full sphere paradox REVIEWED

Paradox 12.4.15 with full sphere limit PhL 12.4.15 Right now bowl doc contains the result of part 1 below. Since V does not contain a, I don't know how to take a limit to maintain fixed R and have a→ 0 with R = a/u0 holding constant. In my Maple plot, R does get very large and is not fixed. I cannot change the meaning of the u coordinate! So I leave this as it is. // 12.31.15 Here is the paradox. 1. On the one hand: I start with my solution V which is this V(ξ,u) = (V0/π) { π - cot-1[] + cot-1[] } (1) I take the limit u0 = 0 and this gives cos(π-u/2) = cos(u/2-π) = -cos(u/2) // for any u cos(2u0-u) → cos(-u) = cosu cos(u0- u/2) → cos(-u/2) = cos(u/2) // limits as u0 → 0 V(ξ,u) = (V0/π) { π - cot-1[] + 1*cot-1[] } = V0 The two terms are equal and opposite it seems for all values of u in the range 2π, and so the two cot-1 terms cancel and the result is V0. 2. On the other hand: I enter exactly the equation (1) above into Maple, Maple then does a scan of ρ and z, and at each point it uses these equations to get ξ and u: ξ = th-1 (2a|ρ| / (a2+ρ2+z2) u = getu(ρ,z) returns a value of u in range (0,2π) and then it just plots the above function versus ρ and z, and it shows a dropoff 1/r outside the bowl, for example, for u0 = .01π/8 3. Paradox is this: Why does Maple show a dropoff, but my limit shows no dropoff. Plan A: I think I know the answer, and Maple helped me out. Notice the scale on the above drawing! I had to increase it a lot to "get outside the bowl". If I don't increase the scale, then the entire plot is basically inside the bowl and you get the constant value V0. So the confusion is related to the fact that the bowl gets infinitely large in this limit! Basically the bowl fills the entire upper half plane in z,ρ space, and so at all points in that half plane you will have V = V0! So I have to regulate the limit. I said it was a "thought experiment only" but in fact it is not so. You have in fact a = R |sinu0| and if you want to keep R fixed, you have to scale down a as shown. But the parameter a does not appear in the potential! So here is the problem: How do you take a limit of the above potential in such a way as to keep R fixed? I could replace sinu0 by a/R and in the end take the limit a→ 0. How would that work? We start with V(ξ,u) = (V0/π) { π - cot-1[] + cot-1[] } (1) Now replace u0 = sin-1(a/R) |sin(u0)| = a/R and then you could write out the above formula in detail. Then cos(2u0-u) = cos(2u0)cos(u) + sin(2u0)sin(u) = [ 1-2sin2(u0)] cos(u) + 2sin(u0)cos(u0)sin(u) = [ 1-2sin2(u0)] cos(u) + 2sign(u0) |sin(u0)|sin(u) = [ 1-2a2/R2] cos(u) + 2 sign(sin(u0))(a/R)sin(u) and you could then install this in the second factor in two places. Similarly you could write cos(u0-u/2) = cos(u0)cos(u/2) + sin(u0)sin(u/2) = cos(u/2) + 2 sign(sin(u0))(a/R) sin(u/2) So the expression is then very messy, but you could present it this way V(ξ,u; u0) = (V0/π) { π - cot-1[] + cot-1[] } (1) where cos(2u0-u) = [ 1-2a2/R2] cos(u) + 2 sign(sin(u0))(a/R)sin(u) cos(u0-u/2) = cos(u/2) + 2 sign(sin(u0))(a/R) sin(u/2) I see no gain by writing this out as a single formula. But what happens now if we take the limit u0→ 0? where cos(2u0-u) = [ 1-2a2/R2] cos(u) + 2 sign(sin(u0))(a/R)sin(u) cos(u0-u/2) = cos(u/2) + 2 sign(sin(u0))(a/R) sin(u/2) we know that |sin(u0)| = a/R so a = R|sin(u0)| so in this limit we have to have a→ 0. In this limit, both second terms vanish and we just get cos(0-u) = [ 1] cos(u) cos(0-u/2) = 1 cos(u/2) and everything is just fine. What happens as we take u0 → 2π? Again we have a→ 0 and we get the same two little equations as above. I think we have to "stay away from the limit" a little bit. So lets write u0 = 0 + ε ε > 0 a << R a = R|sinu0| = Ru0 = Rε a/R = ε Then we have cos(2u0-u) = [ 1-2a2/R2] cos(u) + 2(a/R)sin(u) = [ 1-2ε2] cos(u) + 2εsin(u) ≈ [ 1-2ε2] cos(u) + 2ε [ 1 - (1/2)ε2 ] sin(u) ≈ [ 1-2ε2] cos(u) - 2ε sin(u) /// dropping the ε3 small term ≈ cosu - 2ε sinu and cos(u0-u/2) = cos(u/2) + 2 sign(sin(u0))(a/R) sin(u/2) = cos(u/2) + 2 ε sin(u/2) = cos(u/2) + 2 ε sin(u/2) So lets try going just keeping the leading ε terms, so we have cos(2u0-u) = cosu - 2ε sinu cos(u0-u/2) = cos(u/2) + 2 ε sin(u/2) and we then want to stick these into V(ξ,u) = (V0/π) { π - cot-1[] + cot-1[] } (1) Well we will need chξ - cos(2u0-u) = chξ - [ cosu - 2ε sinu] = chξ -cosu + 2ε sinu Then the argument of the second cot term is this = Then the potential is given by V(ξ,u) = (V0/π) { π - cot-1[] + cot-1[] } (1) I don't see anything useful happening here! Nice try. Take ε → 0 and we have the same paradox problem. Nice try, but no cigar. Plan B. Consider again V(ξ,u; u0) = (V0/π) { π - cot-1[] + cot-1[] } (1) If we take u0→ 0 with a fixed, we get R→ ∞ and the sphere fills the entire upper half plane! This upper half plane is characterized by u in the range (0,π). The lower half plane is (π,2π). You might think that you should be V < V0 in the lower half plane, but the sphere is so big, that you never move any significant distance from its surface in the lower half plane and V remains at V = V0 . So basically you get V = V0 everywhere, and this agrees with what the limit says. Treated as an abstract function, if you consider V(ξ,u; u0) stated above and take u0 → 0, you get V0 and there is nothing you can do about it! You cannot take some kind of limit which keeps the sphere at a constant radius. I think the only way out is to write the above entirely in terms of z,ρ,R and a. THEN maybe we can do something to get our limits. I might use some results from my Jackson limit work. I know for example that + = 2a ch(ξ/2) / r = a / (chξ - cosu) thξ = 2aρ/(ρ2+ z2 +a2) th2ξ = [2aρ/(ρ2+ z2 +a2)]2 = 1 - 1/ch2ξ 1/ch2ξ = 1 - [2aρ/(ρ2+ z2 +a2)]2 Lets define g ≡ 2aρ/(ρ2+ z2 +a2) Then 1/ch2ξ = 1-g2 ch2ξ = 1/(1-g2) 2 ch2ξ/2 = 1+chξ = 1 + 1/ ch2ξ/2 = [ 1 + 1/]/2 chξ/2 = Then I can use this to get + = 2a ch(ξ/2) / = 2a / and this gives this very messy result 1/ = [ + ]/[ 2a ] and the point is that I have written this object entirely in z,ρ coordinates. Now g = 2aρ/(ρ2+ z2 +a2) g2 = [2aρ/(ρ2+ z2 +a2)]2 = 4a2ρ2/ (ρ2+ z2 +a2)2 1-g2 = 1 - 4a2ρ2/ (ρ2+ z2 +a2)2 = [(ρ2+ z2 +a2)2 - 4a2ρ2 ] / (ρ2+ z2 +a2)2 = [ρ4+z4 + a4 + 2ρ2z2 -2ρ2a2 + 2a2z2] / (ρ2+ z2 +a2)2 ≡ f(ρ,z,a) / (ρ2+ z2 +a2)2 = / (ρ2+ z2 +a2) 1/ = (ρ2+ z2 +a2)/ 1 + 1/ = 1 + (ρ2+ z2 +a2)/ = [ + (ρ2+ z2 +a2)] / It is just a real mess. This is not the right Plan. Maple can do it, why can't I do it? Plan C. Recall ξ = tanh-1[2aρ/(ρ2+ z2 +a2)] ρ = (1.3.5) u = tan-1[ 2az/(ρ2+z2- a2)] φ = tan-1(y/x) r2 = ρ2+ z2 Suppose we say that a << ρ,z right from the start. Then we get ξ = tanh-1[2aρ/r2] ≈ 2aρ/r2 a ≈ R u0 u = tan-1[ 2az/r2] ≈ 2az/r2 assumes r >> R so that r2 = z2+ρ2 Then we have simple Cartesian forms for these complicated toroidal parameters. ξ = 2aρ/r2 u = 2az/r2 a << ρ, |z| r >> R So both ξ and u are going to be small in this Plan C. Now once again V(ξ,u; u0) = (V0/π) { π - cot-1[] + cot-1[] } (1) This is the form Maple uses to make the plot which shows 1/r drop, so I am quite sure it is right. I know this is a large r regime. r = a / (chξ - cosu) But when ξ and u are small we have chξ - cosu = ξ2/2 + u2/2 so r = a / [ (ξ2 + u2)/2 or r = 2a/ Now consider since u,u0 are both small, as is their difference, cos(2u0-u) = 1 - (2u0-u)2/2 + O(u4) if both u and u0 are small cos(u0-u/2) = 1 - (u0-u/2)2/2 + O(u4) chξ - cos(2u0-u) = [1 + ξ2/2] - [ 1 - (2u0-u)2/2] = ξ2/2 + (2u0-u)2/2 = chξ - cosu = ξ2/2 + u2/2 = cos(u/2) = (1 - (u/2)2) Now go back to the original toroidal potential V(ξ,u; u0) = (V0/π) { π - cot-1[] + cot-1[] } (1) r1 r2 r3 Ratio r2 is this r2 = = = not small Ratio r3 is this r3 = = = large (we assume u and u0 are small) Ratio r1 is this r1 = = = large Here then is our complete potential in this regime V(ξ,u; u0) = (V0/π) { π - cot-1(r1) + r2 cot-1(r3) } All the time I have been thinking ξ = 2aρ/r2 u = 2az/r2 a << ρ, |z| u0 = a/R so I guess you could say things are expressed fully in Cartesian coordinates. What do I do with a = Rsinu0 = Ru0 ? Well ξ = 2 Ru0ρ/r2 u = 2 Ru0z/r2 a << ρ, |z| So now our potential has no a showing, but it does have R showing! What can I do next? Consider cot-1(r1) = tan-1(1/r1) ≈ 1/r1 cot-1(r3) = tan-1(1/r3) ≈ 1/r3 Then the total potential is V(ξ,u; u0) = (V0/π) { π - 1/r1 + 1/r3 } Now we have 1/r3 = / = and 1/r1 = = small and so V(ξ,u; u0) = (V0/π) { π - + } Now we do have r ≈ 2a/ = 2Ru0 / = 2Ru0/r = (1/) = (1/) 2Ru0/r = Ru0/r ξ2/2+ u2/2 = 2R2u20/r2 So simplifying denominators, we get V(ξ,u; u0) = (V0/π) { π - (1/)2R2u20/r2 + (1/) Ru0/r } = (V0/π) { π - (1/)(u0R/r)[2(u0R/r) - ] } Now (u0R/r) << 1 for large r, and also because u0 is small, so we have = (V0/π) { π - (1/)(u0R/r)[ - ] } = (V0/π) { π + (u0R/r) } = (V0) { 1 + (u0R/πr) } and for the first time I get something that at least looks like a 1/r potential. The result I want is V(ξ,u; u0) = Q/r = (CV0)/r = (RV0)/r = V0 (R/r) In my result if it were correct the first term swamps the second and we get our familiar V = V0 . Summary: I tried to do the case where u,u0 and ξ are all very small, thinking that would simplify things. But in the end, instead of getting 1/r distant behavior, I get V = V0, conflicting with what I see Maple plotting in front of my eyes. The far field potential should be K/r for any value of u0. I should not have to take u0 small. But of course we have to have u≥ u0 so then u can no longer be small. So in this case, what do you say about an expression for r? Well, I need to repair my expression for r in the Section 1! This is tricky, you have to choose your r coordinate. I guess I have it right. It is relative to the origin. So I have r = a / (chξ - cosu) r2 = x2+y2+z2 and this r is NOT with respect to the center of the bowl! Now where is the large r limit? Certainly we need chξ - cosu → 0. Certainly that requires ξ near 0 and u near 0 so I think I have that part correct: Where is large r? It is in the region of ξ ≈ 0 and u ≈ 0 and in that region we have r ≈ 2a/ = 2R|sinu0| / r2 ≡ ρ2+z2 So for fixed u0 we certainly can associate large r with small ξ and u except we need u≥ u0, so you really cant "get to" small u. Idea: Maybe H have to stick with the full expression r = a / (chξ - cosu) and not make any assumptions about "where this is large". I should play with this a bit. r = a / (chξ - cosu) // correct from bowl doc = a shξ / (chξ - cosu) = a shξ / (chξ - cosu) r = a How do you make r be large? You MUST have the denominator small here, so you need these ingredients: ξ ≈ small u ≈ 2π which always lies in the range u0, u0+2π So maybe my u = 0 was the bad turn I took in all the above! I need to fix this wherever I say it in bowl doc. Keeping u as is, we end up with r = a Now let u = 2π + ε ε could have either sign cos(2π+ε) = cos(ε) = 1-ε2/2 Then we have r = a = a = 2a again, r = 2a So "large r" is a certain region of (u,ξ) space where ξ >> ξ2 + ε2 1 >> ξ + ε2/ξ ξ + ε2/ξ << 1 For this last to be true, since both terms are positive, they must both be small. So we need ξ << 1 AND ε2 << ξ or ε2 << ξ << 1 Implications? ξ << 1 ξ2 << 1 ε << 1 ε2 << 1 ξ + ε << 1 ξ2+ ε2 << 1 << 1 So this is something new in this document. ξ << 1 AND (u-2π)2 << ξ This certainly is a very strange region of ξ,u space! (u-2π)2 << ξ << 1 Certainly Maple "finds" this region during its plot. Fact: Here is the region of (u,ξ) space where one finds large r: (u-2π)2 << ξ << 1 ε2 << ξ << 1 u = 2π + ε So if you want to examine the potential for large r, you have to go to this region! Question: do we know that ε << ξ ? Well we know that ε2 << ξ so ε << . Example: Suppose ε = 10-4 and ξ = 10-5 . Then ε2 = 10-8 << 10-5 but ε = 10-4 is NOT << 10-5. And also ε2 = 10-8 is NOT << ξ2 = 10-10. Fact: We do NOT know that ε << ξ so we do NOT know that ε2 << ξ2 . So I guess I could try to study up on the various terms in this select region V(ξ,u) = (V0/π) { π - cot-1[] + cot-1[] } (1) R1 R2 R3 cos(u/2) = cos(π+ε/2) = cos(ε/2+π) = -cos(ε/2) ≈ - [ 1 - (ε/2)2/2] = ε2/8 - 1 cos(u) = cos(2π+ε) = cosε ≈ 1 - ε2/2 chξ - cosu = (1 + ξ2/2 - 1 + ε2/2) = (ξ2+ε2)/2 cannot replace with ξ2/2 !!! ≈ / in my region cos(2u0-u) = cos(2u0- [2π+ε]) = cos(2u0- ε - 2π) = cos(2u0 - ε) chξ - cos(2u0-u) = 1+ξ2/2 - cos(2u0 - ε) = ξ2/2 + [1-cos(2u0 - ε)] = ξ2/2 + 2sin2(u0 - ε/2) = cos(u0- u/2) = cos(u0 - π - ε/2) = - cos(u0-ε/2) // rechecked all on 12/5/15 made fixes Now look at the ratios, R1 = = ≈ - 2/ = large in my region R2 = = ≈ = small = (/2) / |sin(u0- ε/2)| ≈ (/2)/ |sin(u0)| R3 = = ≈ - = - sign(u0) cot(u0) = moderate size say Note: I have made various approximations in getting final results for R1,R2,R3. So here then is my potential in the special region // repairs done!!! V(ξ,u) = (V0/π) { π - cot-1[R1] + R2 * cot-1[R3] } = (V0/π) { π - cot-1[- 2/ ] + (/2)/ |sin(u0)| cot-1[ - sign(u0) cot(u0)] } Now I happily replace π - cot-1(-2/) = cot-1(2/) so we then have V(ξ,u) = (V0/π) { cot-1(2/) + (/2) / |sin(u0)| cot-1[- sign(u0) cot(u0)] } Now assume for the moment that u0 lies in 0 to π so sign(u0) = 1, then we have: V(ξ,u) = (V0/π) { cot-1(2/) + (/2)/ |sin(u0)| cot-1[- cot(u0)] } and one of my "problems" is solved, the first free-standing term is now gone! Now cot-1(2/) ≈ cot-1(+∞) = 0 so we are left with V(ξ,u) = (V0/π) (/2)/ |sin(u0)| cot-1[- cot(u0)] // in my region of interest Now cot-1 ( - cot(u0)) = π - cot-1[cot(u0)] = π - u0 So I then have V(ξ,u) = (V0/π) [ (/2)/ |sin(u0)|] (π-u0) // in my region of interest Now recall that r = 2a // correctly quoted from above (ξ2+ ε2) = 2aξ STOP. Redo r this way sin(u) = sin(2π+ε) in sin(ε) = ε r = a / (chξ - cosu) // correct from bowl doc = a / [ (ξ2+ε2)/2] = 2a / // as I state in bowl doc So here is where we stand right now V(ξ,u) = (V0/π) [ (/2)/ |sin(u0)|] (π-u0) // in my region of interest r = 2a / 0 < u0 < π So now replace = 2a/r to get V(ξ,u) = (V0/π) [ (a/r)/ |sin(u0)|] (π-u0) // in my region of interest But a = R|sinu0| so write again as V(ξ,u) = (V0/π) [ ( R|sinu0| /r)/ |sin(u0)|] (π-u0) // in my region of interest = (V0R/π)(π-u0) (1/r) Side comment: z/ρ = sinu/shξ = ε/ξ and we do NOT know that this is small, problem fixed! Now what? What is the capacitance of our bowl? I have that at the end of section 5 C = (R/π) { (π- u0) + sin(u0) } Then πC/R = (π- u0) + sin(u0) So my result is not quite right, but this is the closest I have ever gotten. The result I expect is this Q = CV0 V = Q/r = V0(C/r) = V0 { (R/π) [ (π - u0) + sin(u0)]}/r = (V0R/π) [ (π-u0) + sin(u0)] / r // What I expected = (V0R/π)[ (π-u0)] / r // What I got This is looking very promising, I seem to have lost a term somewhere!!!! Otherwise very good! Back up to this point and try undoing certain approximations: R1 = = ≈ - (1-ε2/8) 2/ = - 2/ + (1/4) ε2/ recall ε2 << ξ << 1 So here is a candidate "extra term" which may be "finite" . Can I claim that ε2 << ? ε4 << ξ2+ ε2 ? Not obvious to me, so let's see what this particular extra term does π - cot-1[ - (1-ε2/8) 2/] Well, this extra term really has no effect because we still get argument = -∞. So this is not it. Move to R2 which I think is the mostly likely factor which will have that extra term. R2 = = Now write ξ2/2 + 2sin2(u0 - ε/2) = 2sin2(u0 - ε/2) + ξ2/2 = 2sin2(u0 - ε/2) [ 1 + (ξ2/2) / [ 2sin2(u0 - ε/2)] ] = 2sin2(u0 - ε/2) [ 1 + (1/4) ξ2 / sin2(u0 - ε/2)] It would be nice if I had some common order thing, like saying ε = ε and ξ = f ε and then do order ε. Maybe do that on the final presentation. Now consider sin(u0- ε/2) ≈ sin(u0) - (ε/2)cos(u0) 2sin2(u0 - ε/2) ≈ 2 []2