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I have a lot of confusion on special relativity. I am going to use the muon example because that's what was clearest for me to visualize.

A muon generally decays too fast for it to reach the earth from where it is created in the atmosphere, but we detected them here on earth because they go so fast they experience relativistic effects.

From the muons frame, there is length contraction: its going the same speed in a shorter amount of time, which makes enough sense to me. From the muons frame, the change in position between it and the earth between it's creation and detection is smaller than in the frame of the earth. i.e. $$\Delta x_e=\gamma \Delta x_m$$ (I think, sorry if I have the equation wrong).

But the frames were chosen somewhat arbitrarily: why couldn't we say the exact opposite: earth is experiencing time slower than the muon, and the distance traveled is shorter in the earth frame?

I know I have some fundamental misunderstanding about relativity here (and I think similar questions have been asked but I haven't found one with an answer that I really understood), so any help would be greatly appreciated.

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If you want everything to be symmetric then just identify the key events in one frame and use the Lorentz transform to transform them into the other frame.

The length contraction and time dilation formulas are just shortcuts to the Lorentz transform for special cases. They will automatically pop out of the Lorentz transform whenever appropriate. So they really are not necessary nor appropriate for new students.

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The effects of relativity are indeed symmetric. We consider the muon to speeding towards us at nearly the speed of light, and we calculate that its life span is extended by about a factor of ten owing to time dilation. In the muon's frame, the muon is stationary, its life-span is unchanged, the Earth is speeding towards it at nearly the speed of light, and it is time on the Earth that is dilated by a factor of ten.

There is no contradiction between these two viewpoints- they are simply reciprocal perspectives. Suppose you are on a sandy beach and you walk 10m away from the water and then back again. In your frame you have walked 20m altogether. From the point of view of someone passing on the deck of a boat which is going parallel to the shore at the same speed that you are walking, your 20m walk at right angles to the waterline appears to be a zigzag path at forty five degrees to the waterline and about 28m long, so to them it seems longer than it does to you. If they walk 10m across to the other side of their deck then back again, what seems to them to be a 20m walk at right angles to the shore seems to you to be a 28m walk at forty five degrees to it. You each see the reciprocal effect that the other sees.

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You're not the only one with a lot of confusion about special relativity! The thing to focus on is what is happening locally for an observer, vs what is happening far away for that observer. Someone on Earth (say, the middle of a field in rural Pennsylvania) will see events and distances in his immediate neighborhood progressing just as you would expect from any law of physics, including any muons that are stationary with respect to him decaying at the normal rate predicted by QM.

But if he looks far away and observes the muon from space which is traveling very fast with respect to him (if he could somehow actually see the muon's progression) it would be decaying much more slowly. E.g. if instead of a muon it were a clock flying through space that could be observed with a telescope, the seconds would tick more slowly as observed by the terrestrial observer.

Now from the standpoint of a person riding alongside the muon or flying clock, he sees the muon decay at a normal rate, his own clock tick at a normal rate, but he sees the Earth clock and Earth muons taking much longer to proceed in their processes. And furthermore he sees the Earth's lengths contracted. The Earth-man also sees the Space-man's lengths to be contracted, but the original problem did not emphasize this because muons are point particles.

So the situation is completely symmetric.

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  • $\begingroup$ Well, it's almost completely symmetric. Since the muon accelerates towards the earth but not vice versa, it perceives the distance between itself and the earth to be shorter than observers on the earth do. $\endgroup$
    – Gumby The Green
    Commented Mar 30, 2022 at 5:27
  • $\begingroup$ I believe the implicit assumption of the scenario was that the muon, created in a cosmic ray collision in the upper atmosphere, traveled at essentially constant speed to the surface or the Earth. But regardless the acceleration should not break the symmetry. $\endgroup$
    – RC_23
    Commented Mar 31, 2022 at 1:22
  • $\begingroup$ Sure, it travels at essentially constant speed after it accelerates nearly instantaneously. I'm just reiterating one source of asymmetry noted in the question - the fact that the distance the muon travels toward the earth is shorter in its frame than it is in the earth's frame - pointing out that this is enabled by the asymmetry of the forces, causality, and acceleration acting on the muon vs the earth: clearly, the muon is propelled toward the earth and not the other way around. $\endgroup$
    – Gumby The Green
    Commented Mar 31, 2022 at 5:11
  • $\begingroup$ I want to make sure it's clear that the contracted distance between the top of the Earth's atmosphere and the Earth's surface, in the muon's frame, is solely due to the Earth's motion in the muon's frame. No acceleration is needed, and if the muon were traveling from interstellar space at a constant speed all the way to the Earth's surface, the contracted length would be the same. $\endgroup$
    – RC_23
    Commented Mar 31, 2022 at 16:46
  • $\begingroup$ I agree but I would add that it’s due to the motion of the earth and its atmosphere in the muon’s frame. The fact that the atmosphere shares the earth’s frame is the reason why the distance between the earth and the top of its atmosphere (the distance the muon travels) is contracted in the muon’s frame but not in the earth’s. Arguably, this is functionally equivalent in the given scenario to what I was saying earlier but hopefully the point is more clear and technically correct now. $\endgroup$
    – Gumby The Green
    Commented Apr 3, 2022 at 10:30

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