Smythe Problem 38 by Double Inversion Attempt #3

Smythe Problem 38 by Double Inversion Attempt #3 PhL 6.8.10 NOTE: This doc is fully reviewed in "top level review...". Overall, Attempt #3 was just a scratch working effort and there is really nothing in this doc that is now of any value, so don't give it another look! I want to type in this doc while I look at other docs. In Attempt #2 I reviewed things pretty well, and I got the connection between R and r1 cleaned up. What remains is the business with b2-s2 and cosγ' and such things, which are still a complete mess. Plan A. Playing with cosγ'. 1 Something fishy with this bowl result from Attempt #2 (but comes out OK) 4 Compute Jackson's 2.10 Special Case Result 5 Thurs June 10 : Trying harder to relate the two different ρ' distances. 7 Plan A. Playing with cosγ'. First, let's collect information about r3 which is the inversion-1 image of point C. For origin at the inversion origin, we can say r3 = (a2/C2) C = [a2/( d2+ c2)] C For origin at sphere center, we can write r3 = (A,θ3,φ3= 0) where θ3 = 2 tan-1(c/d) Now the magic angle γ' is that between two unit vectors with origin at sphere center: cosγ' = 3 = (cosθcosθ3 + sinθsinθ3cosφ) where r = (A,θ,φ) r3 = (A,θ3,0) // spherical, sphere center Since both r and r3 lie on the sphere, we know that θ = 2α θ3 = 2α3 tanα3 = c/d cosα = R/(2A) so we can say cosγ' = 3 = cos(2α) cos(2α3) + sin(2α) sin(2α3) cosφ and we can try to replace these trig functions with functions of parameters. But that seems unpleasant. But the result I am looking for is also a bit unpleasant. So lets brute force ahead a bit: sin(2α) = 2 sinα cosα = 2 cosα = 2 (R/2A) cos(2α) = 2 cos2α - 1 = 2 (R/2A)2 - 1 R = a2r1/ r12 = ρ'2 + S2 - 2S ρ' cosφ cosφ = (ρ'2 + S2 - r12)/(2Sρ') Comment: Angle α changes as you move your point r on the sphere. In the form above, this is tracked by variable R which in turn is related to r1 and ρ' in the iris world. R2/4A2 = (a4r12/4A2) 1/ [ (d2+S2)(ρ'2+d2) ] so this does connect us to iris world tanα3 = (c/d) sec2α3 = 1 + tan2α3 = 1 + (c/d)2 = 1/cos2α3 cosα3 = 1/ sin2α3 = 1 - cos2α3 = 1 - (1/(1 + (c/d)2) = (c/d)2/ (1 + (c/d)2) sinα3 = (c/d) (1/) sin(2α3) = 2sinα3cosα3 = 2(c/d) (1/) (1/) = 2(c/d) / (1 + (c/d)2) cos(2α3) = 2 cos2α3 - 1 = 2 / (1 + (c/d)2) - 1 = (1 – (c/d)2)/ (1 + (c/d)2) We are building up lots of ugliness. Here are the four pieces so far constructed cosγ' = 3 = cos(2α) cos(2α3) + sin(2α) sin(2α3) cosφ sin(2α) = 2 sinα cosα = 2 cosα = 2 (R/2A) cos(2α) = 2 cos2α - 1 = 2 (R/2A)2 - 1 sin(2α3) = 2sinα3cosα3 = 2(c/d) (1/) (1/) = 2(c/d) / (1 + (c/d)2) cos(2α3) = 2 cos2α3 - 1 = 2 / (1 + (c/d)2) - 1 = (1 – (c/d)2)/ (1 + (c/d)2) The last pair are a function only of (c/d), first pair of (R/2A). I will then have cosγ' = f( (c/d), (R/2A), cosφ) b = B (d2 + S2) / (B2-S2) c = S (B2 + d2) / (B2-S2) R = a2r1/ (c/d) = (S/d) (B2 + d2) / (B2-S2) cosφ = (ρ'2 + S2 - r12)/(2Sρ') The question then is this: if I stuff all this mess in, will I get my cosγ' mess shown in Attempt #2? Here we go: cosγ' = 3 = cos(2α) cos(2α3) + sin(2α) sin(2α3) cosφ = (2 (R/2A)2 - 1) (1 – (c/d)2)/ (1+(c/d)2) + 2 (R/2A) (2(c/d) / (1 + (c/d)2)) (ρ'2 + S2 - r12)/(2Sρ') = {(2 (R/2A)2 - 1) (1 – (c/d)2) + 2 (R/2A) (2(c/d) (ρ'2 + S2 - r12)/(2Sρ') } /(1+(c/d)2) = {(2 (R/2A)2 - 1) (d2 – c2) + 2d2 (R/2A) (2(c/d) (ρ'2 + S2 - r12)/(2Sρ') } /(c2+d2) So I am finding here that (c2+ d2) cosγ' = {(2 (R/2A)2 - 1) (d2 – c2) + 2d2 (R/2A) (2(c/d) (ρ'2 + S2 - r12)/(2Sρ') } But now we have R = a2r1/ = 2Adr1/ (R/2A) = dr1/ (R/2A)2 = d2r12/ (d2+S2)(ρ'2+ d2) so throw this in and we get (c2+ d2) cosγ' = {(2 [d2r12/ (d2+S2)(ρ'2+ d2) ] - 1) (d2– c2) + 2d2 (dr1/) (2(c/d) (ρ'2 + S2 - r12)/(2Sρ') = f(c, d, r1, S, ρ' ) whereas Attempt #2 requires me to show that (c2+ d2) cosγ' ={ (d2+ c2) (ρ'2+ d2) (B2-S2) - (Bd)2 (d2 + S2)2 [ 2 - (1/πr1B)2 (B2-S2)(ρ'2- B2) ] }/D1 = F(c, d, r1, S, ρ', B) D1 = (ρ'2+ d2) (B2-S2)2 I think I am missing a relation involving B! Triangle with two angles "e" is isosceles and says 2e + θb = π. But this is just a special case of my general rule which here says e = (π-θb)/2 as in α = (θ/2). tan(π-θb)/2 = B/d = cot(θb/2) We have a square peg trying to fit into a round hole her. Plan B. Let's just study the result I get for σ, forget about getting Smythe's result for now, and see if mine seems reasonable at all, and whether it agrees with Jackson's claim for center of hole Green's charge. The reasonability test is always a good one! Something fishy with this bowl result from Attempt #2 (but comes out OK) σ(x,y,z) = – (2Ad /π2R3) / ρ' = d = a2/2A x' = (d/2) sec2(θ/2) sinθ cosφ = d tan(θ/2) cosφ R = 2Acos(θ/2) y' = (d/2) sec2(θ/2) sinθ sinφ = d tan(θ/2) sinφ c = d tan(θc/2) (1+ tan2(θb/2))/ (1 - tan2(θc/2) tan2(θb/2)) = (d/2)[ tan(θ1/2) + tan(θ2/2)] b = d tan(θb/2) (1+ tan2(θc/2))/ (1 - tan2(θc/2) tan2(θb/2)) = (d/2)[ tan(θ1/2) – tan(θ2/2)] Define t = tan(θ/2) t1 = tan(θ1/2) t2 = tan(θ2/2) c = (d/2)(t1+ t2) b = (d/2)(t1- t2) b2- ρ'2 = b2 - [ (x'-c)2 + y'2] = b2 - [ x'2 + y'2 + c2 - 2cx'] = (b2- c2) - (x'2+ y'2) + 2cx' = (b+c)(b-c) - (x'2+ y'2) + 2cx' = dt1(-dt2) - (d2t2) + 2(d/2)(t1+ t2) d t cosφ = d2[ -t1t2- t2+ t(t1+ t2) cosφ ] In the plane of the sphere we have φ = 0 so this becomes b2- ρ'2 = d2[ -t1t2- t2+ t(t1+ t2) ] = - d2 [ t2 - (t1+ t2)t + t1t2 ] = - d2(t-t1)(t- t2) = - d2(t-t1)(t+|t2|) = d2(t1-t)( |t2| + t) OK, happy again. If t = 0 so we are on the bowl. we get b2- ρ'2 = +d2t1|t2| > 0 as expected. And as expected this b2- ρ'2 vanishes (for φ = 0) when t = t1 or t = - |t2|. Compute Jackson's 2.10 Special Case Result Our iris charge density in general is this (Attempt #2 page 13) σ'(r') = (r/a)3 σ(r) = (r/a)3 [- (a2/π2R3) / ] r' = (a2/r2) r rr' = a2 But let's take the special case that θc= 0 so our iris point charge is in the center of the hole. Consider then that u = (a2/R2)( sinθc(z-A) - cosθcx) = - (a2/R2)x v = (a2/R2)y = (a2/R2)y w = (a2/R2)( - cosθc(z-A) - sinθcx+A) = - (a2/R2) (z-A) b = d tan(θb/2) (1+ tan2(θc/2))/ (1 - tan2(θc/2) tan2(θb/2)) = d tan(θb/2) B = d /tan(θb/2) => bB = d2 R2 = [ x2 + y2 + (z-A)2 ] - 2A (z-A) cosθc + A2 - 2Ax sinθc = [ x2 + y2 + (z-A)2 ] - 2A (z-A) + A2 For σ work, we know that (x,y,z) like on the sphere in R space, and this sphere is [ x2 + y2 + (z-A)2 ] = A2 => R2 = 2A2 - 2A (z-A) = 2A ( A - z+A) = 2A(2A-z) In our "battle notes" we have z = (a2/r'2) z' = [a2/(ρ'2+ d2)] d = 2Ad2 / (ρ'2+ d2) so that R2 = 2A(2A - 2A d2 / (ρ'2+ d2)) = (2A)2 [ 1 - d2 / (ρ'2+ d2) ] = (2A)2 ρ'2 / (ρ'2+ d2) => R = 2Aρ' / Now go back to these guys, u = - (a2/R2)x = - (a2/R2) [a2/(ρ'2+ d2)] ρ' cosφ' = - a4cosφ/ [ (2A)2ρ' ] = - (cosφ/ρ') (a2/2A)2 = - (d2/ρ') cosφ So we know then that u = - (d2/ρ') cosφ v = - (d2/ρ') sinφ b = d2/B c = 0 // from our general forms for b and c elsewhere so we then have our quantity of major interest b2 - s2 = b2 - (u-c)2 -v2 = b2 - [u2+v2] = (d2/B)2 - (d2/ρ')2 = d4 (1/B2- 1/ρ'2) = d4 (ρ'2-B2)/(B2ρ'2) => 1/ = (Bρ'/d2) (1/) Now go back to our iris charge formula above ( recall charge q = 1) σ'(r') = (r/a)3 [- (a2/π2R3) / ] = - (1/π2a)(r/R)3 / Above we had R = 2Aρ' / r = a2/ => (r/R) = (a2/2Aρ') = (d/ρ') Then we have σ'(r') = - (1/π2a)(r/R)3 / = - (1/π2a) (d/ρ')3 (Bρ'/d2) (1/) = - (Bd/π2a ρ'2) (1/) Now from elsewhere I showed that (applied to our special case) qhole = 1 (rg'/a) = (d/a) So my result is now σ'(r') = - qhole (B/π2 ρ'2) (1/) and FINALLY, this agrees exactly with Jackson page 53 where he has a = B and r = ρ' and qhole = +1. This is the first contact I have had on checking the iris result. Thurs June 10 : Trying harder to relate the two different ρ' distances. (1) In my Attempt #2 section " What do my Inversion Lemmas have to say about s? ", I showed this to be true: ( where s = ρ' of the R' space of the first inversion) s = (AC/R) 2 sin(γ'/2) cosγ' = (cosθcosθ3 + sinθsinθ3cosφ) = 3 |sphere ctrd (2) In the special case that r lies in the plane of paper, I then know that the length s of our charged disk in the first inversion picture is related to other parameters in this way s = (AC/R) 2 sin[(θpre-θ3)/2] > 0 where R = the inversion-origin length of vector r in first-inversion picture C2 = d2+ c2 θ3 = sphere-centered angle of point r3 prior to sphere rotation tan(θ3/2) = c/d θpre = angle of point r on sphere prior to sphere rotation Since we are in our "special case" that φ = 0 we know that R = 2Acos(θpre/2) so we really have s = (AC/[2Acos(θpre/2)]) 2 sin[(θpre-θ3)/2] = (C/[cos(θpre/2)]) sin[(θpre-θ3)/2] = C sin[(θpre-θ3)/2] / cos(θpre/2) C2 = c2 + d2 All these parameters are those of the "first inversion picture" and so involve the pre-rotated sphere (3) On the other hand, if we go to our second-inversion picture, we can "work backwards" and try to relate the distance ρ' on the iris plane to R space parameters of the post-rotated sphere. In my section (5a) I showed for example that ρ'2 = (2Ad)2/r2 - d2 where r = inversion length of r in post-rotation sphere picture All these parameters are those of the "second inversion picture" and so involve the post-rotated sphere. It is easy to show that r = 2Acos(θpost/2) in analogy with R = 2Acos(θpre/2). Thus we have ρ'2 = (2Ad)2/r2 - d2 = (2Ad)2/[2Acos(θpost/2)]2 - d2 = (d)2/[cos(θpost/2)]2 - d2 = d2 { sec2(θpost/2) - d2 } = d2 tan2(θpost/2) // obvious anyway So far then we have s2 = C2 sin2[(θpre-θ3)/2] / cos2(θpre/2) ρ'2 = d2 tan2(θpost/2) I would argue that (see below) θpost = θpre + (π-θc) => θpost/2 = (θpre- θc)/2 + π/2 Then tan(θpost/2) = tan [π/2 + (θpre- θc)/2 ] = – cot[(θpre- θc)/2] We then have s2 = C2 sin2[(θpre-θ3)/2] / cos2(θpre/2) ρ'2 = d2 cot2[(θpre- θc)/2] Now adopt the convention that θ = θpre Then we have' s2 = (c2+ d2) sin2[(θ-θ3)/2] / cos2(θ/2) ρ'2 = d2 cot2[(θ- θc)/2] which is still pretty nasty. We can add these facts: tan(θ/2) = (ρ'/d) tan(θ3/2) = (c/d) tan(θc/2) = S/d I think the following is all just scraps -PhL 6.17.10 *****************************************************************************8 Plan A: Try to relate everything to second-inversion picture parameters. I know how to express R in these parameters already, hold on that a second. What about θ which appears above? I think this is true: θpost = θpre + (π-θc) For example, the right end of the sphere at θpre = 0 in the pre-rotation picture moves CCW by amount (π-θc) in our rotation and so this point ends up at θpost = (π-θc). So, I will now attempt to express "s" entirely in terms of the second-inversion picture: s = (AC/R) 2 sin[(θpre-θ3)/2] We start with θpre = θpost - (π-θc) => θpre-θ3 = θpost - (π-θc) -θ3 = ( θpost - π + θc - θ3 ) = ( θpost + θc - θ3 - π) sin[(θpre-θ3)/2] = sin[(( θpost + θc - θ3 - π)/2] = - sin[ π/2 – (θpost + θc - θ3)/2] = - cos[(θpost + θc - θ3)/2] I will just ignore the "sign issue", somehow this cos must be negative. From #2 notes we also know that R = 2Acos(θpre/2) So then here is what we have s = (2AC/R) | cos[(θpost + θc - θ3)/2 ] | tan(θ3/2) = c/d tan(θc/2) = S/d But now go on with R = 2Acos(θpre/2) = 2A cos[ and we have now achieved our goal of relating s to second-inversion picture parameters. Meanwhile, were I to draw in the various sphere-centered angles in the second-inversion picture, I would find from a right triangle that r2 = (Asinθpost)2 + (A+Acosθpost)2 = 2A2(1 + cosθpos_) = 4A2 cos2(θpost/2) so our first result above becomes ρ'2 = (2Ad)2/r2 - d2 = (2Ad)2/[4A2 cos2(θpost/2)] - d2 = d2 sec2(θpost/2) - d2 = d2[sec2(θpost/2) - 1] = d2 tan2(θpost/2) => ρ' = d tan(θpost/2) => tan(θpost/2) = (ρ'/d) a fact that is totally obvious from the picture so I didn't have to do the last 3" of work, but OK. So here then is where we stand: s = (2AC/R) | cos[θpost + θc - θ3)/2] | with R = a2r1/ ρ' = d tan(θpost/2) Let's now set θpost= θ, our sphere angle in the second-inversion world. We then have s = (2AC/R) |cos[(θ + θc - θ3)/2] | ρ' = d tan(θ/2) with these extra facts R = a2r1/ tan(θ3/2) = c/d tan(θc/2) = S/d b = B (d2 + S2) / (B2-S2) c = S (B2 + d2) / (B2-S2) So this is the best I have done so far in "relating" s to ρ'. We can square both sides to get s2 = (2AC/R)2 cos2[(θ + θc - θ3)/2] = (2AC/R)2 (1/2) [ 1 + cos(θ + θc - θ3) ] ρ'2 = d2 tan2(θ/2) Let's rewrite s2 some more. We know that C2 = c2+ d2 and R = a2r1/ so (2AC/R)2 (1/2) = (1/2)(2A)2(d2+c2) (d2+S2) (d2+ρ'2) /(a2r1)2 = (1/2)(2A)2(d2+c2) (d2+S2) (d2+ρ'2) /(2Adr1)2 = (1/2)(d2+c2) (d2+S2) (d2+ρ'2) /(dr1)2 so we then have s2 = (1/2)(d2+c2) (d2+S2) (d2+ρ'2) /(dr1)2 [ 1 + cos(θ + θc - θ3) ] ρ'2 = d2 tan2(θ/2) Math subdetail. Now we have three analogous relations, and all three fractions have the same denom d : tan(θ/2) = (ρ'/d) tan(θ3/2) = (c/d) tan(θc/2) = S/d Now consider a template case tan(x/2) = f => tan2(x/2) = f2 = (1-cosx)/(1+cosx) => (1+c)f2= 1-c = f2+cf2 => c(1+f2)= 1-f2 => cosx = (1-f2)/(1+f2) => sinx = 2f/(1+f2) // obvious since makes s2 + c2 = 1 and we just assume for now that all three angles and fractions are positive. Now suppose f = y/d. Then we have cos(x) = (1-f2)/(1+f2) = (1-(y/d)2)/(1+(y/d)2) = (d2-y2)/(d2+ y2) sinx = 2f/(1+f2) = 2(y/d)(1+(y/d)2) = 2dy/(d2+y2) Therefore in our three cases we may conclude that cosθ = (d2-ρ'2)/(d2+ ρ'2) sinθ = 2d ρ'/(d2+ρ'2) cosθ3 = (d2-c2)/(d2+ c2) sinθ3 = 2d c/(d2+c2) cosθc = (d2-S2)/(d2+ S2) sinθc = 2d S/(d2+S2) Now define ψ = θc - θ3 sinψ = sin(θc - θ3) = sinθccosθ3 - sinθ3cosθc = [2d S (d2-c2) - 2d c(d2-S2)] / [(d2+ c2) (d2+S2)] =2d [S (d2-c2) - c(d2-S2)] / [(d2+ c2) (d2+S2)] cosψ = cos(θc - θ3) = cosθccosθ3 + sinθ3sinθc = [(d2-S2) (d2-c2) + 2d c 2d S2] / [(d2+ c2) (d2+S2)] = [(d2-S2) (d2-c2) + 4d2 S2] / [(d2+ c2) (d2+S2)] Now go back to our result for s, which write as s2 = (2AC/R)2 cos2[(θ + θc - θ3)/2] = (2AC/R)2 (1/2) [ 1 + cos(θ + θc - θ3) ] = 2 (AC/R)2 [ 1 + cos(θ + ψ) ] We then have cos(θ + ψ) = cosθcosψ - sinθsinψ = (d2-ρ'2)/(d2+ ρ'2) [(d2-S2) (d2-c2) + 4d2 S2] / [(d2+ c2) (d2+S2)] - 2d ρ'/(d2+ρ'2) * 2d [S (d2-c2) - c(d2-S2)] / [(d2+ c2) (d2+S2)] = { (d2-ρ'2) [(d2-S2) (d2-c2) + 4d2 S2] - 4d2 ρ' [S (d2-c2) - c(d2-S2)] } / D1 D1 = (d2+ c2) (d2+S2) (d2+ρ'2) Then we can say 1 + cos(θ + ψ) = { D1 + (d2-ρ'2) [(d2-S2) (d2-c2) + 4d2 S2] - 4d2 ρ' [S (d2-c2) - c(d2-S2)] } / D1 = { (d2+ c2) (d2+S2) (d2+ρ'2) + (d2-ρ'2) [(d2-S2) (d2-c2) + 4d2 S2] - 4d2 ρ' [S (d2-c2) - c(d2-S2)] }/D1 Now define D1 = (d2+ c2) (d2+S2) (d2+ρ'2) F1 = (d2-S2) (d2-c2) + 4d2 S2 F2 = S (d2-c2) - c(d2-S2) then 1 + cos(θ + ψ) = { D1 + (d2-ρ'2) F1- 4d2 ρ' F2 }/D1 The numerator is our main interest N = { D1 + (d2-ρ'2) F1- 4d2 ρ' F2 } At this point I have to use Maple, and things are a big mess, and my Plan A is not doing too well. Plan A Step 2. The two quantities we really want to relate are these b2 - s2 and ρ'2 - B2 where we know that b = B (d2 + S2) / (B2-S2) c = S (B2 + d2) / (B2-S2) s2 = (1/2)(d2+c2) (d2+S2) (d2+ρ'2) /(dr1)2 [ 1 + cos(θ + θc - θ3) ] ρ'2 = d2 tan2(θ/2)