- #1

fluidistic

Gold Member

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## Main Question or Discussion Point

Maxwell's equations with charges can be written as the following (in the cgs system):

[tex]\frac{\partial \vec E}{\partial t} =c \vec \nabla \wedge \vec B-4\pi \vec J[/tex].

[tex]\frac{\partial \vec B}{\partial t} =-c \vec \nabla \wedge \vec E[/tex].

[tex]\vec \nabla \cdot \vec E =4 \pi \rho[/tex].

[tex]\vec \nabla \cdot \vec B =0[/tex].

If I'm right, these equations are valid only where charges are present. It means, almost nowhere! For instance consider a positively charged table. [tex]\rho(t,\vec x)[/tex] is of course not continuous since we're dealing with charges. In other words it is zero everywhere when there's no charge. If we assume the electron has a volume, then the equations would be valid within the volume of the electron. So in an neutral atom, these equations are valid less than 0.001% of the space (only where the proton and the electron are).

I'm just curious if I'm right. In an affirmative case, why do we bother with them and why don't we only use Maxwell's equation in vacuum?

In the latter case, I could imagine having a lot of boundary equations (precisely the places enclosing the charges) and I could only use Maxwell's equation in vacuum, taking into account all the boundary equations not to lose any information about the charges.

[tex]\frac{\partial \vec E}{\partial t} =c \vec \nabla \wedge \vec B-4\pi \vec J[/tex].

[tex]\frac{\partial \vec B}{\partial t} =-c \vec \nabla \wedge \vec E[/tex].

[tex]\vec \nabla \cdot \vec E =4 \pi \rho[/tex].

[tex]\vec \nabla \cdot \vec B =0[/tex].

If I'm right, these equations are valid only where charges are present. It means, almost nowhere! For instance consider a positively charged table. [tex]\rho(t,\vec x)[/tex] is of course not continuous since we're dealing with charges. In other words it is zero everywhere when there's no charge. If we assume the electron has a volume, then the equations would be valid within the volume of the electron. So in an neutral atom, these equations are valid less than 0.001% of the space (only where the proton and the electron are).

I'm just curious if I'm right. In an affirmative case, why do we bother with them and why don't we only use Maxwell's equation in vacuum?

In the latter case, I could imagine having a lot of boundary equations (precisely the places enclosing the charges) and I could only use Maxwell's equation in vacuum, taking into account all the boundary equations not to lose any information about the charges.