I'm trying to prove the formula for curl in curvilinear coordinates and am following the proof in the book "Vector Analysis and Cartesuan Tensors" by Bourne and Kendall.

I got this far:

-rewritting $\nabla \times \textbf{F}=\nabla \times [F_u \textbf{e}_u+F_v \textbf{e}_v+F_w \textbf{e}_w]$

-looking at just one term and expanding $\nabla \times (F_u \textbf{e}_u)=F_u(\nabla \times \textbf{e}_u)-\textbf{e}_u \times (\nabla F_u)$

-using the fact that $\textbf{e}_u=h_1 \nabla u$. The first term disappears because the curl of a grad is zero, and we are left with $\nabla \times (F_u \textbf{e}_u)=h_1 (\nabla F_u)\times (\nabla u)$

However, the proof I'm following demands that $h_1$ is inside the first gradient and it achieves that by performing the substitution sooner. I am concerned of the legality of this move since when a substitution is made shouldn't make a difference. Is there something I'm missing?

  • $\begingroup$ You need to rewrite $\nabla F_u$ in the basis $\{ \mathbf{e}_u, \mathbf{e}_v, \mathbf{e}_w \}$. $\endgroup$ – md2perpe Sep 21 '18 at 19:59

Nabla is a differential operator, a non-constant factor cannot simply be moved outside the nabla: $$\nabla \times \mathbf e_u = \nabla \times (h_1 \nabla u) = (\nabla h_1) \times \nabla u, \\ F_u \nabla \times \mathbf e_u - \mathbf e_u \times \nabla F_u = (F_u \nabla h_1) \times \nabla u + (h_1 \nabla F_u) \times \nabla u = \nabla(h_1 F_u) \times \nabla u.$$


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