Smythe Problem 38 by Double Inversion Attempt #2

Smythe Problem 38 by Double Inversion Attempt #2 PhL 6.7.10 NOTE: See "top level review" for comments on this doc which is now completely obsolete. My Attempt #1 has ballooned out to 40 pages, so hard to see the forest for the trees, the usual reason to start over. The idea is to come up with the Smythe Problem 38 formula for iris charge density with a charge in the hole. My method here is to do this by a sequence of two inversions. I have changed many symbol names to remove painful confusions. The First Inversion Problem. (Sections 1 and 2) 1 The Potential in R space. (Sections 3 and 4) 3 The Charge Distribution on the Bowl (Sections 5, 6, 7) 5 Strategy for Making contact with the iris problem 7 Doing the Bowl Rotation prior to attempting the Second Inversion 8 Setting up for the Second Inversion 12 The Battle to Eliminate Non-Iris Symbols 14 What do my Inversion Lemmas have to say about R? 18 What do my Inversion Lemmas have to say about s? 19 Return to comparing my and Smythe's σ forms 23 OK, take a few shots in the dark now. 26 The First Inversion Problem. (Sections 1 and 2) I related the R' space problem of a charged disk (plus a constant potential) to the R space problem of a spherical bowl (radius A) plus a charge located on the non-bowl part of the bowl's sphere (at the inversion origin). The picture is this: I am making an important "deviation" from the usual inversion notation here. A point r which might be anywhere in R-space (but has a specific in-plane location in the picture above) can be written in two different spherical coordinate ways, depending on one's choice of origin: inv ctr sph ctr r = (R,α,φ) = (r,θ,φ) spherical // note that α =θ/2 only for r on sphere r = (x,y,z) = (X,Y,Z) Cartesian // x = X, y=Y, z=Z+A For the "inversion theory", the form (R,α,φ) is the one that is relevant, because we want to be using "the inversion origin". Therefore, we shall use the symbol R (not r) to denote the length of vector r. We do a similar thing for the point which is the "inverse point" to r, normally called r', so we have r' = (R',α,φ) = (r',θ',φ) spherical θ^ r' = (x',y',z') = (X',Y',Z') Cartesian Notice that r and r' have the same azimuth in both origin coordinate systems, so we don't need φ'. And notice also that r and r' have the same polar angle α in one system, but their θ-type polar angles will be different in the sphere-centered system. This means that our various inversion formulas will have forms like this (inversion origin) Φ(r) = (a/R) Φ'(r') RR' = a2 r' = (a2/R2) r σ(r) = (a/R)3 σ'(r') R = R' = The fact that I refer to the two inversion spaces as the R space and the R' space can be thought of in that R and R' are the lengths (from the inversion origin) of corresponding points r and r' in these two spaces. Section 1 discusses the geometry of this picture. A major simplifying fact is that the two polar angles α and θ are related by α = θ/2, when r lies on the sphere. I write down various simple relations: cosα = R/(2A) = d/R' tanα = h'/d d = a2/2A RR' = a2 where cosα = R/(2A) comes from a right triangle (not shown) inscribed on the diagonal, while cosα = d/R' comes from the obvious triangle that is drawn. My big concern when I started into this was to show that the surface patch was indeed a spherical bowl, but later I derived my Circle Theorem for Inversion which says circles map into circles, so the fact that the perimeter of the disk maps into a circle bounding the patch is completely obvious, so much of the work in the raw notes is superfluous. A major thing to note is that the origin in the above picture is not located at the opposite pole of the bowl because the disk (radius b) is vertically offset by distance c so the bowl is rotated upwards as shown. Before I knew the Circle Theorem, I attempted Plan A to show the patch was a bowl, but I made an incorrect assumption, namely, that a ray connecting the origin to the center of the disk passes through the center of the bowl rc (polar angles αc and θc). Realizing this error, I went on to Plan B where I showed that | r - rc | = constant when r is constrained to lie on the patch boundary. Thus, I showed by brute force that this boundary was a circle. Along the way more "facts" came out, such as tanθc = tan [(θ1/2)+ (θ2/2)] = 2cd / (d2+b2-c2) θc = (θ1 + θ2)/2 In the drawing, θ1 > 0 and θ2 < 0 which allows for θc = 0 as a special case. In Section 2, I just summarize the results of Section 1. The Potential in R space. (Sections 3 and 4) In Section 3 I start on this problem. In R' space we have a disk carrying positive charge Q and to it we add a constant potential -V which causes V = 0 on the disk. We look up the formula in Jackson for the potential of a charged disk, and we thus find that the potential in R' space is this: Φ'(x',y',z') = Φ'(r') = (Q/b) sin-1 [ 2b / ( + ) ] - V where ρ' = and -V = -Q/C with C = (2/π) b, the capacitance of the disk. Since this is in R' space, I use primed coordinates. Our only mod on Jackson's formula is to replace z with z'-d, no rocket science here. The corresponding potential in R space is given by the Inversion Theorem which says Φ(r) = (a/R) Φ'(r') RR' = a2 r' = (a2/R2) r Again, we use unprimed coordinates in R space, and primed ones in R' space. In each space we are using a Cartesian coordinate system whose origin is at the "inversion origin". Thus, our result is this: ΦG(r) = (a/R) Φ'(r') – (aV)/R = (a/R) Φ'(r') – (aQ/C)/R Φ'(r') = (Q/b) sin-1 [ 2b / ( + ) ] ρ' = R = r' = (a2/R2) r This is basically the final answer for the potential in R space. You see how the constant potential term -V in R' space maps into a point charge at the origin in R space. In the raw notes I rewrite this in terms of coordinates X,Y,Z origined at bowl-center and I verify that V = 0 on the bowl. Section 4: Since Q is a free parameter, I can set it to -aQ/C = +1 or Q = -C/a. We then find this result: g(r|ξ) = – (2/πr) sin-1 [ 2b / ( + ) ] + 1/R 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 c = (d/2)[ tan(θ1/2) + tan(θ2/2)] b = (d/2)[ tan(θ1/2) – tan(θ2/2)] where we are now in the sphere-centered coordinates X,Y,Z. We see that the point charge in R' space at the inversion origin is now a positive unit point charge, and therefore we can interpret the total potential as an official "Green's Function" for our bowl where the Green's charge has a specific location on the math sphere outside the bowl. I show that, although the inversion radius "a" appears in the above equations, the potential g(r|ξ) is independent of a, as we know it must be since it is the solution of a Green's problem that does not know about "a". Rather than set a = 1 or some other value, I leave it in the equations. For one thing, it helps showing the dimensions of things. Digression: At this point, I define θb = (θ1 - θ2)/2 as the "half angle" of the spherical bowl, which one could interpret as the polar angle of the circular bowl boundary in a spherical system in which the bowl pole lay on the +z axis. We already have θc = (θ1 + θ2)/2. It is then useful to describe parameters b and c shown above in terms of angles θb and θc . My convention is to regard θ1 as always positive so that regardless of the sign of θ2, we have θb > 0. Angle θc can have either sign. I show that b = d tan(θb/2) (1+ tan2(θc/2))/ (1 - tan2(θc/2) tan2(θb/2)) c = d tan(θc/2) (1+ tan2(θb/2))/ (1 - tan2(θc/2) tan2(θb/2)) It is now a simple matter to replace X,Y,Z with spherical coordinates r,θ,φ origined at sphere center, X = rsinθcosφ Y = rsinθsinφ Z = rcosθ The only effect on the above equations is to change the last two lines to be these: R = r2 = X2+Y2+Z2 x' = (a2/R2) r sinθ cosφ y' = (a2/R2) r sinθ sinφ z' = (a2/R2)( r cosθ +A) The z-axis for this spherical coordinate system is as shown in the picture, origin is sphere center, so θ of the picture becomes the polar angle. We are now 20 pages into the raw notes (but only 5 pages in these META notes). The Charge Distribution on the Bowl (Sections 5, 6, 7) Jackson gives us σ on the disk in R' space which is this: ( I will write θ(b-ρ') once, then be rid of it) (this is the sum of the charge on both sides of the disk, by the way, see page 93 comment pencil) σ'(r') = σ'(x',y',z'=d) = (Q/2πb) θ(b-ρ')/ ρ' = ρ' ≤ b We then use the Inversion Theorem to find σ in R space, σ(r) = (a/R)3 σ'(r') R = r' = (a2/R2) r where our x,y,z are origined at the inversion origin. Thus we have σ(r) = (a/R)3 (Q/2πb) / ρ' = d = a2/2A x' = (a2/R2)x y' = (a2/R2)y z' = d = (a2/R2)z R = Here we are using arbitrary charge Q on the disk, but we can again set Q = -C/a, in which case we get σ(r) = – (a2/π2R3) / ρ' = d = a2/2A x' = (a2/R2)x y' = (a2/R2)y z' = d = (a2/R2)z R = As I did for the potential, I show that σ(r) does not really depend on a. One important point: whereas the potential g(r|ξ) is defined for all x,y,z , the charge density is restricted to the sphere's surface, but that restriction is a little hard to see in these x,y,z coordinates origined at the inversion origin. Notice that we have set z' = d since that is where σ' is located in R' space, and that really gives the restricting condition on x,y,z which is then (a2/r2)z = d. But we really want to move again to sphere-centered X,Y,Z. I do this in the raw notes to get σ(r) = – (a2/π2R3) / ρ' = d = a2/2A x' = (a2/R2)X y' = (a2/R2)Y z' = d = (a2/R2)(Z+A) R = where now X,Y,Z are origined at sphere center and we are Green's Function normalized on Q. Then as with the potential we can convert to spherical coordinates (r,θ,φ) at sphere center. However, since we are now talking σ, we are allowed to simplify things by setting r = A. As a result, the above becomes (along with z' = d which we can ignore) 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)] Inspection shows that this result does not depend on d, so you could set d = 2 (for example) to get a result involving only angles. I then try to consider limits of the above general σ formula, but the only limit that is reasonable is when θc = 0 in which case our Green's point charge is at the bowl opposite pole and we get σ(A,θ,φ) = – (4π2A2)-1 sec3(θ/2) / Remember this is NOT the σ on a charged bowl, it is the σ for our special Green's function problem. In Section 6 I show the main picture above for various sizes of a. My usual pictures imply a > 2A, but any a>0 is viable. In Section 7, I just recap the results for the potential and charge density in our sphere-centered spherical coordinates. We can start with some common equations: d = a2/(2A) ξ = (r,θ,φ) = (A,π,0) = Green's charge loc θ1 = θc + θb θb = bowl spread polar angle θ2 = θc – θb θc = point charge offset polar angle c = (d/2)[ tan(θ1/2) + tan(θ2/2)] = d tan(θc/2) (1+ tan2(θb/2))/ (1 - tan2(θc/2) tan2(θb/2)) b = (d/2)[ tan(θ1/2) – tan(θ2/2)] = d tan(θb/2) (1+ tan2(θc/2))/ (1 - tan2(θc/2) tan2(θb/2)) ρ' = Then we get these official R-space Green's Function and σ results: g(r|ξ) = – (2/πR) sin-1 [ 2b / ( + ) ] + 1/R R = x' = (a2/R2) r sinθ cosφ y' = (a2/R2) r sinθ sinφ z' = (a2/R2)( r cosθ +A) σ(A,θ,φ) = – (d/2)(1/2π2A2) sec3(θ/2) / = – (a2/8π2A2) sec3(θ/2) / R = 2Acos(θ/2) x' = (d/2) sec2(θ/2) sinθ cosφ = d tan(θ/2) cosφ d = a2/(2A) y' = (d/2) sec2(θ/2) sinθ sinφ = d tan(θ/2) sinφ Keep in mind that R is the distance from the inversion origin to an arbitrary point r in R space, and in the case of σ, this point r happens to lie on the sphere. The following picture is useful for a general point r: It shows the same point r in two different spherical coordinate systems, and point r is in general NOT in the plane of paper. The law of cosines for the one triangle shown says R2 = r2 + A2 - 2Ar cos(π-θ) => R2 = r2 + A2 + 2Ar cosθ which confirms our expression above for R. Again, we get α = θ/2 only when r lies on the sphere. Strategy for Making contact with the iris problem We now want to use our knowledge of the above R space Green's Function potential and σ to do the following second inversion problem: In this inversion problem we have "actively" rotated the bowl and its fields, potential, charge distribution to get a new physical situation. In the new situation, the bowl is symmetrical on our new z axis, which puts the Green's point charge off this z axis. We know that V = 0 on the left since this is a Green's function problem, so that makes V = 0 on the iris on the right. Our qG Green's charge on the left then maps into some qhole point charge on the right, and then on the right we have a certain different Green's function involving an iris with a point charge located somewhere in the hole. Smythe and for a special case Jackson tell us σ' on the iris, so finally we have something we can check! Doing the Bowl Rotation prior to attempting the Second Inversion Our first concern, then, is converting our First Inversion problem solution to a form that can be used in our Second Inversion problem. Basically all we have to do is take our solution and rotate it Ry(π-θc) about sphere center so that the bowl pole lies at the new inversion origin. We will actively rotate all points on the sphere (including bowl and Green's charge). If we rotate the physical apparatus, we also rotate the potential it creates and the σ it carries. Unfortunately, this is not a simple azimuthal rotation, but is in the polar dimension. For this reason, it seems easiest to express things in sphere-centered Cartesian coordinates and do the rotation there. So our first step is get everything expressed in sphere-centered X,Y,Z. Since this is always confusing to me, let's just write things out in full detail in such coordinates. But before I do this, I have to reverse process the σ a bit. We had σ(A,θ,φ) = – (d/2)(1/2π2A2) sec3(θ/2) / with R = 2Acos(θ/2) and d = a2/(2A) We need now to get rid of the secant stuff, so we write ( we are Green's unit normalized here) σ(r) = – (d/2) (1/2π2A2) (2A/R)3 / = - (a2/π2R3) / So here then are our results for sphere-centered Cartesians: 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 again we note that for σ only, we get z' = d and X2+Y2+Z2 = A2. These are the only "parts in our machine" that "move" under such a rotation: R = x' = (a2/R2)X y' = (a2/R2)Y z' = (a2/R2)(Z+A) __________________________________________________________________________________ So I am now going to shift from (X,Y,Z) to what I will call (X",Y",Z") where Z" will then be the z axis of the last drawing shown above. Obviously such a rotation does not affect Y, so Y" = Y and we need think only of a 2D rotation. Ry(ψ) = (I will now momentarily lapse into lower case notation, then return soon to upper case. ) Normally we think of Rz acting on , so cyclic says Ry acts on , thus we have so = Ry(ψ) = ψ = π-θc Again, we are actively rotating any point r into point r". We might as well simplify right now and use cosψ = cos(π-θc) = -cos(θc) sin ψ = sin(π-θc) = +sin(θc) so we really have = => x" = - sinθcz - cosθcx y" = y z" = - cosθcz + sinθcx The inverse solution is found by negating the rotation angle ( and I resume upper case) X = sinθcZ" - cosθcX" Y = Y" Z = - cosθcZ" - sinθcX" Our quantities above R = x' = (a2/R2)X y' = (a2/R2)Y z' = (a2/R2)(Z+A) then become ( remember: x",y",z" are sphere-center origined coordinates) R = x' = (a2/R2)( sinθcZ" - cosθcX") y' = (a2/R2)Y" z' = (a2/R2)( - cosθcZ" - sinθcX"+A) For the potential, we know that X"2+ Y"2+ Z"2 = r2. For the σ we have in particular X"2+ Y"2+ Z"2 = A2. __________________________________________________________________________________ Digression: At this juncture, we could transform to sphere-centered spherical coordinates which we would of course refer to as just (r,θ,φ). We would say, as usual, X" = rsinθcosφ Y" = rsinθsinφ Z" = rcosθ r = (r,θ,φ) A hand calculation shows that R2 = r2 + A2 - 2rA cosγ where cosγ ≡ (cosθccosθ + sinθcsinθcosφ) Remember that R is the distance to an observation point r from the Green's point charge -- for example, that is how it appears as the last term in the potential, 1/R. When we rotate the bowl+charge keeping r fixed, this R distance changes! Here is a picture after our rotation, so γ must be the angle shown between two legs of the triangle whose upper right corner lies outside the plane of paper. Other calculations: x' = (a2/R2)( sinθcZ" - cosθcX") = = (a2/R2)( sinθc rcosθ - cosθc rsinθcosφ) z' = (a2/R2)( - cosθcZ" - sinθcX"+A) = (a2/R2)( - cosθcrcosθ - sinθcrsinθcosφ + A) So we then end up with these unpleasant looking results in our sphere-centered sphericals: R = cosγ ≡ (cosθccosθ + sinθcsinθcosφ) x' = (a2r/R2)( sinθc cosθ - cosθc sinθcosφ) y' = (a2r/R2) sinθsinφ z' = (a2r/R2)( - cosθccosθ - sinθcsinθcosφ+A/r) and when talking about σ, we could set r = A. These messy angle combinations are just what you get when you do this somewhat ugly rotation. I don't think this set of results is very useful, so end of this "aside". Well, we will get this: b2 - ρ'2 = b2 - (x'-c)2-y'2 = b2- c2 - x'2 - y'2 +2cx' = (b2- c2) - (x'2+y'2-2cx') = ugly mess This is just an ugly mess, no matter how I examine it. ___________________________________________________________________________________ So let's go back to our earlier results R = x' = (a2/R2)( sinθcZ" - cosθcX") y' = (a2/R2)Y" z' = (a2/R2)( - cosθcZ" - sinθcX"+A) Finally, we make another transform to get to inversion-origin Cartesian coordinates. We will call these final coordinates just x,y,z, so we want to do this: z = Z" + A y = Y" x = X" so our results above become now R = x' = (a2/R2)( sinθc(z-A) - cosθcx) y' = (a2/R2)y z' = (a2/R2)( - cosθc(z-A) - sinθcx + A) b2 - ρ'2 = b2 - (x'-c)2-y'2 and we now have our Green's problem in the required inversion (x,y,z) coordinates. Scratch paper reveals that { I did this twice on separate scratch papers } R2 = [ x2 + y2 + (z-A)2 ] - 2A (z-A) cosθc + A2 - 2Ax sinθc for potential R2 = 2A2 - 2A (z-A) cosθc - 2Ax sinθc for σ = 2A( A - (z-A) cosθc - x sinθc) // since [ x2 + y2 + (z-A)2 ] = A2 neither of which is very simple. Here is where we now stand with our rotated bowl and charge, where I am now replacing symbol ρ' by s in anticipation of a new symbol ρ' appearing in the second inversion problem setup: d = a2/(2A) ξ = Green's charge loc θb = bowl spread polar angle θc = point charge offset polar angle c = d tan(θc/2) (1+ tan2(θb/2))/ (1 - tan2(θc/2) tan2(θb/2)) b = d tan(θb/2) (1+ tan2(θc/2))/ (1 - tan2(θc/2) tan2(θb/2)) s = d = a2/2A g(r|ξ) = – (2/πR) sin-1 [ 2b / ( + ) ] + 1/R σ(r) = - (a2/π2R3) / R2 = [ x2 + y2 + (z-A)2 ] - 2A (z-A) cosθc + A2 - 2Ax sinθc x' = (a2/R2)( sinθc(z-A) - cosθcx) y' = (a2/R2)y z' = (a2/R2)( - cosθc(z-A) - sinθcx+A) This form is, however, unacceptable for the follow reason: When we attempt our second inversion, we are going to have NEW variables like x', y' and z' that are components of r'. We must therefore give new names to our variables of the same name above. The variables x,y,z are of course OK because these are the correct components or r for this second inversion. So replace x', y', z' → u,v,w since none of these symbols has yet been used. I will retain the symbol R and not use any new R's in the second inversion! We then repeat that above with this change at the start of our next section. Setting up for the Second Inversion We have now completed doing our very ugly rotation of the bowl + charge, and here is how things stand for our R space potential and charge distribution: d = a2/(2A) ξ = Green's charge loc θb = bowl spread polar angle θc = point charge offset polar angle c = d tan(θc/2) (1+ tan2(θb/2))/ (1 - tan2(θc/2) tan2(θb/2)) b = d tan(θb/2) (1+ tan2(θc/2))/ (1 - tan2(θc/2) tan2(θb/2)) s = b2 - s2 = b2 - (u-c)2 -v2 g(r|ξ) = – (2/πR) sin-1 [ 2b / ( + ) ] + 1/R σ(r) = - (a2/π2R3) / R2 = [ x2 + y2 + (z-A)2 ] - 2A (z-A) cosθc + A2 - 2Ax sinθc u = (a2/R2)( sinθc(z-A) - cosθcx) v = (a2/R2)y w = (a2/R2)( - cosθc(z-A) - sinθcx + A) where (x,y,z) are based at the inversion origin, and here is the picture for our proposed Second Inversion: Notice that, for r' on the iris plane, we can write r' = (z'= d, ρ', φ') in cylindrical coordinates, and we show ρ' in the picture. This is not the same ρ' we had in our first inversion problem, which is why I renamed that old ρ' to be s. We now have the following battle to face: we like symbols like S, B, ρ' which relate to the iris. We need to rewrite our data above with only these symbols appearing! We must remove the following symbols: x,y,z, θc, θb, A, R, b, c, d To this end, in the raw notes, I show these miscellaneous facts: tan(θb/2) = d/B tan(θc/2) = S/d b = (d2/B) ( 1 + (S/d)2) / (1-(S/B)2) c = S ( 1 + (d/B)2) / (1-(S/B)2) which allow me to remove some of the symbols on my list. Let us now "do" the second inversion from the point only of σ. We get σ'(r') = (r/a)3 σ(r) = (r/a)3 [- (a2/π2R3) / ] r' = (a2/r2) r rr' = a2 and here I am using the conventional notation for r and r', and I use the same a as I used in the first inversion. Before continuing, we can take a peak at the Holy Grail which is Smythe's solution to this problem written in terms of my symbols: σ'(ρ',φ', z' = d)Smythe = - (qhole/2πr12) / r12 = ρ'2+ S2 - 2Sρ'cosφ' r1 = | r' - rhole| Structurally my result has the same general form as Smythe's so we keep the faith a little while longer. The Battle to Eliminate Non-Iris Symbols ( second inversion) This is a frontal brute-force attack which, even if it were to work, is the wrong way to go, but let's just try it anyway hoping it will break through. We already had these from above: (0) tan(θb/2) = d/B tan(θc/2) = S/d b = (d2/B) ( 1 + (S/d)2) / (1-(S/B)2) c = S ( 1 + (d/B)2) / (1-(S/B)2) (1) Let's start by writing the following to eliminate r' wherever it might appear r' = // inversion origin for second inversion (2) Then we can eliminate r as follows r = a2/r' = a2/ = 2Ad / // inversion origin for second inversion (3) We are happy with r' because it is on the iris plane r' = (z'= d, ρ', φ')cyl = (ρ'cosφ', ρ'sinφ',z'=d) x' = ρ' cosφ' y' = ρ' sinφ' z' = d // inversion origin for second inversion (4) Now eliminate x,y,z as follows. Here is one way to state the point inversion r = (a2/r'2) r' Therefore we have 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) (5) Now what about various forms of θc? We already stated this one tan(θc/2) = S/d where we use the thin triangle and the fact that α = θ/2 for points on the sphere. We know Schaum p 16 that tan2(θc/2) = (S/d)2 = (1-cosθc)/(1+cosθc) => cosθc = (d2- S2)/(d2+S2) => sinθc = 2dS/(d2+S2) // you can see that sin2 + cos2 = 1 where I just assume now that θc> 0. (5a) (added 6.10.10 9 AM) Looking at our second inversion picture, for arbitrary r and r' we can use our inversion lemma to conclude that , | r'1 - r'2| = (r2'/r1) | r1 - r2| // template from Jackson META notes | r' - C1| = (C1/r) | r - F | where, in inversion-origin Cartesian coordinates, we have C1 = (0,0,d) C1 = d F = (0,0,2A) so therefore we know that ρ' = (d/r) | r - F | which relates the R' space distance to parameters in R space. If we let ρ' swing around in the iris plane, of course the point r moves around on the sphere, but r does not change. Now, to evaluate the absolute value, we need to select an origin in R space. If we use the sphere-center origin, we have | r - F |2 = A2 + A2 - 2AA cosθ = 2A2(1-cosθ) = 4A2sin(θ/2) | r - F | = 2Asin(θ/2) This continues to be valid for any r on its circle as ρ' swings around. We then we have ρ' = (d/r) 2Asin(θ/2) From (2) above we know that r = 2Ad / But I now realize that this last result by itself really is what I want, so I can forget what I have just done in this little subsection. We have r2 = (2Ad)2 / (ρ'2+ d2) => (ρ'2+ d2) = (2Ad)2/r2 ρ'2 = (2Ad)2/r2 - d2 My goal in this section is simply to relate ρ' of R' space to parameters of the R space, and this really does just that. I will as usual maintain the work just done in case it happens to be needed later. (6) Now is the time to work on R : R2 = [ x2 + y2 + (z-A)2 ] - 2A (z-A) cosθc + A2 - 2Ax sinθc But we only care about r on the sphere (since that puts r' on the iris) so we get the simplified form noted above R2 = 2A( A - (z-A) cosθc - x sinθc) I can then replace: (z-A) = {[a2/(ρ'2+ d2)] d - A } and we find that // recall d = a2/2A R2 = 2A( A - (z-A) cosθc - x sinθc) = 2A( A - {[a2/(ρ'2+ d2)] d - A } (d2- S2)/(d2+S2) - [a2/(ρ'2+ d2)] ρ' cosφ'2dS/(d2+S2)) R2/(2A) = A - {[a2/(ρ'2+ d2)] d - A } (d2- S2)/(d2+S2) - [a2/(ρ'2+ d2)] ρ' cosφ'2dS/(d2+S2) (ρ'2+ d2) R2/(2A) = A(ρ'2+ d2) - {[a2 d - A(ρ'2+ d2)} (d2- S2)/(d2+S2) - [a2 ρ' cosφ'2dS/(d2+S2) (d2+S2) (ρ'2+ d2) R2/(2A) = A(ρ'2+ d2) (d2+S2) - {[a2 d - A(ρ'2+ d2)} (d2- S2) - a2 ρ' cosφ'2dS = A(ρ'2+ d2)[ (d2+S2) + (d2- S2) ] - a2d(d2- S2) - a2 ρ' cosφ'2dS = A2d2(ρ'2+ d2) - a2d(d2- S2) - a2 ρ' cosφ'2dS = a2d (ρ'2+ d2) - a2d(d2- S2) - a2 ρ' cosφ'2dS = a2d (ρ'2) - a2d(- S2) - a2 ρ' cosφ'2dS = a2d ρ'2 + a2dS2 - 2dS a2 ρ' cosφ' = a2d (ρ'2 + S2 - 2S ρ' cosφ') Here is a picture of the iris viewed from the right We can suddenly identify d = a2/2A (ρ'2 + S2 - 2S ρ' cosφ') = r12 = | r' - r'g|2 which we know shows up in Smythe's answer somehow! So we then have just shown by usual brute force that (d2+S2) (ρ'2+ d2) R2/(2A) = a2d r12 R2 = a2d r12 (2A) (d2+S2)-1 (ρ'2+ d2)-1 = a4 r12 (d2+S2)-1 (ρ'2+ d2)-1 R = a2r1/ Some theorem is going on here that I don't know yet to make this a reasonably simple result. I will figure it out later when the dust settles. [ see below! ] Comment: In the case that φ' = π in the above picture, we have ρ' pointing up so S+ρ' = r1. (7) Now is the time to work on s : s = u = (a2/R2)( sinθc(z-A) - cosθcx) v = (a2/R2)y c = S ( 1 + (d/B)2) / (1-(S/B)2) We start with v and have v = (a2/R2) [a2/(ρ'2+ d2)] ρ' sinφ' = (a4/R2) [1 /(ρ'2+ d2)] ρ' sinφ' But we know that (a4/R2) = r1-2 (d2+S2) (ρ'2+ d2) so we find that v = (d2+S2) ρ' sinφ' / r12 Then move on to u where we have d = a2/2A u = (a2/R2)[ sinθc(z-A) - cosθcx]} = (a2/R2)[ 2dS/(d2+S2) * {[a2/(ρ'2+ d2)] d - A} - (d2- S2)/(d2+S2) * [a2/(ρ'2+ d2)] ρ' cosφ' ] (R2/a2) u = [ 2dS/(d2+S2) * {[a2/(ρ'2+ d2)] d - A} - (d2- S2)/(d2+S2) * [a2/(ρ'2+ d2)] ρ' cosφ' ] (ρ'2+ d2) (R2/a2) u = [ 2dS/(d2+S2) * {a2 d - A(ρ'2+ d2)} - (d2- S2)/(d2+S2) * a2ρ' cosφ' ] (d2+S2) (ρ'2+ d2) (R2/a2) u = 2dS (a2d - A(ρ'2+ d2)) - (d2- S2)a2ρ' cosφ' = 2dS (2Ad2 - A(ρ'2+ d2)) - (d2- S2)a2ρ' cosφ' = 2dS (Ad2 - Aρ'2) - (d2- S2)2Adρ' cosφ' = 2Ad { S (d2 - ρ'2) - (d2- S2)ρ' cosφ' } = a2 { S (d2 - ρ'2) - (d2- S2)ρ' cosφ' } => u = (a4/R2) (d2+S2)-1 (ρ'2+ d2)-1 [ S (d2 - ρ'2) - (d2- S2)ρ' cosφ' ] But we found earlier that R2 = a4 r12 (d2+S2)-1 (ρ'2+ d2)-1 or (a4/R2) = r1-2 (d2+S2) (ρ'2+ d2) Thus we really have u = [ S (d2 - ρ'2) - (d2- S2)ρ' cosφ' ] / r12 (ρ'2 + S2 - 2S ρ' cosφ') = r12 It could be worse. Now write c = S ( 1 + (d/B)2) / (1-(S/B)2) = S (B2-d2)/(B2-S2) = { S (B2-d2)/(B2-S2) r12} / r12 We now assembly our wobbly pieces s = s2 = (u-c)2 +v2 = { [ S (d2 - ρ'2) - (d2- S2)ρ' cosφ' ] / r12 - { S (B2-d2)/(B2-S2) r12} / r12 }2 + {(d2+S2) ρ' sinφ' / r12}2 r14 s2 = { [ S (d2 - ρ'2) - (d2- S2)ρ' cosφ' ] - { S (B2-d2)/(B2-S2) r12} }2 + {(d2+S2) ρ' sinφ' }2 (B2-S2)2 r14 s2 = { [ S (d2 - ρ'2) - (d2- S2)ρ' cosφ' ] (B2-S2) - { S (B2-d2) r12} }2 + {(d2+S2) (B2-S2) ρ' sinφ' }2 UNCLE!!! I have had enough blind algebra now. I am obviously still missing some important ingredient. What do my Inversion Lemmas have to say about R? I just went and generalized this in the Jackson inversion meta notes. So let's look at our second inversion picture a bit again An obvious pair of points is shown in this picture! The alternate lemma tells us that | r' - ri'| = (ri'/r) | r - ri| Let the subscript i refer to our Green's charge and its image, so we have | r' - rG'| = (rG'/r) | r - rG | But recall that | r - rG | = R, the distance between point r and the Green's charge (see page 9 above, with little picture). Similarly, from this end-on picture we see that | r' - rG'| = r1. Thus, our alternate lemma is telling us this : r1 = (rG'/r) R => R = (r/rG') r1 From item (2) above we have that r = a2/r' = a2/ // this is the second inversion ρ' and we can see from the first picture that rG' = This our little lemma is telling us that R = ( a2/)/() r1 = a2 r1 / [] which we can compare to the result of our R section (6) above R = a2r1/ r12 = S2 + ρ'2 - 2Sρ' cosφ' Notice how the lemma lets us avoid doing the huge amount of algebra we did in section (6) above. This is a weapon I need to stay aware of! What do my Inversion Lemmas have to say about s? I am now hopeful of applying a lemma of this form to deal with the symbol "s" which was making such a huge mess in section (7) above. We know that s is really ρ' in our first inversion problem, so let's recall that picture again (where I have added point r3 to be discussed below). We can see that s = ρ' = |C - r'| where C and r' are both in R' space. To apply our little lemma, we need to know the image point in R space of C (the disk center). We know this image point (which I shall call r3) lies somewhere on the circle since disk plane maps to sphere. One way to write the answer is of course based on our basic symmetry point fact that r' = (a2/R2) r => r = (a2/R'2) r' => r3 = (a2/C2) C r3 = a2/C where it is pretty clear that C2 = d2+ c2. So there is our answer: r3 = [a2/( d2+ c2)] C and this then is the point I have added to the picture. Can I find the θ angle for this point r3, call it θ3? I think I know the α3: tanα3 = r3x/r3z = Cx/Cz = c/d => tan(θ3/2) = c/d I like how simple this result is. So we can write in sphere-centered spherical coordinates r3 = (A,θ3,φ3= 0) where θ3 = 2 tan-1(c/d) Meanwhile, our lemma says | r' - ri'| = (ri'/r) | r - ri| // template where we take ri to be r3 and ri' to be C, and as usual R is the inversion-origin length of r, | r' - C| = (C/R) | r - r3| So we now know that ρ' = (C/R) | r - r3| r3 = (a2/C2) C C2 = d2+ c2 and we have at least related our distance ρ' to something entirely in the sphere world (R space). Now what happens when we rotate our bowl? We can follow r3 pretty simply. In the rotated R space it ends up here: r3r = Ry(π-θc) r3 = Ry(π-θc) (A,θ3,φ3= 0) = (A, π-θc+θ3, 0) and this is just a specific point on the circle after rotation. So angle(r3r) = π-θc+θ3 ≡ θ3r Now we want to look at our second inversion situation. Here I have added r3r. This should appear just above the origin (which is where rc ends up after rotation), but I intentionally distort the position of r3r so I can get its imagine point r3r' into the picture. Also, recall the ρ' shown here is unrelated to the ρ' = s we are talking about, so don't get confused by that overload. The image point of r3r is at this location, which is on the iris plane, r3r' = (a2/r3r2) r3r or r3r = (a2/r3r'2) r3r' We can then apply our alternate lemma in our second inversion by selecting these two point pairs r,r' and r3r, r3r' to get | r' - r3r'| = (r3r'/R) | r - r3r| This is just an idea I am pursuing, and I am thinking it is not leading anywhere useful. I have so far these facts: ρ' = (C/R) | r - r3| r3 = Ry(-π+θc) r3r r3r = (r3r2/a2) r3r' // I chose the other form here r = (a2/R'2) r' so we can combine these all to get ρ' = (C/R) | (a2/R'2) r' - Ry(-π+θc){ (r3r2/a2) r3r'} | which then let's me relate s = ρ' to objects only in the final R' space. But it does not look very simple and therefore does not look useful. Suppose we think of all vectors as relative to sphere center. Then the Ry means what it says. We also know that r3r = A in that case, so we can rewrite as ρ' = (C/R) | (a2/R'2) r' - (A2/a2) Ry(-π+θc) r3r' | Maybe better to not bother with r3r' since it seems to be nothing intuitive in R' space. So go back to : ρ' = (C/R) | r - r3| = (C/R) | r - Ry(-π+θc) r3r | where the RHS at least has things in our second inversion picture. Something to remember: in these forms, r stays fixed and we are rotating the bowl-sphere and all its points. If we again use sphere centered coordinates we have r = A r3r = A 3r ρ' = (AC/R) | - Ry(-π+θc) 3r | Then | - Ry(-π+θc) 3r |2 Now recall from above that θ3r = π-θc+θ3 3 ≡ Ry(-π+θc) 3r then has angle θ3 I guess I should back up some more: ( treat vectors as sphere-centered) ρ' = (C/R) | r - r3| = (AC/R) | - 3| | - 3|2 = 2 - 23 = 2-2cosφ' = 2(1-cosγ') = 4 sin2(γ'/2) I think I know this: (green Jackson p 68 top) 3 = cosγ' = cosθcosθ3 + sinθsinθ3cos(φ-φ3) But we have φ3 = 0 so 3 = cosγ' = (cosθcosθ3 + sinθsinθ3cosφ) so then s = ρ' =(AC/R) | - 3| = (AC/R) 2 sin(γ'/2) where cosγ' = (cosθcosθ3 + sinθsinθ3cosφ) r = (A,θ,φ) sphere center C2 = d2+ c2 tan(θ3/2) = c/d r = (R,θ/2,φ) inversion center Everything in this section is "first inversion picture". Comment: I could consider just the case φ = 0 where r and r' are in the plane of paper. In this special case, we could conclude that cosγ' = (cosθcosθ3 + sinθsinθ3) = cos(θ-θ3) so γ' = θ-θ3 > 0. Conclusion of this section: the "inversion lemmas" have nothing useful to say about s = ρ'. However, I have obtained a fairly compact expression for s= ρ' by making use of the new point r3 which is the image of the disk center: s = (2AC/R) sin(γ'/2) cosγ' = (cosθcosθ3 + sinθsinθ3cosφ) = 3 We can make these replacements: (where in R of course ρ' and r1 are second inversion parameters) C = R = a2r1/ r12 = S2 + ρ'2 - 2Sρ' cosφ' Return to comparing my and Smythe's σ forms Here is the original Problem 38 quote: I can translate: ( just want to do this one more time to avoid an error) Smythe me q qhole b S a B r r1 c ρ' => σ = -( qhole/2π2r12)/ r12 = ρ'2+ S2 - 2Sρ'cosφ' As a check, if charge is at hole center, we have S = 0 so this becomes σ = -( qholeB/2π2ρ'2) / which is half Jackson's answer on page green 53. So Smythe must be giving just the top or bottom surface. According to Smythe, the factor 1/ is present whether or not the charge is at the center of the hole. Now quoting from my earlier notes "Let us now "do" the second inversion from the point only of σ. We get σ'(r') = (r/a)3 σ(r) = (r/a)3 [- (a2/π2R3) / ] r' = (a2/r2) r rr' = a2 and here I am using the conventional notation for r and r', and I use the same a as I used in the first inversion. Before continuing, we can take a peek at the Holy Grail which is Smythe's solution to this problem written in terms of my symbols: ( I remove Smythe's factor of 2 so his σ' is then the total charge on both sides of the iris, to match what I have done) σ'(ρ',φ', z' = d)Smythe = - (qhole/π2r12) / r12 = ρ'2+ S2 - 2Sρ'cosφ' r1 = | r' - rhole| Structurally my result has the same general form as Smythe's so we keep the faith a little while longer." Smythe's form has a general qhole, but my form is based on the true Green's function, so in my results I will have q'i = qi(a/ri) which says qhole = 1 (a/rg) but I think better to instead use qi = q'i(a/ri') which says 1 = qhole (a/rg') rg'2 = S2+ d2 Therefore qhole = 1 * ( / a) Smythe's answer applied to my situation is then this: σ'(ρ',φ', z' = d)Smythe = - ( / a) (1/2πr12) / whereas my result is this: σ'(r')PL = - (r3/a) [(1/π2R3) / ] r' = (a2/r2) r rr' = a2 so showing these to be equal would require showing this: - (r3/a) [(1/π2R3) / ] = - ( / a) (1/πr12) / (r3) [(1/πR3) / ] = () (1/r12) / r12r3 [(1/R3) / ] = π () ( / ) r3 [(r1/R)3 / ] = π r1() ( / ) Our first inclination is to get rid of R on the left using our above double-confirmed result, R = a2 r1 / [] => (r1/R) = / a2 Then the above hoped-for equality becomes r3 [( / a2)3 / ] = π r1() ( / ) r3 (ρ'2 + d2 )(S2+d2) ( / a6) / = π r1() ( / ) r3 (S2+d2) ()3/ a6) / = π r1 ( / ) r3 (S2+d2) ()/ a2)3 / = π r1 ( / ) But we know that we can replace r = a2/r' = a2/ which then gives (S2+d2) / = π r1 ( / ) (S2+d2) / = π r1b ( / ) // mult both sides by b so at least it is getting "simpler", though still looks rather "distant". Both sides are L1 in units. We also have this data (ρ'2 + S2 - 2S ρ' cosφ') = r12 s = (2AC/r) sin(γ'/2) cosγ' = (cosθcosθ3 + sinθsinθ3cosφ) = 3 b = (d2/B) ( 1 + (S/d)2) / (1-(S/B)2) = (d2B) ( 1 + (S/d)2) / (B2-S2) b = B (d2 + S2) / (B2-S2) Insert this into the above to get (S2+d2) / = π r1(B (d2 + S2) / (B2-S2)) ( / ) 1/ = π r1(B (1 /) ( 1/ ) 1 / = π r1B / [] = ] / (πr1B) = (1/πr1B) So we have now boiled down the entire Smythe mystery to this last putative dimensionless requirement! We do already have these extra facts on hand r12 = ρ'2 + S2 - 2S ρ' cosφ s = (2AC/r) sin(γ'/2) cosγ' = (cosθcosθ3 + sinθsinθ3cosφ) = 3 C2 = d2+ c2 b = B (d2 + S2) / (B2-S2) c = S ( 1 + (d/B)2) / (1-(S/B)2) = S (B2 + d2) / (B2-S2) r = a2/r' = a2/ At this point I am wondering about our quantity d = a2/(2A) . Is this still a free-sliding parameter? I think it is. Double d will make B and S and ρ' and r1 all double. OK, take a few shots in the dark now. Shot A We have cosγ' = (cosθcosθ3 + sinθsinθ3cosφ) Recall from earlier that tan2(θc/2) = (S/d)2 = (1-cosθc)/(1+cosθc) => cosθc = (d2- S2)/(d2+S2) => sinθc = 2dS/(d2+S2) // you can see that sin2 + cos2 = 1 Apply this same idea tan(θ3/2) = c/d so all we have to do is replace S with c. Thus cosθ3 = (d2- c2)/(d2+c2) sinθ3 = 2dc/(d2+c2) Then cosγ' = (cosθcosθ3 + sinθsinθ3cosφ) = ((d2- c2)cosθ + 2dc sinθ cosφ)/ (d2+c2) but this does not seem very useful. What I really need is some clever information about this charge density quantity that comes right from our original charged disk. Shot B. Consider that b2 - ρ'2 = (b+ρ')(b-ρ') which is the product of two distances on the disk plane. For any r' on the disk plane, we can say tanα+ = (b+ρ')/2d tanα- = (b-ρ')/2d In fact, we could define (cylindrical coordinates centered on the inversion origin) r+' ≡ (b+ρ', φ, d) r+'2 = (b+ρ')2 + d2 r-' ≡ (b-ρ', φ, d) r-'2 = (b-ρ')2 + d2 where the distance results come from simple triangles angled out of the plane. We can also express these in spherical coordinates based at the inversion origin: r+' ≡ (r+', α+, φ) r-' ≡ (r-', α-, φ) We could then try to locate the corresponding points on the sphere (spherical origin is at inv origin) r+ = (a/r+')2r+' = (r+, α+, φ) r+ = a2/r+'2 r- = (a/r-')2r-' = (r-, α-, φ) r- = a2/r-'2 The point of this "shot in the dark" is to try to use the inversion theory lemmas. We can say b+ρ' = | r+' - C | = (C/r+) | r+ - r3 | b-ρ' = | r-' - C | = (C/r-) | r- - r3 | Then our "quantity of interest" becomes b2 - ρ'2 = (C2/r+r-) | r+ - r3 | | r- - r3 | If nothing else, we have related our quantity of interest to some distances in the R space world of the first inversion. Earlier we showed that r3 is located on the sphere in our picture plane at angle θ3 where tan(θ3/2) = c/d or tanα3 = c/2 r3 = (a2/C2) C Now we have to choose a coordinate system in which to work. First, try spherical at inv origin. Then we have r+ = (a/r+')2r+' = (r+, α+, φ) r+ = a2/r+'2 r3 = (a2/C2) C = (a2/C2) ( C, α3, 0) = (a2/C, α3, 0) We find that r+ r3 = r+ (a2/C) ( cosα+ cosα3 + sinα+sinα3 cosφ) r- r3 = r- (a2/C) ( cosα- cosα3 + sinα-sinα3 cosφ) and then all we get is a huge mess. It was just an idea. Let's go back to the charge density on the bowl before and after its rotation. Before rotation of the bowl, we have this charge density lying on the bowl σ(r) = – (a2/π2R3) / Question: Do I fully understand this formula for σ(r) ? Suppose the bowl were very small and the disk were also correspondingly small. Then R does not vary (much) on the bowl and the only variation is from the factor. Our bowl is some kind of tilted disk, and we just get a multiple of this factor. On the other hand, if the disk and bowl are "large", then R varies a lot on the bowl, and so does ρ', and so we have both factors involved. Along the way, I wrote σ in the following form 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 where now ALL dependence is explicit! You give me a point (θ,φ) on the bowl (origin at center), I give you the value for σ. I make use of intermediate variables x' and y' here. These in fact are coordinates on the disk, but no matter. We could write b2 - ρ'2 = b2 - (x'-c)2-y'2 = b2 - [ (d tan(θ/2) cosφ - c)2 + (d tan(θ/2) sinφ)2 ] = b2 - [ d2 tan2(θ/2) + c2 - 2dc tan(θ/2) cosφ ] = b2 - c2 - d2 tan2(θ/2) - 2dc tan(θ/2) cosφ = b2 - c2 - d2 tan(θ/2)[ tan(θ/2) + 2(c/d) cosφ] // pre-rotation polars. This result is not simple, but here we have a direct function of (θ,φ) with no intermediaries. But it is hard to "rotate the bowl" directly in spherical coordinates, which is why I used Cartesians when I did this. I think what I did was correct. Dealing more with s and the putative Smythe requirement Here is what I need to show: = (1/πr1B) where S,B,r1 and ρ' are all parameters in the iris picture. Here is what (I think) I know: b = B (d2 + S2) / (B2-S2) s = (2A/a2) sin(γ'/2) L = L cosγ' = (cosθcosθ3 + sinθsinθ3cosφ) = 3 Then (s/b) = (2A/Ba2) [(B2-S2)/ (d2 + S2)] sin(γ'/2) = (1/Bd) [(B2-S2)/ (d2 + S2)] sin(γ'/2) (s/b)2 = (1/Bd)2 (d2+ c2) (ρ'2+ d2) (B2-S2)2 sin2(γ'/2) / (d2 + S2)2 1- (s/b)2 = { (d2 + S2)2 - (1/Bd)2 (d2+ c2) (ρ'2+ d2) (B2-S2)2 sin2(γ'/2) }/(d2 + S2)2 But according to Smythe, he has 1- (s/b)2 = (2/πr1B)2 (B2-S2)(ρ'2- B2) If I set these equal, I get { (d2 + S2)2 - (1/Bd)2 (d2+ c2) (ρ'2+ d2) (B2-S2)2 sin2(γ'/2) }/(d2 + S2)2 = (2/πr1B)2 (B2-S2)(ρ'2- B2) or (d2 + S2)2 - (1/Bd)2 (d2+ c2) (ρ'2+ d2) (B2-S2)2 sin2(γ'/2) = (2/πr1B)2 (B2-S2)(ρ'2- B2) (d2 + S2)2 or (1/Bd)2 (d2+ c2) (ρ'2+ d2) (B2-S2)2 sin2(γ'/2) = (d2 + S2)2 [ 1 - (2/πr1B)2 (B2-S2)(ρ'2- B2) ] or (d2+ c2) (ρ'2+ d2) (B2-S2)2 sin2(γ'/2) = (Bd)2 (d2 + S2)2 [ 1 - (2/πr1B)2 (B2-S2)(ρ'2- B2) ] or (d2+ c2) (ρ'2+ d2) (B2-S2)2 2sin2(γ'/2) = (Bd)2 (d2 + S2)2 [ 2 - (1/πr1B)2 (B2-S2)(ρ'2- B2) ] OK, let's now replace 2sin2(γ'/2) = (1-cosγ') to get (d2+ c2) (ρ'2+ d2) (B2-S2)2 (1-cosγ') = (Bd)2 (d2 + S2)2 [ 2 - (1/πr1B)2 (B2-S2)(ρ'2- B2) ] (1-cosγ') = (Bd)2 (d2 + S2)2 [ 2 - (1/πr1B)2 (B2-S2)(ρ'2- B2) ] / [(d2+ c2) (ρ'2+ d2) (B2-S2)2] cosγ' = 1 - (Bd)2 (d2 + S2)2 [ 2 - (1/πr1B)2 (B2-S2)(ρ'2- B2) ] / [(d2+ c2) (ρ'2+ d2) (B2-S2)2] or cosγ'= { (d2+ c2) (ρ'2+ d2) (B2-S2) - (Bd)2 (d2 + S2)2 [ 2 - (1/πr1B)2 (B2-S2)(ρ'2- B2) ] }/ D where D = [(d2+ c2) (ρ'2+ d2) (B2-S2)2] I have no idea how I would "explain this" identity. Notice that 1-cosγ' = Where is this angle γ' ? Maybe finding it would help. We know that cosγ' = 3 with sphere center as origin of these vectors. Now consider this triangle: I am now at 30 pages and spending lots of timing scrolling around, so I will start Attempt #3!