Green s Function for Problems with Layers

Transcription

1 Chapter 9 Green s Function for Problems with Layers 9. Continuity conditions The medium comprises two or more straight or curved layers. The properties of the medium are constant within each layer; they change discontinuously across the interface separating two layers. Within each layer the phenomena are described by a differential equation; the latter cease to be valid in the interface. There they are replaced by continuity conditions. For example in electrostatics, magnetostatics and electrodynamics the field components must fulfil the following continuity conditions: The tangential components of the electric and of the magnetic field must be continuous as well as the normal components of the dielectric displacement and of the magnetic induction. These conditions are cast into the following formulas: n E E =, Et = E t ; 9. n H H =, Ht = H t ; 9. n D D =, Dt = D t ; 9.3 n B B =, Bt = B t Green s functions for two-layer problems It is convenient though not obligatory to ornate the scalar Green s function with two subscripts: G ik r, r i label of layer containing the field point k label of layer containing the source point This is written out in more detail for a problerm with two layers as: G r, r : r L, r L ; G r, r : r L, r L ; G r, r : r L, r L ; G r, r : r L, r L. 9.5 It is essential to write the differential equations with selfadjoint operators. Even if the parameters describing the physical properties of the layers are constant within the layers, one should write the differential equations as if these parameters were functions of the space coordinates. If there are n layers then the labels i, k run as,,..., n.

2 9.. The electrostatic two-layer problem The Poisson equation for the potential in a problem with a layered dielectric should be written as: div ε rgradφ = ρ r. 9.6 The corresponding differential equation for the Green s function for two layers each consisting of a constant dielectric labelled as ε, ε respectively is: ε i G ik = δik δ r r. 9.7 There is no summation over the repeated subscript i! The pieces of the Green s function having equal subscripts are the solutions of inhomogeneous differential equations, those bearing different subscripts are solutions of homogeneous equations. Still, the latter, too, are determined uniquely by the continuity conditions. In the present case these are eqs.9. and 9.3. We assume that the potential consists also of two pieces, each one valid in the corresponding layer. The boundary conditions then connect these two pieces and their normal derivatives. The relations are then taken over for the pieces of the Green s function. The points of the interface are denoted as r. So we get from the first condition: E i = Φ i, Eit = t Φ i. 9.8 For points in the interface r we find: r r : Eit = t Φ i = E it = t Φ i, Φ = Φ G k = G k, k =,. 9.9 Condition 9.3 gives: Φ n r r : D n = ε Φ ε = ε Φ = D n = ε Φ n, ε G k G k = ε, k =, The source representation In order to derive the source representation, it is most convenient to start where the dependence of the dielectric constant and of the Green s function is taken into account in the dependence on the variables. At the end, the constancy of the dielectric constant in each layer is shown by constants with subscripts. Then also the other quantities are provided with the corresponding labels. We start from Green s second theorem in the generalized form for the primed variable and we identify one function with the potential Φ r the other one with G r, r : G r, r ε r Φ r } {{ } = ρ r r d r Φ r ε r G r, r }{{} = δ r r = df n G r, r ε r Φ r df n Φ r ε r G r, r. d r = We do the integration over the delta-distribution and use the symmetry of the scalar Green s function. Rearranging the resulting equation gives: Φ r = G r, r ρ r d r 9. df ε r G r, r Φ r Φ r G r, r.

3 Now we assume that the charge density ρ is given in a volume V extending over both layers. The surface F is in the interface separating the two layers; it splits the volume V into a part V lying in layer and a part V lying in layer. The surfaces S, S respectively cover the volumes V, V respectively so that S S encloses V ; similarly S S encloses V. Now we specialize the dielectric constants as being constant in each layer. We use the labels belonging to the layers as defined for the Green s function in eqs.9.5 and we attach corresponding labels to Φ and ρ. So we get: Φ r = G r, r ρ r d r G r, r ρ r d r Φ r = V F F V F F df ε G r, r Φ r Φ r G r, r df ε G r, r Φ r Φ r G r, r G r, r ρ r d r V V. G r, r ρ r d r df ε G r, r Φ r Φ r G r, r df ε G r, r Φ r Φ r G r, r 9..3 Two dielectric half-spaces separated by a plane interface The formulas of the previous section are specialized to this case no summation over i ε i G ik = δ ik δ r r. 9.4 G k = G k, 9.5 G k ε z lim G k =, z = G k ε z ; 9.6 lim k =. z 9.7 There are several methods to calculate this Green s function. One uses image charges. Another one uses integral representations including the Sommerfeld integral. Computing the Green s function by image charges The problem is a standard exercise in electrostatics as an example that the method of image charges well-known for ideally conducting planes of infinite extent works also for dielectric half spaces, e.g. 9.]. We introduce cylindrical coordinates We assume a source point location ρ =, z = z. The four pieces of the corresponding Green s function are: G ρ, z, ρ =, z = ε ] ε, 9.8 4πε R ε ε R G ρ, z, ρ =, z = ε ] ε, 9.9 4πε R ε ε R G ij ρ, z, ρ =, z =, i j; 9. πε ε ε R j R, = ρ z z. 9. 3

4 For a source position ρ, φ off the z-axis, one need only replace ρ by ρ ρρ cosφ φ ρ in the last expressions above. Details of the calculations are presented in the notebook quoted in Computing the Green s function by integral representations The Sommerfeld integral 5.3 is used to represent the source term: ρ c = e ζ c J ζρ dζ. 9. Correspondingly, the following integral representations are set up for the four pieces of the Green s function: G ii = J ζρ e ζ z z h ii ζ e ζz] dζ, 9.3 4πε i = 4πε i G ij = 4π J ζρ g ii ζ dζ, 9.4 J ζρ g ij ζ e ζz dζ; g ij ζ = h ij ζ. 9.5 The signs of the exponentials have been chosen to satisfy the boundary conditions at z = ±, eqs.9.7, the minus sign for i =, the plus for i =. These expressions are inserted into the continuity conditions 9.5 and 9.6. The resulting linear equations for the four amplitudes h ij are solved and the amplitudes are inserted into the integrals given above. So we get: G ρ, z, ρ =, z = 4πε G ρ, z, ρ =, z = 4π G ρ, z, ρ =, z = 4π G ρ, z, ρ =, z = 4πε ε ε ε ε dζ J ζρ dζ J ζρ e ζ z z ε ] ε e ζzz, 9.6 ε ε dζ J ζρ e ζz z, dζ J ζρ e ζ z z, 9.9 e ζ zz 9.3 ε ] ε e ζzz. 9.3 ε ε The calculation and the results are presented in the notebook displayed in AnMe4-4-.pdf. All the above integrals may be transformed into closed expressions with the help of the Sommerfeld integral 9.. The resulting expression agree with those given in eqs Green s function for the three-dimensional potential equation with two layers We consider the three-dimensional potential equation for a problem with two plane layers. The thickness of the upper lower layer is p q, s. Fig.9.3. The Green s function is a scalar comprising four pieces distinguished by the two labels i, k. The first second label denotes the layer containing the point of observation the source point. 4

5 =================== z = p: G k = ;! z = :! HG'L k =! G'L k, HGL k = HGL k. G k! G k =================== z = -q < : G k = The Greens functions of the diffusion equation! i D g ik z,zp - k g ik = - dz - zp d ik is computed for a t Figure 9.: An infinite plane condensor comprising two layers with permeability ε and ε. It is bounded problem. by planes of ideal conductivity. The primes of the continuity conditions at z = denote derivations This equation with respect results tofrom z. the potential equation if the transverse coordinates are treated by Fourier transforms. The Green's function consists of 4 pieces denoted by g ik Hz, zpl ; the first subscript b z, The Green s function G the coordinate of the ik r, r point is a solution of the following inhomogeneous differential equation of observation i =, for < z < p, -q < z < ; the second subscript to th : point coordinate zp k ε i G =, ik r, for r = < zp δ< ik p, δ r -q < r zp <. The Green's function is symmetric 9.3 in z and zp.!z is piecewise constant as shown in the figure above. p z, z q, < x, x <, < y, y < The Green s function is represented as a Fourier integral in the transverse directions x, y: G ik r, r = 4π dk x dk x e ikxx x k yy y g ik k x, k y ; z, z Inserting this and the Fourier integral representations of the Delta-distribution in the transverse coordinates gives the following differential equation for the amplitude function g ik k x, k y ; z, z = g ik κ; z, z : d ε i dz κ ḡ ik κ; z, z = δ ik δz z 9.35 with κ = A particular solution of the inhomogeneous equations is: The general solution for each piece of the Green s function is: k x k y ḡ ik κ; z, z = δ ik κε i e κ z z ḡ ik κ; z, z = δ ik κε i e κ z z a ik κ eκz b ik κ e κz These solutions are inserted into the integral representation In addition, plane polar coordinates are introduced for the transverse variables in coordinate and in k-space: x = ρ cos φ, y = ρ sin φ; 9.39 x = ρ cos φ, y = ρ sin φ ; 9.4 k x = κ cos ψ, k y = κ cos ψ. 9.4 R = x x y y = ρ ρρ cosφ φ ρ

6 In place of the integral representation 9.34 we get: G ik r, r = 4π G ik ρ, φ, z; ρ, φ, z = π g ik κ; z, z = δ ik ε i e κ z z π dκ κ dψ e iκr cosφ φ ψḡ ik κ; z, z, 9.43 dκ J κr g ik κ; z, z, 9.44 a ik e κz b ik e κz The integral over ψ is an integral representation of the Bessel function J κr: J κr = π dψ e iκr cosφ φ ψ π If the source is on the z axis then R = ρ. Most of the derivations below will be done for this special case. The transition to the general case just requires the replacement r R An integral representation for the four pieces of the Green s function In preparation Expressions for the cured Green s function In preparation. 9.4 Further examples of Green s functions for problems with several layers The approach to find Green s functions in problems with several layers explained in the first sections of this chapter has been used to treat models of counter configurations comprising up to three layers 9.]. The corresponding Green s functions encounter convergence troubles if the field point is in the same plane as the source point. Similar difficulties arise also from image points. All these problems are cured by subtracting terms in the integral representations to compensate convergence problems. 9.5 Mathematica notebooks 9.5. Point charge q with two dielectrica: Method of images AnMe9-4-.pdf, 6 pages Greens function for two dielectric halfspaces. Method of integral representations Greens function for two dielectric halfspaces. representations of the four pieces. AnMe9-4-.pdf, 4 pages. Computation of the amplitudes in the integral 6

7 9.5.3 Greens function for a condensor with two layers. Convergence acceleration by removing slowly convergent terms in the integral representation In preparation. 9.6 References 9.] J.D. Jackson, Classical Electrodynamics. 3rd. ed., Wiley, ] Th. Heubrandtner, B. Schnizer, C. Lippmann, W. Riegler, Static electric fields in an infinite plane condensor with one or three homogeneous layers. Nucl. Instrum. Methods Phys. Res., A : 489 no.-3, pp = CERN-EP--4 ; Geneva : CERN, 8 Jan 9.3] Th. Heubrandtner, B. Schnizer, C. Lippmann, W. Riegler, Static electric fields in an infinite plane condensor with one or three homogeneous layers. Report CERN-OPEN--74, 3 Oct. This is an extended preprint version of the previous paper 7