TITLE: Frenet-Serret and Vector Fields
QUESTION [8 upvotes]: The well-known Frenet-Serret equations, 
$\dot T(s) = \kappa N(s), \tag 1$
$\dot N(s) = -\kappa(s) T(s) + \tau(s) B(s), \tag 2$
$\dot B(s) = -\tau(s) N(s), \tag 3$
where
$T = \dot \alpha(s), \tag 4$
$\alpha(s)$ being a unit speed curve in $\Bbb R^3$ with arc-length $s$, are most often applied to discover and describe properties of such space curves.
Given an open set
$U \subset \Bbb R^3, \tag 5$
and a vector field
$X \in C^\infty(U, \Bbb R^3) \tag 6$
on $U$, we may of course consider the flow $\phi_X(x, t)$ of $X$; the reader will recall it is defined as, more or less, the entire family of integral curves of the vector field $X$ in the sense that
$\phi_X(x, 0) = x, \; \forall x \in U, \tag 7$
and
$\dfrac{d}{dt}\phi(x, t) = X(\phi(x, t)). \tag 8$
A somewhat natural area of inquiry based upon these two concepts, the Frenet-Serret apparatus and vector fields and their flows, is the relationship 'twixt the Frenet-Serret formulas and the integral curves of $X$; that is, finding the expressions for $T(s)$, $N(s)$, $B(s)$, $\kappa(s)$ and $\tau(s)$ in terms of $X$ and related quantities such as its magnitude $\vert X \vert = \langle X, X \rangle^{1/2}$ and its derivatives $\nabla X$ etc.
The Question then becomes:
Given a (sufficiently smooth) vector field $X$ on an open set $U \subset \Bbb R^3$, find the vector fields $T(s)$, $N(s)$, and $B(s)$ and the scalar quantities $\kappa(s)$ and $\tau(s)$ associated with the integral curves of $X$, expressed in terms of $X$ and it's associated quantities such as $\vert X \vert$ and so forth.
A Few Observations:  Given such an open set $U$ and vector field $X$, of course it is true that the flow $\phi_X(x, t)$ may not exist for all values of $t$, but this is of no consequence here since all calculations are local in nature.  Indeed, for all $x \in U$ the flow $\phi_X(x, t)$ is defined for sufficiently small values of $t$, and this is sufficient for the present purposes.
A Useful Starting Point may be the observation that $X/\vert X \vert$ is a unit vector field, and that in fact
$T(s) = \dfrac{X(\alpha(s))}{\vert X(\alpha(s))\vert} \tag 9$
along the arc-length parametrized integral curve $\alpha(s)$ of $X/ \vert X \vert$.  Of course, we may also adopt and employ the given parametrization of the integral curves of $X$ by $t$, as in (7), (8); in fact we have
$\dfrac{ds}{dt} = \vert X(\alpha(t)) \vert, \; \dfrac{dt}{ds} = \vert X(\alpha(s)) \vert^{-1}, \tag{10}$
which allow the conversion 'twixt $t$ and $s$ via integration:
$s - s_0 = \displaystyle \int_{t_0}^t  \vert X(\alpha(u)) \vert \; du, \; t - t_0 = \displaystyle \int_{s_0}^s  \vert X(\alpha(u)) \vert^{-1} \; du. \tag{11}$
We can also express the unit tangent vector $T$ in terms of the parameter $t$:
$T(t) = \dfrac{X(\alpha(t))}{\vert X(\alpha(t))\vert}. \tag{12}$
In these formulas the reader will recognize that $\alpha(t)$ and $\alpha(s)$ represent the same curves in the geometrical sense, that is, he same paths in $\Bbb R^3$, though they are differently parametrized.

REPLY [3 votes]: The first thing we need to do is construct the unit tangent vector field $T$ to the integral curves of $X$ from $X$ itself; now the magnitude of $X$ is
$\vert X \vert = \langle X, X \rangle^{1/2}; \tag 1$
thus normalizing $X$ yields
$T = \dfrac{X}{\vert X \vert} = \dfrac{X}{\langle X, X \rangle^{1/2}} = \langle X, X \rangle^{-1/2}X, \tag 2$
where $T$ is the unit tangent field to the integral curves of $X$;
$\nabla_{X/\vert X \vert} \left(\dfrac{X}{\vert X \vert} \right) = \nabla_{\langle X, X \rangle^{-1/2}X}  \left( \langle X, X \rangle^{-1/2}X \right) = \langle X, X \rangle^{-1/2}\nabla_X  \left( \langle X, X \rangle^{-1/2}X \right); \tag 3$
$\nabla_X  \left( \langle X, X \rangle^{-1/2}X \right) = \left ( \nabla_X\langle X, X \rangle^{-1/2} \right ) X +  \langle X, X \rangle^{-1/2} \nabla_X X; \tag 4$
$ \nabla_X\langle X, X \rangle^{-1/2} = -\dfrac{1}{2}\langle X, X \rangle^{-3/2} \nabla_X \langle X, X \rangle =  -\dfrac{1}{2}\langle X, X \rangle^{-3/2} \left ( \langle \nabla_X X, X \rangle + \langle X, \nabla_X X \rangle \right )$
$=  -\dfrac{1}{2}\langle X, X \rangle^{-3/2} \left (2 \langle X, \nabla_X X \rangle \right ) = -\langle X, X \rangle^{-3/2} \langle X, \nabla_X X \rangle; \tag 5$
$\nabla_X  \left( \langle X, X \rangle^{-1/2}X \right) = -\langle X, X \rangle^{-3/2} \langle X, \nabla_X X \rangle X +  \langle X, X \rangle^{-1/2} \nabla_X X; \tag 6$
$\nabla_{X/\vert X \vert} \left(\dfrac{X}{\vert X \vert} \right) = \langle X, X \rangle^{-1/2}\nabla_X  \left( \langle X, X \rangle^{-1/2}X \right)$
$=  -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X +  \langle X, X \rangle^{-1} \nabla_X X; \tag 7$
we confirm that this vector is normal to $X$:
$\left \langle \nabla_{X/\vert X \vert} \left(\dfrac{X}{\vert X \vert} \right), X \right \rangle = \langle  -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X +  \langle X, X \rangle^{-1} \nabla_X X, X \rangle$
$= \langle -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X, X \rangle + \langle \langle X, X \rangle^{-1} \nabla_X X, X \rangle$
$=  -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle \langle X, X \rangle + \langle X, X \rangle^{-1} \langle \nabla_X X, X \rangle$
$=  -\langle X, X \rangle^{-1} \langle X, \nabla_X X \rangle  + \langle X, X \rangle^{-1} \langle \nabla_X X, X \rangle = 0; \tag 8$
(7) is thus normal to $X/\vert X \vert$ as well; thus we have $N$, the unit normal to our integral curves, expressible as
$N = \dfrac{ -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X +  \langle X, X \rangle^{-1} \nabla_X X }{\vert -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X +  \langle X, X \rangle^{-1} \nabla_X X  \vert }; \tag 9$
now
$\vert -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X +  \langle X, X \rangle^{-1} \nabla_X X  \vert^2$
$= \langle -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X, -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X \rangle$
$+ 2 \langle   -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X, \langle X, X \rangle^{-1} \nabla_X X \rangle + \langle \langle X, X \rangle^{-1} \nabla_X X, \langle X, X \rangle^{-1} \nabla_X X \rangle$
$= \langle X, X \rangle^{-4} \langle X, \nabla_X X \rangle^2 \langle X, X \rangle - 2 \langle X, X \rangle^{-3} \langle X, \nabla_X X \rangle^2 + \langle X, X \rangle^{-2} \langle \nabla_X X,  \nabla_X X \rangle$
$= \langle X, X \rangle^{-3} \langle X, \nabla_X X \rangle^2 - 2 \langle X, X \rangle^{-3} \langle X, \nabla_X X \rangle^2  + \langle X, X \rangle^{-2} \langle \nabla_X X,  \nabla_X X \rangle$
$= \langle X, X \rangle^{-2} \langle \nabla_X X,  \nabla_X X \rangle -  \langle X, X \rangle^{-3} \langle X, \nabla_X X \rangle^2;  \tag{10}$
thus,
$\vert -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X +  \langle X, X \rangle^{-1} \nabla_X X  \vert$
$= \sqrt{ \langle X, X \rangle^{-2} \langle \nabla_X X,  \nabla_X X \rangle -  \langle X, X \rangle^{-3} \langle X, \nabla_X X \rangle^2}$
$= (\langle X, X \rangle^{-2} \langle \nabla_X X,  \nabla_X X \rangle -  \langle X, X \rangle^{-3} \langle X, \nabla_X X \rangle^2)^{1/2}; \tag{11}$
this quantity (11), being the magnitude of $\nabla_{X / \vert X \vert} (X / \vert X \vert)$, is in fact the curvature of the integral curves of $X$, so we write
$\kappa = (\langle X, X \rangle^{-2} \langle \nabla_X X,  \nabla_X X \rangle -  \langle X, X \rangle^{-3} \langle X, \nabla_X X \rangle^2)^{1/2}, \tag{12}$
which leads to
$\kappa^2 = \langle X, X \rangle^{-2} \langle \nabla_X X,  \nabla_X X \rangle -  \langle X, X \rangle^{-3} \langle X, \nabla_X X \rangle^2$
$= \langle X, X \rangle^{-3} \langle \nabla_X X,  \nabla_X X \rangle\langle X, X \rangle -  \langle X, X \rangle^{-3} \langle X, \nabla_X X \rangle^2$
$= \dfrac{ \langle \nabla_X X,  \nabla_X X \rangle\langle X, X \rangle -  \langle X, \nabla_X X \rangle^2}{\langle X, X \rangle^3} = \dfrac{\vert \nabla_X X \vert^2 \vert X \vert^2 - \langle X,  \nabla_X X \rangle^2}{\vert X \vert^6},  \tag{13}$
and returning once again to
$\kappa = \dfrac{(\vert \nabla_X X \vert^2 \vert X \vert^2 - \langle X,  \nabla_X X \rangle^2)^{1/2}}{\vert X \vert^3},  \tag{14}$
if $\alpha(t)$ is an integral curve of $X$, that is, if $\alpha(t)$ satisfies
$\dot \alpha(t) = X(\alpha(t)), \tag{15}$
then
$\ddot \alpha(t) = \nabla_X X, \tag{16}$
whence
$\kappa = \dfrac{\vert \ddot \alpha(t) \vert^2 \vert \dot \alpha(t) \vert^2 - \langle \dot \alpha(t), \ddot \alpha(t) \rangle^2}{\vert \dot \alpha(t) \vert^3},  \tag{17}$
which is the usual formula for $\kappa$ in terms of the derivatives of a (not necessarily arc-length) parametrized curve $\alpha(t)$. (14) expresses this quantity in terms of the vector field $X$, and hence applies to all the integral curves of this vector field.  We also observe that (13) ensures
$\kappa^2 > 0, \tag{18}$
so that $\kappa$ is well-defined (i.e., real) from (14), (18); this follows from the Cauchy-Schwarz inequality applied to the numerator of (13), that is, to
$\vert \nabla_X X \vert^2 \vert X \vert^2 - \langle X,  \nabla_X X \rangle^2, \tag{19}$
for indeed we have
$\langle X,  \nabla_X X \rangle^2 \le \vert \nabla_X X \vert^2 \vert X \vert^2. \tag{20}$
We have now derived expressions for $T$, $N$ and $\kappa$ in terms of the vector field $X$; we pause to note that $T$ and $N$ are in this case vector fields themselves, and $\kappa$ a scalar field, on any open set $U$ where $X$ is defined; new fields from old.
Having built $\kappa$, $T$ and $N$ from $X$, we proceed to calculate
$B = T \times N; \tag{21}$
by means of (2) and (9) we write
$T \times N = \dfrac{X}{\vert X \vert} \times  \dfrac{ -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X +  \langle X, X \rangle^{-1} \nabla_X X }{\vert -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X +  \langle X, X \rangle^{-1} \nabla_X X  \vert }$
$= \dfrac{X \times (-\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X +  \langle X, X \rangle^{-1} \nabla_X X)}{\vert X \vert\vert -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X +  \langle X, X \rangle^{-1} \nabla_X X  \vert}; \tag{22}$
we evalutate the numerator:
$X \times (-\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X +  \langle X, X \rangle^{-1} \nabla_X X)$
$= -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X \times X +  \langle X, X \rangle^{-1} X \times \nabla_X X = \langle X, X \rangle^{-1} X \times \nabla_X X, \tag{23}$
since $X \times X = 0$; thus (22) becomes
$T \times N = \dfrac{\langle X, X \rangle^{-1} X \times \nabla_X X}{\vert X \vert\vert -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X +  \langle X, X \rangle^{-1} \nabla_X X  \vert}; \tag{24}$
we multiply the numerator and denominator through by $\langle X, X \rangle > 0$ and obtain
$B = T \times N = \dfrac{ X \times \nabla_X X}{\vert X \vert \langle X, X \rangle \vert -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X +  \langle X, X \rangle^{-1} \nabla_X X  \vert}$
$= \dfrac{ X \times \nabla_X X}{\vert X \vert \vert -  \langle X, X \rangle^{-1}\langle X, \nabla_X X \rangle X + \nabla_X X  \vert} = \dfrac{ X \times \nabla_X X}{\vert X \vert \vert \nabla_X X  - \langle X, X \rangle^{-1}\langle X, \nabla_X X \rangle X \vert}; \tag{25}$
Nota Bene:  Whoops!  Looks like I hit that old "Post" button a tad prematurely!  So stay tuned, there is a bit more to come!  End of Note.