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Just a simple question. Does a muon detector on Earth's surface correctly measure the mean lifetime of a muon? I would think the answer is no because most muons detected are created about 15 km above Earth's surface, so on average they will have survived for some amount of time calculated using special relativity before reaching the detector.

The detector I'm talking about is a scintillator, so it detects when the muon enters the detector with a flash, then if the muon decays it detects another flash. The time between flashes is said to be the muon's lifetime.

Is this muon's lifetime the machine giving me the actual lifetime (2.2$\mu s$), or just the lifetime of it on Earth's surface?

here is a plot of my data showing a histogram of decay times

enter image description here

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If you haven't seen it, try – DJohnM Feb 19 '14 at 6:22
Two flashes stands for one muon, not for birth and decay. The construction of the detector is dealt with angle, so particle must pass through both lids to be registered. – Vladimir Feb 19 '14 at 7:24
Nice question. I supervised a muon decay lab course, and when a student asked me this for the first time, I had a really hard time finding the answer. Now I like to ask my students this question to give them something to think about :-). – jdm Feb 19 '14 at 8:21
up vote 15 down vote accepted

Your worry is not necessary.

In the usual experiment the detector measures the distribution of the time between the muon stopping in the detector and the time of it's decay. Then an exponential curve is fit to the data and the lifetime taken from the fit parameters

Muon decay is a processes analogous to radioactive decay, and (like all exponential processes) it has the special property that the lifetime from the fit does not depend on the absolute start time.

An important thing here is what the "lifetime" means. And what it doesn't mean.

It does not mean that muons tend to decay $2.2 \,\mathrm{\mu s}$ after they are created (which would imply that you are looking for a peak and that the absolute start time matters).

It does mean that Given a collection of muons, if you wait $2.2 \,\mathrm{\mu s}$ about 67% of them will have decayed. And if you wait another lifetime about 67% of the one that were left will have decayed. And so on. This means that anytime you fit an exponential to the decay time distribution you will always measure the same lifetime.

A kind of old school way to see this is to note that $y = A \exp \left( -t/\tau \right)$ implies that (writing $A = \exp \alpha$) $$ \ln y = \alpha -\frac{1}{\tau} t \,,$$ which is a straight line if you plot $\ln y$ as a function of $t$. The slope of the line is $-1/\tau$ and does not depend on the range of $t$ over which you measure it.

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Hmmm, I edited my above post to display a graph I got from my data. The times are measured in nanoseconds (x-axis) and you can see that it is peaking in the 200 nanosecond range, while the actual lifetime of a muon is around 2.2 microseconds. Do you know why this would be happening? – Spaderdabomb Feb 19 '14 at 4:04
You're not looking for a peak (and you don't have any that are statistically significant), you want to fit an exponential to that curve. If you don't have a computer fitter available plot the log of each bin and fit a line (that you can do by mark-one eyeball if need be). In any case, just looking at your data I think you are right on. – dmckee Feb 19 '14 at 4:25
Thanks so much! I understand this all much better =)))))))) – Spaderdabomb Feb 19 '14 at 4:49

"The difference between theory and practice is that, in theory, there's no difference, but in practice, there is."

At Colorado State University over the past several years, students using the Teachspin muon lifetime apparatus have been consistently getting muon lifetimes that are slightly but significantly too short. In late 2012 I spent several months collecting my own data, and re-analyzing the student data from spring 2012. My results are explained in (Ignore the (relatively small amount of) material about QTD and Van de Graaff generators; they are irrelevant for your purposes.) There are many possible sources of experimental error, some of which I detail in section 4. I also explain how to correctly analyze a truncated exponential (or geometric) distribution, and give source code for all my analysis software.

Section 3.6 explains that "a property unique to the exponential and geometric distributions is that they are memoryless; the probability of an event occurring does not depend on how long we have already waited for it to occur. Another way of saying this is that they have a constant failure rate. Because muon decay (like nuclear decay) follows an exponential decay pattern, we do not need to be concerned with the history of the muon before it arrives in our apparatus. Indeed, the entire experiment would not be possible without this property, since we do not measure the entire lifetime of the muon from its creation in a pion decay while still traveling at nearly lightspeed, but only its lifetime after coming to rest in our detector." These distributions are also the maximum-Shannon-entropy distributions given only the constraint of having a particular mean lifetime, which means that assuming any other distribution of decays is equivalent to claiming that you know something special about the internal decay process, that allows you to predict it better.

After the above paper, I was able to demonstrate that a part of the error was due to contamination by shorter-lived particles like pions (26 nS) and K_longs (51 nS). Since CSU is at about 1500 m altitude, a few more of these should be expected than at sea level, but the data indicates even more than that. The best guess at the moment is that the excess were "structural" pions generated by cosmic rays colliding with the steel and concrete of the physics building; these wouldn't have to travel very far. They show up as an increased number of events in the first few decay bins.

However, pion contamination is insufficient to explain the whole discrepancy. Much of the remainder is still a mystery.

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I'm sure I don't need to tell you, but for future readers: sums of exponentials with different lifetimes (i.e. $e^{-t/\tau} + e^{-t/\sigma}$ with $\tau \ne \sigma$) don't have the memoryless property if you fit a single exponential. This implies that if you see a changing of the lifetime depending on what part of the data you fit you may have contamination with a second lifetime. – dmckee Feb 19 '14 at 14:44
"but only its lifetime after coming to rest in our detector." How do you ensure that? Through going muons will leave two flashes; anticoincidence ? – anna v Sep 21 '15 at 15:59
@annav One way I've seen it done is to use a short enough tank of scintillator that the through-going deposit is much less than $51 \,\mathrm{MeV}$ and select on the Michel electron peak as a delayed coincidence. – dmckee Dec 7 '15 at 23:05
@dmckee Right, to estimate the shorter-lived particles I had to fit a sum of exponentials. I don't consider this analysis to be completely reliable because it is unstable under bootstrap Monte Carlo resampling; I probably need more data to improve that. – Howard A. Landman Jan 27 at 15:24
@annav Because the lifetime is short (2.2 microseconds) and the rate of muon arrival is relatively slow, most pairs of flashes indicate a decay event. There is a significant noise background from a pair of different particles passing through, but this noise has an approximately flat distribution while the decays follow a steeply declining exponential. Thus we fit to an exponential plus a constant and expect the exponential to represent the decays and the constant to represent the noise. See the paper for more details, especially subsections 4.1, 4.7, and 4.8 and section 5. – Howard A. Landman Jan 27 at 15:46

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