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We discuss them further in Chap. 7. 65 α such More generally, there exist solutions having a complex wavevector k = β − jα that β , α are not collinear. , will be different. 10. 4) β − jα α)·E0 = 0, the magnetic field is computed from With E0 chosen to satisfy k · E0 = (β β − jα α)×E0 /ωμ. Eq. 2), H0 = k × E0 /ωμ = (β Let us look at an explicit construction. We choose β , α to lie on the xz plane of ˆ βx and α = ˆ ˆ αx . Thus, Fig. 1) do not preclude a complex-valued wavevector k. For example, if the medium is lossy, we must replace {η, k} by {ηc , kc }, where kc = β − jα, resulting from a complex effective permittivity c .

2), with the right-hand side of Amp` ere’s law given by Jtot = J + jωD = jω c E . The assumption of uniformity (∂x = ∂y = 0), will imply again that the fields E, H are transverse to the direction ˆ z. 4) cE These may be written in a more convenient form by introducing the complex wavenumber kc and complex characteristic impedance ηc defined by: kc = ω μ The sense of rotation may be determined from the cross-product E (0)× E (t). If the rotation is counterclockwise, this vector will point towards the positive z-direction, and otherwise, it will point towards the negative z-direction.

4) 68 2. Uniform Plane Waves These two expressions are equivalent to first-order in v/c. This follows from the Taylor series approximation (1 +x)−1 1 −x, which is valid for |x| 1. More generally, to first order in va /c and vb /c, Eq. 1) does depend only on the relative velocity. 8) 1 − β2 Setting kz = k cos θ, kx = k sin θ, with k = ω/c, and similarly in the S frame, kz = k cos θ , kx = k sin θ , with k = ω /c, Eqs. 7) Using the first-order Taylor series expansion (1 + x)±1/2 = 1 ± x/2, one can show that Eq.