paweld
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I wonder why we always assume that electromagnetic field energy density is
w = \frac{1}{2}(E^2+B^2). In fact energy of the filed can be any
function w for which there exists such vectror field \vec{S} that
continuity equation is fulfilled:
\vec{E} \vec{J}+\frac{\partial w}{\partial t}+\nabla \vec{S}=0.
Where \vec{E} \vec{J} is a work donw by the filed in unit volume and time.
Obviously we can choose: w = \frac{1}{2}(E^2+B^2) and
S =\vec{E}\times \vec{B} but there are also other options.
Because \frac{\partial \rho}{\partial t}+\nabla \vec{j}=0 (\rho - density
of charge, \vec{j} - density of current) we can take for example:
w = \frac{1}{2}(E^2+B^2) + \alpha \rho and S =\vec{E}\times \vec{B} + \alpha\vec{j}.
This change doesn't affect the continuity equation and of course both \rho and \vec{j} can by express
by means of Maxwell equation in terms of \vec{E} and \vec{B}
so that w and \vec{S} depend only on the field.
Are there any physical arguments for w = \frac{1}{2}(E^2+B^2) or it's
only our traditional choice and others are as well.
w = \frac{1}{2}(E^2+B^2). In fact energy of the filed can be any
function w for which there exists such vectror field \vec{S} that
continuity equation is fulfilled:
\vec{E} \vec{J}+\frac{\partial w}{\partial t}+\nabla \vec{S}=0.
Where \vec{E} \vec{J} is a work donw by the filed in unit volume and time.
Obviously we can choose: w = \frac{1}{2}(E^2+B^2) and
S =\vec{E}\times \vec{B} but there are also other options.
Because \frac{\partial \rho}{\partial t}+\nabla \vec{j}=0 (\rho - density
of charge, \vec{j} - density of current) we can take for example:
w = \frac{1}{2}(E^2+B^2) + \alpha \rho and S =\vec{E}\times \vec{B} + \alpha\vec{j}.
This change doesn't affect the continuity equation and of course both \rho and \vec{j} can by express
by means of Maxwell equation in terms of \vec{E} and \vec{B}
so that w and \vec{S} depend only on the field.
Are there any physical arguments for w = \frac{1}{2}(E^2+B^2) or it's
only our traditional choice and others are as well.