Show Cart EOM give Ang EOM

Show that Cartesian EOM give Angular EOM PhL 1.13.17 In Section F.5 I use the torque equations to obtain angular equations of motion as follows + sinθcosθ(3ω2 - 2) + ωsinθcosφ (ω cosθ cosφ - 2sinθ ) = 0 // + 2 cotθ – ωcosφ (ωsinφ -2) = 0 // (*) (F.5.7) In Section F.6 I use force equations to verify each of the above equations. In Section F.8 I derive two EOM directly in Cartesian coordinates, they are 1 y - x = - ω2xy - 2ωx // when converted to θ,φ this matches (F.5.7) 2 z - x = (ω2b3/r'13)bx - ω2(b+z)x + 2ωx where r'12 = r12 + b2 + 2bz (**) In Maple "show cart EOM equals ang EOM.mws" I show the first equation of (**) replicates the second equation of (*), giving then a triple derivation of this equation. I think that the first equation of (*) is a certain linear combination of the two equations in (**). When I have Maple apply this linear combination, I obtain the first equation of (*) BUT the red term is missing. This term is part of the "constant" in that equation. I spent about 2 hours yesterday trying to find the error but failed, so today I will resume. One bedtime idea is to write all equations assuming velocities are all 0, which should bring a focus on the constant terms! So start with (F.8.1) and examine only constant terms: Feff,1 = m1 a1 (F.6.5) (F.8.1) ≈ - (GMEm1/r'13)r'1 - T 1 – m1S' – m1ω x (ω x r1) – 2m1 ω x v1 – m1 x r1 . 1 2 3 4 5 6 Term 1: - (GMEm1/r'13)r'1 = - (GMEm1/r'13)(b + r1) = - (GMEm1/r'13) [ b + x + y + z ] = - (GMEm1/r'13) [ x + y + (b+z) ] (F.8.3) Term 2: - T 1 = -(T/r1)r1 = -(T/r1) [ x + y + z] (F.8.4) Term 3: – m1S' = – m1 x b - m1 ω x (ω x b) // (F.6.3) = - m1ω x (ω x b) // satellite in circular orbit, = 0 = - m1(ωb)ω + m1ω2b //- A x (A x C) = -(AC)A + A2C = m1ω2b = m1ω2b (F.8.5) Term 4: – m1ω x (ω x r1) = -m1(ωr1)ω + m1ω2r1 // identity shown above = -m1ω2(r1) + m1ω2r1 = -m1ω2(x) + m1ω2 [ x + y + z] = m1ω2 [ y + z] (F.8.6) Terms 5 and 6 make no contribution. So unfortunately MOST terms contribute to the constant part. I now assemble the equations Feff,1 = m1 a (F.8.1) ≈ - (GMEm1/r'13)r'1 - T 1 – m1S' – m1ω x (ω x r1) – 2m1 ω x v1 – m1 x r1 1 2 3 4 5 6 : m1 ax = - (GMEm1/r'13) x - (T/r1)x ok 1 2 : m1 ay = - (GMEm1/r'13) y - (T/r1)y + m1ω2y + 2m1ωvz ok 1 2 4 : m1 az = - (GMEm1/r'13) (b+z) - (T/r1)z + m1ω2b + m1ω2z - 2m1ωvy ok (F.8.9) 1 2 3 4 I then rewrite the three equations : = - [(ω2b3/r'13) + (T/m1r1)]x ok = - [(ω2b3/r'13) + (T/m1r1) - ω2]y + 2ωvz ok = - [(ω2b3/r'13) - ω2](b+z) - (T/m1r1)z - 2ωvy ok I then define A ≡ (ω2b3/r'13) B ≡ (T/m1r1) // rescaled tension and equations become 1 = - (A + B)x ok 2 = - (A + B - ω2)y + 2ωvz ok 3 = - (A - ω2)(b+z) - Bz - 2ωvy ok I eliminate B from the first two equations and end up with y - x = -ω2xy - 2ωzvz When this equation is "converted to θ,φ" I get exactly the second equation of (F.5.7) so I don't have doubts about the above elimination. I then eliminate B from the first and third equations 1*z z = - (A + B)xz ok 3*x x = - (A - ω2)(b+z)x - Bxz - 2ωxvy ok = -Abx -Azx +ω2(b+z)x - Bxz- 2ωxvy ok = - (A + B)xz - Abx +ω2(b+z)x- 2ωxvy ok Subtract so that the -(A + B)xy terms cancel, z - x = Abx - ω2(b+z)x + 2ωxvy (F.8.14) At this point then the two Cartesian equations are 1 y - x = - ω2xy // when converted to θ,φ this matches (F.5.7) 2 z - x = (ω2b3/r'13)bx - ω2(b+z)x + 2ωxvy A supporting equation is this r'12 = (r1+b)2 = r12 + b2 + 2 r1 b = r12 + b2 + 2 r1 [b] = r12 + b2 + 2bz . ok (F.8.11) I then do the far approximation for the first term in 2 : r'12 = r12 + b2 + 2bz = b2 [ 1 + (r1/b)2 + 2(z/b)] ≈ b2[ 1 + 2(z/b) ] ok r'1-2 ≈ b-2 [1-2(z/b)] ok r'1-3 ≈ b-3 [1-2(z/b)]3/2 ≈ b-3 [ 1 +(3/2)(-2(z/b)] = b-3 [ 1 - 3(z/b)] ok so (b/r'1)3 = 1 - 3(z/b) ok Equation 2 above then becomes z - x = (ω2b3/r'13)bx - ω2(b+z)x+ 2ωxvy ok = ω2[1 - 3(z/b) ]bx - ω2(b+z)x+ 2ωxvy ok = ω2bx - 3(z/b)ω2bx - ω2bx - ω2zx+ 2ωxvy ok = - 3(z/b)ω2bx - ω2zx+ 2ωxvy ok = - 3ω2xz - ω2zx+ 2ωxvy ok = - 4ω2xz+ 2ωxvy ok My two equations in the far approximation are then 2 y - x = - ω2xy - 2ωzvz // when converted to θ,φ this matches (F.5.7) 1 z - x = - 4ω2xz + 2ωxvy // renumber equations here! I now see that I must include the velocity terms after all because they will contribute when I convert to the θ and φ coordinates! These come only from Term 5. Term 5: -2m1 ω x v1 = -2m1 [ω] x ( vx + vy + vz) = -2m1ω [ vy - vz] (F.8.7) so these add terms to the z and y equations above. Go install these now in blue. Done. I now go to "show cart EOM....mws" and I enter equations which will do the conversion of these EOM's from Cart to Spher coordinates: I then use these objects in my entry of the two equations of interest. Here is how I process equation 2 2 y - x = - ω2xy - 2ωzvz First you see it properly entered as eq2. I divide by r12 , make a subs I want to compare this equation to + 2 cotθ – ωcosφ (ωsinφ -2) = 0 For comparison, multiply the above equation by -sin2θ -sin2θ - 2 cosθsinθ + sin2θωcosφ (ωsinφ -2) = 0 or -sin2θ - 2 cosθsinθ + ω2sin2θcosφ sinφ -sin2θωcosφ 2 = 0 Now reorder the terms - 2 cosθsinθ -sin2θ + ω2sin2θcosφ sinφ - 2ωsin2θcosφ = 0 This agrees exactly with the last blue Maple equation above, so I conclude that the Cart equation 2 is exactly the same as the angular equation 2. Now look at eq1. (1) You can see that I have properly entered my equation 1 which is this 1 z - x = - 4ω2xz + 2ωxvy (2) Here is the equation I am aiming for, taken from (*) above, + sinθcosθ(3ω2 - 2) + ωsinθcosφ (ω cosθ cosφ - 2sinθ ) = 0 (3) This target equation (3) has no term, but Maple (*) for eq1 does have such a term. This tells me that in order to get (3), I will have to do a linear combination of Maple eq2 and eq1 so as to cancel the term. So at this point in Maple we have eq1 := -2r12cos2θ sinφ + stuff eq2: = -2r12sinθcosθ + stuff Consider now eq1 * sinθ = -2r12sinθ cos2θ sinφ + stuff eq2 * cosθsinφ = -2r12sinθcos2θ sinφ + stuff Therefore this should cancel if I make this definition eq3 := sinθ * lhs(eq1) - cosθsinφ * lhs(eq2) I will now try this linear combination in Maple : Looking through the terms we see that there is no term, so the desired cancellation has been achieved. A guiding term here is sinθ cosφ . Now we need to simplify getting rid of cos3(θ): Next divide out the leading factor sinθcosφ and then do another subs: I will now manually translate the final result of 6 terms -sinθcosθ 2 + + 4ω2sinθcosθ - 2ωcosφ + 2ωcos2θ cosφ - ω2sinθcosθsin2φ and I now reorder these terms - sinθcosθ 2 + 4ω2sinθcosθ - ω2sinθcosθsin2φ +2ωcosφ (cos2θ - 1) or - sinθcosθ 2 + 4ω2sinθcosθ - ω2sinθcosθsin2φ - 2ωcosφsin2θ A = Maple Now the target here is this from (*) above + sinθcosθ(3ω2 - 2) + ωsinθcosφ (ω cosθ cosφ - 2sinθ ) = 0 or - sinθcosθ 2 + 3ω2sinθcosθ + ω2sinθcosθcos2φ - 2ωsin2θcosφ B = target ok ok ok If I can show that the "constant terms" are the same, then the game is won! 4ω2sinθcosθ - ω2sinθcosθsin2φ = 3ω2sinθcosθ + ω2sinθcosθcos2φ ? 4ω2cosθ - ω2cosθsin2φ = 3ω2cosθ + ω2cosθcos2φ ? 4ω2cosθ - ω2cosθ(1-cos2φ) = 3ω2cosθ + ω2cosθcos2φ ? 4ω2cosθ - ω2cosθ + ω2cosθcos2φ = 3ω2cosθ + ω2cosθcos2φ ? 3ω2cosθ + ω2cosθcos2φ = 3ω2cosθ + ω2cosθcos2φ yes !!!!!!! So finally it works, hurray hurray !!!!!! Now how can I do all this more efficiently! here again is the target equation: + sinθcosθ(3ω2 - 2) + ωsinθcosφ (ω cosθ cosφ - 2sinθ ) = 0 + 3ω2sinθcosθ -sinθcosθ 2 + ω2sinθcosθcos2φ - 2ωsin2θcosφ - sinθcosθ 2 + [3ω2sinθcosθ + ω2sinθcosθcos2φ ] - 2ωsin2θcosφ - sinθcosθ 2 + ω2sinθcosθ[3 + cos2φ ] - 2ωsin2θcosφ - sinθcosθ 2 - 2ωsin2θcosφ + ω2sinθcosθ(3 + cos2φ) - sinθcosθ 2 +ωsinθ( -2sinθcosφ + ωcosθ(3+cos2φ)) Nothing is very elegant.