E dot B equals 0 v1 REVIEWED

E dot B = 0 text for lines doc PhL 1.28.14 This is the third version of this doc in which I attempt to come up with a lower bound for ω above which one will have E B = 0 in a transmission line. I could never get happy with my estimates. At one point below I get a result that the bound is about f = 1 Hz, but who knows. I gave up and left it as a Reader Exercise! I really have no idea what that lower limit is. One could compute it perhaps for the Chapter 6 worked out example, but much work would be required. I added that as part of the Exercise. Appendix M: Estimate of lowest frequency for which E•B = 0 at the conductor surface. This is a surprisingly complicated estimate. We shall assume rectangular conductors with cross section as shown in Fig C.*, where we know that E•H ≠ 0 at ω = 0. Our interest will be the nature of the perpendicularity of the fields just outside the conductor surface in the cross section picture. (a) Estimate of Hz / Hφ We assume our conductors are widely spaced so we can crudely use certain m = 0 results from Appendix D (even though Appendix D is for a round wire). The main result of interest from Appendix D comes from (D.6.1) which estimates the ratio of the longitudinal H field to the transverse H field. We assume μd = μc= μ so that H is continuous at the conductor boundary, so the (D.6.1) results then apply just outside the surface. The estimate implied by (D.6.1) is this: | Hz / Hφ | ~ | βd/βc | . This ratio appears in the form (βd/β') but β' ≈ β = βc for the conductor. We assume the conductor is copper with a large σc, and for the dielectric we assume σd = 0 and ε = ε0. Also, μd = μc = μ0. Then from (1.5.1), βc2 = -jωμ0σc βd2 = ω2μ0ε0 => | βd/βc |2 = = ω (ε0/σc) ≈ ω (10-11/108) ≈ ω 10-18 => | Hz / Hφ | ≈ | βd/βc | ≈ 10-9 // SI units Certainly for any reasonable frequency this ratio is very small. It seems likely this ratio estimate is valid for any non-singular point on the cross section of a conductor of any shape. (b) Estimate of Er/Hφ For a round wire and m = 0, Appendix D gives the following surface charge density [ (D.1.5a) and (D.2.31)] n(φ) = N0 = (βd/2πωa) I The transverse electric field at the conductor surface is then, from (1.1.47), Er = n/ε0 = (βd/2πωaε0) I . dim OK Since our assumed transmission line has widely spaced conductors, the m = 0 estimate is reasonable. For non-round conductors (at non-singular points on the periphery) we use this same formula, with the understanding that a is some transverse dimension of the conductor. Inserting ** for βd we get Er2 ≈ (βd/2πωaε0)2 I2 = βd2 (I/2πωaε0)2 = ω2μ0ε0 I2/ (4π2ω2a2ε02) = (μ0/ε0) I2/ (2πa)2 so Er ≈ I/(2πa) ≈ (377/2π) I/a ≈ 60Ω I/a dim OK For a round wire we also know from Ampere's law the value of Hφ at the wire surface, 2πa Hφ = I => Hφ = I/(2πa) The ratio then of Er to Hφ is given by Er/ Hφ = = 2π 60Ω = 377Ω which replicates the plane-wave result that E/H = the impedance of free space. (c) Examination of E H Let us review our proof from (***) that E H = 0. First, Maxwell's curl H equation (1.1.1) says, using both D = εE and J = σE, curl H = (jωD + J ) = (jωεE + σE ) = (jωε + σ)E Then E H = (jωε + σ)-1 curl H H . In cylindrical coordinates we have H = Hr + Hθ + Hz curl H = [ r-1∂θHz - ∂zHθ] + [∂zHr - ∂rHz] + [ r-1∂r(rHθ) - r-1∂θHr ] so curl H H = Hr [ r-1∂θHz - ∂zHθ] + Hθ[∂zHr - ∂rHz] + Hz [ r-1∂r(rHθ) - r-1∂θHr ] If we set Hz ≡ 0 and Hr ≡ 0 with the argument that Hθ is so large, this becomes curl H H = 0 [ r-1∂θ0 - ∂zHθ] + Hθ[∂z0 - ∂r0] + 0 [ r-1∂r(rHθ) - r-1∂θ0 ] = 0 and from this we concluded that E H = 0. But in reality we don't have Hz and Hr both vanishing, so we then back up to write |E| |H| cosα = (jωε + σ)-1 { Hr [ r-1∂θHz - ∂zHθ] + Hθ[∂zHr - ∂rHz] + Hz [ r-1∂r(rHθ) - r-1∂θHr ] } Maybe better to do this with the other curl equation to get' |E| |B| cosα = (ωε)-1 | Er [ r-1∂θEz - ∂zEθ] + Eθ[∂zEr - ∂rEz] + Ez [ r-1∂r(rEθ) - r-1∂θEr ] | Since we are just above the conductor surface, using the local cylindrical coordinate system of Fig 3.3 we argue that Eθ = 0 because the surface is an equipotential and we are just above it. So, |E| |B| cosα = (ωε)-1 | Er [ r-1∂θEz] + Ez [- r-1∂θEr ] | |E| |B| cosα = (ωε)-1 | Er [- ∂zEθ] + Eθ[∂zEr - ∂rEz] + Ez [- r-1∂r(rEθ) ] | Next, we replace ∂z derivatives by jβd based on the travelling wave on the line, so |E| |B| cosα = (ωε)-1 | Er [- jβdEθ] + Eθ[jβdEr - ∂rEz] + Ez [ -∂rEθ - r-1Eθ ] | Now we make a wild guess that we can replace ∂r by (1/a) where a is a generic transverse dimension of the conductor, and we also set r = a, so then |E| |B| cosα = (ωε)-1 | Er [- jβdEθ] + Eθ[jβdEr -(1/a)Ez] + Ez [-(1/a)Eθ - a-1Eθ ] | = (ωε)-1 | -jβdErEθ + jβd EθEr - (1/a)Ez Eθ - (2/a)EzEθ | = (ωε)-1 | (3/a)EzEθ | Notice that EzEθ is doubly small here! We then get cos α ≈ (ωε)-1 | (3/a)EzEθ | / (ErBθ) Above we found that Er ≈ 60Ω I/a Hθ ≈ I/(2πa) Bθ ≈ μ0I/(2πa) How can you have Eθ ??? *********************************8 Next we stare at (D.6.1) to conclude that, since Ez and Eφ are continuous at the boundary, being tangential E fields, we can neglect Eφ ( = Eθ) relative to Ez since Eφ/Ez ≈ (βd/βc). Then |E| |B| cosα = (-jωε)-1 { Er [- jβdEθ] + Eθ[jβdEr - ∂rEz] + Ez [∂rEθ + r-1Eθ ] } From Appendix D since maybe I know a little more about the E fields. Then very roughly keep just the Er term, |E| |B| cosα ≈ (-jωε)-1 { Er [ r-1∂θEz - ∂zEθ] Since the conductor surface is an equipotential surface and we are just above it, we set ∂θEz = 0. And from the wave nature of the fields ∂zEθ = jβdEθ so then we have |E| |B| cosα ≈ (-jωε)-1 { Er [- jβdEθ] } The next required ingredient is an estimate for the ratio of the transverse E field to the transverse H field. For this estimate we We assume a wide spacing of the conductors so that we can crudely use our m = 0 field results of Appendix D to es Hz/Hφ We shall now try to estimate a lower limit for ω such that we expect E•H = 0. There is probably a simple way to do this, but we proceed as best we can. For a ballpark estimate, let us assume that σ = 0 in the dielectric, so the neglected terms in (3.7.3) are these cosα |neglected = { ( ∂xHy - ∂yHx)Hz + ( ∂yHz)Hx + (- ∂xHz)Hy } / {(jωε) |E| |H| } . Although the numerator terms involving Hz are small, their effect is amplified by dividing by ω when ω is small. A reasonable condition for the applicability of our claim above that E•H = 0 would be this | cosα |neglected | << 1 or | ( ∂xHy - ∂yHx)Hz + ( ∂yHz)Hx + (- ∂xHz)Hy | / {(ωε) |E| |H| } << 1 . As part of our ballpark estimate, we let Hx "represent" the transverse action (probably cylindrical coordinates would be better here for which Hx → Hφ) and set Hy = 0 to get | (- ∂yHx)Hz + ( ∂yHz)Hx | / {(ωε) |E| |H| } << 1 or | Hz2 ∂y(Hx/Hz) | / {(ωε) |E| |H| } << 1 We next assume that ∂y(Hx/Hz) ~ (Hx/Hz)/D where D is some transverse dimension of the transmission line in question. Then | Hz2 (Hx/Hz)/D | / {(ωε) |E| |H| } << 1 or | Hz Hx /D | / {(ωε) |E| |H| } << 1 But since |H| ≈ |Hx| in our ballpark scenario, this says | Hz/D | / {(ωε) |E| } << 1 . From the Appendix D study of a round wire, using the m = 0 results shown in (D.6.1) we find that | Hz / Hx | ~ | Hz / Hφ | = | Bz / Bφ | ~ | β/βc | where βc is for the conductor and β for the dielectric. Although this result applies to the H field inside the round wire, if μ = μc then H is continuous at the boundary, so the result then applies for H just outside the wire as well. From (1.5.1) this ratio is roughly | β/βc |2 = | ω2με / (-jωμcσc) | ≈ (ε/σc) ω μ ≈ μc Then, replacing Hx by Hφ which is the main transverse field in cylindrical coordinates, | Hz|2 ≈ | Hφ|2 | β/βc |2 = | Hφ|2 (ε/σc) ω Squaring our condition *** gives | Hz |2 << D2(ωε)2 |E|2 or | Hφ|2 (ε/σc) ω << D2ε2ω2 |E|2 or | Hφ|2 (1/σc) << D2ε ω |E|2 or ω >> ( |Hφ|/ |E|)2 (1/D2σcε0) ε ≈ ε0 The next task is to estimate the ratio |Hφ|/ |E| for a transmission line. Roughly we can estimate |E| = V(z)/D where V(z) is the potential difference of Chapter 4 and D is a transverse dimension of the line. For a round conductor of radius a, we know that just outside the conductor surface 2πa Hφ ≈ i(z) where i(z) is the conductor current. Then: |Hφ|/ |E| ≈ (1/2πa) i(z) / [V(z)/D] = (1/2π)(D/a) i(z)/V(z) But V(z)/i(z) is Z0 for the transmission line. Even for a 60 Hz power transmission line, one is in the "high frequency limit" in terms of the formula (4.4.16) Z0 ≈ (K /) 30Ω εrel ≡ ε/ε0 . (4.4.16) The reason is that for such a transmission line, G is extremely small and ωL dominates R which is very low as well in order to reduce ohmic losses in the line. Setting εrel ≈ 1 we then have |Hφ|/ |E| ≈ (1/2π)(D/a) i(z)/V(z) ≈ (1/2π)(D/a) / Z0 ≈ (1/2π)(D/a)(1/K)(1/30Ω) The dimensions on the left are (amps/m) / (volts/m) = (amps/volts) = 1/ohms, in agreement with the right where only the last factor has dimensions. Then ** becomes ω >> ( |Hφ|/ |E|)2 (1/D2σcε0) = (1/2π)2(D/a) 2(1/K)2(1/30Ω)2 / (D2σcε0 ) or ω >> (1/2πa K 30Ω)2 / (σcε0 ) Another dimension check for the right side [looking as usual at (1.1.28) and (1.1.29)], dim(RHS) = m-2 ohm-2 / [ohm-1 m-1 * farad m-1] = ohm-1/farad = sec-1 Then ω >> 1/(60π)2 1/K2 (1/a)2 / (σcε0 ) Using ε0 ~ 10-11 and σc ~ 108 we have ω >> 1000/(60π)2 * 1/K2 (1/a)2 For a typical transmission line with a = 1/2 cm and wire separation 1 m, we find from *** that K = 4 ln(b/a) = 4 ln( 1m/ 0.5*10-2m) = 4 ln(20) = 12 which corresponds to Z0 = K 30Ω = 360Ω ( normally 400-600 Ω for a power line). Then ω >> .03 1/K2 (1/a)2 = .03 (1/144) (200)2 = 8 Then 2πf = 8 says f ~ 1 Hz !! *******************************************************88 I am getting nowhere with all this work, so I think I will bail out and just make a comment. Here then is a rewrite of Fact 4: Fact 4: In a cross sectional sketch of a transmission line, the E and B field lines are perpendicular at every point in the dielectric. Proof: From Maxwell's curl E equation (1.1.2) in the ω domain we have curl E = - jωB . Then B E = (-jω)-1 curl E E = (-jω)-1 [ ( ∂xEy - ∂yEx)Ez + ( ∂yEz - ∂zEy)Ex + ( ∂zEx - ∂xEz)Ey ] . To the extent that Ez << Ex , Ey we set Ez ≈ 0 to get B E = (-jω)-1 [- (∂zEy)Ex + (∂zEx)Ey ] + (-jω)-1 [ correction terms ] where "correction terms" accounts for the fact that the dismissed terms are not exactly zero. Then B E = (-jω)-1 [Ey2 ∂z(Ex/Ey)] + (-jω)-1 [ correction terms ] (3.7.1) However, we argued in Fact 3 that the shape of fields does not vary with z. Thus, the ratio of two components like Ex/Ey cannot vary with z. Thus ∂z(Ex/Ey) = 0 so B E = (-jω)-1 [ correction terms ] . Letting α be the angle between B and E, we can write |B| |E| cosα = (ω)-1 | correction terms | so then cosα = (ω)-1 | correction terms | / (|B| |E|) . Since the correction terms involve Ez , they are small compared to (|B| |E|) and we then conclude that cosα ≈ 0 and α ≈ π/2 and then BE = 0 . Caveat: However, for sufficiently small ω the right side of ** blows up and our conclusion is no longer valid. A condition for validity would then be cosα << 1 or (ω)-1 | correction terms | / (|B| |E|) << 1 or ω >> | correction terms | / (|B| |E|) In particular, this condition is violated when ω = 0. This then explains why one does not see BE = 0 at the surface of rectangular conductor operating at low frequency as shown in Fig C.2. _________________________________________________________________________ older stuff: Recall these two equations from earlier sections curl B = μ (jωε E + J ) = μ (jωε E + σE ) = μ(jωε + σ)E (2.2.1) β2 = ω2 μ [ ε + σ/jω] = ω2 μ( jωε + σ) /jω = ωμ( jωε + σ) /j . (1.5.1) If follows from these equations that, curl B = j(β2/ω) E ≡ C E . (3.7.1) To show that the E and B fields are perpendicular, we will show that E•B = 0. We have from (3.7.1), C E•B = curl B B = ( ∂xBy - ∂yBx)Bz + ( ∂yBz - ∂zBy)Bx + ( ∂zBx - ∂xBz)By . Since Bz ≈ 0 ( see estimate in previous section), we are left with only two terms C E•B = By(∂zBx) - Bx(∂zBy) = By2 ∂z (Bx/By) . However, we argued in Fact 3 that the shape of fields does not vary with z. Thus, the ratio of two components like Bx/By cannot vary with z. Thus, E•B = 0 so the E and B lines are perpendicular everywhere in the dielectric. ( We also assume Ez ≈ 0 since it is small in the dielectric).