By Tommaso Dorigo | December 28th 2009 10:55 AM | 32 comments | Print | E-mail | Track Comments

I am an experimental particle physicist working with the CMS experiment at CERN. In my spare time I play chess, abuse the piano, and aim my dobson...

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Two days ago I offered you three problems in experimental particle physics, of varied complexity. Three readers tried answering the test in the comments thread: a rather underwhelming turnaround, but what did I expect - we are deep in Christmas vacations after all.

I will give below my own answers to the questions, and then comment some of those I received. For ease of reading, I paste here again the three questions.

1. The charged pion is a particle made by a  pair of quarks: for the positive one we have the composition $\pi^+ = (u \bar d)$, and for its antiparticle $\pi^- = (\bar u d)$. The charged pion decays in about ten nanoseconds into a muon and its neutrino, as in $\pi^- \rightarrow \mu \bar \nu_\mu$.
In the decay, the quark-antiquark pair "annihilate" into a negative W boson, which then materializes the muon-antineutrino pair. The question is: why does the pion prefer to decay to a muon-antineutrino pair instead than in an electron-antineutrino pair ? Note, weak interactions, which are responsible for the decay, are flavor-democratic: there is a thing called "lepton universality of the weak currents" which demands that the intensity of the couplings to all leptons be the same. Also note, the pion has a mass of 139 MeV, the muon a mass of 105 MeV, the electron a mass of 0.5 MeV, and the neutrino is virtually massless. How come the pion prefers the heavier muon, if there would be more "phase space" to yield an electron instead ?
Hint: helicity is the key. Okay, you don't know what helicity is, so go to wikipedia, damn it.

2. In the famous 1964 experiment that demonstrated the violation of CP invariance and won Cronin and Fitch their Nobel prize in Physics, the four experimenters (Christenson, Cronin, Fitch, and Turlay) allowed a beam of neutral kaons to decay inside a large bag
filled with Helium. The observation of the decay of two charged pions of the neutral kaons constituted indisputable proof of the fact that these particles had violated the combined symmetry of charge-conjugation and parity. In fact, the beam of kaons had been
treated in a way that only one of the two possible quantum states had survived.
The question is, why Helium ? What properties has this gas that were suitable for the experiment ? Mind you, this is a tricky one.
Hint: the answer has only in part something to do with Physics...

3. You have a b-tagging algorithm which is capable of properly identifying jets as being originated from b-quark hadronization. You measure two numbers: the probability
P(btag|bjet)=0.5 as the probability of tagging a jet if it was originated by b-quark hadronization; and the probability P(btag|not bjet)=0.01 of incorrectly btagging a non-b-jet. From those, you then get P(no-btag|bjet) = 1-P(btag|bjet), and similarly P(no-btag|no-bjet) = 1-P(btag|no-bjet).
The question is: given a sample of 1000 jets tagged as b-jets by the algorithm, what fraction of them are b-jets ? In other words, what is P(bjet|btag) ?
Hint: think Bayesian!

1. The charged pion is a spin-zero particle which needs to decay into two fermions -there is only neutrinos, electrons, and muons available with a smaller mass, and they are all fermions. Fermions in the standard model are left-handed: they carry a half unit of spin, and when we say "left-handed" we mean that their spin has direction aligned opposite to their momentum vector. Further, the anti-particle of a left-handed fermion will have to be right-handed, i.e. its spin will have to be aligned along the momentum.
Now, only massless fermions need to be perfectly left-handed: for a massive particle, helicity is not a good quantum number because the massive particle travels at a speed smaller than the speed of light: for an observer who traveled faster and overtook it, the spin of the fermion would flip its direction! For this reason, the less relativistic a fermion is, the less strict is its obedience of left-handedness. Weak interactions, in a way, are increasingly willing to waive the rule, the slower the fermion is traveling.
Now, in the pion rest frame, it is clear that one of the two fermions will have to get a half-unit of spin aligned in a direction opposite to that of its own liking. Take a negative pion decaying into a lepton-antineutrino pair: the antineutrino will be forced to be right-handed, and will have the spin pointing along its direction of flight; the other particle, which travels away in the opposite direction, will then have to also be right-handed, to balance to zero the total spin of the final state. A right-handed electron is thus much more unlikely than a right-handed muon, because it is emitted with a much higher velocity, being less massive. This is the cause of the strong suppression factor that makes pions decay primarily into muon-antineutrino pairs.

2. What the experimenters tried to minimize with the large helium bag was the regeneration of neutral kaons with the wrong CP eigenstate. Regeneration is a phenomenon that occurs when a quantum system which is the superposition of states gets scrambled up by the different interactions that its constituents undergo. A perfect example of this phenomenon happens when the long-lived component of neutral kaons traverses matter, because weak interactions act differently on the $(d \bar s)$and $(s \bar d)$ quark doublets of which long-lived neutral kaons are made. Let me explain this in some more detail. The physics of neutral kaons is absolutely fascinating!

Neutral kaons are made by a $K^0 = (d \bar s)$combination of quarks; their antiparticles are instead a $\bar K^0 = (s \bar d)$combination. These two states "mix together", and what we observe to decay is a short-lived component, called $K^0_S$, which may decay into a $\pi^+ \pi^-$pair, and a long-lived component, which decays into pion triplets ( $K^0_L \to \pi^+ \pi^- \pi^0$), with a lifetime 600 times longer. It turns out that the long-lived component is in a quantum state of CP = -1, and so it cannot decay to two pions, a state of CP= +1. We can write the short- and long-lived components as a superposition of $K^0$and $\bar K^0$:
$K^0_S = (K^0 + \bar K^0)$
$K^0_L = (K^0 - \bar K^0)$
The Nobel winners created a beam of neutral kaons and allowed it to drift in vacuum until the short-lived, two-pion-decaying component had died away completely. They then studied the decays of these pure $K^0_L$ particles: observing the forbidden two-pion decay would be proof of CP violation -unless a few $K^0_S$ had popped out in the beam by a regeneration in matter!

The regeneration phenomenon "remixes", in a way, the soup of K0 and anti-K0 which make up the $K^0_L$: by suppressing more the anti-K0 component (which may produce hyperons when colliding with nucleons) the quantum state acquires a small component of K0_S, spoiling the experiment! Check it out below:

1. The beam is made purely by $K^0_L = K^0 + \bar K^0$

2. Matter reduces differently the K0 and anti-K0 components: $K^0 \to f_1 K^0$, $\bar K^0 \to f_2 \bar K^0$

3. The new quantum state can be written $f_1 K^0 + f_2 \bar K^0$or more properly as$[(f_1+f_2)/2] K^0_L + [(f_1-f_2)/2] K^0_S$

4. If, as we stated, f1 and f2 are different (and they are, due to the willingness of the s-quark in anti-K0 to bind with a (ud) pair from a nucleon to create a (uds) combination, something which anti-s quarks cannot do when interacting with ordinary matter) the interaction of $K^0_L$with matter regenerates a non-zero $[(f_1-f_2)/2]K^0_S$component!

The Nobel winners were very careful to minimize the impurity of the beam of kaons, such that the observation of the two-pion decay would be a true smoking gun of the CP violating interaction.  Now, the best choice to avoid the regeneration phenomenon would of course be a large vacuum vessel, but that was impractical to handle and consequently costly. Hydrogen, being the lightest element, was the next-best choice, but the explosive properties of the substance made it very dangerous to deal with in the experimental setup. So Helium was the best compromise. Below you can see a sketch of the experimental setup: the beam of kaons enters from the left between the two collimators, and the detectors "observe" the region within the Helium bag.

3. This was a tricky question, but the hint was there as a warning: if one really applied Bayesian reasoning, one would have started asking oneself the right question straight away: what is the prior probability of the examined jets to be originated by b-quark hadronization in the sample ? In other words, what is P(bjet) before anybody pulls out a b-tagging device ?
If one knows that number, one can determine the fraction of b-tag jets that are true b-jets in the sample; if one doesn't, there is no solution.
In fact, Bayes theorem provides the solution fairly quickly if we know P(bjet). Let us take P(bjet) to be 0.05, which is not too different from what typically happens at a hadron collider. In that case we can use Bayes formula:

P(bjet|btag)=P(btag|bjet)*P(bjet)/P(btag).

Notice that at the denominator we have P(btag). This can be computed from the data at hand: from a generic sample of jets, 5% of which are b-jets, the probability to b-tag one of them is

P(bjet)*P(btag|bjet)+(1-P(bjet))*P(btag|nobjet),

which turns out to be 0.05*0.5*0.95*0.01=0.0345. This means that in a sample which has 5% of true b-jets, our 50% efficient, 1% fake rate algorithm will tag a total of 3.45% of them.

Now we have all the data we need: We use Bayes formula to get

P(bjet|btag)=P(btag|bjet)*P(bjet)/P(btag) = 0.5*0.025/0.0345 = 0.724.

In a sample of 1000 b-tagged jets, we might thus expect that 724 of them are true b-jets.
Now that we have worked through the example above, which fixed P(bjet) at 5%, it is immediately clear why the data in the original problem was insufficient: we had no "prior" to use as an input. The answer is: "There is not enough information to determine the fraction of b-jets in the 1000 b-tagged sample."

Now that I in turn did my own homework -that is, writing down the solution in a sort-of-understandable fashion- let me have my share of fun. I will comment the answers given by the few courageous readers who took the challenge:

Lubos Motl:
1) Correct, if a bit too long. But I do like long answers, they show the true reasoning of  the student and allow for a better evaluation.
2) Wrong! Lubos said the reason for Helium was to minimize interactions with the pions, i.e. the final state of the decay. The answer, as I discussed at length above, was to avoid the regeneration phenomenon as much as possible.
3) Correct, although Lubos is very cautious ("I think the dependence on P(bjet) does not disappear"). He correctly explains how the result depends on the prior.
Result: 2/3.

SLW:
1) not quoted
2) Partly correct! The cost factor is in fact the driving reason for choosing a Helium bag.
3) I did not check the formula, but I suppose that not reaching a definite value means that SLW did realize it was not solvable. So I would give two-thirds of a point for this answer.
Result: 1.4/3.

Tulpoeid:
1) Basically correct, although not discussed in enough detail.
2) Wrong! Tulpoeid is also misled by the decay rather than by the nature of the phenomenon to be constrained.
3) Correct, although living on the shoulders of the previous solver.
Result: 1.5/3.

This was fun. I will have more questions for the second half of the vacations!

Hi Tommaso, fine but you haven't actually explained why it doesn't matter whether the pions interact with the medium - to distinguish whether there are 2 pions or 3 pions etc.
I wrote it was it was to minimize some interactions (because the helium is inert, neutral, spin 0, all kinds of vanishing charges like that) - and rating such morally correct answer as 80% worse than the answer that "everything is about money" is a bit bizarre. ;-)

Helium costs from \$3.25 to \$15.00 per liter,

so it is more expensive than Coke. If the guy hasn't explained why Coke hasn't be used instead, it just can't be a correct answer. ;-)

Best wishes
Lumo
Thanks a lot for this and I'm looking forward to the next set
-- but seriously, the third question was the easy one, Lubos and I worked independently and he just published first!!

Lubos, I agree - helium has some attractive properties (especially with respect to coke or toilet paper) also for tracing the decay products, although this is really not crucial in the experiment. I was more strict with you because you certainly came across the regeneration phenomenon in your studies and apparently forgot it.

Eleni, yes, okay... The third question was the easiest, although it could be confusing. I will have a few more questions, avrio h methavrio.

Cheers,
T.
Dear Tommaso, based on your explanation, I have no doubts that the phenomenon exists and it is the key to the helium decision even though it was a messy problem to calculate the interactions of K0L and K0S with different materials. It doesn't take much time to agree that the interactions of the K before the decay is more dangerous than the interactions of the decay products.
But I've never heard the term "regeneration phenomenon" in my life, it sounds like Italian English to me, and the only time I met helium in action was when the experimental physics professors in Prague were demonstrating their helium dwarf voice in a show dedicated to the freshmen like myself 17 years ago. ;-)

You know, orthodox high-energy physicists usually don't care about more helium nuclei than one or a couple - the behavior of macroscopically many helium atoms is wisely and generously left to low-energy and other low-brow physicists. :-)

True, "regeneration" is very much like in Italian, rigenerazione. However, it is the way it is called, and I did not make it up... See http://en.wikipedia.org/wiki/Kaon .

So I learn that I have some misconceptions about the courses that would-be particle theorists end up taking before they become high-brow stringers :)

Cheers,
T.
I confirm it's called this way in Wikipedia. We have "regenerace" in Czech, too. That wouldn't mean much by itself.
Isn't it obvious that theoretical physicists are not learning this stuff? The closest thing of this kind I remember - about the interactions of high-energy particles with materials - was an undergrad course on the basics of nuclear physics, including the health hazards of radiation, nuclear power plants etc.

This is clearly a topic for high-energy experimenters, and in some sense, only high-energy experimenters because this is a completely special kind of phenomena relevant for particle detection engineering etc. I suspect that condensed matter physicists are not routinely taught regeneration of kaons, either.

There exist mathematically similar if not isomorphic situations that can be taught elsewhere, and the logic is surely not difficult. But the phenomenon including the particular context is clearly relevant for particle experimenters only.
Let me say a few words what the theorists learn about your tasks. In short, almost nothing - and none of the tasks is taught "completely". And for good reasons...

...

First, theoretical particle physicists surely learn lots of things about the vanishing of amplitudes because of the spin conservation and helicity and chirality constraints. But they rarely look at them in the context of the pion decays. So obviously, they may be expected to solve such a problem but it's a dirty advanced QCD in some sense - or an old-fashioned nuclear physics, if you look at it from a different angle - so it's just not a part of the education "from Newton to the state-of-the-art physics". Pion decays simply don't belong to this path these days.
I have explained that the regeneration of kaons is not taught although mathematically similar or isomorphic problems are of course taught.

The Bayesian inference is not really taught, either. Of course, it's a part of culture that theoretical high-energy physicists do learn elsewhere (it's a part of science that is used in practice in all other disciplines), and I agree that they should, but it's not a part of the path "from Newton to state-of-the-art physics", either. The probabilistic reasoning needed on this path appears in statistical physics (and thermodynamics) and various predictions of processes (probabilities as inherent objects in quantum mechanics etc.).

But in all these cases, one naturally deals with the frequentist interpretation of the probabilities. I think that I have discussed this issue many times. The reason why I consider the frequentist probability more conceptually solid stems from the fact that theorists like myself make predictions, i.e. their probabilities make sense, at least in principle, by a sufficient repetition of the same experiments. Theorists may always assume that they have a well-defined theory to start with - which predicts clear frequentist probabilities - and they can also assume that experiments are repeated infinitely many times, at least in principle.

Experimenters, on the contrary, don't know anything about the world - I mean that they know no theory yet - so they're using their observations to infer at least something about the world. The logic is the opposite. And no "sufficient number of copies of the experiments" is guaranteed, either. So they kind of have to leave all the inference to theorists who start with theories and falsify them against the evidence, or they have to use some Bayesian inference which is much more conceptually shaky, arbitrary, and affected by uncontrollable conventions and priors than the frequentist probability, but it's often useful in the dirty world of an experimenter who knows nothing about the theoretical framework of our world and who only has a couple of observations.
Hi Lubos,
you mention several things I would like to comment on, but I have time only for a mention to the phenomena of strangeness oscillation and k0 regeneration: these are textbook examples of quantum mechanics and they should be included in any serious book on QM. So it is not something that only experimentalists should get exposed to... Just my two cents.
Cheers,
T.
Dear Tommaso, I wasn't officially exposed to Feynman's Lectures in Physics - they were not used anywhere officially - but of course it was a pretty favorite book of mine that I became familiar with.
...

I say it because its Volume 3 is the only major quantum mechanics textbook that deals with two-state QM systems and oscillations as with the key problem. In my opinion, this is a wise approach because it builds QM "independently" from classical physics and focuses on the inherently QM aspects which is probably more important than to repeat some aspects in which it resembles classical physics (which leads to lots of misunderstandings of the people later).

http://en.wikipedia.org/wiki/The_Feynman_Lectures_on_Physics#Volume_3._Q...

The main examples of the two-state systems, which are given special sections, are precession of spin-1/2 particles, the ammonia maser, the hyperfine splitting of Hydrogen, and [in the many-state generalization], propagation in a crystal lattice.

As far as I remember, long-lived and short-lived kaons also appear somewhere in the "other examples" but Feynman doesn't discuss regeneration. There are many other detailed issues to consider - concerning th states in which the kaon is produced etc. The interaction with an additional environment is an overly complex bonus.

Moreover, the Feynman lectures probably come from the very same year as the Cronin-Fitch experiment - 1964. So it was too new if not causally impossible. Other people later just didn't include it into any standard introductions, so it didn't ever become standard as a material to learn.

If you protested that I discuss undergraduate texts, then: yes, mentally - by the difficulty of the material, both mathematically and conceptually - this regeneration *is* an undergraduate material. So if it doesn't occur in the undergrad era, it's unreasonable for it to appear in graduate basic courses etc. because it doesn't give the theory student any mathematical or conceptual technology he didn't know yet.
Re: "a rather underwhelming turnaround". This is the second time in a month or two that I've opened up this blog and found multiple articles apparently posted simultaneously. Now I've been somewhat bored, and have been RELIGIOUSLY checking in (as befits the season), so I'm quite sure that I need to do something like "refresh".

P.S. Wish me luck at PRL.

I think scientificblogging should do smth about this sooner or later... it's also been happening to me and other people for many months now, the only solution I've found is click on one of the visible articles and then check the "More Articles" list on the right. Plus, the number of comments appearing below the title is definitely not to be trusted.

Since I got started with my wishlist, the site is by far the heaviest I visit (might take around a minute to fully load) and I'd appreciate a button to tag "inappropriate content" for ads so that they don't appear again... one can't show colleagues an article and rest assured that half-naked girls won't appear (and no half-naked guys at all?), or show non-scientists an article and rest assured that it's free of ads about paranormal activity. (Yes I'm aware of the two logical jokes-sort-of lurking in the previous sentence.)

And of course, happy 2010 to everyone involved!

Hi Eleni,

I know, I am forwarding this comment to Hank in the hope he will do something about it -I am blogging from the mountains where my connection is not very wide-band, and every second time I load the page the browser gets stuck trying to load whoknowswhat from "www.cruiseflorida.com". I think this is nonsensical, but I have no power other than moral suasion to change it.

As for the ads, this is unfortunately part of the game. Typically ads here are not inappropriate, and I doubt anything that has to do with paranormal activities has ever appeared, or I would really start a strike ;-)

Best wishes to you too
Cheers,
T.
Hi Carl,

good luck with PRL. As for the problems with this blog, I can only reiterate a request for a check to the blog owner...

Cheers,
T.
A follow-up to the first question, or rather to Lubos's answer: the argument that, if the electron were massless, the decay would violate angular momentum conservation looks like an exact, nonperturbative statement. But Lubos said: 'In the twistor space, the massless electron amplitude is "more than maximally helicity violating, MHV" amplitude and vanishes.' The vanishing of beyond-MHV amplitudes is true only at tree level, in general. So: is it true that a pion decay to a massless electron and neutrino would be exactly zero, nonperturbatively? If so, what is the essential difference between a process like this and a process like the ++++...+ n-gluon amplitude, which has a finite one-loop contribution? If not, what's the loophole in the argument that it vanishes?

I think I know roughly how the answer goes, but I don't have time to think it through right now, so I leave it as a follow-up exercise and perhaps a further opportunity for Lubos to show off ;-)

Hi Anon, great issues to consider. First, your arguments are much more robust than you modestly indicate. If a process is forbidden by the violation of the angular momentum, it is forbidden exactly, nonperturbatively, at all loops and beyond. As a matter of principle, that much is true both for neutrinos and for gluons in N=4 SYM.
But one must be careful what exact process actually violates the angular momentum conservation.

In the case of the gluon scattering, none of the processes you find disputable violates the angular momentum conservation. Indeed, the one-loop contribution is nonzero, so it is allowed.

In the case of weak decays, the process that would violate the angular momentum conservation is a decay of a spin-0 particle to (essentially) a *massless* particle-antiparticle pair which have a fixed helicity (i.e. chirality, too) - a left-handed particle and a right-handed antiparticle going in the opposite direction.  Because of the different helicity *and* different direction, their angular momenta coincide, and their sum is therefore nonzero.

But this is only relevant for strictly massless, chiral fermions - which is why I didn't find this "nonperturbatively exact" statement any useful to write a concise solution.

Both electroweak reality as well as the gluon-gluino dynamics are different.

In the electroweak case, the electrons and muons are 4-component spinors, rather than 2-component spinors, so the decay is clearly possible because the helicity is not strictly linked to chirality for massive fermions. Indeed, this possible decay product arises "from the coupling of the old 2-component spinor with the new 2-component spinor which is previously decoupled". That's why the amplitude depends on the mass - the coupling between the two 2-component spinors - and the decay rate is proportional to the square of the mass.

Once the charger fermion mass is nonzero, the decay is possible - even at the tree level.

There is an essential difference between the massless chiral fermions and the N=4 super Yang-Mills, too. The gauge theory, because of the high supersymmetry, is innately non-chiral, left-right symmetric, so the left-handed and right-handed particles always come together, and both are allowed to interact at the same time. That's why no nonperturbative arguments for the decays can be constructed. The helicities of gluons and gluinos may always be chosen so that the process is allowed by the angular momentum. That's why the vanishing of some amplitudes is a "twistor miracle" and only works at the tree level (or a particular number of loops).

The only reason why we could ever obtain an exact non-perturbative vanishing of the amplitudes in the massless electroweak case was that some particles had constraints on their chirality/helicity - but such constraints exist neither in the world of massive Dirac fermions, nor in the world of non-chiral N=4 gauge theory.
Nice post, Tommaso. A couple of comments on the first question.

for an observer who traveled faster and overtook it, the spin of the fermion would flip its direction!
Actually, when overtaking a moving object, what flips direction is the object's momentum, not spin. Since momentum flips, and spin does not, helicity flips its sign as well.

Another way of putting the issue of the muon preference is that weak currents can only reach a right-handed charged lepton through a mass term.

Mmmmhh...!!! Does one really need to invoke twistors to understand pion decay? Odd...

It's not that someone *needs* them. What's fascinating is that the twistors *can* be used to calculate related processes (especially gluon scattering amplitudes) and to shed light on their structure, miraculous vanishing, and other aspects. I meant it partly as a joke for the pions but it is no joke for the processes where it works.
Whoever is not intrigued by such a thing is a barbarian who has no talent to think about the physical world.
Hi Anon,
of course! Helicity flips. Thanks for spotting it... There is always more to explain than I can manage, it's so hard to not give anything for granted!
Cheers,
T.
Lubos said:
"... Feynman's Lectures in Physics ... was a pretty favorite book ... its Volume 3 is the only major quantum mechanics textbook that deals with two-state QM systems and oscillations as with the key problem. In my opinion, this is a wise approach because it builds QM "independently" from classical physics and focuses on the inherently QM aspects ...".

The QM independent of classical physics approach of Feynman is indeed good, and Lubos's approach is indeed wise,
and
it is in fact the basis of the book Collective Electrodynamics - Quantum Foundations of Electromagnetism" by Carver Mead (MIT 2000), which Carver Mead says "... came about as a direct result of ... interactions with Richard Feynman. ...
Vol. II of Feynman Lectures on Physics ... separated science into two worlds: quantum and classical. For [Feynman] the vector potential was primary in the quantum world, whereas E and B wre necessary for the classical world ...
Why didn't he use his knowledge of quantum electrodynamics to "take the vector and scalar potentials as fundamental quantities in a set of equations that replace the Maxwell equations ... ?
...
A superconductor ... Chapter 21 in Volume III of The Feynman Lectures in Physics ... is a quantum system on a classical scale ...
electrical current ... is made up of moving electrons ... that ... have momentum ... that ... do not correspond in the most direct way to our expectations from classical mechanics ...
Collective quantum systems do not have a classical correspondence limit.
...
For almost all problems involving currents in wires, the electron density is sot high, and the requirement for charge neutrality enforced so strongly, that he momentum of the collective, interacting system is overwhelmingly larger than that calculated by adding the momenta of the free particles moving at the same velocity ...".

I think that if such books were more widely used in physics education, it would (as Lubos indicated) prevent "lots of misunderstandings of the people later".

Tony Smith

"Whoever is not intrigued by such a thing is a barbarian who has no talent to think about the physical world."

I cannot be held responsible for anything, I'm just following orders.

I didn't hold you responsible for anything. I just kindly pointed out why you were a barbarian - which is a property independent of someone's being a hired gun. Most barbarians have always been hired guns, anyway. ;-)
Tommaso,

Question 1 was the easiest. And the only one I got, so forgive me if I pick nits at your solution. :-)
"Fermions in the standard model are left-handed" - Not true, of course! There are both left- and right-handed fermions even in the standard model. However a valid remark is that they have different interactions.
".. for an observer who traveled faster and overtook it, the spin of the fermion would flip its direction" - No again! The spin would remain the same and the momentum would flip direction.
Mostly my unease with what you say lies in assigning blame to the particles themselves (do electrons really 'want' to be left-handed??) whereas IMHO the properties of the weak interaction are to blame. It is all about coupling, and only that. The W boson interacts with fermion pairs via a V-A current, which always couples to both left- and right-handed fermions. However the proportions vary, and as the particle becomes more and more relativistic the coupling to the left-handed part dominates. So the coupling to the (less relativistic) right-handed muon is stronger than it would be for a right-handed electron. N'est-ce pas?

"The W boson interacts with fermion pairs via a V-A current, which always couples to both left- and right-handed fermions"

No, it doesn't. V-A means precisely that it only couples to left-handed fermions (and right-handed antifermions). The coupling to right-handed fermions comes only through the mass term, and that's why the muon wins: it's heavier than the electron. (By the way, velocity doesn't have anything to do with this, there is no non-relativistic approximation involved at all.)

A final comment from my part... I had no idea what the answer to the 2nd question was, so I found Tommaso's answer quite educating. I wonder, from the figure it seems that the helium bag is about 2 meters across, is it really so difficult to make a vacuum chamber of that size? (In case it's not obvious enough: I have no idea how to make a vacuum chamber...)

Wow Anon, I do not know much myself, but the walls of a 8-cubic-meter vacuum chamber have to hold a considerable pressure. This would mean problems for the detection of the pions and their measurement, but I think the important point is the cost of the whole thing. Definitely overkill, when there is such a simple alternative as a helium bag.

Interesting to note is an anecdote retold by Cronin (I think). Shortly after the diffusion of their result, somebody at a dinner started arguing that a fly within the helium bag could have been a source of regeneration of K_S if it had been in the right spot. They ended up computing, on the back of a towel, the atomic number of such a fly to account for the observed phenomenon, and it turned out that it had to be a transuranic fly :D .

Cheers,
T.
"The W boson interacts with fermion pairs via a V-A current, which always couples to both left- and right-handed fermions"

"No, it doesn't. V-A means precisely that it only couples to left-handed fermions (and right-handed antifermions)."

Sorry, I meant, ".. which always couples to particles of both positive and negative helicity." Handedness is defined in terms of the chirality operator gamma5 while helicity is defined in terms of momentum dot spin.

"(By the way, velocity doesn't have anything to do with this, there is no non-relativistic approximation involved at all.)"

People often make the ultrarelativistic approximation in which handedness and helicity are the same.

Hi all,

I am quite happy to stand corrected in my explanations, for two reasons. One, because it shows that my readers are alert. And two, because to explain these complicated things it is necessary to make a few concessions to correctness. To make an example other than the "left-handed fermions" or the "spin flipping direction" trivializations, take the answer to question two. I totally disregard to 1) normalize the quantum states; 2) put the bra's around the particle symbols; 3) explain that we are in fact dealing with K_1 and K_2 states (due to the small but nonzero epsilon); 4) explain that the phenomenon of oscillation is what is behind the whole thing.
See, if I had to be careful and accurate it would have taken me twice as long to write the article above; I would have been checking things, and then I would have been explaining irrelevant details. I am happy to be inaccurate, because we do need to step down a bit if we want to popularize science.

But please, continue to stress my mistakes! It is an important part of the whole process...
Cheers,
T.
Bill K said:
"... There are both left- and right-handed fermions even in the standard model ...
Handedness is defined in terms of the chirality operator gamma5 while helicity is defined in terms of momentum dot spin. ... People often make the ultrarelativistic approximation in which handedness and helicity are the same. ...
Mostly my unease with what you say lies in assigning blame to the particles themselves (do electrons really 'want' to be left-handed??) ...".

In his book Leptons and Quarks (North-Holland 1982) L. B. Okun said:
"... A particle is said to possess ... left-handed helicity ... if its spin is directed ... opposite to that of its momentum ...
The concept of helicity is not Lorentz invarian if the particle mass is non-zero. The helicity of such a particle depends upon the motion of the observer's frame of reference. ... Overtaking a particle is the more difficult, the higher its velocity, so that helicity becomes a better quantum number as velocity increases. It is an exact quantum number for massless particles. ...
This means ... that at v [approaching] 1 [speed of light],
paricles have only left-handed helicity,
and antiparticles only right-handed helicity. ...".

So, in Okun's view, YES, electrons really 'want' to be left-handed.

Tony Smith

When you spill water on a hot surface, like a stove top, why does the water move outward from the center of the spill, then bubble up at the edge of the burner surface?

I think the outward motion is driven by the temperature gradient, but admittedly I have never really tried to think at this problem from a quantitative point of view.

I will be happy to hear others ponder on this interesting practical issue.

Cheers,
T.
I will be happy to hear others ponder on this interesting practical issue.

Leidenfrost effect - no pondering required.
Ah thanks Patrick!
Cheers,
T.