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Here’s what I don’t understand about time dilation. Alice is at rest and Bob is moving with velocity V with respect to Alice. Let’s say Alice measures two events separated in time by $\Delta t_A$ according to her own watch, and she measures those events in the same place, so she has measured her own proper time.

By applying LTs, we get that $\Delta t_B’ = \gamma \Delta t_A$. What does this time mean? Is this the proper time elapsed between the two events as measured by another observer Bob who is measuring time with his own clock at the same place? I mean, what we compute by using LTs, is Bob’s proper time or is it something different? And if it IS something different, how can we compute Bob’s proper time?

Also, since $\Delta t_B’ > \Delta t_A$, I don’t understand why people usually state that “moving clocks tick slower’. Alice being at rest thinks her proper time $\Delta t_A$ is less than $\Delta t_B’$. For examples, Alice thinks she has measured one hour and Bob measured 2 hours, right? If I were Alice, I would conclude that Bob’s moving clock is running faster, rather than slower.

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Let’s say Alice measures two events separated in time by ΔtA according to her own watch, and she measures those events in the same place, so she has measured her own proper time.

Technically, the proper time is only defined along the worldline of an observer, so for the time between two events to be Alice's proper time it is necessary not only that they be at the same place, but also that place must be Alice's position. (This is a nit-picking requirement that is not important in SR but becomes important in GR). So I will assume that the two events are indeed on Alice's worldline so that it is indeed her proper time.

Is this the proper time elapsed between the two events as measured by another observer Bob who is measuring time with his own clock at the same place? I mean, what we compute by using LTs, is Bob’s proper time or is it something different?

It cannot be Bob's proper time because Bob is moving relative to Alice, so at most one of those events could be on both Bob and Alice's worldline. So it is instead Bob's coordinate time, meaning the time as shown on a lattice of clocks at rest relative to Bob and synchronized using the Einstein synchronization convention.

I don’t understand why people usually state that “moving clocks tick slower’.

It is true that this phrasing is a bit confusing. Since whether a clock is moving or not depends on the reference frame and since no frame is specified, it is ambiguous.

What is unambiguous is the following: $\Delta t_A$ is measured by a single clock which was present at both events, but $\Delta t'_B$ is measured by a pair of synchronized clocks each of which was present at only one event. The time difference for the single clock is always less than the time difference for the pair of synchronized clocks. "Moving clocks tick slower" means that whenever you are comparing a single clock to a lattice of synchronized clocks (in the lattice clock's frame) then the single clock will be slower whenever it is moving (in the lattice frame).

For examples, Alice thinks she has measured one hour and Bob measured 2 hours, right? If I were Alice, I would conclude that Bob’s moving clock is running faster, rather than slower.

But in that comparison Alice is not looking at a single Bob clock. Any one Bob clock will run slow compared to Alice's lattice of synchronized clocks. But Bob's clocks are not synchronized in Alice's frame due to the relativity of simultaneity. So subtractions of times on different Bob clocks is essentially meaningless for Alice.

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  • $\begingroup$ I really like your answer, but I don’t exactly understand what you mean when you talk about “lattice of clocks”. I think what I think is a moving clock, you consider to be a bunch of rest clocks forming like a net, is that it? However, I don’t think thats strictly necessary for me to get your point. You say that the time difference for a single clock is always less than the time difference for a pair of clocks. But why is that so? You state it, but I don’t know how you came to that conclusion. I might as well just need some more time to pause and ponder on what you’ve written. $\endgroup$
    – Rainbow
    Jun 27, 2022 at 16:57
  • $\begingroup$ This is just cosmetic objection, but I think it would be less convoluted to just say that the time $\Delta t'_B$ is Bob's proper time between events A' and B' on his worldline that are simultaneous (according to Bob) with original events. $\endgroup$
    – Umaxo
    Jun 27, 2022 at 18:17
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    $\begingroup$ @Rainbow I have added a pictorial representation of a lattice of clocks. Basically, that is part of the definition of an inertial reference frame in SR. As far as why the time difference for a single clock is always less than the time difference for a pair of clocks, that is because of time dilation. It can be derived directly from the Lorentz transform $\endgroup$
    – Dale
    Jun 28, 2022 at 2:23
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$\Delta t_{B}'$ is the time the moving observer (Bob) measures between the events for which Alice measured $\Delta t_{A}$. Do not conflate this with proper time though. Proper time refers to the measurement in the POV of an observer who is being the subject of that time-like separation of events. Here Alice and Bob observe a third phenomenon, with the difference being that Bob is in a moving frame of reference. If the subject of observation was Bob himself, then you could speak of Bob's proper time. Of course in that case Bob would be the static observer from his POV (since he sees himself stationary in his own frame of reference), so his measured proper time would be smaller than the time Alice measured.

A example of this is the famous muon paradox. There - without accounting for special relativity- muons even with velocities very close to $c$ would in their vast majority decay within a short distance from the upper levels of Earth's atmosphere where they are generated due to cosmic rays. Yet we are able to detect a large number of them. From our perspective, their time is dilated (we are Alice) since they are moving and thus manage to reach Earth's surface. From their POV where they are stationary (they are Bob), time is not dilated i.e. it is the muons' proper time. What amends this is that from their own frame of reference, it is the Earth that is moving towards them, meaning the distance they cover is subject to relativistic length contraction.

The concept of clocks ticking slower is precisely in reference to time dilation with regards to the subject of observation being the second observer as well. The moving muons (Bob) in their frame check their clock and notice no changes. The scientists on Earth (Alice) check the former's clock and find it has moved less than their own between two events. Hence the moving clock "ticks more slowly".

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  • $\begingroup$ Alright, so “moving clocks tick more slowly” only makes sense when stated by an observer S which is at rest with respect to a moving observer S’, the moving observer being what is observed by the observer at rest S. Is that it? This does makes sense because here t’<t does allow S to come to the conclusion that the moving clock runs slows, since the time t’ it shows is less than t. Also, thanks a lot for answering :). $\endgroup$
    – Rainbow
    Jun 27, 2022 at 17:12
  • $\begingroup$ Yes, that's more or less the idea. $\endgroup$
    – rhomaios
    Jun 27, 2022 at 17:42
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The statement that 'moving clocks tick more slowly' is misleading in two important ways.

Firstly, if while you sit at your desk reading this you are passed by two spaceships in line, one ahead of the other, and you look at a clock aboard each spaceship as it passes, the time difference between the two clocks will be greater than the time difference by your watch, ie the moving clock on the second spaceship will appear to have been ticking faster, not slower, than your watch. In that context the phrase 'moving clocks tick more slowly' is literally false.

Secondly, time dilation in SR reflects the fact that the time difference between two events that occur in the same place in one inertial frame will always be less than the time between them in any other inertial frame in which they occur in two separate places. That is a property of the geometry of spacetime, which clocks measure. For example, an interval between two events might be five seconds in one frame and eight seconds in another; the fact that accurate clocks in each frame report the interval as being longer in one frame than in the other is not because the moving clocks are ticking more or less slowly, but because the intervals are actually different. The phrase 'moving clocks tick more slowly' gives the impression that as a consequence of movement a clock under-reports time in some way, which is simply incorrect: an accurate clock will still record a second as a second.

The time dilation formula in SR applies where you have one clock which is passed by two others (or, equivalently, moves between two others) which are stationary relative to each other. The duration measured by the single clock will be less than the duration according to the time difference between the two clocks.

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Relativity can indeed be confusing, when you have to figure out what is what. Time dilation is a joint effort of Einstein and Newton. You can see a short video on youtube if you search 'time dilation Newton'. Time dilation is the result of Newtonian momentum conservation, while Einstein famous formula (E = mc²) is a mass increase for objects in motion. The video will carry you through the math (simple algebra), so it is not too complicated.

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  • $\begingroup$ "Einstein famous formula" is just incorrect. The correct versions are 1) $E_0 = mc^2$, and 2) $E=\sqrt{(mc^2)^2 + p^2 c^2}$. $\endgroup$
    – warlock
    Sep 15, 2022 at 7:19

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