appendix C fiddle

Idea PhL 3.26.05 Here I was just wondering if you could warp some or all of the grid lines in curvilinear x'-space to make all the angles be right. I did not finish it, does not seem at all useful. Appendix C framework: 'θ 'ρ = e'θ e'ρ / (h'θh'ρ) = eθ eρ / (h'θh'ρ) = 'θρ / (h'θh'ρ) = [b2-a2]ρ sin(θ)cos(θ)/ {ρ} = [b2-a2] sin(θ)cos(θ)/ {} = cosψ ψ(θ) = cos-1[[b2-a2] sin(θ)cos(θ)/ {}] Can I show that |'θ 'ρ| ≤ 1? | eθ eρ | < (h'θh'ρ) ? 'θρ < (h'θh'ρ) ('θρ)2< (h'θh'ρ)2 ('θρ)2 < 'θθ'ρρ det(g') > 0 ? Yes, we do know this from Section 5 (d). So this suggests a rather elaborate way to draw the θ-ρ picture. It is drawn in Cartesian space such that at every point, the angle between the two sets of lines is angle ψ(ρ,θ). The lines intersecting are lines of constant ρ and the lines of constant θ. These would be curves satisfying dρ/dθ = tan[ψ(θ)] Now secψ = / {[b2-a2] sin(θ)cos(θ)} sec2ψ = (a2sin2θ + b2cos2θ)( a2cos2θ + b2sin2θ) / {[b2-a2]2 sin2(θ)cos2(θ)} tan2ψ = sec2ψ - 1 = (a2sin2θ + b2cos2θ)( a2cos2θ + b2sin2θ) / {[b2-a2]2 sin2(θ)cos2(θ)} – {[b2-a2]2 sin2(θ)cos2(θ)} / {[b2-a2]2 sin2(θ)cos2(θ)} NUM = (a2sin2θ + b2cos2θ)( a2cos2θ + b2sin2θ) - [b2-a2]2 sin2(θ)cos2(θ) = a4s2c2+ a2b2c4 + a2b2s4 + b4s2c2 - b4s2c2 - a4s2c2 + 2a2b2s2c2 = [a4 +b4 - b4 -a4 + 2a2b2] s2c2 + a2b2c4+ a2b2s4 = 2a2b2 s2c2 + a2b2 (s4+c4) = a2b2 [s4+c4 + 2s2c2] = a2b2(s2+c2)2 = a2b2 Thus we find that tan2ψ = a2b2 / {[b2-a2]2 sin2(θ)cos2(θ)} Therefore, tanψ = ab / {|b2-a2|sin(θ)cos(θ)} = [ 2ab/|b2-a2| ] csc(2θ) much simpler than I thought. Our curves for the wavy vertical constant θ lines are then set by dρ/dθ = tan[ψ(θ)] = [ 2ab/|b2-a2| ] csc(2θ) dρ = [ 2ab/|b2-a2| ] csc(2θ)dθ ρ = [ 2ab/|b2-a2| ] { -(1/2)ln [csc(2θ) + cot(2θ)] + constant } //Maple At θ = π/4 we find that csc(2θ) + cot(2θ) = 1 so ρ(π/4) = [ 2ab/|b2-a2| ] { -(1/2)ln [1] + constant } = [ 2ab/|b2-a2| ] * constant At θ = 0+ we find that csc(2θ) + cot(2θ) = +∞ so ρ = -∞ + constant. I give up, this all makes no sense.