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I'm trying to understand how the Frenet frame is formed from the normal and tangent on a curve in $\mathbb{R}^3$.

For a curve $\gamma (s)$ in $\mathbb{R}^3$ parameterised by arc length let $T(s) = \gamma' (s)$ be the unit tangent.

From a previous questions, I understand that from $T . T = 1 $ we get $T . T' = 0$. Then this gives us that $T'(s) = \kappa (s) N(s)$ for $\kappa(s) \in \mathbb{R}$ for $N(s)$ the unit normal.

BUT surely in $\mathbb{R}^3$ there are infinitely many normals as the vector can just 'rotate' round the curve whilst remaining perpendicular to the tangent and so I can't my head around how we can deduce $T'(s) = \kappa N(s)$ just from that fact $T . T' = 0$ ?

Thanks!

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By definition $N(s)=T'(s)/\Vert T'(s)\Vert$ –  user8268 Dec 17 '12 at 10:04
    
Then surely $\kappa$ is always $1$ if $T$ is the unit tangent? –  user53076 Dec 17 '12 at 10:18
    
no; $\kappa=\Vert T'\Vert$ –  user8268 Dec 18 '12 at 21:22

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Let's briefly talk about the case of plane curves first. In that case, there are some "canonical" ways of getting normal vector $N(s)$ from a unit tangent vector $T(s)$ - one way would be rotating the tangent vector by 90 degrees anti-clockwisely. If we define normal vector $N(s)$ this way, then it did not a priori depend on $T'(s)$, but then a calculation would show that $T'(s)$ and $N(s)$ lies on the same line, and curvature can be expressed as the ratio of their lengths.

For space curves, like you said there's no "canonical" way of defining normal vector by rotation, since there are infinitely many directions to rotate. So we do NOT define normal vectors by this method. Instead, we use the clue in the case of plane curves, and define $N(s) = \frac{T'(s)}{\|T'(s)\|}$. This way we can at least get a unit vector that is always perpendicular to the unit tangent vector $T(s)$.

Would $\kappa$ be always 1 then? NO. It should actually be $\|T'(s)\|$, assuming that you started with arc length parametrization. Again this is just the ratio of lengths of $T'(s)$, and $N(s)$.

We do lose something in this case though. In the case of plane curves, $N(s)$ may not have the same sign as $T'(s)$, so the sign of curvature has meaning - whether it is locally convex or concave. This does not carry over to the case of space curves.

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In the three-dimensional case there is a small proviso, namely the condition $T'(s)\ne0$ for the considered $s$ values. –  Christian Blatter Dec 17 '12 at 13:08

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