Analytic solutions for Dp branes in SFT
aa r X i v : . [ h e p - t h ] J un Preprint typeset in JHEP style - HYPER VERSION
SISSA/29/2010/EP hep-th/yymm.xxxx
Analytic solutions for Dp branes in SFT
L.Bonora
International School for Advanced Studies (SISSA)Via Bonomea 265, 34136 Trieste, Italy, and INFN, Sezione di Trieste, Italy;E-mail: [email protected] , S.Giaccari
International School for Advanced Studies (SISSA)Via Bonomea 265, 34136 Trieste, Italy, and INFN, Sezione di Trieste, Italy;E-mail: [email protected] , D.D.Tolla
Department of Physics and University College, Sungkyunkwan University, Suwon440-746, South KoreaE-mail: [email protected]
Abstract:
This is the follow-up of a previous paper [ArXiv:1105.5926], where we calcu-lated the energy of an analytic lump solution in SFT representing a D24-brane. Here wedescribe an analytic solution for a D p -brane, for any p , and compute its energy. Keywords:
String Field Theory. ontents
1. Introduction 12. A D23-brane solution 2
3. The ǫ -regularization 64. The energy of the D –brane 85. D(25-p) brane solutions 12
1. Introduction
This paper is a continuation and extension of [1]. Recently, following an earlier suggestionof [2], a general method has been proposed, [3], to obtain new exact analytic solutions inWitten’s cubic open string field theory (OSFT) [4], and in particular solutions that de-scribe inhomogeneous tachyon condensation. There is a general expectation that an OSFTdefined on a particular boundary conformal field theory (BCFT) has classical solutionsdescribing other boundary conformal field theories [7, 8]. Previously analytic solutionswere constructed describing the tachyon vacuum [5, 6, 9, 10, 11, 12, 13, 14, 15, 16, 17,18, 19, 20, 21] and of those describing a general marginal boundary deformations of theinitial BCFT [22, 23, 24, 25, 26, 27, 28, 29, 30, 31], see also the reviews [32, 33]. In thispanorama an element was missing: the solutions describing inhomogeneous and relevantboundary deformations of the initial BCFT were not known, though their existence waspredicted [7, 8, 34]. In [2, 3] such solutions were put forward, in [1, 35] the energy of aD24-brane solution was calculated for the first time. Here we wish to extend the methodand the results of [1] to describe analytic SFT solutions corresponding to D(25- p )-branesfor any p . The extension is nontrivial because new aspects and problems arise for p > ψ u ,u for a D23brane, compute its energy functional, study its UV and IR behaviour and verify that thevalue of its energy functional depends on the parameter v = u /u . Next, in section 3,in analogy with [1], we introduce the ǫ -regularized solutions ψ ǫu ,u , which represents thetachyon condensation vacuum. Then we verify that the difference ψ u ,u − ψ ǫu ,u , whichis a solution to the equation of motion over the vacuum represented by ψ ǫu ,u , has theexpected energy of a D23-brane. At this point the extension to a generic D(25- p )-brane isstraightforward and we summarize it in section 5.– 1 – . A D23-brane solution Let us briefly recall the technique to construct lump solutions by incorporating in SFTexact renormalization group flows generated in a 2D CFT by suitable relevant operators.To start with we enlarge the well-known
K, B, c algebra defined by K = π K L | I i , B = π B L | I i , c = c (cid:18) (cid:19) | I i , (2.1)in the sliver frame (obtained by mapping the UHP to an infinite cylinder C of circum-ference 2, by the sliver map f ( z ) = π arctan z ), by adding a state constructed out of a(relevant) matter operator φ = φ (cid:18) (cid:19) | I i . (2.2)with the properties [ c, φ ] = 0 , [ B, φ ] = 0 , [ K, φ ] = ∂φ, (2.3)such that Q has the following action: Qφ = c∂φ + ∂cδφ. (2.4)One can show that ψ φ = cφ − K + φ ( φ − δφ ) Bc∂c (2.5)does indeed satisfy the OSFT equation of motion Qψ φ + ψ φ ψ φ = 0 (2.6)In order to describe the lump solution corresponding to a D24-brane in [3, 1] we used therelevant operator, [2], φ u = u (: X : +2 log u + 2 A ) (2.7)defined on C , where X is a scalar field representing the transverse space dimension, u isthe coupling inherited from the 2D theory and A is a suitable constant.In the case of a D φ ( u ,u ) = u (: X : +2 log u + 2 A ) + u (: X : +2 log u + 2 A ) (2.8)where X and X are two coordinate fields corresponding to two different space directions.There is no interaction term between X and X in the 2D action.Then we require for φ u the following properties under the coordinate rescaling f t ( z ) = zt f t ◦ φ ( u ,u ) ( z ) = 1 t φ ( tu ,tu ) (cid:16) zt (cid:17) . (2.9)– 2 –he partition function corresponding to the operator (2.8) is factorized, [36, 37]: g ( u , u ) = g ( u ) g ( u ) g ( u i ) = 1 √ π √ u i Γ(2 u i ) e u i (1 − log 2 u i ) (2.10)where in (2.10) we have already made the choice A = γ − π . This choice implieslim u ,u →∞ g ( u , u ) = 1 (2.11)With these properties all the non-triviality requirements of [3, 1] for the solution ψ ( u ,u ) ≡ ψ φ ( u ,u are satisfied. Therefore we can proceed to compute the energy. To this end wefollow the pattern of Appendix D of [3], with obvious modifications. So, for example, h X ( θ ) X ( θ ′ ) i Disk = h X ( θ ) i Disk h X ( θ ′ ) i Disk = Z ( u ) h u Z ( u ) h u (2.12)and so on.Going through the usual derivation one gets that the energy functional is given by E [ ψ ( u ,u ) ] = − h ψ ( u ,u ) ψ ( u ,u ) ψ ( u ,u ) i (2.13)where E [ ψ ( u ,u ) ] = 16 Z ∞ dt dt dt E ( t , t , t ) g ( u T, u T ) · ( u G u T ( 2 πt T ) G u T ( 2 π ( t + t ) T ) G u T ( 2 πt T )+8 u G u T ( 2 πt T ) G u T ( 2 π ( t + t ) T ) G u T ( 2 πt T )+ u (cid:16) − ∂ u T g ( u T, u T ) g ( u T, u T ) (cid:17) + 2 u u (cid:16) − ∂ u T g ( u T, u T ) g ( u T, u T ) (cid:17)! · (cid:16) G u T ( 2 πt T ) + G u T ( 2 π ( t + t ) T ) + G u T ( 2 πt T ) (cid:17) + u (cid:16) − ∂ u T g ( u T, u T ) g ( u T, u T ) (cid:17) + 2 u u (cid:16) − ∂ u T g ( u T, u T ) g ( u T, u T ) (cid:17)! · (cid:16) G u T ( 2 πt T ) + G u T ( 2 π ( t + t ) T ) + G u T ( 2 πt T ) (cid:17) + u (cid:16) − ∂ u T g ( u T, u T ) g ( u T, u T ) (cid:17) + u (cid:16) − ∂ u T g ( u T, u T ) g ( u T, u T ) (cid:17)! ) (2.14)When writing ∂ u T g ( u T, u T ) we mean that we differentiate (only) with respect to thefirst entry, and when ∂ u T g ( u T, u T ) (only) with respect to the second. This can be– 3 –ritten also as E [ ψ ( u ,u ) ] = 16 Z ∞ dt dt dt E ( t , t , t ) g ( u T, u T ) · ( u G u T ( 2 πt T ) G u T ( 2 π ( t + t ) T ) G u T ( 2 πt T )+8 u G u T ( 2 πt T ) G u T ( 2 π ( t + t ) T ) G u T ( 2 πt T )+ (cid:16) − u ∂ T g ( u T, u T ) g ( u T, u T ) (cid:17) · (cid:16) G u T ( 2 πt T ) + G u T ( 2 π ( t + t ) T ) + G u T ( 2 πt T ) (cid:17) + (cid:16) − u ∂ T g ( u T, u T ) g ( u T, u T ) (cid:17) · (cid:16) G u T ( 2 πt T ) + G u T ( 2 π ( t + t ) T ) + G u T ( 2 πt T ) (cid:17) + (cid:16) − ∂ T g ( u T, u T ) g ( u T, u T ) (cid:17) ) (2.15)where now ∂ T g ( u T, u T ) means differentiation with respect to both entries. A furtheruseful form is the following one E [ ψ ( u ,u ) ] = 16 Z ∞ ds s Z dy Z y dx π sin πx sin πy sin π ( x − y )g( s, vs ) (2.16) · ( G s (2 πx ) G s (2 π ( x − y )) G s (2 πy )+ v G vs (2 πx ) G vs (2 π ( x − y )) G vs (2 πy ) − (cid:16) ∂ s g( s, vs )g( s, vs ) (cid:17)(cid:16) G s (2 πx ) + G s (2 π ( x − y )) + G s (2 πy ) (cid:17) − (cid:16) v ∂ s g( s, vs )g( s, vs ) (cid:17)(cid:16) G vs (2 πx ) + G vs (2 π ( x − y )) + G vs (2 πy ) (cid:17) + (cid:16) − ∂ s g( s, vs )g( s, vs ) (cid:17) ) (2.17)where s = 2 u T, v = u u and, by definition, g( s, vs ) ≡ g ( s/ , vs/
2) = g ( u T, u T ). Thederivative ∂ s in ∂ s g( s, vs ) acts on both entries. We see that, contrary to [3], where the u dependence was completely absorbed within the integration variable, in (2.17) there is anexplicit dependence on v . First of all we have to find out whether E [ ψ ( u ,u ) ] is finite and whether it depends on v .To start with let us notice that the structure of the x, y dependence is the same asin [1]. Therefore we can use the results already found there, with exactly the same IR– 4 – s → ∞ ) and UV ( s ≈
0) behaviour. The differences with [1] come from the various factorscontaining g or derivatives thereof. The relevant IR asymptotic behaviour is g ( s, vs ) ≈ v v s (2.18)for large s ( v is kept fixed to some positive value). The asymptotic behaviour does notchange with respect to the D24-brane case (except perhaps for the overall dominant asymp-totic coefficient, which is immaterial as far as integrability is concerned), so we can concludethat the integral in (2.17) is convergent for large s , where the overall integrand behavesasymptotically as 1 /s .Let us come next to the UV behaviour ( s ≈ G s . We have14 π s g ( s, vs ) (cid:18) ∂ s g ( s, vs ) g ( s, vs ) (cid:19) = −
116 ( π √ v ) s − π √ vs (cid:0) (1 + v )(1 + 2 γ ) + 2 log 2 + 2(1 + v ) log s + 2 v log(2 v ) (cid:1) + O ((log s ) )(2.19)The double pole in zero is to be expected. Once we integrate over s we obtain a behaviour ∼ s near s = 0. This singularity corresponds to ∼ δ (0) ∼ V , which can be interpretedas the D25 brane energy density multiplied by the square of the (one-dimensional) volume,see Appendix C of [1]). In order to extract a finite quantity from the integral (2.17) wehave to subtract this singularity. We proceed as in [1] and find that the function to besubtracted to the LHS of (2.19) is h ( v, s ) = (cid:16) −
116 ( π √ v ) s + 116 π √ vs − π √ vs (cid:0) (1 + v )(1 + 2 γ ) + 2 log 2 + 2(1 + v ) log s + 2 v log(2 v ) (cid:1)(cid:17) · e ss − (cid:0) s − s + 2 s + s (cid:1) ( − s ) (2.20)in the interval 0 ≤ s ≤ v -dependent.As for the quadratic terms in G s and G vs the overall UV singularity is −
316 ( π √ v ) s − v )8 ( π √ v ) s + O ((log s ) ) (2.21)and the corresponding function to be subtracted from the overall integrand is h ( v, s ) = − e ss − (cid:0) s − s + 2 s + s (cid:1) (1 + s + 2 sv )16 π s ( s − √ v (2.22)in the interval 0 ≤ s ≤ v dependent.Finally let us come to the cubic term in G s and G vs . Altogether the UV singularitydue to the cubic terms is −
18 ( π √ v ) s + ( γ + log s )(1 + v ) + log 2 + v log(2 v )4 π √ vs + O ((log s ) ) (2.23)– 5 –he overall function we have to subtract from the corresponding integrand is h ( v, s ) = 216 π √ vs e ss − (cid:0) s − s + 2 s + s (cid:1) ( s − · (cid:0) − s + 2 s (1 + v )( γ + log s ) + s log 4 + 2 sv log(2 v ) (cid:1) (2.24)for 0 ≤ s ≤ v dependent.As explained in [1] the result of all these subtractions does not depend on the particularfunctions h , h , h we have used, provided the latter satisfy a few very general criteria.After all these subtractions the integral in (2.17) is finite, but presumably v depen-dent. This is confirmed by a numerical analysis. For instance, for v = 1 and 2 we get E ( s ) [ ψ ( u ,u ) ] = 0 . ( s ) means UVsubtracted. It is clear that this cannot represent a physical energy. This is not surprising.We have already remarked in [1] that the UV subtraction procedure carries with itself acertain amount of arbitrariness. Here we have in addition an explicit v dependence thatrenders this fact even more clear. The way out is the same as in [1]. We will compare the(subtracted) energy of ψ ( u ,u ) with the (subtracted) energy of a solution representing thetachyon condensation vacuum, and show that the result is independent of the subtractionscheme.
3. The ǫ -regularization As we did in section 8 of [1], we need to introduce the ǫ -regularization and the ǫ -regularizedsolution corresponding to (2.8). We recall the general form of such solution ψ φ = c ( φ + ǫ ) − K + φ + ǫ ( φ + ǫ − δφ ) Bc∂c (3.1)where ǫ is an arbitrary small number. In the present case φ ≡ φ ( u ,u ) = u (: X : +2 log u + 2 A ) + u (: X : +2 log u + 2 A ) (3.2)It is convenient to split ǫ = ǫ + ǫ and associate ǫ to the first piece in the RHS of (3.2)and ǫ to the second. We will call the corresponding solution ψ ǫ ( u ,u ) . After the usual– 6 –anipulations the result is E [ ψ ǫ ( u ,u ) ] = 16 Z ∞ dt dt dt E ( t , t , t ) g ( u T, u T ) e − ǫT · ( u G u T ( 2 πt T ) G u T ( 2 π ( t + t ) T ) G u T ( 2 πt T )+8 u G u T ( 2 πt T ) G u T ( 2 π ( t + t ) T ) G u T ( 2 πt T )+ u (cid:16) ǫ u − ∂ u T g ( u T, u T ) g ( u T, u T ) (cid:17) + 2 u u (cid:16) ǫ u − ∂ u T g ( u T, u T ) g ( u T, u T ) (cid:17)! · (cid:16) G u T ( 2 πt T ) + G u T ( 2 π ( t + t ) T ) + G u T ( 2 πt T ) (cid:17) + u (cid:16) ǫ u − ∂ u T g ( u T, u T ) g ( u T, u T ) (cid:17) + 2 u u (cid:16) ǫ u − ∂ u T g ( u T, u T ) g ( u T, u T ) (cid:17)! · (cid:16) G u T ( 2 πt T ) + G u T ( 2 π ( t + t ) T ) + G u T ( 2 πt T ) (cid:17) + u (cid:16) ǫ u − ∂ u T g ( u T, u T ) g ( u T, u T ) (cid:17) + u (cid:16) ǫ u − ∂ u T g ( u T, u T ) g ( u T, u T ) (cid:17)! ) (3.3)or E [ ψ ǫ ( u ,u ) ] = 16 Z ∞ dt dt dt E ( t , t , t ) g ( u T, u T ) e − ǫT · ( u G u T ( 2 πt T ) G u T ( 2 π ( t + t ) T ) G u T ( 2 πt T )+8 u G u T ( 2 πt T ) G u T ( 2 π ( t + t ) T ) G u T ( 2 πt T )+ u (cid:16) ǫ − ∂ T g ( u T, u T ) g ( u T, u T ) (cid:17)! · (cid:16) G u T ( 2 πt T ) + G u T ( 2 π ( t + t ) T ) + G u T ( 2 πt T ) (cid:17) + u (cid:16) ǫ − ∂ T g ( u T, u T ) g ( u T, u T ) (cid:17)(cid:17) · (cid:16) G u T ( 2 πt T ) + G u T ( 2 π ( t + t ) T ) + G u T ( 2 πt T ) (cid:17) + (cid:16) ǫ − ∂ T g ( u T, u T ) g ( u T, u T ) (cid:17) ) (3.4)– 7 –nd finally E [ ψ ǫ ( u ,u ) ] = 16 lim ǫ → Z ∞ ds s Z dy Z y dx E (1 − y, x ) g( s, vs ) e − ηs (3.5) · ( G s (2 πx ) G s (2 π ( x − y )) G s (2 πy )+ v G vs (2 πx ) G vs (2 π ( x − y )) G vs (2 πy )+ 12 (cid:16) η − ∂ s g( s, vs )g( s, vs ) (cid:17)(cid:16) G s (2 πx ) + G s (2 π ( x − y )) + G s (2 πy ) (cid:17) + 12 v (cid:16) η − ∂ s g( s, vs )g( s, vs ) (cid:17)(cid:16) G vs (2 πx ) + G vs (2 π ( x − y )) + G vs (2 πy ) (cid:17) + (cid:16) η − ∂ s g( s, vs )g( s, vs ) (cid:17) ) where E (1 − y, x ) = π sin πx sin πy sin π ( x − y ) and η = ǫ u . It is worth remarking that theresult (3.5) does not depend on the splitting ǫ = ǫ + ǫ .The integrand in (3.5) has the same leading singularity in the UV as the integrandof (2.17). The subleading singularity on the other hand may depend on ǫ . Thus it mustundergo an UV subtraction that generically depends on ǫ . We will denote the correspondingsubtracted integral by E ( s ) [ ψ ǫ ( u ,u ) ]. The important remark here is, however, that in thelimit ǫ → e − ηs appearing in the integrand of (3.5) changes completely its IR struc-ture. It is in fact responsible for cutting out the contribution at infinity that characterizes(2.17) and (modulo the arbitrariness in the UV subtraction) makes up the energy of theD23 brane.In keeping with [1], we interpret ψ ǫ ( u ,u ) as a tachyon condensation vacuum solutionand E ( s ) [ ψ ǫ ( u ,u ) ] the energy of such vacuum. This energy is actually v - (and possibly ǫ )-dependent. We will explain later on how it can be set to 0.
4. The energy of the D –brane As explained in [1], the problem of finding the right energy of the D23 brane consists inconstructing a solution over the vacuum represented by ψ ǫ ( u ,u ) (the tachyon condensationvacuum). The equation of motion at such vacuum is Q Φ + ΦΦ = 0 , where Q Φ = Q Φ + ψ ǫ ( u ,u ) Φ + Φ ψ ǫ ( u ,u ) (4.1)One can easily show that Φ = ψ ( u ,u ) − ψ ǫ ( u ,u ) (4.2)is a solution to (4.1). The action at the tachyon vacuum is − hQ Φ , Φ i − h Φ , Φ Φ i . (4.3)– 8 –hus the energy is E [Φ ] = − h Φ , Φ Φ i = − (cid:2) h ψ ( u ,u ) , ψ ( u ,u ) ψ ( u ,u ) i − h ψ ǫ ( u ,u ) , ψ ǫ ( u ,u ) ψ ǫ ( u ,u ) i− h ψ ǫ ( u ,u ) , ψ ( u ,u ) ψ ( u ,u ) i + 3 h ψ ( u ,u ) , ψ ǫ ( u ,u ) ψ ǫ ( u ,u ) i (cid:3) . (4.4)Eq.(4.2) is the lump solution at the tachyon vacuum, therefore this energy must be theenergy of the lump.The two additional terms h ψ ǫ ( u ,u ) , ψ ( u ,u ) ψ ( u ,u ) i and h ψ ( u ,u ) , ψ ǫ ( u ,u ) ψ ǫ ( u ,u ) i aregiven by h ψ ǫ ( u ,u ) , ψ ( u ,u ) ψ ( u ,u ) i = − lim ǫ → Z ∞ ds s Z dy Z y dx e − ηs E (1 − y, x ) e ηsy g( s, vs ) · ((cid:16) η − ∂ s g( s, vs )g( s, vs ) (cid:17)(cid:16) − ∂ s g( s, vs )g( s, vs ) (cid:17) (4.5)+ G s (2 πx ) G s (2 π ( x − y )) G s (2 πy ) + v G vs (2 πx ) G vs (2 π ( x − y )) G vs (2 πy )+ 12 (cid:16) η − ∂ s g( s, vs )g( s, vs ) (cid:17)(cid:16) G s (2 π ( x )) + v G vs (2 π ( x )) (cid:17) + 12 (cid:16) − ∂ s g( s, vs )g( s, vs ) (cid:17)(cid:16) G s (2 πy ) + G s (2 π ( x − y )) + v (cid:16) G vs (2 πy ) + G vs (2 π ( x − y )) (cid:17)(cid:17)) . and h ψ ( u ,u ) , ψ ǫ ( u ,u ) ψ ǫ ( u ,u ) i = − lim ǫ → Z ∞ ds s Z dy Z y dx e − ηs E (1 − y, x ) e ηsx g( s, vs ) · ((cid:16) η − ∂ s g( s, vs )g( s, vs ) (cid:17) (cid:16) − ∂ s g( s, vs )g( s, vs ) (cid:17) (4.6)+ G s (2 πx ) G s (2 π ( x − y )) G s (2 πy ) + v G vs (2 πx ) G vs (2 π ( x − y )) G vs (2 πy )+ 12 (cid:16) η − ∂ s g( s, vs )g( s, vs ) (cid:17)(cid:16) G s (2 πx ) + G s (2 πy ) + v (cid:16) G vs (2 πx ) + G vs (2 πy ) (cid:17)(cid:17) + 12 (cid:16) − ∂ s g( s, vs )g( s, vs ) (cid:17)(cid:16) G s (2 π ( x − y )) + v G vs (2 π ( x − y )) (cid:17)) . Now we insert in (4.4) the quantities we have just computed together with (2.17) and (3.5).We have of course to subtract their UV singularities. As we have already remarked above,such subtractions are the same for all terms in (4.4) in the limit ǫ →
0, therefore theycancel out. So the result we obtain from (4.4) is subtraction-independent and we expect itto be the physical result.In fact the expression we obtain after the insertion of (2.17,3.5,4.5) and (4.6) in (4.4)looks very complicated. But it simplifies drastically in the limit ǫ →
0. As was noticedin [1], in this limit we can drop the factors e ηsx and e ηsy in (4.5) and (4.6) because ofcontinuity . What we cannot drop a priori is the factor e − ηs . It is useful to recall that the limit ǫ → e ηsx or e ηsy is convergent. This is true for the x and y integration, but it is not the casefor instance for the integral (4.9) below. – 9 –ext it is convenient to introduce ˜g( s, vs ) = e − ηs g( s, vs ) and notice that η − ∂ s g( s, vs )g( s, vs ) = − ∂ s ˜g( s, vs )˜g( s, vs ) (4.7)Another useful simplification comes from the fact that (without the e ηsx or e ηsy factors)upon integrating over x, y the three terms proportional to G s (2 πx ), G s (2 πy ) and G s (2 π ( x − y )), respectively, give rise to the same contribution. With this in mind one can easily realizethat most of the terms cancel and what remains is E [Φ ] = 16 lim ǫ → Z ∞ ds s Z dy Z y dx E (1 − y, x ) ( g( s, vs ) (1 − e − ηs ) · " G s (2 πx ) G s (2 π ( x − y )) G s (2 πy ) + v G vs (2 πx ) G vs (2 π ( x − y )) G vs (2 πy )+ 12 (cid:16) − ∂ s g( s, vs )g( s, vs ) (cid:17)(cid:16) G s (2 πx ) + G s (2 π ( x − y )) + G s (2 πy ) (cid:17) + 12 v (cid:16) η − ∂ s g( s, vs )g( s, vs ) (cid:17)(cid:16) G vs (2 πx ) + G vs (2 π ( x − y )) + G vs (2 πy ) (cid:17) + (cid:16) − ∂ s g( s, vs )g( s, vs ) (cid:17) +˜g( s, vs ) (cid:16) ∂ s ˜g( s, vs )˜g( s, vs ) − ∂ s g( s, vs )g( s, vs ) (cid:17) ) (4.8)The term proportional to 1 − e − ηs vanishes in the limit ǫ → E [Φ ] = 16 lim ǫ → Z ∞ ds s Z dy Z y dx E (1 − y, x ) g( s, vs ) e − ηs η = 14 π lim ǫ → Z ∞ ds s g( s, vs ) η e − ηs . (4.9)where g( s, vs ) = g( s )g( vs ). In [1] the analog of this was the coefficient α that determinesthe energy of the solution. This contribution comes from the ǫ term in the last line of(3.4). If one knows the asymptotic expansion of the integrand for large s , it is very easyto extract the exact ǫ → s → ∞ . In fact splitting the s integration as 0 ≤ s ≤ M and M ≤ s < ∞ ,where M is a very large number, it is easy to see the the integration in the first intervalvanishes in the limit ǫ →
0. As for the second integral we have to use the asymptoticexpansion of g( s, vs ): g( s, vs ) ≈ v v s + . . . . Integrating term by term from M to ∞ ,the dominant one gives 14 π e − ηM (2 + 2 M η + M η ) (4.10)– 10 –hich, in the ǫ → π . The other terms are irrelevant in the ǫ → E [Φ ] = 12 π . (4.11)We recall that 1 in the numerator on the RHS is to be identified with lim s →∞ g ( s, vs ).We conclude that T = 12 π (4.12)This is the same as T , so it may at first be surprising. But in fact it is correct becauseof the normalization discussed in App. C of [1]. Compare with eqs.(C.1) and (C.7) there:when we move from a Dp -brane to a D ( p − π (remember that α ′ = 1), but simultaneously we have to divide by 2 π because the volumeis measured with units differing by 2 π (see after eq. (C.6)).In more detail the argument goes as follows (using the notation of Appendix C of[1]). The volume in our normalization is V = 2 π V , where V is the volume in Polchinski’stextbook normalization, [38], see also [39]. The energy functional for the D24 brane isproportional to the 2D zero mode normalization (which determines the normalization ofthe partition function). The latter is proportional to V . Since V = 2 π V , normalizing withrespect to V is equivalent to multiplying the energy by 2 π . This implies that T D = 12 π T D (4.13)where T represents the tension in Polchinski’s units. The energy functional in (2.16) de-pends linearly on the normalization of g( s, vs ), which is the square of the normalizationof g ( s ), so is proportional to V . Therefore the ratio between the energy with the twodifferent zero mode normalizations is (2 π ) . Consequently we have T D = 1(2 π ) T D (4.14)Since, from Polchinski, we have T D = 2 π T D = (2 π ) T D = 2 (4.15)eq.(4.12) follows.We end this section with two comments. The first is about E ( s ) [ ψ ǫu ,u ]. This isinterpreted as the energy at the tachyon condensation vacuum after the UV subtraction,which represents itself the energy of the tachyon condensation vacuum. Therefore it shouldvanish. In fact it does not vanish and its value is ( v, ǫ )-dependent. The reason it does notvanish is that the subtraction itself is ( v, ǫ )-dependent and this is due to the arbitrarinessof the subtraction scheme. However we can always fix E ( s ) [ ψ ǫu ,u ] to zero by subtractinga suitable constant. Of course we have to subtract the same constant from E ( s ) [ ψ u ,u ].The second comment concerns the dependence on ǫ of (4.4). The result we have derivedin this section makes essential use of the limit ǫ →
0, but we believe that it should hold for– 11 –ny ǫ . In [1] it was in fact argued that this should be so, based on the ǫ –independence ofthe UV subtractions. In this paper the UV subtractions are generically ǫ -dependent and wecannot use the same simple argument. However there is no reason to believe that the RHSof (4.4) is ǫ dependent, although it is more complicated to prove it. Such complication hasto do only with the technicalities of the ǫ -regularization. It is possible to envisage otherregularizations in which the UV subtractions are independent of the regulator. We willpursue this point elsewhere.
5. D(25-p) brane solutions
The previous argument about D-brane tensions can be easily continued and we always findthat the value to be expected is T − p = 12 π , ∀ p ≥ φ u = p X i =1 u i (: X i : +2 log u i + 2 A ) (5.2)where X i will represent the transverse direction to the brane and u i the corresponding 2Dcouplings. Since the u i couplings evolve independently and linearly, the partition functionwill be g ( u , . . . , u p ) = g ( u ) g ( u ) . . . g ( u p ).The derivation of the energy of such solutions is a straightforward generalization ofthe one above for the D23-brane and we will not repeat it. The final result for the energyabove the tachyon condensation vacuum is E [Φ ] = 16 lim ǫ → Z ∞ ds s Z dy Z y dx E (1 − y, x ) g( s, v s, . . . , v p − s ) e − ηs η = − π lim ǫ → Z ∞ ds s g( s, v s, . . . , v p − s ) η e − ηs . (5.3)where v = u u , v = u u , . . . . It is understood that the UV singularity has been subtractedaway from the integral in the RHS, therefore the only contribution comes from the regionof large s . Since, again lim s →∞ g( s, v s, . . . , v p − s ) = 1, we find straightaway that E [Φ ] = 12 π . (5.4)from which (5.1) follows. References [1] L. Bonora, S. Giaccari and D. D. Tolla,
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