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Scattering Cross Section - At high energies

  1. Feb 24, 2015 #1
    1. The problem statement, all variables and given/known data

    Why does ## \frac{ e^+ + e^- \rightarrow \mu^+ + \mu^- }{e^+ + e^- \rightarrow \tau^+ + \tau^- } \rightarrow 1## at high energies?

    Would it be the same if it was ## \frac{ e^+ + e^- \rightarrow \mu^+ + \mu^- }{e^+ + e^- \rightarrow e^+ + e^- }##?

    2. Relevant equations


    3. The attempt at a solution

    For the first reaction, propagator factor for the photon is ##\frac{1}{(2E)^2 - m_\gamma^2} = \frac{1}{4E^2}##. Two vertex factors are ##g_{EM}^2##. Thus ## |M_{fi}|^2 = |\frac{g_{EM}^2}{4E_e^2}|^2##.

    Fermi's golden rule is given by:
    [tex]\Gamma = 2\pi |M_{fi}|^2 \frac{dN}{dE_f} [/tex]

    Differential cross section is number of particles scattered per unit solid angle per unit time divided by flux:

    [tex]\frac{d\sigma}{d\Omega} = \frac{1}{2\nu_e} \Gamma = \frac{1}{2\nu_e} 2\pi |M_{fi}|^2 \frac{dN}{dE_f} [/tex]

    The density of states is given by: ##dN = \frac{p^2_\mu}{(2\pi)^3} d\Omega dp_\mu## and substituting gives:

    [tex] d\sigma = \frac{1}{2\nu_e} 2\pi |M_{fi}|^2 \frac{p^2}{(2\pi)^3} \frac{dp}{dE_0} d\Omega [/tex]

    At high energies, ##E_\mu \approx \frac{1}{2} E_0##, so ## \frac{dp_\mu}{dE_0} = \frac{1}{2} \frac{dp_\mu}{dE_\mu} = \frac{E_\mu}{p_\mu}##.
     
    Last edited: Feb 24, 2015
  2. jcsd
  3. Feb 24, 2015 #2
    At high energies, ##v_\mu \approx c##, so ##\frac{dp_\mu}{dE_\mu} = \frac{1}{2} \frac{E_\mu}{p_\mu} = \frac{1}{2c} = \frac{1}{2}##, letting c=1. This gives:

    [tex] d\sigma = \frac{1}{2\nu_e} 2\pi \left( \frac{g_{EM}^2}{4E_e^2} \right)^2 \frac{p^2}{(2\pi)^3} \frac{1}{2} d\Omega [/tex]

    [tex] \sigma = \frac{1}{2\nu_e} 2\pi \left( \frac{g_{EM}^2}{4E_e^2} \right)^2 \frac{p^2}{(2\pi)^3} \frac{1}{2} 4\pi [/tex]

    [tex] = \frac{1}{64} \frac{1}{\nu_e} \frac{1}{\pi} \frac{g_{EM}^2}{E_e^4} p_\mu^2 [/tex]

    Using ##\frac{g_{EM}^2}{4\pi} = \alpha##,

    [tex]\sigma = \frac{\pi}{4} \frac{1}{\nu_e} \frac{p_\mu^2}{E_e^4} \alpha^2 [/tex]

    For highly relativistic case, ##p_\mu \approx E_\mu = E_e## and ##\nu_e \approx c = 1##.

    [tex]\sigma = \frac{\pi}{4} \frac{1}{E^2} = \pi \frac{\alpha^2}{(2E)^2} = \pi \frac{\alpha^2}{s^2} [/tex]
    where ##s=2E## is the centre of mass energy.
     
    Last edited: Feb 24, 2015
  4. Feb 24, 2015 #3
    Since the scattering cross section only depends on:

    a) Centre of mass energy
    b) ##g_{EM}##

    Wouldn't both ##\rightarrow 1## at high energies (##p >> m##)?
     
  5. Feb 28, 2015 #4

    ChrisVer

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    I find the 2nd question a little tricky...
    For the first you dealt only for one possible channel : the s-channel ..
    For the second case however, the electrons can also "not interact" at all... apart from annihilating and giving again e^+/- pair..
     
  6. Mar 16, 2015 #5
    I think I got it.

    Part (a)
    The cross section is proportional to ##|M_{fi}|^2##.

    First process is given by:

    [tex]e^+ + e^- \rightarrow \gamma^* \rightarrow \mu^+ + \mu^-[/tex]

    The propagator factor for this process is ## \frac{1}{P \cdot P - m_\gamma^2} = \frac{1}{P \cdot P}## since ##m_\gamma = 0 ##.

    Second process is given by:

    [tex]e^+ + e^- \rightarrow Z^0 \rightarrow \tau^+ + \tau^-[/tex]

    The propagator factor for this process is ## \frac{1}{P \cdot P - m_Z^2} ##.

    At high energies, this propagator factor ##\approx \frac{1}{P \cdot P}##

    Thus they approach 1 at high energies.

    Part (b)

    Wouldn't this always be true? Given that the propagator factor for both are equal to ##\frac{1}{P \cdot P}##.
     
  7. Mar 16, 2015 #6

    ChrisVer

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    The part a is clear for you I think... Even for photon interactions alone, the ratio you give approaches 1 for high energies...it can also be done for the Z0.

    For part b, you don't have to think of propagators alone...
    you can have:
    [itex]e^- e^+ \rightarrow \gamma^* \rightarrow e^- e^+[/itex] (s-channel)
    as well as
    [itex] e^- e^+ \rightarrow e^- e^+[/itex] ...by t-channel...they don't have to annihilate to photon and produce the pairs again as you have to do for the case of muon products.
     
  8. Mar 16, 2015 #7

    ChrisVer

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  9. Mar 16, 2015 #8
    Their vertex factors are equal, since they the muons and electrons have equal charge.

    But what about their propagator factors?
     
  10. Mar 16, 2015 #9

    ChrisVer

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    For high energies, the propagators of the weak and electromagnetic interaction are the same. It goes as [itex]\frac{1}{s}[/itex]
    The difference comes by the 2nd diagram that is possible for the electron/positron products interaction and doesn't exist for the muons.
     
  11. Mar 16, 2015 #10
    Which process dominates? Which is the "s-channel" and which is the "t-channel"?
     
  12. Mar 16, 2015 #11

    ChrisVer

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    In my figure the diagram with muons and the same one with electrons are the s-channel processes.
    The extra diagram for the [itex]e^- \rightarrow e^-[/itex] and [itex] e^+ \rightarrow e^+[/itex] which exists only for the [itex]e^-e^+ \rightarrow e^+ e^-[/itex] process is the t-channel....

    I don't understand what you mean by "dominates". I am explainging why:
    [itex] \frac{e^+e^- \rightarrow \mu^+\mu^-}{e^+e^- \rightarrow e^+e^-}[/itex] shouldn't be 1.
     
  13. Mar 16, 2015 #12
    All three are in the same picture.

    By "dominate" I mean which is most frequent/most likely?
     
  14. Mar 16, 2015 #13

    ChrisVer

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  15. Mar 16, 2015 #14
    Got it. For electron-positron scattering, is the s-channel more likely or the t-channel more likely?
     
  16. Mar 16, 2015 #15

    ChrisVer

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    Both are the same (I think), since the interaction is symmetric under the interchange [itex]t \leftrightarrow s[/itex]
    But nevertheless, how can this "help" you?
     
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