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Recap: A spaceship is heading towards the earth at a relativistic velocity, if the earth in its frame of reference measures some distance between itself and the spaceship, would the spaceship in its own frame of reference at that time, measure the same distance between itself and the earth?

My thoughts on this are that in both frames of reference, the other is heading towards them with the same velocity and neither frame of reference can see the two objects (Earth and spaceship) as stationary and both see the other moving toward them at the same speed, so would I be right in saying they both see the same distance between themselves at all times and neither see some sought of length contraction in their inertial reference frames? (Until collision)

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  • $\begingroup$ Length contraction would occur when a length measured in a frame of reference for which two points are stationary is compared to a length measured in a reference frame for which the two points are not stationary. But in both reference frames, the distance between the observer, stationary in their reference frame, and the other observer moving toward them, is constantly changing, as both observers see the other one moving towards them at the same velocity. I still can't see how one observer on earth or the spaceship could measure a different distance. $\endgroup$ Oct 27, 2022 at 6:31
  • $\begingroup$ We would measure a distance with way markers and so would the ship... but the ship's way markers would move relative to us and ours would move relative to the ship. The situation is fully symmetric. So you could ask if either observer sees the distances between the other observer's way markers contracted and they would. $\endgroup$ Oct 27, 2022 at 6:39

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The problem is that the observers on Earth and the observers on the space ship are not measuring the same thing when they measure times. So you can say:

At time $t$ observers on Earth measure the distance to the spaceship to be $x$
At time $t'$ observers on the spaceship measure the distance to Earth to be $x'$

but since the Earth and spaceship time axes are different you have to define how you are going to relate the times $t$ and $t'$.

There is an obvious way to do this if you don't mind a slight modification to your experiment. Suppose the ship passes the Earth and at the moment it passes both the Earth and ship observers reset their clocks to read zero. Then for both observers their clocks measure the time elapsed since passing. We can then ask how the observers find the Earth-ship distance changes with the time shown by their clocks.

This simplifies the situation because in both frames the relative speed is the same i.e. the Earth and ship observers measure the velocity to be $+v$ and $-v$ respectively. Then for both observers the separation is just given by:

$$ s = v \tau $$

where $\tau$ is the time shown by their clocks. If we define the separation this way it is the same for both observers.

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  • $\begingroup$ Thank you, that's very helpful, but what if the Observer on earth notes some event at the spaceship when the spaceship is a certain distance from earth according to the earthly observer, would the observer on the spaceship see the earth as being equally far away for the event noted by the earthly observer in the spaceship's frame of reference? $\endgroup$ Oct 27, 2022 at 11:21
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To answer your original question and the follow-up you added to John's answer, the key point you must bear in mind is that when you talk about how far away a moving object is, you must specify a time. An observer on Earth might wonder how far away the space-ship is 'now', while the observer on the spaceship might wonder how far away the Earth is 'now', and since they both mean different things by 'now' they get different answers.

So let's imagine that the spaceship jettisons an empty fuel tank at noon Earth time. For the Earth people, that happens when the time on Earth is noon, so when they talk about how far away the spaceship was when it jettisoned the tanks, they mean how far away was the spaceship when it was noon at the spaceship and noon on Earth. However, for the people on the spaceship who jettisoned the tank, the current time on Earth is not noon (it is either earlier or later than noon, depending upon the direction in which the spaceship is headed), so the distance to the Earth 'now' is actually the distance to the Earth at some earlier or later time than noon on Earth.

So hopefully you see that length contraction is a result of measuring the distance to a moving object at different times in two different frames, with the natural result that you will get different answers because the position of the object changes between the two measurements.

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