Smythe Problem 38 by Double Inversion Attempt #4

Smythe Problem 38 by Double Inversion Attempt #4 PhL 6.10.10 NOTE: This doc is reviewed in "top level review...". doc I now spend day after day trying to make the Smythe result work, but it refuses to work! I thrashed around in Attempt #3 but got nowhere, but I want to keep that algebra for possible later use, but I want to clear the decks for a cleaner start, so we are now in Attempt #4. Study the distance s. 2 Bigger contradiction: 2 Resolution: 3 Study the distance s, Version 2 start review here 4 Now I am perhaps coming into another mistake I may have made. 4 Question: How are (θ,φ) and (θ",φ") related to each other? 7 Restatement from above about b2 - s2 in terms of post rotation angles: 8 Starting Over with an "Active" View of the Bowl Rotation: Rotating points and fields! 9 Study the distance s, Version 3 11 Why we have two different θ coordinates which are not the same. 16 Doing Things Right, Version 1 18 Doing Things Right, Version 2: The Basic Theory Equations 21 Doing Things Right, Version 2: Apply these Equations 24 As outlined in Attempt #2, in order to reconcile my result with that of Smythe, I have to show the following fact to be true: (S2+d2) (1/ ) = r1 ( 1/ ) " putative" My problem is that I simply do not see why anything like this should be true ! In this expression, we know that (ρ'2 + S2 + 2S ρ' cosφ") = r12 φ" = second picture azimuth so we are talking about an arbitrary point r, not just a point in the plane of our drawings. The equality above is so "simple", there must be an easy and elegant way to show it, without going off into hundreds of lines of algebra. In the above equation we have B,S,r1,ρ' all parameters in the iris space b,s parameters in the charged disk space d the oddball scaling parameter I can move b to the iris space using this fact b = B (d2 + S2) / (B2-S2) // is this valid? [ YES. ] so imagine we use this to replace b. Then we have B,S,r1ρ' all parameters in the iris space s parameter in the charged disk space d the oddball scaling parameter So you would think that all we have to do is figure out how to relate the distance s to parameters in the iris space, then we should have it! Sounds easy. Study the distance s. From #2 "What do my Inversion Lemmas have to say about s?" after "back up some more" we know that s = ρ' =(AC/R) | - 3| = (AC/R) 2 sin(γ'/2) R = 2A cos(θ/2) so we have this fairly simple result: s = C sin(γ'/2)/ cos(θ/2) (*) γ and θ are in the first-inversion picture _____________________________ error and recovery ________________________________ As we let s "rotate around" on the charged disk, point r rotates around on the sphere, but angle γ' does not change. The reason is that, during this rotation, the arrow r from sphere center describes a cone of angle γ' relative to the line from sphere center to fixed point r3. For a general position of r we know that cosγ' = (cosθcosθ3 + sinθsinθ3cosφ) and as we do this motion, both θ and φ vary, but γ' stays fixed. But now we have a contradiction: if γ' stays fixed, and θ changes, then s must change from (*). My expression for s must therefore be wrong! I am completely blocked now until this contradiction is resolved, this will take maybe 6 hours. Bigger contradiction: Imagine ρ' = s rotating around on the disk. As it rotates, the point r rotates in some manner. In Attempt #1 I argued that as ρ' goes around the perimeter of the disk, r goes around a circle on the sphere and this circle bounds a "spherical cap" and the center of the spherical cap (lying on the cap) is at point rc which is at angle θc. But intuitively (perhaps wrongly) you would think then that as you reduced ρ' in length, as it circles around C, you would think you just have a smaller spherical bowl still centered at rc. But then in the limit that ρ' = 0, the bowl should be at rc. But that seems to imply that rc = r3 which I have argued elsewhere is not the case. So my entire universe is tumbling down now. I am unable to do simple geometry. Resolution: The circular locus which is the perimeter of the disk (radius b) maps into a circular locus on the sphere. This circular locus defines a "bowl" whose "pole" rc is located here at angle θc in the plane of paper, where tanθc = 2cd / (d2+b2-c2) We could consider a set of concentric circles on the disk. Suppose one has radius ρ'. Then for such a circle, the pole is at this angle tanθc = 2cd / (d2+ρ'2-c2) which is a function of ρ'. Thus, the "bowl pole" in fact MOVES as you work your way in on this set of c concentric circles on the disk. In the limit of ρ' = 0, we get tanθc = 2cd / (d2-c2) => tan(θc/2) = (c/d) = tan(θ3/2) So the "bowl pole" moves as we reduce ρ'. I think I did not realize this fact, and it may have led to mistakes in my 75 pages of algebra! I now have to go scan everything and make repairs! Scan Attempt #1: Everywhere in this entire doc I always use only ρ' = which is still correct and is unaffected by the above doc. Scan Attempt #2: Through page 12, this is still the case, even when I write s = . Still clean through page 18. Finally on page 21 we get to the section " What do my Inversion Lemmas have to say about s?" I add the point r3 which is the back map of point C. I then ignore everything till the second "go back" in red, so to page 24. We are OK through all of that section. Page 25 starts the Smythe compare, and that section is OK, we are to page 28. OK to page 31. OK, scan done. Scan Attempt #3. I just did a quickie on this, since mostly trash anyway. Now that we have understood this confusion, let's start over: ________________________ resume ____________________________________________ Study the distance s, Version 2 start review here From #2 "What do my Inversion Lemmas have to say about s?" after "back up some more" we know that s = ρ' =(AC/R) | - 3| = (AC/R) 2 sin(γ'/2) R = 2A cos(θ/2) so we have this fairly simple result: [ θ is a first-inversion picture angle] s = C sin(γ'/2)/ cos(θ/2) (*) cosγ' = (cosθcosθ3 + sinθsinθ3cosφ) = 3 As we let s "rotate around" on the charged disk at some radius s ≤ b, point r rotates around on the sphere. In fact it rotates on a cone with a central point that lies somewhere between rc and r3. This cone center point aligns with rc when s = b, and aligns with r3 when s = 0. The cone angle only agrees with γ' in this last limiting case. So we now see this fact: as s rotates around on the disk at some radius s > 0, the point r moves on the sphere in a circular locus, the angles (θ,φ) of r vary, and γ' varies as well, because we are not just rotating on a γ' cone. So, all three variables θ,φ,γ' are varying. So, when we look at equation (*) above, we see that the LHS stays fixed during our rotation, but both trig functions on the RHS vary. They vary in a way that the variations offset and we have a constant s on the LHS. Now, as we said, we want to somehow relate distance s to parameters in the iris world. But this expression for s is not very handy because it is a messy function of (θ,φ) of the pre-rotation sphere world. We can regard C and θ3 as constants, they are no problem. Now I am perhaps coming into another mistake I may have made. I kept saying in some sense that φ = φ', but when we do our bowl rotation, we can regard that as a change from θpre, φpre to θpost, φpost and the two φ's are not going to be the same! [ that is correct, they are not the same 6.13.10] Below I will refer to these two coordinate pairs as θ,φ and θ",φ". Let's go back instead to our "Cartesian" approach where we had (I quote changing ρ' to s) b2 - s2 = b2 - (x'-c)2-y'2 = b2- c2 - x'2 - y'2 +2cx' = (b2- c2) - (x'2+y'2-2cx') = "ugly mess" R = x' = (a2/R2)X y' = (a2/R2)Y z' = (a2/R2)(Z+A) where r = (X,Y,Z) is a point on the sphere in the first inversion picture, sphere-centered. If we actively rotate this point into r" = (X",Y",Z") by angle β = (π-θc), we are then saying r" = Ry(β) r which is the same as r = Ry(-β) r" . On page 9 of Attempt #2A we did R" = Ry(-β) R so I will steal the results there and modify them to get X = -cosθcX" - sinθcZ" = A( -cosθcsinθ"cosφ" - sinθccosθ") Y = Y" = A(sinθ"sinφ") Z = sinθcX" - cosθcZ" = A(sinθcsinθ"cosφ" - cosθccosθ") Z + A = A(sinθcsinθ"cosφ" - cosθccosθ" + 1) We can insert these into the above expressions then to get x' = (a2/R2)X = (a2/R2)[ -cosθcX" - sinθcZ"] = (a2/R2)A( -cosθcsinθ"cosφ" - sinθccosθ") y' = (a2/R2)Y = (a2/R2)[ Y"] = (a2/R2)A (sinθ"sinφ") z' = (a2/R2)(Z+A) = (a2/R2){ sinθcX" - cosθcZ"} + A} = (a2/R2)A{ (sinθcsinθ"cosφ" - cosθccosθ") + 1} Further, we have R = = A At this point, we can tell Maple to compute the inside of this radical: ("f1 f2 f3 R thing.mws") and we find then that R2 = 2A2 ( 1 - cosθccosθ" + sinθcsinθ"cosφ") So now we can express things in terms of the coordinates θ",φ" of the actively rotated point r": b2 - s2 = b2 - (x'-c)2-y'2 = b2- c2 - x'2 - y'2 +2cx' = (b2- c2) - (x'2+y'2-2cx') = "ugly mess" R2 = 2A2 ( 1 - cosθccosθ" + sinθcsinθ"cosφ") x' = (Aa2/R2)( -cosθcsinθ"cosφ" - sinθccosθ") y' = (Aa2/R2) (sinθ"sinφ") z' = (Aa2/R2) (sinθcsinθ"cosφ" - cosθccosθ" + 1) What is the meaning of R here? It is a certain distance in the pre-rotation picture. The pre-rotation result is simple: R = 2Acos(θ/2) which follows from the usual development based on the pre-rotation picture R2 = (A+Acosθ)2 + (Asinθ)2 I now want to say more about the meaning of R. We can write R = | r - O | where O means the origin in the pre-rotation space. After the rotation, R is the same number, but we can write it as R = | Ry(π-θc)r - Ry(π-θc)O | ≡ | r" - O" | where r" = Ry(π-θc) r O" = Ry(π-θc) O O" = Ry(π-θc) O = Ry(β) O = O = O = = = // agrees with pic #2 The location O" is where I put the Green's charge in the post-rotation picture, and r" is a vector whose sphere-centered spherical coordinates are (θ",φ"). [ Good! ] If I drew an accurate picture, r" would lie a bit below the origin of the post-rotation picture. However, I am allowed to draw it anywhere I like, as long as I understand what it is. What seems to be missing is a direct relationship between (θ,φ) and (θ",φ"). I have sort of danced around it, but have not written it down yet. Question: How are (θ,φ) and (θ",φ") related to each other? In this section, we are talking about taking a vector at θ,φ and actively rotating into a vector at θ",φ", all this being done in the first inversion space whose coordinates we always call θ,φ. I am not talking here about any coordinate system which has spherical angles θ",φ". Consider a vector on the sphere before and after the rotation; V = (A,θ,φ) // pre-rotation If I "actively" rotate this vector by Ry(π-θc), I get some new vector in the same space: V" = (A,θ",φ") = (1,θ,φ) " = (1,θ",φ") " = Ry(π-θc) Alternatively, I can let S" be a coordinate system that is rotated backwards relative to S and I can ask what the coordinates of the vector V are in this rotated coordinate system. (6.13.10) I can do the math right here, stealing stuff from "effect of transformations..." . In that doc I used R" = R1-1R, but here I am using R" = R1R for an active rotation so I will have to modify the result by negating the rotation angle. My result in that doc for R1 = Ry(β) was this ("euler stuff.mws") " = = To apply this to the current situation, I would set β = - (π-θc). This means we replace cos(β) = cos(π-θc) = -cosθc sin(β) = -sin(π-θc) = -sinθc giving this result " = = Ry(π-θc) = Ry(π-θc) = which is three equations: // THIS shows how (θ",φ") is related to (θ,φ) !!! sinθ"cosφ" = -cosθcsinθcosφ+ sinθccosθ sinθ"sinφ" = sinθsinφ cosθ" = -sinθcsinθcosφ - cosθccosθ The last one tells us θ" and the first two give us φ" in (0,2π). So, if we take a charge point on the pre-rotation sphere located at θ,φ, when we do our active rotation, that same charge point ends up at location θ",φ". The main point: in general we end up with θ",φ" ≠ θ,φ Special case φ = 0: φ = 0: cosφ = 0 sinφ = 0 sinθ"cosφ" = -cosθcsinθ+ sinθccosθ = sin(θc-θ) sinθ"sinφ" = 0 cosθ" = -sinθcsinθ - cosθccosθ = - cos(θc-θ) = cos( π - (θc-θ)) But θ" is a polar angle and is restricted then to (0,π). The angles θ and θc are also polar angles and also lie in this range. Therefore (θc-θ) lies in (-π,π) so that π - (θc-θ) lies in (0,2π). Aside: Consider then the abstract problem cosθ" = cosx where x can lie in the range (0,2π) but we require θ" in (0,π). If x is in the lower range half, θ" = x, but if x lies in the upper half, then θ" = 2π-x (draw some pictures). Then we conclude that sinθ" = sinx for x in (0,π), but sinθ" = sin(2π-x) = - sinx for x in (π,2π). Can write this as sinθ" = (-1)Int(x/π) sinx where Int means integer part of. So continuing the above, we get #3 => sinθ" = (-1)Int([π - (θc-θ]/π) sin (π - (θc-θ)) = (-1)Int([π - (θc-θ]/π) sin(θc-θ) #1 => cosφ" = (-1)Int([π - (θc-θ]/π) = (-1)Int([π+θ-θc]/π) If as in my picture θ > θc, then Int(π+θ-θc) = 1, cosφ" = -1, φ" = π. Else φ" = 0. So: φ = 0 => θ" = φ" = Restatement from above about b2 - s2 in terms of post rotation angles: A few sections above we wrote these results which express b2 - s2 in terms of the location of the post-rotated point r", b2 - s2 = b2 - (x'-c)2-y'2 = b2- c2 - x'2 - y'2 +2cx' = (b2- c2) - (x'2+y'2-2cx') = "ugly mess" R2 = 2A2 ( 1 - cosθccosθ" + sinθcsinθ"cosφ") x' =(Aa2/R2)( -cosθcsinθ"cosφ" - sinθccosθ") y' =(Aa2/R2) (sinθ"sinφ") z' = (Aa2/R2) (sinθcsinθ"cosφ" - cosθccosθ"+ 1) and then we learned that we can connect post rotation angles to pre rotation angles this way: sinθ"cosφ" = -cosθcsinθcosφ+ sinθccosθ sinθ"sinφ" = sinθsinφ cosθ" = -sinθcsinθcosφ - cosθccosθ As a quick check, let's insert these last results into our R2 expression above. R2 = 2A2 [ 1 - cosθccosθ" + sinθcsinθ"cosφ" ] = 2A2 [ 1 - cosθc(-sinθcsinθcosφ - cosθccosθ) + sinθc(-cosθcsinθcosφ+ sinθccosθ) ] = 2A2 [ 1 + cosθc(cosθccosθ) + sinθc(sinθccosθ) ] = 2A2[ 1 + cosθ] = 2A2cos2(θ/2) => R = 2A cos(θ/2). and this agrees with our traditional first-inversion picture computation of R2 ! review check good to here Starting Over with an "Active" View of the Bowl Rotation: Rotating points and fields! I am now rereading my section " Doing the Bowl Rotation prior to attempting the Second Inversion" in Attempt #2. I don't want to rotate any coordinate systems, I want to actively rotate the bowl and its potential and charge distribution. I want to have only coordinates called (X,Y,Z) for sphere-centered Cartesians. How do I DO such an active rotation? Let's start with the Green's charge which is at point O prior to rotation. In my new picture I will have rnew,Green's charge = Ry(π-θc) O = Ry(π-θc) (-A, 0,0) I use the word "new" when discussing points in the apparatus after doing this active rotation. So that is all well and good for rotating physical points on an apparatus. But how do I rotate a potential or electric field or charge? These things are all mathematical "fields", call them generically f(X,Y,Z). Both the potential and charge σ are "scalar fields" and I expect the rule to be this fnew(r) = f(R-1r) Let's try the simplest possible example, the potential of an electric dipole moment p = p : V = (pr) / r3 = pz/r3 = pcosθ/r2 We usually draw "contours" by setting V = constant so we then have r2 = (p/V)cosθ = a cosθ = a z/r => r3 = az In Cartesians this becomes (looking just in the x,z plane say) (x2+ z2)3/2 = az => (x2+ z2)3 = a2z2 which is a 6th order curve that makes the usual "dumbbell" which has a positive value for z>0. Here is the + side of the dumbbell: So here is a pair of contours for our dipole potential, where I draw in x,y space: (see left pic) Suppose we now rotate the dipole moment actively so that pnew = Rz(θ1) p . I expect to get the picture in the middle. Now let's see if our conjecture is correct. We have V(r) = V(r,θ) = pcosθ/r2 Vnew(r) =?= V(R-1r) = V(r,θ-θ1) = pcos(θ-θ1)/r2 So the new potential in the middle has the contour maximum when θ = θ1. So Vnew is indeed the correct "rotated potential", and our general rule is validated in this simple example. We can then simply redraw the middle picture as on the right and introduce a new coordinate system called S' with coordinates (x',y') = r' where r' = R-1r. We could then say V'new(r') = V(r) = V(Rr') which says that the new potential viewed in the new coordinates is the same as the original potential viewed in the original coordinates, and you see a vertical dumbbell in either case, left vs right pics. Now, suppose we have a more general situation V(r) = V(r,θ,φ) Vnew(r,θ,φ) = V(R-1r) which is how we "rotate the field by rotation R". Suppose we have this active rotation in Cartesian coordinates, R = Rz(φ1) Ry(θ1) = = R(θ1, φ1) If we applied this to our little p dipole, it would take it to down off the z axis, then rotate azimuth, so our p would end up at angles (θ1,φ1). We can then compute R-1 = RT to get a matrix for R-1. Our general question then is this. Suppose we write r' ≡ R-1r = R-1(θ1, φ1) R(θ,φ) r = r Ry(-θ1) Rz(-φ1) Rz(φ) Ry(θ) = r Ry(-θ1) Rz(φ-φ1)Ry(θ) = (r,θ',φ') Find expressions for θ' and φ'. Fri Jun 11. OK, I have gone off and written a transformation doc, I have created Attempt #2A with some things clarified, but I don't think anything major changed, maybe a few signs. What is clearer is that my active bowl rotation makes sense, and I know how the charge and fields rotate in a conceptual sense as well as in the sense of having exact expressions (which I mostly already had). So let's therefore back up and try again: Study the distance s, Version 3 From #2 "What do my Inversion Lemmas have to say about s?" after "back up some more" we know that s = ρ' =(AC/R) | - 3| = (AC/R) 2 sin(γ'/2) R = 2A cos(θ/2) Review 1: s/ |r-r3| = primed/unprimed = (C/R), simple application of the alternate lemma. But |r-r3|2 = A2+ A2 - 2AAcosγ' = 2A2(1-cosγ') = 4A2sin2(γ'/2). So s = 2Asin(γ'/2) (C/R) Review 2: R2 = (Asinθ)2 + (A+Acosθ)2 = A2 + A2 + 2A2cosθ = 2A2(1+cosθ) = 4A2 cos2(θ/2). so we have this fairly simple result: [ angles are first-inversion picture angles ] s = C sin(γ'/2)/ cos(θ/2) (*) cosγ' = (cosθcosθ3 + sinθsinθ3cosφ) = 3 As we let s "rotate around" on the charged disk at some radius s ≤ b, point r rotates around on the sphere. In fact it rotates on a cone with a central point that lies somewhere between rc and r3. This cone center point aligns with rc when s = b, and aligns with r3 when s = 0. The cone angle only agrees with γ' in this last extreme case. So we now see this fact: as s rotates around on the disk at some radius s > 0, the point r moves on the sphere in a circular locus, the angles (θ,φ) of r vary, and γ' varies as well (θ3 is a constant), because we are not just rotating on a γ' cone. So, all three variables θ,φ,γ' are varying. So, when we look at equation (*) above, we see that the LHS s stays fixed during our rotation, but both trig functions on the RHS vary. They vary in a way that the variations offset and we have a constant s on the LHS. Now, as we said, we want to somehow relate distance s to parameters in the iris world. But this expression for s s = C sin(γ'/2)/ cos(θ/2) (*) cosγ' = (cosθcosθ3 + sinθsinθ3cosφ) = 3 tan(θ"/2) = (ρ'/d) // θ" means Pic 2 angle is not very "handy" because it is a messy function of (θ,φ). We can regard C and θ3 as constants, they are no problem. Nevertheless, if Smythe's result is correct, (S2+d2) (1/ ) = π r1 ( 1/ ) (*) checked !!! Then I should at least be able to verify it by "brute force". Let's gather up all the potentially useful pieces, assuming I did them correctly: (Items I use will be blue, and then red after checking) C2 = d2+ c2 tan(θb/2) = d/B // since tan([π-θb]/2) = B/d = cot(θb/2) from picture b = B (d2 + S2) / (B2-S2) c = S (B2 + d2) / (B2-S2) r' = // inversion origin for second inversion r = a2/r' = a2/ = 2Ad / x = (a2/r'2) x' = [a2/(ρ'2+ d2)] ρ' cosφ' y = (a2/r'2) y' = [a2/(ρ'2+ d2)] ρ' sinφ' z = (a2/r'2) z' = [a2/(ρ'2+ d2)] d = 2Ad2 / (ρ'2+ d2) 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) tan(θ3/2) = c/d cos(θ3/2) = d/ sin(θ3/2) = c/ sinθ3 = 2dc/(d2+c2) cosθ3 = (d2- c2)/(d2+c2) ρ' = (d/r) 2Asin(θ/2) ρ'2 = (2Ad)2/r2 - d2 r = 2Acos(θ/2) R = a2r1/ (ρ'2 + S2 + 2S ρ' cosφ') = r12 Notice that only two things in (*) depend on azimuth angle φ = φ': r1 and s. Therefore, in order for the above to be true. we must somehow have r1 = independent of φ or r12(b2- s2) = independent of φ So as usual, we start with s s2 = C2 sin2(γ'/2)/ cos2(θ/2) = (d2+ c2) sin2(γ'/2) (ρ'2+d2)/d2 = d-2 (d2+ c2) (ρ'2+d2) sin2(γ'/2) so the whole ball of wax hinges on this γ' thing. We have sin2(γ'/2) = (1/2)(1 - cosγ') s2 = (1/2d2) (d2+ c2) (ρ'2+d2) (1 - cosγ') Just as an aside, we know that b2 = B2 (d2 + S2)2 / (B2-S2)2 so that b2 - s2 = B2 (d2 + S2)2 / (B2-S2)2 – (1/2d2) (d2+ c2) (ρ'2+d2) (1 - cosγ') Now back to the main act: cosγ' = (cosθcosθ3 + sinθsinθ3cosφ) = (d2- ρ'2)/(d2+ ρ'2) * (d2- c2)/(d2+c2) + 2d ρ'/(d2+ ρ'2)* 2dc/(d2+c2)* cosφ = [(d2- ρ'2) (d2- c2) + 2d ρ'* 2dc* cosφ]/[(d2+ ρ'2) (d2+c2)] = [(d2- ρ'2) (d2- c2) + 4d2cρ' cosφ]/[(d2+ ρ'2) (d2+c2)] which is "not too bad". Then we can compute 1 - cosγ' = {(d2+ ρ'2) (d2+c2) - [(d2- ρ'2) (d2- c2) + 4d2cρ' cosφ] } / [(d2+ ρ'2) (d2+c2)] = {(d2+ ρ'2) (d2+c2) - (d2- ρ'2) (d2- c2) - 4d2cρ' cosφ } / [(d2+ ρ'2) (d2+c2)] But we have (d2+ ρ'2) (d2+c2) - (d2- ρ'2) (d2- c2) = = (d4 + ρ'2d2 + d2c2 + ρ'2 c2) - (d4 - ρ'2d2 - d2c2 + ρ'2 c2) = ( d4 + ρ'2d2 + d2c2 + ρ'2 c2 - d4 + ρ'2d2 + d2c2 - ρ'2 c2) = (ρ'2d2 + d2c2 + ρ'2d2 + d2c2) = 2 d2c2 + 2 ρ'2d2 So we then have 1 - cosγ' = { 2 d2c2 + 2 ρ'2d2 - 4d2cρ' cosφ } / [(d2+ ρ'2) (d2+c2)] = 2d2( c2 + ρ'2 - 2cρ' cosφ ) / [(d2+ ρ'2) (d2+c2)] which says (d2+ ρ'2) (d2+c2)( 1 - cosγ')/ (2d2) = ( c2 + ρ'2 - 2cρ' cosφ ) but this then tells us that s2 = (1/2d2) (d2+ c2) (ρ'2+d2) (1 - cosγ') or s2 = ( c2 + ρ'2 - 2cρ' cosφ ) which is an amazing fact that begs for an interpretation. The three distances s, c, ρ' don't exist in the same space, yet we seem to have some kind of triangle rule here. [ However, in the inversion-one R space we DO have this triangle condition: r22 = c2 + s2 + 2cs cosφ ] Now we can check our claim about φ dependence. We have r12 = S2 + ρ'2 + 2Sρ' cosφ // sign corrected I think So we would need this to be true: r12(b2- s2) = independent of φ which says (S2 + ρ'2 + 2Sρ' cosφ) [b2 - (c2 + ρ'2 - 2cρ' cosφ ) ] = independent of φ But this thing is a polynomial of degree 2 in cosφ, so how can it be independent of φ ? Quadratic term: (2Sρ') (- 2cρ') cos2φ ≠ 0 Linear term: [2Sρ' (b2- c2 - ρ'2) - 2cρ' (S2+ ρ'2) ] cosφ ≠ 0 So there is no need to keep going, we are already in trouble, the Smythe requirement above cannot be true. When S = 0 we have c = 0 so then these two terms DO vanish and we have a chance (and of course we also know it works). So I have now proven that my Smythe condition is NOT true! (S2+d2) (1/ ) ≠ π r1 ( 1/ ) Either I am wrong or Smythe is wrong (any bets? ) Let's go back now to our new-found expression for s and try to disprove or prove it: s2 = ( c2 + ρ'2 - 2cρ' cosφ ) But wait, I just had an idea. Just as there are two different ρ' variables, maybe there are two different θ variables. I argued that I actively rotated the bowl, and that the (θ,φ) before and after were the same, for a point r in the space (origin at bowl center). Why we have two different θ coordinates which are not the same. Consider, side by side, our two inversion pictures: Had I done this accurately, both spheres would be the same size, the lower bowl would have a small bowl angle (same as the upper bowl), and the iris hole radius would be quite large, perhaps 8" high in the lower picture (hence, hard to draw and still see sphere details, which is why I distorted it). Also, we set the two distances d the same in both pictures. Imagine I had done this. Then consider in the upper picture the sphere point r shown, in the plane of paper, with θ = 45 degrees. The inversion angle is θ/2 and at this angle we can see that the σ' at some point r' on the disk maps into some σ on the bowl at r. That is just fine. Now suppose we go to the lower picture and consider θ = 45 degrees in that picture. The corresponding point r is not on the bowl! And its mapping r' is not on the iris! Now in the upper picture we might say that tan(θ/2) = (s+c)/d and in the lower picture that tan(θ/2) = ρ'/d. If we regard θ = θ as the same angle (after all, θ = θ = 45 degrees, how can you say they are not the same?), we get ρ' = s+c. Here, s corresponds to a point on the disk, while ρ' corresponds to a point NOT on the iris. So you ask: what is the meaning of a relation obtained in this manner, for example, of the claim that ρ' = s + c ? We certainly know that the charge density on the disk at s will NOT be related to the zero charge density that will be present at ρ' = s+c which in the hole of the iris. Let's try reductio ad absurdam. Our pictures can be simplified: If someone says both these pictures are valid, you would correctly conclude that ρ' = s+c. But if the ρ' you actually want goes up to the dot in the second picture, then you would say yes, ρ' = s+c, but although that is the s I want to be talking about, that is not the ρ' I want to be talking about, so the equation is of no use to me, although true. My conclusion is that you cannot treat the angle θ in my two pictures as if it were the same angle. If you do, then the things it marks don't "correspond", and we are trying to trace σ → σ → σ for corresponding points. So I suspect in my work above I have identified the two angles. For example, I make use of both these "facts" : R = 2A cos(θ/2) where θ is from the first-inversion picture cos(θ/2) = d/ where θ is from the second-inversion picture By equating these cos(θ/2) objects, I end up with a garbage equation, similar to the ρ' = s+c just mentioned. So that is easily a good reason why my result comes out "wrong" (disagrees with Smythe). The right way out of this problem is to "be done" with the first picture and extract from it the charge σ on the bowl in terms of parameters involving the bowl. Then rotate this charge. Then draw the second picture and go ahead and use θ in that picture if you like. I think I have essentially done this, so let's try to track it. Doing Things Right, Version 1 (1) Compute Φ and σ on the bowl in the first picture prior to bowl rotation. Result is this: g(r|ξ) = – (2/πR) sin-1 [ 2b / ( + ) ] + 1/R σ(r) = - (a2/π2R3) / ρ' = R = x' = (a2/R2)X d = a2/2A y' = (a2/R2)Y ξ = (0,0,-A) = Green's charge loc z' = (a2/R2)(Z+A) θc = (θ1 + θ2)/2 where (X,Y,Z) are sphere-centered Cartesian coordinates. The pre-rotated bowl and its orientation is fully characterized by parameters A, θ1 and θ2. That is to say, the bowl metal is on the right side of the sphere, and in the planar picture its angle θ runs from θ2 up to θ1. We could alternatively use θc = (θ1 + θ2)/2 // angle bowl is "tilted up" in the pre-rotation picture θb = (θ1 - θ2)/2 // bowl polar angle (in any picture) and we could make these replacements c = (d/2)[ tan(θ1/2) + tan(θ2/2)] b = (d/2)[ tan(θ1/2) – tan(θ2/2)] At this point, then, we could imagine that our pre-rotation bowl's Φ and σ are fully described by (X,Y,X) along with parameters A, θb, θc, a, d , where these last two we know are just dummy variables and nothing depends on them. Note: If we select (X,Y,Z) on the metallic part of the sphere, then ρ' will lie on the metallic disk, so we will find that ρ' ≤ b. (2) Rotate this bowl (and the point charge). After doing this, Φ and σ for the new post-rotated bowl are given by: (I have renamed R to be H) gpost(R; ξ',σ') = g (R"; ξ,σ) = – (2/πH) sin-1 [ 2b / ( + ) ] + 1/H σpost(R) = σ(R") = - (a2/π2H3) / H = H(R") = H(X",Y",Z") = | R" - (0,0,-A| ρ' = H = = H(X",Y",Z") x' = (a2/H2)X" d = a2/2A y' = (a2/H2)Y" ξ = (0,0,-A) = Green's charge loc z' = (a2/H2)(Z"+A) θc = (θ1 + θ2)/2 where the meaning of the dummy temporary symbols X" Y" and Z" is this: X" = -cosθcX - sinθcZ Y" = Y Z" = sinθcX - cosθcZ Thus, we now have gpost and σpost expressed as functions of (X,Y,Z) sphere centered. This rotated bowl's parameters again can be regarded as just A, θb, θc, a, d. (3) Convert this to inversion coordinates appropriate for the second-inversion picture, then we have: (where I now remove the post labels from g and σ) g(R; ξ',σ') = – (2/πH) sin-1 [ 2b / ( + ) ] + 1/H σ(R) = - (a2/π2H3) / H = H(R") = H(X",Y",Z") = | R" - (0,0,-A| s = ρ' = H = = H(X",Y",Z") x' = (a2/H2)X" d = a2/2A y' = (a2/H2)Y" ξ = (0,0,-A) = Green's charge loc z' = (a2/H2)(Z"+A) θc = (θ1 + θ2)/2 where the meaning of the dummy temporary symbols X" Y" and Z" is this: X" = -cosθcx - sinθc(z-A) Y" = y Z" = sinθcx - cosθc(z-A) and (x,y,z) are the inversion-centered Cartesian coordinates. Again, the bowl parameters can be regarded as A, θb, θc, a, d. Notice that we don't have any angles θ,φ in this section (3). (4) Now draw the full second-inversion picture where (x,y,z) are the inversion coordinates. We may now introduce (θ,φ) coordinates for this picture if we like, in the obvious manner. We may not equate these to angles of the same name in the first-inversion picture, however. We can proceed to write various equations which relate to the geometry of the second-inversion picture, some of which involve θ,φ. (5) We shall be interested in the quantity s. We see that s = x' = (a2/H2)X" y' = (a2/H2)Y" H = X" = -cosθcx - sinθc(z-A) Y" = y What does this all mean? If we pick a location r = (x,y,z) in our second-inversion picture which lies on the sphere in that picture, then that point on the sphere will "correspond" to this value of s in the first picture. If s is a point on the disk, then (x,y,z) will be a point on the bowl of the rotated sphere. Point (x,y,z) corresponds to (X", Y", Z") of section 2 above which is a point on the rotated sphere, and this corresponds to point (X,Y,Z) on the unrotated sphere. (6) So let's go back to (X", Y", Z") which is a point on the rotated sphere. We could choose to represent this point in (θ,φ) coordinates of the 2nd picture. So go back to (2) above and write: gpost(R; ξ',σ') = g (R"; ξ,σ) = – (2/πH) sin-1 [ 2b / ( + ) ] + 1/H σpost(R) = σ(R") = - (a2/π2H3) / H = H(R") = H(X",Y",Z") = | R" - (0,0,-A| ρ' = H = = H(X",Y",Z") x' = (a2/H2)X" d = a2/2A y' = (a2/H2)Y" ξ = (0,0,-A) = Green's charge loc z' = (a2/H2)(Z"+A) θc = (θ1 + θ2)/2 where the meaning of the dummy temporary symbols X" Y" and Z" is this: X" = Asinθcosφ Y" = Asinθsinφ Z" = Acosθ If we pick θ,φ such that we have a metallic point on the rotated sphere, then ρ' we get working backwards will be ρ' < b so we have a metallic point on the charged disk. If this is all correct, we then have STOP, try another approach. Doing Things Right, Version 2: The Basic Theory Equations Let's imagine four worlds: disk, pre-rotated sphere, post-rotated sphere, iris. We want to trace a "point" through these four worlds. (1) In Disk World, we have a point rdisk at some arbitrary location in Disk World space, not necessarily on the disk. The origin for rdisk is our inversion origin. (2) We do our first inversion, and we find the corresponding point in Pre World to be this rpre = (a/rdisk)2 rdisk rdisk = | rdisk | where rPre is relative to the same inversion origin. However, we can construct a vector rPre,sc , which means this same vector but relative to sphere-center in Pre World. We find that rpre = A + rpre,sc => rpre,sc = rpre – A (3) To get from Pre World to Post World, we do a certain active rotation R1 about the sphere center. Let's just keep this arbitrary for now and call it R1. We then obtain rpost,sc = R1 rpre,sc so now we happily know our corresponding point in Post World, but so far it is relative to sphere center there. But we can convert rpost,sc to Post World inversion coordinates like so rpost = A + rpost,sc (4) next, we take rpost into Iris World as follows: riris = (a/rpost)2rpost = rpost= | rpost | where we assume we use the same scaling factor "a" for our second inversion. (5) Comments: we have done the trace rdisk → rpre → rpost → riris . Presumably, if we start with a point on the charged disk itself in Disk World, we will end up at the "corresponding point" on the iris in Iris World. (6) We can combine things together, so let's work from the start: rpre = (a/rdisk)2 rdisk rdisk = | rdisk | Nothing fancy so far. We presumably know rdisk and hence its length rdisk. Next, rpre,sc = rpre – A = (a/rdisk)2 rdisk – A So far so good. Next, rpost,sc = R1 rpre,sc = R1 [(a/rdisk)2 rdisk – A ] and again, everything is known, no mysteries here. Next rpost = A + rpost,sc = A + R1 [(a/rdisk)2 rdisk – A ] At this point, we have to do some work (but we know how to do it) and we compute rpost= | rpost | Then we go to the last step riris = (a/rpost)2 rpost = (riris/a)2 rpost and we are done. (7) Now that we have figured out the sequence of corresponding points, let's consider what happens to potential. I could assume some completely strange initial Φ in Disk World, something very general. So we have some Φdisk(rdisk) which we are given at the start. To get to Pre World, we have to do our first inversion. I find in my Jackson META notes that one form we can use is φ(r) = (a/r) φ'(r'). I will think of the prime as being Disk World and unprimed as being Pre World, so we can say Φpre(rpre) = (a/rpre) Φdisk(rdisk) = (rdisk/a ) Φdisk(rdisk) which seems pretty simple and doable. The next step I think is pretty simple as well: Φpost(rpost) = Φpre(rpre) The potential at the rotated point is the post world is the same as that at the pre rotated point in the Pre World. There is no scale factor involved here as there is with a rotation. Now we can do our inversion to the iris. I select φ'(r') = (a/r') φ(r) and now identify prime with the iris and unprimed with the post world. So Φiris(riris) = (a/riris) Φ(rpost) = (rpost/a) Φ(rpost) So now we have traced the surface charge density through the four worlds, and we should be able to combine or telescope this down. We have Φiris(riris) = (rpost/a) Φ (rpost) = (rpost/a) Φpre(rpre) = (rpost/a) (rdisk/a ) Φdisk(rdisk) So this is a final result I want to "work with" : Φiris(riris) = (rpost/a) (rdisk/a ) Φdisk(rdisk) (8) Now let's repeat all the above for our charge density σ. Start with some σdisk(rdisk) in Disk World. As we go through the first inversion, I will use σ(r) = (a/r)3 σ'(r') = (r'/a)3 σ'(r') to get σpre(rpre) = (rdisk/a)3 σdisk(rdisk) pre = noprime disk=prime Then of course we have σpost(rpost) = σpre(rpre) and again there is no scaling factor. For the last inversion I use σ'(r') = (r/a)3 σ(r) = (a/r')3 σ(r) so σiris(riris) = (rpost/a)3 σpost(rpost) post = noprime iris=prime So here is the whole ball of wax: σiris(riris) = (rpost/a)3 σpost(rpost) = (rpost/a)3 σpre(rpre) = (rpost/a)3 (rdisk/a)3 σdisk(rdisk) Here then our the complete results of this section: (a) riris = (a/rpost)2 rpost where rpost = A + R1 [(a/rdisk)2 rdisk – A ] (b) Φiris(riris) = (rpost/a) (rdisk/a ) Φdisk(rdisk) (c) σiris(riris) = (rpost/a)3 (rdisk/a)3 σdisk(rdisk) This is the first time I have stated these rather elegant equations. Doing Things Right, Version 2: Apply these Equations We start by looking at our first-inversion picture and saying, for a point ON the disk, because we are first going to worry about charge distribution: ( For the potential we have to replace d with general z. ) rdisk = (c + s cosφ, s sinφ, d) // inversion origin Cartesian, check rdisk2 = s2 + c2 + d2 + 2cscosφ Let's be careful at this very early stage not to get messed up. If we look at our disk from the left, facing the +z axis, here is what we see: So φ = 0 when our rdisk is at the highest point in the plane of paper or our first-inversion picture. We might as well install our correct R1 = Ry(π-θc) rotation. Then we have rpost = A + Ry(π-θc) [(a/rdisk)2 (c + s cosφ, s sinφ, d) – A ] = A + (1/rdisk2) Ry(π-θc) [ a2 (c + s cosφ, s sinφ, d) – rdisk2 (0,0,A) ] = A + (1/rdisk2) Ry(π-θc) [(a2c + a2s cosφ, a2s sinφ, a2d) –(0,0, rdisk2A) ] = A + (1/rdisk2) Ry(π-θc) [(a2c + a2s cosφ, a2s sinφ, a2d - rdisk2A) ] Now we go write Ry(π-θc) = = where we know that cos(π-θc) = - cosθc sin(π-θc) = + sin θc Then we compute Ry(π-θc) [...] to be: = Now define a temp symbol ε = a2/rdisk2 We then have (1/rdisk2) Ry(π-θc)[...] = Now finally we want to add A and we get our final result: rpost = ε = a2/rdisk2 So, for s ≤ b, this rpost should be a point on the post-rotation metal part of the sphere. Our next step is this: riris = (a/rpost)2 rpost = (riris/a)2 rpost and it seems easier maybe to choose the second form here. Then we have riris = (riris/a)2 Looking now at our second-inversion drawing, we know we can write this as riris = (ρ' cosφ', ρ' sinφ', d) riris2 = ρ'2 + d2 with this picture, again we look from the left out the +z axis, I am NOT making any claim that φ = φ' !! I can see from the pictures, however, that φ = 0 => φ' = π for the way I have defined my angles. A point high on the charged disk maps to a point near the θ1 edge of the bowl, and after rotation this ends up on the lower left section of the rotated bowl, and this then maps down to the lower part of the iris! Everything stays in the plane of paper! Similarly, φ = π => φ' = 0. So we have ended up with the following fact: ((ρ'2 + d2) / a2) = which gives us three equations, and where ε = a2/rdisk2 = a2/ (s2 + c2 + d2 + 2cscosφ) The three equations are these: // all are L3 = L3 (ρ'2 + d2) [ -ε(c+scosφ)cosθc + (εd - A)sinθc] = a2 ρ'cosφ' (ρ'2 + d2) [ε s sinφ] = a2ρ' sinφ' (ρ'2 + d2) [ε(c+scosφ)sinθc – (εd - A)cosθc + A] = a2d ε = a2/ (s2 + c2 + d2 + 2cscosφ) I would now like to "test" these equations somehow, a sanity check if you will. So let's set φ = 0 which we know goes with φ' = π and see what they say: ε = a2/ (s2 + c2 + d2 + 2cs) = a2/ [ d2 + (s+c)2 ] (ρ'2 + d2) [ -ε(c+s)cosθc + (εd - A)sinθc] = a2 ρ'(-1) (ρ'2 + d2) [ε s 0] = a2ρ' 0 (ρ'2 + d2) [ε(c+s)sinθc – (εd - A)cosθc + A] = a2d I think I can steal these earlier results which I think are still correct, tan(θc/2) = S/d cos(θc/2) = d/ sin(θc/2) = S/ sinθc = 2dS/(d2+S2) cosθc = (d2- S2)/(d2+S2) Let's then focus hard on the third equation which seems to make a testable statement. We rewrite it: {ε(c+s)sinθc – (εd - A)cosθc + A} = a2d/ (ρ'2 + d2) Mult through by [ d2 + (s+c)2 ] to get {a2(c+s)sinθc – (a2d - [ d2 + (s+c)2 ]A)cosθc + [ d2 + (s+c)2 ]A} = [ d2 + (s+c)2 ] a2d/ (ρ'2 + d2) Now install the trig facts and we have {a2(c+s) 2dS/(d2+S2) – (a2d - [ d2 + (s+c)2 ]A) (d2- S2)/(d2+S2) + [ d2 + (s+c)2 ]A} = [ d2 + (s+c)2 ] a2d/ (ρ'2 + d2) Mult through by (d2+S2): {a2(c+s) 2dS – (a2d - [ d2 + (s+c)2 ]A) (d2- S2) + (d2+S2) [ d2 + (s+c)2 ]A} = (d2+S2) [ d2 + (s+c)2 ] a2d/ (ρ'2 + d2) Maybe {..} will simplify, but Maple says no. But I could replace a2 = 2Ad at least. According to Maple, we have ("attempt4scratch.mws") so that {...} = 2A d2 ( 2Sc + 2Ss + S2 + c2+ 2cs + s2 ) = h Then our third equation above boils down to 2A d2 ( 2Sc + 2Ss + S2 + c2+ 2cs + s2 ) = (d2+S2) [ d2 + (s+c)2 ] a2d/ (ρ'2 + d2) 2A d2 ( 2Sc + 2Ss + S2 + c2+ 2cs + s2 ) = (d2+S2) [ d2 + (s+c)2 ] 2Ad2/ (ρ'2 + d2) (ρ'2 + d2) ( 2Sc + 2Ss + S2 + (s+c)2 ) = (d2+S2) [ d2 + (s+c)2 ] (ρ'2 + d2) ( 2S(s+c) + S2 + (s+c)2 ) = (d2+S2) [ d2 + (s+c)2 ] This looks like a quadratic equation which you could solve for s + c as f(ρ', d, S), something that would certainly be interesting to me. Keep rewriting: (ρ'2 + d2) ( 2S(s+c) + S2 + (s+c)2 ) – (d2+S2) [ d2 + (s+c)2 ] = 0 A (s+c)2 + B (s+c) + C = 0 A = (ρ'2 + d2) - (d2+S2) = (ρ'2- S2 ) B = 2S (ρ'2 + d2) C = (ρ'2 + d2) S2 - (d2+S2) d2 = (ρ'2 ) S2 - (d2) d2 = ρ'2S2 - d4 So our solution will involve B2 - 4AC = 4S2 (ρ'2 + d2)2 - 4 (ρ'2- S2 )( ρ'2S2 - d4) = 4 ρ'2 (d2+ S2)2 this last according to Maple, ("attempt4scratch.mws" continued) Therefore our solution is this: (s+c) = [ -2S (ρ'2 + d2) ± 2ρ'(d2+ S2)] / [ 2 (ρ'2- S2 ) ] = [ -S (ρ'2 + d2) ± ρ'(d2+ S2)] / (ρ'2- S2 ) which looks certainly promising! But why are there two solutions? Let's write each one (s+c)+ = (ρ' - S)(d2-ρ'S)/ (ρ'2- S2 ) = (d2-ρ'S)/(ρ'+S) (s+c)- = – (ρ' + S)(d2+ρ'S)/ (ρ'2- S2 ) = – (d2+ρ'S)/(ρ'-S) Now s, ρ', d, S must be positive, but c could be negative. I don't really know what to make of this result of my look at the third equation. If I pick some ρ' on the lower part of the iris (in plane), this equation tells me the value of s on the disk. I am now very discouraged by this "new approach" (yet another approach). I liked the idea that everything seems so simple in this little summary: (a) riris = (a/rpost)2 rpost where rpost = A + R1 [(a/rdisk)2 rdisk – A ] (b) Φiris(riris) = (rpost/a) (rdisk/a ) Φdisk(rdisk) (c) σiris(riris) = (rpost/a)3 (rdisk/a)3 σdisk(rdisk) This seems to say that if you start with a potential and charge on the disk, it is an easy matter to find the potential and charge on the iris. But the mapping between riris and rdisk seems very messy. Well, maybe not all THAT bad. Let's review: rdisk = (c + s cosφ, s sinφ, d) // inversion origin Cartesian, check rdisk2 = s2 + c2 + d2 + 2cscosφ riris = (riris/a)2 ε = a2/rdisk2 = a2/ (s2 + c2 + d2 + 2cscosφ) riris = (ρ' cosφ', ρ' sinφ', d) riris2 = ρ'2 + d2 => ((ρ'2 + d2) / a2) =