Confusion regarding divergencies of the Kretchmann scalar for the time dependent Schwarzschild metric

Question

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?

Answer 1

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.

Answer 2

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 $K$.