jim delta 1_14_11 11AM

the doc is: jim delta 1_14_11 11AM.doc Jim, Here is some more evidence for limε→0 [(1/r)∂r(r/) ] = + 2π δ(2)(r) = + 2π[ δ(r)/2πr] = δ(r)/r Suppose we multiply both sides by r. Then this ought to be true limε→0 [∂r(r/) ] = δ(r) (1) and this is then an n=1 case of Stakgold's theorem. Let's see if we can verify equation (1) . We first do the usual Maple calculation to find that ∂r(r/) = ε2 / (r2+ ε2)3/2 = ε2 / [ ε2 ( (r/ε)2 + 1)] 3/2 = ε2 / [ ε3 ( (r/ε)2 + 1) 3/2] = ε-1 / ( (r/ε)2 + 1) 3/2 Therefore, adding a factor of K to both sides, we have shown that K ∂r(r/) = ε-1 { K / ( (r/ε)2 + 1) 3/2 ) So define g(x) = K/(x2 +1)3/2 and we then have K ∂r(r/) = ε-1 g(r/ε) Stakgold's theorem for general n=1 says this (see Appendix below) !Syntax Error, Idr g(r) = 1/2 Sn(1) = "area of 1 dimensional sphere" = 2 limε→0 [ ε-1 g(x/ε) ] = δ(x) on R1 which is the real axis x in (-∞,∞) limε→0 [ ε-1 g(r/ε) ] = (1/2)δ(r) for r in (0,∞) // See Appendix We determine K from the above normalization condition, 1/2 = !Syntax Error, Idr g(r) = !Syntax Error, Idr { K/(x2 +1)3/2} = K !Syntax Error, Idr/ (r2 +1)3/2 = K * 1 = K which then says K = 1/2, so we then have (1/2) ∂r(r/) = ε-1 g(r/ε) limε→0 [(1/2) ∂r(r/)] = limε→0[ε-1 g(r/ε)] = (1/2)δ(r) Therefore we have shown that limε→0 [ ∂r(r/)] = δ(r) which is the claim of equation (1) above. Appendix on Stakgold's Theorem for n = 1 Stakgold's theorem for n = n says !Syntax Error, Idr rn-1 g(r) = 1/Sn(1) Sn(1) = area of n dimensional sphere limε→0 [ ε-n g(r/ε) ] = δ(n)(r) on Euclidean space Rn Stakgold's theorem for n = 1 says !Syntax Error, Idr g(r) = 1/2 Sn(1) = "area of 1 dimensional sphere" = 2 limε→0 [ ε-1 g(x/ε) ] = δ(x) on R1 which is the real axis x in (-∞,∞) (a) First of all, the normalization 1/2 arises simply in this manner ∫R1 f(x) dx = !Syntax Error, If(x) dx = 1 If f(x) = f(|x|) = g(r), this becomes 1 = !Syntax Error, I f(|x|) dx = = 2 !Syntax Error, I f(x) dx = 2 !Syntax Error, I g(r) dr so we get !Syntax Error, I g(r) dr = 1/2 Thus, we don't have to worry about interpreting "the area of a 1 dimensional sphere". (b) Second of all, we need to modify the conclusion of the n=1 theorem in this way limε→0 [ ε-1 g(x/ε) ] = δ(x) x in the range (-∞,∞) limε→0 [ ε-1 g(r/ε) ] = (1/2) δ(r) r in the range (0,∞) Here is an explanation. In the first case we have, for a test function φ(x) !Syntax Error, Idx φ(x) δ(x) = φ(0) and we imagine this as (think of δ(x) as a thin gaussian with two halves of equal area) !Syntax Error, Idx φ(x) { δlefthalf(x)θ(-x) + δrighthalf(x)θ(x) ) = φ(0) Then we would claim that in the following integral only the right half contributes, so !Syntax Error, Idx φ(x) δ(x) = !Syntax Error, Idx φ(x) δrighthalf(x) = (1/2) φ(0) (2) But people usually think of a radial delta function as doing this !Syntax Error, Idr φ(r) "δ(r)" = φ(0) (3) Multiply both sides of (3) by 1/2 to get !Syntax Error, Idr φ(r) (1/2) "δ(r)" = (1/2) φ(0) (4) Comparing (2) and (4) shows that (1/2) "δ(r)" = δ(x) So our Stakgold theorem says limε→0 [ ε-1 g(r/ε) ] = δ(x) = (1/2) "δ(r)" But in the radial coordinate context for r in (0,∞), nobody writes a double quote. They instead say !Syntax Error, Idr φ(r) δ(r) = φ(0) // defines what is meant by the δ(r) symbol in this context and so the Stakgold n=1 theorem says this limε→0 [ ε-1 g(r/ε) ] = (1/2) δ(r)