What is the figure of 6 years representing in resolving the Twin Paradox?

[mrausum]

Resolving the "Twin Paradox"

http://mentock.home.mindspring.com/twins.htm

I've been trying to follow this unusual explanation to resolve twin paradox, which uses the lorentz relativistic velocity transformation equation to get the speed bob zooms off after Ann at,15/17 C. I can understand up to that point of the explanation and get the same value, however why does it take him 4 years to catch up with Ann at this speed? It's not explained. Maybe it's obvious? Help!

(please no explanations involving the relativistic dopler shift equation).

 

[mrausum]

Nevermind, i resolved the problem :D. For anyone interested, I worked out the time it took for bob to catch up in the inertial bob frame, then divided this by the new gamma between bobs returning frame and bobs inertial frame (from v = 15/17 C), to get 4 years.

 

[JesseM]

Nevermind, i resolved the problem :D. For anyone interested, I worked out the time it took for bob to catch up in the inertial bob frame, then divided this by the new gamma between bobs returning frame and bobs inertial frame (from v = 15/17 C), to get 6 years.

What is the figure of 6 years supposed to represent? In your original post you were asking how they derived the figure of 4 years for Bob to catch up with Ann again, in the frame where Ann is moving at 3/5c and Bob is moving at 15/17c after Bob's acceleration.

Anyway, here's one way of deriving the 4 year figure. Before Bob accelerates, in this frame Bob is at rest while Ann is moving away at 3/5c, and this lasts for 4 years, so when Bob accelerates the distance between Ann and himself will be (3/5c)*(4 years) = 12/5 light years. Then after he accelerates, he'll be moving towards her at 15/17c while she continues to move away at 3/5c, so the "closing velocity" between them in this frame (the rate at which the distance between them is shrinking in this frame, which is different from the velocity of Ann in Bob's rest frame) will be 15/17c - 3/5c = 75/85c - 51/85c = 24/85c. So, if the distance between them is initially 12/5 light years at the moment Bob accelerates, the time in this frame for Bob to catch up will be (12/5 ly)/(24/85c) = 17/2 years. This is a lot longer than 4 years! But note that Bob's clock is running slow during the return journey, by a time dilation factor of sqrt(1 - (15/17)^2) = sqrt(289/289 - 225/289) = sqrt(64/289) = 8/17. So, during the 17/2 years it takes in this frame for Bob to catch up with Ann after accelerating, Bob's clock only ticks forward by (8/17)*(17/2) = 8/2 = 4 years.

 

[mrausum]

What is the figure of 6 years supposed to represent? In your original post you were asking how they derived the figure of 4 years for Bob to catch up with Ann again, in the frame where Ann is moving at 3/5c and Bob is moving at 15/17c after Bob's acceleration.

Anyway, here's one way of deriving the 4 year figure. Before Bob accelerates, in this frame Bob is at rest while Ann is moving away at 3/5c, and this lasts for 4 years, so when Bob accelerates the distance between Ann and himself will be (3/5c)*(4 years) = 12/5 light years. Then after he accelerates, he'll be moving towards her at 15/17c while she continues to move away at 3/5c, so the "closing velocity" between them in this frame (the rate at which the distance between them is shrinking in this frame, which is different from the velocity of Ann in Bob's rest frame) will be 15/17c - 3/5c = 75/85c - 51/85c = 24/85c. So, if the distance between them is initially 12/5 light years at the moment Bob accelerates, the time in this frame for Bob to catch up will be (12/5 ly)/(24/85c) = 17/2 years. This is a lot longer than 4 years! But note that Bob's clock is running slow during the return journey, by a time dilation factor of sqrt(1 - (15/17)^2) = sqrt(289/289 - 225/289) = sqrt(64/289) = 8/17. So, during the 17/2 years it takes in this frame for Bob to catch up with Ann after accelerating, Bob's clock only ticks forward by (8/17)*(17/2) = 8/2 = 4 years.

Sorry, i meant 4 years for this example. I'm writing this solution to the paradox as a short first year essay/report and so I'm using different values.

Thanks for the explanation, that's exactly what i did, except you explained it a lot clearer. Good to know I'm using the right method.