App D.9 Paradox of the Day 10.5.14 REVIEWED

App D Paradox of the Day 10.5.14 PhL 10.5.14 Retrograde motion on Appendix D.9 ! I had it right long ago, then I changed it , and now I change it back. It is a tricky notational issue really, k → k' called k, I think the new D.9 clarifies this tricky issue and I did several tests to make sure it is now right. 1. On the one hand, I have argued in Appendix D.9 that when we change from G = 0 to G > 0, the Appendix D solutions change simply by the addition of the factor (ξ2/ε2) to all the E and B fields. For example, when G = 0 the E field solutions are, Second summary of the E field solutions : (D.2.33) 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 Eθ(r,m) = (1/4) ηm I Rdc (ak) hm hm = [ - ] where I = 2πa (ω/k) N0 = 2πa (ω/k) (1/2πa) q = (ω/k) CV . (D.2.31c) Thus, for G = 0 the E fields are Ez(r,m) = (1/4) ηm (ω/k) CV Rdc (aβ') fm Er(r,m) = (j/4) ηm (ω/k) CV Rdc (ak) gm Eθ(r,m) = (1/4) ηm (ω/k) CV Rdc (ak) hm and for G > 0 the fields are Ez(r,m) = (ξ2/ε2) (1/4) ηm (ω/k) CV Rdc (aβ') fm Er(r,m) = (ξ2/ε2) (j/4) ηm (ω/k) CV Rdc (ak) gm Eθ(r,m) = (ξ2/ε2) (1/4) ηm (ω/k) CV Rdc (ak) hm The entire change can be encapsulated by saying I = (ω/k) CV → I = (ω/k) CV(ξ2/ε2) which we can write as IG>0 / IG=0 = (ξ2/ε2) 2. On the other hand, I know that 1/Z0 = where for low ω we have R = Rdc2 . I also know that I = V/Z0 Therefore I = V G > 0 For G = 0 this reads I = V G = 0 The ratio by which I increases is this: IG>0 / IG=0 = / = 3. The Paradox: Comparing these two results, we must have (ξ2/ε2) = But (ξd/εd) = (εd - jσd/ω)/εd = 1 - j (σd/εd) (1/ω) = 1 - j (G/C) (1/ω) = 1 + G/(jωC) Thus, it must be true that 1 + G/(jωC) = or = So THAT is the paradox. 4. Think source is in Appendix D.9 (d) I do remember dealing with this square root issue once before in Appendix D.9, then I got rid of the square root. So I guess to resolve this I need to do yet another review of that section. I see a comment in red on this in "old App D with betad.doc". Plan A [ ignore this, go to Plan B ] At large ω, we know that that for G = 0, k2 = βd02 = ω2μdεd . But when G > 0, we have k2 = βd2 = ω2μdξd. [ wrong! ] Are these claims true, or not? The values for k2 come from the usual k(ω) formula in the large ω limit, as for example in (Q.3.5), Re(k) ≈ (ω/vd) vd = 1/ ≈ 1/ Im(k) ≈ - ω tanL /(2vd) where it says vd = 1/ ≈ 1/ . Why does this not depend on whether Gdc = 0 or Gdc > 0 ? Well, the model of (Q.1.9) says G(ω) = C (ωd + tanL ω ) ωd ≡ (σd/εd) Gdc ≡ ωd C where ωd = Gdc/C. But when ω is high enough, you ignore ωd and so the result is independent of Gdc. That is the reason. Meanwhile, LeC = μdεd = 1/vd2 (4.1219) It seems that Gdc has no effect whatsoever on the value of k for large ω. βd02 = ω2μdεd = ω2/vd2 βd0 = ω/vd so only βd0 is involved at large ω. If you "turn on" the dielectric conductivity at high ω, there is no change in the value of k (assuming you have not altered tanL in this turn on). OK, for high ω this is the case. Plan B. Start here: Nm → (ξd/εd) Nm everywhere . (D.9.23) For example, for the total current we have from (D.2.31a) that I = 2πa (ω/k) N0 , so here is the change in symbol I caused by turning on the conducting dielectric, I = 2πa {N0} → 2πa {(ξd/εd)N0} . We must now treat a subtle point regarding k in order to obtain the correct "rule" for how turning on the dielectric DC conductivity alters the E and B fields presented earlier in Appendix D. Although k is treated as a generic constant in Appendix D, we will eventually be setting it to the specific value k(ω), k = k(ω) ≡ -j = -j (5.3.6) where the four parameters are as given in the simple model of (Q.1.9). In particular, C(ω) = C, a constant, whereas G(ω) = Gdc + C tanLω. so we may write k = -j ≈ -j which shows the traditional dependence on Gdc and C. Thus we can write k(Gdc> 0) = -j k(Gdc= 0) = -j . The point is that the value of k changes when Gdc is turned on. The k ratio is then = . We can then rewrite the above theory alteration for turning on the dielectric conductivity in this manner I(Gdc= 0) = 2πa {N0} I(Gdc> 0) = 2πa {(ξd/εd)N0} . The current ratio is then = (ξd/εd) = (ξd/εd) . But from (1.5.1a) we know that ξd ≡ εd - jσd/ω => (ξd/εd) = 1 - j (σd/εd)(1/ω) Then using (4.4.10) C/Gdc = εd/σd (4.4.10) we find that (ξd/εd) = 1 - j (Gdc/C)(1/ω) = 1 + Gdc/(jωC) = . Therefore our current ratio becomes, = (ξd/εd) = = = Here then is our rule: The effect of moving from Gdc = 0 to Gdc > 0 on the Appendix D fields is that one should take I → I. In other words, all the E fields become "larger" by the factor . Since the B fields are obtained from -jωB = curlE as in Section D.4, it follows that the B fields also become larger by this same factor . Another way to express this rule is this I = 2πω (a/k) N0 → I' = 2πa (ω/k') (ξd/εd) N0 (D.2.31a) or I = (ω/k) CV → I' = (ω/k') CV (D.2.31c) where I and k are for Gdc= 0, and I' and k' are for Gdc > 0. ************************************ Now let's start over on our Paradox! 1A. On the one hand, I have argued in Appendix D.9 that when we change from G = 0 to G > 0, the Appendix D solutions change simply by the addition of the factor (ξ2/ε2) to all the E and B fields and changing k to k'. For example, when G = 0 the E field solutions are, Second summary of the E field solutions : (D.2.33) 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 Eθ(r,m) = (1/4) ηm I Rdc (ak) hm hm = [ - ] where I = 2πa (ω/k) N0 = 2πa (ω/k) (1/2πa) q = (ω/k) CV . (D.2.31c) Thus, for G = 0 the E fields are Ez(r,m) = (1/4) ηm (ω/k) CV Rdc (aβ') fm Er(r,m) = (j/4) ηm (ω/k) CV Rdc (ak) gm Eθ(r,m) = (1/4) ηm (ω/k) CV Rdc (ak) hm and for G > 0 the fields are Ez(r,m) = (ξ2/ε2) (1/4) ηm (ω/k') CV Rdc (aβ') fm Er(r,m) = (ξ2/ε2) (j/4) ηm (ω/k') CV Rdc (ak) gm Eθ(r,m) = (ξ2/ε2) (1/4) ηm (ω/k') CV Rdc (ak) hm where k' is given by the usual formula but with G > 0, whereas k is the same formula with G = 0. The entire change can be encapsulated by saying I = (ω/k) CV → I = (ω/k') CV(ξ2/ε2) As shown above, the current ratio for making this change is: IG>0 / IG=0 = = (*) 2A. On the other hand, I know that 1/Z0 = where for low ω we have R = Rdc2 . I also know that I = V/Z0 Therefore I = V G > 0 For G = 0 this reads I = V G = 0 The ratio by which I increases is this: IG>0 / IG=0 = / = = (**) 3A. The Paradox Resolved: Comparing these two results (*) and (**), we see that they are exactly the same and thus there is no paradox! Comments: this business with the symbol I is quite confusing. It appears in my second summary of the E fields in (D.2.33) for example. The problem is that k also appears in these formulas, and you have to remember to take k → k'. so the fields in general do not do a simple scaling! For example,. Second summary of the E field solutions : Rdc = β'2 = β2 - k2 (D.2.33) 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 Eθ(r,m) = (1/4) ηm I Rdc (ak) hm hm = [ - ] There is a k hidden inside the fm factors for example, and k → k'. There are also some explicit k values sitting there, and they too must do k→ k'. NOW, if we think of I = (ω/k) CV then there is no N0 appearing, so our only rule is to take k→k' I guess. That in itself creates the factor in going I → I', since ω and V and C don't change. But the problem is that I always just write k as k, and its value then goes in through the formula when I eventually apply the k(ω) formula. So this is all very confusing and needs to be cleaned up !!! It is true that the actual physical current in the wire increases by the factor . Next day 10.6.14. I am now thinking about how to rewrite the text after (D.9.24). Several possibilities: 1. If I don't distinguish the change in k as k', then (D.9.24) is extremely unclear. You get the impression that the current changes by the full factor (ξd/εd). I could say I = 2πaN0(ω/k) → I' = 2πaN0(ω/k') (ξd/εd) . (D.9.24) where k applies to the G = 0 situation and k' to the G>0 situation. But then I have to segue to the notion that k is not a fixed constant, but k = k(ω) = our formula, and then k' ≠ k. The prime might be too lightweight.