<p>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.</p>

<p>The <a href="https://en.wikipedia.org/wiki/Parameterized_post-Newtonian_formalism" rel="nofollow">parametrized post-Newtonian formalism</a>, 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.</p>

<p>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.</p>

<hr>

<p>References:</p>

<ol>
<li>Costa, L. F. O, Natário, J. "Gravito-electromagnetic analogies". <em>Gen. Rel. Grav.</em> <strong>46</strong>, 1792 (2014)[arXiv:<a href="http://arxiv.org/abs/1207.0465" rel="nofollow">1207.0465</a>]</li>
</ol>

CC BY-SA 3.0
2015-03-10T22:34:27.017Z
2015-03-10T22:34:27.017Z
Stan Liou
<p>In general relativity the gravitational field of rotating mass is described by <a href="http://en.wikipedia.org/wiki/Kerr_metric" rel="nofollow"><em>Kerr metric</em></a>. It can be written as:</p>

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

<p>The last term is responsible for the so called <a href="http://en.wikipedia.org/wiki/Lense%E2%80%93Thirring_precession" rel="nofollow"><em>Lense-Thiring Effect</em></a>, a graviomagnetic effect as already mentioned by Stan Liou. A nice description can be found in the mission description of <a href="https://einstein.stanford.edu/MISSION/mission1.html" rel="nofollow">Gravity Probe B</a>. So if I look at your image I think your intuition is correct.</p>

CC BY-SA 3.0
2015-03-11T19:56:37.260Z
2015-03-11T21:29:53.130Z
2015-03-11T21:29:53.130Z
Sjoerd222888
<p>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.</p>

<p>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:</p>

<p><img src="https://i.sstatic.net/edkbq.png" alt="enter image description here"></p>

<p>to this?</p>

<p><img src="https://i.sstatic.net/nIiin.jpg" alt="enter image description here"></p>

CC BY-SA 3.0
2015-03-16T15:27:11.697Z
Were you looking for equations or a visual answer?
Stan Shunpike
CC BY-SA 3.0
2015-03-19T06:33:25.697Z
so, it means that gravity is changing in every point.
Darius Miliauskas
<gravity><rotation>
Does rotation affect gravitational lines of force
Robert Cartaino
# Does rotation affect gravitational lines of force

(Asked by: username_0 on 2015-03-10)

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](https://i.sstatic.net/edkbq.png)

to this?

![enter image description here](https://i.sstatic.net/nIiin.jpg)

**Tags:** gravity, rotation
> **Comments:**
> **username_1**: Were you looking for equations or a visual answer?  
> **username_2**: so, it means that gravity is changing in every point.  

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## 2 answers

### Answer by username_3 on 2015-03-10. Score: 2

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](https://en.wikipedia.org/wiki/Parameterized_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](http://arxiv.org/abs/1207.0465)]


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### Answer by username_4 on 2015-03-11. Score: 2

In general relativity the gravitational field of rotating mass is described by [*Kerr metric*](http://en.wikipedia.org/wiki/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*](http://en.wikipedia.org/wiki/Lense%E2%80%93Thirring_precession), a graviomagnetic effect as already mentioned by Stan Liou. A nice description can be found in the mission description of [Gravity Probe B](https://einstein.stanford.edu/MISSION/mission1.html). So if I look at your image I think your intuition is correct.


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