Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cross Section for e+e- in EM interaction - Is it the same?

  1. Apr 25, 2015 #1
    I was studying my notes and specifically for the ##e^+e^- \rightarrow \mu^+ \mu^-## process, cross section is given by
    [tex]\sigma = \frac{4\pi}{3} \left( \frac{\alpha \hbar c}{W} \right)^2 [/tex]
    where ##\alpha = \frac{g_{EM}^2}{4\pi}## and ##W## is the centre of mass energy.

    Is this the same for ##e^+e^- \rightarrow \tau^+ \tau^-## and ##e^-e^+ \rightarrow e^-e^+## process? I know the vertex factor is the same as they have the same charge.
     
  2. jcsd
  3. Apr 25, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    This is for high energies? Otherwise phase-space is an issue.
    For taus you get the same cross-section, but for electrons you have more diagrams to include which changes the answer.
     
  4. Apr 25, 2015 #3
    I know that for the second process ##e^+e^- \rightarrow e^+e^-## you can either have electron-electron scattering, or annihilation-production. So is the cross section for the second process simply twice of the first?

    And yeah I think it's reasonable to assume that ##W >> m_\mu c^2## so the mass of the products don't matter.
     
  5. Apr 25, 2015 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    I would be surprised if it is exactly twice (you have to add amplitudes, not probabilities), but I don't know the factor. You can look it up.
     
  6. Apr 25, 2015 #5
    I think the relation is given by
    2012_B4_Q3.png
    Calculation is given in http://physics.weber.edu/schroeder/feynman/feynman3.pdf
     
  7. Apr 25, 2015 #6

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    Your intuition is right. It's not a factor of two. It's infinite. You can have arbitrarily low momentum transfer in the electron case (so the cross-section blows up) but not in the muon case.
     
  8. Apr 25, 2015 #7
    Why is phase space an issue if it is not "high energies"? The two calculations are the same for mu and tau surely (just use m_lepton). Of course that's assuming stable final state.

    Practically for electron scattering you use a small photon mass?
     
  9. Apr 25, 2015 #8

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Well, a trivial case is an energy like ##\sqrt{s}=1\, GeV##, where tau production is not possible at all. And even above that, as long as the leptons are not highly relativistic the cross-section looks different.

    The photon mass is zero. You need the additional approximation of small lepton masses for this simple formula.
     
  10. Apr 25, 2015 #9
    So the formula for the cross section changes depending on if "the leptons are not highly relativistic"?

    I am aware that the photon mass is zero. Is the cross section for electron (or quark scattering for that matter)? No. In qcd you define the final state in terms of ir safe observables - jets. My question was, so you do the same in qed, and I guessed this could be done by introducing a small photon mass, as is regularly done for 1-loop integrals with collinear divergences...
     
  11. Apr 25, 2015 #10

    ChrisVer

    User Avatar
    Gold Member

    I don't know your background, but why don't you try to extract the cross section for this process? [itex]e^- e^+ \rightarrow \mu^- \mu^+[/itex] and see by yourself?
     
  12. Apr 25, 2015 #11
    Missed an *infinite* there
     
  13. Apr 25, 2015 #12
    Well, the Feynman rules don't change, the 2-body phase space doesn't change, and the incoming flux is independent of whether it is muons or taus.

    So no, the formula doesn't change.
     
  14. Apr 25, 2015 #13
    Feynman calculations are out my syllabus, so we take a rather holistic approach towards feynman diagrams.
     
  15. Apr 26, 2015 #14

    ChrisVer

    User Avatar
    Gold Member

    Well OK, I think then some people have already answered you.
    The thing is that there are different amplitudes, which for the muon and tau look the same.
    The electron-positron has one additional diagram, which corresponds to scattering rather than annihilation.

    The differential cross section (I found it here, but I didn't look through the calculations http://www.physics.usu.edu/Wheeler/QFT/PicsII/QFT10Feb23Muon.pdf) for the electron positron to muon-antimuon is:
    [itex]\frac{d \sigma}{d \Omega} =\frac{e^4}{64 \pi^2 s} \frac{\sqrt{1- \frac{4m_\mu^2}{s}}}{\sqrt{1-\frac{4m_e^2}{s}}} \Big[ 1+ \frac{4}{s} (m_e^2 +m_\mu^2) + \Big( 1- \frac{4m_e^2}{s} \Big) \Big( 1- \frac{4m_\mu^2}{s} \Big) \cos^2 \theta \Big] [/itex]

    Where [itex]s = W^2[/itex] and if you want to put taus as a last product, you have to replace the [itex]m_\mu[/itex] with [itex]m_\tau[/itex].

    Now if you use that [itex]m_e, m_\mu \ll W =\sqrt{s} < 90~GeV[/itex] and carry out the integration of the differential cross section you will get to the cross section you posted. But the above limit must be taken and that's why everyone said that the leptons should be relativistic .. I think they avoid to write an [itex]W=90~GeV[/itex] to avoid the creation of [itex]Z^0[/itex] resonance (where the decay via [itex]Z^0[/itex] will become important).

    As for the electron positron scattering, you would have a similar term which would correspond to your annihilation diagram squared, and two additional terms: 1 for the scattering and 1 for the interference of amplitudes.
    I think this is given by Bhabha cross section.
     
  16. Apr 26, 2015 #15

    ChrisVer

    User Avatar
    Gold Member

    Yup, the ultrarelativistic Bhabha differential cross section is:
    [itex] \frac{d \sigma}{d \Omega}=\frac{e^4}{32 \pi^2 s} \Big[ \frac{1}{2} (1+\cos^2 \theta) + \frac{1+ \cos^4 (\theta/2)}{sin^4 (\theta/2)} - \frac{2 \cos^4 (\theta/2)}{\sin^2 (\theta/2)} \Big] [/itex]

    If you try to integrate it (using right cuts for the angles to avoid the explosion of the cross section at the singular points) [itex] \theta_0 \le \theta \le \pi [/itex], you will get:

    [itex] \sigma= \frac{4 \pi \alpha^2}{s} \Big[ \frac{1}{\sin^2(\theta_0/2)} + \log \sin^2 (\theta_0/2) + \frac{4}{3} \Big][/itex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Cross Section for e+e- in EM interaction - Is it the same?
  1. Cross section (Replies: 7)

  2. Cross section (Replies: 0)

Loading...