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save old App H Theorem 3 REVIEWED

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A short archived Word note from Phil's transmission-lines Appendix H, holding the old last section (Theorem 3), which now lives in Appendix I and is marked for eventual deletion. It proves that the 2D Laplacian of ln(1/r) is -2πδ(r) by editing the 3D proof of Theorem 1. The proof shows the Laplacian vanishes for r>0, then uses the divergence theorem on a small circular hole to get an integral of 1.

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Save old App H last section which is now in App I This is just archive, it is in Appendix I now, so throw this out some day. Theorem 3: 2[ln(1/r)] = - 2πδ(r) in two dimensions (H.1)' Our 2D proof here will be an "copy, paste and edit" version of the 3D proof of Theorem 1. We shall use the word "volume" with symbol V to denote area. A "spherical cavity" now means a circular hole. It should be understood that 2 = ∂x2+∂y2 and δ(r)= δ(x)δ(y). We indicate all edits in red and equation numbers are the same as for Theorem 1 with a prime added. Proof: Let volume V be all of 2D space. Carve out from V a small spherical cavity of radius a centered at r = 0. If we call this spherical volume Va and then V' = V - Va is the original volume with the spherical cavity. In order to show that some function g(r) = δ(r), one has to show that lima→0 ∫V' dV g(r) = 0 (H.2a)' lima→0 ∫Va dV g(r) = 1 (H.2b)' This is basically the definition of δ(r). Since δ(r) has units L-2, g(r) has units L-2. Comment: When any differential operator like or 2 is applied to ln(r0/r), the result is independent of r0 so we can always take r0 = 1. For example, ∂x [ln(r0/r)] = ∂x [ lnr0 + ln(1/r)] = ∂x ln(1/r). In what follows, ln(r) and ln(1/r) are always acted upon by differential operators, so we can interpret these objects as dimensionless quantities ln(r/r0) and ln(r0/r). Then it is clear below that dim [g(r)] = L-2. Our candidate function of interest is g(r) = - (1/2π) 2[ln(1/r)] = +(1/2π) 2 [ ln(r)] (H.3)' Using 2 in polar coordinates acting on a function of r, one finds that, since ∂r(1) = 0, 2[ln(r)] = (1/r)∂r(r∂r) [ln(r)] = 0 r > 0 (H.4)' so that g(r) = - (1/2π) 2[ln(1/r)] = 0 r > 0 . (H.5)' Thus, condition (H.2a)' is trivially satisfied since r > 0 everywhere in volume V'. It remains to verify condition (H.2b)'. Consider the integral appearing in the left side of (H.2b)' ∫Va dV g(r) = - (1/2π) ∫Va dV 2[ln(1/r)] = - (1/2π) ∫Va dV [ln(1/r)] . (H.6)' The divergence theorem says ∫V dV div F = ∫S dS F (H.7)' where V is any closed volume whose surface is S, and dS points out. Using V = Va and F = [ln(1/r)] = ∂r(ln(1/r)) = - ∂r(lnr) = [ -r-1] we find that [ here dS = dS where dS = adθ = a piece of circumference of the circle bounding Va ] LHS (H.7) = ∫Va dV div [ln(1/r)] = ∫Va dV 2[ln(1/r)] = ∫Va dV [-2πg(r)] = -2π ∫Va dV g(r) RHS (H.7) = ∫S dS [ ln(1/r)] = ∫dθ [a ] [ ln(1/r)]|r=a = ∫dθ [a ] [-a-1] = -2π which tells us that ∫Va dV g(r) = 1 for any a. Thus, lima→0 ∫Va dV g(r) = 1 and we have then verified (H.2b)'. Therefore we conclude that - (1/2π)2[ln(1/r)] = δ(r) (H.8)' or 2[ln(1/r)] = - 2πδ(r) which is (H.1)'. QED