Special relativity muon example

[Kara386]

Homework Statement

Here's a standard example of special relativity in action:

The mean lifetime of the muon as measured in a laboratory is about 2µs (rounded to 1 s.f.). Thus, the typical distance traveled by a muon should be about $3\times 10^8ms^{-1}\times 2\times 10^6s = 600m$. The atmosphere is about 20 km thick, so the fraction reaching Earth should be about $e^{\frac{-20km}{0.6km}} = e^{-33}$ ≈ 0. However, we detect ∼ 1% at sea level! How can we explain this?

Homework Equations

$l = \frac{l_0}{\gamma}$
$t = \gamma t_0$

The Attempt at a Solution

Here $\gamma$ is about 7. This is probably just me being really stupid, but this is a worked example we've been given and in the solution, it says that the length according to the muon in the muon's frame is $\frac{20km}{7} = 3km$.

I thought that in the equation $l = \frac{l_0}{\gamma}$, $l_0$ was the 'proper length' - the length measured in the rest frame of the muon. Well, that would make the length the muon sees $20 \times 7$. wouldn't it? That's obviously wrong, because length in the moving frame contracts. So do I just have the definitions of $l$ and $l_0$ the wrong way round? Or worse, do I have them the right way round but their definitions wrong?

 

[Doc Al]

I thought that in the equation $l = \frac{l_0}{\gamma}$, $l_0$ was the 'proper length' - the length measured in the rest frame of the muon.

Yes, $l_0$ is the proper length, but proper length means the length of something as measuring from its rest frame. Here we are talking about the "length" of the atmosphere, so 20 km is the proper length and $l$ is the distance measured from the muon's moving frame.

 

[Kara386]

Yes, $l_0$ is the proper length, but proper length means the length of something as measuring from its rest frame. Here we are talking about the "length" of the atmosphere, so 20 km is the proper length and $l$ is the distance measured from the muon's moving frame.

So the proper length of the atmosphere is the same in lots of different frames, but proper time is specific to each frame? So for the muon, proper times are times measured in its frame, but if we chose Earth's frame, while the proper length of the atmosphere is the same, proper times are different, right?

 

[Doc Al]

So for the muon, proper times are times measured in its frame, but if we chose Earth's frame, while the proper length of the atmosphere is the same, proper times are different, right?

Clocks always measure 'proper time'. Here, the muon itself acts like a clock, so from the Earth's viewpoint the decaying muon is a clock that runs slow.