# Special Relativity - Time dilation

1. Sep 20, 2008

### kehler

1. The problem statement, all variables and given/known data
Assume a round trip with constant proper acceleration. It would have four segments: acceleration away from Earth, deceleration, acceleration towards Earth and deceleration to reach Earth. The time for each segment is the same, so the total trip time is four times that for a segment. Assume each segment is 5 years of proper time, so the total trip lasts 20 years for the traveller.
Assume the hyper-drive accelerates the ship at 1g=10m/s^2 which is approximately 1.06c/year. (1year=3.16x10^7s)

Calculate how much time passes on Earth. Use the time dilation formula. Note that the time on Earth for a segment can be written as 5= .
The following integral should be useful:

3. The attempt at a solution
I figured I just had to substitude numbers into the formula. I put '5' in place of the 'T's in the formula, and used alpha=1.06c/year. I then multiplied my answer by 4.
I got the answer to be 19.99981719. I find that kinda weird cos shouldn't more time pass on Earth than for the ship?? Did I do something wrong? :S

2. Mar 23, 2009

### cincirob

I don't visit this forum very often and I ran across this in a search. You don't have the integral corrrect. Here is the proper analysis:

t' = t(1-(v/c)^2)^.5

v=at

t' = Int[t(1-(at/c)^2)^.5]dt

t' = (-a/c)Int[t(t^2 - (c/a)^2)]dt

Int[x(x^2 - a^2)^.5]dx = 1/3(x^2 - a^2)^1.5

t' = (-c/a)^2(1/3)(t^2 - (c/a)^2)^1.5

t' =(1/3) (-a/c)^2(t^2 - (c/a)^2)^1.5