# Muon experiment

I am new to special relativity and I'm finding it really hard to understand concepts such as time dilation. The example I'm trying to understand is that of a decaying muon. The muon starts falling from height $$h$$ down to the Earth, at speed $$v = 0.99c$$. What I want is to obtain the time ellapsed during that process as measured in two different reference frames: the Earth's and the muon's reference frame.

Here's how I've tried to find those times ellapsed. First I define $$S$$ to be the Earth's reference frame, which I assume to be stationary. Then I define $$S'$$ to be the reference frame moving at velocity $$v$$ with respect to $$S$$. In $$S'$$ the muon is at rest. I also define two events (1) 'starting to fall' and (2) 'reaching Earth's surface'. Thus we should have:

According to $$S$$: $$x_1 \neq x_2$$, $$\Delta x = h$$, $$t_1 \neq t_2$$, $$\Delta t = \Delta x /v$$.

According to $$S'$$: $$x_1' = x_2'$$, $$\Delta x' = 0$$, $$t_1' \neq t_2'$$, $$¿ \Delta t' ?$$.

I thought that in order to find $$\Delta t'$$ we could use LTs as follows:

$$t_1' = \gamma (t_1 - \frac{v}{c^2}x_1)$$ $$t_2' = \gamma (t_2 - \frac{v}{c^2}x_2)$$

And substracting those we get:

$$\Delta t' = \gamma (\Delta t - \frac{v}{c^2} \Delta x)$$

But there must be something wrong with my reasoning because the right answer should be $$\Delta t' = \Delta t / \gamma$$. Could anyone explain to me what's wrong?

• I go through exactly this calculation here. Jun 5, 2022 at 10:49
• @JohnRennie Thanks for answering, I wasn’t aware my question has already been covered. However there is something I don’t quite understand about your calculation. Why do you identify the proper time $t_{\mu}$ with an unprimed $t$ instant? Isn’t that time measured according to $S’$, which is the muon reference frame moving with velocity $v$ respect to $S$? Jun 5, 2022 at 10:59
• @JohnRennie This is so confusing. Even if you decided to name $S$ what I named $S'$ and to name $S'$ what I named $S$, shouldn't I obtain the same result? The decision of naming your reference frames shouldn't influence the result, right? Jun 5, 2022 at 12:24

$$\Delta t' = \gamma (\Delta t - \frac{v}{c^2} \Delta x)$$
But there must be something wrong with my reasoning because the right answer should be $$\Delta t' = \Delta t / \gamma$$. Could anyone explain to me what's wrong?
Nothing is wrong, you just stopped too soon. If you continued by substituting in for $$\Delta x$$ and simplifying then you would get:
$$\Delta t' = \gamma \left(\Delta t - \frac{v}{c^2} \Delta x\right)$$ $$= \gamma \left(\Delta t - \frac{v}{c^2} v \Delta t \right)$$ $$= \gamma \left( 1-\frac{v^2}{c^2}\right) \Delta t$$ $$=\Delta t/\gamma$$