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G.7 Alternate Maple calculation of (v) for orthogonal systems ???? 4/23/15 For an orthogonal coordinate system, the transformation from x-space to the space of the orthogonal coordinates (call it x"-space) is x" = FM(x) as described in Section ***. Under FM, the tensor transformation rule is given in (E.9.20), A" ijk... = Mii'Mjj'Mkk'...... A i'j'k'... (E.8.20) where M = HR. H is a diagonal matrix of scale factors, and R is the R-matrix for x' = F(x), and M is a simple rotation matrix, as one would expect for going from Cartesian to Unit-Vector curvilinear basis vectors. Therefore, an alternate calculation of the components of (Δv) can be done as follows: [(v)())]ij = (v)"ij = MiaMjb(v)ab = MiaMjb∂bva Comment: This would be useful if you wanted [(v)())]rθ in terms of ∂xvy stuff. But nobody ever wants that, so we continue. But we know that ∂b = Rcb ∂'c va = Rda v'd => ∂bva = Rcb ∂'c(Rda v'd) and then [(v)())]ij = MiaMjb Rcb ∂'c(Rda v'd) = MiaMjb Rcb [ Rda∂'c( v'd) + (∂'cRda)v'd] Conclusion: But this is really no simpler than my existing result [(v)()]ij = h'i-1 h'j-1 [(∂'jv'i) + Rim(∂'jRcm) v'c] (G.5.1) so my x"-space does not provide an easy silver bullet for finding the [(v)()]ij ! It's real payoff is in the covariance of equations. What equation similar to (E.8.20) above links x" space directly to x' space? A" ijk... = Hii'Hjj'Hkk'...... A' i'j'k'... But this says nothing really, and you still have to compute the A' i'j'k'... so again there is no silver bullet.