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This question grew out of a research project in which I am an active participant, a project which curiously enough requires a hefty dose of the theory of curves and surfaces in $\Bbb R^3$. I needn't go into more detail here.

One of my colleagues in said endeavor noticed that a previous answer of mine, given in response to a question by Helen Waters, presented results which were very illuminating to his own work. The question to which I refer is:

Representing a unit speed curve on a sphere in terms of its Frenet Frame.

In point of fact, similar questions have been posed more than once here on math.stackexchange.com; but this particular one became the object of my colleague's attention, discovered by googling around.

In the course of our work it became clear that a generalization of this question and its answer is also of significant importance to us. I pose the generalized question here, and present my answer below.

A unit speed curve $\alpha(s)$, where $s$ as usual denotes arc-length, which lies on a sphere of radius $r$ centered at a point $c \in \Bbb R^3$ satisfies an equation of the form

$(\alpha(s) - c) \cdot (\alpha(s) - c) = r^2; \tag 1$

if we differentiate this equation with respect to $s$ and recall that the unit tangent vector $T(s)$ to $\alpha(s)$ is given by

$T(s) = \dot \alpha(s), \tag 2$

we obtain

$T(s) \cdot (\alpha(s) - c) = 0; \tag 3$

successive differentiation of this formula with respect to $s$ yields the results of the answer to the cited question; the engaged reader my consult it to see the details.

As indicated in the title, I seek here to broaden the result given in the above link to more general, not-necessarily spherical surfaces. To this end we note that, just as a patch on the sphere of radius $r$ centered at $c \in \Bbb R^3$ may be represented as as a $2$-parameter vector function $\mathbf r(u, v) \in \Bbb R^3$ such that

$(\mathbf r(u, v) - c) \cdot (\mathbf r(u, v) - c) = r^2, \tag 4$

so a general surface patch $\mathcal S$ in $\Bbb R^3$ may be represented by a vector function $\mathbf r(u, v)$, but sans the constraint (4). When (4) no longer applies, we may consider the role of $\delta(s)$, where

$\delta^2(s) = \mathbf r(u(s), v(s)) \cdot \mathbf r(u(s), v(s)) = \alpha(s) \cdot \alpha(s) \tag 5$

is the squared magnitude of $\mathbf r(s)$, i.e., $\delta(s)$ is the distance of $\alpha(s)$ from the coordinate origin $O$; here $(u(s), v(s))$ is the path $\alpha(s)$ takes in terms of the patch coordinates $u$ and $v$.

Seen from this point of view, what I wish to ask becomes:

The Question: Given a surface patch $\mathcal S$ specified by a vector function $\mathbf r(u, v)$ of two parameters $u$ and $v$, and a unit speed curve $\alpha(s)$ with differentiable non-vanishing curvature and non-vanishing torsion in $\mathcal S$ , express

$\alpha(s) = \mathbf r(u(s), v(s)) \tag 6$

in terms of the Frenet Frame of $\alpha(s)$ and the function $\delta(s)$.

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This looks pretty ugly to me, but here goes. We write $$\alpha = \lambda T + \mu N + \nu B \tag{$\star$}$$ for some scalar functions $\lambda, \mu, \nu$. From $\|\alpha\|^2 = \delta^2$, we get $\lambda = \alpha\cdot T = \delta\delta'$. On the other hand, as we've shown numerous times in posts to which you referred, we differentiate ($\star$) and use the Frenet equations to get $$ T = (\lambda'-\mu\kappa)T + (\lambda\kappa + \mu' - \nu\tau)N + ((\mu\tau+\nu')B,$$ and so \begin{align*} \lambda' &= 1 + \mu\kappa \\ \nu' &= -\mu\tau \\ \mu' &= \nu\tau - \lambda\kappa. \end{align*} This gives us \begin{align*} \mu &= \frac1{\kappa}(\delta\delta'-1) \\ \nu &= \frac1{\tau}(\mu'+\lambda\kappa) = \frac1{\tau}\left(\Big(\frac{(\delta\delta'-1}{\kappa}\Big)' + \delta\delta'\kappa\right). \end{align*} And the final equality gives the constraint ODE $$0=\nu'+\mu\tau = \left(\frac1{\tau}\left(\Big(\frac{(\delta\delta'-1}{\kappa}\Big)' + \delta\delta'\kappa\right)\right)' + \frac{\tau}{\kappa}(\delta\delta'-1).$$ Yuck.

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  • $\begingroup$ I agree 'tis not the most elegant question ever posed, but thanks for answering, +1, endorsed! $\endgroup$ Nov 20 '19 at 17:18
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My answer to the question I posted above is as follows:

We begin with the observation that the relation

$\alpha(s) \cdot \alpha(s) = \mathbf r(u(s), v(s)) \cdot \mathbf r(u(x), v(s)) = \delta^2(s) \tag 1$

may be differentiated with respect to $s$,

$2\alpha(s) \cdot \dot \alpha(s) = 2\mathbf r(s) \cdot \dot{\mathbf r}(s) = 2\delta \dot \delta, \tag 2$

or

$\dot \alpha(s) \cdot \alpha(s) = \dot{\mathbf r}(s) \cdot \mathbf r(s) = \delta \dot \delta; \tag 3$

here of course $\dot{\mathbf r}(s)$ is the total derivative of $\mathbf r$ with respect to $s$, i.e.

$\dot{\mathbf r}(s) = \mathbf r_u \dot u(s) + \mathbf r_v \dot v(s), \tag 4$

where

$\mathbf r_u = \dfrac{\partial \mathbf r}{\partial u}, \tag 5$

etc. Since

$\alpha(s) = \mathbf r(s) \tag 6$

is a unit-speed curve, we have the unit tangent vector to $\alpha(s)$,

$T(s) = \dot \alpha(s) = \dot{\mathbf r}(s); \tag 7$

in light of this we have from (3)

$T(s) \cdot \mathbf r(s) = \delta \delta_s, \tag 8$

which gives the component of $\mathbf r(s)$ along $T(s)$; we next differentiate (8) again with respect to $s$ to find

$\dot T(s) \cdot \mathbf r(s) + T(s) \cdot \dot{\mathbf r}(s) = \delta_s^2 + \delta \delta_{ss}, \tag 9$

which by virtue of (7) becomes

$\dot T(s) \cdot \mathbf r(s) + T(s) \cdot T(s) = \delta_s^2 + \delta \delta_{ss}; \tag{10}$

since $T(s)$ is a unit vector,

$\dot T(s) \cdot \mathbf r(s) + 1 = \delta_s^2 + \delta \delta_{ss}; \tag{11}$

next, the Frenet-Serret equation

$\dot T(s) = \kappa(s) N(s) \tag{12}$

may be used to transform (11) to

$\kappa(s) N(s) \cdot \mathbf r(s) + 1 = \delta_s^2 + \delta \delta_{ss}; \tag{13}$

since we have assumed that

$\kappa(s) \ne 0, \tag{14}$

we may re-arrange (13) into the form

$N(s) \cdot \mathbf r(s) = \dfrac{\delta_s^2 + \delta \delta_{ss} - 1}{\kappa(s)}, \tag{15}$

which gives the compoent of $\mathbf r(s)$ along $N(s)$; we may in fact continue in this direction, and take the $s$-derivative of (13):

$\dot \kappa(s) N(s) \cdot \mathbf r(s) + \kappa(s) \dot N(s) \cdot \mathbf r(s) + \kappa(s) N(s) \cdot \dot{\mathbf r}(s) = (\delta_s^2 + \delta \delta_{ss})_s; \tag {16}$

in accord with (7) we see that

$ N(s) \cdot \dot{\mathbf r}(s) = N(s) \cdot T(s) = 0, \tag{17}$

whence

$\dot \kappa(s) N(s) \cdot \mathbf r(s) + \kappa(s) \dot N(s) \cdot \mathbf r(s) = (\delta_s^2 + \delta \delta_{ss})_s; \tag {18}$

as for the right-hand side of this equation,

$(\delta_s^2 + \delta \delta_{ss})_s = (\delta_s^2)_s + (\delta \delta_{ss})_s = 2\delta_s \delta_{ss} +\delta_s \delta_{ss} + \delta\delta_{sss} = 3\delta_s \delta_{ss} + \delta_s \delta_{sss}; \tag{19}$

thus,

$\dot \kappa(s) N(s) \cdot \mathbf r(s) + \kappa(s) \dot N(s) \cdot \mathbf r(s) = 3\delta_s \delta_{ss} + \delta_s \delta_{sss}; \tag {20}$

we now exploit the so-called second Frenet-Serret equation,

$\dot N(s) = -\kappa(s) T(s) + \tau(s) B(s) \tag{21}$

by substituting it into (20):

$\dot \kappa(s) N(s) \cdot \mathbf r(s) + \kappa(s) ( -\kappa(s) T(s) + \tau(s) B(s)) \cdot \mathbf r(s) = 3\delta_s \delta_{ss} + \delta_s \delta_{sss}, \tag {22}$

or

$\dot \kappa(s) N(s) \cdot \mathbf r(s) - \kappa^2(s) T(s) \cdot \mathbf r(s) + \kappa(s) \tau(s) B(s) \cdot \mathbf r(s) = 3\delta_s \delta_{ss} + \delta_s \delta_{sss}, \tag {23}$

in which the term containing $B(s) \cdot \mathbf r(s)$ may be isolated, thus:

$\kappa(s) \tau(s) B(s) \cdot \mathbf r(s) = \kappa^2(s) T(s) \cdot \mathbf r(s) -\dot \kappa(s) N(s) \cdot \mathbf r(s) + 3\delta_s \delta_{ss} + \delta_s \delta_{sss}, \tag {24}$

from which

$B(s) \cdot \mathbf r(s) = \dfrac{\kappa(s)}{\tau(s)} T(s) \cdot \mathbf r(s) - \dfrac{1}{\kappa(s) \tau(s)} \dot \kappa(s) N(s) \cdot \mathbf r(s) + \dfrac{1}{\kappa(s) \tau(s)} (3\delta_s \delta_{ss} + \delta_s \delta_{sss}); \tag {25}$

we can now use (8) and (15) to eliminate $T(s) \cdot \mathbf r(s)$ and $N(s) \cdot \mathbf r(s)$ from this expression, to find:

$B(s) \cdot \mathbf r(s) = \dfrac{\kappa(s)}{\tau(s)} \delta \delta_s - \dfrac{1}{\kappa(s) \tau(s)} \dot \kappa(s) (\dfrac{\delta_s^2 + \delta \delta_{ss} - 1}{\kappa(s)}) + \dfrac{1}{\kappa(s) \tau(s)} (3\delta_s \delta_{ss} + \delta_s \delta_{sss}). \tag {26}$

Now the expansion of the vector $\mathbf r(s)$ in terms of the Frenet Frame $T(s)$, $N(s)$, $B(s)$ is

$\mathbf r(s) = (\mathbf r(s) \cdot T(s)) T(s) + (\mathbf r(s) \cdot N(s))N(s) + (\mathbf r(s) \cdot B(s)) B(s); \tag{27}$

using (8), (15) and (26) we arrive at

$\mathbf r(s) = (\delta \delta_s) T(s) + \left (\dfrac{\delta_s^2 + \delta \delta_{ss} - 1}{\kappa(s)} \right )N(s) $ $+ \left ( \dfrac{\kappa(s)}{\tau(s)} \delta \delta_s - \dfrac{1}{\kappa(s) \tau(s)} \dot \kappa(s) \left (\dfrac{\delta_s^2 + \delta \delta_{ss} - 1}{\kappa(s)} \right ) + \dfrac{1}{\kappa(s) \tau(s)} (3\delta_s \delta_{ss} + \delta_s \delta_{sss}) \right ) B(s). \tag{28}$

By virtue of this equation we may express $\vert \mathbf r(s) \vert$, the distance of the point $\mathbf r(s)$ from the origin, as

$\vert \mathbf r(s) \vert^2 = \mathbf r(s) \cdot \mathbf r(s) = (\delta \delta_s)^2 + \left (\dfrac{\delta_s^2 + \delta \delta_{ss} - 1}{\kappa(s)} \right )^2$ $+ \left ( \dfrac{\kappa(s)}{\tau(s)} \delta \delta_s - \dfrac{1}{\kappa(s) \tau(s)} \dot \kappa(s) \left (\dfrac{\delta_s^2 + \delta \delta_{ss} - 1}{\kappa(s)} \right ) + \dfrac{1}{\kappa(s) \tau(s)} (3\delta_s \delta_{ss} + \delta_s \delta_{sss}) \right )^2; \tag{29}$

we observe that in the case

$\delta = \text{constant} \tag{30}$

all the derivatives

$\delta_s, \; \delta_{ss}, \; \delta_{sss} = 0, \tag{31}$

and (28) reverts to

$\mathbf r(s) = -\dfrac{1}{\kappa(s)} N(s) + \dfrac{ \dot \kappa(s)}{\kappa^2(s) \tau(s)} B(s), \tag{32}$

whilst (29) becomes

$\vert \mathbf r(s) \vert^2 = \dfrac{1}{\kappa^2(s)} + \dfrac{ (\dot \kappa(s))^2}{\kappa^4(s) \tau^2(s)}; \tag{33}$

formulas (32) and (33) agree with the answers I gave in https://math.stackexchange.com/questions/2504918/representing-a-unit-speed-curve-on-a-sphere-in-terms-of-its-frenet-frame which treats the case of constant $\delta = r$, that is, when the surface in which $\alpha(s)$ lies is a sphere.

In the above discussion, we have focused exclusively on surfaces given in parametric form $\mathbf r(u, v)$, and we relied extensively on the fact that for such surfaces the quantity $\delta = \vert \mathbf r(u, v) \vert$ is the Euclidean distance in $\Bbb R^3$ 'twixt the origin $O = (0, 0, 0)$ and the point $\mathbf r(u, v)$. However, it appears this question may also be addressed by representing surfaces implicitly, that is by specifying a (sufficiently smooth)

$f: \Omega \subset \Bbb R^3 \to \Bbb R \tag{34}$

$\Omega$ open, with

$\nabla f(x, y, z) \ne 0, \; (x, y, z) \in \Omega; \tag{35}$

since specifying a surface implicitly does not give us a direct handle on the distance of points from the origin, we might expect any formulas we derive along such lines to exhibit significant points of difference from (28)-(29). Finally, it is also possible to represent the $\mathbf r$ in polar form, that is, writing

$\mathbf r = r\mathbf e_r, r = \vert \mathbf r \vert = \delta \ne 0, \; \vert \mathbf e_r \vert = 1, \tag{36}$

and perhaps carrying out calculations similar to the above but using $\mathbf r$ in this form; e.g., (7) becomes

$T(s) = \dot \alpha(s) = \dfrac{d}{ds} (r(s)\mathbf e_r(s)) = \dot r(s) \mathbf e_r(s) + r(s) \dot{\mathbf e}_r(s); \tag{37}$

if this formulation is followed through, the contributions of changes in distance m$O$ ($r(s)$ direction $(\mathbf e_r(s)$ of $\alpha(s)$ to the coefficients of $T$, $N$, and $B$ may perhaps be quantified. But this post is quite long enough as is, so I will leave such undertakings for later.

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