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I was reading about rotating pulsars and other more-massive objects spinning at absurd speeds (hundreds of revolutions per second), and there was one point of which I was not clear.

Does a rotation of a mass change the gravitational lines of force? In other words, does a body which is rotating change the direction of gravitational pull from this:

enter image description here

to this?

enter image description here

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  • $\begingroup$ Were you looking for equations or a visual answer? $\endgroup$ – Stan Shunpike Mar 16 '15 at 15:27
  • $\begingroup$ so, it means that gravity is changing in every point. $\endgroup$ – Darius Miliauskas Mar 19 '15 at 6:33
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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 $$\mathrm{d}s^2 = -(1+2\Phi)\,\mathrm{d}t^2 + 2\mathcal{A}_j\,\mathrm{d}t\,\mathrm{d}x^j + (1-2\Phi)\delta_{ij}\,\mathrm{d}x^i\,\mathrm{d}x^j$$ where $\Phi\equiv -U$ is essentially the Newtonian gravitational potential, while $\mathcal{A}_j\equiv-\tfrac{7}{4}V_j - \tfrac{1}{4}W_j$ in terms of the other PPN potentials. For the four-velocity $U^\alpha\equiv{\mathrm{d}x^\alpha}/{\mathrm{d}\tau} = (U^0,\vec{U})$, the geodesic equation for time-independent $\Phi$ and $\mathcal{A}$ becomes, to linear order in $\Phi$, $$\frac{\mathrm{d}\vec{U}}{\mathrm{d}\tau} = U^0(\vec{G} + \vec{U}\times\vec{H})\text{,}$$ where the gravitoelectric field is $\vec{G} = -\nabla\Phi$ and the gravitomagnetic field is $\vec{H} = \nabla\times\vec{\mathcal{A}}$, 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 $\vec{G} = -\nabla\Phi - \partial_t\vec{\mathcal{A}}$, 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]
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In general relativity the gravitational field of rotating mass is described by Kerr metric. It can be written as:

$$ ds^2 = ds^2_{\text{Schwarzschild}} + \frac{4GJ}{c^2r} \sin^2 \theta dt d\phi $$

The last term is responsible for the so called Lense-Thiring Effect, a graviomagnetic effect as already mentioned by Stan Liou. A nice description can be found in the mission description of Gravity Probe B. So if I look at your image I think your intuition is correct.

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