jim delta 1.13.11 5PM

The name of this doc is: " jim delta 1.13.11 5PM.doc " . PhL 1.13.11 Jim, here are some of the details the I omitted from the previous doc. I am doing my best here to be detailed and accurate, but I know that I always make mistakes. At least you can show me where (so I put equation numbers this time in some places). Contents ( you can do shift-click on any of these items to go to that section) 0. The Stakgold book theorem says this: 1 1. Showing that ∂r2 (r [1/r]) = – 2π δ(2)(r) . 2 2. Showing that (1/r) ∂r(r [1/r]) = + 2π δ(2)(r) . 3 3. Confirmation from the 2D Divergence Theorem. 4 4. 3D Divergence Theorem. Why does this give a different kind of result? 6 5. How do we apply the Stakgold theorem to the 3D case? 7 6. Attempt the same 3D problem using 1/(r+ε) regulation (this section added 1.17.11) 8 0. The Stakgold book theorem says this: _________________________________________________________________ Theorem n: Any non-negative locally integrable function g(r) for which this is true !Syntax Error, Idr rn-1 g(r) = 1/Sn(1) Sn(1) = area of n dimensional sphere has the property that limε→0 [ ε-n g(r/ε) ] = δ(n)(r) Note: = δ(n)(r) = δ(x1)δ(x2)....δ(xn) _________________________________________________________________ Theorem for n=2: Any non-negative locally integrable function g(r) for which this is true !Syntax Error, Idr r g(r) = 1/2π has the property that limε→0 [ ε-2 g(r/ε) ] = δ(2)(r) Note: = δ(2)(r) = δ(x)δ(y) __________________________________________________________________________ Detail: Function g(r) is "locally integrable" means that ∫K dr |g(r)| < ∞ where K is any bounded subset of the open interval (0,∞ ). So for example g(r) = 1/[r(r2+1)3/2] meets this requirement even though the integral diverges at the lower endpoint. As shown below, the normalization integral !Syntax Error, Idr r g(r) is finite. 1. Showing that ∂r2 (r [1/r]) = – 2π δ(2)(r) . Consider ∂r2(r[1/r]), where we define as a shorthand notation [1/r] ≡ limε→0 (1/) (1.1) so the operator is ∂r2(r..) and the thing operated on is [1/r] ≡ limε→0 (1/). Then we have ∂r2 (r [1/r]) = ∂r2 (r limε→0 (1/)) = limε→0[ ∂r2 (r/) ] (1.2) Maple tells us that ∂r2 (r/) = -3 rε2 / (r2+ ε2)5/2 (1.3) But we can write rε2 = ε3(r/ε) {(r2+ε2)}5/2 = {ε2 [(r/ε)2+1 ]}5/2 = ε5 [(r/ε)2+1] 5/2 => -3 rε2 / (r2+ ε2)5/2 = -3 ε3(r/ε) / (ε5 [(r/ε)2+1]5/2) = ε-2 { -3(r/ε)/ [(r/ε)2+1]5/2 } (1.4) Therefore, adding a factor of K to both sides, we have shown that K ∂r2 (r/) = ε-2 { -3K(r/ε)/ [(r/ε)2+1]5/2 } (1.5) If we now define g(x) = -3Kx/[(x2+1)5/2] (1.6) Then the above becomes K ∂r2 (r/) = ε-2 g(r/ε) (1.7) We now want to find the value of K which provides our desired normalization !Syntax Error, Idx x g(x) = 1/2π. So (1/2π) = !Syntax Error, Idx x { -3Kx/[(x2+1)5/2] } = -3K { !Syntax Error, Idx x2 / (x2+1)5/2 } = -3K {1/3} = -K (1.8) where Maple evaluates the integral as shown. Therefore, K = - (1/2π). We then have -(1/2π) ∂r2 (r/) = ε-2 g(r/ε) (1.9) and then according to Stakgold's theorem with n = 2, limε→0 [-(1/2π) ∂r2 (r/)] = limε→0 [ε-2 g(r/ε)] = δ(2)(r) (1.10) which says limε→0 [∂r2 (r/) ] = – 2π δ(2)(r) (1.11) __________________________________________________________________________ 2. Showing that (1/r) ∂r(r [1/r]) = + 2π δ(2)(r) . Consider (1/r) ∂r(r [1/r]), where we define as a shorthand notation [1/r] ≡ limε→0 (1/) (2.1) so the operator is (1/r) ∂r(r..) and the thing operated on is [1/r] ≡ limε→0 (1/). Then we have (1/r) ∂r(r [1/r]) = (1/r)∂r(r limε→0 (1/)) = limε→0[ (1/r)∂r(r/) ] (2.2) Maple tells us that (1/r)∂r(r/) = (1/r) ε2 / (r2+ε2)3/2 (2.3) But we can write (1/r) ε2 = ε (ε/r) = ε / [ (r/ε) ] {(r2+ε2)}3/2 = {ε2 [(r/ε)2+1 ]}3/2 = ε3 [(r/ε)2+1] 3/2 => (1/r) ε2 / (r2+ε2)3/2 = ε / ( (r/ε) ε3 [(r/ε)2+1] 3/2) = ε-2 {1/ ( (r/ε) [(r/ε)2+1] 3/2) } (2.4) Therefore, adding a factor of K to both sides, we have shown that K (1/r)∂r(r/) = ε-2 { K/ [ (r/ε) [(r/ε)2+1] 3/2] } (2.5) If we now define g(x) = K/[x(x2+1)3/2] (2.6) Then the above becomes K (1/r)∂r(r/) = ε-2 g(r/ε) (2.7) We now want to find the value of K which provides our desired normalization !Syntax Error, Idx x g(x) = 1/2π. So (1/2π) = !Syntax Error, Idx x { K/[x(x2+1)3/2]} = K { !Syntax Error, Idx / (x2+1)3/2 } = K { 1 } (2.8) where Maple evaluates the integral as shown. Therefore, K = (1/2π). We then have (1/2π) (1/r)∂r(r/) = ε-2 g(r/ε) (2.9) and then according to our theorem with n = 2, limε→0 [(1/2π) (1/r)∂r(r/) ] = limε→0 [ε-2 g(r/ε)] = δ(2)(r) (2.10) which says limε→0 [(1/r)∂r(r/) ] = + 2π δ(2)(r) (2.11) __________________________________________________________________________ 3. Confirmation from the 2D Divergence Theorem. In the last two sections we have argued that, limε→0 [∂r2 (r/) ] = – 2π δ(2)(r) (3.1) = (1.11) limε→0 [(1/r)∂r(r/) ] = + 2π δ(2)(r) (3.2) = (2.11) Now let's consider the sum of the two objects on the left and see if we can show that they add up to 0 by some other method. In our compact notation we have for this sum ∂r2(r [1/r]) + (1/r)∂r(r [1/r]) = [ ∂r2 + (1/r)∂r] (r [1/r]) (3.3) We can write [ ∂r2 + (1/r)∂r] F(r) = (1/r)∂r(r ∂r F(r)) = 22D,radial F(r) = 22D F(r) (3.4) Thus, the sum of the two objects on the left above is ∂r2(r [1/r]) + (1/r)∂r(r [1/r]) = 2(r [1/r]) (3.5) where by 2 we mean the full 2D Laplacian 2 = (1/r)∂r(r ∂r) + (1/r2)∂φ2 = ∂x2 + ∂y2 (3.6) Now since we have 2 floating around, we try to apply the divergence theorem. In N dimensions this says ∫V dV A = ∫S dSA (3.7) where V is an N-dimensional volume, and S is an N-1 dimensional surface enclosing that volume. We can apply this theorem to A = ψ(r) where we are still in N dimensions. We find that ∫V dV ψ = ∫S dSψ = ∫S (dS ) ψ = ∫S dS ∂nψ (3.8) where is normal to the surface patch dS and ∂n is the normal derivative. So far then we have ∫V dV 2ψ = ∫S dS ∂nψ valid in N dimensions N = 2,3,4..... (3.9) If the volume is a hypersphere in N dimensions this says ∫sphere dV 2ψ = ∫S dS ∂rψ (3.10) since is now the normal direction. Applying this version of the divergence theorem for N = 2, we find ∫disk dV 2ψ = ∫circle dl ∂rψ (3.11) If the disk has radius "a", then dl = adθ and our theorem says ∫disk dV 2ψ = a ∫dθ (∂rψ)|r=a (3.12) Now we apply this to our special function, using our same compact notation, ψ = (r [1/r]) (3.13) We know that at r = a, [1/r] = (1/r) since we are away from the problem point r = 0. Therefore (∂rψ)|r=a = (∂r{ r [1/r]} )|r=a = (∂r{ r (1/r)} )|r=a = (∂r{ 1} )|r=a = 0 (3.14) Therefore we have shown that ∫disk dV 2(r [1/r]) = 0 (3.15) We then take the limit of this result as a → 0 and conclude that 2(r [1/r]) = 0 (3.16) This is consistent with the sum of equations (3.1) and (3.2) above, 2(r [1/r]) = ∂r2(r [1/r]) + (1/r)∂r(r [1/r]) = - 2π δ(2)(r) + 2π δ(2)(r) = 0 __________________________________________________________________________ 4. 3D Divergence Theorem. Why does this give a different kind of result? The claim in 3D is this 3D([1/r]) = - 4πδ(3)(r) The divergence theorem above in N dimensions stated ∫dV 2ψ = ∫S dS ∂nψ For N = 3 dimensions, and for the volume being a sphere of radius "a", this becomes ∫dV 2ψ = ∫S a2dΩ (∂rψ)|r=a If ψ = ψ(r) only, this becomes ∫dV 2ψ = a2 ∫ dΩ (∂rψ)|r=a = 4π a2 (∂rψ)|r=a We apply this now to ψ = [1/r] to get ∫dV 2[1/r] = 4π a2 (∂r[1/r])|r=a = 4π a2 (∂r(1/r))|r=a = 4πa2(-1/r2) |r=a = -4π Therefore we have shown that ∫dV 2[1/r] = -4π We conclude taking the limit a→ 0 that 2[1/r] = -4π δ(3)(r) Comparison: 23D([1/r]) = -4π δ(3)(r) 22D(r [1/r]) = 0 __________________________________________________________________________ 5. How do we apply the Stakgold theorem to the 3D case? For n = 3 the theorem says !Syntax Error, Idr r2 g(r) = 1/4π // normalization integral limε→0 [ ε-3 g(r/ε) ] = δ(3)(r) // delta function generator Consider then 23D([1/r]) = (1/r2)∂r(r2 ∂r [1/r] ) As usual the function acted upon is this [1/r] ≡ limε→0 (1/) So we consider 23D([1/r]) = 2[limε→0 (1/)] = limε→0 (1/r2)∂r[ r2 ∂r(1/) ] where for the rest of this section, 2 refers to the full 3D Laplacian which in sphericals is this 2 = (1/r2)∂r(r2 ∂r + (1/r2){ cscθ∂θ(sinθ∂θ) + csc2θ ∂φ2 } As usual, we call upon Maple to compute (1/r2)∂r[ r2 ∂r(1/) ] = -3ε2 / (r2+ε2)5/2 But we can write (as we did below equation 1.3 above) {(r2+ε2)}5/2 = {ε2 [(r/ε)2+1 ]}5/2 = ε5 [(r/ε)2+1] 5/2 so we get (again, adding constant K to both sides) K(1/r2)∂r[ r2 ∂r(1/) ] = ε-3 { -3K/ [(r/ε)2+1] 5/2 } If we now define (now there is no numerator factor x in g(x) as there was in 1.6) g(x) = -3K/[(x2+1)5/2] then the above becomes K(1/r2)∂r[ r2 ∂r(1/) ] = ε-3 g(r/ε) We now want to find the value of K which provides our desired normalization !Syntax Error, Idx x2 g(x) = 1/4π. So (1/4π) = !Syntax Error, Idx x2 { -3K/[(x2+1)5/2] } = -3K { !Syntax Error, Idx x2 / (x2+1)5/2 } = -3K {1/3} = -K and this in fact is the same integral we encountered in (1.8) above. Therefore K = - (1/4π) and we have then shown that - (1/4π) (1/r2)∂r[ r2 ∂r(1/) ] = ε-3 g(r/ε) and then according to our theorem with n = 3, limε→0 [- (1/4π) (1/r2)∂r[ r2 ∂r(1/) ] = limε→0 [ε-3 g(r/ε)] = δ(3)(r) which says limε→0 [ (1/r2)∂r[ r2 ∂r(1/) ] = limε→0 [ε-3 g(r/ε)] = -4π δ(3)(r) so we have then shown, using Stakgold's little theorem, that 23D([1/r]) = (1/r2)∂r[ r2 ∂r([1/r]) ] = -4π δ(3)(r) 6. Attempt the same 3D problem using 1/(r+ε) regulation (this section added 1.17.11) This will be a copy, paste and edit from the preceding section. For n = 3 the theorem says !Syntax Error, Idr r2 g(r) = 1/4π // normalization integral limε→0 [ ε-3 g(r/ε) ] = δ(3)(r) // delta function generator Consider then 23D([1/r]) = (1/r2)∂r(r2 ∂r [1/r] ) Now assume the function acted upon is this [1/r] ≡ limε→0 (1/(r+ε)) So we consider 23D([1/r]) = 2[limε→0 (1/(r+ε))] = limε→0 (1/r2)∂r[ r2 ∂r(1/(r+ε)) ] where for the rest of this section, 2 refers to the full 3D Laplacian which in sphericals is this 2 = (1/r2)∂r(r2 ∂r + (1/r2){ cscθ∂θ(sinθ∂θ) + csc2θ ∂φ2 } As usual, we call upon Maple to compute [ for visual comparison: (1/r2)∂r(r2 ∂r [1/r] ) ] which tells us that (1/r2)∂r[ r2 ∂r(1/(r+ε)) ] = -2ε / [r(r+ε)3] = -2ε / [ {[r/ ε]ε}{ ε3([r/ ε]+1)3 }] = -2ε-3 / [ {[r/ ε]}{ ([r/ ε]+1)3 }] = -2ε-3 / ( [r/ ε] ([r/ ε]+1)3] ) Therefore we have -K(1/r2)∂r[ r2 ∂r(1/(r+ε)) ] = ε-3 { 2K g(r/ ε) } where g(x) = 2K x/(x+1)3 ~ 1/x2 - K(1/r2)∂r[ r2 ∂r(1/) ] = ε-3 { 3K/ [(r/ε)2+1] 5/2 } where g(x) = 3K /(x2+1)5/2 ~ 1/x5 where the second line in blue is the result from the preceding section, put there for comparison. Notice that the r2 in causes a much greater denominator power at large x. We now want to find the value of K which provides our desired normalization !Syntax Error, Idx x2 g(x) = 1/4π. So (1/4π) = !Syntax Error, Idx x2 {2K x/(x+1)3] } = 2K !Syntax Error, Idx x2 x/(x+1)3] ~ !Syntax Error, Idx 1 = linearly divergent (1/4π) = !Syntax Error, Idx x2 { -3K/[(x2+1)5/2] } = -3K { !Syntax Error, Idx x2 / (x2+1)5/2 } = -3K {1/3} = -K where on the second line I show in blue the convergent corresponding line from the previous section. Conclusion: This method of regulation fails! Therefore constant * limε→0 [(1/r2)∂r[ r2 ∂r(1/)] = δ(3)(r) = δ(r)/[4πr2] constant' * limε→0 [(1/r2)∂r[ r2 ∂r(1/(r+ε))] >> δ(3)(r) = δ(r)/[4πr2] The second sequence is "too big" in that it produces something whose 3D integral instead of being 1 is in fact infinite. If we multiply the above by r2 on both sides we restate as constant" * limε→0 [∂r[ r2 ∂r(1/)] = δ(r) constant''' * limε→0 [∂r[ r2 ∂r(1/(r+ε))] >> δ(r) Here is a plot of ∂r(r2 ∂r [1/r] ) for the three cases black green red [1/r] = (1/r) 1/ 1/(r+ε) The black curve is just 0 everywhere since ∂r[ r2 ∂r(1/(r))] = 0. The green curve is approaching a delta function, while the red curve is approaching something that looks like a delta function, but whose integral is becoming infinite in the limit ε→0. The green curve has a finite integral in this limit. Notice that the limit here is the "radial δ(r)" discussed elsewhere which exists only for r ≥ 0 since r is a radial coordinate, and for which !Syntax Error, Idr δ(r) f(r) = f(0) As ε→0, the red and green curves jam themselves against the wall r = 0. All three curves have the value 0 at the location r = 0: Here is the same plot for ε = .01: