# Does time expand along with space?

1. Apr 12, 2015

### jcap

The flat FRW metric is given by:
$$ds^2=-c^2dt^2+a(t)^2dr^2$$
If we take $dt=0$ then we get:
$$ds=a(t)\ dr$$
The proper distance between co-moving points scales with $a(t)$. Thus we find that space expands.

If we take $ds=0$ to find the null geodesic followed by a light beam we get:
$$c\ dt=a(t)\ dr$$
Surely this implies that cosmological time intervals expand in the same way as space intervals?

2. Apr 12, 2015

### wabbit

This implies $dt=\frac{c}{a(t)}dr$, which is a useful relation but says nothing of "time expansion".
If $\tau$ is the proper time of a comoving observer, and $\delta$ the distance in his reference frame to another comoving object situated at comoving distance $r$,
$\quad\tau=t$ (no time expansion)
$\quad\delta=a(t)r$ ("space expansion")

But you may be referring to the time dilation effect when this observer looks at a comoving source. Light emitted by that source at $t$ and $t+dt$ is received at $t'$ and $t'+\frac{dt}{a(t)}$ because the second pulse was emitted from a more distant source than the first - this is actually the same thing as the redshift ot a distant source due to expansion.

Last edited: Apr 12, 2015
3. Apr 12, 2015

### Staff: Mentor

No, it implies that the wavelength of light expands with $a(t)$--i.e., that the light gets redshifted as the universe expands, as wabbit said.