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?

  • $\begingroup$ 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
    Commented Jul 12, 2015 at 13:07

1 Answer 1


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 $(1 + \kappa s)^{-1}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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