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?? :)
