ring source

1 Ring source in Toroidal Coordinates PhL 12.31.13 1. Ring source in cylindrical coordinates In a potential theory problem (Laplace equation) an d in electrostatics language, a ring source of total charge q in cylindrical coordinates with radius ρc and at z = 0 would be written as source = q δ (z) δ(ρ-ρc)/[2πρ] . The reason for the 2 πρ is that we need the 3D integral of the source to be q: ∫dφ ∫ρdρ ∫dz q δ(z) δ(ρ-ρc)/[2πρ] = 2π ∫ρdρ ∫dz q δ(z) δ(ρ-ρc)/[2πρ] = q ∫dρ ∫dz δ(z) δ(ρ-ρc) = q The problem then is to express this ri ng source in toroidal coordinates. 2. Digression on the equations for th e toroidal coordinate level curves (I have derived these equations elsewhere and they probably appear somewhere in M&F or elsewhere. ) horizontal circles (tori cross sections, let ρ = horizontal coordinate, z = vertical coordinate) (ρ - acothξ)2 + z2 = a2/sh2ξ ρc = acothξ R = a/|sh ξ| ( Historically (200BC), the horizontal circles are called the "circles of Appollonius" and they have some very simple properties, notably, | r - a | / | r + a | = e-ξ. ) vertical circles (when truncated at z=0, they give the bowls) ρ2 + (z - acotu)2 = a2/sin2u z c = a cot(u) R = a/|sinu| The picture going with these equations showing the various circles is this 2 Remember that you rotate this bipolar 2D system around the vertical z axis by φ to get toroidals. 3. Geometry associated with the ring source Now consider the following picture, where on the right is one of our "horizontal circles" being a cross section of a gray toroid, We know at once from the previous section that the center of the circle is at ρc = a coth( ξ0). We also know that R = a/|sh ξ0| . Then ρc2 - R2 = a2 coth2(ξ0) - a2/(shξ0)2 = a2[coth2(ξ0) - csch2(ξ0)] = a2 ρc/R = a coth( ξ0)* shξ0/a = ch(ξ0) / / 0 < ξ0 < ∞ in normal usage We have now derived this set of equations which appeared in my earlier email and which appear in the bowl paper, ρ c = a cothξ0 R = a /sh ξ0 => ρc2 - R2 = a , ρc/R = chξ0 (10.9) 3 Also, we can get an expression for cylindrical ρ expressed in toroidals as follows (these x,y,z equations are those which in fact define toroidal coordinates ( ξ,u,φ) ) x = a cos φ shξ/(chξ - cosu) = cosφ hφ y = a sin φ shξ/(chξ - cosu) = sin φ hφ z = a sinu/(ch ξ - cosu) = sinu h ξ we find that ρ 2 = x2+y2 = a2sh2ξ / (chξ - cosu)2 => ρ = a shξ/(chξ - cosu) . The inverse equations are given by ξ = tanh -1[2aρ/(a2+ ρ2+ z2)] tan(u) = [ -2az/(a2-ρ2-z2)] // note that z = 0 => u = 0 as one possibility tanφ = y/x However, one really needs to execute the following algorithm to find compute ξ ,u,φ from x,y,z : 1. Compute φ from the last equation above, in range (0,2 π) say. Then sin φ and cosφ are known. 2. Compute ξ from the unambiguous formula shown above involving tanh -1. 3. Compute cosu from the " x = " formula above. 3. Compute sinu from the "z = " formula above. 4. Then compute u manually from these values. The scale factors are given by h ξ = hu = a/[ chξ - cosu] h φ = a shξ/[ chξ - cosu] = sh ξ hμ hξ hu hφ = shξ hξ3 = a3 shξ /[ chξ - cosu]3 which will be used below. Note: M&F deal with this stuff on page 1210. Instead of using our angle u, they use an angle θ. Here is the connection between their θ and our u: The relationship between u and θ is this: u = (θ+π) mod (2π) 4 or u = θ+π for θ in (0,π ) u = θ-π for θ in (π,2π) or u = θ + π sign(π-θ) For any θ in (0,2π) we know the following to be true: sinθ = - sinu cosθ = - cosu tanθ = + tanu . 4. Jacobians and final exp ression for the ring source We know in spherical coordinates that the Jac obian tells how volume elements transform, dxdydz = J(r,θ ,φ)drdθdφ = r 2sinθ drdθdφ J(r, θ,φ) = r2sinθ . Here J is the Jacobian for these curvilinear coordinates and for orthogonal coordinates ξ 1,ξ2,ξ3 we have J(ξ 1,ξ2,ξ3) = h1h2h3 . // product of the scale factors In cylindrical coordinates we have dxdydz = J( ρ,z,φ)dρdzdφ = ρdρdzdφ J ( ρ,z,φ) = ρ In toroidal coordinates we have instead dxdydz = J( ξ,u,φ)dξdudφ J( ξ,u,φ) = h 1 h2 h3 = shξ h13 = shξ a3 / [chξ - cosu]3 Thus, dxdydz = a 3 shξ (chξ - cosu)-3 dξdudφ . We can then get this connection between cylindrical and toroidal coordinates using dxdydz = dxdydz" ρdρdzdφ = a3 shξ (chξ - cosu)-3 dξdudφ . // note that dimensionally correct But above we had ρ = a shξ (chξ - cosu) -1 so then a shξ (chξ - cosu)-1dρdzdφ = a3 shξ (chξ - cosu)-3 dξdudφ 5 or dρdzdφ = a2 (chξ - cosu)-2 dξdudφ or dρdz = a 2 (chξ - cosu)-2 dξdu . Since delta functions work as the inverse of volume elements, we then know that δ(ρ-ρ1)δ(z-z1) = a-2(chξ - cosu)2 δ(ξ-ξ1)δ(u-u1) where ξ 1 and u1 are the toroidal coordinates matching ρ1 and z1 in cylindrical coordinates. We are more interested in [ from above we know that 1/ ρ = (chξ - cosu)/(ash ξ) ] δ(z-z1) δ(ρ-ρ1)/ [2πρ] = a-2(chξ - cosu)2 δ(ξ-ξ1)δ(u-u1) * (chξ - cosu)/(2 πashξ) = ( 1 / 2 π) a -3(chξ - cosu)3 δ(ξ-ξ1)δ(u-u1) /shξ For the desired ring source at the cross hairs of the above drawing, we have ρ 1 = ρc // ρc = a cothξ0 z1 = 0 u 1 = 0 // the ring source lies on the "iris" limit of the bowl ξ1 = 2ξ0 . To verify this last equation fro m our inversion formula above, ξ1 = tanh-1[2aρ/(a2+ ρ12+ z12)] = tanh-1[2aρc/(a2+ ρc2)] = t a n h -1[2a a coth ξ0/(a2+ a2 coth2ξ0)] = tanh-1[2 cothξ0/(1 + coth2ξ0)] = t a n h -1[2 /(tanhξ0 + cothξ0)] = tanh-1[2 tanhξ0 /(tanh2ξ0 +1)] = t a n h -1[ tanh(2ξ0)] = 2 ξ 0 . Is this really true? To check, compute ρ 1 using formula above: ρ 1 = a shξ1/(chξ1 - cosu1) = a sh(2 ξ0)/( ch(2ξ0) - 1) = a 2 sh(ξ 0)ch(ξ0) / [2sh2(ξ0)] = a ch( ξ0)/ sh(ξ0) = a coth( ξ0) which is correct. Then we have shown that 6 δ(z-z1) δ(ρ-ρ1)/ [2πρ] = (1/2 π) a-3(chξ - cosu)3 δ(ξ-ξ1)δ(u-u1) /shξ = ( 1 / 2 π) a -3(chξ1 - cosu1)3 δ(ξ-ξ1)δ(u-u1) /shξ1 = ( 1 / 2 π) a-3(chξ1 - 1)3 δ(ξ-ξ1)δ(u) /shξ1 / / u 1 = 0 = ( 1 / 2 π) a-3(2sh2(ξ1/2))3 δ(ξ-ξ1)δ(u) /[2sh( ξ1/2)ch(ξ1/2)] = ( 1 / 2 π) a -3(8sh6(ξ1/2)) δ(ξ-ξ1)δ(u) /[2sh(ξ 1/2)ch(ξ1/2)] = ( 1 / 2 π) a -3(4sh5(ξ1/2)) δ(ξ-ξ1)δ(u) /[ch(ξ1/2)] = ( 1 / 2 π) a -3(4sh5(ξ0)) δ(ξ-2ξ0)δ(u) /[ch(ξ0)] = ( 1 / 2 π) a -3 4sh4(ξ0) th(ξ0) δ(ξ-2ξ0)δ(u) = ( 2 / π) a -3sh4(ξ0) th(ξ0) δ(ξ-2ξ0)δ(u) . Then the ring source of total charge q expr essed in toroidal coordinates is given by source = q δ (z) δ(ρ-ρc)/[2πρ] or source = q (2/ π) a-3sh4(ξ0) th(ξ0) δ(ξ-2ξ0) δ(u) I guess the main fact is that source = constant δ(ξ-2ξ 0) δ(u) where the details are only needed for normalization w ith respect to q. I am a little surprised at the 2 ξ 0 , maybe it is wrong. I originally thought it would just be δ(ξ-ξ0). The work above is just a "quick shot" and is not fully checked!!