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The r=0 BC idea for Ephi REVIEWED

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A short reviewed note by Phil dated 11.29.13 from the Chapter 2 round wire material. It records an attempt to argue that the second boundary condition in Appendix D(c) is Eφ=0 at r=0 rather than at r=a, and discusses Er and Eφ at the singular origin of polar coordinates. It works through Bessel function small-x limits to test the constants a_m, finding a0 and a1 vanish but m>1 undetermined. A header says the idea was later rejected as nonsense.

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The r=0 BC idea for Ephi PhL 11.29.13 This was my attempt to say that the second boundary condition of App D (c) shown below is that Eφ vanishes at r = 0 instead of r = a. This idea later got completely rejected as non-sense, and the r = 0 BC was then reinforced in text at the start of App D. Question: If you have fields Er(r,m) and Eφ(r,m) as functions of r, what can you say about these functions when they are evaluated at r = 0 ? Think of this as a polar coordinates question. This is a singular point of the coordinate system of course. If you had Er ≠ A > 0 at r=0, say, then somehow Er points in all directions at once. I want to argue that therefore Er = 0 at such an origin. But if you take a uniform E field and plop down an origin somewhere, that does not mean Er= 0. (c) Application of the Boundary Conditions We have two boundary conditions to impose: Er(r=a,m) = (jω/σ) Nm (D.2.18) Eφ(r=0,m) = 0 (D.2.19) The first is the radial charge pumping condition shown in (D.1.23) above, while the second a statement of the fact that, since there is no "vortex line" of infinite B field at r = 0, Eφ(r=0) cannot point in any direction and must therefore be 0. These two boundary conditions serve to determine the two constants am and Km, though a bit of algebra is required. The first step is to use (D.2.11) for Er and (D.2.15) for Eφ to write out the two boundary conditions as am xa-1 Jm(xa) + Jm+1(xa) = (jω/σ) Nm (1) limx→0 [- am (1/x) Jm(x) + ( + ) Jm+1(x)] = 0 . (2) Now we know that J0(x) ≈ 1 Jm(x) ≈ (x/2)m/m! Jm+1(x) = (x/2)m+1/(m+1)! (1/x)Jm(x) ≈ 2-m/m! * xm-1 For any m ≥ 0, the second term in (2) vanishes so we then have limx→0 [- am (1/x) Jm(x)] = 0 . (2)' For m = 0, a0 must vanish, otherwise the limit is infinite. For m > 0 we have limx→0 [- am (1/x) (x/2)m/m!] = 0 . (2)' or limx→0 [- am (x/2)m-1/m!] = 0 . (2)' For m = 1 we find limx→0 [- a1 (x/2)0/1!] = 0 . (2)' which says a1 = 0. For m > 1, this limit yields no conclusion about am. Status: This means we don't have enough information to determine the constants for m > 1.