Why galaxies with a redshift > 2 have a lower velocity now, compared to when the light was emmited?

Question

I was playing around with astropy cosmology data, and I made the following graph:

On the x axis I plotted Redshift, and on the y axis the speed of light as 1c, 2c etc. Until a redshift of ~ 2, my reasoning is like this:

Since the light was emitted from a galaxy (Z=1), the universe expanded, accelerating, and the velocity 'now' is greater than velocity 'then'.

But, why after a redshift of ~2 the velocity 'now' is lower than in the past?

Update: Indeed the Hubble parameter was much bigger in the past!

Answer 1

The velocity of a galaxy at redshift $z$ today is $$v_0 = H_0 D(z) \ ,$$ where $D(z)$ is the comoving radial distance, and $H_0$ today's value of the Hubble parameter.

The distance of the same galaxy at emission of the photon we received today is given by the angular diameter distance, which is - in a flat universe - given by $$\frac{D(z)}{1+z} \ .$$ Hence, the velocity of the galaxy in the past is given by $$v_{\text{past}} = H(z)\frac{D(z)}{1+z} \ ,$$ where we have to take into account that the Hubble parameter is a function of $z$.

From the 1st Friedman equation, we obtain for a flat universe filled with matter and vacuum energy: $$H(z) = H_0 \sqrt{\Omega_m (1+z)^3 + \Omega_{\Lambda}} \ .$$

Solving for $v_0 = v_{\text{past}}$ with $\Omega_m = 0.3$ and $\Omega_{\Lambda}=0.7$ yields $z \simeq 2$ (and $z=0$). For $z>2$, we find indeed that the recession velocity in the past is bigger than today.

Answer 2

The recession speed of an object that moves with the Hubble flow, as a function of time, is a constant times $\dot a$ (not $H$).

$\dot a$ decreased in the past ($\ddot a<0$) and is now increasing ($\ddot a>0$), so there was a time $t<t_0$ at which $\dot a(t)=\dot a(t_0)$, and at all $t'<t$, $\dot a(t')>\dot a(t_0)$.