The Shankar Connection

The Shankar Connection PhL 1.18.11 Summary: In Sec 1 I review how the problem arises, Ku=μu. In Sec 2 I repeat the Rothwell expansion idea and conclude there is no Lxy(1/r) = δ solution for L which matches at each point in kt space. I can only come up with non-linear L that sort of work, but non-linear is no good. In Sec 3 I show explicitly that constant coefficient Lxy cannot work due to "analyticity". In Sec 4 I repeat Rothwell for δ(r-r') and get same no-go conclusion. In Sec 5 I report on my 1/4/11 emails with Shankar in which he tersely claims no solution because |q| is not analytic, which I interpreted to mean |kt| of my Rothwell thing. I reply that I think he is only ruling out constant coefficient solutions but he does not respond to that. In Sec 6 I note that Jim later in fact found two solutions, and they do not have constant coefficients. In Sec 7 I ask why the Shankar/Rothwell method misses these solutions, and the answer is clear: In Sec 8 I take one of Jim's solutions and show that it in fact solves the Rothwell expansion, but only globally where you include the d2k integral, there is not a match locally at each point in k space. Sec 9 states my conclusions. __________________________________________________________________________________ 1. Why was I interested in Lxy [ 1/r] = δ(2)(r). 1 2. Using the Rothwell/Cloud expansion for 1/r : no solution seems possible. 1 3. Trying for Lxy with constant coefficients: no solution is possible. 2 4. Rothwell/Cloud also gives no solution to Lxy(1/R) = 2πδ(2)(r-r') 3 5. Report of my exchange with Shankar 3 6. But despite the Shankar method failure, Jim finds two solutions. 4 7. Why did these solutions not show up in the Shankar analysis? 5 8. Brute force calculation of the Rothwell integral for one of the Jim solutions. 6 9. Conclusions 8 1. Why was I interested in Lxy [ 1/r] = δ(2)(r). In the v3 doc I am wondering about this integral equation ∫∫S dx'dy' 1/( ) u(x',y') = μ u(x,y) Ku = μu (0) though I don't seem to write this anywhere in v3 in Cartesian coordinates (but I do in my doc to R Price later on). Looking at the above, I am wondering if there is some operator Lxy such that Lxy [ 1/( ] = δ(x-x')δ(y-y') (1) If there does exist such an Lxy , then the following would have to also be true Lxy [ 1/r] = δ(x)δ(y) = δ(2)(r) (2) Probably when I first did this, I thought (2) (1) but in fact that is not true. But (1) (2) and therefore we know that !(2) !(1). So if (1) was to be true, then (2) would have to be true, and so I focused at this point only on (2). 2. Using the Rothwell/Cloud expansion for 1/r : no solution seems possible. In Section 3 of the v3 doc I remembered a fact from the Rothwell/Cloud Cartesian work I did where I showed that (either sign works) 1/r = / = (1/π) ∫d2kt exp(±ikt r) /(2kt) (3) So I thought maybe I could expand the LHS and RHS of (2) to see what happens. That gave ∫d2kt (1/2kt) Lxy [exp(±ikt r)] = ∫d2kt exp(±ikt r) (4) At this point, you cannot remove the d2kt integral from both sides because you don't have a clean set of basis functions because you don't know what Lxy [exp(±ikt r)] is going to do. BUT, IF you could find an Lx such that Lxy [exp(±ikt r)] = 2kt exp(±ikt r) (5) then you would have a complete set, and then this equation would be justified. I came up with some nonlinear operators that would have this behavior Lxy [f] = 2f | f | 1 Lxy [f] = 2 2 (6) Lxy [f] = 2 f 3 = 2 ( - ∂x2 - ∂y2)1/2 f Operators 1 and 2 are I nonlinear operators, while operator 3 does not exist in any sense that I know about. If you try a binomial expansion, you get negative powers of the derivative. You could somehow try continuing off the set of positive integral exponents to the value (1/2). There may be some sense to this in the world of fractional calculus, but that is not what I am looking for here. So I dismiss the last operator 3 as making no sense at all, and I have to dismiss the first two since they are non-linear. This is because, if you want to solve (0) by using (1). Lxy has to pass through a sum = integral and thus must be linear for things to work. So I have no solution! 3. Trying for Lxy with constant coefficients: no solution is possible. I then asked this sort of related question: "Is there an Lxy with constant coefficients that solves (5)? " I answered this question on the whiteboard which I now transcribe. We would have to have {Σn=0∞ Σm=0∞ anm ∂xn ∂ym } [exp(-ikt r)] = 2kt exp(-ikt r) which says {Σn=0∞ Σm=0∞ anm (-ikx)n (-iky)m } [exp(-ikt r)] = 2 exp(-ikt r) which says Σn=0∞ Σm=0∞ anm (-ikx)n (-iky)m = 2 = 2 | kt| = 2kt (7) But there is no solution to this equation, and there are several ways to explain why: if you use the binomial expansion for the RHS, you get negative powers of one or the other ki terms (a+b)n = an +n an-1b + (n,2) an-2b2 + ..... use with n = 1/2 but the left hand side has only positive powers, so no go. In other words, there is no solution to this Σn=0∞ Σm=0∞ anm xnym = Due to the range of our Fourier expansions, we have to have (7) be true for kx,ky in the full ranges (-∞,∞), so if a double series existed, it would have to converge for all kx,ky. The series on the LHS of (7), for fixed y, must then be analytic in the entire kx complex plane. But the series on the RHS clearly has branch points at kx = ±i ky and so is not analytic in the kx complex plane. So the bottom line is that I have no solution to (5) and probably none exists. 4. Rothwell/Cloud also gives no solution to Lxy(1/R) = 2πδ(2)(r-r') I think I can redo the above analysis more generally this way. I write 1/R = / = (1/π) ∫d2kt exp(±ikt (r-r')) /(2kt) (3)' where I have just replaced r by r-r' on both sides . Then the next equation becomes ∫d2kt (1/2kt) Lxy [exp(±ikt (r-r'))] = ∫d2kt exp(±ikt (r-r')) (4)' And now we seek a solution which says Lxy [exp(±ikt (r-r')] = 2kt exp(±ikt (r-r')) (5)' But now we just divide both sides by exp(±ikt (-r')) and we are back to equation (5) having to be true. So my inability to find Lxy in this context applies to the more general case. The constant coefficients argument goes through as well. 5. Report of my exchange with Shankar (sent 1/5/11) His reply is this In momentum space such an operator would have to be |q| which, because it is not analytic cannot be written in terms of any simple differential operator. S This agrees I think with my general analysis above, where I think I have made clearer what variable we are talking about for analyticity. I then replied and then issued an open Torrey invitation to which he replied he would try to exploit that invitation. So that then is the Shankar connection. 6. But despite the Shankar method failure, Jim finds two solutions. I then entered my little project with Jim, and we in fact found two solutions to the equation: L(1/r) = 2π δ(2)(r) Here are the solutions. Lxy(3)([1/r] ) = (1/r)∂r(r [1/r] ) = (1/) [ x∂x + y∂y + 1] () [1/] Lxy(1) ([1/r] ) = – ∂r2(r[1/r]) = – (1/) [ x2 ∂x2 + y2∂y2 +2xy∂x∂y + 2x ∂x + 2y∂y ] [1/] These are NOT solutions to Lxy(1/R) = 2π δ(2)(r-r') but OK. 7. Why did these solutions not show up in the Shankar analysis? They don't have constant coefficients, so that part is OK. But what about Lxy [exp(-ikt r)] = 2kt exp(-ikt r) (5 with -) This came from an integral which swept over the entire kt space and we can regard r as fixed. For a particular value of kt there will be some angle θ such that kt r = ktr cosθ. Then we have Lxy [exp(-irktcosθ)] = 2kt exp(-irktcosθ)) (5 with -) Now lets define kt = k and a = -icosθ so we want to consider Lxy eαkr = 2k eakr and we now try one of our known solutions – ∂r2(r eakr ) = 2k eakr ? But Maple says – ∂r2(r eakr ) = [ -2ak - ra2k2] eakr = -ak(2+akr) eakr so obviously this is not going to work. In the Shankar thing, we looked for a solution which made the equality in (4) ∫d2kt (1/2kt) Lxy [exp(±ikt r)] = ∫d2kt exp(±ikt r) (4) such that there was equality at every point in k-space. This is a special kind of solution and it seems that the solutions we found are more global in nature. In fact, this must be true ∫d2kt (1/2kt) {– ∂r2(r [exp(-ikt r)])} = (2π)2 δ(2)(r) 8. Brute force calculation of the Rothwell integral for one of the Jim solutions. Let's now work on the LHS: LHS = ∫d2kt (1/2k) { -ak(2+akr) eakr } where a = -icosθ Now write ∫d2kt = ∫k dk ∫dθ so we then have LHS = ∫k dk ∫dθ (1/2k) { -ak(2+akr) eakr } = (1/2) ∫ dk ∫dθ { -(-icosθ)k(2-icosθkr) e-icosθkr } = (i/2) ∫ dk k !Syntax Error, Idθ { cosθ (2-icosθkr) e-icosθkr } = i ∫ dk k !Syntax Error, Idθ { cosθ (2-icosθkr) e-icosθkr } Let's define z = cosθ so this becomes dz = -sinθdθ = - dθ = i ∫ dk k !Syntax Error, Idz/ {z (2-izkr) e-izkr } (*) There are two terms. The first is proportional to !Syntax Error, Idz/* z e-izkr = !Syntax Error, Idz/* z ( -i sin(krz) ) = -i!Syntax Error, Idz/* z sin(krz) = -2i !Syntax Error, Idz/* z sin(krz) = (-2i) (π/2)J1(kr) and we looked up to find in GR7 The second term is proportional to !Syntax Error, Idz/* z2 e-izkr = !Syntax Error, Idz/* z2 (cos(krz) ) = 2 !Syntax Error, Idz/* z2 cos(krz) = 2 !Syntax Error, Idz/* z2 (-1)(1/rz)2∂k2 cos(krz) = -2/(r)2 * ∂k2 !Syntax Error, Idz/ * cos(krz) = - 2/(r)2 * (π/2) ∂k2 J0(kr) = - (π/r2) ∂k2 J0(kr) and we looked up to find in GR7 What a nightmare! So we now have for our LHS = LHS1 + LHS2, LHS1 = i ∫ dk k (2) (-2i) (π/2)J1(kr) = i (2) (-2i) (π/2) ∫ dk k J1(kr) = i (-2i) (π) (1/r2) // Maple says the integral is just 1/r2 = + (2π/r2) LHS2 = i ∫ dk k (-ikr) (-π/r2) ∂k2 J0(kr) = i (-π/r2))(-ir) ∫ dk k2 ∂k2 J0(kr) = - π/(r) ∫ dk k2 ∂k2 J0(kr) Wow. Maple says that ∂k2 J0(kr) = -r2 [ J0(kr) - (1/rk)J1(rk) ] so we then have LHS2 = - π/(r) (-r2) ∫ dk k2 [ J0(kr) - (1/rk)J1(rk) ] But we just saw from above that ∫ dk k J1(rk) = (1/r2) so writing LHS2 = LHS2a + LHS2b we have LHS2b = - π/(r) (-r2) (-1/r) ∫ dk k J1(rk) = - π/(r) (-r2) (-1/r) (1/r2) = -π/(r) (1/r) = -π/r2 LHS2a = - π/(r) (-r2) ∫ dk k2 J0(kr) = - π/(r) (-r2) (-1/r3) = - π/(r2) // Maple int So barring errors, I have shown that LHS = + (2π/r2) -π/r2 - π/(r2) = 0 Hurray!!! All the steps above are certainly valid for r > 0 and this is just the result we wanted to get in this case. But the δ(r) is somehow hidden. We would have to do everything with regulation, but I am ready to stop now. 9. Conclusions (1) The Jim solutions did not show up in the Shankar analysis because the latter required equality at each point in k-space, Lxy [exp(±ikt r)] = 2kt exp(±ikt r), and doing it that way, we get no solutions. (2) The Jim solutions involve the global k-space integration. (3) For r > 0, I explicitly computed ∫d2kt (1/2kt) Lxy [exp(±ikt r)] for one of Jim's solutions, and rather amazingly, you really do get 0, which is consistent with the required result (2π)2 δ(2)(r). But things are extremely tangled in this approach and it would be pretty hard I think to ferret out the δ(r) part, but I know it must work because the Jim solutions really are solutions. (4) Maybe I will send Shankar these results.