Conformal or Cosmological Time?

1. Jun 18, 2013

johne1618

Let us assume a flat FRW metric
$$ds^2=-dt^2+a(t)^2(dx^2+dy^2+dz^2).$$
where $t$ is cosmological time, $x,y,z$ are comoving space coordinates, the speed of light $c=1$ and $a(t_0)=1$ at the present cosmological time $t_0$.

Imagine a light beam travelling in the x-direction. It travels on a null geodesic $ds=0$ therefore its path obeys the relation
$$a(t)dx=dt$$
Therefore at the present time $t_0$ during an interval of cosmological time $dt$ the light beam travels a proper distance $a(t_0)dx=dx$.

Now imagine a time $t$ in the future when the Universe has expanded by a factor $a(t)$.

During the same interval of cosmological time $dt$ the light beam now travels a proper distance $a(t)dx$.

Thus, in the future, the light beam travels further in the same interval of cosmological time and therefore its speed seems to have increased according to an observer at the present time $t_0$.

I think this paradox is resolved if the time interval the later observer at time $t$ measures expands by the same factor of $a(t)$ according to the present observer.

Let us assume that observers actually measure time in units of conformal time $d\tau$ such that
$$dt = a(t) d\tau$$
Then for the later observer at cosmological time $t$ we have
$$\frac{a(t) dx}{dt} = \frac{a(t) dx}{a(t) d\tau} = \frac{dx}{d\tau} = 1$$
This agrees with the speed of light measured by the present observer at cosmological time $t_0$
$$\frac{a(t_0)dx}{dt}=\frac{dx}{a(t_0)d\tau}=\frac{dx}{d\tau}=1$$
Thus if we assume that both observers measure conformal time $\tau$ rather than cosmological time $t$ then both will agree with the other's measurement of the speed of light.

Last edited: Jun 18, 2013
2. Jun 18, 2013

marcus

Here's a picture showing conformal time (in the bottom frame). You've probably seen it many times but's still worth glancing at now and then I find. Jorrie has it in his signature along with Lightcone calculator.

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