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If there is no particular absolute choice of frame of reference, the guy who sits on Earth is also moving away from the guy on spaceship perspective and hence time on Earth should also dilate when viewed from the guy on spaceship perspective. But why does the guy moving on spaceship look younger in twin paradox? What am I misunderstanding terribly?

ADDED:: Is time dilation symmetric? If one frame of reference are moving with constant velocity w.r.t other, we have to transformation relations

$$\Delta t = \frac{\Delta t'}{\sqrt{1 - \frac{v^2}{c^2}}}$$ Does each one of them see clock on other tick slowly than their own?

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  • $\begingroup$ We are not comparing two inertial reference frames here. The guy in the spaceship has to accelerate and decelerate to get back to earth and compare who has aged more. The guy who stays on earth doesn't go through all these velocity changes. $\endgroup$ – Johannes Aug 5 '13 at 16:15
  • $\begingroup$ The guy on the spaceship has to jump from one inertial frame to another in order to return to Earth. This breaks the symmetry. See wikipedia. $\endgroup$ – Pulsar Aug 5 '13 at 16:16
  • $\begingroup$ @Johannes is it due to acceleration? also to add to my question ... if one frame is moving with constant velocity, does both of them measure time dilation for each other? $\endgroup$ – hasExams Aug 5 '13 at 16:17
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    $\begingroup$ See also: physics.stackexchange.com/q/2554 $\endgroup$ – Johannes Aug 5 '13 at 16:21
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    $\begingroup$ @hasExams: they have to come back to the same position to compare their ages to each other. In the most commonly phrased version of the paradox, one of them turns around and comes back to meet. If you draw the thing out on a spacetime diagram, you'll see that the turnaround event is where they sync up their notions of "who is younger". If the "moving one" says moving at constant velocity forever, the effect is symmetric. $\endgroup$ – Jerry Schirmer Aug 5 '13 at 16:36
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Here's how I think about it.

First of all, twins A and B have identical clocks consisting of two mirrors, where a photon bounces back and forth between them. That's what you call a "tick" and "toc", and it's a nice way to build a clock because the speed of light is always constant no matter who measures it. What's more, everything depends on it, like a person's age, which is just a certain number of ticks and tocs.

They are not so far apart or going so fast that they can't watch each other's clock as well as their own.

enter image description here

OK, as far as A is concerned, he's standing still, but B is moving. Since B is moving, the photon has to travel a longer distance between the mirrors, which takes more time. So as far as A is concerned, B's clock is running slower, ergo B is aging more slowly.

Now switch to B's frame of reference, and he sees A's clock running more slowly.

So you see, it's all relative :)

Of course, the paradox is resolved when one undergoes acceleration to turn around and come back. That one will actually be younger.

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I assume the inertial twin is A ,while the one in the spacecraft is B

The relative velocity between A and B is v.

The distance B has to travel is from x0 to x1 as seen by A observer= x1-x0.

The same distance is k*(x1-x0) as seen by B observer,where k =(1-(v^2/c^2))^0.5 because of length contraction.

So from the point of view of the observer A, B has to travel a distance equals to 2*(x1-x0). The time measured in A's frame = 2*(x1-x0)/v.

While from the point of view of the observer B,

The point x1 is travelling towards him at speed of v.

the initial distance between B and x1 = k*(x1-x0).

x1 reaches B and reverses course to move away from him and x0 reaches B again.

The total distance measured in B's frame = 2*k*(x1-x0).

The time measured in the B's frame of reference for this journey = 2*k*(x1-x0)/v.

The whole thing is down to the initial assumption that the points x0,x1 are inertial to the A observer.

This is embedded in the problem.

In fact we can design simpler thought experiments without acceleration ,deceleration and yet the same result will still hold.

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  • $\begingroup$ You may want to use Latex to improve your answer. $\endgroup$ – Ali Aug 8 '13 at 2:48

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