Gravity and Strings

322 The Kaluza–Klein black hole
and keep only the zero mode −1/( √ 2πℓk0). The replacement of the δ function by its
constant zero mode gives us the z-independent harmonic functions and metric, which can
be immediately rewritten in terms of d-dimensional fields that we express both in the KK
frame and in the modified Einstein frame for the interesting, asymptotically flat d > 4
cases:
ds 2 KK = H −1 dt 2 − d x 2
d−1 ,
d ˜s 2 E
= H − d−3
d−2 dt 2 − H 1
d−2 d x 2
d−1 ,
Ãt = α(H −1 − 1), ˜k = H 1 2 , α =±1,
H = 1 +
h
, h =
|xd−1|
d−3
16πG ( ˆd)
N pz
2πℓk2 0 (d − 3)ω(d−2)
.
(11.131)
This is the d-dimensional extreme electric KK BH solution. As expected, it describes a
massive, electrically charged object that should be a KK mode. It does not have a regular
horizon. It is clear that, had we started from the general family of pp-wave solutions
Eqs. (10.42), we would have obtained a family of solutions of the same form but with arbitrary
harmonic functions. Thus, we can construct solutions of the KK action Eq. (11.39)
with several of these objects with charges of the same sign in static equilibrium by the
standard procedure. Now, the equilibrium is more difficult to describe because a third interaction,
mediated by the KK scalar k, comes into play. On the other hand, in the reduction of
the ERN solution we also found a solution describing a massive object charged with respect
to a vector field and with a non-trivial scalar, but different from this one. The reason is that
they obey different equations of motion, the difference being the strength with which the
KK scalar couples to the vector field. We will study these dilaton “BHs” in more detail in
Section 12.1.
We can calculate the mass and charge of the above solutions to check that they do indeed
correspond to those of a KK mode. From
and the definition of the mass M,
d−3
− d − 3 h
˜gE tt = H d−2 ∼ 1 − , (11.132)
d − 2 |xd−1|
d−3
˜gE tt ∼ 1 − 16πG(d)
N M
(d − 2)ω(d−2)
1
, (11.133)
|xd−1|
d−3
we find M = pzk −1
0 ,asexpected.
The electric charge can be calculated using the definition in Eqs. (11.63), finding first
d−1
˜k
2 d−2 ⋆ ˜F =±(d − 3)h d d−2 , (11.134)
where dd−2 is the unit (d − 2)-sphere volume form, whose integral over the sphere just
gives ω(d−2) (see Appendix C). The final result is ˜q =±pzk −1
0 (pz was taken to be positive),
also as expected.