I'm following Schutz's General Relativity book and I am confused about his description and derivations of a spherically symmetric spacetime. I searched online and found that using Killing vectors is a common approach. It is unfortunately not (yet) part of Schutz' reasoning, so I don't yet quite understand that methodology.
Instead, his argument begins by reminding us that that line element of a 2-sphere in flat Minkowski space is $$ds^2 = r^2(d\theta^2 + \sin^2\theta d\phi^2) = r^2d\Omega^2.$$ This is a familiar result from multivariable calculus. He then makes the claim verbatim that a
"... a spacetime is spherically symmetric can now be made more precise; it implies that every point of spacetime is on a two-surface... whose line element is $dl^2=f(r',t)d\Omega^2$ where $f(r',t)$ is an unknown function of the other two coordinates of our manifold, $r'$ and $t$."
Firstly, what is $r'$? Besides being another coordinate, $r'$ came out of no where. Is it a transformation of the radial coordinate $r$?
Secondly, why do we need $f(r', t)$ in the first place? By the definition, every point must be on some surface of a two-surface. At a given origin, can't we can describe all points a distance $r$ away as having $dl^2 = r^2d\Omega^2$? Is the fuss with $f(r', t)$ because not all points in spacetime can be described as points on a sphere centred at my chosen origin? So I would need to do some coordinate transformation $r'\leftarrow r$ to get to another sphere's centre? Is this a correct interpretation?
Schutz attached a picture of a wormhole as an example. Clearly, there is circular symmetry about a vertical axis going through the centre of the throat. What would $r$ and $r'$ be here? Would $r$ be the radius of each of the cocentric circles? Then $r'$ would be the displacement from my origin to another point on the centre axis?
Lastly, he then defines the radial coordinate $r$ such that $f(r', t) = r^2$. This makes me even more confused why we had to introduce $f$ in the first place; it looks like we're back where we started! What does this mean here?
