Can somebody explain what does it mean when a solution is "asymptotically flat"? like the schwarzschild metric which is asymptotically flat solution to vacuum Einstein equations.
1 Answers
The definition of asymptotic flatness varies from text to text, but there are in general, 3 conditions for a solution of Einstein's field equations to be considered asymptotically flat. Consider a spacetime $(\hat{M}, \hat{g}_{ab})$, with the following 3 properties:
There exists a function $\omega \geq 0$, such that $g_{ab} = \omega^2 \hat{g}_{ab}$,
On the boundary, one has $\omega = 0$, and $\omega_{,a} \neq 0$,
Every null geodesic intersects the boundary at two points.
Such spacetimes are called asymptotically simple. But, now if the metric tensor $g_{ab}$ is indeed a solution of the Einstein field equations, the spacetime then is considered asymptotically flat.
For the Schwarzschild solution, it is easiest to see this using null coordinates, $u,v$, which the metric then takes the form:
$ds^2 = -dudv \left(1 - \frac{2M}{r}\right) + r^2 \left(d\theta^2 + \sin^2 \theta d\phi^2\right)$, where $r = r(u,v)$. So, make the further coordinate transformation:
$\tan U = u, \tan V= v$,
then one can show that the conformal metric takes the form:
$ds^2 = \cos^{-2}U \cos^{-2}V \left(1- \frac{2M}{r}\right) dU dV + r^2 \left[d\theta^2 + \sin^2 \theta d\phi^2\right]$.
Now, if you let $\omega = \cos U \cos V$, this can be written as:
$ds^2 = \omega^{-2}\left[ \left(1- \frac{2M}{r}\right) dU dV + r^2 \omega^2 \left(d \theta^2 + \sin^2 \theta d\phi^2\right)\right]$
Now, at the boundary where $\omega = 0$, we have clearly that $U = \pm \pi/2$ or that $V = \pm \pi/2$, i.e., we have "mapped the infinities $u,v = \pm \infty$ to finite values. Since clearly $r \omega \neq 0$ when $\omega = 0$ then, the boundary consists of two points being intersected.
This is an idea of how one shows that the Schwarzschild metric is asymptotically flat.
If you want more details on this, I suggest you look at Wald's classic text.
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