~rprice/area51/documents/roswell.tex ROSWELL COORDINATES FOR TWO CENTERS As on April 16, 00 no changes since Dec 4. I. Definitions of coordinates We define the Roswell coordinates χ, Θ. A better name will be found? coordinate, and Θ. y r P Here χ is a radial-like θ r 1 θ 1 x In two dimensions, for two points on the Cartesian x axis, at distance a from the origin, the Roswell coordinates are defined by the diagram above and by: χ = r 1 r = {[ x a + y ] [ x + a + y ]} 1/4 1 Θ = 1 θ 1 + θ = 1 xy tan 1. x a y It is relatively straightforward to invert these relationships and find the Cartesian coordinates x, y in terms of χ, Θ: [ 1 ] x = a + χ cos Θ + a 4 + a χ cos Θ + χ 4 3 [ 1 ] y = a χ cos Θ + a 4 + a χ cos Θ + χ 4 4 There are two interesting limits to these expressions. When χ 0, and Θ < π/, then r a and χ a r 1 and Θ θ 1 /. A similar solution applies to Θ > π/, with 1 and interchanged. The second limit is for large χ. In this case, χ x + y and Θ tan 1 y/x. The Roswell coordinates therefore aside from a factor of here and there correspond to polar coordinates very close to the two point sources, and correspond to polar coordinates far from both points. 1
II. PDE in Roswell coordinates Let ϕ In cartesian coordinates, x = r cos ϕ, y = r sin ϕ, we have Our symmetry amounts to the replacement thus in cartesians our linear operator becomes: where LΨ = Ψ t = With a great deal of tedious algebra we find: ϕ = x y. 5 t = Ω ϕ, 6 + Ω x y. 7 Ψ = Ψ, χχ + Ψ,ΘΘ Θ + Ψ,χΘ Θ χ + ΨΘ Θ + Ψ χ χ, 8 χ χ + χ + and so forth. Note that both χ adn Θ are harmonic, so that χ = Θ = 0. We will leave these terms in now, to enhance clarity. In a similar manner we find that where Ψ ϕ = Āχ, θ Ψ + Bχ, θ Ψ + Cχ, θ Ā y + x xy B y + x xy C y + x Ψ + Dχ, θ Ψ Ψ + Ēχ, θ. 9 xy D y χ + χ x xy χ y x Ē y Θ + Θ x xy Θ y x + 10 11 1 13 14
with When these expressions are used we get: LΨ = A Ψ + B Ψ + C Ψ + D Ψ + E Ψ. 15 A = χ Ω Ā 16 B = Θ Ω B 17 C = χ Θ Ω C 18 D = χ Ω D 19 E = Θ Ω Ē. 0 With the use of Eqs. 1,, 3, 4, we have the coefficients A, B, C, D, E that appear above, as explicit functions of χ, Θ. II. Expansion of D problem Take our problem to be LΨ = Sources. 1 We can add a term like λψ 3 as an effective source. For more general nonlinearities, there will be depdences on Ψ and on its derivatives in the coefficients A, E. For now we will focus on the linear equation. We can use finite differencing on Eq. 1, along with Eqs. 15... 16. To use an expansion we write Ψχ, Θ and the sources as Ψχ, Θ = N and our equation becomes [ α n χ, Θ a n χ Here the coefficients are a n χe inθ. Sources = σ n χe inθ. + β n χ, Θ a ] nχ a nχ + γ n χ, Θ a n χ e inθ = σ n χe inθ 3 α n χ, Θ = A = χ Ω Ā 4 β n χ, Θ = inc + D = in χ Θ + χ Ω D Ω D + in C 5 γ n χ, Θ = ine n B = in Θ n Θ Ω inē n B 6 3
We now take a multipole of this equation, by multiplying by e imθ and integrating over Θ, then dividing by π. This eliminates the appearance of Θ in the equation, and leads to: where α mn a n χ α mn χ = 1 π β mn χ = 1 π + β mn a n χ + γ mn a n χ = σ m χ, 7 e mθ α n χ, Θe nθ dθ 8 e mθ β n χ, Θe nθ dθ 9 γ mn χ = 1 e mθ γ n χ, Θe nθ dθ 30 π We can now solve the problem in Eq. 7 30 by imposing a grid on the χ axis. This means we choose discrete values χ k and solve α mn a n χ k k + β mn a n χ k k + γ mn a n χ k = σ m χ k. 31 We have P values of χ k, with k = 1... P, and n has the N + 1 values n = 0, ±1, ±.... We therefore have P N + 1 unknowns a n χ k. We also have this many equations, since Eq. 31 is P equations for every value of m, and there are N + 1 values of m. Note: Of the P N + 1 unknowns and equantions, P equations and P unknowns are real, and NP equations and unknowns are complex. III. Warm up problem Let us start with the simplest possible problem: we will try to solve for the potential due to two equal point charges, in 3D. The idea is to interpret our x axis as the z axis of a cylindrical coordinate system, and our y coordinate is to be the ρ coordinate of cylindrical coordinates. With this interpretation, the laplace operator takes the form + + 1 y. 3 This means that we are looking for a solution of the equation: Ψ, χχ + Ψ,Θχ Θ χ + Ψ,ΘΘ Θ + Ψ,χ χ + 1 y + Ψ, Θ Θ + 1 y = 0. 33 This can be simplified somewhat. Since Θ is harmonic, and the level surfaces of Θ and χ are orthogonal, it follows that Θ = 0 and Θ χ = 0. Note that χ is not harmonic; it is logχ that is harmonic. With these simplifications, our equation becomes Ψ, χχ + Ψ,ΘΘ Θ + Ψ,χ 4 χ + 1 y 1 + Ψ, Θ y = 0. 34
The particular partial derivatives needed are listed in the Appendix. Of course we know the solution we want: It is 1 Ψ exact = x a + y + 1 x + a + y. 35 In terms of the notation of Fig. 1, this means 1/r 1 + 1/r. From this we infer the boundary conditions we need. At r 1 0 our condition is Ψ 1 + 1 a r 1, and hence Ψ 1 + a. At large χ we have Ψ. a χ χ III. Outgoing wave condition In standard polar coordinates, the outgoing condition is The extra ϕ derivative serves to kill the monopole term. We must express this condition in Roswell coordinates using r = Ψ,rϕ = ΩΨ,ϕϕ. 36 r + r. With such relations, the outgoing condition in Eq. 36 becomes Appendix I Needed expressions for the warm up 3D problem: Q a 4 + a χ cosθ + χ 4 37 1 y = a + Q χ 3 38 1 Q a y = a + χ cos Θ + Q χ 6 sin Θ 39 Θ = Q χ 4 40 = Q χ 41 χ = Q χ 3 4 5
These expressions have been checked by evaluating Eq. 34 for Ψ = 1/r 1 + 1/r and Ψ = 1/ x + y, and showing that both are zero. Additional checks have shown that logχ and log x + y vanish for the D Laplacian in Roswell coordinates. The D Laplacian is the expression in Eq. 34 with the 1/y terms missing. Appendix II Needed expressions for D waves: C y Ā y B y + x + x xy + x xy xy + = a4 sin Θ. 43 χ = a cos Θ + χ 44 χ 4 = a sin Θ a cos Θ + χ 45 χ 3 D y χ + x χ xy χ y x = a a + 3 a cos Θ + χ cos Θ χ 3 46 Ē y Θ + x Θ xy Θ y x = a a cos Θ + χ sin Θ χ 4 47 6