Gravity and Strings

280 The Taub–NUT solution
The electrically charged Taub–NUT solution was found by Brill in [184] and is
ds 2 = f (r)(dt + 2N cos θ dϕ) 2 − f −1 (r)dr 2 − r 2 + N 2 d 2 (2) ,
Ftr = 4q(r 2 − N 2 )
r 2 + N 2
, ( ⋆F)tr = 8qNr
(r 2 + N 2 ,
) 2
f (r) = (r − r+)(r − r−)
r 2 + N 2
,
r± = M ± r0, r 2 0 = M2 + N 2 − 4q 2 .
(9.56)
It reduces to the RN solution when we set the NUT charge to zero. It is trivial to generalize
these solutions to the magnetic and dyonic cases.
In contrast to the Taub–NUT solution, the charged Taub–NUT solution does have an
extremal limit M 2 + N 2 = 4q 2 in which the extremality parameter r0 vanishes and the two
zeros of the metric function f (r) coincide. In this case, by shifting the radial coordinate to
ρ = r − M and defining Cartesian coordinates such that ρ =|x3|,wefind a simple form of
the solution, 8
ds 2 =|H| −2 (dt + A) 2 −|H| 2 d x 2
3 ,
At = 2Re(eiαH), Ãt = 2Im(eiαH), M + iN
H = 1 + ,
|x3|
A = Aidx i , ɛijk∂i A j =±Im(H∂kH).
(9.57)
As in some of the other “extreme” solutions that we have found so far, 9 it turns out that
we obtain a solution for any complex harmonic function H(x3). Byabsorbing the complex
phase e iα into H, wecan write the general solution in this form:
ds 2 =|H| −2 (dt + A) 2 −|H| 2 d x 2
3 ,
At = 2Re H, Ãt =−2Re(iH),
A = Aidx i , ɛijk∂i A j =±Im(H∂kH),
∂i∂iH = 0.
(9.58)
Metrics of the above form are known as conformastationary metrics [640]. Observe that
the integrability condition of the equation for the 1-form A is the Laplace equation for H.
8 Here we are actually taking the extreme limit of the dyonic solution, which indeed has a simpler form. The
information on the electric and magnetic charges is contained in the SO(2) electric–magnetic-duality phase
e iα .
9 But not in all of them. In particular, not in the Kerr BH.