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Attraction between two charges in QED

  1. Sep 19, 2010 #1
    Attraction between nuclei and electron in QED is the most simply described with Feynman's diagram and calculation, when an electron fastly flow close to a nuclei. (?? scattering)

    1. But how, in QED, to describe attraction between almost standstill electron and nuclei.
    In classical physics this gives F=e^2/r^2.
    Is this constant flow of virtual photons, which gives this attraction?

    2. What is difference between classical and QED scattering of an electron on the nuclei?

    I read some Feynman books, so you can give me its references.

    Thanks in advance.
  2. jcsd
  3. Sep 26, 2010 #2
    In feynman's "Feynman Lecture on Gravitation" eq. 3.2.8 is divided in two parts.

    = (j4' j4)/k^2+1/(w^2-k^2)(j1' j1+j2' j2)
    j4 part is instantaneous, other tho parts are retarded.

    j4 part means instantaneously acting Couloumb's potential.

    But, Couloumb's potential in classical electrodynamics is not instantaneous.
    For instance if sun suddenly dissapears, its gravity on earth will disapear only after 8 minuts. The same is at one charge (on smaller distance).
    How to explain this? Is maybe instantaneous part only approximation, or?
  4. Oct 10, 2010 #3
    Carlip http://arxiv.org/abs/gr-qc/9909087
    wrote about how speed of planet compensates (centralise force) so that planet behaves similarly that gravitational interaction is instantenous. But how this connect with Feynman's instantenous force?
  5. Oct 10, 2010 #4


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    It dependas how you formulate the theory.

    In high energy physics you insist on explicit covariante and chose a covariant gauge. In this gauge scattering processes are "easy", whereas bound states are "hard".

    But you can chose a different gauge, e.g. the Coulomb gauge (which is preferred as only in this gauge static charges do not radiate). If you eliminate A°(x) which is unphysical and solve the Gauss law contraint you find a Hamiltonian with the Coulomb term

    [tex]V_C = e^2 \int d^2x \int d^3y \frac{\rho(x) \rho(y)}{|x-y|}[/tex]

    describing the Coulomb interaction between the fermionic charge density operators.

    I hope this makes it clear that QED contains ordinary electrostatic interaction as well. It's only in certain gauges that this isn't obvious.
  6. Oct 11, 2010 #5


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    There have been plentiful of articles in the "American Journal of Physics" explaining how the instantaneous Coulombic part is modified by the photonic part so as to yield a retarded interaction.
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