the curl E problem REVIEWED

The curl E problem PhL 8.18.14 Reviewed on 9.10.14. In both Appendix D and Plan C Revised (reflection), I compute B fields from Maxwell's equation B = - (1/jω) curl E // curl E = -∂tE I am then interested in the limit of things as ω→0. For any physical situation including at DC, the B fields must be finite. Therefore, the following fact must be true. Theorem 1 Suppose as ω→ 0 that [curl E]i → ωs [stuff]i where stuff is non-infinite. Note that Bi = +(j/ω) [curl E]i = j [stuff]i ωs-1 . In this case one of the following must apply: (1) if s > 1, then Bi = 0 (2) if s = 1, then Bi = j [stuff]i which could be either finite or 0 (3) if s < 1, then Bi = ∞ and we have a non-physical situation. Now, the general equation for the curl is as follows: (cylindrical coordinates, partial waves, and e-jkz assumption on all field components) curl E(r,θ,z) = [ r-1∂θEz - ∂zEθ] + [∂zEr - ∂rEz] + [ r-1∂r(rEθ) - r-1∂θEr ] curl E(r,m,z) = [ r-1jmEz+ jkEθ] + [-jkEr - ∂rEz] + [ r-1∂r(rEθ) - r-1jmEr ] or [curl E(r,m,z)]r = r-1jmEz+ jkEθ [curl E(r,m,z)]θ = -jkEr - ∂rEz = A ω + B = [ B + A] = B [curl E(r,m,z)]z = r-1∂r(rEθ) - r-1jmEr Let's first look at our Appendix D E fields written out for the case of small ω, Ez(r,m) = (1/2) ηm I Rdc (r/a)|m| (|m|+1) Er(r,m) = (j/4) ηm I Rdc (ak) [(r/a)|m|+1 + (r/a)|m|-1] (D.11.7) Eθ(r,m) = (1/4) ηm I Rdc (ak) [(r/a)|m|+1 - (r/a)|m|-1] Ez(r,0) = I Rdc Er(r,0) = (j/2) I Rdc (ak) (r/a) Eθ(r,0) = 0 // low ω E fields where things are expressed in terms of current I. We know that at z = 0 we have I = V/Z0 where Fact 8: In the low frequency limit with G = 0, both components diverge as 1/: (Q.8) Re(Z0) ≈ (1/) 1/ Im(Z0) ≈ - (1/) 1/ The above has been replaced by the following, which is in fact exactly the same : Fact 8: The small ω limit for Z0(ω), assuming ωd = 0, is given by (Q.8.6) Re(Z0) ≈ 1/ + (1/2) [Ldc/] + O(ω3/2) Im(Z0) ≈ - 1/ + (1/2) [Ldc/] + O(ω3/2) This says that Z0 = (1/) 1/(1-j) = 1/ e-jπ/4 1/Z0 = e+jπ/4 and thus, I = V/Z0 = V e+jπ/4 Note Added: In lines doc we have I = 2πω (a/k) N0. Using k as given just below we get I = 2πω (a/k) N0 = 2πωaN0[e+jπ/4 / ()] = 2πaN0 e+jπ/4/ Comparing this to the above expression for I gives V e+jπ/4 = 2πaN0 e+jπ/4/ or V = 2πaN0 / or V = 2πaN0 or V C = 2πaN0 or q = 2πaN0 and this agrees with (D.1.8), where N0 = <n(θ)> Moreover, we also have Fact 4: In the low frequency limit with G = 0 , (Q.4) Re(k) ≈ ω1/2 + ω3/2 Im(k) ≈ - ω1/2 + ω3/2 ω << R/L The above has been replaced by the following, which is again exactly the same : Fact 4: The small ω limit for k(ω), assuming ωd = 0, is given by (Q.4.9) Re(k) ≈ + ( 1 - tanL/2) + O(ω3/2) Im(k) ≈ - ( 1 + tanL/2) + O(ω3/2) Thus we can write k = ω1/2(1-j) = e-jπ/4 I = V/Z0 = V e+jπ/4 Notice then that Ik = V e+jπ/4 * e-jπ/4 = ω V C. We can then rewrite our small ω field expressions as Ez(r,m) = (1/4) ηm V Rdc {2 e+jπ/4 (r/a)|m| (|m|+1) } Er(r,m) = (1/4) ηm V Rdc {(Ca) j [(r/a)|m|+1 + (r/a)|m|-1] ω} (D.11.7) Eθ(r,m) = (1/4) ηm V Rdc {(Ca) [(r/a)|m|+1 - (r/a)|m|-1] ω} Ez(r,0) = V Rdc { e+jπ/4 } Er(r,0) = V Rdc {C (j/2) r ω} Eθ(r,0) = 0 // low ω E fields Notice that all E field components are non-infinite for any ω ≥ 0. We can simplify the above by writing things this way Ez(r,m) = Az(r,m) Er(r,m) = Ar(r,m) ω (D.11.7) Eθ(r,m) = Aθ(r,m) ω Ez(r,0) = Az(r,0) Er(r,0) = Ar(r,0) ω Eθ(r,0) = Aθ(r,0) ω Aθ(r,0)= 0 it happens // low ω E fields where all the coefficients are finite (non-zero and non-infinite). So now we reconsider our three curl E components stated earlier, where we replace k = e-jπ/4 [curl E(r,m,z)]r = r-1jmEz+ j e-jπ/4Eθ [curl E(r,m,z)]θ = -j e-jπ/4Er - ∂rEz [curl E(r,m,z)]z = r-1∂r(rEθ) - r-1jmEr Now let's look at our three curl E components, one at a time. First, [curl E(r,m,z)]r = r-1jmEz+ j e-jπ/4Eθ = r-1jm Az(r,m)+ j e-jπ/4 Aθ(r,m) ω = Aω1/2 + Bω3/2 = ω1/2[ A + Bω] = Aω1/2 For m = 0 we have instead, since Aθ(r,0) = 0, [curl E(r,0,z)]r = 0 + 0 = 0 Thus, according to Theorem 1, this component of curl E gives Br = ∞ for m > 0, unphysical. s = 1/2. Next, we have [curl E(r,m,z)]θ = -j [ e-jπ/4]Er - ∂rEz // matches (D.4.7) for Bθ = -j e-jπ/4 Ar(r,m) ω - ∂r Az(r,m) = A ω3/2 + B ω1/2 = B ω1/2 = in violation [curl E(r,0,z)]θ = -j e-jπ/4Er - ∂rEz = -j e-jπ/4 Ar(r,0) ω - ∂r Az(r,0) But it happens that Az(r,0) = V Rdc { e+jπ/4} = constant, so ∂r Az(r,0) = 0 and we get = A ω3/2 = NOT in violation Finally, we have [curl E(r,m,z)]z = r-1∂r(rEθ) - r-1jmEr = r-1∂r(r Aθ(r,m) ω) - r-1jm Ar(r,m) ω = ω [r-1∂r(r Aθ(r,m) ) - r-1jm Ar(r,m)] and [curl E(r,0,z)]z = r-1∂r(rEθ) - r-1j0Er = r-1∂r(rEθ) = ω [r-1∂r(r Aθ(r,0) ] = ω 0 So this component is OK for m ≥ 0. Conclusion: For Appendix D fields we find that as ω → 0. m > 0 m = 0 [curl E(r,m,z)]r Br → ∞ Br → 0 [curl E(r,m,z)]θ Bθ → ∞ Bθ → 0 [curl E(r,m,z)]z Bz → finite Bz → 0 Example: Bθ(r,m) = +(j/ω) [curl E]θ = (j/ω) B ω1/2 = j B ω-1/2 → ∞ I would say this is a huge problem with Appendix D that I failed to notice until recently!! These two fields are infinite in the ω→0 limit: Br(r,m) and Bθ(r,m) for m > 0 Review of 9/10/14 is fine from start down to here, I agree with everything said so far. Discussion. Here are the assumptions that go into Appendix D: 1. e-jkz dependence of everything on z (ansatz) 2. the two boundary conditions (ignoring Debye term) [ see separate doc re-studying Debye ] 3. the expressions for k(ω) and Z0 as obtained from the TL equations. This is done in the Appendix K network model and in Chapter 4. Note that network model => TL equations => Z0 = and k = -j physics model => TL equations => Z0 = and k = -j but in the physics model, we had to do "averaging" and this makes things hazy. In App D we obtain solutions for the E fields which satisfy the E Helmholtz equations and div E = 0 and the two boundary conditions. When the B fields are obtained, we satisfy ALL four of Maxwell's equations. So OK, is there some physical reason that Bθ(r,m) → ∞ for all r ≤ a as ω→ 0 in my infinite transmission line made from two cylinders? You would think just the opposite, that as I→0, the Bθ field would surely vanish! Since Jz → 0 for all r (as ) where would Bθ be coming from?? In fact ALL Ei fields are going to 0. All currents are then Ji = 0. How can you have Bθ = ∞ if there are no currents to create the Bθ field. Rephrase this question: What causes [curl E]θ → I as ω→ 0? [curl E(r,m,z)]θ = -jk Er - ∂rEz with Er(r,m) = (j/4) ηm I Rdc (ak) [(r/a)|m|+1 + (r/a)|m|-1] ~ ω (D.11.7) Ez(r,m) = (1/2) ηm I Rdc (r/a)|m| (|m|+1) ~ ω1/2 The problem arises from the second term -∂rEz which has I ~ dependence on ω, and thus we get that - ∂rEz ~ causing [curl E(r,m,z)]θ ~ causing Bθ ~ ω-1/2. Can I see this problem in the non-small-ω limit expressions? [ yes, as shown below] Ez(r,m) = (1/4) ηm I Rdc (aβ') fm fm = [ - ] x = β'r Er(r,m) = (j/4) ηm I Rdc (ak) gm gm = [ + ] xa = β'a I want to ponder the quantity -∂rEz . It already has from the I, and another from β', but what happens with ∂rfm(r) ? Should be easy to do: ∂rfm = dx/dr d/dx f(x) = β' f'm(x) = β' [ - ] * J'm(x) = (β' /2) [ - ] { Jm-1(x) - Jm+1(x)} . Now for small x and xa the leading term in each bracket is the one selected below, ≈ (β' /2) [] { Jm-1(x) } ~ β' (xa)-m-1 xm-1 = β' (x/xa)m-1 xa-2 ~ β' β'-2 ~ 1/β' ~ 1/ . Then I have shown that -∂rEz ~ 1/ ~ which agrees with the small ω field expression results, and which results in Bθ → ∞. Comment 1. The problem arises because Bθ ~ (1/ω) ∂rEz and Ez ~ I rm ~ rm. At small β, the E Helmholtz equations are (2 + β2)E = 0 and in particular (2 + β2) Ez = 0 which is then 2 Ez ≈ 0 which is in fact the 3D scalar Laplace equation. If we assume e-jkz then ∂z → -jk and we then end up with a 2D scalar Laplace equation involving polar variables r and θ. We know from Stakgold that the solutions of such an equation always have the form rm ejmθ, and therefore we expect to have Ez(r,m) ~ rm and that is exactly what we are getting as just shown on the first line of this comment. Our issue then is not with the rm part, it is with the coefficient. We solved the (2 + β2)E = 0 Helm equations in App D with the ansatz e-jkz and got the general form Ez(r,m) = - j (β'/k) Jm(x) x = β'r (D.1.27) Er(r,m) = am x-1 Jm(x) + Jm+1(x) β'2 = β2 - k2 (D.2.11) jEθ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) (D.2.15) in terms of two coefficients am and Km. Since Jm(x) = (x/2)m/m! and x = β' r, we see at once Ez(r,m) = - j (β'/k) (x/2)m/m! ~ (β'/k) β' m rm ~ (β')m+1 (1/k) rm There is the expected rm factor just explained. In order to get Bθ(r,m) finite, coefficient Km just do exactly the right thing. But from my two boundary conditions I get = (jω/2σ) Nm [ – ] ~ ω ~ ω (β'a)-m-1 ~ ω (β')-m-1 and then we get Ez(r,m) = (β')m+1 (1/k) rm = (β')m+1 (1/k) ω (β')-m-1 rm = (1/k) ω rm = (ω/k)rm and once again you have (ω/k) ~ and so Ez(r,m) ~ rm which causes Bθ = ∞. Therefore, the only way to "fix" this problem is to change Km to some other form, and that means one must change at least one of the boundary conditions. Perhaps that answer is that the ansatz e-jkz is not valid for small ω. But then why should we expect it to be valid for larger ω? In partial waves where ∂θ → jm and also ∂z → -jk, this equation reads [r2∂r2 + r ∂r - m2 + r2 β'2] Ez(r,m) = 0 . (D.1.15) Review of 9/10/14 is fine from start down to here, I agree with everything said so far. Update 9.10.14. Appendix Q has been rewritten, but it makes no difference for the curl E problem since the super low ω limits are the same. Of course I didn't really know that till I did it Here is the problem in a nutshell: I am proposing adding the following verbatim to Section D.11. Another low-frequency anomaly and its implications. We consider here a sequence of steps which leads to a contradictory result. The scenario here is a transmission line one of whose conductors is a round wire, but it is helpful just to think of both conductors being round wires as studied in Chapter 6. In this case we know for example that ηm = -ηm where ηm are the partial wave moments of a round wire surface charge density. 1. For monochrome fields, Maxwell tells us that (in any coordinate system), Bi = +(j/ω) [curl E]i 2. The low ω and low k limit of the Appendix D partial wave E field is this: Ez(r,m) = (1/2) ηm I Rdc (r/a)|m| (|m|+1) m>0 Er(r,m) = (j/4) ηm I Rdc (ak) [(r/a)|m|+1 + (r/a)|m|-1] (D.11.7) Eθ(r,m) = (1/4) ηm I Rdc (ak) [(r/a)|m|+1 - (r/a)|m|-1] These expressions incorporate the ansatz assumption that E(r,θ,z) = E(r,θ,0) e-jkz which is assumed at the very start of Appendix D. 3. In cylindrical coordinates, curl E is given by curl E(r,θ,z) = [ r-1∂θEz - ∂zEθ] + [∂zEr - ∂rEz] + [ r-1∂r(rEθ) - r-1∂θEr ] . With the above mentioned ansatz we get ∂z → -jk. If at the same time we convert to partial waves where ∂θ → jm, the above curl expression becomes, curl E(r,m,z) = [ r-1jmEz+ jkEθ] + [-jkEr - ∂rEz] + [ r-1∂r(rEθ) - r-1jmEr ] or [curl E(r,m,z)]r = r-1jmEz+ jkEθ [curl E(r,m,z)]θ = -jkEr - ∂rEz [curl E(r,m,z)]z = r-1∂r(rEθ) - r-1jmEr . For example, inserting the low-ω E field expressions from above gives, [curl E(r,m,z)]r = r-1jmEz+ jkEθ = r-1jm [(1/2) ηm I Rdc (r/a)|m| (|m|+1)] + jk [(1/4) ηm I Rdc (ak) [(r/a)|m|+1 - (r/a)|m|-1] ] = j (1/4) ηm Rdc { 2 I r-1m (r/a)|m| (|m|+1) + a k2I [(r/a)|m|+1 - (r/a)|m|-1] } . 4. If we assume that the k appearing above for the internal round wire E field is the same as the k which appears in the network model of Appendix **, and in the large ω theory of Chapters 4 and 5, namely, k = -j = -j , then using the parameter model of Appendix Q we know that, for Gdc = C ωd = 0, Fact 4: The small ω limit for k(ω), assuming ωd = 0, is given by (Q.4.9) Re(k) ≈ + ( 1 - tanL/2) + O(ω3/2) Im(k) ≈ - ( 1 + tanL/2) + O(ω3/2) We might as well assume a vacuum dielectric, so than both Gdc = 0 and tanL = 0. In this case, we find that. as ω→ 0, k = ω1/2(1-j) = e-jπ/4 as ω→ 0 so k2 = RdcC e-jπ/2ω = -jRdcCω . // as also in (D.11.2) Now from (D.2.31) we can determine how I depends on ω for small ω ( N0 = <n(θ)> ), I = 2πω (a/k) N0 = 2πωaN0[e+jπ/4 / ()] = 2πaN0 e+jπ/4/ which says I ~ for small ω. As expected, I→ 0 since Z0 → ∞ as ω→ 0. Then also, k2I = -jRdcCω * 2πaN0 e+jπ/4/ = constant * ω3/2. 5. Installing these expressions for k2 and k2I into [curl E(r,m,z)]r gives: (all for m > 0) [curl E(r,m,z)]r = j (1/4) ηm Rdc { 2 I r-1 m (r/a)|m| (|m|+1) + a k2I [(r/a)|m|+1 - (r/a)|m|-1] } = m a(r,|m|) + b(r,|m|) ω3/2 . // ηm = η-m = η|m| for Ch 6 TL 6. From Step 1 we then find that, as ω→0, Br(r,m) = +(j/ω) [curl E]r = (j/ω) [m a(r,|m|) + b(r,|m|) ω] = j m a(r,|m|) / + j b(r,|m|) ≈ j m a(r,|m|) / as ω→ 0 → ∞ for m ≠ 0 so in the DC limit of the theory, we end up with an infinite radial magnetic field! More generally we find, m > 0 m = 0 [curl E(r,m,z)]r Br → ∞ Br → 0 [curl E(r,m,z)]θ Bθ → ∞ Bθ → 0 [curl E(r,m,z)]z Bz → finite Bz → 0 7. Do the +m and - m terms cancel? Br(r,θ) = Br(r,m) ejmθ + Br(r,-m) e-jmθ // m and -m terms only = j a(r,|m|)/ * [ m ejmθ + (-m) e-jmθ] = j m a(r,|m|)/* [ 2jsin(mθ)] → ∞ still So cancellation does not save the day. 8. Conclusions: In looking for traveling wave solutions of the form ej(ωt-kz) going down a transmission line, the assumption that k = -j = -j for ultra-low ω leads to the contradiction of infinite B fields as ω→ 0 (the DC limit of the theory). The assumption k = -j follows from the transmission line equations (4.11.14/15) or (K.5/6) and from the requirement that the interior and exterior wave solutions must match at the round wire surfaces. We conclude that the transmission line equations are not valid at ultra-low frequencies. These equations were derived in Chapters 4 and 5 only for "large" ω, although the network model of Appendix K supports the transmission line equations all the way down to ω → 0. We conclude therefore that the network model is invalid at ultra-low frequencies as a representation of the physics of an actual transmission line. Redo everything for the G> 0 case. Start with Ez(r,m) = (1/2) ηm I Rdc (r/a)|m| (|m|+1) Er(r,m) = (j/4) ηm I Rdc (ak) [(r/a)|m|+1 + (r/a)|m|-1] (D.11.7) Eθ(r,m) = (1/4) ηm I Rdc (ak) [(r/a)|m|+1 - (r/a)|m|-1] Ez(r,0) = I Rdc Er(r,0) = (j/2) I Rdc (ak) (r/a) Eθ(r,0) = 0 // low ω E fields And write these as Ez(r,m) = Am I (r/a)|m| Er(r,m) = Bm I k [(r/a)|m|+1 + (r/a)|m|-1] (D.11.7) Eθ(r,m) = Bm I k [(r/a)|m|+1 - (r/a)|m|-1] Ez(r,0) = A0 I Er(r,0) = B0 I k Eθ(r,0) = 0 // low ω E fields Rewrite using I k = 2πω a N0 Ez(r,m) = Am I (r/a)|m| Er(r,m) = Bm 2πω a N0 [(r/a)|m|+1 + (r/a)|m|-1] (D.11.7) Eθ(r,m) = Bm 2πω a N0 [(r/a)|m|+1 - (r/a)|m|-1] Ez(r,0) = A0 I Er(r,0) = B0 2πω a N0 Eθ(r,0) = 0 // low ω E fields Now use I = 2π(ω/k) a N0, so once again Ez(r,m) = Am' (ω/k) (r/a)|m| Er(r,m) = Bm 2πω a N0 [(r/a)|m|+1 + (r/a)|m|-1] (D.11.7) Eθ(r,m) = Bm 2πω a N0 [(r/a)|m|+1 - (r/a)|m|-1] Ez(r,0) = A0' (ω/k) Er(r,0) = B0 2πω a N0 Eθ(r,0) = 0 // low ω E fields The dependences are the same for m = 0 or m > 0 really. Meanwhile, [curl E(r,m,z)]r = r-1jmEz+ jkEθ = (ω/k) + kω [curl E(r,m,z)]θ = -jkEr - ∂rEz = kω + (ω/k) [curl E(r,m,z)]z = r-1∂r(rEθ) - r-1jmEr . = ω + ω So if k→k0, then all these things are proportional to ω and we do avoid all infinite B fields Br = constant Bθ = constant Bz = constant