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I am reading this paper Formation and evaporation of non-singular black holes by S.A. Hayward.

It deals with the possible metric of a non singular black-hole. The autor consider the metric $$ds^2=-F(r)dt^2+\dfrac{1}{F(r)} dr^2+r^2d\omega^2.$$ In order to find the non singular metric, the autor imposes two conditions:

1) $F(r)\rightarrow 1-\dfrac{2M}{r}$ for $r\rightarrow \infty$

2) a flatness conditon: $F(r)\rightarrow 1-\dfrac{r^2}{l^2}$ for $r\rightarrow0$

What does the second condition mean?

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Mathematically, it just means that the metric is forced (by this construction) to go to Minkowski space at the origin in a smooth way since $F(0)=1$.

Whether that has any physical significance is a question for the theory to answer in the context of experimental data. There's some wiggle room here because anything outside the event horizon cannot see the inside structure, but many of these types of approaches are either without compelling physical motivation or modify the metric enough that there is no event horizon and the "trick" becomes exposed to observers "outside".

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  • $\begingroup$ Would it be the same using $F(r)\rightarrow 1-\dfrac{r}{l}$? $\endgroup$ – mattiav27 Nov 14 '19 at 14:40
  • $\begingroup$ Also note that the $r^2/l^2$ term is there to provide for a possible non-zero energy density near the origin (which works out to be $3/(8 \pi l^2)$ assuming Einstein's equation holds.) $\endgroup$ – Michael Seifert Nov 14 '19 at 14:41
  • $\begingroup$ @MichaelSeifert so $F(r)\rightarrow 1-\dfrac{r}{l}$ is not an option? $\endgroup$ – mattiav27 Nov 14 '19 at 14:42
  • $\begingroup$ You need a certain amount of smoothness at the origin. At least two derivatives for the Einstein equations. That will limit your options, @mattiav27. $\endgroup$ – Brick Nov 14 '19 at 14:43
  • $\begingroup$ @mattiav27: I'm pretty sure that in that case the curvature tensor would not be smooth at the origin. For example, if you take a metric $$ds^2 = \left(1 - \frac{\sqrt{x^2 + y^2 + z^2}}{l}\right)(dx^2 + dy^2 + dz^2)$$ (which is what you're proposing), then its first derivatives are not continuous at the origin because of the square root, which means that the second derivatives (and hence the curvature) are not well-defined there. $\endgroup$ – Michael Seifert Nov 14 '19 at 14:44

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