the search for T

The Search for T PhL 1.12.17 Question: How can I get an estimate of T as I did in the angular world in (F.6.29)? There I had three equations in θ,φ and I used the first radial one to estimate T. In Cartesians I have this = - [(ω2b3/r'13) + (T/m1r1)]x = - [(ω2b3/r'13) + (T/m1r1) - ω2]y + 2ω = - [(ω2b3/r'13) - ω2] (b+z) - (T/m1r1)z - 2ω x2+y2+z2 = r12 (F.8.10) From (F.6.25) I know that (b/r'1)3 ≈ 1 - 3(r1/b)cosθ ≈ [1 - 3z/b] so equations become = - [(ω2[1 - 3z/b]) + (T/m1r1)]x = - [(ω2[1 - 3z/b]) + (T/m1r1) - ω2]y + 2ω = - [(ω2[1 - 3z/b]) - ω2] (b+z) - (T/m1r1)z - 2ω First equation says ( I am looking for a result T = 3m1 ω2r1cos2θ = (3m1 ω2/r1)r12cos2θ = (3m1 ω2/r1)z2 m1r1 = - m1r1[(ω2[1 - 3z/b]) + Tx m1r1 = - m1r1ω2 + m1r1ω23(z/b) + Tx I don't see my result coming out of this equation! I do know that x2+y2+z2 = r12 x +y +z = 0 x + 2 + y +2 + z + 2 = 0 but that does not seem to help much. Suppose I drop the z/b terms in all three equations above: = - [ω2 + (T/m1r1)]x = - [ω2 + (T/m1r1) - ω2]y + 2ω = - (T/m1r1)z - 2ω again unclear what I can do here. Go back again to = - [(ω2b3/r'13) + (T/m1r1)]x = - [(ω2b3/r'13) + (T/m1r1) - ω2]y + 2ω = - [(ω2b3/r'13) - ω2] (b+z) - (T/m1r1)z - 2ω x2+y2+z2 = r12 (F.8.10) What happens if I eliminate the (ω2b3/r'13) terms instead of eliminate T? 1 = - (A + B)x 2 = - (A + B - ω2)y + 2ω 3 = - (A - ω2) (b+z) - Bz - 2ω 4 x2+y2+z2 = r12 (F.8.12) When I try to eliminate A from 1 and 2, B also goes away and that gives what I already know y - x = -ω2xy - 2ωx (F.8.13) Let's now try to eliminate A from 1 and 3 1*(b+z): (b+z) = - (A + B)x(b+z) 3*x: x = - (A - ω2) (b+z)x - Bzx - 2ω x = -A (b+z)x + ω2 (b+z)x - Bzx - 2ω x Now subtract to get (b+z) - x = - (A + B)x(b+z) +A (b+z)x - ω2 (b+z)x + Bzx + 2ω x = -Ax(b+z) - Bx(b+z) +A (b+z)x - ω2 (b+z)x + Bzx + 2ω x = - Bx(b+z) - ω2 (b+z)x + Bzx + 2ω x = - Bxb - Bxz - ω2 (b+z)x + Bzx + 2ω x = - Bxb - ω2 (b+z)x + 2ω x and so B manages to survive. I then have (b+z) - x = - Bxb - ω2 (b+z)x + 2ω x or b + z - x = - Bxb - ω2 (b+z)x + 2ω x (*) But I also know that z - x = Abx - ω2(b+z)x + 2ωx so equation (*) becomes Abx - ω2(b+z)x + 2ωx + b = - Bxb - ω2 (b+z)x + 2ω x or Abx + b = - Bxb or Ax + = - Bx or = - (A+B)x around in a circle of course. So unlike in spherical coordinates, there seems to be no easy want to extract an approximate expression for T because it seems you need the radial equation to do that! What happens if I just insert my known result for T into these equations? = - [(ω2b3/r'13) + (T/m1r1)]x = - [(ω2b3/r'13) + (T/m1r1) - ω2]y + 2ω = - [(ω2b3/r'13) - ω2] (b+z) - (T/m1r1)z - 2ω x2+y2+z2 = r12 (F.8.10) T = 3m1 ω2r1cos2θ = (3m1 ω2/r1)r12cos2θ = (3m1 ω2/r1)z2 dimT = M T-2L = ma correct (T/m1r1) = 3ω2(z/r1)2 T-2 = T-2 Go back instead to = - [(ω2[1 - 3z/b]) + (T/m1r1)]x = - [(ω2[1 - 3z/b]) + (T/m1r1) - ω2]y + 2ω = - [(ω2[1 - 3z/b]) - ω2] (b+z) - (T/m1r1)z - 2ω and insert (T/m1r1) = 3ω2(z/r1)2to get = - [(ω2[1 - 3z/b]) +3ω2(z/r1)2]x = - [(ω2[1 - 3z/b]) + 3ω2(z/r1)2 - ω2]y + 2ω = - [(ω2[1 - 3z/b]) - ω2] (b+z) - 3ω2(z/r1)2z - 2ω Now drop the z/b terms where appropriate = - [(ω2[1]) +3ω2(z/r1)2]x = - [ - 3ω2z/b + 3ω2(z/r1)2]y + 2ω = - [ - 3ω2z/b] (b+z) - 3ω2(z/r1)2z - 2ω or = - [(ω2 +3ω2(z/r1)2]x = - [ - 3ω2z/b + 3ω2(z/r1)2]y + 2ω = - [ - 3ω2z/b] b- 3ω2(z/r1)2z - 2ω or = - ω2[(1 +3(z/r1)2]x = - 3ω2[ -z/b + (z/r1)2]y + 2ω = 3ω2z - 3ω2(z/r1)2z - 2ω or = - ω2[(1+3(z/r1)2]x = - 3ω2(z/r1)2y + 2ω = 3ω2z[1 - (z/r1)2] - 2ω We know that x + 2 + y +2 + z + 2 = 0 but if velocities are small we can say x + y + z ≈ 0 Insert now from above (with velocities ignored) to check this result x + y + z = x{- ω2[(1+3(z/r1)2]x} + y{ - 3ω2(z/r1)2y} + z{3ω2z[1 - (z/r1)2]} = - ω2[(1+3(z/r1)2]x2 - 3ω2(z/r1)2y2 + 3ω2[1 - (z/r1)2] z2 = -ω2x2 - 3ω2(z/r1)2x2 - 3ω2(z/r1)2y2 + 3ω2z2 - 3ω2 (z/r1)2z2 = -ω2x2 + 3ω2z2 - 3ω2(z/r1)2[ x2+y2+z2] = -ω2x2 + 3ω2z2 - 3ω2(z/r1)2[r12] = -ω2x2 + 3ω2z2 - 3ω2z2 = -ω2x2 // same result I got last time. so it almost works! Very unclear. x2+y2+z2 = r12 x +y +z = 0 x + 2 + y +2 + z + 2 = 0