Muon experiment

Question

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?

Answer 1

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?

Nothing is wrong, you just stopped too soon. If you continued by substituting in for $\Delta x$ and simplifying then you would get:

$$ \Delta t' = \gamma \left(\Delta t - \frac{v}{c^2} \Delta x\right)$$ $$ = \gamma \left(\Delta t - \frac{v}{c^2} v \Delta t \right)$$ $$= \gamma \left( 1-\frac{v^2}{c^2}\right) \Delta t$$ $$=\Delta t/\gamma$$