1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

E+e- collison Resonance peaks cross-section partial widths

  1. Mar 22, 2015 #1
    Question:

    The Breit-Wigner cross-section for a resonance R is ## \sigma_{i \to f} =12\pi\frac{\Gamma_{R\to i} \Gamma_{R\to f}}{(s-M^{2})^{2}+M^{2}_{R}\Gamma^{2}_{R total}}## [1],

    where ##s## is the com energy squared, ##M_{R}## is the mass of the resonance , ##\Gamma_{R total}## is the total width of the particle, and ##\Gamma_{R\to i,f}## is the partial width for the decay of the particle into the initial and final state particles respectively.

    Assuming that ##\Gamma_{R \to e+e-}=\Gamma_{R \to \mu+\mu-}=\Gamma_{R \to \tau+\tau-}## use the plot (attached) of the cross-section near the peak to obtain ##\Gamma_{R total}, \Gamma_{R \to \mu+\mu-}, \Gamma_{R \to hadrons}## and the mass of the resonance?

    (The cross-section for ##e+e- \to \mu+\mu-## has been multiplied by ten in the figure).

    Solution:

    I'm fine with attaining these figures:

    ##M_{R}=91.2 Gev##
    ##\sigma_{hadrons}=39.5 nb ##
    ##\sigma_{\mu+\mu-}=2.0nb##

    I understand that the formula reduces to ## \sigma_{i \to f} =12\pi\frac{\Gamma_{R\to i} \Gamma_{R\to f}}{M^{2}_{R}\Gamma^{2}_{R total}}## around the peak.

    I don't understand:

    1) Why ##\Gamma_{total}=(92.5-90.0)=2.5GeV##

    So i see this has been read of from the hadron peak, but I thought that this figure corresponds to ##\Gamma_{R \to hadrons}##.

    I think i'm still struggling with these definitions of initial widths, final width and total width. I know that the product has many decay routes, and ##\Gamma_{total}=\Sigma\Gamma_{i}## . I'm unsure of what is defined as an initial state and what is defined as an final state.

    Maybe when I understand these better I'll understand why ##2.5GeV## is not ##\Gamma_{R \to hadrons}##.

    2) Why ##\sigma_{\mu+\mu-}=\frac{12\pi\Gamma^{2}_{\mu+\mu-}}{M^{2}\Gamma_{total}^{2}}##.

    As in I don't understand why ## \Gamma_{R \to i}= \Gamma_{R \to f}## here.

    So when computing ##\sigma_{hadrons}##, ##\Gamma_{R \to i}= \Gamma_{\mu+\mu-}## and ##\Gamma_{R \to f}= \Gamma_{hadrons}## Again I'm unsure.

    Any one who can help shed some light on this, greatly appreciated !!!!!!!
     
  2. jcsd
  3. Mar 22, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    There is no attachment which makes your questions hard to understand.

    1) The width of the peak is always the total width because it comes from the denominator only (the numerator does not depend on s).

    2) This is the peak value? See where the cross-section gets maximal, plug it into the formula.

    ##\Gamma_{R \to i}= \Gamma_{R \to f}## for muons because it got defined above (the process has the same width for electrons and muons).
     
  4. Mar 22, 2015 #3
    Apologies, attached here.
    (can no longer edit).
     

    Attached Files:

  5. Mar 24, 2015 #4
  6. Mar 24, 2015 #5
    [QUOTE="mfb, post: 5050415, member: 405866] (the process has the same width for electrons and muons).[/QUOTE]

    I dont know how you know which of electrons or muons is the initial/final decay mode?
    What's meant by these terms? I cant find a clear definition on the internet, thanks appreciate it.
     
    Last edited: Mar 24, 2015
  7. Mar 24, 2015 #6

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    You have electron/positron collisions -> ee is your initial state.
    You consider the decay to muons -> µµ is your final state.

    Decay and production give the same factor, it does not matter which one is which.
     
  8. Mar 25, 2015 #7
    Ah thanks . so for ##\sigma_{hadrons}, \Gamma_{i}=\Gamma_{e+e-} ## and ##\Gamma_{f}=\Gamma_{hadrons}##, (For## \Gamma_{i}## we've just used ##\Gamma_{e+e-}=\Gamma_{\mu+ \mu-}##

    I'm still confused with ##\Gamma_{total}=\Sigma \Gamma_{i}##
    Isn't the only initial state e+e- here?

    The solution gives it as ##\Gamma_{total}=\Gamma_{hadrons}+\Gamma_{u+u-}+\Gamma_{\tau+ \tau-}+\Gamma_{unseen decays}##

    Thanks.
     
  9. Mar 25, 2015 #8

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    That should be the sum over all partial decay widths.
     
  10. Mar 25, 2015 #9
    Ah thanks. Apologies it had ##\Gamma_{total}=3\Gamma_{\mu+ \mu-}+\Gamma_{hadrons}+\Gamma_{unseen} ##
    So its took ##\Gamma_{e+e-}## into account - I don't understand why, this isn't a decay route?

    I'm still unsure why you read ##\Gamma_{total}## from the hadron peak, why not the ##\mu+ \mu- ## peak say?
     
  11. Mar 27, 2015 #10
    So ##\Gamma_{total}## is the total width of the resonance peak,
    But there are two resononances - one from the ##\mu+\mu-## and one from the ##e+e-## , how do you know which to consider?
     
  12. Mar 27, 2015 #11

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    e+ e- is a possible decay.

    You can get the total width from all peaks. Shouldn't matter which one you take (you can even use multiple together). Hadrons are more frequent.
    What do you mean with two resonances?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: E+e- collison Resonance peaks cross-section partial widths
  1. Ee ->e+e+ scattering (Replies: 9)

Loading...