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Uniform Plane Waves 38 2. Uniform Plane Waves Because also ∂zEz = 0, it follows that Ez must be a constant, independent of z, t. Excluding static solutions, we may take this constant to be zero. Similarly, we have 2 = Hz 0. Thus, the fields have components only along the x, y directions: E(z, t) = xˆ Ex(z, t)+yˆ Ey(z, t) Uniform Plane Waves (transverse fields) (2.1.2) H(z, t) = xˆ Hx(z, t)+yˆ Hy(z, t) These fields must satisfy Faraday’s and Amp`ere’s laws in Eqs. (2.1.1). We rewrite these equations in a more convenient form by replacing and μ by: 1 η 1 μ = ,μ= , where c = √ ,η= (2.1.3) ηc c μ Thus, c, η are the speed of light and characteristic impedance of the propagation medium. Then, the first two of Eqs. (2.1.1) may be written in the equivalent forms: ∂E 1 ∂H ˆz × =− η 2.1 Uniform Plane Waves in Lossless Media ∂z c ∂t (2.1.4) ∂H 1 ∂E The simplest electromagnetic waves are uniform plane waves propagating along some η ˆz × = ∂z c ∂t fixed direction, say the z-direction, in a lossless medium {, μ}. The assumption of uniformity means that the fields have no dependence on the The first may be solved for ∂zE by crossing it with ˆz. Using the BAC-CAB rule, and transverse coordinates x, y and are functions only of z, t. Thus, we look for solutions noting that E has no z-component, we have: of Maxwell’s equations of the form: E(x, y, z, t)= E(z, t) and H(x, y, z, t)= H(z, t). ∂E ∂E ∂E ∂E † = ˆz × × ˆz = (ˆz · ˆz)−ˆz ˆz · = Because there is no dependence on x, y, we set the partial derivatives ∂x 0 and ∂z ∂z ∂z ∂z ∂y = 0. Then, the gradient, divergence, and curl operations take the simplified forms: where we used ˆz · ∂zE = ∂zEz = 0 and ˆz · ˆz = 1. It follows that Eqs. (2.1.4) may be replaced by the equivalent system: ∂ ∂Ez ∂E ∂Ey ∂Ex ∇∇=ˆz , ∇∇·E = , ∇∇×E = ˆz × =−xˆ + yˆ ∂z ∂z ∂z ∂z ∂z ∂E 1 ∂ =− (ηH × ˆz) Assuming that D = E and B = μH , the source-free Maxwell’s equations become: ∂z c ∂t (2.1.5) ∂ 1 ∂E ∂E ∂H (ηH × ˆz)=− ∂H ˆz × =−μ ∂z c ∂t ∇∇×E =−μ ∂z ∂t ∂t ∂H ∂E Now all the terms have the same dimension. Eqs. (2.1.5) imply that both E and H × = ∂E ˆz satisfy the one-dimensional wave equation. Indeed, differentiating the first equation ∇∇×H = ∂z ∂t ∂t ⇒ (2.1.1) with respect to z and using the second, we have: ∂Ez ∇∇· = = 0 2 2 E 0 ∂z ∂ E 1 ∂ ∂ 1 ∂ E =− (ηH × ˆz)= or, ∂z2 c ∂t ∂z c2 ∂t2 ∇∇·H = 0 ∂Hz = 0 ∂z ∂2 1 ∂2 − E(z, t)= 0 (wave equation) (2.1.6) An immediate consequence of uniformity is that E and H do not have components ∂z2 c2 ∂t2 along the z-direction, that is, E = H = 0. Taking the dot-product of Amp`ere’s law z z and similarly for H. Rather than solving the wave equation, we prefer to work directly with the unit vector ˆz, and using the identity ˆz · (ˆz × A)= 0, we have: with the coupled system (2.1.5). The system can be decoupled by introducing the so- called forward and backward electric fields defined as the linear combinations: ∂H ∂E ∂Ez ˆz · ˆz × = ˆz · = 0 ⇒ = 0 ∂z ∂t ∂t 1 E+ = (E + ηH × ˆz) 2 † ∂ The shorthand notation ∂x stands for . (forward and backward fields) (2.1.7) ∂x 1 E− = (E − ηH × ˆz) 2 2.1. Uniform Plane Waves in Lossless Media 39 40 2. Uniform Plane Waves Component-wise, these are: Inserting these into the inverse formula (2.1.9), we obtain the most general solution of (2.1.5), expressed as a linear combination of forward and backward waves: 1 1 Ex± = (Ex ± ηHy), Ey± = (Ey ∓ ηHx) (2.1.8) 2 2 E(z, t) = F(z − ct)+G(z + ct) We show next that E+(z, t) corresponds to a forward-moving wave, that is, moving 1 (2.1.12) towards the positive z-direction, and E−(z, t), to a backward-moving wave. Eqs. (2.1.7) H(z, t) = ˆz × F(z − ct)−G(z + ct) η can be inverted to express E, H in terms of E+, E−. Adding and subtracting them, and using the BAC-CAB rule and the orthogonality conditions ˆz · E± = 0, we obtain: The term E+(z, t)= F(z − ct) represents a wave propagating with speed c in the positive z-direction, while E−(z, t)= G(z+ct) represents a wave traveling in the negative E(z, t) = E+(z, t)+E−(z, t) z-direction. (2.1.9) 1 To see this, consider the forward field at a later time t + Δt. During the time interval H(z, t) = ˆz × E+(z, t)−E−(z, t) η Δt, the wave moves in the positive z-direction by a distance Δz = cΔt. Indeed, we have: In terms of the forward and backward fields E±, the system of Eqs. (2.1.5) decouples E+(z, t + Δt) = F z − c(t + Δt) = F(z − cΔt − ct) into two separate equations: ⇒ E+(z, t + Δt)= E+(z − Δz, t) E+(z − Δz, t) = F (z − Δz)−ct = F(z − cΔt − ct) ∂E+ 1 ∂E+ =− ∂z c ∂t This states that the forward field at time t + Δt is the same as the field at time t, (2.1.10) ∂E− 1 ∂E− but translated to the right along the z-axis by a distance Δz = cΔt. Equivalently, the =+ ∂z c ∂t field at location z + Δz at time t is the same as the field at location z at the earlier time t − Δt = t − Δz/c, that is, Indeed, using Eqs. (2.1.5), we verify: E+(z + Δz, t)= E+(z, t − Δt) ∂ 1 ∂ 1 ∂E 1 ∂ (E ± ηH × ˆz)=− (ηH × ˆz)∓ =∓ (E ± ηH × ˆz) ∂z c ∂t c ∂t c ∂t Similarly, we find that E−(z, t + Δt)= E−(z + Δz, t), which states that the backward field at time t + Δt is the same as the field at time t, translated to the left by a distance Eqs. (2.1.10) can be solved by noting that the forward field E+(z, t) must depend Δz. Fig. 2.1.1 depicts these two cases. on z, t only through the combination z − ct (for a proof, see Problem 2.1.) If we set E+(z, t)= F(z − ct), where F(ζ) is an arbitrary function of its argument ζ = z − ct, then we will have: ∂E+ ∂ ∂ζ ∂F(ζ) ∂F(ζ) = F(z − ct)= = ∂z ∂z ∂z ∂ζ ∂ζ ∂E+ 1 ∂E+ ⇒ =− ∂E+ ∂ ∂ζ ∂F(ζ) ∂F(ζ) ∂z c ∂t = F(z − ct)= =−c ∂t ∂t ∂t ∂ζ ∂ζ Vectorially, F must have only x, y components, F = xˆFx + yˆFy, that is, it must be transverse to the propagation direction, ˆz · F = 0. Similarly, we find from the second of Eqs. (2.1.10) that E−(z, t) must depend on z, t through the combination z + ct, so that E−(z, t)= G(z + ct), where G(ξ) is an arbitrary (transverse) function of ξ = z + ct. In conclusion, the most general solutions for the Fig. 2.1.1 Forward and backward waves. forward and backward fields of Eqs. (2.1.10) are: The two special cases corresponding to forward waves only (G = 0), or to backward E+(z, t) = F(z − ct) (2.1.11) ones (F = 0), are of particular interest. For the forward case, we have: E−(z, t) = G(z + ct) E(z, t) = F(z − ct) with arbitrary functions F and G, such that ˆz · F = ˆz · G = 0. 1 1 (2.1.13) H(z, t) = ˆz × F(z − ct)= ˆz × E(z, t) η η 2.1. Uniform Plane Waves in Lossless Media 41 42 2. Uniform Plane Waves This solution has the following properties: (a) The field vectors E and H are perpen- In the general case of Eq. (2.1.12), the E/H ratio does not remain constant. The dicular to each other, E · H = 0, while they are transverse to the z-direction, (b) The Poynting vector and energy density consist of a part due to the forward wave and a part three vectors {E, H, ˆz} form a right-handed vector system as shown in the figure, in the due to the backward one: sense that E × H points in the direction of ˆz, (c) The ratio of E to H × ˆz is independent PP=E × H = c ˆz |F |2 − |G |2 of z, t and equals the characteristic impedance η of the propagation medium; indeed: 1 1 (2.1.18) = | |2 + | |2 = | |2 + | |2 1 w E μ H F G H(z, t)= ˆz × E(z, t) ⇒ E(z, t)= ηH(z, t)׈z (2.1.14) 2 2 η Example 2.1.1: A source located at z = 0 generates an electric field E(0,t)= xˆ E0 u(t), where The electromagnetic energy of such forward wave flows in the positive z-direction. u(t) is the unit-step function, and E0, a constant. The field is launched towards the positive With the help of the BAC-CAB rule, we find for the Poynting vector: z-direction.
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