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!

Resonance production of charmonium

  1. Feb 27, 2013 #1
    The Question:

    The cross section for the scattering of two particles with spins sa and sb via a resonance with
    spin J is:
    [itex]\sigma(E)=\frac{\pi\lambda^{2}(2J+1)}{(2S_{a}+1)(2s_{b}+1)} \frac{\Gamma_{i}\Gamma_{j}}{(E-E_{0})^{2}+\frac{\Gamma^{2}}{4}}[/itex]
    with [itex]\lambda=1/p[/itex], E is the centre-of-mass energy, E0 is the rest
    mass energy of the resonance, [itex]\Gamma[/itex] is the total width of the resonance and [itex]\Gamma_{i, f}[/itex] are the partial widths for the decay of the resonance into the initial and final states, respectively.

    The cross section for the production of the J/ψ resonance in e+e− collisions,
    followed by its decay into e+e−, integrated over the centre-of-mass energy is:
    [itex]\int \sigma(E)dE=\frac{3\pi^{2}}{2}\lambda^{2}B^{2}_{J/\psi\rightarrow e^+e^-\Gamma}[/itex]
    with
    [itex]B_{J/\psi\rightarrow e^+e^-}=\frac{\Gamma_{J/\psi\rightarrow e^+e^-}}{\Gamma}[/itex]
    and we took the limits on the integral to be [itex]\pm \infty [/itex] and lambda is constant

    The cross section for the production of (a) hadrons, (b) μ+μ− and (c)
    e+e− is shown at http://imgur.com/94OYTaM. The measurements were made during a scan of the beam energies at the SPEAR
    storage ring at SLAC using beams of e+ and e− circulating in opposite directions with the same
    energy.
    The observed width of the peak is due to the energy spread of the beams at each point in
    the scan, the actual J/ width is much smaller than the observed width of the distributions.
    However the relative centre-of-mass energies are known to about 1 part in 104.

    At each scan point the beam energy spread produces a spread in the centre-of-mass energies
    E′ distributed about the average centre-of-mass energy with a probability distribution
    f (E − E′). Show that the measured area under the resonance peak is the same as the true
    area under the peak, i.e.

    [itex]\int \sigma_{int}=\int \sigma_{measured}dE[/itex]


    My attempt:

    Measured is wider than the actual resonance peak. So for the area to be the same, the actual peak has to be taller. All I can think of is that! How can I prove that mathematically?

    [For some reason, the equations haven't come out well so please use this to view them http://www.codecogs.com/latex/eqneditor.php] [Broken]

    Thanks
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 27, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    The equations will be fine if you use correct tags: "[/itex]" instead of "[\itex]".

    I think this can be solved with a simple manipulation of a double-integral (with E and E' as variables).
     
  4. Feb 27, 2013 #3
    Thanks for pointing the [/itex] out.

    But how would you involve the f(E-E')?
     
  5. Feb 27, 2013 #4
    Is [itex]\int\int \sigma f(E,E')dEdE' = \int\int \sigma dEdE' [/itex] correct then?
     
  6. Feb 27, 2013 #5

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    σ depends on E, f(E,E') depends on the difference between those only, and I don't know what you mean with d.
    In addition, some explanation or even steps how to get your integral would be useful.
     
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: Resonance production of charmonium
  1. Nuclear resonance (Replies: 8)

  2. What is a resonance? (Replies: 2)

  3. Resonance in phase (Replies: 2)

  4. Δ Resonance (Replies: 0)

Loading...