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1.Why is the observer at the bottom measuring the proper time rather than the coordinate time?

2.How do we go from time dilation to $g_{00}$ or from $g_{00}$ to time dilation?

I'm trying to understand the following passages.

In relativity, proper time along a timelike world line is defined as the time as measured by a clock following that line.

Coordinate time is the time between two events as measured by an observer using that observer's own method of assigning a time to an event.

-https://en.wikipedia.org/wiki/Proper_time

A common equation used to determine gravitational time dilation is derived from the Schwarzschild metric, which describes space-time in the vicinity of a non-rotating massive spherically symmetric object. The equation is

${\displaystyle t_{0}=t_{f}{\sqrt {1-{\frac {2GM}{rc^{2}}}}}=t_{f}{\sqrt {1-{\frac {r_{s}}{r}}}}=t_{f}{\sqrt {1-{\frac {v_{e}^{2}}{c^{2}}}}}=t_{f}{\sqrt {1-\beta _{e}^{2}}}<t_{f}}$

where $t_{0}$ is the proper time between two events for an observer close to the massive sphere, i.e. deep within the gravitational field.

Are the events also close to the observer and massive object? It seems so.

$t_{f}$ is the coordinate time between the events for an observer at an arbitrarily large distance from the massive object (this assumes the far-away observer is using Schwarzschild coordinates, a coordinate system where a clock at infinite distance from the massive sphere would tick at one second per second of coordinate time, while closer clocks would tick at less than that rate).

-https://en.wikipedia.org/wiki/Schwarzschild_metric

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It's most important to focus on question 2. I'll work with $c=1$.

Time dilation is computed as $\sqrt{ds^2/dt^2}$ on suitable assumptions of $dx^i/dt$. Special relativity in Minkowski space gives time dilation due to motion, say with Cartesian coordinates $x^i$ satisfying $dx^i=\beta^i dt$; using this in $ds^2=dt^2-dx^2$, $\sqrt{ds/dt}=1/\gamma$. In general relativity, time dilation can happen even to motionless bodies with $dx^i=0$, so $ds^2=g_{\mu\nu}dx^\mu dx^\nu\implies\sqrt{ds^2/dt^2}=\sqrt{g_{00}}$.

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  • $\begingroup$ Thanks. I understand time dilation with respect to different observers. But this seems to be time dilation of curved spacetime with respect to the flat space that would be there in the absence of the massive object. Which I guess means with respect to far away from the massive object. I think I'm starting to understand. $\endgroup$
    – user288901
    Feb 15 at 8:30
  • $\begingroup$ @ExpertNonexpert also note the Schwarzschild metric is asymptotically flat $\endgroup$
    – Eletie
    Feb 15 at 9:51
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    $\begingroup$ @ExpertNonexpert: the main point is that coordinate time is useful only as a label for spacetime points. All clocks ever measure is proper time along different curves. The use for the coordinate time is definiting exact spacetime points so that two different observers can definie exact intervals over spacetime to conduct experiments where they can say "I am observing time dilation, because our proper time measurements of these spacetime intervals disagree" $\endgroup$ Feb 15 at 17:53

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