Contents

326 21 Non-empty stationary axisymmetric solutionssolution. They obtained the new complex null tetradk i =(1, 0,a/∆, (r 2 + a 2 )/∆), l i = 1 2 (r2 + a 2 cos 2 θ) −1 (−∆, 0,a,r 2 + a 2 ),(21.23)m i =2 −1/2 (0, 1, i/ sin θ, ia sin θ)/(r +ia cos θ), ∆ ≡ r 2 + a 2 + e 2 − 2mr(x 1 = r, x 2 = θ, x 3 = ϕ, x 4 = t), which represents a solution of theEinstein–Maxwell equations. The corresponding metric is given by( drds 2 =(r 2 + a 2 cos 2 2 )θ)∆ +dθ2 −(1 −[+sin 2 θ r 2 + a 2 +a2 sin 2 θr 2 + a 2 cos 2 θ (2mr − e2 )−2a sin2 θr 2 + a 2 cos 2 θ (2mr − e2 )dϕdt.)2mr − e2r 2 + a 2 cos 2 dt 2θ]dϕ 2 (21.24)For e = 0, it goes over into the Kerr metric (20.34). The Kerr–Newmansolution (21.24) may describe the exterior gravitational field of a rotatingcharged source and contains three real parameters: m (mass), e (charge)and a (angular momentum per unit mass).The only non-vanishing components of the Weyl and Maxwell tensors,with respect to the null tetrad (21.23), are√m(r +ia cos θ) − e2Ψ 2 = −(r − ia cos θ) 3 (r +ia cos θ) , κ02 Φ 1 =e2(r − ia cos θ) 2 . (21.25)In the metric (21.24) the complex scalar potentials Φ and E, with respectto the Killing vector ξ = ∂ t , are given byΦ=er − ia cos θ , E =1− 2mr − ia cos θ . (21.26)E is a linear function of Φ. As can be seen from Φ, the magnetic dipolemoment vanishes if the source is either non-rotating (a = 0, Reissner–Nordström solution) or uncharged (e = 0, Kerr solution).Ernst and Wild (1976) applied the Harrison transformation (34.12)E ′ =Λ −1 E, Φ ′ =Λ −1 (Φ − BE), Λ=1− 2BΦ+B 2 E (21.27)(B being a real parameter) to obtain the solution for a Kerr–Newmanblack hole immersed in a homogeneous magnetic field. Note that theErnst potentials in (21.27) are formed with respect to the Killing vectorη = ∂ ϕ . For a discussion of the Ernst–Wild solution and further