# Is this a black hole?

I came across this metric definition:

$ds^2=-\left(1-\frac{r_s}{r}\right)^2dt^2 + \left(1-\frac{r_s}{r}\right)^{-2}dr^2+r^2d\Omega^2$

I was trying to figure out if it describes a black hole, but I was unable to find the proper definition.

My hints so far:

• The time coordinate does not change its sign when r crosses $r_s$
• There are two singularities, one at $r=r_s$ which I guess arises from coordinate singularity and another at $r=0$, which looks like the center of a black hole?

What do you think?

• This is the Schwarzschild metric, which is indeed a black hole if the spacetime is defined for $r < r_s$. Commented Feb 15, 2018 at 15:37
• please note the second power , which is different from schwarchild metric . Commented Feb 15, 2018 at 15:38
• The metric (v1) is not the Schwarzschild metric because of some powers of 2. One may check that the metric does not satisfy the vacuum EFEs, even in the presence of a cosmological constant. Commented Feb 15, 2018 at 15:40
• then it is fairly simple : there's only one static black hole that is spherically symmetric, and that is the Schwarzschild metric. Unless there is some coordinate transformation that can turn one into the other (I don't think that's the case), it is not a black hole. Commented Feb 15, 2018 at 15:41
• @Slereah only one vacuum static spherically symmetric black hole, and that's in general relativity. Filling spacetime with stuff and/or going to alternative theories gives you more black hole solutions; whether a given metric is or is not a black hole doesn't depend on the theory. Commented Feb 15, 2018 at 15:50

We need to be careful about terminology because the phrase black hole tends to be used to mean a vacuum solution. The only two vacuum solutions are the Schwarzschild and Kerr metrics, so in everyday use black hole means either the Schwarzschild or Kerr metric.

But if we relax the requirement for the geometry to be a vacuum solution then we can get other forms of black hole. For example if we add a cosmological constant we get the de Sitter-Schwarzschild solution and if we add null dust we get the Vaidya metric. Your geometry is also a non-vacuum geometry.

In principle we can tell it's not a vacuum solution by calculating the Einstein tensor from your metric, because this gives us the stress-energy tensor and that will turn out not to be zero. However the GR Mathematica notebook I have does not include a routine for calculating the Einstein tensor and calculating it by hand would be something of an ordeal. I did calculate the Ricci tensor and it seems fairly obvious from a glance that this is going to give a non-zero Einstein tensor.

But I would guess your question is really whether this geometry has a horizon. In general finding horizons is a surprisingly hard thing to do because we can only find them by considering the entire spacetime. However in a simple, static, geometry like this one we can simply look at the coordinate speed of light and see if it's zero. We do this by noting that for a light ray $ds = 0$. If we choose a radial trajectory, so $d\Omega=0$ and make this substitution we get:

$$0=-\left(1-\frac{r_s}{r}\right)^2dt^2 + \left(1-\frac{r_s}{r}\right)^{-2}dr^2$$

Giving us:

$$\frac{dr}{dt} = \pm\left(1-\frac{r_s}{r}\right)^2$$

And since this is zero at $r=r_s$ this is indeed an event horizon.

Courtesy of AccidentalFourierTransform we can calculate the Ricci and Kretschmann scalars. The Ricci scalar turns out to be zero everywhere, which isn't much help, but the Kretschmann scalar is:

$$K = \frac{8r_s^2 ( 6 r^2 - 12 r r_s + 7 r_s^2)}{r^8}$$

This remains finite at $r=r_s$, but goes to infinity as $r \to 0$. So the former is a coordinate singularity while the latter is a true curvature singularity.

• It would be interesting to see the components of the stress-energy tensor in this case, for $r < r_s$ and $r > r_s$. What kind of matter produces that metric ? Does it satisfy all the energy conditions ? It shouldn't be too much difficult to find out, using the tetrad formalism : $\boldsymbol{e}^0 = (1 - r_s/r) \, \boldsymbol{d}t$, $\boldsymbol{e}^1 = (1 - r_s/r)^{-1} \, \boldsymbol{d}r$, ..., and I would recommend a simple radial coordinate change.
– Cham
Commented Feb 15, 2018 at 17:38
• Hmm, the radial coordinate change I was suggesting is ugly, actually : $u = r + r_s \ln{(r/r_s - 1)}$ (for $r > r_s$). Not obvious to change $r$ as a function of $u$.
– Cham
Commented Feb 15, 2018 at 17:48
• @J mathematica code to compute the Einstein tensor: mathematica.stackexchange.com/a/138841/34893 the result is $$G_{\mu\nu}=\left( \begin{array}{cccc} \frac{(r-R_s)^2 R_s^2}{r^6} & 0 & 0 & 0 \\ 0 & -\frac{R_s^2}{r^2 (r-R_s)^2} & 0 & 0 \\ 0 & 0 & \frac{R_s^2}{r^2} & 0 \\ 0 & 0 & 0 & \frac{{R_s}^2 \sin ^2(\theta )}{r^2} \\ \end{array} \right)$$ Commented Feb 15, 2018 at 17:52
• @AccidentalFourierTransform aha, thanks. That's identical to the Ricci tensor, which makes sense because I had already computed the Ricci scalar by hand and got $R=0$. So the energy density is sane but it has a negative radial pressure component. Commented Feb 15, 2018 at 18:01
• That is weird. It means that some energy conditions are violated, apparently. Since $R = 0$, this implies some massless radiation.
– Cham
Commented Feb 15, 2018 at 19:02