# How far can you travel until you can't get back to where you started?

I searched for this question but couldn't find it, so I'm asking. Lets just ignore turning into a liquid at high accelerations and the fact that an object with mass cannot travel at the speed of light (I say that last one because as your velocity approaches the speed of light, we could just say it is the speed of light, and then just say its less than the answer you get from the math). So since the distances in the universe change over time, and the velocity it takes to travel those distances also changes, there exists places that are impossible to reach, as it would require faster than light travel. My question then becomes, if I intended to return to my starting point, how far could I travel until returning would become impossible?

• The speed of light is the same for all observers regardless of their velocity relative to another's. The speed of light is 299,792,458 meters per second. If you are traveling at, say, 299,792,457 m/s (99.99% c), you will still observe a beam of light travel at 299,792,458 m/s. – BillDOe Dec 27 '17 at 20:10
• You're asking about cosmic expansion and how the size of the observable universe affects round-trips, right? – Nathan Tuggy Dec 27 '17 at 20:47
• This is cool question. I don't have the chops to figure it out, but at some distance from Earth, the expansion of the Universe is faster than the speed of light, so you can't back. It would have to be billions of light years, because we can see stuff out that far, so light from there can reach us. – zeta-band Dec 27 '17 at 21:29
• This turns out to be a very interesting question, although I'm not sure it is on topic here. There is an excellent answer (somewhere) in Physics stackexchange explaining that the expansion of space is so fast and distances are so large that light from some parts will never, ever be able to reach other parts. There are in fact situations where the Maine humor adage You can't get there from here applies (as does this). So if light can't do it, the rocket can't either! – uhoh Dec 27 '17 at 21:37
• I believe the seed for this starts in the extended discussion in comments below this answer to the question Am I attracting Pluto? However I'll keep looking for the reference to the rate of expansion and distant points. – uhoh Dec 27 '17 at 21:51

In the expanding universe the interval is, $$ds^2=c^2 dt^2-a^2(t) d\vec{x}^2$$

The interval along lightlike trajectory is zero, so from that we get, $$\frac{dx}{dt}=\frac{c}{a(t)},\quad \Delta x(t_0,t_1)=c\int_{t_0}^{t_1}\frac{dt}{a(t)}$$ If you as usually done choose that $a$ at present time ($t=0$) is equal to $1$. the coordinate $x$ will tell you how far this is at the present. To learn how far it is at $t=t_1$ you should multiply it on $a(t_1)$.

Now the equation above require you to know how exactly scale factor $a(t)$ will evolve in the future which is determined by the matter. Luckily the problem is much simplified as we entered the dark energy dominated era with dark energy most likely being just a cosmological constant. We then assume that the Hubble parameter is approximately constant (actually it will gradually drop by about 20% as matter will disperse and its contribution will decrease), $$H\equiv\frac{1}{t_H}\equiv\frac{\dot{a}}{a}\simeq \mathrm{const},\Rightarrow a(t)=e^{t/t_H}$$ With $t_H\approx 14.4 Gy$ (where Gy stands for gigayear i.e. billion years) Then, $$\Delta x(t_0,t_1)=ct_H\Big(e^{-t_0/t_H}-e^{-t_1/t_H}\Big)$$ Note that if we take $t_0=0$ (i.e. shoot the light pulse now) the fartherst point it may get is $c t_H$. This is the cosmological event horizon - anything further can't be reached because of the universe expansion. The point of no return for light is defined in such a way that when light reaches it the point of departure is exactly on its cosmologial event horizon.

In general case the cosmological horizon depends on $t_0$. However when $H=\mathrm{const}$ we get highly symmetric de Sitter spacetime for which $a(t+\Delta t)=a(t)a(\Delta t)$ and $a(t_0)\Delta x(t_0,t_1)=\Delta x(0,t_1-t_0)$. So to find the point of no return we simply take, $$a(t_{NR})\Delta x(0,t_{NR})=ct_H$$ This gives us $t_{NR}=t_H\ln{2}\approx 10 Gy$. At the present this point is at $\Delta x=\frac{ct_H}{2}\approx 7.2 Gly$ ($7.2$ billion light-years) from us.

To put those numbers into perspective the age of the universe is 13.8 Gy, the radius of the observable universe is 45.7 Gly.

UPD: By the way there's a very simple way to understand that for any FRLW cosmology with event horizon the sphere of no return will always have the radius equal to half of the horizon radius. To do so you need to use conformal time, $$\tau=\int \frac{dt}{a(t)},\quad ds^2=a^2(\tau)\Big(c^2 d\tau^2-d\vec{x}^2\Big)$$ The nice thing about the conformal time is that lightcones look very simple, just like in the flat spacetime $\Delta x=c \Delta \tau$. But how cosmological horizons then appear, what prevents lightcones from intersecting with some wordline $x=\mathrm{const}$? This happens when $a$ grows sufficiently fast, then $\tau$ happens to be restricted by some finite $\tau_{max}$ where $a\rightarrow+\infty$. So for lightcones this looks just like a Minkowski spacetime where you cut off everything after $\tau_{max}$. If the intersection lies exactly at $\tau_{max}$ this is a horizon, if it happens after that then no signal can pass.

From that it's obvious that the cosmological horizon for $x=0$ is the past lightcone originating at $\tau_{max}$ i.e. $|\vec{x}_H|=c(\tau_{max}-\tau)$ To find the point of no return we send lightsignal $x=c(\tau-\tau_0)$ and look where it intersects with the horizon lightcone. It's pretty obvious that the intersection will happen at $x_{NR}=x_H/2=\frac{c}{2}(\tau_{max}-\tau_0)$. For better understanding I add the illustration. Again I remind you that this is true for any $a(t)$ growing sufficiently fast in time.

• Can you put 7.2 Gly in perspective somehow? How does that compare to the present age of the universe for example, or the present "size" of the universe? This isn't a distance unit most people work with on a daily basis! :-) – uhoh Jan 2 '18 at 15:07
• @uhoh: A little more than half the size of the observable universe. Actually, once you reach that point you'll be the Hubble Radius (14GLy) away from Earth; a bit more than the Observable Universe radius - thanks to space expansion providing about half of the distance covered (plain old movement through space being the other half). - the point that NOW is 7.2GLy away will be at Hubble Radius then. Then, to return, you must begin the arduous climb against the slippage of space expansion to return eventually, in an infinitely distant future. 7.2GLy there, $\infty$ GLy back. – SF. Jan 2 '18 at 15:26
• @S.F. I'm afraid you underestimate the size of the observable universe. It's much larger than 13.8 Gly again thanks to the expansion of the universe. – OON Jan 2 '18 at 15:49
• @OON Just clarifying, so the non-returnable distance is 7.2 Gly or half that distance? As in, if i went 7.2 giga-light-years away, can i come back, or would I have to go half that distance in order to travel the other half back? – Terran Jan 2 '18 at 16:59
• ^ And a final clarification: the e^ - e^ equation when halved is based on the hyperbolic sin but diverges with a same signed exponent because of the two time variables or due to some other reason? I recognized the equation from hyperbolic geometry and thought I would ask :) – Terran Jan 2 '18 at 17:16

Nobody knows for sure.

Recent WMAP experiments suggest a planar Universe in this, the answer for your question is infinite. If the Universe has a spherical space-like geometry, its radius seems to be at least 300billion light years.

Note, this is only a theoretical thing, because even if you are going with $c$, the Universe expands faster. It is not known, what will happen to it on the long term. Currently an accelerating expansion is visible, but also decelerating expansion and exponential acceleration were already happened.

I'm very much not an expert on the subject, but my understanding of the problem is this:

The universe, to our knowledge expands at a steady rate of 0.007% per million years. All of it.

Doesn't sound like much. but ask a financier about the value of compound interest.. essentially, the more money you have, the better 0.007% sounds. If I have 7000 dollarpounds in my bank account, that'll be an extra dollarpound per million years. But if I have \$£7,000,000, then I'll have an extra 1000 dollarpounds.

So, if it's 2.5 million lightyears from here to Andromeda. then in a million years that distance will have grown by 17,500 lightyears. (ignoring that andromeda is in reality moving towards us under the effects of mutual gravity)

Measure a wide enough region and the overall growth of that region per million years will exceed a million lightyears at a time.

That's your limit. you can't go faster than light, but the universe's expansion is not limited by that. so eventually every body in the universe will pass that limit relative to one another and be isolated. forever.

By my math, Assuming you can travel at the speed of light, the furthest away an object can be and still be reachable at all is 142,857,142.857 Lightyears away.

Anything further away than that is accelerating away faster than you can ever catch up to.

Provided that your destination will still be within 142.86 million lightyears of your starting point when you get there, you should still be able to get home after the 284 million year round-trip.

This is however thrown off kilter by things like your own acceleration, I'm assuming you can accelerate instantly and decelerate just as fast.

What worries me is that I've come to a substantially smaller conclusion to all these much more knowledgable people who are also answering the question. I think I'm missing something important.

• I would guess that it has something to do with spacetime, how velocity and acceleration is impacted from multiple observers over the course of some amount of time. So maybe it requires that math you did combined with Lorentz transformations or something... I think you and I are in the same boat on this one, while I do believe that the answer I chose is the correct one (it has the math I have seen before), just understanding it at a fundamental level seems to be the disconnect... – Terran Jan 4 '18 at 20:52
• Personally I am getting close, and I would like to share with you what I have done... I worked backwards from time dilation to get escape velocity (by accident), then from that I got Lorentz transformations and rapidity. Now I am working on minkowski space... So I guess a good start is to look at time dilation equations, and I at least saw a pattern, it seems that they can be rewritten to "ratio"=sin^-1(v/c), and time dilated=time initial * cos(ratio)^-1 ... and just work backwards? Thats what I did at least :/ btw it has something to do with hyperbolic geometry... – Terran Jan 4 '18 at 20:58
• Surely time-dilation only actually matters if you care about how long the subjective flight time is going to take.Your ship will take X time to get where it's going, but the crew will experience a much shorter period at higher velocities. I'm not sure how time-dilation will affect the journey itself.. it presumably still takes the same objective time to get there. we're already looking at an Inertialess Drive with a limit of the speed of light. It might actually be easier to model it as a Laser of infinite accuracy. Does light suffer from time-dilation? – Ruadhan2300 Jan 5 '18 at 15:55
• I would like to say that it does, but you have many perspectives of said time dilation... From the light's perspective there is no time, the time perceived rate is infinite. From anyone else's perspective of the light, its time rate should be zero. The reason I was setting the maximum limits in my question (light speed, no acceleration or deceleration) is because that would encompass all possible answers (the maximum answer), if anything was less it would fall under the answer..... wait.... it just hit me... If you traveled at the speed of light... there is no time. I'm gonna ask OON. lol – Terran Jan 5 '18 at 17:13