Appendix H REVIEWED

Appendix H : Calculation of 2[1/r] , 2[h(r)/r] and 2[e-jkr/r] This has been installed, no need to keep long term. __________________________________________________________________________________ Theorem 1: 2[1/r] = - 4πδ(r) (H.1) Proof: Let V be some volume that contains the origin point r = 0. Carve out from V a small spherical cavity of radius a surrounding 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). Our 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 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, (H.2a) is trivially satisfied since r > 0 everywhere in volume V'. It remains to show (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], ∫Va dV g(r) = - (1/4π) ∫Sa dS [1/r] = - (1/4π) [4πa2 ] [1/r] = -a2 [-r-2]r=a = a2 a-2 = 1 . 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/R) = 4π δ(x-x') (H.15) Thus we may identify e-jkR/R as the Helmholtz equation Green's Function.