TOPIC V BLACK HOLES IN STRING THEORY

TOPIC V BLACK HOLES IN STRING THEORY

Lecture notes Making black holes How should we make a black hole in string theory? A black hole forms when a large amount of mass is collected together. In classical general relativity, the resulting gravitational field then deforms spacetime to create a horizon. To make a black hole in string theory, we must use the elementary excitations gravitons, strings, branes etc. present in the theory. We should put a large number of these excitations together, and see if we get a black hole. The simplest excitation is the graviton. But the graviton is massless, and so moves at the speed of light. While a black hole can be made from a sufficient number of massless particles, it is easier to start with excitations that are massive, since these can be made to stay at a given location. String theory has extended objects like strings and branes. In their lowest energy state, the force of tension makes such objects collapse to a point, and we again get massless excitations like gravitons and gauge fields. To get a massive excitation, we should somehow stretch these objects to a nonzero size. The simplest way to do this is to compactify some directions of space. We can then wrap the extended object on these compact directions, causing it to have a nontrivial extent, and thus a nonzero mass. Let us now follow this path, and see if we get black hole.. Wrapped strings We start as follows: (i) String theory lives in 9+ spacetime dimensions. We compactify p directions to circles; let these be the directions x,... x p. This leaves 9 p space directions noncompact. Thus we will get a black hole in (9 p) + spacetime dimensions. (ii) We wrap a string along one of the compact directions, say x. In the noncompact directions, we let this string be at the origin: {x p+,... x 9 } = 0. Let the length of the x circle be L. Then the energy of the wrapped string is E = T NS L (.) From the viewpoint of the noncompact directions, this gives a point mass m = T NS L (.2) 2

.. WRAPPED STRINGS 3 Figure.: (a) The horizontal direction represents the noncompact space directions, while the vertical direction represents the compact directions. A string is wrapped on a compact circle. (b) From the viewpoint of the dimensionally reduced theory, we see only the noncompact directions. The string now appears as an object with some mass m. (a) (b) Figure.2: (a) If we take many separately wound strings, then we would be describing many separate masses. (b) We take one multiwound string; this is a bound state, and so corresponds to one massive object. at the location {x p+,... x 9 } = 0 (fig..). (iii) We want a large amount of mass to make a big black hole. Thus we take n strings wrapped as above, with n (.3) (iv) If we have n separately wrapped strings (as in fig..2(a)), then we would be trying to make n separate tiny black holes. We wish to make one large black hole. Thus we need to consider a bound state of these n strings. We have already seen that such a bound state has a simple form: the string wraps n times around x before closing. From the viewpoint of the noncompact directions, this gives an object of mass at the location {x p+,... x 9 } = 0 (fig..2(b)). m = n T NS L (.4) Now we ask: has this construction resulted in a black hole? To answer this question, we must compute the metric produced by the above mentioned strings.

4 LECTURE NOTES. MAKING BLACK HOLES.. The metric produced by strings Let us take type IIA string theory for concreteness, and for the moment let all directions be noncompact. It will be convenient to begin by looking at the string metric g S µν, though later we will consider the Einstein metric g E µν as well. Let us separate a direction x along which we will presently place our strings. Introducing polar coordinates in the directions transverse to x, the flat spacetime metric is ds 2 S = dt 2 + dx 2 + dr 2 + r 2 dω 2 7 (.5) Now consider n string stretching along the direction x, placed at the location r = 0 in the transverse space. The metric produced by these strings has the form [] ds 2 S = H [ dt 2 + dx 2 ] + dr 2 + r 2 dω 2 7 (.6) where H = + Q r 6 (.7) Metrics of strings and branes will be crucial in our study, so let us analyze this metric in some detail: (i) The string has a 2-dimensional worldsheet spanning the directions t, x. We see that the metric is Lorentz invariant in this t, x space. This invariance reflects the fact that the action of the string is just given by the area of the worldsheet. (ii) The coefficient function H is a harmonic function in the 7-dimensional space transverse to the brane: H 8 2 x 2 i=2 i H = 0 (.8) This is analogous to the fact that in the Schwarzschild metric we get g tt = ( 2M r ), which is harmonic in 3-dimensional space. (iii) The quantity Q is proportional to the number of strings n Q n (.9) We assume that n to get strong sources whose metric can be well described by a classical approximation. The string is a charged object, and radiates the gauge field B ab. The coupling of the string to the gauge field has the form of an integral along the worldsheet: BdA. The worldsheet of the string lies along t, x. Thus we expect to generate a gauge field B tx. We have B tx = H = + Q r 6 (.0)

.. WRAPPED STRINGS 5 This gauge field is analogous to the field produced by a point charge q in electromagnetism. In the electromagnetic case the worldline is along t, and the coupling has the form: q Adl. This results in a gauge field component A t = q r, which is just the scalar potential produced by a stationary point charge. This scalar potential is harmonic: A t = 0. Similarly, B tx is harmonic in the transverse space x 2,... x 9. Note that we can add a constant to B tx by a gauge transformation. Thus we can make B tx 0 at infinity, and this potential then looks entirely analogous to the electromagnetic case. The gravitational solution produced by the string has one more nontrivial component: the dilaton φ. Recall that from an M-theory perspective, the string is given by a M2 brane wrapped on the direction x. The tension of the M brane tends to squeeze this direction to a smaller value near the location of the branes. But the quantity e φ reflects the size of the x direction. Thus we expect that e φ will approach smaller values as we approach r = 0. We have We see that as r 0 e φ = H = ( + Q r 6 ) (.) e φ r6 Q 0 (.2) so φ and the x circle gets pinched to zero length. Let us summarize the gravitational solution produced by elementary strings lying along x ds 2 S = H [ dt 2 + dx 2 ] + dr 2 + r 2 dω 2 7 B tx = H e 2φ = H (.3) with H = + Q r 6 (.4) a harmonic function in the transverse 8-dimensional space. What is remarkable is that we can replace H by any harmonic function, and still get a solution to the field equations. Thus we can take q H = + x x 6 + q 2 x x 2 6 + + q k x x k 6 (.5) where x, x 2,... x k are points in the 8-dimensional transverse space. This solution corresponds to string sources with strengths q i at locations x i, with all strings stretching along the direction x...2 Compactifying directions In the above discussion we had 9+ noncompact dimensions. Let us now see how we can compactify p of these directions and move towards the black hole solution we desire.

6 LECTURE NOTES. MAKING BLACK HOLES Compactifying x The string above stretched for an infinite length along the x direction. But a black hole should look like a localized object in spacetime. Thus the direction x must be among the directions we compactify. We see that the solution (.3) is independent of x. Thus it remains a solution if we assume that 0 x < L (.6) and that we have the identification x = 0 x = L (.7) Thus compactification along the direction of the string does not change the algebraic form of the solution. Compacitfying a transverse direction x 2 Now supposed we wish to compactify a transverse direction like x 2 on a circle of length L 2. A solution with such a compactification would be periodic under the shift x 2 x 2 + L 2 (.8) Unlike the situation with x, the solution (.3) is not periodic under shifts of x 2. To get a solution with the required periodicity, we take a -dimensional array of string sources along the x 2 direction, placed at locations Thus the harmonic function H has the form where H = + x 2 = nl 2, < n < (.9) n= r 2 = Q (r 2 + (x 2 nl 2 ) 2 ) 3 (.20) 9 x 2 i (.2) Bu construction, the functions appearing in the solution (.3) are now periodic under (.8), and we have compactified the transverse direction x 2. We will actually be interested in a useful approximation of the above solution. We normally think of the compact directions as having a fixed, small size, perhaps of order planck length. The length scale set by Q, on the other hand will be large, since Q n, and we make a large black hole by taking n. Thus we wish to consider the limit i=3 L 2 Q 6 (.22)

.. WRAPPED STRINGS 7 In this limit the sum in (.20) can be replaced by an integral n= Thus H takes the form Q (r 2 + (x 2 nl 2 ) 2 ) 3 n= Q (r 2 + (x 2 nl 2 ) 2 ) 3 = 3π 8L 2 r 5 (.23) H = + Q r 5 (.24) which is a harmonic function in the noncompact space x 3,... x 9, which is now 7-dimensional. Proceeding in this way, we find that if we compactify x and p other directions, then the solution has the form (.3) with H = + Q r (7 p) (.25) where r is the radial coordinate in the noncompact directions. We will be interested in the case where we compactify x and 4 other directions. In that case the harmonic function has the form..3 Looking for a horizon H = + Q r 2 (.26) In the above discussion we have worked with the string metric gab S. The physics of black holes was, on the other hand, is naturally described in the Einstein metric gab E. For example the entropy formula S bek = A 4G (.27) has the area of the horizon A measured using the Einstein metric. Thus we start by converting the solution (.3) to the Einstein metric. We have g E ab = e φ 2 g S ab (.28) This gives ds 2 E = H 3 4 [ dt 2 + dx 2 ] + H 4 [dr 2 + r 2 dω 2 3] + H 4 4 dx i dx i (.29) i= To look for a horizon, we consider a surface of constant r, and take the limit r 0. A constant r surface is a cylindrical 8-dimensional surface, with the following structure:

8 LECTURE NOTES. MAKING BLACK HOLES (i) In the transverse directions, we have a S 3 of radius H 8. The area of this sphere is A S 3 = (2π 2 )H 3 8 (.30) where is the volume of a 3-sphere of unit radius. (ii) The direction x has the length A Ω 3 = 2π 2 (.3) L x = H 3 8 L (.32) (iii) The directions along the T 4 have a volume V 4 = H 2 V4 (.33) Thus the overall area of the surface at radius r = r 0 is A H = ((2π ) ( ) ) 2 )H 3 8 r 3 H 3 8 L (H 2 V = 2π 2 LV H 2 r 3 2π 2 LV Q 2 r 2 (.34) We see that in the limit r 0 0, we get A H 0 (.35) and we have not succeeded in getting a black hole. It is not hard to see the reason for our failure. The strings wrap around the direction x. Their tension pinches this circle, making it have zero length at the location of the strings. We will now see that our failure to get a nonzero A H here is a good thing, since a nonzero A H would have led to an immediate problem for string theory. The entropy of the NS solution Since A H = 0, we find that the Bekenstein entropy is S bek = A H 4G = 0 (.36) Let us compare this to the microscopic entropy S micro of our system. We have taken a bound state of n strings, with no excitations on these strings. We have seen in (??) that the degeneracy of such a bound state is N = 256 (.37) Thus S micro = ln[256] 0 (.38)

.2. THE NS-P SOLUTION 9 The symbol 0 here means that S micro is a fixed number, rather than a number that grows with n. We take n to make a good classical solution fro our string source. If A H was nonzero for this classical solution, then it would be something that increased with n, and this would contradict (.38). As things have turned out, we can write and string theory has saved itself from a problem. S bek = S micro 0 (.39).2 The NS-P solution The tension of the NS strings had pinched the circle x, making the horizon area zero. What should we do to make the x direction not pinch? To get a idea of what we need, consider the energy of the wrapped strings without, for the moment, considering their backreaction on the metric. The energy is E NS = n T NS L (.40) This energy is minimized for L = 0, which is why the tension of the strings pinches the x circle to zero at the location of the strings. We thus need to add something whose energy will increase when L is decreased. Consider a graviton carrying n p units of momentum along the x. The energy of this momentum mode will be E P = 2πn p L (.4) This energy increases when L is decreased. Assuming that the directions x,... x 4 and y are compactified, one finds that the metric of such a graviton mode is ds 2 string = H [ dt 2 + dx 2 + K(dt + dx ) 2 ] + [dr 2 + r 2 dω 2 3] + where and the gauge field and dilaton vanish 4 dx i dx i (.42) i= K = Q p r 2 (.43) B µν = 0, e 2φ = (.44) The quantity Q p is proportional to the number of units of momentum Q p = g2 α 4 V R 2 n p (.45)

0 LECTURE NOTES. MAKING BLACK HOLES Now we take both NS and P charges. The solution is ds 2 string = H [ ( dt 2 + dx 2 ) + K(dt + dx ) 2] + [dr 2 + r 2 dω 2 3] + B tx = H 4 dx i dx i i= e 2φ = H (.46).2. Microscopic entropy of the NSP bound state We have a bound state of n strings and n p units of momentum. We have seen that this bound state is degenerate: there are N e 2 2π n n p (.47) states with the same mass and charge. These different states correspond to different ways in which the momentum n p can be carried along the multiwound string as travelling waves. Thus the microscopic entropy is S micro = ln N = 2 2π n n p (.48) This time we have an entropy that increases with the charges n, n p. Let us now compute the area of the horizon..3 The entropy of a gas of vibrations Let a direction y be compactified to a circle of length L. Consider a string wrapped n w on this circle; thus the total length of the string is L T = n w L. Let the string carry n p units of momentum along this string, say in the positive y direction. At this moment we put no excitations traveling in the negative y direction, though we will do that as well later. The total energy and momentum on the string must have the form E = P = 2πn p L = 2πn wn p L T (.49) The individual excitations carrying thus momentum can be in various harmonics k on the string with length L T.. An excitation in the kth harmonic has energy and momentum e k = p k = 2πk L T (.50) Let there be m k excitations in the harmonic k. Then we must have k m k = n w n p (.5) k=

.3. THE ENTROPY OF A GAS OF VIBRATIONS Our goal is to find the number N of sets {n k } which satisfy this relation; this number N gives the degeneracy of states with the quantum numbers (.49), and the entropy of the vibrating string will then be given by.3. The partition function S micro = ln N (.52) The partition function of a system is defined as Z[β] = e βestate (.53) states where β = (.54) T and T is the temperature. We first consider the partition function for a single boson, then for a single fermion, and finally, for the case where we have f B flavors of bosons and f F flavors of fermions. The partition function for a single boson Consider just one bosonic fourier mode k. Each excitation of this mode has energy e k = 2πk/L T. Since the number of excitations can be m k = 0,, 2,..., summing over various numbers of these excitations gives the contribution Z B k = m k =0 e βm ke k = e βe k (.55) Since the various k harmonics describe independent sets of excitations, the corresponding contributions Z k need to be multiplied together. We can consider the log of Z, where we find ln Z B = ln ZK B = ln[ e βe k ] (.56) k= For large values of n w, n p, the values of k are peaked at k, where we can approximate the sum over k by an integral ln Z B 0 dk ln[ e βe k ] = L T 2π k= The partition function for a single fermions 0 de k ln[ e βe k ] = L T π 2 2πβ 6 (.57) A fermion mode can have only two possible occupation numbers m k = 0,. Thus in place of (.55) we get Z F k = e βm ke k = + e βe k (.58) m k =0

2 LECTURE NOTES. MAKING BLACK HOLES Again taking the log of Z F and approximating the sum over k by an integral, we get log Z F L T de k ln[ + e βe k ] = L T π 2 (.59) 2π 2πβ 2 0 We see that ln Z F = 2 ln ZB, so a fermions counts as half a boson for the purposes of its contribution to the partition function. Several flavors of bosons and fermions Suppose we have f B bosonic degrees of freedom and f F fermionic degrees of freedom. Since each degree of freedom gives independent excitations, the corresponding partition functions Z are multiplied together, which leads to a sum over the corresponding logarithms log Z = f B log Z B + f F log f F = (f B + 2 f F ) πl T 2β c(πl T 2β ) (.60) where we have defined c = f B + 2 f F (.6) The quantity c is called the central charge and gives a measure of the effective degrees of freedom of the system..3.2 The thermodynamics of string vibrations Let us now use the above partition function to compute various thermodynamics quantities describing the vibrating string. General relations The average energy is given in terms of the partition function by Applying this to (.60), we find E = Z ( β)z = β ln Z (.62) E = cπl T 2β 2 (.63) Thus if we put an energy E on the string, and assume that it is distributed thermally among all possible excitations, then this thermal distribution will be characterized by a temperature T = β = 2E πl T c (.64)

.3. THE ENTROPY OF A GAS OF VIBRATIONS 3 The entropy of the distribution is S = ln N, where N is the average number of states that contribute to the partition function. Thus we can write Z = states e βestate e S βe (.65) This gives S = ln Z + βe (.66) Applying this to (.60), we find S = cπl T 6β = cπlt E 3 (.67).3.3 The string with charges n w, n p Let us now apply these general relations to our case of the 2-charge extremal system the string with winding n w and momentum n p. Thermodynamic quantities The string has 8 transverse directions in which it can vibrate, so f B = 8. By supersymmetry, there are a corresponding number of fermionic flavors, so f F = 8. Thus c = f B + 2 f F = 8 + 4 = 2 (.68) We have noted that E = 2πn wn p L T, L T = n w L (.69) Then (.64) gives T = 2 L np n w (.70) and (.67) gives S = 2 2π n n p (.7) Qualitiative picture of the vibrating string From the above computation we can extract a few other details. In a thermal distribution, the average energy of an excitation is ē T n n p L T (.72) From (.50) we see then that that the generic quantum is in a harmonic k n n p (.73)

4 LECTURE NOTES. MAKING BLACK HOLES on the multiwound string. Given that the total energy is (.69), we find that the number of quanta is m n n p (.74) From (.55) we can find the average occupation number of a bosonic energy level e k m k = β Z B k = e k e βe (.75) k Z B k so for the generic quantum with e k β we have m k (.76) For fermionic levels, m k = Z F k e k β Z F k = e βe k + (.77) so for the generic quantum with e k β we again have m k (.78) To summarize, there are a large number of ways to partition the energy into different harmonics. One extreme possibility is to put all the energy into the lowest harmonic k = ; then the occupation number of this harmonic will be m = n n p (.79) At the other extreme we can put all the energy into a single quantum in the harmonic n n p ; i.e. k = n n p, m k = (.80) But the generic state which contributes to the entropy has its typical excitations in harmonics with k n n p. There are n n p such harmonic modes; and the occupation number of each such mode is < m k >. These details about the generic state will be important to us later..3.4 The Bekenstein entropy of the NSP state The Einstein metric is ds 2 E = H 3 4 [ dt 2 +dx 2 +K(dt+dx ) 2 ]+H 4 [dr 2 +r 2 dω 2 3]+H 4 In the limit r 0, we get ds E r 3 2 [ dt 2 +dx 2 ]+ Q p (dt+dx Q 3 4 r 2 Q 3 ) 2 + Q 4 4 r 2 4 dx i dx i (.8) i= dr 2 +Q 4 r 3 2 dω 2 3 + Q 4 r 2 4 dx i dx i i= (.82)

.3. THE ENTROPY OF A GAS OF VIBRATIONS 5 Now look at the hypersurface t = 0. The surface at a given value of r has the following components to its area: (i) The direction along x has a length L x = Q 2 p L (.83) r 4 Q 3 8 (ii) The directions along the torus give V 4 = Q 2 r V (.84) (iii) The angular sphere gives V Ω3 = 2π 2 Q 3 8 r 9 4 (.85) Thus the area of the horizon is ( ) ( ) Q 2 p Q 2 ( ) A H = L r 4 Q 3 8 r V 2π 2 Q 3 8 r 9 4 = 2π 2 LV Q 2 p Q 2 r (.86) Once again, we find that as we take r 0, we get A H 0 (.87) So we still do not seem to have a black hole. This time however, there is a difference: we do have, in fact a nonzero horizon area; its just that in our present approxmation we are not able to see it..3.5 The NS-NS5-P black hole The metric produced by such NS5 branes is ds 2 string = [ dt 2 + dy 2 ] + H 5 [dr 2 + r 2 dω 2 3] + H 5 dx i dx i (.88) where H 5 = + Q 5 r 2 (.89) Here Q 5 is proportional to the number n 5 of NS5 branes Q 5 = α n 5 (.90) We get a dilaton e 2φ = H 5 (.9)

6 LECTURE NOTES. MAKING BLACK HOLES We also get a B µν field. Recall that the NS5 brane is the magnetic dual of the NS brane. Thus the B µν field lies along directions of the angular sphere S 3 B φψ = sin 2 θ (.92) We can now write down the solution with NS-NS5-P charges ds 2 string = H [ ( dt 2 + dx 2 ) + K(dt + dx ) 2] + H 5 [dr 2 + r 2 dω 2 3] + B tx = H, B φψ = sin 2 θ 4 dx i dx i e 2φ = H 5 H (.93) i= The horizon area has the following components (i) The angular directions have the area A ω3 = 2π 2 H 3 2 5 r 3 (.94) (ii) The torus directions have the volume A T 4 = V (.95) (iii) The x direction has a length dominated by the term L x = K 2 H 2 L (.96) Thus the overall area of the horizon in the string metric is ( ) ( A string H = 2π 2 H 3 2 5 r 3) K 2 (V ) L = 2π 2 V Lr 3 H 3 2 H 5 H 2 K 2 (.97) 2 The 0-D Einstein metric g E µν is related to the string metric g S µν by g E µν = e φ 2 g S µν = ( H H 5 ) 4 (.98) Thus ( ) 4 g E A E µν = = H A string = 2π 2 V Lr 3 H 2 5 H H 2 K 2 2π 2 V L Q Q 5 Q p 5 g S µν (.99)

.3. THE ENTROPY OF A GAS OF VIBRATIONS 7 Recalling that and that we find that Q = g2 α 3 V n, Q 5 = α n 5, Q p = g2 α 4 V R 2 n p (.00) G = 8π 6 g 2 α 4 (.0) S bek = A H 4G = 2π n n 5 n p (.02) The nonextremal gravity solution We continue to use the compactification M 9, M 4, T 4 S. We have charges NS, NS5, P as before, but also extra energy that gives nonextremality. The metric and dilaton are [?] ds 2 string = H [ dt2 +dy 2 + r2 0 r 2 (cosh σdt+sinh dr 2 4 σdy)2 ]+H 5 [ ( r2 0 r ) +r2 dω 2 3]+ dz a dz a 2 a= (.03) where e 2φ = H 5 H (.04) H = + r2 0 sinh 2 α r 2, H 5 = + r2 0 sinh 2 γ r 2 (.05) The integer valued charges carried by this hole are The energy (i.e. the mass of the black hole) is ˆn = V r2 0 sinh 2α 2g 2 α 3 (.06) ˆn 5 = r2 0 sinh 2γ 2α (.07) ˆn p = R2 V r 2 0 sinh 2σ 2g 2 α 4 (.08) E = RV r2 0 2g 2 (cosh 2α + cosh 2γ + cosh 2σ) (.09) α 4 The horizon is at r = r 0. From the area of this horizon we find the Bekenstein entropy The Hawking temperature is S Bek = A 0 4G 0 = 2πRV r3 0 g 2 α 4 cosh α cosh γ cosh σ (.0) T H = [( S E )ˆn,ˆn 5,ˆn p ] = 2πr 0 cosh α cosh γ cosh σ (.)

8 LECTURE NOTES. MAKING BLACK HOLES The extremal limit: Three large charges, no nonextremality The extremal limit is obtained by taking while holding fixed r 0 0, α, γ, σ (.2) r 2 0 sinh 2 α = Q, r 2 0 sinh 2 γ = Q 5, r 2 0 sinh 2 σ = Q p (.3) This gives the extremal hole we constructed earlier. For this case we have already checked that the microscopic entropy agrees with the Bekenstein entropy (??). It can be seen that in this limit the Hawking temperature is T H = 0. Two large charges + nonextremality We now wish to move away from the extremal 3-charge system, towards the neutral Schwarzschild hole. For a first step, we keep two of the charges large; let these be NS, NS5. We will have a small amount of the third charge P, and a small amount of nonextremality. The relevant limits are r 0, r 0 e σ r 0 e α, r 0 e γ (.4) Thus σ is finite but α, γ. We are close to the extremal NS-NS5 state, so we can hope that the excitations will be a small correction. The excitations will be a dilute gas among the large number of ˆn, ˆn 5 charges and a simple model for these excitations might give us the entropy and dynamics of the system. The BPS mass corresponding to the ˆn NS branes is M BP S = Rˆn α = RV r2 0 2g 2 α 4 sinh 2α = RV r2 0 2g 2 α 4 (cosh 2α e 2α ) RV r2 0 2g 2 cosh 2α α 4 (.5) The BPS mass corresponding to the ˆn 5 NS5 branes is M5 BP S = RV ˆn 5 g 2 α 3 = RV r2 0 2g 2 α 4 sinh 2γ = RV r2 0 2g 2 α 4 (cosh 2γ e 2γ ) RV r2 0 2g 2 cosh 2γ α 4 (.6) Thus the energy (.09) can be written as The momentum is Note that E = M BP S + M5 BP S + E, E RV r2 0 2g 2 cosh 2σ (.7) α 4 P = ˆn p R = RV r2 0 2g 2 sinh 2σ (.8) α 4 E + P RV r2 0 2g 2 α 4 e2σ, E P RV r2 0 2g 2 α 4 e 2σ (.9)

.3. THE ENTROPY OF A GAS OF VIBRATIONS 9 We wish to compute the entropy (.0) in this limit. Note that ˆn = V r 2 0 2g 2 α 3 sinh 2α V r2 0 g 2 α 3 cosh2 α (.20) ˆn 5 = r2 0 2α sinh 2γ r2 0 α cosh2 γ (.2) We then find S Bek 2π ˆn ˆn 5 [ R R 2 ( E + P ) + ( E P ) ] 2 (.22) Let us now look at the microscopic description of this nonextremal state. The NS, NS5 branes generate an effective string as before. In the extremal case all the excitations were right movers (R) on this effective string, so that we had the maximal possible momentum charge P for the given energy. For the nonextremal case we will have momentum modes moving in both R,L directions. Let the right movers carry n p units of momentum and the left movers n p units of (oppositely directed) momentum. Then (ignoring any interaction between the R,L modes) we will have E = R (n p + n p ), P = R (n p n p ) (.23) Since we have ignored any interactions between the R,L modes the entropy S micro of this gas of momentum modes will be the sum of the entropies of the R,L excitations. Thus using (??) we write S micro = 2π ˆn ˆn 5 n p + 2π ˆn ˆn 5 n p (.24) But using (.23) in (.22) we find S micro = 2π ˆn R R ˆn 5 [ 2 ( E + P ) + ( E P ) ] (.25) 2 Comparing to (.22) we find that S micro S Bek (.26) We thus see that a simple model of the microscopic brane bound state describes well the entropy of this near extremal system.

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