Suppose f is a unit speed curve. Assuming a and b are constants, what can be said about the Frenet Serret apparatus for f?

Question

Suppose $f(s)$ is a unit speed curve. We know that $({\bf T}, {\bf N}, {\bf B})$ is an orthonormal basis for $R^3$. As such, we can write $f(s)$ in the form: $$f(s) = a(s) {\bf T} + b(s) {\bf N} + c(s) {\bf B}$$ Assume $a(s)$ and $b(s)$ are nonzero constants. What can be said about $a$, $b$, $c$, $\kappa$, and $\tau$?

I've made the observation that $a(s)$ is the inner product of $f(s)$ with $\bf T$, which I will write $\langle f(s), {\bf T} \rangle$.

Similarly, $b(s) = \langle f(s), {\bf N} \rangle$ and $c(s) = \langle f(s), {\bf B} \rangle$.

I think that $b(s) = -1 / \kappa$. Am I on the right track?

Answer 1

Hint Differentiating the relation $$f = a {\bf T} + b {\bf N} + c {\bf B}$$ and using the Frenet-Serret Identities gives \begin{align} {\bf T} &= a {\bf T}' + b {\bf N}' + c' {\bf B} + c {\bf B}' \\ &= a (\kappa {\bf N}) + b (-\kappa {\bf T} + \tau {\bf B}) + c' {\bf B} + c (-\tau {\bf N}) \\ &= -b \kappa {\bf T} + (a \kappa - c \tau) {\bf N} + (b \tau + c') {\bf B} . \end{align} Comparing like coefficients gives the system \begin{align} 1 &= - b \kappa \\ 0 &= a \kappa - c \tau \\ 0 &= b \tau + c' . \end{align} The first equation gives the suspected (constant) value $\kappa = -1 / b$. After substituting we can solve the remaining two equations for $\tau$ and (by integrating), $c$.

Answer 2

With

$f(s) = a T + b N + c(s) B, \tag 1$

and

$\Vert f(s) \Vert = 1, \tag 2$

it follows that $s$ is the arc-length along $f(s)$; thus

$T = \dot f(s) = a \dot T + b \dot N + \dot c(s) B + c(s) \dot B; \tag 3$

now introducing the Frenet-Serret equations

$\dot T = \kappa N, \tag 4$

$\dot N = -\kappa T + \tau B, \tag 5$

$\dot B = -\tau N, \tag 6$

and inserting them into (3) we find

$T = a(\kappa N) + b(-\kappa T + \tau B) + \dot c(s) B -c(s) \tau N$ $= -b\kappa T + (a\kappa - c(s) \tau)N + (b\tau + \dot c(s))B; \tag 7$

comparing the coefficients of the three orthonormal vectors $T$, $N$, and $B$ yields

$-b\kappa = 1, \tag 8$

$a\kappa - c(s) \tau = 0, \tag 9$

$b\tau + \dot c(s) = 0; \tag{10}$

we thus see, in accord with our OP Gary Zoldiek, that

$\kappa = -\dfrac{1}{b}, \; \text{a constant}, \tag{11}$

which, since we conventionally assume $\kappa >0$, implies that

$b < 0; \tag{12}$

from (9), (10) (11) we also have

$c(s) \tau = a\kappa = -\dfrac{a}{b}, \tag{13}$

and

$\dot c(s) = -b\tau; \tag{14}$

now since

$a \ne 0 \ne b, \tag{15}$

(13) implies that

$c(s) \ne 0 \ne \tau, \tag{16}$

whence we may write

$\tau = -\dfrac{a}{bc(s)}, \tag{17}$

hence from (14),

$\dot c(s) = -b\tau = \dfrac{a}{c(s)}; \tag{18}$

thus,

$\dfrac{d}{ds} (\dfrac{1}{2} c^2(s)) = c(s) \dot c(s) = a, \tag{19}$

or

$\dfrac{d}{ds} (c^2(s)) = c(s) \dot c(s) = 2a; \tag{20}$

we integrate 'twixt $s_0$ and $s$:

$c^2(s) - c^2(s_0) = 2a(s - s_0); \tag{21}$

$c^2(s) = c^2(s_0) - 2a(s - s_0), \tag{22}$

$c(s) = \pm \sqrt{c^2(s_0) - 2a(s - s_0)}, \tag{23}$

as long as the signs allow; it then follows from (17) that

$\tau(s) = \mp \dfrac{a}{b \sqrt{c^2(s_0) - 2a(s - s_0)}}. \tag{24}$

So, what have we learned about $a$, $b$, $c$, $\kappa$ and $\tau$?

Well, $b < 0$, $c(s)$ is given by (23), $\kappa = -1/b$ is constant, and $\tau$ is as in (24); apparently $a$ and $c(s_0)$ are the two free parameters upon which all else swings. We should also observe that $c(s)$ and $\tau(s)$ will become singular when $s$ passes through

$s = s_0 + \dfrac{c^2(s_0)}{2a} \tag{25}$

where $c(s) = 0$; depending on the sign of $a$, the curve $f(s)$ may be extended arbitrarily far in the direction of either increasing or decreasing $s$.

I think there are many more intriguing facts about $f(s)$ which will be revealed by further scrutiny of these results, but for now I haven't the time to say more.