(Hyper)Surface of Simultaneity How can I determine the surfaces of simultaneity if I know the metric? In particular, what are the surfaces of simultaneity for rotating disk with Langevin metric:
$$
ds^2=-(1-\omega^2r^2)dt^2+2r^2\omega d\phi dt+dr^2+r^2d\phi^2+dz^2
$$
where $\omega$ is a constant angular velocity?
 A: Recall how there are multiple surfaces of simultaneity in special relativity? Same thing in general relativity.
Recall how in special relativity you can pick one vector somewhere and find a single surface of simultaneity that has all its tangent vectors be spacelike everywhere and be orthogonal to that one single timelike vector at that one event?
That one isn't true in general relativity. Multiple surfaces could have all spacelike tangent vectors everywhere and be orthogonal to that one timelike vector at that one event.
A: In this particular case, a simple change of variable on the angular coordinate can separate the time variable and show that the (local) surface of simultaneity is always the 3d Euclidian space, or else, always has a 3d Euclidian metric:
Define $\phi = \phi' - \omega t$, such that
$$
d\phi = d\phi' - \omega dt
$$
and substitute in the original expression defining the metric:
$$
ds^2 = -(1 - r^2\omega^2)dt^2 + 2r^2 \omega ( d\phi' - \omega dt) dt + dr^2 + r^2 ( d\phi' - \omega dt)^2 +dz^2 =\\
= -dt^2 + r^2\omega^2 dt^2 + 2r^2 \omega d\phi' dt - 2r^2 \omega dt^2 + dr^2 + r^2 d\phi' ^2 + r^2\omega^2 dt^2 - 2r^2 \omega d\phi' dt + dz^2 =\\
= -dt^2 + dr^2 + r^2 d\phi' ^2 + dz^2 = -dt^2 + d\sigma^2
$$
where $d\sigma^2 =  dr^2 + r^2 d\phi' ^2 + dz^2$ is obviously the Euclidian metric in cylindrical coordinates.
In other words, the metric describes 3d space as seen from a locally rotating frame given by $\phi' = \phi + \omega t$.
