gravity - Does rotation affect gravitational lines of force

An spherical electric charge has the same electric field lines whether spinning or not. The difference between those two cases is entirely in the magnetic field. Thus, one should expect as similar thing to happen for gravity.

The parametrized post-Newtonian formalism, weak-field GTR has the metric
$$mathrm{d}s^2 = -(1+2Phi),mathrm{d}t^2 + 2mathcal{A}_j,mathrm{d}t,mathrm{d}x^j + (1-2Phi)delta_{ij},mathrm{d}x^i,mathrm{d}x^j$$
where $Phiequiv -U$ is essentially the Newtonian gravitational potential, while $mathcal{A}_jequiv-tfrac{7}{4}V_j - tfrac{1}{4}W_j$ in terms of the other PPN potentials. For the four-velocity $U^alphaequiv{mathrm{d}x^alpha}/{mathrm{d}tau} = (U^0,vec{U})$, the geodesic equation for time-independent $Phi$ and $mathcal{A}$ becomes, to linear order in $Phi$,
$$frac{mathrm{d}vec{U}}{mathrm{d}tau} = U^0(vec{G} + vec{U}timesvec{H})text{,}$$
where the gravitoelectric field is $vec{G} = -nablaPhi$ and the gravitomagnetic field is $vec{H} = nablatimesvec{mathcal{A}}$, here $nabla$ being used in the ordinary sense of $nabla_i = partial_i$, i.e. with respect to the Euclidean metric $delta_{ij}$. If $Phi$ and $mathcal{A}$ are not time-independent, then $vec{G} = -nablaPhi - partial_tvec{mathcal{A}}$, paralleling electromagnetism, but there will be an extra term in the analogue of Lorentz force that has no electromagnetic counterpart.

Much the same thing can be done in any stationary spacetime, including the rotating Kerr black hole. See also Costa and Natário (2014) for a much more general treatment of several gravito-electromagnetic analogies.

---

References:

1. Costa, L. F. O, Natário, J. "Gravito-electromagnetic analogies". Gen. Rel. Grav. 46, 1792 (2014)[arXiv:1207.0465]