bowl bug 11_27_15 REVIEWED

Bug found with σ on the bowl. PhL 1.27.15 I had trouble here with the disk limit of σin and this problem is explored more in v2 of this doc with the same title. I failed to realize that you have to handle R→∞ at the same time. I then went off pondering whether my whole understanding of σ is correct! All resolved now. Here is the problem. The first question is OK, but not the second. Question 1: I say that σ is infinite at the bowl edge. How does that appear in my result? I never checked on that. σ+ = σin = ( V0/4π2R) [ (B/A) – tan-1(B/A)] A = cos(u0/2) and B = At the bowl edge, we are at the focal point so we have ξ → +∞. Then B → ∞ and σin(near edge) = ( V0/4π2R) [ (B/A) – π/2 ] ≈ ( V0/4π2R) (B/A) Now B/A = (chξ)1/2/ [cos(u0/2)] so the answer close to the edge is σin(near edge) = ( V0/4π2R) (chξ)1/2/ [cos(u0/2)] which does indeed blow up as claimed as ξ → ∞. Question 2. How about at the bowl center? There ξ → 0 and we get B = = . Then B/A = / [cos(u0/2) ] Now = sin(u0/2) B/A = sin(u0/2)/ [cos(u0/2) ] = tan(u0/2) and then we have σin(at center) = ( V0/4π2R) [ tan(u0/2) – tan-1(tan(u0/2))] = ( V0/4π2R) [ tan(u0/2) – u0/2] Consider the sequence of upper bowls from large to small so u0 → π from below. This says that σin (and σout which differs by a constant) blows up for the flat disk at disk center! This result I am sure is wrong. So we have a new problem. Time to check my calculation of σ starting with (4.1). I need to show that it replicates the flat disk charge distribution Question 3: Let's start with the full σin result and take it to the flat disk limit. Here is my full result σ+ = σin = ( V0/4π2R) [ (B/A) – tan-1(B/A)] A = cos(u0/2) and B = The flat disk has u0 = π and then A = cos(π/2) = 0 B = The limit then says σ+ = σin = ( V0/4π2R) [ (B/0) – tan-1(B/0)] = ( V0/4π2R) [ ∞ – π/2] So something is WRONG with my σ result! I should have checked this limit when I wrote the doc. Stupid. [ again I overlook the fact that R → ∞ in the disk limit ] Check the calculation of σ I start with V(ξ,u) = V0 !Syntax Error, Idτ Piτ-1/2(chξ) ch[(π–u0)τ] ch[(π+u0-u)τ]/ ch2(πτ) (2.8) σ+ = – (1/4π) (1/hu) ∂uV(ξ,u)|u=u0 1/hu = (chξ - cosu)/a Where do I get this second line from? No reference is given. Is this result in any of my web docs? I do have this in SI units, Dn1 - Dn2 = nfree or [ε1E1n - ε2E2n] = nfree (1.1.47) Consider a boundary like this Suppose 2 = metal and 1 = vacuum. We then have [ε1E1n - ε2E2n] = nfree → ε0E1n = nfree Translate this to say σ = ε0En SI units So far so good. Meanwhile, E = - grad V We look this up in tensor doc where we have [grad f](x) = (∂if ) ei [grad f](x) = (∂if ) ei = (1/hi2) (∂if ) ei [grad f](x) = hi(∂if ) i = (1/hi) (∂if ) // M&S 1.05 (10.1.17) or [grad V](x) = (∂iV ) ei [grad V](x) = (∂iV ) ei = (1/hi2) (∂iV ) ei [grad V](x) = hi(∂iV ) i = (1/hi) (∂iV ) i // M&S 1.05 (10.1.17) where italic V means it is a function of the curvilinear coordinates. Assume that i is a unit vector in curvilinear coordinates which is normal to our surface of interest, which would be u for our case study. Then [grad V](x) = (1/hu) (∂uV ) u Then E = - grad V Euu = - (1/hu) (∂uV ) u Eu = - (1/hu) (∂uV ) and then finally σ = ε0Eu = - ε0 (1/hu) (∂uV ) and then divide by 4πε0 to get cgs units σ = ε0Eu = - (1/4π) (1/hu) (∂uV ) and this is what I used. Question: But there is something unclear to me here. Go back to [ε1E1n - ε2E2n] = nfree Are these E-field component covariant or contravariant !!! Go back to the derivation (where I have fixed the bad reference which was 1.1.13), div D = ρfree ∫V ρfree dV = ∫S D dS (1.1.32) Now how do you do this in covariant notation? D dS = DidSi = DidSi My lines doc does not address this question. In the n-piped discussion I do mention area with covariant index stuff, and there fore example I have dA'n = J2(Πi≠ndx'i) e'n since (e'n)i = δni (7.18.1) (8.4.c.5) where we are in x'-space. I show in toroidals.doc that J = hξhuhφ = If I then set n = 2 for the u coordinate, we get dA'n = J2(Πi≠ndx'i) e'n dAu = J2 dξdφ eu dSu = J2 dξdφ eu Then we have D dS = ( Du eu) ( J2 dξdφ eu) = Du J2 dξdφ so it is the contravariant component ! Then we really have [ε1En1 - ε2En2] = nfree [ε1Eu1 - ε2Eu2] = σ εoEu1 = σ We now trace through the above discussion. [grad V](x) = (∂iV ) ei [grad V](x) = (∂iV ) ei = (1/hi2) (∂iV ) ei [grad V](x) = hi(∂iV ) i = (1/hi) (∂iV ) i // M&S 1.05 (10.1.17) E = - grad V Euu = - (1/hu) (∂uV ) u Eu = - (1/hu) (∂uV ) Eu = - (1/hu) (∂uV ) Then σ = εoEu1 = - ε0 (1/hu)(∂uV ) // with an upper index! So my document should say not this σ+ = – (1/4π) (1/hu) ∂uV(ξ,u)|u=u0 1/hu = (chξ - cosu)/a σ– = +(1/4π) (1/hu) ∂uV(ξ,u)|u=u0+2π // cgs units selected here (4.1) but this σ+ = – (1/4π) (1/hu) ∂uV(ξ,u)|u=u0 1/hu = (chξ - cosu)/a σ– = +(1/4π) (1/hu) ∂uV(ξ,u)|u=u0+2π // cgs units selected here (4.1) So this is the first thing I have found to be done wrong. Now we have ∂i = Σj gij∂j = gii∂i For i = 2 = u this says ∂u = guu∂i = hu2 ∂u ∂u = hu-2∂u Then I would seem to get σ+ = – (1/4π) (1/hu)3 ∂uV(ξ,u)|u=u0 1/hu = (chξ - cosu)/a σ– = +(1/4π) (1/hu)3 ∂uV(ξ,u)|u=u0+2π // cgs units selected here (4.1) Basically this adds an extra factor of (1/hu)2 to my charge densities. Note that 1/hu = (chξ - cosu)/a (1/hu)2 = (chξ - cosu)2/a2 = B4/a2 Then my new charge density results (if everything ELSE was done correctly) would be σ+ = σin = ( V0/4π2R) [ (B/A) – tan-1(B/A)] (B4/a2) A = cos(u0/2) and B = B2 = chξ - cosu0 How does this change the limit u0 ≠ π ? The problem remains unrepaired! A → 0 and σ → ∞ for any ξ. I have some major problems here. I will respond to the guy first, and then tackle these problems. This might result in a major rewrite of bowl doc!