A study of (R)-1 expansions in cylindrical coordinates

A study of 1/R expansions in cylindrical coordinates PhL 5.22.10 I have just finished doing this for spherical coordinates, so let's try to replicate things here for the cylindrical case. From "cylindrical atoms.doc" there are two atomic systems often used, (a) e±kz [ Jm(kρ), Nm(kρ)] e±imφ // expo in z (b) [ Km(kρ), Im(kρ)] e±ikz e±imφ // expo in ρ As usual, I assume oscillatory in φ. So I am perhaps going to have to do everything twice. Overview (4 pages, added 9.9.10). 1 Part A: Green's Function for Infinite cylinder using Jm functions 4 Pre 0. Introduction: Green's function for an infinite cylinder of radius b. 4 Pre 1. Use the pillbox condition to find the Amn coefficients for the infinite cylinder. 5 Part B: 1/R Green's Function using Jm functions 8 0. Introduction: Free space Green's function in cylindrical coordinates 8 1. Use the pillbox condition to find the Amn coefficients. 9 2. Development of the Addition Theorem 12 Part C: 1/R Green's Function using Km functions 15 0. Introduction: Free space Green's function in cylindrical coordinates 15 1. Use the pillbox condition to find the Amn coefficients. 15 2. Development of the Addition Theorem 18 Part D: Summary of All Results 20 Part A : 20 Part B, including the Qm-1/2 expansion : 20 Part AB: 22 Part C : 22 Part D (spherical results): 23 ___________________________________________________________________________________ Overview (4 pages, added 9.9.10). In Part A I seek the J-J Green's Function for an infinite cylinder of radius b using the Jm atomic form. This problem is not really the 1/R expansion, but is interesting in its own right. In Part Pre 0 I state the Smythian form as follows, where m is all integers and Jm(knb) = 0 so the "k" sum is over kn : V(r | r') = Σmk Amk exp(-k|z-z'|) Jm(kρ) eimφ I write r = (ρcosφ, ρsinφ, z) for the observation point, and r' = (ρ'cosφ',ρsinφ',z') for the Green's charge location, and I derive the famous results that R2 = ρ2 + ρ'2 + (z-z')2 - 2ρρ'cos(φ-φ') ' =[ ρρ'cos(φ-φ') + zz']/[ ] = cos(γ) Then in part Pre 1 I do the usual pillbox method (perp to z, since ρ and φ are oscillatory) to find Amk. This involves the usual φ orthogonality and the Fourier-Bessel ρ orthogonality, and the result is Amn = q e=imφ' Jm(knρ') [b Jm+1(bkn)/]-2 / kn where Jm(knb) = 0 so that our Green's Function is then [ I call this a J-J form since the product of two J's appears ] V(r | r') = q [b /]-2 Σmn [Jm+1(bkn)]-2 (1/kn) exp(-kn|z-z'|) Jm(knρ) Jm(knρ') eimφ e-imφ' which is manifestly symmetric. At this point, I review Stak's general proof that a Green's Function must in fact be symmetric, and this also implies what is called the reciprocity principle of electrostatics. I was not able to find verification of my result for V above but will keep my eye out for it. It should be noted that this cylinder Green's problem can be solved using the I-K atomic form rather than the J-J form. I have not done the problem this way, but based on the 1/R analysis in Part C below, we will get the sum over kn replaced by an integral over positive k. I have seen this result in the literature, and I think this is the way people usually do such a problem since the Fourier-Bessel is a bit messy. See Part C below. [ One could say you use k = imaginary in the Mellin-Barnes sense for your J-J solution, and this is analogous to using the conical/toroidal/Mehler functions in spherical atomic forms. Someday I hope to carry out this little program and show that you pick up the series by closing the Barnes contour.] In Part B I seek the (J-J) 1/R expansion using Jm functions. In Part 0 I repeat work from Part A, but now the spectrum of k is the continuous range (0,∞), that is the key difference. I wanted to make Part A and Part B "look similar" which is why I used the Σk notation in Part A (k = kn etc) and ∫dk in Part B. The Smythian form for 1/R this time is claimed to be: f(r | r') = Σm ∫dk Amk exp(-k|z-z'|) Jm(kρ) eimφ In Part 1 I redo the pillbox analysis (again perp to z) to find the Amk. This time of course we use the Hankel Transform for our second orthogonality, and the result is found to be simply this: Amk = q e-imφ' Jm(kρ') // Amk is really Amk(r') but r' is suppressed so our Green's function 1/R is then given by (Σm over all integers in first form, and set q=1) V(r | r') = 1/R = Σm !Syntax Error, Idk exp(-k|z-z'|) Jm(kρ) Jm(kρ') eimφ e-imφ' = Σm=0∞εm cos[m(φ-φ')] !Syntax Error, Idk exp(-k|z-z'|) Jm(kρ) Jm(kρ') // J form (*) which is again manifestly symmetric and which may be compared to the cylinder result above from Part A. I found verification of the above 1/R expansion in MF. In Part 2 I derive the addition theorem. I first let r' = (0,0 ,z') to obtain the special case result of (*) above, 1/R~ = 1/ = !Syntax Error, I dk exp(-k|z-z'|) J0(kρ) which I verify in GR. Since the above is true for any value of ρ2, I replace ρ2 by ρ2 + ρ'2 - 2ρρ'cos(φ-φ') to obtain the result fully general result [ since R2 = ρ2 + ρ'2 + (z-z')2 - 2ρρ'cos(φ-φ') ] 1/R = !Syntax Error, Idk exp(-k|z-z'|) J0(k) (**) This, then, is our 1/R expansion analogous to the famous sphericals result (note m=0 everywhere) 1/R = Σn (r<)n(r>)-n-1 Pn(cosθcosθ' + sinθsinθ'cos(φ-φ')) If we now equate the 1/R form (**) with the J-form for 1/R shown above (*), we find that J0(k) = Σm=0∞εm cos[m(φ-φ')] Jm(kρ) Jm(kρ') which is the "addition theorem" [ verified GR7 page 940] analogous to the spherical addition theorem Pn[cosθcosθ' + sinθsinθ'cos(φ-φ')] = Σm=0∞ εm cos[m(φ-φ')] f(n,-m) Pnm(cosθ) Pnm(cosθ')] I am sure the cylindrical addition theorems have group theoretic underpinnings as the sphericals do, but I have never pursued this subject. The group is of course continuous but different, not SO(3). In Part C I repeat all of the above 1/R expansion work using the I-K form instead of the J-J form. We start with this Smythian form in Part 0, f(r | r') = q/R = Σm!Syntax Error, I dk Amk cos[k(z-z')] Im(kρ<) Km(kρ>) eimφ Then in Part 1 we do the pillbox to get the Amk . But now z and φ are oscillatory, so we do the pillbox perp to ρ instead of z. We use the usual φ orthogonality, but this time we use the Cosine Integral Transform for our z orthogonality, and we find this simple result: Amk = (2/π)q e–imφ' which then tells us our resulting 1/R expansion 1/R = (2/π) Σm!Syntax Error, I dk cos[k(z-z')] Im(kρ<) Km(kρ>) eimφ e–imφ' = (2/π) Σm=0∞ εm cos[m(φ-φ')]!Syntax Error, I dk cos[k(z-z')] Im(kρ<) Km(kρ>) which is verified page 86 (3.148). In Part 2 I develop the addition theorem for this case. As before, I start with r' = (0,0 ,z') to get 1/R~ = 1/ = (2/π) !Syntax Error, I dk cos[k(z-z')] K0(kρ) or: 1/ = (2/π) !Syntax Error, I dk cos(kz) K0(kρ) which I then verify in Jackson and GR7. Doing the same addition theorem trick as above, we find the 1/R expansion and the addition theorem, both of which appear in : 1/R = (2/π) !Syntax Error, I dk cos[k(z-z')] K0(k) // 1/R expansion K0(k) = Σm=0∞ εm Im(kρ<) Km(kρ>) cos[m(φ-φ')] // addition theorem In Part D I summarize all the results of Parts A,B,C, but I also add some new results concerning the Qn-1/2 function. First, we have the 1/R expansion which appears in the "Selvaggi" paper 1/R = 1/(π) Σm=0∞εm cos[m(φ-φ')] Qm-1/2(β) β = [ ρ2 + ρ'2 + (z-z')2 ]/(2ρρ') and which I derive in another of my docs called "integral doc" (see also toroidal 1/R doc). If I then equate this to the J-J expansion for 1/R, I conclude that Qm-1/2(β) / (π) = !Syntax Error, Idk exp(-k|z-z'|) Jm(kρ) Jm(kρ') which I then verify in GR. I now have no doubt but that the above 1/R Selvaggi expansion is really just the 1/R expansion in toroidal coordinates, but in my toroidal 1/R doc I don't quite show that this is in fact the case, more work is needed there. [ that work is now done!] The Selvaggi expansion is the toroidal 1/R expansion but some parts of it are expressed in cylindrical rather than toroidal coordinates so it is a bit of a hybrid thing. ___________________________________________________________________________________ Part A: Green's Function for Infinite cylinder using Jm functions Pre 0. Introduction: Green's function for an infinite cylinder of radius b. (I did this problem accidentally, but since I finished it, I will keep it). Our point charge q is located at some point r' = (ρ',φ',z') inside a cylinder of radius b. We write the following atomic form for the potential V(r | r') = Σmk Amk exp(-k|z-z'|) Jm(kρ) eimφ We know this z stuff as follows. Draw a z-plane at z' which passes through our point charge. On either size of this plane, we must have expo decay as we go to large |z|. There is some discrete spectrum of k determined presumably by the zeros of Jm on our cylindrical boundary, but we leave it just as = Notice that we are parameterizing our two points r and r' this way r = (ρcosφ, ρsinφ, z) => = (ρcosφ, ρsinφ, z)/ r' = (ρ'cosφ',ρsinφ',z') => ' = (ρ'cosφ',ρ'sinφ',z')/ => ' =[ ρρ'(cosφcosφ' + sinφsinφ') + zz']/[ ] = cos(γ) => ' =[ ρρ'cos(φ-φ') + zz']/[ ] = cos(γ) so that γ is the angle directly between our two vectors r and r'. But for our cylindrical case, we are really more interested (later) in these facts, simpler than the above, rr' = ρρ'cos(φ-φ') + zz' r2 = ρ2 + z2 r'2 = ρ'2 + z'2 R2 = |r-r'|2 = r2 + r'2 - 2 rr' = ρ2 + z2 + ρ'2 + z'2 - 2ρρ'cos(φ-φ') -2 zz' = ρ2 + ρ'2 + (z-z')2 - 2ρρ'cos(φ-φ') At this point there are several different things we can do. Pre 1. Use the pillbox condition to find the Amn coefficients for the infinite cylinder. The general pillbox condition is this ( units) ∂q1Vi - ∂q1Vo = 4πq (h1/h2h3) δ(q2-q2') δ(q3-q3') where q1 is the dimension perp to our partition of space into two regions at the point charge. I will assume for the moment that z' > 0 and that "outer" means z > z' which is farther away from the origin, etc. We then have (that is to say, the "inner" region contains the origin) Vi = Σmk Amk exp(+k(z-z')) Jm(kρ) eimφ z < z' Vo = Σmk Amk exp(-k(z-z')) Jm(kρ) eimφ z > z' So q1 = z. We can then take q2 = ρ and q3 = φ and use these facts h1 = hz = 1 h2 = hρ = 1 h3 = hφ = ρ I take these facts from M&M page 178. I really should start with the three Cartesian equations and run my little program to compute the metric tensor, but OK for now. Our pillbox condition is then this: ∂zVi - ∂zVo = 4πq (1/ρ) δ(ρ-ρ') δ(φ-φ') We then need to compute the derivatives ∂zVi = Σmk Amk k exp(+k(z-z')) Jm(kρ) eimφ z < z' ∂zVo = Σmk Amk (-k) exp(-k(z-z')) Jm(kρ) eimφ z > z' then evaluate both on the bounding surface between the two regions (which means z = z' ) ∂zVi = Σmk Amk k Jm(kρ) eimφ ∂zVo = –Σmk AmkkJm(kρ) eimφ so ∂zVi - ∂zVo = 2 Σmk Amk k Jm(kρ) eimφ = 4πq (1/ρ) δ(ρ-ρ') δ(φ-φ') We will use two orthogonalities to find the Amn. First, apply ∫dφ e-im'φ to both sides and use (1/2π) !Syntax Error, I dφ eimφ e-im'φ = δmm' // orthogonality The LHS becomes 2 Σmk Amk k Jm(kρ) ∫dφ e-im'φ eimφ = 2 Σmk Amk k Jm(kρ) 2πδmm' = 4π Σk Am'k k Jm'(kρ) and the RHS becomes 4πq (1/ρ) δ(ρ-ρ') e–im'φ' so we then have (replace m' with m) 4π Σk Amk k Jm(kρ) = 4πq (1/ρ') δ(ρ-ρ') e-imφ' Now we reach into our prepared "transforms doc" and pull out this rabbit, f(ρ) = Σn=1∞ An,ν Jν(knρ) // expansion An,ν = 2/[b2Jν+1(bkn)2] !Syntax Error, Idρ ρ f(ρ) Jν(knρ) // projection [b Jν+1(bkn)/]-2!Syntax Error, Idρ ρ Jν(knρ) Jν(kn'ρ) = δn,n' // orthogonality Σn=1∞ [b Jν+1(bkn)/]-2 Jν(knρ) Jν(knρ') = δ(ρ-ρ')/ρ // completeness where we now assume that the kn quantized values are set by Jm(knb) = 0 so the radius of whatever BC cylinder we have is ρ = b. So let's first rewrite our above equation showing the spectral kn ( we change then from Amk to Amn) 4π Σn Amn kn Jm(knρ) = 4πq (1/ρ') δ(ρ-ρ') e-imφ' where n = 1,2,3... just enumerate the zeros of Jm(knb), and we note that the kn depend on m, but we don't want to over clutter our notation to show this fact right now. So, the next step is to apply !Syntax Error, Idρ ρ Jm(kn'ρ) to both sides of the above equation. LHS = 4π Σn Amn kn !Syntax Error, Idρ ρ Jm(kn'ρ) Jm(knρ) = 4π Σn Amn kn δn,n'[ b Jm+1(bkn)/]2 = 4π Amn' kn' [b Jm+1(bkn')/]2 RHS = 4πq (1/ρ') e-imφ' !Syntax Error, Idρ ρ Jm(kn'ρ) δ(ρ-ρ') = 4πq (1/ρ') e-imφ' ρ' Jm(kn'ρ') = 4πq eimφ' Jm(kn'ρ') The result is this (change n' to n) 4π Amn kn [b Jm+1(bkn)/]2 = 4πq e-imφ' Jm(knρ') Amn = q e=imφ' Jm(knρ') [b Jm+1(bkn)/]-2 / kn So we now have our coefficients! We can install these into our expansion above V(r | r') = Σmn Amn exp(-kn|z-z'|) Jm(knρ) eimφ = Σmn { q e-imφ' Jm(knρ') [b Jm+1(bkn)/]-2 / kn } exp(-kn|z-z'|) Jm(knρ) eimφ = q [b /]-2 Σmn [Jm+1(bkn)]-2 (1/kn) exp(-kn|z-z'|) Jm(knρ) Jm(knρ') eimφ e-imφ' We notice that the result is symmetric under r ↔ r' . The reason this is true is not as obvious as it is in our 1/R free space problem where R = |r-r'|. If r is close to the cylinder and r' close to the center line, it is not obvious that switching these points should give the same result! Ie, that V(r | r') = V(r' | r). There is no obvious transformation of space that swaps the two points and leaves the cylinder where it is. But we have just proven it is symmetric in this case, and we know from our Stakgold reading that all Green's Functions have this symmetry. This is discussed in Chapter 6, and I will quote my entire little note section on this topic, it is worth rereading right now since we have an actual application ______________________________________________________________ Theorem 2: The function g(x|ξ) is symmetric. This is not as obvious, so let's do the work he outlines. -2g(x|ξ) = δ(x-ξ) g = 0 on σ // definitions -2g(x|η) = δ(x-η) g = 0 on σ - g(x|η)2g(x|ξ) = g(x|η)δ(x-ξ) // process - g(x|ξ)2g(x|η) = g(x|ξ)δ(x-η) - g(x|η)2g(x|ξ) + g(x|ξ)2g(x|η) = g(x|η)δ(x-ξ) - g(x|ξ)δ(x-η) // subtract ∫R dx { g(x|ξ)2g(x|η) – g(x|η)2g(x|ξ)} = g(ξ|η) - g(η|ξ) // integrate over R The purpose is to get the RHS as shown, which we would like to claim vanishes. Use Green #2 on the LHS and it becomes LHS = ∫σ dS { g(x|ξ) ∂n g(x|η) – g(x|η) ∂n g(x|ξ) } = 0 since both g's vanish on σ, and since we assume no singular behavior of ∂n g(x|-) on σ. QED. But if g(x|ξ) = g(ξ|x), then this must be true as well for v(x,ξ) in 6.75, since it is obviously true for E. So if we put unit charge at B, we get a certain v(A,B). But if we put the unit charge at A, then at point B we see v(B,A). These numbers are the same, v is symmetric. This fact is not really obvious when surface σ is present (on which g = 0). The fact is called the reciprocity principle of electrostatics. ______________________________________________________________ We now would like some verification of the above formula. I downloaded a paper that does this, but it uses sin in z and uses the K and I function approach. I think my result above is correct, but no one does it that way because it is easier to use the I and K method since orthog is simpler. So I guess I will just state my result with no verification. It was "the wrong problem" anyway. Now on to the right problem. Part B: 1/R Green's Function using Jm functions 0. Introduction: Free space Green's function in cylindrical coordinates The potential of a point charge at the origin is q/R with R = r. If we move the point charge to some other origin r', the potential is q/R where R = |r - r'|. In either case, the potential is a solution of the Laplace equation (away from the point charge) and can therefore be written in terms of the "atoms" of whatever coordinate system we want to use. For example, if we think of R as f(r) we know we can expand it this way in cylindricals (the notation here means that r' is a parameter, r is the variable) f(r | r') = Σm ∫dk Amk exp(-k|z-z'|) Jm(kρ) eimφ The k spectrum is continuous because there is no finite cylindrical surface on which we have V = 0. The sum Σm is over all integers, not just non-negative integers. We know the z stuff as follows. Draw a z-plane at z' which passes through our point charge. On either size of this plane, we must have expo decay as we go to large |z|. There is some discrete spectrum of k determined presumably by the zeros of Jm on some cylindrical boundary, but we leave it just as We are doing one of the "two ways" mentioned above. Notice that we are parameterizing our two points r and r' this way r = (ρcosφ, ρsinφ, z) => = (ρcosφ, ρsinφ, z)/ r' = (ρ'cosφ',ρsinφ',z') => ' = (ρ'cosφ',ρ'sinφ',z')/ => ' =[ ρρ'(cosφcosφ' + sinφsinφ') + zz']/[ ] = cos(γ) => ' =[ ρρ'(cos(φ-φ') + zz']/[ ] = cos(γ) so that γ is the angle directly between our two vectors r and r'. But for our cylindrical case, we are really more interested (later) in these facts, simpler than the above, rr' = ρρ'cos(φ-φ') + zz' r2 = ρ2 + z2 r'2 = ρ'2 + z'2 R2 = |r-r'|2 = r2 + r'2 - 2 rr' = ρ2 + z2 + ρ'2 + z'2 - 2ρρ'cos(φ-φ') -2 zz' = ρ2 + ρ'2 + (z-z')2 - 2ρρ'cos(φ-φ') At this point there are several different things we can do. 1. Use the pillbox condition to find the Amn coefficients. The general pillbox condition is this ( units) ∂q1Vi - ∂q1Vo = 4πq (h1/h2h3) δ(q2-q2') δ(q3-q3') where q1 is the dimension perp to our partition of space into two regions at the point charge. I will assume for the moment that z' > 0 and that "outer" means z > z' which is farther away from the origin, etc. We then have (that is to say, the "inner" region contains the origin) Vi = Σm∫dk Amk exp(+k(z-z')) Jm(kρ) eimφ z < z' Vo = Σm∫dk Amk exp(-k(z-z')) Jm(kρ) eimφ z > z' So q1 = z. We can then take q2 = ρ and q3 = φ and use these facts h1 = hz = 1 h2 = hρ = 1 h3 = hφ = ρ I take these facts from M&M page 178. I really should start with the three Cartesian equations and run my little program to compute the metric tensor, but OK for now. Our pillbox condition is then this: ∂zVi - ∂zVo = 4πq (1/ρ) δ(ρ-ρ') δ(φ-φ') We then need to compute the derivatives ∂zVi = Σm∫dk Amk k exp(+k(z-z')) Jm(kρ) eimφ z < z' ∂zVo = Σm∫dk Amk (-k) exp(-k(z-z')) Jm(kρ) eimφ z > z' then evaluate both on the bounding surface between the two regions (which means z = z' ) ∂zVi = Σm∫dk Amk k Jm(kρ) eimφ ∂zVo = –Σm∫dk AmkkJm(kρ) eimφ so ∂zVi - ∂zVo = 2 Σm∫dk Amk k Jm(kρ) eimφ = 4πq (1/ρ) δ(ρ-ρ') δ(φ-φ') We will use two orthogonalities to find the Amn. First, apply ∫dφ e-im'φ to both sides and use (1/2π) !Syntax Error, I dφ eimφ e-im'φ = δmm' // orthogonality The LHS becomes 2 Σm∫dk Amk k Jm(kρ) ∫dφ e-im'φ eimφ = 2 Σm∫dk Amk k Jm(kρ) 2πδmm' = 4π Σ∫dk Am'k k Jm'(kρ) and the RHS becomes 4πq (1/ρ) δ(ρ-ρ') e–im'φ' so we then have (replace m' with m) 4π ∫dk Amk k Jm(kρ) = 4πq (1/ρ') δ(ρ-ρ') e-imφ' Now we reach into our prepared "transforms doc" and pull out this rabbit, (Hankel Transform) f(ρ) = !Syntax Error, Idk k Jν(kρ) Fν(k) // expansion Fν(k) = !Syntax Error, Idx x Jν(kρ) f(ρ) // projection !Syntax Error, Idρ ρ Jν(kρ) Jν(k'ρ) = δ(k-k')/k // orthogonality !Syntax Error, Idk k Jν(kρ) Jν(kρ') = δ(ρ-ρ')/ρ // completeness So, the next step is to apply !Syntax Error, Idρ ρ Jm(k'ρ) to both sides of the above equation. LHS = 4π ∫dk Amk k!Syntax Error, Idρ ρ Jm(k'ρ) Jm(kρ) = 4π ∫dk Amk k δ(k-k')/k = 4π Amk' RHS = 4πq (1/ρ') e-imφ' !Syntax Error, Idρ ρ Jm(k'ρ)δ(ρ-ρ') = 4πq (1/ρ') e-imφ' ρ' Jm(k'ρ') = 4πq e-imφ' Jm(k'ρ') The result is this (change k' to k) 4π Amk = 4πq e-imφ' Jm(kρ') Amk = q e-imφ' Jm(kρ') So we now have our coefficients! We can install these into our expansion above V(r | r') = q/R = Σm ∫dk Amk exp(-k|z-z'|) Jm(kρ) eimφ = Σm ∫dk q e-imφ' Jm(kρ') exp(-k|z-z'|) Jm(kρ) eimφ = q Σm ∫dk exp(-k|z-z'|) Jm(kρ) Jm(kρ') eimφ e-imφ' where Σm is over all integers. So here then is our interesting result: 1/R = Σm ∫dk exp(-k|z-z'|) Jm(kρ) Jm(kρ') eimφ e-imφ' We know that the result is real, so the imaginary part vanishes, and this can also be written as 1/R = Σm cos[m(φ-φ')] ∫dk exp(-k|z-z'|) Jm(kρ) Jm(kρ') We can at this point do the usual "fold over" operation. We know J-m(z) = (-1)mJm(z) so everything is even in m. Thus, the negative sum will equal the positive sum, and we get at once 1/R = Σm=0∞εm cos[m(φ-φ')] ∫dk exp(-k|z-z'|) Jm(kρ) Jm(kρ') This of course has the expected symmetric form under r ↔ r' . We now want some verification of the above formula. on page 86 gives 1/R "the other way" which I am not doing right now, so we need to look elsewhere. Smythe? On page 194 he does the I and K version, similar to . Does not give the above. MF? On page 1263 they have it! I copy: Hurray! By the way, the toroidal Q function version is mentioned here http://en.wikipedia.org/wiki/Green%27s_function_for_the_three-variable_Laplace_equation 2. Development of the Addition Theorem Now, what happens if we put our point charge on the z axis? Then r' = (0,0 ,z') We have then that Jm(kρ') = Jm(0) = δm,0 [ p 72 or Schaum p 136 ] so we have 1/R = Σm=0∞εm cos[m(φ-φ')] ∫dk exp(-k|z-z'|) Jm(kρ) δm,0 = ∫dk exp(-k|z-z'|) J0(kρ) which is a pretty simple result indeed. But in this case we also know that 1/R = 1/. So we have this interesting result 1/ = !Syntax Error, I dk exp(-k|z-z'|) J0(kρ) (*) which is better stated as 1/ = !Syntax Error, I dk exp(-k|z|) J0(kρ) Can I verify this result? GR p 694 says So set ν = 0 and x = k and β = ρ and α = |z| and only the denominator survives, QED. This result is similar to p 86 3.150. Now I need something like my little Theorem 1 of the spherical 1/R document? Well, here it is simpler than that. Our equality (*) is valid for any quantity ρ>0 you want. To stress that point, write 1/ = !Syntax Error, I dk exp(-k|z-z'|) J0(ka) Now recall from above that we have R2 = ρ2 + ρ'2 + (z-z')2 - 2ρρ'cos(φ-φ') Therefore, if we set a2 = ρ2 + ρ'2 - 2ρρ'cos(φ-φ') // which is also a geometric law of cosines we then have = R so we then have 1/R = !Syntax Error, Idk exp(-k|z-z'|) J0(k) This, then, is our result analogous to the sphericals result 1/R = Σn (r<)n(r>)-n-1 Pn(cosθcosθ' + sinθsinθ'cos(φ-φ')) But now consider that our more general 1/R result from above is this: (one form of it) 1/R = Σm=0∞εm cos[m(φ-φ')] ∫dk exp(-k|z-z'|) Jm(kρ) Jm(kρ') Setting things equal we must have Σm=0∞εm cos[m(φ-φ')] ∫dk exp(-k|z-z'|) Jm(kρ) Jm(kρ') = !Syntax Error, I dk exp(-k|z-z'|) J0(k) Although exp(-k|z-z'|) functions are not really a complete set, we suspect these things can only really be equal for all values of z and z' if the following is true J0(k) = Σm=0∞εm cos[m(φ-φ')] Jm(kρ) Jm(kρ') Verification for this addition theorem comes from GR7 page 940, with The variable names are very skewed, but we see it is exactly the same. The connection is this: GR7 me m k ρ ρ r ρ' k m φ φ-φ' This addition theorem is similar to 3.151, but his is for the I K world. I will be doing that soon. I am sure this addition theorem also has group theory support, but I have never read that stuff. On the list! In the spherical case, we had Pn(cosγ) where γ was the angle between r and r'. Here we have in place of that our J0(ka) where a is the distance between the projections of r and r' onto the same z plane. No doubt this is an invariant of some symmetry operation the underlies the group interpretation. Part C: 1/R Green's Function using Km functions 0. Introduction: Free space Green's function in cylindrical coordinates In this case, our two spaces will be an inside cylinder, and the complementary outside space. The boundary is a cylindrical surface which contains our Green's charge. The atomic form is this f(r | r') = q/R = Σm!Syntax Error, I dk Amk cos[k(z-z')] Im(kρ<) Km(kρ>) eimφ where m is over all integers. This is a "Smythian form" in that I have built in the known symmetry for both z,z' and ρ,ρ'. Since even under swap symmetry, we cannot have a sin[k(z-z')] term. And of course I is the function good near ρ = 0, etc. Notice that we are parameterizing our two points r and r' this way r = (ρcosφ, ρsinφ, z) => = (ρcosφ, ρsinφ, z)/ r' = (ρ'cosφ',ρsinφ',z') => ' = (ρ'cosφ',ρ'sinφ',z')/ => ' =[ ρρ'(cosφcosφ' + sinφsinφ') + zz']/[ ] = cos(γ) => ' =[ ρρ'(cos(φ-φ') + zz']/[ ] = cos(γ) so that γ is the angle directly between our two vectors r and r'. But for our cylindrical case, we are really more interested (later) in these facts, simpler than the above, rr' = ρρ'cos(φ-φ') + zz' r2 = ρ2 + z2 r'2 = ρ'2 + z'2 R2 = |r-r'|2 = r2 + r'2 - 2 rr' = ρ2 + z2 + ρ'2 + z'2 - 2ρρ'cos(φ-φ') -2 zz' = ρ2 + ρ'2 + (z-z')2 - 2ρρ'cos(φ-φ') At this point there are several different things we can do. 1. Use the pillbox condition to find the Amn coefficients. The general pillbox condition is this ( units) ∂q1Vi - ∂q1Vo = 4πq (h1/h2h3) δ(q2-q2') δ(q3-q3') where q1 is the dimension perp to our partition of space into two regions at the point charge. In our current application, q1 = ρ so we make these choices h2 = hz = 1 h1 = hρ = 1 h3 = hφ = ρ I take these facts from M&M page 178. I really should start with the three Cartesian equations and run my little program to compute the metric tensor, but OK for now. Our pillbox condition is then this: ∂ρVi - ∂ρVo = 4πq (1/ρ) δ(z-z') δ(φ-φ') We then need to compute the derivatives Vi = Σm!Syntax Error, I dk Amk cos[k(z-z')] Im(kρ) Km(kρ') eimφ Vo = Σm!Syntax Error, I dk Amk cos[k(z-z')] Im(kρ') Km(kρ) eimφ ∂ρVi = Σm!Syntax Error, I dk k Amk cos[k(z-z')] Im(kρ)' Km(kρ') eimφ ∂ρVo = Σm!Syntax Error, I dk k Amk cos[k(z-z')] Im(kρ') Km(kρ)' eimφ then evaluate both on the bounding surface between the two regions (which means ρ' = ρ ) ∂ρVi = Σm!Syntax Error, I dk k Amk cos[k(z-z')] Im(kρ)' Km(kρ) eimφ ∂ρVo = Σm!Syntax Error, I dk k Amk cos[k(z-z')] Im(kρ) Km(kρ)' eimφ so ∂ρVi - ∂ρVo = Σm!Syntax Error, I dk k Amk cos[k(z-z')] W[Km(kρ), Im(kρ)] eimφ = Σm!Syntax Error, I dk k Amk cos[k(z-z')] (kρ)-1 eimφ // p 86 = (1/ρ) Σm!Syntax Error, Idk Amk cos[k(z-z')] eimφ The pillbox condition is therefore, (1/ρ) Σm!Syntax Error, Idk Amk cos[k(z-z')] eimφ = 4πq (1/ρ) δ(z-z') δ(φ-φ') or Σm!Syntax Error, Idk Amk cos[k(z-z')] eimφ = 4πq δ(z-z') δ(φ-φ') LHS = RHS We will use two orthogonalities to find the Amn. First, apply ∫dφ e-im'φ to both sides and use (1/2π) !Syntax Error, I dφ eimφ e-im'φ = δmm' // orthogonality The LHS becomes Σm!Syntax Error, Idk Amk cos[k(z-z')] ∫dφ e-im'φ eimφ = Σm!Syntax Error, Idk Amk cos[k(z-z')] 2π δmm' = 2π !Syntax Error, Idk Am'k cos[k(z-z')] and the RHS becomes 4πq δ(z-z') e–im'φ' so we then have (replace m' with m) 2π !Syntax Error, Idk Amk cos[k(z-z')] = 4πq δ(z-z') e–imφ' !Syntax Error, Idk Amk cos[k(z-z')] = 2q δ(z-z') e–imφ' or !Syntax Error, Idk Amk cos(kz) = 2q δ(z) e–imφ' LHS = RHS Now we reach into our prepared "transforms doc" and pull out this rabbit, (Cosine Integral Transform) Summary Case 2: For z in (-∞,∞) and k in (0,∞) where f(z) is even in z: f(z) = !Syntax Error, Idk fk cos(kz) // expansion pair 2 fk = !Syntax Error, Idz f(z) cos(kzx) // projection !Syntax Error, Idz cos(kz) cos(k'z) = π δ(k-k') // orthogonality !Syntax Error, Idk cos(kz) cos(kz') = (π/2)[ δ(z-z') + δ(z+z')] // "completeness" So, the next step is to apply !Syntax Error, Idz cos(k'z) to both sides of the above equation LHS = RHS: LHS = !Syntax Error, Idk Amk !Syntax Error, Idz cos(k'z) cos(kz) = !Syntax Error, Idk Amk π δ(k-k') = π Amk' RHS = 2q !Syntax Error, Idz cos(k'z)δ(z) e–imφ' = 2q e–imφ' The result is this (change k' to k) π Amk = 2q e–imφ' Amk = (2/π)q e–imφ' So we now have our coefficients! We can install these into our expansion above f(r | r') = q/R = Σm!Syntax Error, I dk Amk cos[k(z-z')] Im(kρ<) Km(kρ>) eimφ = (2/π)q Σm!Syntax Error, I dk cos[k(z-z')] Im(kρ<) Km(kρ>) eimφ e–imφ' with the result then that 1/R = (2/π) Σm!Syntax Error, I dk cos[k(z-z')] Im(kρ<) Km(kρ>) eimφ e–imφ' where as noted above m is summed over all integers. We get immediate verification of this result from page 86 (3.148). Bateman page 5 shows that K-ν = Kν for general ν (page 5), and I-m = Im for integer m only. This means that, if we first replace the expos with cos[m(φ-φ')], we can reflect the negative side of the series in the usual way, and we then get this alternate form: 1/R = (2/π) Σm=0∞ εm!Syntax Error, I dk cos[k(z-z')] Im(kρ<) Km(kρ>) cos[m(φ-φ')] and this appears as (3.149). As usual, εm is the Neumann factor which is 2 for m > 0 else 1. 2. Development of the Addition Theorem Now, what happens if we put our point charge on the z axis? Then r' = (0,0 ,z') Since ρ' = 0, surely ρ' = ρ< so we have in this case 1/R = (2/π) Σm=0∞ εm!Syntax Error, I dk cos[k(z-z')] Im(0) Km(kρ) cos[m(φ-φ')] and as before we get Im(0)= δm0 so this thing shrinks down to 1/R = (2/π) !Syntax Error, I dk cos[k(z-z')] K0(kρ) which is a pretty simple result indeed. But in this case we also know that 1/R = 1/. So we have this interesting result 1/ = (2/π) !Syntax Error, I dk cos[k(z-z')] K0(kρ) (*) which is better stated as 1/ = (2/π) !Syntax Error, I dk cos(kz) K0(kρ) Can I verify this result? has it on page 86 as (3.150). Also GR7 page 719 and there it is. Our equality (*) is valid for any quantity ρ>0 you want. To stress that point, write 1/ = (2/π) !Syntax Error, I dk cos[k(z-z')] K0(ka) Now recall from above that we have R2 = ρ2 + ρ'2 + (z-z')2 - 2ρρ'cos(φ-φ') Therefore, if we set a2 = ρ2 + ρ'2 - 2ρρ'cos(φ-φ') // which is also a geometric law of cosines we then have = R so we then have 1/R = (2/π) !Syntax Error, I dk cos[k(z-z')] K0(k) This, then, is our result analogous to the sphericals result 1/R = Σn (r<)n(r>)-n-1 Pn(cosθcosθ' + sinθsinθ'cos(φ-φ')) But now consider that our more general 1/R result from above is this: (one form of it) 1/R = (2/π) Σm=0∞ εm!Syntax Error, I dk cos[k(z-z')] Im(kρ<) Km(kρ>) cos[m(φ-φ')] Setting things equal we must have (2/π) !Syntax Error, I dk cos[k(z-z')] K0(k) = (2/π) Σm=0∞ εm!Syntax Error, I dk cos[k(z-z')] Im(kρ<) Km(kρ>) cos[m(φ-φ')] Although cos[k(z-z')] functions are maybe not really a complete set, we suspect these things can only really be equal for all values of z and z' if the following is true K0(k) = Σm=0∞ εm Im(kρ<) Km(kρ>) cos[m(φ-φ')] and this is our addition theorem, verified as p 86 (3.151). Finally, in Part D I summarize all results of this doc, for Parts A,B and C. Part D: Summary of All Results Part A : The Green's Function for a point charge q located at r' inside an infinitely long conducting cylinder of radius b can be written as, g(r | r') = q (2/b2) Σm Σn [Jm+1(bkn)]-2 (1/kn) exp(-kn|z-z'|) Jm(knρ) Jm(knρ') eimφ e-imφ' The literature usually provides an alternate formula using I and K functions which I could, but did not, derive in this doc. I did the harder way. βn(m) = bkn are the zeros of Jm(knb) = 0 which are enumerated by the index n = 1,2... so we have really kn = kn(m) = βn(m)/b . The m sum is over all integers. Using methods shown above, this result can be written by replacing the two exponentials with cos[m(φ-φ')]. Once this is done, these same methods show that the negative half of the series can be reflected and we get the "all real" result: g(r | r') = q (2/b2) Σm=0∞ εm cos[m(φ-φ')] Σn [Jm+1(bkn)]-2 (1/kn) exp(-kn|z-z'|) Jm(knρ) Jm(knρ') Because I did it the hard way, I was unable to verify these results against any other sources. The results do in fact show the required symmetry Part B, including the Qm-1/2 expansion : The free-space 1/R Green's function for a unit point charge can be written as 1/R = Σm eimφ e-imφ'!Syntax Error, Idk exp(-k|z-z'|) Jm(kρ) Jm(kρ') where the m sum is over all integers, and the symmetry is manifest. As in part A, we can write this in its real form as follows: 1/R = Σm=0∞εm cos[m(φ-φ')] !Syntax Error, Idk exp(-k|z-z'|) Jm(kρ) Jm(kρ') This is verified in MF as noted above. The related addition theorem formulas are these: 1/ = !Syntax Error, Idk exp(-k|z|) J0(kρ) 1/R = !Syntax Error, Idk exp(-k|z-z'|) J0(k) J0(k) = Σm=0∞εm cos[m(φ-φ')] Jm(kρ) Jm(kρ') The first and third are verified in GR7, while the second is a restatement of the first and must be true as well, so I regard it as verified. In passing, I want to mention the Selvaggi result which I derived in my "integral doc" which is this: 1/R = 1/(π) Σm=0∞εm cos[m(φ-φ')] Qm-1/2(β) β = [ ρ2 + ρ'2 + (z-z')2 ]/(2ρρ') This result was derived (by me) by doing a direct Cosine Fourier Series expansion of 1/R 1/R ≡ 1/|r-r'| = 1/ So we can compare these two results for 1/R: 1/R = Σm=0∞εm cos[m(φ-φ')] !Syntax Error, Idk exp(-k|z-z'|) Jm(kρ) Jm(kρ') 1/R = 1/(πρρ') Σm=0∞εm cos[m(φ-φ')] Qm-1/2(β) β = [ ρ2 + ρ'2 + (z-z')2 ]/(2ρρ') The following would seem then to be true: !Syntax Error, Idk exp(-k|z-z'|) Jm(kρ) Jm(kρ') = Qm-1/2(β) / (π) Lo and behold, this result appears in GR page 696 where x = k, β = ρ, γ = ρ', ν = m, α = |z-z'|. This Qm-1/2 expansion is still in terms of cylindrical coordinates, we don't see any toroidal coordinates appearing here (except φ). Maybe at some point I will understand why a toroidal harmonic appears in this cylindrical problem. An opportunity will come soon when I do 1/R for toroidal coordinates, on the menu. : Suppose we wanted to know the potential of just the induced charge in our Part A cylinder problem. We could find this by subtracting out the potential of the point charge. Thus, V(r|r')due to induced charge = (2/b2) Σm=0∞εm cos[m(φ-φ')] Σn=1∞ [Jm+1(bkn)]-2 (1/kn) exp(-kn|z-z'|) Jm(knρ) Jm(knρ') – Σm=0∞εm cos[m(φ-φ')] ∫dk exp(-k|z-z'|) Jm(kρ) Jm(kρ') The two terms cannot be very will combined as they can in other problems, here because one is discrete and the other continuous. Part C : The free-space 1/R Green's function for a unit point charge can be written as 1/R = (2/π) Σm!Syntax Error, I dk cos[k(z-z')] Im(kρ<) Km(kρ>) eimφ e–imφ' where the m sum is over all integers, and the symmetry is manifest. As in part A, we can write this in its real form as follows: 1/R = (2/π) Σm=0∞ εm!Syntax Error, I dk cos[k(z-z')] Im(kρ<) Km(kρ>) cos[m(φ-φ')] Both these forms are verified in page 86. The related addition theorem formulas are these: 1/ = (2/π) !Syntax Error, I dk cos(kz) K0(kρ) 1/R = (2/π) !Syntax Error, I dk cos[k(z-z')] K0(k) K0(k) = Σm=0∞ εm Im(kρ<) Km(kρ>) cos[m(φ-φ')] These results are also verified on green page 86. Part D (spherical results): Let's just add here the corresponding spherical coordinate results. The Green's function for a unit point charge is given by 1/R = Σn=0∞ Σm f(n,-m) (r<)n(r>)-n-1 Pnm(z) Pnm(z') eimφ e-imφ' where Σnm is the conventional spherical double sum. One can regard the m sum as being over all integers with the cutoff being provided by the P functions. This result is confirmed by as shown in my 1/R spherical doc when converted to his Y functions. In the usual way, we can reflect the negative m contributions and write this as 1/R = Σn=0∞ Σm=0∞ εm f(n,-m) (r<)n(r>)-n-1 Pnm(z) Pnm(z') cos[m(φ-φ')] I don't know where this is stated in this manner but feel certain it is correct. The addition theorem related equations are these: 1/R = Σn (r<)n(r>)-n-1 Pn(cosθcosθ' + sinθsinθ'cos(φ-φ')) Pn[cosθcosθ' + sinθsinθ'cos(φ-φ')] = Σm=0∞ εm f(n,-m) Pnm(cosθ) Pnm(cosθ') cos[m(φ-φ')] The first line is verified page 62. The second from p 69, also Bateman p 169 and GR7.