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What is momentum transfer cross section ?

  1. Nov 17, 2005 #1

    vanesch

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    What is "momentum transfer cross section" ?

    Hi all,
    I'm reading a book on electron-molecule interactions and I'm puzzled by a quantity (well, by the difference in definition of two quantities). The book is, for your information, "fundamental electron interactions with plasma processing gases" by L. G. Christophorou.
    One considers the elastic cross sections of electron-molecule interactions for a given molecule in a given state (say, ground state), with the differential cross section:
    [tex] dN_s = \sigma_{e,diff}(\epsilon,\theta) N_e N_t d\Omega [/tex]
    where a beam of N_e electrons per cm^2 per second strike a target gas containing N_t molecules, and dN_s is the number of electrons which are elastically scattered per second in the solid angle dOmega under an angle theta.
    (Formula 2.1 in the book).
    He then defines the total elastic scattering cross section (eq. 2.2):
    [tex] \sigma_{e,t}(\epsilon) = \int_0^{2 \pi} \int_0^{\pi}\sigma_{e,diff}(\epsilon,\theta) \sin \theta d\theta d\phi [/tex]
    which is of course clear,
    but he also defines:
    [tex] \sigma_{m}(\epsilon) = \int_0^{2 \pi} \int_0^{\pi}\sigma_{e,diff}(\epsilon,\theta)(1-\cos \theta) \sin \theta d\theta d\phi [/tex]
    in equation (2.3) and calls it the "momentum transfer cross section".
    Anybody an idea what that stands for ?
     
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  3. Nov 17, 2005 #2

    Gokul43201

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    Here's a plausibility (read "hand-waving") argument, not a derivation ...

    Consider two kinds of scattering events : one where the electron deflects by small angles ([itex]\theta \rightarrow 0[/itex]) and a second, where the electron deflects by large angles [itex]\theta \rightarrow \pi[/itex]). Which of these two events involves a greater momentum transfer from the electron to the molecule ?

    The [itex]1-cos\theta [/itex] is something like a weight that says that large angle scattering is more important for momentum transfer than small angle scattering. You can also think of it as the change in momentum of the electron (along the incident direction), which, in the elastic limit is the momentum added to the molecule in this "forward" direction.

    To elaborate a wee bit : If [itex]\theta[/itex] is the angle through which the electron is scattered, the final momentum of the electron (non-relativistic) along the incident direction is simply [itex]p_i cos\theta [/itex], as long as it is being scattered by something much heavier than itself.

    Disclaimer : I have not read the book or studied electron-molecule scattering, so take the above as merely my opinion.
     
    Last edited: Nov 17, 2005
  4. Nov 17, 2005 #3

    vanesch

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    Yes, that's what I more or less guessed after having pondered a bit more over it. When you calculate the horizontal momentum component (in the axis of the incoming electron) that is *transferred* to the target species, you find that it will have a theta dependence in [itex] 1 - \cos \theta [/itex]. So if you want to calculate a quantity that will depend on that momentum transfer averaged over all scattering events, you will have to calculate in the end the integral given by the momentum transfer cross section.

    All NON-horizontal momentum transfer will be cancelled out on average, so I guess that this momentum transfer cross section is the useful quantity to find out how much momentum is transferred from an electron beam to the medium (how much horizontal wind you create by the electron beam).
    That's indeed a usage I can find for that expression - something that occured to me and a collegue I asked - and apparently you too.

    But it is funny to call this a "cross section" and I wondered whether it had some explicit meaning.
     
  5. Nov 18, 2005 #4

    Gokul43201

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    Well, it has dimensions of scattering cross-section! I've come across less compelling reasons for the way things get named. IMO, the shine-Gordon equation takes the cake !

    Zz may have more to say about this when he gets back, but people in cond mat see momentum scattering cross sections all the time, because that is the quantity that is relevant to transport (think momentum ~ current).
     
  6. Nov 18, 2005 #5

    vanesch

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    Well, I encountered the concept for the first time, and noticed (that's why I took up the book!) that you shouldn't include them when summing different types of scattering cross sections to obtain the total scattering cross section :redface:
     
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