# Relativity - time dilation

I'm learning about relativity and I'm having some issues with it and the twin paradox. I found many questions and answers on this subject but they did not answer my specific problem.

In my thought experiment I added a third twin - triplet. So here it goes:

Let's have triplets $A$,$B$ and $C$. sitting still at same point in space.

They want to test relativity so they do this experiment..

They all reset their clocks, and $B$ and $C$ start moving together away from $A$ at some ridiculous speed $v_{365}$ which coincidentaly has lorentz factor $365$. they move at uniform motion in only one direction (let's say acceleration time nears 0).

State 1: \begin{align}t(A)=t(B)=t(C)&=0 \\ v(A,B) = v(A,C) &= v_{365}; \\ v(B,C) &= 0\end{align}

$t(A)$ being A's clock, $v(A,B)$ being velocity at which $A$ and $B$ are moving away from each other, etc..

Now after some time, let's say when $t(A) = 1\text{ year}$, brother $B$ starts to decellerate to full stop. It also shows up that brother $C$ was sleeping through whole first part of the experiment. But now he wakes up and sees that his brother $B$ is accelerating away from him in the opposite direction than was planned but that doesn't bother him.

State 2: \begin{align}t(A)&=1\text{ year} \\ t(B)=t(C) &\stackrel{?}{=} 1\text{ day} \\ v(A,B) &= 0 \\ v(A,C)=v(B,C) &= v_{365}\end{align}

Now after another while lets say when $t(A)=2\text{ years}$, brother $C$ sees that brother $B$ isnt stopping any time soon so he decides to match his speed.. (brother $C$ stops now from point of $A$)

State 3: \begin{align}t(A)&=2\text{ years} \\ t(B)&= 1\text{ year }1\text{ day} \\ t(C)&= 2\text{ days} \\ v(A,B)=v(A,C)=v(B,C)&=0\end{align}

That's it.. now they stand still some light year away from each other, but they can communicate and compare their clocks..

here are my questions:

1] Is my assesment of time dilation effects correct in this scenario? if not which state contains first error and why?

2] If my assesment is correct, then from point of brother $C$ it was brother $B$ who was accelerating away, yet it was also brother $B$ who aged when they got in same speed - therefore it seems to me its always the one who is matching speed who doesnt age - is this a general rule?

3] If answer to questions 1 and 2 is yes, then can this effect be used to gather infomation that would take much longer to extract in normal speeds?

Regards

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Your problem is with the notion of simultinaeity. You are implicitly using $A$'s clock to define all of the "now"s. You need to factor in the times required to send radio messages so that everyone can compare notes. An experiment like this is best done with a spacetime diagram, first from $A$'s perspective, and then from $C$'s perspective. –  Jerry Schirmer Aug 4 '12 at 0:11
The fact i defined changes by times on clock of A should not be an problem, it can happen incidentaly at that time from point of A, it should not change physics. As to comparing their notes, they dont compare their times until they are all standing still to each other. There should be no problem in compensating for their distance, assuming they know speed of light. –  code Aug 4 '12 at 6:44

To develop a reliable SR intuition, you should become comfortable with spacetime diagrams and the distinction between coordinate time $t$ and proper time $\tau$.

Choose a reference frame (coordinate system) in which A remains at the spatial origin; the coordinate time and A's proper time (A's "wristwatch" time) can be chosen to have the same value.

Let's denote the coordinate time in this frame $t_A$ and the proper time of A, B and C, $\tau_A, \tau _B$ and $\tau_C$ respectively.

Now, in the set-up you describe, we have:

$t_A = 0$: $\ \ \tau_A = \tau_B = \tau_C = 0$.

$t_A = 1$ year: $\ \ \tau_A = 1$ year, $\ \tau_B = \tau_C = 1$ day.

$t_A = 2$ years: $\ \ \tau_A = 2$ years, $\ \tau_B = 1$ day $+ 1$ year, $\ \tau_C = 2$ days.

The coordinate time is read directly from the spacetime coordinates of the events. The proper time is calculated along the world lines in the spacetime diagram.

While the world line for A is along the time axis of frame A, the world lines for B and C have kinks where the accelerations occur.

Importantly, there is no inertial reference frame in which the world lines for B & C do not have kinks. This is crucial to understanding why the elapsed proper times for B and C are less than A's proper time.

A remains in one inertial frame while B & C make "jumps" (boosts) to a frame moving with respect to the A frame and then back. The lines of simultaneity for B & C rotate at each kink.

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Thats exactly what im trying to do- develop SR intuition. Could you answer the question 2? Is the info who accelerates unimportant? Does it always depend only on who is decelerating ? –  code Aug 4 '12 at 7:05
Honestly, your question 2 is so poorly phrased, it's not clear what you're asking. –  Alfred Centauri Aug 4 '12 at 11:05

1] Is my assesment of time dilation effects correct in this scenario? if not which state contains first error and why?

It seems fine to me. I am assuming $t$ refers to self-time everywhere, and that simultaneous events are defined by observer A.

2] If my assesment is correct, then from point of brother C it was brother B who was accelerating away, yet it was also brother B who aged when they got in same speed - therefore it seems to me its always the one who is matching speed who doesnt age - is this a general rule?

Yes. The initial acceleration of B away from C is not important: it just sets up the initial conditions in which B and C are moving at different velocities. At this point they both agree that they are of the same age.

What is important is who can trust their special-relativity calculations regarding the other guy, after some time has elapsed. C sees B moving away and he can compute B's time using dilation. But once C decelerates he can throw his calculations in the trash because special-relativity does not apply to him: he needs general-relativity. B's calculations about C are fine though, all the way to the end.

Perhaps you already understand this, but I want to emphasize that in general-relativity (i.e. if your space was not flat) you would be more restricted, and would generally have to get everyone to meet in the same place in order to compare their clocks. In special-relativity you can get away with what you did, because two observers at the same velocity can work out their relative age by sending light signals back and forth.

3] If answer to questions 1 and 2 is yes, then can this effect be used to gather infomation that would take much longer to extract in normal speeds?

I don't see how.

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If you are down voting, it will be nice if you explain your reason for doing so. –  Guy Gur-Ari Aug 4 '12 at 1:43
To clarify, I did not downvote your answer, someone else did. To explain question 3: Imagine computer chip running some brute force code-breaking alghorithm, which has really low chance of success in close future. Now consider we got the technology and accelerate it to lorentz factor 10^6, and then after say a month we send another chip after this one, at which moment the first chip should seem 10^6 months older to the second chip, now the first one sends signal to the second with the result of his calculation.. and the second sends result to earth. :D :D haha i started laughing at my own idea –  code Aug 4 '12 at 6:58
You have to be careful about simultaneity, which is not a frame-independent concept. In your original question, you defined simultaneous events in the frame of observer A (whether you intended to or not). As long as you stick with a frame you're fine. But now you launch the second chip in one frame, and want to discuss what it looks like in the first chip's frame, so you have to be careful. For example, in the first chip's frame the new launch does not happen when he is $10^{-6}$ months old, but after that! You have to work it out carefully. –  Guy Gur-Ari Aug 4 '12 at 22:51
On more general grounds, as someone on earth you cannot gain a speed advantage by doing a calculation on a moving ship (e.g. a ship orbiting the earth). In our frame a moving computer will always take longer to complete the calculation than one staying on earth. This is even without accounting for the time it takes to transmit the result back to earth. –  Guy Gur-Ari Aug 4 '12 at 22:54