Cosmological redshift

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

Etherington reciprocity

$D_L/(1+z)=D_C=D_A(1+z)$

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

</aside>

Looking into the past

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:

$$ d_c=c \eta=c\int^{t_{obs}}{t{emit}}\frac{1}{a(t)}\mathrm{d}t $$

This allows us to calculate the comoving distance to an object given when the light from it was emitted (or vice versa!). 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