📂Geometry

Frenet-Serret Formulas: Curvature, Tangent, Normal, Binormal, Torsion

Definition ¹

Let’s say $\alpha$ is a unit speed curve.

1. The speed $\kappa (s) := \left| T^{\prime}(s) \right|$ of the tangent $T(s) = \alpha^{\prime} (s)$ is called the curvature $\alpha (s)$.
2. The function obtained by dividing the velocity $T^{\prime}(s)$ of the tangent of $\alpha$ by the curvature $\kappa (s)$, namely, defined as $N$, is referred to as the Normal vector field. $$ N(s) := {{ T^{\prime}(s) } \over { \left| T^{\prime}(s) \right| }} = {{ T^{\prime}(s) } \over { \kappa (s) }},\qquad \kappa (s) \ne 0 $$
3. The Binormal vector field is defined as follows: $$ B(s) := T(s) \times N(s) $$
4. The scalar function defined as follows $\tau$ is called Torsion. $$ \tau (s) := - \left< B^{\prime}(s) , N (s) \right> $$
5. The collection of tangent, normal, binormal, curvature, and torsion as defined above is called the Frenet-Serret Apparatus. $$ \left\{ T(s), N(s), B(s), \kappa (s), \tau (s) \right\} $$

Explanation

• Do not waste time worrying about the differences between speed and velocity in the definitions. If confused, it’s better to simply remember it as speed.
• Especially, the orthogonal set $\left\{ T(s), N(s), B(s) \right\}$ is called the Frenet-Serret Frame.
• Although $\alpha$ is a unit speed curve in the definition, it can be generalized simply.

Curvature

While those studying the original text may be familiar with the pronunciation “curvature,” it’s more natural to write it as curvature in text.

In formulas, curvature simply represents how much the tangent (the direction) changes. The more a curve’s direction changes implies how curved it is, thus it’s a valid definition. Of course, if $\kappa (s) = 0$ then the curve is a straight line.

The definition in the case of non-unit speed $\alpha$ is as follows:

$$ \begin{equation} \kappa (t) := \dfrac{\left| T^{\prime}(t) \right|}{\left| \alpha^{\prime}(t) \right|} \end{equation} $$

Normal

As usual, Normal is not translated as ‘regular’ but rather naturalized as ‘perpendicular’. Though it applies to Normal, considering it together with Binormal shows that this might not be the best translation. In the context of geometry, understanding Normal as ‘perpendicular’ tends to be better for mental health. In this sense, being ‘Normal’ could be interpreted as being ‘perpendicular’ to something.

The magnitude of the normal vector is always fixed as $1$, and since $\alpha^{\prime} = T$, $N$ corresponds to the second derivative of $\alpha$. Considering that torsion involves up to $B^{\prime} = T^{\prime} \times N + T \times N^{\prime}$, it can be seen that to deal with the Frenet-Serret Apparatus, $\alpha$ must be differentiable at least three times.

It’s also referred to as the Principle Unit Normal Vector.

Binormal

Binormal is naturalized as ‘Secondary Normal,’ suggested by the name that it follows the normal. Following the interpretation of ’normal’ mentioned above, Bi (2, two) attached to Binormal gives the sense of ‘being perpendicular to both the tangent and normal’.

This becomes clearer with formulations. The very definition of Binormal being the cross product of tangent and normal, from the name alone, it appears to have been decided to be perpendicular to both tangent and normal.

Torsion

Torsion translates directly to ’twisting’ and even in Japanese incorporates 天地² with Hiragana mixed in. Such translations attempt to preserve the meaning of Torsion, all being somewhat unclear, and the term ’twisting’ itself isn’t quite appropriate as an academic term, so it might be more comfortable to just call it [Torsion] as is.

It’s not particularly important whether minus $-$ is included; it’s merely convention.

Looking at the formula, it’s incredibly difficult to understand why torsion is defined this way, and more study is needed to accept this formulaically. For now, it can be understood as ‘defining it this way makes many formulas neat,’ but this definitely isn’t a good attitude when studying geometry. Let’s try to intuitively understand it, even if it seems forced.

Following $\alpha (s)$, the plane that is perpendicular to $B(s)$ is called the Osculating Plane³. If $B$ is constant, that is, always the same, then the osculating plane of the curve $\alpha$ won’t change, hence it will lie within a plane as in $\mathbb{R}^{3}$.

Meanwhile, as mentioned earlier, since the speed of the normal vector is always $|N| = \left| T^{\prime} / \left| T^{\prime} \right| \right| = 1$, the magnitude of the torsion $\tau$ is proportional to the magnitude of $B^{\prime}$. Considering this with the above osculating plane, the magnitude of the torsion $\left| - \left< B^{\prime}(s) , N (s) \right> \right|$ can be seen as a measure of ‘how much the curve attempts to deviate from the osculating plane’. As previously mentioned, if $B$ is constant and $B^{\prime} = 0$, it means the curve has no intention of leaving its (osculating) plane.

Physical Meaning⁴

The curve $\alpha$ can be seen as a position in physics. $\alpha^{\prime}$ is called velocity because $\alpha$’s variable is considered as time, and the function’s value as position, in fact, this concept is a mathematical abstraction of the curve in differential geometry. Now, let’s assume a body is moving along the curve $\alpha$. In physics, position is expressed as $\mathbf{r}$.

$$ \alpha (t) = \mathbf{r}(t) $$

Since velocity is the derivative of position, $\mathbf{v} = \dfrac{d \mathbf{r}}{dt} = \dfrac{d \alpha}{d t}$ is obtained. By the definition of tangent, $T = \dfrac{\alpha^{\prime}}{\left| \alpha^{\prime} \right|} = \dfrac{\mathbf{v}}{v}$ is obtained.

$$ \mathbf{v} = v T $$

Now, let’s differentiate both sides by $t$. The derivative of velocity is acceleration, so

$$ \mathbf{a} = \dfrac{d \mathbf{v}}{dt} = v^{\prime} T + v T^{\prime} $$

By $(1)$, we obtain the following:

$$ \kappa = \dfrac{\left| T^{\prime} \right|}{\left| \mathbf{r}^{\prime} \right|} = \dfrac{\left| T^{\prime} \right|}{v} \implies \left| T^{\prime} \right| = v\kappa $$

By the definition of the normal, we obtain the following:

$$ N = \dfrac{T^{\prime}}{\left| T^{\prime} \right|} = \dfrac{T^{\prime}}{v\kappa} \implies T^{\prime} = v\kappa N $$

Therefore, acceleration can be expressed as follows:

$$ \mathbf{a} = v^{\prime}T + v^{2}\kappa N $$

This shows that acceleration is a vector in the Osculating Plane and is represented as a linear combination of $T$ and $N$.