If we send a photon away from earth, it would eventually reach a comoving distance of about 16.5 billion light years (the universe's event horizon). I'm interested in how far (in comoving distance) particles moving below $c$ can eventually travel and at what velocity they would arrive if they were to reach a distant comoving galaxy near the end of this journey.

A simple answer to both would be that a particle moving at $kc$ (where $0<k<1$) could eventually travel $k$ times as far as light could. This is what I get if I assume $v_{peculiar}$ is always equal to $kc$ and $v_{recessional} = H(t) \cdot distance(t)$ (both in proper coordinates) and take the integral of velocity over time to get the distance.

However, I've also heard that the momentum of a particle reduces as it travels, resulting in redshift for photons (since they can't lose velocity) and in velocity loss for other particles. It seems that this could affect the above calculations. For example, this textbook says $\frac{\dot{p}}{p}=-H$.

Does this mean that a particle launched at $kc$ would instead travel less than $k$ times 16.5 billion light years and arrive at a galaxy near that limit with a relative (proper) velocity near zero?

(Bonus points if you could derive an equation for the the maximum distance as a function of $k$.)

  • $\begingroup$ Thanks. I hadn't seen that question, but the answer (if correct) would answer the first half of my question. I'm a bit doubtful of whether it is correct though, and would want to understand why the simple argument I suggested doesn't work. $\endgroup$
    – Toby Ord
    Commented Nov 27, 2015 at 16:12

1 Answer 1


The cosmological event horizon is the furthest distance light can travel and is also the upper bound to how far a massive particle can travel. A free falling massive particle will asymptotically approach a point which is receding from the point which it started with the same recessional velocity as its original peculiar velocity.

I believe the correct expression would be:

$$s = \int^{\infty}_{t_0} \frac{kdt}{a(t)}$$

Where $s$ is the total comoving distance traveled by the particle, $t$ is cosmological time, $t_0$ is the time the particle set off, $k$ is the peculiar speed of the particle at $t_0$ and $a(t)$ is the scale factor.

  • $\begingroup$ Regarding the first paragraph, I'm not sure that is right, since it would imply that something travelling at the speed of light would approach the Hubble radius, whereas it actually approaches the event horizon, which is a bit further out (14.4 billion light years vs 16.5). $\endgroup$
    – Toby Ord
    Commented Nov 27, 2015 at 16:14
  • $\begingroup$ Your maths may well be right. It limits to the event horizon, which is the correct answer at light speed, and it gives my 'simple answer' of $k$ times as far as the event horizon. However, it is a different result to the answer suggested in the potential duplicate question, and I don't know which to believe. $\endgroup$
    – Toby Ord
    Commented Nov 27, 2015 at 16:16
  • $\begingroup$ @TobyOrd, on your first point: yes, but the Hubble horizon and event horizon asymptotically coincide (i.e. they will be arbitrarily close together after an arbitrarily long amount of time). Or in other words I am talking about the current recessional velocity of the point, but the final recessional velocity. $\endgroup$
    – John Davis
    Commented Nov 27, 2015 at 16:50
  • $\begingroup$ thanks for the clarification. I hadn't thought of it that way. Re my second point, do you think the graph in the answer to the possible duplicate question is wrong? It seems to me that it could be. $\endgroup$
    – Toby Ord
    Commented Nov 27, 2015 at 17:40
  • $\begingroup$ On your second point, look at the 3 diagrams in Figure 1 on this paper: arxiv.org/pdf/astro-ph/0310808v2.pdf, now note that in the third diagram the trajectory of free-falling particles are just straight lines whose slopes are decided by their current speeds. I hesitate a little though as I would be interested to see why my answer differs from @Jim 's. $\endgroup$
    – John Davis
    Commented Nov 27, 2015 at 17:51

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