- #1
paweld
- 255
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I wonder why we always assume that electromagnetic field energy density is
[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.
[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.