I’m trying to get a better grasp on cosmological horizons and have a question regarding recessional velocity. In particular:

What is the average recessional velocity of a cosmological object (irrespective of distance) an observer would expect to observe if measured at some time in the future?

Here is my attempt:

If the current cosmic event horizon $D_{eh}$ is the maximum proper distance from which light emitted now can ever reach the observer in the future, then to calculate the average recessional velocity $\bar v_{rec}$ of an observable that could potentially be detected by the observer at some time between now $t_0$ and $t=\infty$, one can calculate the recessional velocity of the current event horizon $v_{eh}$ and divide by two—i.e., as recessional speed is proportional to distance, this is the average between $v_{eh}$ and the observer’s recessional velocity, $v_{ob}=0$. Therefore, \begin{align} \bar v_{rec} = \frac{1}{2}v_{eh}(t_0) = \frac{1}{2}H_0 D_{eh}. \end{align} If the Hubble constant $H_0$ is about $70$ km/s/Mpc, and assuming a standard $\Lambda$CDM-model, $D_{eh}$ is about 5 Gpc, then \begin{align} \bar v_{rec} \approx 0.58c. \end{align}

Does this reasoning make sense? Thanks for your help.

  • 1
    $\begingroup$ I think you are implicitly assuming that the density of “cosmological objects” (what are they ? Stars ? Galaxies ?) is the same at all distances. I am not sure this assumption is correct. $\endgroup$
    – gandalf61
    Aug 7, 2022 at 10:43
  • $\begingroup$ Right, I’m making that assumption. Although, I’m not sure why that should affect the average recession velocity. $\endgroup$ Aug 7, 2022 at 11:40
  • 2
    $\begingroup$ The question about the average recession velocity doesn't make sense since you'd need a finite universe to determine it, but it would make sense to ask about the average en.wikipedia.org/wiki/Peculiar_velocity#Cosmology $\endgroup$
    – Yukterez
    Aug 9, 2022 at 15:34
  • $\begingroup$ Is "average recession velocity" truly a meaningful quantity? What can you calculate with it? It seems like trying to define an "average value" for the function $y = e^x$. Yes, you can calculate one, but a flat line running through that graph at some value tells you little or nothing about the function itself. $\endgroup$
    – RC_23
    Aug 12, 2022 at 3:08

1 Answer 1


If I understand your assumption that all of the matter objects are similar, and all on a large scale have similar densities, then what determines one's velocity from us is determined by the equation:

$$V = D * H_0. $$

$V$ is velocity from us, $D$ is distance from us, and $H_0$ is the Hubble constant with the reciprocal value: $1/H_0 = 14.4 \, \text {Gyrs}$.

So, what needs to be calculated is the average of distance A of points within a sphere from its center. The radius of the sphere R will be

$$R = c * 14.4 \, \text {Gyrs}. $$ (Note: c = speed of light.)

I think you that should have the first try to calculating A. If you can't solve for A, then I will offer some more help.

The reason for this value of R (also known as the event horizon) is that the distance R is where anything further away travels away from the center faster than the speed of light.


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