old handedness section

old handedness section PhL 2.26.12 Today in tensor doc I replaced the old handedness section with a better version, and the old one is now parked here. (i) Handedness of the en and the sign of det(S). Handedness As noted in Section 3 (a), since transformation F is invertible, the tangent basis vectors en are linearly independent and form a basis in x-space. In general the en are non-orthogonal, and one might ask what it means for the set of basis vectors { e1 , e2.... eN } to comprise a "right handed coordinate system". For N=3, we shall define right and left-handed coordinate systems in this way, e2 x e3 e1 > 0 right handed system e2 x e3 e1 < 0 left handed system Section 5 (j) showed that |en| = ≡ h'n > 0 "scale factor n" One can then define unit-vector versions of the en n = en/h'n and then the above definition of right-handedness can equivalently be written, 2 x 3 1 > 0 If {1,2,3} were basis vectors for an orthogonal coordinate system with 2 x 3 = 1 (and cyclic), one would find 2 x 3 1 = 1 > 0, so the above definition agrees with the usual meaning of a right handed system in the orthogonal case. For a skewed system as shown in this picture, it will still be true that 2 x 3 1 > 0 but one will find 2 x 3 1 < 1. This picture illustrates the notion that an orthogonal system can be deformed into a non-orthogonal one in such a way as to remain right handed. Now returning to e2 x e3 e1 > 0, one can write the LHS as e2 x e3 e1 = [ e2 x e3]k (e1)k = εkij (e2)i(e3)j (e1)k = εkij Si1Sj2 Sk3 = det(S) Therefore, for N=3 the vectors {1,2,3} form a right-handed coordinate system if det(S) > 0. An example is provided by spherical coordinates where {, , } = {1, 2, 3 } form a right-handed orthogonal system and det(S) = r2sinθ > 0 (Section 3 Example 2). The above discussion can be reframed in terms of the reciprocal base vectors En along with the en. Appendix A shows that Ek ≡ det(R) (-1)k-1 e1 x e2 x ......x eN // ek missing; N > 2 so that for N = 3 one has E1 ≡ det(R) e2 x e3 The rule for right-handedness given above e2 x e3 e1 > 0 can then be restated as E1 e1 / det(R) > 0 But E1 e1 = 1 from section (b) above, so this just says det(S) > 0 This is the same conclusion reached earlier for a right-handed system, but this second method shows the way to generalizing the idea to N > 3. Working backwards, det(S) > 0 Ek ek / det(R) > 0 { det(R) (-1)k-1 e1 x e2 x ......x eN } ek / det(R) > 0 (-1)k-1 [ e1 x e2 x ......x eN] ek > 0 // ek missing in cross product If this last inequality is true for any k in the range (1,2..N), it is true for all k, so one could just test the case k = 1, [e2 x e3......x eN] e1 > 0 If this is true, then we call { e1 , e2.... eN } a right-handed coordinate system and det(S) > 0. Otherwise the coordinate system is left-handed and det(S) < 0. In various locations we have used the symbol σ = sign[det(S)]. Conclusion. Let bn be a complete basis in x-space. These basis forms a right-handed basis if the following is true det (b1, b2.....bn ) = εabc..x (b1)a (b1)a.... (bN)x = b1 [b2 x b3......x bN] > 0 In x-space there exist a complete set of basis vectors en determined by transformation S, in fact (en)i = Sin. The sign of det(S) For a given ordering of the x'i coordinates, det(S) will have a certain sign. By changing the x'i ordering to any odd permutation of the original ordering (for example, swap two coordinates), det(S) will negate because two columns of Sik(x') ≡ (∂xi/∂x'k) will be swapped. In the curvlinear coordinates application it is therefore always possible to select the ordering of the x'i coordinates to cause det(S) to be positive. One always starts with a right-handed Cartesian system for x-space, and det(S)>0 then guarantees that the en will form a right-handed system there as well. Since the underlying transformation F is assumed invertible, one cannot have det(S)=0 anywhere in the domain x (or range x') of x' = F(x), and therefore det(S) cannot change sign anywhere in the space of interest. For graphical reasons, we have selected coordinates in the "wrong order" in both the polar coordinates examples (called Example 1) and in the elliptic polar system studied in Appendix C, which is why detS < 0 for both these systems.