PeterDonis
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Ich said:In my opinion, the coordinate time is relevant only if is connected with something physically interesting
In Schwarzschild coordinates, it is: the timelike Killing vector is ##\partial / \partial t##, i.e., the ##t## coordinate is aligned with the timelike Killing vector. That means that if the reference Killing vector (in your terminology) is at infinity, the coordinate time dilation in Schwarzschild coordinates is exactly what you're calling the time dilation relative to the reference Killing vector.
Ich said:The time dilation of a moving object would then naturally be the product of its four-velocity with the respective Killing vector.
This can't be quite right as a general statement, because the dot product between two 4-vectors is an invariant, but time dilation is coordinate-dependent. However, per my comments above, if we use Schwarzschild coordinates, we can ignore that complication.
Having said that, let's compute the dot product explicitly. It is ##u \cdot k = g_{\mu \nu} u^{\mu} k^{\nu}##, where ##u^{\mu}## is the 4-velocity and ##k^{\nu}## is the Killing vector. As I noted above, the Killing vector is just ##\partial / \partial t##, so its components are ##(1, 0, 0, 0)##; that means the only term in the dot product is the ##t## term, so we have ##u \cdot k = g_{tt} u^t = g_{tt} dt / d\tau = g_{tt} / \left( d\tau / dt \right)##.
In other words, the dot product ##u \cdot k## is not the same as what is usually called the "time dilation", which is ##d\tau / dt##. It's not even the same as the reciprocal of the time dilation, because of the factor ##g_{tt}##. But it just so happens that, for a static observer, since ##d\tau / dt = \sqrt{g_{tt}}##, the dot product becomes ##u \cdot k = g_{tt} / \sqrt{g_{tt}} = \sqrt{g_{tt}}##, which happens to be equal to ##d\tau / dt##. So for a static observer, the relationship you came up with does happen to hold; but unfortunately it doesn't generalize to a non-static observer.