- #1

paweld

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[tex]w = \frac{1}{2}(E^2+B^2) [/tex]. In fact energy of the filed can be any

function [tex]w[/tex] for which there exists such vectror field [tex] \vec{S} [/tex] that

continuity equation is fulfilled:

[tex] \vec{E} \vec{J}+\frac{\partial w}{\partial t}+\nabla \vec{S}=0[/tex].

Where [tex] \vec{E} \vec{J}[/tex] is a work donw by the filed in unit volume and time.

Obviously we can choose: [tex]w = \frac{1}{2}(E^2+B^2) [/tex] and

[tex]S =\vec{E}\times \vec{B} [/tex] but there are also other options.

Because [tex] \frac{\partial \rho}{\partial t}+\nabla \vec{j}=0 [/tex] ([tex]\rho [/tex] - density

of charge, [tex] \vec{j}[/tex] - density of current) we can take for example:

[tex] w = \frac{1}{2}(E^2+B^2) + \alpha \rho[/tex] and [tex]S =\vec{E}\times \vec{B} + \alpha\vec{j}[/tex].

This change doesn't affect the continuity equation and of course both [tex]\rho [/tex] and [tex] \vec{j}[/tex] can by express

by means of Maxwell equation in terms of [tex] \vec{E}[/tex] and [tex] \vec{B}[/tex]

so that [tex]w[/tex] and [tex]\vec{S}[/tex] depend only on the field.

Are there any physical arguments for [tex]w = \frac{1}{2}(E^2+B^2) [/tex] or it's

only our traditional choice and others are as well.