Eddington-Finkelstein coordinates not well-defined?

Question

Consider the Schwarschild solution $$d s^{2}=-\left(1-\frac{2 m}{r}\right) d t^{2}+\frac{d r^{2}}{1-\frac{2 m}{r}}+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \varphi^{2}\right) $$

and the radial null geodesics (in Schwarschild coordinates): $$t=\pm(2 m \ln |r-2 m|+r)+\text { constant }. $$

The advanced Eddington Finkelstein (EF) coordinates are defined as ($\bar{t}$,r,θ,φ) with $$\bar{t}=t+2 m \ln (r-2 m) $$

The metric in EF coordinates has the form: $$d s^{2}=-\left(1-\frac{2 m}{r}\right) d \bar{t}^{2} -\frac{4m}{r}d \bar{t} dr-\left(1+\frac{2m}{r}\right)dr^2+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \varphi^{2}\right) $$ which is not singular at $r=2m$ (but still singular at $r=0$). The incoming radial geodesics (corresponding to $-$) become: $$\bar{t}=-r+\text{constant} $$

and the outgoing (corresponding to $+$): $$\bar{t} = 4m\ln(r-2m) +r +\text{constant} .$$ My understanding is that the above definition for $\bar{t}$ is only valid for $r>2m$. However we use the solution in EF coordinates in all $r>0$ and when we draw the geodesics we extend them to $r=0$, even thought $\bar{t}$ is not defined for $r<2m$ since the quantity in $\ln$ is negative. The mathematical treatment seems a bit imprecise, what's up? I'd appreciate a more rigorous approach to this.

Answer 1

When you choose $(\bar{t}, r \ldots)$ instead of $(t, r \ldots)$ and notice that inside $r<2m$ the EF metric is a valid and regular solution, what you find is an extension (like an analytical extension) of the solution in the region $r>2m$ into a region below the event horizon. It was not there before because of the coordinate singularity in $r=2m,$ but now your new coordinates allow you to see that there is nothing strange for incoming geodesics in $r=2m.$ This is because there was no true singularity there but just a failure of the Schwarzschild original coordinates to describe the region close to the horizon. I believe some textbooks discuss this, D'Inverno, Ryder, etc. talk about it. But in short, usually physicist extend the solution of one region where it works into another.