Q:

How can I calculate the time dialation with constant acceleration of g, starting at velocity 0 and ending at velocity .99C and then deccelerating at constant g back to velocity 0?

- Anthony (age 34)

Holmdel, NJ

- Anthony (age 34)

Holmdel, NJ

A:

For other readers, this means that the acceleration experienced by the traveler is g, not the acceleration as seen by the one who stays home.

There are two different times here, one according to the traveler who sets out and accelerates back, the other according to the stay-at-home. these are standard exercises, but the beauty of doing this site is that it gives an excuse to exercise a creaky old brain.

Here's what I get:

traveler time: 2(c/g)*arctanh(0.99)

home time: 2(c/g)*0.99/sqrt(1-0.99^{2}).

c/g is about one year. Evaluating the functions then gives for the times

~5.2 years and ~ 14 years. You might want to plug in the numbers to get that a little more accurately.

Mike W.

There are two different times here, one according to the traveler who sets out and accelerates back, the other according to the stay-at-home. these are standard exercises, but the beauty of doing this site is that it gives an excuse to exercise a creaky old brain.

Here's what I get:

traveler time: 2(c/g)*arctanh(0.99)

home time: 2(c/g)*0.99/sqrt(1-0.99

c/g is about one year. Evaluating the functions then gives for the times

~5.2 years and ~ 14 years. You might want to plug in the numbers to get that a little more accurately.

Mike W.

*(published on 09/05/2012)*