Curl of the curl?

  • Thread starter magnifik
  • Start date
  • #1
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given curl E = -1/c*([tex]\partial[/tex]H/[tex]\partial[/tex]t)
div E = 0
div H = 0
curl H = 1/c*([tex]\partial[/tex]E/[tex]\partial[/tex]t), find

[tex]\nabla[/tex] x ([tex]\nabla[/tex] x E)

how do i take [tex]\nabla[/tex] x curl E? i tried to do it by determinants, but i'm not sure which values correspond to the i, j, and k. so my next assumption is that there is some property that i can take advantage of to solve the problem. please help. thanks.
 

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
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[tex]
\nabla \times E = -\frac{1}{c} \frac{\partial H}{\partial t}
[/tex]


you could do it directly
[tex]
\nabla \times (\nabla \times E)
=\nabla \times ( -\frac{1}{c} \frac{\partial H}{\partial t})
=-\frac{1}{c}(\nabla \times \frac{\partial H}{\partial t})
[/tex]

and i think it should be ok to take the time derivative outside the curl, though you may want to confirm that...
[tex]
-\frac{1}{c}(\nabla \times \frac{\partial H}{\partial t})=-\frac{1}{c}\frac{\partial}{\partial t}(\nabla \times H)
[/tex]

and it should follow, otherwise, if you know the original field you could make use of the equality
[tex]
\nabla \times (\nabla \times E)
=\nabla (\nabla \cdot E) - \nabla^2 E
[/tex]

which should simplify with some of the other info
 
Last edited:

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