We modify and generalize the known solution for the electromagnetic field when a vacuum, stationary, axisymmetric black hole is immersed in a uniform magnetic field to the case of nonvacuum black holes (of modified gravity) and determine all linear terms of the vector potential in powers of the magnetic field and the rotation parameter.
The magnetic field problemA Killing vector ξ μ in vacuum (no stress-energy T μν ≡ 0) is endowed with the property of being parallel, that is, proportional, to some vector potential A μ that solves the sourceless (no currentsMaxwell field equations. So, ξ μ is itself a solution to the same source-less Maxwell field equations. In Ref.[1], this property was employed as an ansatz to determine the electromagnetic field of a vacuum, stationary, axisymmetric, asymptotically flat black hole placed in a uniform magnetic field that is asymptotically parallel to the axis of symmetry. The ansatz stipulates that the vector potential of the solution be in the plane spanned by the timelike Killing vector ξ μ t = (1, 0, 0, 0) and spacelike one ξ μ ϕ = (0, 0, 0, 1) of the stationary, axisymmetric black hole,Since ξ μ t and ξ μ ϕ are pure geometric objects, they do not encode information on the applied magnetic field B; such information is encoded in the coefficients (C t , C ϕ ). Here B is taken as a test field, so the metric of the stationary, axisymmetric black hole too does not encode any information on the applied magnetic field.In this work, a spacetime metric has signature (+, −, −, −) and andrespectively, 1 where B and Q are seen as perturbations, that is, if the metric of the background black hole is that of Kerr, then Qξ t μ /(2M) is, up to an additive constant, the one-form A μ dx μ = −Qr(dt − a sin 2 θ dϕ)/ρ 2 of the Kerr-Newman black hole (with ρ 2 = r 2 + a 2 cos 2 θ ). It is important to emphasize this point: the potential given by (1) and (3) is not an exact solution to the source-less Maxwell equations,if the background metric is that of the charged black hole itself. Rather, it is a solution to (4) if the background metric is that of the corresponding uncharged black hole. For instance, in the Kerr background metric, the potential given by (1) and (3) is a solution to (4), but in the KerrNewman background metric the nonvanishing electric charge density J t and the ϕ current density J ϕ expand in powers of Q aswhich are zero to first order only. Even if rotation is suppressed (a = 0), J t is still nonzero:and its integral charge is also nonzero. Where does this electric charge density come from (the only existing electric 1 The sign "+" in (3) in front of Q is due to our metric-signature choice.123