Assuming the spatially flat FRW metric for simplicity:(adsbygoogle = window.adsbygoogle || []).push({});

$$ds^2=c^2dt^2-a(t)^2(dx^2+dy^2+dz^2)$$

where ##t## is cosmological time, ##a(t)## is the scaling factor and ##x,y,z## are co-moving Cartesian co-ordinates.

Light freely propagating along the ##x##-direction follows the null geodesic, with spacetime interval ##ds=0##, so that we have the relationship:

$$\frac{c\ dt}{a(t)}=dx\ \ \ \ \ \ \ \ \ \ \ \ (1)$$

The co-moving interval ##dx## in this expression is constant which implies that the cosmological time interval ##dt## must scale like ##a(t)##.

Now let us imagine that we have a clock that consists of a rigid ruler of fixed proper length ##dl## with an optical fiber attached to it.

Each tick of the clock consists of a light pulse sent down the optic fibre which takes a constant time interval ##d\tau##:

$$d\tau = \frac{dl}{c}\ \ \ \ \ \ \ \ \ \ \ \ (2)$$

Now the co-moving interval ##dx## in the null geodesic of freely propagating light, equation ##(1)##, and the proper length of the rigid ruler ##dl## are both constant. Without loss of generality let us arrange for them both to be equal in magnitude so that we have:

$$dx=dl\ \ \ \ \ \ \ \ \ \ \ \ (3)$$

By substituting equations ##(2)## and ##(3)## into the null geodesic equation ##(1)## we obtain the relationship:

$$\frac{dt}{a(t)}=d\tau\ \ \ \ \ \ \ \ \ \ \ (4)$$

Thus each tick of our clock ##d\tau## does not measure an interval of cosmological time ##dt## as one might expect but instead it measures an interval ##dt/a(t)## which is in fact an interval of conformal time.

Is this reasoning correct?

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# Do clocks measure conformal time?

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