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Curl of the Polarization (Electrostatics)

  1. Jul 24, 2013 #1
    I've been reading Griffith's "Introduction to Electrodynamics" and I've got to this part where it says:

    "When you are asked to compute the electric displacement, first look for symmetry. If the problem exhibits spherical, cylindrical, or plane symmetry, then you can get [itex]\vec{D}[/itex]directly from Gauss's equation (for the displacement) in integral form. (Evidently in such cases ∇x[itex]\vec{P}[/itex] is automatically zero, but since symmetry alone dictates the answer you're not really obliged to worry about the curl.)"

    Now, why is it that the curl of the polarization is always zero in those cases where there is symmetry?

    Is it just because in every case that exhibits symmetry the polarization is perpendicular to the boundary?
     
    Last edited: Jul 24, 2013
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  3. Jul 24, 2013 #2

    Andy Resnick

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    Are you considering electrostatics or electrodynamics?
    In what sense do you mean 'symmetry': a symmetric homogenous body, or some sort of crystal symmetry?
     
  4. Jul 24, 2013 #3
    Electrostatics. Symmetry meaning the first one, a symmetric homogeneous body.
     
  5. Jul 24, 2013 #4

    Andy Resnick

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    Since ∇x E = -∂B/∂t = 0 for statics and D = E + P, the result immediately follows.
     
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