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Answer to the muon decay experiment question June 5, 2008

Posted by dorigo in personal, physics, science.
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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 pictorially, by pasting here four graphs I obtained this evening with a simple root macro. The macro perform a simple simulation of the experiment and -save some (unlikely) mistake on my part, it allows to obtain a histogram of counts in the 256 bins of a simulated multichannel analyzer (MCA), mapping the time frame 0 - 20 \mu s after a start has been detected by the experimental setup. The macro also allows the user to vary the number of bins of the MCA, the fraction of background, the exposure -but we need not be concerned with that.

A word of warning: the simulation is crude – I did not bother to insert the geometry of the apparatus (which may change from one setup to another), the energy loss of muons and electrons in Aluminum, or other details: I just simulated at random muon decays and background, recording whichever produces a stop in the MCA counts. Also, note that only positive muons are simulated here: as I warned in the post-scriptum to my post of yesterday, and as is further discussed in the thread to that post, negative muons get captured by Aluminum nuclei in the K-shell, and they decay much faster than positive ones -about 2.5 times more so. The decay produces a nuclear recoil which gets recorded in the scintillator as if it were an electron, causing in the end the MCA to record not one, but two different exponential distributions (with time constants of 2.2 and 0.88 microseconds for positive and negative muons, respectively) on top of random backgrounds. I ask the reader to forget about this detail in the following.

In the plots you see the stop time distribution resulting from a million starts, in a situation when there is a large amount of background, and when a realistic fraction of positrons get detected by the apparatus (roughly 30%). The first plot shows the MCA distribution fit with a single exponential (answer #1). This is a very wrong modeling of what is happening, of course: in the presence of background -caused by particles randomly crossing scintillators B or C during the 20 microseconds of exposure after a real muon start- the multichannel will record both a muon decay signal with a time structure, and casual hits at random times.

The second plot shows how the data is interpreted with a functional form representing answer #2: a falling exponential plus a constant. Here things look definitely better, but a keen eye will spot the problem in the fit: the green curve stays above, then below, and then above the data points as we move left to right. In fact, in the upper right corner root informs us that the fit probability is zero! There is thus something wrong with answer #2.

The third plot shows how the data is interpreted with a functional form representing answers #3 and #4: two falling exponentials with different time constants. The fit now is extremely good! The fit probability is a whooping 67.5%, indicating that our model of the data is correct, within the statistical power of the analyzed data. What makes options #3 and #4 different is that in the former case one expects one of the two exponentials to have a decay constant equal to the muon lifetime -which in the simulation was set to 2.0 microseconds for simplicity, while option #4 predicts that the lifetime measurement will be biased. Well, by looking at the parameters of the fitting function (which is p_0 e^{p_1 t} + p_2 e^{p_3 t}) one observes that neither p_1 nor p_3 have a value close to -1/0.000002=-500,000 (remember, the decay law is e^{-t/\tau_\mu} so the fit parameter describing the exponential slope has to be the inverse of the opposite of the muon lifetime).

The fourth plot -at this point, only academic- shows the results of a constant fit. Of course, there is a lot of time dependence in the data, and not just a flat distribution!

Now let me comment the result. Why are we finding that the background -which is uncorrelated with the time of arrival of stopped muons- shows a time structure ? And why on earth is the muon lifetime coming out biased low (the fit in the third plot has a value p_0 larger than 500,000) ?

The answer is not quite simple, but one can arrive at it by intuition, as I mentioned in the former post. First of all, we have to note that while the muon causing the start of our clock has one and only one chance to yield a signal -with the exponential law we discussed above-, backgrounds, however small, have multiple, albeit tiny, chances of providing counts in the time window: only the first arriving background count stops our clock and records an entry in the MCA, though. So, imagine you have not just one (and in our experimental setup, there is only a 20% chance that a background hit occurs in the 20 microseconds), but many background hits distributed at random throughout your 20 microseconds time window: they are overall flat -no time information from them- and yet, our data acquisition system records the first one! No wonder we get an exponential for the background as well!

So, the background produces an exponentially falling distribution, as showed by the fit in picture three. But why, however, do we get a wrong lifetime from the hits really coming from muon decays ?

The reason is the same as above. Muons decaying early on have a smaller chance of being beaten on time by a random background hit than muons decaying later on. So, the time distribution of real muon decay stops is also biased towards smaller values. We get a time constant smaller than the muon decay lifetime, that is.

Not convinced ? I can provide the root macro, and (by downloading the root package from cern) you should be able to test the system by yourself in minutes. Drop me a line if you wish to.

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.