special relativity - Muon demonstration of time dilation

Looking at the second paragraph of page 120 here. Is the half life given measured in the rest frame of the muon? Then the calculation shows that the time to reach sea level from the muon's perspective is less than its half life measured in its rest frame. This explains why so many reach sea level.

But, could we have drawn the same conclusion by instead working out the half life of the muon in our frame? Then $\tau$ would be multiplied by a factor of $\gamma=10$ in our frame to give a value more than $t=7\times10^{-6}$, equally explaining why so many reach sea level.

Am I right in thinking it would be clearer if $\tau$ was labelled $\tau'$ in the text instead, since my understanding is that it's the half life measured in the muon's frame?

Answer

The text you've linked is indeed misleading but correct. There are two ways to solve this problem and they both give the same answer:

1. 
   

   Time dilation. This says that the lifetime of the particle in its rest frame is a "proper time" $\tau'$, which gets dilated for us as $\tau = \gamma ~ \tau'.$ The number of lifetimes that the particle actually experiences while going through the atmosphere is therefore $t / \tau = t / (\gamma \tau'),$ and we expect $e^{-t/\tau}$ of the particles to persist after we observe that time $t$.

   
   


2. 
   
   

   Length contraction. This says that the lifetime of the particle in its rest frame is a "proper time" $\tau'$, and our time $t$ should instead be turned into a distance $L = t~v$ in our frame of reference. The particle does not see this distance directly but actually sees the length-contracted distance $L/\gamma$ which it traverses in a time $t' = (L / \gamma) / v$ (since in relativity if A is moving with speed $v$ away from B in B's coordinates, then also B is moving with speed $v$ away from A in A's coordinates). The number-of-lifetimes is therefore $t' / \tau' = (t v) / (\gamma v \tau') = t / (\gamma \tau').$

   
   

The work that you're looking at is doing analysis (2) but phrasing it as a time dilation problem. This is possible due to the equivalence above, but it will build up a false intuition, because it's not clear to the student why you are dividing $t$ by $\gamma$ rather than multiplying it: relativity of course says that the ground-frame is moving at speed $v$ relative to the particle, why don't we dilate the time even more?

This in turn builds up a false intuition that the ground frame is special, and that relativity is complicated. It is actually very simple: you multiply proper times (source and detector in the same place in the reference frame) by $\gamma$; anything else requires factoring in the effect of the relativity of simultaneity to truly extract the time taken.