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Range of validity of Maxwell's equations with charges

  1. Mar 19, 2010 #1

    fluidistic

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    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.
     
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  3. Mar 19, 2010 #2

    Matterwave

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    Maxwell's equations are valid even when no charges are present. In that case, you can set [tex]\rho=0[/tex] and you can see that the equations are still there, and the E and B fields also interact.

    We use this formulation because it's more general than the formulation without charges.

    Because the equations are differential equations, even if some terms equal zero, the E and B fields need not equal zero. Just like if dx/dt=0 it doesn't mean x=0 for all time.
     
  4. Mar 19, 2010 #3

    fluidistic

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    Ah ok, thanks.
    I'm getting it. I feel I'm learning a lot, it's incredibly nice!
     
  5. Mar 19, 2010 #4

    Pengwuino

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    You need to consider that charges create fields throughout your space, not simply where the charges are. Infact, in classical EM, the fields diverge at the actual charges. For example your coulomb field blows up as your field point approaches your charges.

    When you formulate a problem, you define a space you're working in. Maxwell's equations will be valid in the entire space. You don't have a new equation for every infinitesimal volume in the space, it's simply one set of equations for the whole space given some charge distribution inside.

    Think of a gravitational potential. It's defined for your entire volume of space, not just where the mass is.
     
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