Waves can be guided not only by conductors, but by dielectrics. Fiber optics cable of silica has nr varying with radius. Simplest: core radius a with n = n 1, surrounded radius b with n = n 0 < n 1. Total internal reflection if α > α c = sin 1 n 0 /n 1 Equivalently θ < θ max = cos 1 n 0 /n 1. This is geometrical optics. Needs λ a. Two kinds of : α α multimode, a λ, treat with geometrical optics. Typically a 25 µm, b 75 µm, λ 0.85 µm. single mode, a λ, treat as wave guide. Typically a 2 µm. θ a b nalysis with in Normal
Define = n2 1 n2 0 2n 2 1 n 0, typically about 0.01. 1 n 1 cos θ max 1 1 2 θ2 max = 1, so θ max 2. How many modes can propagate? Uncertainty principle: only one mode can fit per unit dpdq D volume in phase space, N = for each 2π mode in D dimensions. Here D = 2, the cross section has coordinate integral d 2 q = πa 2. s k k z tan θ max = k z 2, d 2 p = 2 d 2 k = 2π 2 kz. 2 There are two polarizations, so 1 N = 2 πa 2 2π 2 2πkz 2 = 1 2 V 2, wherev := ka 2 is called the fiber parameter. nalysis with in Normal
Problem with simple fiber t each angle θ < θ max, light travels indefinitely down the fiber. But to go a large distance L down the fiber, θ it travels a different distance L sec θ, so light from different θ s arrive with different phases, interfere! Fix: make several transitions to lower n. In fact, for homework Jackson 8.14 you will θ find a perfect fix, using n varying continuously with radius. nalysis with in Normal
nx varying with radius x Consider dielectric with ɛ x varying smoothly, µ = µ 0 as silica is not magnetic. ssume single frequency ω. Maxwell gives So E ɛ E = 0 = ɛ E + ɛ E E = µ 0 H t = iµ 0ω H H = ɛ E = iωɛe t H = 0. = 2 E + E = iµ 0 ω H nalysis with in Normal = µ 0 ω 2 ɛ E = 2 E 1 ɛ ɛ E
The same for H: H = 2 H + H = iω Thus 2 H = iω ɛ E iωɛ E = iω ɛ E + µ 0 ω 2 ɛh. 2 E + µ0 ω 2 ɛe + 1 ɛ ɛ 2 H + µ0 ω 2 ɛ H iω ssume ɛ varies slowly compared to λ, ɛ ɛ λ = ɛω c. E ɛe = 0 ɛ E = 0 Other terms are ω 2 /c 2 times E or H, but ɛ terms are ɛ/ɛλ times E, λ 2 = ω 2 /c 2, so they can be ignored. Both E H satisfy 2 + ω2 c 2 n2 r ψ r = 0. nalysis with in Normal
ψ oscillates rapidly on scale λ. Take this away by defining the eikonal S r, with so 2 ψ = iω = ψ r = e iωs r/c Se c iωs r/c [ iω c 2 S i ω c = ω 2 n 2 /c 2 e iωs/c 2 S 2 ] e iωs r/c nalysis with in Normal n 2 r S S = i c ω 2 S. Now c/ω λ while S varies with n r, much more slowly, so we can set the r.h.s to zero, for the approximation: S S = n 2 r.
The approximation: S S = n 2 r doesn t give the direction S changes, but does give the rate. Define ˆk r so S = n rˆk r. Near a point r 0, ψ r e iω S r 0 + r r 0 S /c = e iωs r 0/c e iωˆk r r 0 n r/c, so it is locally a plane wave with k = ωn r/c. Consider an integral curve, that is, a ray following S, let s be the distance along that curve. Then d r/ds = ˆk, n rd r/ds = S, so d ds n r d r ds = d S ds = ds ds = n r. 1 Γ nalysis with in Normal Meridional rays pass through axis m = 0 as waves skew rays do not, travel helically m 0 modes.
We will treat only meridional, effectively in xz plane, with x a radial direction z along the fiber. We assume n r = nx independent of z. Take x z components of 1 : d ds n r d r ds = n r d dnx nx sin θ = ds dx, d dn r nx cos θ = = 0. ds dz So nx cos θ = constant. With θ0 < θ max ray reaches a maximum radius x max with n := n0 cos θ0 = nx max. dz ds = cos θ = n nx, d ds = n d nx dz, so the x component of 1 gives dn dx = n d n dx nx = n d nx dz nx dz nx dz so n dx, dz n 2 d2 x dz 2 = nxdnx dx = 1 d 2 dx n2 x. nalysis with in Normal
n 2 d2 x dz 2 = 1 d 2 dx n2 x Looks like ma = dv/dx with potential 1 2 n2 x time z, so as for Newton, multiply by velocity dx/dz, to get 1 2 n2 d dz dx 2 = 1 dz 2 d dz n2 x = n 2 dx 2 dz }{{} =0 at x max = n 2 x n 2. The distance travelled along z in getting from the axis to x is x dz x dx zx = dx = n dx n 2 x n, 2 0 the distance from one axis crossing to the next is xmax dx Z = 2 n n 2 x n. 2 0 0 nalysis with in Normal
The optical distance nxds between axis crossings is xmax L opt = 2 0 xmax = 2 = 2 0 xmax 0 nx ds dz dz dx dx nx nx n n 2 x n 2 x n 2 dx. n n 2 x n dx 2 Over a long distance L, many axis crossings L/Z, total phase change is proportional to L L opt Z. It is ideal if L opt Z is independent of n, for otherwise different rays will destructively interfere. You will find the ideal in problem 8.14. Signals will also degrade with distance if there is dispersion over the bwidth of the signal. There is also some absorption in real dielectrics. These two issues for silica favor using λ 1.4µm. nalysis with in Normal
We will not cover single-mode, or normal modes in. So we are skipping section 8.11. Now we will discuss sources of electromagnetic fields in conducting waveguides. Next time, we will begin discussing sources more generally, We will first cover spherical waves of Jackson 9.6, then come back to the beginning of chapter 9. nalysis with in Normal
of Waves in We discussed waves propagation without sources in waveguides. Now consider a given distribution of charges currents in a localized region of the waveguide. We assume the charge motion is otherwise determined, ignoring back reaction of the fields on the charges. Then Maxwell s equations are still linear inhomogeneous in the fields, with boundary conditions still time-independent, so fourier transform in time will give frequency components independently in terms of frequency components of the source distribution. way from the sources, waves as before, with the general fields superpositions of normal modes with z dependence given by k = ± ω 2 /c 2 γλ 2. nalysis with in Normal
k = ± ω 2 /c 2 γλ 2. We need to consider not only right-moving k real, > 0 left-moving k real, k < 0 modes, but also the damped modes, ω < cγ λ, with k imaginary. Far from the sources, only the real k modes will matter, but we need all modes for a complete set of states. Exp our fields in normal modes, indexed by λ, which includes a type TE or TM or TEM as well as indices quantum numbers defining the mode. Each mode λ has two k values, a positive one, k > 0 real, or k imaginary, Im k > 0, a negative one, k < 0 real, or k imaginary, Im k < 0. For each λ let k λ be the positive value i.e. positive real or imaginary with positive imaginary part. nalysis with in Normal
For the positive mode part of the fields, we have [ ] E + λ x, y, z = Eλ x, y + ẑe z λ x, y [ ] H + λ x, y, z = Hλ x, y + ẑh z λ x, y e ik λz e ik λz where E λ x, y H λ x, y are purely transverse, are determined by E z H z as in lecture 6 Jackson 8.26. The negative modes are found by z z, which involves a parity transformation, under which E z changes sign but the transverse part doesn t. But the magnetic field is a pseudovector, so under parity it behaves the opposite way, H z doesn t change sign but Hx, y does. Thus [ ] E λ x, y, z = Eλ x, y ẑe z λ x, y e ik λz H λ [ x, y, z = H ] λ x, y + ẑh z λ x, y e ik λz nalysis with in Normal
The normal modes are determined by solutions of the Helmholtz equation 2 t + γλ 2 ψ = 0, with ψ Γ = 0 for TM modes or Neumann conditions on Γ for the TE modes. With two more indices, λ TM mn give a complete set of functions on the cross section with ψ Γ = 0, λ TE mn give a complete set of functions on the cross section with ψ = 0. Γ n Note that if ψ λ ψ µ are solutions to 2 t + γ 2 ψ = 0 with γ = γ λ γ = γ µ respectively, [ ] t ψ λ t ψ µ = t ψ λ t ψ µ ψ λ 2 ψ ψ µ = ψ λ Γ n ψ λ 2 t ψ µ = 0 + γλ 2 ψ λ ψ µ where the vanishing of the Γ holds if ψ λ satisfies Dirichlet boundary conditions ψ λ Γ = 0 or ψ µ satisfies Neumann conditions ψ µ / n Γ = 0. nalysis with in Normal
Reversing µ λ, we see that if both satisfy the same condition, t ψ λ t ψ µ = γλ 2 ψ λ ψ µ = γµ 2 ψ λ ψ µ, so if γ λ γ µ, ψ λψ µ = 0. If there are several solutions with the same γ, with γ 2 real, we may choose them all to be real or have the same phase, in which case ψ λψ µ is a real symmetric matrix which can be diagonalized, so we may choose basis functions ψ λ to be orthonormal under integration over. nalysis with in Normal
It is convenient to chose the basis functions so that the transverse electric field at z = 0 is real normalized. For TM modes, E = i k λ /γ 2 λ t ψ, E z = ψ, so we chose ψ to be imaginary for k λ real real for k λ imaginary, with Then ψ λ ψ µ = γ2 λ kλ 2 δ λµ. E λ E µ = k λk µ γλ 2γ2 µ = k λk µ γ 2 λ γ2 µ = k λk µ γ 2 µ γ 2 λ t ψ λ t ψ µ γ 2 λ k 2 λ ψ λ ψ µ δ λµ = δ λµ. nalysis with in Normal
For TE modes, H λ = i k λ /γ 2 λ t ψ λ = 1 Z λ ẑ E, where Z λ = Z 0 k/k λ, k := ω/c. Here we choose ψ imaginary as k λ Z λ is real with ψ λ ψ µ = Then for two TE modes E λ E µ = k λk µ Z λ Z µ = k λk µ γ 2 λ γ2 µ = k λk µ γ 2 λ γ2 µ H zλ H zµ = γ2 λ kλ 2 δ λµ. Z2 λ γ 2 λ γ2 µ ψ µ ψ λ Γ n 0 + γ 2 λ t ψ λ t ψ µ ψ λ ψ µ ψ λ 2 t ψ µ nalysis with in Normal = k λk µ γ 2 µ γλ 2 kλ 2 δ λµ = δ λµ.
Finally, suppose λ is a TM mode with E λ = i k λ /γ 2 λ t ψ λ, µ is a TE mode, with E µ = i k µ Z µ /γ 2 µ ẑ t ψ µ. Then E λ E µ = k λk µ Z µ γ 2 λ γ2 µ = k λk µ Z µ γλ 2 ẑ γ2 µ The integral t ψ λ t ψ µ = = t ψ λ ẑ t ψ µ Γ t ψ λ t ψ µ t ψ λ t ψ µ ψ λ t ψ µ dl = 0 by Stokes theorem the fact that ψ λ vanishes on the boundary. nalysis with in Normal
Thus we have shown or chosen that the transverse electric fields for the normal modes satisfy E λ E µ = δ λµ for all the modes, TE TM. s we have shown E λ E µ = δ λµ, as H λ = Z 1 λ ẑ E λ, we have H λ H µ = 1 δ λµ, in Z 2 λ calculating the time average power flow P = 1 E 2 H ẑ to the right, we can use Eλ H µ ẑ = 1 Z λ δ λµ nalysis with in Normal
For a rectangular wave guide If the cross section is 0 x a 0 y b, the equation separates in x y, modes are labelled by integers m n, the number of half wavelengths in each direction, TM waves: ψ S = 0 E z mn = ψ = 2iγ mn k λ ab sin E x mn = E y mn = mπx a 2πm mπx γ mn a ab cos a 2πn mπx γ mn b ab sin a sin cos sin nπy b nπy b nπy b,,, nalysis with in Normal
TE waves: where ψ n = 0 S H z mn = ψ = 2iγ mn k λ Z λ ab cos E x mn = E y mn = mπx nπy cos a b 2πn mπx nπy γ mn b ab cos sin, a b 2πm mπx nπy γ mn a ab sin cos, a b γ 2 mn = π 2 m 2 a 2 + n2 b 2. The overall constants are determined from the normalization E2 x + E 2 y = 1, except that for TE modes, we need an extra factor of 1/ 2 for each n or m which is zero, as cos 2 mπx/a = a1 + δ m0 /2., nalysis with in Normal
Expansion of Free Waves Except where there are sources, the fields are an expansion in terms of normal modes, divided into positive negative components: with E = E + + E, H = H + + H, E ± = λ ± λ E ± λ, H± = λ ± λ H ± λ, Coefficients ± λ are uniquely determined by transverse E H along any cross section. For example, at z = 0 E has expansion coefficients + λ + λ while H has coefficients + λ λ. From the orthonormality properties we find ± λ = 1 E 2 E λ ± Zλ 2 H H λ. nalysis with in Normal
Now consider a source J xe iωt confined to some region z [z, z + ]. Consider cross sections S S +, with all sources between S z S + them. so at S + there is no amplitude for any mode with negative k or with i k, which would represent left-moving waves or exponential blow up as z +. The reverse is true at S, so E = + λ E + λ, H = + λ H + λ at S + λ λ E = λ E λ, H = λ H λ at S λ λ nalysis with in Normal
In between, we have the full Maxwell equations with sources, E = B t = iωµ 0H, H = J+ɛ0 E t = J iωɛ 0E, while the normal modes obey Maxwell equations without sources: H ± λ = iωɛ 0 E ± λ, E ± λ = iωµ 0 H ± λ. If we apply the identity B = B B, we find E H ± λ E ± λ H = E H ± λ E H ± λ E ± λ H + E ± λ H = iωµ 0H H ± λ + iωɛ 0E E ± λ iωµ 0H ± λ H + E ± λ J iωɛ0e = J E ± λ nalysis with in Normal
If we integrate this over the volume between S S +, using Gauss theorem the boundary condition that E = 0 at the conductor s surface, E H ± λ E ± λ H ˆn = J E ± λ, S V where S consists of S + with ˆn = ẑ, S with ˆn = ẑ. Let s take the upper sign. The contribution from S + is can be reduced to an integral over at z = 0: + E + λ λ H + λ E + λ H + λ λ S + z = + Eλ λ H λ E λ H λ λ = λ + λ S + Eλ E λ Z 1 λ ẑ E λ z eik λ+k λ z Z 1 λ ẑ E λ z eik λ+k λ z nalysis with in Normal
+ E + λ λ H + λ E + λ H + λ λ S + z = 1 + λ Eλ Z E λ 1 λ λ Z λ = λ + λ δ λλ 1 Z λ 1 Z λ E λ E λ e ik λ+k λ z = 0. On the other h, the contribution from S is E λ H + λ E + λ H λ ẑ λ λ S = λ λ S Eλ H λ + E λ H λ = λ λ 2 Z λ δ λλ = 2 Z λ λ so λ = Z λ 2 V J E + λ. e ik λ+k λ z ẑ e ik λ k λ z nalysis with in Normal
The same argument for the lower sign, as spelled out in the book, gives the equation with the superscript signs reversed, so both are ± λ = Z λ 2 V J E λ. In addition to sources due to currents, we may have contributions due to obstacles or holes in the conducting boundaries. These can be treated as additional surface terms in Gauss law by excluding obstacles from the region of integration V, but this requires knowing the full fields at the surface of the obstacles or the missing parts of the waveguide conductor. This is treated in 9.5B, but we won t discuss it here. So finally we are at the end of Chapter 8. nalysis with in Normal