Torsion coefficient

Question

We went over the Frenet-Serret formulas today in class and the professor wrote $$d\mathbf{\hat{B}}/dt=-\tau{}\mathbf{\hat{N}}.$$ He said that the coefficient is always negative (so that tau is always positive). I don't understand why this has to be and would quite like an explanation.

Answer 1

In my experience with the Frenet-Serret apparatus it is the cuvrature which must always be positive; it is the torsion which may change sign along a regular curve.  To explain:

Let $\alpha(s) \subsetneq \Bbb R^3$ be a unit-speed curve; $s$ is the arc-length along $\alpha$; then as is well-known the unit tangent vector to $\alpha(s)$ is

$T(s) = \dot \alpha(s); \tag 1$

since

$T(s) \cdot T(s) = 1, \tag 2$

it follows in the usual manner that

$T(s) \cdot \dot T(s) = 0; \tag 3$

assuming that

$\dot T(s) \ne 0, \tag 4$

we may set

$\dot T(s) = \kappa(s) N(s), \tag 5$

where $N(s)$ is also a unit vector,

$\Vert N(s) \Vert^2 = N(s) \cdot N(s) = 1; \tag 6$

and by definition

$\kappa(s) > 0; \tag 7$

we have,

$\kappa(s) = \kappa(s) \Vert N(s) \Vert = \Vert \kappa(s)N(s) \Vert = \Vert \dot T(s) \Vert > 0; \tag 8$

$\kappa(s)$ is the magnitude of $\dot T(s)$; $N(s)$ is the direction.

We proceed to examine $\dot N(s)$; since

$T(s) \cdot N(s) = 0, \tag 8$

we have its derivative

$\dot T(s) \cdot N(s) + T(s) \cdot \dot N(s) = 0, \tag 9$

which yields, via (5), (6) and (8),

$\kappa(s) + T(s) \cdot \dot N(s) = \kappa(s) N(s) \cdot N(s) + T(s) \cdot \dot N(s)$ $= \dot T(s) \cdot N(s) + T(s) \cdot \dot N(s) = 0, \tag{10}$

and thus

$T(s) \cdot \dot N(s) = -\kappa(s); \tag{11}$

that is, the component of $\dot N(s)$ along $T(s)$ is $-\kappa(s)$.

Now

$\alpha(s) \subsetneq \Bbb R^3, \tag{12}$

and therefore we cannot assume that $\dot N(s)$ is collinear with $T(s)$; in general we must allow for a component of $\dot N(s)$ normal to both $N(s)$ and $T(s)$; we note however that (6) implies $\dot N(s)$ has no component along $N(s)$ itself, since it yields upon differentiation

$2N(s) \cdot \dot N(s) = 0 \Longrightarrow N(s) \cdot \dot N(s) = 0; \tag{13}$

bearing these facts in mind, we define the binormal vector

$B(s) = T(s) \times N(s), \tag{14}$

and noting that

$T(s) \cdot B(s) = N(s) \cdot B(s) = 0, \tag{15}$

and by virtue of (8) that

$B(s) \cdot B(s) = 1, \tag{16}$

we define the torsion $\tau(s)$ of $\alpha(s)$ as the component of $\dot N(s)$ along $B(s)$, so that

$\dot N(s) = -\kappa(s) T(s) + \tau (s) B(s), \tag{17}$

and we may thus also write

$\tau(s) = \tau(s) B(s) \cdot B(s) = \dot N(s) \cdot B(s); \tag{18}$

we now have, since (14) implies

$T(s) \times B(s) = -N(s), \tag{19}$

$\dot B(s) = \dot T(s) \times N(s) + T(s) \times \dot N(s)$ $= \kappa(s) N(s) \times N(s) + T(s) \times \dot N(s)$ $= \tau(s) T(s) \times B(s) = -\tau(s) N(s). \tag{20}$

In the above derivation of the Frenet-Serret equations (5), (17) and (20), we have nowhere found reason to stipulate

$\tau(s) > 0; \tag{21}$

indeed, it is well-known that

$\tau(s) = 0, \; \dot B(s) = 0, \tag{22}$

when $\alpha(s)$ is a planar curve; see this question and my answer for a detailed explanation.

We may in fact reverse the sign of the torsion of $\alpha(s)$ via the transformation

$\alpha(s) \to -\alpha(s); \tag{23}$

then

$T(s) = \dot \alpha(s) \to -T(s), \tag{24}$

and

$N(s) \to -N(s), \; \dot N(s) \to -\dot N(s); \tag{25}$

however,

$B(s) \to (-T(s)) \times (-N(s))$ $= T(s) \times N(s) = B(s); \tag{26}$

that is, $B(s)$ remains invariant under (23), hence also

$\dot B(s) \to \dot B(s); \tag{27}$

it now follows from

$\dot B(s) = -\tau(s) N(s) \tag{28}$

that

$\tau(s) \to -\tau(s) \tag{29}$

under (23); therefore the sign of $\tau(s) \ne 0$ may always be reversed by taking the transformation (23); it follows that a curve of negative torsion exists if and only if a curve of positive torsion also exists; thus (21) is in fact false in general.

For more on this topic, the reader may consult the answer I gave to this question.

Torsion coefficient