# Twin paradox without length contraction

1. Feb 22, 2010

### ed2288

Hi all, apologies if this has been posted a million times before....

I'm trying to explain the twin paradox without getting involved with length contraction.

One way to think of it is Twin A remains at rest on Earth then twin B goes off at 4c/5 to Alpha centauri 4 light years away, then comes back again having aged 6 years, 3/5 as much as twin A.

Alternatively, can I think about it this way? Twin B remains at rest, while twin A and Alpha centauri speed off at 4c/5 for 5 years in the rest frame of B. When Alpha centauri reaches twin B, he suddenly accelerates to 40c/41 (relativistically added 4c/5 and 4c/5) and catches Twin A up. The lorentz factor of twin B becomes 41/9...and from here I struggle to get the nice and simple Twin A = 10 years, Twin B = 6 years like I did before.

2. Feb 22, 2010

### grav-universe

Well, you've it from the perspective of A, so one way to look at it from B's perspective using only times is that 3 years passes on B's clock away and back, so B sees a time dilation of .6 for A while do so also, so sees only 1.8 years pass on A's clock away and back, so 3.6 years in all. However, at the point of turn-around, the simultaneity between A and B shifts from past-back to future-forward as B conceptualizes the reality of what is taking place for A's clock, for a time difference of 2 y (v / c)^2 t using the time that B measured for the one way journey after initially synchronizing with A's clock before the journey began. So when B turns around, he says that A's clock fast forwards to an additional time reading of 2 y (v / c)^2 t = 6.4 years, which added to the original time dilation read upon A's clock, will give 10 years for A and 6 years for B.

Last edited: Feb 22, 2010
3. Feb 22, 2010

### JesseM

If you want to consider the frame where twin B is at rest during the outbound leg of the trip, you do need to take into account length contraction relative to the twin A frame, since in this frame the distance between Earth and Alpha Centauri is not 4 light years but only (3/5)*4=2.4 light years. Other than that, this method should work fine for calculating the time for each twin. In this frame the time for twin B to get from Earth to AC will be 2.4/(4/5) = 3 years, then on the return trip you have to take into account that both twin B and Earth are moving, so the time in this frame will not just be 2.4/(40/41), instead it'll take longer since the Earth is moving in the same direction as twin B but at a smaller speed. Are you having trouble figuring out the length of the return trip in this frame or does the problem lie elsewhere?

4. Feb 22, 2010

### JesseM

I don't think this answer is really addressing ed2288's question, the question was about figuring things out from the perspective of a single inertial frame where twin B is at rest during the outbound leg and moving at (40/41)c during the inbound leg, not about using a single non-inertial frame where B is at rest during both phases, or combining the results of two inertial frames (one where B is at rest during the outbound leg and another where B is at rest during the inbound leg), as you seem to be doing.

Last edited: Feb 22, 2010
5. Feb 22, 2010

### grav-universe

Oh right, thanks JesseM. :) Okay, so using only times and relative speeds while viewing things from the same inertial rest frame, then, A takes off at 4/5 c for 3 years, not 5 by the way, according to the rest frame of B. So 3 years passes for B and 1.8 years passes for A according to the rest frame of B. Then B takes off toward A at 40/41 c. Now since we are still observing the clocks from the original rest frame, we must find the time that it takes for B to catch up to A, which is v_A t / (v_B - v_A) = (4/5 c) (3 years) / (40/41 c - 4/5 c) = 13.6666 years according to a clock that remains in the original rest frame, where A travels away at v_A for a time of t, then B catches up with a difference in speeds of v_B - v_A. However, during this time, A's clock has time dilated by .6, so a time of 8.2 years will pass for A, making 1.8 years + 8.2 years = 10 years in all for A, while B's clock time dilates by 9/41, so a total time of 3 years + 3 years = 6 years passes for B when A and B meet back up.

Last edited: Feb 22, 2010
6. Feb 22, 2010

### ed2288

Thanks for the replies, I'm understanding this much more. In response to JesseM, yes it is this return length that I'm having difficulty calculating. Here's my tentative method which gives an answer, is this the right way of doing it?

Find the difference in speed between twin A and B by relativistically subtracting 3/5 c from 40/41 c to get 77/85 c. At this speed the lorentz factor is 85/36 meaning the Earth-AC distance contracts to (4*36/85) = 144/85 light years. So travelling at 77/85 c Twin B takes exactly 2 years to catch up with twin A. Given twin B waited 4 years to start his journey, he will have aged 4+2=6 years as expected. CORRECTION: I've just noticed 144/77 does in fact NOT equal 2! Sorry for that...

To me this seems a more intuitive way to do the calculation, I'm just confused as to why that simple subtraction formula holds with having to do it relativistically?

Last edited: Feb 22, 2010
7. Feb 22, 2010

### grav-universe

That should be relativistically subtract 4/5 c from 40/41 c to get 4/5 c back since this is what you relativistically added before.

[EDIT after JesseM's next post] - That is, this is the relativistic difference in speeds if you are trying to find the speed A and B now observe between themselves, otherwise just subtract them directly to find the difference in speeds according to B's original rest frame.

We don't need to include length contraction or simultaneity shifts since everything is done from an inertial frame, but we still need time dilation and the relativistic addition of speeds in order to transfer to a different inertial frame for the same scenario to take place.

Last edited: Feb 22, 2010
8. Feb 22, 2010

### JesseM

Since both 3/5 c and 40/41 c are from the perspective of the same frame (the frame where twin B was at rest during the outbound leg), the difference in speed in this frame is just (40/41 - 3/5)c. You only use the formula for relativistic velocity addition/subtraction when you're comparing different frames, like if you know A is going at 0.8c in B's frame, and you know B is going at 0.5c in C's frame, and you want to know how fast A is going in C's frame.