-1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.