Serret-Frenet equations in ${ℝ}^2$

Given a plane curve, we may associate to each point on the curve an orthonormal basis consisting of the unit normal tangent vector and the unit normal. In general, different points will have different bases associated to them, so we may ask how the basis depends upon the choice of point. The Serret-Frenet equations answer this question by relating the rte of change of the basis vectors to the curvature of the curve.

Suppose $I$ is an open interval and $c:I\to {ℝ}^2$ is a twice continuously differentiable curve such that $\parallel c^{\prime }\parallel =1$. Let us then define the tangent vector and normal vector as

$𝐓$ $=$ $c^{\prime },$

$𝐍$ $=$ $J\cdot 𝐓,$

where is the rotational matrix that rotates the plane $90$ degrees counterclockwise.

Curvature

Differentiating $⟨c^{\prime },c^{\prime }⟩=1$ yields $⟨{𝐓}^{\prime },𝐓⟩=0$, so ${𝐓}^{\prime }$ is in the orthogonal complement of $𝐓$, which is $1$-dimensional. Since $J\cdot 𝐓$ is also in the orthogonal complement, it follows that there exists a function $\kappa :I\to ℝ$ such that

$${𝐓}^{\prime }=\kappa J\cdot 𝐓.$$

Furthermore, $\kappa$ is uniquely determined by this equation. We define this unique $\kappa$ to be the curvature of $c$. Explicitly,

$$\kappa =⟨{𝐓}^{\prime },J\cdot 𝐓⟩.$$

Serret-Frenet equations

By the definition of curvature

${𝐓}^{\prime }$ $=$ $\kappa J\cdot 𝐓=\kappa 𝐍$

and so

${𝐍}^{\prime }$ $=$ $J\cdot {𝐓}^{\prime }=\kappa J𝐍=-\kappa 𝐓$

since $J^2=-\mathrm{I}$. These are the Serret-Frenet equations

Title	Serret-Frenet equations in ${ℝ}^2$
Canonical name	SerretFrenetEquationsInmathbbR2
Date of creation	2013-03-22 15:16:57
Last modified on	2013-03-22 15:16:57
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	8
Author	rspuzio (6075)
Entry type	Definition
Classification	msc 53A04
Related topic	SerretFrenetFormulas