# Representing a unit speed curve on a general surface in terms of its Frenet Frame

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:

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 $$\delta(s)$$.

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.

• I agree 'tis not the most elegant question ever posed, but thanks for answering, +1, endorsed! Nov 20 '19 at 17:18

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.