# What does asymptotically flat solution mean?

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.

• A nice open-access presentation is Frauendiener, link.springer.com/article/10.12942/lrr-2004-1 . What level of presentation are you looking for? A full definition is pretty technical. – Ben Crowell Feb 4 '18 at 3:53
• Are you asking for a mathematical definition (which is technical) or just an explanation in words of what it's about? – Ben Crowell Feb 4 '18 at 22:14

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:

1. There exists a function $\omega \geq 0$, such that $g_{ab} = \omega^2 \hat{g}_{ab}$,

2. On the boundary, one has $\omega = 0$, and $\omega_{,a} \neq 0$,

3. 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.

• But, now if the metric tensor gab is indeed a solution of the Einstein field equations, the spacetime then is considered asymptotically flat. I assume you're sketching a definition of asymptotic flatness for vacuum spacetimes, and this refers to the vacuum field equations? – Ben Crowell Feb 4 '18 at 3:39
• I think this sketch could be clarified by explaining what you mean by "the boundary." As written, it gives the impression that manifolds automatically come equipped with a boundary, or that the boundary exists in the physical manifold, not just the unphysical manifold. You never say that $g$ is the metric on an unphysical manifold, or that the idealized boundary points are points in this unphysical manifold. I think the physical content that you're expressing is that an asymptotically flat space is one for which there exists a conformally equivalent unphysical manifold having [...] – Ben Crowell Feb 4 '18 at 3:49
• [...] a boundary consisting of spacelike infinity $i^0$ and null infinity $\mathscr{I}^+$ and $\mathscr{I}^-$. – Ben Crowell Feb 4 '18 at 3:51