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Re: Relative Velocity Time Dilation:

Pioneers have travelled 4.2 light years to Alpha Centauri Cb in 9 months. Handwavium tech can transport mass in this timeframe. Upon their arrival to Cb, they are 9 months older, but how much time has passed on Earth?

Can you math-heads give me a rough idea of what this calculation would be?

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    $\begingroup$ There seems to be two questions. Can you edit out one, then ask it in its own thread. (The 3 weeks "massless transmission" would seem to be up to you anyhow as it breaks the speed of light and thus known physics too, but why wouldn't it be the same both-ways?) $\endgroup$ Commented May 18 at 18:14
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    $\begingroup$ if this is within real physics, the first one is a straight up time dilation calculation. the second one is not how physics works. if it's a fictional ftl tech, then you decide the rules. $\endgroup$
    – ths
    Commented May 18 at 18:18
  • $\begingroup$ Ty @Escapeddentalpatient. - "Mass Transportation" also breaks the speed of light, but given my lack of physics knowledge I haven't the slightest idea how to go about even loosely applied basic laws to influence the theory of this. $\endgroup$ Commented May 18 at 19:14
  • $\begingroup$ @Escapeddentalpatient. I wonder, am I understanding this correctly that it would be a dilation/difference between what the "data" experienced in time passage, versus the static observers of the sender or the recipient? And seeing how the "data" doesn't age, then it doesn't really matter what the dilation is? $\endgroup$ Commented May 18 at 19:18
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    $\begingroup$ Regarding the faster-than light information transmission: That's a matter for the next question, but, no-one can answer that except you - since it breaks known physics. You're making the rules up about how it works. We can certainly help you figure out consistent rules, but you'd need to specify them in the first place. $\endgroup$ Commented May 18 at 19:48

3 Answers 3

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A little bit more than 4.2 years. For any large time dilation factor, the astronauts spend most of the journey very close to the speed of light in the Earth frame, so you don't need to do a hard calculation unless you care about getting several decimal places of accuracy. Just use the amount of time it takes for light to get there, add a few percent, and mention that the course isn't a perfectly straight line to assuage any especially pedantic readers.

However, it can't be done the way you want with any technology permitted by real science if we further constrain the problem to keep regular humans alive during the trip. If you want squishy humans onboard to stay to stay alive, there are maximum acceleration limits, and you're way past them.

For ship time $t'_\text{max}$, ship observer measuring constant acceleration $a$, earth time $t_\text{max}$, with a flip halfway to the target point so you leave Earth at 0 and arrive at the target point at 0, speed of light $c$, total distance $X$

$t'_\text{max} = 2\frac{c}{a} \sinh^{-1}(\frac{a}{ct_\text{max}}) = 2\frac{c}{a} \sinh^{-1}(\frac{a}{c} \sqrt{\frac{(0.5Χ)^2}{c^2} + \frac{Χ}{a}})$

(the 2 is for the flip, we're covering one distance twice in a row)

The total distance measured by an observer on Earth (or Alpha Centauri) from Earth to the target point is related to time by the ship's velocity as measured by Earth and Earth's ordinary time, but the coordinate acceleration measured by Earth is not constant.

Plugging in a few test values I see that to get $t'_\text{max} = .75 yr$ with $X = 4.2 LY$ we need $a \approx 14g$, which will kill everyone onboard within a few minutes.

Referencing this article by Philip Gibbs.

Try deciding how big of an acceleration you want your futuristic medical tech to be able to help the astronauts survive, then plugging in $a$ and $X$ to get the ship time at arrival.

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    $\begingroup$ With suitable equipment available on Earth right now, humans can easily sustain 14g, I answered a question on max acceleration here a while ago: worldbuilding.stackexchange.com/questions/247019/… $\endgroup$
    – quarague
    Commented May 19 at 15:18
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    $\begingroup$ @Fattie Well, the ESA (and the physics stack exchange also linked in my answer) disagrees with you. You will not be squished against the tank wall, this does work. $\endgroup$
    – quarague
    Commented May 19 at 15:44
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    $\begingroup$ @Fattie If the fluid has the same density as the person, there will be no buoyant force at any acceleration, although there will be internal forces in the human because of the human's variable density (especially in the sinuses and lungs, since the difference in density between air and body is much bigger than the difference between fat and bone). $\endgroup$
    – g s
    Commented May 19 at 15:57
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    $\begingroup$ You can do this experiment. Make some thin simple syrup by adding a heap of sugar to scalding hot water. Get a clear jug and fill it most of the way with lukewarm water. Put a grape in the jug. Add syrup until the grape begins to float. Fill the bottle the rest of the way up with water until it starts to spill, then lid the bottle. Hold the jug up so that you can observe the motion of the grape, then violently swing it back and forth as hard as you can. $\endgroup$
    – g s
    Commented May 19 at 16:00
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    $\begingroup$ @Fattie If you had perfectly matching densities you would not notice arbitrarily high densities. Now humans don't have perfectly homogenous densities but their density is approximately that of water, so it partially works, much better than humans in air. The ESA made experients to figure out how well it works and their result was humans in water with air breathing can sustain up to 24g. $\endgroup$
    – quarague
    Commented May 20 at 6:04
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I am not sure if I follow other answers. Let's ignore time dilation before we hit the desired travel speed for simplicity.

At 3g acceleration and deceleration we can reach the speed of light in 4 months. We would also travel for 2 light months of distance during that time. We also need to decelerate to 0 which will take another 4 months and will cover 2 light months. We are left with ~46 months to cover at close to speed of light so that it would take a month for the travelers. This roughly translates to 0.999765c. So in total, roughly 4.5 years.

You could decrease that to 2.6g if you are willing to hit almost 1c. But I am guessing it would be much more difficult to do that.

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    $\begingroup$ This is a great answer. $\endgroup$
    – Fattie
    Commented May 19 at 15:38
  • $\begingroup$ If you boost at 3g (ship frame proper acceleration) for 4 months (Earth frame) you end up at about 0.7c and the journey takes about 3 years. If you boost at 3g (Earth frame coordinate acceleration) for 4 months, you flatten yourself with hundreds of gravities of proper acceleration along the way. $\endgroup$
    – g s
    Commented May 19 at 21:40
  • $\begingroup$ If you want to read up, you're looking for the difference between velocity and rapidity. $\endgroup$
    – g s
    Commented May 19 at 21:51
  • $\begingroup$ So I shouldn't ignore relativity, but since this is not even science based, I am guessing this answer could be used to cheat it. $\endgroup$ Commented May 20 at 13:50
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    $\begingroup$ @reiterwriter 4.5 for Earth observer. 9 months for those in the ship. $\endgroup$ Commented May 21 at 6:21
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Assuming you have inertial dampeners or functional equivalent, so that we don't need to worry about squishing the squishy humans as in g s's answer, there's an easy way to do the calculation. It looks like Pythagoras' theorem but with a minus sign instead of a plus sign:

$$ (\text{ship time})^2 = (\text{Earth time})^2 - (\text{distance})^2. $$

You have to measure time in years and distance in lightyears (or time in seconds and distance in light-seconds etc.) for it to work.

In your case you want the ship time to be 9 months = 0.75 years and the distance is 4.2 light-years, so

$$ (0.75)^2 = (\text{Earth time})^2 - (4.2)^2 $$ or $$ \text{Earth time} = \sqrt{(0.75)^2+(4.2)^2} \approx 4.27\,\text{years}. $$

This is assuming flat space-time (a reasonable approximation in this case) and travel at a constant speed (may or may not be reasonable depending on the properties of handwavium).

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  • $\begingroup$ I think this answer could be improved by using (s/c)^2 instead of s^2 so you don't need c=yr=1 units. $\endgroup$
    – g s
    Commented May 19 at 7:34
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    $\begingroup$ @gs I suppose that's a matter of taste - I was going for making it as easy to understand as possible and I don't think the choice makes it any less correct (it arguably makes it more correct). OP is using light-years and years anyway. $\endgroup$
    – N. Virgo
    Commented May 19 at 7:53
  • $\begingroup$ No one's fault but my own, as I don't grasp "why" it's true, but it looks like the time that passes on Earth would still equate to around 4.2 light years for observers watching our 9-month handwavium travellers. That is very interesting, and could add some complexity to my intended plot. Thank you for this simple answer that accounts for my handwavium =) $\endgroup$ Commented May 20 at 17:11

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