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How exactly to obtain Frenet Frame via GramSchmidt process? 
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#1
Aug1612, 09:45 AM

P: 108

I have a regular curve, [itex]\underline{a}(s)[/itex], in ℝ^{N} (parameterised by its arc length, [itex]s[/itex]).
To a running point on the curve, I want to attach the (Frenet) frame of orthonormal vectors [itex]\underline{u}_1(s),\underline{u}_2(s),\dots, \underline{u}_N(s)[/itex]. However, looking in different books, I find different claims as to how these should be obtained. Specifically, some books suggest that GramSchmidt should be applied to:[tex]\underline{a}^{\prime}(s), \underline{a}^{\prime \prime}(s), \dots , \underline{a}^{(N1)}(s)[/tex]while another book suggests that [itex]\underline{u}_{k+1}(s)[/itex] is obtained by applying GramSchmidt to [itex]\underline{u}_k^{\prime}(s)[/itex]. Which should I use? 


#2
Aug1612, 09:47 AM

P: 108

To add a little more detail.....
Since [itex]s[/itex] is an invariant parameter, I start with:[tex]\underline{u}_1(s) = \underline{a}^{\prime}(s)[/tex] Then, using [itex]\underline{a}^{\prime \prime}(s) = \underline{u}_1^{\prime}(s)[/itex] as the next linearly independent vector for GramSchmidt gives:[tex]\underline{u}_2 = \frac{ \underline{u}_1^{\prime}  (\underline{u}_1^T\underline{u}_1^{\prime}) \underline{u}_1}{\Vert numerator \Vert}[/tex] However, for [itex]\underline{u}_3, \underline{u}_4, \dots [/itex] the two approaches appear to become different. 


#3
Aug1912, 10:47 AM

Sci Advisor
P: 1,716

In three dimension a curve parameterized by arc length has acceleration perpendicular to the tangent. The cross product of the unit tangent with the normalized acceleration is perpendicular to both and this gives you the third vector in the frame. I do not believe that the Frenet frame can include other vectors so starting with an arbitrary basis and GramSchmiditfying will not work.
In higher dimesions there are many frames that extend the unit tangent and normalized acceleration. Gram Schmitt would work on a given basis but I am not sure what its geometric meaning would be. 


#4
Aug1912, 11:50 AM

P: 108

How exactly to obtain Frenet Frame via GramSchmidt process?
Thanks for your response.



#5
Aug1912, 02:58 PM

Sci Advisor
P: 1,716

Rather than defining a frame this way in terms of the motion along the curve. one could just pick some basis at every points and Gram Schmitify it to get an orthonormal frame. This would not be the Frenet frame in general. 


#6
Aug1912, 05:32 PM

P: 108

So, going back a few steps, is the following roughly true? If I zoom in super close to the curve, it looks like a straight line pointing in the direction of [itex]\underline{u}_1(s)[/itex]. I zoom out a bit, and actually the curve looks like a little circular arc lying in the plane of [itex]\underline{u}_1(s),\underline{u}_2(s)[/itex] and with radius equal to the inverse of the first curvature. Then I zoom out a bit more, and see that the curve actually lifts out from the plane of [itex]\underline{u}_1(s),\underline{u}_2(s)[/itex] in the direction of [itex]\underline{u}_3(s)[/itex]... so (locally) the curve looks like a piece of helix (?) (... and so on in N dimensions...) 


#7
Aug1912, 05:40 PM

P: 108

(I may have abused my notation there... I mean in the vicinity of some specific [itex]s[/itex], rather than the general definition of [itex]s[/itex] as arc length)



#8
Aug2312, 03:05 PM

P: 108

Hmm, it seems like this thread has dried up. Perhaps I can rephrase the original question.
Do the following two approaches yield the same result?[tex]\underline{u}_k(s)=\frac{\underline{a}^{(k)}(s)  \sum\limits_{m=1}^{k1}\left(\underline{u}_m^T(s)\underline{a}^{(k)}(s)\right) \underline{u}_m(s)}{\Vert numerator \Vert}[/tex]... suggested in, for example, [1, p. 13] (link) and [2] (link). [tex]\underline{u}_k(s)=\frac{\underline{u}_{k1}^{\prime}(s)  \sum\limits_{m=1}^{k1}\left(\underline{u}_m^T(s)\underline{u}_{k1}^{\prime}(s) \right) \underline{u}_m(s)}{\Vert numerator \Vert}[/tex]... suggested in, for example, [3, p. 159]. In other words, is the subspace spanned by [itex]\left\{\underline{a}^{\prime}, \underline{a}^{ \prime \prime}, \dots, \underline{a}^{(k)}\right\}[/itex] the same as the subspace spanned by [itex]\left\{\underline{u}_1, \underline{u}_2, \dots, \underline{u}_{k1}, \underline{u}_{k1}^{\prime} \right\}[/itex]? References: [1] W. Kühnel, "Differential Geometry: Curves  Surfaces  Manifolds". [2] Wikipedia, "Frenet–Serret formulas". [3] H. W. Guggenheimer, "Differential Geometry", McGraw Hill (or Dover Edition), 1963 (1977). 


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