question on Fourier + parallel plates Cartesian

Question on Fourier; Parallel Plate Cartesian Green's PhL 7.17.10 0. Overview ( 2 pages, added 9.8.10) 2 1. Verification of Fourier 3D 1/R expansion. 3 2. Do we have a new atomic form for Cartesians? 4 3. Examining the Rothwell/Cloud parallel plate Green's Function Solution 4 4. Rederive the 1/R expansion using the 3D Fourier Transform Notation 6 5. How would you solve the parallel-plate problem using a "transverse Fourier Series"? 7 (a) do the Fourier Transform only in x and y, not z. 7 (b) Solve the 1D Helmholtz Green's Function problem in z which arises. 8 (c) Resumption of Main Flow; Statement of the Parallel Plate Cartesian Green's Function 11 (d) Transverse Fourier Expansion for 1/R. 12 (e) Subtracting off the point charge potential 14 (f) Errata Search. 15 Preliminary Question: (overview starts down 1/2 page) How does a 3D Fourier expansion (part of the usual transform) relate to the subject of Cartesian atoms and Smythian forms? For example, consider this 3D Fourier expansion of 1/R which appears in MF 1/R = (1/2π2)∫d3k exp[ ik(r- r1) ] / k2 In the notes below, I show that the integrand here is in fact NOT a Cartesian atom, yet the integral still solves Laplace. The usual atoms reappear when the above is rewritten in its 2D "transverse form" 1/R = (1/π) ∫d2kt exp[ ±ikt(r-r1) exp(-kt|z-z1|)/(2kt) Appendix A A.55 as discussed below using a contour integration. Meanwhile, the solution to the parallel plate problem in Cartesian coordinates comes out being: ( kt = (kx,ky) and kt = κz = ) G(r;r1) = (2π)-2 !Syntax Error, I!Syntax Error, I d2kt (exp[-i kt(r-r1)] /kt) sinh[kt(d -z>)] sinh(ktz<) /sinh(ktd) z> = max(z,z1) z< = min(z,z1) 0 < z1 < d (between the plates) where again in this Smythian form you see our usual Cartesian atomic form (written in exponentials) 1' [exp(kxx), exp(-kxx)], [exp(kyy), exp(-kyy)], [exp(κzz), exp(-κzz) ] kz = imaginary = iκz where x and y are oscillatory and z is radial/expo. ________________________________________________________________________________ 0. Overview ( 2 pages, added 9.8.10) In Section 1 I shows that if we assume that 1/R = (1/2π2)∫d3k exp[ ik(r- r1) ] / k2 we find that 2 (1/R) = - 4π q δ3(r-r1) which is the correct Green's equation for a unit point charge (q=1) at r = r1. In Section 2 I comment that the integrand exp[ ik(r- r1) ] / k2 is itself NOT a Cartesian atomic form, since 2 exp[ ik(r- r1) ]/k2 = - exp[ ik(r- r1) ] ≠ 0 and thus does not satisfy Laplace. Yet everywhere away from r = r1 the integral of this integrand does satisfy Laplace. In Section 3 I quote the Rothwell/Cloud solution to our problem of the Green's Function for parallel plates treated in Cartesian coordinates, and I was lucky to find this solution in a web search. I first show that it is in fact a superposition of valid Cartesian atomic forms, but I don't at this point verify their solution which, by the way, is only that part of the solution due to the induced charge on the plates, which they call the "secondary" potential in a problem like this. In Section 4 I start with -2u(r;r1) = 4πδ(r-r1) and do a 3D Fourier to find U = F3D(u) is given by U(k; r1) = 4π (2π)-3e+ikr1/k2 and this then is just another way to confirm the 1/R expansion shown above. Section 5 has several lettered parts, to wit: In Part (a) I start again with -2u(r;r1) = 4πδ(r-r1) but now I do only a 2D Fourier expansion which is essentially doing a Stakgold "partial eigenfunction expansion" to solve this parallel plates problem. You are then left with this 1D Green's function problem in coordinate z, where now U = F2D(u) , (kt2 - ∂z2) U(kt; z; r1) = 4π (2π)-2 δ(z-z1) exp(iktr1) U(kt; z=0; r1) = 0 U(kt; z=d; r1) = 0 // potential vanishes on the plates! Then in Part (b) I solve this 1D Green's function problem (lots of detail) and I get g(z; z1) = -(a/k) sinh[k(z>-d)] sinh(kz<) /sinh(kd) z> = max(z,z1) where g(z; z1) is shorthand for U(kt; z; r1) and k is really kt and a = exp(iktr1)/π . In Part (c) I write this out as U(kt; z; r1) = - exp(iktr1) (1/πkt) sinh[kt(z>-d)] sinh(ktz<) /sinh(ktd) and I insert this into the 2D Fourier expansion to get the final result G(r;r1) = (2π)-2 !Syntax Error, I!Syntax Error, I d2kt (exp[-i kt(r-r1)] /kt) sinh[kt(d -z>)] sinh(ktz<) /sinh(ktd) = (1/π2)!Syntax Error, I!Syntax Error, Idkxdky cos[kx(x-x1)] cos[ky(y-y1)] sinh[κz(d -z>)]sinh(κz z<)/[ κzsinh(κzd)] z> = max(z,z1) z< = min(z,z1) 0 < z1 < d (between the plates) κz = . which is the Green's function for a unit positive point charge located at r1 = (x1, y1, z1) between parallel plates located at z=0 and z=d, when observed at point r = (x,y,z) , and where kt = (ktx, kty, 0) and we are in green Jackson cgs units. Either form for G is just a linear combination of Cartesian atoms, and I note the manifest r ↔ r1 symmetry. In order to verify Rothwell and Cloud's claimed result for the potential due just to the charge induced on the plates, I need to subtract out the contribution of the point charge to the above G(r;r1) result. Since the above G involves a d2kt integration, it is very useful to have a similar integral for the point charge 1/R. So in Part (d) I develop this 1/R expansion, and show that it agrees with Appendix A.55 of R&C, 1/R = (1/π) ∫d2kt exp[ ±ikt(r-r1) exp(-kt|z-z1|)/(2kt) Appendix A A.55 which requires playing with a contour integral, something familiar to me from "propagator theory". Finally then in Part (e) I subtract out this 1/R contribution and I get this result : Gind(r;r1) = -(2π)-2∫d2kt{2sinh[kt(z>-d)]sinh(ktz<) /sinh(ktd) + exp(-kt|z-z1|) } exp[i kt(r-r1)]/(2kt) where I now show a direct comparison with the R&C result. It is not obvious that the two results agree, but, using Maple to do the algebra, I show that they do in fact agree exactly, except R&C have an error in their exponential argument -- they should really have exp(-j kρ[r'-r]). Mr. Rothwell confirmed this erratum and I have copied the email comm at the end of this doc. ________________________________________________________________________________ 1. Verification of Fourier 3D 1/R expansion. First of all, how would we verify this? 2 |r- r1|-1 = 2 (1/R) = 12 { (1/2π2)∫d3k exp[ ik(r- r1) ] / k2} = (1/2π2)∫d3k k-2 2 exp[ ik(r- r1) ] = (1/2π2)∫d3k k-2 i2 k2 exp[ ik(r- r1) ] = - (1/2π2)∫d3k exp[ ik(r- r1) ] = - 4π (1/2π)3 (2π)3 δ(r-r1) = - 4π q δ3(r-r1) // q = 1 The Poisson equation says -2 V = 4πρ For a point charge q located at position r1 this says -2 V = 4πq δ3(r-r1) But we know that the potential of a point charge is just V = q |r- r1|-1 and we know it is unique. Thus, the above expansion must be equal to 1/R. The 1/R expansion shown above is the solution of this Poisson equation which we can regard as just a 3D Green's Function problem with no BC's, the Stak "fundamental solution". 2. Do we have a new atomic form for Cartesians? Notice that 1/R is a solution of Laplace as long as you stay away from the point charge. Staying away, then, we seem to have a "separated" atomic form as follows, where all three terms are oscillatory, exp[ ik(r- r1) ]= exp[ikx(x-x1)] exp[iky(y-y1)] exp[ikz(z-z1)] But this product of atoms does NOT satisfy Laplace, since each term makes the same sign contribution to the curvature, 2 exp[ ik(r- r1) ] = -k2 exp[ ik(r- r1) ] ≠ 0 However, we can construct with this atomic form a function which, in a very large region of space, IS a solution of Laplace, and that is the 1/R shown above. Rather than having the ki2 components add up to zero for each term, we have global phase cancellation at all points except at the point charge. Since the atom itself does not satisfy Laplace, I don't think I would classify it as one of my "atomic forms". 3. Examining the Rothwell/Cloud parallel plate Green's Function Solution I found the following solution in a book Electromagnetics by Rothwell and Cloud (I have it). Here is the picture and the solution: (this really is a 3D problem, by the way) And here while we're at it is a related follow-on problem: In this book, the authors use the word "secondary" to mean the potential due to induced charge, and "primary" is the exciting charge, here a point charge. The notation used above is certainly hazy, but if we look back in the text we see this info: In the notes below, I derive the expansion 3.77 which is in fact correct. I also derive the parallel plate solution shown above, and I did find a significant erratum which I sent off to the author (namely, that the expo in 3.212 should be as it appears in 3.77). As for 3.78, it is a viable "atomic form" for this simple reason: If you apply 2, you get the sum of two terms which are from the [...] and from the expo factor. The [...] contributes +kρ2 and the expo contributes - kρ2, so indeed you get 2Φ2(r) = 0. How would I enter this in my list of "atoms" ? One way is this: [ exp(kzz), exp(-kzz) ] [cos(kxx + kyy) ] which can be written as [ exp(kzz), exp(-kzz) ] [cos(kxx)cos(kyy) - sin(kxx)sin(kyy) ] This is a subset of the more general atomic form [ exp(kzz), exp(-kzz) ] [sin(kxx), cos(kxx) ] [sin(kyy), cos(kyy) ] so I don't think (3.78) is as fully general as my atomic form. In any event, you see that we are expo in z, and oscillatory in x and y, and we integrate over kx and ky, and (0,∞) would be general enough I think for the spectra. So I don't think I have anything new to add to my atomic forms for Cartesians. My PDF Rothwell and Cloud book has no page numbers and is perhaps an early draft? I think it matches their 2001 First Edition however. 4. Rederive the 1/R expansion using the 3D Fourier Transform Notation Nothing new relative to section 1, above, just doing it in a different notation. Suppose you start with a raw Green's equation [ true Green's would be G = u/(4π) ] -2u(r;r1) = 4πδ(r-r1) // Jackson convention with unit cgs charge You could Fourier-expand the δ and the u solution: δ(r-r1) = (2π)-3 ∫d3k e-ik(r-r1) u(r;r1) = ∫d3k e-ikr U(k; r1) -2u(r;r1) = ∫d3k e-ikr k2 U(k; r1) We then have the following k-space equation k2 U(k; r1) = 4π (2π)-3e+ikr1 and we have a solution to our problem in k-space, which is just this: U(k; r1) = 4π (2π)-3 e+ikr1/k2 We then want to transform this solution back to r-space, u(r;r1) = ∫d3k e-ikr U(k; r1) = ∫d3k e-ikr {4π (2π)-3e+ikr1/k2 } = 4π (2π)-3 ∫d3k e-ik(r-r1)/k2 = (1/2π2) ∫d3k e-ik(r-r1)/k2 = 1/R where in the last step we use our result above. So this is another way to obtain that result, if you will. 5. How would you solve the parallel-plate problem using a "transverse Fourier Series"? (a) do the Fourier Transform only in x and y, not z. Think of our plates in z, and we decide to transform only the x and y directions. We start again with -2u(r;r1) = 4πδ3(r-r1) and we make these expansions: δ3(r-r1) = δ(z-z1) (2π)-2 ∫d2kt exp[-i kt(r-r1)] u(r;r1) = ∫d2kt exp[-i ktr] U(kt; z; r1) where kt is a 2-component transverse k vector, but can think as 3D vector with ktz = 0. We next compute the effect of 2 on u : -2 u(r;r1) = -(t2 + ∂z2) u(r;r1) = = ∫d2kt (-t2exp[-i ktr]) U(kt; z; r1) + ∫d2kt exp[-i ktr] (-∂z2 U(kt; z; r1) ) = ∫d2kt exp[-i ktr]) kt2U(kt; z; r1) + ∫d2kt exp[-i ktr] (-∂z2 U(kt; z; r1) ) = ∫d2kt exp[-i ktr]) { (kt2 - ∂z2) U(kt; z; r1) } Our Green's equation then reads: ∫d2kt exp[-i ktr]) { (kt2 - ∂z2) U(kt; z; r1) } = 4π ∫d2kt exp[-i kt(r-r1)] { δ(z-z1) (2π)-2} and thus in kt space we have this equation to solve: (kt2 - ∂z2) U(kt; z; r1) = 4π (2π)-2 δ(z-z1) exp(iktr1) We can now regard this, for a particular value of kt, as a 1D Helmholtz Green's Function problem. But we now want to impose our boundary conditions U(kt; z=0; r1) = 0 U(kt; z=d; r1) = 0 (b) Solve the 1D Helmholtz Green's Function problem in z which arises. So let's simplify the notation and write k = kt (k2 - ∂z2) g(z; z1) = a δ(z-z1) a = 4π (2π)-2 exp(iktr1) = exp(iktr1)/π The homo solutions of this equation are sh(kz) and ch(kz) due to the relative minus sign. So we construct our Green's function like this: g(z; z1) = Ash(kz) 0 ≤ z ≤ z1 g(z; z1) = Bsh(kz) + Cch(kz) z1 ≤ z ≤ d The first line meets the BC at z = 0, and we need to work on the second one: Bsh(kd) + Cch(kd) = 0 => C = -B th(kd) So we now have it down to two coefficients: g(z; z1) = Ash(kz) 0 ≤ z ≤ z1 g(z; z1) = B[ sh(kz) - th(kd)ch(kz) ] z1 ≤ z ≤ d The function must be continuous at z1 so we have Ash(kz1) = B[ sh(kz1) - th(kd)ch(kz1) ] Then we have our jump condition to worry about. I will just rederive that right here rather than look it up. Start with (k2 - ∂z2) g(z; z1) = a δ(z-z1) Now integrate both sides around z1 to get !Syntax Error, I(k2 - ∂z2) g(z; z1) = a The k2 term gives nothing, so we have !Syntax Error, I∂z2 g(z; z1) = -a The integrand is a perfect differential, so we have derived our jump condition, ∂zg(z; z1)+ - ∂zg(z; z1)- = -a But from above we have g(z; z1) = Ash(kz) 0 ≤ z ≤ z1 g(z; z1) = B[ sh(kz) - th(kd)ch(kz) ] z1 ≤ z ≤ d ∂z g(z; z1) = A k ch(kz) 0 ≤ z ≤ z1 ∂z g(z; z1) = B k [ ch(kz) - th(kd)sh(kz) ] z1 ≤ z ≤ d So our jump condition is this: ∂zg(z; z1)+ - ∂zg(z; z1)- = -a B k [ ch(kz1) - th(kd)sh(kz1) ] – A k ch(kz1) = -a So let's now assemble our continuity and our jump conditions: Ash(kz1) = B[ sh(kz1) - th(kd)ch(kz1) ] Ach(kz1) = B[ ch(kz1) - th(kd)sh(kz1) ] + a/k We want then to solve these equations for A and B, then we will have found our Green's 1D solution. This is just a Cramer's Rule problem which Maple is happy to solve for us: where it puts everything into expo notation. Manual fiddling then shows that A = - (a/k) sinh[k(z1-d)] / sinh(kd) B = -(a/k) sinh(kz1) coth(kd) I then verified this with Maple by entering these things as AA and BB and then subtracting. So, we now know our 1D Green's function which meets the parallel plates BC's: g(z; z1) = Asinh(kz) 0 ≤ z ≤ z1 g(z; z1) = B[ sinh(kz) - tanh(kd)cosh(kz) ] z1 ≤ z ≤ d g(z; z1) = - (a/k)sinh(kz) sinh[k(z1-d)] / sinh(kd) 0 ≤ z ≤ z1 g(z; z1) = -(a/k) [ sinh(kz) - tanh(kd)cosh(kz)] sinh(kz1) coth(kd) z1 ≤ z ≤ d We can simplify the last expression a little: -(a/k) [ sinh(kz) - tanh(kd)cosh(kz)] sinh(kz1) coth(kd) = -(a/k) [ sinh(kz) - sinh(kd)cosh(kz)/cosh(kd)] sinh(kz1) cosh(kd)/sinh(kd = -(a/k) [ sinh(kz)cosh(kd) - sinh(kd)cosh(kz)] sinh(kz1) /sinh(kd) = -(a/k) [ sinh[k(z-d)] sinh(kz1) /sinh(kd) Then write g(z; z1) = -(a/k) sinh[k(z1-d)] sinh(kz) / sinh(kd) 0 ≤ z ≤ z1 g(z; z1) = -(a/k) sinh[k(z-d)] sinh(kz1) /sinh(kd) z1 ≤ z ≤ d which is our familiar form g(z; z1) = -(a/k) sinh[k(z>-d)] sinh(kz<) /sinh(kd) z> = max(z,z1) We can confirm that: (1) both these g forms are linear combinations of sh and ch and solve our homo ODE (2) the first g = 0 on the lower plate, the second g=0 on the upper plate (c) Resumption of Main Flow; Statement of the Parallel Plate Cartesian Green's Function Now let's go back to our original problem where a = exp(iktr1)/π and k2 = kt2 and we have shown then that U(kt; z; r1) = - exp(iktr1) (1/πkt) sinh[kt(z>-d)] sinh(ktz<) /sinh(ktd) Now we can insert this back into our transverse Fourier transform to get u(r;r1) = ∫d2kt exp[-i ktr] U(kt; z; r1) = ∫d2kt exp[-i ktr] { - exp(iktr1) (1/πkt) sinh[kt(z>-d)] sinh(ktz<) /sinh(ktd) } = -(1/π) ∫d2kt (exp[-i kt(r-r1)] /kt) sinh[kt(z>-d)] sinh(ktz<) /sinh(ktd) This is the complete solution to this problem, it includes the effect of the point charge and of the induced charge. If we wanted a true Green's function, we divide by 4π and say G(r;r1) = – (2π)-2 ∫d2kt (exp[-i kt(r-r1)] /kt) sinh[kt(z>-d)] sinh(ktz<) /sinh(ktd) where z> = max(z,z1). I think it is more logical to change the sinh argument so it is positive, then G(r;r1) = (2π)-2 !Syntax Error, I!Syntax Error, I d2kt (exp[-i kt(r-r1)] /kt) sinh[kt(d -z>)] sinh(ktz<) /sinh(ktd) z> = max(z,z1) z< = min(z,z1) 0 < z1 < d (between the plates) ********* which is the Green's function for a unit positive point charge located at r1 = (x1, y1, z1) between parallel plates located at z=0 and z=d, when observed at point r = (x,y,z) , and where kt = (ktx, kty, 0) and we are in green Jackson cgs units. Note that the symmetry of G is manifest: swapping r ↔ r1 causes z ↔ z1 which switch has no effect on the z< and z> part of the solution (eg, max(z,z1) = max(z1,z)). It causes the phase to change sign which makes no difference. Note also that, as per the general Stak theorem, G is positive everywhere. Let's try a rewrite of the above. We replace the expo with cos, so we have exp[-i kt(r-r1)] → cos[kt(r-r1)] = cos[ktx(x-x1) + kty(y-y1)] = cos[ktx(x-x1)] cos[kty(y-y1)] – sin[ktx(x-x1)] sin[kty(y-y1)] But the sin sin term vanishes because, for example, sin[ktx(x-x1)] is odd in ktx but the integration range is even, and the rest of the integrand is even since kt = . So we drop the sin term and get G(r;r1) = (2π)-2!Syntax Error, I!Syntax Error, I d2kt(cos[ktx(x-x1)] cos[kty(y-y1)] /kt) sinh[kt(d -z>)] sinh(ktz<) /sinh(ktd) We can then fold those two integrals to get this final form: G(r;r1) = (1/π2)!Syntax Error, I!Syntax Error, Id2kt cos[ktx(x-x1)] cos[kty(y-y1)] sinh[kt(d -z>)] sinh(ktz<) /[ktsinh(ktd)] G(r;r1) = (1/π2)!Syntax Error, I!Syntax Error, Idkxdky cos[kx(x-x1)] cos[ky(y-y1)] sinh[κz(d -z>)]sinh(κz z<)/[ κzsinh(κzd)] z> = max(z,z1) z< = min(z,z1) 0 < z1 < d (between the plates) ********* where κz = . Here we see one of our standard Cartesian atomic forms in action! 1 [sin(kxx), cos(kxx)], [sin(kyy), cos(kyy)], [exp(κzz), exp(-κzz) ] kz = imaginary = iκz so there is nothing new under the sun regarding atoms! (d) Transverse Fourier Expansion for 1/R. If we want, we can get the potential due just to the induced charge by subtracting out (from u) 1/R = (1/2π2)∫d3k exp[ ik(r- r1) ] / k2 but we want to get this into a 2D Fourier form, 1/R = (1/2π2)∫d2kt dkz exp[ ikt(r-r1) exp[ ikz(z-z1) ] / (kt2 + kz2) = (1/2π2)∫d2kt exp[ ikt(r-r1) !Syntax Error, I dkz exp[ ikz(z-z1) ] / (kt2 + kz2) = (1/2π2)∫d2kt exp[ ikt(r-r1) !Syntax Error, I dkz cos[ kz(z-z1) ] / (kt2 + kz2) = (1/π2)∫d2kt exp[ ikt(r-r1) !Syntax Error, I dkz cos[ kz(z-z1) ] / (kt2 + kz2) We have from (GR7 p 435) so let β = kt and a = |z-z1| and our integral is then !Syntax Error, I dkz cos[ kz(z-z1) ] / (kt2 + kz2) = Ko[kt |z-z1|] // modified Bessel function So we then have 1/R = (1/π2)∫d2kt exp[ ikt(r-r1)] Ko[kt |z-z1|] This is no doubt correct, but I think there is a simpler result available. We had earlier that = (1/2π2)∫d2kt exp[ ikt(r-r1) !Syntax Error, I dkz exp[ ikz(z-z1) ] / (kt2 + kz2) Suppose we treat the rightmost integral as a contour integral, where we have poles at ±ikt in the kz plane. Maybe write it like this: I = !Syntax Error, I dw exp[ iw(z-z1) ] / (kt2 + w2) = !Syntax Error, I dw exp[ iw(z-z1) ] / [(w-ikt) (w+ikt)] If z>z1, we close the contour up top since exp[ i(i|w|(z-z1) ] = exp[ -|w|(z-z1) ] → 0. We then pick up the residue of the pole at w = +ikt and we get I = (2πi) exp[ i(ikt)(z-z1)]/ (2ikt) With the other sign choice, we close bottom and get (leading minus since wrong circulation sense) I = - (2πi) exp[ i(-ikt)(z-z1)]/ (-2ikt) = (2πi)-1 exp[ i(ikt)(z1-z)]/ (2ikt) We combine these results then to have I = (2πi) exp[ i(ikt)|z-z1|]/ (2ikt) = (2π) exp(-kt|z-z1|)/(2kt) Then we find that 1/R = (1/2π2)∫d2kt exp[ ikt(r-r1) !Syntax Error, I dkz exp[ ikz(z-z1) ] / (kt2 + kz2) = (1/2π2)∫d2kt exp[ ikt(r-r1) I = (1/2π2)∫d2kt exp[ ikt(r-r1) {(2π) exp(-kt|z-z1|)/(2kt)} = (1/π) ∫d2kt exp[ ±ikt(r-r1) exp(-kt|z-z1|)/(2kt) and this result agrees with Rothwell and Cloud Appendix A A.55. I suspect I have gone down this same road many times in my prior life, perhaps with Kittel, perhaps with some wave stuff, but fine, here it is again. I show explicitly that either sign is OK in the phase exponent since 1/R is real. (e) Subtracting off the point charge potential So the a/R above is what we want to subtract off to get just the potential of the induced charge, uind(r;r1) = -(1/π) ∫d2kt (exp[i kt(r-r1) /kt) sinh[kt(z>-d)] sinh(ktz<) /sinh(ktd) - (1/π) ∫d2kt exp[ ikt(r-r1) exp(-kt|z-z1|)/(2kt) = - (1/π) ∫d2kt (exp[i kt(r-r1)]/(2kt)) { 2 sinh[kt(z>-d)] sinh(ktz<) /sinh(ktd) + exp(-kt|z-z1|) } = - (1/π) ∫ d2kt { 2 sinh[kt(z>-d)] sinh(ktz<) /sinh(ktd) + exp(-kt|z-z1|) } exp[i kt(r-r1)]/(2kt) Now recall the comment above that "true Green's would be G = u/(4π)" . So our true Green's function then is this (that is, G satisfies the ODE driven by a δ with no 4π ) Gind(r;r1) = -(2π)-2∫d2kt{2sinh[kt(z>-d)]sinh(ktz<) /sinh(ktd) + exp(-kt|z-z1|) } exp[i kt(r-r1)]/(2kt) where I show their solution for comparison. I think they have a typo in the exponential where they don't show r. Let's evaluate the {..} factor in the two cases: { 2 sinh[kt(z>-d)] sinh(ktz<) /sinh(ktd) + exp(-kt|z-z1|) } z > z1 = 2 sinh[kt(z-d)] sinh(ktz1) /sinh(ktd) + exp(-kt(z-z1)) = [ 2 sinh[kt(z-d)] sinh(ktz1) + sinh(ktd)exp(-kt(z-z1)) ]/ sinh(ktd) So make this agree with their result, I would have to show that 2 sinh[kt(z-d)] sinh(ktz1) + sinh(ktd)exp(-kt(z-z1)) = exp(-kt(d-z)) sinh(ktz1) + exp(-ktz) sinh(kt(d-z1)) Maple confirms that this is in fact an equality! Now, what happens with the other sign? { 2 sinh[kt(z>-d)] sinh(ktz<) /sinh(ktd) + exp(-kt|z-z1|) } z < z1 = 2 sinh[kt(z1-d)] sinh(ktz) /sinh(ktd) + exp(-kt(z1-z)) = [2 sinh[kt(z1-d)] sinh(ktz) + sinh(ktd)exp(-kt(z1-z))]/ sinh(ktd) To make this agree with their result, I would have to show that 2 sinh[kt(z1-d)] sinh(ktz) + sinh(ktd)exp(-kt(z1-z)) = exp(-kt(d-z)) sinh(ktz1) + exp(-ktz) sinh(kt(d-z1)) and Maple verifies this as well! ("question on Fourier 2") Therefore I have shown this fact: - [...] = {....} so our answers differ only inasmuch as they have omitted r in their expo! (f) Errata Search. Google books shows the 2001 edition having the error. (this book of course has page numbers). I look at Edward Rothwell's website and see that a 2nd edition is out, October 28, 2008. I don't have any download on this new book so cannot check errata. I sent him an email about this error, I hope to get some kind of response, probably that it was fixed in 2nd Ed. // I did in fact get an email response, and Mr. Rothwell confirmed the error, saying they found it when they wrote their book of solutions! _________________________________________________________________________ Hi Ed, For a very obscure reason, I was looking at your first edition Electromagnetics book and noted a small erratum that you have no doubt fixed in your 2nd edition. Since I don't have access to that 2nd edition, I don't know if it has been fixed or not, so I report it anyway. It involves a certain problem in the book, This result is exactly correct (by my painful calculation) except the exponential factor should be Without this correction, the Green's Function appears to be independent of r ! I looked a bit for errata on your two editions but could not find any online. I have not read much in your book but it looks very good to me! I was a Jackson student circa 1971 (green Jackson), and have recently been reading my two old volumes of Ivar Stakgold on BV problems which enlarge the playing field somewhat. Best regards, -Phil Lucht Salt Lake City _____________________________________________________________________ Hi Phil: You're absolutely right. We didn't discover that typo until we wrote a solution manual for the 2nd ed. We cleaned up a lot of typos and minor errors, but I'm sure there are still a few lurking in there. I took "physics" EM out of Jackson as well. My red copy has duct tape holding it together :-) The biggest difference between the second and first edition of our book is that we added a chapter on integral equations and numerical solutions. We also added a lot more problems and, as I mentioned, tried to clear up the errata. If you're not into integral equations, there's probably no reason to get the second edition. Thanks for letting me know about the error, and if you spot any more please pass them along. Take care, Ed _____________________________________________________________________