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## Main Question or Discussion Point

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

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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

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tom.stoer

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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.

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.

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