c τ (0) NT D3 4r 2 = r4 c zeroth order EM tensor τ (0) = NT D3m(G 1 ) second order large r correction to the metric h (2) ( ) = r4 c 1 4r 2 c c (mg 1 ) + O r 6 plug τ (0) (2) and h into ove sol. of linearized Einstein eq. NT D3 r 4 ( ) c 1 4r 2 c c (mg 1 ) + O r 6 = r4 c τ (2) r 4 r 4 NT D3 c 4r 2 c c (mg 1 ) second order EM tensor τ (2) = 0 Born-nfeld is not corrected at second order Andreev, Tseytlin (1988)

28 Fourth order EM tensor? solution of linearized Einstein eq. h (4) τ (4) r 4 + r4 c c c τ (2) NT D3 4r 2 = r4 c zeroth and second order EM tensors NT D3 16 log r c c d d τ (0) r4 c τ (0) = NT D3m(G 1 ) τ (2) = 0 fourth order large r correction to the metric h (4) plug τ (0), τ (2) (2) and h rc 4 16 log r c c d d (mg 1 ) +O NT D3 ( ) = r4 c 1 16 log r c c d d (mg 1 ) + O r 4 into ove sol. of linearized Einstein eq. ( ) 1 r 4 = r4 c τ (4) rc 4 r 4 + NT D3 16 log r c c d d (mg 1 )

29 Outline 1 Overview 2 Born-nfeld side 3 Gravity side 4 Conclusions

30 Conclusions explored a new duality between B theory and 5d gravity open/closed string duality for D3-branes in a slowly varying background Kalb-Ramond field F world-volume derivative expansion to build the gravity solution for D3 with a slowly varying F dual to B theory (including derivative corrections) proved that the B action is not corrected up to the second order in the derivative expansion what out the fourth order?

31 Extra slides

32 Duality chain generating constant F D3 solution start with the D3 solution h = η 1 + r4 c r 4 e η = 1 + r4 c r 4 φ = χ = 0 A = B = 0 T-duality along x 1 D2-branes smeared along x 1 rotation in the 12-plane + T-duality along x 1 D1-D3 bound state with D-strings along x 3

33 Duality chain... cont d D1-D3 bound state with D-strings along x 3 T-duality along x 3 D0-D2 bound state smeared along x 3 rotation in the 23-plane + boost along x 3 + T-duality along x 1 F1-D1-D3 bound state with F-strings along x 3 and D-strings in the 13-plane

34 Second order solution Eqs that determine Y (2) H m, (0) Y (2) +H m, (1) particular solution Y (2) = (2) ry +H m, (2) 2 r Y (2) = K m,, a b Y (0) +L m,,j ay (0) b Y (0) n=0 C (2),n r 2+4n solution of the homogeneous part δy = general second order solution Y (2) + δy n=1 C (2),n r 4n demand δy to be a small perturbation of Y (0) and whole solution to be asymptotically flat δy = 0