Gravitational time dilation at a given radius $r$ around a mass looks equal to the special‐relativistic time dilation you’d get if you moved at the local escape velocity. For an object freely falling from rest at infinity down to radius $r$, does the gravitational and velocity time dilation double up, or do they somehow cancel? In other words, from the perspective of an observer at infinity who is stationary with respect to the mass, what is the net time dilation factor for the free‐falling object?
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nir asked: "For an object freely falling from rest at infinity down to radius $r$, does the gravitational and velocity time dilation double up, or do they somehow cancel?"
From the perspective of the stationary bookkeeper it multiplies while from the perspective of the free falling observer cancels (exactly if $\rm v=±c\sqrt{r_s/r}$, therefore $\rm g^{tt}=1$ in raindrop coordinates where the local clocks and rulers fall with the escape velocity).
The gravitational component of the time dilation is absolute while the kinematic component is relative. For the (positive or negative) escape velocity they are equal in magnitude, so such an object has the gravitational time dilation squared $\rm dt/d\tau=1 \div \sqrt{(1-r_s/r) \times (1-v^2/c^2)}$ from the external perspective and no time dilation $\rm d\tau/dt=\sqrt{(1-r_s/r)\div(1-v^2/c^2)}$ relative to the external observer from its own perspective.
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