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Condition of continuity of E field at a boundary

  1. Dec 28, 2012 #1
    I am trying to understand the derivation of Snell's law using Maxwell's equation and got stuck.

    My text book says that "the E field that is tangent to the interface must be continuous" in order to consider refraction of light.
    If it were static E field I understand this is true because in electrostatics

    rotE = 0

    holds. However Snell's law describes how electromagnetic waves change their direction of propagation when going through an interface of two mediums. Since our E filed is changing dynamically, we should use the equation

    rotE = -∂B/∂t

    in stead. To me it is not obvious why this equation leads to the continuity condition.
    How does the continuity condition in Snell's law appears from Maxwell's equations?
    Last edited: Dec 28, 2012
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  3. Dec 28, 2012 #2

    Simon Bridge

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  4. Dec 29, 2012 #3

    Meir Achuz

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    The continuity of E tangential comes from applying Stokes' theorem to rotE = -∂B/∂t.
    The area for [tex]\int{\bf dS}\partial_t{\bf B}[/tex] shrinks to zero.
    Last edited: Dec 29, 2012
  5. Dec 29, 2012 #4
    Stokes's theorem for rotE is
    [tex]\int{ rot\bf{E}}{\bf dS}= \oint _{∂S} \bf Edx = \oint _{∂S} \bf \partial_t B dx [/tex]

    How does this lead to the continuity condition?
  6. Dec 29, 2012 #5


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  7. Dec 30, 2012 #6
    OK I see. It seems like the continuity condition is something to do with the fact that the interface has zero volume and the planar surface is sufficiently large.
    The path of line integration must be an infinitely thin rectangular when the area of the box approaches to 0.
    The shape of the box is the key because it will allow the line integration to become 0 even if rotE is non-zero.
    Thanks for all the replies.

    Will the same rule apply if there is a gradient layer between the two phases?
    Let's say the geometry is no longer flat but curved, and the curvature of radius is comparable to the thickness of gradient layer.
    I'm pretty sure that the parallel component of the E field strength will still be continuous at any point.
    But will the phase still be the same?
  8. Jan 1, 2013 #7

    Simon Bridge

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    You know what the answer has to be already - what usually happens to Snell's Law when the surface is curved or the interface is not sharp?

    You could try working it out for a simple setup - like a spherical interface (par-axial) - and see if the general boundary conditions give you the appropriate equations.
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