3. Doing the Second Inversion

Doing the Second Inversion PhL 6.15.10 See top level doc for an overview of what is going on in the 33 pages below! It is just amazing how complicated it all is, but it got me to the right Problem 38 σ answer, along with a potential Φ to boot. I later had good evidence that the Φ answer is correct as well, when I discovered the simpler disk-to-disk single inversion way to solve this problem. Overview: This is just a very long, pure nuts and bolts, grind it out calculation. I give a detailed overview in the top level review doc. That overview by itself is 11 pages long, more of a "meta" doc. If you want a quick overview, just read every line of the contents below. 1. Preliminaries. 1 2. Do the second inversion, formally. 2 3. Evaluation of σiris(r') 2 4. Does the z' = d condition have anything to say? 7 5. Continue processing the b2 - s2 nightmare. 8 6. Install b2- s2 result and compare with Smythe 10 7. Do the second inversion of the potential and process it. 11 (1) Starting Point. 11 (2) Major Change In Notation. 12 (3) Jackson's potential inversion rule 13 (4) Express in iris coordinates. 14 (5) Convert to iris center coordinates. 14 (6) Convert to cylindrical iris coordinates 15 (7) It will probably help to move some factors around: 15 (8) Correct for charge size. 16 (9) Combine the two terms inside {...} 16 (10) Get rid of A. 16 (11) Process the T factor: 17 (12) Process the s variable. 19 (13) Using the ellipse theorem and get final result for Φ. 20 (14) Show that Φ = 0 on the iris metal 24 (15) Can we perhaps use the ellipse theorem a second time ? 25 (16) Compute the charge density? 27 (a) Review of input information and strategy: 27 (b) Compute facts about n. 28 (c) Compute facts about m. 29 (d) Compute facts about m+n 29 (e) Compute ∂ 29 (f) Compute . 30 (g) Compute ∂Zi Φiris(r')|Zi=0 31 1. Preliminaries. Here is where we stand at this point: (1) I have done the first inversion in the old documents, to get σ and Φ for the bowl from the disk. (2) In doc "How to rotate the bowl", I have derived expressions for σpost and Φpost for the post-rotated bowl. (3) In doc " Parameter facts for Pic 1 and Pic 2.doc" I have, from scratch, written down the various geometric and inversion facts for our two "pictures". 2. Do the second inversion, formally. We shall now do "the second inversion" right here. First, here is how we left things after the first inversion and after the subsequent bowl rotation (from "how to rotate"): σpost(r") = σ(r) = – (a2/π2R3) / d = a2/2A R = 2A[sinθc sinθ"cosφ" - cosθc cosθ" + 1] s = x' = (a2A/R2) [-cosθc sinθ"cosφ" - sinθc cosθ"] y' = (a2A/R2) [sinθ"sinφ" z' = (a2A/R2) [sinθc sinθ"cosφ" - cosθc cosθ" + 1] // = d Let's forget the potential for right now and concentrate on σ. Here is the basic second inversion claim: σ'(r') = (R"/a)3 σpost(r") // see Jackson META page 8 where we are of course in Pic 2 now. Of course the point r" on the sphere corresponds to point r' on the iris. From Pic 2 fact (d) we have R" = 2A cos(θ"/2) So here is the charge density on the iris: σiris(r') = (R"/a)3 σpost(r") = (2A/a)3 cos3(θ"/2) σpost(r) = – (2A/a)3 cos3(θ"/2) (a2/π2R3) / = – (8A3/π2aR3) cos3(θ"/2) / dim = 1/L2 where, of course, s and R are given by the complicated expressions above. 3. Evaluation of σiris(r') We have σiris(r') = – (8A3/π2aR3) cos3(θ"/2) / with R = 2A[sinθc sinθ"cosφ" - cosθc cosθ" + 1] s = x' = (a2A/R2) [-cosθc sinθ"cosφ" - sinθc cosθ"] y' = (a2A/R2) [sinθ"sinφ"] We want to get rid of angle functions in favor of iris parameters! To this end, I call upon these facts: tan(θc/2) = S/d cos(θc/2) = d/ sin(θc/2) = S/ sinθc = 2dS/(d2+S2) cosθc = (d2- S2)/(d2+S2) tan(θ"/2) = (ρ'/d) cos(θ"/2) = d/ sin(θ"/2) = ρ'/ sinθ" = 2d ρ'/(d2+ ρ'2) cosθ" = (d2- ρ'2)/(d2+ ρ'2) B = d cot(θb/2) r12 = S2 + ρ'2 + 2Sρ' cosφ" So let's now try blindly "working on" some of our angle functions. (sinθ"sinφ") = [2d ρ'/(d2+ ρ'2)] sinφ" Next, (cosθcsinθ"cosφ" + sinθccosθ") = (d2- S2)/(d2+S2)* 2d ρ'/(d2+ ρ'2) * cosφ" + 2dS/(d2+S2)* (d2- ρ'2)/(d2+ ρ'2) = [(d2- S2) 2d ρ' cosφ" + 2dS (d2- ρ'2)] / [(d2+S2) (d2+ ρ'2)] = 2d [(d2- S2) ρ' cosφ" + S (d2- ρ'2)] / [(d2+S2) (d2+ ρ'2)] Next, [ 1 - cosθccosθ" + sinθcsinθ" cosφ" ] = 1 - (d2- S2)/(d2+S2)*(d2- ρ'2)/(d2+ ρ'2) + 2dS/(d2+S2) 2d ρ'/(d2+ ρ'2) cosφ" = 1 - [(d2- S2) (d2- ρ'2) - 2dS2d ρ' cosφ"] / [(d2+S2) (d2+ ρ'2)] = { (d2+S2) (d2+ ρ'2) - (d2- S2) (d2- ρ'2) + 2dS2d ρ' cosφ"}/ [(d2+S2) (d2+ ρ'2)] Now as I have done several times before, we go off and find that (d2+S2) (d2+ ρ'2) - (d2- S2) (d2- ρ'2) = (d4 + d2S2 + d2ρ'2 + S2ρ'2) - (d4 - d2S2 - d2ρ'2 + S2ρ'2) = (d2S2 + d2ρ'2) - (- d2S2 - d2ρ'2) = 2(d2S2 + d2ρ'2) = 2d2(S2+ρ'2) So we then have [] = { 2d2(S2+ρ'2) + 2dS2d ρ' cosφ"}/ [(d2+S2) (d2+ ρ'2)] = { 2d2(S2+ρ'2) + 4d2S ρ' cosφ"}/ [(d2+S2) (d2+ ρ'2)] = 2d2 { (S2+ρ'2) + 2S ρ' cosφ"}/ [(d2+S2) (d2+ ρ'2)] = 2d2 { S2+ρ'2 + 2S ρ' cosφ"}/ [(d2+S2) (d2+ ρ'2)] But happily from above we have r12 = S2 + ρ'2 + 2Sρ' cosφ" So we then end up with [ 1 - cosθccosθ" + sinθcsinθ"cosφ" ] = 2d2 r12 / [(d2+S2) (d2+ ρ'2)] which I think is a key result. This also tells us that R2 = 2A2[ 1 - cosθccosθ" + sinθcsinθ"cosφ" ] = 4A2 d2 r12 / [(d2+S2) (d2+ ρ'2)] = (2Adr1)2 / [(d2+S2) (d2+ ρ'2)] R = 2Adr1/ // agrees with previous work in Attempt #2A Here then is a summary of our angle functions (with a common denom) and R: (cosθc sinθ"cosφ" + sinθc cosθ") = 2d [(d2- S2) ρ' cosφ" + S (d2- ρ'2)] / [(d2+S2) (d2+ ρ'2)] (sinθ"sinφ) = [2d ρ'/(d2+ ρ'2)] sinφ" = [2d ρ' (d2+S2) sinφ" ] / [(d2+S2) (d2+ ρ'2)] (1 - cosθc cosθ" + sinθc sinθ"cosφ" ) = 2d2 r12 / [(d2+S2) (d2+ ρ'2)] R2 = (2Adr1)2 / [(d2+S2) (d2+ ρ'2)] We continue now, recalling from above, s = x' = – (Aa2/R2) (cosθc sinθ"cosφ" + sinθc cosθ") y' = (Aa2/R2) (sinθ"sinφ") Notice that (Aa2/R2) = (Aa2/ (2Adr1)2) [(d2+S2) (d2+ ρ'2)] = (Aa2/4A2d2r12) [(d2+S2) (d2+ ρ'2)] = (2dA/4Ad2r12) [(d2+S2) (d2+ ρ'2)] = (1/2dr12) [(d2+S2) (d2+ ρ'2)] = (1/2d) (1/r12) [(d2+S2) (d2+ ρ'2)] Thus we have x' = – (Aa2/R2) (cosθc sinθ"cosφ" + sinθc cosθ") = – (1/2d) (1/r12) [(d2+S2) (d2+ ρ'2)]) * 2d [(d2- S2) ρ' cosφ" + S (d2- ρ'2)] / [(d2+S2) (d2+ ρ'2)] = – (1/2d) (1/r12) 2d [(d2- S2) ρ' cosφ" + S (d2- ρ'2)] = – (1/r12) [(d2- S2) ρ' cos φ" + S (d2- ρ'2)] // not too bad Next we do y' = (Aa2/R2) (sinθ"sinφ") = (1/2d) (1/r12) [(d2+S2) (d2+ ρ'2)] * [2d ρ' (d2+S2) sinφ" ] / [(d2+S2) (d2+ ρ'2)] = (1/2d) (1/r12) [2d ρ' (d2+S2) sinφ" ] = (1/r12) ρ' (d2+S2) sinφ" Now I need something nice for c. But we have something, which is this (param facts Pic 2) b = B (d2+ S2) / (B2- S2) c = S (B2+ d2) / (B2- S2) and above x' = – (1/r12) [ρ' (d2- S2) cosφ" + S (d2- ρ'2)] y' = (1/r12) [ρ' (d2+S2) sinφ" ] So now we can insert things to obtain: s = s2 = (x'-c)2 + y'2 Let's remember now what we are trying to do ! We want this: σiris(r') = – (8A3/π2aR3) cos3(θ"/2) / = - (2A/R)3 (1/π2a) cos3(θ"/2) / and we have something nice for R, and we have been "working on" the s2 nightmare. We can simplify the above by inserting R and cos(θ"/2) as follows cos(θ"/2) = d/ R = 2Adr1/ => (2A/R) = / (dr1) So we have (2A/R)3 = (d2+S2)3/2 (d2+ ρ'2)3/2 / d3r13 Then we get σiris(r') = – (d2+S2)3/2 (d2+ ρ'2)3/2 (1/ π2a d3r13) cos3(θ"/2) / = – (d2+S2)3/2 (d2+ ρ'2)3/2 (1/ π2a d3r13) d3 (ρ'2+ d2)-3/2 / = – (d2+S2)3/2 (1/ π2a r13) / We are now down to this endgame situation: σiris(r') = – (d2+S2)3/2 (1/ π2a r13) / // qhole = / a s2 = (x'-c)2 + y'2 x' = – (1/r12) [ρ' (d2- S2) cosφ" + S (d2- ρ'2)] y' = (1/r12) [ρ' (d2+S2) sinφ" ] b = B (d2+ S2) / (B2- S2) c = S (B2+ d2) / (B2- S2) Admittedly, this is for the situation in which qhole = (/ a) If we wanted the result for some arbitrary charge q1 in the hole, we would divide by qhole to get our unit charge result, then multiply by q1 . Then we get σiris(r') = – q1 (a/) (d2+S2)3/2 (1/ π2a r13) / = – (q1/ π2 r13) (d2+S2) / // dim = Q/L2 correct Before going on, we can take a look at Attempt #2's interpretation of Smythe's claim for this result σSmythe = – ( q1/π2r12)/ where I have doubled Smythe's result to get sum of charges on top and bottom of iris since this is the way I started things on the disk with Jackson. These results are enticingly similar. Smythe's result has the φ" dependence all in the r12 factor. To show our two results agree, I would have to show that (d2+S2) / = r1 / (d2+S2)2 / (b2-s2) = r12 (B2- S2)/ (ρ'2- B2) (b2-s2) = (d2+S2)2 (ρ'2- B2) / [r12 (B2- S2)] with the cosφ" appearing only in that r1 factor. 4. Does the z' = d condition have anything to say? Although we did not need to use it, we have this condition floating around in our σ analysis, you see it at the start of this doc, z' = d = (a2/R2)(Z^+A) = (Aa2/R2) (sinθc sinθ"cosφ" - cosθc cosθ" + 1) which I write as (Aa2/R2) (sinθc sinθ"cosφ" - cosθc cosθ" + 1) = d LHS = RHS Let's just evaluate LHS and see what happens. We will use our result from above. I just treat this like another angular factor: (sinθc sinθ"cosφ" - cosθc cosθ" + 1) = =2dS/(d2+S2)* 2d ρ'/(d2+ ρ'2) * cosφ" - (d2- S2)/(d2+S2)* (d2- ρ'2)/(d2+ ρ'2) + 1 = 1 + [ 2dS 2d ρ' * cosφ" - (d2- S2) (d2- ρ'2)] / [(d2+S2) (d2+ ρ'2)] = {(d2+S2) (d2+ ρ'2) + 2dS 2d ρ' * cosφ" - (d2- S2) (d2- ρ'2) } / [(d2+S2) (d2+ ρ'2)] The two factors without the angle are the same as I dealt with above, where I found, (d2+S2) (d2+ ρ'2) - (d2- S2) (d2- ρ'2) = 2d2(S2+ρ'2) So we then have (sinθc sinθ"cosφ" - cosθc cosθ" + 1) = {2d2(S2+ρ'2) + 2dS 2d ρ' * cosφ" } / [(d2+S2) (d2+ ρ'2)] = {2d2(S2+ρ'2) + 4d2S ρ' * cosφ" } / [(d2+S2) (d2+ ρ'2)] = 2d2{S2+ρ'2 + 2Sρ' cosφ" } / [(d2+S2) (d2+ ρ'2)] = 2d2r12 / [(d2+S2) (d2+ ρ'2)] So we are now left with this LHS (we expect this to come out being d) LHS = (Aa2/R2) (sinθc sinθ"cosφ" - cosθc cosθ" + 1) R2 = 4A2d2r12 / [(d2+S2) (d2+ ρ'2)] => LHS = (Aa2/4A2d2r12) [(d2+S2) (d2+ ρ'2)] * 2d2r12 / [(d2+S2) (d2+ ρ'2)] = (Aa2/4A2d2r12) * 2d2r12 = (a2/2A) = d So this was just an algebra check and it came out right. 5. Continue processing the b2 - s2 nightmare. Above I arrived at this situation s2 = (x'-c)2 + y'2 = (c-x')2 + y'2 x' = – (1/r12) [ρ' (d2- S2) cosφ" + S (d2- ρ'2)] y' = (1/r12) [ρ' (d2+S2) sinφ" ] b = B (d2+ S2) / (B2- S2) c = S (B2+ d2) / (B2- S2) If the two trig terms had the same factor, at least sinφ" would go away and I would at least end up with s2 being a function only of φ" which I physically know must be true (and of course it is true in Smythe's answer). So I guess I will try to hunt down a possible sign error here! // I cannot find a sign error, so just proceed anyway: s2 = {c-x'}2 + y'2 = { S (B2+ d2) / (B2- S2) + (1/r12) [ρ' (d2- S2) cosφ" + S (d2- ρ'2)] }2 + (1/r14) ρ'2 (d2+S2)2 sin2φ" = (1/r14) [ { r12 S (B2+ d2) / (B2- S2) + [ρ' (d2- S2) cosφ" + S (d2- ρ'2)] }2 + ρ'2 (d2+S2)2 sin2φ" ] = (1/[r14(B2- S2)2]) * [ { r12 S (B2+ d2) + (B2- S2) [ρ' (d2- S2) cosφ" + S (d2- ρ'2)] }2 + (B2- S2)2 ρ'2 (d2+S2)2 sin2φ" ] = (1/[r14(B2- S2)2] ) * ( f12 + f2) This evaluation is a job for Maple. We also have r12 = S2 + ρ'2 + 2Sρ' cosφ" cosφ" = ( r12 - S2 - ρ'2 )/2Sρ' b2- s2 = b2 - (1/[r14(B2- S2)2] * f = (b2 [r14(B2- S2)2] - f) /[r14(B2- S2)2] But b2(B2- S2)2 = B2 (d2+ S2)2 / (B2- S2)2 *(B2- S2)2 = B2 (d2+ S2)2 so we have b2- s2 = (B2 (d2+ S2)2 r14 - f) / /[r14(B2- S2)2] = num/den I entered this in Maple and got these results: ("inversion Smythe algebra.mws" This, my result is this: b2 - s2 =(B2- ρ'2) (d2+ S2)2 / [ r12(S2- B2)] = [ (d2 + S2)2/ r12 ] (ρ'2- B2) / (B2-S2) which is looking very good!!! 6. Install b2- s2 result and compare with Smythe My result from above was this σiris(r') = – q1 (a/) (d2+S2)3/2 (1/ π2a r13) / = – (q1/ π2 r13) (d2+S2) / // dim = Q/L2 correct and I now install my new finding b2 - s2 = [ (d2 + S2)2/ r12 ] (ρ'2- B2) / (B2-S2) = [ (d2 + S2)/ r1 ] / 1/ = r1 / [(d2 + S2) ] which then gives my final result σiris(r') = – (q1/ π2 r13) (d2+S2) / = – (q1/ π2 r13) (d2+S2) r1 / [(d2 + S2) ] = – (q1/ π2 r12) / The Smythe result quoting from above is this: σSmythe = – ( q1/π2r12)/ FINALLY I have derived Smythe's result. Take note: 7:30 PM June 14, 2010. At some point I will report just how LONG I have been working on getting his result! His result is correct for the charge on one of the surfaces as in his book, and I have doubled his result to match my Jackson. For the first time ever I believe Smythe's result and I have validated the inversion/rotation/inversion method of deriving it from the σ on a charged isolated disk. Now that I know it is right, I can try to understand how to do it without all the messy algebra that only Maple could figure out! 7. Do the second inversion of the potential and process it. Now let's try to write the corresponding iris potential. (1) Starting Point. The potential after rotation and before second inversion is given by this (from how to rotate bowl) Φpost(r") = Φ(r) = g(r|ξ) = – (2/πQ) sin-1 [ 2b / ( + ) ] + 1/Q σpost(r") = σ(r) = – (a2/π2Q3) / Q = s = u = (a2/Q2)[ -cosθcx" - sinθc(z"-A)] v = (a2/Q2)y" w = (a2/Q2)( sinθcx" - cosθc(z"-A) +A) where r" = (x",y",z") are Cartesians based at the inversion origin. We are now evaluating at a general point r", so it is no longer on the sphere, and the corresponding point r' is not on the iris. Here is a picture: (2) Major Change In Notation. I now want to "do" the second inversion for the potential using the canonical r and r' vector notation. So I will now replace r" by r, adjusting all our expressions and creating a new picture. Φpost(r) = – (2/πQ) sin-1 [ 2b / ( + ) ] + 1/Q σpost(r) = – (a2/π2Q3) / Q = s = u = (a2/Q2)[ -cosθcx - sinθc(z-A)] v = (a2/Q2)y w = (a2/Q2)( sinθcx - cosθc(z-A) +A) where r = (x,y,z) are inversion origin Cartesians. So we now carry on with the second inversion! (3) Jackson's potential inversion rule is this ( I select one of the rules) ( look at Pic 2) φ'(r') = (a/r') φ(r) which we apply in our situation to get Φiris(r') = (a/r') Φpost(r) so we get Φiris(r') = (a/r') { – (2/πQ) sin-1 [ 2b / ( + ) ] + 1/Q } Q = Q2 = (x2+ y2+ (z-A)2 + 2A[sinθcx - cosθc(z-A)] + A2 s = u = (a2/Q2)[ -cosθcx - sinθc(z-A)] v = (a2/Q2)y w = (a2/Q2)( sinθcx - cosθc(z-A) +A) (4) Express in iris coordinates. But naturally we want to see our result in terms of the coordinates of point r', not r. We know that r = (a/r')2r' x = (a/r')2x' y = (a/r')2y' z = (a/r')2z' so we can install these things to get Φiris(r') = (a/r') {– (2/πQ) sin-1 [ 2b / ( + ) ] + 1/Q } Q2 = ((a/r')4 x'2+ (a/r')4 y'2+ ((a/r')2z'-A)2 + 2A[sinθc(a/r')2x' - cosθc((a/r')2z' - A)] + A2 s = u = (a2/Q2)[ -cosθc(a/r')2x' - sinθc((a/r')2z'-A)] v = (a2/Q2) (a/r')2y' w = (a2/Q2)( sinθc(a/r')2x' - cosθc[(a/r')2z'-A] +A) (5) Convert to iris center coordinates. But we would rather have our Cartesian coordinates centered on the iris center, so define x' = Xi r'2 = Xi2 + Yi2 + (Zi+d)2 = ρ'2 + (Zi+d)2 y' = Yi z' = (Zi+d) // If Zi = 0 for on iris plane, then z' = d and we write it again Sam, Φiris(r') = (a/r') { – (2/πQ) sin-1 [ 2b / ( + ) ] + 1/Q } Q2 = ((a/r')4 Xi 2+ (a/r')4 Yi2+ ((a/r')2(Zi+d)-A)2 + 2A[sinθc(a/r')2 Xi - cosθc((a/r')2(Zi+d) -A)] +A2 s = u = (a2/Q2)[ -cosθc(a/r')2 Xi - sinθc((a/r')2(Zi+d)-A)] v = (a2/Q2) (a/r')2 Yi w = (a2/Q2)( sinθc(a/r')2 Xi - cosθc[(a/r')2(Zi+d)-A] +A) Although it is pretty messy, this gives our iris potential in terms of iris coordinates! (6) Convert to cylindrical iris coordinates with the strange history names r' = (ρ',Zi,φ). This used to be φ", notice that both r and r' have the same φ. then we have' Xi = ρ'cosφ Yi = ρ'sinφ Zi = Zi So write it again Sam #2: Φiris(r') = (a/r') { – (2/πQ) sin-1 [ 2b / ( + ) ] + 1/Q } Q2 = ((a/r')4 ρ'2+ ( ((a/r')2(Zi+d)-A)2 + 2A[sinθc(a/r')2 ρ'cosφ - cosθc((a/r')2(Zi+d) -A)] + A2 s = r'2 = ρ'2 + (Zi+d)2 u = (a2/Q2)[ -cosθc(a/r')2 ρ'cosφ - sinθc( (a/r')2(Zi+d) - A)] v = (a2/Q2) (a/r')2 ρ'sinφ w = (a2/Q2){ sinθc(a/r')2 ρ'cosφ - cosθc[ (a/r')2(Zi+d) - A] +A} (7) It will probably help to move some factors around: Q2 = (a/r')4 ( ρ'2 + ( Zi+d-A(r'/a)2 )2 + 2A (r'/a)2 { sinθcρ'cosφ - cosθc [ Zi+d -A(r'/a)2] } + (r'/a)4A2 ) = (a/r')4 T2 // define T2 u = (a/r')2(a2/Q2)[ -cosθc ρ'cosφ - sinθc( (Zi+d) - A(r'/a)2)] v = (a/r')2 (a2/Q2) ρ'sinφ w = (a/r')2 (a2/Q2)( sinθc ρ'cosφ - cosθc[ (Zi+d) - A(r'/a)2] +A(r'/a)2) We notice now that (a/r')2(a2/Q2) = (r'2/T2) Q = (a/r')2 T so things get a little better, where I finally factor the Q out on the Φ line, Φiris(r') = (a/r') (1/Q){ – (2/π) sin-1 [ 2b / ( + ) ] +1 } = (r'/a) (1/T){ – (2/π) sin-1 [ 2b / ( + ) ] +1 } T2 = ( ρ'2 + ( Zi+d-A(r'/a)2 )2 + 2A(r'/a)2 { sinθcρ'cosφ - cosθc [ Zi+d -A(r'/a)2] } + (r'/a)4A2 ) u = (r'2/T2) [ -cosθc ρ'cosφ - sinθc{ (Zi+d) - A(r'/a)2}] v = (r'2/T2) ρ'sinφ Q = (a/r')2 T w = (r'2/T2))( sinθc ρ'cosφ - cosθc[ (Zi+d) - A(r'/a)2] +A(r'/a)2) (8) Correct for charge size. Now recall this from our σ work above, "Admittedly, this is for the situation in which qhole = (/ a) If we wanted the result for some arbitrary charge q1 in the hole, we would divide by qhole to get our unit charge result, then multiply by q1 ." Doing this process, we get Φiris(r') = q1 (a/ )(r'/a) (1/T){ – (2/π) sin-1 [ 2b / ( + ) ]+1 } = q1 (r'/ ) (1/T){ – (2/π) sin-1 [ 2b / ( + ) ]+1 } (9) Combine the two terms inside {...}: write {} = (1 - (2/π) sin-1X = (2/π) [ π/2 - sin-1X ] = (2/π) cos-1X Proof: let sin-1X = θ, some angle. then [ π/2 - sin-1X ] = [ π/2 - θ] cos[ π/2 - θ] = sinθ => [ π/2 - θ] = cos-1 [ sinθ] = cos-1 [ X] => [ π/2 - sin-1X ] = cos-1X Then we can write Φiris(r') = q1 (r'/ ) (1/T)(2/π)cos-1 [ 2b / ( + ) ] T2 = ( ρ'2 + ( Zi+d-A(r'/a)2 )2 + 2A(r'/a)2 { sinθcρ'cosφ - cosθc [ Zi+d -A(r'/a)2] } + (r'/a)4A2 ) u = (r'2/T2) [ -cosθc ρ'cosφ - sinθc{ (Zi+d) - A(r'/a)2}] v = (r'2/T2) ρ'sinφ Q = (a/r')2 T w = (r'2/T2))( sinθc ρ'cosφ - cosθc[ (Zi+d) - A(r'/a)2] +A(r'/a)2) (10) Get rid of A. At this point let's replace A(r'/a)2 = (r'2/2d) and write it again Φiris(r') = (2q1/πT) (r'/ ) cos-1 [ 2b / ( + ) ] T2 = ( ρ'2 + ( Zi+d-(r'2/2d) )2 + 2(r'2/2d) { sinθcρ'cosφ - cosθc [ Zi+d -(r'2/2d)] } + (r'2/2d)2 ) s = r'2 = ρ'2 + (Zi+d)2 u = (r'2/T2) [ -cosθc ρ'cosφ - sinθc{ (Zi+d) - (r'2/2d)}] v = (r'2/T2) ρ'sinφ Q = (a/r')2 T w = (r'2/T2))( sinθc ρ'cosφ - cosθc[ (Zi+d) - (r'2/2d)] +(r'2/2d)) r'2 = ρ'2 + (Zi+d)2 sinθc = 2dS/(d2+S2) cosθc = (d2- S2)/(d2+S2) b = B (d2+ S2) / (B2- S2) c = S (B2+ d2) / (B2- S2) and now we are completely rid of both a and A, and everything is in terms of iris-center cylindrical coordinates. It took an amazing amount of work to arrive at this point! Still, the following facts are not clear: How do you show potential is independent of d ? How do you show potential is symmetric under Zi→ -Zi? I think we are getting closer to some Maple work now. Surely there are many errors in what I have done above, could be in any of those 10 steps! Here is one hopeful fact: (scratch paper) r'2 = ρ'2 + (Zi+d)2 [ Zi+d -(r'2/2d)] = (d2- Zi2- ρ'2)/(2d) // rechecked on scratch so here finally is something symmetric under Zi→ -Zi. Continue here next time! It is 10 PM 6.19.10. (11) Process the T factor: Let's now insert the θc trig functions and the above "hopeful fact" T2 = ( ρ'2 + (d2- Zi2- ρ'2)2/(2d)2 + 2(r'2/2d) { 2dS/(d2+S2) * ρ'cosφ - (d2- S2)/(d2+S2) (d2- Zi2- ρ'2)/(2d) } + (r'2/2d)2 ) Mult through by (2d)2 [ I now group the last term with the first two ] (2d)2 T2 = ( (2d)2ρ'2 + (d2- Zi2- ρ'2)2 + r'4 +2r'2 { (2d)2S/(d2+S2) * ρ'cosφ - (d2- S2)/(d2+S2) * (d2- Zi2- ρ'2)} ) Now mult through by (d2+S2) // r'2 = ρ'2 + (Zi+d)2 (2d)2 (d2+S2) T2 = ( (d2+S2) [ (2d)2ρ'2 + (d2- Zi2- ρ'2)2 + r'4 ] + 2r'2 { (2d)2Sρ'cosφ - (d2- S2)(d2- Zi2- ρ'2)} ) We throw this into Maple as follows ( "Smythe38phiwork.mws") where the last line is (2d)2 (d2+S2) T2 . Therefore, (2d)2 (d2+S2) T2 = (2d)2 [ ρ2 + (Zi+d)2 ] ( ρ2 + S2 + Zi2 + 2Sρcosφ ) = (2d)2 r'2 ( ρ2 + S2 + Zi2 + 2Sρcosφ ) Now the (...) quantity is our Φ analogue of the Smythe r1 factor, just that now we are off the plane so we have this extra Z2 term. So let's redefine r12 = ( ρ2 + S2 + Zi2 + 2Sρcosφ ) r'2 = ρ'2 + (Zi+d)2 where r' is the distance from inversion origin to point r' , as shown in the above figure. Then we have (2d)2 (d2+S2) T2 = (2d)2 r'2 r12 (d2+S2) T2 = r'2 r12 T = r' r1 1/T = / (r'r1) and this is the kind of result I like to see!!! Our "factor out front" in Φ is then (2q1/πT) (r'/ ) = (2q1/π) (r'/ )( / (r'r1) ) = (2q1/πr1) So we now have Φiris(r') = (2q1/πr1) cos-1 [ 2b / ( + ) ] where r12 = ( ρ'2 + S2 + Zi2 + 2Sρcosφ ) So two comments! (1) the leading factor is symmetric in Zi (2) the leading factor contains no d's . So things are looking very promising. (12) Process the s variable. We have s = where [ we include w for later use ] u = (r'2/T2) [ -cosθc ρ'cosφ - sinθc{ (Zi+d) - (r'2/2d)}] v = (r'2/T2) ρ'sinφ w = (r'2/T2){ sinθc ρ'cosφ - cosθc[ (Zi+d) - (r'2/2d)] +(r'2/2d) } sinθc = 2dS/(d2+S2) cosθc = (d2- S2)/(d2+S2) 1/T = / (r'r1) First, notice that (r'2/T2) = r'2 (S2 + d2)/(r'2r12) = (S2 + d2)r1-2 Now throw in this and the trig functions to get u = (S2 + d2)r1-2 [ -(d2- S2)/(d2+S2) * ρ'cosφ - 2dS/(d2+S2) *{ (Zi+d) - (r'2/2d)}] v = (S2 + d2)r1-2 ρ'sinφ w = (S2 + d2)r1-2 { 2dS/(d2+S2)* ρ'cosφ - (d2- S2)/(d2+S2) [ (Zi+d) - (r'2/2d)] +(r'2/2d) } u = r1-2 [ -(d2- S2)ρ'cosφ - S{ 2d(Zi+d) - r'2} ] v = r1-2 ρ'sinφ (S2 + d2) w = r1-2 { 2dSρ'cosφ - (d2- S2) [ (Zi+d) - (r'2/2d)] +(S2 + d2) (r'2/2d) } Multiply through by (2d)r12 (2d) r12u = 2d [ -(d2- S2)ρ'cosφ - S{ 2d(Zi+d) - r'2} ] ≡ u1 (2d) r12v = 2d ρ'sinφ (S2 + d2) ≡ v1 (2d) r12w = { (2d)2 Sρ'cosφ - (d2- S2) [ 2d(Zi+d) - r'2] +(S2 + d2) r'2 } ≡ w1 Note in passing: I should have made use of this fact, but did not do so: { 2d(Zi+d) - r'2} = 2d(Zi+d) - ρ'2 - (Zi+d)2 = (Zi+d)(2d - Zi - d) - ρ'2 = (Zi+d)(d-Zi) - ρ'2 = (d + Zi)(d-Zi) - ρ'2 = d2 - Zi2 - ρ'2 Now we are ready to look at s. s = s2 = (u-c)2 + v2 [(2d) r12]2s2 = ([(2d) r12]u-[(2d) r12]c)2 + ([(2d) r12]v)2 = (u1 - [(2d) r12]c)2 + v12 Now install c = S (B2+ d2) / (B2- S2) to get [(2d) r12]2s2 = (u1 - [(2d) r12]c)2 + v12 = ( u1 - [(2d) r12] S (B2+ d2) / (B2- S2) )2 + v12 [ (B2- S2) (2d) r12 ]2s2 = ≡ s1s = [ (B2- S2)u1 - [(2d) r12] S (B2+ d2) ]2 + [ (B2- S2) v1 ]2 // L12 The result for s1s is a big mess, so let's check for errors. My Maple entry is good. Well, I took this mess and I replaced cosφ and sinφ with the r1 expression, then it is "not so bad" s1s =[ (B2- S2) (2d) r12 ]2s2 = = (2d)2 (d2 + S2)2 h where I use h as a symbol for this set of 10 terms, and Maple now knows P10. The good news is that the d-dependence is entirely in the first two factors, which is promising. We have [ (B2- S2) (2d) r12 ]2s2 = (2d)2 (d2 + S2)2 h [ (B2- S2) r12 ]2s2 = (d2 + S2)2 h What does this tell us (13) Using the ellipse theorem and get final result for Φ. At this point, I find that s is not a perfect square (since the above 10 terms do not factor), and I really did not expect it would come out that way. I now call upon the famous ellipse theorem from ellipses.doc which says (1/2) ( + )2 = (x2 + y2 + a2) + ( + )2 = 2(s2 + (w-d)2 + b2) + 2 Now recall that so far we have Φiris(r') = (2q1/πr1) cos-1 [ 2b / ( + ) ] So the denom of the cos-1 arg can be replaced by the ellipse theorem above (with a sqrt), and s only appears squared on the RHS. So let's write this as dac2 = 2(s2 + (w-d)2 + b2) + 2 where dac2 means "denominator of arc cosine squared". Our ingredients at this point are these: [ (B2- S2) r12 ]2s2 = (d2 + S2)2h (2d) r12w = { (2d)2 Sρ'cosφ - (d2- S2) [ 2d(Zi+d) - r'2] +(S2 + d2) r'2 } ≡ w1 b = B (d2+ S2) / (B2- S2) As our first step, multiply dac2 by (B2- S2)2r14 ≡ f2 : f2 dac2/2 = (f2s2 + f2(w-d)2 + f2b2) + Next, multiply both sides by (2d)2 r14 ≡ g2 f2 g2 dac2/2 = (f2 g2s2 + f2(gw-gd)2 + g2f2b2) + Next, multiply both sides by (B2- S2)2 ≡ k2 f2 g2 k2 dac2/2 = (f2 g2 k2s2 + f2 k2 (gw-gd)2 + g2f2 k2b2) + Now we make these identifications f2s2 = (B2- S2)2r14 s2 = (d2 + S2)2 h gw = (2d) r12 = w1 kb = (B2- S2) B (d2+ S2) / (B2- S2) = B(d2+ S2) We then have f2 g2 k2 dac2/2 = (g2 k2 (d2 + S2)2 h + f2 k2 (w1-gd)2 + g2f2 B2(d2+ S2)2) + where f = (B2- S2)r12 g = (2d) r12 k = (B2- S2) w1 = as already in Maple h = as already in Maple My Big Hope is that the argument of the radical will be a perfect square, but not totally necessary. Call it argrad = [g2 k2(d2 + S2)2 h - f2 k2 (w1-gd)2- f2 g2 B2(d2+ S2)2]2 + 4g2 k4(d2 + S2)2 h f2 (w1- gd)2 To my great amazement, Maple gives me this relatively simple result argrad = It is very close to a perfect square! Let's write it out by hand argrad = 42r112d4 (d2+S2)4 (B2- S2)6 [ (ρ'-B)2+Zi2] [ (ρ'+B)2+Zi2] // checked I just did a backscan looking for sign errors in linear ρ' terms. The u,v,w is always -.+.+. I don't see an error! So I guess the above is correct and we don't have a perfect square. Let's just forge ahead then. We had, f2 g2 k2 dac2/2 = (g2 k2 (d2 + S2)2 h + f2 k2 (w1-gd)2 + g2f2 B2(d2+ S2)2) + For the LHS we know, f2 g2 k2 dac2/2 = (B2- S2)2r14 (2d)2 r14 (B2- S2)2 dac2/2 = (B2- S2)4r18(2d)2 dac2/2 leaving us with (B2- S2)4r18(2d)2 dac2/2 = (g2 k2 (d2 + S2)2 h + f2 k2 (w1-gd)2 + g2f2 B2(d2+ S2)2) + Let's go now and compute the non-radical stuff on the right, which I have not done yet. j ≡ (g2 k2 (d2 + S2)2 h + f2 k2 (w1-gd)2 + g2f2 B2(d2+ S2)2) Maple gives us a pretty good result for this: j = And I will call the 7 term factor by the name m, so which I point out is symmetric in Z and has no d's appearing anywhere! So we have j = (2d)2r16 (S2-B2)2 (d2+S2)2 m Installing this "j" thing above we get (B2- S2)4r18(2d)2 dac2/2 = j + = (2d)2r16 (S2-B2)2 (d2+S2)2 m + Now back to our result argrad = 42r112d4 (d2+S2)4 (B2- S2)6 [ (ρ'-B)2+Zi2] [ (ρ'+B)2+Zi2] L = 4 r16 d2 (d2+S2)2 (B2- S2)3 Then we have (B2- S2)4r18(2d)2 dac2/2 = (2d)2r16 (S2-B2)2 (d2+S2)2 m + 4 r16 d2 (d2+S2)2 (B2- S2)3 and we can cancel a few things ( each term is L20) (B2- S2)2r12 dac2/2 = (d2+S2)2 m + (d2+S2)2 (B2- S2) Then we have dac2 = 2 (B2- S2)-2r1-2 (d2+S2)2 [ m + (B2- S2) ] and we now take the square root dac = (B2- S2)-1r1-1 (d2+S2) which is at least manageable. Our cos-1 argument is 2b/dac with b = B (d2+ S2) / (B2- S2) so arg of cos-1 = (Br1) 1/ and our potential is now Φiris(r') = (2q1/πr1) cos-1 [ 2b / ( + ) ] = (2q1/πr1) cos-1 [ arg of cos-1 ] = (2q1/πr1) cos-1 [ ( Br1/ ] with r12 = ( ρ'2 + S2 + Zi2 + 2Sρ'cosφ ) The good news is : (1) it is brief enough that I can actually write it down. (2) it is a function of Z2 and this has the looked-for symmetry for Z → -Z; (3) there are no appearances of symbol "d" in the result. (4) arc cos argument is dimensionless. All these facts are good evidence for a lack of major errors in the algebra. I am able to write m this way m = (ρ'2+Zi2)(S2-B2) + B2(B2- S2 + 2r12) // rechecked, this is correct. so here is my final result Φiris(r') = (2q1/πr1) cos-1[ (Br1) * 1/] where r12 = ( ρ'2 + S2 + Zi2 + 2Sρ'cosφ ) => 2r1dr1 = 2ZidZi (14) Show that Φ = 0 on the iris metal. Obviously we need to run some checks on this result! The first check is to see if we get 0 on the iris! So set Zi= 0 and see what happens: Φiris(r'; Zi= 0) = (2q1/πr1) cos-1[ (Br1) * 1/] where r12 = ( ρ'2 + S2 + 2Sρ'cosφ ) Let's examine the argument of the radical here: argrad = (ρ'2)(S2-B2) + B2(B2- S2 + 2r12) + (B2- S2) = m + n Clearly r1 is finite, so in order to get Φ = 0 we are going to need arg of cos-1 = 1 so cos-1 = 0. So we are going to need this: (Br1)/ = 1 B2r12 2 = argrad So we are going to want to see this (ρ'2)(S2-B2) + B2(B2- S2 + 2r12) + (B2- S2) = B2r122 We might imagine that = ± (ρ'-B)(ρ'+B) = ± (ρ'2-B2) so we might then have (ρ'2)(S2-B2) + B2(B2- S2 + 2r12) ± (B2- S2) (ρ'2-B2) = B2r122 (ρ'2)(S2-B2) + B2(B2- S2 + 2r12) ± ρ'2(B2-S2)∓B2(B2-S2) = B2r122 Now if we go with the upper sign, we get some cancellation (ρ'2)(S2-B2) + B2(B2- S2 + 2r12) + ρ'2(B2-S2)– B2(B2-S2) = B2r122 B2( 2r12) = B2r122 and there you have it!!! This says that (Br1)/ = 1 and then cos-1(1) = 0 and Φ = 0. I just knew it was going to come out this way, just as it does for the disk. So go back to the sign issue again. We have = + (ρ'-B) > 0 when ρ' > B which means we are "on the iris" If we are in the hole, then we have = + (B-ρ') > 0 = -(ρ'-B) and this is the lower sign situation. So in the hole we do NOT get Φ = 0. What DO we get in the hole? Well, from above we had this situation argrad = (ρ'2)(S2-B2) + B2(B2- S2 + 2r12) ± ρ'2(B2-S2)∓B2(B2-S2) and if we are in the hole, we want to take the lower signs, so we have argrad = (ρ'2)(S2-B2) + B2(B2- S2 + 2r12) - ρ'2(B2-S2)+B2(B2-S2) = - 2ρ'2(B2-S2) + B2 [ 2(B2-S2) + 2r12] = - 2ρ'2(B2-S2) + B2 [ 2(B2-S2) + 2( ρ'2 + S2 + 2Sρ'cosφ )] = - 2ρ'2(-S2) + B2 [ 2(B2-S2) + 2( S2 + 2Sρ'cosφ )] = - 2ρ'2(-S2) + B2 [ 2(B2) + 2( 2Sρ'cosφ )] = 2ρ'2S2 + 2B2 [ B2 + 2Sρ'cosφ] So we could insert this into our Φiris and get our result for potential in the hole Φiris(r'; Zi= 0) = (2q1/πr1) cos-1[ (Br1) / ] As a special case, we could ask about the potential in the hole at ρ' = S, which is the circle which contains our point charge. We get argrad = 2ρ'2S2 + 2B2 [ B2 + 2Sρ'cosφ] r12 = ( S2 + S2 + 2S2cosφ ) = 2S4 + 2B2 [ B2 + 2S2cosφ] Now as a special case of this special case, let's look at φ = π, which is the location of the point charge. We then have argrad = 2S4 + 2B2 [ B2 - 2S2] = 2 [ S4 + B4 - 2B2S2] = 2(B2-S2)2 r1→ 0 We then are faced with cos-1((Br1) / [(B2-S2) ]) = cos-1[Br1/(B2-S2)] = cos-1(0) = π/2 The potential at this point is then Φ = (2q1/πr1) π/2 = q/r1 as expected, and it of course blows up, despite that fact that the iris is an infinite object. (15) Can we perhaps use the ellipse theorem a second time ? (1/2) ( + )2 = (x2 + y2 + a2) + which says (1/) ( + ) = We have this "quantity of interest" Can we identify this somehow with the RHS of the previous line, for some choice of the three arguments a, x and y ? Proportional would be OK. Consider [ (ρ'-B)2+Zi2] [ (ρ'+B)2+Zi2] = (ρ'-B)2(ρ'+B)2 + Zi2(ρ'+B)2 + Zi2(ρ'-B)2 + Zi4 = (ρ'2-B2)2 + 2Zi2(ρ'2+B2) + Zi4 = ρ'4 + B4 - 2ρ'2B2 + 2Zi2(ρ'2+B2) + Zi4 = ρ'4 + B4 + Zi4 - 2ρ'2B2 + 2Zi2(ρ'2+B2) (x2 - y2- a2)2 + 4x2y2 = x4 + y4 + a4 + 2x2y2 - 2a2x2- 2a2y2 = f(x,y) It sure seems close, but the signs are wrong. Secondly, there is no φ dependence here, but there is inside the r1 factor in the rest of the radical which has to be (x2+y2+ a2) . So I am afraid I cannot make this "fit". Try one more time: (x2 + y2+ a2) = [(ρ'2+Zi2)(S2-B2) + B2(B2- S2 + 2r12)]/ (B2-S2) = [- (ρ'2+Zi2) + B2 + 2r12(B2/ (B2-S2) ] (x2 + y2+ a2)2 = [- (ρ'2+Zi2) + B2 + 2r12(B2/ (B2-S2) ]2 (x2 - y2- a2)2 + 4x2y2 = [(ρ'2-B2)2 + 2Zi2(ρ'2+B2) + Zi4] Subtract to get 2x2(y2+a2) + 4x2y2 = [- (ρ'2+Zi2) + B2 + 2r12(B2/ (B2-S2) ]2 - [(ρ'2-B2)2 + 2Zi2(ρ'2+B2) + Zi4] Even if it works, it is a complete mess, so forget it! (16) Compute the charge density? Φiris(r') = (2q1/πr1) cos-1[ (Br1) * 1/] where r12 = ( ρ'2 + S2 + Zi2 + 2Sρcosφ ) => 2r1dr1 = 2ZidZi We expect to find that [ E = 4πσ in Jackson units, p 12 ] [ put in Smythe for one side ] σ = - ( qhole/2π2r12)/ ∂Zi Φiris(r')|Zi=0 = -4πσSmythe = - 4π [ - ( q1/2π2r12)/ ] = ( 2q1/πr12)/ ] One little thing to note is this: the "outside factor" has Zi dependence through r1. But, the contribution from this term to the derivative will be 0 because cos-1 = 0 on the iris, as we already showed. So only dependence inside the cos-1 will contribute. We might abbreviate u = (Br1)/ Φiris(r') = (2q1/πr1) cos-1(u) ∂Zi Φiris(r')|Zi=0 = (2q1/πr1) [ -1/ ] ∂Ziu where u2 = 2B2r12/(m+n) We know that at Zi = 0 we will have m+n = (Br1)2 as we already showed. So we have u2 = 2B2r12/ 2B2r12 = 1 so = 0 in this case, and we must get cancellation from something in ∂Ziu. It turns out that we will want to keep the first order in Zi term in both ∂Ziu up top, and in down below. Here then is the complete calculation which I hate to admit took me more than a whole day to get correct. (a) Review of input information and strategy: u = (Br1)/ 1 = (Br1o)/ u2 = 2B2r12/(m+n) 1 = 2B2r1o2/(m0+n0) r12 = ( ρ'2 + S2 + Zi2 + 2Sρcosφ ) r1o2 = ( ρ'2 + S2 + 2Sρcosφ ) m = (ρ'2+Zi2)(S2-B2) + B2(B2- S2 + 2r12) m0 = - (B2- S2)(ρ'2-B2) + 2B2r1o2 n = (B2- S2) n0 = + (B2- S2)(ρ'2-B2) ρ' > B Φiris(r') = (2q1/πr1) cos-1(u) Note that m0+ n0 = 2B2r1o2 ∂Zi Φiris(r')|Zi=0 ≈ (2q1/πr1) [ -1/ ] ∂Ziu 1-u2 = [ (m+n) - 2B2r12]/(m+n) ∂Zir1 = (Zi/r1) ∂Zir12 = 2r1∂Zir1 = 2r1 (Zi/r1) = 2Zi - 2u∂Ziu = 2B2 [ r12∂Zi(m+n) - (m+n) 2Zi] / (m+n)2 Our interest will be in the ratio ∂Ziu/ which is a 0/0 situation for Zi= 0, so we shall keep the lowest terms for both in Zi as we do our calculation. (b) Compute facts about n. As a start, let's expand n for small Zi: ( a2 + b2)1/2 = a + (1/2)a-1b2 + ... assume ρ' > B ≈ (ρ'-B) + (1/2) Zi2/ (ρ'-B) = (ρ'-B)[ 1+ (1/2)Zi2/ (ρ'-B)2] ≈ (ρ'+B) + (1/2) Zi2/ (ρ'+B) = (ρ'+B)[ 1+ (1/2)Zi2/ (ρ'+B)2] ≈ (ρ'2- B2) [ 1 + (1/2)Zi2 { 1/(ρ'-B)2 + 1/(ρ'+B)2 } ] = (ρ'2- B2) [ 1 + Zi2 (ρ'2+B2)/ (ρ'2- B2)2 ] So we then have, for small Zi n = (B2- S2) ≈ (B2- S2) (ρ'2- B2) [ 1 + Zi2 (ρ'2+B2)/ (ρ'2- B2)2 ] ∂Zin = (B2- S2) (ρ'2- B2) 2Zi(ρ'2+B2)/ (ρ'2- B2)2 = (B2- S2) 2Zi(ρ'2+B2) / (ρ'2- B2) n = n0 + (B2- S2) Zi2 (ρ'2+B2)/ (ρ'2- B2) (c) Compute facts about m. Next we have m = (ρ'2+Zi2)(S2-B2) + B2(B2- S2 + 2r12) ∂Zim = 2Zi(S2-B2) + 2B2 ∂Zir12 = 2Zi(S2-B2) + 4B2 Zi = 2Zi [ (S2-B2) + 2B2] = 2Zi(S2+B2) m = m0 + Zi2(S2-B2) + 2B2Zi2 = m0 + Zi2(S2+B2) (d) Compute facts about m+n m+n = {m0 + Zi2(S2+B2)} + {n0 + (B2- S2) Zi2(ρ'2+B2) / (ρ'2- B2) } = m0 + n0 + Zi2 { (S2+B2) + (B2- S2) (ρ'2+B2) / (ρ'2- B2) } = m0 + n0 + Zi2 { (S2+B2) (ρ'2- B2) + (B2- S2) (ρ'2+B2) }/(ρ'2- B2) = m0 + n0 + Zi2 { 2B2(ρ'2-S2) }/(ρ'2- B2) = m0 + n0 + 2 B2Zi2 (ρ'2-S2)/(ρ'2- B2) = 2B2r1o2 + 2 B2Zi2 (ρ'2-S2)/(ρ'2- B2) Then we know that ∂Zi(m+n) = 4 B2Zi (ρ'2-S2)/(ρ'2- B2) (e) Compute ∂Ziu . We start with this from above, u2 = 2B2r12/(m+n) 2u∂Ziu = 2B2 [ (m+n) ∂Zi(r12) - r12 ∂Zi(m+n) ] /(m+n)2 ∂Ziu = B2 [ (m+n) ∂Zi(r12) - r12 ∂Zi(m+n) ] /(m+n)2 where we just set u = 1 on the LHS. We also replace m+n in the denom with 2B2r1o2 because its correction would be at least order Z3 which we are going to ignore. ∂Ziu = B2 [ (m+n) ∂Zi(r12) - r12 ∂Zi(m+n) ] /(2B2r1o2)2 Now, our two derivatives are linear in Z, so we shall only keep linear terms in this numerator difference and ignore higher order terms. Therefore, we can write the above as ∂Ziu = B2 [(2B2r1o2) ∂Zi(r12) - r1o2 ∂Zi(m+n) ] /(2B2r1o2)2 = [(2B2) ∂Zi(r12) - ∂Zi(m+n) ] /(2Br1o)2 We now install our derivatives computed above, namely ∂Zi(m+n) = 4 B2Zi (ρ'2-S2)/(ρ'2- B2) ∂Zir12 = 2Zi to get ∂Ziu = [(2B2) ∂Zi(r12) - ∂Zi(m+n) ] /(2Br1o)2 = [(2B2) 2Zi - 4 B2Zi (ρ'2-S2)/(ρ'2- B2) ] /(2Br1o)2 = 4ZiB2 [ 1 - (ρ'2-S2)/(ρ'2- B2) ] /(2Br1o)2 = Zi [ 1 - (ρ'2-S2)/(ρ'2- B2) ] /(r1o)2 = Zi [(ρ'2- B2) - (ρ'2-S2) ] /[r1o2(ρ'2- B2)] = Zi [S2-B2 ] /[r1o2(ρ'2- B2)] = - Zi (B2-S2) /[r1o2(ρ'2- B2)] (f) Compute . Start with this 1-u2 = 1 - 2B2r12/(m+n) = [ (m+n) - 2B2r12]/ (m+n) We anticipate that the difference here will have Zi2 terms and higher, so we can replace the denominator with 2B2r1o2 since the correction would be 4th order. So 1-u2 =[ (m+n) - 2B2r12]/ (2B2r1o2) Into this we install two facts already found, m+n = 2B2r1o2 + 2 B2Zi2 (ρ'2-S2)/(ρ'2- B2) r12 = r1o2 + Zi2 The leading terms cancel and we are left with 1-u2 = [ 2 B2Zi2 (ρ'2-S2)/(ρ'2- B2) - 2B2Zi2]/ (2B2r1o2) = Zi2 B2 [ (ρ'2-S2)/(ρ'2- B2) - 1]/ (B2r1o2) = Zi2 [ (ρ'2-S2)/(ρ'2- B2) - 1]/ (r1o2) = Zi2 [ (ρ'2-S2) - (ρ'2- B2)]/ [r1o2(ρ'2- B2)] = Zi2 [ B2-S2]/ [r1o2(ρ'2- B2)] Therefore we have found that = Zi / [r1o ] 1/ = r1o / [Zi ] (g) Compute ∂Zi Φiris(r')|Zi=0 From our starting info we have, ∂Zi Φiris(r')|Zi=0 ≈ (2q1/πr1o) [ -1/] ∂Ziu and we now install our two pieces from the last two sections above, ∂Ziu = - Zi (B2-S2) /[r1o2(ρ'2- B2)] 1/ = r1o / [Zi ] to get ∂Zi Φiris(r')|Zi=0 ≈ (2q1/πr1o) [ -1/] ∂Ziu = (2q1/πr1o) [ 1/] (-∂Ziu) = (2q1/πr1o) [r1o / [Zi ]] Zi (B2-S2) /[r1o2(ρ'2- B2)] = (2q1/πr1o2) (/ ) Our expected answer was this: ( from section 16 above.) ∂Zi Φiris(r')|Zi=0 = ( 2q1/πr1o2)/ and we have (finally after 2 days of algebra screwing up) the expected and known-correct result. So this is a powerful test of our potential!