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which gives = 1 gs2 (14.40)
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CHAPTER 14 Black Holes
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Recalling that the entropy of an excited string is proportional to the product of its mass and length [Eq. (14.34)], this allows us to write the entropy in terms of the mass of the black hole and the gravitational constant using [Eqs. (14.39) and (14.37)]. This gives a result proportional to Eq. (14.33):
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D D S = m B 3 G D 3
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(14.41)
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This calculation was not formal by any means. It just relied on some basic considerations from string theory, but it gave the correct result modulo a constant. Now let s turn to the ve-dimensional black hole example. The structure of the ve-dimensional geometry is as follows. We take a circular dimension of radius R denoted by S 1 and a four torus T 4 so that T 5 = T 4 S1 As mentioned above, the black hole actually has two string components. One is an actual string (a D1-brane) which wraps around S 1 and so has winding modes which will contribute to its mass. The D5-brane wraps around S 1 and also has KaluzaKlein modes quantized on the circle T 5 . The starting point is the metric given by
2 ds 2 = 2 /3 dt 2 + 1/3 ( dr 2 + r 2 d 3 )
(14.42)
where:
3 r 2 = 1 + i r i =1
(14.43)
A result from superstring theory that we simply take as a given because it s beyond the scope of our discussion that the BPS condition is satis ed. What this means for us is that charges are additive. The upshot of this is we can write the mass of the black hole as M = M1 + M 2 + M3
String Theory Demysti ed
Now the calculation of the entropy is actually quite straightforward. From the metric in Eq. (14.42), there are three radii associated with the horizon. Using Eq. (14.9) with D = 5 , we can write each of these as ri2 = 16 mG5 gs2 8 s = m 3 3 RV i (14.44)
where R is the radius of the circular dimension S 1 and V is the volume of the torus. The individual masses can be calculated from string considerations. The rst two masses are due to winding modes. First, the string winding around radius R gives m1 = Q1 R gs 2 s (14.45)
For the D5-brane, rst we have the winding mode which wraps around the circle and torus: m2 = Q5 RV gs 6 s (14.46)
Then we have a third mass, due to the Kaluza-Klein excitation of the D5-brane along the circular dimension: m3 = n R (14.47)
Now let s calculate each of the radii:
r1 =
RV gs
m1 =
(14.48)
V (14.49)
r2 =
RV gs
m2 = gs
r3 =
m3 =
(14.50)
CHAPTER 14 Black Holes
Now, the area in ve dimensions is A = 2 2 r1r2 r3 = 2 2 The entropy is S= A 4G5 gs2 8 s Q1Q5 n RV
(14.51)
where G5 = ( 2 )5 G10 RV = ( 2 )5 8 6 gs2 8 . Putting everything together we obtain s the result: S = 2 Q1Q5 n (14.52)
Summary
One of the recent successes of string theory has been its ability to count up the microscopic states of a black hole to calculate its entropy. The result obtained in this manner agrees with the semiclassical expressions, providing strong support for string theory as a quantum theory of gravity.
Quiz
1. Find the temperature of a D = 4 charged black hole. 2. The Hagedorn temperature is the temperature above which multiple strings would coalesce into a single string. Take the density of string states to be n = exp( 4 m ) and write down the partition function. What condition is necessary for the partition function to be nite This gives the Hagedorn temperature. 3. Suppose that the sun were to collapse to a black hole. What would be its temperature 4. Estimate the lifetime of a six solar mass black hole that is evaporating from the Hawking process.
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