# Triplet paradox: where's the error in my logic?

1. Nov 14, 2011

### gespex

Consider the following scenario:
There is a triplet of persons A, B and C. Person A stays on earth, while person B and C both go onto two different space missions, the directions parallel to each other. Person B travels at 0.45*c, person C at 0.9*c. The space missions are both set to take exactly one earth-year, that is, person B and C will be back after 1 year having elapsed for person A.
How much time elapsed for person B and C when they come back?
For person B: sqrt(1 - 0.45^2) ~ 0.8930 years
For person C: sqrt(1 - 0.9^2) ~ 0.4359 years

Now person B calculates how much time elapsed for person C. As they are inertial reference frames, wouldn't time for C speed up by a factor of:
1/sqrt(1 - (v2 - v1)^2/c^2)
Where v2 and v1 are the two velocities, 0.45c and 0.9c? That is, it would slow down by a factor of:
sqrt(1 - (0.9*c - 0.45*c)^2/c^2) = sqrt(1 - 0.45^2)

Since the time elapsed for person B is sqrt(1 - 0.45^2), would he not calculate the time elapsed for person C as:
sqrt(1 - 0.45^2)*sqrt(1 - 0.45^2) = 1 - 0.45^2

So wouldn't person B consider person C (1 - 0.45^2) ~ 0.7975 years older, even though person A would consider person C 0.8930 - 0.4359 ~ 0.4571 years older.

What's the error in my logic here?

2. Nov 14, 2011

### gordonj005

if you're considering the frame of reference of B, from his point of view he is stationary. that is, A is travelling away from him at -0.45c and C is travelling away from him at 0.45c. so from B's point of view, the time elapsed is 1 year, while on A it seems to him, has elapsed 0.8930 of a year, and on C it seems to him that 0.8930 of a year has elapsed for him. taking the reference frame of C, it seems to him on A 0.435 of a year has elapsed and on B 0.8930 of a year has elapsed, while on C a full year has passed by.

So in terms of age:

Frame of A: B is 0.107 years younger, C is 0.565 years younger
Frame of B: A is 0.107 years younger, C is 0.107 years younger
Frame of C: B is 0.107 years younger, A is 0.565 years younger

So we note that comparing A and C, they both feel like the other is 0.565 years younger, comparing A and B they each feel that the other is 0.107 years younger, and comparing B and C they each feel as though the other is 0.107 years younger.

Is your question: Who is right?

3. Nov 14, 2011

### Staff: Mentor

Neither B nor C are inertial. The standard time dilation formulas don't apply.

Pick any inertial frame you like and you will get the same results.

4. Nov 14, 2011

### Staff: Mentor

No, this is wrong. Since A, B, and C all come back together at the end, the time elapsed for each of them is an invariant and doesn't depend on whose frame you use to calculate it. From B's point of view, 0.8930 years have elapsed when he and A and C come back together. Similarly, from C's point of view, 0.4359 years have elapsed when all three come back together.

What the OP is missing is the fact that B and C do not stay in the same inertial frame for their entire trip, while A does. (We are assuming that we neglect the Earth's gravity and the fact that it is orbiting the Sun; it would be better to set the thing up so that A stays in the same space station, floating freely far away from all gravitating bodies, for the whole time, while B and C go out and come back, to avoid complications.) So the calculations can't be done from B's or C's "point of view" the same way they can be done from A's point of view.

5. Nov 14, 2011

### PAllen

Just to add that another error is believing B sees C traveling at .45c. Velocities are not additive in relativity. Due to velocity addition formula, B sees C traveling at .7563c. However, this does not account most of the resolution of the 'paradox'. The main issue, as pointed out, is that B and C path's are not inertial. One thing that can be stated is what B would visually see:

For the first half of B's time between separation and uniting, B would see C's clock run slow, consistent with relative speed of .7563. However, for the second half of B's separated time, they would see C's clock run at a much slower rate than the first half, accounting for the discrepancy.

[EDIT: correction: B would visually see 3 different rates on C's clock. First, as expected for .7563 inertial relative recession speed; then extremely slow; then faster than the initial period. This is raw observation, no corrections for doppler. Corrected for doppler, the beginning and end rates would both be consistent with .7563 relative speed, but the intermediate slower rate accounts for the discrepancy (compared to treating B as if they were inertial)]

Last edited: Nov 14, 2011
6. Nov 14, 2011

### gespex

Good points, guys... Thanks for the replies!

7. Nov 14, 2011

### PAllen

And one more thing ... when B sees C's clock running very slow, it is as if C were moving .96c away from B. This is because for some time after B has turned around (note, the turnaround itself is not relative, it is an absolute phenomenon), they still see C receding, not yet turned around. This is just another application of velocity addition rule.

8. Nov 15, 2011

### M1keh

Ok. Sorry to jump in on this one, but ...

If you have three staging posts in a row, with A in the middle and the other two an hour away at 0.8c, according to A. All 3 are stationery with respect to A, so all can agree on the time. If B is approaching one way at 0.8c and C is approaching in the opposite direction at 0.8c, both pass the outposts at the same time, according to A, who is in constant contact with the outposts. Both B & C sync times with the outposts, and therefore with A. Now B & C both take an hour to reach A, but their times only show that 36 minutes have passed. However, both B & C show the same time when they pass each other at A. When the 3 are at A, they can all agree on the event of them checking the time as they're all at the same location.

Now there's no acceleration, so relativity does apply, but B & C agree on the time taken to travel from the outpost to A, despite the fact that there's a massive difference in their speed.

How ?

9. Nov 15, 2011

### PAllen

This example gets at relativity of simultaneity. A thinks B and C passing the posts and synchronizing are simultaneous events, and that B and C clocks are running at the same slow rate from there. B thinks C past the its post well after B past its post, an that C clock runs slow from this point. Similarly, C thinks B past its post 'late', and its clock runs slow. The affects all balance so that when B,and C are momentarily co-located, they see their clocks agree - but each has a different explanation of the sequence of events, and rates of clocks, accounting for this.

Last edited: Nov 15, 2011
10. Nov 15, 2011

### M1keh

Yes. I'd forgotten that one. I'll take it away and have a thunk. :-)

Thanks.

11. Nov 15, 2011

### M1keh

Actually ...

If both B & C are not single observers, but observers coming in waves of 3. In A's frame all 3 B's are (1hr@0.8c) apart and all C's are (1hr@0.8c) apart, then as B2 & C2, reach A, C1 & B3 are at one outpost and C3 & B1 are at the other.

Assuming B1, B2 & B3 and C1, C2 & C3 can agree on the times between themselves and C2 & B2 can now agree the time that they passed A, how can C1 & B3 and C3 & B1 disagree on the time that they pass the outposts ?

12. Nov 15, 2011

### PAllen

The key to understanding this is unchanged. According to each observer (An, Bn, Cn) there are two separate effects they see in the other's clocks: slower rates, and offsets between each clock. That is, if Bn sets their B1,B2,B3 clocks so they read the same (according to Bn), then according to both A and C, they never read the same. This is again due to relativity of simultaneity - A and C are seeing different events in the B1, B2, B3 world lines as simultaneous than B sees as simultaneous. Once this is understood, nothing about my prior explanation needs to change.

13. Nov 15, 2011

### D H

Staff Emeritus
Even though shouting (or its textual equivalent) is normally rude, some points are so important that they just need to be shouted out. PAllen didn't shout. I'll shout it out for him.

The twin paradox was constructed to be as simple as possible and still show some rather bizarre consequences of relativity. Adding a triplet, or putting in waypoints, will not make the paradox easier to understand. It only adds complications, unneeded complications at that.

14. Nov 16, 2011

### M1keh

At the risk of angering the gods. Adding complication in the questions is only necessary because of the complications added to the theory ? Pulling out the 'relativity of simultaneity' card complicates the stage we're working on and requires more complex questions to help with our (or at least my) understanding of how this all works ?

Am I missing something ... besides a few marbles ... the 'Twins Paradox' isn't a paradox at all under Special Relativity. It's just an example used to help explain the affects of time dilation ?

PAllen : Thanks for the patient reply, but ...

If B1,B2,B3 agree on the time and C1,C2,C3 agree on the time, B1 & C3 are at the same location and must agree on events at that location, similarly with B3 & C1. A time displayed at the outposts must be agreed on by all 6 ? B3 & C1 see it's 6pm, B3 tells B1 & B2, C1 tells C2 & C3 and all agree that B3 & C1 were at the outpost when the outpost's clock struck 6pm ?

15. Nov 16, 2011

### PAllen

If B1,B2,B3 clocks ever look the same to B observer, they will look offset to C observer. This is a fact, a direct consequence of relativity of simultaneity. Further, if the 3 passing events are simultaneous for A they will occur in a different order for B, and another different order for C. Each passing event is a single event, but each thinks the 3 passing events occurs in a different order.

16. Nov 16, 2011

### D H

Staff Emeritus

They are instead paradoxical in the sense that they appear to run against the grain of everything we thought (and oftentimes still think) to be true. That thinking, which is based on experiences limited to very low speeds compared to light, is flat out wrong when relative velocities become very high. One point of the paradoxes is to illustrate situations where our Newtonian/Aristotelian thinking is just wrong.

17. Nov 16, 2011

### M1keh

Ok. May as well have A1,A2,A3 on the three staging posts.

If A2 sees the Bn coming, all travelling at (1hour*0.8c) intervals, A2 knows that the Bn will all hit An at the same time (in An frame). A2 sends a signal to A1 & A3 advising them that hostilities have broken out and all Bn are to be destroyed. A1 & A3 succeed in destroying B1 & B3, but A2 fails and B2 gets away. B2 sends a signal to B1 & B3 to avoid An. One of B1 & B3 wont yet have reached An (in Bn frame) and could receive the message before reaching the staging post and avoid being destroyed.

I know 0.8c probably doesn't give enough time for the signal to reach B1 or B3, but a smaller value presumably would. How can A1 & A3 destroy B1 & B3 when one of them doesn't get there ?

18. Nov 16, 2011

### Staff: Mentor

This is a perfect example of what PAllen and DH were saying. Adding complications doesnt aid understanding, it just distracts from the point which has already been answered. Here, for no reason (because you have already been correctly answered) you are asking a very tedious question. You have 3 A objects, 3 B objects, and 3 messages, for a total of 9 worldlines. Since you have two reference frames that is a total of 18 expressions with 24 intersections. All for the purpose of repeating an answer that you can get simply by going back and re-reading it.

If you really want this question answered you should work out the 18 expressions and 24 intersections yourself. Here is the equation to transform between frames:
http://en.wikipedia.org/wiki/Lorent...ormation_for_frames_in_standard_configuration

Within each frame each worldline will be of the form $x_n=v_nt+x_{0n}$. If you get stuck please post your work and we can help you get unstuck.

19. Nov 16, 2011

### M1keh

Fair point, if a bit harshly put. I'll take the formulas away and see if I can plug some numbers in, to see what I get.