Transit time for a round trip to alpha centuri..

[oracle99]

TL;DR

A simple relativity question.

Just a simple little question as my first post here..

Suppose you have a space craft that has an engine that can exert an acceleration of 1G. You fly this to Alpha Centauri accelerating, then halfway turn around and decelerate at 1G, so you have gravity for the whole trip.
Once you arrive, you turn around and fly back home the same way.

How long would appear to elapse for the traveler, and how long would appear to elapse for people on Earth? What is the formulea for different accelerations?

 

[Ibix]

Using Earth time and starting from rest, distance travelled in time $t$ at constant acceleration $a$ is $s=\frac 1a(\sqrt{1+a^2t^2/c^2}-1)$.

Note that if you use light years for distance and years for time then $c=1$ and 1g is within 5% of 1 light year per year squared, so $s\approx\sqrt{1+t^2}-1$ and each two light year phase of the journey takes about $\sqrt{8}\approx 2.8$ years. So 5.7 years all the way there, and 11.3 for the round trip.

Ship board time, $\tau$, is related to Earth time, $t$, by $\tau=\frac ca\sinh^{-1}(at/c)$. Again setting $c=1$ and $a\approx 1$, $\sinh^{-1}(2.8)\approx 1.76$ for each two light year phase, so 3.5 years to Alpha Centauri and 7 years for the round trip.

 

[FactChecker]

Using Earth time and starting from rest, distance travelled in time $t$ at constant acceleration $a$ is $s=\frac 1a(\sqrt{1+a^2t^2/c^2}-1)$.

I assume the acceleration here is measured in the spaceship time and distance, since the OP wants it to feel like Earth's 1g to the travelers.

 

[Ibix]

I assume the acceleration here is measured in the spaceship time and distance, since the OP wants it to feel like Earth's 1g to the travelers.

Yeah, $a$ is proper acceleration here.

 

[Ibix]

You can't maintain a Minkowski coordinate acceleration of 1g for more than a year(ish) anyway, even in principle, since you'd reach lightspeed at that point.

 

[oracle99]

So if you travel 3 years you get
sqrt(3*3+1)-1= 2.16 light years, about half way to Alpha Centauri, so decelerate is the same, so approx 6 years out, 6 years back, 12 years, from the travelers perspective?
oh I see you updated it, that would be earth dwellers time. Thanks!

Another little point, there is often discussion about traveling faster than the speed of light - you can, but the people back home won't see it that way!

 

[phyzguy]

Another little point, there is often discussion about traveling faster than the speed of light - you can, but the people back home won't see it that way!

No, you can't.

 

[PeterDonis]

you can, but the people back home won't see it that way!

You can't outrun a light ray that's moving in the same direction as you are. That's not a matter of who is "seeing" it; it's an invariant fact.

There are ways to calculate a "speed" that will give a result greater than $c$, but given the above fact, describing any such calculation as showing that anything is "moving faster than light" is problematic.