## Experimenter for one day – a subtlety in the muon lifetime measurement June 4, 2008

Posted by dorigo in personal, physics, science.
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I wish to offer you, my dear reader, the chance to play the part of the experimenter and confront you with a simple measurement setup, wherein a subtlety is hidden -one the experimenter needs to realize if he is to perform his experiment correctly. I will describe the setup and the equipment, and give maybe too much detail, as per my typical style. Then I will formulate a question, giving you one day to ponder on it. I am going to try to make this post so simple that you are not expected to know anything about particle physics in order to participate. All is needed is the knowledge that muons decay, with a lifetime of about two microseconds, producing an energetic electron; but you just acquired that knowledge by reading the last sentence.

The experiment seeks to determine the muon lifetime by stopping muons from cosmic rays inside a thick bar of aluminum, and then measuring the time it takes for an electron to emerge from it. A simplified view of the apparatus is as shown below.

There is a 8-inch shield of lead on top (the blue rectangle), which stops the so-called soft component of cosmic rays (three thin lines ending inside it are shown). This is due to protons, pions, electrons, and photons – all secondaries produced by the high-energy interaction of a cosmic proton in the upper atmosphere or subsequent decays in flight. Those particles are largely absorbed by the lead, while a good fraction of muons -also secondary products of the cosmic ray- pass it rather easily.

Muons are special: they can traverse large amounts of matter with only minor modifications of their momentum. Those that manage to punch through the lead cross a pair of scintillation detectors (read out by photomultipliers labeled A,B at one end), and then enter an aluminum bar (in red). While most muons also exit the aluminum at the bottom, leaving a signal in the other scintillation counter (cyan layer read out by photomultiplier C), a small fraction of them stop in the aluminum, where they stay and then decay with a typical exponential law.

The stopping of muons can be identified by requiring that there is no signal in the lower counter in coincidence with the two on top. This anti-coincidence also determines the start of our clock $t_{start}$: we want to see how long it takes for the stopped muons to produce an electron, which yields a delayed signal in one of the counters surrounding the aluminum bar (the one labeled C in the plot above). The delayed signal is the stop of the clock $t_{stop}$: so the time the muon sat in the aluminum bar is simply determined as $\Delta t = t_{stop} - t_{start}$. Once a stop signal is seen, the value of $\Delta t$ is stored, and the experiment is reset to its initial state, looking for another start. If instead 20 microseconds pass without a stop signal occurring, the event is discarded and the experiment reset, getting ready to catch another stopped muon.

In the experimental setup, you work with a multichannel analyzer (MCA) which converts the time interval $\Delta t$ into a number T from 1 to 256. T=1 corresponds to $20 \mu s /256 = 0.08$ microseconds, while T=256 corresponds to 20 microseconds. Every time a start and a stop occur, the histogram is filled with an entry corresponding to the bin $T=256 \times \Delta t /20 \mu s$. After a long enough exposure, one should obtain a histogram showing the typical exponential decay law, proportional to the function $N(t)= e^{-t/\tau_\mu}$, where $\tau_\mu$ is the decay constant, the muon lifetime we want to measure. Are you still with me ? Well, the math is basically over.

Now, all this is nice and simple, but the Devil created backgrounds to keep physicists busy. In fact, during those 20 microseconds while our apparatus waits for the muon to spit out an electron, our counters may be crossed by another particle. This will cause a stop, and the MCA will be filled with a random entry. This entry, it is important to stress, has absolutely no correlation with the arrival time of the muon ($t_{start}$). So we may expect them to be distributed with a totally flat distribution between 0 and 20 microseconds: each one of our 256 channels will be ridden with some background hits, after any given exposure.

Now the question for you is the following. Given that muons produce stops with a falling exponential law – whose decay constant is equal to the muon lifetime (2 microseconds), and thus much smaller than the 20 microseconds during which we keep our system receptive of a stop signal- while background hits produce stops at random times, what is the kind of histogram one expects after a long enough exposure if the overall number of decay and random stops is roughly equal? I give you five options below, and one day to think about it.

1. A falling exponential distribution with time constant equal to the muon lifetime
2. A falling exponential distribution with time constant equal to the muon lifetime, on top of a flat distribution
3. Two falling exponential distributions one on top of each other, one with time constant equal to the muon lifetime, the other with a longer time constant
4. Two falling exponential distributions one on top of each other, neither of which with a time constant equal to the muon lifetime
5. A flat distribution

Of course, the problem can be solved analytically, but the math needed is not totally trivial, and I bet you have rather use your intuition. Good luck! Please type your answer in the comment box before reading what others may have answered, to make things more interesting!

Post-scriptum: just before going to bed, I realized that some of the more knowledgeable among you might be misled by the fact that positive and negative muons do quite different things when they stop in aluminum or other materials. Please disregard this detail – or rather, keep it in mind, we will discuss it tomorrow; it has no bearing on the answer of the question I posed above.

PPS: And the answer is here.

1. embarrassed - June 5, 2008

I have to say “none of the above”. I believe the answer is e^-(rm+rb)t, where rm is the muon decay rate (1/tau_mu) and rb is the background particle arrival rate. Detector C can’t distinguish between a muon decay and a background particle, so this is the addition of two Poisson processes, which is the addition of the exponents, not the addition of two separate exponentials, i.e. e^(a+b) not e^a + e^b. I’m not a physicist, so it’s not too embarrassing to have gotten this wrong and I look forward to learning the correct answer.

2. embarrassed - June 5, 2008

Hmmm, maybe I got the right answer after all. If I interpret “one on top of each other” as meaning multiplication of the individual distributions instead of addition then my answer matches choice 3.

3. dorigo - June 5, 2008

Just a clarification embarassed: “on top of” means addition, as entries get added to other entries in each bins of the histogram.

I will discuss the solution later today.

Cheers,
T.

4. Chase - June 5, 2008

Of course the “multiple guess” format constrains the thinking…. I learned a lot by just reading the 5 possible answers. I think the answer #4 is the best. I am a little confused by the description of a “totally flat distribution” of background counts… I assume you mean they arrive a constant rate, producing an exponential histogram. You could also have some kind of noise which really had a flat distribution across the channels… but I am going with a background that is real signals from other particles hitting the third scintillator.

Then I imagine that we would get an exponential distribution of random events across the histogram and an exponential distribution of real muon hits, but they interrupt each other and we can’t tell the difference between a muon stop and random stop. Therefore, neither represents the muon lifetime (when the time constant for muon decay is comparable to the rate constant for the arrival of random particles). Thanks for the puzzle… now I’ll read the comments and see how wrong I am!

5. Enrique - June 5, 2008

I think the 1st answer is correct and the experiment is well-designed. If at t=0 you get signal in A and B but not in C you get a Tstart. Then if you get a signal in C between 1/256 microseconds and 20 microseconds you interpret it as a muon. But if you get a signal in C not correlated with your Tstart at T=0 that signal has to have a correlated signal in A and B at the same time (say Tsignal)and then it’s one of the cases with signal in A, B and C at the same time (Tsignal) so they do not open a Tstart and should not be counted by the detector as a muon.

6. embarrassed - June 5, 2008

I now agree with Enrique. My mistake was thinking that I was only allowed to use data from C to determine T_stop.

7. Luboš Motl - June 5, 2008

The correct answer is clearly #2, Tommaso essentially told us.

The number of decays per second,

dN/dt = C exp(-t/t0) / t0 + C’

where t0 is the lifetime around 2 microseconds. It’s that simple only because the transitive property of the exponential function.

C is the overall normalization and C’ is the background strength.

Right? I notice that the competition says it is either #4 or #6 haha.

8. Joe - June 5, 2008

I think Ans 1 is the right one. There are two contributions to the histogram, one being the real muon decays, the other being the false stops. The real ones contribute a exponential decay curve with time constant equal to muon lifetime. The subtlety is in the false ones.

The false ones are selected. False stops with larger Delta Ts are more rare because they are more likely to be pre-empted by the real decay events. The frequency of false stops of a certain Delta T is proportional to the area under the exponetial decay curve beyond said Delta T, which is again the same exponential curve.

However, there seem to be simple ways by which one can eliminate the false stops entirely. If you spaces A and B adequately, you can tell the direction any passing particle is moving in, and suppress all comers after each Tstart to wait only for goers.

9. Luboš Motl - June 5, 2008

I would actually agree that Joe’s would be a very good point (and the answer would be #1; and it is important to distinguish the probability that “at least one stop” is observed after “t” from “exactly one stop” is observed after “t”) if it were possible to guarantee that all stopped muons behave in the simple fashion – that they wait for a moment and then decay with the standard falling exponential time dependence.

But I find it likely that some muons will be lost “forever” or for much longer timescales. A negative muon can be captured into a K-shell (the internal one) of an atom and then be eaten by the nucleus (muon plus quark goes to the isospin-paired quark plus an invisible muon neutrino). If that happens, the long-time probability that we observe the “stop” is dominated by the flat background again.

Analogously, the positive muon can annihilate and I assume that the scintillators don’t see photons. So qualitatively, the result works for both mu+ and mu- which is probably why Tommaso says that it doesn’t matter. The punch line is that there is a still a flat background left (as a combination of the muons that are not stopped and those that are stopped and survive). There are all kinds of processes so that it is difficult to imagine that you can get rid of it entirely.

Let me keep #2.

10. Luboš Motl - June 5, 2008

The words “don’t see photons” above should be “don’t see neutrinos” because the annihilation I meant was a t/u-channel exchange of W bosons, mu+ + e- goes to muon antineutrino plus electron neutrino.

Some of these processes is likely to generate “lost” (anti)muons in any environment. When the original muon is lost, the first observed “stop” is likely to come from the background again – the “stop” is then uncorrelated to the “start” and we obtain a flat distribution. #2

11. dorigo - June 5, 2008

Hey, this thread has diverged. I had tried to avoid it by post-scribing, to no avail…

Lubos, negative muons do cause a signal. Their lifetime in Al k-shells is about 0.88 microseconds – 40% of positive ones; the decay yields a neutrino, but the nucleus recoils and is highly ionizing: many will be absorbed, but quite a few make it to the scintillator, causing another exponential with smaller time constant. However, I wanted to avoid this detail… I will discuss the answer to the question later.

Cheers,
T.

12. Answer to the muon decay experiment question « A Quantum Diaries Survivor - June 5, 2008

[…] Ok, one day has passed, and a few readers have decided to leave a comment with their answer to the question I posed in the former post, along with some additional considerations. I will first show the answer to the question […]

13. dorigo - June 5, 2008

So, now that I published the answer to my question (see here), I can declare the winner. Chase got it right: background is a real nightmare in this experiment. In order to reduce it, the best setup consists in scintillators read out by TWO photomultipliers at the two opposite ends. The stop signal is thus a coincidence of the light output of the scintillation counter, strongly reducing noise hits (from photomultiplier dark currents). A further reduction of backgrounds coming from randomly crossing cosmic rays not stopped by the lead shiedl is provided by an outer veto – a cage of scintillators which the decay positron cannot reach, and which have to stay silent while counters B or C record a stop signal (a so-called anticoincidence, or veto).

This experiment, although quite simple, is a joy to perform and to study. I did it during my undergraduate course, and I was lucky enough to have a lot of equipment available -our setup ran with TEN photomultipliers reading out six scintillation counters, and there were three different aluminum bars. I will describe it in another post.

Congratulations to all who participated, anyway: the bias discussed in these posts is one nobody had thought about, and the experiment had been performed a dozen times by students before I came by in 1990…

Cheers all,
T.

14. carlbrannen - June 7, 2008

Oh come on now, surely Lubos knows all about these things.

15. dorigo - June 7, 2008

Sorry – I should have specified I was referring to the former students who did the experiment in Padova during the eighties. Of course the mechanism by which a distribution is biased by a selective data-taking is widely known.

Cheers,
T.

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