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Given a Frenet-Serret frame $(\vec T(t), \vec N(t), \vec B(t))$ defined by a curve $\vec \gamma(t)$ with $$\begin{array}{rcl} |\tfrac{d}{dt}\vec\gamma(t)| &\equiv& 1, \\ \vec T(t) &:=& \tfrac{d}{dt}\vec\gamma(t), \\ \kappa(t) &:=& |\tfrac{d}{dt}\vec T(t)|, \\ \kappa(t)\vec N(t) &:=& \tfrac{d}{dt}\vec T(t), \\ \vec B(t) &:=& \vec T(t)\times\vec N(t), \\ \tfrac{d}{dt}\vec B(t) &=:& -\tau\vec N(t) \end{array}$$ one can define a local coordinate system $$\vec x(t,n,b) := \vec\gamma(t) + n\vec N(t) + b\vec B(t)$$ (where the range of $n,b$ is defined such that this yields an injective function, assuming $\kappa,\tau\neq0$). $(\vec T,\vec N ,\vec B)$ are assumed to be independent of $n,b$, i.e. for a fixed $t$ the $\vec x(t,n,b)$ span a 2D plane perpendicular to $\vec T(t)$ with Cartesian coordinates $(n,b)$.

Then what are the derivatives of these unit vectors, especially

what are $(\vec a\vec\nabla)\vec T$ (for a vector $\vec a$, i.e. directional derivative of $\vec T$) and $\Delta\vec T$ (i.e. Laplacian of $\vec T$) expressed in these coordinates?

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My first impulse was simply claiming $\vec a\vec\nabla = \partial_t$ and $\Delta\vec T = \partial_t^2\vec T$ and then using the Frenet-Serret Formulas, but since the coordinate system is curved this is probably wrong. I wonder if it's something similar to spherical coordinates, i.e. $\Delta\vec T=\kappa^2\partial_t(\frac1{\kappa^2}\partial_t\vec T)$ etc. – Tobias Kienzler Jun 17 '13 at 14:55
Be careful. You seem to be thinking of $t$ as arclength, but then your formula for $\vec T$ is overly complicated. Do you mean to compute these derivatives only along $n=b=0$? You have coordinates in a neighborhood of $\vec\gamma$, but do you have vector fields on that neighborhood? – Ted Shifrin Jun 17 '13 at 15:49
@TedShifrin $t$ is only the arclength if $|d/dt\,\vec\gamma|\equiv1$, though that could be assumed for simplicity. Concerning your second point, I forgot to mention that $\vec T,\vec N,\vec B$ are assumed to be independent of $n,b$, i.e. for a fixed $t$ the $\vec x(t,n,b)$ span a 2D plane perpendicular to $\vec T(t)$ with Cartesian coordinates $(n,b)$. Yet due to parallel transport $\Delta\vec T(t)$ might not be independent of $n$ or $b$ I think... – Tobias Kienzler Jun 17 '13 at 15:56
I'm just commenting that in the remainder of your formulas you did assume $t$ to be arclength. I'll think about the rest in a bit. – Ted Shifrin Jun 17 '13 at 16:03
I don't see any subtleties in your original question. What might be more interesting is to instead think of the Frenet frame of the curve $\gamma_{n,b}(t)=x(t,n,b)$ as defining the framing $T,N,B$. – Ted Shifrin Jun 17 '13 at 17:32
up vote 0 down vote accepted

Alright, let's just follow the Wikipedia article on curvilinear coordinates:

From $\vec x(t,n,b) = \vec\gamma(t) + n\vec N(t) + b\vec B(t)$ one obtains

$$\begin{array}{rcl} \vec h_t &=& \partial_t\vec x = \underbrace{\vec\gamma'}_{=\vec T} + n\underbrace{\vec N'}_{=-\kappa\vec T + \tau\vec B} + b\underbrace{\vec B'}_{=-\tau\vec N} \\ &=& (1-\kappa)\vec T - n\tau\vec N + b\tau\vec B \\\Rightarrow h_t &=& \sqrt{(1-\kappa)^2+(n^2+b^2)\tau^2}, \\\vec h_n &=& \partial_n\vec x = \vec N \Rightarrow h_n=1, \\\vec h_b &=& \partial_b\vec x = \vec B \Rightarrow h_b=1. \end{array}$$

Note how for non-vanishing torque $\tau\neq0$ this yields $\vec h_t\,\not\|\,\vec T$. Then, following these formulas, one obtains

$$(\vec a\vec\nabla)\vec T = \frac{\vec a\vec h_t}{h_t^2}\underbrace{\partial_t\vec T}_{=\kappa\vec N}$$


$$\begin{array}{rcl} \Delta\vec T &=& \left(\frac1{h_t}\partial_t\right)^2 \vec T = \frac1{h_t}\partial_t\left(\frac{\kappa(t)}{h_t}\vec N\right) \\ &=& \frac1{h_t}\left(\frac\kappa{h_t}\right)'\vec N + \frac{\kappa}{h_t^2}\left(-\kappa\vec T + \tau\vec B\right) \\ \text{with}\quad h_t' &=& \frac{(1-\kappa)\kappa'+(n^2+b^2)\tau\tau'}{h_t} \\ \Rightarrow \frac1{h_t}\left(\frac\kappa{h_t}\right)' &=& \frac{\kappa' h_t^2 - \kappa[(1-\kappa)\kappa' + (n^2+b^2)\tau\tau']}{h_t^4} \\ &=& \frac{\kappa'[(n^2+b^2)\tau+2\kappa^2-3\kappa+1] - \tau'\kappa\tau(n^2+b^2)}{h_t^4} \end{array}$$

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No complaints? Then I'll have to accept my own answer... – Tobias Kienzler Jun 26 '13 at 6:31

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