Is it possible to assign a physical radius to a black hole? The Schwarzschild metric is given by:
$$c^2d\tau^2 = \left(1-\frac{r_s}{r}\right)c^2 dt^2-\left(1-\frac{r_s}{r}\right)^{-1}dr^2 - r^2 \left(d\theta^2 + \sin^2 \theta \, d\varphi^2\right).$$
The Schwarzschild radius $r_s$ is often referred to as 'the' black hole radius. But the coordinate distance only coincides with the physical distance infinitely away from the black hole. If I wanted to estimate a physical radius for the black hole, I would try something like:
$$R=\int_0^{r_S}\frac{dr}{\sqrt{1-r_s/r}}.$$
Which gives nonsensical results. Is it possible to make sense of it?
 A: $(1-r_s/r)^{-1/2}dr$ is the radial distance measured by a (local) observer at fixed $r$. However it is only possible to remain at fixed $r$ when outside the Schwarzschild radius: $r>r_s$. (The reason it is "nonsensical" is that this attempts to measure a proper time, not a proper distance, so you should add a minus sign before taking the square root.)
To measure inside the Schwarzschild radius, we must first decide on which observers to use: in other words, choose an allowable velocity. A simple choice is observers who fell from rest, far away from the black hole ("raindrops"). These measure the radial proper distance to be exactly $dr$, so for them $r$-coordinate intervals are exactly what their rulers measure. See Taylor & Wheeler, Exploring Black Holes (2000, $\S B.3$).
So yes, it is possible to assign a physical radius to a black hole, and $r$ seems the best choice. (Another reason is $r$ is the "reduced circumference" or "areal radius" -- look it up if you like.) But don't forget distance is relative to the observer.
