4
$\begingroup$

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

|cite|improve this question
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Jerry Schirmer Aug 13 '18 at 19:11
0
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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 '15 at 9:51
  • 2
    $\begingroup$ @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 '15 at 13:53
  • $\begingroup$ @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. $\endgroup$ – John Rennie Nov 26 '15 at 16:40
  • $\begingroup$ @javier: Can I drop one of the space components instead, say $\phi$ ? $\endgroup$ – imranal Nov 27 '15 at 23:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.