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In the lorentz tranformation, I have got the following equation for time: $$t' = \dfrac{t - \dfrac{ux}{c^2}}{\sqrt{1-\dfrac{u^2}{c}}}$$

The $S'$ system is moving relative to $S$ system with speed $u$, and in the following context I will denote $\gamma = \dfrac{1}{\sqrt{1-\dfrac{u^2}{c}}} (\gamma > 1 \ \text{for} \ u < c)$. I will maily focusing on the $t$ term and supposing $x = 0$ (the $\dfrac{ux}{c^2}$ term is irrelevant to my problem). So I get: $$t' = \gamma t \quad \text{or} \quad \Delta t' = \gamma \Delta t\tag{1}$$ The thing which confuses me is the following:

I have read the Feynman Lectures on Physics, and he gives an example about muons (I have summarized the content):

A muon have a life time about $2.2 \ \mu \text{s}$, When it comes to earth with relative speed $u$, in the muon point of view, it lives $2.2 \ \mu \text{s}$, and to us (stationary observer on earth), the muon lives longer than $2.2 \ \mu \text{s}$. The factor by which the time is increased is $\dfrac{1}{\sqrt{1-\dfrac{u^2}{c}}}$.

In my understaning about the above example, the muon is in $S'$ system, which means $\Delta t' = 2.2 \ \mu \text{s}$. The observer on earth is in $S$ system, and the time measured will be denoted by $\Delta t$, So according to the example:

$$\Delta t = \dfrac{\Delta t'}{\sqrt{1-\dfrac{u^2}{c}}} =\gamma \Delta t' \tag{2}$$

We will conclude that $\Delta t$ is longer than $\Delta t'$ from $(2)$, which is familiar to us (time is slower in $S'$). I will make a Simple example on my own:

System $S'$ is moving relative to system $S$ with speed $u$ which makes $\gamma = 2$.In $S'$, if a clock elapsed $1 \ \text{s}$, then the observer in $S$ will get $\Delta t = \gamma \Delta t' = 2 \times 1 = 2 \ \text{s}$ on his own clock(which made from same factory with the clock in $S'$ and they have synchronized the two clock when $S'$ is stationary), this result is in consistent with my thought. But go back to the formula $(1)$, which is quite the opposite of $(2)$.

I can't find why the example's conclusion contradicts the Lorentz transformation of time.

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2 Answers 2

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In the Muon decay problem, it is more convenient to choose Muon frame as the $S$ system in your formula. Note there is a term depending on space coordinate $x$ that cannot be ignored. When we consider the question "what is the life time of Muon?", we are considering two events: the creation and decay of the Muon. The two events have spacetime coordinates $(t_1, x_1), (t_2, x_2)$ in $S$ system and coordinates $(t_1', x_1'), (t_2', x_2')$ in the $S'$ system. Now apply the formula (for clarity setting $c=1$) to both events 1 and 2 $$t'= \gamma t - \gamma u x$$ and take the difference ($\Delta t = t_2-t_1$ etc.) $$\Delta t' = \gamma \Delta t - \gamma u \Delta x$$ Since $S$ system is relatively stationary to the Muon, the two events happen at the same place, so $\Delta x= 0$ and we finally conclude $\Delta t' = \gamma \Delta t$, i.e. the earth frame ($S'$ system) measures a longer lifetime. As an exercise, you can choose the earth frame as the $S$ system, in this case $\Delta x \neq 0$. Using the formula above you can reach the same conclusion.

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  • $\begingroup$ Thanks for your answering. I have check the thing you have mentioned carefully. I find that the mistake I made exactly is that I have missing the second term in my equation, namely $\frac{ux}{c^2}$.When I substitute $\Delta x = u \Delta t$ in the first equation I have posted, I get the right answer:$\Delta t = \gamma \Delta t '$. I also realized that I am not understand the second term physically(even though equation seems to me get the right result). I will be appreciate if you can recommend me a book which can describe this thing in more detail and more physically. $\endgroup$
    – zjp
    Jun 13, 2022 at 15:43
  • $\begingroup$ Additionally, another thing I have found is that when $\Delta x = 0$(a clock A in a stationary man A's hand elapsed $\Delta t$), The man B who is moving will get $\Delta t' = \gamma \Delta t > \Delta t$(The man B may think the clock A is moving backwards). The result is that man B thinks that man A's clock is slower. When $\Delta x = u \Delta t$(a clock B in moving man B's hand elapsed $\Delta t'$), $\Delta t' = \frac{\Delta t}{\gamma} < \Delta t$.What happens to A? The man A thinks that the clock of B is slower. this result is seems bizarre to me. $\endgroup$
    – zjp
    Jun 13, 2022 at 16:03
  • $\begingroup$ @zjp the second term in the equation embodies the relativity of simultaneity. Supppose in $S$ frame two events happen at the same time $t_1=t_2=t$ but at different locations $x_1\neq x_2$. Then in $S'$ frame it would appear the two events happen at different times $t_1'\neq t_2'$. For your additional question, yes, both observers would think the clock of the other is running slow. But relativity tells us that perception of space and time would be different in different frames which is exactly what happens here. You can read more in twin paradox $\endgroup$
    – Dayan
    Jun 16, 2022 at 4:45
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There is no contradiction. Time dilation is symmetrical, which means that if a clock in the frame of the muon seems to us to be dilated by a factor 𝛾 then a clock in our frame seems dilated by the same factor 𝛾 in the frame of the muon.

The reason why you might be confused is that you are using the symbols Δt′and Δt in two different ways in equations 1) and 2).

In one scenario, where t' is the time on the moving clock, then Δt'<Δt.

In the other scenario t is the time on the moving clock, so Δt<Δt'.

So you have Δt'<Δt in one scenario and Δt<Δt' in the other. You think these are contradictions, but that is because you are not understanding that they represent two quite separate but complementary circumstances.

The time dilation formula applies when you have one clock moving between two other which are stationary relative to each other. That is the essential point- you are not comparing one clock with one other. Time dilation arises because in the frame of the moving clock the two stationary clocks are out of synch with each other.

So if you compare the time on a single clock in one frame with the time on two other clocks it passes in another frame, you will always find that the elapsed time on the single clock is less than the elapsed time according to the two clocks that have passed.

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  • $\begingroup$ Thanks for yours answering, when I made some calculation, I have got your point, just like the comments I post above, but I am not sure it is right or not(It refers to the bizarre thing I mentioned above). If it is right, I really really need a explanation! :) $\endgroup$
    – zjp
    Jun 13, 2022 at 16:06
  • $\begingroup$ I will expand my answer to try to explain why you might be confused. $\endgroup$ Jun 14, 2022 at 5:53

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