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Starting from a simplified radial Freidman Walker metric we have $$ds^2 = -c^2 dt^2 + a(t)^2 dr^2 $$ How does one measure one's proper time operationally?

One times a light beam along an element of proper distance $ds$. Thus an element of proper time $d\tau$ is given by $$d\tau = \frac{ds}{c}$$ From the metric above we have an element of proper distance $ds$ (where $dt=0$) given by $$ds = a(t) dr$$ Thus $$d\tau = \frac{a(t)dr}{c}\ \ \ \ \ \ \ \ \ \ \ \ (1)$$ Now light travels on a null geodesic where $ds=0$ so from the above metric we also have $$\frac{a(t)dr}{dt} = c\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ Substituting Equation (2) into Equation (1) we find $$d\tau = dt$$ Thus an interval of our proper time $d\tau$ is the same as an interval of cosmological time $dt$ provided we measure the time light takes to travel an expanding distance $ds$. But our clocks don't expand with the Universe. They are rigid like we are. Thus if we measure our proper time interval $d\tau$ we would use $$d\tau = \frac{dr}{c}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$$ where $dr$ is a distance interval that does not expand with the Universe. If we now substitute the light-path expression (2) into Equation (3) we find $$d\tau = \frac{dt}{a(t)}$$ This seems to imply our local proper time will speed up as the Universe expands. Is this right?? :)

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Why was this downvoted?. Looks like a perfectly reasonable misconception . +1 to reverse the downvote . –  Dimensio1n0 Aug 2 '13 at 14:24
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1 Answer 1

First, it is Friedmann not Freidman. And the metric is the Friedmann–Lemaître–Robertson–Walker metric or FLRW metric.

Second, proper time $d\tau=ds/c$ is not valid for null geodesics except as formal meaningless parameter (i.e., it cannot be interpreted as time).

Third, the FLRW metric describes a homogeneous Universe, therefore time is the same everywhere and given by $t$. There is not such a thing as a cosmological time and a different local time.

Fourth, you are combining mutually incompatible equations and obtaining inconsistent results.

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