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Maxwell's equations form

  1. Jul 30, 2008 #1
    Easy question for those who know, I expect. It would help me understand though.

    Generally, wherever I look for information the equation

    curl E = - dB/dt

    is given, but in some areas I see the equation

    curl E(r,t) = j*omega*u0*H(r,t)

    where j is the imaginary unit, omega is angular freq., u0 is permeability of free space.

    B=u0*H so that's ok, but does this imply that dB/dt = -j*omega*B ?

    Also, curl H(r,t) = -j*omega*e0*e*E(r,t) is given where e0 is permittivity of free space, e is a dielectric tensor. I assume this can be explained in the same way.

    I'm missing a chunk of understanding as you probably notice ;)
     
  2. jcsd
  3. Jul 30, 2008 #2

    Andy Resnick

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    Usually, when trying to write down solutions to Mawell's equations, people assume two things about the field:

    1) The spatial part and the temporal part can be separated: E(r,t) = R(r)T(t).
    2) The temporal part can be written as T(t) = Ae[itex]^{i \omega t}[/itex].

    The first assumption is fairly basic to solving differential equations, and separable functions are a very important class of solutions- I can't give a consise reason why, but for now, it makes it possible to analytically solve the equations.

    The second assumption just means that the temperal part oscillates like a sine wave. It's written that way to be more general (and actually, the full expression is T(t) = Ae[itex]^{i \omega t}[/itex] + BAe[itex]^{-i \omega t}[/itex] ). There's good reasons for this assumption as well, which I don't need to get into now.

    Anyhow, hopefully you can see where the j[itex]\omega[/itex] comes from now- electrical engineers use 'j' instead of 'i' becasue 'i' is current density.

    The other part is the conversion of E to D, and B to H. But you seem to have a handle on that part.
     
  4. Jul 30, 2008 #3
    Ah, yes I see now.

    Thanks very much!
     
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