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School of Mathematics, Trinity College, Dublin 2, Ireland.

arXiv:hep-th/9303140v2 31 Aug 1993TCD–3–93March 1993Tachyon Back Reactionon d = 2 String Black HoleS. Kalyana RamaSchool of Mathematics, Trinity College, Dublin 2, Ireland.email: kalyan@maths.tcd.ieABSTRACT.

We describe a static solution for d = 2 criticalstring theory including the tachyon T but with its potential V (T)set to zero. This solution thus incorporates tachyon back reac-tion and, when T = 0, reduces to the black hole solution.

WhenT ̸= 0 we find that (1) the Schwarzschild horizon of the aboveblack hole splits into two, resembling Reissner-Nordstrom hori-zons and (2) the curvature scalar develops new singularities atthe horizons. We show that these features will persist even whenV (T) is nonzero.

We present a time dependent extension of ourstatic solution and discuss some possible methods for removingthe above singularities.1

1. IntroductionIt is a challenging task to resolve the puzzles of gravitating systems suchas the nature of singularities, the end effect of Hawking radiation, and in-formation loss(?) inside the black hole.

Recently there has been a renewedeffort [1, 2] to solve these problems in the simpler context of two dimensional(d = 2) systems, using the string inspired toy models for quantum gravity,the inspiration having come from the discovery of black holes in d = 2 criticalstrings [3, 4, 5].The d = 2 black hole for graviton-dilaton system was discovered inan SL(2, R)/U(1) gauged Wess-Zumino-Witten model [3]; as a solution ofO(α′) β-function equations [6, 7] for critical string theory with d = 2 targetspace-time [4]; and in other forms [5]. Similar toy models in d = 2 space-timefor quantum gravity including matter were studied first by [1] and then bymany others [2] with a view of solving various puzzles of gravitating sys-tems in this simpler context.

Most of these models have mainly studied thegraviton-dilaton system.For a more complete story, one should also include tachyon, the onlyremaining low energy degree of freedom for d = 2 strings (and which isa massless excitation for d = 2). Though important, its inclusion resultsin non linear equations which are solved only in a few asymptotic cases[8, 2].The solution of β-function equations for d = 2 strings includingtachyons is not known.

However, solving for tachyon in the d = 2 string blackhole background leads to singular behaviour [3, 4]. Thus it is important tounderstand the back reaction of tachyons for d = 2 critical strings, especiallybecause of its importance as a model for d = 2 quantum gravity.In this work we describe a static solution, recently announced in [9], of theβ-function equations for the low energy d = 2 critical string theory includingtachyon T but not its potential V (T).

This solution incorporates tachyonback reaction. The solution has, beside the black hole mass parameter, a newparameter ǫ which is a measure of tachyon strength.

We find that: (1) TheSchwarzschild horizon of the previous black hole splits into two, resemblingReissner-Nordstrom horizons. However, the solution cannot be analyticallycontinued in between the two “horizons”.

(2) The curvature scalar R has apoint singularity, as for ǫ = 0, but when ǫ > 0 develops new singularities atthe horizons. Hence this solution is not a black hole solution in the usual2

sense, though for ǫ = 0 it is. Though we needed to set V = 0 to obtain anexplicit solution, we show later that the features of this solution will persisteven when V is correctly taken into account.

Thus this shows that whentachyon back reaction is included, the d = 2 string does not admit any staticsolution which can be interpreted as a black hole in the usual sense. Wealso present a time dependent extension of our solution which, however, doesnot alter the features of the static solution, and then discuss critically somepossible methods of removing the singularities.In section 2 we present β-function equations in various forms and obtaina static solution of these equations in section 3 assuming that the tachyonpotential vanishes.

We discuss the features of this solution in section 4 andshow, in section 5, that these features will persist even when the tachyonpotential is nonzero. In section 6 we present a time dependent extension ofour static solution and then conclude with a critical discussion of possiblemethods of removing the singularities, atleast the new ones.2.

β-function EquationsIn this section we give β-function equations in various forms for the lowenergy d = 2 critical string theory including graviton (Gµν), dilaton (φ), andtachyon (T) fields. We consider mostly the “static” case.

The sigma modelaction of the d = 2 critical string theory for Gµν, φ, and T, in a notationsimilar to that of [4], is given bySsigma =18πα′Zd2˜x√g(Gµν∇xµ∇xν + α′Rφ + 2T)where ˜x are the world sheet coordinates and xµ, µ = 0, 1 are the targetspace coordinates. The conformal invariance of the above action requires thefollowing β-function equations to be satisfied:Rµν + ∇µ∇νφ + ∇µT∇νT=0R + (∇φ)2 + 2∇2φ + (∇T)2 + 4γK=0∇2T + ∇φ∇T −2γKT=0(1)where γ = −2α′ , K = 1 + V4γ, V = γT 2 + O(T 3) is the tachyon potential and3

KT ≡dKdT . These equations also follow from the target space effective actionS =Zd2x√G eφ (R −(∇φ)2 + (∇T)2 + 4γK) .

(2)Note from equation (2) that the dilaton field e−φ2 acts as a coupling constant.Consider now the target space conformal gauge ds2 = eσdu dv where u =x0 + x1 and v = x0 −x1. In this gauge the only nonzero connection (Γκµν)and curvature (Rµν) components areΓuuu = ∂uσ , Γvvv = ∂vσ ; Ruv = ∂u∂vσ(3)and equations (1) become∂2uφ −∂uσ∂uφ + (∂uT)2=0∂2vφ −∂vσ∂vφ + (∂vT)2=0∂u∂vσ + ∂u∂vφ + ∂uT∂vT=0∂u∂vφ + ∂uφ∂vφ + γKeσ=02∂u∂vT + ∂uφ∂vT + ∂vφ∂uT −γKTeσ=0 .

(4)Defining new coordinates ξ = uv, χ = u/v, and, for convenience eΣ = γξeσ,equations (4) can be written as(φ11 −Σ1φ1 + T 21 ) + (φ00 −Σ0φ0 + T 20 )=0(Σ11 + φ11 + T 21 ) −(Σ00 + φ00 + T 20 )=02φ01 −Σ1φ0 −Σ0φ1 + 2T0T1=0(T11 + φ1T1 −12eΣKT) −(T00 + φ0T0)=0(φ11 + φ21 + eΣK) −(φ00 + φ20)=0(5)where (· · ·)0 = (χ ddχ)(· · ·) and (· · ·)1 = (ξ ddξ)(· · ·). In the following we firstspecialise to the static case where the fields depend only on ξ.

Then equations(5) becomeΣ11 + φ1Σ1 = Σ11 + φ11 + T 21=0T11 + φ1T1 −12eΣKT = φ11 + φ21 + eΣK=0 . (6)4

To proceed further, we first note that the fields can be expected to evolvedepending on the local metric given by Σ and hence look for their solutionsin terms of X(ξ) ≡Σ1. Letting F(X) = φ1, (· · ·)′ =ddX (· · ·), and notingthat (· · ·)1 = (· · ·)′X1, equations (6) giveX1 + XF = 0,T 21 −XF(1 + F ′)=0XFF ′′ + (XF ′ −F)(1 + F ′) + eΣT1KT(XF)−1=0 .

(7)The curvature scalar is given by R = −4γe−ΣXF. Dividing the last equationabove by XF(1 + F ′) and changing the independent variable from ξ to Xthe β-function equations for the static case can be written asFΣ′ + 1=0Xφ′ + 1=0XFT ′2 −(1 + F ′)=0F ′′1 + F ′ + F ′F −1X −T ′KTeΣXF(1 + F ′)=0(8)where X and ξ are related by X1 = −XF.

Equations (8) can be simplifiedfurther, for example, as follows. LetL′ =T ′KTeΣXF(1 + F ′) .

(9)In this case the last equation in (8) can be integrated to giveF(1 + F ′) = a2XeL(10)where a2 is a constant. This immediately gives T 21 = a2X2eL from which weobtainT ′ = −aF −1eL2 .

(11)Substituting these expressions in (9) and replacing L by a new field H = aeL2the β-function equations becomeFΣ′ + 1=0FT ′ + H=01 + F ′ −XHF=0H′ + KT eΣ2X2F=0(12)5

andXF + F 2 −X2H2 + KeΣ = 0 . (13)The above equation forms an integral of the set of differential equations (12).We present one more version of the β-function equations which appearssimpler but is just as hard to solve as other versions.

However, this form willbe useful later. Note that equations (12) are invariant under the scalingX→λXF→λFΣ→Σ + 2 ln λ(14)where λ is a constant.

Hence define a set of scale invariant variables [10]x = −X2e−Σ , f = FX , s = xΣ′ . (15)In terms x and f equations (12) and (13) becomesf + 1=0x(1 + 2f)dTdx + H=0x(1 + 2f) dfdx + Kx=0x(1 + 2f)dHdx −KT2x=0(16)andf + f 2 −H2 −Kx = 0 .

(17)The curvature scalar is given by R = 4γxf.We now consider the β-function equations (1) in the Schwarzschild gaugeds2 = −G(dx0)2 + G−1(dx1)2.For the static case we have G = G(x1).Equations (1) then becomeG′′ + G′φ′=0φ′′ + T ′2=0Gφ′′ + φ′(G′ + Gφ′) + 4γK=0GT ′′ + T ′(G′ + Gφ′) −2γKT=0(18)6

where in equations (18) and (19) (· · ·)′ denotesddx1(· · ·) and γ, K, and KTare as defined before. These equations appear to have no connection withequations (6) in the conformal gauge.

However, if we define G and ψ byG = 4γe−G , ψ′ = φ′ −G′ ,(19)then equations for G, ψ, and T become exactly the same as equations (6) forΣ, φ, and T respectively. Thus this implies that restricting the fields Σ, φ,and T to depend only on ξ corresponds to the static case.

Henceforth, inthis paper we will work only in the conformal gauge.Finally, let us briefly discuss the residual transformations which preserveboth the conformal gauge ds2 = eσdudv and the static nature of the fields.Such transformations are given byu = ˜ub, v = ˜vb(20)where b is an arbitrary constant. Under transformations (20) we haveξ=˜ξb(· · ·)˜1=b(· · ·)1˜σ=σ + (b −1) ln ˜ξ + 2 ln b˜Σ=Σ + 2 ln b˜X=bX˜F=bF(21)where (· · ·)1 denotes (ξ ddξ)(· · ·) as before.

Also equations (6) remain invariantunder these transformations. Thus, for example, one may scale both X and Fby an arbitrary constant using the residual conformal gauge transformations(20).3.

Static SolutionsWe now consider the solutions of β-function equations described in lastsection. First let T = 0.

In this case, the general solution of equations (1)is given by the two dimensional black hole solution presented in [4] whichcorresponds to the static case. This solution can be written as follows.

T = 07

implies that H = KT = 0 in equation (12). We then obtain F = 1 −Xwhere, using transformations (20), the integration constant is set equal to1.Equation (13) then gives eΣ = X −1.From X1 = −XF it followsthat X = α(ξ + α)−1 where α is a constant related to the black hole massparameter a of ref.

[4] by a = −γα (see section 4 also). This constitutes thegeneral solution of equation (1) when T = 0.However, by setting T = 0 we loose all information about tachyon backreaction.

It is important to incorporate this back reaction particularly inview of the hints of instability found in various works [3, 4, 8] when tachyonis included. The main difficulty when T ̸= 0 is that the resulting equationsare hard to solve explicitly.

In this section we present a static solution whenT ̸= 0 thus including the tachyon back reaction.Notice that equations (6) can be solved explicitly if one assumes that thetachyon potential vanishes, i.e. , V (T) = 0.

The solution obtained in thisapproximation incorporates tachyon back reaction and exhibits new featureswhich, we show later, will persist even when V (T) ̸= 0.When V = 0, K = 1 and (12) implies that H = constant. Let H2 =ǫ(1 + ǫ) where ǫ ≥0 (equivalently ǫ can be ≤−1) so that T 21 = X2H2 ≥0.We then have F(1 + F ′) = ǫ(1 + ǫ)X.

Note that this equation is invariantunder X →λX and F →λF where λ is a constant. This invariance suggeststhe substitution [11] X = es, F = es ˜F.

Integrating the resulting equationgives(F −ǫX)ǫ (F + (1 + ǫ)X)1+ǫ = constant . (22)Note that for ǫ = 0 (hence T = 0) in the above expression we get the blackhole solution of [4].In principle, this forms the complete solution.

However it is difficult tounderstand this solution — to integrate X1 = −XF, to understand how thevarious fields Σ, φ, T, and the curvature scalar R behave, etc. .

Thereforewe choose a different parametristaion for the above solution. LetF −ǫX = lBτ −1; F + (1 + ǫ)X = Bτ δ−1(23)where τ ≥0 is a new parameter, l = ±1, δ = (1 + 2ǫ)(1 + ǫ)−1, ǫ ≥0, B =A(1+2ǫ) and A is a constant.

Note that for l = −1 the constant in equation(22) need not be real and hence our choice of l constitutes minimal analyticcontinuation. Equations (23) giveX = Aτ −1(τ δ −l); F = Aτ −1(ǫτ δ + l(1 + ǫ))(24)8

from which it follows that (1 + ǫ)τ ˙X = F where˙(· · ·) = d(···)dτ . Equations (12)or (6) then give˙φ=−X−1 ˙X˙Σ=−((1 + ǫ)τ)−1˙T=−√δ −1τ −1τ1=−(1 + ǫ)τX(25)which can be integrated to obtaineφ=β0τ(τ δ −l)−1T=−√δ −1 ln τeΣ=−lB2τ −1(1+ǫ)Z τ0 dτ(τ δ −l)−1=A(1 + ǫ) ln( αmξ )(26)where β0 and α are constants and m = ±1 is the sign of ξ.

The curvaturescalar R is given byR = 4γ(1 + 2ǫ)−2τ −δ(τ δ −l)(ǫτ δ + l(1 + ǫ)) . (27)Equations (26) and (27) form the solution of equations (6) when V = 0.

Usingthe transformations (20) we can set B = 1. If ǫ = 0 and τ is eliminated, theabove expressions reduce to those given in the beginning of this section.For future use we also write down the expressions for x and f defined in(15).

For ǫ = 0 we havex =X21 −X ,f = 1 −XX(28)where X = α(ξ + α)−1. For ǫ ̸= 0 we get from (26)x=l(1 + 2ǫ)−2τ −δ(τ δ −l)2f=ǫτ δ + l(1 + ǫ)τ δ −l.

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4. Features of the static solutionsLet us now consider in detail the static solution given by equations (26)and (27).

The τ-integration in (26) cannot be done in a closed invertibleform except when ǫ = 0 or ∞. However to understand the solution andits salient features it is not necessary to do this integration.

The integrand(τ δ −l)−1 is regular for ǫ ≥0 and is well defined for τ ≥0. We only needfrom the equation relating τ and ξ the value of ξ for a given τ.

Since theintegrand varies monotonically with respect to ǫ, this information is easy toobtain qualitatively for generic ǫ.ǫ = 0This gives the solution obtained in [4]. In this case we get from (26)ln(τ −l) = ln( αmξ )(30)where τ ≥0, α is a constant and we have set B (= A) = 1.

Using theexpression for eΣ in (26) and its definition eΣ ≡γξeσ where γ = −2α′ we seethat l = sign(ξ) so that eσ ≥0. We first take l = m = 1 and writeτ −1 = αξ .

(31)For ǫ = 0 the above equation can be considered to be valid for −∞≤τ ≤∞.We then obtain, with β0 a constant,eφ=β0α (ξ + α)e−σ=−γ(ξ + α)(32)which is the solution of [4] if we set a = −γα = β0 where a is the blackhole mass parameter of [4]. This solution is asymptotic to flat space timeat ξ = ∞, has a horizon at ξ = 0, and a singularity at ξ = −α where thecurvature scalar R becomes divergent.

For more details see [4].However, in obtaining the above solution we let τ take negative valuesalso. This is alright for ǫ = 0.

But when ǫ > 0,τ δ in the integrand is not10

well defined for τ < 0. Therefore we always restrict τ to be ≥0.

We thenproceed as follows. Let ξ ≥0.

Then l = m = 1 so that the metric is welldefined as explained above. The range ∞≥ξ ≥0 outside the horizon thencorresponds to 1 ≤τ ≤∞.

We call this Branch I. When ξ ≤0 we havel = m = −1 and the range −α ≥ξ ≥0 inside the horizon corresponds to0 ≤τ ≤∞.

We call this Branch II. These two branches describe the solutiongiven above when ǫ = 0.

We assume that these are also the relevent brancheswhen ǫ > 0.We now make a few comments on the mass of the black hole, Mbh, whichcan be calculated using the behaviour of the metric in the asymptotically flatregion following the methods of Witten [3] or those of, for example, [12]. Wefind it convenient to use the formula given by T. Tada and S. Uehara in [12]which, in our notation, reads asMbh = (−γ)12eφ(e−Σφ21 + 1)(33)where (· · ·)1 = (ξ ddξ)(· · ·) as before.

Using (32) we obtainMbh = (−γ)12β0 . (34)Note further that Mbh as given in (33) is conserved, i.e.

,M1 = 0(35)as can be easily verified. (ii) ǫ > 0This case includes the tachyon back reaction.

Consider branch I, i.e. , τ ≥1and l = m = 1.

The integralR τ0 dτ(τ δ −1)−1 tends to −∞as τ →1+ andtends to a finite positive semidefinite value (= A(1 + ǫ) ln αξ+, by definition)as τ →∞. Thus we have ∞≥ξ ≥ξ+ > 0 as 1 ≤τ ≤∞.

Similarly forbranch II, the integralR τ0 dτ(τ δ + 1)−1 tends to a finite positive semidefinitevalue (= A(1 + ǫ) ln αξ−, by definition) as τ →∞.It then follows that−α ≤ξ ≤−ξ−< 0 as 0 ≤τ ≤∞. Moreover ξ± →0 as ǫ →0.

This impliesthat the horizon which was located at ξ = 0 when ǫ = 0 now splits into twolocated at ξ = ±ξ±. Thus the horizon resembles Schwarzschild horizon forǫ = 0 and Reissner-Nordstrom horizon for ǫ > 0.11

That τ = ∞does indeed correspond to the location of horizon can alsobe seen by the zeroes of the metric Gµν in Schwarzschild coordinates (r, t)defined by dξ = 2(−γ)12ξe−Σdr and dχ = 2(−γ)12χdt. The metric is given byds2 = eΣdt2 −e−Σdr2(36)where eΣ = γξeσ and γ = −2α′.

Equation (26) implies that eΣ →0 as τ →∞for any ǫ ≥0. Hence τ = ∞corresponds to a horizon.

Note also that inSchwarzschild coordinates the fields are indeed static, i.e , independent oftime coordinate t and that the asymptotically flat region is given by r →∞.Though for ǫ > 0 the metric components Gµν have the right zeroes andpoles at τ →∞, it is not really a horizon in the usual sense because Gµνcannot be made regular by transforming coordinates. Thus τ = ∞is notjust a coordinate singularity.

This can be seen by evaluating the curvaturescalar R at τ = ∞. We see from equation (27) that R →∞as τ →∞with astrength proportional to ǫ for small values of ǫ.

Thus these new horizons havecurvature singularities. Also, unlike in the case of Reissner-Nordstrom blackhole, the above solutions cannot be extended analytically into the region−ξ−≤ξ ≤ξ+, retaining their correct limiting behaviour as ǫ →0.

Henceour solution is not really a black hole solution, though for ǫ = 0 it is.The curvature scalar R has another singularity in branch II at τ = 0 ascan be seen from equation (27). This is a singularity inside the inner horizonand is present even when ǫ = 0, i.e.

, T = 0. This is the singularity presentin the blackhole solutions of [3, 4].Note further that at τ = ∞the dilaton field eφ →τ −ǫ1+ǫ (→0).

Since e−φ2acts as a coupling constant in equation (2), it follows that one is dealing witha strong coupling regime near the horizon when ǫ > 0, i.e , when tachyonback reaction is included.Consider now the “black hole” mass Mbh calculated from the behaviourof the fields in our solution in the asymptotically flat region. It can be easilyseen that in the asymptotically flat region (branch I, τ →1+) the tachyonfield is negligible.

(This is true even when the tachyon potential V ̸= 0. Seesection 5 and, for example, [8]).

Therefore, equation (33) can still be appliedwith negligible error to calculate Mbh from the fields in the asymptoticallyflat region. We obtainMbh = (−γ)12β01 + 2ǫ.

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Thus the behaviour of the fields in the asymptotically flat region is not givingany indication of the new singularities that arise as a result of the tachyonback reaction. However, the formula (33) cannot be relied upon since it is notconserved when one uses the equations of motion (6) which include tachyonalso.

That is,(Mbh)1 ̸= 0 . (38)It is not clear if any conserved quantity exists when tachyons are included.We could not find any, except the trivial ones given in the equations of motion(6) or their linear combinations.5.

Effect of tachyon potential, V (T)In this section we consider the tachyon potential, V . It was necessary toassume that V = 0 in deriving our solution.

We show below that even whenV ̸= 0 the features we found in our solution will persist.It is convenient to view the tachyon equation in (6) as an equation for an(anti)damped oscillator where the couplings to graviton and dilaton providethe damping force and the tachyon potential V provides the restoring force.This would mean that as one moves in from the asymptotically flat region(branch I, τ →1+) towards the horizon (τ →∞), the tachyon will interactwith gravitational field acquiring a large kinetic energy. Then the tachyonpotential can be neglected in comparison to its kinetic energy.

That this isactually what happens can be seen by calculating the kinetic ((∇T)2) andpotential (V ) energy terms in the action (2) (or, equivalently, the damping(φ1T1) and the potential (eΣKT) terms in the tachyon equation in (6) ). Usingthe expressions (∇T)2 = 4γe−σT 21 , T = −√δ −1 ln τ, φ1T1 =qǫ(1 + ǫ)XF,and eΣKT = eΣ( T2 + O(T 2)) and the solution given in equation (26) onesees that tachyon kinetic energy indeed dominates its potential energy awayfrom the asymptotically flat region.

Hence, neglecting tachyon potential is avalid approximation far inside from the asymptotically flat region (branch I,τ >> 1).Asymptotically, however, the kinetic and potential enrgy of the tachyonare of the same order and hence V cannot be neglected1. But it is reasonable1Indeed, in the asymptotically flat region τ →1+, TV =√T where TV is the correcttachyon solution including its potential V (T ) = γT 2.

This relation can be worked out13

to expect that the correct asymptotically flat solution when V ̸= 0 can bematched at some point to (26) which becomes more and more valid as onenears the horizon. That this is likely to be the case can be seen by performingone iterative calculation when V ̸= 0 as follows.

Near the asymptotically flatregion, let τ = 1+y where y > 0 is small. Denoting˙(· · ·) =ddτ (· · ·) =ddy(· · ·),the dilaton and tachyon equations in (6) become¨φ + τ −1 ˙φ + (1 + T 24 )eΣτ 21=0¨T + τ −1 ˙T −TeΣ4τ 21=0(39)where we have taken V (T) = γT 2.

Note that τ −1 ˙T and −TeΣ4τ 21 act like effectivedamping and restoring forces respectively in the tachyon equation. For y →0the solutions (26) become, with B = 1,τ1=−y + · · ·eΣ=−1 +y1 + ǫ + · · ·φ=−ln y + constant +2 + ǫ2(1 + ǫ) y + · · ·(40)where · · · denote higher order terms in y.

The restoring force is given by−TeΣ4τ 21= T4y2(1 −y + · · ·) . (41)It follows from the tachyon equation in (39) thatT = a0√y + · · ·(42)where a0 is a constant.

Note that the above solution for the tachyon with itspotential included is indeed the square root of the tachyon solution given ineasily by substituting the expressions for graviton and dilaton that follow from equation(26) into the tachyon equation given in (6) and solving for tachyon near τ →1+ as we willshow. Or, it can also be seen easily from a calculation in “linear dilaton vacuum” similarto that of de Alwis and Lykken in ref.

[8].14

equation (26) when V = 0. Substituting this value of tachyon in the dilatonequation in (39) one obtains the correction to the restoring force given by−TeΣ4τ 21= T4y2(1 −y −a20y + · · ·) .

(43)What the above expression implies is that the tachyon back reaction reducesthe restoring force as one moves inside away from the asymptotically flatregion. Thus it indicates that the restoring force (and hence the tachyonpotential) is less and less important as one moves inside and that it caneventually be neglected, thus validating our assumption.

Furthermore, thepotential V (T) = γT 2 itself neglects O(T 3) corrections given, e.g. as in [13],byV = γ(T 2 −T 324 + O(T 4)) .

(44)Thus the correction term implies that the potential is less steep, suggestingthat the potential energy can eventually be neglected (compared to the ki-netic energy). Also, various recent works [8, 2, 3, 4] find instability of thed = 2 string black hole solution when tachyon is included.

Since our solutioncan be viewed as arising from a similar instability, it seems that neglect-ing tachyon potential is reasonable and that the qualitative features of oursolution will persist even when V (T) is properly taken into account.We now show by a perturbative analysis near the horizon when V ̸= 0that the horizon still develops a curvature singularity. This is seen by ananalysis of equations (16).We first show that x →0 ( ∞) in the asymptotically flat region (nearhorizon).

Recall that x = −X2e−Σ where X = Σ1 and note that (· · ·)1 =12(−γ)12eΣ ddr(· · ·) in (r, t) coordinates described in section 4. Since the metricin (r, t) coordinates is given by equation (36), it follows that in the asymp-totically flat region Σ →0 and also X →0.

Hence, x →0 there. Similarlynear the horizon eΣ →0 by definition and hence Σ →−∞.

Since X = Σ1, itfollows that X ̸= 0 and therefore x →∞near the horizon. This behaviourof x tending to zero in the asymptotically flat region and to infinity near thehorizon remains true irrespective of the nature of tachyon potential and canbe seen explicitly from equations (28) and (29) for special cases T = 0 andV = 0 respectively.Consider the β-function equations in the form given in (16) and (17).

Letus call the last three equations in (16) as T, f, and H equation respectively.15

To find the solution near x = 0 or ∞one first starts with an ansatz for thetachyon field and then solves f, H, and T equations, in that order, uptoleading corrections. The equation (17) will be used to fix constants.

Theansatz is considered correct if the solution of T equation gives the same formfor T, to the leading order, as originally assumed.Consider first x →0 and the ansatz T = a0xn for the tachyon. Let n > 0so that T will not be divergent in the asymptotically flat region x →0.

Aftersolving f, H, and T equations one gets, for V = γT 2,f + f 2=x−1 + f 20 −14 + · · ·H=−a02 x−14 + H0 + · · ·T=a0x14 −H0√x + · · · . (45)Here and in the following a0, f0, and H0 are constants.

For V = 0 one getsH = H0 + · · · and T = a0√x + · · · and f as above.Next consider the ansatz T = a0xn ln x for the tachyon as x →0 (withn > 0 for the same reasons as above). After solving f, H, and T equationsone gets, for V = γT 2,f + f 2=x−1 + f 20 −14 + · · ·H=−a02 x−14(ln x + 4) + · · ·T=a0x14 ln x + · · · .

(46)The above ansatz is not valid for V = 0. In the above equations · · · denotethe subleading terms in powers of x which can be obtained by performingmore iterations.

Note that for both the ansatzes above, the curvature scalarR (= 4γxf) tends to zero as x →0 as expected and the tachyon behaviouragrees with that found in [8].We now consider the region near horizon where x →∞and V ̸= 0. Wewould like to know how the solution behaves near the horizon, i.e.

, x →∞,and in particular, whether the curvature scalar R is divergent or not whenV ̸= 0. Let us first consider the ansatz T = a0xn + · · ·, where · · · denotesubleading terms.

The exponent n can be positive or negative and a0 is aconstant. After solving f, H, and T equations one gets n ≤12 andT ≈constant + ln x + xn−1(47)16

schematically, omitting the coefficients. This is in contradiction to the orig-inal ansatz which, therefore, is incorrect.Next we take the ansatz T =a0 ln x + · · · .

Solving f, H, and T equations we getf + f 2=f 20 −14 + a204 x−1(ln2 x + 2 ln x + 2) −x−1 + · · ·H=−2a0f0 −a08f0x−1(ln x + 1) + · · ·T=a0 ln x + x−1[ a308f 20(ln2 x + 4 ln x + 6)−a016f 20(ln x + 2) + a02f 20] + · · ·(48)where · · · denote O(x−2 ln∗x) terms for some integer ∗. Note that this solu-tion of T agrees to the leading order with our starting ansatz which, therefore,is correct.

We also get from equation (17) thatf 20 −14 = 4a20f 20(49)which implies that f +f 2, and hence f, is nonzero as long as a0, and hence, Tis nonzero. Note that the solution given in (26) for V = 0 satisfies the aboveequation, as it should.

One can continue the above iteration to obtain thesubleading terms of O(x−2 ln∗x) but they are not necessary for our purpose.If the tachyon is not identiaclly zero, i.e. a0 ̸= 0, then as x →∞, fapproaches a nonzero constant and hence the curvature scalar given by R =4γxf diverges near the horizon where x →∞.

Note also that the leadinglogarithmic behaviour of T near the horizon when V ̸= 0 agrees with oursolution (26) obtained assuming V = 0.Thus the curvature scalar R is divergent at the horizon even when thetachyon potential V (T) ̸= 0. Hence it follows that when tachyon back reac-tion is included, the d = 2 string does not admit any static solution whichcan be interpreted as a black hole in the usual sense.6.

DiscussionsIn this section we first briefly present a time dependent extension of oursolutions and then discuss possible methods of removing the singularities17

found in previous sections. Consider the ansatz for the fields Σ, φ, and TΣ(ξ, χ)=Σ(ξ) + ˜Σ(χ)φ(ξ, χ)=χ(ξ) + ˜φ(χ)T(ξ, χ)=T(ξ) + ˜T(χ)(50)which is the nonlinear analog of seperation of variables.

The variables ξ andχ are defined in section 2. The equations for the ξ dependent fields are thesame as in (6) with eΣ understood as eΣ(ξ,χ) and those for the χ dependentfields are given byΣ00 + φ0Σ0=0Σ00 + φ00 + T 20=0Σ1φ0 + Σ0φ1 −2T0T1=0T00 + φ0T0=0φ00 + φ20=0(51)where we have omitted the tildes over the χ dependent fields.As before(· · ·)0 ≡(χ ddχ)(· · ·) and (· · ·)1 ≡(ξ ddξ)(· · ·).

Note that the above equationsare the same as those given in (6) with γ = 0 and ξ replaced by χ. From thefirst, fourth, and fifth equations above it immediately follows thatΣ0 = a1φ0,T0 = a2φ0(52)where a1 and a2 are constants.

The second equation in (51) givesa22 = 1 + a1(53)while the third equation implies, using (26) for Σ(ξ), φ(ξ), and T(ξ), thata1 = 0, a22 = (4ǫ(1 + ǫ))−1 . (54)The splitting of the β-function equations (5) into (6) and (51) and the useof the solutions (26) of equations (6), which involve eΣ terms, is justified aposteriori since a1 = 0 and hence Σ = Σ(ξ).

The equation for φ(χ) can beeasily solved and we getφ(χ)=ln(b1 ln Aχ)T(χ)=ln(b2 ln Aχ)(55)18

where A, b1, and b2 are constants. Note however that this time dependentextension does not alter any of the features of the previous static solution(26).Nor is it likely to represent the general solution of the β-functionequations (5).We now discuss some possible interaction terms that may remove thesingularities found in previous sections, atleast the new ones:(1) Higher order α′ corrections: Tseytlin [14] had shown that black holesolutions of [3, 4, 5] survive these corrections.

Very likely, the solution givenhere will also survive these corrections since the tachyon can be thought ofas an (anti)damped oscillator gaining energy by gravitational interactions —so that it would have grown strong before one reaches the region of strongcurvature where α′ corrections are deemed important. Thus these correctionsmay not remove the singularities.

(2)Higher massive modes : For similar reasons as above higher massivemodes An are of no help. In fact, taking their effective action as given in [13]with zero potential, it is easily seen that An’s have solutions similar to thatof tachyon T and hence do not remove the singularities.

(3) Antisymmetric tensor, H (indices on H suppressed) : This field is notthere for d = 2 space-time. However, for d = 2 toy models of a D dimensionalspace-time, as considered in [1, 2] for example, the resulting equations thatinclude quantum effects will be similar to the β-function equations for d = 2strings.

H field interactions present in such cases may possibly remove thesingularities. Moreover, H fields invariably arise in any space-time obtainedfrom string theory, so it is natural to include them.

(4) Supersymmetry : This symmetry introduces fermions which may pro-vide enough repulsive force to avoid the formation of the singularities. Onthe other hand, the fermions might instead form attractive condensates andnot remove the singularities.

(5) Other approaches : It will also be interesting to find an analog ofour solution, which includes tachyon back reaction, in the context of gaugedWess-Zumino-Witten models such as considered in [3] or in the context ofmatrix model where tachyon equation in the black hole background has beenfound recently [15]. Perhaps in these contexts one may get a better insightinto the singularities and methods of avoiding them.The above (and other) possibilities are worth pursuing.

It is important toresolve this problem of new singularities — to understand them better andto remove them if possible. This issue is particularly relevent for the string19

inspired toy models of d = 2 quantum gravity that may answer the puzzlesof d = 4 space-time. The removal of the singularities seen here might alsosuggest new interactions that could be important.I thank S. Sen for encouragement and S. R. Das for pointing out the lastpaper in [8].

This work is supported by EOLAS Scientific Research ProgramSC/92/206.Notes Added:(1) N. E. Mavromatos has kindly brought ref. [16] to our attention wheretime dependent perturbations of the black hole solution are considered per-turbatively in the asymptotically flat region.Notes Added in Proof:(1) In ref.

[17] Mann et al. consider 1 + 1 dimensional black hole for-mation in various models, for the first time to our knowledge by two sidedgravitational collapse.

For the string inspired models they find that blackhole formation requires dilaton surface charges, interpreted as arising from atachyon potential.We comment on two recent papers [18, 19] which appeared after our work:(2) In ref. [18] Marcus and Oz calculate tachyon effects in d = 2 stringblack hole and find none of the features presented here.

This is not surprisingsince they work with the tachyon equation in black hole background, thusnot taking into account the effects of backreacted graviton and dilaton ontachyon. Therefore tachyon back reaction is not completely accounted for intheir work.

(3) In ref. [19] Peet et al.

point out that besides the singular solutiondescribed in this paper, the d = 2 string admits another solution whichis regular at the horizon.It can be obtained by taking the ansatz T =const. + .

. .

for the tachyon near the horizon in equations (16) and (17).This solution is trivial (that is, T = const.) when the tachyon potentialV = 0 and has infinite energy when V ̸= 0 (see [19]).

These authors furtherargue that a singular configuration cannot be dynamically formed startingfrom a regular one. We find their argument inconclusive and will addressthis issue elsewhere [20].20

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[10] This transformation is an extension of a similar one given in NonlinearOrdinary Differential Equations and Their Applications, P. L. Sachdev,p. 61 (Vol.

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