Time in the negative mass Schwarzschild solution

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I have read that for the Schwarzschild metric solution with $M<0$, something odd happens with the time coordinate. For the constants of motion, $dt/d\tau=e(1 – 2GM/r)^{-1}$ with $M<0$ and $e$ a constant of motion, at $r\to\infty$, $dt/d\tau=e$ and as $r$ decreases, so does $dt/d\tau$, reaching zero at $r=0$, which is the opposite behaviour to the $t$ coordinate in the ordinary Schwarzschild
metric.

What is happening with the time coordinate in the negative mass Schwarzschild solution?

Source: Schwarzschild Metric With Negative Mass Author: Glenn Rowe

Best Answer

I am not sure that I properly understand the question. However, looking at the paper in the link you see that the relevant equations are two $$\frac{dt}{d\tau}=e(1 - 2GM/r)^{-1}$$ and $$\frac{dr}{d\tau} = \pm \sqrt{e^2 -\left(1 + \frac{C}{r}\right)\left(1 + \frac{\ell^2}{r^2}\right)}$$ where $e$ and $\ell$ are constants.

In the paper, an issue is made explicit:

"Because the time coordinate behaves differently in this case, is it still correct to identify it with the proper time of an observer at infinity? "

Yes it is: If a particle stays at rest at infinity then both $r\to +\infty$ and $\frac{dr}{d\tau}\to 0$ there. That is possible only if $e=0$ from the second equation. The first equation implies that $t= \tau + constant$. Hence, again $t$ is the proper time of a particle at rest very far from the singularity.

Regarding the fact that negative mass "speeds up" (proper) time, I think that the meaning of that statement is as follows. Look at the plots for $x>1$. The red line is $\frac{d\tau}{dt}$ for positive mass, the blue one is the analog for negative mass. Here Schwarzschild radius is at $x=1$ as $2GM =1$. The two plots can be actually compared only in the "positive-mass external" region $x>1$. You see that the "speed of the proper time vs (Killing) time" $\frac{d\tau}{dt}$ increases while approaching $x=1$ and diverges around the singularity $x=0$ along the blue line (negative mass). Conversely, it decreases along the red line (positive mass).

However, far from away the singular region, i.e., for $x\to +\infty$ the two curves coincide.

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