# Correction to the field energy due to the existence of discrete charges

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In the classical electromagnetic field theory, the field density of energy is given by:

$$u = (\epsilon/2)E^2 + (\mu/2)H^2$$
One of the differences between the classical electromagnetic theory and the real world is that in classical EM all charge and current density, (ρ(r), J(r)), is indistinguishable and every point of charge interacts with the rest through the generated EM field. On the other hand, in the real world we have discrete charged particles that do not interact electromagnetically with themselves (at least not directly), an example of this statement is the Hydrogen's electron Hamiltonian, the potential we see is the one created by the proton but there is no contribution from the electron itself.

The overall EM field energy changes with the interaction with charges as:

$$\partial_t u = - E · J + Flow term$$
Where there is no charge, the overall field energy does not change, it only flows.

If we have an infinitesimal density of charge isolated ρ(r)dr3, the energy of its field is order dr5, that means this energy is negligible compared with the charge’s mass energy.

We can conclude then that when "infinitesimal" charges are far away from each other, the EM field does not have energy but it earns it when the charges get closer. But if we accept that discrete charges do not interact with themselves through EM field, then it is evident that there is no work done in bringing together the discrete particle charge. A system with only one discrete particle would bring no energy to the EM field but it would have an electric field, so according to the classical formula, the field would have energy. This fact makes necessary some correction should be done on the formula of EM field energy.

I would like to know if some of my assumptions is wrong or if this correction is explained in some theory.

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