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How exactly to obtain Frenet Frame via Gram-Schmidt process?

  1. Aug 16, 2012 #1
    I have a regular curve, [itex]\underline{a}(s)[/itex], in ℝN (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 Gram-Schmidt should be applied to:[tex]\underline{a}^{\prime}(s), \underline{a}^{\prime \prime}(s), \dots , \underline{a}^{(N-1)}(s)[/tex]while another book suggests that [itex]\underline{u}_{k+1}(s)[/itex] is obtained by applying Gram-Schmidt to [itex]\underline{u}_k^{\prime}(s)[/itex].

    Which should I use?
  2. jcsd
  3. Aug 16, 2012 #2
    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 Gram-Schmidt 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.
  4. Aug 19, 2012 #3


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    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 Gram-Schmiditfying 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.
  5. Aug 19, 2012 #4
    Thanks for your response.
    Could you expand on what you mean? Is it possible/likely that both methods I mentioned are valid?

    What I'd like to do is define the Cartan matrix (containing curvature, tortion, ..., general curvatures) and so develop expressions relating the moving frame, [itex]\underline{u}_1(s),\underline{u}_2(s),\dots, \underline{u}_N(s)[/itex], and its derivative, [itex]\underline{u}_1^{ \prime}(s),\underline{u}_2^{ \prime}(s),\dots, \underline{u}_N^{ \prime}(s)[/itex]. I find that this is straightforward using the second method I mentioned in my original post ("while another book suggests that..."). However, before I had checked in books, I had thought the first method would be the way to go...
  6. Aug 19, 2012 #5


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    The second method seems right because it is defined by the motion along the curve. Each successive vector in the frame points in the direction that the hyperplane spanned by the previously defined vectors is moving. It is possible though that there will be points on the curve where these derivatives are zero.

    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.
  7. Aug 19, 2012 #6
    Ah, this is very insightful. In fact, I hadn't realised that this is probably the whole point of what I'm trying to do.

    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...)
  8. Aug 19, 2012 #7
    (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)
  9. Aug 23, 2012 #8
    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}^{k-1}\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}_{k-1}^{\prime}(s) - \sum\limits_{m=1}^{k-1}\left(\underline{u}_m^T(s)\underline{u}_{k-1}^{\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}_{k-1}, \underline{u}_{k-1}^{\prime} \right\}[/itex]?

    [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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