I have a question which I believe will greatly help my understanding of the Hubble sphere and what it means physically.

Say we have a flat ($k=0$) Universe with no dark energy ($\Lambda=0$) during the matter-dominated era such that the first Friedmann equation is

\begin{equation} \Big(\frac{\dot{a}}{a}\Big)^2 = \frac{8\pi G \rho_m}{3}, \end{equation}

where $\rho_m$ is the energy density of a perfect fluid and $a$ is the scale radius of the Universe.

If I send out a photon today, at time $t_0$ when the Hubble constant has value $H_0$ and the scale radius has value $a_0$, what is the maximum physical distance over which I can 'communicate' using this photon?

That is, if an observer at this distance sends me back a photon immediately after receiving my photon, what is the maximum physical distance between me and the observer which will allow their photon to reach me at some finite time?

It would also be great to have a comparison between this case and the case for an accelerating Universe, highlighting the major differences.


1 Answer 1


After answering several similar questions on Physics SE I have realised I can answer this question myself using the concepts of the (comoving) Particle Horizon, the (comoving) Event Horizon and the Comoving Hubble Sphere.

If I want to know the maximum distance I can communicate in an any amount of future time, then I should consider the Event Horizon, which gives the greatest comoving distance from which an observer at time $t_2$ can receive a signal emitted at time $t_1$, as

\begin{equation} \chi_{\text{ev}} = \int^{t_2}_{t_1} \frac{dt}{a(t)}. \end{equation}

This is the distance at which the rate of expansion of the emitter at $t_1$ is just under the speed of light $c$, so that the photon can $just$ 'outrun' the Universe's expansion at this point.

I must then equate this to the maximum comoving distance I can send a photon between times $t_0$ and $t_1$ which is given by the exactly analogous concept of the Particle Horizon

\begin{equation} \chi_{\text{ph}} = \int^{t_1}_{t_0} \frac{dt}{a(t)}, \end{equation}

which gives the maximum comoving distance over which the observer at $t_1$ will be able to receive my photon. Completing the example for the matter-dominated era as in my question gives (cancelling multiplicative constants)

\begin{equation} \int^{t_1}_{t_0} t^{-2/3} \; dt = \int^{t_2}_{t_1} t^{-2/3} \; dt \end{equation}

\begin{equation} \implies t_2^{1/3} = 2t_1^{1/3}-t_0^{1/3}. \end{equation}

Assuming we know the time $t_0$ we sent out the original signal and the deadline $t_2$ by which we want to get it back, we can then calculate $t_1$ and hence the comoving distance $\chi_{ph}$ to which we may send the signal.

Accelerating Universe: In the above example for the MD era (and in fact also for the RD era) we can put $t_2 \rightarrow \infty$ and see that given an infinite amount of future time we could communicate to any distance. But in an accelerating Universe there is a finite limit even when we have infinite time as I mentioned at the end of my answer to this question. During the $\Lambda$-dominated era this would be given by

\begin{equation} \int^{t_1}_{t_0} e^{-Ht} \; dt = \int^{\infty}_{t_1} e^{-Ht} \; dt \end{equation}

\begin{equation} \implies t_1 = t_0 + \ln{2}, \end{equation}

if we assume $H \sim$ const. (note that this is not a good assumption, but is included so I can solve analytically for illustrative purposes).

Want to get the signal back within the next Hubble time: In this case we should replace the Particle and Event Horizons by Comoving Hubble Spheres for me and the other observer. This is the distance over which comoving particles can travel in the course of one Hubble time.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.