According to Dodelson, Modern Cosmology (p.146)

There is a subtle distinction between the comoving horizon $\eta$ and the comoving Hubble radius $(aH)^{-1}$. If particles are separated by distances greater than $\eta$, they never could have communicated with one another; if they are separated by distances greater than $(aH)^{-1}$, they cannot talk to each other now.

if two observers are separated by distances greater than the Hubble radius, they cannot talk to each other now.

However, it's stated in the first answer of this post that "a galaxy on the Hubble radius can still send signals to us (or we to them)".

I'm not sure which one is the correct explanation here, the one in the post makes more sense to me.


2 Answers 2


The textbook is just wrong.

In the second edition (which was published after this question), that language has been changed to

The comoving Hubble radius is the approximate distance over which light can travel in the course of one expansion time, i.e., the time in which the scale factor increases by a factor of $e$. It provides a yardstick to assess whether particles can, at the given epoch, communicate within one $e$-fold of expansion.

which is more sensible.


Saying that two such observers cannot talk to each other "now" is a weird way of putting it, but here's how I understand Dodelson.

Consider two observers, A and B, separated by a distance greater than the Hubble radius, $d_H$. At some time, A emits a photon towards B. Of course, locally (relative to the expansion), the photon travels at the speed of light ($v_{pec} = c$), but since it is emitted at a distance from B greater than $d_H$, Hubble's law tells us that the photon's recession velocity relative to B is greater than the speed of light, $v_{rec} >c$. Its total velocity is $v_{tot} = v_{rec} - c$, in a direction away from B. Perhaps this is what Dodelson means by "now" -- when first emitted, the light is actually moving away from B. But, with time, since $d_H$ grows (as long as $\dot{\rho} <0$), eventually the Hubble radius "catches up" to the photon, at which point it has zero velocity relative to B. Once $d_H$ overtakes the photon, it begins moving towards B, faster and faster until $v_{tot} = c$ in B's local frame.

More generally, the rate of change of the recession velocity is easily found from Hubble's law to be $\dot{v}_{rec}=-H^2 q$, where $q = -1+\dot{H}/H^2$ is the deceleration parameter. As long as the universe is decelerating (such that $q>0$) all observers, however distant, will be able to exchange light signals in the future.


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