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Tau leptonic decay - Lifetimes and modes

  1. Apr 12, 2015 #1
    1. The problem statement, all variables and given/known data

    (a) Explain lepton universality.
    (b) Explain why decay mode is forbidden and find hadronic branching ratios.
    (c) Find the lifetime of tau lepton.
    (d) What tau decay mode would be suitable?
    (e) Find the precision.
    (f) How do you improve the results?
    (g) Why is it much harder to measure lifetime of muon?

    leptondecay.png


    2. Relevant equations


    3. The attempt at a solution

    Part(a)

    For weak interactions, the coupling of leptons to gauge bosons are the same for all leptons. Thus different decay modes have the same vertex factor.

    Part(b)
    It is forbidden because for weak charged interactions, a tau neutrino must be produced? By universality, branching ratio to electronic decay is 18%. Thus branching ratio to hadronic decay is 64%.

    Part(c)
    Given that ##\Gamma \propto G_F^2 m_\tau ^5##, let us consider
    [tex]\frac{\Gamma_{\left( \tau^- \rightarrow e^- + \bar {\nu_e} + \nu_\tau \right)}}{\Gamma_{\left( \mu^- \rightarrow e^- \bar{\nu_e} + \nu_\mu \right)}} = \left(\frac{m_\tau}{m_\mu}\right)^5 [/tex]
    [tex]\frac{t_\mu}{t_\tau} = \left(\frac{m_\tau}{m_\mu}\right)^5[/tex]
    [tex]t_\tau \approx 2 \times 10^{-12} s [/tex]

    For lepton decay of tau lepton, only ##18%## decays to electronic mode.
    For lepton decay of muon, it only electronic mode is possible.

    Factoring in branching ratio, ##t_\tau \approx 3 \times 10^{-13} s##.

    I have a feeling that this is a roundabout way to calculate the tau lepton's lifetime. Is there a more straightforward way to calculate this?

    Part(d)
    To achieve better tracks in a cloud chamber, the muonic decay is preferred over the electronic decay, since mass of the electron is much smaller than that of a muon. Is this right?

    Part(e)
    Not sure how to find the uncertainty, as we are not given the speed. Can we assume a speed of like ##\approx 100m s^{-1}##? Then the uncertainty would simply be ##10^{-8}s##.

    I'm not sure on the last 2 parts..
     
  2. jcsd
  3. Apr 13, 2015 #2

    mfb

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    How is the different mass an argument?
    Both decays have a serious issue: how do you find the point where the tau decayed?

    You can calculate the speed of the tau - the Z is produced at rest in the detector frame.
    Where does 100m/s come from? In general, all particles in high-energy physics move at speeds that are significant fractions of the speed of light.

    Cloud chambers mainly have historic value - there are much better detectors available now.
     
  4. Apr 18, 2015 #3
    Part(d)
    Since the decay position vertex is reconstructed from trajectories of charged particles, an electron is stable compared to a muon (lifetime of ##2\mu s##) so we should use electronic decay?

    Part(e)
    Ok, then how can I find the uncertainty without knowing the speed. Can I assume that the speed is roughly ##c##? Then uncertainty would be ##\frac{10 \times 10^{-6}}{c} \approx 10^{-14}##.

    Part (f)
    I looked it up - there's something called hadronic calorimeters that can detect such processes?

    Part (g)
    I suppose the pp process undergoes strong interaction while the electronic process undergoes weak interaction? And for strong interactions to be detectable, deep scattering is required which makes it significantly harder.
     
    Last edited: Apr 18, 2015
  5. Apr 18, 2015 #4

    mfb

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    Muons are stable enough to neglect their decays, but how do you reconstruct a vertex with a single charged particle? How do you find out where the particle was produced?

    The speed is roughly c but you have to consider time dilation. You'll need the energy (or velocity, or momentum) of the tau. Think of the Z decay and energy/momentum conservation.

    Calorimeters don't do tracking. There are other detector types for this purpose.

    The strong interaction cannot produce taus, but the particles created from the strong interaction are important.
     
  6. Apr 19, 2015 #5
    I suppose there is some kind of experimental signature involved? Do we use muons or electrons then?


    Since energy and momentum must be conserved at every vertex, the energy is spread to the muon/electron, its neutrino and radiation? To include the effect of time dilation, i multiply by ##\gamma##?

    If both the pp and e+e- are weak interactions, why is it harder to measure any one over the other?
     
  7. Apr 19, 2015 #6

    mfb

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    For lifetime measurements: no, not at all. You use decays with at least two charged particles (which means at least three due to charge conservation). That allows to calculate a vertex.
    I don't understand that question. The energy of the tau can be computed based on the Z decay, this has nothing to do with the tau decay.
    You probably want to divide by that.
    That question does not make sense at all.
     
  8. May 10, 2015 #7
    Part (d)
    I made some progress. The electronic mode is helicity suppressed due to the electron's mass. So the ##\tau \rightarrow \nu_\tau + \mu + \nu_\mu## is preferred.

    Part (e)
    Still not sure on this bit to find the uncertainty. Can I use the uncertainty priniple ##\Delta E \Delta t \sim \hbar## to give ##E \sim \Delta E \sim 3 \times 10^{-22}## somehow?

    Final part (f)
    For production of tau muon, this process requires pulling quark-antiquark pairs out of the vacuum during the pp collision process in the LHC. This involves dissolving the inner proton structure by pulling the quarks apart to increase the exchange gluon energy. Given that many other products are produced alongside, detecting the specific tau muon makes it significantly harder.
     
  9. May 10, 2015 #8

    mfb

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    They have nearly the same branching fraction, but both are unsuitable. You cannot measure the point where the tau decayed with just a single charged track. You need at least two charged tracks to reconstruct a vertex - and due to charge conservation this means you need at least three charged particles.

    No, the uncertainty comes from the resolution of the flight distance, a purely classical effect.

    If a Z at rest decays to two tau, what is the energy of each tau?
    If a tau lives for one lifetime, how far does it travel before it decays? You'll need special relativity.
    What is the relative uncertainty on this value if your measurement has 10µm uncertainty?

    What is a tau muon?
    The other collision products are an issue, yes. The unknown tau energy is another one (you have to measure it, with its own uncertainty).
     
  10. May 10, 2015 #9
    That's right, I was confusing pion decay with tau decay. I looked up wikipedia for tau decays, I think ##\tau \rightarrow \nu_\tau + \pi^{+} + \pi^{-} + \pi^{-}## is possible?

    The energy of the tau would be half of Z's mass, ##45 594 MeV/c^2## which gives ##\gamma = 25.7##. Due to length contraction, ##l = \frac{10^{-5}}{\gamma} = 3.9 \times 10^{-7} m##. Tau is moving close to speed of light, so ##\Delta t_\tau = \frac{l}{c} = 1.3 \times 10^{-15} s##, which is about 1% of the lifetime of tau.


    I meant tau sorry.

    I think an entire summary of the ##e+e-## collider is
    2014_B4_Q1_2.png
     
    Last edited: May 10, 2015
  11. May 10, 2015 #10

    mfb

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    Yes.

    Right.
    That does not work. For the tau the detector is length contracted, for us the tau lifetime is dilated (longer). Not the other way round.

    That is a possible process, right.
     
  12. May 10, 2015 #11
    Ok so just working on timescales, we factor in time dilation once.
    [tex] \delta t = \gamma \frac{\delta l}{c} = 9 \times 10^{-13} s [/tex]
    This is three times the lifetime!
     
  13. May 10, 2015 #12

    mfb

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    Somehow you always put the gamma factor in the wrong place.
     
  14. May 10, 2015 #13
    Ok. So the error in length is ##\delta l = 10^{-5}m##. Speed of tau is ##c##. Time in our frame is ##\delta t = \frac{\delta l}{c}##. But we want to find the time in the tau's rest frame, so ##\delta t_0 = \frac{1}{\gamma} \delta t = \frac{l}{\gamma c} = 1.3 \times 10^{-15}##. That's what I got before..
     
  15. May 10, 2015 #14

    mfb

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    But now you got it with the right argument.
     
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