I think I may have some fundamental misunderstanding about what $dt, dx$ are in general relativity.
As I understand it, in special relativity, $ds^2=dt^2-dx^2$, we call this the length because it is a quantity that is invariant under Lorentz boosts. If a ball is moving in space and I want to calculate the $ds$ for the ball to travel from point A to point B, then $d\tau=ds$ (where $\tau$ is proper time) because according to the ball, $dx=0$, since point A and point B are at the same place in its own frame.
Assuming this is correct, this is all fine with me. Now fast forward to general relativity. $ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$…so if a ball is travelling in spacetime under whatever metric, then to me it would seem like, in the ball's frame, we should set $dx^i=0$, and then we get $ds^2=g_{00}d\tau^2$…however, from what I am told and from what I have read, $ds^2=d\tau^2$. When I look at the Schwarzschild metric, for example, $g_{00}$ doesn't appear to be 1 in a travelling ball's frame.
What have I misunderstood about the interval $ds$?. Is it that $ds$ only purely has to do with geometry of spacetime, and doesn't quite represent the distance between events?
Best Answer
The problem arose when you wrote $ds^2 = g_{00} d\tau^2$. Generally one of your coordinates $x^\mu$ will be timelike, and the others spacelike, but the timelike one is not in general the proper time of someone whose spatial coordinates are not changing. That is, $t \neq \tau$. Using your sign convention,1 $d\tau^2 = ds^2$, so the (arbitrarily large) lapse in proper time between two events A and B is $$ \Delta\tau = \int\limits_\text{path} \sqrt{ds^2} = \int_{\sigma_\mathrm{A}}^{\sigma_\mathrm{B}} \sqrt{g_{\mu\nu} \frac{\mathrm{d}x^\mu}{\mathrm{d}\sigma} \frac{\mathrm{d}x^\nu}{\mathrm{d}\sigma}} \ \mathrm{d}\sigma. $$ Here $\sigma$ is any parameter that parametrizes the specific path (required to be timelike) taken from A to B, and your location along the path at parameter $\sigma$ has coordinates $x^\mu(\sigma)$.
Suppose $x^i \equiv 0$ along the path. Also call $x^0$ by the name $t$. Then most of the terms in the sum vanish and we have $$ \Delta\tau = \int_{t_\mathrm{A}}^{t_\mathrm{B}} \sqrt{g_{00}} \ \mathrm{d}t. $$ In fact we could have gotten this just by examining $$ d\tau^2 = g_{\mu\nu} \mathrm{d}x^\mu \mathrm{d}x^\nu. $$
Now often in GR we do call that timelike coordinate $t$, as I have done. Usually when this is done, as in the Schwarzschild metric $$ ds^2 = \left(1 - \frac{2M}{r}\right) \mathrm{d}t^2 - \left(1 - \frac{2M}{r}\right)^{-1} \mathrm{d}r^2 - r^2 \left(\mathrm{d}\theta^2 + \sin^2(\theta) \ \mathrm{d}\phi^2\right), $$ that $t$ becomes arbitrarily close to the $\tau$ of a local observer "at rest" in these coordinates as you move into the flat regime. In this case, as you move away from the mass at small $r$, $t \to \tau$. In other words, you can only neglect the $g_{00}$ term in some asymptotic cases (which you'll note correspond to $g_{00} \to 1$). In your SR example, you happened to use coordinates that matched those of the observer whose proper time you cared to measure, but you could have boosted to a new frame, in which case the $\mathrm{d}t^2$ part of the metric would have had some prefactor involving $\gamma$.
1 I will point out that most often in GR proper, $ds^2$ is negative for timelike separations, with $d\tau^2 = -ds^2$. The convention you adopted is more common in particle physics.