<p>In the first example, you certainly could use the calculation of special relativity to measure the difference in the time shown on a clock which was making the high speed journey and one which was at rest. The details may make the calculation trickier, but we can use the flat minowsky space of the clock at rest to do the calculation, so it is relatively straightforward. [see what I did there?]. It would be a factor of about 1.6 so the journey that would take a little over 67 years would, on board ship, seem to take 67*1.6 = 41 years. (I have ignored the time spent accelerating and decelerating, in order to keep the maths simple. Earth would observe the ship leaving 30years after it did (light distance) and arriving after 67 years, however, since the light must travel further, the ship would appear to arrive after 67+50-30 years = 87 years.</p>

<p>For the multiple journey the calculations are just longer. The direction of travel doesn't affect time dialation, only the speed. So it would again be straightforward to calculate the total time dialation.</p>

<p>It would be quite possible for Earth to communicate with the ship. But those the messages would not arrive for up to 50 years. Messages would be certainly out of date.</p>

CC BY-SA 3.0
2016-03-22T14:59:23.967Z
So, if I used the correct formula for the first leg of the trip and we determine the 1.6 multiplying factor just for ease of thought then would the following scenario be correct? Earth Departure: January 1, 2112 / Star 1 Arrival: Ship time: January 1, 2162, Earth time: January 1, 2198/ Star 2 Arrival: Ship time January 1, 2172, Earth time 2214/ Earth return arrival: January 1 2222, Earth Time: 2300.
Jack R. Woods
CC BY-SA 3.0
2016-03-22T23:32:17.423Z
Yes, you can do the maths in the Earths frame, and it keeps the calculations simple.  Remember that even after the ship arrives at star 1, it would take 50 years for the light to reach Earth. So Earthlings would observe the ship arriving at star one in 2248, and so on.
James K
CC BY-SA 3.0
2016-03-21T00:42:04.333Z
2016-03-21T00:42:04.333Z
James K
<p>I'm once again confused by the twin paradox. Let's say I am on an interstellar starship flying at 0.6c from a star 30 light years away from Earth to a star 50 light years away from Earth and 40 light years from each other. How would you estimate the difference between the time on the ship when you reach the destination versus the time a person on Earth thinks it should be? Now let's say I travel back to the original star (In both circumstances, I accelerate at 1g until reaching 0.6c.) I would assume that I could use the twin paradox formula to determine the difference in elapsed time on the ship from that measured by a clock left at the first star.</p>

<p>Now, let's say I'm traveling in the same manner (1g acceleration until reaching 0.6c and 1g deceleration for the same duration until arrival) beginning at Earth, traveling to 16 stars and returning to Earth. I am thinking that the time difference between the ship clock (date) and Earth time would be based on the distance and acceleration/ deceleration segments of each leg of the journey independent of the components of the travel vectors in the direction (toward or away from the Earth). Am I right??</p>

<p>Would Earth send messages to the ship based on conventional measurements (ie Physics 101)? I would want to determine the difference between the time on Earth when the message is sent and the ship time when the message is received, considering of course the time of transit of the message (at light speed).</p>

CC BY-SA 3.0
2016-02-22T02:15:56.100Z
@barrycarter No, not intentional, I kind of picked the distances at random, but maybe calculations might be easier in this special case. Actually my question is more in the second and third paragraphs. I am trying to write a book series about a multi-generational ship doing something similar to what is mentioned above. The starship will communicate with Earth and I want to establish time frames both on Earth and the ship as well as elapsed time upon return.
Jack R. Woods
CC BY-SA 3.0
2016-02-28T05:49:02.473Z
@barrycarter If you take v(0) = 0 and x(0) = 0, then the classical motion equations (a good approximation up to even .6c) would be:                        t = v(final)/a = (0.6c)/g and x(final) = 1/2 x a x tSQ = 1/2 x g x tSQ. Approximating g = 10 m/s/s and c = 3 x 10E8 m/s, then t =
Jack R. Woods
CC BY-SA 3.0
2016-02-28T05:59:39.070Z
@barrycarter continuing the above paragraph (i had to run);
Jack R. Woods
CC BY-SA 3.0
2016-02-28T06:15:33.583Z
@barry carter t = .6c/g = 1.8 x 10E7 sec (about 7 months) and               x = .5g x tSQ = 1.6 x 10E15 meters (about 1/3 light year)                        I find it fascinating that a spaceship traveling at 1g acceleration (artificial gravity) takes very close to one year to approach the speed of light. Coincidence? It's meant to be!!
Jack R. Woods
CC BY-SA 3.0
2016-02-28T06:25:06.740Z
@barrycarter thanks for the link/ it does take a little longer using relativistic equations, but still ballpark/ all of these sights just show one destination though/ i want to travel to multiple stars and then return to earth
Jack R. Woods
<special-relativity><interstellar-travel>
Special Relativity as Applied to an Interstellar Starship?
Jack R. Woods
# Special Relativity as Applied to an Interstellar Starship?

(Asked by: username_0 on 2016-02-18)

I'm once again confused by the twin paradox. Let's say I am on an interstellar starship flying at 0.6c from a star 30 light years away from Earth to a star 50 light years away from Earth and 40 light years from each other. How would you estimate the difference between the time on the ship when you reach the destination versus the time a person on Earth thinks it should be? Now let's say I travel back to the original star (In both circumstances, I accelerate at 1g until reaching 0.6c.) I would assume that I could use the twin paradox formula to determine the difference in elapsed time on the ship from that measured by a clock left at the first star.

Now, let's say I'm traveling in the same manner (1g acceleration until reaching 0.6c and 1g deceleration for the same duration until arrival) beginning at Earth, traveling to 16 stars and returning to Earth. I am thinking that the time difference between the ship clock (date) and Earth time would be based on the distance and acceleration/ deceleration segments of each leg of the journey independent of the components of the travel vectors in the direction (toward or away from the Earth). Am I right??

Would Earth send messages to the ship based on conventional measurements (ie Physics 101)? I would want to determine the difference between the time on Earth when the message is sent and the ship time when the message is received, considering of course the time of transit of the message (at light speed).

**Tags:** special-relativity, interstellar-travel

> **Comments:**
> **username_0**: @barrycarter No, not intentional, I kind of picked the distances at random, but maybe calculations might be easier in this special case. Actually my question is more in the second and third paragraphs. I am trying to write a book series about a multi-generational ship doing something similar to what is mentioned above. The starship will communicate with Earth and I want to establish time frames both on Earth and the ship as well as elapsed time upon return.  
> **username_0**: @barrycarter If you take v(0) = 0 and x(0) = 0, then the classical motion equations (a good approximation up to even .6c) would be:                        t = v(final)/a = (0.6c)/g and x(final) = 1/2 x a x tSQ = 1/2 x g x tSQ. Approximating g = 10 m/s/s and c = 3 x 10E8 m/s, then t =  
> **username_0**: @barrycarter continuing the above paragraph (i had to run);  
> **username_0**: @barry carter t = .6c/g = 1.8 x 10E7 sec (about 7 months) and               x = .5g x tSQ = 1.6 x 10E15 meters (about 1/3 light year)                        I find it fascinating that a spaceship traveling at 1g acceleration (artificial gravity) takes very close to one year to approach the speed of light. Coincidence? It's meant to be!!  
> **username_0**: @barrycarter thanks for the link/ it does take a little longer using relativistic equations, but still ballpark/ all of these sights just show one destination though/ i want to travel to multiple stars and then return to earth  

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## 1 answers

### Answer by username_1 on 2016-03-21. Score: 1 | Accepted answer

In the first example, you certainly could use the calculation of special relativity to measure the difference in the time shown on a clock which was making the high speed journey and one which was at rest. The details may make the calculation trickier, but we can use the flat minowsky space of the clock at rest to do the calculation, so it is relatively straightforward. [see what I did there?]. It would be a factor of about 1.6 so the journey that would take a little over 67 years would, on board ship, seem to take 67\*1.6 = 41 years. (I have ignored the time spent accelerating and decelerating, in order to keep the maths simple. Earth would observe the ship leaving 30years after it did (light distance) and arriving after 67 years, however, since the light must travel further, the ship would appear to arrive after 67+50-30 years = 87 years.

For the multiple journey the calculations are just longer. The direction of travel doesn't affect time dialation, only the speed. So it would again be straightforward to calculate the total time dialation.

It would be quite possible for Earth to communicate with the ship. But those the messages would not arrive for up to 50 years. Messages would be certainly out of date.

> **Comments:**
> **username_0**: So, if I used the correct formula for the first leg of the trip and we determine the 1.6 multiplying factor just for ease of thought then would the following scenario be correct? Earth Departure: January 1, 2112 / Star 1 Arrival: Ship time: January 1, 2162, Earth time: January 1, 2198/ Star 2 Arrival: Ship time January 1, 2172, Earth time 2214/ Earth return arrival: January 1 2222, Earth Time: 2300.  
> **username_1**: Yes, you can do the maths in the Earths frame, and it keeps the calculations simple.  Remember that even after the ship arrives at star 1, it would take 50 years for the light to reach Earth. So Earthlings would observe the ship arriving at star one in 2248, and so on.  

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