The Kerr metric expressed in terms of polar coordinates $r,\theta,\phi$, such that $x = r\sin(\theta)\cos(\phi)$, $y = r\sin(\theta)\sin(\phi)$, $z = r\cos(\theta)$. Then the Kerr metric is given as \begin{align*} ds^2 = &-\left(1 - \frac{2GMr}{r^2+a^2\cos^2(\theta)}\right) dt^2 + \left(\frac{r^2+a^2\cos^2(\theta)}{r^2-2GMr+a^2}\right) dr^2 + \left(r^2+a^2\cos(\theta)\right) d\theta^2\\ &+ \left(r^2+a^2+\frac{2GMra^2}{r^2+a^2\cos^2(\theta)}\right)\sin^2(\theta) d\phi^2 - \left(\frac{4GMra\sin^2(\theta)}{r^2+a^2\cos^2(\theta)}\right) d\phi\, dt \end{align*} where $a \equiv S/M$ is the object's angular momentum per unit mass, and $G$ is the gravitational constant. This is an exact solution for the empty-space Einstein equation.

Say, If we are to consider the metric for a constant time, $t_0$. Is it then possible to define the Kerr metric on a submanifold of spacetime, say only in space? If so how can I accomlish this? Is it as simple as dropping the time dependent terms, i.e \begin{align*} ds^2 = & \left(\frac{r^2+a^2\cos^2(\theta)}{r^2-2GMr+a^2}\right) dr^2 + \left(r^2+a^2\cos(\theta)\right) d\theta^2\\ &+ \left(r^2+a^2+\frac{2GMra^2}{r^2+a^2\cos^2(\theta)}\right)\sin^2(\theta) d\phi^2 \end{align*} or do I need to use the induced metric to describe the metric on the submanifold?

Edit : I solved the geodesic differential equations using a "time independent" Kerr metric, with a = 0 (i.e this reduces Kerr metric to the Schwarzschild metric), and the Schwarzschild radius to define the other parameters :

Geodesics for a time independent Kerr metric

Most plots I got spiraled around a singularity at the origo.

Here is a plot where I set $\phi$ to a constant, the z-axis becomes the "time" :

Geodesics for $phi$ independent Kerr metric

Update : I have found the following figure which seem to verify my first figure. enter image description here

Strategies for Direct Visualization of Second-Rank Tensor Fields by Werner Benger and Hans-Christian Hege

  • $\begingroup$ At heart, something like this has to be properly understood as a 3+1 decomposition of the spacetime, which can be understood using the ADM decomposition: en.wikipedia.org/wiki/ADM_formalism Note that different choices for the time coordinate will give you radically different 3-geometries. $\endgroup$ Aug 13, 2018 at 19:11

1 Answer 1


The metric is telling you how to calculate the proper time along a path of your choosing. If you select a path where the time is everywhere constant then as you integrate along that path $dt = 0$ and any terms involving $dt$ disappear. It is as simple as that.

  • $\begingroup$ Its nice to know that. That simplifies things a lot for me. I intend to visualize the metric by solving the geodesic differential equations. And it makes things much easier if I can simply consider the problem in 3D. $\endgroup$
    – imranal
    Nov 26, 2015 at 9:51
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    $\begingroup$ @imranal: I don't think you can actually do that. Geodesics in space are not the same as geodesics in spacetime. $\endgroup$
    – Javier
    Nov 26, 2015 at 13:53
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    $\begingroup$ @imranal: I think what you're describing is a bit different to what I thought you meant. The hypersurface of constant time is a Riemannian manifold, with a metric obtained by setting $ds=0$. You could solve the geodesic equation for this manifold and maybe this is a good way to understand the shape of the manifold. However the curves you get have no physical relevance in the sense that they are not physically meaningful trajectories. $\endgroup$ Nov 26, 2015 at 16:40
  • $\begingroup$ @javier: Can I drop one of the space components instead, say $\phi$ ? $\endgroup$
    – imranal
    Nov 27, 2015 at 23:32

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