[Physics] Does a muon detector on Earth’s surface correctly measure the mean lifetime of a muon

data analysisexperimental-physicsparticle-detectorsparticle-physicsspecial-relativity

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

Best Answer

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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