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Electromagnetism question

  1. Nov 7, 2008 #1
    Is [tex]\vec{\nabla^{2}}[/tex] [tex]\vec{E}[/tex] a dot or a vector product?
     
    Last edited: Nov 7, 2008
  2. jcsd
  3. Nov 7, 2008 #2

    gabbagabbahey

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    Do you mean [itex]\nabla ^2 V[/itex]?
     
  4. Nov 7, 2008 #3
  5. Nov 7, 2008 #4

    gabbagabbahey

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    Whenever you take the Laplacian of a vector, its shorthand for

    [tex]\hat{x} \frac{\partial ^2 E_x}{\partial x}+\hat{y} \frac{\partial ^2 E_y}{\partial y}+\hat{z} \frac{\partial ^2 E_z}{\partial z}[/tex]

    in Cartesian coordinates and the definition in curvilinear coordinates is similar.
     
  6. Nov 7, 2008 #5
    How can I then use Maxwell's equations to derive the wave equations for EM waves?
     
  7. Nov 7, 2008 #6

    gabbagabbahey

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    There is no need to use the above definition for that....

    Just use the rules for vector second derivatives and look at

    [tex]\vec{\nabla} \times (\vec{\nabla} \times \vec{E})[/tex]

    and

    [tex]\vec{\nabla} \times (\vec{\nabla} \times \vec{B})[/tex]


    ...you will need to use maxwell's equations and the continuity equation.
     
  8. Nov 7, 2008 #7
    [tex]\vec{\nabla} \times (\vec{\nabla} \times \vec{E})[/tex] = ?
     
  9. Nov 7, 2008 #8

    gabbagabbahey

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    There is a rule that will help you determine that; it should be in your text somewhere (possibly even inside the front cover!)
     
  10. Nov 7, 2008 #9
    Thanks!

    And one more question:

    Why can [tex]\vec{\nabla} \times (-\partial B / \partial t)[/tex] be written [tex](-\partial/ \partial t)\vec{\nabla}\times B[/tex]
     
    Last edited: Nov 7, 2008
  11. Nov 7, 2008 #10

    gabbagabbahey

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    Because [itex]\vec{\nabla}[/itex] represents a spatial derivative, and since space and time are assumed to be independent for classical E&M, it doesn't matter whether you take the time derivative before the spacial derivative, or vice versa.
     
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