I was simply thinking that the gradient in the Frenet-Serret coordinate at a particular point is similar to the gradient in the Cartesian coordinate. I simply assumed that Frenet space is an orthonormal space and the the only nonvanishing term of the gradient $\phi$ will be $\frac{\partial\phi}{\partial s}\hat{e}_s$ where $s$ is the arc-length (tangent) coordinate, along the reference curve. However, when I read this paper, I realized for the scalar field $\phi$ $$\nabla\phi=\frac{\partial\phi}{\partial x}\hat{e_x}+\frac{\partial\phi}{\partial y}\hat{e_y}+\frac{1}{h}\frac{\partial\phi}{\partial s}\hat{e_s}$$ where the so-called scale factor $h=1+\kappa(s)x$ in which $\kappa(s)$ is local curvature of the reference curve. How the curvature of the reference curve shows up in the gradient written in Frenet frame? That's the part I don't understand. Where am I doing anything wrong?
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$\begingroup$ Minor comment to the post (v3): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$– Qmechanic ♦Jul 12, 2015 at 13:07
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I believe you and the author are referring to different "gradients". You mean the gradient along the manifold defined by the particle trajectory; since this is a one dimensional manifold, the gradient would indeed have only one component. The authors are referring to the full three-space gradient. Using the formulae $$ r = \frac{1}{\kappa}+x \\ \phi = \kappa s $$ it is easy to show that the scale factor for $s$ would be $(1 + \kappa s)^{-1}$.
