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Let us have the familiar Lorentz transformations, given by :

$$\Delta t_{s'}=\gamma\left(\Delta t_s-\frac{v\Delta x_s}{c^2}\right)$$ $$\Delta t_{s}=\gamma\left(\Delta t_{s'}+\frac{v\Delta x_{s'}}{c^2}\right)$$

I want to derive the concept of time dilation from here. My problem is, different sources do it differently. For example, what books like Griffiths, or Dr. Brian Greene's lecture on youtube does, is state the following :

We are looking at a clock in motion. Let $S$ be our frame at rest, and $S'$ be the frame of the moving clock. The clock is at rest in it's own moving frame, and so $\Delta x_{s'}=0$. Plugging this into the second equation, we have :

$$\Delta t_s=\gamma \Delta t_{s'}$$

This clearly shows that time elapsed on our stationary clock between any two events is more than the time elapsed on the moving clock. This shows that the moving clock is running slow, from our perspective.

However, other books like Kleppner and even the Wikipedia article does this slightly differently :

The state $\Delta x_s=0$, and use the first relation to show :

$$\Delta t_{s'}=\gamma\Delta t_s$$

This is exactly opposite of what predicted by our first relation. Hence I'm inclined to believe that this new relation doesn't relate the time elapsed on the moving watch and the watch at rest. Instead it relates something else, like the time period of the moving watch against that of the watch in rest. Since, the moving watch runs slow, it should take a longer time in the perspective of the watch at rest to complete one second, or one full turn.

My question is, how does setting $\Delta x_{s'}$ or $\Delta x_s$ to be equal to $0$, change our interpretation of what $\Delta t_s$ and $\Delta t_{s'}$ actually represent. In the first case, they represent the elapsed time on each clock, but in the second case, they are measuring the time period of each clock ? How and why does the interpretation of these terms suddenly change ?


The second confusion is, when we say a moving clock runs slower, what we actually mean is that, when 'we the observer at rest' looks at a moving clock, it seems to tick slowly compared to the watch which is at rest with us. Suppose, when our clock ticks $5$ times, the clock in motion ticks once, between two events $A$ and $B$. But now suppose, someone else is moving with the clock in motion. That person encounters the same two events. However, he is at rest with respect to the moving clock. How many ticks of the clock would he record?

The answer is supposed to be $1$ tick since the moving clock is ticking slower. But this is something that I'm having difficulty wrapping my head around. The moving clock ticks once, between $A$ and $B$, when we at rest are looking at it. Why will it also tick once, for someone moving along with the clock, and looking at it ? Is this because, according to the moving observer, the distance between the two events has been length contracted ?

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My question is, how does setting $\Delta x_{s'}$ or $\Delta x_s$ to be equal to $0$, change our interpretation of what $\Delta t_s$ and $\Delta t_{s'}$ actually represent.

Setting $x'=0$ in the formula $$\Delta t_{s}=\gamma\left(\Delta t_{s'}+\frac{v\Delta x_{s'}}{c^2}\right)$$ means that the clock rests in S', so $\Delta t_{s'}$ is the "proper time" of that clock, which is time dilated with respect to "coordinate time" $\Delta t_s=\gamma \Delta t_{s'}$ in S.

Setting $x=0$ in the other formula $$\Delta t_{s'}=\gamma\left(\Delta t_s-\frac{v\Delta x_s}{c^2}\right)$$ means that we talking about another clock which is at rest in S, so $\Delta t_{s}$ is the "proper time" of that clock, which is time dilated with respect to "coordinate time" $\Delta t_{s'}=\gamma\Delta t_s$ in S'.

Why will it also tick once, for someone moving along with the clock, and looking at it ? Is this because, according to the moving observer, the distance between the two events has been length contracted ?

Yes, the distance is length contracted. It's precisely what happens in the muon-atmosphere time dilation experiments. In the atmosphere rest frame the decay rate of the muons is time dilated so that they can hit Earth's surface, while in the muon rest frame the atmosphere is length contracted so that the muons can hit Earth's surface also in this frame.

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I'll just address your 'second confusion', since the first part of your question has been fully answered.

It is misleading to think that a clock moving past you ticks more slowly than your own watch. In its rest frame the clock continues to tick at the same rate as your watch- however, the geometry of spacetime is such that the clock is moving along a rising plane of time in your frame, in which all clocks in your frame appear to the moving clock to be progressively running ahead of each other- it is that which causes the appearance of time dilation.

To illustrate the principle with simple example, suppose you start walking down a corridor at exactly ten O'clock, according to both your watch and a person standing still next to you holding a clock. After ten seconds on your watch you reach another person standing with a clock- their clock is running a second ahead of the first person's so it reads 10:00:11, while your watch reads 10:00:10. You walk for another ten seconds and find a third person with a clock, that clock being set a second ahead of the last one, so it reads 10:00:22, while your watch reads 10:00:20. And so on. You will see that your watch seems to be losing a second at each new clock you reach, ie to be time dilated. In fact, your watch and all the clocks you pass are running at exactly the same rate, but your watch appears to be running slow because each of the clocks you pass is progressively set ahead of the last one.

The effect I have described is directly analogous to the cause of time dilation, which arises because a plane of simultaneity in the rest frame of a clock is a rising plane of time in any frame through which the clock is moving, time being progressively ahead in the clock's direction of motion. It is that which causes the apparent slow running of the clock, whereas in fact the clock always ticks at the same rate.

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  • $\begingroup$ Do you think it is a bit misleading to say that 'Moving clocks run slow', as often it becomes difficult to see, which clock is moving exactly. Shouldn't a better statement be 'proper time' is less than 'coordinate time', if we define proper-time to be the elapsed time, in a frame where the events happen at the same location ? $\endgroup$ Commented Nov 12, 2021 at 15:39
  • $\begingroup$ In my example, $\Delta x = 0$. Hence, time in S frame, which is supposed to be at rest, is less than time in S' frame which is supposed to be moving. This is rather misleading if someone blindly follows the mantra, that moving clocks tick slower. From the initial wording, one might be led to thinking that the S' clock is moving, since that is the moving frame here. $\endgroup$ Commented Nov 12, 2021 at 15:42
  • $\begingroup$ Yes, I think it is very misleading to say simply that moving clocks run slow. It suggests that a clocks rate of ticking changes when it moves, which is quite wrong. Unfortunately our language is not sufficiently precise to deal with the nuances of the physics. $\endgroup$ Commented Nov 12, 2021 at 16:03

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