Deriving Coulomb's Law with Virtual Photon Exchange Theory: Complete Explanation

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 6K views
sharma_satdev
Messages
33
Reaction score
0
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
 
Physics news on Phys.org
http://www.scribd.com/doc/70796478/97/The-Coulomb-Potential
 
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]

with

[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: