Experimenter for one day – a subtlety in the muon lifetime measurement June 4, 2008Posted by dorigo in personal, physics, science.
Tags: experiment, muon lifetime, question
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 : 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 : so the time the muon sat in the aluminum bar is simply determined as . Once a stop signal is seen, the value of 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 into a number T from 1 to 256. T=1 corresponds to 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 . After a long enough exposure, one should obtain a histogram showing the typical exponential decay law, proportional to the function , where 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 (). 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.
- A falling exponential distribution with time constant equal to the muon lifetime
- A falling exponential distribution with time constant equal to the muon lifetime, on top of a flat distribution
- 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
- Two falling exponential distributions one on top of each other, neither of which with a time constant equal to the muon lifetime
- 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.