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How long would it take to reach the current edge of the reachable portion of the universe, with the following bounds in mind:

I would like to know the time the traveler experiences, and the time that everyone else (e.g. those on earth) experience.

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  • $\begingroup$ I think this is probably more suited to the space.stackexchange.com forum anyway, but I dont think its feasible to say it would need to accelerate for the whole journey (+/-) because the speed of light is finite and you cant accelerate past it. $\endgroup$ – Dean Apr 16 '16 at 14:51
  • $\begingroup$ I realize that the observer would not go past the speed of light, but length contraction and time dilation takes place at near relativistic speeds. Relativity is weird, but amazing! I wonder if it could be moved to the space.stackexchange.com if it is deemed necessary? $\endgroup$ – Jonathan Apr 16 '16 at 15:16
  • $\begingroup$ Since this journey is completely infeasible, but can be answered with physics for the hypothetical case, I think it should stay here. $\endgroup$ – pela Apr 16 '16 at 16:54
  • $\begingroup$ I think Pela provided a better refined answer, feel free to look at the implications I propose in the end of my answer as well, I think they are well worth taking a look at. $\endgroup$ – Jonathan Apr 16 '16 at 17:48
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Jonathan's answer is essentially correct, but as Rob Jeffries comments, he doesn't take into account that the Universe is expanding during the journey.

The edge of the observable Universe is 47 billion lightyears (Gly) away. Even if you are a lightbeam, you cannot reach that point. The farthest you can go if departing today is roughly 5 Gpc, or 17 Gly, but this journey would of course take infinitly long (or else it wouldn't be "the farthest you can go"). This distance is probably what the linked article is referring to (I didn't read the article; it's very, very long).

So, in order for the answer to be any fun, you have to freeze the Universe, using magic, which is what Jonathan's calculator is doing. Here I'll just provide the analytical solution: In that case, the proper time $\tau$ (i.e. the time experienced by the traveler) to reach a distance $x$ when traveling at a constant acceleration $a$ is $$ \tau = \frac{c}{a} \cosh^{-1} \left( \frac{ax}{c^2} +1 \right), $$ where $c$ is the speed of light. If you wish to decelerate after having reached halfway, you just divide $x$ by $2$ and multiply the result by $2$.

If you plug in the $x=15\,\mathrm{Gly}$ you request, you get roughly 45 years. To get to the edge of the Universe at 47 Gly actually only takes a few years more. The reason for this is simply that traveling at 1G gets you to (almost) the speed of light in only a couple of years, and hence you experience (almost) no time, no matter how far you go.

The time experienced for the Earthlings for the traveler at constant acceleration is given by $$ t(\tau) = \frac{c}{a} \sinh \left( \frac{a \tau}{c} \right), $$ which works out to 15 Gyr for the 15 Gly, and 47 Gyr for the observable Universe. The reason is simply that the traveler, from the point of view of the Earthlings, extremely fast reaches a speed which is almost the speed of light.

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  • $\begingroup$ Thanks for providing the formula, revision, and for a well thought out answer! Could you please specify what all the variables are in the formula? $\endgroup$ – Jonathan Apr 16 '16 at 17:18
  • $\begingroup$ @Jonathan: I think it's there already: "…the proper time $\tau$ to reach a distance $x$ when traveling at $a$". Or which variables are you thinking of? $\endgroup$ – pela Apr 16 '16 at 17:33
  • $\begingroup$ ah yes, I don't know how I missed that (unless it was edited in). Thank you! $\endgroup$ – Jonathan Apr 16 '16 at 17:35
  • $\begingroup$ So how does it work out if you do consider the expansion of space time. As your approach C length contracts and at the same time space time is expanding. Does the traveler therefore observe a greatly increased expansion rate one which dominates the length contraction he is also observing. Preventing him from reaching the edge of the observable universe? The implications of this is that speeds are not relative and that there is actually a universal reference frame which you could discover by adjusting your speed until you minimize the observed expansion rate of the universe. This seem wrong. $\endgroup$ – trampster Feb 20 '18 at 1:10
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    $\begingroup$ If instead the expansion rate is the same to all observers regardless of velocity then the universe wont expand much in those 45 years from his point of view and therefore he can make it well past the edge of the observable universe. $\endgroup$ – trampster Feb 20 '18 at 1:14
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I welcome everyone else's input on both this answer and possible different solutions! This is what I came up with:

First off, providing a 1G thrust the whole way (switching to decelerating half way there) would be a great way to provide "artificial gravity". I found a site that actually has scripts to help with such calculations! Assuming the site is correct, here is what we come up with: http://www.cthreepo.com/lab/math1/ Using the Long Relativistic Journeys calculation:

  • Enter acceleration: 1G
  • Enter light years to travel: 15000000000
  • Time experienced by observer: Infinity (incorrect, and the site points out that due to a limitation in the java script, such large numbers do not calculate properly)
  • Time experienced on Earth: 15010384285.855611

However, I noticed something interesting that might still get us there:

  • For a trip of 15 light years, it shows the observer experiences about 5.54 years, and the earth time experienced is about 16.84 years.

  • For a trip of 150 light years, it shows 9.80 years for the observer, and 152.03 years for earth

  • For a trip of 1500 light years, it shows 14.24 years for the observer, and 1502.97 years for earth

  • For a trip of 15000 light years, it shows 18.70 years for the observer, and 15012.32 years for earth

I am noticing a pattern here... The time for the observer increases by about 4.5 years every time the distance is multiplied by 10. The time experienced by earth grows closer to the number of light years as distance is increased too. Now, let's see where the calculation breaks down, and see if this pattern is still happening before that point.

  • It breaks down at 1.5 billion light years, so lets take a few steps back and see if we still see the pattern.

  • For a trip of 1,500,000 light years, it shows 27.63 years for the observer, and 1501040.36 years for earth. Note that commas were added to easier see what number we are talking about, when entering numbers into the calculation, don't use commas.

  • For a trip of 15,000,000 light years, it shows 32.10 years for the observer, and 15010386.22 years for earth.

  • For a trip of 150,000,000 light years, it shows 35.61 years for the observer, and 150103844.77 years for earth.

Yes, the pattern still largely holds. It seems be about 3.5 - 4.5 years for each additional factor of 10 for distance (and seems to be getting less time "near the breaking point"). Let's assume based on this that each factor of 10 increase produces an approximate 4 year increase on the observer's time, and that the time experienced on earth is very close to the number of light years travelled.

  • For a trip of 1.5 billion light years, this would mean 39.61 years experienced for the observer, and slightly over 1.5 billion years experienced on earth.

  • Now for the big one and our answer, a 15 billion light year trip should take about 43.61 years for the observer, while slightly over 15 billion years would be experienced on earth!

This has some astonishing implications:

  • With such a ship, we could theoretically travel to the end of the reachable universe within a human lifetime (though 15 billion years would have passed on earth)!
  • We could also theoretically reach any point within this area in less time (so we could travel practically anywhere, to a neighboring galaxy, to a galaxy off in the distance, to the center of our galaxy, etc...) within a human lifetime (in less than 44 years)!
  • We could time travel into the distant future by going out somewhere very distant at relativistic speeds, then coming back to earth! Note: going back in time still hasn't been figured out yet as far as we know. For large distances, the time travelled would be slightly more than double the distance in light years travelled (accounting for the return trip).
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  • $\begingroup$ This appears to be a standard twin-paradox solution; but how have you accounted for the expansion of the universe? $\endgroup$ – Rob Jeffries Apr 16 '16 at 16:04
  • $\begingroup$ Good point, I did not account for the expansion of the universe taking place here. That's why I am looking for additional feedback. $\endgroup$ – Jonathan Apr 16 '16 at 16:54
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    $\begingroup$ @RobJeffries and Jonathan: But if you take into account expansion, the answer is almost given in the question: You can go no farther than 5 Gpc, which would take $\infty$ as seen from the Earthlings, and only a few years as seen from the traveler, since she very quickly reaches (almost) the speed of light and hence experiences (almost) no time to go anywhere. $\endgroup$ – pela Apr 16 '16 at 17:00
  • $\begingroup$ @pela, Does this imply that in 15 billion years (roughly 3 * 5 Gpc = 15 G light years), something like a big rip is going to happen to the universe? $\endgroup$ – Jonathan Apr 16 '16 at 17:10
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    $\begingroup$ @Jonathan: No, these calculations assume a cosmological constant, i.e. no big rip. But the acceleration still means that the recession velocity of any object not gravitationally bound to us will increase without bounds, eventually making them unreachable. $\endgroup$ – pela Apr 16 '16 at 17:29

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