A time dilation paradox

Question

We know from special relativity if an airplane is moving with some large speed away from us, its clock will run slower relative to us. now I have another scenario: imagine two airplanes which are moving away from each other with equal velocities, like this:

now the question is: which airplane's clock is running slower? you may say both clocks are synchronous but we know one of the airplanes has a relative speed to the other one so we must have time dilation. if you chose one of the airplanes then I also disagree because both are the same and the problem has symmetry.

Answer 1

which airplane's clock is running slower?

Depends on which is the frame of reference you are considering.

In A's frame of reference, B's clock is running slower than clock A.
In B's frame of reference, A's clock is running slower than clock B.

In the frame of reference of someone on the ground, both A and B's clocks are ticking slower than the ground clocks AND ticking at the same rate.

Answer 2

Each plane observes the clock on the other plane running slow when compared to its own. There is no paradox as there is no abolute "time", only the proper time of each each observer.

Answer 3

This is a do the Lorentz transform situation.

We have two moving frames: $S'_{\pm}$, moving at $\pm v_0$. And a stationary ATC frame, $S$. They're all at the same spatial point at $t=t'_+=t'_-=0$.

In $S$, $t$ ticks away as the planes move along the world lines:

$$x_{\pm}(t) = (t,x_{\pm}) = (t, \pm v_0 t)$$

Let's transform $x_{\pm}(t)$ in $S'_{\pm}$.

First the position:

$$ x'_{\pm}=\gamma(x_{\pm}-(\pm v_0 t))$$ $$ x'_{\pm}=\gamma(\pm v_0 t -(\pm v_0 t))$$ $$ x'_{\pm} =0 $$

Meaning, the planes remain at the origin of their coordinate system in which they are rest. Good.

Now the time coordinate:

$$ t'_{\pm} =\gamma_0\big(t -(\pm v_0)x_{\pm})\big)$$ $$ t'_{\pm} =\gamma_0\big(t -(\pm v_0)(\pm v_0 t)\big)$$ $$ t'_{\pm} =\gamma_0(t - v^2_0t))$$ $$ t'_{\pm} =t \times \gamma_0(1 - v^2_0))=t\gamma_0\gamma_0^{-2}$$ $$ t'_{\pm} =t/\gamma_0 $$

So both plane clocks tick slower than the ATC clock, being dilatedly $\gamma_0 = 1/\sqrt{1-v_0^2}$.

Now we pick a plane and have it look at the other plane. The relative velocity of $S'_{\pm}$ in $S'_{\mp}$ is:

$$ v_1^{\pm} = \frac{\pm v + \pm v}{1+(\pm v)(\pm v)}$$ $$ v_1^{\pm} = \frac{\pm 2v_0}{1+v_0^2}$$

Now take the coordinate of the other $\mp$ plane as seen from $S'_{\pm}$ at time $t'_{\pm}$:

$$ x'_{\pm}(t'_{\pm}) = (t'_{\pm}, \mp v_1 t'_{\pm}) $$

and transform it to $S'_{\mp}$.

I'll leave that as an exercise. The answer is:

$$ t'_{\mp} = t'_{\pm}/\gamma_1 $$

showing that each plane sees the other's clock running slower.

This is really is the first paradox of time dilation, and the resolution to the apparent contradiction (of course there is no actual contradiction), is that when $S'_{\pm}$ is looking at $S'_{\mp}$'s clock, it's not the same time that $S'_{\mp}$ is looking at $S'_{\mp}$'s clock. The relativity of simultaneity matters.

Answer 4

Time is a lot more like space than you would expect. We don't see this because we move so slowly.

Light travels at $3 \times 10^8$ m/s. We are comfortable at $3$ m/s. We have difficulty understanding relativistic physics.

Consider a world where the fastest motion is $3 \times 10^{-8}$ m/s. This is about 1 m/year, the speed of a glacier. It is not much faster than $1$ cm/year, the speed of continental drift. We can learn about our difficulties by looking at the difficulties glacier world physicists have with everyday physics.

In classical glacier world physics, each object has a fixed, intrinsic property called position. Every observer agrees on the position of a given object. Position can be used as the identity of the object.

However, precise measurements or measurements over long time intervals show that position changes with time. This leads to the counter-intuitive concepts of "velocity" and the "failure of sameplaceity".

These can usually be ignored. But observers traveling at everydayistic velocities would see strange effects. Bob and Alice both agree that they both have position $0$ at time $t_0$. At $t_1$, Bob says he has position $x_0$, just as one would expect. Likewise, Alice says all is normal with her. But Bob says Alice is at $x_1$ and Alice says Bob is at $-x_1$.

This is not only counter-intuitive. It is a paradox. Not only does an object have different positions at different times, but different observers don't agree on what the positions are. They see a difference growing in opposite directions.

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Our conceptual difficulties are much the same. We are used to different positions at different times, and two observers measuring different positions for the same object.

We think of position and time as the identify of an event. We think of all events at a given time as identifying the state of the universe. We think all observers should agree on the time measurement of a given event, even if they disagree on position. We do not expect the velocity of the observer to affect the measurement of time.

This understanding of time is incorrect. Instead, we see on a space-time diagram that one observer's present includes another observer's past and future in a velocity dependent way. This leads to all the counter-intuitive concepts and paradoxes of special relativity.

Length contraction is one example. An observer sees a rod at rest. He measures the position of the two ends at the same time. From this he gets the length.

Another moving observer measures the positions of the two ends at his version of the same time. Since he is using different times than the first observer, he is seeing the ends in different places. It is not surprising that he gets a different length.