new appendix H REVIEWED

11.14.13 Draft of App H. I split old App H into two appendices, one for 2D and one for 3D. Appendix H : The Poisson and Helmholtz Free-Space Propagators in 3D Note: Appendix I deals with these propagators in 2D rather than 3D. __________________________________________________________________________________ H.1 Overview and Meaning of Free-Space Propagators This appendix proves the following Facts: Fact 1 : -2[1/4πr] = δ(r) (H.2.1) (H.1.1) Fact 2 : -2[h(r)/r] = 4π h(0) δ(r) - h"(r)/ r (H.3.1) (H.1.2) Fact 3 : - (2+k2) (e-jkr/4πr) = δ(r) (H.3.5) (H.1.3) Throughout, 2 is the usual 3D Laplacian operator 2 = ∂x2 + ∂y2 + ∂z2. In the first and last results above, if one replaces r → r-r' (a simple translational shift of origin) ones finds -2[1/4πR] = δ(r-r') R = | r - r' | (H.1.4) - (2+k2) [e-jkR/4πR] = δ(r-r') δ(r-r') = δ(x-x') δ(y-y') δ(z-z') (H.1.5) The quantities in brackets are known as free-space Green's Functions (Green Functions) or propagators, or as "fundamental solutions": 1/4πR = the Poisson 3D free-space propagator (H.1.6) e-jkR/4πR = the Helmholtz 3D free-space propagator (H.1.7) The significance of these propagators is the following: -2 f(x) = s(x) => f(x) = ∫d3x' [1/4πR] s(x') + homogeneous solutions The Poisson Equation (H.1.8) - (2+k2) f(x) = s(x) => f(x) = ∫d3x' [e-jkR/4πR] s(x') + homogeneous solutions The Helmholtz Equation (H.1.9) The equations on the left are inhomogeneous partial differential equations driven by source function s(x). If one is careful to include in s(x) all source contributions (such as those on boundary surfaces), one generally does not have to add any homogeneous solutions on the right. A homogeneous solution refers to -2 fh(x) = 0, for example. The solutions shown on the right above can be instantly verified as follows: f(x) = ∫d3x' [1/4πR] s(x') + fh(x) -2 f(x) = ∫d3x' (-2 [1/4πR] ) s(x') -2 fh(x) = ∫d3x' δ(r-r') s(x') - 0 = s(x) (H.1.10) and similarly for - (2+k2) f = g. A "free space" Green's Function gF in general is a solution of D gF(r, r') = δ(r-r'), gF(r, r') → 0 as r → ∞ (H.1.11) where D is some differential operator. The condition on the right says gF must vanish on the Great Sphere. More generally one can write D g(r, r') = δ(r-r'), g(r, r') = 0 for r on some closed surface enclosing a region of interest (H.1.12) In this second form, the ∫d3x' is over the volume inside that closed surface. We shall not make use of this more general form in this document. George Green (1793-1841), by the way, was an English grain miller. Looking at f(x) = ∫d3x' [1/4πR] s(x') = ∫ gF(x,x') [s(x') d3x'], one can say that the kernel Green's Function gF(x,x') "propagates" a tiny piece of "source" [s(x')d3x'] from location x' to location x so that the solution f(x) is then a sum of all such propagated contributions as the source ranges over the entire volume of interest, which for us is all 3D space where the source is non-vanishing. __________________________________________________________________________________ H.2 Fact 1: -2[1/r] = 4πδ(r) (H.2.1) 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 carved out: In order to show that some function g(r) = δ(r), one has to show that lima→0 ∫V' dV g(r) = 0 (H.2.2a) lima→0 ∫Va dV g(r) = 1 (H.2.2b) 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.2.3) 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.2.4) so that g(r) = - (1/4π) 2[1/r] = 0 r > 0 . (H.2.5) Thus, condition (H.2.2a) is trivially satisfied since r > 0 everywhere in volume V'. It remains to verify condition (H.2.2b). Consider the integral appearing in the left side of (H.2.2b) ∫Va dV g(r) = - (1/4π) ∫Va dV 2[1/r] = - (1/4π) ∫Va dV [1/r] . (H.2.6) The divergence theorem says ∫V dV div F = ∫S dS F (H.2.7) 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.2.7) = ∫Va dV div [1/r] = ∫Va dV 2[1/r] = ∫Va dV [-4πg(r)] = -4π ∫Va dV g(r) RHS (H.2.7) = ∫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.2.2b). Therefore we conclude that the candidate g(r) of (H.2.3) is in fact the same as δ(r) so - (1/4π)2[1/r] = δ(r) (H.2.8) or 2[1/r] = - 4πδ(r) (H.2.9) which is (H.2.1). QED __________________________________________________________________________________ H.3 Fact 2: 2[h(r)/r] = - 4π h(0) δ(r) + h"(r)/ r (H.3.1) Proof: Start with this vector identity, 2(φψ) = φ2ψ + ψ2φ + 2 φ ψ . (H.3.2) This identity is valid in any number of dimensions (implied sum on i from 1 to N) , ∂i2(φψ)= ∂i[ (∂iφ)ψ + ψ(∂iφ)] = (∂i2φ)ψ + (∂iφ) (∂iψ) + φ(∂i2ψ) + (∂iφ) (∂iψ) . So apply (H.3.2) to the case φ = h and ψ = r-1, 2(h r-1) = h2(r-1) + r-12h + 2 h (r-1) = - h 4π δ(r) + r-12h + 2 [ h' (-r-2) ] // using (H.2.9) = - 4π h(0) δ(r) + r-12h - 2 r-2 h'(r) . (H.3.3) Algebra shows that, using spherical coordinates, 2h = (1/r2)∂r(r2∂r)h(r) = h"(r) + (2/r)h'(r) (H.3.4) 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.3.1). QED Fact 3: - (2+k2) (e-jkr/4πr) = δ(r) (H.3.5) This Fact is just an application of Fact 2 to the case h(r) = e-jkr : h = e-jkr h(0) = 1 h' = -jk e-jkr h" = -k2 e-jkr 2[h(r)/r] = - 4π h(0) δ(r) + h"(r)/ r (H.3.1) so 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) or - (2+k2) (e-jkr/4πr) = δ(r) as claimed.