Gravity and Strings

318 The Kaluza–Klein black hole
ˆd it is unnecessary to sum the infinite series and then calculate the zero mode [712]: it is
possible to approximate the infinite sum by an integral. First we change variables,
and we have
with
un =
H = 1 +
∼ 1 +
z − 2πnRz
|x ˆd−2 |
2πnRz 2π(n + 1)Rz
, un ∈ ,
|x ˆd−2 | |x ˆd−2 |
, (11.118)
h
ˆd−3 |x ˆd−2 |
h
ˆd−3 |x ˆd−2 |
n=+∞
n=−∞
1
2π Rz
|x ˆd−2 |
1
(1 + u 2 n
+∞
−∞
ˆd−3
) 2
du
(1 + u2 ) ˆd−3
2
h
= 1 +
′
, (11.119)
ˆd−4 |x ˆd−2 |
h ′ = hω ( ˆd−4)
. (11.120)
2π Rzω ( d−5) ˆ
It is clear that this approximation is valid if |x ˆ
d−2 | >> Rz and for ˆd ≥ 5. For ˆd = 4 the
series does not converge. In fact, defining now
we have
H = 1 + h
n=+∞
|x2| n=−∞
= 1 + lim
v→+∞
un = (z − 2πnRz) ∈ [2πnRz, 2π(n + 1)Rz], (11.121)
1
(|x2| 2 + u2 n ) 1 ∼ 1 +
2
h
2π Rz
⎧
h
⎨
v
ln
⎩|x2|
+
⎫
2 v
⎬
1 +
|x2| ⎭
π Rz
+∞
−∞
du
(|x2| 2 + u 2 ) 1 2
∼− h
ln |x2|+D, (11.122)
π Rz
where D is a divergent constant. The solution to this problem [712] is to redefine each term
in the H series with a constant chosen so as to cancel D out:
n=+∞
H = h
n=−∞
1
(|x2| 2 + (z + 2πnRz) 2 |) 1 2
n=+∞
− 2h
n=1
1
. (11.123)
2πnRz
The solution is not asymptotically flat, but this is to be expected on physical grounds.