general relativity - Do clocks measure conformal time (new argument)?

Assuming the spatially flat FRW metric for simplicity:

$$ds^2=c^2dt^2-a(t)^2(dx^2+dy^2+dz^2)$$ where $t$ is cosmological time, $a(t)$ is the scaling factor and $x,y,z$ are co-moving spatial Cartesian co-ordinates.

Light freely propagating along the $x$-direction follows the null geodesic with spacetime interval $ds=0$. Therefore for small finite co-ordinate intervals $\Delta t$ and $\Delta x$ along the null geodesic we have the relationship:

$$\frac{c\ \Delta t}{a(t)}=\Delta x\tag{1}$$ The co-moving interval $\Delta x$ in this expression must be constant as it is the distance between co-moving spatial co-ordinates. This implies that the cosmological time interval $\Delta t$ must scale like $a(t)$.

Now let us imagine a model of a clock that consists of a rigid ruler of fixed proper length $\Delta l$ with an optical fiber attached along it.

Let us assume that the rigid ruler is co-moving with the expansion of the Universe.

Let us suppose that each tick of the clock consists of a light pulse sent down the optic fiber (assuming refractive index 1 for simplicity) which takes a constant time interval $\Delta\tau$ given by:

$$\Delta \tau = \frac{\Delta l}{c}\tag{2}$$ Now the co-moving interval $\Delta x$ in the null geodesic of freely propagating light, equation $(1)$, and the proper length of the rigid ruler $\Delta l$ are both constant. Without loss of generality let us arrange for them both to be equal in magnitude so that we have: $$\Delta x=\Delta l\tag{3}$$ By substituting equations $(2)$ and $(3)$ into the null geodesic equation $(1)$ we obtain the relationship: $$\frac{\Delta t}{a(t)}=\Delta \tau\tag{4}$$ Thus each tick of our clock $\Delta \tau$ does not measure an interval of cosmological time $\Delta t$ as one might expect but instead it measures an interval $\Delta t/a(t)$ which is in fact an interval of conformal time.

Is this reasoning correct?

Answer

Firstly, notice that the Frequency of light used to measure time will not remain constant. I wont use this but do take note.

Secondly notice that If the object is rigid, then and One end is fixed at x = 0, then the other end in the co-ordinate systems used will be at

$\Delta x = \frac{\Delta L}{a(t)}$

So Now light using equation 1 from your question we get

$\Delta t = a(t)*\frac{\Delta x}{c} = \frac{\Delta L}{c}$

So both clocks reproduce cosmological time as expected.

I hope I understood your arguments correctly. I think the flaw in the argument is you cannot allow for both the rod to be rigid and its ends to be co-moving at the same time.