second Mie scattering attempt1

Second Attempt at Mie Scattering Coefficients PhL 2.15.03 Attempt to set up the scattering problem in the N and M world, Version 2 Here is Krugel's setup of the problem: Ep = E [ M(1)o1 - i N(1)e1 ] E = i // incident plane wave Es = E [ - b M(3)o1 + i a N(3)e1 ] *** // incident plane wave, (3) means h(1) (kr) Ei = E [ c M(1)o1 - i d N(1)e1 ] // incident plane wave, (1) means j(kr) Now we write the H equations using our above 3 rules from going from E to B noted above: " Summary: To get B from E, for each term use the nature of, and -i times the coeff of, the cross term. " H'p = H [ - M(1)e1 - i N(1)o1 ] H = i H's = H [ a M(3)e1 + i b N(3)o1 ] H'i = H [ - d M(1)e1 - i c N(1)o1 ] Now use H' = H and = 1/ = v/c and 1/ = = n, the index. Hp = no H [ - M(1)e1 - i N(1)o1 ] H = i // OK only if = 1 Hs = no H [ a M(3)e1 + i b N(3)o1 ] Hi = ni H [ - d M(1)e1 - i c N(1)o1 ] / i = inside Now the boundary conditions are of the general form Xi = Xp + Xs evaluated at r=a, because both the plane wave and the scattered wave are on the "outside", while the internal is on the "inside". So we can write our boundary conditions as r x Ei = r x Ep + r x Es r x Hi = r x Hp + r x Hs In each expansion, we can then remove the sum and ignore the constant X overall factor and we end up with inside plane scattered r x [ c M(1)o1 - i d N(1)e1 ] = r x [ M(1)o1 - i N(1)e1 ] + r x [ - b M(3)o1 + i a N(3)e1 ] r x [- ni d M(1)e1 - i ni c N(1)o1 ] = r x [ - no M(1)e1 - no i N(1)o1 ] + r x [ no a M(3)e1 + i no b N(3)o1 ] Balance the M and N terms separately, and use these results derived elsewhere: r x M(1)o1 = f * - r2 o r x N(1)o1 = (rf)' / n * (1/k) r x o // k here on right is /c Here are the results of this balance, c ji = 1 jo - b ho - i d (rji)' /ni = - i (rjo)' /no + i a (rho)' /no - ni d ji = - no jo + no a ho - i ni c (rji)' /ni = -i no (rjo)' /no + i no b (rho)' /no First stage simpify, and use n = ni / no c ji = jo - b ho - d (rji)'= - n (rjo)' + n a (rho)' - nd ji = - jo + a ho - c (rji)'= - (rjo)' + b (rho)' Rewrite for Maple: c* ji = jo - b* ho - d* rjip = - n*rjop + n*a* rhop - n*d *ji = - jo + a* ho - c* rjip = -rjop + b* rhop Put these into Maple: The result is this: 2 -rjop ho + rhop jo ji rjop - jo rjip n ji rjop - jo rjip n (-rjop ho + rhop jo) {c = ------------------, b = ------------------, a = ---------------------, d = ----------------------} -rjip ho + rhop ji -rjip ho + rhop ji 2 2 -rjip ho + n rhop ji -rjip ho + n rhop ji As usual, this result is wrong, and is even more wrong than last time! The n2 factors are back! Apepndix: Hypothesis #1 again. Suppose I now omit the H to H' factors. Tha means to remove explicit n's from last two equations. Go back above and start with c ji = 1 jo - b ho - i d (rji)' /ni = - i (rjo)' /no + i a (rho)' /no - ni d ji = - no jo + no a ho - i ni c (rji)' /ni = -i no (rjo)' /no + i no b (rho)' /no OK Now remove those n factors from last 2 equations c ji = 1 jo - b ho - i d (rji)' /ni = - i (rjo)' /no + i a (rho)' /no - d ji = - jo + a ho - i c (rji)' /ni = -i (rjo)' /no + i b (rho)' /no And now simplify some more c ji = jo - b ho - d (rji)' = - n(rjo)' + a n (rho)' - d ji = - jo + a ho - c (rji)' = - n (rjo)' + b n (rho)' And our new Maple equation set is (first 2 are same) c* ji = jo - b* ho - d* rjip = - n*rjop + n*a* rhop - d *ji = - jo + a* ho - c* rjip = -n*rjop + n*b* rhop This gives Maple result, n (rjop ho - jo rhop) n (rjop ho - jo rhop) n ji rjop - jo rjip n ji rjop - jo rjip {d = - ---------------------, c = - ---------------------, b = -------------------, a = -------------------} %1 %1 %1 %1 %1 := -rjip ho + n rhop ji which of course is still wrong. So this hypothesis does not fix things!