## The twin paradox: the crucial bits

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### The twin paradox: the crucial bits

It is commonly known that in relativity that a moving clock ticks more slowly than a stationary one. This statement has been challenged as being inconsistent with the central tenets of relativity and is called the twin paradox. The name comes from a pair of twins, one of whom stays on Earth, while the other moves at high velocity to a distant star and returns. When the twins are reunited, the homebody is older than the traveler.

One of the reasons that the twin paradox is called a paradox is that the relativity says that any inertial observer can call themselves stationary. It would seem that the travelling twin could say that he was stationary while the Earth zoomed away and then zoomed back.

A common explanation of this paradox is that the travelling twin experienced acceleration to slow down and reverse velocity. While it is clearly true that a single person must experience this acceleration, you can show that the acceleration is not crucial. What is crucial is that the travelling twin experienced time in two reference frames, while the homebody experienced time in one. We can demonstrate this by a modification of the problem. In the modification, there is still a homebody and a person travelling to a distant star. The modification is that there is a third person even farther away than the distant star. This person travels at the same speed as the original traveler, but in the opposite direction. The third person's trajectory is timed so that both of them pass the distant star at the same time. As the two travelers pass, the Earthbound person reads the clock of the outbound traveler. He then adds the time he experiences travelling from the distant star to Earth to the duration experienced by the outbound person. The sum of these times is the transit time.

Note that no acceleration occurs in this problem...just three people experiencing relative inertial motion.

The following figure illustrates the situation.

Relativity_demo

What follows is a derivation of the problem.

Start with three inertial observers: A, B, C. This is different from the usual way to do this, in which there are two observers, one at rest and the other zooms out to some distance L and returns.

In this case, at time 0 in A's frame, A is at rest at the origin. B is at the origin, moving to the right with velocity v. Observer C is at a distance of 2L in A's frame, moving to the left at velocity v.

In A's frame, what happens is B moves to point 2 (taking time L/v to do it). C also moves to point 2 in the same amount of time. As B passes C, C reads B's digital clock. B continues to the right, while C continues to the left. After the same amount of time (L/v), C is now at A's position and B is at 2L. We always consider the right to be the positive x direction.

So, as far as A is concerned, there are three important space/time coordinates, denoted 1, 2, 3. 1 is the start, 2 is when B and C meet at distance L, and 3 is when C returns to A. Mathematically, we could say:

$\Large (x,t)_{1A} = (0, 0)$
$\Large (x,t)_{2A} = (L, \frac{L}{v})$
$\Large (x,t)_{3A} = (0, \frac{2L}{v})$

Hopefully this is clear.

Now let us remember the Lorentz frame transformation equations:

$\Large x' = \gamma (x - vt)$
$\Large t' = \gamma (t - vx/c^2)$

With this, we can essentially trivially find the (x,t) in both the B & C frame. Specifically:

$\Large (x,t)_{1B} = (0, 0)$
$\Large (x,t)_{2B} = (0, \frac{1}{\gamma} \frac{L}{v})$
$\Large (x,t)_{3B} = (-2 \gamma L, 2 \gamma \frac{L}{v})$

and

$\Large (x,t)_{1C} = (0, 0)$
$\Large (x,t)_{2C} = (2 \gamma L, \gamma \frac{L}{v} (1 + \frac{v^2}{c^2}))$
$\Large (x,t)_{3C} = (2 \gamma L, 2 \gamma \frac{L}{v})$

Now to get the time passage of C from 2 to 3, we need to subtract their times:

$\Large \Delta t_{3,2 C} = t_{3C} - t_{2C} = \frac{1}{\gamma} \frac{L}{v}$.

So the total time experienced by two, inertial observers (as compared to A) is

$\Large t_{2B} + \Delta t_{3,2 C} = \frac{2}{\gamma} \frac{L}{v}$.

So, what we see is the following.

The observers inhabiting multiple frames experience a shortening of time. I did not explain the acceleration histories of the various observers. You could imagine three objects in a co-moving frame and then have several options of acceleration resulting in the three inertial frames I've mentioned.

Given the straightforward methodology described here, you can show that the observers in multiple frames experience the time dilation. [Hint, if you pick B or C as the "at rest" frame, you have to properly calculate the velocities.]

This really is left as an exercise for the student.

Lincoln
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 RussellRamp, Hendrick Laursen, ElectraQT liked this post

### Re: The twin paradox: the crucial bits

Note that a thorough explanation of how the ticking of moving clocks depends on both location and velocity can be found in another similar post.

viewtopic.php?f=84&t=8856

This shows another interesting facet of relativity which shows how two people in motion actually agree with one another when one takes into account both the relative location and motion.

Lincoln
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