2
$\begingroup$

I am a novice in mathematics in general and even more so in differential geometry. Currently, I am looking to generalize the Frenet-Serret formulas to $n$ dimensions. At the moment, I am interested in the generalization of curvature to n dimensions. On this Wikipedia page, a formula for generalized curvature is proposed, which is as follows: $$ X_i(s)=\frac{\langle e'_i(s),e_{i+1}(s)\rangle}{\|r'(s)\|} $$ On the same page, $$e_j(s) = \frac{\bar{e}_j(s)}{\|\bar{e}_j(s)\|}$$ with $$\bar{e}_j(s) = r^{(j)}(s) - \sum_{i=1}^{j-1} \langle r^{(j)}(s),e_i(s)\rangle e_i(s)$$.

The problem, for me, is that both $X_i(s)$ and $e_j(s)$ are parameterized by the arc length $s$. What I am looking for is rather than these expressions being expressed as a function of time $t$.

My question is, how do I transition from $X_i(s)$ and $e_j(s)$ to $X_i(t)$ and $e_j(t)$?

$\endgroup$
6
  • 4
    $\begingroup$ Where did you see that they are treating $s$ as arclength rather than an arbitrary parameter? If $s$ were indeed arclength, then one would have $\|r’(s)\|=1$. $\endgroup$
    – Ted Shifrin
    Commented Apr 4 at 15:07
  • $\begingroup$ Conceptually, the natural approach is to compute the first $n$ derivatives of the path, then (assuming the derivatives comprise a linearly independent set at each point) perform Gram-Schmidt on the ordered set of derivatives to get an orthonormal basis. That is (more-or-less precisely) what the formulas for the $\bar{e}$ accomplish. $\endgroup$
    – Andrew D. Hwang
    Commented Apr 4 at 15:21
  • 1
    $\begingroup$ they are not parametrized by arc length $\endgroup$
    – Masacroso
    Commented Apr 4 at 20:02
  • 1
    $\begingroup$ One issue with the wikipedia concept is that different people may author different parts of the same article, and there is no editorial overview to ensure consistency between the parts. Whoever added the $n$-dimensions section gave formulas applicable to a general parametrization, even though they used the variable $s$ which conventionally denotes arc-length parametrization. $\endgroup$
    – Paul Sinclair
    Commented Apr 5 at 17:05
  • 1
    $\begingroup$ If $\mathbf r(t)$ is a curve parametrised by $t$, then we can define $s(t) = \int_0^t \|\dot{\mathbf r}(\sigma)\|\,d\sigma$ as in the article. Then the arclength parametrization $\mathbf q(s)$ of the same curve satisfies $\mathbf q(s(t)) = \mathbf r(t)$. And $$\dot{\mathbf r}(t) = \mathbf q'(s(t))\dot s(t)$$ by the chain rule, where the dot denotes differentiation with respect to $t$ and the prime denotes differentiation with respect to $s$. $\dot s(t)$ is a scalar and thus pulls out of all the vector calculations as a multiplier. $\endgroup$
    – Paul Sinclair
    Commented Apr 5 at 17:13

0

Reset to default

You must log in to answer this question.