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## 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 travelled 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?