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Using Frenet-Serret formulas how to find positions of curves in 3-space when calculating the two space curves:

$ \kappa= 1, \tau=1; $

$ \kappa= \cos (s/a), \tau= \sin(s/a). $

Next, when $ \ Q = \left [ \begin{array}{c} T \\ N \\ B \end{array} \right ] $ , can we treat $ T,N,B$ of the frame in the same way separately as three scalar functions to integrate the three coupled equations: $ { Q^{'}}= \left [ \begin{array}{} 0 & \kappa & 0 \\ -\kappa & 0 & \tau \\ 0 & -\tau & 0 \end{array} \right ] \cdot Q \,? $

Narasimham
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1 Answers1

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Regarding the first question, Frenet-Serret formulas don't tell anything about the initial position, or the initial orientation, of the curve they determine locally. This is exactly the difference between an ODE and an IVP; without initial conditions one can only consider the (existence and uniqueness) of the general solution. In order words, if one only knows $\kappa$ and $\tau$ of a curve in $\mathbb{R}^3$, one can't determine where the curve is, or if it's coming closer to a predetermined destination.

Regarding the second question, as Chappers commented one can transform the Frenet-Serret ODE with the unknown taking values in $M_3(M_{3\times 1}(\mathbb{R}))$ (where for $V$ a vector space $M_{n\times m}(V)$ denotes the vector space of $n\times m$ matrices whose entries are vectors in $V$) to an ODE with the unknown taking values in $M_{9\times 1}(\mathbb{R})$, as $M_3(M_{3\times 1}(\mathbb{R}))\cong M_{9\times 1}(\mathbb{R})$ as vector spaces. Note that here there is some choice involved; e.g. the isomorphism

$$\phi: \begin{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\\\begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix}\\\begin{pmatrix}z_1\\z_2\\z_3\end{pmatrix}\end{pmatrix}\mapsto \begin{pmatrix}x_1\\x_2\\x_3\\y_1\\y_2\\y_3\\z_1\\z_2\\z_3\end{pmatrix}$$

transforms the Frenet-Serret ODE into

$$\left(\begin{array}{c}T_1\\T_2\\T_3\\\hline N_1\\N_2\\N_3\\\hline B_1\\B_2\\B_3\end{array}\right)'= \left(\begin{array}{ccc|ccc|ccc} 0 & 0 & 0 & \kappa & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \kappa & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \kappa & 0 & 0 & 0 \\\hline -\kappa & 0 & 0 & 0 & 0 & 0 & \tau & 0 & 0 \\ 0 & -\kappa & 0 & 0 & 0 & 0 & 0 & \tau & 0 \\ 0 & 0 & -\kappa & 0 & 0 & 0 & 0 & 0 & \tau \\\hline 0 & 0 & 0 & -\tau & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & -\tau & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -\tau & 0 & 0 & 0 \end{array}\right) \left(\begin{array}{c}T_1\\T_2\\T_3\\\hline N_1\\N_2\\N_3\\\hline B_1\\B_2\\B_3\end{array}\right).$$

However note that it is more natural to consider the Frenet-Serret equation as an equation for triples for $3$-dimensional vectors, as this equation really is an ODE on $SO(3,\mathbb{R})$, which is exactly the symmetry/ ambiguity group associated to the first problem (up to translations in $\mathbb{R}^3$).

Alp Uzman
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