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Say we have my friend blasting off in a spaceship to Planet Physics and I’m sitting here on earth. I think it’s quite obvious that I will measure the proper length between the planets (because I could use a meter ruler and measure it all) and that he will measure the contracted length (because his length is squished).

What I don’t understand is why does he measure the proper time of the journey and I don’t. I can appreciate that my friend should measure proper time because “he is stationary relative to both events (launch and touchdown)” but when I sit on earth I think I’m stationary to the launch and touch down too. So, why can I not measure the proper time it takes to reach planet physics? Can you explain why I am not stationary relative to touch down (I always see the planet staying still in the sky!)

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    $\begingroup$ There is no such thing as being "stationary relative to an event". $\endgroup$ – Emilio Pisanty Sep 21 at 7:06
  • $\begingroup$ @robjeffries : I once unthinkingly posted a comment essentially identical to yours and was glad to be corrected. I hope you are too. :) $\endgroup$ – WillO Sep 21 at 13:20
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The proper time is by definition the time as measured by an inertial observer who is present at both events. That's your friend and not you.

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    $\begingroup$ You mean "the proper time interval between two events". Proper time is just time measured along a particular observer's world line. $\endgroup$ – Rob Jeffries Sep 21 at 19:21
  • $\begingroup$ @RobJeffries : This is of course a question of vocabulary, not of physics, and perhaps different people speak different dialects. But I'm fairly sure that in standard usage, "proper time" is time measured along an inertial observer's worldline, and therefore measures the proper time interval between any two events on that worldline. The proper time interval between two events is (as I'm sure we agree) uniquely defined,and "proper time" --- at least if I'm right about standard usage --- refers to essentially just that. $\endgroup$ – WillO Sep 22 at 1:53
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    $\begingroup$ Semantics or not, I disagree. The first thing I check does not have your boldface addition and matches my description precisely. en.wikipedia.org/wiki/Proper_time Alfred's answer is correct. $\endgroup$ – Rob Jeffries Sep 22 at 7:56
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Assuming flat spacetime for simplicity, there is a proper time $\Delta\tau$, associated with two events with time-like interval, related to the Lorentz invariant interval $\Delta s^2 = (c\Delta t)^2 - (\Delta r)^2$:

$$c\Delta\tau = \sqrt{\Delta s^2}$$

This proper time is equal to the elapsed time (between the events) given by an inertial clock with world line through both events. Put another way, an unaccelerated clock co-located with both events, will measure the proper time associated with the events.

But in general, there is the proper time $\tau$ along the world line of an accelerated observer through two events (which will always be less than the proper time associated with the two events). This proper time is defined as the elapsed time between the events given by a clock with a world line through both events.

The clock on the spacecraft records the proper time along the accelerated world line from the launch event to the landing event. While the world line of your clock is through the launch event, it is not through the landing event.

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Special relativity is not special. It is simply a process where an observer A measures observer B when they are in relative motion. There is no preferred frame of reference. Thus observer A cannot say they are stationary (preferred frame) while observing B. They both move away from each other. That is what relativity is. Both observers see each other’s “clocks” run slower. When they return their clocks are identical. This is the result of using a method of observation with a particle or wave that has a finite speed. Special relativity also applies to sound waves. In fact any wave. Submarines that use sonar actually observe the other moving submarine shorter than its length when both are stationary or moving in tandem. It is the result of the finite speed of the sound wave. Lengths do not physically shrink. The just appear to shrink. It is an optical illusion only.

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    $\begingroup$ Welcome New contributor John ruffolo! I've downvoted your answer for the "not useful" reason - your answer doesn't even mention the notion of proper time, the primary focus of the OP's question. $\endgroup$ – Alfred Centauri Sep 21 at 17:30
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As other have pointed out, the concept of "proper time" is what Einstein discarded in special relatively. You each experience time and distance differently but in a mutually consistent way and reconcilable way.

You scenario is however not covered by special relativty.

If your friend was making a fly-by Earth follwed by a fly-by of Plant Physics having maintained a constant course and speed, (e.g. without accelerating and decelerating) then you would bot be in "inertial frames" with a constant relative velocity between you and him, which is what special relativity covers.

Because you and your friend start and end with no or little relative velocity but he has had undergone high and sustained acceleration on the space ship while you have not, then your experiences are not symmetrical, there is no "paradox" with him having aged less than you and when he sends you a message to say he arrived, you may be long dead of old age.

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    $\begingroup$ Proper time is perfectly well defined in SR, the scenario is perfectly well described by SR, any acceleration is totally irrelevant to the question at hand, and the OP did not ask about the "paradox" you're gratuitously trying to debunk. You're averaging one major error per paragraph. $\endgroup$ – WillO Sep 21 at 13:25
  • $\begingroup$ Thanks - Fair point - I had forgotten the specific meamomg rather than him thinking it is some kind of absolute time. But anyway its space-time path-dependent isn't it? So acceleration changes the path you follow. $\endgroup$ – Duke Bouvier Sep 21 at 15:24
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    $\begingroup$ We can take the accelerations at the beginning and end to be instantaneous, so they have no bearing on the relevant part of the traveler's path, and he is inertial for the relevant part of the journey. And even if there were relevant accelerations, SR would have no problem dealing with them. $\endgroup$ – WillO Sep 21 at 15:49

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