Time Dilation with Two Spacecraft

1. Jul 27, 2014

Art_Vandelay

I want to write a science-fiction story concerning the rapid increase in technology affecting the speed at which spacecraft travel and passengers age (relative to each other).

Theoretically, would it be possible that a son leaves a certain time (a number of years) before his father to travel to an exoplanet, then his father departs on a spacecraft that travels near light speed. Would his father age significantly less than his son, to the point where the son is older than his father. And if so, how far away would the planet have to be and how fast would the respective space craft need to travel?

Thanks!

2. Jul 28, 2014

ghwellsjr

Yes, it is possible and not hard to calculate what the age difference is for any given scenario. By trial and error, you can hone in on a scenario that you like.

I'm going to assume that the father arrives at the exoplanet at the same time as his son and that the exoplanet is not moving with respect to the earth.

First you would calculate how much the son ages for his trip. Given a distance between the earth and the exoplanet and a speed for the son, you would first calculate the time according to the earth/planet rest frame that his trip took. Let's call that t1. Then, based on the speed, you multiply that time by the inverse of gamma. We'll call that t2. You also do this for the father yielding t3 for the earth/exoplanet time for his trip and t4 for how much time he aged. Next you have to calculate how much time the father has to wait before taking off. This would be t1-t3. We have to add to that the amount that the father aged during his trip which would be t1-t3+t4, his total aging from the time his son left him until he rejoins his son.

Now I've already done a bunch of trials and the one I like puts the exoplanet 48 light-years away from earth. The son travels at 0.6c, getting there in 48/0.6=80 years of earth time. So t1=80. The inverse of gamma at 0.6c is √(1-0.62) = √(1-0.36) = √(0.64) = 0.8 so t2 = 80(0.8) = 64.

The father travels at 0.96c, getting there in 48/0.96=50 years of earth time. So t3 = 50. The inverse of gamma at 0.96c is √(1-0.962) = √(1-0.9216) = √(0.0784) = 0.28 so t4 = 50(0.28) = 14.

The amount of time that the father waits before leaving is t1-t3 = 80-50 = 30. The total amount of time that the father ages is 30+14 = 44.

So the son ages 64 years while the father ages 44. As long as the father was less than 20 years old when he had his son, then he will be younger when he arrives at the exoplanet with his son.

Here is a spacetime diagram to illustrate the scenario showing the moment when the son departs to the moment when they both reunite on the exoplanet. The son is represented in blue and the father in red. The dots mark off one-year increments of time:

Although this particular scenario makes the son quite elderly by the time he meets up with his father, I don't think you can find another one that would be much better.

Attached Files:

• FatherSonTrips1.PNG
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3. Jul 28, 2014

Orodruin

Staff Emeritus
You can make the difference in aging arbitrarily close to the aging of the son. However, if you want to fix the aging difference to the difference in age between the two, this will imply very long travel distances.

Example: Son aging 10 times more implies that his speed is 0.98c and that the father's is very close to c. This would imply a gamma factor of about 5 and the son would need to travel about 100 light years to age by 20 years while the father would remain behind for about two years and not age significantly during the trip.

4. Jul 28, 2014

ghwellsjr

Good point. Here is a spacetime diagram to illustrate your scenario. The speed of the son is actually 0.9806c and the father is 0.99995c:

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