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The FRW metric at the origin $r=0$ is simply given by: $$ds^2 = -dt^2 + a^2(t)\ dr^2$$ Setting $dt=0$ gives us an element of proper distance $ds$ given by: $$ds = a(t)\ dr$$ Thus we get the well known result that space expands with the scale factor.

Setting $dr=0$ gives us the relationship between an element of proper time $d\tau$ and an element of co-ordinate time $dt$ (using $d\tau^2 = - ds^2$): $$d\tau = dt$$

Thus we get the well known result that for a co-moving observer proper time is the same as co-ordinate time.

However there is one more relation one can derive. We can set $ds=0$. We then obtain a relationship between an element of co-ordinate time $dt$ and an element of co-ordinate separation $dr$ : $$dt = a(t)\ dr$$

Surely this relationship implies that elements of co-ordinate time $dt$ (and thus proper time $d\tau$ for a co-moving observer) expand with the scale factor in the same way that elements of proper distance expand with the scale factor?

John Eastmond
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  • With $ds=0$ you find the equation of the light cone. So $\frac{dr}{dt}=\frac{1}{a(t)}$ – Antonio Ragagnin Jun 16 '14 at 10:02
  • True. However as the speed of light is constant for all observers, including co-moving observers, then time intervals must expand along with space intervals. – John Eastmond Jun 16 '14 at 10:36
  • I think the co-moving co-ordinate $r$ does not depend on $t$ by definition. Therefore the equation $dr/dt=1/a(t)$ is false. One can only assert $ds/dt=1$ where $ds=a(t)dr$. – John Eastmond Jun 16 '14 at 11:54

1 Answers1

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As Antonio says in his comment, by setting $ds = 0$ all you are doing is writing the equation of a null geodesic, i.e. the trajectory of a ray of light. And you deduce that (again as Antonio has pointed out):

$$ \frac{dr}{dt}=\frac{1}{a(t)} $$

At first glance this looks as if the speed of light falls as the universe expands, but you need to remember that the $r$ coordinate is comoving distance not proper distance. The speed of light is indeed falling in comoving coordinates, but the speed you and I measure with our rulers and clocks is the proper distance per second not the comoving distance per second, and that stays constant at $c$.

Response to comment: if we call the change in proper distance $dr_p$ then we have:

$$ dr_p = a(t) dr $$

and substituting this in the above equation for the speed of light we get:

$$ \frac{\tfrac{1}{a(t)}dr_p}{dt}=\frac{1}{a(t)} $$

or:

$$ \frac{dr_p}{dt}=1 $$

as we expect. We don't need $dt$ to be changing to keep the speed of light at $c$.

John Rennie
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  • But as I say in my above comment a co-moving observer measures proper distance per proper time (same as co-ordinate time). Proper distance is expanding so proper time must expand in order to ensure that c=1. – John Eastmond Jun 16 '14 at 10:47
  • @JohnEastmond: no, because the speed of light in comoving coordinates is falling. The expansion in the proper distance balances this out and keeps the speed of light at $c$. The proper time doesn't need to change. – John Rennie Jun 16 '14 at 10:49
  • But a truly co-moving observer could only measure time with an expanding light clock - he would not have access to fixed atomic distances like we do. Therefore his time intervals would expand with proper distances. – John Eastmond Jun 16 '14 at 11:00
  • @JohnEastmond: I'm not sure I see your point. The speed of light is invarient for a comoving observer so they can measure distance in light years and the distance they measure will be the proper distance (at least locally). The expansion just means the number of light years they measure to some other distant comoving observer increases with time (as the universe expands). – John Rennie Jun 16 '14 at 11:07
  • As I understand it if $r$ is a co-moving co-ordinate then it cannot depend on $t$. Therefore the equation $dr/dt=1/a(t)$ is wrong? One can only assert $ds/dt=1$. – John Eastmond Jun 16 '14 at 11:28
  • @JohnEastmond: you derived that equation by requiring that $ds = 0$. It's not some general equation for calculating how comoving distance changes with time. It's an equation that tells you how comoving distance must change with time to keep the line element equal to zero. – John Rennie Jun 16 '14 at 12:07
  • I would say $dr=\mbox{const}$ and $ds=0$ so that we must have $dt \propto a(t)$. – John Eastmond Jun 16 '14 at 13:34
  • Let us continue this discussion in chat. – John Eastmond Jun 16 '14 at 20:20