Muon demonstration of time dilation

Question

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 1

Note thanks to dmckee: What the text calls half-life should really be mean lifetime.

Mean lifetimes are always given in the particle's rest frame, because what other frame would you use? If I tell you the lifetime I measured in my frame but I don't tell you the relative motion between the particle and my lab, the information is useless. And since $\tau$ is the symbol most commonly used for mean lifetime, it's standard notation to say that the muon's lifetime is $\tau = 2\times 10^{-6}\,\text{s}$.

Of course we could have used any other frame. As dmckee pointed out in the comments, we can measure the fraction of muons that make it to the ground (though we need to know how many muons were produced in the atmosphere) to obtain the lifetime in our frame and then multiply by the time dilation factor (assuming we know the muons' speed) to find the lifetime in the muons' rest frame, which is the number you will find in tables everywhere. In practice this is probably not a very good way to measure lifetimes; I imagine we can't really know the number of muons produced in the upper atmosphere very well.

The primes are a bit of a problem, and I think the author should have used primes for the ground frame, and no primes for the muon's frame. The reason is that if the muon frame is the primed one (as it is in the text), we might be tempted, as you say, to call its lifetime $\tau'$, and the lifetime as measured in our frame $\tau$. But this would be quite confusing to a reader who was just skimming the text looking for the lifetime. They would probably look for something called $\tau$, when the actual lifetime was in fact $\tau'$.

Answer 2

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.

Answer 3

Looking at the second paragraph of page 120 [...]

[Muons] decay [...] with a half-life $\tau \approx 2 \times 10^{-6}~\text{s}$. Muons are created when cosmic rays hit the atmosphere, and subsequently rain down on Earth. Yet to make it down to sea level, it takes about $t \approx 7 \times 10^{-6}~\text{s}$ [...] the muons are travelling at a speed $v \approx 0.99~\text{c}$, giving $\gamma~\approx~10$. From the muon's perspective, the journey only takes $t' = t / \gamma \approx 7 \times 10^{-6}~\text{s}$

Is the half-life given [...]

The value $2 \times 10^{-6}~\text{s}$ is actually both fairly close to the known mean life duration (or for short: the "mean life") of muons, which has been measured as $\tau \approx 2.197 \times 10^{-6}~\text{s}$,
and also fairly close to the corresponding half-life $\approx~\text{Ln}[~2~] \times 2.197 \times 10^{-6}~\text{s} \approx 1.523 \times~10^{-6}~\text{s}$.

measured in the rest frame of the muon?

The mean life duration (and correspondingly the half-life) is of course a mean values of a sample of many individual life duration values of individual muons; in recent practice a sample of billions of individual values; cmp http://muon.npl.washington.edu/exp/MuLan/TalksAndPresentations/MuLan2004_PRL_Submit.pdf

In order to meaningfully compare these individual values, and put them in relation to each other (such as to determine their mean value) it is not required that all these individual muons had been members of one common rest frame. (Although, of course the comparison is simplified if they are, in practice, at least approximately.)

Indeed, in order to compare the life duration of any one muon to the duration of $1~\text{second}$ (or to the duration of some particular number of oscillation periods of certain ceasium atoms), or in order to compare the individual life duration values of any two muons to each other, it is not even necessary that any one muon had been a member of a rest frame. Surely muons circlating in a storage ring can be individually attributed a life duration value, too.

What can be said is that the life duration value of an individual muon is attributed specificly to this muon, as its duration from having indicated its production until having indicated its decay; it characterizes this muon properly.

Am I right in thinking it would be clearer if $\tau$ was labelled $\tau'$ in the text instead

If the prime ($'$) is supposed to denote quantities which are properly attributed to the muon (rather than, say, to constituent of the atmosphere, or to constituent of the Earth's surface) then: yes.