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hw6solns - HW 6 solns Phys 6561 1 Jackson 7.10 1 B c t 1 D...

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HW 6 solns. Phys 6561 October 16, 2011 1 Jackson 7.10 (a) No free currents, non permeable media ∇ × E = 1 c B t ∇ × B = 1 c D t (1) Thus, ∇ × ∇ × E = 1 c 2 2 D t 2 (2) Take E = E 0 exp ( i ( k · r ω t )) and D = · E 0 exp ( i ( k · r ω t )) = D 0 exp ( i ( k · r ω t )) . Substituting these definitions into (1) we get ∇ × ∇ × E 0 e i ( k · r ω t ) = 1 c 2 2 t 2 D 0 e i ( k · r ω t ) (3) Using ∇ × ( ψ a ) = ( ψ ) × a + ψ ∇ × a , we get k × k × E 0 + ω 2 c 2 D 0 = 0 (4) (b) Use a × ( b × c ) = ( a · c ) b ( a · b ) c : k 2 [( n · E 0 ) n E 0 ] + ω 2 c 2 D 0 = 0 (5) where k = k n . Use a coordinate frame in which is diagonal and Diag [ ] = { 1 , 2 , 3 } . Also k 2 / ω 2 = 1 /v 2 ; i /c 2 = 1 /v 2 i . This gives n 2 1 v 2 1 v 2 + 1 v 2 1 n 1 n 2 v 2 n 1 n 3 v 2 n 1 n 2 v 2 n 2 2 v 2 1 v 2 + 1 v 2 2 n 2 n 3 v 2 n 1 n 3 v 2 n 2 n 3 v 2 n 2 3 v 2 1 v 2 + 1 v 2 3 E 1 E 2 E 3 = 0 (6) 1
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Condition for a non trivial solution is Det [ M ] = 0 , Det [ M ] = n 2 1 v 2 1 v 2 + 1 v 2 1 n 2 2 v 2 1 v 2 + 1 v 2 2 n 2 3 v 2 1 v 2 + 1 v 2 3 n 2 2 v 2 n 2 3 v 2 n 2 1 v 2 1 v 2 + 1 v 2 1 n 2 1 v 2 n 2 2 v 2 n 2 3 v 2 1 v 2 + 1 v 2 3 n 2 3 v 2 n 2 1
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