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:


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:


A light beam is described by substituting $ds=0$ in the metric to get the null geodesic:


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.

Light Clock Cosmological

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:


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:


Therefore in conformal co-ordinates we have:


On integrating:

$$\rho = \tau$$

Thus in conformal co-ordinates the light clock is described by the diagram below:

Light Clock Conformal

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:



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 $t$:


using $d\tau=dt/a(t)$ we find


Integrating we find the expected 45 degree light cones in the $\tau$,$r$ spacetime diagram:


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.


1 Answer 1


No, the argument is not correct. The spatial "conformal" coordinate $R$ in which, together with the conformal time $\tau$, the angle of the light rays is 45 degrees is not $\rho$ but nothing else than $r$: $$ ds^2 = -dt^2 + a(t)^2 dr^2 = a(t)^2 (-d\tau^2 + dr^2) $$ If you want a diagram with $\tau$ on the vertical axis where the light rays are drawn at 45 degrees, the horizontal axis must be $r$. And because the proper length of the mirror is kept constant, the right end of the mirror will be at $r=L / a(t)$, just like you wrote at the beginning.

So in this diagram, $\Delta \tau$ between the two ticks will be equal to $2r_{\rm right}$ i.e. like $2L/a(t)$ as well, and because $\Delta \tau = 2L/a(t)$, then $\Delta t = 2L$, assuming that $L\ll R_{\rm Hubble}$ and so on. So just like one may determine with common sense, it's the cosmological time that is measured by the number of ticks, not the conformal time.

The two lines starting with "The transformed radial co-ordinate $\rho$ is given by" are incorrect or at least misleading. One may transform coordinates in any way he wants but only some coordinates complete $\tau$ to conformal coordinates, and it is $r$, not $\rho$.

  • $\begingroup$ Instead of introducing $\rho$ maybe I should have just introduced $r(\tau)=r(t)dt/d\tau$. If the end of the light clock travels on the path $r(t)=1/a(t)$ then that is also $r(\tau)=(1/a)*a=1$. $\endgroup$ Apr 15, 2015 at 18:16
  • $\begingroup$ Sorry, this equation is completely wrong. If you have $r$ as a function of $\tau$, the correct formula is still $r(\tau)=r(\tau(t))$, i.e. $r=r$. There is no $dt/d\tau$ here. The latter factor only appears if you express the derivative $r'$ - $dr/dt$ and $dr/d\tau$ differ by this factor. $\endgroup$ Apr 16, 2015 at 5:11
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    $\begingroup$ Ok - I see what i've done wrong now! $\endgroup$ Apr 16, 2015 at 17:27

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