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Electromagnetic Expressions

  1. Oct 27, 2011 #1
    Hello all!
    I am getting into the electromagnetic waves section in my optics class and in some of these derivations, they are using expressions that I cannot remember how they were derived or the context of them. My book from my E&M class I took awhile back is currently at my house and I am at my school so I can not look it up. The expressions I am talking about are..

    [tex]E=cB[/tex]
    Where E and B are the magnitudes of the electric and magnetic fields respectively.

    Energy Density in the Electric field
    [tex]u_{E}=\frac{1}{2}ε_0E^2[/tex]

    Energy Density in the Magnetic field
    [tex]u_{B}=\frac{1}{2} \frac{1}{μ_0}B^2[/tex]

    Can anyone show where these expressions come from? It would help my understanding a lot! Thanks!
     
  2. jcsd
  3. Oct 28, 2011 #2

    vanhees71

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    Everything in classical electromagnetic theory follows from Maxwell's equations or, more elegantly, from the corresponding action principle.

    Your first expression is valid for plane waves or in the far-field limit, i.e., at distances from any sources (i.e., charge and current distributions) much larger than the typical wavelength of your em. wave.

    To prove it take the Maxwell equations for the free field and look for plane-wave solutions of the form

    [tex]\vec{E}(t,\vec{x})=\vec{E}_0 \exp(-\mathrm{i} \omega t+\mathrm{i}\vec{k} \cdot \vec{x}),[/tex]
    [tex]\vec{B}(t,\vec{x})=\vec{B}_0 \exp(-\mathrm{i}\omega t+\mathrm{i}\vec{k} \cdot \vec{x}).[/tex]

    The other expressions do not make much sense taken separated from each other. There is only an electromagnetic field not an electric and a separated magnetic field. It depends on the frame of reference of the observer, what are electric and magnetic components of the field. In relativity the field-strength components are components of an antisymmetric 2nd-rank tensor.

    You find the expression for the energy density of the em. fields with help of Noether's theorem or less systematically by taking the time derivative of the expression,

    [tex]\epsilon=\frac{\epsilon_0}{2} \vec{E}^2+ \frac{1}{2 \mu_0} \vec{B}^2[/tex]

    and use Maxwell's equation to derive the energy-conservation law in form of a continuity equation. You'll also find that the energy-current density is given by something proportional to the Poynting vector [itex]\vec{S}=\vec{E} \times \vec{B}[/itex].
     
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