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 1J a c k s o n 7 . 1 0 (a) No free currents, non permeable media ∇× E = 1 c B ∂t B = 1 c D (1) Thus, ∇×∇× E = 1 c 2 2 D 2 (2) Take E = E 0 exp ( i ( k · r ωt )) and D = ° · E 0 exp ( i ( k · r )) = D 0 exp ( i ( k · r )) . Substituting these deFnitions into (1) we get E 0 e i ( k · r ) = 1 c 2 2 2 D 0 e i ( k · r ) (3) Using ( ψ a )=( ψ ) × a + ψ a ,weget 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 (5) where k = k n . Use a coordinate frame in which ° is diagonal and Diag [ ° ]= { ° 1 2 3 } .A l so k 2 2 =1 /v 2 ; ° i /c 2 /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 (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 v 2 ° n 2 2 v 2 1 v 2 + 1 v 2 2 ± +2 n 2 1 v 2 n 2 2 v 2 n 2 3 v 2 (7) This can be further simpliFed to Det [ M n 2 1 v
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hw6solns - HW 6 solns Phys 6561 1 Jackson 7.10 1 B c t 1 D...

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