# Do clocks measure conformal time?

1. May 23, 2015

### jcap

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 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?

2. May 23, 2015

### Staff: Mentor

This doesn't make sense. These intervals are differentials; you have to integrate them to get something meaningful.

No, they aren't. The scale factor $a(t)$ is the ratio of proper length to coordinate length; so at a given instant of time $t$, we have $dl = a(t) dx$. But you are considering a null geodesic that travels for a finite coordinate time $t$, so $a(t)$ changes while it travels. That means the ratio of $dl$ to $dx$ changes.

No. See above.

3. May 23, 2015

### jcap

Ok so I should use small finite intervals in my argument instead of differentials i.e

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

The intervals are assumed to be small enough so that $a(t)$ does not change appreciably.

I accept that the scale factor $a(t)$ is the ratio of proper length to coordinate length so that for the ruler with constant proper length $\delta l$ one can write:

$\delta l = a(t) \delta x$

But the coordinate length of the ruler, $\delta x$, is a different coordinate length than the one I use in the null geodesic equation for freely propagating light:

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

The co-ordinate length of the ruler $\delta x$ changes with time so it cannot be a static co-moving spatial interval as required by the null geodesic equation. Only the proper length of the ruler $\delta l$ is constant which can be compared to the constant co-moving interval $\Delta x$ in the null geodesic equation.

Last edited: May 23, 2015
4. May 23, 2015

### Staff: Mentor

Then you are in a local inertial frame, in which everything works like flat spacetime; you just rescale all your spatial coordinates by the current value of $a(t)$ and the metric is the Minkowski metric. So you aren't proving anything about the metric you started with.

To put it another way, your original purpose was to show something about how changes in $a(t)$ affect other things. But now you're removing all changes in $a(t)$ from consideration. So you obviously aren't going to accomplish anything towards your original purpose.

5. May 23, 2015

### Staff: Mentor

I suppose that, rather than allow this discussion to prolong itself, I should just go ahead and show how null geodesics are correctly dealt with in your original metric. Start with your equation (1):

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

Now observe that the proper length $dl$ of the ruler is related to its coordinate length $dx$ as follows:

$$\frac{dl}{a(t)} = dx$$

Obviously we can equate these two expressions for $dx$ and then cancel the factors of $a(t)$ to obtain

$$c dt = dl$$

It should be obvious that this means the light travels proper distance $dl$ in proper time interval $dt$--i.e., that coordinate time intervals in this metric are also proper time intervals for observers at rest in these coordinates.

6. May 23, 2015

### Staff: Mentor

Once again, you are misinterpreting that equation. It is not telling us about an absolute coordinate interval $dx$. It is only telling us the relationship between coordinate intervals along a null geodesic--i.e., if we pick any coordinate interval $dt$, it tells us the corresponding interval $dx$ along the geodesic, or vice versa. The coordinate length of the ruler is irrelevant to that relationship.