Where is the observer in AdS-Schwarzschild coordinates?

Question

for an AdS-Schwarzschild black hole in 4d, the metric is

$$ ds^2 = -f(r)dt^2 + \frac{dr^2}{f(r)} + r^2d\Omega^2 $$

where $f(r) = 1 + r^2/l^2 - C/r$. $l$ is the AdS length scale and $C$ is some positive constant.

My question is: Where is the observer whose metric is the one above?

For normal Schwarzschild, the observer is always positioned at radial infinity, and in pure AdS the observer lies in the center, i.e. $r=0$. But in this case, since the center is inside the black hole and being infinitely far away requires infinite amount of energy (if you are massive), then Im confused where the observer would be.

My initial guess would be the radial position $r_0$ where $f(r_0)=1$ so that the space around the observer is locally flat, i.e. equivalence principle. But this would require a constant acceleration away from the black hole to maintain the position.

Answer 1

Where is the observer whose metric is the one above?

I do not think that question makes much sense in general. In GR, we have coordinate maps and atlases, not observers and frames. Coordinate map is arbitrary homeomorphism from some open set in spacetime to $\mathbb{R}^4.$

Moreover, metric is geometric object. There is no observer associated with it. You can associate observer with some coordinate system. But not for general coordinates. You are not at all guaranteed, that the coordinates you have written down can have nice interpretation as you would like.