Let us assume that an observer is stationary at the origin in expanding space. We assume the FRW metric near the origin is given by:
$$ds^2=-dt^2+a(t)^2dr^2$$
Let us assume that the observer measures time by bouncing a light beam at a mirror that is at a constant proper unit distance away from him.
By substituting $dt=0$ in the metric we find that an interval of proper distance $ds$ is given by:
$$ds=a(t)\ dr$$
Integrating we find the proper distance $S$ to an object at co-moving co-ordinate $r$ is given by:
$$S=a(t)\ r$$
Thus if the mirror has constant proper distance $S=1$ then it follows a path in co-moving co-ordinates given by:
$$r=\frac{1}{a(t)}$$
A light beam is described by substituting $ds=0$ in the metric to get the null geodesic:
$$dr=\frac{dt}{a(t)}$$
Integrating we get the path of a light beam:
$$r = \int \frac{dt}{a(t)}$$
The spacetime diagram below shows the observer attempting to measure cosmological time $t$ using the light clock.
One can see that as the mirror gets closer in co-moving co-ordinate $r$ the period of the light clock gets smaller and smaller. Thus the light clock is getting faster relative to cosmological time $t$.
Now let us make a transformation to conformal time $\tau$ given by the relationship:
$$d\tau=\frac{dt}{a(t)}$$
The transformed radial co-ordinate $\rho$ is given by:
$$\rho\ d\tau = r\ dt$$
Therefore the path of the mirror in conformal co-ordinates is given by:
$$\rho = r\ \frac{dt}{d\tau}$$
$$\rho = \frac{1}{a(t)} \cdot a(t) = 1$$
The path of a light beam in co-moving co-ordinates is given by:
$$dr=\frac{dt}{a(t)}$$
Therefore in conformal co-ordinates we have:
$$d\rho=d\tau$$
On integrating:
$$\rho = \tau$$
Thus in conformal co-ordinates the light clock is described by the diagram below:
One can now see that the light clock period is constant - the clock is working properly.
Thus a light clock measures conformal time $\tau$ rather than cosmological time $t$.
Is this argument correct?
P.S. In answer to Lubos's objection I should not have introduced a new co-ordinate $\rho$. Instead I should keep the co-moving co-ordinate $r$ but change the time variable that it depends on from cosmological time $t$ to conformal time $\tau$.
Therefore I should express the transformation as:
$$r(\tau)d\tau=r(t)dt$$
$$r(\tau)=r(t)\frac{dt}{d\tau}$$
As the end of the light clock travels on the path $r(t)=1/a(t)$ and an interval of conformal time is $d\tau=dt/a(t)$ we find that, in terms of conformal time $\tau$, the co-moving co-ordinate of the end of the light clock, $r(\tau)$, is constant i.e.:
$$r(\tau)=\frac{1}{a(t)}\cdot a(t) = 1$$
Similarly the light beam geodesic is initially in terms of $t$:
$$dr(t)=\frac{dt}{a(t)}$$
using $d\tau=dt/a(t)$ we find
$$dr(\tau)=d\tau$$
Integrating we find the expected 45 degree light cones in the $\tau$,$r$ spacetime diagram:
$$r(\tau)=\tau$$.
PS I now realise that my idea is wrong.
The co-moving distance light travels in a small interval of cosmological time $\delta t$ is: $$r = \frac{c \delta t}{a(t)}$$. As we have $r \propto 1/a(t)$ then we get: $$\frac{c \delta t}{a(t)} \propto \frac{1}{a(t)}$$ $$\delta t \propto 1$$ Therefore the light clock does measure cosmological time correctly.
