304
Black holes in string theory
world volume (X 2 , X 3 , X4, x 5 ), and the final factor on the remining dimensions
(y6, Y 7, y8. y9) that are transverse to both. This symmetry forbids rigid branes
from carrying linear or angular momentum. So the question arises as to what
degrees of freedom do carry the momentum n/ R. An obvious possibility is the
massless states of the open strings that begin and end on D-branes. (The massive
excitations of the O-branes have masses proportional to g; I and. hence, do not
play a role when the coupling is weak.) The I-I states, in which the open string
begins on a 0 I-brane and ends on a 0 I-brane, generate a vector supermultiplet in
the adjoint representation of the U(ql) gauge group. similarly for the 5-5 states
with gauge group U(qs). In geometrical terms, the VEVs of the scalar fields
in these supermultiplets correspond to separations of the individual 01- and 05branes from each other. This takes us away from the black-hole state, which
has maximal degeneracy. So instead we consider the 1-5 and 5-1 states. These
generate hypermultiplets in the (fl' 9s)+(91' qs) of U(ql) x U(qs), which gives
a total of 4qlqS (scalar) bosons and an equal number of (Weyl) fermions. The
vacuum expectation values of the scalars are associated with the 4 coordinates
(transverse to the 0 I-branes) of each 0 I-brane giving its position relative to each
05-brane. This configuration must be made to carry P = n/ R momentum in
the x I-direction. With the four coordinates x 2 , x 3 , x4, x S compactified on a torus
whose size is small compared to that of the circle on which x I is compactified
(V ;/4 « R), we effectively have a two-dimensional field theory on the world
volume (xO.xl) of the D-string. The Hamiltonian is H = n/R and this
has to be distributed among the 4qlqS bosons and fermions. Apart from the
(minor) complication introduced by having fermions. this is precisely the problem
discussed in section 10.4 in which we estimated the number of (bosonic) string
states having mass n (in string units).
As before. the problem may be solved using a generating function (the
partition function)
00
Z(w) == trw N = Ldnw n
(10.150)
n::{)
where N == PR and d n is the (required) number of states having eigenvalue n of
N (and. therefore. right-moving momentum n/ R). Then
00
trwN = n (1 + w'" )4qIQS
(10.151)
w m
m=1 1 -
•
The fermions give the terms in the numerator and the bosonic contribution in
the denominator is derived precisely as in (10.81). As in section 10.4, we may
estimate d n for large values of n (exercise 12) with the result [22]
dn ... exp (21r ,Jqlqsn).
(10.152)
Hence.
lndn ... 27r,Jqlqsn = Sbh
(10.153)
Black holes in string theory
world volume (X 2 , X 3 , X4, x 5 ), and the final factor on the remining dimensions
(y6, Y 7, y8. y9) that are transverse to both. This symmetry forbids rigid branes
from carrying linear or angular momentum. So the question arises as to what
degrees of freedom do carry the momentum n/ R. An obvious possibility is the
massless states of the open strings that begin and end on D-branes. (The massive
excitations of the O-branes have masses proportional to g; I and. hence, do not
play a role when the coupling is weak.) The I-I states, in which the open string
begins on a 0 I-brane and ends on a 0 I-brane, generate a vector supermultiplet in
the adjoint representation of the U(ql) gauge group. similarly for the 5-5 states
with gauge group U(qs). In geometrical terms, the VEVs of the scalar fields
in these supermultiplets correspond to separations of the individual 01- and 05branes from each other. This takes us away from the black-hole state, which
has maximal degeneracy. So instead we consider the 1-5 and 5-1 states. These
generate hypermultiplets in the (fl' 9s)+(91' qs) of U(ql) x U(qs), which gives
a total of 4qlqS (scalar) bosons and an equal number of (Weyl) fermions. The
vacuum expectation values of the scalars are associated with the 4 coordinates
(transverse to the 0 I-branes) of each 0 I-brane giving its position relative to each
05-brane. This configuration must be made to carry P = n/ R momentum in
the x I-direction. With the four coordinates x 2 , x 3 , x4, x S compactified on a torus
whose size is small compared to that of the circle on which x I is compactified
(V ;/4 « R), we effectively have a two-dimensional field theory on the world
volume (xO.xl) of the D-string. The Hamiltonian is H = n/R and this
has to be distributed among the 4qlqS bosons and fermions. Apart from the
(minor) complication introduced by having fermions. this is precisely the problem
discussed in section 10.4 in which we estimated the number of (bosonic) string
states having mass n (in string units).
As before. the problem may be solved using a generating function (the
partition function)
00
Z(w) == trw N = Ldnw n
(10.150)
n::{)
where N == PR and d n is the (required) number of states having eigenvalue n of
N (and. therefore. right-moving momentum n/ R). Then
00
trwN = n (1 + w'" )4qIQS
(10.151)
w m
m=1 1 -
•
The fermions give the terms in the numerator and the bosonic contribution in
the denominator is derived precisely as in (10.81). As in section 10.4, we may
estimate d n for large values of n (exercise 12) with the result [22]
dn ... exp (21r ,Jqlqsn).
(10.152)
Hence.
lndn ... 27r,Jqlqsn = Sbh
(10.153)