Origin: https://arxiv.org/pdf/0808.1081.pdf
$$ \textrm{Define:} \quad a=\frac{D_{then}}{D_{now}} $$
$$ 1+z \equiv \frac{\lambda_{observed}}{\lambda_{emitted}}=\frac{1}{a} \approx 1+\frac{v_{rs}}{c}\textrm{ if } v_{rs} \ll c $$
$$ H \equiv \frac{\dot{a}}{a} \implies v_{rec}=HD \quad (\because v=\dot{D}) $$
Proper distance: actual distance at current time, which changes with the expansion of the universe
Comoving distance: distance at current time but does not change with the expansion of the universe
$D_L/(1+z)=D_C=D_A(1+z)$
where $D_L$ is the luminosity distance, $D_C$ is the comoving distance and $D_A$ is the angular diameter distance
<aside> 💡 PRACTICE: Explain Etherington reciprocity
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Conformal time: useless on its own
$$ \eta=\int^{t_{obs}}{t{emit}}\frac{1}{a(t)}\mathrm{d}t $$
However, when multiplied by the speed of light, you get its comoving distance:
$$ d_c=c \eta=\int^{t_{obs}}{t{emit}}\frac{c}{a(t)}\mathrm{d}t $$
This allows us to calculate the comoving distance to an object given when the light from it was emitted (or vice versa!). Intuitively, the stuff in the integral is the infinitesimal proper distance travelled. Multiplying the comoving distance by the scale factor at any time gives the proper distance to the object at any time.
By using $t_{emit}=0$ and $t_{obs}=t_0$, comoving distance obtained is the particle horizon, which is basically the size of the universe
Replacing $c$ with the speed of a rocket yields the distance a rocket travels given its emitted time and observed time (assuming the rocket moves at that speed relative to CMBR)