Question about gravitational time dilatation For $g_{00} = 1 - 2GM/c^2r$ , the time interval $\Delta T$ measured in a stationary frame at a distance $r$ from the source and the time interval $\Delta t$ measured by a frame at $r= \infty$ are related via
$$\Delta T = \sqrt{1 - \frac{2GM}{rc^2}} \Delta t$$
But, the invariant quantity $\Delta \tau^2$ in both frames should be the same:
$$\Delta \tau^2 = \left( 1 - \frac{2GM}{c^2r} \right) (\Delta T)^2 = (\Delta t)^2$$
which gives the reverse relation between $\Delta T$ and $\Delta t$. What am I doing wrong?
 A: The eigen time of a stationary frame at distance $r$ from a gravitational source is
$$(\Delta T)^2 =\frac{(\Delta s)^2}{c^2} = \left(1-\frac{2GM}{c^2 r}\right)(\Delta t)^2$$
i.e. the time measured in the stationary frame at distance $r$.
The quantity $\Delta t$  is the coordinate time.
In order to know the coordinate time one can go to a frame in a region where there is no gravitational effect (i.e. an area with a Minkowski metric), formally achieved by putting $r\rightarrow \infty$. In such a frame we would have as eigen time (since there we have $(\Delta x)=(\Delta y)= (\Delta z)=0$.):
$$\frac{(\Delta s)^2_{r\rightarrow \infty}}{c^2}  = (\Delta t)^2 -[(\Delta x)^2 -(\Delta y)^2 -(\Delta z)^2]/c^2 = (\Delta t)^2$$
In such a frame the eigen time corresponds to the coordinate time $\Delta t$.
So one can relate the eigen time measured at distance $r$ with the eigen time measured at $r\rightarrow \infty$  by
$$(\Delta T)^2_{r}  = \left(1-\frac{2GM}{c^2 r}\right)(\Delta t)^2_{r\rightarrow \infty}$$
Therefore there is no contradiction at all.
