Smythe Problem 38 by Double Inversion Attempt #1

Smythe Problem 38 by Double Inversion Attempt #1 PhL 6.3.10 NOTE: This obsolete document is fully reviewed at the start of Attempt #2A. There is no need to reread Attempt #1 ever again, but I keep it since this is where things started. If you want to know what #1 is about, go read the review at the start of #2A. 1. Lemma: Show that the image on the sphere of the disk perimeter is a circle. 1 Plan A [ fails, but some algebra is useful] 4 Plan B: [ succeeds] 7 2. Review of the above work. 10 3. The potential Φ(r) in R space 12 4. Green's Function (restricted case) for a Spherical Bowl 15 Identity for tangent sum and differences. 17 5. Charge distribution on the bowl 20 Some limits for σ. 24 6. Comments on the size of the inversion sphere radius a. 26 7. Summary of Potential and Charge Distribution in R space: 27 8. Changing to a rotated spherical coordinate system. 28 9. One more change to the coordinate system: rotate about y by π. 32 10. Repeat last steps but in Cartesian coordinates. 34 11. Doing a second inversion to get the iris with charge in its hole. 37 1. Lemma: Show that the image on the sphere of the disk perimeter is a circle. Comment: I now realize this is trivial since inversion always maps circles into circles as well as spheres into spheres, my Circle Theorem. But I develop here all the algebra needed for later sections. I suppose it was useful to explicitly show the lemma to be true in this specific case. Consider this picture: Although the disk is in R'-space and the sphere patch is in R-space, we draw everything in the same picture with the same origin. We have a right-handed x,y,z system as shown. We shall for the moment consider a spherical coordinate system whose origin is the picture's origin. In this system, a vector r lies on the spherical patch and has polar angle α and some azimuth φ not shown. The vector r' lies on the disk and is at the same polar and azimuthal angles as r. So we have r = (r,α,φ)s r' = (r',α,φ)s with rr' = a2 We can also write these vectors in Cartesian coordinates as follows: r = (Rsinθcosφ, Rsinθsinφ, R+Rcosθ)c r' = (h'cosφ, h'sinφ,d)c with h'/d = Rsinθ/(R+Rcosθ) = sinθ/(1+cosθ) = tan(θ/2) = tanα where the last ratio comes from similar triangles. And we know that α = θ/2. From the Cartesian form we can compute that r2 = R2( sin2θ + (1+cosθ)2) = R2(2 + 2cosθ) = 2R2(1 + cosθ) = 2R2 2cos2(θ/2) = 4R2cos2α r'2 = h'2 + d2 = d2( (h'/d)2+ 1) = d2( tan2α + 1) = d2 sec2α So we have shown that r = 2Rcosα r' = d secα => rr' = 2Rd = a2 which just serves to "verify" our Cartesian coordinates from the picture. Our Cartesian coordinates can also be written this way: r = R(sinθcosφ, sinθsinφ, 1+cosθ)c r' = d (tanα cosφ, tanα sinφ,1)c Question: how do we know that point r lies on the sphere shown? One way would be to show this: | r - (0,0,R)c|2 = R2 or R2| (sinθcosφ, sinθsinφ, 1+cosθ)c - (0,0,1)c|2 = R2 or | (sinθcosφ, sinθsinφ, 1+cosθ)c - (0,0,1)c|2 = 1 or | (sinθcosφ, sinθsinφ, cosθ)c |2 = 1 QED So far we have not said much about the disk location other than it lies in the z = d plane. We shall now assume that the disk starts off centered on the z axis with radius b, but is then shifted upward by an amount c so that the disk center has y = 0 and thus lies in the plane of paper. The center of the disk is then this: C = (c,0,d)c C2 = c2 + d2 and the locus of points on the disk is given by | r' - C | ≤ b with z' = d or r'2 + C2 - 2r' C ≤ b2 with z' = d We may compute r' C = d (tanα cosφ, tanα sinφ,1)c (c,0,d)c = d c tanα cosφ + d2 so our condition is then: (a2/r)2 + c2 + d2 - 2(d c tanα cosφ + d2) ≤ b2 (a2/r)2 + c2 - d2 - 2d c tanα cosφ ≤ b2 where recall that r = (r,α,φ)p . Thus, we have arrived at a condition on r which puts r on our spherical patch, although this fact is not obvious. So we shall now move to a new spherical coordinate system with origin and sphere center, where polar angle is θ, azimuth still φ. Rewrite the above condition as r'2 + c2 - d2 - 2d c tanα cosφ ≤ b2 d2 sec2α + c2 - d2 - 2d c tanα cosφ ≤ b2 For example, if r' = C , disk center, we would have φ = 0 and tanα = c/d so the LHS is this: LHS = d2 sec2α + c2 - d2 - 2d c (c/d) 1 = d2 sec2α + c2 - d2 - 2c2 = d2 (sec2α-1) - c2 = d2 (tan2α) - c2 = d2 (c2/d2) - c2 = c2- c2 = 0 as expected. On the other hand, if we are on the disk perimeter, we must have d2 sec2α + c2 - d2 - 2d c tanα cosφ = b2 Our Big Question is this: what locus of points on the sphere is described by the above equation? We can rewrite it like this: d2 sec2(θ/2) + c2 - d2 - 2d c tan(θ/2) cosφ = b2 or d2 ( sec2(θ/2) - 1) + c2 - 2d c tan(θ/2) cosφ = b2 or d2 tan2(θ/2) + c2 - 2d c tan(θ/2) cosφ = b2 where r = (R,θ,φ)s' where s' means spherical coordinates centered at sphere center. I would like to claim that this locus is in fact a circle lying on the sphere, but how do I show that? Plan A [ fails, but some algebra is useful] Let's assume that our patch is in fact a spherical cap. If so, the center of this cap probably lies on the ray from the origin to point C [ this assumption is wrong, causing Plan A to fail.] . So let's call this cap center rc. Then we know that r = (a2/r'2) r' // inversion equation rc = (a2/C2) C = (a2/ [ c2 + d2]) (c,0,d)c rc2 = (a2/ [ c2 + d2])2 (c2 + d2) = a4/ (c2+ d2) rc = a2/ The Plan A is then to try to show this to be true: | r - rc | = constant where rc = (a2/ [ c2 + d2]) (c,0,d)c r = R(sinθcosφ, sinθsinφ, 1+cosθ)c subject to d2 sec2(θ/2) + c2 - d2 - 2d c tan(θ/2) cosφ = b2 So let's then compute: | r - rc |2 = r2 + rc2 - 2 r rc r = 2Rcosα = 2Rcos(θ/2) r rc = R(sinθcosφ, sinθsinφ, 1+cosθ)c (a2/ [ c2 + d2]) (c,0,d)c = (R a2/ [ c2 + d2]) (sinθcosφ, sinθsinφ, 1+cosθ)c (c,0,d)c = (R a2/ [ c2 + d2]) [ c sinθcosφ + d(1+cosθ) ] = (R a2/ [ c2 + d2]) [ c 2 sin(θ/2) cos(θ/2) cosφ + d 2 cos2(θ/2)] = (2 R a2/ [ c2 + d2]) [ c sin(θ/2) cos(θ/2) cosφ + d cos2(θ/2)] Then we have | r - rc |2 = (2Rcos(θ/2))2 + a4/ (c2+ d2) - (4 R a2/ [ c2 + d2]) [ c sin(θ/2) cos(θ/2) cosφ + d cos2(θ/2)] We want to show that this is a constant if our condition above is true, so mult through by (c2+d2) , want to show that: (c2+ d2) (2Rcos(θ/2))2 + a4 - (4 R a2) [ c sin(θ/2) cos(θ/2) cosφ + d cos2(θ/2)] = const So process this expression's LHS LHS = (c2+ d2) 4R2cos2(θ/2) + a4 - 4 R a2[ c sin(θ/2) cos(θ/2) cosφ + d cos2(θ/2)] = cos2(θ/2) { (c2+ d2) 4R2 - 4 R a2 d } - 4 R a2 c sin(θ/2) cos(θ/2) cosφ Now cancel a 4R from all three terms to get = cos2(θ/2) { (c2+ d2) R - a2 d } - a2 c sin(θ/2) cos(θ/2) cosφ = cos2(θ/2) [ { (c2+ d2) R - a2 d } - a2 c [sin(θ/2) / cos(θ/2)] cosφ ] = cos2(θ/2) [ { (c2+ d2) R - a2 d } - a2 c tan(θ/2) cosφ ] Now multiply through by the constant 2d to get = cos2(θ/2) [2d { (c2+ d2) R - a2 d } - a2 2 d c tan(θ/2) cosφ ] We want to show this is a constant using our condition above. So try this: d2 tan2(θ/2) + c2 - 2d c tan(θ/2) cosφ = b2 => - 2d c tan(θ/2) cosφ = {b2 - c2 - d2 tan2(θ/2)} Then we want to show this is constant: K = cos2(θ/2) [2d { (c2+ d2) R - a2 d } - a2 {b2 - c2 - d2 tan2(θ/2)} ] = cos2(θ/2) 2d { (c2+ d2) R - a2 d } - cos2(θ/2) a2 {b2 - c2 - d2 tan2(θ/2)} = cos2(θ/2)[ 2d { (c2+ d2) R - a2 d } - a2 {b2 - c2} ] + sin2(θ/2) [ a2(c2+ d2)] = cos2(θ/2)[ 2d { (c2+ d2) R - a2 d } - a2 {b2 - c2} ] +(1 - cos2(θ/2)) [ a2(c2+ d2)] We now drop the constant term and the rest is this: cos2(θ/2) [2d { (c2+ d2) R - a2 d } - a2 {b2 - c2} - [ a2(c2+ d2)] ] We will be successful if we can show that [2d { (c2+ d2) R - a2 d } - a2 {b2 - c2} - [ a2(c2+ d2)] ] = 0 [2d { (c2+ d2) R - a2 d } - a2 {b2 } - [ a2( d2)] ] =?= 0 [{ (c2+ d2) 2d R -2 a2 d2 } - a2b2 - a2d2 ] =?= 0 [ { (c2+ d2) a2 -2 a2 d2 } - a2b2 - a2d2 ] =?= 0 // using 2R = a2/d [ (c2+ d2) -2 d2 } - b2 - d2 ] =?= 0 [ c2+ d2 -2 d2 - b2 - d2 ] =?= 0 [ c2- 2 d2 - b2 ] =?= 0 Not quite. We are stuck with b dependence which means it cannot work. So: Plan B: [ succeeds] Let's instead try to use those angles like θ2 to locate the center of the spherical cap, assuming we have a spherical cap. We have: tanα1 = (c+b)/d = (b+c)/d = tan(θ1/2) > 0 tanα2 = (c-b)/d = tan(θ2/2) < 0 The spherical cap pole must be located at angle θc = (θ1 + θ2)/2 = (θ1/2)+ (θ2/2) Then tanθc = tan [(θ1/2)+ (θ2/2)] = { tan(θ1/2) + tan(θ2/2) } / { 1 – tan(θ1/2) tan(θ2/2)} = { (b+c)/d + (c-b)/d } / { 1 – (b+c)/d * (c-b)/d } = { (b+c)d + (c-b)d } / { d2 – (b+c) *(c-b)} = { (2cd } / { d2 + (b2-c2)} = 2cd / (d2+b2-c2) // valid when things are as shown in the figure If c = 0, we get the expected result θc = 0. Notice that as c gets very large, our cap moves to the top left of the sphere and we get θ1 and θ2 both approaching π, so (θ1 + θ2)/2 can approach π and can at least be larger than π/2, and in this realm, tanθc < 0 as Schaum page 14 graph shows. Meanwhile, since the disk is centered on the y = 0 plane, our spherical patch must be a mirror image in the y=0 plane, so we think then that the bowl center point has φc = 0. So, our spherical cap postulated center is at this location (spherical cap center lies on the spherical cap) rc = (R,θc,0)s' or rc = R(sinθc, 1+cosθc)c The Plan B is then to try to show this to be true: | r - rc | = constant where rc = R(sinθc, 0, 1+cosθc)c tanθc = 2cd / (d2+b2-c2) cotθc = (d2+b2-c2)/(2cd) r = R(sinθcosφ, sinθsinφ, 1+cosθ)c subject to d2 tan2(θ/2) + c2 - 2d c tan(θ/2) cosφ = b2 As in Plan A, we proceed along by brute force, | r - rc |2 = r2 + rc2 - 2 r rc r = 2Rcos(θ/2) rc = 2Rcos(θc/2) r rc = R2 [ sinθc sinθ cosφ + (1+cosθc)( 1+cosθ) ] so we then have | r - rc |2 = r2 + rc2 - 2 r rc = 4R2cos2(θ/2) + 4R2cos2(θc/2) - 2 R2 [ sinθc sinθ cosφ + (1+cosθc)( 1+cosθ) ] and we want then to show this is a constant. We therefore throw out terms we already know are constants such as 4R2cos2(θc/2) and we are left with = 4R2cos2(θ/2) - 2 R2 [ sinθc sinθ cosφ + (1+cosθc)( 1+cosθ) ] = 4R2cos2(θ/2) - [2R2sinθc sinθ cosφ + 2R2 (1+cosθc)( 1+cosθ) ] = 4R2cos2(θ/2) - [2R2sinθc sinθ cosφ + 2R2 (1 + cosθc + cosθ + cosθc cosθ) ] Again, we throw out constant terms to get = 4R2cos2(θ/2) - [2R2sinθc sinθ cosφ + 2R2 ( cosθ + cosθc cosθ) ] = 2cos2(θ/2) - [sinθc sinθ cosφ + cosθ (1+cosθc) ] = 2cos2(θ/2) - sinθc sinθ cosφ – cosθ (1+cosθc) ] = 2cos2(θ/2) - sinθc 2sin(θ/2) cos(θ/2) cosφ – (2 cos2(θ/2) - 1) (1+cosθc) ] = 2cos2(θ/2) - sinθc 2sin(θ/2) cos(θ/2) cosφ – (2 cos2(θ/2) ) (1+cosθc) ] + (1+cosθc) Drop the last constant term and keep plugging along = 2cos2(θ/2) - sinθc 2sin(θ/2) cos(θ/2) cosφ – (2 cos2(θ/2) ) (1+cosθc) ] = cos2(θ/2) - sinθc sin(θ/2) cos(θ/2) cosφ – cos2(θ/2) (1+cosθc) ] = – cos2(θ/2) cosθc - sinθc sin(θ/2) cos(θ/2) cosφ = – cos2(θ/2) [ cosθc + sinθc sin(θ/2) / cos(θ/2) cosφ ] = – cos2(θ/2) [ cosθc + sinθc tan(θ/2) cosφ ] Multiply through by 2dc = – cos2(θ/2) [2dc cosθc + sinθc 2dc tan(θ/2) cosφ ] Now let's make use of our "disk perimeter" condition which is this: d2 tan2(θ/2) + c2 - 2d c tan(θ/2) cosφ = b2 => 2d c tan(θ/2) cosφ = d2 tan2(θ/2) + c2 - b2 Then our hopefully constant expression becomes = – cos2(θ/2) [2dc cosθc + sinθc {2dc tan(θ/2) cosφ} ] = – cos2(θ/2) [2dc cosθc + sinθc { d2 tan2(θ/2) + c2 - b2} ] = – cos2(θ/2) (2dc cosθc) – sinθc d2 sin2(θ/2) – cos2(θ/2) sinθc (c2 - b2) = cos2(θ/2) { –sinθc (c2 - b2) – 2dc cosθc} – sin2(θ/2) { sinθc d2 } = cos2(θ/2) { – sinθc (c2 - b2) – 2dc cosθc} –(1-cos2(θ/2)) { sinθc d2 } Again throw out constant term and left with = cos2(θ/2) { – sinθc (c2 - b2) – 2dc cosθc + sinθc d2} At least now θc depends on b so we might get a cancellation. Keep going = cos2(θ/2) sinθc { –(c2 - b2) – 2dc [cotθc] + d2} = cos2(θ/2) sinθc { – (c2 - b2) – 2dc [(d2+b2-c2)/(2cd)] + d2} = cos2(θ/2) sinθc { – (c2 - b2) – (d2+b2-c2) + d2 } = cos2(θ/2) sinθc { – (c2 - b2) – (b2-c2) } = - cos2(θ/2) sinθc { (c2 - b2) + (b2-c2) } = - cos2(θ/2) sinθc { 0 } = 0 So finally we have shown that | r - rc | = constant if we assume that our spherical cap center is at the place indicated above, and if we assume the disk perimeter condition. This was a very ugly brute force method, but at least we know now that we do in fact have a "spherical cap" in this problem. [ The elegant method of showing this "lemma" is noted in the review below, and is based on the simple fact that circles map into circles under inversion as well as conformal maps! ] Just out of curiosity, if we run a ray through the spherical cap center, where does it hit the disk? We had rc = R(sinθc, 0, 1+cosθc)c tanθc = 2cd / (d2+b2-c2) cotθc = (d2+b2-c2)/(2cd) r' = d (tanα cosφ, tanα sinφ,1)c = d (tanα , 0 ,1)c = (d tan(θc/2), 0, d) So this ray hits the disk plane at this location up the x axis: h'c = d tan(θc/2) = d { 2 tan(θc)/(1-tan2(θc) ) = d { 2 2cd / (d2+b2-c2) / [ 1 - {2cd / (d2+b2-c2)}2] = d { 4 cd (d2+b2-c2) / [(d2+b2-c2)2 - {2cd }2] = 4 cd2 (d2+b2-c2) / [(d2+b2-c2)2 - 4c2d2] = c {4 d2 (d2+b2-c2) / [(d2+b2-c2)2 - 4c2d2] } In general, this point doesn't even lie on the disk! For example, if the spherical cap is on the top of the sphere, we get θc = π/2 which corresponds to the denominator vanishing and h'c = ∞, yet the disk would be at some nice finite location. So my Plan A assumption that hc' = c was quite wrong indeed. 2. Review of the above work. The picture of interest is this: The vertically shifted disk appears on the right and maps via "inversion" onto a spherical cap on the sphere, marked with a heavy curve in this slice drawing. Various facts are described above, which I will now summarize here: (0) The disk has radius b, and is vertically offset by distance c above the z axis. (1) We can write out the corresponding points r and r' in Cartesian coordinates (with origin at the inversion origin) as follows: r = R (sinθ cosφ, sinθ sinφ, 1+cosθ) // on the bowl cap r' = d (tan(θ/2) cosφ, tan(θ/2) sinφ, 1 ) // on the disk where φ is an azimuthal angle around the z axis. It turns out that the vector lengths are these r = 2R cos(θ/2) r' = d sec(θ/2) => rr' = 2Rd = a2 as required by inversion (2) The patch on the sphere is in fact a "spherical cap", a fact that is not really obvious from staring at the picture, and that was the point of this whole "lemma". You could imagine that the patch might have an elliptical boundary rather than a circular one, but this is not the case, it turns out. [ After doing all this work, I added and proved a "Circle Theorem" to my Jackson inversion notes, which theorem claims circles map into circles under inversion, so the fact that a disk maps into a spherical cap becomes completely obvious. ] We found that the center rc of the spherical cap is at location θ = θc and φ = 0 where θc is the angle noted below, ( this being a point on the sphere) rc = R (sinθc , 0, 1+cosθc) where θc = (θ1 + θ2)/2 tanθc = 2cd / (d2+b2-c2) where the indicated angles are shown in the picture, so tan(θ1/2) = (b+c)/d tan(θ2/2) = (c-b)/d As the picture is drawn, θ1 > 0 and θ2 < 0, but if the disk is moved up a lot, both these angles become positive. Of course if c = 0 we find that θ2 = - θ1 and θc = 0, which is the classic bowl situation. (3) The perimeter of the disk maps (by inversion) into this curve which lies on the sphere and bounds the spherical patch d2 tan2(θ/2) + c2 - 2d c tan(θ/2) cosφ = b2 r = R where r = (r,θ,φ) are spherical coordinates relative to the center of the sphere. It is certainly not obvious that this curve is in fact a circle. But I know it is a circle because I was able to show that for points r which satisfy these two equations, we have | r - rc | = constant where rc is the center of the spherical cap as shown above. I did not compute the location of the circle center, but that would be easy to do: rcc = (r1 + r2)/2 = R (sinθ1 + sinθ2 , 0, 2 + cosθ1 + cosθ2)/2 (4) One might think that the center of the spherical cap would lie on a ray from the origin to the center of the disk, but this is in fact not the case except when c = 0. The ray through the center of the cap in fact lies at this distance above the z axis on the plane of the disk, h'c = d tan(θc/2) = d { 4 cd (d2+b2-c2) / [(d2+b2-c2)2 - 4c2d2] whereas the center of the disk lies at h' = c. If the cap lies on top of the sphere, for example, you find that this h'c = ∞ whereas c is not infinite. 3. The potential Φ(r) in R space Here I just quote some Jackson results. On page 92 he gives the potential (I have derived it elsewhere) of a disk in the z=0 plane in cylindrical coordinates which is centered at the z,y origin and has radius a. If we offset this disk as shown in our picture above, we can define C = (c,0,d) // center of the disk ρ = (x-c, y) // a vector whose magnitude is cyl coord ρ ρ2  = (x-c)2 + y2 We can then translate Jackson's 3.178 as follows (our radius is b instead of a) Φ'(x,y,z) = (Q/b) sin-1 [ 2b / ( + ) ] where ρ = If we think of this as being in R' space, we then want to translate as follows: Φ'(x',y',z') = Φ'(r') = (Q/b) sin-1 [ 2b / ( + ) ] where ρ' = According to our inversion theorem, we can use this to compute the potential in R' space from that in R space as follows: Φ'(r) = (a/r) Φ(r') rr' = a2 r' = (a2/r2) r But this is also true if we completely swap the R and R' spaces, so we can say Φ(r') = (a/r') Φ'(r) r'r = a2 r = (a2/r'2) r' But it is still true if we now retain our spaces and just swap the points r ↔ r' , so we get Φ(r) = (a/r) Φ'(r') rr' = a2 r' = (a2/r2) r Now we know that r2 = x2+ y2+ z2 So our potential in R space is then this: Φ(r) = (a/r) Φ'(r') Φ'(r') = (Q/b) sin-1 [ 2b / ( + ) ] ρ' = r = r' = (a2/r2) r which we could process a bit if we wanted (I may do this below). Now, suppose our R' space disk carries a charge Q, has capacitance C and potential V = Q/C which we shall assume is positive. We then want to add in R' space a constant potential -V so that our disk in R' space will then be at zero potential (instead of potential V). We know that in R space this is reflected in the appearance of a point charge at the origin of magnitude q = a(-V). Let's assume we have done this. The result is that our spherical bowl is now at zero potential ( V=0 to V=0 theorem), and the situation then in R space is the Green's Function for a spherical bowl in the presence of charge q = -aV located at the origin. Our Green's function potential in R space is then 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 d = a2/2R If we "do the math", we should find that ΦG(r) = 0 on our hemispherical bowl, but ΦG(r) ≠ 0 on the math sphere complement of the bowl ! Note that the (x,y,z) above have their origin at the inversion origin, as shown in the figure. We could shift to another Cartesian system (X,Y,Z) with origin at sphere center. Then x = X, y = Y, and z = Z + R. We can then rewrite all of the above as: 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) (X,Y,Z+R) d = a2/2R Question: How can we show explicitly that this vanishes on the bowl? Well, on the entire math sphere surface we know that X2+ Y2+ Z2 = R2, so we can write r = = r' = a2 [ 2R(R+Z)]-1 (X,Y,Z+R) x' = a2 [ 2R(R+Z)]-1X y' = a2 [ 2R(R+Z)]-1Y z' = a2 [ 2R(R+Z)]-1(Z+R) =a2/2R = d We then get this tricky business that z'-d = 0 (which I have pondered elsewhere). We then have Φ'(r') = (Q/b) sin-1 [ 2b / ( + ) ] = (Q/b) sin-1 [ 2b / ( + ) ] If ρ' > b, which means we are off the disk in R' space, then we get = (q/b) sin-1 [ 2b / ( (ρ'-b) + (ρ'+b) ) ] = (q/b) sin-1 [ 2b / 2ρ' ] = (q/b) sin-1 [ b / ρ' ] which is not a constant. But if we are on the disk with ρ' < b, we get a sign reversal in the first square root and we then have Φ'(r') = (Q/b) sin-1 [ 2b/2b ] = (Q/b) sin-1 [ 1 ] = (Q/b)(π/2) which IS a constant. Jackson page 92 reminds us that the disk capacitance is this: C = (2/π) b so we have found that when ρ' < b, meaning we are on the disk in R' space, we find that Φ'(r') = (Q/b)(π/2) = Q/C = V We then add to this our point charge potential in R space to get ΦG(r) = (a/r) Φ'(r') – (aV)/r = (a/r) Q/C – (aQ/C)/r = 0 and we have thus verified that the potential on the entire spherical cap is 0. On the complementary math sphere region, we instead get ΦG(r) = (a/r) Φ'(r') – (aV)/r = (a/r) (q/b) sin-1 [ b / ρ' ] – (a Q/C)/r which is NOT a constant. So, we have in effect found the Green's Function for a spherical bowl, but only in the special case that the Green's point charge lies somewhere on the complementary part of the sphere containing the bowl. Looking at our picture above, the point charge lies at polar angle θc above the bowl's opposite pole. So let's go for a formal statement of our result. 4. Green's Function (restricted case) for a Spherical Bowl If a Green's point charge of size q = - aQ/C is placed at angle θc away from the opposite pole of a spherical bowl on a sphere of radius R (the charge must not lie on the bowl), then the Green's function for the bowl is given by the following, where (X,Y,Z) are coordinates of a Cartesian coordinate system which has its origin at the bowl-sphere center, and such that the Green's point charge lies at the point (X,Y,Z) = (0,0,-R) : ΦG(r) = (a/r) Φ'(r') – (aV)/r = (a/r) Φ'(r') – (aQ/C)/r Φ'(r') = (Q/b) sin-1 [ 2b / ( + ) ] ρ' = r = x' = (a2/r2)X d = a2/2R C = (2/π) b y' = (a2/r2)Y z' = (a2/r2)(Z+R) But we really want to have a unit positive Green's point charge, so let's scale the above result by multiplying our entire solution by -1/q = -C/(aQ). Then for our unit positive point charge, our true Green's function is then this: g(r|ξ) = -C/(aQ) (a/r) Φ'(r') – (aQ/C)/r * -C/(aQ) ξ = (0,0,-R) = Green's charge loc = -C/(Q) (1/r) Φ'(r') + 1/r = – (C/Qr) Φ'(r') + 1/r = – (C/Qr) (Q/b) sin-1 [ 2b / ( + ) ] + 1/r = – (C/br) sin-1 [ 2b / ( + ) ] + 1/r = – ((2/π) b /br) sin-1 [ 2b / ( + ) ] + 1/r = – (2/πr) sin-1 [ 2b / ( + ) ] + 1/r So our full result is now: g(r|ξ) = – (2/πr) sin-1 [ 2b / ( + ) ] + 1/r ρ' = r = x' = (a2/r2)X d = a2/2R y' = (a2/r2)Y ξ = (0,0,-R) = Green's charge loc z' = (a2/r2)(Z+R) But we have constants a, c, b floating around in our result which seem to have nothing to do with our spherical bowl problem! So we have more work to do. The bowl itself is described by angles θ1 and θ2 as noted above, so we write tan(θ1/2) = (b+c)/d tan(θ2/2) = (c-b)/d Assuming we can find a and hence d by some other method, we can regard these two equations as determining values for b and c, removing those two constants from our result. tan(θ1/2) + tan(θ2/2) = 2c/d => c = (d/2)[ tan(θ1/2) + tan(θ2/2)] tan(θ1/2) - tan(θ2/2) = 2b/d => b = (d/2)[ tan(θ1/2) – tan(θ2/2)] Now what about a ? If we double a, then d becomes 4x larger so b and c also become 4x larger. Also, the variables x', y' and z' become 4x larger. This means that ρ' becomes 4x larger. Then each square root in the denominator of g becomes 4x larger, but the numerator b is also 4x larger, so g does not change. In other words, if we were to write everything out, we would find that "a" cancels out completely from our result. I originally set a = 1 arbitrarily here, but I think now in retrospect I want to keep showing a. So here then is our Green's function: ( note that "a" appears in four places, shown in red) g(r|ξ) = – (2/πr) sin-1 [ 2b / ( + ) ] + 1/r ρ' = r = x' = (a2/r2)X d = a2/2R y' = (a2/r2)Y ξ = (0,0,-R) = Green's charge loc z' = (a2/r2)(Z+R) θc = (θ1 + θ2)/2 c = (d/2)[ tan(θ1/2) + tan(θ2/2)] b = (d/2)[ tan(θ1/2) – tan(θ2/2)] Things are still however a little hazy with the three angles. The full "bowl angle" is this θ1- θ2, but we would usually describe a spherical bowl by its half angle, sort of the polar angle. So we have θc = (θ1 + θ2)/2 // offset angle of Green's charge from opposite bowl pole θb = (θ1 - θ2)/2 // bowl polar angle Then we have θc + θb = θ1 θc – θb = θ2 So more algebra now: tan(θ1/2) ± tan(θ2/2) = tan([θc + θb] /2) ± tan([θc - θb]/2) Identity for tangent sum and differences. tan(A+B) ± tan(A-B) = (tanA +tanB)/(1-tanA tanB) ± (tanA -tanB)/(1+tanA tanB) = [ (tanA +tanB) (1+tanA tanB) ± (tanA -tanB) (1-tanA tanB) ] / (1 - tan2A tan2B) = [ (α +β) (1+α β) ± (α -β) (1- α β) ] / (1 - α2β2) Let's do the two signs separately. First the + sign N+ = (α +β) (1+α β) + (α -β) (1- α β) = α + β + α2β+ αβ2 + α - β - α2β + αβ2 = α + αβ2 + α + αβ2 = 2α + 2αβ2 = 2α(1+β2) N- = (α +β) (1+α β) - (α -β) (1- α β) = α + β + α2β+ αβ2 - α + β + α2β - αβ2 = β + α2β + β + α2β = 2β + 2βα2 = 2β(1+α2) Therefore: tan(A+B) + tan(A-B) = 2 tanA (1+ tan2B)/ (1 - tan2A tan2B) tan(A+B) - tan(A-B) = 2 tanB (1+ tan2A)/ (1 - tan2A tan2B) Now set A = (θc/2) and B = (θb/2) and we then find that tan(θ1/2) + tan(θ2/2) = tan([θc + θb] /2) + tan([θc - θb]/2) = 2 tan(θc/2) (1+ tan2(θb/2))/ (1 - tan2(θc/2) tan2(θb/2)) etc so tan(θ1/2) + tan(θ2/2) = 2 tan(θc/2) (1+ tan2(θb/2))/ (1 - tan2(θc/2) tan2(θb/2)) tan(θ1/2) - tan(θ2/2) = 2 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)) b = d tan(θb/2) (1+ tan2(θc/2))/ (1 - tan2(θc/2) tan2(θb/2)) So OK, this just shows how to compute b and c directly from θc and θb. I don't think anything was really gained by this, so let's restate in a more practical manner (where r = (X,Y,Z) ) g(r|ξ) = – (2/πr) sin-1 [ 2b / ( + ) ] + 1/r ρ' = r = x' = (a2/r2)X d = a2/2R y' = (a2/r2)Y ξ = (0,0,-R) = Green's charge loc z' = (a2/r2)(Z+R) θ1 = θc + θb θ2 = θc – θb c = (d/2)[ tan(θ1/2) + tan(θ2/2)] b = (d/2)[ tan(θ1/2) – tan(θ2/2)] But now the capital letters (X,Y,Z) are a bit confusing, so replace them with lower case, and rewrite everything in "computational order", Also, I replace the sphere's radius R by letter A, and then I rename the variable r above to be R, which is the usual Green's distance R, to avoid confusion with a new r which could be a spherical coordinate with origin at sphere center. Thus we have, d = a2/(2A) ξ = (0,0,-A) = 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)] b = (d/2)[ tan(θ1/2) – tan(θ2/2)] R = x' = (a2/R2)x y' = (a2/R2)y z' = (a2/R2)(z+A) ρ' = g(r|ξ) = – (2/πR) sin-1 [ 2b / ( + ) ] + 1/R At some point I will try to get verification of this result from other sources. Normally any kind of Green's function like this is some horrible double sum of coordinate system harmonics, usually P and Q functions, so this result by that measurement is extremely simple, there are no sums at all, everything is in closed form with elementary functions. Surely this is the problem Kelvin attacked and maybe I could confirm my result against his results. I have not done the "charged bowl problem" here, I have done a special case of the Green's function. In terms of spherical coordinates we could write x = r sinθ cosφ y = r sinθ sinφ z = r cosθ so that R2 = x2 + y2 + (z+A)2 = r2 + a2 + 2Az = r2 + A2 + 2Ar cosθ so here is a restatement, where I now give up on computation order and put things in logical order g(r|ξ) = – (2/πR) sin-1 [ 2b / ( + ) ] + 1/R where ρ' = and: 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)] b = (d/2)[ tan(θ1/2) – tan(θ2/2)] R = x' = (a2/R2) r sinθ cosφ y' = (a2/R2) r sinθ sinφ z' = (a2/R2)( r cosθ +A) where we have spherical coordinates (r,θ,φ) centered at the sphere center. 5. Charge distribution on the bowl It is pretty clear from the above that the potential created by the induced charge on the bowl is this: V(r|ξ) = – (2/πR) sin-1 [ 2b / ( + ) ] since the other term is the potential just of the point charge. Using the above spherical coordinates centered at bowl center we could compute ∂rV to obtain the charge density on the bowl's surface, which would certainly be a check on our result, since we have a more direct path to this charge density from the general inversion formalism, which I will now examine using the rule σ'(r') = (r/a)3 σ(r) 2D surface charge density or σ(r) = (a/r)3 σ'(r') rr' = a2 r' = (a2/r2) r In R' space, the charge distribution on the disk is given Jackson page 93. This charge density lies in the plane z = d and of course ρ ≤ b, so in that region we have σ(x,y,z=d) = (Q/2πb) θ(b-ρ)/ ρ = ρ ≤ b but in our application this disk is in in R' space so we write this as σ'(r') = σ'(x',y',z'=d) = (Q/2πb) θ(b-ρ')/ ρ' = ρ' ≤ b We then compute the charge in R space to be σ(r) = (a/r)3 σ'(r') r = r' = (a2/r2) r So install the various pieces and we get σ(r) = (a/r)3 (Q/2πb) θ(b-ρ')/ ρ' = d = a2/2R x' = (a2/r2)x y' = (a2/r2)y z' = d = (a2/r2)z r = and let's not forget about these facts θ1 = θc + θb θ2 = θc – θb c = (d/2)[ tan(θ1/2) + tan(θ2/2)] b = (d/2)[ tan(θ1/2) – tan(θ2/2)] { Check: if θc= 0, get θ1 = θb= -θ2 so b = d tan(θb/2) and c = 0. Then ρ' = . The bowl edge in this case should then be at ρ'2 = b2 which says x'2 + y'2 = b2 which says (a2/r2)2 (x2+ y2) = b2. But if we then go into spherical coordinates with origin at inversion origin, x = r sin(θ/2) cosφ etc so x2+ y2 = r2sin2(θ/2). We then have (a2/r2)2 r2 sin2(θ/2) = b2 = d2 tan2(θb/2) which says a4sin2(θ/2)= r2d2 tan2(θb/2). But from way back we know that r = 2R cos(θ/2) and that d = a2/(2R) so we have then a4sin2(θ/2)= [2R cos(θ/2)]2[ a2/(2R)]2 tan2(θb/2) = a4cos2(θ/2) tan2(θb/2) => sin2(θ/2) = cos2(θ/2) tan2(θb/2) tan2(θ/2) = tan2(θb/2) => θ = θb } This then is the charge distribution on our spherical bowl in Cartesian coordinates centered at the inversion origin. We have assumed that the charge on the disk is Q. We know from above that in this case, the Green's point charge has size q = - aQ/C with C = (2/π) b. As we did with the potential, we want now to scale down the charge density by factor -1/q = -C/(aQ) so that our problem then has a unit point charge for a Green's charge. Our scaled charge density then becomes this σ(r) = [-C/(aQ)] (a/r)3 (Q/2πb) θ(b-ρ')/ where I won't change notation for σ(r). We then have q = +1 so we have 1 = - aQ/C and Q = -C/a. So after this scaling, our disk in R' space holds a Q which is negative, causing σ' on the disk to be negative, and so σ on our spherical bowl is also negative, and the Green's charge is positive. We can combine the constants a bit using a2 = 2Rd and C = (2/π) b, [-C/(aQ)] (a/r)3 (Q/2πb) = – (2/π) b * 1/(aQ) * a3/r3 * Q/(2πb) = – (2/π) * 1/(a) * a3/r3 * 1/(2π) = – (1/π) * 1/(1) * a2/r3 * 1/(π) = – (1/π2) * a2/r3 = – (a2/π2r3) and then ( note that σ has dimensions 1/L2 σ(r) = – (a2/π2r3) θ(b-ρ')/ ρ' = d = a2/2R x' = (a2/r2)x y' = (a2/r2)y z' = d = (a2/r2)z r = c = (d/2)[ tan(θ1/2) + tan(θ2/2)] b = (d/2)[ tan(θ1/2) – tan(θ2/2)] Question: does our R space charge density depend on parameter "a" ? If we double a, we scale up x' and y' and z' and d and c and b by 4x. Thus ρ' increases by 4x. Then increases by 4x in the denominator, but the leading factor contains a2 which increases by 4x, so σ(r) is independent of a. Thus, we can set a to whatever constant we want. Knowing this, I will continue to keep "a" in the various equations, at least for now. Now as before, we would like to get this into better coordinates. Right now, x,y,z are Cartesian relative to the inversion origin. Let's first shift to the same cap Cartesians we used above, x = X, y = Y, and z = Z + R, so then we have σ(X,Y,Z) = – (a2/π2r3) θ(b-ρ')/ ρ' = d = a2/2R x' = (a2/r2)X y' = (a2/r2)Y z' = d = (a2/r2) (Z+R) r = But now the capital letters (X,Y,Z) are a bit confusing, so replace them with lower case. Also, I replace the sphere's radius R by letter A, and then I rename the variable r above to be R, which is distance from a point r to the inversion origin, to avoid confusion with a new r which will soon be a spherical coordinate with origin at sphere center. Thus we have, σ(x,y,z) = – (a2/π2R3) θ(b-ρ')/ 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 Cartesians with origin at sphere center. We then go to the same spherical coordinates used above x = r sinθ cosφ r2 = x2 + y2 + z2 d = a2/2A y = r sinθ sinφ z = r cosθ R2 = x2 + y2 + (z+A)2 = r2 + A2 + 2Az = r2 + A2 + 2Ar cosθ x' = (a2/R2) r sinθ cosφ y' = (a2/R2)r sinθ sinφ z' = (a2/R2) (r cosθ + A) c = (d/2) [ tan(θ1/2) + tan(θ2/2)] b = (d/2) [ tan(θ1/2) – tan(θ2/2)] But we know that our charge in R space is all located at r = A, so if we are talking about the R space charge distribution, we can replace r by A and the above becomes R2 = r2 + A2 + 2Ar cosθ = A2 + A2 + 2A2 cosθ = 2A2 (1+cosθ) = 4A2cos2(θ/2) R = 2Acos(θ/2) x' = (a2/R2)x = (a2/(4A2cos2(θ/2)))x = (a2/(4A2cos2(θ/2))) A sinθ cosφ = (a2/4A) sec2(θ/2) sinθ cosφ = (d/2) sec2(θ/2) sinθ cosφ similarly for y', but z' = (a2/R2)(z+A) = (a2/(4A2cos2(θ/2))) (A cosθ+A) = (a2/4A) sec2(θ/2) (cosθ+1) = (d/2) sec2(θ/2) (cosθ+1) = (d/2) sec2(θ/2) 2 cos2(θ/2) = d // as expected While were at it, we can write (a2/π2R3) = (2Ad/π2) (1/8A3cos3(θ/2)) = (d/4π2A2)sec3(θ/2) = (d/2)(1/2π2A2) sec3(θ/2) Thus we have these simplified results: σ(A,θ,φ) = – (d/2)(1/2π2A2) sec3(θ/2) θ(b-ρ')/ ρ' = x' = (d/2) sec2(θ/2) sinθ cosφ = d tan(θ/2) cosφ y' = (d/2) sec2(θ/2) sinθ sinφ = d tan(θ/2) sinφ z' = d c = (d/2) [ tan(θ1/2) + tan(θ2/2)] b = (d/2) [ tan(θ1/2) – tan(θ2/2)] d = a2/2A Suppose we now extract our factor (d/2) from everything and define some cap letter quantities, x' = (d/2)X' etc Then we get ρ' = = (d/2) = (d/2) σ(A,θ,φ) = – (d/2)(1/2π2A2) sec3(θ/2) θ(b-ρ')/ = – (d/2)(1/2π2A2) sec3(θ/2) θ(b-ρ') (d/2)-1 / = – (1/2π2A2) sec3(θ/2) θ(b-ρ') / So here then is our final result: ( dimensions charge/A2are correct) σ(A,θ,φ) = – (1/2π2A2) sec3(θ/2) θ(b-ρ')/ X' = sec2(θ/2) sinθ cosφ = 2tan(θ/2) cosφ X'2 + Y'2 = 4 tan2(θ/2) Y' = sec2(θ/2) sinθ sinφ = 2tan(θ/2) sinφ C = [ tan(θ1/2) + tan(θ2/2)] B = [ tan(θ1/2) – tan(θ2/2)] θ1 = θc + θb θ2 = θc – θb { Another way to get these results is just set d = 2 in our previous results, which means a2 = 2Ad = 4A. } This then is the surface charge distribution on the spherical bowl in the Green's Function of a spherical bowl of radius A and polar angle θb with a unit positive Green's point charge located at a position θc away from the opposite bowl pole. I have left in the theta function just to show where the charge ends. In our spherical coordinates this is some complicated function of θ and φ and θ1 and θ2. Some limits for σ. There are certainly some limits of this result we can use to check things. Limit #1. Suppose we make θ1 = θ2 be very small, so the bowl approaches a point charge. Then basically we will have θ ≈ 0 on the entire bowl, cosθ ≈ 1 and sec(θ/2) ≈ 1 and we have X' = Y' = 0 Z' = 1 C = θ1/2 B = 0 = = → 0 so we fine that σ(A,θ,φ) = ∞ in this case. That is to say, if you put a Green's point charge a foot away from a tiny sphere, that sphere will need an infinite induced surface charge as it get's infinitely small. We could take a different limit where we regard the bowl then as a disk, and we get the charge induced on a disk, etc etc, but let's not do that. Limit #2. Suppose we let the bowl become the entire sphere? This limit is also not very well defined. B Limit #3: Suppose θc = 0 so we are right at the pole with our Green's charge. Then θ1 = θb = -θ2, so we then have C = [ tan(θ1/2) + tan(θ2/2)] = 0 B = [ tan(θ1/2) – tan(θ2/2)] = 2tan(θb/2) B2-(X'-C)2 - Y'2 = B2- (X'2 + Y'2) = 4 tan2(θb/2) – 4 tan2(θ/2) = 2 σ(A,θ,φ) = – (2π2A2)-1sec3(θ/2) θ(b-ρ')/ [2] = – (4π2A2)-1 sec3(θ/2) θ(θb-θ) / This shows that, as expected, the σ blows up in a characteristic way at the bowl edge θ = θb. This then is a very simple result and there is no φ dependence. I have simplified the Heaviside function in the obvious manner shown. If we further assume that θβ = π/2 so we have a hemispherical bowl, then tan(θb/2) = 1 and we have σ(A,θ,φ) = – (4π2A2)-1 sec3(θ/2) / where I don't bother with Heaviside. This result is so simple, I can probably find a web check somewhere. Comments: (1) This charge density is again expressed without sums in closed form with just elementary functions, so in that respect, it is a simple result. Rewrite the bowl density a few more times please. At one point above we had this: σ(x,y,z) = – (a2/π2R3) θ(b-ρ')/ ρ' = d = a2/2A R = x' = (a2/R2)x y' = (a2/R2)y z' = (a2/R2)(z+A) = d // this line is not necessary and will be dropped in next write in our Cartesian system with origin at sphere center. But then I was able to simplify since we know that the quantity r = A when talking about σ. That is to say, we found that R2 = r2 + A2 + 2Ar cosθ = A2 + A2 + 2A2 cosθ = 2A2 (1+cosθ) = 4A2cos2(θ/2) => R = 2Acos(θ/2) so we then have σ(x,y,z) = – (a2/π2R3) θ(b-ρ')/ ρ' = d = a2/2A R = 2Acos(θ/2) x' = (a2/R2)x y' = (a2/R2)y We can at this point go to spherical coordinates and we get σ(x,y,z) = – (a2/π2R3) θ(b-ρ')/ ρ' = d = a2/2A R = 2Acos(θ/2) x' = (a2/R2) r sinθ cosφ y' = (a2/R2) r sinθ sinφ So here is our grand summary of the charge density on the bowl in sphericals: σ(x,y,z) = – (a2/π2R3) θ(b-ρ')/ ρ' = d = a2/2A 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))) R = 2Acos(θ/2) = A x' = (a2/R2) A sinθ cosφ = (d/2) sec2(θ/2) sinθ cosφ = d tan(θ/2) cosφ y' = (a2/R2) A sinθ sinφ = (d/2) sec2(θ/2) sinθ sinφ = d tan(θ/2) sinφ 6. Comments on the size of the inversion sphere radius a. So far, all my pictures have been drawn like so: The inversion sphere is dashed and passes between the bowl and the disk, and r and r' are mirror points inverted through this inversion sphere. In this particular picture, we have roughly d ≈ 4A = a2/2A so that a2 = 8A2 and a ≈ 2.8 A. You can see that as long as a > 2A, the picture has the rough form shown above. The limiting picture with a = 2A looks like this: where now the disk in R's space touches the bowl in R space. But of course there is no reason we cannot have a be any size we want, so here we show what happens for a small choice of a: The disk now lies to the left of the bowl and lies inside the inversion sphere. So fine, the picture is different, but the math is all the same. 7. Summary of Potential and Charge Distribution in R space: (a) The bowl Green's function potential for a unit Green's point charge is this: g(r|ξ) = – (2/πR) sin-1 [ 2b / ( + ) ] + 1/R where ρ' = and: 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)] b = (d/2)[ tan(θ1/2) – tan(θ2/2)] R = x' = (a2/R2) r sinθ cosφ y' = (a2/R2) r sinθ sinφ z' = (a2/R2)( r cosθ +A) where we have spherical coordinates (r,θ,φ) centered at the sphere center. (b) The bowl charge corresponding to this Green's function problem is given by σ(A,θ,φ) = – (d/2)(1/2π2A2) sec3(θ/2) θ(b-ρ')/ ρ' = c = (d/2) [ tan(θ1/2) + tan(θ2/2)] b = (d/2) [ tan(θ1/2) – tan(θ2/2)] d = a2/2A x' = (d/2) sec2(θ/2) sinθ cosφ = d tan(θ/2) cosφ y' = (d/2) sec2(θ/2) sinθ sinφ = d tan(θ/2) sinφ in the same spherical coordinates. Notice that this charge density only exists at r=A on a portion of the sphere which we call "the bowl". The constants b,c,d are of course the same as for the potential. 8. Changing to a rotated spherical coordinate system. Consider a coordinate system S with coordinate r = (x,y,z). Imagine some other coordinate system S' in which the same point is represented as r' = (x',y',z'). If we get from S to S' by some rotation R, we know that is the same as staying in S and changing to r' = R-1r . We have already expressed our potential and charge distribution in a Cartesian coordinate system with coordinates above which we called r = (x,y,z). This system has its origin at sphere center. Here is a simplified picture: So this is coordinate system S. I now want to think about (red) coordinate system S' which causes the bowl to be centered on the right, and causes the point charge to be lifted up by angle θc. We might say S' = Ry(θc) S. What this means for example is that ' = Ry(θc) . In general, then, we can work in S and actively rotate the bowl (and everything else) such that r' = Ry(-θc) r . Copying from our matrix doc we have Ry(-) = = This tells us that x' = r sinθ'cosφ' = cosθc x - sinθc z = cosθc r sinθ cosφ - sinθc r cosθ y' = r sinθ' sinφ' = y = r sinθ sinφ z' = r cosθ' = sinθc x + cosθc z = sinθc r sinθ cosφ + cosθc r cosθ We cancel the r's to get sinθ'cosφ' = cosθc sinθ cosφ - sinθc cosθ sinθ'sinφ' = sinθ sinφ cosθ' = sinθc sinθ cosφ + cosθc cosθ We would like to know θ' and φ' in terms of θ and φ. The third line tells us cosθ' but since we know θ' is in the range (0,π), this gives θ' without ambiguity. Let's rewrite these three lines as cosθ' = sinθc sinθ cosφ + cosθc cosθ cosφ' = (cosθc sinθ cosφ - sinθc cosθ)/sinθ' sinφ' = sinθ sinφ/sinθ' and then we know φ' precisely from the last two lines. As an example, suppose θ = θ and φ = 0. Then cosθ' = sinθc sinθ cosφ + cosθc cosθ = sinθc sinθ + cosθc cosθ = cos(θ-θc) cosφ' = (cosθc sinθ - sinθc cosθ)/sinθ' = sin(θ-θc)/sinθ' sinφ' = sinθ sinφ/sinθ' = 0 From the last equation we know that sinφ' = 0 and from the second that cosφ' = sin(θ-θc)/sinθ' = sin(θ-θc) / where the root is positive because sinθ' is always positive since θ' is a polar angle in (0,π). Then cosφ' = sin(θ-θc) / = sin(θ-θc) / |sin(θ-θc)| Since θ is a polar angle, it ranges (0,π), but we could have θc anywhere in (-π,π). This means that (θ-θc) can have either sign, so sin(θ-θc) can have either sign. If sin(θ-θc) > 0, we find φ' = 0, but if sin(θ-θc) < 0 then we instead find φ' = π. To summarize what we have just done: we can change from the (r,θ,φ) coordinates to (r,θ',φ') coordinates with the same origin, such that in the primed coordinates, the Green's point charge will be at r'green = Ry(-θc) rgreen = Ry(-θc) (0,0,-A) = (Asinθc, 0, -Acosθc) which has a positive x' component in our example where θc ≈ + π/10 say. As for the bowl, in this new system it is "centered" on the z' axis with bowl polar angle being θb. We can "invert" all our results above by taking θc→ -θc and then we get the results we will actually need. Doing this we find that, sinθcosφ = cosθc sinθ' cosφ' + sinθc cosθ' sinθsinφ = sinθ' sinφ' cosθ = - sinθc sinθ' cosφ' + cosθc cosθ' If we look at our expression for the bowl potential shown above, we have to deal with R = = x' = (a2/R2) r sinθ cosφ = (a2/R2) r (cosθc sinθ' cosφ' + sinθc cosθ') y' = (a2/R2) r sinθ sinφ = (a2/R2) r sinθ' sinφ' z' = (a2/R2)( r cosθ +A) = (a2/R2)[ r (- sinθc sinθ' cosφ' + cosθc cosθ') +A] But now that we have done all this, we will drop the primes and write our potential in the spherical coordinates which cause the bowl to be centered on the z axis, and which cause the Green's charge to be up at angle θc: g(r|ξ) = – (2/πR) sin-1 [ 2b / ( + ) ] + 1/R where ρ' = and: d = a2/(2A) ξ = (Asinθc, 0, -Acosθc) = 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))) R = x' = (a2/R2) r (cosθc sinθ cosφ + sinθc cosθ) y' = (a2/R2) r sinθ sinφ z' = (a2/R2)[ r (- sinθc sinθ cosφ + cosθc cosθ) +A] Note that objects like x' are coordinates in the R' system and are not x in the rotated S' system or anything like that. We again have some overloading here of the prime. Here is the picture which goes with the above equations: So if you tell me θc and θb, and bowl radius A, I will tell you the potential at all points in space. We are still free to set a to any value we like, since the potential does not depend on it, although it appears to. What about the surface charge σ on the bowl? The quantities we need there are these: R = 2Acos(θ/2) = A x' = (a2/R2) r sinθ cosφ = (d/2) sec2(θ/2) sinθ cosφ = d tan(θ/2) cosφ y' = (a2/R2) r sinθ sinφ = (d/2) sec2(θ/2) sinθ sinφ = d tan(θ/2) sinφ But these are the same quantities as above with the simplification that r = A, so just take that results above R = x' = (a2/R2) r (cosθc sinθ cosφ + sinθc cosθ) y' = (a2/R2) r sinθ sinφ and set r = A. The first line becomes R = = = A Therefore our σ result is this: (same b and c as the potential result) σ(x,y,z) = – (a2/π2R3) θ(b-ρ')/ ρ' = d = a2/2A R = A x' = (a2/R2) A (cosθc sinθ cosφ + sinθc cosθ) y' = (a2/R2) A sinθ sinφ 9. One more change to the coordinate system: rotate about y by π. There is now just one more transformation I want to do on this picture. First, rotate everything about the y axis just to get a new picture: The expression given above is still the potential for this picture. I now want to go to yet another coordinate system I will call θ",φ" which we get to by doing r" = Ry(-π) r where in effect I have now set θc = π just in a local sense for the math I am about to do. We then get our results above with this replacement, so we have sinθ"cosφ" = cosθc sinθ cosφ - sinθc cosθ = - sinθ cosφ sinθ"sinφ" = sinθ sinφ cosθ" = sinθc sinθ cosφ + cosθc cosθ = - cosθ which we restate as sinθ"cosφ" = - sinθ cosφ sinθ"sinφ" = sinθ sinφ cosθ" = - cosθ The last line really tells us that θ" = (π-θ) which gives cosθ" = - cosθ sinθ" = + sinθ so our first two lines above are now cosφ" = - cosφ sinφ" = + sinφ and we now see how all trig quantities transform under this rotation. So write R = x' = (a2/R2) r (cosθc sinθ cosφ + sinθc cosθ) y' = (a2/R2) r sinθ sinφ z' = (a2/R2)[ r (- sinθc sinθ cosφ + cosθc cosθ) +A] → R = x' = (a2/R2) r (-cosθc sinθ" cosφ" - sinθc cosθ") y' = (a2/R2) r sinθ" sinφ" z' = (a2/R2)[ r (+ sinθc sinθ" cosφ" - cosθc cosθ") +A] So we now have a new expression for the potential, and a new picture to go along with it, where we now drop the double primes : g(r|ξ) = – (2/πR) sin-1 [ 2b / ( + ) ] + 1/R where ρ' = and: d = a2/(2A) ξ = (-Asinθc, 0, Acosθc) = 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))) R = x' = (a2/R2) r (-cosθc sinθ cosφ - sinθc cosθ) y' = (a2/R2) r sinθ sinφ z' = (a2/R2)[ r (+ sinθc sinθ cosφ - cosθc cosθ) +A] The corresponding bowl result uses the R,x' and y' lines with r = A, so now R = A and we get σ(x,y,z) = – (a2/π2R3) θ(b-ρ')/ ρ' = d = a2/2A R = A x' = (a2/R2) A (-cosθc sinθ cosφ - sinθc cosθ) y' = (a2/R2) A sinθ sinφ I want the potential and charge in this form because the above picture is soon going to be inverted to give the iris with a charge in the hole, off center, which is the problem I am aiming at. 10. Repeat last steps but in Cartesian coordinates. I wanted things in sphericals centered at bowl sphere center, because this lets you check things most easily. But now that I am about to do another inversion transformation to get the iris, I really want things in Cartesian coordinates centered at the inversion origin! Let's start by recalling previous results. For the potential, I go back to a very early result before we did scaling of the point charge: ΦG(r) = (a/r) Φ'(r') – (aV)/r = (a/r) Φ'(r') – (aQ/C)/r d = a2/2R Φ'(r') = (Q/b) sin-1 [ 2b / ( + ) ] ρ' = r = r' = (a2/r2) r σ(r) = (a/r)3 (Q/2πb) θ(b-ρ')/ where b and c are stated elsewhere and R is the sphere radius again. I have added the corresponding charge density. Our first step is to do the small Ry(-θc) rotation as per above, which resulted in x' = cosθc x - sinθc z y' = y z' = sinθc x + cosθc z where primes here mean rotated coordinates (not R' space coordinates). The inverted equations are of course these x = cosθc x' + sinθc z' y = y' z = - sinθc x' + cosθc z' We then want to do another Ry(-π) rotation to get to an S" Cartesian coordinate system which has the effect of negating x' and z', so we have x = - cosθc x" - sinθc z" y = y" z = sinθc x" - cosθc z" Notice that r"2 = r2 since we are just doing rotations. So our results are then (the primes appearing below are the original R' space ones) ΦG(r) = (a/r) Φ'(r') – (aV)/r = (a/r) Φ'(r') – (aQ/C)/r d = a2/2R Φ'(r') = (Q/b) sin-1 [ 2b / ( + ) ] ρ' = r = r' = (a2/r2) r σ(r) = (a/r)3 (Q/2πb) θ(b-ρ')/ where x = - cosθc x" - sinθc z" y = y" z = + sinθc x" - cosθc z" Here (x", y", z") are a Cartesian system centered at the middle of the bowl and here is the corresponding picture: I really need now to get rid of the double primes and primes and clean other things up. Start with ΦG(r) = (a/r) (Q/b) sin-1 [ 2b / ( + ) ] – (aQ/C)/r σ(r) = (a/r)3 (Q/2πb) / We need to compute ρ' so we can get rid of it: ρ'2 = (x'-c)2 + y'2 = [(a2/r2)x - c]2 + [(a2/r2)y]2 = (a2/r2)2 { [x - c (r2/a2)]2 +y2 } = (a2/r2)2 { [(- cosθc x" - sinθc z") - c (r2/a2)]2 +y"2 } ≡ w2 z' = (a2/r2)z = (a2/r2) (sinθc x" - cosθc z") ≡ t where I am now making up new symbol names. We then have ΦG(r) = (a/r") (Q/b) sin-1 [ 2b / ( + ) ] – (aQ/C)/r" σ(r) = (a/r")3 (Q/2πb) / r" = w2 = (a2/r"2)2 { [(- cosθc x" - sinθc z") - c (r"2/a2)]2 +y"2 } t = (a2/r"2) (sinθc x" - cosθc z") Now finally we can remove all the double primes and adjust our picture, so we have ΦG(r) = (a/r) (Q/b) sin-1 [ 2b / ( + ) ] – (aQ/C)/r σ(r) = (a/r)3 (Q/2πb) / r = d = a2/2R w2 = (a2/r2)2 { [(- cosθc x - sinθc z) - c (r2/a2)]2 +y2 } t = (a2/r2) (sinθc x - cosθc z) 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))) where the size of our Green's Charge is qG = – (aQ/C) with C = (2/π) b. 11. Doing a second inversion to get the iris with charge in its hole. The above picture gets us all set up to so this next step. As noted earlier, in the first phase the inversion radius "a" is arbitrary and neither the potential nor σ depends on it. For our second phase we could use a new inversion radius, perhaps a1 and likely we will later set a1 = a, but let's keep it separate for a while. Question: What is the radius B of the iris? Consider the upper dotted ray to the iris edge. Let the angle at the origin be called αu. We know from very early work that αu = (π-θb)/2. The horizontal edge of this triangle has length d1 = a12/2R. If the iris has radius B, then we have tan[(π-θb)/2] = B/d1 => B = d1 tan[(π-θb)/2] = d1 cot(θb/2) Question: what is radius S of the charge in the iris hole? We could draw a corresponding triangle going through this hole charge, the angle would be αc = θc/2 and we would say tan(θc/2) = S/d1 => S = d1 tan(θc/2) Question: what is the size of the charge in the hole? Jackson notes say q'i = qi(a/ri). The distance ri must be determined from a triangle to the Green's charge from the inversion origin. Angle is αc as noted just above. So tanαc = rise/run = Rsinθc/ (R + Rcosθc) = sinθc/(1+cosθc) = tan(θc/2) so we are at least consistent. The hypotenuse of our triangle is ri so we have ri2 = (R + Rcosθc)2 + (Rsinθc)2 = R2 { 1 + 1 + 2 cosθc} = 2 R2 (1 + cosθc) = 4R2 cos2(θc/2) So here then is the size of our in-the-hole charge qhole = (a1/[ 2Rcos(θc/2)]) qG = (a1/[ 2Rcos(θc/2)]) [– (aQ/C)] with C = (2/π) b Question: what is the charge density on the iris? Quoting from far above, we have σ'(r') = (r/a1)3 σ(r) = (r/a1)3 (a/r)3 (Q/2πb) / r' = (a12/r2) r Now at once we have excellent motivation to set a1 = a. Then we have σ'(r') = (Q/2πb) / r' = (a2/r2) r which is a mighty simple result, which I now need to process. w2 = (a2/r2)2 { [(- cosθc x - sinθc z) - c (r2/a2)]2 +y2 } 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)) d = a2/2R We know from just above that tan(θb/2) = d/B tan(θc/2) = S/d Then we have b = d (d/B) ( 1 + (S/d)2) / ( 1 - (S/d)2(d/B)2 ) = (d2/B) ( 1 + (S/d)2) / (1-(S/B)2) c = d (S/d) ( 1 + (d/B)2)/ (1-(S/B)2) = S ( 1 + (d/B)2)/ (1-(S/B)2) Restate b = (d2/B) ( 1 + (S/d)2) / (1-(S/B)2) c = S ( 1 + (d/B)2)/ (1-(S/B)2) The relative simplicity of these two results provide encouragement to the "hopeful". But w does not look simple enough. In our σ equation, we want r' to lie on the iris and that means r must lie on the bowl. But for points r on the bowl, we know that x2 + y2 + (z-R)2 = R2 Thus we could write w2 = (a2/r2)2 { [( cosθc x + sinθc z) + c (r2/a2)]2 +[ R2- x2- (z-R)2] } so if you pick x,z with each less than R, you will get some w. Does this simplify? {} = ( cosθc x + sinθc z)2 + c2(r2/a2)2 + 2c (r2/a2) ( cosθc x + sinθc z) + R2 - x2 - (z-R)2 = x2(cos2θc - 1) + z2(sin2θc - 1) + 2xz cosθc sinθc + x 2c (r2/a2) cosθc + z 2c (r2/a2) + R2- R2 + 2Rz + c2(r2/a2)2 = - sin2θc x2 - cos2θc z2 + 2xz cosθc sinθc + x 2c (r2/a2) cosθc + z 2[c (r2/a2) + R] + c2(r2/a2)2 Answer: no, it does not simplify. I am suddenly wondering about this little Lemma I proved in the Jackson meta notes for inversion | r' - ri| = (ri/r) | r - r'i| This just seems to be a pretty powerful fact that maybe wants to get used. In our iris situation, we certainly can talk about our one pair of charges: ri = 2R cos(θc/2) = inversion origin to Green's charge qG ri' = // Pythagorean theorem for triangle We could then apply this to any pair (r,r') we wanted. Now recall Smythe's answer to this problem which I am trying to get to: σ(r',θ') = - (q/2πr12) / r12 = r'2 + b2 - 2br' cosθ' I can translate these Smythe parameters to my current context: Smythe Me q qhole = (a1/[ 2Rcos(θc/2)]) qG b S a B θ φ' r' ρ' r12 | ρ' - ri|2 = ρ'2+ S2 - 2ρ'S cosφ' Then the Smythe result is this σ(ρ',φ', z' = d)Smythe = - (qhole/2πr12) / which I can compare with my result σ'(r') = (Q/2πb) / r' = (a2/r2) r b = (d2/B) ( 1 + (S/d)2) / (1-(S/B)2) w2 = (a2/r2)2 { [( cosθc x + sinθc z) + c (r2/a2)]2 +[ R2- x2- (z-R)2] } I don't see agreement, but at least we seem to be playing in the same ball park. Obviously it should not be taking 40 pages of work to get his result, but I am a novice and novices have to pay heavily for being so. I have now been working on this today Sunday 6.6.10 for about 11 hours, so let's give it a rest. After all, this problem has been pending for maybe 2-3 months now, it can wait for tomorrow.