# What rules out a negative radial coordinate $r<0$ for the Schwarzschild metric?

In deriving the Schwarzschild metric, there's nothing mathematical that rules out negative values of $r$, as far as I can tell. That is, as long as $r = 0$ and $r \ne r_s$ (where $r_s$ is the (positive) Schwarzschild radius), the metric satisfies the Einstein field equations.

So why don't we consider the region $r < 0$ as a possible spacetime? Or, equivalently, why don't we consider metrics of the form

$$ds^2 = -(1 + r_s/r) \, dt^2 + (1+r_s/r)^{-1} \, dr^2 + r^2 \, d\Omega^2$$ for $r > 0$?

One physical argument would be that this spacetime would have a naked singularity at $r = 0$. But that seems somewhat arbitrary. Is there a better reason to rule it out?

• One physical argument would be that this spacetime would have a naked singularity at r=0. The Schwarzschild metric can be used to describe the external field of a spherically symmetric body, and so can the negative-mass Schwarzschild metric. The earth's external field can be modeled by the Schwarzschild metric, for example.
– user4552
Jan 13, 2018 at 16:01
• Apr 10 at 8:08

Well, for one, $r$ is considered a kind of distance from the central point mass - namely it's a generalization of the polar coordinate $r$ which is a distance from the origin, and distances are never negative, so you could argue that geometrically a negative $r$ does not make any sense. It's like asking what number has a negative absolute value. (You can try to create such numbers like how the imaginary number $i$ was created, but the system is not very mathematically interesting, I believe. And there's zero obvious way to give it any kind of sensible physical interpretation.) It doesn't represent distance in an obvious way because of the warped geometry - moreover, there is no non-arbitrary "now" surface on which to measure such a distance - but it generalizes one.
Nonetheless there is a case which, although it does not appear to be actually realizable in our universe, the metric equation you give would make conceptual, geometric sense at least: it is equivalent to taking the mass $M$ of the point mass to be negative, so a point concentration of negative mass, at non-negative radii. This would be more reasonable - the only trouble is as far as we can tell, matter with negative mass does not exist. All existing particles have positive mass. There is something like negative mass (and so also, negative energy) in quantum mechanics, but due to a phenomenon known as "quantum interest" it does not appear to be isolatable from a (larger!) quantity of positive mass.
• Thanks! I did see the negative mass solution floating around, but that seems more speculative than considering a metric with $1 + r_s/r$ instead of $1 - r_s/r$... Jan 13, 2018 at 7:32