saved old Appendix H REVIEWED

Saved old Appendix H PhL 11.14.13 I replaced this old App H with a pair of new appendices, App H and App I. They subsume all the results developed in this old App H. So this is just for archive, can be tossed eventually. Appendix H : Calculation of 2[1/r] , 2[h(r)/r] and 2[e-jkr/r], and 22D[ln(1/r)] __________________________________________________________________________________ Theorem 1: -2[1/r] = 4πδ(r) (H.1) Proof: Let volume V be all of 3D 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-3, g(r) has units L-3. Our candidate function of interest is g(r) = - (1/4π) 2[1/r] (H.3) Using 2 in spherical coordinates acting on a function of r, one finds that, since ∂r(1) = 0, 2[1/r] = (1/r2)∂r(r2∂r) [1/r] = 0 r > 0 (H.4) so that g(r) = - (1/4π) 2[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) = - (1/4π) ∫Va dV 2[1/r] = - (1/4π) ∫Va dV [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 = [1/r] = ∂r(1/r) = -r-2 we find that LHS (H.7) = ∫Va dV div [1/r] = ∫Va dV 2[1/r] = ∫Va dV [-4πg(r)] = -4π ∫Va dV g(r) RHS (H.7) = ∫S dS [1/r] = ∫dΩ [a2 ] [1/r]|r=a = ∫dΩ[a2 ] [-a-2] = -4π 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 - (1/4π)2[1/r] = δ(r) (H.8) or 2[1/r] = - 4πδ(r) which is (H.1). QED __________________________________________________________________________________ Theorem 2: 2[h(r)/r] = - 4π h(0) δ(r) + h"(r)/ r (H.9) Proof: Start with this vector identity 2(φψ) = φ2ψ + ψ2φ + 2 φ ψ (H.10) so that 2(h r-1) = h2(r-1) + r-12h + 2 h (r-1) = - h 4π δ(r) + r-12h + 2 [ h' (-r-2) = - 4π h(0) δ(r) + r-12h - 2 r-2 h'(r) (H.11) Algebra shows that, 2h = (1/r2)∂r(r2∂r)h(r) = h"(r) + (2/r)h'(r) (H.12) so then 2(h r-1) = - 4π h(0) δ(r) + r-1 [h"(r) + (2/r)h'(r) ] - 2 r-2 h'(r) = - 4π h(0) δ(r) + h"(r)/ r which is the claim of (H.9). QED __________________________________________________________________________________ Example of Theorem 2: h(r) = e-jkr h = e-jkr h(0) = 1 h' = -jk e-jkr h" = -k2 e-jkr 2(e-jkr/r) = - 4π 1 δ(r) + [-k2 e-jkr ] / r = -4πδ(r) - k2(e-jkr/r) Thus, ( 2+k2) (e-jkr/r) = - 4πδ(r) (H.13) or - (2+k2) (e-jkr/4πr) = δ(r) (H.14) If we translate the origin from 0 to x' and define R = |r - x'| and rename r to be x, (H.13) becomes - (2+k2) (e-jkR/4πR) = δ(x-x') (H.15) Thus we may identify e-jkR/4πR as the Helmholtz equation Green's Function. In the case k = 0 we get - 2(1/4πR) = δ(x-x') (H.16) and we may identify (1/4πR) as the Poisson equation Green's Function as shown in (A.0.3) and (A.0.4). __________________________________________________________________________________ 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