Question 4 Sec 8.8 tide error RESOLVED

Question 4 PhL 2.27.17 This resulted in a rewrite of the section in question within Section 8.8. It is all done and installed. I made a stupid error with the rotations as noted below due to lack of clarity on the Passive View concept. How do I clean up my claims around (8.8.44) which I think are wrong! I say : We have now redefined the Earth frame on the right to be Frame S (formerly it was Frame S') and we have drawn new x,y,z axes for this new Frame S so the z axis points away from mass M1. In Frame S then the angle θ is the usual spherical-coordinates polar angle. We now assume that the Earth rotates about some axis ' (new Frame S') which is obtained by rotating the axis by angles θ1 and φ1 as follows (see (E.2.2) for the matrix), ' = Rz(φ1) Ry(θ1) ≡ R1 = = = . // Frame S' components (8.8.44) I think this is correct. In more detail, I could say this (',',') = Rz(φ1)Rx(θ1)Rz(ψ1) (,,) which is exactly (G.5.15). One of these equations indeed then is ' = Rz(φ1) Ry(θ1) where the Rz(ψ1) has no effect on . Now given this rule for basis vectors being back-rotated, I know that for any Kinematic Vector like r, (r)' = R1-1 r In the Passive View, this gives the coordinates of the vector r in Frame S'. The corrected version of this is then (r)' = = (Frame S' coordinates of vector r) = R1-1 r R1 = Rz(φ1)Rx(θ1) = R1-1 = R1T = so then (r)' = = R1-1 r = new and r = = R1(r)' = new (8.8.45) Suppose we define spherical coordinates in both Frame S and Frame S' we then have = and = THEN we have = and from this I can write (I did this totally wrong, amazing! cosθ = cosθ1cosφ1sinθ'cosφ' - sinθ1sinθ'sinφ' + sinθ1cosφ1 cosθ' cos2θ = 2cos2θ - 1 . new (8.8.48) Comment: Here I am so much tempted to say that, since ' = Rz(φ1) Ry(θ1) ≡ R1 , I can apply this not just to the vector but to any vector r to get r' = R1r. But I just wrote Section 1.3 where I just got done saying that in the Passive View, the basis vectors back-rotate e'n = R-1en, then Kinematic Vectors stay put and must do (V)' = RV. I say this directly in (1.3.3). So if e'n = R1en , then (V)' = R1-1V . So I think my corrected results above are right. Let's now continue with the corrected equations: First we get : Recall the equation of the water surface from (8.8.33), r(θ) = R2 + a cos2θ . a > 0 ok (8.8.33) This implies a tide height of h(θ) = a cos2θ . ok (8.8.49) Therefore on our idealized Earth which rotates about an axis (θ1,φ1) relative to Fig (8.8.43) we obtain the following tide height during the day h(t) = a cos2θ(t) = a [ 2cos2θ(t) - 1 ] = a [ 2 (cosθ1cosφ1sinθ'cosωt - sinθ1sinθ'sinωt + sinθ1cosφ1 cosθ' )2 - 1 ] . new (8.8.50) Here θ' indicates the line of latitude at which our Observer is positioned. I then do Example 1: Example 1: (' = ) Suppose the Earth's rotation axis were in the direction in Fig (8.8.43) (pointing out of the plane of paper). In that case one has θ1= π/2 and φ1= π/2, since ' = Rz(π/2) Ry(π/2) = = = . (8.8.51) so far this seems right. But now continue with corrections Then from (8.8.50), h(t) = a [ 2(cosθ1cosφ1sinθ'cosωt - sinθ1sinθ'sinωt + sinθ1cosφ1 cosθ')2 - 1 ] = a [ 2( - sinθ'sinωt )2 - 1 ] = a [ 2( sinθ'sinωt )2 - 1 ] = a [ 2sin2θ'sin2ωt - 1 ] . // same answer as wrong cosθ gave!!! (8.8.52) OK. I will now go for a full rewrite of this frames doc section! It was a good catch! It means all my plots are wrong. The code for this section was in rotation matrices.mws and I will now make a new version of it with a 1 suffix. Got this done today 2.27.17.