A Simple Theorem for Orthogonal Coordinate Systems
DOCX · 18.7 KB
Open DOCX file
A brief note by Phil (dated 1.22.12, updated 2.11.12) on orthogonal curvilinear coordinates. It states a theorem expressing Cartesian unit vectors in terms of the tangent base vectors and unit vectors with scale factors h'i, defining Q with Q Q^T = 1. The proof uses results from his tensor document, shows det Q = ±1 via the Jacobian, and notes a possible parity inversion.
AI-written summary; may contain errors. This description is approximate.
Extracted text (machine-read; may contain errors)
A Simple Theorem for Orthogonal Coordinate Systems PhL 1.22.12
updated 2.11.12
When I first did this, it seemed strange and complicated. Now it is all on 2/3 of a page.
Assumptions: x-space is Cartesian with unit vectors un, and x' are orthogonal coordinates. The usual tangent base vectors are en and the corresponding unit vectors are n , all as in tensor doc.
Theorem: un = Σi h'i-2 Rin ei
un = Σi h'i-1 Rin i = Σi Qin i
Qin ≡ h'i-1 Rin QQT= 1 and det(Q) = 1 Q = rotation
Proof:
(a) I know from tensor doc that in full generality
un = Ren = Σi Rin ei = Σi Rin h'i i and by the way ej un = Rjn
[ but I did not know this until recently and then I added it to Section 3 (b) of tensor doc ! ]
Then for g = 1 and orthogonal this becomes
= Σi Rin h'i i = Σi giiRin h'i i = h'i-1 Rini => Qin = h'i-1 Rin QED.
(b) Meanwhile then Qin = h'i-1 Rin = h'i Rin = H'ii' Ri'n so , using tensor doc facts,
detQ = detH' detR = (Πih'i) detR = (Πih'i) J-1
But (Πih'i)2 = det(') = J2 so J = ±(Πih'i) and thus detQ = ±1 and we really cannot rule out the - sign because Q could in principle be a rotation and a parity inversion together.
(c) Finally, since Q = H'R in this sense just shown above, we have Q-1 = R-1H'-1 = SH'-1 and writing this out gives Q-1in = Sin h'n-1 = Rni h'n-1 = Qni = QTin and therefore Q-1 = QT so QQT= 1.