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Homework Help: 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]
    [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
    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]

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


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    2017 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


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    2017 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.
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