old Appendix D REVIEWED

Old Appendix D 10.13.13 I had my new Appendix D in a separate file for a long time, but today I decided to install it into the main doc since I guess it is reasonably stable. Here I store the App D that got replaced. _______________________ Appendix D: The General Electric Field in a Round Wire In this Appendix we present a somewhat lengthy calculation of the electric field (and therefore the current) in a round wire without assuming that such fields are symmetrical about the axis. The conclusions one can draw from the analytic solution given below are supportive of the general discussion elsewhere in this monograph: 1. The Er and Eφ field components are very small. In fact, they are smaller than Ez by the factor (a/λ), where a = wire radius, and λ = wavelength of the wave on a transmission line containing the wire. This fraction is always assumed small in any analysis of a transmission line. 2. It is possible to maintain the condition Eφ = 0 on the wire surface. This condition is consistent with the idea that the wire surface is an electrical equipotential at any constant z. 3. We get an explicit formula for the surface impedance. We find that it is non-uniform around the boundary of the wire cross section, and that Jz is non-uniform inside the wire. This result is true even in the DC limit of a transmission line. 4. We suggest a method for computing the E fields in the low frequency limit of a transmission line consisting of round wires. First, solve the "electrostatic" problem illustrated in Chapter 6. Knowing the potential φ, compute the radial electric field at the surface of the conductors. From this compute the surface charge density n(φ) as a function of azimuth around the wire Then, as described below, compute the moments Nm (or ηm) of this charge distribution, and use the formulas below to find the electric field. This is a highly technical appendix. The reader is invited to inspect the boxed results and comments at the end. D.1 The General Method and Solution for Ez The starting point for the calculation is the Helmholtz equation (1.6.2) for the E field inside the wire, ( 2 + β2 ) E(r,φ z, t) = 0 . (D.1.1) Here we use cylindrical coordinates (r,φ,z), as appropriate for a straight round wire. The first step is to expose the assumed t and z dependence, and in doing so, define the field E(r,φ) : E(r,φz,t) = ej(ωt-βz) E(r,φ) . (D.1.2) Here, symbol βd stands for the value that parameter β = ω takes in the dielectric. We reserve the symbol β with no subscript to mean β inside the metal of our wire. The form shown in (D.1.2) is a simple wave traveling down a transmission line in the +z direction. We assume that one of the conductors of this transmission line is our round wire, while the other conductor is (other conductors are) unspecified. One can regard (D.1.2) as an assumed variable-separated form for a solution, an "ansatz". Our second ansatz is that Eφ(r=a,φ) = 0 which means at any z = constant slice, the circular wire perimeter is an equipotential. If a consistent solution to the field equations can be found with these assumptions, they are justified de facto. (a) Partial Wave Expansions The next step is to do a "partial wave expansion" (that is, a complex Fourier series expansion) of E(r,φ) in terms of "azimuthal harmonics" eimφ, so that the variable φ is replaced with the partial wave index m: E(r,φ) =!Syntax Error, I E(r,m) ejmφ E(r,m) = (1/2π) !Syntax Error, Idφ E(r,φ) e-jmφ (D.1.3) Since the usual complex part has been extracted in (D.1.2), we assume that E(r,φ) is real so the right euqation above says that E(r,-m) = E(r,m)* . We can then re-express the above as, E(r,φ) = E(r,0) +!Syntax Error, I[ E(r,m)ejmφ + E(r,-m)e-jmφ] = E(r,0) +!Syntax Error, I[ E(r,m)ejmφ + [E(r,m)ejmφ]*] = E(r,0) + 2!Syntax Error, IRe{ E(r,m)ejmφ } = E(r,0) + 2!Syntax Error, I[ Re(E) + j Im(E)] [ cos(mφ) + j sin(mφ)] = E(r,0) + 2!Syntax Error, I[Re(E)cos(mφ) - Im(E) sin(mφ)] + 2j!Syntax Error, I[ Im(E)cos(mφ) + Re(E) sin(mφ)] Since E(r,φ) is assumed real, the second sum must vanish for all r and φ, giving this final result, E(r,φ) = E(r,0) + 2!Syntax Error, I[Re{E(r,m)}cos(mφ) - Im{E(r,m)} sin(mφ)] . (D.1.4) In analogy with (D.1.2) and the discussion of Section 1.7 regarding complex functions, we define a complex surface (r=a) charge density n(φ,z,t) which has the following ansatz variable-separated form, n(φ,z,t) = ej(ωt-βz) n(φ) (D.1.5) where n(φ) is real. We then expand n(φ) as in (D.1.3) and (D.1.4), n(φ) = !Syntax Error, I Nm ejmφ Nm = (1/2π) !Syntax Error, Idφ n(φ) e-jmφ . (D.1.6) n(φ) = N0 + 2!Syntax Error, I[Re{Nm}cos(mφ) - Im{Nm} sin(mφ)] (D.1.7) Since n(φ) is real, Nm = N-m*. These Nm are the "moments" of the surface charge distribution, and their complex nature keeps track of cos(mφ versus sin(mφ) components. (b) Charge Pumping Boundary Condition The reason we are so interested in n(φ) is that it acts as a driving source of the radial electric field in the wire. Recall the equation of continuity (1.1.8), divJ = -jωρ ∫S dSJ = -jω ∫V dV ρ . // divergence theorem (D.1.8) When applied to a thin box of radial area dS straddling the wire surface, one finds that ∫S dSJ = -Jr(r=a,φ)dS and ∫V dV ρ = n(φ) dS so that Jr(r=a,φ) = jω n(φ) . (D.1.9) We assume that there is no current outside the wire to get this result. Since J = σE, this is really a boundary condition on the radial electric field, Er(r=a,φ) = (jω/σ) n(φ) . (D.1.10) Since we have parallel partial wave expansions for both sides, we can rewrite (D.1.10) in m space as Er(r=a,m) = (jω/σ) Nm . (D.1.11) Thus, the radial electric field must have a certain value at the r=a boundary in each partial wave. And the value it must have is determined by the moment of the charge distribution. By way of interpretation, the surface charge of a transmission line is "pumped" by the radial current in the wire. Since divJ = 0 inside the wire, this radial current is accompanied by the usual longitudinal current one expects to find inside the conductors of a transmission line. (c) The Ez Solution Here is the Helmholtz equation (D.1.1) for Ez as in (D.1.2) (cylindrical coordinates r,φ,z ) [∂r2 + (1/r) ∂r + (1/r2) ∂φ2 + ∂z2 + β2 ] ej(ωt-βz)Ez(r,φ) = 0 . (D.1.12) Inserting the expansion (D.1.3) and moving the m sum to the left gives !Syntax Error, I [∂r2 + (1/r) ∂r + (1/r2) ∂φ2 + ∂z2 + β2 ] ej(ωt-βz)Ez(r,m) ejmφ = 0 . (D.1.13) We can then make the obvious replacements ∂z = -jβd and ∂φ = +jm to get, !Syntax Error, I { [∂r2 + (1/r) ∂r -m2 (1/r2) – βd2 + β2 ] ej(ωt-βz)Ez(r,m) } ejmφ = 0 . (D.1.14) Due to the completeness of functions ejmφ on the interval (-π.π), we conclude that { } = 0, or [∂r2 + (1/r) ∂r -m2 (1/r2) – βd2 + β2 ] ej(ωt-βz)Ez(r,m) = 0 . (D.1.15) [ I.e., apply !Syntax Error, Idφ e-jm'φ to both sides of (D.1.14) and use !Syntax Error, Idφ ej(m-m')φ = 2π δm,m'. ] Next, multiply both sides of (D.1.15) by r2 e-j(ωt-βz) to get, [r2∂r2 + r ∂r - m2 + r2 ( β2- βd2)] Ez(r,m) = 0 or [r2∂r2 + r ∂r + (r2 β'2 - m2)] Ez(r,m) = 0 (D.1.16) where β'2 = β2 - βd2 . (D.1.17) In a conductor, β is huge compared to the dielectric βd, so we could ignore the distinction between β and β'. Setting x = β'r we find ∂r = β'∂x and then r∂r = x∂x and so on so that (D.1.16) reads [x2∂x2 + x ∂x + (x2 - m2)] Ez(x/β',m) = 0 . x = β'r (D.1.18) This is Bessel's equation [ Spiegel 24.1] and the solution subject to the condition that Ez be finite at r = 0 is Ez(x/β',m) = Czm Jm(x) or Ez(r,m) = Czm Jm(β'r) (D.1.19) where Czm is an arbitrary constant for each partial wave m. Equation (D.1.19) is consistent with (2.1.22). In Section 2.1 we dealt only with the m=0 partial wave, which embodies the symmetrical part of the problem. Also, we assumed constant z behavior so βd = 0 and the question of β versus β' never arose. D.2 The Solutions for Er and Eφ In Cartesian coordinates, the Helmholtz equation (D.1.1) can be written ( 2 + β2 )Ei = 0 for i = 1,2,3 meaning x,y,z. The point is that (2E)i = 2(Ei). We quietly used this fact for Ez in the previous section. For non-Cartesian coordinates like r and φ things are more complicated (see for example Morse and Feshbach Vol I p 116, Moon and Spencer p 139, or do a web search on "vector Laplacian"; the author's Tensor Analysis document, Sections 13, 14 and 15, derives these results for arbitrary coordinate systems): E = Er + Eφ + Ez (D.2.1) (2E)r = 2Er - (2/r2) ∂φEφ - (1/r2) Er (D.2.2a) (2E)φ = 2Eφ + (2/r2) ∂φEr - (1/r2) Eφ (D.2.2b) (2E)z = 2Ez (D.2.2c) where 2 = (1/r)∂r(r∂r) + (1/r2)∂φ2 + ∂z2 = ∂r2 + (1/r)∂r + (1/r2)∂φ2 + ∂z2 . (D.2.3) Here is a summary of vector differential operators in cylindrical coordinates taken from Morse and Feshbach above, where the last line corresponds to the above discussion: (D.2.4) (a) The Er Solution As noted above, it is a characteristic of curvilinear coordinate systems that the Laplacian operator applied to a vector quantity involves a cross coupling between field components which does not occur in Cartesian coordinates. In our case, this affects the Er and Eφ field components, but not Ez. Here is the Helmholtz equation (D.1.1) for Er(r,φ,z) using (D.2.2a), 2Er - (2/r2) ∂φEφ - (1/r2) Er + β2 Er = 0 . (D.2.5) The fact that Eφ is cross-coupled with Er is a minor inconvenience. We eliminate it by using the condition that divE = 0, which in cylindrical coordinates appears as in (D.2.4) times r, ∂r (r Er) + ∂φEφ + r ∂zEz = 0 . (D.2.6) Expanding the 2 operator as in (D.2.3), and eliminating ∂φEφ from (D.2.6) , we get: ∂r2 + (1/r)∂r + (1/r2)∂φ2 + ∂z2Er + (2/r2) [∂r (r Er) + r ∂zEz] - (1/r2) Er + β2 Er = 0 or ∂r2 Er + (3/r) ∂rEr + (1/r2) Er + (1/r2) ∂φ2 Er + ∂z2 Er + β2Er = (-2/r) ∂zEz . (D.2.7) We got rid of Eφ, but now we are stuck with Ez on the right. Actually, we are happy to have this factor on the right, because this is how the longitudinal current in the wire gets coupled into the radial current which feeds the surface charge. Recall that E(r,φz,t) = ej(ωt-βz) !Syntax Error, I E(r,m) ejmφ . (D.1.2), (D.1.3) As in the previous section, we insert into (D.2.7) the above expansion for the various E components, replace ∂φ2 = -m2 and ∂z = -jβd, use completeness of the ejmφ, install (D.1.19) for Ez(r,m), and multiply by r2 -- with β'2 as in (D.1.17) -- to get, [r2∂r2 + 3r∂r + (1-m2) + r2 β'2] Er(r,m) = 2j βd r Czm Jm(β'r) . (D.2.8) In order to get the left side into something recognizable, we define Er(r,m) = x-1 fm(x) (D.2.9) where x is a dimensionless radial variable which will play a major role in the following, x ≡ β'r and xa ≡ β' a . (D.2.10) Then (D.2.8) becomes [x2∂x2 + 3x∂x + (1-m2) + x2] { x-1 fm(x)} = 2j (βd/β') xCzm Jm(x) or x [x2∂x2 + 3x∂x + (1-m2) + x2] { x-1 fm(x)} = K x2 Jm(x) (D.2.11) where Km ≡ 2j (βd/β') Czm . (D.2.12) Ever eager, Maple computes the left side of (D.2.11), so that (D.2.11) becomes [ x2 ∂x2 + x ∂x + (x2-m2)] fm(x) = Km x2 Jm(x) . (D.2.13) The left side of (D.2.13) is the normal Bessel operator [ Spiegel 24.1] , but the equation is also driven by a power times a Bessel function. The solution to the equation is the homogeneous solution of the Bessel equation plus the particular solution which is the response to the driving function on the right hand side. The homogeneous solution is the usual linear combination of Jm(x) and Ym(x), but we must reject Ym(x) since it blows up at x=0 and thereby causes the field Er to be singular, which it cannot be, smack in the middle of a wire. The particular solution is not very obvious and required some hunting to find. It is this fm(x)particular = (1/2) Km [ x Jm+1(x) ] . (D.2.14) a Maple confirms, continuing the above code, Therefore, we now have this full solution for fm(x) fm(x) = fm(x)particular + fm(x)homogeneous = (1/2) Km [ x Jm+1(x) ] + am Jm(x) and then from (D.2.9) the full solution Er Er(r,m) = am x-1 Jm(x) + Jm+1(x) . (D.2.15) For each value of m, there are two as-yet undetermined constants, am and (Km/2). However, looking at (D.2.10), we see that, since J0(x) ≈ 1 for small x, we must have a0 = 0 (D.2.16) to keep Er finite at r = 0. We shall obtain expressions for am and (Km/2) below. (b) The Eφ Solution Start with div E = 0 as stated in (D.2.6), ∂r (r Er) + ∂φEφ + r ∂zEz = 0 . (D.2.6) Repeat the steps (D.1.12) through (D.1.17) to convert the above equation to m space. The rules are that ∂φ → jm and ∂z → -jβd and Ei(r,φ,z,t) → Ei(r,m), so that ∂r (r Er(r,m)) -jm Eφ(r,m) + r ( -jβd )Ez(r,m) = 0 . Then insert (D.1.19) for Ez(r,m) to get, ∂r (r Er(r,m)) + jmEφ(r,m) - jβd r Czm Jm(β'r) = 0 or ∂r (r Er(r,m)) + jmEφ(r,m) - (1/2) [2j(βd/β') Czm ] β'r Jm(β'r) = 0 or ∂r (r Er(r,m)) + jmEφ(r,m) - (1/2) Km x Jm(x) = 0 // x = β'r as earlier so jmEφ(r,m) = (Km/2) x Jm(x) - ∂x (x Er(r,m)) . (D.2.17) Next, insert (D.2.15) Er(r,m) so jmEφ(r,m) = (Km/2) x Jm - ∂x (x { am x-1 Jm + Jm+1}) = (Km/2) x Jm - ∂x (am Jm + x Jm+1) = - am Jm' + [ - x Jm+1' - Jm+1 + xJm] . (D.2.18) At this point we invoke the recurrence relations AS 10.6.2 to write Jm+1' = Jm - (m+1)x-1Jm+1 first relation with ν = m+1 Jm' = -Jm+1 + (m/x)Jm second relation with ν = m (D.2.19) Insert these into (D.2.18) to get jmEφ(r,m) = - am Jm' + [ - x Jm+1' - Jm+1 + xJm] = - am {-Jm+1 + (m/x)Jm } + [ - x { Jm - (m+1)x-1Jm+1} - Jm+1 + xJm] = am Jm+1 - am (m/x)Jm + [ - x Jm + (m+1) Jm+1 - Jm+1 + xJm] = am Jm+1 - am (m/x)Jm + [ m Jm+1] = - am (m/x)Jm + ( m + am ) Jm+1 Dividing by m then gives the final solution: jEφ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) . (D.2.20) We have already noted that a0 = 0. For m=0, (D.2.20) is invalid because we divided (D.2.18) by m. In fact, we really know nothing at all about the field Eφ(r,m=0) from our solution method, because the Eφ fields were obtained from ∂φEφ in the div E = 0 equation (D.2.6). Since Eφ(r,m=0)ej0φ does not depend on φ, this simply gives 0, but does not tells us what Eφ(r,m=0) might be. Presumably we could find it from the φ Helmholtz equation which we have so far ignored. Conjecture: It seems likely that Eφ(r,m=0) = 0. In the m=0 partial wave where everything is symmetrical in azimuth, it is hard to imagine a constant (in φ) field Eφ circulating around the axis of the wire. Thus, we interpret the quantity (+ ) in (D.2.20) as being 0 when m=0. (c) Application of the Boundary Conditions We have two boundary conditions to impose: Er(r=a,m) = (jω/σ) Nm (D.2.14) Eφ(r=a,m) = 0 (D.2.15) The first is the radial charge pumping condition shown in (D.1.9) above, while the second is our ansatz assumed earlier that any z = constant circle on the wire surface be an equipotential. These two boundary conditions serve to determine the two constants am and Km, though a bit of algebra is required. The first step is to use ** and ** to write out the two boundary conditions as am xa-1 Jm(xa) + Jm+1(xa) = (jω/σ) Nm (1) - am xa-1 Jm(xa) + ( + ) Jm+1(xa) = 0 (2) Addition and subtraction of these equations gives two new equations, Jm+1(xa) + ( + ) Jm+1(xa) = (jω/σ) Nm (3) 2 am xa-1 Jm(xa) - Jm+1(xa) = (jω/σ) Nm (4) the second of which may be immediately solved for am = (jω/σ) Nm . which agrees with (D.2.16) Using the recursion relation 2m x-1 Jm = [Jm+1 + Jm-1] , equation (2) may be solved to get ( + ) = (jω/2σ) Nm [ + ] Finally, subtracting ** from ** we find = (jω/2σ) Nm [ – ] To summarize, am = m(jω/σ) Nm m≥0 = (jω/2σ) Nm [ – ] m≥ 0 (+ ) = (jω/2σ) Nm { + } m>0 (D.2.16) ok to here The last equation of the above set should not be used for m=0, due to the = condition, but the other two are correct. Since J-n = (-1)nJn , for m = 0 we have J-1 = - J1 so K0 becomes just K0 = 2 (jω/σ) N0 (D.2.17) We have now completely solved the problem. To review, here are the three fields gathered together in one spot. For Ez we use (D.2.8) to replace Czm . jEφ(r,m) = - am x-1 Jm(x) + (+ ) Jm+1(x) Er(r,m) = am x-1 Jm(x) + Jm+1(x) Ez(r,m) = -j(β'/βd) Jm(x) verify all 3 (D.2.18) D.3 Statement of the Results Although we have a complete solution to our problem, it is useful to normalize the results to some well defined absolute scale, rather than the factors Nm as is done above. The first step in doing this is to compute the total current I in the wire, I = !Syntax Error, Idφ !Syntax Error, Ir dr Jz(r,m) ejmφ (D.3.1) It is clear from (D.3.1) that only the m=0 partial wave can make any contribution, so we write I = !Syntax Error, I2πrdr σEz(r,m=0) = 2πσ [-j(β'/βd ) ] !Syntax Error, Ir dr J0(x) = -2πσj {} !Syntax Error, Idx x J0(x) = -2πσj [(jω/σ) N0 { }] xaJ1(xa) = -2πσj N0 a2 (D.3.2) At this point it is convenient to introduce the DC resistance per unit length of our wire, Rdc = (D.3.3) along with a new symbol to indicate the relative surface charge moment, ηm ≡ (D.3.4) Using (D.3.3) and (D.3.4) in (D.3.2) gives (jω/σ) Nm = (j/2) (aβd) ηm I Rdc (D.3.5) We can use (D.3.5) to get our coefficients in (D.2.16) into a form that is then normalized to the total current I. Putting all the pieces together, we get: E fields in a Round Wire, m > 0 jEφ(r,m) = (j/4) (aβd) ηm I Rdc [ - + { + } ] Er(r,m) = (j/4) (aβd) ηm I Rdc [ + + { - } ] Ez(r,m) = (1/4) ηm I Rdc [ - ] (D.3.6) E fields in a Round Wire, m = 0 jEφ(r,m) = 0 Er(r,m) = (j/2) (aβd) I Rdc [ ] Ez(r,m) = (1/2) I Rdc [ ] (D.3.7) Comments about the solution: (1) Notice that the components Eφ and Er are smaller than Ez by factor (βda). This is the dimensionless smallness parameter which defines the "transmission line limit", see Chapter 4. (2) If there exist moments Nm of the charge distribution on the wire with m > 1, then the corresponding ηm ≠ 0, and it is clear that Ez and hence Jz is non-uniform across the surface of the wire. The non-uniformity is not "small" but has the full strength of ηm. Of course we only expect to get significant moments of charge density n(φ) when conductors are "fat and close". It seems likely in this case that the largest contribution will come from m=1. (3) Because Ez is non-uniform, we conclude that the surface impedance Zs = Ez/I will also be non-uniform around the boundary of the wire. In fact, here is our explicit formula for the surface impedance of a round wire. At r=a, it is a function only of azimuth angle φ : Zs (φ) = (1/2)Rdc { [ ] + } (D.3.8) D.4 The Low Frequency Limit If we assume that ω is small enough that xa = β'a << 1, we can greatly simplify the above results. Another way to state this limit is that (a/δ) << 1, which means the skin depth is much larger than the radius of the wire. All limits come from the leading term of Jn(x) which is Jn(x) ≈ xn/ (2nm!) must be Jn(x) ≈ xn/ (2nn!) (D.4.1) In the formulas for Eφ and Er in the box (D.3.6), the leading term is the first term, the other terms may be neglected. We find that, ≈ (r/a)m-1 (D.4.2) and the first factor in Ez is its leading term, ≈ 2(m+1) (r/a)m (D.4.3) For the m=0 fields we notes these facts, [ ] ≈ (r/a) [ ] ≈ 2 (D.4.4) So here is a complete summary of the low frequency results: E fields in a Round Wire, m > 0, Low Frequency Limit jEφ(r,m) = (j/4) (aβd) ηm I Rdc [ - (r/a)m-1 ] Er(r,m) = (j/4) (aβd) ηm I Rdc [ + (r/a)m-1 ] Ez(r,m) = (1/2) ηm I Rdc [ (m+1) (r/a)m ] (D.4.5) E fields in a Round Wire, m = 0, Low Frequency Limit jEφ(r,0) = 0 Er(r,0) = (j/2) (aβd) I Rdc (r/a) Ez(r,0) = I Rdc (D.4.6) In this very last equation, we recover the fact that a piece of wire really does act like a resistor at low frequencies. However, there is still a small linear radial field which serves to feed the symmetric surface charge in the m=0 partial wave. In fact, if you integrate the radial current over the surface of a long piece of wire of length λ/ 2, taking into account the z dependence, you get 2I. See Section 3.7, Figure 2 for a drawing of the current we have just computed. D.5 What about φ, A and B ? Inside the conductor, we know that φ satisfies (D.1.1), and we know that φ is constant at r=a. This means that φ is similar to our Ez solution in the m=0 partial wave, and vanishes for all higher partial waves. Thus we write φ(r,m) = δm,0 φ0 J0(x)/J0(xa) x = β'r (D.5.1) where φ0 is the value of the potential on the surface at r=a. Since we know E(r,m) from Section 3 (6), we can solve for the vector potential A as follows: -jωA = E + grad φ (D.5.2) Converted to partial waves, this says -jωAr(r,m) = Er(r,m) + δm,0 ∂rφ(r,0) -jωAφ(r,m) = Eφ(r,m) - (1/r) jm δm,0 φ(r,0) = Eφ(r,m) -jωAz(r,m) = Ez(r,m) - jβd δm,0 φ(r,0) (D.5.3) Thus, for m≠0 we have -jωA = E. For m = 0 there are extra pieces as shown for Ar(r,0) and Az(r,0). Since A(r,m) is known, A(r,φ,z) is also known, and then so too is B = curl A. This curl can also be performed in each partial wave if desired by replacing ∂φ = -jm and ∂z = -jβd as usual. Since -jωA = E in the higher partial waves, we may conclude that the transverse components of A are small compared to the longitudinal component, just as is the case for E, see (D.3.6). For m=0 we know that Aφ = Eφ = 0, but Ar is a combination of Er and ∂r φ which is not small compared to Az, due to the ∂rφ contribution. Thus, we arrive at the conclusion that, although Ar is negligible in the dielectric, it cannot be ignored inside the conductor. This explains, incidentally, the problem one encounters with the gauge condition (1.5.5) divA = -j(β2/ω)φ. Since φ is continuous at the boundary, and β2 takes a jump of many orders of magnitude (from βd to β'), something on the left side must change violently. But Az is also continuous. It is the m=0 Ar inside the conductor that takes up the slack. In fact, taking the difference inside minus outside we conclude that ∂rAr (r=a-ε) = -j(β'2/ω)φ0 (D.5.4) from which we can determine φ0The potential Ar is discontinuous at r=a.