On D-brane interaction & its related properties
aa r X i v : . [ h e p - t h ] F e b USTC-ICTS-19-11
On D-brane interaction & its related properties
Qiang Jia, J. X. Lu, Zihao Wu and Xiaoying Zhu
Interdisciplinary Center for Theoretical StudyUniversity of Science and Technology of China, Hefei, Anhui 230026, ChinaandPeng Huanwu Center for Fundamental Theory, Hefei, Anhui 230026, China
Abstract
We compute the closed-string cylinder amplitude between one Dp brane andthe other Dp ′ brane, placed parallel at a separation, with each carrying a generalconstant worldvolume flux and with p − p ′ = 0 , , , p ≤
6. For the p = p ′ ,we show that the main part of the amplitude for p = p ′ < p = p ′ = 5 or 6 case. For all other p − p ′ = 2 , , p = p ′ case.Combined both, we obtain the general formula for the amplitude, which is valid foreach of the cases considered and for arbitrary constant worldvolume fluxes. Thecorresponding general open string one-loop annulus amplitude is also obtained bya Jacobi transformation of the general cylinder one. We give also the general openstring pair production rate. We study the properties of the amplitude such as thenature of the interaction, the open string tachyonic instability, and the possibleopen string pair production and its potential enhancement. In particular, in thepresence of pure magnetic fluxes or magnetic-like fluxes, we find that the natureof interaction is correlated with the existence of potential open string tachyonicinstability. When the interaction is attractive, there always exists an open stringtachyonic instability when the brane separation reaches the minimum determinedby the so-called tachyonic shift. When the interaction is repulsive, there is no suchinstability for any brane separation. We also find that the enhancement of openstring pair production, in the presence of pure electric fluxes, can occur only for the p − p ′ = 2 case. Introduction
Computing the interaction amplitude between one Dp and the other Dp ′ , placed parallelat a separation transverse to the Dp brane, with each carrying a general constant world-volume flux (We consider only constant worldvolume flux(es) in this paper) and with p − p ′ = 0 , , , p ≤
6, has its own interest by itself. As we will see, the amplitudeitself exhibits many interesting properties. For example, the contribution from the so-called NS-NS sector or R-R sector has a nice form, determined by the certain propertiesof the worldvolume background fluxes relevant to the amplitude, and can be expressed interms of certain θ -functions and the Dedekind η -function. The total amplitude can alsobe expressed in terms of a certain θ -function, usually the θ -function, using a special formof the more general identity relating various different θ -functions obtained from the con-tributions of the NS-NS and R-R sectors after the so-called Gliozzi-Scherk-Olive (GSO)projection, and the Dedekind η -function, so exhibiting the expected modular property ofthe amplitude.A Dp brane carrying no worldvolume flux is a non-perturbative stable Bogomol’ny-Prasad-Sommereld (BPS) solitonic extended object in superstring theories (for example,see [2]), preserving one half of the spacetime supersymmetries. It has its tension and car-ries the so-called RR charge. When we place two such Dp branes parallel at a separation,the net interaction between the two actually vanishes due to the 1/2 BPS nature of thissystem. We can check this explicitly by computing the lowest order stringy interactionamplitude in terms of the closed string tree-level cylinder diagram. We have here the so-called NS-NS contribution, due to the brane tension, which is attractive, and the so-calledR-R contribution, due to the RR charges, which is repulsive. The BPS nature of eachDp brane identifies its tension with its RR charge in certain units and as such the sumof the two gives an expected zero net interaction by making use of the usual ‘abstruseidentity’ [3]. This same amplitude can also be computed via the so-called open stringone-loop annulus diagram. The same conclusion can be reached. For a system with p − p ′ = κ = 0 , , ,
6, we have as usual NN = p ′ + 1 for which the two ends ofopen string obey the Neumann boundary conditions, ND = κ for which one end of the open string obeysthe Dirichlet boundary conditions while the other the Neumann ones, and DD = 9 − p ′ − κ for which thetwo ends of open string obey the Dirichlet boundary conditions. In general, placing an infinitely extended Dp in spacetime will cause it to curve. For our purpose,we try to avoid this to happen at least to the probe distance in which we are interested. For this, we needto limit our discussion in this paper to p ≤ g s ≪ α ′ / ≫ r ≫ g / (7 − p ) s α ′ / as discussed in section 2 of [1]. ′ with p > p ′ and p ≤ .For the p − p ′ = 2 case, we have an attractive interaction while for the p − p ′ = 6 case wehave a repulsive one. As such, the underlying spacetime supersymmetries are all broken.However, for the p − p ′ = 4 case, the net interaction vanishes and the underlying systemis still BPS, preserving 1/4 of spacetime supersymmetries. Each of these, regarding thesupersymmetry breaking or preservation, can be checked explicitly following [5, 6], forexample.When the brane worldvolume fluxes are present, we now expect in general a non-vanishing interaction. Except for the ( p = 6 , p ′ = 0) case mentioned above and the( p = 6 , p ′ ≤
6) cases to be considered later in this paper, the long-range interactionbetween the Dp and Dp ′ for other cases is in general attractive when the electric and/ormagnetic fluxes on the Dp ′ are parallel to the corresponding ones on the Dp, respectively.The reason for this is simple since only different constituent branes contribute to thislong-range interaction and each contribution is from the respective NS-NS sector and isattractive. For example, if we have both electric and magnetic fluxes present on Dp andDp ′ , the F-strings (see footnote (4)) within Dp ′ have no interaction with their parallelF-strings within Dp but have a long-range attractive interaction with D(p - 2) branes(see footnote (4)) within Dp. However, as indicated above for p = 6 and p ′ ≤
6, thelong-range interaction can be repulsive in the presence of certain types of fluxes. This hasbeen demonstrated in the simplest possible cases in [18] when p − p ′ = 2. We will spellout the condition in general for this to be true later in this paper.For certain type of fluxes (to be specified later on), the nature of the interaction atsmall brane separation (attractive or repulsive) remains unclear if it is computed in termsof the closed string tree-level cylinder amplitude. In general, this implies new physics to For the p − p ′ = 2 case, the long-range interaction is attractive since the contribution from either theexchange of massless dilaton or the exchange of massless graviton is attractive while for the p − p ′ = 6case the long-range interaction is repulsive since the contribution from the exchange of massless dilatonis repulsive and exceeds the attractive contribution from the exchange of massless graviton. However,for the p − p ′ = 4 case, the repulsive contribution from the exchange of massless dilaton just cancels theattractive one from the exchange of massless graviton and this gives a net vanishing interaction. Eachof these can be checked explicitly, for example, see [4]. Each of these remains to be true for any braneseparation as we will show later in this paper. The electric flux on a Dp-brane stands for the presence of F-strings, forming the so-called (F, Dp)non-threshold bound state [7–14], while a magnetic flux stands for that of co-dimension 2 D-branes insidethe original Dp brane, forming the so-called (D(p-2), Dp) non-threshold bound state [15–17], from thespacetime perspective. These fluxes are in general quantized. We will not discuss their quantizations inthe text for simplicity due to their irrelevance for the purpose of this paper. p − p ′ = 2),we find that there is a correlation between the nature of the interaction between the Dp andthe Dp ′ and the potential open string tachyonic instability. If the interaction is attractive,the open string connecting the two D-branes has a tachyonic shift to its spectrum [19, 20].We have then the onset of tachyonic instability when the brane separation reaches theminimum determined by the shift. Once this instability develops, the attractive braneinteraction diverges. We have then the so-called tachyon condensation and as such aphase transition occurs, releasing the excess energy of this system. For example, for p = p ′ , this process restores not only the gauge symmetry from U (1) × U (1) → U (2) butalso the supersymmetry from none to half of the spacetime supersymmetries [21]. In theso-called weak field limit, the corresponding instability is just the analog of the Nielsen-Olesen one [22] of non-abelian gauge theories such as the electroweak one and the gaugesymmetry restoration was considered in [23, 24]. On the other hand, if the interaction isrepulsive, we don’t have this tachyonic shift and therefore have no tachyonic instabilityto begin with.When we have only worldvolume electric fluxes present, the underlying system is ingeneral no longer 1/2 BPS and breaks all its supersymmetries, therefore unstable. Thismanifests itself by the appearance of an infinite number of simple poles in the integrandof the integral representation of the open string one-loop annulus amplitude, implyingthat the interaction amplitude has an imaginary part. Each of these simple poles actuallyindicates the corresponding open string pair production under the action of the appliedelectric fluxes [21,25,26]. The imaginary amplitude just reflects the decay of the underlyingsystem via the so-called open string pair production, releasing the excess energy of thesystem until it reaches the corresponding 1/2 BPS stable one. This is the analog ofthe Schwinger pair production in quantum electrodynamics (QED) [27]. For unorientedbosonic string and type I superstring, this was pursued a while ago in [28, 29]. When theapplied electric flux reaches its so-called critical field determined by the fundamental stringtension, the open string pairs are produced cascadingly and there is also an instabilitydeveloped.When both electric and magnetic fluxes are present in a certain way, the open stringpair production has an enhancement, uncovered recently in [6, 18, 21, 26, 30, 31].As explained in [21, 31], there is no open string pair production for a single Dp branein Type II string theories even if we apply an electric flux on the brane, unless it reaches4ts critical value . The simple reason for this is that each of the virtual open strings inthe pairs from the vacuum has its two ends carrying charge + and − , respectively, givinga net zero-charge of the open string, called the charge-neutral open string, and the netforce acting on the string vanishes under the applied constant electric field. So the electricfield can only stretch each open string but cannot separate the virtual open string andthe virtual anti-open string in each pair to give rise to the pair production. This can alsobe explained by the fact that a Dp brane carrying a constant worldvolume electric flux isactually a 1/2 BPS non-threshold bound state [11, 14], therefore it is stable and cannotdecay via the open string pair production.In order to produce the pair production in Type II string theories, a possible choice isto let the two ends of the open string experience different electric fields since the charge-neutral nature of the open strings cannot be altered. The above mentioned two Dp-branesystem is probably the simplest one for this purpose. We compute this pair productionrate [6, 25, 26, 30] and find it indeed non-vanishing. However, for any realistic electricfield applied, the rate is in general vanishingly small and it has no practical use. Butwhen an additional magnetic flux is added in a certain way, the rate can be enhancedenormously [6, 30] and the largest possible rate is for the system of two D3 branes whenthe electric and magnetic fields are collinear [30, 31]. This enhanced pair production mayhave the potential to be detected [21].For this to occur in practice, we need to assume one of the D3 branes to be relevant toour 4-dimensional world and the other D3 to be invisible (hidden or dark) to us. For thissimple system, it appears that there is a possibility for the detection of the pair productionbut there is an issue if one carefully examines the underlying physics as discussed in detailin [21]. The mass spectrum of the open string connecting the two D3 at a separation y has a mass shift m = y/ (2 πα ′ ) at each mass-level. That the corresponding modes ateach given mass-level all have this same shift is due to the underlying supersymmetryin the absence of worldvolume fluxes. For example, the lowest-mass eight bosons andeight fermions all have the same mass y/ (2 πα ′ ) which becomes massless at y = 0 andthe underlying system is 1/2 BPS in the absence of worldvolume fluxes. In general, thelaboratory electric and magnetic fields are much smaller than the string scale and theweak field limit holds. So the contribution to the pair production is due to the above 8bosonic (8 B ) and 8 fermionic (8 F ) lowest-mass charged modes of the open string. Fromthe worldvolume viewpoint, these 8 B + 8 F massive charged modes, in the absence of When the applied electric field reaches its critical value, it will break the virtual open strings fromthe vacuum and the pair production is due to the breaking of each of these open strings, not to theseparation of the pair of the virtual open string and the virtual anti open string under the electric field. U (2) → U (1) × U (1) with oneof scalars taking its expectation value ∼ y but the underlying 16 supersymmetries remainintact, giving rise to the 4-dimensional N = 4 massive gauge theory with one massivecharged vector (W-boson), 5 massive charged scalars and their corresponding fermionicsuper partners, all with mass m = y/ (2 πα ′ ). In the presence of worldvolume fluxes andif the brane interaction is non-vanishing, the underlying supersymmetries are all broken.The presence of practical magnetic fluxes can also give a tiny mass shift to the massivecharged vector [20].We therefore naturally expect the mass scale m = y/ (2 πα ′ ) no less than a few TeV(since no supersymmetry has been found in LHC yet) and this requires the electric andmagnetic fields 21 orders of magnitude larger than the current laboratory limit to makethe detection possible. The other choice is to take the other D3 as a dark one and forthis, we don’t have a priori knowledge of the mass scale m . If it happens to be no largerthan the electron mass, we may have an opportunity to detect the open string rate if theQED Schwinger pair production becomes feasible. Even so, we still have to explain whythe other charged fermions other than the one identified with the electron, the chargedscalars and the charged vector, all having the same mass m ∼ m e = 0 .
51 MeV, are notthe Standard model particles.All these issues, one way or the other, are due to that the 16 (8 B + 8 F ) relevant modesall have the same mass m = y/ (2 πα ′ ) from the underlying supersymmetry. In addition,the currently available laboratory electric and magnetic fields are too small. The naturalquestion is: does there exist a possibility that we can get around these issues in practice?A while ago, one of the present authors along with his collaborator considered a systemof one Dp and one Dp ′ , placed parallel at a separation transverse to the Dp brane, with p − p ′ = 2 and with each carrying only one flux [18], and found that whenever there isan electric flux present along the NN-directions, there is an open string pair productionenhancement even in the absence of a magnetic flux. The novel feature found in [18] isthat the Dp ′ -brane plays effectively as a magnetic flux of stringy order (see footnote (4)).In other words, if our D3 brane has a nearby D-string, for example as a cosmic string, thisD-string appears effectively as a stringy magnetic field. This field can give rise to the pairproduction enhancement, which can hardly possible with a laboratory magnetic field, ifour D3 carries an electric flux along the D-string direction. In addition, the underlyingsystem breaks all supersymmetries intrinsically. So this consideration may provide asolution to the above question raised. This is the other line motivating us to consider thebrane interaction in general between one Dp and the other Dp ′ with p − p ′ = k = 0 , , , ′ , placed parallel at a separation transverseto the Dp, with each carrying a general worldvolume flux and with p − p ′ = 0 , , , p ≤
6. We will show that the key part of the amplitude in terms of the θ -functions andthe Dedekind η -function for each of the p = p ′ < p = p ′ = 5 or 6 case. We further demonstrate that the amplitude for p − p ′ = 2 , , p = p ′ case by choosing specific magneticfluxes along the 2, 4, 6 ND-directions, a trick greatly simplifying the computations . Wecompute first the closed-string cylinder amplitude using the closed string boundary staterepresentation of D-brane [32–36], which has the advantage of holding true for a generalworldvolume constant flux [14]. By a Jacobi transformation of this, we can obtain thecorresponding open-string annulus amplitude. We will also compute the open string pairproduction rate if any and discuss the relevant analytical structures of the amplitude. Wewill explore the nature (attractive or repulsive) of the interaction at large brane separationand small brane separation, respectively, and study various instabilities such as the onsetof tachyonic one at small brane separation. In particular, we find that there is a correlationbetween the nature of interaction being attractive and the existence of tachyonic shift,which can give rise to the onset of tachyonic instability when the brane separation reachesthe minimum determined by the tachyonic shift. We will determine at which conditionsthere exists the open string pair production and its possible enhancement. We will alsospeculate possible applications of the enhanced open string pair production for practicalpurpose.Before we move to discuss the plan of this paper, we would like to point out a fewthings which are relevant and will be mentioned only briefly here due to the scope ofthis paper . Note that the added fluxes on either Dp or Dp ′ , considered in this paper,are general, with each electric one standing for F-strings [7–14] and each magnetic onestanding for co-dimension 2 D-branes [15–17] inside the Dp or Dp ′ , so these fluxes representdifferent intersecting D-branes and/or F-strings. We know that the magnetized D-branesare related to the corresponding intersecting D-branes via T-dualities [15, 36–40] (see [41]for a more complete list of references, in particular for phenomenological applications).We also know that the electrified D-branes are related to moving D-branes again viaT-dualities [36, 42]. We limit ourselves in this paper to consider the Dp and the Dp ′ to We will also provide the physical rationale for this. This paragraph is added to address the comments/questions by one of the referees. We thank theanonymous referee for these which help to clarify certain issues not addressed in the early version of thispaper.
7e placed parallel, not oblique, at a separation only for simplicity and for the purposeof seeking the open string pair production enhancement. The extension of the presentconsideration to the oblique Dp and Dp ′ may provide more feasibilities in giving rise tothe open string pair production in practice. For example, we can use the intersectingbranes to give a lower string scale and/or to give light stringy states [43]. In this paperwe only use the so-called no-force condition to discuss whether the underlying systempreserves certain supersymmetry but this can also analyzed in detail following that forintersecting branes [5, 15, 37–40, 44–46] (see [41] for a rather complete list of referenceson phenomenological applications to supersymmetry breaking for intersecting branes).One of primary purposes of this paper is to seek a D-brane system containing a D3 asour own world which can give rise to the earthbound laboratory testable open stringpair production without the need of string compactifications. We also know that withstring compactfications the intersecting D-branes in Type II string theory, for examplethe intersecting D6 branes, provide phenomenologically interesting models [41, 47–49]. Inthis paper, we consider only the system of a single Dp and a single Dp ′ with each carryinga general flux. The extension of this to N and N ′ branes with each stack carrying therespective overall U(1) flux in a similar setting looks straightforward. For example, thetwo corresponding amplitudes differ by an overall factor N N ′ . We can understand this bynoting that the closed string cylinder or the open string annulus interaction between eachgiven brane in the first stack and that in the second stack is the same as we compute inthis paper and total counting factor is N N ′ . Also note that the closed string cylinder orthe open string annulus amplitude is the lowest order stringy interaction for the systemconsidered. This multiplicity factor N N ′ can also be understood from the long-rangeinteraction between the two stacks computed from the effective field theory, for example,see [4].The paper is organized as follows. In section 2, we give a brief review of the closed-string boundary state representation of Dp-brane carrying a general constant worldvolumeflux and set up conventions for latter sections. We give also a general discussion oncomputing the closed-string cylinder amplitude between a Dp and a Dp ′ , placed parallelat a separation transverse to the Dp, with each carrying a general constant worldvolumeflux and with p − p ′ = 0 , , , p ≤
6. In section 3, we first compute the closed-stringcylinder amplitude for each of the 0 ≤ p = p ′ ≤ θ -functionsand the Dedekind η -function. We study the nature of interaction and find that therepulsive interaction can only be possible for p = p ′ = 6 and for certain purely magnetic8uxes present. For all other cases, the long-range interaction is attractive. We also findthe correlation between the nature of interaction being attractive and the existence oftachyonic shift, which will give rise to the onset of tachyonic instability when the braneseparation reaches the minimum determined by the tachyonic shift. We compute thedecay rate of the underlying system and the corresponding open string pair productionrate when they exist and discuss their potential enhancement. In section 4, we move tocompute the amplitude for each of p − p ′ = 2 , , p ≤
6, respectively, using the known p = p ′ one with a specific choice of the magnetic fluxes on the Dp ′ , along now the 2, 4,6 ND-directions. Basically, the amplitude for each p − p ′ = 2 , , p ≤ p = p ′ one computed in section 3 by a special choiceof magnetic fluxes along the 2, 4, 6 ND-directions on the Dp ′ brane. We also provide theunderlying physical reason for this. Similar properties of the amplitude as discussed insection 3 are also given. We discuss and conclude this paper in section 5. In this section, we give a brief review of Dp brane boundary state carrying a generalconstant worldvolume flux, following [36]. We also give a general discussion in computingthe closed-string cylinder amplitude between a Dp brane and a Dp ′ brane, placed parallelat a separation transverse to the Dp, with each carrying a general constant worldvolumeflux and with p − p ′ = 0 , , , p ≤ | B, η i , with η = ± . However, only the combinations | B i NS = 12 [ | B, + i NS − | B, −i NS ] , | B i R = 12 [ | B, + i R + | B, −i R ] , (1)are selected by the Gliozzi-Scherk-Olive (GSO) projection in the NS-NS and R-R sectors,respectively. The boundary state | B, η i for a Dp-brane can be expressed as the productof a matter part and a ghost part [34–36], i.e. | B, η i = c p | B mat , η i| B g , η i , (2)where | B mat , η i = | B X i| B ψ , η i , | B g , η i = | B gh i| B sgh , η i (3)9nd the overall normalization c p = √ π (cid:0) π √ α ′ (cid:1) − p . As discussed in [14, 36], the operator structure of the boundary state holds true evenwith a general constant worldvolume flux and is always of the form | B X i = exp( − ∞ X n =1 n α − n · M · ˜ α − n ) | B X i , (4)and | B ψ , η i NS = − i exp( iη ∞ X m =1 / ψ − m · M · ˜ ψ − m ) | i (5)for the NS-NS sector and | B ψ , η i R = − exp( iη ∞ X m =1 ψ − m · M · ˜ ψ − m ) | Bη i (6)for the R-R sector. The ghost boundary states are the standard ones as given in [34],independent of the fluxes, which we will not present here. The M-matrix , the zero-modes | B X i and | B, η i encode all information about the overlap equations that thestring coordinates have to satisfy. They can be determined respectively [14, 32, 36] as M = ([( η − ˆ F )( η + ˆ F ) − ] αβ , − δ ij ) , (7) | B X i = [ − det( η + ˆ F )] / δ − p ( q i − y i ) Y µ =0 | k µ = 0 i , (8)for the bosonic sector and | B ψ , η i = ( C Γ Γ · · · Γ p η Γ η U ) AB | A i| ˜ B i , (9)for the R-R sector. In the above, the Greek indices α, β, · · · label the world-volumedirections 0 , , · · · , p along which the Dp brane extends, while the Latin indices i, j, · · · label the directions transverse to the brane, i.e., p + 1 , · · · ,
9. We define ˆ F = 2 πα ′ F with F the external worldvolume field. We also have denoted by y i the positions of the D-branealong the transverse directions, by C the charge conjugation matrix and by U the matrix U ( ˆ F ) = 1 q − det( η + ˆ F ) ; exp (cid:18) −
12 ˆ F αβ Γ α Γ β (cid:19) ; (10) We have changed the previously often used symbol S to the current M to avoid a possible confusionwith the S-matrix in scattering amplitude. | A i| ˜ B i stands for the spinor vacuum of the R-Rsector. Note that the η in the above denotes either sign ± or the worldvolume Minkowskiflat metric and should be clear from the content.We now come to compute the closed-string tree-level cylinder amplitude between aDp and a Dp ′ as stated earlier viaΓ = h B p ′ ( ˆ F ′ ) | D | B p ( ˆ F ) i , (11)where D is the closed string propagator defined as D = α ′ π Z | z |≤ d z | z | z L ¯ z ˜ L . (12)Here L and ˜ L are the respective left and right mover total zero-mode Virasoro generatorsof matter fields, ghosts and superghosts. For example, L = L X + L ψ + L gh0 + L sgh0 where L X , L ψ , L gh0 and L sgh0 are the respective ones from matter fields X µ , matter fields ψ µ ,ghosts b and c , and superghosts β and γ , and their explicit expressions can be found inany standard discussion of superstring theories, for example in [35], therefore will not bepresented here. The above total amplitude has contributions from both NS-NS and R-Rsectors, respectively, and can be written as Γ p,p ′ = Γ NSNS + Γ RR . In calculating eitherΓ NSNS or Γ RR , we need to keep in mind that the boundary state used should be the GSOprojected one as given in (1).For this, we need to calculate first the amplitude Γ( η ′ , η ) = h B ′ , η ′ | D | B, η i in eachsector with η ′ η = + or − , B ′ = B p ′ ( ˆ F ′ ) and B = B p ( ˆ F ). Actually, Γ( η ′ , η ) dependsonly on the product of η ′ and η , i.e., Γ( η ′ , η ) = Γ( η ′ η ). In the NS-NS sector, this givesΓ NSNS ( ± ) ≡ Γ( η ′ , η ) when η ′ η = ± , respectively. Similarly we have Γ RR ( ± ) ≡ Γ( η ′ , η )when η ′ η = ± in the R-R sector. We then haveΓ NSNS = 12 [Γ
NSNS (+) − Γ NSNS ( − )] , Γ RR = 12 [Γ RR (+) + Γ RR ( − )] . (13)Given the structure of the boundary state, the amplitude Γ( η ′ η ) can be factorized asΓ( η ′ η ) = c p ′ c p α ′ π Z | z |≤ d z | z | A X A bc A ψ ( η ′ η ) A βγ ( η ′ η ) . (14)In the above, we have A X = h B ′ X || z | L X | B X i , A ψ ( η ′ η ) = h B ′ ψ , η ′ || z | L ψ | B ψ , η i ,A bc = h B gh || z | L gh0 | B gh i , A βγ ( η ′ η ) = h B sgh , η ′ || z | L sgh0 | B sgh , η i . (15)11he above ghost and superghost matrix elements A bc and A βγ ( η ′ η ), both independent ofthe fluxes and the dimensionalities of the branes involved, can be calculated to give, A bc = | z | − ∞ Y n =1 (cid:0) − | z | n (cid:1) , (16)and in the NS-NS sector A βγ NSNS ( η ′ η ) = | z | ∞ Y n =1 (cid:0) η ′ η | z | n − (cid:1) − , (17)while in the R-R sector A βγ RR ( η ′ η ) = R0 h B sgh , η ′ | B sgh , η i | z | ∞ Y n =1 (cid:0) η ′ η | z | n (cid:1) − , (18)where R0 h B sgh , η ′ | B sgh , η i denotes the superghost zero-mode contribution which requiresa regularization along with the zero-mode contribution of matter field ψ in this sector.We will discuss this regularization later on.With the above preparation, we are ready to compute the closed string tree-levelcylinder amplitudes for the systems under consideration. We first compute the closed-string tree-level cylinder amplitude for the case of p = p ′ . This serves as the basis forcomputing the amplitude for each of the p = p ′ cases. The general steps follow those givenin section 2 of [26] but with a few refinements. Once the closed string tree-level cylinderamplitude is obtained, we use a Jacobi transformation to obtain the corresponding openstring one-loop annulus amplitude. We will have these done in the following two sectionsone by one. We also discuss the properties of the respective amplitude in each case. p = p ′ case As indicated already in the previous section, the computations of the amplitude boildown to computing the matrix elements of matter part, i.e, A X and A ψ given in (15).For this, the following property of the matrix M given in (7) can be used to simplify theircomputations greatly, M µ ρ ( M T ) ρ ν = ( M T ) µ ρ M ρ ν = δ µ ν , (19)where T denotes the transpose of matrix. Following [26], for a system of two Dp branes,placed parallel at a separation y , with one carrying flux ˆ F ′ and the other carrying flux ˆ F ,we can then have, A X = V p +1 h det( η + ˆ F ′ ) det( η + ˆ F ) i (2 π α ′ t ) − p e − y πα ′ t ∞ Y n =1 (cid:18) − | z | n (cid:19) − p p Y α =0 − λ α | z | n , (20)12nd in the NS-NS sector A ψ NSNS ( η ′ η ) = ∞ Y n =1 (cid:0) η ′ η | z | n − (cid:1) − p p Y α =0 (cid:0) η ′ ηλ α | z | n − (cid:1) , (21)while in the R-R sector A ψ RR ( η ′ η ) = R h B ′ ψ , η ′ | B ψ , η i | z | ∞ Y n =1 (cid:0) η ′ η | z | n (cid:1) − p p Y α =0 (cid:0) η ′ ηλ α | z | n (cid:1) , (22)where R h B ′ ψ , η ′ | B ψ , η i denotes the zero-mode contribution in this sector which, whencombined with the zero-mode contribution from the superghost, needs a regularizationmentioned earlier. We will present the result of this regularization later on. In the above, | z | = e − πt , V p +1 denotes the volume of the Dp brane worldvolume, λ α are the eigenvaluesof the matrix w (1+ p ) × (1+ p ) defined in the following W = M M ′ T = w (1+ p ) × (1+ p ) I (9 − p ) × (9 − p ) ! , (23)where the matrix M or M ′ is the one given in (7) when the corresponding flux is ˆ F orˆ F ′ , respectively, and I stands for the unit matrix. We can also express the matrix w interms of matrix s and s ′ as w α β = ( ss ′ T ) α β = h ( I − ˆ F )( I + ˆ F ) − ( I + ˆ F ′ )( I − ˆ F ′ ) − i α β , (24)where s α β = [( I − ˆ F )( I + ˆ F ) − ] α β , (25)and similarly for s ′ but with ˆ F replaced by ˆ F ′ . Note that the two factors ( I − ˆ F ) and( I + ˆ F ) − in s are inter-changeable and this remains also true for the s ′ . In the above ‘ I ’stands for the (1 + p ) × (1 + p ) unit matrix. For matrix s , we have s α γ ( s T ) γ β = δ α β . Thisholds also for the matrix s ′ and matrix w . The above orthogonal matrix W , satisfying W W T = W T W = I × , (26)can be obtained from a redefinition of the certain oscillator modes, say ˜ a nν , which is atrick used in simplifying the evaluation of the matrix elements of matter part from thecontribution of oscillator modes. Let us take the following as a simple illustration forobtaining the matrix W . In obtaining A X , we need to evaluate, for given n >
0, thefollowing matrix element, h | e − n α µn ( M ′ ) µ ν ˜ α nν | z | α τ − n α nτ e − n α ρ − n ( M ) ρ σ ˜ α − nσ | i = h | e − n α µn ( M ′ ) µ ν ˜ α nν e − | z | nn α ρ − n ( M ) ρ σ ˜ α − nσ | i , (27)13here | i stands for the vacuum. Purely for simplifying the evaluation of the matrixelement on the right of the above equality, we first define ˜ α ′ µ = ( M ′ ) µ ρ ˜ α ρ where we haveomitted the index n since this works for both n > n <
0, due to the matrix M ′ being real. Note that the commutation relation [ ˜ α ′ n µ , ˜ α ′ m ν ] = η µν δ n + m, continues to hold,using the property of matrix M ′ as given in (19). With this property of matrix M ′ , wecan have ˜ α µ = ( M ′ T ) µ ν ˜ α ′ ν . Substituting this into (27) for n < α ′ , we have (27) as h | e − n ˜ α µn α nµ e − | z | nn α ρ − n W ρ σ ˜ α − nσ | i , (28)where W is precisely the one given in (23). Since W is an unit matrix in the absence offluxes, we expect that it can be diagonalized with the deformation of adding fluxes usingthe following non-singular matrix V , V = v (1+ p ) × (1+ p ) I (9 − p ) × (9 − p ) ! , (29)such that W = V W V − . (30)In the above, W = λ λ . . . λ p I (9 − p ) × (9 − p ) , (31)and v is a (1 + p ) × (1 + p ) non-singular matrix. We prove (29), (30) and (31) to holdtrue in general in Appendix A. We now further define , for n > α ′ nµ = ( V − ) µ ν α nν and α ′ µ − n = α ν − n V ν µ , and ˜ α ′− nµ = ( V − ) µ ν ˜ α − nν and ˜ α ′ µn = ˜ α νn V ν µ . Note that ˜ α ′ µn α ′ nµ = ˜ α µn α nµ .The matrix element (28) becomes h | e − n ˜ α ′ µn α ′ nµ e − | z | nn λ ρ α ′ ρ − n ˜ α ′− nρ | i . (32)We have now the commutator relations [ α ′ nµ , α ′ ν − m ] = nδ νµ δ n,m and [ ˜ α ′ µn , ˜ α ′− mν ] = nδ µν δ n,m when n, m >
0. We still have α ′ nµ | i = ˜ α ′ µn | i = 0 and h | α ′ µ − n = h | ˜ α ′− nµ = 0. The This purely serves the purpose of simplifying the evaluation of the matrix element (28). For this, wekeep the annihilation operator α ′ nµ with a lower Lorentz index µ while the creation operator α ′ ν − n withan upper Lorentz index ν . It will be opposite for the corresponding oscillators with tilde. p + 1 eigenvalues λ α with α = 0 , · · · p are not all independent and can actually bedetermined by the given worldvolume fluxes. First from the given property of w , we havedet w = 1, which gives p Y α =0 λ α = 1 . (33)The eigenvalue λ satisfies the following equationdet (cid:0) λ δ α β − w α β (cid:1) = 0 , (34)as well as the equation det (cid:0) λ − δ α β − ( w − ) α β (cid:1) = 0 . (35)The last one can also be written asdet (cid:0) λ − δ α β − w α β (cid:1) = 0 , (36)where we have used ( w − ) α β = ( w T ) α β = η ββ ′ w β ′ α ′ η αα ′ . In other words, for everyeigenvalue λ of w , its inverse λ − is also an eigenvalue. So the p + 1 eigenvalues λ α arepairwise. When p = even, this must imply that one of the eigenvalues is unity. Given thisproperty of λ α , the equation (33) satisfies automatically. For convenience, we now relabelthe eigenvalues pairwise as λ α and λ − α with α = 0 , · · · [( p − /
2] and keep in mind thatthere is one additional unity eigenvalue, i.e., λ = 1, when p = even. Here [( p − / p − /
2. For example, for p = 6, it gives aninteger 2.For a general p ≤
6, we need at most the following three equations to determine thecorresponding eigenvalues λ α , λ − α with α = 0 , , · · · , [( p − /
2] plus λ = 1 if p = even.For p = even, we have λ = 1 and1 + [ p − ] X α =0 (cid:0) λ α + λ − α (cid:1) = tr w, p − ] X α =0 (cid:0) λ α + λ − α (cid:1) = tr w , p − ] X α =0 (cid:0) λ α + λ − α (cid:1) = tr w , (37)while for p = odd, we have instead[ p − ] X α =0 (cid:0) λ α + λ − α (cid:1) = tr w, [ p − ] X α =0 (cid:0) λ α + λ − α (cid:1) = tr w , [ p − ] X α =0 (cid:0) λ α + λ − α (cid:1) = tr w . (38)In the above, the w is given in (24) in terms of fluxes ˆ F and ˆ F ′ . Concretely, we list therespective equations needed to determine the corresponding eigenvalues in Table 1 for15 Equation(s) for eigenvalue(s)0 λ = 11 λ + λ − = tr w λ + λ − = tr w − λ = 13 P α =0 ( λ α + λ − α ) = tr w, P α =0 ( λ α + λ − α ) = tr w P α =0 ( λ α + λ − α ) = tr w − , P α =0 ( λ α + λ − α ) = tr w − , λ = 15 P α =0 ( λ α + λ − α ) = tr w, P α =0 ( λ α + λ − α ) = tr w , P α =0 ( λ α + λ − α ) = tr w P α =0 ( λ α + λ − α ) = tr w − , P α =0 ( λ α + λ − α ) = tr w − , P α =0 ( λ α + λ − α ) = tr w − λ = 1Table 1: The equations needed to determine the corresponding eigenvalues for p ≤ p ≤
6. We actually don’t need to solve the eigenvalues from the equations given in Table1 for the matrix elements given in (20), (21) and (22), respectively, for each case. Let ususe one particular example for p = 3 to illustrate this. For example, the following productin (20) can be expressed in terms of tr w, (tr w ) and tr w as (cid:0) − λ | z | n (cid:1) (cid:0) − λ − | z | n (cid:1) (cid:0) − λ | z | n (cid:1) (cid:0) − λ − | z | n (cid:1) = 1 − X α =0 (cid:0) λ α + λ − α (cid:1) | z | n + (cid:2) (cid:0) λ + λ − (cid:1) (cid:0) λ + λ − (cid:1)(cid:3) | z | n − X α =0 (cid:0) λ α + λ − α (cid:1) | z | n + | z | n = 1 − tr w | z | n + 12 (cid:2) (tr w ) − tr w (cid:3) | z | n − tr w | z | n + | z | n . (39)We are now ready to express the amplitude (14) given in the previous section in the NS-NS or R-R sector in a more compact form. For the NS-NS sector, using (16), (17), (20)and (21) for the contributions from the ghost bc , superghost βγ , the matter X and ψ ,respectively, we have the NSNS-amplitude asΓ NSNS ( η ′ η ) = V p +1 q det( η + ˆ F ′ ) det( η + ˆ F )(8 π α ′ ) p Z ∞ dtt − p e − y πα ′ t | z | − × ∞ Y n =1 (cid:18) η ′ η | z | n − − | z | n (cid:19) δ p, even − p [ p − ] Y α =0 (1 + η ′ ηλ α | z | n − ) (1 + η ′ ηλ − α | z | n − )(1 − λ α | z | n ) (1 − λ − α | z | n ) , (40)16nd similarly for the R-R sector, using (16), (18), (20) and (22), we haveΓ RR ( η ′ η ) = V p +1 q det( η + ˆ F ′ ) det( η + ˆ F )(8 π α ′ ) p h B ′ , η ′ | B, η i Z ∞ dtt − p e − y πα ′ t × ∞ Y n =1 (cid:18) η ′ η | z | n − | z | n (cid:19) δ p, even − p [ p − ] Y α =0 (1 + η ′ ηλ α | z | n ) (1 + η ′ ηλ − α | z | n )(1 − λ α | z | n ) (1 − λ − α | z | n ) . (41)In obtaining the above, we have used the following relations c p π (2 π α ′ ) − p = 1(8 π α ′ ) p , Z | z |≤ d z | z | = 2 π Z ∞ dt, (42)where the explicit expression for c p as given right after (3) has been used and | z | = e − πt as given earlier. The above zero-mode contribution h B ′ , η ′ | B, η i ≡ h B sgh , η ′ | B sgh , η i × h B ′ ψ , η ′ | B ψ , η i (43)can be computed for the general fluxes ˆ F and ˆ F ′ , using the expression for the R-R sectorzero-mode (9) along with (10) and following the regularization scheme given in [34, 50],as h B ′ , η ′ | B, η i = − δ ηη ′ , + q det( η + ˆ F ) det( η + ˆ F ′ ) × [ p +12 ] X n =0 (cid:18) n n ! (cid:19) (2 n )! ˆ F [ α β ··· ˆ F α n β n ] ˆ F ′ [ α β ··· ˆ F ′ α n β n ] . (44)From (40) and (41) as well as (13), we have the GSO projected NSNS-amplitude,Γ NSNS = 12 [Γ
NSNS (+) − Γ NSNS ( − )]= V p +1 q det( η + ˆ F ′ ) det( η + ˆ F )2 (8 π α ′ ) p Z ∞ dtt − p e − y πα ′ t | z | − " ∞ Y n =1 A n (+) − ∞ Y n =1 A n ( − ) , (45)and the GSO projected RR-amplitudeΓ RR = 12 [Γ RR (+) + Γ RR ( − )]= V p +1 q det( η + ˆ F ′ ) det( η + ˆ F )2 (8 π α ′ ) p h B ′ , η ′ | B, η i Z ∞ dtt − p e − y πα ′ t ∞ Y n =1 B n (+) , (46)17here A n ( ± ) = (cid:18) ± | z | n − − | z | n (cid:19) δ p, even − p [ p − ] Y α =0 (1 ± λ α | z | n − ) (1 ± λ − α | z | n − )(1 − λ α | z | n ) (1 − λ − α | z | n ) ,B n (+) = (cid:18) | z | n − | z | n (cid:19) δ p, even − p [ p − ] Y α =0 (1 + λ α | z | n ) (1 + λ − α | z | n )(1 − λ α | z | n ) (1 − λ − α | z | n ) . (47)In obtaining (46), we have used the property of the zero-mode (44) which has the onlycontribution from ηη ′ = +.To proceed, we express the eigenvalues λ α = e πi ν α with α = 0 , · · · [( p − / ν α takes either real or purely imaginary value. In the former case, it looks that we couldtake ν α ∈ [0 , /
2] since only the combination of λ α + λ − α = 2 cos 2 πν α appears in theamplitude and ν α = 0 corresponds to the absence of fluxes. However, taking accountof the zero-mode contribution in the R-R sector as discussed in Appendix B (see (50)given below), we still need to take ν α ∈ [0 , ν α ’s can take imaginary value (see Appendix A) and we can choose thisparticular ν = ν = i ¯ ν with ¯ ν ∈ (0 , ∞ ) since λ + λ − = 2 cosh 2 π ¯ ν . This is actuallythe consequence of the matrix w (24) being a general Lorentz transformation. It happenswhenever the applied electric fluxes cannot be eliminated by a Lorentz transformation.We can now express the NSNS-amplitude (45) in terms of the θ -functions and theDedekind η -function as (see, for example, [51] for their definitions)Γ NSNS = 2[ p +12 ] V p +1 q det( η + ˆ F ′ ) det( η + ˆ F )2 (8 π α ′ ) p Z ∞ dtt − p e − y πα ′ t [ η ( it )] − [ p +12 ] [ p − ] Y α =0 sin πν α θ ( ν α | it ) × [ θ (0 | it )] − [ p +12 ] [ p − ] Y α =0 θ ( ν α | it ) − [ θ (0 | it )] − [ p +12 ] [ p − ] Y α =0 θ ( ν α | it ) . (48)Similarly for the RR-amplitude (46), we haveΓ RR = 2[ p +12 ] − V p +1 q det( η + ˆ F ′ ) det( η + ˆ F )2 (8 π α ′ ) p h B ′ , η ′ | B, η i Z ∞ dtt − p e − y πα ′ t [ η ( it )] − [ p +12 ] × [ θ (0 | it )] − [ p +12 ] [ p − ] Y α =0 θ ( ν α | it ) θ ( ν α | it ) [ p − ] Y α =0 sin πν α cos πν α . (49)In Appendix B, we show that the zero-mode contribution (44) can be written in terms of18 The θ -terms in the square bracket in (51) and their simplification0 θ (0 | it ) − θ (0 | it ) − θ (0 | it ) = 2 θ (0 | it ) = 01 or 2 θ (0 | it ) θ ( ν | it ) − θ (0 | it ) θ ( ν | it ) − θ (0 | it ) θ ( ν | it ) = 2 θ (cid:0) ν (cid:12)(cid:12) it (cid:1) θ (0 | it ) θ ( ν | it ) θ ( ν | it ) − θ (0 | it ) θ ( ν | it ) θ ( ν | it ) − θ (0 | it ) θ ( ν | it ) θ ( ν | it )= 2 θ (cid:0) ν + ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν (cid:12)(cid:12) it (cid:1) θ (0 | it ) θ ( ν | it ) θ ( ν | it ) θ ( ν | it ) − θ (0 | it ) θ ( ν | it ) θ ( ν | it ) θ ( ν | it )5 or 6 − θ (0 | it ) θ ( ν | it ) θ ( ν | it ) θ ( ν | it )= 2 θ (cid:0) ν + ν + ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν + ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν + ν − ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν − ν (cid:12)(cid:12) it (cid:1) Table 2: The θ -terms in the square bracket in (51) and their simplification.the ν α when ν α ∈ [0 ,
1) as h B ′ , η ′ | B, η i = − δ ηη ′ , + [ p − ] Y α =0 cos πν α . (50)We would like to stress that the above zero-mode expression plays a key role in writingthe total amplitude in a nice form as given later in (54).With the above, we can have the total interaction amplitude for 0 ≤ p = p ′ ≤ p,p = Γ NSNS + Γ RR = 2[ p +12 ] V p +1 q det( η + ˆ F ′ ) det( η + ˆ F )2 (8 π α ′ ) p Z ∞ dtt − p e − y πα ′ t [ η ( it )] − [ p +12 ] [ p − ] Y α =0 sin πν α θ ( ν α | it ) × [ θ (0 | it )] − [ p +12 ] [ p − ] Y α =0 θ ( ν α | it ) − [ θ (0 | it )] − [ p +12 ] [ p − ] Y α =0 θ ( ν α | it ) − [ θ (0 | it )] − [ p +12 ] [ p − ] Y α =0 θ ( ν α | it ) . (51)We would like to stress that in spite of its appearance, each term in the above squarebracket is the product of four theta-functions of the same type. For convenience, we listthe three terms in the square bracket for each case in Table 2. In this Table, we also usethe following identity for the θ -functions from [52] to simplify the formula2 θ ( w | τ ) θ ( x | τ ) θ ( y | τ ) θ ( z | τ ) = θ ( w ′ | τ ) θ ( x ′ | τ ) θ ( y ′ | τ ) θ ( z ′ | τ )+ θ ( w ′ | τ ) θ ( x ′ | τ ) θ ( y ′ | τ ) θ ( z ′ | τ ) − θ ( w ′ | τ ) θ ( x ′ | τ ) θ ( y ′ | τ ) θ ( z ′ | τ )+ θ ( w ′ | τ ) θ ( x ′ | τ ) θ ( y ′ | τ ) θ ( z ′ | τ ) , (52)19here w ′ , x ′ , y ′ and z ′ are related to w, x, y, z as2 w ′ = − w + x + y + z, x ′ = w − x + y + z, y ′ = w + x − y + z, z ′ = w + x + y − z. (53)Note that θ (0 | τ ) = 0. So we have computed the closed string cylinder amplitude foreach of the 0 ≤ p = p ′ ≤ θ -function terms in the square bracketin (51) now given by the simplified term containing only the θ -function listed in Table2. Carefully examining the integrand of the amplitude for each case, in particular the θ -function factor, we observe that the amplitude can be expressed by a universal formulafor each of 0 ≤ p = p ′ ≤ ν α . Concretely, the general amplitude isΓ p,p = 2 V p +1 h det( η + ˆ F ′ ) det( η + ˆ F ) i (8 π α ′ ) p +12 Z ∞ dtt − p e − y πα ′ t η ( it ) × θ (cid:0) ν + ν + ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν + ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν + ν − ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν − ν (cid:12)(cid:12) it (cid:1) θ ( ν | it ) θ ( ν | it ) θ ( ν | it ) Y α =0 sin πν α = 2 V p +1 h det( η + ˆ F ′ ) det( η + ˆ F ) i (cid:2)P α =0 cos πν α − Q α =0 cos πν α − (cid:3) (8 π α ′ ) p +12 × Z ∞ dtt − p e − y πα ′ t ∞ Y n =1 C n , (54)where C n = ˜ C n (1 − | z | n ) Q α =0 [1 − | z | n cos 2 πν α + | z | n ] , (55)with˜ C n = (cid:2) − | z | n cos π ( ν + ν + ν ) + | z | n (cid:3) (cid:2) − | z | n cos π ( ν − ν + ν ) + | z | n (cid:3) × (cid:2) − | z | n cos π ( ν + ν − ν ) + | z | n (cid:3) (cid:2) − | z | n cos π ( ν − ν − ν ) + | z | n (cid:3) (56) or = (cid:2) − | z | n e iπν cos π ( ν + ν ) + e πiν | z | n (cid:3) × (cid:2) − | z | n e iπν cos π ( ν − ν ) + e πiν | z | n (cid:3) × (cid:2) − | z | n e − iπν cos π ( ν + ν ) + e − πiν | z | n (cid:3) × (cid:2) − | z | n e − iπν cos π ( ν − ν ) + e − πiν | z | n (cid:3) . (57)20n the first line of equality in (54), the following factor in the integrand θ (cid:0) ν + ν + ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν + ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν + ν − ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν − ν (cid:12)(cid:12) it (cid:1) θ ( ν | it ) θ ( ν | it ) θ ( ν | it ) Y α =0 sin πν α (58)is actually for p = p ′ = 6 or 5. But it will reduce to the p = p ′ = 4 or 3 case if we takethe limit ν = 0, to the p = p ′ = 2 or 1 if we take the limits ν = 0 , ν = 0 and to the p = p ′ = 0 case if we take ν = 0 , ν = 0 , ν = 0. We will explain this nice feature of theamplitude later on.This same amplitude can also be expressed in terms of the open string one-loop annulusone via the Jacobi transformation t → t ′ = 1 /t , from the first equality in (54), asΓ p,p = − i V p +1 h det( η + ˆ F ′ ) det( η + ˆ F ) i Q α =0 sin πν α (8 π α ′ ) p +12 Z ∞ dt ′ t ′ p − e − y t ′ πα ′ η ( it ′ ) × θ (cid:0) ν + ν + ν it ′ (cid:12)(cid:12) it ′ (cid:1) θ (cid:0) ν − ν + ν it ′ (cid:12)(cid:12) it ′ (cid:1) θ (cid:0) ν + ν − ν it ′ (cid:12)(cid:12) it ′ (cid:1) θ (cid:0) ν − ν − ν it ′ (cid:12)(cid:12) it ′ (cid:1) θ ( ν it ′ | it ′ ) θ ( ν it ′ | it ′ ) θ ( ν it ′ | it ′ )= 2 V p +1 h det( η + ˆ F ′ ) det( η + ˆ F ) i (8 π α ′ ) p +12 Z ∞ dtt p − e − y t πα ′ Y α =0 sin πν α sinh πν α t × " X α =0 cosh πν α t − Y α =0 cosh πν α t − ∞ Y n =1 Z n , (59)where in obtaining the first equality in (59), we have used the following relations for the θ -function and the Dedekind η -function, η ( τ ) = 1( − iτ ) / η (cid:18) − τ (cid:19) , θ ( ν | τ ) = i e − iπν /τ ( − iτ ) / θ (cid:18) ντ (cid:12)(cid:12)(cid:12) − τ (cid:19) , (60)in the second equality we have dropped the prime on t , and Z n = ˜ Z n (1 − | z | n ) Q α =0 [1 − | z | n cosh 2 πν α t + | z | n ] , (61)with˜ Z n = (cid:2) − | z | n cosh π ( ν + ν + ν ) t + | z | n (cid:3) (cid:2) − | z | n cosh π ( ν − ν + ν ) t + | z | n (cid:3) × (cid:2) − | z | n cosh π ( ν + ν − ν ) t + | z | n (cid:3) (cid:2) − | z | n cosh π ( ν − ν − ν ) t + | z | n (cid:3) (62) or = (cid:2) − | z | n e − πν t cosh π ( ν + ν ) t + e − πν t | z | n (cid:3) × (cid:2) − | z | n e − πν t cosh π ( ν − ν ) t + e − πν t | z | n (cid:3) × (cid:2) − | z | n e πν t cosh π ( ν + ν ) t + e πν t | z | n (cid:3) × (cid:2) − | z | n e πν t cosh π ( ν − ν ) t + e πν t | z | n (cid:3) . (63)21n what follows, we will discuss each of the four cases: 1) p = p ′ = 6 or 5, 2) p = p ′ = 4or 3, 3) p = p ′ = 2 or 1, and 4) p = p ′ = 0, separately. p = p ′ = 6 or case For the respective general worldvolume fluxes ˆ F and ˆ F ′ , from the second equality of (54),we have the interaction amplitude for p = p ′ = 6 or 5 case asΓ p,p = 2 V p +1 (cid:2)P α =0 cos πν α − Q α =0 cos πν α − (cid:3) q det( η + ˆ F ′ ) det( η + ˆ F )(8 π α ′ ) p +12 × Z ∞ dtt − p e − y πα ′ t ∞ Y n =1 C n , (64)where C n is given in (55). For large brane separation y , the amplitude has its contributionmostly from the large t integration for which C n ≈
1. We have thereforeΓ p,p V p +1 ≈ (cid:2)P α =0 cos πν α − Q α =0 cos πν α − (cid:3) q det( η + ˆ F ′ ) det( η + ˆ F )2 − (8 π α ′ ) p +12 × Z ∞ dtt − p e − y πα ′ t = (cid:2)P α =0 cos πν α − Q α =0 cos πν α − (cid:3) q det( η + ˆ F ′ ) det( η + ˆ F )2 p − π p +12 (2 πα ′ ) p − y − p Z ∞ dtt − p e − t = (cid:2)P α =0 cos πν α − Q α =0 cos πν α − (cid:3) q det( η + ˆ F ′ ) det( η + ˆ F )2 p − π p +12 (2 πα ′ ) p − y − p Γ (cid:18) − p (cid:19) , (65)where in the second equality, we have rescaled the integration variable t , and Γ(7 − p/ /
2) = √ π for p = 6 and Γ(1) = 1 for p = 5, respectively.According to our conventions, Γ p,p > p,p < F = P α =0 cos πν α − Q α =0 cos πν α − ν , ν and ν vanishes, it isnon-negative. For example, taking ν = 0, we can write it as F = (cos ν − cos ν ) ≥ ν = i ¯ ν with ¯ ν ∈ (0 , ∞ ), and ν , ν ∈ (0 , F = sinh π ¯ ν + cos πν + cos πν − π ¯ ν cos πν cos πν = sinh π ¯ ν + cos πν + cos πν − (cid:16) π ¯ ν (cid:17) cos πν cos πν = 4 (cid:16) cosh π ¯ ν − cos πν cos πν (cid:17) sinh π ¯ ν πν − cos πν ) > . (66)In other words, the large brane separation is always attractive. Case 2): all ν , ν , ν ∈ (0 , p = 6 since it needs at least six worldvolume spatialdirections. For this case, in order to determine the condition for the sign of F , we rewriteit as F = 2 sin π ν + ν + ν π ν − ν + ν π ν + ν − ν π ν − ν − ν . (67)Note that F is symmetric under the exchange of any two of ν ’s. So without loss ofgenerality, we can assume ν ≤ ν ≤ ν . Given this, we have three subcases to consider dueto the allowed range for each of the ν ’s: a) 0 < ν + ν + ν <
2, b) ν + ν + ν = 2 and c) 2 <ν + ν + ν <
3. For subcase a), it is clear that the first two factors sin π ( ν + ν + ν ) / > π ( ν − ν + ν ) / > π ( ν − ν − ν ) / <
0. So the signof F is determined by that of third factor sin π ( ν + ν − ν ) /
2. If ν + ν > ν , then F < ν + ν < ν , we have F > F = 0 when ν + ν = ν .In other words, if the possible largest one of the three ν ’s is smaller than the sum ofthe remaining two, the long-range interaction is repulsive. Otherwise, it is attractiveor vanishes. For subcase b), F = 0 and so we have the force vanishing, implying theunderlying system to be supersymmetric. For subcase c), we have both the first factorsin π ( ν + ν + ν ) / < π ( ν − ν − ν ) / < π ( ν − ν + ν ) / >
0. For this subcase, we must have ν + ν > ν given each of the ν α ∈ (0 ,
1) and we therefore have the third factor sin π ( ν + ν − ν ) / >
0. Hence we havenow F >
0, therefore the interaction is again attractive. In other words, the long-rangeinteraction can only be repulsive when the three ν ’s are non-vanishingly real and satisfy ν + ν + ν <
2, and when the possible largest one of the three ν ’s is smaller than the sumof the remaining two. The above discussion actually remains true for any brane separationas we will show this later on. For ν + ν + ν <
2, as will be shown, when the possiblelargest one of the three ν ’s is smaller than the sum of the remaining two, the interactionis repulsive for any brane separation and there is no tachyonic instability. However, whenthe largest one is larger than the sum of the remaining two, the corresponding interactionremains attractive at least until the brane separation reaches the minimum determinedby the so-called tachyonic shift and after that there is a tachyonic instability to occur.23his tachyonic instability occurs whenever the interaction is attractive and the braneseparation reaches the minimum. When the possible largest one equals to the sum of theremaining two or ν + ν + ν = 2, the interaction vanishes for any brane separation andthis no-force implies the underlying system to be supersymmetric.We now come to explain the above. It is well-known that when a D2 or D4 brane isplaced parallel to a D6 at a separation transverse to the D6, the interaction between themis zero or attractive while by the same token, the interaction between the D6 and a D0is repulsive. Each of these cases can be examined easily in the following section when weconsider the case of p − p ′ = 2 , , p ≤
6. For the above, we need to have p = p ′ = 6along with ν α ∈ (0 ,
1) for α = 0 , ,
2. In other words, at least one of the two D6 carries aconstant magnetic flux, for example, ˆ F , with non-vanishing components ˆ F , ˆ F and ˆ F ,which can give rise to the above three ν α ∈ (0 , T Z (cid:16) C − n ∧ ˆ F n (cid:17) , (68)where T is the D6 brane tension, C p +1 is the potential minimally coupled with Dp, andˆ F n = ˆ F ∧ · · · ∧ ˆ F stands for the wedge product of n ˆ F . So n = 0 gives the couplingof D6 with the R-R potential C , n = 1 gives the coupling of D4 with C , n = 2 givesthe coupling of D2 with C and n = 3 gives the coupling of D0 with C . Given whatwe have for the non-vanishing components of ˆ F , we can have n = 0 , , ,
3. So this givesan explanation of the delocalized D0, D2, D4 within D6 mentioned above. So when thepossible largest one of the three ν ’s is smaller than the sum of the remaining two for thecase ν + ν + ν <
2, the above cylinder amplitude shows that the repulsive interactionbetween the other D6 and the delocalized D0 on this D6 overtakes the attractive onesbetween the other D6 and the delocalized D2 or D4. Otherwise, the attractive interactionovertakes the repulsive one. The net interaction vanishes when the two equals. This alsoexplains that when one of the three ν ’s is imaginary or vanishes, the net interaction isattractive since we must have one of the ˆ F , ˆ F and ˆ F being zero which implies theabsence of D0 branes on the D6.For small brane separation, the small t integration in (64) becomes important and hasto be considered. For small t , C n in (55) can be large. In particular, C n blows up when t → − | z | n ) in its denominator. So overall we have a blowingup factor Q ∞ n =1 (1 − | z | n ) − for t → e − y / (2 πα ′ t ) in the integrand becomes vanishinglysmall when t →
0. So there is a competition between the two and one expects a potentially24nteresting physics to occur when t →
0. This will become manifest when we transformthe closed string cylinder amplitude to the corresponding open string one-loop annulusone and it is a potential open string tachyonic instability. For now, the nature of Γ p,p depends on that of the parameters ν , ν and ν .Following the previous discussion for large y , we have that the interaction is attractivewhenever the three ν ’s are all real and the possible largest one of these is larger thanthe sum of the remaining two when ν + ν + ν < < ν + ν + ν < y . Note that the C n (55) is still positive even for small t andthis can be easily checked. Each factor in the numerator ˜ C n (56) is positive, for example,the first factor [1 − | z | n cos π ( ν + ν + ν )+ | z | n ] > − | z | n + | z | n = (1 −| z | n ) >
0. Bythe same token, each factor in the denominator is also positive. In other words, the signof Γ p,p is still determined by that of F given earlier. For this, the attractive interactionacting between the two D6 has a tendency to move the two towards each other and tomake the brane separation smaller. Therefore the exponential factor e − y / (2 πα ′ t ) will makeits suppressing less important and one expects that the diverging effect from C n at small t will become to dominate at a certain point. So we expect then a potential instabilitymentioned above to occur. On the other hand, when the possible largest one of three ν ’s is less than the sum of the remaining two when ν + ν + ν <
2, the interaction isrepulsive and as such has a tendency to move the two D6 apart further. So this makes thesuppression of the exponential factor e − y / (2 πα ′ t ) in the integration more important anddisfavors the instability to occur. So this appears to provide a correlation between thenature of interaction and the existence of potential tachyonic instability. We will showlater that this is indeed true.For small t , there appears a new feature when one of three ν ’s takes an imaginaryvalue. For example, we choose ν = i ¯ ν with ¯ ν ∈ (0 , ∞ ). Now C n can be negative.By the same token as given in the previous paragraph, each factor in ˜ C n (57) continuesto be positive, for example, the third factor [1 − | z | n e π ¯ ν cos π ( ν + ν ) + e π ¯ ν | z | n ] > − | z | n e π ¯ ν + e π ¯ ν | z | n = (1 − e π ¯ ν | z | n ) >
0. However, the factor [1 − | z | n cosh 2 π ¯ ν + | z | n ] ≈ − cosh 2 π ¯ ν ) <
0, in the denominator of C n , for small t . Since there are aninfinite number of C n appearing as product in the integrand, the sign of Γ p,p will then beindefinite. So for small y , the nature of the interaction becomes obscure for the case underconsideration and this indicates that there should exist new physical process occurringin addition to the potential tachyonic instability mentioned above for small t . This newphysics is actually the decay of the underlying system via the so-called open string pairproduction under the action of applied electric fluxes which makes ν become imaginary.25ll these will become manifest when the interaction is expressed in terms of the openstring variable as the open string one-loop annulus amplitude (59) for p = 6 which weturn next. Note that this consideration applies to both p = p ′ = 6 and p = p ′ = 5 cases.The open string one-loop annulus amplitude for p = p ′ = 5 ,
6, respectively, can beread from the second equality of (59) asΓ p,p = 2 V p +1 h det( η + ˆ F ′ ) det( η + ˆ F ) i (8 π α ′ ) p +12 Z ∞ dtt p − e − y t πα ′ Y α =0 sin πν α sinh πν α t × " X α =0 cosh πν α t − Y α =0 cosh πν α t − ∞ Y n =1 Z n , (69)where Z n is given in (61). Note that the closed string t-variable and the open stringt-variable are inversely related to each other. So small t in closed string case implies larget in open string one. So the potential open string tachyonic instability should show upfor large t in the integrand of the above amplitude if it exists at all. Let us examine thisin detail.For large t, note that n ≥ Z n ≈ ν ’s being real with ν α ∈ (0 ,
1) and ν + ν + ν < ν α are real with ν α ∈ (0 ,
1) and 2 < ν + ν + ν < Z ≈ − e π ( ν + ν + ν − t and Z n ≈ n ≥ ν ’s are real (only valid for p = p ′ = 6), we once again assume ν ≤ ν ≤ ν without loss of generality since the amplitude is symmetric under the exchange of anytwo of the three ν , ν and ν . From the discussion in the closed string variable, we knowthat when ν + ν > ν for ν + ν + ν <
2, the interaction is repulsive and one expectsno tachyonic instability. Let us check this explicitly here. From (69), the terms in thesquare bracket can be expressed as2 sinh π ( ν + ν + ν ) t π ( ν − ν + ν ) t π ( ν + ν − ν ) t π ( ν − ν − ν ) t . (70)So from this, one sees for large t and ν + ν > ν with ν + ν + ν < ∼ − e − y t πα ′ e π ( ν + ν + ν ) t sinh πν t sinh πν t sinh πν t t →∞ −→ − e − y t πα ′ , (71)which shows no tachyonic shift and therefore no potential tachyonic instability. This isconsistent with our anticipation. However, when ν + ν < ν , we do expect to see the26otential tachyonic instability. Now (70) gives ∼ e πν t and the integrand is positive andbehaves like ∼ e − y t πα ′ e πν t sinh πν t sinh πν sinh πν t t →∞ −→ e − y t πα ′ e π ( ν − ν − ν ) t , (72)where we have a so-called tachyonic shift ( ν − ν − ν ) / > m = y (2 πα ′ ) − ν − ν − ν α ′ , (73)which becomes tachyonic if y < π p ν − ν − ν ) α ′ . Let us now move to the case2 < ν + ν + ν <
3. For this, we must have ν + ν > ν . Then (70) gives, for large t, ∼ − e π ( ν + ν + ν ) t , a negative infinity, and the integrand in (69) is still positive and behavesas − e − y t πα ′ e π ( ν + ν + ν ) t sinh πν t sinh πν sinh πν t Z t →∞ −→ e − y t πα ′ e π ( ν + ν + ν − t , (74)where we have used Z ∼ − e π ( ν + ν + ν − t and Z n ≈ n ≥ ν + ν + ν − / m = y (2 πα ′ ) − ν + ν + ν − α ′ , (75)which becomes tachyonic when y < π p ν + ν + ν − α ′ . Whenever this happens,the integrand blows up for t → ∞ and this reflects the onset of tachyonic instability.Then we will have a phase transition via the so-called tachyon condensation. Once again,this is consistent with our expectation. So this confirms our earlier assertion that thereis indeed a correlation between the attractive nature of interaction and the existence of atachyonic instability.The tachyonic shift and the appearance of tachyon mode can also be understood fromthe spectrum of the open string connecting the two D6 carrying magnetic fluxes whichgive rise to the ν , ν and ν following [19,20]. Let us use an explicit example for p = p ′ = 6to demonstrate this. For this purpose, we choose the following magnetic flux for ˆ F ˆ F = − ˆ g ˆ g − ˆ g ˆ g − ˆ g ˆ g , (76)27nd for ˆ F ′ we just replace each g ’s in ˆ F with a prime on it. Following the prescriptiongiven earlier, we have λ α + λ − α = 2 (1 − ˆ g α )(1 − ˆ g ′ α ) + 4ˆ g α ˆ g ′ α (1 + ˆ g α )(1 + ˆ g ′ α ) , (77)where α = 0 , , λ α = e πiν α with ν α ∈ [0 , πν α = | ˆ g α − ˆ g ′ α | g α ˆ g ′ α . (78)We have also now the amplitude (69) with h det( η + ˆ F ′ ) det( η + ˆ F ′ ) i / = Y α =0 (cid:2) (1 + ˆ g α )(1 + ˆ g ′ α ) (cid:3) / . (79)Type I superstring in a single magnetic background, say, the magnetic field being in 56-directions, has been discussed in [20]. Here we have three magnetic fields, the first in 12-directions, the second in 34-directions and the third in 56-directions. The generalizationof the discussion given there to the present case in the R-sector is straightforward andthe conclusion remains the same even if we exclude the contribution of y / (2 πα ′ ) tothe energy square. In other words, unlike the case in the NS-secor which we will turnnext, there is no possibility for the existence of tachyonic shift in the R-sector. TheGSO-projected R-sector ground states (the eight fermions 8 F ) have masses no less than y/ (2 πα ′ ). In what follows, we focus here only on the generalization to the present NS-sector. As before, without loss of generality, we assume once again ν ≤ ν ≤ ν . Theenergy spectrum is now α ′ E = y (2 π ) α ′ + X α =0 h (2 N α + 1) ν α − ν α S α i + L freeNS , (80)where N α = b + α, b α, , S α = ∞ X n =1 (cid:0) a + α,n a α,n − b + α,n b α,n (cid:1) + ∞ X r =1 / (cid:16) d + α,r d α,r − ˜ d + α,r ˜ d α,r (cid:17) ,L freeNS = X α =0 ∞ X n =1 n ( a + α,n a α,n + b + α,n b α,n ) + ∞ X r =1 / r ( d + α,r d α,r + ˜ d + α,r ˜ d α,r ) −
12 + L ⊥ freeNS . (81)In the above, N α defines the corresponding Landau-level for α = 0 , ,
2, respectively, S α isthe spin operator and L ⊥ freeNS is the part contributing to the zero-mode Virasoro generatorfrom the 0, 7, 8, 9-directions. One can check when ν + ν < ν and ν + ν + ν <
2, the28ossible lowest energy state is from the GSO-projected ground state d +2 , / | i NS and forthis we have α ′ E = y (2 π ) α ′ − ν − ν − ν ,S = 1 , S = S = 0 , N = N = N = 0 , L freeNS = 0 . (82)Here the first equation is exactly the same as (73). In other words, when ν > ν + ν and ν + ν + ν < ν /
2, rather than the smallerone ( ν − ν − ν ) /
2, in the absence of the other two fluxes. In other words, in order tohave the largest tachyonic shift, we need to choose to apply the largest magnetic one butno more. This largest tachyonic shift is also responsible for the largest open string pairproduction enhancement discussed in [21, 31]. We will also address this later when wediscuss the open string pair production in the presence of electric fluxes.Similarly, when 2 < ν + ν + ν < ν + ν > ν , the lowest energy state is theGSO-projected one d +0 , / d +1 , / d +2 , / | i NS which gives S = S = S = 1 and L freeNS = 1 andfor this we have α ′ E = y (2 π ) α ′ − ν + ν + ν − ,S = S = S = 1 , N = N = N = 0 , L freeNS = 0 . (83)The lowest energy is nothing but the effective mass given in (75) with the same tachyonicshift ( ν + ν + ν − /
2. So the tachyonic instability can be discussed exactly the sameand will not be repeated here.We now move to the case when one of three ν ’s is imaginary, say, ν = i ¯ ν , and both ν and ν are real. So this applies to both p = p ′ = 5 and p = p ′ = 6. Now the openstring annulus amplitude (69) becomesΓ p,p = 2 V p +1 h det( η + ˆ F ′ ) det( η + ˆ F ) i (8 π α ′ ) p +12 Z ∞ dtt p − e − y t πα ′ sinh π ¯ ν sin πν sin πν sin π ¯ ν t sinh πν t sinh πν t × (cid:20) (cosh πν t − cosh πν t ) + 4 sin π ¯ ν t (cid:18) cosh πν t cosh πν t − cos π ¯ ν t (cid:19)(cid:21) × ∞ Y n =1 Z n , (84) Both the state d +0 , / | i NS and d +1 , / | i NS have their respective energy higher than that of d +2 , / | i NS . Z n , from (61), becomes Z n = ˜ Z n (1 − | z | n ) [1 − | z | n cos 2 π ¯ ν t + | z | n ] Q α =1 [1 − | z | n cosh 2 πν α t + | z | n ] , (85)with ˜ Z n , from (63), as˜ Z n = (cid:12)(cid:12)(cid:2) − | z | n e − iπ ¯ ν t cosh π ( ν + ν ) t + e − iπ ¯ ν t | z | n (cid:3)(cid:12)(cid:12) × (cid:12)(cid:12)(cid:2) − | z | n e − iπ ¯ ν t cosh π ( ν − ν ) t + e − iπ ¯ ν t | z | n (cid:3)(cid:12)(cid:12) > . (86)Note that in the denominator of Z n , the factor 1 − | z | n cos π ¯ ν t + | z | n > (1 − | z | n ) > α = 1 ,
2, 1 − | z | n cosh πν α t + | z | n = (1 − | z | n e πν α t )(1 − | z | n e − πν α t ) > n ≥ ν α ∈ [0 , Z n >
0. Note also that every other factor in theintegrand, except for the sin π ¯ ν t in the denominator, is also positive. The interestingphysics shows up precisely due to this factor sin π ¯ ν t . It gives an infinite number ofsimple poles of the integrand at t k = k/ ¯ ν with k = 1 , , · · · along the positive t-axis.This implies that the interaction amplitude has an imaginary part, indicating the decayof the underlying system via the so-called open string pair production. By saying this, wefirst need to note that the integral has no singularity when we take t →
0. Secondly, weneed to have y > π p | ν − ν | α ′ to avoid a potential tachyonic instability and to validateour amplitude computations since otherwise the integrand blows up for large t as ∼ e − y t πα ′ e π | ν − ν | t = e − πt (cid:20) y π )2 α ′ − | ν − ν | (cid:21) , (87)and as such a phase transition, called tachyon condensation, occurs. The decay rate ofthe underlying system per unit volume of Dp brane via the open string pair productioncan be computed, following [28], as the sum of the residues of the simple poles of theintegrand in (84) times π and is given as W p,p = − V p +1 = 2 h det( η + ˆ F ′ ) det( η + ˆ F ) i sinh π ¯ ν sin πν sin πν ¯ ν (8 π α ′ ) p +12 ∞ X k =1 ( − ) k +1 (cid:16) ¯ ν k (cid:17) p − e − ky π ¯ ν α ′ × (cid:16) cosh kπν ¯ ν − ( − ) k cosh kπν ¯ ν (cid:17) sinh kπν ¯ ν sinh kπν ¯ ν Z k (¯ ν , ν , ν ) , (88)where Z k = ∞ Y n =1 [1 − − ) k | z k | n cosh kπ ( ν + ν )¯ ν + | z k | n ] [1 − − ) k | z k | n cosh kπ ( ν − ν )¯ ν + | z k | n ] (1 − | z k | n ) [1 − | z k | n cosh kπν ¯ ν + | z k | n ][1 − | z k | n cosh kπν ¯ ν + | z k | n ] , (89)30ith | z k | = e − kπ/ ¯ ν . Note that when ¯ ν → ∞ , we have Z k → ∞ for k = odd while Z k → k = even due to | z k | →
1. For the rate, the odd-k terms, each is blowing up andpositive, are dominant over the almost vanishing negative even-terms, and so the rateblows up. This gives another singularity. As we will see, this is due to the electric fieldreaching its critical value. The open strings break under the action of the critical fieldand their production cascades.According to [54], the rate (88) should be more properly interpreted as the decay oneof the underlying system rather than the pair production one. The pair production rateis just the leading k = 1 term in the above rate as W (1) p,p = 2 h det( η + ˆ F ′ ) det( η + ˆ F ) i sinh π ¯ ν sin πν sin πν ¯ ν (8 π α ′ ) p +12 ¯ ν p − e − y π ¯ ν α ′ × (cid:16) cosh πν ¯ ν + cosh πν ¯ ν (cid:17) sinh πν ¯ ν sinh πν ¯ ν Z (¯ ν , ν , ν ) . (90)So the pair production simply cascades when ¯ ν → ∞ . Since W (1) is symmetric to ν and ν , without loss of generality and for convenience, we assume ν ≥ ν for thefollowing discussion. Given ν ≥ ν ∈ [0 , Z , from (89), increasesas ¯ ν increases. When ¯ ν ≫ ν , the factor (cid:16) cosh πν ¯ ν + cosh πν ¯ ν (cid:17) / sinh πν ¯ ν sinh πν ¯ ν ∼ ν / ( π ν ν ) increases also when we increase ¯ ν . Since p ≥
5, all other factors have anoverall increase when we increase ¯ ν . This holds true at least for the most interestingcases with a large enhancement of the rate and is also expected since ¯ ν is related to theapplied electric fluxes and increases when any of them increases, which will be explicitlydemonstrated in an example given later. When ¯ ν ∼ ν ≥ ν , this same factor will notplay important role for the rate. The rate is now controlled by the other factors and stillincreases with the ¯ ν . If ¯ ν ≪ ν ≤ ν , this implies ¯ ν ≪
1. So Z (¯ ν , ν , ν ) ≈
1. Thepair production rate is W (1) p,p ≈ π h det( η + ˆ F ′ ) det( η + ˆ F ) i sin πν sin πν (8 π α ′ ) p +12 ¯ ν p − e − y π ¯ ν α ′ e π ( ν − ν ν , (91)where the factor e π ( ν − ν ) / ¯ ν ≫
1, a large enhancement of the rate in the presence ofmagnetic fluxes. If y > π p ν − ν ) α ′ (for avoiding the tachyonic instability), therate still increases when we increase ¯ ν . For the purpose of illustration, we consider the31 = p ′ = 5 case and take the following simple flux for ˆ F asˆ F = − ˆ f ˆ f − ˆ g ˆ g − ˆ g g , (92)where ˆ f stands for the electric flux along 01-directions while ˆ g , ˆ g are the magnetic onesalong 23- and 45-directions, respectively. Similarly for ˆ F ′ but denoting every quantitywith a prime. We can then determine the eigenvalues as λ + λ − = 2 (1 + ˆ f )(1 + ˆ f ′ ) − f ˆ f ′ (1 − ˆ f )(1 − ˆ f ′ ) , λ a + λ − a = 2 (1 − ˆ g a )(1 − ˆ g ′ a ) + 4ˆ g a ˆ g ′ a (1 + ˆ g a )(1 + ˆ g ′ a ) , (93)where a = 1 ,
2. By setting λ α = e iπν α with α = (0 , a ), we havetanh π ¯ ν = | ˆ f − ˆ f ′ | − ˆ f ˆ f ′ , tan πν a = | ˆ g a − ˆ g ′ a | g a ˆ g ′ a , (94)where we have set ν = i ¯ ν . Note that | ˆ f | , | ˆ f ′ | < | ˆ g a | , | ˆ g ′ a | < ∞ . As explainedearlier, we always have ¯ ν ∈ (0 , ∞ ) and ν a ∈ [0 ,
1) for the amplitude and the rate. Itis clear that ¯ ν increases when we increase ˆ f or ˆ f ′ as mentioned earlier. Now the factordet( η + ˆ F ′ ) det( η + ˆ F ) = (1 − ˆ f )(1 − ˆ f ′ ) Q a =1 (1 + ˆ g a )(1 + ˆ g ′ a ). The open string pairproduction rate can now be expressed as W (1) p,p = 2 | ˆ f − ˆ f ′ || ˆ g − ˆ g ′ || ˆ g − ˆ g ′ | ¯ ν (8 π α ′ ) p +12 ¯ ν p − e − y π ¯ ν α ′ (cid:16) cosh πν ¯ ν + cosh πν ¯ ν (cid:17) sinh πν ¯ ν sinh πν ¯ ν Z (¯ ν , ν , ν ) , (95)where ¯ ν and ν a with a = 1 , p = p ′ = 6case. The earlier general discussion on how the pair production rate depends on theapplied electric fluxes or ¯ ν for fixed ν a with a = 1 , ν a for fixed non-vanishing ¯ ν . The rate for vanishing magnetic fluxes can be obtainedfrom (95) as W (1) p,p (ˆ g a = ˆ g ′ a = 0) = 2 | ˆ f − ˆ f ′ | (8 π α ′ ) p +12 ¯ ν p − e − y π ¯ ν α ′ Z (¯ ν , ν a = 0) . (96)32e have then W (1) p,p (ˆ g a , ˆ g ′ a = 0) W (1) p,p (ˆ g a = ˆ g ′ a = 0) = | ˆ g − ˆ g ′ || ˆ g − ˆ g ′ | ν (cid:16) cosh πν ¯ ν + cosh πν ¯ ν (cid:17) sinh πν ¯ ν sinh πν ¯ ν Z (¯ ν , ν a = 0) Z (¯ ν , ν a = 0) . (97)For non-vanishing ν a ∈ (0 ,
1) and | ˆ g a − ˆ g ′ a | ∼ O (1), if ¯ ν is not too small, the presence ofmagnetic fluxes will not give a significant enhancement of the rate as can be seen fromthe above. However, if instead ν a / ¯ ν ≫ | ˆ g a − ˆ g ′ a | ≥ ν a / ¯ ν ≪ g a , ˆ g ′ a are large, with a = 1 ,
2, the rate has a significant enhancement. Let us consider thelater case for which all ˆ g a and ˆ g ′ a are large. Now ν a are small. So from (94), we haveˆ g a ˆ g ′ a πν a = | ˆ g a − ˆ g ′ a | . The ratio of (97) becomes W (1) p,p (ˆ g a , ˆ g ′ a = 0) W (1) p,p (ˆ g a = ˆ g ′ a = 0) ≈ ˆ g ˆ g ′ ˆ g ˆ g ′ ≫ , (98)much enhanced. In the above, we have used Z (¯ ν , ν a ≪ ≈ Z (¯ ν , ν a = 0).Unless we consider relevant physics in string scale, the fluxes ˆ f , ˆ f ′ , ˆ g a and ˆ g ′ a are ingeneral small in terms of string scale. In other words, | ˆ f | ≪ , | ˆ f ′ | ≪ , | ˆ g a | ≪ | ˆ g ′ a | ≪
1. We then have π ¯ ν = | ˆ f − ˆ f ′ | ≪ , πν a = | ˆ g a − ˆ g ′ a | ≪
1. The rate (95) becomes W (1) p,p = 8 π ν ν (8 π α ′ ) p +12 ¯ ν p − e − y π ¯ ν α ′ (cid:16) cosh πν ¯ ν + cosh πν ¯ ν (cid:17) sinh πν ¯ ν sinh πν ¯ ν , (99)where Z (¯ ν , ν , ν ) ≈
1. In the weak field limit, we showed in [31] that adding magneticfluxes ˆ g and ˆ g ′ , assuming ν ≤ ν , in general diminishes rather than enhances the rate.This can also be understood via the tachyonic shift discussed earlier. So for the purposeof enhancing the rate via adding magnetic fluxes, we merely need to add only the possiblelargest fluxes ˆ g and ˆ g ′ . In other words, for given ¯ ν ≪ ν , thecorresponding largest possible rate is W (1) p,p = 8 π ¯ ν ν (8 π α ′ ) p +12 ¯ ν p − e − y π ¯ ν α ′ (cid:16) cosh πν ¯ ν + 1 (cid:17) sinh πν ¯ ν . (100)This rate formula is actually valid for p ≥ ν ≪ ν ≪
1, it is clearthat the smallest p = 3 case gives the largest rate [21, 31]. The enhancement due to theadded magnetic fluxes is W (1) p,p (¯ ν , ν = 0) W (1) p,p (¯ ν , ν = 0) = πν ¯ ν h πν ¯ ν i πν ¯ ν , (101)33hich is always greater than unity for πν / ¯ ν >
0. One can check this numerically. Inparticular, when ν / ¯ ν ≫
1, this ratio becomes W (1) p,p (¯ ν , ν = 0) W (1) p,p (¯ ν , ν = 0) = πν ν e πν ν ≫ . (102) p = p ′ < cases Given the discussion for p = p ′ = 5 , p = p ′ < p = p ′ < M -matrix (7), the bosonic zero-mode (8) in thebosonic sector and the fermionic zero-mode (9) in the R-R sector in the matter part. Therest are independent of this dimensionality. Let us first examine carefully the M -matrix(7) which we rewrite here for convenience, M = ([( η − ˆ F )( η + ˆ F ) − ] αβ , − δ ij ) , (103)where α, β are along the brane directions while i, j are along the directions transverse tothe brane. For example, let us first consider the D6 brane. In other words, α, β = 0 , , · · · i, j = 7 , ,
9. For any other Dp with even p <
6, we denote their α ′ , β ′ = 0 , , · · · p along its brane directions and i ′ , j ′ = p + 1 , · · · − p as directions transverse to this brane.Its corresponding M p -matrix with a general worldvolume flux ( ˆ F p ) α ′ β ′ can be taken as aspecial case of the D6 brane, namely M , in the following sense. For the D p with even p <
6, we have M p = ([( η p − ˆ F p )( η p + ˆ F p ) − ] α ′ β ′ , − δ i ′ j ′ ) . (104)We now extend α ′ , β ′ = 0 , , · · · , p to α, β = 0 , , · · · F p to ˆ F taking the followingspecial form, ( ˆ F ) αβ = ( ˆ F p ) α ′ β ′ g − ˆ g g − p − ˆ g − p × . (105)34ith this special choice, the M turns out to give just M p when we take the specialmagnetic fluxes ˆ g k → ∞ with k = 1 , · · · , (6 − p ) /
2. Let us see this in detail. With thespecial flux (105), we have( M ) αβ = ( M p ) α ′ β ′ − ˆ g g g g g g − ˆ g g . . . − ˆ g − p g − p g − p g − p g − p g − p − ˆ g − p g − p , (106)which becomes ( M ) αβ = (( M p ) α ′ β ′ , − δ k ′ l ′ ) with k ′ , l ′ = p + 1 , · · · g k → ∞ with k = 1 , · · · (6 − p ) /
2. So we have M = (( M p ) α ′ β ′ , − δ i ′ j ′ ) = M p for the above specialchoice of the flux ˆ F (105) when we take ˆ g k → ∞ with k = 1 , · · · (6 − p ) /
2. In otherwords, M p is just a special case of M when the worldvolume flux of D6 takes a specialchoice as indicated above. This same discussion applies to the odd p < p = 5.This same applies to the R-R zero-mode contribution (50) to the amplitude. We dis-cuss this in great detail in Appendix B and refer there for detail. These two considerationsexplain the following part of the integrand in the amplitude (54), θ (cid:0) ν + ν + ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν + ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν + ν − ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν − ν (cid:12)(cid:12) it (cid:1) θ ( ν | it ) θ ( ν | it ) θ ( ν | it ) η ( it ) Y α =0 sin πν α , (107)which is valid in general for p = p ′ = 5 or 6 but will reduce to the corresponding expectedone for p = p ′ < η p + ˆ F p )] / , the other part of the zero mode has nothingto do with the applied worldvolume flux and therefore this same trick as used for the M -matrix and the RR zero-mode does not apply here. This zero-mode contribution tothe amplitude gives essentially the other part of the integrand as q det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) V NN (cid:0) π α ′ t (cid:1) − DD2 e − y πα ′ t , (108)where V NN = V p ′ +1 denotes the volume of the Dp ′ worldvolume following the conventionsgiven in footnote (1) and DD denotes the DD-directions. Here DD = 9 − p with ourconventions. It is obvious that the t − (9 − p ) / -factor in the amplitude (54) for p = p ′ < p = p ′ = 5 or 6 even we choose the respective special fluxes andtake the corresponding limits.We therefore give an explanation to the cylinder amplitude (54) for the case of p = p ′ <
5. Given the extended flux (105) for ˆ F (or ˆ F ) and similarly for ˆ F ′ (or ˆ F ′ ) but withnow ˆ g ′ k with a prime, we have now at leasttan πν = (cid:12)(cid:12)(cid:12)(cid:12) ˆ g − p − ˆ g ′ − p (cid:12)(cid:12)(cid:12)(cid:12) g − p ˆ g ′ − p for even p < , (cid:12)(cid:12)(cid:12)(cid:12) ˆ g − p − ˆ g ′ − p (cid:12)(cid:12)(cid:12)(cid:12) g − p ˆ g ′ − p for odd p < , (109)which gives ν → g (6 − p ) / → ∞ and ˆ g ′ (6 − p ) / → ∞ (or ˆ g (5 − p ) / →∞ and ˆ g ′ (5 − p ) / → ∞ ). So with a vanishing ν = 0, we have θ ( ν | it ) → η ( it ) sin πν and the closed string tree-level cylinder amplitude (54) now becomesΓ p,p = 2 V p +1 h det( η p + ˆ F ′ p ) det( η p + ˆ F p ) i (8 π α ′ ) p +12 Z ∞ dtt − p e − y πα ′ t η ( it ) θ (cid:0) ν + ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν (cid:12)(cid:12) it (cid:1) θ ( ν | it ) θ ( ν | it ) × Y α =0 sin πν α = 2 V p +1 h det( η p + ˆ F ′ p ) det( η p + ˆ F p ) i (cos πν − cos πν ) (8 π α ′ ) p +12 Z ∞ dtt − p e − y πα ′ t ∞ Y n =1 C n , (110)where C n from (55) becomes C n = ˜ C n (1 − | z | n ) Q α =0 [1 − | z | n cos 2 πν α + | z | n ] , (111)with ˜ C n from (56) or (57) as˜ C n = (cid:2) − | z | n cos π ( ν + ν ) + | z | n (cid:3) (cid:2) − | z | n cos π ( ν − ν ) + | z | n (cid:3) (112) or = (cid:2) − | z | n e iπν cos πν + e πiν | z | n (cid:3) (cid:2) − | z | n e − iπν cos πν + e − πiν | z | n (cid:3) . (113)36he corresponding open string one-loop annulus amplitude, from (59), is nowΓ p,p = − V p +1 h det( η p + ˆ F ′ p ) det( η p + ˆ F p ) i Q α =0 sin πν α (8 π α ′ ) p +12 Z ∞ dtt p − e − y t πα ′ η ( it ) × θ (cid:0) ν + ν it (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν it (cid:12)(cid:12) it (cid:1) θ ( ν it | it ) θ ( ν it | it )= 2 V p +1 h det( η p + ˆ F ′ p ) det( η p + ˆ F p ) i Q α =0 sin πν α (8 π α ′ ) p +12 Z ∞ dtt p − e − y t πα ′ × (cosh πν t − cosh πν t ) sinh πν t sinh πν t ∞ Y n =1 Z n , (114)where Z n , from (61), becomes Z n = ˜ Z n (1 − | z | n ) Q α =0 [1 − | z | n cosh 2 πν α t + | z | n ] , (115)with ˜ Z n , from (62) or (63), as˜ Z n = (cid:2) − | z | n cosh π ( ν + ν ) t + | z | n (cid:3) (cid:2) − | z | n cosh π ( ν − ν ) t + | z | n (cid:3) (116) or = (cid:2) − | z | n e − πν t cosh πν t + e − πν t | z | n (cid:3) (cid:2) − | z | n e πν t cosh πν t + e πν t | z | n (cid:3) . (117)As before, the large y interaction can be obtained from (110) with the large t -integrationas Γ p,p V p +1 ≈ (cos πν − cos πν ) q det( η p + ˆ F ′ p ) det( η p + ˆ F p )2 p − π p +12 (2 πα ′ ) p − y − p Γ (cid:18) − p (cid:19) , which is always non-negative, therefore implying an attractive interaction in general. Thisis consistent with what has been discussed for the p = p ′ = 5 or 6 case given in the previoussubsection. In other words, whenever p = p ′ <
6, the long-range interaction is alwaysnon-repulsive. The possible long-range repulsive interaction for p = p ′ can occur only for p = p ′ = 6 with all three ν , ν , ν real and non-vanishing. As discussed in the previoussubsection, it actually happens when the possible largest of the three ν α ’s is smaller thanthe sum of the remaining two when ν + ν + ν <
2. When ν is imaginary, given as ν = i ¯ ν with ¯ ν ∈ (0 , ∞ ), once again the integrand of the open string one-loop amplitude(114) has an infinite number of simples poles at t k = k/ ¯ ν with k = 1 , , · · · along thepositive t -axis, indicating the decay of the underlying system via the open string pair37roduction. The decay rate and the corresponding open string pair production rate canbe computed as before, respectively, as W p,p = 2 h det( η p + ˆ F ′ p ) det( η p + ˆ F p ) i sinh π ¯ ν sin πν ¯ ν (8 π α ′ ) p +12 ∞ X k =1 ( − ) k +1 (cid:16) ¯ ν k (cid:17) p − e − ky π ¯ ν α ′ × (cid:16) cosh kπν ¯ ν − ( − ) k (cid:17) sinh kπν ¯ ν Z k (¯ ν , ν ) , (118)where Z k (¯ ν , ν ) = ∞ Y n =1 h − − ) k | z k | n cosh kπν ¯ ν + | z k | n i (1 − | z k | n ) h − | z k | n cosh kπν ¯ ν + | z k | n i , (119)with | z k | = e − kπ/ ¯ ν , and W (1) p,p = 2 h det( η p + ˆ F ′ p ) det( η p + ˆ F p ) i sinh π ¯ ν sin πν ¯ ν (8 π α ′ ) p +12 ¯ ν p − e − y π ¯ ν α ′ × (cid:16) cosh πν ¯ ν + 1 (cid:17) sinh πν ¯ ν Z (¯ ν , ν ) . (120)The above two rates can be obtained from (88) and (90), respectively, by taking the limit ν →
0. We now discuss the cases for p = p ′ < The p = p ′ = 3 or case: The p = p ′ = 3 can be obtained from the p = p ′ = 5while p = p ′ = 4 can be obtained from the p = p ′ = 6 in the sense described in thepresent subsection given above. In either case, the worldvolume flux can be extended thefollowing way,ˆ F αβ = ˆ F α ′ β ′ g − ˆ g , ˆ F ′ αβ = ˆ F ′ α ′ β ′ g ′ − ˆ g ′ . (121)We then have here tan πν = | ˆ g − ˆ g ′ | g ˆ g ′ , (122)which gives ν → g → ∞ and ˆ g ′ → ∞ . The general closed stringtree-level cylinder amplitude is just given by (110) while the corresponding open stringone-loop annulus one is given by (114). The respective physics of these amplitudes such38s the nature of the interaction, the relevant instabilities and the potential open stringpair production and its enhancement can be similarly discussed in general following whatwe have done for the respective p = p ′ = 5 and p = p ′ = 6 cases. So we will not repeatthe same discussion here. For example, one typical interesting case is the p = p ′ = 3 onefor the following choice of fluxes,ˆ F α ′ β ′ = f − ˆ f g − ˆ g × , ˆ F ′ α ′ β ′ = f ′ − ˆ f ′ g ′ − ˆ g ′ × . (123)With the above fluxes, we havetanh π ¯ ν = | ˆ f − ˆ f ′ | − ˆ f ˆ f ′ , tan πν = | ˆ g − ˆ g ′ | g ˆ g ′ . (124)The closed string cylinder amplitude, the open string annulus amplitude, the decay rateand open string pair production rate of this system can be directly read from (110),(114), (118) and (120), respectively, for the present case. Their explicit expressions willnot be written down here. Their analysis, in particular the open string pair productionand its enhancement along with their potential applications, has been discussed in greatdetail in [21, 30, 31]. Again we will not repeat it here and refer there for detail. For the p = p ′ = 3 case, the discussion with the most general worldvolume fluxes is given explic-itly in a forthcoming paper by one of the present authors [55] and the basic conclusionremains the same. For example, the interaction amplitude can be given in terms of sixLorentz invariants constructed from the fluxes. The p = p ′ = 1 or case: The p = p ′ = 1 can be obtained from the p = p ′ = 5 while p = p ′ = 2 can be obtained from the p = p ′ = 6 again in the sense described in thepresent subsection given earlier. In either case, the worldvolume flux can be extended thefollowing way asˆ F αβ = ˆ F α ′ β ′ g − ˆ g g − ˆ g , ˆ F ′ αβ = ˆ F ′ α ′ β ′ g ′ − ˆ g ′ g ′ − ˆ g ′ . (125)We then have tan πν = | ˆ g − ˆ g ′ | g ˆ g ′ , tan πν = | ˆ g − ˆ g ′ | g ˆ g ′ , (126)39here both ν → ν → g k → ∞ , ˆ g ′ k → ∞ with k = 1 ,
2. Theclosed string cylinder amplitude can be obtained from (110) along with (111) and (112)or (113) by taking ν → ν → p = p ′ = 5 or 6 as well as in theprevious work [6, 21, 26, 30], which requires p = p ′ ≥ p = p ′ = 2 case, which also occurs for 2 ≤ p = p ′ ≤
6, as discussed in [26]by one of the present authors, when the added fluxes satisfy certain conditions. We referthis to [26] for detail.
The p = p ′ = 0 case: This is a trivial one and can be obtained from the p = p ′ = 6,similarly as above, by setting ν → ν → , = 0. For this system, there are no fluxes which canbe added to the worldvolume and therefore this system remains still as a 1/2 BPS one.The Γ , = 0 is just the usual “no-force” condition.In summary, in this section, we compute the closed string cylinder as well as thecorresponding open string one-loop annulus amplitude for the system of two Dp branes,placed parallel at separation, with 0 ≤ p = p ′ ≤
6, carrying the most general respectiveconstant worldvolume fluxes. We observe that the θ -function factor (107) in the integrandof the closed string cylinder amplitude for p = p ′ = 6 or 5 works also for the correspondingfactor for p = p ′ < ν → ν → , ν → ν , ν , ν → ≤ p = p ′ ≤ M (7) and the various zero-modes in the matter sector ofthe closed string boundary state representation of Dp-brane. This nice feature, as wewill see, provides us a trick to compute the closed string cylinder amplitude (as well asthe open string annulus amplitude) for the system of Dp and Dp ′ with p − p ′ = 2 , , p = p ′ .This will greatly simplify the computations for p = p ′ cases given in the following section.40sing the computed amplitudes in this section, we give also a general discussion on theproperties of the amplitudes such as the nature of the interaction, the onset of potentialtachyonic instability which is associated with the added worldvolume magnetic fluxes andthe open string pair production when an electric flux is added. In particular, we find thatthe interaction can be repulsive only for p = p ′ = 6 and when the added fluxes are allmagnetic with the possible largest one of three ν α ’s smaller than the sum of the remainingtwo when ν + ν + ν <
2. Otherwise, it is attractive or vanishes. The later case occurseither for the largest one equaling the sum of the remaining two or ν + ν + ν = 2.Whenever this happens, it implies that the underlying system is supersymmetric. Wealso find that the nature of the interaction is correlated with the existence of a potentialtachyonic instability. When the interaction is repulsive, there is no tachyonic instability.While the interaction is attractive, there is one. We give also a detail discussion on theopen string pair production enhancement following the line given in [6, 21, 26, 30]. p = p ′ case In this section, we move to compute the closed string tree-level cylinder amplitude betweenone Dp and the other Dp ′ , placed parallel at a separation transverse to the Dp, with p − p ′ = κ = 2 , , p ≤ p > p ′ . The discussion given in theprevious section for computing the cylinder amplitude for p = p ′ case makes it easierto carry out the computations in the present section. Once the cylinder amplitude isobtained, we can again use a Jacobi transformation to obtain the corresponding openstring one-loop annulus amplitude.The trick used in the subsection 3.2 helps us here in obtaining the amplitude for p = p ′ from that of p = p ′ if we extend the general flux ˆ F ′ p ′ on the Dp ′ to the ˆ F ′ p on a Dp in asimilar fashion as we did in extending a Dp brane flux for p < F ′ p ′ on the Dp ′ as,ˆ F ′ αβ = ( ˆ F ′ p ′ ) α ′ β ′ g ′ − ˆ g ′ g ′ κ − ˆ g ′ κ ( p +1) × ( p +1) , (127)where α, β = 0 , , · · · , p and α ′ , β ′ = 0 , , · · · p ′ . Here κ = 2 , ,
6. As before, at the end41f computations, we need to send ˆ g ′ k → ∞ with k = 1 , · · · , κ/
2. As discussed in thesubsection 3.2, this extension will not change anything about the corresponding matrix M ′ p ′ given in (7) for the present case and the RR zero-mode contribution in the matter partto the amplitude so long the above limit is taken at the end of computations. Moreover,unlike the extension given there, we have here a bonus for the extension of the bosoniczero-mode contribution to the amplitude so long things are taken care of properly. Let usexplain this in detail. If one computes the bosonic zero-mode contribution in the matterpart to the amplitude for the present case, as already given in (108), it is q det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) V NN (cid:0) π α ′ t (cid:1) − DD2 e − y πα ′ t . (128)In the present context, we have V NN = V p ′ +1 and DD = 9 − p . If we use the trick mentionedabove, we have then the following q det( η p + ˆ F ′ p ) det( η p + ˆ F p ) V p+1 (cid:0) π α ′ t (cid:1) − − p e − y πα ′ t . (129)Comparing the two, the nice thing here is that the t-dependent part is the same and theirdifference occurs only in the t-independent part. From (127), we have det( η p + ˆ F ′ p ) =(1 + ˆ g ′ ) · · · (1 + ˆ g ′ κ/ ) det( η p ′ + ˆ F ′ p ′ ) = ˆ g ′ · · · ˆ g ′ κ/ det( η p ′ + ˆ F ′ p ′ ) when we take ˆ g ′ k → ∞ with k = 1 , · · · κ/
2. So we have q det( η p + ˆ F ′ p ) det( η p + ˆ F p ) = ˆ g ′ · · · ˆ g ′ κ/ q det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) . (130)Note also that V p +1 = V p ′ +1 V κ . For a Dp brane with the flux (127), following the discussionof (68), we have the following coupling among others, T p Z ( C p ′ +1 ∧ ˆ F ′ ∧ ˆ F ′ · · · ∧ ˆ F ′ ) p +1 , (131)where the number of ˆ F ′ ’s is ( p − p ′ ) / C p ′ +1 is the ( p ′ +1)-form RR potential whichcan couple with Dp ′ brane. It is clear that when we take all ˆ g ′ k → ∞ with k = 1 , · · · , κ/ T p V κ ˆ g ′ · · · ˆ g ′ κ/ Z C p ′ +1 , (132)where we have now p − p ′ = κ and the coefficient in front of the coupling denotes thequantized charge N of Dp ′ brane in terms of its tension. In other words, we have N T p ′ = T p V κ ˆ g ′ · · · ˆ g ′ κ/ , (133)42hich gives V κ = N ˆ g ′ · · · ˆ g ′ κ/ T p ′ T p . (134)With the above considerations, now (129) becomes q det( η p + ˆ F ′ p ) det( η p + ˆ F p ) V p+1 (cid:0) π α ′ t (cid:1) − − p e − y πα ′ t = ˆ g ′ · · · ˆ g ′ κ/ q det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) V p ′ +1 N ˆ g ′ · · · ˆ g ′ κ/ T p ′ T p (cid:0) π α ′ t (cid:1) − − p e − y πα ′ t = N c p ′ c p q det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) V p ′ +1 (cid:0) π α ′ t (cid:1) − − p e − y πα ′ t , (135)where we have used the relation T p ′ /T p = c p ′ /c p with the normalization c p = √ π (2 π √ α ′ ) − p for the boundary state given right after (3). This factor c p ′ /c p is the one just needed toconvert the factor c p , which is used to compute the cylinder amplitude when p = p ′ (forexample, see (14)), to c p × c p ′ /c p = c p c p ′ , the correct one for the present amplitude. Thelarge integer N here implies what has been computed using the trick described is actu-ally between one single Dp and N Dp ′ branes (not a single Dp ′ ). To obtain the wantedamplitude with a single Dp ′ , we need to divide so obtained amplitude by N . Given whathas been discussed, the closed string tree-level cylinder amplitude for p − p ′ = κ = 0 canbe obtained from (54) asΓ p,p ′ = Γ p,p N = 2 V p +1 h det( η p + ˆ F ′ p ) det( η p + ˆ F p ) i Q α =0 sin πν α N (8 π α ′ ) p +12 Z ∞ dtt − p e − y πα ′ t η ( it ) × θ (cid:0) ν + ν + ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν + ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν + ν − ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν − ν (cid:12)(cid:12) it (cid:1) θ ( ν | it ) θ ( ν | it ) θ ( ν | it )= c p ′ c p V p ′ +1 h det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) i Q α =0 sin πν α (8 π α ′ ) p +12 Z ∞ dtt − p e − y πα ′ t η ( it ) × θ (cid:0) ν + ν + ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν + ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν + ν − ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν − ν (cid:12)(cid:12) it (cid:1) θ ( ν | it ) θ ( ν | it ) θ ( ν | it )= 2 V p ′ +1 h det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) i Q α =0 sin πν α κ (8 π α ′ ) p ′ +12 Z ∞ dtt − p e − y πα ′ t η ( it ) × θ (cid:0) ν + ν + ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν + ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν + ν − ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν − ν (cid:12)(cid:12) it (cid:1) θ ( ν | it ) θ ( ν | it ) θ ( ν | it )43 2 V p ′ +1 h det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) i (cid:2)P α =0 cos πν α − Q α =0 cos πν α − (cid:3) κ (8 π α ′ ) p ′ +12 × Z ∞ dtt − p e − y πα ′ t ∞ Y n =1 C n , (136)where in the first equality the Γ p,p is the cylinder amplitude (54) for the extended flux ˆ F ′ p given in (127), in the second equality we have used V p +1 = V p ′ +1 V κ and (135), in the thirdequality we have used the explicit expression for c p = √ π (2 π √ α ′ ) − p , and C n continues tobe given by (55) and the extension trick discussed in section 3 for ν α ’s still applies here.It is clear that the basic structure of the above cylinder amplitude is the same as thatfor the p = p ′ case discussed in the previous section. So we expect the same properties ofthe amplitude as discussed there such as the nature of the interaction and the potentialinstabilities. So we will not repeat this discussion here. Moreover, we also expect somespecial features to arise here which will be discussed later in this section.Once we have the closed string tree-level cylinder amplitude (136), the correspondingopen string one-loop annulus amplitude can be obtained from the next to the last equalityof this amplitude above by a Jacobi transformation following the standard prescriptiongiven earlier in the previous section. This open string one-loop annulus amplitude is thenΓ p,p ′ = − i V p ′ +1 h det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) i Q α =0 sin πν α κ (8 π α ′ ) p ′ +12 Z ∞ dtt p − e − y t πα ′ η ( it ) × θ (cid:0) ν + ν + ν it (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν + ν it (cid:12)(cid:12) it (cid:1) θ (cid:0) ν + ν − ν it (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν − ν it (cid:12)(cid:12) it (cid:1) θ ( ν it | it ) θ ( ν it | it ) θ ( ν it | it )= 2 V p ′ +1 h det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) i κ (8 π α ′ ) p ′ +12 Z ∞ dtt p − e − y t πα ′ Y α =0 sin πν α sinh πν α t × " X α =0 cosh πν α t − Y α =0 cosh πν α t − ∞ Y n =1 Z n , (137)where Z n is still given by (61). The use of this open string one-loop annulus amplitudeis for analyzing the small y behavior such as the onset of tachyonic instability and thatwhen ν is imaginary, the underlying system will decay via the so-called open string pairproduction. So we will give here a general discussion of both.When all three ν , ν and ν are real and non-vanishing, once again without loss ofgenerality and for convenience, we assume ν ≤ ν ≤ ν . Let us first focus on the case ν + ν + ν <
2. If ν + ν > ν , the interaction (136) is repulsive and there is no44otential tachyonic instability which can be checked from the integrand of (137) for large t . If ν + ν = ν , the amplitude vanishes and this indicates the preservation of certainsupersymmetry. On the other hand, if ν + ν < ν , the interaction is attractive and forlarge t , it behaves as ∼ e − y t πα ′ e π ( ν − ν − ν ) t = e − πt (cid:18) y π )2 α ′ − ν − ν − ν (cid:19) , (138)which blows up if y < π p ν − ν − ν ) α ′ , indicating the onset of tachyonic instability.For the case ν + ν + ν = 2, the interaction again vanishes and the underlying systempreserves certain supersymmetry. For 2 < ν + ν + ν <
3, we can only have ν + ν > ν and the interaction is also attractive. For large t, the amplitude behaves as ∼ e − y t πα ′ e π ( ν + ν + ν − t = e − πt (cid:18) y π )2 α ′ − ν ν ν − (cid:19) , (139)which blows up if y < π p ν + ν + ν − α ′ , indicating again the tachyonic instability.So everything here is consistent with what has been discussed in the previous section for p = p ′ case.If ν = i ¯ ν , i.e., imaginary, with ¯ ν ∈ (0 , ∞ ), so the factor sin πν / sinh πν t in theintegrand of (137) becomes sinh π ¯ ν / sin π ¯ ν t , indicating the appearance of an infinitenumber of simple poles of the integrand at t k = k/ ¯ ν with k = 1 , , · · · . So this impliesthat the amplitude has an imaginary part, indicating the decay of the underlying systemvia the so-called open string pair production. The decay rate per unit volume of Dp ′ brane worldvolume can be computed as before as W p,p ′ = − V p ′ +1 = 2 − κ h det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) i sinh π ¯ ν sin πν sin πν ¯ ν (8 π α ′ ) p ′ +12 ∞ X k =1 ( − ) k +1 (cid:16) ¯ ν k (cid:17) p − × (cid:16) cosh kπν ¯ ν − ( − ) k cosh kπν ¯ ν (cid:17) sinh kπν ¯ ν sinh kπν ¯ ν e − ky πα ′ ¯ ν Z k (¯ ν , ν , ν ) , (140)where Z k = ∞ Y n =1 (cid:20)(cid:16) | z k | n − − ) k | z k | n cosh kπν ¯ ν cosh kπν ¯ ν (cid:17) − | z k | n sinh kπν ¯ ν sinh kπν ¯ ν (cid:21) (1 − | z k | n ) (cid:16) − | z k | n cosh πkν ¯ ν + | z k | n (cid:17) (cid:16) − | z k | n cosh πkν ¯ ν + | z k | n (cid:17) , (141)45ith | z k | = e − kπ/ ¯ ν . As before, the open string pair production rate is given by the k = 1term of the above as W (1) p,p ′ = 2 − κ h det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) i sinh π ¯ ν sin πν sin πν (8 π α ′ ) p ′ +12 ¯ ν p − e − y πα ′ ¯ ν × (cid:16) cosh πν ¯ ν + cosh πν ¯ ν (cid:17) sinh πν ¯ ν sinh πν ¯ ν Z (¯ ν , ν , ν ) . (142)In the above, we assume p ≥
5. The p = 3 , ν → p = 1 , ν , ν → p − p ′ = 2, p − p ′ = 4 and p − p ′ = 6 one by onein the following.Before we move to that, let us check one thing mentioned in the previous section.It is that the interaction between a Dp and a Dp ′ for 6 ≥ p > p ′ , placed parallel at aseparation and without any brane flux present, is attractive when p − p ′ = 2, vanisheswhen p − p ′ = 4 and is repulsive when p − p ′ = 6 ( p = 6 , p ′ = 0). Let us check each ofthem explicitly here. For p = 6 and p ′ = 0, we have here ν = ν = ν = 1 / , = − V π α ′ ) Z ∞ dtt e − y πα ′ t ∞ Y n =1 (1 + | z | n ) (1 − | z | n ) (1 + | z | n ) , (143)where we have used (55) for C n . It is indeed repulsive since Γ , < y . For p − p ′ = 2, we have one of three ν , ν , ν being half and the remaining two being zerowhile for p − p ′ = 4, we have two of them being half and the remaining one being zero.For the former case, the cylinder amplitude (136) is now,Γ κ =2 = 2 V p ′ +1 (8 π α ′ ) p ′ +12 Z ∞ dtt − p e − y πα ′ t ∞ Y n =1 (1 + | z | n ) (1 − | z | n ) (1 − | z | n ) , (144)where we have used (55) for C n . It is indeed attractive since Γ κ =2 > y . For thelatter case, the amplitude Γ κ =4 simply vanishes due to the constant factor [ P α =0 cos πν α − Q α =0 cos πν α −
1] in the amplitude (136) being zero for the present case. All are asexpected. We now discuss each separate case mentioned earlier. p − p ′ = 2 case In this subsection, we will focus on the p − p ′ = 2 case, specifically. We will discuss hereeach of p = 6 or 5; p = 4 or 3 and p = 2, separately. Let us begin with p = 6 or 5 case.46 he p = 6 or case: For either of these two cases, we can extend the flux ˆ F ′ p ′ to ˆ F ′ p ,prescribed in (127), as ( ˆ F ′ p ) αβ = ( ˆ F ′ p ′ ) α ′ β ′ g ′ − ˆ g ′ , (145)where we will take ˆ g ′ → ∞ at the end of computations. For illustration purpose, weconsider the flux ˆ F p on the Dp brane the following formˆ F αβ = ( ˆ F p ′ ) α ′ β ′ g − ˆ g , (146)where g is finite. We can then determine ν astan πν = | ˆ g ′ − ˆ g | g ′ ˆ g , (147)which gives tan πν = 1 / ˆ g when we take ˆ g ′ → ∞ limit. For a given fixed ˆ g , the discussiongoes exactly the same as we did for the p = p ′ = 6 case given in the previous section. Forthis reason, we will not repeat it here. We here focus on vanishingly small ˆ g (in practiceˆ g ≪
1) for which we have ν → /
2. From (136), we have the closed string cylinderamplitude for the present case asΓ p,p ′ = 2 V p ′ +1 h det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) i (cid:0) cos πν − sin πν (cid:1) (8 π α ′ ) p ′ +12 Z ∞ dtt − p e − y πα ′ t ∞ Y n =1 C n , (148)where C n , from (55), is now C n = ˜ C n (1 − | z | n ) Q α =0 [1 − | z | n cos 2 πν α + | z | n ] , (149)with˜ C n = (cid:2) (1 + | z | n ) − | z | n sin π ( ν + ν ) (cid:3) (cid:2) (1 + | z | n ) − | z | n sin π ( ν − ν ) (cid:3) (150) or = (cid:2) (1 + e πiν | z | n ) − | z | n e iπν sin πν (cid:3) × (cid:2) (1 + e − πiν | z | n ) − | z | n e − iπν sin πν (cid:3) . (151)When p = 6, we have two choices: 1) ν , ν ∈ [0 ,
1) and 2) ν = i ¯ ν with ¯ ν ∈ (0 , ∞ ) and ν ∈ [0 , πν > sin πν , vanishes if47os πν = sin πν , and is repulsive if cos πν < sin πν . Let us discuss each of these indetail. For the attractive interaction, when ν < /
2, we have, from cos πν > sin πν ,cos πν > sin πν which gives ν + ν < / ν . In other words, the largest ν is largerthan the sum of the remaining two and this is consistent with what we have achieved inthe previous section since now ν + ν + ν < / <
2. While for ν > / ν cannot be 1 / − cos πν > sin πν which gives ν > ν + 1 / ν + ν and again the largest ν is larger than the sum of the remaining two. The interactionvanishes when ν + ν = 1 / ν if ν ≤ / ν = ν + 1 / ν + ν if ν > / ν + ν > / ν if ν ≤ / ν ≤ / ν < ν + 1 / ν + ν if ν ≤ / ν > / ν < ν + 1 / ν + ν if ν > / ν ≤ / < ν + ν < / ν > / ν > / ν > ν we willhave ν < ν + 1 / ν + ν or if ν > ν we will have ν < ν + ν . Note that for therepulsive interaction, we have ν + ν + ν <
2. Everything here is consistent with whathas been discussed in the p = p ′ = 6 case in the previous section if we take the present ν = 1 /
2. So here is just a special case with ν = 1 / p = p ′ = 6 in the previous section.For small y , the small t integration becomes important for the amplitude. This givesalso a potential singularity of this amplitude since we have two potential sources for this.One is from the t − (9 − p ) / factor in the integrand and the other comes from infinite productof C n , each of which has a factor (1 − | z | n ) ∼ t for small t in the denominator of C n .Both of them blow up for small t . The discussion will again be the special case of whathas been discussed in the previous section for the p = p ′ cases and will not be repeatedhere.For the second case, the sign of the integrand becomes again indefinite for small t andthe discussion goes the same as we did for the p = p ′ case and will not be repeated here.The underlying physics for either of these two cases will become clear if we examine itfrom the corresponding open string one-loop annulus amplitude. This annulus amplitudecan be read from (137) for the present case asΓ p,p ′ = 2 V p ′ +1 h det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) i (8 π α ′ ) p ′ +12 Z ∞ dtt p − e − y t πα ′ sinh π ¯ ν sin πν sin π ¯ ν t sinh πν t sinh πt × "(cid:18) cosh πν t − cosh πt (cid:19) + 4 sin π ¯ ν t (cid:18) cosh πν t cosh πt − cos π ¯ ν t (cid:19) ∞ Y n =1 Z n , (152)48here Z n can be read from (61) as Z n = [1 − | z | n cos 2 π ¯ ν t + | z | n ] − ˜ Z n (1 − | z | n ) [1 − | z | n cosh πt + | z | n ] [1 − | z | n cosh 2 πν t + | z | n ] , (153)with˜ Z n = "(cid:18) − | z | n cos π ¯ ν t cosh π (cid:18) ν + 12 (cid:19) t + | z | n (cid:19) +4 | z | n sin π ¯ ν t sinh π (cid:18) ν + 12 (cid:19) t (cid:21) "(cid:18) − | z | n cos π ¯ ν t cosh π (cid:18) ν − (cid:19) t + | z | n (cid:19) +4 | z | n sin π ¯ ν t sinh π (cid:18) ν − (cid:19) t (cid:21) > . (154)For large t , Z n ≈ ∼ e − y t πα ′ e π | − ν | t = e − πt (cid:20) y π )2 α ′ − | / − ν | (cid:21) , (155)which blows up when y < π p | − ν | α ′ , indicating the onset of tachyonic instabilitymentioned above. Once again, the factor sin π ¯ ν t in the denominator of the integrand ofthe amplitude gives an infinite number of simple poles along the positive t-axis (note thatthe integrand is regular as t →
0) at t k = k/ ¯ ν with k = 1 , , · · · . This implies that theamplitude has an imaginary part, indicating the decay of the underlying system via theso-called open string pair production. The decay rate per unit p ′ -brane volume can becomputed as before to give W p,p ′ = − V p ′ +1 = 2 h det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) i sinh π ¯ ν sin πν ¯ ν (8 π α ′ ) p ′ +12 ∞ X k =1 ( − ) k +1 (cid:16) ¯ ν k (cid:17) p − e − ky πα ′ ¯ ν × (cid:16) cosh kπν ¯ ν − ( − ) k cosh kπ ν (cid:17) sinh kπν ¯ ν sinh kπ ν Z k (¯ ν , ν ) , (156)where Z k (¯ ν , ν ) = ∞ Y n =1 h − − ) k | z k | n cosh kπ ¯ ν (cid:0) ν + (cid:1) + | z k | n i (1 − | z k | n ) h − | z k | n cosh kπ ¯ ν + | z k | n i × h − − ) k | z k | n cosh kπ ¯ ν (cid:0) ν − (cid:1) + | z k | n i h − | z k | n cosh kπν ¯ ν + | z k | n i , (157)49ith | z k | = e − kπ/ ¯ ν . The open string pair production rate is the k = 1 term of the aboveand it is W (1) p,p ′ = 2 h det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) i sinh π ¯ ν sin πν (8 π α ′ ) p ′ +12 ¯ ν p − e − y πα ′ ¯ ν × (cid:16) cosh πν ¯ ν + cosh π ν (cid:17) sinh πν ¯ ν sinh π ν Z (¯ ν , ν ) . (158)One can check easily that the above decay rate or the open string pair production rateis just the special case of (140) or (142) for ν = 1 / κ = 2, respectively. There isan interesting enhancement of the pair production rate even in the absence of magneticflux for which we have ν = 0 for small ¯ ν . This rate can be obtained from the above bytaking ν → W (1) p,p ′ = 2 h det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) i sinh π ¯ ν (8 π α ′ ) p ′ +12 ¯ ν p − e − y πα ′ ¯ ν (cid:16) π ν (cid:17) sinh π ν Z (¯ ν , , (159)where Z (¯ ν ,
0) = ∞ Y n =1 h | z | n cosh π ν + | z | n i (1 − | z | n ) h − | z | n cosh π ¯ ν + | z | n i . (160)We would like to remark here that given the form of flux ˆ F p (146), the above decay orpair production rate is also valid for p ≥ ν = 0. For illustration, let ushave a consideration of the following special choice of fluxes ˆ F p and ˆ F ′ p ′ for p = 5 asˆ F = f − ˆ f · · · × , ˆ F ′ = f ′ − ˆ f ′ × , (161)where there is no magnetic flux present. This gives the ˆ g = 0 in (146) and so we have ν = 1 /
2. With this special choice of fluxes, we have ν = 0 andtanh π ¯ ν = | ˆ f − ˆ f ′ | − ˆ f ˆ f ′ , (162)50he pair production rate (159) becomes W (1) p,p ′ = 2 | ˆ f − ˆ f ′ | (8 π α ′ ) p ′ +12 ¯ ν p − e − y πα ′ ¯ ν (cid:16) π ν (cid:17) sinh π ν Z (¯ ν , , (163)where p = 5 and p ′ = 3. As pointed out above, this rate is also valid for p = 3 and p ′ = 1.For small ¯ ν , Z (¯ ν , ≈ e π/ (2¯ ν ) ≫ p = p ′ case. This large enhancement was also considered by one of thepresent authors in [18] and it is essentially due to the Dp ′ brane acting effectively as astringy magnetic flux. The p = 4 or case: For this case, we need to set ν = 0 from the outset. Now the roleof ν in the above p = 6 or 5 case is replaced by that of ν in the present one. By thesame token, we extend the flux ˆ F ′ p ′ to ˆ F ′ p the following way,( ˆ F ′ p ) αβ = ( ˆ F ′ p ′ ) α ′ β ′ g ′ − ˆ g ′ , (164)where we will take ˆ g ′ → ∞ at the end of computations. For illustration purpose, weconsider the flux ˆ F p on the Dp brane the following formˆ F αβ = ( ˆ F p ′ ) α ′ β ′ g − ˆ g , (165)where ˆ g is finite. We can then determine ν astan πν = | ˆ g ′ − ˆ g | g ′ ˆ g , (166)which gives tan πν = 1 / ˆ g when we take ˆ g ′ → ∞ . For a general ˆ g , the present discussionis not different from its correspondence in the p = p ′ case in the previous section and wewill not repeat it here. We here also focus on the small or vanishing ˆ g for which we have ν → /
2. The closed string cylinder amplitude can be read from the last equality in(136) asΓ p,p ′ = 2 V p ′ +1 h det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) i cos πν (8 π α ′ ) p ′ +12 Z ∞ dtt − p e − y πα ′ t ∞ Y n =1 C n , (167)51here C n can be read from (55) as C n = (1 + | z | n ) (1 + 2 | z | n cos 2 πν + | z | n ) (1 − | z | n ) (1 − | z | n cos 2 πν + | z | n ) . (168)It is clear that this interaction can only be attractive which is consistent with what wehave achieved in the previous section. This interaction vanishes if ν = 1 /
2. This caneasily be understood as follows. The ν = 1 / ′ carries no flux or from that the Dp carriesno flux but the Dp ′ carries such a magnetic flux. In the former case, the contribution tothe interaction from the Dp is actually dominated by the infinitely large magnetic fluxwhich gives an infinite many of D(p - 2) branes whose dimensionality is the same as thatof the Dp ′ with p ′ = p − ′ behaves effectively asinfinitely many D(p ′ - 2) branes. So now the interaction is between one Dp and infinitelymany D(p - 4)-branes, placed parallel at a separation, which vanishes since there does notexist any interaction between D-branes whose dimensionality differs by 4. Given what hasbeen said, the two cases are still different in that the former case preserves 1/2 spacetimesupersymmetries while the later preserves only 1/4.Again the small y physics can be best described in terms of the corresponding openstring one-loop annulus amplitude (137). For the present case, it isΓ p,p ′ = 2 V p ′ +1 h det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) i sin πν (8 π α ′ ) p ′ +12 Z ∞ dtt p − e − y t πα ′ × (cid:0) cosh πν t − cosh πt (cid:1) sinh πν t sinh πt ∞ Y n =1 Z n , (169)where Z n can be read from (61) as Z n = h(cid:0) | z | n − | z | n cosh πν t cosh πt (cid:1) − | z | n sinh πν t sinh πt i (1 − | z | n ) (1 − | z | n cosh πt + | z | n )(1 − | z | n cosh 2 πν t + | z | n ) . (170)When ν = 1 /
2, once again the amplitude vanishes and we have explained this in thecylinder amplitude. For p = 4, we have in general ν ∈ [0 , t , we have Z n ≈ ∼ e − y t πα ′ e π | − ν | t = e − πt (cid:20) y π )2 α ′ − | − ν | (cid:21) , (171)52hich blows up when y < π p | − ν | α ′ , indicating again the onset of tachyonic instabil-ity. We now consider an imaginary ν = i ¯ ν with ¯ ν ∈ (0 , ∞ ). We can use the followingspecific fluxes ˆ F p and ˆ F ′ p ′ to give a representative discussion,( ˆ F p ) = − ( ˆ F p ) = ˆ f , ( ˆ F ′ p ′ ) = − ( ˆ F ′ p ′ ) = ˆ f ′ , (172)where the rest components of both ˆ F p and ˆ F ′ p ′ are zero and we have also taken the ˆ g = 0given in (165). With this choice, we havetanh π ¯ ν = | ˆ f − ˆ f ′ | − ˆ f ′ ˆ f . (173)We have now the amplitude (169) asΓ p,p ′ = 2 V p ′ +1 | ˆ f − ˆ f ′ | (8 π α ′ ) p ′ +12 Z ∞ dtt p − e − y t πα ′ (cid:0) cos π ¯ ν t − cosh πt (cid:1) sin π ¯ ν t sinh πt ∞ Y n =1 Z n , (174)where Z n continues to be given by (170) but with ν = i ¯ ν . This amplitude has now atachyonic instability when y < π √ α ′ . In addition, the sin π ¯ ν t factor in the denominatorof the integrand of the above amplitude gives again an infinite number of simple poles at t k = k/ ¯ ν with k = 1 , , · · · and therefore the amplitude has an imaginary part, indicatingthe decay of the underlying system via the so-called open string pair production. Thedecay rate per unit volume of Dp ′ brane can be computed to give W p,p ′ = − V p ′ +1 = 4 | ˆ f − ˆ f ′ | ¯ ν (8 π α ′ ) p ′ +12 ∞ X k =1 ( − ) k +1 (cid:16) ¯ ν k (cid:17) p − e − ky πα ′ ¯ ν h cosh kπ ν − ( − ) k i sinh kπ ν Z k (¯ ν , ν = 1 / , (175)where Z k (¯ ν , ν = 1 /
2) = ∞ Y n =1 h − − ) k | z k | n cosh kπ ν + | z k | n i (1 − | z k | n ) (1 − | z k | n cosh kπ ¯ ν + | z k | n ) , (176)with | z k | = e − kπ/ ¯ ν . The open string pair production rate is given by the first k = 1 termof the above and it is W (1) p,p ′ = 4 | ˆ f − ˆ f ′ | (8 π α ′ ) p ′ +12 ¯ ν p − e − y πα ′ ¯ ν h cosh π ν + 1 i sinh π ν Z (¯ ν , ν = 1 / , (177)53here Z (¯ ν , ν = 1 /
2) = ∞ Y n =1 h | z | n cosh π ν + | z | n i (1 − | z | n ) (1 − | z | n cosh π ¯ ν + | z | n ) . (178)This pair production rate is, as expected, the same as that given in (163). The samediscussion applies here, too. Note also that the decay rate (175) and the pair productionrate (177) are just special cases of (140) and (142) when we take ν = 0 and ν = 1 /
2, asexpected.
The p = 2 case: This is the last case we will discuss in this subsection. The D0 branecannot carry any worldvolume flux. However, by the same token as before, we can havethe following extension as ˆ F ′ = g ′ − ˆ g ′ , (179)where we will set ˆ g ′ → ∞ at the end of computations. For this case, we will consider themost general D2 worldvolume flux as an example. This flux can be expressed asˆ F = f ˆ f − ˆ f g − ˆ f − ˆ g . (180)Using (24), we have We here take this simple case as a direct check of the trick used. For the general flux (180) on D2, wecan compute its M-matrix using (7) as M = ( M α β , − δ i j ) with α, β = 0 , , i, j = 3 , , · · ·
9, where M α β = (1 + ˆ g − ˆ f − ˆ f ) − g + ˆ f + ˆ f −
2( ˆ f + ˆ f ˆ g ) 2( ˆ f ˆ g − ˆ f )2( ˆ f ˆ g − ˆ f ) 1 − ˆ g + ˆ f − ˆ f
2( ˆ f ˆ f − ˆ g ) −
2( ˆ f + ˆ f ˆ g ) 2( ˆ f ˆ f + ˆ g ) 1 − ˆ g − ˆ f + ˆ f , (181)while for D0 brane, we have M ′ = (1 , − δ i ′ j ′ ) with i ′ , j ′ = 1 , , · · ·
9. So we have W = M M ′ T = w α β I × ! , (182)where w α β = (1 + ˆ g − ˆ f − ˆ f ) − g + ˆ f + ˆ f
2( ˆ f + ˆ f ˆ g ) −
2( ˆ f ˆ g − ˆ f )2( ˆ f ˆ g − ˆ f ) − (1 − ˆ g + ˆ f − ˆ f ) −
2( ˆ f ˆ f − ˆ g ) −
2( ˆ f + ˆ f ˆ g ) −
2( ˆ f ˆ f + ˆ g ) − (1 − ˆ g − ˆ f + ˆ f ) , (183)which is nothing but the limit of w α β given in (184) when we set ˆ g ′ → ∞ . This confirms that the trickused in obtaining the corresponding eigenvalues of w works indeed. α β = g + ˆ f + ˆ f g − ˆ f − ˆ f −
2( ˆ f + ˆ f ˆ g )(1 − ˆ g ′ ) −
4( ˆ f − ˆ f ˆ g )ˆ g ′ (1+ˆ g − ˆ f − ˆ f )(1+ˆ g ′ ) −
4( ˆ f + ˆ f ˆ g )ˆ g ′ +2( ˆ f − ˆ f ˆ g )(1 − ˆ g ′ )(1+ˆ g − ˆ f − ˆ f )(1+ˆ g ′ ) −
2( ˆ f − ˆ f ˆ g )1+ˆ g − ˆ f − ˆ f (1 − ˆ g + ˆ f − ˆ f )(1 − ˆ g ′ )+4(ˆ g − ˆ f ˆ f )ˆ g ′ (1+ˆ g − ˆ f − ˆ f )(1+ˆ g ′ ) 2(1 − ˆ g + ˆ f − ˆ f )ˆ g ′ − g − ˆ f ˆ f )(1 − ˆ g ′ )(1+ˆ g − ˆ f − ˆ f )(1+ˆ g ′ ) −
2( ˆ f + ˆ f ˆ g )1+ˆ g − ˆ f − ˆ f g + ˆ f ˆ f )(1 − ˆ g ′ ) − − ˆ g − ˆ f + ˆ f )ˆ g ′ (1+ˆ g − ˆ f − ˆ f )(1+ˆ g ′ ) (1 − ˆ g − ˆ f + ˆ f )(1 − ˆ g ′ )+4(ˆ g + ˆ f ˆ f )ˆ g ′ (1+ˆ g − ˆ f − ˆ f )(1+ˆ g ′ ) . (184)One can check explicitly that the above w has one eigenvalue unity and the other two λ and λ − satisfy λ + λ − = 2(ˆ g + ˆ f + ˆ f − g − ˆ f − ˆ f , (185)where we have taken ˆ g ′ → ∞ . Setting λ = e πiν , we havetan πν = q − ˆ f − ˆ f ˆ g . (186)We have two cases to consider: 1) ˆ f + ˆ f <
1, 2) 1 < ˆ f + ˆ f < g . For the firstcase, ν ∈ [0 , g is finite, the discussion goes the same as the pure magnetic case of p = p ′ = 2 discussed in the previous section and we will not repeat it here and refer therefor detail. If there is no magnetic flux, i.e., ˆ g = 0, we have then ν = 1 /
2. The closedstring cylinder amplitude can be obtained from (136) with ν = ν = 0 and ν = 1 / , = 2 V q − ˆ f − ˆ f (8 π α ′ ) Z ∞ dtt e − y πα ′ t ∞ Y n =1 (1 + | z | n ) (1 − | z | n ) (1 − | z | n ) , (187)where we also use (55) for C n . The integrand of this amplitude has a potential divergencebut has no sign ambiguity for small t , indicating a potential tachyonic instability but noopen string pair production, even though there exist applied electric fluxes. To see bothof these clearly, we need the corresponding open string one-loop annulus amplitude whichcan be read from (137) asΓ , = 2 V q − ˆ f − ˆ f (8 π α ′ ) Z ∞ dtt e − y t πα ′ (cid:0) cosh πt − (cid:1) sinh πt × ∞ Y n =1 [1 − | z | n cosh πt + | z | n ] (1 − | z | n ) [1 − | z | n cosh πt + | z | n ] , (188)where we have used (61) for Z n . For large t , the integrand of the above behaves like ∼ e − y t πα ′ e πt = e − πt (cid:20) y π )2 α ′ − (cid:21) , (189)55hich blows up when y < π √ α ′ , indicating the onset of tachyonic instability. The inte-grand is regular at t → − ˆ f − ˆ f = ǫ → + . With ˆ g = 0, from (185), we alwayshave ν = 1 / − ˆ f − ˆ f = ǫ → + . Due to thetachyonic instability when y < π √ α ′ , we need to have y > π √ α ′ to validate the amplitudecomputations. Once this holds, the effective tension on the virtual open strings is lessthan the critical one even if we take 1 − ˆ f − ˆ f → + . So this limiting tension cannotbreak the open strings and therefore there is no open string pair production.We now move to the second case for which we cannot set ˆ g vanish. So we haveˆ g < ˆ g + ˆ f + ˆ f − < g and 0 < ˆ g + 1 − ˆ f − ˆ f < ˆ g . From (185), this must implythat ν is imaginary, i.e., ν = i ¯ ν with ¯ ν ∈ (0 , ∞ ). This can also be seen directly from(186) and it is now tanh π ¯ ν = q ˆ f + ˆ f − | ˆ g | . (190)The present closed string cylinder amplitude can be read from (136) with ν = ν = 0and ν = i ¯ ν asΓ , = 2 V q g − ˆ f − ˆ f (cosh π ¯ ν − (8 π α ′ ) Z ∞ dtt e − y πα ′ t × ∞ Y n =1 [1 − | z | n cosh π ¯ ν + | z | n ] (1 − | z | n ) (1 − | z | n cosh 2 π ¯ ν + | z | n ) , (191)where we have used (55) for C n . As before, the large separation interaction is obviouslyattractive but the integrand for small t has an ambiguity of its sign in addition to apotential singularity. The sign ambiguity implies a decay of the underlying system via theopen string pair production while the potential singularity implies a potential tachyonicinstability. To check both of these explicitly, we need to examine the corresponding open56tring one-loop annulus amplitude which can be read from (137) asΓ , = 2 V q ˆ f + ˆ f − π α ′ ) Z ∞ dtt e − y t πα ′ (1 − cos π ¯ ν t ) sin π ¯ ν t × ∞ Y n =1 [1 − | z | n cos π ¯ ν t + | z | n ] (1 − | z | n ) [1 − | z | n cos 2 π ¯ ν t + | z | n ] , (192)where we have used (61) for Z n . For large t , the integrand of this annulus amplitude doesnot have a blowing up behavior and therefore there is no potential tachyonic singularity.However, the integrand does have an infinite number of simple poles at t k = (2 k − / ¯ ν with k = 1 , , · · · , indicating the decay of the system via the open string pair production.The decay rate and the open string pair production rate can be read from (140) and (142),respectively, with ν = ν = 0, as W p,p ′ = 16 ¯ ν q ˆ f + ˆ f − π α ′ ) ∞ X k =1 (cid:18) ¯ ν k − (cid:19) p − e − (2 k − y πα ′ ¯ ν ∞ Y n =1 (1 + | z k − | n ) (1 − | z k − | n ) , (193)where we have used (141) for Z k and | z k | = e − kπ/ ¯ ν , and W (1) p.p ′ = 16 q ˆ f + ˆ f − π α ′ ) ¯ ν p − e − y πα ′ ¯ ν ∞ Y n =1 (1 + | z | n ) (1 − | z | n ) . (194)Both of the rates blow up when ¯ ν → ∞ for which ˆ f + ˆ f − → ˆ g , the critical limit.In the above, we have an interesting thing happening. Note that the above discussionfor ˆ f + ˆ f < g = 0. For given ˆ g = 0, there is a potential openstring tachyonic instability but no open string pair production if ˆ f + ˆ f < f + ˆ f > .For the former, the electric fluxes representing the respective delocalized fundamentalstrings (see footnote (4)) have no interaction with the D0 brane [4]. So their presencejust gives certain modifications of the pure magnetic case of the underlying system butnot its characteristic behavior, as discussed above. So a potential tachyonic instabilityis expected when the brane separation reaches the distance determined by the tachyonicshift. For the latter, we have to admit that we don’t have a better explanation of itexcept for the following observation. Before that, we would also like to point out thatwhen ˆ f + ˆ f = 1, the above amplitudes and rates computed all vanish. Note that in both cases we need to have ˆ f + ˆ f < g and for the former case, it satisfies trivially. w given in (184) while keeping ˆ g ′ large. Note thatthe tr w is a D2 worldvolume Lorentz invariant and the eigenvalue equation (185) is nowreplaced by λ + λ − = tr w − g ˆ g ′ ) − (ˆ g − ˆ g ′ ) + (1 + ˆ g ′ )( ˆ f + ˆ f )(1 + ˆ g ′ )(1 + ˆ g − ˆ f − ˆ f ) , (195)which gives (185) if we send ˆ g ′ → ∞ . For the present purpose, we keep ˆ g ′ large and takethe limit ˆ g ′ → ∞ only at the end of the discussion. If we set λ = e πiν , we have fromthe above, cos πν = 1 + ˆ g ˆ g ′ q (1 + ˆ g ′ )(1 + ˆ g − ˆ f − ˆ f ) , sin πν = q (ˆ g − ˆ g ′ ) − (1 + ˆ g ′ )( ˆ f + ˆ f ) q (1 + ˆ g ′ )(1 + ˆ g − ˆ f − ˆ f ) . (196)Note that we have a few Lorentz invariants of D2 brane worldvolume: ˆ F αβ ˆ F αβ = 2[ˆ g − ( ˆ f + ˆ f )] , ˆ F ′ αβ ˆ F ′ αβ = 2ˆ g ′ , ˆ F αβ ˆ F ′ αβ = 2ˆ g ˆ g ′ . Therefore the numerator on the right side ofsin πν in (196) is also Lorentz invariant. In other words, the ν is a Lorentz invariant. Letus now examine this numerator which can be rewritten as [ˆ g ′ (1 − ˆ f − ˆ f ) + ˆ g − g ′ ˆ g − ( ˆ f + ˆ f )] / . Due to the ˆ g ′ -factor in the denominator, we need to have a ˆ g ′ -factor in thenumerator to give a non-vanishing ν and the numerator becomes [(1 − ˆ f − ˆ f )ˆ g ′ ] / whenwe take ˆ g ′ → ∞ . Note that (1 − ˆ f − ˆ f )ˆ g ′ is also a Lorentz invariant of D2 worldvolumesince it is related to ( ǫ αβγ ǫ αβδ + ˆ F α β ǫ α β δ ˆ F α β γ ) ˆ F ′ γτ ˆ F τδ = 4ˆ g ′ (1 − ˆ f − ˆ f ). So it isclear now that the sign of 1 − ˆ f − ˆ f determines the nature of ν , real or imaginary!When ˆ f + ˆ f < ν is real and the underlying system with ˆ g ′ → ∞ resembles a puremagnetic case. When ˆ f + ˆ f = 1, ν vanishes. While ˆ f + ˆ f >
1, the ν is imaginaryand the underlying system with ˆ g ′ → ∞ has a long-range attractive interaction but atsmall brane separation the amplitude has a sign ambiguity, indicating a decay via theopen string pair production as described above. p − p ′ = 4 case For p ≤
6, we have only three cases to consider in this subsection: 1) p = 6 , p ′ = 2; 2) p = 5 , p ′ = 1 and 3) p = 4 , p ′ = 0. The extension of ˆ F ′ p ′ on the Dp ′ brane to ˆ F ′ p given in58127) in the present context takes the following form( ˆ F ′ p ) αβ = ( ˆ F ′ p ′ ) α ′ β ′ g ′ − ˆ g ′ g ′ − ˆ g ′ , (197)where we need to take both ˆ g ′ → ∞ and ˆ g ′ → ∞ at the end of relevant computations. The p = 6 case: For a general flux ˆ F on D6, even with the above extension (197)for ˆ F ′ on D2, the characteristic behavior of the closed string cylinder amplitude or thecorresponding open string one-loop annulus amplitude is similar to that for the p = p ′ = 6discussed in the previous section. We here specify the ˆ F to the following form along withthe extension of ˆ F ′ as,ˆ F αβ = ( ˆ F ) α ′ β ′ g − ˆ g g − ˆ g , ˆ F ′ αβ = ( ˆ F ′ ) α ′ β ′ g ′ − ˆ g ′ g ′ − ˆ g ′ . (198)We have therefore tan πν = 1ˆ g , tan πν = 1ˆ g , (199)where we have taken ˆ g ′ → ∞ and ˆ g ′ → ∞ . For general ˆ g and ˆ g , the discussion continuesto be the same as that of the p = p ′ = 6 case. We further specify to the case of bothˆ g = 0 and ˆ g = 0 for which ν = ν = 1 /
2. Now the closed string cylinder amplitude canbe read from (136) with ν = ν = 1 / , = − V h det( η + ˆ F ′ ) det( η + ˆ F ) i sin πν (8 π α ′ ) Z ∞ dtt e − y πα ′ t × ∞ Y n =1 [(1 + | z | n ) − | z | n cos πν ] (1 − | z | n ) (1 + | z | n ) (1 − | z | n cos 2 πν + | z | n ) , (200)where we have used (55) for C n . The amplitude vanishes for ν = 0 when ˆ F = ˆ F ′ = 0.This is consistent with the fact that there is no interaction between a Dp and a Dp ′ ,placed parallel at a separation, with p − p ′ = 4. When ν ∈ (0 , p = p ′ = 6 in the previous sectionsince we have here ν + ν > ν if ν ≤ / ν + ν > ν if ν > /
2, i.e., when thepossible largest one among the three ν , ν , ν is less than the sum of the remaining two.We therefore don’t expect to have a potential open string tachyonic instability which isobvious from the above amplitude.When ν is imaginary, i.e., ν = i ¯ ν with ¯ ν ∈ (0 , ∞ ), the large separation interactionbecomes attractive. For small y , the small t becomes important in the integration. Butfor small t , the factor (1 − | z | n cosh 2 π ¯ ν + | z | n ) in the denominator of the infiniteproduct in the integrand can be negative and this gives the ambiguity about the sign ofthe integrand. As before, we expect the decay of the underlying system via the so-calledopen string pair production.Either of the above will become manifest if we look from the corresponding open stringone-loop annulus amplitude which can be read from (137) in the present context asΓ , = 4 V h det( η + ˆ F ′ ) det( η + ˆ F ) i sin πν (8 π α ′ ) Z ∞ dtt e − y t πα ′ × sinh πν t (cid:0) cosh πν t − cosh πt (cid:1) sinh πν t sinh πt ∞ Y n =1 Z n , (201)where Z n can be read from (61) as Z n = [1 − | z | n cosh πν t + | z | n ] (1 − | z | n ) (1 − | z | n cosh πt + | z | n ) × h (1 + | z | n − | z | n cosh πν t cosh πt ) − | z | n sinh πν t sinh πt i (1 − | z | n cosh 2 πν t + | z | n ) . (202)Since ν ∈ [0 , t , Z n ≈ ∼ e − y t πα ′ , (203)which indicates no tachyonic instability as expected. When ν = i ¯ ν with ¯ ν ∈ (0 , ∞ ), theintegrand has an infinite number of simple poles occurring at t k = k/ ¯ ν with k = 1 , , · · · ,indicating the decay of the underlying system via the so-called open string pair produc-tion. The decay rate and the open string pair production rate are given by the (140) and(142), respectively, with ν = ν = 1 /
2, and are not given here explicitly. In particular,we would like to point out that there is no open string enhancement here even for small60 ν since it is in general given by e π | ν − ν | / ¯ ν which is unity here. However, for generalnon-vanishing fluxes ˆ g and ˆ g , this enhancement can still be significant. The p = 5 case: Here p ′ = 1. The extension of a general flux ˆ F ′ on D1 to ˆ F ′ , following(127), is ( ˆ F ′ ) αβ = f ′ − ˆ f ′ g ′ − ˆ g ′ g ′ − ˆ g ′ , (204)where as before we take both ˆ g ′ → ∞ and ˆ g ′ → ∞ at the end of computations. Fora general ˆ F on D5, the discussion goes the same as what has been discussed for the p = p ′ = 5 case in the previous section. An example of the following flux on D5( ˆ F ) αβ = f − ˆ f g − ˆ g g − ˆ g , (205)along with the extension (204) corresponds just to a special case of what has been dis-cussed in great detail in [31]. So we refer there for detail and will not repeat it here. Forthis, it is also essentially the same as the ν being the imaginary case of the D6-D2 systemdiscussed above. The p = 4 case: Here p ′ = 0. The extension of no flux on D0 to ˆ F ′ , following (127), as( ˆ F ′ ) αβ = g ′ − ˆ g ′ g ′ − ˆ g ′ , (206)where once again we take both ˆ g ′ → ∞ and ˆ g ′ → ∞ at the end of computations. For ageneral flux on D4, the discussion will go the same as that for the p = p ′ = 4 case givenin the previous section. We could give some sample discussion for the present extended61ux (206) and some special choice of flux on D4 but this will not give anything new. Theclosed string cylinder amplitude, the open string annulus one, the potential decay rateand the potential open string pair production rate can all be read from the correspondingfrom (136), (137), (140) and (142), respectively, for the present consideration. So we omitto write each of them explicitly here. p − p ′ = 6 case This is the last case to be considered in this section. For p ≤
6, we have only one case toconsider, namely, p = 6 , p ′ = 0. The extension of no flux on D0 to ˆ F ′ , following (127), as( ˆ F ′ ) αβ = g ′ − ˆ g ′ g ′ − ˆ g ′ g ′ − ˆ g ′ , (207)where similarly we need to take ˆ g ′ → ∞ , ˆ g ′ → ∞ and ˆ g ′ → ∞ at the end of computations.As before, for a general flux ˆ F on D6, the relevant discussion goes more or less the sameas that for the p = p ′ = 6 case discussed in the previous section and we will not repeat ithere. We could give a sample discussion for the following flux on D6,( ˆ F ) αβ = f ˆ f − ˆ f g − ˆ f − ˆ g g − ˆ g g − ˆ g , (208)and this is still a rather general case of the more general discussion for the p = p ′ = 6case mentioned above but now withtan πν = q − ˆ f − ˆ f ˆ g , tan πν = 1ˆ g , tan πν = 1ˆ g , (209)where we have taken ˆ g ′ → ∞ , ˆ g ′ → ∞ and ˆ g ′ → ∞ . If we further set ˆ g = ˆ g = 0, wehave ν = ν = 1 /
2. For this special case, the closed string cylinder amplitude can be62ead from (136) asΓ , = − V q g − ˆ f − ˆ f sin πν π α ′ ) Z ∞ dtt e − y πα ′ t × ∞ Y n =1 [(1 + | z | n ) − | z | n cos πν ] (1 − | z | n ) (1 + | z | n ) (1 − | z | n cos 2 πν + | z | n ) , (210)where we have used (55) for C n . Except for the overall constant factor, this amplitudelooks essentially the same as the corresponding one for the p = 6 , p ′ = 2 case discussed insubsection 4.2. For real and non-vanishing ν , which requires ˆ f + ˆ f < ν + ν > ν if ν ≤ / ν + ν > ν if ν > /
2, i.e. the possible largest one is less than the sumof the remaining two as discussed for the p = p ′ = 6 case in the previous section. When ν = 0 for which ˆ g = 0 and ˆ f + ˆ f = 1, the amplitude vanishes but this is different fromthe p = 6 , p ′ = 2 case for which the the fluxes on D6 and D2 all vanish (or the fluxes on D6are vanishing except for the ones along the D2 directions which are identical to those on theD2). The explanation for the present vanishing interaction goes like this. The interactionbetween a D0 and a D6 carrying no flux is repulsive. The magnetic flux ˆ g stands fordelocalized D4 within D6 which has no interaction with D0 since their dimensionalitydiffers by four. The electric flux ˆ f , ˆ f stand for the delocalized fundamental strings withinD6 which have attractive interaction with the D0. So the vanishing of this amplitudemust imply the cancellation of the repulsive interaction between the D6 and the D0 withthe attractive one between the D0 and the fundamental F-strings within the D6 whenˆ f + ˆ f = 1. This is also consistent with the general conclusion reached for the p = p ′ = 6case in the previous section that ν + ν = ν for which the amplitude vanishes.For real non-vanishing ν , given what we learned earlier in this paper, we expect noopen string tachyonic instability. Let us check this explicitly by examining the corre-sponding open string one-loop annulus amplitude which can be read from (137) asΓ , = V q − ˆ f − ˆ f (8 π α ′ ) Z ∞ dtt e − y t πα ′ sinh πν t (cid:0) cosh πν t − cosh πt (cid:1) cosh πν t sinh πt ∞ Y n =1 Z n , where Z n , read from (61), is Z n = [1 − | z | n cosh πν t + | z | n ] (1 − | z | n ) (1 − | z | n cosh πt + | z | n ) × (cid:2) (1 − | z | n cosh πν t cosh πt + | z | n ) − | z | n sinh πν t sinh πt (cid:3) (1 − | z | n cosh 2 πν t + | z | n ) . (211)63or large t , Z n ≈ ∼ − e − y t πα ′ , (212)which vanishes for all y = 0, therefore no tachyonic divergence as expected.Let us consider ν to be imaginary which requires ˆ g = 0 and 1 < ˆ f + ˆ f < g .We now set ν = i ¯ ν with ¯ ν ∈ (0 , ∞ ). We have now from (209)tanh π ¯ ν = q ˆ f + ˆ f − | ˆ g | , (213)in addition to ν = ν = 1 / g = ˆ g = 0. The open string one-loop annulusamplitude is now, from (211) with ν = i ¯ ν ,Γ , = V q ˆ f + ˆ f − π α ′ ) Z ∞ dtt e − y t πα ′ sin π ¯ ν t (cid:0) cosh πt − cos π ¯ ν t (cid:1) cos π ¯ ν t sinh πt ∞ Y n =1 Z n , (214)where Z n continues to be given by (211) but now with ν = i ¯ ν . Now this amplitude hasan infinite number of simples poles of its integrand occurring at t k = (2 k − / ¯ ν with k = 1 , , · · · , giving an imaginary part of the amplitude. This further indicates the decayof the underlying system via the so-called open string pair production. The decay rateand the open string pair production rate are given, respectively, by (140) and (142) forthe present case with ν = ν = 1 / W = 4 q ˆ f + ˆ f − ν (8 π α ′ ) ∞ X k =1 (cid:18) ¯ ν k − (cid:19) cosh k − π ¯ ν sinh k − π ¯ ν e − (2 k − y πα ′ ¯ ν Z k − (¯ ν , / , / , (215)where Z k (¯ ν , / , /
2) = ∞ Y n =1 (1 + | z k | n ) (cid:16) | z k | n cosh kπ ¯ ν + | z k | n (cid:17) (1 − | z k | n ) (cid:16) − | z k | n cosh kπ ¯ ν + | z k | n (cid:17) , (216)with | z k | = e − kπ/ ¯ ν . As before, the open string pair production rate is given by the k = 1term of the above as W (1) = 4 q ˆ f + ˆ f − π α ′ ) ¯ ν / cosh π ¯ ν sinh π ¯ ν e − y πα ′ ¯ ν Z (¯ ν , / , / . (217)We would like to point out that with the choices of fluxes (207) and (208), the aboveamplitudes and rates share qualitatively the same properties as their correspondences,64espectively, in the case of p = 2 , p ′ = 0 discussed in subsection (4.1), even though thedetails are different. For example, both of the rates blow up when ¯ ν → ∞ which occursas ˆ f + ˆ f − → ˆ g , reaching the so-called critical field. For small ¯ ν ≪ Z ≈ p = 2 , p ′ = 0 case.For a general ¯ ν , these two rates are different. Note that we don’t have the exponentialenhancement of the rate for small ¯ ν , either. We compute, in this paper, the closed string cylinder amplitude between a Dp and aDp ′ , placed parallel at a separation along the directions transverse to the Dp, with eachcarrying their general constant worldvolume fluxes and with p − p ′ = κ = 0 , , , p ≤
6. We find that the amplitude for each of the p − p ′ = κ = 0 cases can be obtained asjust a special case of the corresponding amplitude for the p = p ′ case based on the relatedphysical consideration presented in the previous sections. As such, we find a universalformula for this closed string cylinder amplitude, valid for all cases specified above, asΓ p,p ′ = 2 V p ′ +1 h det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) i Q α =0 sin πν α κ (8 π α ′ ) p ′ +12 Z ∞ dtt − p e − y πα ′ t η ( it ) × θ (cid:0) ν + ν + ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν + ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν + ν − ν (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν − ν (cid:12)(cid:12) it (cid:1) θ ( ν | it ) θ ( ν | it ) θ ( ν | it )= 2 V p ′ +1 h det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) i (cid:2)P α =0 cos πν α − Q α =0 cos πν α − (cid:3) κ (8 π α ′ ) p ′ +12 × Z ∞ dtt − p e − y πα ′ t ∞ Y n =1 C n , (218)where C n continues to be given by (55). The amplitude for each given pair of p and p ′ andthe corresponding given worldvolume fluxes can be obtained from the above as a specialcase as prescribed in the previous two sections. The corresponding open string one-loopannulus universal amplitude can be obtained from the above via the Jacobi transformation t → t ′ = 1 /t along with the relations for the θ -function and the Dedekind η -functiongiven in (60) as 65 p,p ′ = − i V p ′ +1 h det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) i Q α =0 sin πν α κ (8 π α ′ ) p ′ +12 Z ∞ dtt p − e − y t πα ′ η ( it ) × θ (cid:0) ν + ν + ν it (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν + ν it (cid:12)(cid:12) it (cid:1) θ (cid:0) ν + ν − ν it (cid:12)(cid:12) it (cid:1) θ (cid:0) ν − ν − ν it (cid:12)(cid:12) it (cid:1) θ ( iν t | it ) θ ( iν t | it ) θ ( iν t | it )= 2 V p ′ +1 h det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) i κ (8 π α ′ ) p ′ +12 Z ∞ dtt p − e − y t πα ′ Y α =0 sin πν α sinh πν α t × " X α =0 cosh πν α t − Y α =0 cosh πν α t − ∞ Y n =1 Z n , (219)where we have dropped the prime on the open string variable t and Z n continues to begiven by (61). If one of three ν α , say ν , is imaginary, the underlying system decays viathe open string pair production. The general decay rate is W p,p ′ = 2 − κ h det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) i sinh π ¯ ν sin πν sin πν ¯ ν (8 π α ′ ) p ′ +12 ∞ X k =1 ( − ) k +1 (cid:16) ¯ ν k (cid:17) p − × (cid:16) cosh kπν ¯ ν − ( − ) k cosh kπν ¯ ν (cid:17) sinh kπν ¯ ν sinh kπν ¯ ν e − ky πα ′ ¯ ν Z k (¯ ν , ν , ν ) , (220)where Z k is given by (141). The corresponding open string pair production rate is givenby the leading k = 1 term of the above as W (1) p,p ′ = 2 − κ h det( η p ′ + ˆ F ′ p ′ ) det( η p + ˆ F p ) i sinh π ¯ ν sin πν sin πν (8 π α ′ ) p ′ +12 ¯ ν p − e − y πα ′ ¯ ν × (cid:16) cosh πν ¯ ν + cosh πν ¯ ν (cid:17) sinh πν ¯ ν sinh πν ¯ ν Z (¯ ν , ν , ν ) . (221)With the above, we have studied various properties of the amplitudes for each of thesystems considered such as the nature of the interaction, the open string tachyonic insta-bility, and the open string pair production if it exists and the associated enhancement. Inparticular, we find that the interaction can be repulsive for p ′ ≤ p = 6 with all three pa-rameters ν , ν , ν being real and non-vanishing, i.e., ν α ∈ (0 ,
1) with α = 0 , ,
2. Since theamplitude is symmetric with respect to the three ν , ν , ν , we can assume ν ≤ ν ≤ ν without loss of generality. The repulsive interaction occurs indeed when ν + ν > ν and66 + ν + ν <
2. In other words, whenever the sum of two smaller ν ’s (here ν and ν )is larger than the largest ν (here ν ) and ν + ν + ν <
2, the underlying interactionis repulsive. The reason for the above requirements is simple. The repulsive inter-braneinteraction occurs, in the absence of the worldvolume fluxes, only for the system of p = 6and p ′ = 0 for which we have ν = ν = ν = 1 / ν + ν > ν and ν + ν + ν = 3 / < p and p ′ , the inter-braneinteraction, in the absence of the worldvolume fluxes, is either attractive or vanishing.So to have a potential repulsive-interaction, we first need to have the presence of D6 andsecondly we need to have D0 which can be realized in general for p ′ ≤ p = 6 with allthree ν , ν , ν ∈ (0 , ν , ν , ν ∈ (0 , ν ≤ ν ≤ ν , we have shown in section 3 and checked for each caseconsidered later that whenever ν + ν > ν and ν + ν + ν <
2, the net interactionis repulsive. The interaction vanishes whenever ν + ν = ν or ν + ν + ν = 2. Theinteraction is attractive whenever ν + ν < ν and ν + ν + ν < ν + ν > ν and2 < ν + ν + ν < ν , ν , ν is imaginary, the underlying system is unstableand decays via the so-called open string pair production. This is reflected in that the openstring one-loop amplitude has an imaginary part. Again without loss of generality, wechoose ν = i ¯ ν with ¯ ν ∈ (0 , ∞ ). This is related to the applied electric flux(es). Whenthe applied electric flux reaches its critical one, we have ¯ ν → ∞ , the open string pairproduction rate diverges and the pair production cascades, giving rise also to the otherinstability of the system. We have also studied the potential enhancement of the pairproduction rate in the presence of magnetic fluxes and our findings here are consistent withour previous studies on this. The enhancement is determined by the so-called tachyonic67hift which can be given in general as | ν − ν | / ν , ν ∈ [0 , ν , ν , ν are small. We have that the larger the shift is, the larger the open string pairproduction enhancement. For this purpose, we prefer to have the presence of the largerof ν and ν while turning off the smaller one such that the enhancement is larger. Forexample, we keep ν while drop ν . So the question is: can we realize the largest shiftwhich is ν / /
2? This is one of the motivations for this paper as mentioned in theIntroduction. Though we cannot realize ν / / ν / /
4, almostequally as good, and it can come from the system of p − p ′ = 2 without adding anyworldvolume magnetic fluxes given the above consideration. This is due to that the Dp ′ brane acts effectively as a magnetic field which can give rise to ν = 1 / ν occurs for p = 3 , p ′ = 1 with purely added electric fluxes along the D1-directions. This system has ν = 1 /
2, giving the possible largest enhancement. This may have a potential applicationin practice which we would like to pursue in the near future.One last thing, we have briefly mentioned in the introduction, is the relationship ofthe present discussion with that for a system of Dp and Dp ′ ( p ≥ p ′ ) with the two branesnot at rest but with a constant relative motion transverse at least to the Dp ′ and/ora rotation between certain transverse directions and the brane directions. As discussedin [36], a Dp brane carrying a constant electric flux along certain spatial direction isequivalent to a boosted and delocalized D(p - 1) brane along this direction. They arerelated by a T-duality along this direction and the boost velocity is determined by theelectric flux. By a similar token, a Dp brane carrying a magnetic strength F ij with i < j (also called magnetized brane), for example, is equivalent to intersecting D(p - 1) branesat an angle between the spatial i -direction and the spatial j-direction and determinedby the magnetic flux, for example, by a T-duality along j-direction [15, 36–40]. Here themagnetized Dp brane and the intersecting D(p - 1) branes are related by a T-dualityalong the j-direction and the rotation is determined by the magnetic flux. Since theresulting D(p - 1) brane(s) in either case is delocalized along certain directions transverseto the brane, it is probably much easier and much more straightforward to compute thesame interaction between two such D branes using their equivalent ones carrying fluxes atrest as discussed in this paper even though computations of the interaction for localizedsuch objects are known (for example, see [36, 46, 56]). In this paper we only use theso-called no-force condition to discuss whether the underlying system preserves certainsupersymmetry but for intersecting branes the underlying supersymmetry can be analyzedin detail following [5, 15, 37–40, 44–46] (see [41] for a rather complete list of references onphenomenological applications to supersymmetry breaking for intersecting branes).68 cknowledgments The authors would like to thank the anonymous referees for their suggestions and questionsraised which help to improve the present manuscript and acknowledge support by grantsfrom the NSF of China with Grant No: 11775212, 11947301 and 11235010.
Appendix A
In this Appendix, we first give a general discussion of the eigenvalues of the matrix w (24). Note that s = ( η − ˆ F )( η + ˆ F ) − , s ′ = ( η − ˆ F ′ )( η + ˆ F ′ ) − , w = ss ′ T , (222)and each of them satisfies the same relation, e.g., w α γ ( w T ) γ β = δ α β . (223)This is actually a relation satisfied by a Lorentz transformation in (1 + p) dimensions.In other words, either s or s ′ is a general Lorentz transformation since either flux ˆ F or ˆ F ′ counts the number of independent parameters of SO (1 , p ) as ( p + 1) p/
2. Thisholds also true for w since it is the product of s and s ′ . In addition, we have det s =det[( η − ˆ F )( η + ˆ F ) − ] = det( η − ˆ F ) det( η + ˆ F ) − = det( η − ˆ F ) det( η − ˆ F ) − = 1 where wehave used ( η + ˆ F ) T = ( η − ˆ F ) in the third equality since ˆ F is antisymmetric. This sameholds for s ′ , too. So we have det w = det s det s ′ T = 1 as well. Note also here η αβ = ( − , , · · · , α, β = 0 , , · · · p. (224)Given the above, it is clear that a purely electric flux gives a Lorentz boost while a purelymagnetic one gives only a rotation of SO ( p ). When both ˆ F and ˆ F ′ are each purely electric,the resulting w is in general not a pure Lorentz boost unless the two electric fluxes arecollinear. However, the resulting w is a rotation when both ˆ F and ˆ F ′ are each purelymagnetic. If w is indeed a pure Lorentz boost, it can always be brought to the followingform by a SO ( p ) rotation R , w = r T ˜ wr, (225)where ˜ w = γ γvγv γ I ( p − × ( p − , r = R p × p ! , (226)69ith γ = (1 − v ) − / and v <
1. The rotation just brings the velocity along the ‘1’-direction. So it is clear that a general Lorentz boost has only one non-trivial pair realeigenvalues of λ ± = γ (1 ± v ) with λ + λ − = 1 and the rest are all unity. If w is insteada pure SO(p) rotation, e.g. w = r with r given above, we have its eigenvalues 1 and therest in pairs as λ α , λ − α with λ α = e πiν α and α = 0 , , · · · ( p − / p = even or1, 1, and the rest in pairs as before but now with α = 0 , , · · · , ( p − / p = odd.Here ν α are all real with each ν α ∈ [0 , /
2] for the amplitude as discussed in Section 3.For a general w , we need to put some extra efforts to figure out the nature of itseigenvalues. For this, since w is a Lorentz matrix with det w = 1, we can set w = e K , (227)where from ( w − ) α β = ( w T ) α β we have K α β = − K β α . (228)Now solving the eigenvalue problem of w is transformed to that of K . In other words, wehave f ( ρ ) = det (cid:0) ρ δ α β − K α β (cid:1) = 0 . (229)Since K T has the same eigenvalues as K , we have also f ( ρ ) = det (cid:2) ρ δ α β − ( K T ) α β (cid:3) = 0 , (230)which implies whenever ρ is an eigenvalue so is − ρ from (228). In other words, theeigenvalues appear in pairs. When p = even, we have always one zero-eigenvalue sincedet K α β = − det K αβ = 0, giving the unity eigenvalue of w discussed in Section 3, andthe rest are in pairs.Let us discuss the p = even case first. For the zero-eigenvalue, we can choose thecorresponding eigenvector as e such that K α β e β = 0. We have two sub-cases to consider: e · e = − e · e = 1 where we have normalized each as indicated . For the first We choose here the same conventions as used in Section 3. For certain choice of the fluxes, we may have the eigenvector being light-like. This can always be takenas certain limit of the time-like or space-like limit as discussed. When this happens, the corresponding K matrix can still be diagonalized. Here we use a simple example for p = 2 to illustrate this. Now themost general K can be expressed as K α β = f f f gf − g , (231) e = e such that { e , e , · · · , e p } forms a complete normalized basisof the eigenvector space, giving η αβ e α ¯ α e β ¯ β = η ¯ α ¯ β or η αβ e α ¯ α e β ¯ β = η ¯ α ¯ β , (234)where the α, β indices are raised or lowered using η αβ or η αβ and similarly for the ¯ α, ¯ β indices. So e α ¯ α or e α ¯ α is also a Lorentz transformation. We take α = (0 , a ) ( ¯ α = (0 , ¯ a ))from now on with a = 1 , , · · · p (¯ a = 1 , , · · · , p ). We have now K α β e β = 0 and from(228) we have also ( e T ) β K β α = 0. With these two, we have, from (234),¯ K ¯ α ¯ β ≡ ( e T ) ¯ α α K α β e β ¯ β = K ¯ a ¯ b ! , (235)where ¯ K ¯ a ¯ b = ¯ K ¯ a ¯ b = − ¯ K ¯ b ¯ a is real and antisymmetric from the property of K given in(228). So it can be diagonalized by a unitary matrix u of the following form with itspurely imaginary eigenvalues in pairs as ( ρ ¯ a , − ρ ¯ a ) with ¯ a = 1 , · · · , p/ K = U ¯ K U + = u p × p ! K ) p × p ! u + p × p ! , (236)where ( ¯ K ) p × p = ρ . . . − ρ . . . . . . ρ p/
00 0 . . . − ρ p/ . (237) where we assume f , f , g are all non-negative without loss of generality. This matrix has three expectedeigenvalues: 0 , ρ , − ρ with ρ = p f + f − g . It can be diagonalized as K = V ¯ K V − , (232)where the diagonal matrix ( ¯ K ) α β = (0 , ρ , − ρ ) and the non-singular matrix V = 1 ρ g − f f f g − f ρ g − f gρ − f f g − f − f g − f ρ gρ + f f f − g , (233)with its det V = 2. For the 0-eigenvalue, the corresponding eigenvector can be taken as e = ( g, f , − f ) T ,giving K α β e β = 0. Here e = η αβ e α e β = f + f − g which can be either time-like or space-like ingeneral. However, it becomes null when f + f = g , which can be taken as the corresponding limit ofeither time-like or space-like case. So long this is taken as a limiting case, we don’t have a problem todiagonalize the matrix K since det V = 2 is non-singular. K is diagonalized as K = V ¯ K V − , (238)where V = eU with ( e − ) ¯ α β = ( e T ) ¯ α β and U − = U + . Here ¯ K gives the expectedeigenvalues: one zero and the rest being purely imaginary in pairs as indicated in (237).For the second subcase, i.e. e · e = 1, we take now e p = e such that { e , e , · · · , e p } forms a complete normalized basis of the eigenvector space. The following discussion goesexactly the same line as above and end up with now¯ K ¯ α ¯ β ≡ ( e T ) ¯ α α K α β e β ¯ β = ¯ K ¯ α ′ ¯ β ′
00 0 ! , (239)where ¯ α ′ , ¯ β ′ = 0 , , · · · p −
1. The diagonalisation of the matrix ¯ K ¯ α ′ ¯ β ′ follows exactly thesame as K α β for odd p . So we turn now to the odd p case.For this case, since the eigenvalues are in pairs as ( ρ α , − ρ α ) with α = 0 , , · · · ( p − / f ( ρ ) from (229) must be even in power of ρ when we expand the determinantand is f ( ρ ) = ρ p +1 + c ρ p − + c ρ p − + · · · + c p − ρ + det K α β . (240)Note that det K α β = − det K αβ = − (pf( K αβ )) < K αβ = − K βα from (228) andpf(K) denoting the Pfaffian of antisymmetric matrix K αβ . So we have f (0) = det K α β < ρ >
0, the highest power of ρ dominates and so we have f ( ρ ) >
0. Thereforewe must have at least one pair ( ρ , − ρ ) with ρ positive real satisfying the eigenvalueequation f ( ± ρ ) = 0. If det K α β = 0, we will have a pair of zero eigenvalues unless theabove c p − = 0 and if this is case we can do what we have done for the above even p case.If we have more zero eigenvalues, we just repeat this process until we have non-zero ones.For now, we assume det K α β <
0, so we have ρ = 0. The corresponding eigenvectors,denoting as x and x , satisfy their respective equations as K α β x β = ρ x α , K α β x β = − ρ x α . (241)From the first one, we have ρ x · x = ( x ) α K α β x β = x α K αβ x β = 0 where we haveused the property of K from (228). This must imply x · x = 0 since ρ = 0. In otherwords, x is a null vector. By the same token, we have also x as a null vector. We nowshow x · x = 0. Since both are null vectors, without loss of generality, we can alwayschoose x = (1 , , , · · · ,
0) and x = ( | ~a | , ~a ). If x · x = 0, we must have | ~a | = a > x = ( a , a , , · · · ,
0) = a x . This contradicts the fact that the two eigenvectors72re independent since they correspond to different eigenvalues. Therefore, we must have x · x = 0. For convenience, we choose to have x · x = −
2. With this, we define e = 12 ( x + x ) , e = 12 ( x − x ) , (242)such that ( e ) = −
1, ( e ) = 1 and e · e = 0. We now construct an orthogonal basis { e , e , · · · e p } satisfying the same relations as those given in (234). Note that K α β e β = − ( e T ) β K β α = ρ e α and K α β e β = − ( e T ) β K β α = ρ e α . So we have¯ K ¯ α ¯ β = ( e T ) ¯ α α K α β e β ¯ β = ρ ρ K ¯ a ¯ b , (243)where ¯ K ¯ a ¯ b is a ( p − × ( p −
1) antisymmetric matrix ( ¯ K ¯ a ¯ b = ¯ K ¯ a ¯ b = − ¯ K ¯ b ¯ a ) and can bediagonalized, as before, by a ( p − × ( p −
1) unitary matrix u , with its purely imaginaryeigenvalues in pairs as ( ρ ¯ c , − ρ ¯ c ) with ¯ c = 1 , , · · · , ( p − /
2. Note that the symmetricsub-matrix in ¯ K ¯ α ¯ β can be diagonalized by a 2 × R as ρ ρ ! = R ρ − ρ ! R − , (244)where specifically R = 1 √ −
11 1 ! , R − = 1 √ − ! . (245)In other words, the matrix ¯ K ¯ α ¯ β can be diagonalized by the unitary matrix rU as¯ K = r ρ − ρ ¯ K ( p − × ( p − r − = rU ¯ K U + r T = ( rU ) ¯ K ( rU ) − , (246)where r = R I ( p − × ( p − ! , U = I × u ! , (247)and the diagonal matrix ¯ K = ( ρ , − ρ , ρ , − ρ , · · · , ρ ( p − / , − ρ ( p − / ). Given the aboveand from (243), we have K = e ¯ Ke T = erU ¯ K ( rU ) + e T = ( erU ) K ( erU ) − , (248)73here we have used e − = e T . So we prove that in general K has a pair of real eigenvalue( ρ , − ρ ) and the rest are all imaginary and given in pairs as ( ρ c , − ρ c ) with c = 1 , , · · · ( p − / p = even, we have two cases: 1) one zero eigenvalue and therest are all imaginary and given in pairs as ( ρ c , − ρ c ) with c = 1 , , · · · , p/
2; 2) one zeroeigenvalue, a pair of real eigenvalues ( ρ , ρ ) and the rest are all imaginary and given inpairs as ( ρ c , − ρ c ) with c = 1 , , · · · , ( p − /
2. For p = odd, we have in general a pairof real eigenvalues ( ρ , − ρ ) and the rest are all imaginary and given as ( ρ c , − ρ c ) with c = 1 , , · · · , ( p − / ρ = 2 πν and the imaginary eigenvalues ρ c =2 πiν c , from (227) we then obtain the same eigenvalues of w as discussed in Section 3. Appendix B
The zero-mode contribution to the amplitude in the RR sector for p − p ′ = ν = 0 , , , p ≤ h B ′ , η ′ | B, η i ≡ h B sgh , η ′ | B sgh , η i
0R 0R h B ′ ψ , η ′ | B ψ , η i = − δ ηη ′ , + q det( η p + ˆ F ) det( η p ′ + ˆ F ′ ) [ p ′ +12 ] X n =0 [2( n + ν )]!2 n + ν n !( n + ν )! × ˆ F [ α ′ β ′ ··· ˆ F α ′ n β ′ n ˆ F ( p ′ +1)( p ′ +2) ··· ˆ F ( p ′ + ν − p ′ + ν )] ˆ F ′ [ α ′ β ′ ··· ˆ F ′ α ′ n β ′ n ] , (249)where the indices inside the square bracket denote their anti-symmetrization. As indicatedalready in the previous sections, the above zero-mode matrix element for lower p and p ′ cases can be obtained from either p = 6 or p = 5 case depending on p and p ′ being evenor odd if their worldvolume fluxes are extended in a specific way which we turn next. Letus take two explicit examples to demonstrate this. The first one is for p = 5 and p ′ = 3and we will show that the corresponding zero-mode matrix element can be obtained from p = p ′ = 5 case if we extend the p ′ = 3 worldvolume flux ˆ F ′ α ′ β ′ to the p ′ = p = 5worldvolume flux ˆ F ′ αβ the following way,ˆ F ′ αβ = ˆ F ′ α ′ β ′ g ′ − ˆ g ′ , (250)74here we take the magnetic flux ˆ g ′ > p = 5 and p ′ = 3 h B ′ , η ′ | B, η i = − δ ηη ′ , + q det( η + ˆ F ) det( η + ˆ F ′ ) (cid:18) ˆ F + 32 ˆ F [ α ′ β ′ ˆ F ˆ F ′ α ′ β ′ + 158 ˆ F [ α ′ β ′ ˆ F α ′ β ′ ˆ F ˆ F ′ [ α ′ β ′ ˆ F ′ α ′ β ′ ] (cid:19) . (251)We now show that this can also be obtained from the p = p ′ = 5 case but with ˆ F ′ αβ givenby (250). We have now from (249) h B ′ , η ′ | B, η i = − δ ηη ′ , + q det( η + ˆ F ) det( η + ˆ F ′ ) (cid:18) F αβ ˆ F ′ αβ + (cid:18) (cid:19)
4! ˆ F [ α β ˆ F α β ] ˆ F ′ [ α β ˆ F ′ α β ] + (cid:18) (cid:19)
6! ˆ F [ α β ˆ F α β ˆ F α β ] ˆ F ′ [ α β ˆ F ′ α β ˆ F ′ α β ] ! . (252)Note that det( η αβ + ˆ F ′ αβ ) = (1 + ˆ g ′ ) det( η α ′ β ′ + ˆ F ′ α ′ β ′ ) which gives ˆ g ′ det( η α ′ β ′ + ˆ F ′ α ′ β ′ ) forˆ g ′ → ∞ . To have a finite contribution at ˆ g ′ → ∞ , we need each term in the bracket tohave a factor ˆ F ′ = ˆ g ′ . For this,ˆ F αβ ˆ F ′ αβ = 2ˆ g ′ ˆ F + · · · , ˆ F [ α β ˆ F α β ] ˆ F ′ [ α β ˆ F ′ α β ] = 4ˆ g ′ ˆ F [ α β ˆ F ˆ F ′ α β + · · · , ˆ F [ α β ˆ F α β ˆ F α β ] ˆ F ′ [ α β ˆ F ′ α β ˆ F ′ α β ] = 6ˆ g ′ ˆ F [ α β ˆ F α β ˆ F ˆ F ′ [ α β ˆ F ′ α β ] + · · · , (253)where the · · · terms are independent of g ′ . With these and taking ˆ g ′ → ∞ , we can checkeasily that (252) is exactly the same as (251).The second example is to take p = p ′ = 4 and we will show that the zero-mode matrixelement can be obtained from the p = p ′ = 6 case by taking the worldvolume fluxes,respectively, as ˆ F αβ = ˆ F α ′ β ′ g − ˆ g , ˆ F ′ αβ = ˆ F ′ α ′ β ′ g ′ − ˆ g ′ , (254)where α, β = 0 , , · · · , α ′ , β ′ = 0 , , · · ·
4. Note that we need to take ˆ g, ˆ g ′ → ∞ atthe end of computations. For p = p ′ = 4, the zero-mode matrix element from from (249)75s h B ′ , η ′ | B, η i = − δ ηη ′ , + q det( η + ˆ F ) det( η + ˆ F ′ ) (cid:18) F α ′ β ′ ˆ F ′ α ′ β ′ + 38 ˆ F [ α ′ β ˆ F α ′ β ′ ] ˆ F ′ [ α ′ β ′ ˆ F ′ α ′ β ′ ] (cid:19) . (255)For p = p ′ = 6 with the respective worldvolume fluxes given in (254), we have thecorresponding zero-mode matrix element from (249) as h B ′ , η ′ | B, η i = − δ ηη ′ , + q det( η + ˆ F ) det( η + ˆ F ′ ) (cid:18) F αβ ˆ F ′ αβ + 4!2 ˆ F [ α β ˆ F α β ] ˆ F ′ [ α β ˆ F ′ α β ] + 6!2 (3!) ˆ F [ α β ˆ F α β ˆ F α β ] ˆ F ′ [ α β ˆ F ′ α β ˆ F ′ α β ] (cid:19) . (256)We will show that the above is actually the same as that given in (255) for ˆ g, ˆ g ′ → ∞ .For this, note that q det( η αβ + ˆ F αβ ) det( η αβ + ˆ F ′ αβ ) = ˆ g ˆ g ′ q det( η α ′ β ′ + ˆ F α ′ β ′ ) det( η α ′ β ′ + ˆ F ′ α ′ β ′ ) , (257)where we have used (254) and taken ˆ g, ˆ g ′ → ∞ . To have a finite limit, only those termsin the bracket proportional to ˆ g ˆ g ′ in (256) survive. We haveˆ F αβ ˆ F ′ αβ = 2ˆ g ˆ g ′ + · · · , ˆ F [ α β ˆ F α β ] ˆ F ′ [ α β ˆ F ′ α β ] = 2
4! ˆ g ˆ g ′ ˆ F α ′ β ′ ˆ F ′ α ′ β ′ + · · · , ˆ F [ α β ˆ F α β ˆ F α β ] ˆ F ′ [ α β ˆ F ′ α β ˆ F ′ α β ] = (3!) g ˆ g ′ ˆ F [ α ′ β ′ ˆ F α ′ β ′ ] ˆ F ′ [ α ′ β ′ ˆ F ′ α ′ β ′ ] + · · · , (258)where the · · · terms are independent of ˆ g ˆ g ′ . Plugging the above terms to (256) and takingˆ g, ˆ g ′ → ∞ , we get exactly (255).In summary, for various RR zero-mode contributions given in (249), they each canbe obtained from the p = p ′ = 6 or the p = p ′ = 5 case by choosing the correspondingworldvolume fluxes in a way as indicated in the above examples. So we only need to focuson these two cases. They each can be given from (249) as h B ′ , η ′ | B, η i = − δ ηη ′ , + q det( η p + ˆ F ) det( η p + ˆ F ′ ) × X n =0 (2 n )!2 n ( n !) ˆ F [ α β ··· ˆ F α n β n ] ˆ F ′ [ α β ··· ˆ F ′ α n β n ] , (259)76here p = 5 or 6. For either case, let us write the zero-mode contribution as h B ′ , η ′ | B, η i = − δ ηη ′ , + q det( η p + ˆ F ) det( η p + ˆ F ′ ) S, (260)where S = X n =0 (2 n )!2 n ( n !) ˆ F [ α β ··· ˆ F α n β n ] ˆ F ′ [ α β ··· ˆ F ′ α n β n ] . (261)In what follows, we will show( h B ′ , η ′ | B, η i ) = 2 − p δ ηη ′ , + det( I + w ) , (262)where w is given in (24) and for convenience we rewrite it explicitly here w = ( I − ˆ F )( I + ˆ F ) − ( I + ˆ F ′ )( I − ˆ F ′ ) − , (263)with I the ( p + 1) × ( p + 1) unity matrix. Note thatdet( I + w ) = det[( I + ˆ F )( I + w )( I − ˆ F ′ )]det( I + ˆ F ) det( I − ˆ F ′ )= 2 p +1 det( I − ˆ F ˆ F ′ )det( I + ˆ F ) det( I + ˆ F ′ ) . (264)With this, for (262) to hold, we need to show S = det( I − ˆ F ˆ F ′ ) . (265)For this, we represent the S in terms of the following Grassmannian integration, S = ( − ) p +1 Z p Y γ =0 (cid:2) dθ γ dθ ′ γ (1 + θ γ θ ′ γ ) (cid:3) e − ˆ F αβ θ α θ β e ˆ F ′ ¯ α ¯ β θ ′ ¯ α θ ′ ¯ β , (266)where θ γ and θ ′ γ are all real Grassmannian variables. With this, we have S = Z p Y γ =0 dθ γ dθ ′ γ d ˜ θ γ d ˜ θ ′ γ (1 + θ γ θ ′ γ )(1 + ˜ θ γ ˜ θ ′ γ ) e − ˆ F αβ ( θ α θ β +˜ θ α ˜ θ β ) e ˆ F ′ ¯ α ¯ β ( θ ′ ¯ α θ ′ ¯ β +˜ θ ′ ¯ α ˜ θ ′ ¯ β ) . (267)Now we change the integration variables as θ γ = η γ + η ∗ γ √ , ˜ θ γ = η γ − η ∗ γ i √ , θ ′ γ = η ′ γ + η ′∗ γ √ , ˜ θ ′ γ = η ′ γ − η ′∗ γ i √ . (268)77here ∗ denotes the complex conjugate. In terms of η γ , η ∗ γ , η ′ γ and η ′∗ γ , we have S = Z p Y γ =0 dη ′∗ γ dη γ dη ′ γ dη ∗ γ (1 + η γ η ′∗ γ )(1 + η ∗ γ η ′ γ ) e − ˆ F αβ η α η ∗ β e ˆ F ′ ¯ α ¯ β η ′ ¯ α η ′∗ ¯ β . (269)The evaluation of the integral can be simplified if we do the following integration first I = Z p Y γ =0 dη ′ γ dη ∗ γ (1 + η ∗ γ η ′ γ ) e − ˆ F αβ η α η ∗ β e ˆ F ′ ¯ α ¯ β η ′ ¯ α η ′∗ ¯ β = e − ( ˆ F ˆ F ′ ) α ¯ α η α η ′∗ ¯ α . (270)With this, we have S = Z p Y γ =0 dη ′∗ γ dη γ (1 + η γ η ′∗ γ ) I = Z p Y γ =0 dη ′∗ γ dη γ e [ I − ( ˆ F ˆ F ′ )] α ¯ α η α η ′∗ ¯ α = det( I − ˆ F ˆ F ′ ) . (271)In other words, (262) holds indeed. In Appendix A, we have shown w = V w V − with w diagonal with eigenvalue 1 and others in pairs as ( λ α , λ α ) with α = 0 , , p = 6 orwith eigenvalues in pairs as ( λ α , λ α ) with α = 0 , , p = 5. We then have from (262)for p = 5 or 6( h B ′ , η ′ | B, η i ) = 2 − p δ ηη ′ , + det( I + w )= 2 − p δ ηη ′ , + det( I + w )= 2 − p δ ηη ′ , + (1 + δ p, ) Y α =0 (1 + λ α )(1 + λ − α )= 2 δ ηη ′ , + 2 Y α =0 cos πν α , (272)where we have used λ α = e iπν α . This gives the expected result h B ′ , η ′ | B, η i = − δ ηη ′ , + 2 Y α =0 cos πν α , (273)where we have used the known result h B ′ , η ′ | B, η i = − δ ηη ′ , + in the absence of fluxesfor which ν α = 0 and ν α ∈ [0 ,
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