# Taking into account scale expansion for the path taken by a photon

The Robert-Walkerson metric is given by

$$\left ( {\rm d}s \right )^{2}=\left ( cdt \right )^{2}-R^{2}\left ( t \right )\left ( dl \right )^{2}$$

where

$$\left ( dl \right )^{2}=R^{2}\left [ \frac{\left ( dr \right )^{2}}{1-kr^{2}}+r^{2}\left ( \left ( d \theta \right )^{2}+\sin^{2} \theta\left ( d\phi \right )^{2} \right ) \right ]$$

and $R=R\left ( t \right )$ is the expansion scale factor of the Universe with respect to time and $$\left ( dl \right )^{2} =\left [ \frac{\left ( dr \right )^{2}}{1-kr^{2}}+r^{2}\left ( \left ( d \theta \right )^{2}+\sin^{2} \theta\left ( d\phi \right )^{2} \right ) \right ]$$ gives the comoving path of a say, photon, on some geometry in 3-D.

Now, for a photon travelling in a radial path towards us from $r=0$ to $r=r_{1}$ in an EDS Universe $\left ( k=0 \right )$, the FLRW metric reduces to

$$\left ( cdt \right )^{2}=R^2\left ( t \right )\left ( dl \right )^{2}$$

which further reduces to

$$r_{1}=c\int_{t_{1}}^{t_{0}}\frac{dt}{R\left ( t \right )}\equiv d_{comoving}\tag{1}$$

for $t_{1} < t_{0}$.

Here is where I fail to understand what follows. The above FLWR metric already takes into account the Expansion scale factor of the Universe giving the proper distance $d_{proper}$. Yet, from proper distance, Equation (1) gives only the comoving distance from the equation and so the proper distance is

$$d_{proper}=R\left ( t \right )d_{comoving}=R\left ( t \right )r_{1}$$

Why do we take into account the expansion scale factor twice?

## 1 Answer

As you point out ${\rm d}l$ represents a comoving path. So you'd need to multiply it by $R$ to get physical distances. That is

$${\rm d}l = c\frac{{\rm d}t}{R(t)} ~~\Rightarrow~~~ l_1 -\underbrace{l_0}_{=0} = \int_{t_0}^{t_1}c\frac{{\rm d}t}{R(t)}$$

such that $R(t_1)l_1$ is the actual length in physical coordinates.

• Thank you very much. I think I now understand. The link en.wikipedia.org/wiki/… help a lot by mentioning about the reduced circumference $r$ where it was, otherwise, not mentioned in my notes. Since dr would then be the differential reduced circumference which is really just a comoving distance, multiplying it by the expansion scale factor of the Universe converts this to its respective proper distance. – Physkid Nov 13 '17 at 4:28