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If a rocket was moving away from earth at say 0.5c, why do we consider the earth as stationary object and the rocket is moving? Can we assume that the rocket is stopped but the earth is moving backwards? If yes why doesn't the people in rocket age faster and people on earth age slowly?

Mohammad Mizanur Rahaman
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    Does this answer your question? https://physics.stackexchange.com/questions/383248/how-can-time-dilation-be-symmetric Velocity is relative and symmetric, as are functions of velocity like Lorentz factor. – g s Feb 05 '23 at 19:35
  • "If yes why doesn't the people in rocket age faster and people on earth age slowly?" In the rocket frame, the people in the rocket DO age faster than the people on earth, so your question has no basis. – WillO Feb 06 '23 at 01:11

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You can certainly assume that the rocket is stopped while the earth is moving backwards. But the result would not change, here's why.

This is how time dilation is proven:

  • We consider two different frames of reference separated by a Lorentzian boost i.e. one frame is moving with respect to the other at a constant speed. We use unprimed and primed coordinates to distinguish the frames.
  • Then, we consider two points in spacetime and measure the time difference between them in these frames. In addition, we assume that these two events spatially coincide in at least (and of course, at most) one of the two frames.
  • Finally, using the Lorentzian boost, we find the equation for time dilation.

Let's consider the two scenarios implied by the italicized text.

  • If we suppose that the events spatially coincide in the primed coordinates, we have the following: $$\Delta x' = \gamma(\Delta x -\beta c \Delta t)=0$$ $$\Delta x = \beta c \Delta t$$ $$\Delta t' = \gamma \left(\Delta t - \frac{\beta}{c} \Delta x \right) = \gamma (1 - \beta^2)\Delta t = \frac{\Delta t}{\gamma}$$

  • On the other hand, if we suppose that the events spatially coincide in the unprimed coordinates, we have the following: $$\Delta x = \gamma(\Delta x' +\beta c \Delta t')=0$$ $$\Delta x' = -\beta c \Delta t'$$ $$\Delta t = \gamma \left(\Delta t' + \frac{\beta}{c} \Delta x' \right) = \gamma (1 - \beta^2)\Delta t' = \frac{\Delta t'}{\gamma}$$

These two results are not contradictory as the choice of events makes a difference. What is not important, as you may have noticed, is which of these frames is the one we are considering to be in motion! The situation is symmetric until you choose your spacetime events.

So, for instance, when you consider twins, twin $A$ being on Earth, twin $B$ on the rocket. If one twin (does not matter which!), say $A$, observes $B$ and measures the duration between two consecutive birthday cakes on the rocket, his own clock will measure more than one year. And this is reversible: if $B$ observes $A$ and measures the duration between two consecutive birthday cakes on Earth, his clock will measure more than one year.

There is no "faster clock", it depends on the frame chosen.