We know that hubble expansion increases the distance between points in space, and that the cosmological event horizon represents the distance from the observer at which objects are receding faster than the speed of light. We also know that objects at a given proper distance recede over time, and thus traveling toward them will result in the traveler arriving at a point further away than the object's original distance due to the expansion which occurred in the time it took for the traveler to catch up to it.

So is there an equation by which one can know (for a receding object's given proper distance in Giga-lightyears and a traveler's velocity in c) if:

  1. a traveler can arrive at an object within a finite amount of time

  2. the object will still be inside of the takeoff point's cosmological event horizon when the traveler arrives

  3. a round trip at the original approach velocity could occur within a finite amount of time

In order of importance - 2, 3, 1.

Of less pressing importance, but still of some interest to me would be an equation which could yield how long it would take a traveler to arrive at a distant object receding due to hubble expansion when traveling at some fraction of c.

  • 6
    $\begingroup$ Good question, but a small correction: The edge of the "Hubble sphere", not the event horizon, represents the distance from the observer at which objects are receding faster than the speed of light. Outside the Hubble sphere, galaxies recede at v > c, but we easily-ish see them (because light from them has been carried toward us faster than c). The cosmic event horizon is the max distance from where light has had the time to reach us the the Big Bang, and is ~3 times farther away than the Hubble sphere edge, roughly 46 Glyr vs roughly 16 Glyr. $\endgroup$
    – pela
    Commented Mar 6, 2023 at 17:20
  • $\begingroup$ An equation which could yield how long it would take a traveler to arrive at a distant object receding due to hubble expansion when traveling at some fraction of c? Answer to the current title is sphere of influence. $\endgroup$
    – Mazura
    Commented Mar 7, 2023 at 22:38
  • $\begingroup$ What are you actually Asking, please? Hubble expansion increases the distance between points in space. The cosmological event horizon represents the distance from the observer at which objects are receding at more then light speed. Objects do recede over time, so traveling toward them will result in the traveler arriving at a point further away than the object's original distance… The object will still be inside the takeoff point's cosmological event horizon when the traveler arrives $\endgroup$ Commented Mar 7, 2023 at 23:36

2 Answers 2



Let $a(t)$ be the cosmic expansion factor. Let $x$ be a comoving coordinate, so two objects (like galaxies) that are separating solely due to the expansion of the universe have a constant separation in $x$. Since comoving distances are $a$ times physical distances, light covers comoving distance $x$ at the rate $\mathrm{d}x/\mathrm{d}t=c/a(t)$, where $c$ is the speed of light. So between times $t_1$ and $t_2$, light covers comoving distance $$ \Delta x = c\int_{t_1}^{t_2}\frac{\mathrm{d}t}{a(t)}. \tag{1}\label{eq1} $$ Note that this is essentially the same as the calculation of the particle horizon, just for arbitrary starting and ending times. We can also rewrite this expression as $$ \Delta x = c\int_{a_1}^{a_2}\frac{\mathrm{d}a}{a^2 H(a)}, \tag{2}\label{eq2} $$ where $a_1=a(t_1)$ and $a_2=a(t_2)$. This form is usually more convenient because according to the concordance model of cosmology, $$ H(a) = H_0\sqrt{\Omega_m a^{-3}+\Omega_r a^{-4}+\Omega_\Lambda}, \tag{3}\label{eq3} $$ where $H_0\simeq 68~\mathrm{km/s/Mpc}$, $\Omega_m\simeq 0.31$, $\Omega_r\simeq 9.1\times 10^{-5}$, and $\Omega_\Lambda\simeq 0.69$ (e.g. Planck 2018). Here we adopt the convention that $a=1$ today. To convert between expansion factors $a$ and times $t$, you simply use $$ t = \int_0^a \frac{\mathrm{d}a^\prime}{a^\prime H(a^\prime)}. \tag{4}\label{eq4} $$

The scenario

We now have all the necessary tools to carry out the calculation you seek. You want to travel to a distant point that is receding with the Hubble flow of the expanding universe and return. This means you are traveling to and from an object that lies at a constant comoving distance. Let's say that distance is $d$. Assuming your journey starts today ($a=1$), you will arrive at the target when $a=a_d$, where $$ d = c\int_{1}^{a_d}\frac{\mathrm{d}a}{a^2 H(a)}. \tag{5}\label{eq5} $$ Here we assume you travel at essentially light speed. To make the round trip, just double the distance $d$. The corresponding time is straightforwardly evaluated using equation \eqref{eq4}.

Note that you will need to integrate and solve these expressions numerically. For sufficiently large $d$, you will find that equation \eqref{eq5} has no solution. That means the target is so far that the journey is impossible.

Analytic approximation

You can safely neglect radiation ($\Omega_r=0$) when considering the present time and the future. This lets you integrate equations \eqref{eq2} and \eqref{eq4} analytically in terms of known special functions (e.g. hypergeometric functions). However, you still have to solve them numerically.

To do things fully analytically, you want to also neglect matter ($\Omega_m=0$). Then $H(a)=\sqrt{\Omega_\Lambda}H_0\simeq (17~\text{billion years})^{-1}$, so equations \eqref{eq2} and \eqref{eq4} become $$ \Delta x = \frac{c}{H_0\sqrt{\Omega_\Lambda}} \left(\frac{1}{a_1}-\frac{1}{a_2}\right) \tag{7}\label{eq7} $$ and $$ t = t_0 + \frac{\log a}{H_0\sqrt{\Omega_\Lambda}}, \tag{8}\label{eq8} $$ respectively, where $t_0$ is the time today. It then takes time $$ \Delta t(d) = \frac{1}{H_0\sqrt{\Omega_\Lambda}}\log\left(\frac{1}{1-H_0\sqrt{\Omega_\Lambda}d/c}\right) \tag{9}\label{eq9} $$ to cover the comoving distance $d$. Notice how this time diverges as $d$ approaches $c/(H_0\sqrt{\Omega_\Lambda})\simeq 17$ billion light years. If $d\geq c/(H_0\sqrt{\Omega_\Lambda})$, then the trip is impossible.


As an example, suppose you want to travel to a galaxy 5 billion light years away. To get there, we set $d=5$ billion light years, which implies $H_0\sqrt{\Omega_\Lambda}d/c \simeq 5/17$. Thus the trip takes $$ \Delta t\simeq (17~\text{billion years})\log\left(\frac{1}{1-5/17}\right)\simeq 6~\text{billion years}. $$ To get there and return, we instead set $d=10$ billion light years, which implies $H_0\sqrt{\Omega_\Lambda}d/c \simeq 10/17$. Then the round trip takes $$ \Delta t\simeq (17~\text{billion years})\log\left(\frac{1}{1-10/17}\right)\simeq 15~\text{billion years}, $$ meaning the return trip alone takes $15-6=9$ billion years.

Final notes

These are times in the frame of a comoving observer, like (approximately) the Earth or the target. In the frame of the traveler, arbitrarily little time can pass if they travel arbitrarily close to light speed.

Also, if the traveler always moves at some fixed slower-than-light speed with respect to their immediate surroundings, then just replace $c$ with that speed. Note that this kind of motion is accelerated, since the motion of the immediate surroundings changes as you move (due to cosmic expansion). The acceleration required is typically tiny, though. To maintain a speed of $0.99c$, only about $10^{-8}~\mathrm{m/s^2}$ of acceleration is needed. Uniform $9.8~\mathrm{m/s^2}$ acceleration would maintain a speed that is slower than $c$ by about one part in $10^{21}$.

  • 1
    $\begingroup$ Thank you for the response! would it be possible to use an example value to illustrate how the equations would be used? For instance, for a distance of 1 giga light-year? $\endgroup$
    – user52978
    Commented Mar 7, 2023 at 0:08
  • 2
    $\begingroup$ @user52978 Sure, added an example (I picked a somewhat longer distance so that the result would be interesting) $\endgroup$
    – Sten
    Commented Mar 7, 2023 at 1:18

sten's answer is excellent and beautifully analytical; I just wanted to illustrate the journey and show that neglecting matter (and radiation) is a good approximation.

Spacetime diagram

To do this, I solved the full Friedmann equation numerically (well, Python's astropy did it for me) and created a spacetime diagram in comoving coordinates. A spacetime diagram show distance $d$ on the $x$ axis and time $t$ on the $y$ axis. "Comoving coordinates" means that distance is scaled with the expansion of the Universe, so that the comoving distance between objects that don't move stay the same, despite expansion. Furthermore, I scaled time such that one year on the $y$ axis is the same length in the plot as one lightyear on the $x$ axis (this is called conformal time).

A detailed explanation follows below the diagram.


The blue lines

A point in a spacetime diagram consists of a position and a time, together called "an event". The path that something (a galaxy, a spaceship, a light ray, or whatever) takes through space and time is called its worldline. If you don't travel, you will follow a vertical line; "ours" is the blue line called "Here". If you do travel, you will deviate from vertical. Because I show distance in comoving giga-lightyears (Glyr) and time in giga-years (Gyr), and because I set the aspect ratio of the plot to [Universe-size]:[Universe-age], light rays travel on lines of 45º, so this is the maximum deviation from vertical you can do. In general, the worldline of an object moving with velocity $v$ wrt. a comoving ("stationary") object will have an angle $\theta=\tan^{-1}(v/c)$, where $c$ is the speed of light.

At any point in time, everything that happens, happens on a horizontal line. The blue horizontal line called "Now" is what happens today, so Right here, right now is where the blue lines cross.

The green line

At any point in time, light that has been able to reach us since the Big Bang must come from the region bounded by the green lines, the particle horizon (PH). Space inside the PH is called the observable Universe. Today, the PH is some 46 Glyr away, and in the future it will increase to ~63 Glyr. Remember, this is in comoving coordinates. In physical coordinates the distance will increase without bounds, but this means that we will never be able to see anything that is currently farther aways than 63 Glyr.

The red and orange lines

Everything that we see consists of events lying on the "surface" of our past light cone, given by the red line, and everything that we have ever been able to see must have been inside that, marked by the orange region. In the future, the spacetime volume of our past will increase, asymptotically reaching the event horizon (EH), marked by the orange line.

At the moment, the distance to the EH is ~16.6 Glyr. In comoving coordinates this distance will decrease. In other words, even though our observable Universe increases (because light from increasingly distant regions reaches us), the region from which we can receive information send today, decreases. In the far future, we therefore won't be able to send signals to / receive signals from any galaxies outside our local, gravitationally bound local group of galaxies.

The purple lines

The recession speed $v_\mathrm{rec}$ of galaxies increases linearly with distance. That's Hubble's law for you, $v_\mathrm{rec} = H_0 d$, with the Hubble constant $H_0$ being the constant of proportionality. This means that all galaxies inside a region of radius $d_\mathrm{H} \equiv c/H_0$ recede slower that the speed of light, while everything outside recedes faster. The region is called the Hubble sphere, and really has no particular meaning. Farther away, galaxies recede at several times $c$, marked by the purple lines.

The journey from sten's example

Assuming that you can travel arbitrarily close to the speed of light (you can't because you'd get blasted by the near-speed-of-light wind of interstellar and -galactic particles, but that's a minor detail), a journey to "There and back again" will look like an "isosceles right triangle" (45º-45º-90º) with one corner at "Here & Now", apex at "There & in the future", and the last corner at "Here & even farther future". In the spacetime diagram I illustrated the journey from sten's example with a black, dashed line that has its apex at 5 Glyr. The upper corner returns to the Milky Way at $t=29.6\,\mathrm{Gyr}$, i.e. $15.8\,\mathrm{Gyr}$ from now (since the current age of the Universe is $13.8\,\mathrm{Gyr}$). This is slightly faster than sten's $17\,\mathrm{Gyr}$, and the difference is due to my calculations taking matter into account which retards the expansion somewhat for a little while yet (and possibly due to the cosmology used; I used a Planck 2018 cosmology).

Here's a zoom-in of the journey. The thinner, green, dashed line shows a similar journey, but with $v=0.9c$, so the angle with Milky Way's worldline is $\theta=42^\circ$. In this case, to go to the same galaxy 5 Glyr away and back will take longer 24 Glyr.


Maximum distance

The most distant galaxy you can visit if you want to return is found by extending the dashed line all the way to the EH (the orange line). They cross at $d=8.3\,\mathrm{Glyr}$, and this happens $t=11.8\,\mathrm{Gyr}$ from now. The journey back again then takes almost infinitely long (by assumption) and there will be no-one left to tell about your journey, so you might as well skip the return.

If you do skip the return, you might as well try and reach an even more distant galaxy. The most distant galaxy you can visit is one which is at a comoving distance equal to the end of our future light cone, i.e. where the dashed red line hit the top of the diagram. By symmetry, this galaxy is currently located at our EH, i.e. where the horizontal blue line crosses the orange line. This distance is $16.6\,\mathrm{Glyr}$.

  • 3
    $\begingroup$ I expected the Right here right now to go elsewhere $\endgroup$
    – Mormegil
    Commented Mar 8, 2023 at 12:59
  • 1
    $\begingroup$ But to the point: You write “At the moment, the distance to the event horizon is some 63 Glyr” but I see the orange event horizon line cross the blue now line at the lower-mentioned 16.6 Glyr? So what other “distance to the event horizon” is meant? $\endgroup$
    – Mormegil
    Commented Mar 8, 2023 at 13:12
  • 2
    $\begingroup$ @Mormegil Agreed, that's a better reference :D And agreed: What I wrote was nonsense, I mixed up some lines and colors. I edited the text now, but might rephrase later, because I don't think it's so clearly written. Thanks! $\endgroup$
    – pela
    Commented Mar 8, 2023 at 16:03
  • 2
    $\begingroup$ @user52978 The angle of your worldline (i.e. your path through spacetime) in this diagram where light travels on 45º is given by $\tan^{-1}(v/c)$. So as your speed $v$ decreases, the (horizontal) "height" of the black dashed triangle decreases. So the difference in max. dist. compared to going at light speed is equal to the difference in height of the triangle with 45º degree corners, compared to a triangle with, say ~43.5º and 42º for $v=0.95c$ and $v=0.9c$, respectively. If I've done my trigonometry right, you'd reach a distance that's 97.4% and 94.6% of the light speed max, respectively. $\endgroup$
    – pela
    Commented Mar 8, 2023 at 22:48
  • 2
    $\begingroup$ @user52978 I edited the answer to show a subluminal journey :) $\endgroup$
    – pela
    Commented Mar 9, 2023 at 10:07

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