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2.4 Frenet-Serret formulae

From (2.20) and (2.44), we found that

$\displaystyle {\bf t}' = \kappa {\bf n}\;,$			(2.53)
$\displaystyle {\bf b}' = -\tau {\bf n}\;.$			(2.54)

From these equations we deduce

$\displaystyle {\bf n}' = ({\bf b}\times {\bf t})' = {\bf b}'\times {\bf t} + {\... ...\times (\kappa{\bf n}) = -\kappa {\bf t} + \tau {\bf b}\;. % Cho \nonumber \\$

(2.55)

In matrix form we can express the differential equations as

$\displaystyle \begin{pmatrix} {\bf t}' \\ {\bf n}' \\ {\bf b}' \end{pmatrix}=... ...u& 0 \end{pmatrix}\begin{pmatrix} {\bf t} \\ {\bf n}\\ {\bf b} \end{pmatrix}.$

(2.56)

Thus, ${\bf t}$ , ${\bf n}$ , ${\bf b}$ are completely determined by the curvature and torsion of the curve as a function of parameter . The equations $\kappa = \kappa(s)$ , $\tau = \tau(s)$ are called intrinsic equations of the curve. The formulae (2.56) are known as the Frenet-Serret formulae and describe the motion of a moving trihedron ( ${\bf t,\; n,\; b}$ ) along the curve. From these ${\bf t}$ , ${\bf n}$ , ${\bf b}$ the shape of the curve can be determined apart from a translation and rotation. For arbitrary speed curve the Frenet-Serret formulae are given by

$\displaystyle \begin{pmatrix} \dot{\bf t} \\ \dot{\bf n} \\ \dot{\bf b} \end{... ...u& 0 \end{pmatrix}\begin{pmatrix} {\bf t} \\ {\bf n}\\ {\bf b} \end{pmatrix},$

(2.57)

where $v=\frac{ds}{dt}$ is the parametric speed.

Example 2.4.1 As shown in Example 2.3.1 the intrinsic equations of circular helix are given by $\kappa(s)=\frac{a}{c^2}$ , $\tau(s)=\frac{b}{c^2}$ , where $c=\sqrt{a^2+ b^2}$ . In this example we derive the parametric equations of circular helix from these intrinsic equations. Substituting the intrinsic equations into the Frenet-Serret equations we obtain

$\displaystyle \frac{d{\bf t}}{ds}=\frac{a}{c^2}{\bf n},\;\;\;\;\;\; \frac{d{\bf... ...\frac{b}{c^2}{\bf b}, \;\;\;\;\;\; \frac{d{\bf b}}{ds}=-\frac{b}{c^2}{\bf n}\;.$

We first differentiate the first equation twice and the second equation once with respect to , which yield

$\displaystyle \frac{d^2 \mathbf{t} }{ds^2} = \frac{a}{c^2} \frac{d \mathbf{n}}{... ...c{d^2{\bf n}}{ds^2}=-\frac{a}{c^2}\frac{d{\bf t}}{ds}-\frac{b^2}{c^4}{\bf n}\;,$

where the third equation is used to replace $\frac{d{\bf b}}{ds}$ . Eliminating ${\bf n}$ , ${\frac{d{\bf n}}{ds}}$ , ${\frac{d^2{\bf n}}{ds^2}}$ and recognizing that $\mathbf{t} = \frac{d \mathbf{r}}{ds}$ , we obtain the fourth order differential equation

$\displaystyle \frac{d^4 \mathbf{r} }{ds^4}+ \frac{1}{c^2}\frac{d^2 \mathbf{r}}{ds^2}=0\;.$

The general solution to this differential equation is given by

$\displaystyle {\bf r}(s) = {\bf C}_1 + {\bf C}_2s + {\bf C}_3\cos\frac{s}{c} + {\bf C}_4\sin\frac{s}{c}\;,$

where ${\bf C}_1$ , ${\bf C}_2$ , ${\bf C}_3$ and ${\bf C}_4$ are the vector constants determined by the initial conditions. In this case we assume the following initial conditions

$\displaystyle {\bf r}(0)=(a, 0, 0)^T,\;\;\;\;\;\;{\bf r}'(0)=\left(0,\frac{a}{c... ...c^2},0,0\right)^T,\;\;\;\;\;\; {\bf r}'''(0)=\left(0,-\frac{a}{c^3},0\right)^T,$

which yield

$\displaystyle {\bf C}_1=(0,0,0)^T,\;\;\;\;\;\; {\bf C}_2=\left(0,0,\frac{b}{c}\right)^T, {\bf C}_3= (a,0,0)^T,\;\;\;\;\;\;{\bf C}_4=(0,a,0)^T\;,$

thus, we have ${\bf r}(s) = \left(a \cos \frac{s}{c}, a \sin \frac{s}{c}, \frac{bs}{c}\right)^T$ .

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December 2009