Advanced Electrodynamics Exercise 11 Guides

Transcription

1 Advanced Electrodynamics Exercise 11 Guides Here we will calculate in a very general manner the modes of light in a waveguide with perfect conductor boundary-conditions. Our derivations are widely independent of the cross-section, i.e. the profile of the waveguide. In the case of a straight waveguide we will arrive at a rather simple differential equation for the transverse magnetic (TM) modes which we will solve explicitly for the rectangular and the circular waveguide. By using the hodge star and his property of interchanging E and B we will get the transverse electric (TE) modes with only little further calculations. Next we will consider the case of a slightly bent waveguide. Again we can do the calculations independent of the form of the cross-section and derive more involved differential equations. However, with the afore derived solutions of the straight waveguide we can set up a perturbation scheme to calculate the new TM and TE modes. 1 Straight waveguide A typical waveguide is assumed infinitely extended along a specific axis. Possible waveguide-forms are the circular or the rectangular ones. In a more abstract language we say the waveguide is a manifold Ω with boundary Ω. The light-wave is traveling in the volume Ω and is reflected at the surface Ω. Assume a problem adopted coordinate system (x 0 = ct, x, y, z) with the Figure 1: Waveguide-forms ( 1

2 associated Minkowski-metric g µν = 0 g xx g yy where g xx = g xx (x, y) and g yy = g yy (x, y) do not depend on the z-coordinate. The waveguide extends infinitely along the z-axis. In the case of a circular waveguide we will use cylindrical coordinates and in the rectangular case we will choose Cartesian ones. The boundary conditions for a perfect conductor read E ( Ω) = B ( Ω) = 0 where E ( Ω) is parallel and B ( Ω) perpendicular to the boundary Ω of the waveguide. Hence, the parallel electric field at the boundary of our waveguide is zero as well as the perpendicular magnetic field. 1.1 TM modes We will start with the Ansatz for the 4-potential A TM = e i(ωt kz) ( ψ(x, y)dz + φ(x, y)dx 0), where ψ and φ are functions and ω and k are constants. The wave travels along the z-axis with the wavenumber k and frequency ω but is a standing one perpendicular to the propagation direction z. a) Show that this Ansatz leads to TM-modes, i.e. it has no magnetic field component along the propagation direction. By using the Lorentz gaugecondition deduce that d A TM = 0 φ(x, y) = kc ψ(x, y). ω b) Deduce that the perfect conductor boundary conditions lead to Dirichlet boundary conditions for the yet unknown function ψ(x, y), i.e. ψ( Ω) = 0. In order to guarantee the boundary conditions for the magnetic field use that ψ( Ω) = 0 implies dψ is normal to the boundary. Then interpret the wedge-product geometrically, i.e. has the same properties for the covectors as for the vectors. 2

3 Figure 2: Rectangular waveguide ( c) Calculate Maxwell s equations df TM = 0 d F TM = 0 for the given Ansatz. You should find that there is only one non-trivial equation left, i.e. [ ( ) ( )] ψ ψ x x gxx g + y y gyy g = (k 2 ω2 c 2 )ψ g, (1) where g = det(g µν ). Until now we did not specify the cross-section of the waveguide, hence we have derived a very general result. In general we assume that this differential equation admits eigenfunctions ψ mn with eigenvalues ω 2 mn. In a next step we will solve the above equation for two simple cases. d) Solve equation (1) for a rectangular waveguide. Use Cartesian coordinates with 0 x a and 0 y b. You should find eigenfunctions ψ mn with eigenvalues where m, n N. (1 point) ω 2 mn = ω2 c 2 k2 = ( mπ ) 2 ( nπ + a a e) Solve equation (1) for a circular waveguide. Use cylindrical coordinates with 0 r a and 0 φ < 2π. (Hint: You can separate (1) into an equation depending only on φ and one which only depends on r. The differential equation for the r-coordinate can be recast as a Bessel differential equation.) You should find eigenfunctions ψ mn with eigenvalues ω 2 mn = ω2 c 2 k2 = x2 mn a 2, ) 2, where x mn are zeros of the Bessel function of the first kind. 3

4 Figure 3: Circular waveguide ( 1.2 TE modes In the case of TE modes the electric field is perpendicular to the propagation direction z. We could now start with an appropriate Ansatz similar to the case above. However, here we will use a special property of the Hodge-star operator. Namely that it interchanges E and B fields (see for example equation (116) in lecture notes 8). With this we will construct the TE modes from the above calculated TM modes. Therefore we define F TE = F TM. f) Show that the above defined F TE solves Maxwell s equations by using the fact that F TM does. (1 point) g) Show that the 4-potential A TE = e i(ωt kz) gc iω ( ψ y gyy dx ψ ) x gxx dy, leads to the field F TE if ψ is a solution of equation (1). Although in the TE case we can solve the same differential equation as in the TM case, we have to use different boundary conditions. One can proceed in an analogous manner as in b) and interpret the covectors geometrically to find that we need to use the Neumann boundary-conditions for ψ, i.e. n ψ( Ω) = 0 with n is the normal vector to the boundary Ω. h) Solve the rectangular and the circular case for equation (1) with the Neumann boundary-conditions. 4

5 2 Bent waveguide In the case of a bent waveguide the equations become more complicated. We do not pursue a full solution of that problem here but rather derive an perturbative scheme to find approximate solutions for slightly bent waveguides. Under certain assumptions we can use the solutions of the straight waveguide as a starting point. In this case the problem adopted coordinate-system (x 0, u, v, w) has the metric g µν = g uu g vv g ww, (2) where g uu = g uu (u, v), g vv = g vv (u, v) and g ww = g ww (u, v) do not depend on the w-coordinate. For example in the circular case we could use toroidal coordinates similar to (1) in lecture notes 5 (though we have renamed the coordinates here), i.e. R h(u, v) cos(w/r) ϕ 1 (u, v, w) = R h(u, v) sin(w/r) b ζ sin(v) with h = (1 + δζ cos(v)), δ = a/r, ζ = u/a and 0 u a, 0 v < 2π. Here R is the major radius and we only look at a part of the whole torus, i.e. 0 w ɛr with 0 < ɛ < 2π. For the rectangular case we could use R h(u) cos(w/r) ϕ 1 (u, v, w) = R h(u) sin(w/r) v with h = (1 + δζ) and 0 u a, 0 v b, 0 w < ɛr. Here R would be the distance between the origin of the coordinate system and the boundary of the bent waveguide. In both examples we have g ww = h 2 and h has the form h = 1 + δf(u, v). If we now look at the limit of an unbent waveguide, i.e. R while the cross-section of the waveguide stays the same, we see that δ 0. Hence we have a metric similar to the one we assumed before. Therefore we will try the previous Ansatz to deduce an approximate solution of the bent waveguide. i) Use the previous Ansatz for the 4-potential ( A TM = e i(ωt kw) ψ(u, v)dw kc ) ψ(u, v)dx0, ω to derive Maxwell s equations. (3 points) 5

6 You immediately see that only in the case of δ = 0 we can solve the resulting equations exactly. j) Nevertheless, use these results to derive the following differential equation ( ) ω 2 c 2 k2 ψ ( ) ( ) ψ ψ g + u u guu g + v v gvv g [ ( ) f ψ = δ u u guu g + f ( )] ψ 2 v v gvv g h (1 point) With the assumption ˆψ 0 mn ˆψ 0 m n = du dv g uu g vv ( ˆψ0 mn (u, v)) ˆψ0 m n (u, v) = δ mn,m n, where ˆψ 0 mn is a normalized eigenfunctions of (1) to the eigenvalues ω 2 mn, we want to derive a perturbation scheme in δ. We make the Ansatz k 2 = k δk ψ mn = ψ 0 mn + δψ 1 mn +... k) Derive with the operators 1 [ u g uu g uu g vv u + v g vv ] g uu g vv v = (2) guu g vv and [ 2 f h u guu u + f ] v gvv v = L the first order perturbation equation [ (2) + ω 2 mn] ˆψ1 mn = (k L) ˆψ 0 mn, where ω 2 mn = ω2 c 2 k0 2. Then use the expansion ˆψ 1 mn = kl c mn,kl ˆψ0 kl to derive ˆψ 1 mn = kl mn ˆψ 0 kl L ˆψ 0 mn ω 2 mn ω 2 kl ˆψ 0 kl. and k

7 With this we could calculate the first order perturbation of the bent waveguide as ˆψ mn = ˆψ 0 mn + kl mn ˆψ 0 kl L ˆψ 0 mn ω 2 mn ω 2 kl ˆψ 0 kl. Maximal points: 20 Hand in Monday at latest 18:00 o clock - mailbox of Robert van Leeuwen, second floor Nanoscience Building 7