More Physics Questions
    By Tommaso Dorigo | December 29th 2009 10:29 AM | 14 comments | Print | E-mail | Track Comments
    About Tommaso

    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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    The turnaround of the three physics questions I offered a few days ago, to stimulate your neurons and extract you from the chocolate and alcohol flood caused by the usual string of Christmas parties and dinners, was rather scarce. Despite that, I wish to repeat the offer today, making some adjustments to reach a wider public. The questions I offer here are easier but still not accessible to everybody. However, my plans are that at least the answers I will give in a couple of days will be understandable. Further, anyone can try the bonus question I ask at the bottom of this piece...

    So below are three more questions on experimental high-energy physics, a bit more accessible than the previous ones. Please try them and report your answers below. The time you spend on the test will not be wasted!

    1) The LHC experiments will search for Z bosons in their early 2010 data. The Z decay to muon pairs, in particular, provides a means to verify the correct alignment of tracking detectors and the precise modeling of the magnetic field inside the solenoid, which bends charged tracks traversing its volume.
    If the signal cross section equals 50 nanobarns, and only three decays in a hundred produce muon pairs, calculate the integrated luminosity required to identify 100 candidates, assuming that the efficiency with which a muon is detected is 70%.
    Hint: you will need to use the formula , and all the information provided above.

    2) In the decay of stopped muons, , the produced electron is observed to have an energy spectrum peaking close to the maximum allowed value. What is this maximum value, and what causes the preferential decay to energetic electrons ?
    Hint: you might find inspiration in the answer to the first question I posted on Dec. 26th.

    3) The decay to an electron-neutrino pair of the W boson occurs one-ninth of the time, because the W may also decay to the other lepton pairs and to light quarks, and the universality of charged weak currents guarantees an equal treatment of all fermion pairs. The question is: if the W boson had a mass of 300 GeV, what would the rate of electron-neutrino decays ?
    Hint: the top quark would play a role...

    Bonus question: What do you get if you put together three sexy red quarks ?

    Answers in a couple of days... Take your thinking hat and start working now!


    Okay, this time I was able to see the post, but only by clicking around on the "more articles" link.

    I'll try part 1: 100 / (0.70 x 0.70 x0.05 nb)

    First Answer:

    You want to identify 100 Z bosons by detecting 2 muons (u+ and u-) for each Z decay. However since your system is lossy you have to generate more Z bosons then you detect. The number of Z decays you want to detect is N. I'll call the number of Z decays actually required Np.

    Only 3/100 Z decay events will produce the desired muon pair. This factor I'll call ev_eff.

    You have to detect both muons from the decays. The probability of detecting 1 muon is .7 (pm) and the probability of detecting both from a single decay is (pm * pm).

    So if N is then number of detected events and Np is the total number of events then

    N = Np * ev_eff * pm *pm

    and Np = N/(ev_eff * pm * pm)

    Cross section is a = 50 nanobarns

    Required Luminosity L is derived from Np = a*L => L = Np/a


    Np = 100/(.7 * .7 * .03) = 6802.72 Z decays required

    L = 6802.72/50 = 136.05 particles per nanobarn

    Number 2 partial answer...

    e- => electron
    ve => electron neutrino
    u- => muon
    vu => muon neutrino
    W- => W- boson

    The decay process is

    u- -> vu + W- => W- -> e- + ve

    Spins 1/2 -> -1/2 + 1 => 1 -> 1/2 + 1/2

    Note that the muon neutrino and electron neutrino have opposite spins. Since the weak interaction only allows left handed spins the only way for this to be possible is for the e- particle momentum to be large enough so that in it's rest frame it will see a left handed spin on the vu particle. It has to have enough momentum to overtake the vu particle. Since the vu particle has a very small rest mast this translates into a large kinetic energy for the e- particle relative to the original u- particle rest frame which by definition is at zero energy. All the system momentum vectors have to sum to 0.

    I think the e- kinetic energy is going to be E = sqr((me/mvu)^2 * c^4) where

    me = mass of electron
    mvu = mass of muon nuetrino

    but it seems like the minimum energy instead of the maximum.

    Go easy on me, I don't do this much...

    Some opportunity for the other readers to copy. ;-) I have 10 minutes before the badminton.
    The integrated luminosity needed is probably
    100 / 0.03 / 0.7 / 0.7 / 50 nanobarn = 136 inverse nanobarns, which is very fast (unless wrong).

    Maximum energy of the electron means its maximum momentum. Because in the muon rest frame, the total momentum is zero, the neutrinos must carry the opposite momentum, and the length of their total momentum vector must be maximized, too. That's achieved if they go in the same direction - the opposite direction that the electron's momentum.

    So the total energy of the neutrinos is approximately m(muon)/2, and so is the electron energy. That's the approximation where the electron is massless - relatively to the muon. (And of course, so are the neutrinos.) More precisely, for the total energy, we have in c=1 units

    m(muon) = p + sqrt(m(electron)^2+p^2)

    The solution is  p = m(muon)/2 - m(electron)^2/(2m(muon)). From that and the simple dispersion relation for the electron, energy = sqrt(mass^2+momentum^2), we have

    E(electron,max) = sqrt [ m(muon)^2/4 + m(electron)^4/4m(muon)^2 + me^2/2 ]

    Small one-line update after badminton: it seems I can simplify the square root again, giving me 

    E(electron,max) = m(muon)/2 + m(electron)^2 / 2m(muon).

    There's no problem with the helicities, I think. At least for slightly different directions, there can't be any vanishing. But even at the extreme, I think that the electron can just inherit the muon spin, and the neutrino and antineutrino can have mutually cancelling spin (the same direction, opposite types).

    The 1/9 fraction is probably obtained because there are 9 equally good channels:

    electron-antineutrino, muon-antineutrino, tau-antineutrino,

    and red,green,blue version of the two

    up-antidown, charm-antistrange decays

    3+6 equals 9. For a very heavy W boson, the top-antibottom 3 decays (red green blue) would become possible, too. So 9 would be raised to 12, so 1/9 would become 1/12. It's not exact because the top quark phase space would be different because the top quark mass is a substantial fraction of the new W mass. But if you want to know the exact fraction, screw you, Ken.

    Although I was fascinated by this whole thing, I also think this set of questions is better suited for people not involved with particle physics. Or for theorists. But I think the theorists have to try again on one of the answers.
    I certainly wish I could give an answer to the bonus question ... but nothing better than various puns with hadrons, infrared slavery etc crosses my mind and I still can't pin it down!

    I still can't see this post on the front page. Instead it shows the "Answers to Physics Questions" post.

    Not sure what you mean.   Here is a screen cap:
    Hi Hank,
    I know that what Carl writes must sound crazy. And you may think it's just crazy, especially if you take his first name, last name, and their shared record into account. ;-)

    Except that I am sure he knows what he is talking about because I have experienced the same behavior of this blog - and no other blog - a few weeks ago. Roughly for a week, the new posts were not being added to Tommaso's blog at all. I mean the front page of the Dorigo blog,

    Right now, I can see everything but it wasn't always the case. I suspect that some cache, proxy, or caching settings lead the browsers of some people, under certain circumstances, to reuse the obsolete version of the blog's front page.

    Best wishes

    That one we figured out.   There is a 'sticky' setting that can be added to posts, to make something stay on top, like an announcement about an upcoming talk or something, and one of the moderators promoted his articles to Featured and clicked it by mistake.

    So we removed it, since no one really used it except by mistake.    I think it should be fixed now but Tommaso mentioned an issue yesterday so I have been monitoring the comments and his column and checking every time to be sure.

    We do use caching for anonymous people, because our traffic can be quite high.  But if it isn't clearing every 5 minutes that is bad.    It sounds like something we can track down this morning if it is a caching problem.

    As always, the best thing about this community is people let us know if they are not getting what they want.   It sure beats people who are quietly frustrated and never say anything.
    I'm still caught in the same time loop. Meanwhile, PRL efficiently kicked out my paper, with the comment that it has no physics content, but is only a mathematics paper suitable for a more specialist journal.

    The paper is about parameterizations of unitary matrices. I figured unitary matrices are fairly useful to people working in quantum mechanics. There are quite a few papers giving less elegant parameterizations in the literature, so I suppose I'll lengthen it a bit and submit to one of the journals that publish that sort of thing.

    And Lubos, I think it's too late for IJMPD to pull my paper announcing a new theory of gravity, so I'll be a published author about this time tomorrow or so. The only thing worse than a crazy amateur is a crazy professional.

    My answers:
    1) L*50*0.03*0.7*0.7=100, then L=136nb-1=0.136pb-1
    2) take M_mu=105.6MeV, M_e=0.511MeV, E_max=52.8MeV
    The reason for energetic electrons: neutrinos are left-handed,anti-neutrinos are right-handed. In order to get the 1/2 from muon spin, nu_e_bar and nu_mu tend to be aligned thus giving electron the almost maximum energy above.
    3) 1/12
    Bonus question: you will finally get three hadrons!

    About question 2),
    I'm endorsing *lawed answer. Electron takes maximum momentum when one of the two neutrinos gets the minimum energy, so a null momentum: i.e., the electron goes one side with momentum p_e, one neutrinos the other with p_nu. In the muon frame, pe=-p_nu. The total energy (quadratic sum of rest mass and momentum) also has to be conserved: so, m_mu = sqrt(m_e^2+p_e^2)+p_nu, assuming massless neutrinos. Solving for p_e, you have the (text book!) formula p_e = (m_mu^2-m_e^2)/(2m_mu)=52.8 MeV. If electron kinetic energy is what you want, use p_e to compute the total energy and then subtract electron rest mass: sqrt(p_e^2+m_e^2)-m_e = 52.3 MeV.
    ...Thanks Tommaso for proposing this useful gymnastic!

    Let me try this again with a clearer head...

    Answer to question 2:

    The initial muon is stopped so has momentum 0

    The muon decays to one neutrino, one anti-neutrino and one electron

    The decay used the weak interaction and allows only left handed spin on the neutrino and right handed spin on the anti-neutrino

    The center of momentum must remain the same before and after the decay so if the electron is at maximum energy (momentum) then the other decay products must travel at the same absolute momentum in the opposite direction to the electron.

    mass of muon is mu
    mass of electron is me
    mass of both neutrinos is mn
    kinetic energy of electron = ke

    Total available energy = mu
    Available kinetic energy after decay = mu - (me +mn)

    Maximum kinetic energy of the electron ke = (mu - (me + mn))/2

    Is the decay
    u- -> vu + W- ===> W- -> e- + ve
    allowed when e- is at maximum energy?

    Check Spins...

    Assume Right handed spin on muon
    u- vu W- W- e- ve
    1/2 -> -1/2 +1 ===> 1 -> 1/2 1/2
    R L R R R R
    This doesn't work because the W- doesn't do right handed spins

    Assume left handed spin on muon
    u- vu W- W- e- ve
    -1/2 -> 1/2 -1 ===> -1 -> -1/2 -1/2
    L R L L L L
    This works

    I had to cheat aaah... google a little bit to see what this means...

    The two neutrinos travel in almost the same direction when the electron has maximum energy and they have opposite spins so their spins cancel out. To conserve the original system (muon) spin the electron has to have the same spin and direction as the muon. Because of weak decay parity restrictions it must be opposite to the electron momentum vector.

    I not claiming I understand this, but the reason this is the preferred decay mode is governed by a Michel parameter which has a cos(theta) term in the numerator for negative muon. Theta is the angle between the spin direction and the emitted electron. Since it is cos(theta), it is a maximum when the angle (theta) is either 0 or 180 degrees to the spin direction. The 0 direction is out because of the weak parity restriction so the maximum probability for the path of emission is opposite to the muon spin direction which is going to be aligned to the electron emission direction.

    Answer 3

    If the W boson mass is 300 Gev it will allow decays to the top and bottom quarks.

    The decay paths then would be...

    e+ve u/d c/d t/d
    u+vu u/s c/s t/s
    t+vt u/b c/b t/b

    There are now a total of 12 channels so the rate of the e+ve path would be 1/12.