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A textbook I’m using gives an example regarding relativity that I don’t understand. The exact calculations are not that important, I’m more after the conceptual understanding. It is explicitly stated that no acceleration/deceleration happens so that general relativity does not factor into the example.

It says a spaceship travels from Earth to a star. From the Earth/star frame, the distance is 20ly, and the rocket speed is 0.80c. The example says that from the Earth/star frame this will take 25 years, but from the spaceship’s frame it will take 15 years. It also states that from the spaceship’s frame it will see that a clock on earth will only have measured 9 years. I understand everything up to this point.

It then says that from the spaceship’s perspective, a clock on Earth and a clock at the star will be out of sync by 16 years; because from the spaceship’s perspective earth will have measured 9 years and by the time it reaches the star it will have been 25 years for the trip, which is what will be displayed on the star clock.

My problem is, if the star is in the same frame as the earth, then from the spaceship’s perspective wouldn’t both clocks on the Earth and the star have the same reading? Even if the space ship entered the star’s frame, which the example says doesn’t happen, then won’t it also have entered the Earth’s frame; since the Earth and the star are in the same frame of reference? From what I understand, both the Earth and the star will show 9 years have passed from the point of view of the spaceship. How is there a 16 year difference between the earth and the star from the spaceship’s perspective?

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    $\begingroup$ Remember, simultaneity is relative in SR. If the clocks show the same time in the earth/star frame, you can be sure they won't show the same time in the ship frame. I think they are equally out of sync at the start of the journey as at the end, though (but the ship won't notice until it gets there) $\endgroup$ Jan 21, 2021 at 7:21
  • $\begingroup$ @Kristoffer is right. In the ship frame, the earth and star clocks tick equally fast (or equally slow!)).. But the star clock starts out (and remains) 16 years ahead of the earth clock. (Of course in the earth/star frame, the earth and star clocks are synchronized.) $\endgroup$
    – WillO
    Jan 21, 2021 at 7:58
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    $\begingroup$ Can you draw a position vs time graph (also called a Spacetime diagram) showing (on the same graph) the positions of the earth, the star , and the spaceship during their motions. Imagine each object had a wristwatch attached to it. What events (all such points on the graph) are at “t=0 according to the earth”, “... the star”, and “... the [traveling] spaceship”? (For each, you will draw a line.) Try it. $\endgroup$
    – robphy
    Jan 21, 2021 at 14:13

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If I reply as a physicist than I will say that your question doesn't make any sense. Cause, past, present and future exists simultaneously.

$$t = \frac{t`}{\sqrt{1-\frac{c^2}{v^2}}}$$

You can put those values inside above equation than you will get your answer.

Suppose, we(you and me) are twin I am inside earth you are traveling away.And, your average velocity is 0.4c. You came back to me after 30 years. Now, I wanna calculate your age.

$$t = \frac{30years}{\sqrt{1-\frac{c^2}{(0.4c)^2}}}$$

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You are confusing yourself by thinking you "enter" frames, as if you visit them. A frame is a fictitious box you draw. Drawing it around a star and Earth is quite a big one, but it still works for theoretical discussions. But you never "go out" and "revisit" the frame. In this case the spaceship's frame is just a different one, where the clock on the star and Earth will nearly always have different readings. The ship's acceleration can influence the difference between these readings.

The only frame for which a clock on the star and a clock on Earth generally show the same reading, is a photon's, that travels at c. It "sees" both of these at the same time.

For all other frames, there are very limited points in time and space where information from these two clocks will show the same reading at the same time in that frame.

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