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First of all, I know the Vaidya metric exists and I know the properties of the Ricci scalar/Ricci tensor for this metric. But The metric I use is the following: \begin{equation} ds^2=\left( 1-\frac{r_s(t)}{r} \right)dt^2-\left( 1-\frac{r_s(t)}{r} \right)^{-1}dr^2-r^2d\Omega^2 \end{equation} I expected some divergencies of the Ricci tensor's components at the horizon, and it is the case. Together with the Ricci scalar, it gives a nonvanishing stress-energy tensor that diverges at the singularity and at the horizon.

Now, I've calculated the Kretschmann scalar and it has the following expression: \begin{align} K=\frac{12r_s^2}{r^6}+r^4 \left( \partial_0 \frac{\dot{r}_s}{(r-r_s)^3} \right)^2+\frac{2r_s}{r}\partial_0\frac{\dot{r}_s}{(r-r_s)^3}-\frac{18}{r^2}\left( \partial_0 \ln\left( 1-\frac{r_s}{r} \right) \right)^2 \end{align} and gives the right scalar when $r_s(t)=r_s$. But, this scalar diverges at the horizon, and the Kretschmann scalar is said to diverge only at the true singularities. The usual Schwarzschild metric has coordinates singularities at the horizon and one can suppress them by a simple change of coordinate. But in this case it is not true. Are at first sight my calculations wrong or is the fact that one can change the coordinate system to remove the coordinate singularities, a simple mathematical coincidence?

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  • $\begingroup$ That is not the Vaidya metric. $\endgroup$
    – TimRias
    Jul 16, 2021 at 8:24
  • $\begingroup$ I said «I know the vaidya metric exists [...] but...». I thought I was clear. I do not work with the vaidya metric. $\endgroup$ Jul 16, 2021 at 9:28
  • $\begingroup$ Then this just a case GIGO. $\endgroup$
    – TimRias
    Jul 16, 2021 at 12:29

2 Answers 2

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As already mentioned in the comments, the metric isn't a vacuum solution to the Einstein field equations, and in fact we have the off diagonal equation $$ G_{01}= \frac{\dot{r_s}}{r^2 - r r_s} \ . $$ This means that for some $T_{\mu \nu}$ we need some momentum flux, whereas a vacuum solution only exists for $r_s(t)=$ const. I'm not sure of what physical interest the proposed metric has.

Another point is you mention (in the comments) looking at a limit $r = r_s(t)$, but the function $r_s(t)$ is a function of time not the radial coordinate $r$. The Einstein tensor itself is singular if you assume that the radial coordinate and the function $r_s(t)$ coincid, but again, you should solve the Einstein field equations for a given $T_{\mu \nu}$ first before looking at things like the Kretschmann scalar.

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  • $\begingroup$ Thank you for your answer. With this metric I just wanted to know what would be the consequence of the variation of the mass of the blackhole with time. Indeed it is not a vacuum solution, and this shows that the mass of a black-hole cannot vary in time without matter. Why can't I say that this is a solution to EFE for the particuliar stress-energy tensor given by the einstein tensor? $\endgroup$ Jul 16, 2021 at 9:36
  • $\begingroup$ @JeanbaptisteRoux I mean, you can use a stress tensor with off-diagonal components, but then you have to really work out if this describes a blackhole-like solution or something different entirely. Just saying that you define some $T_{\mu \nu}$ to be $G_{\mu \nu}$ doesn't give you anything physical to work with, because you need to then work out what matter distribution you're working with (which could be completely nonphysical) and also apply the continuity equation. $\endgroup$
    – Eletie
    Jul 16, 2021 at 9:47
  • $\begingroup$ But, as black-holes shrink over time due to Hawking radiations, and at any instant the metric is Schwarzschild-like, isn't the metric I propose to describe a black-hole whose radius is changing over time correct? (for instance decreasing if I assume the hawking radiations) $\endgroup$ Jul 16, 2021 at 9:56
  • $\begingroup$ @JeanbaptisteRoux I'm not familiar with this metric, but I took a quick look in the literature and found arxiv.org/abs/1902.07827. The requirements on the stress tensor definitely seem to imply this isn't as simple as the usual Schwarzschild black hole losing mass over time - it can't be because there's no vacuum. (They also don't mention a black hole losing mass via hawking radiation, but instead the formation of dark energy stars, because of the need for negative pressure I suppose. There are additional singularities but it also violates different energy conditions). $\endgroup$
    – Eletie
    Jul 16, 2021 at 10:10
  • $\begingroup$ Thank you for your answers and for the article, I will take a look. It does not exactly answer my question on whether the non-singular behavior of the Kretschmann at the horizon was a mathematical coincidence or not, but I think this was perhaps a bad question after all. I will accept your answer. $\endgroup$ Jul 16, 2021 at 10:21
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  1. If you set $r_s(t)$ to be a constant $r_s$, $K$ becomes $$ K=\frac{180 r^2 r_s^6-240 r^3 r_s^5+180 r^4 r_s^4-72 r^5 r_s^3+12 r^6 r_s^2+12 r_s^8-72 r r_s^7}{r^6 \left(r-r_s\right)^6}; $$
  2. Then taking the limit $r\to r_s$ gives us $$ \lim_{r\to r_s} K= \frac{12}{r_s^4} $$ which is not divergent. In other word, $r=r_s$ is not the divergent point of $K$.
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  • $\begingroup$ Yes of course it behaves well for a constant Schwarzschild radius, but when it is not the case, the Kretschmann scalar seems to be divergent at $r=r_s(t)$. You say that $r=r_s$ is not a divergent point of $K$ but right before you imposed $r_s=c^{ste}$. $\endgroup$ Jul 16, 2021 at 7:55
  • $\begingroup$ First, what do you mean by $r_s(t)=r_s$? Second, the metric which you wrote down is not Schwarzschild metric and it is not the vacuum solution of Einstein equation, thus that the Kretschmann scalar has more than one singularities does not surprise me. $\endgroup$
    – user142288
    Jul 16, 2021 at 8:02
  • $\begingroup$ I mean by this that $r_s$ does not depend on time anymore. The rest of your comment is useless to me. $\endgroup$ Jul 16, 2021 at 9:30

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