Thus we define a set of adjoint or conjugate unit

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Thus, we define a set of adjoint or conjugate unit vectors by means of the relationships ( k ) k = 1; ( k ) t ( σ ) ( ( k ) = 0; ( t ( σ ) ( ) k ) ) k = 0; ( t ( σ ) ( ) k ) ) t ( σ ') ( ( k ) = δ σσ ' [ V-4 ] Thus, with r k = k ) k = k * ( k (so that r k r k = k 2 ) the representation for r D ( r k ) in Equation [ V-3 ] automatically satisfies Equation. [ I-8c ]. A key problem is the representation of the ubiquitous vector operation
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O N C LASSICAL E LECTROMAGNETIC F IELDS P AGE 44 R. Victor Jones, December 19, 2000 t ( r r ) r F ( r r ) = r ∇× r ∇× r F ( r r ) ( 29 = r r ∇⋅ r F ( r r ) -∇ 2 r F ( r r ) [ V-5 ] which appears in Equation [ V-1 ]. For plane wave representation of the operator becomes t ( r k ) r k r k ( 29 t 1 - r k r k = k 2 t 1 - ( k ) k { } . [ V-6 ] Since t 1 = ) k ( k + ) t (1) ( ( k ) ( t (1) ( ) k ) + ) t (2) ( ( k ) ( t (2) ( ) k ) [ V-7 ] the plane wave representation of the operator simplifies to t ( r k ) = k 2 ) t (1) ( ( k ) ( t (1) ( ) k ) + ) t (2) ( ( k ) ( t (2) ( ) k ) { } [ V-8] Using this representation of the t ( r k ) operator and the representation for r D ( r k ) in Equation. [ V-1 ], the generalized Helmholz equation -- i.e. Equation [ V-1 ] -- can be written k 2 {D (1) ( r k ) ) t (1) ( ( k ) ( t (1) ( ) k ) t ε - 1 ) t (1) ( ( k ) [ ] + D (1) ( r k ) ) t (2) ( ( k ) ( t (2) ( ) k ) t ε - 1 ) t (1) ( ( k ) [ ] + D (2) ( r k ) ) t (1) ( ( k ) ( t (1) ( ) k ) t ε - 1 ) t (2) ( ( k ) [ ] + D (2) ( r k ) ) t (2) ( ( k ) ( t (2) ( ) k ) t ε - 1 ) t (2) ( ( k ) [ ] } = ϖ 2 μ 0 D (1) ( r k ) ) t (1) ( ( k ) + D (2) ( r k ) ) t (2) ( ( k ) { } [ V-9 ] Crucial point : To obtain an eigenvalue equation we need to choose the eigenvectors ) t ( σ ) ( ( k ) so that ( t (1) ( ) k ) t ε - 1 ) t (2) ( ( k ) = ( t (2) ( ) k ) t ε - 1 ) t (1) ( ( k ) 0 [ V-10 ] If we can find eigenvectors defined in this way, then Equation [ I-9 ] becomes an eigenvalue equation with eigenvalues -- i.e., the inverse refractive indices -- given by n 1 ( r k ) [ ] - 2 = k 0 2 k 2 = ε 0 ( t (1) ( ) k ) t ε - 1 ) t (1) ( ( k ) [ V-11a ]
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O N C LASSICAL E LECTROMAGNETIC F IELDS P AGE 45 R. Victor Jones, December 19, 2000 n 2 ( r k ) [ ] - 2 = k 0 2 k 2 = ε 0 ( t (2) ( ) k ) t ε - 1 ) t (2) ( ( k ) [ V-11b ] These results -- i.e., Equations [ I-10 ] and [ I-11 ] -- are a complete formal solution of the problem. However, they are difficult to apply in the general case and an addition relationship -- viz., the Fresnel equation of wave normals -- is found to be extremely useful as the starting point for actual computations. For plane waves, Equation [ V-1 ] can be rewritten as k 2 t 1 - k 0 2 ε 0 ( 29 t ε [ ] r E r k ( 29 = r k r k r E r k ( 29 [ V-12 ] Multiplying this equation through by r k k 2 t 1 - k 0 2 ε 0 ( 29 t ε [ ] - 1 we obtain r k r E r k ( 29 = r k k 2 t 1 - k 0 2 ε 0 ( 29 t ε [ ] - 1 r k { } r k r E r k ( 29 [ V-13a ] or 1 = r k k 2 t 1 - k 0 2 ε 0 ( 29 t ε [ ] - 1 r k [ V-13b ] From this result we may develop two important relationships. Using the principal axes coordinates of the dielectric tensor, we can write k 2 t 1 - k 0 2 ε 0 ( 29 t ε [ ] = k 2 - k 0 2 ε 0 ( 29 ε λ ( 29 ( 29 ) a λ ( a λ λ= a c [ V-14a ] and k 2 t 1 - k 0 2 ε 0 ( 29 t ε [ ] - 1 = k 2 - k 0 2 ε 0 ( 29 ε λ ( 29 ( 29 - 1 ) a λ ( a λ λ= a c [ V-14b ] Therefore, Equation [ I-14b ] becomes ) k ( a a ( 29 ( k ) a a ( 29 n 2 - n a 2 + ) k ( a b ( 29 ( k ) a b ( 29 n 2 - n b 2 + ) k ( a c ( 29 ( k ) a c ( 29 n 2 - n c 2 = 1 n 2 [ V-15a ]
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O N C LASSICAL E LECTROMAGNETIC F IELDS P AGE 46 R. Victor Jones, December 19, 2000
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