# Relative size of galaxy calculation

So this is a past exam problem, however I am confused to a question which is related to this graph and the questions states:

Question: On the graph, one galaxy is labelled A. Determine the size of the universe, relative to its present size, when light from the galaxy labelled A was emitted.

In the mark scheme it says: $$z=\frac{v}{c}=\frac{4.6 \times10^4\times 10^3}{3.00\times10^8}=0.15$$ $$\frac{R}{R_0}=z+1$$ $$\frac{R_0}{R}=\frac{1}{z+1}$$ My question is, why is it suddenly calculating z the redshift value, and why is $$\frac{R}{R_0}=z+1$$

The graph is attached below for reference, I will appreciate the clarification.

The relationship between the scale factor and the redshift can be written as,

$$\frac{a(t_{observed})}{a(t_{emitted})}=z+1$$ Derivation of this equation is a bit long but I am sure there are some online sources that you can find (or in your textbook)

Now we know that the proper distance is defined as $$R=a(t)r$$ where $$R$$ is the proper distance and $$r$$ is the comoving distance.

In this case we can re-write the above equation in terms of the proper distance

$$R_{o}=a(t_{o})r$$ and $$R_{e}=a(t_{e})r$$, Hence

$$\frac {a(t_o)}{a(t_e)}=\frac{ \frac {R_o} {r}} {\frac {R_e} {r}}=\frac {R_o} {R_e}=z+1$$

By calculating the $$z$$ we can find the ratios of the proper distances between two events.

Or in this case, the ratios of the size of the observable universe when the light is observed (now) and when the light is emitted from the galaxy A. Since we know the current radius of the observable universe, we can calculate the value when the light is emitted from galaxy A.

Another point: $$\frac{a(t_{observed})}{a(t_{emitted})}=z+1$$ In this equation we can take $$a(t_o)=1$$ as a convetion, so the equation becomes

$$\frac{1}{a(t_{emitted})}=z+1$$ As we can see, $$z$$ is just a tool that we use to determine the scale factor for the time when the light is emitted, with respect to the current scale factor.