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I am learning general relativity, I understand that the metric tensor has a coordinate $t$ corresponding to time. But I know also that time depends on gravity and so the time can change from point to point in a region depending the presence of mass.

To which clock does the time $t$ in the metric tensor correspond to, in which position?

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It seems that be that you are automatically using a fixed set of coordinates, which is not how you should be thinking about the metric. Mathematically by construction, the metric, $g$, is a geometrical object, specifically a 2 rank tensor. As such we are free to choose which local coordinates we want to use to represent it.

In the context of general relativity one of the requirements is that locally (at a space-time point and a neighborhood of it) we should be able to recover Minkowski space, that is the metric should be able to be put in ${\rm diag}(-1,1,1,1)$ form via some coordinate transformation. However the coordinates for the local patch before such transformation might be complicated and in certain solutions one might not even be able to recognize (have) a time-like killing vector, that is a direction that plays the role of time in a consistent manner.

Having mentioned the points above, we have to settle first which coordinates are you using in what scenario. So that one is able to state what are they describing. There are plenty of coordinates and metrics that have the label $t$, so as such the question needs to give more details around that.

I will make a guess and assume you are referring to a FLRW-type of metric which are particularly symmetric. This are typically used to model cosmic evolution and in this case the $t$ represents a cosmic time (see the wiki).

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As in special relativity, there is no concept of absolute time in general relativity. Time is observer-dependent. However, there is still the notion of proper time, which is measured by an observer's clock.

For each observer, it is possible to choose coordinates that locally look like flat Minkowski space at one point, although derivatives of the metric are non-vanishing and the curvature can be non-zero. If the observer does not move with respect to this coordinate system, the four-dimensional line element can be written as $ds^2=-d\tau^2$. In this case $\tau$ corresponds to the time that the observer measures.

In the Schwarzschild geometry $ds^2 = (1-\frac{2M}{r})dt^2+\frac{dr^2}{1-\frac{2M}{r}}+r^2 d\theta^2+r^2\sin^2\theta d\phi^2$, the metric reduces in the limit $r\rightarrow \infty$ to the flat Minkowski metric. Far away from the source, one approximately finds a flat spacetime. Therefore, the time coordinate $t$ in the Schwarzschild metric is usually referred to as the time of a distant observer. This time differs from the time of an observer that falls freely inside the Schwarzschild geometry. As a consequence, a distant observer would not observe his friend crossing the event horizon of a black hole within a finite amount of time, whereas the friend (using Lemaitre coordinates to describe his fall) would cross the event horizon within a finite amount of time.

In the geometry of an expanding universe, $ds^2 = -dt^2 +a(t)^2 (dx^2+dy^2+dz^2)$ the coordinate $t$ has the meaning of a global cosmological time. This time, however differs from the time that an observer measures, when he is at the surface of a massive object.

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    $\begingroup$ You seem to be confusing the concept of co-moving coordinates and an observer at rest, you may want to edit that part with more details. $\endgroup$ – ohneVal May 27 at 8:07
  • $\begingroup$ Good point, thanks. Hopefully, now it's better. $\endgroup$ – p6majo May 27 at 19:20

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