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The finite size of the nucleon

  1. Sep 21, 2014 #1
    Hi,

    I read in article: to incorporate the effects of the finite size of the nucleon, we considered an exponential form factor.
    I want to know what does "the finite size" mean?

    thank you
     
  2. jcsd
  3. Sep 21, 2014 #2

    vanhees71

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    In the context of quantum theory "finite size" means that an object is not described in terms of an irreducible quantum field, i.e., it's not an elementary particle.

    This is, however, a bit too strict in practice, because for many problems in atomic physics it is a good approximation to consider the proton as a point particle and then its mass is much larger than the one of the electron and you can abstract simply as an external Coulomb field. This gives the hydrogen spectrum to quite a high accuracy.

    Now, if you do electron proton scattering this holds true only for not too large collision energies. If the collision energy becomes large enough so that you resolve the "finite size" of a proton, you'll realize that the Coulomb potential is not a good enough approximation anymore, and you must use a form factor. You can e.g., as a first approximation use an extended charge uniform charge distribution.

    At even higher energies, you'll figure out that the proton has a pretty complicated substructure. Then another approximation is at place, the socalled parton model: You describe the proton as bound state of three constituent quarks. This was in fact the historical way, how quarks were discovered (look for Bjorken Scaling in the literature).

    As you see, the necessary level of description of "extended" objects like a nucleon depends on the typical energies involved in the setup, the energy scale. At low energies, where you don't resolve the inner structure of the extended object (in scattering experiments a typical estimate is whether the de Broglie wavelength of the scatterer is large or small compared to the extension of the scattering center), you can describe it as a "point particle" (i.e., an elementary quantum field with the given spin of the particle). The larger the scattering energy gets the more details of the structure of the extended (composed) object you resolve, and you have to change the level of description. In this sense the particle colliders are "microscopes" delivering higher and higher resolution with which you can study the inner structure of particles like the nucleon.

    On the other hand there are also particles which are considered "elementary", the leptons and quarks. In the usual slang we say they are "point particles". This of course only means that they are described with elementary quantum fields (in this case Dirac fields). This, of course, only means that we haven't found any deviations in the cross sections that describe the scattering of such particles from the Standard Model that describes them with elementary quantum fields. In other words, if they have an "extension", i.e., if they are bound states of some other constituents, this extension must be much smaller than we can resolve with the hitherto largest collision energies (available at the LHC).
     
  4. Sep 21, 2014 #3
    Thank you for your help and I appreciate your valuable time.
     
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