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

Coulombs law derivation

  1. Apr 16, 2012 #1
    How the virtual photon exchange theory be used to derive completely Coulombs Law related to electrostatic field? Complete derivation means derivation involving charges and distance
  2. jcsd
  3. Apr 17, 2012 #2
  4. Apr 17, 2012 #3


    User Avatar
    Science Advisor

    You don't need any virtual photons but you can derive Coulombs law w/o using perturbation theory. It's most transparent in
    a) Coulomb gauge or in
    b) A°=0 gauge plus fixing of the residual symmetry of time-independent transformations respecting A°=0.

    In both cases it boils down to construct the inverse of the Laplacian 1/Δ; in Coulomb gauge this is due to the Poisson equation

    [tex]\Delta A^0 = \rho[/tex]


    [tex]\Delta^{-1} \to k^{-2} [/tex]

    in k-space and

    [tex]\Delta^{-1} \to |x|^{-1} [/tex]

    in x-space

    This results in an interaction term

    [tex]V \sim \int d^3x\,d^3y\,\frac{\rho(x)\,\rho(y)}{|x-y|} [/tex]

    Of course there are other interaction terms involving physical (transversal) photons as well.
    Last edited: Apr 17, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook