# Time Dilation v-v Space Travel

I have read two different explanations of time dilation in the context of space travel.

Given: round trip v-v Alpha Centauri, assumed distance 4.37 light years one way. Speed of the space craft: 0.9999C

Explanation 1: From the reference plane of the space ship, the trip would take 8.7409 years. From the reference plane of the earth, the space ship would have been gone 618.0888 years. That's the one I learned in college in the 1960's.

Explanation 2: From the reference plane of the space ship, the trip would appear instantaneous. From the reference plane of the earth, the space ship would have been gone for 8.7409 years.

The implications of these two approaches is obvious: #1 pretty much eliminates space travel for humans. #2 makes space travel trivial; easier even than in STNG.

I was only a physics major for 2 years.........I'm hoping we have some PhD's on the site who can give me the reality on this. Thanks in advance for your info.

• You must be misremembering what you learned in college. If the spaceship is moving at 0.9999c relative to the Earth, and as measured by the Earth the distance it travels is about 8.74 light years, then the time as measured by the Earth is clearly just a little over 8.74 years. Commented Aug 19, 2021 at 20:37
• explanation 1 is wrong, and 2 is very accurate (thought the trip is not instantaneous, it is pretty short)
– user65081
Commented Aug 19, 2021 at 21:05
• First, I want to thank you not only for your answers, but for the non-punishing way in which you presented them. It is refreshing... and appreciated. Commented Aug 19, 2021 at 22:29
• lol...........don't hit "enter" to start a new paragraph..........anyway.......clearly I had a senior moment. Your responses led me to explore the entire area in more detail and that was a pleasant byproduct. It's always nice to learn and it's always important to have inaccurate information corrected. Thank you. Commented Aug 19, 2021 at 22:31
• could we take the discussion a bit further, please. Still Alpha Centauri. Spaceship is now at 1,000,000mph. So, (assuming instant acceleration) the ship will return to earth in around 5700 years. How can I calculate the time that would have passed from the reference plane of the ship? At this low speed, it's unlikely to be instantaneous. There must be a formula to relate speed to "on ship" time. Commented Aug 19, 2021 at 22:52

Let's consider just the 'going' part of the trip. In the reference frame of the earth, the ship is travelling at some speed $$v = 0.9999c$$ and has to travel a distance $$L = 4.37 \, \text{LightYear}$$. This means that for us, standing on earth, the trip would of course take $$t_{earth} = \frac{L}{v} \simeq 4.37044 \, \text{Years}$$ just a bit above the time light would take.
On the ship's frame, however, Alpha centauri is moving towards the origin(the sip, in this case) at a speed of $$0.9999c$$, but only has to travel a distance $$\frac{L}{\gamma}$$ where $$\gamma = \frac{1}{\sqrt{1-(v/c)^{2}}}$$ is the Lorentz's factor. (I am using space contraction as a given here, but it is easiliy verified through Lorentz's transformation). This gives a time observed by the spaceship $$t_{ship} = \frac{L}{\gamma v} \simeq 0.062 \, \text{Years}$$ If we ignore acceleration effects that happen when the ship mask a u-turn to get back, the time for the whole travel is $$2\cdot t_{earth} \simeq 8.7408 \, \text{Years}$$ according to citizens on earth and $$2 \cdot t_{ship} \simeq 0.12 \, \text{Years}$$ for those on the ship.