Why is the angular part of the Schwarzschild metric is not affected by function of $r$ and $t$ as opposed to the first two?

Question

Consider the Schwarzschild metric $$ds^2 = A(r)dt^2 −B(r)dr^2 −C(r)r^2(d\theta^2 +\sin\theta^2d \phi^2),$$ where $A,B,C$ are some functions of $r$. There is no loss of generality if one takes $C = 1$. Why is this? A mathematical explanation that allows one to visualize the situation would be nice.

Answer 1

This is the way the $r$ coordinate is defined.

The $r$ coordinate is not a radial distance i.e. it is not the distance that would be measured if you let down a measuring tape towards the black hole. Instead the $r$ coordinate is defined as the circumference of a circle, $s$, centred on the black hole divided by $2\pi$:

$$ r = \frac{s}{2\pi} \tag{1} $$

In flat space this is of course just the radius of the circle, but in curved space it is not. Instead we use equation (1) to define what we mean by the $r$ coordinate.

From this it follows immediately that the function $C(r)$ in your expression is equal to unity. If we consider a circle centred on the black hole in the equatorial plane ($\theta=\pi/2$) then $dt = dr = d\theta = 0$ and the metric simplifies to:

$$ ds^2 = C(r) r^2d\phi^2 \tag{2} $$

To get the circumference of the circle we integrate $ds$ from $\phi=0$ to $2\pi$:

$$ s = \sqrt{C(r)} r \int_0^{2\pi} d\phi = \sqrt{C(r)} 2\pi r $$

and since we have defined the $r$ coordinate by $s = 2\pi r$ this means $C(r)=1$.

Answer 2

$\let\rho=\varrho$ There's no need to think of geometrical meaning of $r$. Simply define $$\rho = r\,\sqrt{C(r)}$$ and define $D(\rho)$ such that $r=D(\rho)$ - the inverse function of $r\sqrt{C(r)}$. Then $$dr = D'(\rho)\,d\rho$$ and $$ds^2 = A(D(\rho))\,dt^2 - B(D(\rho))\,{D'}^2(\rho)\,d\rho^2 - \rho^2 (d\theta^2 + \sin^2\!\theta\,d\phi^2).$$