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The flat FRW metric is given by:

$$ds^2=-c^2dt^2+a(t)^2dr^2$$

If we take $dt=0$ then we get:

$$ds=a(t)\ dr$$

Thus we find that space expands.

If we take $ds=0$ to find the null geodesic followed by a light beam we get:

$$c\ dt=a(t)\ dr$$

Does this imply that cosmological time expands along with space?

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  • $\begingroup$ Isn't the appropriate parallel to taking $dt = 0$, and concluding space is expanding, to take $dr = 0$ and then $ds^2 = -(cdt)^2$ and conclude that time is not expanding? $\endgroup$ – Alfred Centauri Apr 10 '15 at 21:37
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    $\begingroup$ Duplicate of Does cosmological time expand along with space? $\endgroup$ – ACuriousMind Apr 10 '15 at 22:09
  • $\begingroup$ I believe this has been asked and answered here: physics.stackexchange.com/a/83619/9887 $\endgroup$ – Alfred Centauri Apr 10 '15 at 23:08
  • $\begingroup$ @AlfredCentauri If one takes $dr=0$ then one just finds that proper time for a co-moving observer is the same as cosmological time. But a co-moving observer would be expanding with the Universe whereas we do not. He would be measuring time with an expanding light clock whereas our light clocks have a fixed length. $\endgroup$ – John Eastmond Apr 11 '15 at 3:42
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Well, yes. "Expansion" of time as you call it is the cosmological age. The "direction" of progression of the time defines cosmological arrow of time.

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  • $\begingroup$ I'm arguing that intervals of cosmological time $c\ dt=a(t)dr$ expand in exactly the same way as intervals of space $ds=a(t)dr$. $\endgroup$ – John Eastmond Apr 11 '15 at 3:31

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