bug with small beta limit in Ch 4 REVIEWED

Bug with small beta limit in Ch 4 PhL 1.3.14 My old Chapter 4 did have a big flaw. I want to be talking small β, but my main expansion assumed large β as you see in (4.2.2) quoted below. This resulted in full rework of Chapter 4 and it used to have an Appendix G for support, but that is now gone and replaced with old Appendix M. Nothing like a brand new and fatal bug to deal with after working 3+ months on this monster. Plan A 1 Plan B. 6 Plan C 8 Plan D 11 Plan E 14 Plan A. My expression for V(z) is this, V(z) = !Syntax Error, I q(2m)(z) * { !Syntax Error, Idx' dy' a1(x',y') [ s1m Km (jβs1) - s2m Km (jβs2)] - !Syntax Error, Idx' dy' a2(x',y') [ s1m Km (jβs1) - s2m Km (jβs2)] } (4.2.2) where si = | xi - x' | and x1 is a point on C1 and x2 a point on C2. I am trying to discuss the small β limit of this expression and trying to show that beyond the leading term, things are order β2. But as I examine things, this seems completely wrong. If I look at any of the "four terms" in the above, say for m > 0, I see this Km (jβs1) ≈ 2m-1(m-1)! z-m = 2m-1(m-1)! (jβs1)-m = 2m-1(m-1)! = 2m-1(m-1)! s1-m So the larger m is, the more this thing blows up in my face! It's as if the expansion I used is completely wrong. [ yup ] Let's back up a bit to this point, where I have one of my "four terms" : φ1(x,y,z) = !Syntax Error, I(1/n!) q(n)(z) !Syntax Error, Idx' dy' a1(x',y') !Syntax Error, Idz' (z'-z)n . (4.1.7) R = It seems quite clear that as β→0, the dz' integral blows up something like this !Syntax Error, Idz' (z'-z)n-1 and that of course is what my exact integration shows. !Syntax Error, Idz' (z'-z)n = sn fn(βs) = (G.10) My expansion of q(z) just makes things worse, bringing in those powers (z'-z)n. Is it possible that my Km(jβs) limit is wrong? I have to allow that β can be completely real in a good dielectric, so we really are going right ON the imaginary axis with the argument. According to green Jackson page 75, we can write this in terms of Hm(1)(-βs). Do I know anything about that function for small β ? It seems that Ym(-βs) will diverge. So we still have our power divergence for small z no matter how you write it. It must be that you should not do this dz' integral at this point in the calculation! King deals with this question on his page 16, and I thought I was cleverly generalizing his work. First of all, King only deals with 2 terms instead of my four terms, because he assumes "a balanced line". That is to say, he assumes φ2 = -φ1 so that then V = φ1-φ2 = 2φ1 as in his top page 15. Of course he is already "into" his simple example. Suppose I hold off on the expansion of q(z) so that, for some observation point x in the dielectric, φ1(x,y,z) = !Syntax Error, Idx' dy' a1(x',y') !Syntax Error, Idz' q1(z') . (4.1.5) φ2(x,y,z) = !Syntax Error, Idx' dy' a2(x',y') !Syntax Error, Idz' q2(z') . (4.1.5) Now assume only that q1 = q and q2 = -q1 = -q so we then have φ1(x,y,z) = !Syntax Error, Idx' dy' a1(x',y') !Syntax Error, Idz' q(z') . (4.1.5) φ2(x,y,z) = !Syntax Error, Idx' dy' a2(x',y') !Syntax Error, Idz' [-q(z')] . where R = |x-x'| but of course x' takes different values for the different integrals. I am really stuck at this point, I don't see how to do anything with the subtraction. Maybe slide dz' integration to the left, φ1(x,y,z) = !Syntax Error, Idz' q(z')!Syntax Error, Idx' dy' a1(x',y') . (4.1.5) φ2(x,y,z) = !Syntax Error, Idz' [-q(z')]!Syntax Error, Idx' dy' a2(x',y') . Maybe I can say something about the transverse integral in the transmission line limit where λ >> d. φ1(x,y,z) = !Syntax Error, Idz' q(z')!Syntax Error, Idx1' dy1' a1(x1',y1') . R12 = (x-x1')2 + (y-y1')2 + (z-z1')2 φ2(x,y,z) = !Syntax Error, Idz' [-q(z')]!Syntax Error, Idx2' dy2' a2(x2',y2') . R12 = (x-x2')2 + (y-y2')2 + (z-z2')2 Both integrations have a common z' coordinate, so no need for z1' and z2'. Now it is true that, in the TL limit, we will have βR1 << 1 and βR2 << 1 so we could do expansions: [ we can do these expansions even if βR is NOT << 1 because ∞ disk of convergence ] φ1(x,y,z) = !Syntax Error, Idz' q(z')!Syntax Error, Idx1' dy1' a1(x1',y1') R1-1 Σn=0∞ (-jβR1)n/n! φ2(x,y,z) = !Syntax Error, Idz' [-q(z')]!Syntax Error, Idx2' dy2' a2(x2',y2') R2-1 Σn=0∞ (-jβR2)n/n! . φ1(x,y,z) = !Syntax Error, Idz' q(z') Σn=0∞ (-jβ)n!Syntax Error, Idx1' dy1' a1(x1',y1') R1n-1 φ2(x,y,z) = - !Syntax Error, Idz' q(z') Σn=0∞ (-jβ)n!Syntax Error, Idx2' dy2' a2(x2',y2') R2n-1 Now I claim that, !Syntax Error, Idx1' dy1' a1(x1',y1') R1n-1 = f1n(x,y,z; z') which is a function only of geometry and certainly does not depend on β. What exactly does this integral represent? Here is a picture, where we are integrating the red dot around the red boundary for each φi If we regard a1(x1',y1') as a normalized weighting function, it is like a probability and then what we have is !Syntax Error, Idx1' dy1' a1(x1',y1') R1n-1 = <R1n-1> = expected value of R1n-1 = average value of R1n-1 You can see that if the two points have a very large z separation compared to other separations (that is, if the angle R1 makes is close to parallel with the conductor) then R12 = (x-x1')2 + (y-y1')2 + (z-z1')2 ≈ (z-z1')2 => R1 ≈ |z-z1| <R1n-1> ≈ |z-z1|n-1 So then we have φ1(x,y,z) = !Syntax Error, Idz' q(z') Σn=0∞ (-jβ)n f1n(x,y,z; z') φ2(x,y,z) = - !Syntax Error, Idz' q(z') Σn=0∞ (-jβ)n f2n(x,y,z; z') At least we seem to have a small-β expansion now. Then, φ12 = φ1-φ2 = !Syntax Error, Idz' q(z') Σn=0∞ (-jβ)n [ f1n(x,y,z; z') - f2n(x,y,z; z') ] We can write out some of the terms: φ12 = !Syntax Error, Idz' q(z') [ f10(x,y,z; z') - f20(x,y,z; z') ] + !Syntax Error, Idz' q(z') Σn=2∞ (-jβ)n[ f1n(x,y,z; z') - f2n(x,y,z; z') ] which has the form φ12 = !Syntax Error, Idz' q(z') [ f10(x,y,z; z') - f20(x,y,z; z') ] + order(β2) Write the first two non-vanishing terms only as φ12 = !Syntax Error, Idz' q(z') [ f10(x,y,z; z') - f20(x,y,z; z') ] + !Syntax Error, Idz' q(z') (1/2) (-jβ)2[ f12(x,y,z; z') - f22(x,y,z; z') ] Now look back at !Syntax Error, Idx1' dy1' a1(x1',y1') R1n-1 = f1n(x,y,z; z') For large z-z', we have roughly that R12 = (z-z1')2 so this becomes f1n(x,y,z; z') for large |z-z'| ≈ !Syntax Error, Idx1' dy1' a1(x1',y1') |z-z'|n-1 = |z-z'|n-1 f2n(x,y,z; z') for large |z-z'| ≈ !Syntax Error, Idx2' dy2' a1(x2',y2') |z-z'|n-1 = |z-z'|n-1 This really means we can write φ12 = φ1-φ2 = !Syntax Error, Idz' q(z') Σn=0∞ (-jβ)n [ f1n(x,y,z; z') - f2n(x,y,z; z') ] where perhaps Λ ≥ 100 d where d is the diameter of a tube which can contain both conductors. The main point is that each term in the n sum has a convergent dz' integral AND we have things in a power series expansion in β. The first two series terms are then φ12 = !Syntax Error, Idz' q(z') [ f10(x,y,z; z') - f20(x,y,z; z') ] - (1/2) β2 !Syntax Error, Idz' q(z') [ f12(x,y,z; z') - f22(x,y,z; z') ] Jan 4, 2014 Plan B. Let's go back to φ12 = φ1-φ2 = !Syntax Error, Idz' q(z') Σn=0∞ (-jβ)n [ f1n(x,y,z; z') - f2n(x,y,z; z') ] or φ12 = φ1-φ2 = !Syntax Error, Idz' q(z') Σn=0∞ (-jβ)n [<R1n-1> - <R2n-1> ] !Syntax Error, Idx1' dy1' a1(x1',y1') R1n-1 = f1n(x,y,z; z') = <R1n-1> R12 = (x-x1')2 + (y-y1')2 + (z-z')2 = s12 + (z-z')2 R22 = (x-x2')2 + (y-y2')2 + (z-z')2 = s22 + (z-z')2 and maybe do something with this difference function. s1 s2 Let's try an expansion for where |z-z'| is large. R12 = (z-z')2 + s12 = (z-z')2 [ 1 + s12/ (z-z')2] R1 = |z-z'| ( 1 + s12/ |z-z'|2)1/2 R1n-1 = |z-z'|n-1 ( 1 + s12/ |z-z'|2)(n-1)/2 // (1+x)n = 1n x + n 1n-1x + .... or R1n-1 ≈ |z-z'|n-1 ( 1 + ) R2n-1 ≈ |z-z'|n-1 ( 1 + ) Then write <R1n-1> - <R2n-1> = !Syntax Error, Idx1' dy1' a1(x1',y1') R1n-1 - !Syntax Error, Idx2' dy2' a2(x2',y2') R2n-1 = !Syntax Error, Idx1' dy1' a1(x1',y1') { |z-z'|n-1 ( 1 + ) } - !Syntax Error, Idx2' dy2' a2(x2',y2') { |z-z'|n-1 ( 1 + ) } Now the "1" contributions exactly cancel due to the normalization of a1 and a2 , so that is the main cancellation idea realized yesterday. We then have <R1n-1> - <R2n-1> ≈ !Syntax Error, Idx1' dy1' a1(x1',y1') { |z-z'|n-1 () } - !Syntax Error, Idx2' dy2' a2(x2',y2') { |z-z'|n-1 () } = |z-z'|n-3 { !Syntax Error, Idx1' dy1' a1(x1',y1') s12 - !Syntax Error, Idx2' dy2' a2(x2',y2') s22 } This appears to be a step in the right direction. I could write more terms in this expansion, but let's hold off on that for the moment. What is the meaning of the integrals appearing here? For a given point x in the dielectric, the first integral is a mean value of s12 from the line specified by x,y as z varies. So I have then shown that <R1n-1> - <R2n-1> ≈ |z-z'|n-3 [<s12> - <s22> ] Then φ12 = φ1-φ2 = !Syntax Error, Idz' q(z') Σn=0∞ (-jβ)n [<R1n-1> - <R2n-1> ] ≈ !Syntax Error, Idz' q(z') Σn=0∞ (-jβ)n |z-z'|n-3 [<s12> - <s22> ] ≈ [<s12> - <s22> ] Σn=0∞ (-jβ)n !Syntax Error, Idz' q(z')|z-z'|n-3 OK, at least I have done something. But for larger n, the dz' integral still diverges, so what went wrong? Plan C Well, suppose we "back up" even more like so: φ1(x,y,z) = !Syntax Error, Idz' q(z')!Syntax Error, Idx1' dy1' a1(x1',y1') . φ2(x,y,z) = !Syntax Error, Idz' [-q(z')]!Syntax Error, Idx2' dy2' a2(x2',y2') . And supposed to DON'T expand the exponential yet. Then φ12(x) = φ1(x) - φ2(x) = !Syntax Error, Idz' q(z'){ !Syntax Error, Idx1' dy1' a1(x1',y1') – !Syntax Error, Idx2' dy2' a2(x2',y2') } Now let's try an expansion of . As above we write, R1 = |z-z'| ( 1 +)1/2 = exp[ -jβ |z-z'| ( 1 +)1/2 ] / |z-z'| ( 1 +)1/2 ≈ exp[ -jβ |z-z'| ( 1 + ) ] / |z-z'| ( 1 + ) ≈ exp[ -jβ |z-z'| -jβ ) ] / |z-z'| ( 1 + ) ≈ exp[ -jβ |z-z'| ] exp[ -jβ ) ] / |z-z'| ( 1 + ) = * (*) If I do this as well for , then the "leading term" exactly cancels in φ12 , as happened in my last attempt. Let's go ahead and write it all out φ12 ≈ !Syntax Error, Idz' q(z') !Syntax Error, Idx1' dy1' a1(x1',y1') [ * ] - !Syntax Error, Idz' q(z') !Syntax Error, Idx2' dy2' a2(x2',y2') [ * ] or φ12 ≈ !Syntax Error, Idz' q(z') !Syntax Error, Idx1' dy1' a1(x1',y1') [] - !Syntax Error, Idz' q(z') !Syntax Error, Idx2' dy2' a2(x2',y2') [] I think there is a finite β = 0 term sitting here: φ12β=0 = !Syntax Error, Idz' q(z') { !Syntax Error, Idx1' dy1' a1(x1',y1') [] - !Syntax Error, Idz' q(z') !Syntax Error, Idx2' dy2' a2(x2',y2') [] or φ12β=0 ≈ !Syntax Error, Idz' q(z') { !Syntax Error, Idx1' dy1' a1(x1',y1') [1 - ] - !Syntax Error, Idz' q(z') !Syntax Error, Idx2' dy2' a2(x2',y2') [1 - ] The leading terms cancel due to ai normalization and we are left with φ12β=0 ≈ (- ) !Syntax Error, Idz' q(z') { <s12> - <s22> } = (- ) { <s12> - <s22> } !Syntax Error, Idz' q(z') Now suppose we put in a power series for q(z'), q(z') = q(z) + q'(z) (z'-z) + .. Then the leading term diverges at the end near z' = 0,' !Syntax Error, Idz' q(z) = q(z) !Syntax Error, Idz' = q(z) !Syntax Error, Idx = 2 q(z) !Syntax Error, Idx x-3 = diverges This I guess is due to my expansion for large |z-z'| being bad for small |z-z'|. Ignoring this for the moment: OK, now maybe is the time to do our exponential expansion exp[ -jβ ] = Σn=0∞ (1/n!) (jβ )n = 1 + Σn=1∞ (1/n!) (jβ )n But since I only expanded two terms in (*) above, the higher terms here are not too meaningful. So maybe only keep the first two terms: exp[ -jβ ] ≈ 1 + (1/2) jβ Then we get ≈ [1 + (1/2) jβ ] [1 - ] ≈ 1 + (jβ/4) + order() Then put pieces together again φ12 ≈ !Syntax Error, Idz' q(z') { !Syntax Error, Idx1' dy1' a1(x1',y1') [1 + (jβ/4)] - !Syntax Error, Idz' q(z') { !Syntax Error, Idx2' dy2' a2(x2',y2') [1 + (jβ/4)] As already noted, the leading terms cancel and we then have φ12 ≈ !Syntax Error, Idz' q(z') { !Syntax Error, Idx1' dy1' a1(x1',y1') [ (jβ/4)] - !Syntax Error, Idz' q(z') { !Syntax Error, Idx2' dy2' a2(x2',y2') [(jβ/4)] = (jβ/4) !Syntax Error, Idz' q(z') { <s12>t - <s22>t } or φ12(x,y,z) ≈ (jβ/4) { <s12>t - <s22>t }!Syntax Error, Idz' q(z') Now the dz' integral seems a little more convergent. But the leading term seems order β and not β2 ? What result am I eventually expecting? V(z) = q(z){ !Syntax Error, Idx' dy' a1(x',y') ln(s2/s1) -!Syntax Error, Idx' dy' a2(x',y') ln(s2/s1) } (4.2.5) This is of course another step of differencing, but it has no β dependence. Plan D Start again, φ12(x) = φ1(x) - φ2(x) = !Syntax Error, Idz' q(z'){ !Syntax Error, Idx1' dy1' a1(x1',y1') – !Syntax Error, Idx2' dy2' a2(x2',y2') } R12 = (x-x1')2 + (y-y1')2 + (z-z')2 = s12 + (z-z')2 R22 = (x-x2')2 + (y-y2')2 + (z-z')2 = s22 + (z-z')2 Again, for x somewhere in the middle, this thing has to be finite for any β. If a1 = a2 as in King's example, then you get the difference function ( - ) which is how he deals with things. His symmetry then allows consideration of only 2 terms instead of 4 terms for V(x). What happens if I start off with the expansion q(z') = q(z) + q'(z) (z'-z) + .. This will only be valid when z' is close to z, so why is this useful? One argument is this: when z' is far from z, contributions to the above integrals are small because the two Ri are large. So the main contribution comes from the region where z' is near z, and then this expansion is reasonable. then φ12(x) = φ1(x) - φ2(x) = q(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' a1(x1',y1') – !Syntax Error, Idx2' dy2' a2(x2',y2') } + q'(z) !Syntax Error, Idz' (z'-z) { !Syntax Error, Idx1' dy1' a1(x1',y1') – !Syntax Error, Idx2' dy2' a2(x2',y2') } The problem is then to show that this second line is small compared to the first line in the transmission line limit. But if I make the ansatz that q(z) = q(z0)e-jβz for z dependence then q'(z) = -jβq(z) and then we have φ12(x) = φ1(x) - φ2(x) = q(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' a1(x1',y1') – !Syntax Error, Idx2' dy2' a2(x2',y2') } -jβq(z)!Syntax Error, Idz' (z'-z) { !Syntax Error, Idx1' dy1' a1(x1',y1') – !Syntax Error, Idx2' dy2' a2(x2',y2') } and then the second term is down by a factor of β (at least, maybe it will be more) Idea: in this second term, each Ri is even in (z'-z). Thus, is even in (z'-z) as is . Thus, the entire integrand {...} is even in (z'-z), and thus the entire second integral vanishes exactly!! If we then do the next term in the power series q(z') = q(z) + q'(z)(z'-z) + (1/2)q"(z) (z'-z)2 + .... q(z) = q(z0)e-jβz q'(z) = (-jβ)q(z) q"(z) = (-jβ)2q(z) Then the next term will be (-jβ)2q(z) !Syntax Error, Idz' (z'-z)2 {!Syntax Error, Idx1' dy1' a1(x1',y1') – !Syntax Error, Idx2' dy2' a2(x2',y2') } and this term is then order β2 in agreement with King. So at this point, although pieces are not all there, we have shown that if we drop β2 and higher terms, we end up with φ12(x) = q(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' a1(x1',y1') – !Syntax Error, Idx2' dy2' a2(x2',y2') } So then I only have this thing to deal with, but of course I have to get to the four terms. V(z) = φ12(x1) - φ12(x2) = potential between conductors. When things are symmetric as in King's example, then we get V(z) = φ12(x1) - φ12(x2) = 2 φ12(x1) or V(z) = q(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' a1(x1',y1') – !Syntax Error, Idx2' dy2' a2(x2',y2') } where now R12 = (x1-x1')2 + (y1-y1')2 + (z-z')2 = s112 + (z-z')2 R22 = (x1-x2')2 + (y1-y2')2 + (z-z')2 = s122 + (z-z')2 Now I think if the wires are thin, and we take our touch points to be the nearest points, and we assume the average value for x1' is the wire center, then we get R12 ≈ a2 + (z-z')2 R22 ≈ (x1-x2)2 + (y1-y2)2 + (z-z')2 ≈ b2 + (z-z')2 where d is the separation of the thin wires. I think in this case, the transverse integration is 1 and we have V(z) = q(z) !Syntax Error, Idz' { – } Now the leading term with β = 1 would be V(z) = q(z) !Syntax Error, Idz' { – } = q(z) !Syntax Error, Idz' { - } = q(z) !Syntax Error, Idx { - } According to Maple, so this integral is !Syntax Error, Idz' { – } = !Syntax Error, Idx { - } = ln(b2/a2) = 2ln(b/a) in agreement with King, and then we get V(z) = q(z) 2ln(b/a) = q(z) ln(b/a) = Q/C' C' = πξ/ln(b/a) Taking the ε part we get C = πε/ln(b/a) // agrees with King (30b), and so on for R. So at least I see how the King thing works, vaguely. He makes a lot of approximations, I hoped to do better. Plan E I now assume that higher terms are β2 by the argument given in Plan D, and now I concentrate on evaluation of the main term. Here is my potential at an arbitrary central point, φ12(x) = φ1(x) - φ2(x) = q(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' a1(x1',y1') – !Syntax Error, Idx2' dy2' a2(x2',y2') } But if I am doing β = 0, this is really φ12(x) = φ1(x) - φ2(x) = q(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' a1(x1',y1') – !Syntax Error, Idx2' dy2' a2(x2',y2') } I am done with all expansion things, and I only have to compute this term. I think I should draw a picture that shows all the R's that will be involved. φ12(x1) = q(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' a1(x1',y1') – !Syntax Error, Idx2' dy2' a2(x2',y2') } R112 = (x1-x1')2 + (y1-y1')2 + (z-z')2 = s112 + (z-z')2 s112 = (x1-x1')2 + (y1-y1')2 R122 = (x1-x2')2 + (y1-y2')2 + (z-z')2 = s122 + (z-z')2 s122 = (x1-x2')2 + (y1-y2')2 φ12(x2) = q(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' a1(x1',y1') – !Syntax Error, Idx2' dy2' a2(x2',y2') } R222 = (x2-x2')2 + (y2-y2')2 + (z-z')2 = s222 + (z-z')2 s222 = (x2-x2')2 + (y2-y2')2 R212 = (x2-x1')2 + (y2-y1)2 + (z-z')2 = s212 + (z-z')2 s212 = (x2-x1')2 + (y2-y1)2 R222 = | r2 - r2'|2 Then finally we have V(z) = φ12(x1) - φ12(x2) = q(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' a1(x1',y1') – !Syntax Error, Idx2' dy2' a2(x2',y2') } - q(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' a1(x1',y1') – !Syntax Error, Idx2' dy2' a2(x2',y2') } = q(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' a1(x1',y1')( - ) - !Syntax Error, Idx2' dy2' a2(x2',y2') (- ) } Now what??? I could move the dz' integrations to the right. Then I need, for example, !Syntax Error, Idz' ( - ) = !Syntax Error, Idz' ( - ) But I think I know this integral! Recall from above that !Syntax Error, Idx { - } = ln(b2/a2) = 2ln(b/a) So then we should have !Syntax Error, Idz' ( - ) = !Syntax Error, Idz' ( - ) = 2ln(s21/s11) and then !Syntax Error, Idz' ( - ) = 2ln(s22/s12) and then I arrive here: V(z) = q(z) { !Syntax Error, Idx1' dy1' a1(x1',y1') ln(s21/s11) - !Syntax Error, Idx2' dy2' a2(x2',y2') ln(s22/s12) } I think then this is a more precise version of what I have in my lines doc. Here is a picture to go with the original equations = q(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' a1(x1',y1')( - ) - !Syntax Error, Idx2' dy2' a2(x2',y2') (- ) } The last equation has a different picture all in a constant z plane: V(z) = q(z) { !Syntax Error, Idx1' dy1' a1(x1',y1') ln(s21/s11) - !Syntax Error, Idx2' dy2' a2(x2',y2') ln(s22/s12) } s112 = (x1-x1')2 + (y1-y1')2 s212 = (x2-x1')2 + (y2-y1)2 s122 = (x1-x2')2 + (y1-y2')2 s222 = (x2-x2')2 + (y2-y2')2 Let's now try the two circles: STOP! On the next line, I assume incorrectly that the surface charge is uniform around the round wires, but that is of course only valid for d >> b1, b2 so THAT is my error!! That is why the result I get below disagrees with King, because it makes a false assumption right at the start So that above equation is valid in the sense that, if you KNEW a1(x1',y1') and a2(x1',y1'), you then have a result. So what good is this whole approach since we DONT know these things except in the large separation case? V(z) = q(z) { !Syntax Error, Idx1' dy1' δ(r - b1)/(2πb1) ln(s21/s11) - !Syntax Error, Idx2' dy2' δ(r - b2)/(2πb2) ln(s22/s12) } = q(z) { !Syntax Error, I dr1' !Syntax Error, Idθ'1 δ(r1' - b1) ln(s21/s11) - !Syntax Error, I dr2' !Syntax Error, Idθ'2 δ(r2' - b2) ln(s22/s12) } = q(z) { !Syntax Error, Idθ'1 ln(s21/s11)|r1'=b1 - !Syntax Error, Idθ'2 ln(s22/s12)| r2'=b2 } Now s112 = (x1-x1')2 + (y1-y1')2 = b12 + b12 - 2b1b1cos(θ1) = 2b12 [ 1-cos(θ1)] s122 = (x1-x2')2 + (y1-y2')2 = (d+b2)2 + b22 - 2(d+b2) b2cos(π-θ2) s212 = (x2-x1')2 + (y2-y1)2 = (d+b1)2 + b12 - 2(d+b1) b1cos(θ1) s222 = (x2-x2')2 + (y2-y2')2 = 2b22( 1+cos(θ1)) seem OK Inserting these, we get V(z) = q(z) { !Syntax Error, Idθ'1 ln(s21/s11)|r1'=b1 - !Syntax Error, Idθ'2 ln(s22/s12)| r2'=b2 } = q(z) (1/2) * (T1 + T2 + T3 + T4) T1 = + !Syntax Error, Idθ'1 ln(s212) = !Syntax Error, Idθ'1 ln [(d+b1)2 + b12 - 2(d+b1) b1cos(θ1')] T2 = - !Syntax Error, Idθ'1 ln(s112) = - !Syntax Error, Idθ'1 ln [2b12 ( 1-cos(θ1'))] T3 = - !Syntax Error, Idθ'2 ln(s222) = - !Syntax Error, Idθ'2 ln [2b22( 1+cos(θ2')) ] T4 = + !Syntax Error, Idθ'2 ln(s122) = !Syntax Error, Idθ'2 ln[(d+b2)2 + b22 + 2(d+b2) b2cos(θ2')] This seems more complicated than the example I did in lines doc. Well, I allow arbitrary b1, b2 and d in this example, whereas I did not in the text. now I need to know this integral !Syntax Error, Idx ln (A + Bcosx) = 2π ln[(1/2)(A + )] following from GF7 p531 Let's assume the condition is met for all four terms, check this later. Then T1: A = (d+b1)2 + b12 B = - 2(d+b1) b1 A2-B2 = [(d+b1)2 + b12]2 - 4(d+b1)2 b12 = (d+b1)4 + b14 - 2(d+b1)2 b12 = [(d+b1)2 - b12]2 Then T1 = 2πln[(1/2)( (d+b1)2 + b12) + (d+b1)2 - b12)] = 2π ln[(d+b1)2] = 4π ln(d+b1) _________________ T2: A = 2b12 B = -2b12 A2-B2 = 0 Then T2 = - 2πln[b12] = - 4π ln(b1) _______________________ T3: A = 2b22 B = 2b22 A2-B2 = 0 Then T3 = - 2π ln[b22] = -4π ln(b2) _____________________ T4: A = (d+b2)2 + b22 B = 2(d+b2) b2 T4 = + 4π ln(d+b2) _____________________ To summarize: T1 = 4π ln(d+b1) T2 = -4π ln(b1) T3 = -4π ln(b2) T4 = 4π ln(d+b2) Then we get V(z) = q(z) (1/2) * (T1 + T2 + T3 + T4) = q(z) (1/2)* (ln(d+b1) + ln(d+b2) - ln(b1) - ln(b2) = q(z) ln [] d = inner touch distance bi = radii For very large d this says V(z) = q(z) ln [] If we change to diameters instead of radius this becomes V(z) = q(z) ln [] The capacitance is then 1/C = ln [] How can I verify this result? It says that C = 2πε/ ln [] Suppose we had equal radii. Then we would have C = 2πε/ ln [] = 2πε/ ln [] = πε / ln(d/b) and this does agree with King p 17 (30b). So at least I have a verification for equal radii which means my overall constant is right. Now go back to my more general result: V(z) = q(z) ln [] Let's change symbols on my result by using B = distance between centers = d + b1 + b2 d + b1 = B-b2 d + b2 = B - b1 Then my result becomes V(z) = q(z) ln [] The capacitance is then 1/c = ln [] c = 2πε 1/ ln [] This agrees with King p 28 if I can show that: (ch-1ψ1 + ch-1ψ2) = ln [] He has, converted to my notation, ψ1 = (B2+b22 -b12)/(2Bb1) ψ2 = (B2+b12 -b22)/(2Bb2) Let's try some numbers B = 9 b1 = 2 b2 = 3 d = 4 Result: My result disagrees with King! No need to go any further. A third source is that I do this problem myself later in lines where I find that K = f(x1) - f(x2) = B1 - B2 = ln [(d/a1) + ] + ln [(d/a2) + ] where ai are the radii and 2d is the distance between the foci. My drawing is very unclear. I am doing the potential 2D method, so my result should be same as King's. My b is distance between centers. I claim that d = (1/2b) (6.3.7) But things then look pretty messy for the above K. But my result has no verification, so it is just another floating calculation, not useful to me right now! Matick has a result on page 317 for two equal radii with S as the center separation. So let's compare the three results for that case: King: ψ1 = ψ2 = B/2b so then [ ψ1 + ] [ ψ2 + ] = [ ψ1 + ]2 = ψ12 + ψ12- 1 + 2 ψ1 = 2ψ12 - 1 + 2 ψ1 = 2(B/2b)2 - 1 + 2(B/2b) So King's result is then related to ln [2(B/2b)2 - 1 + 2(B/2b) ] Me: ln [] = ln [] =2 ln [] Matick: ln So we have three different results! I will put in my numbers B = 9 and b = 2, and this shows that King and Matick I think agree, because I find numerically that ()2 = 2(B/2b)2 - 1 + 2(B/2b) which is the same analytically as [(B/2b) + ]2 = 2(B/2b)2 - 1 + 2(B/2b) I predict LHS = RHS. Eval LHS = (B/2b)2 + (B/2b)2 - 1 + 2(B/2b) = 2 (B/2b)2 - 1 + 2(B/2b) // yes So I think King and Mat are compatible. It is MY result that is clearly wrong, it is "too simple". Then (ch-1ψ1 + ch-1ψ2) = ln [ ψ1 + ] + ln [ ψ2 + ] = ln { [ ψ1 + ] [ ψ2 + ] } So what I want to show is this: ln { [ ψ1 + ] [ ψ2 + ] } = ln [] which is to say, I want to show that [ ψ1 + ] [ ψ2 + ] = But numerically I have already shown above that this does not work! MY result after my huge calculation is wrong somehow. There will be many things to check, Manana! ___________________________________________________________________________ The capacitance result is not stated in my Radio Eng book, but page 592 shows a result for Z which has some relevance. Meanwhile, King does this problem and gets a result on page 28 which indicates 1/c = (1/2πε) [cosh-1 + cosh-1] where D = distance between centers = d + b1+b2 . Theorem: cosh-1x = ln[x + ] ≡ ln(y) Proof: Show that cosh(ln[x + ) = x Evaluate the LHS : cosh(ln[x + ) = cosh(ln(y)) = (1/2) [ elny + e-lny] = (1/2) [y +1/y] = (1/2) [x + + x - ] = x QED This is Spiegel 8.56, duh huh Now I just showed the following on scratch paper: y = ch-1ψ1 + ch-1ψ2 => y = ln [ ψ1ψ2 + ψ2 + ψ1 + ] so at least I can make some connection to logs. Just out of curiosity, ψ1 = ψ12 - 1 = [(D2-b22+b12)2 - (2b1D)2 ] / (2b1D)2 Consider (x2-y2+z2)2 - 4 z2x2 Then (x2-y2+z2)2 - 4 z2x2 = [ (x+y)2 - z2] [ (x-y)2 - z2] Thus suggests that we get this not-very-simple expression, ψ12 - 1 = [ (D+b2)2 - b12] [ (D-b2)2 - b12] / (2b1D)2 = / (2b1D) Then I guess = / (2b2D) ψ1ψ2 + ψ2 + ψ1 + = (ψ1+)( ψ2+ ) ψ1+ = + / (2b1D) = [ D2-b22+b12 + ] / (2b1D) Does this simplify? Web Scan. It is not easy to find what I want. I search on "unequal diameters" and get a hit http://www.rfcafe.com/references/electrical/transmission-lines.htm This seems different from King's result. Why don't I say anything about "impedance" of a line? I just added some notes above the start of lines 4.6 and concluded there that at large ω Z0 = 1/C For my two round wire line with large spacing I get C = πε / ln ( b/) (4.6.5) That would imply that Z0 = (1/πε) ln ( b/) = (1/π) ln ( b/) Now assume μ = μ0 and ε = (ε/ε0)ε0 so then Z0 = (1/π) ln ( b/) = (1/π) ln ( b/) = (1/π) ln ( b/) = (1/π) 376.73032 ohms * ln ( b/) = 119.9169852 ln ( b/) = 119.9169852 ln ( 2b/) ≈ 120 ln ( 2b/) (**) Now y = lnx ey = elnx = x z = logx 10z = 10logx = x 10z = ey z = log(ey) = ylog(e) y = ln(ey) = ln(10z) = z ln(10) logx = log(e)lnx lnx = ln(10)logx log(e) = 0.434294482 ln(10) = 2.302585093 These are inverses of each other. It is hard to find this simple fact on the web! lnx = 2.302585093 logx // write this down somewhere for next time. Continuing the above Z0 = (1/π)* 376.73032 ohms * 2.302585093 log10 ( b/) = 276.1190626 log10 ( b/) Here ai are radius values. Write a1 = d1/2 to get = /2 and then I get Z0 =276.1190626 ohms log10 ( 2b/) This agrees with radio book page 592! Finally something agrees with something! That book clearly states that it is assuming b is >> the radii. Now how does this fit into the picture, For large D, the claim is that N = 2D2/d1d2 and then they claim that Z0 = 60 cosh-1(2D2/d1d2) Now consider y = cosh-1(x) x = cosh y x = large and y = large x = (1/2)[ ey + e-y] ≈ (1/2)ey ln(2x) = y y = ln(2x) So their result becomes Z0 = 60 ln(4D2/d1d2) = 120 ln( 2D/) This agrees with my (**) above, so now we have two agreements.