# Differential Geometry/Torsion

Consider a curve of class of at least 3, with a nonzero curvature.

We shall now consider the rate of change of the osculating plane. Of course, when the curve is a plane curve, the osculating plane is the same as the plane of the curve, so it does not change, and consequently, the binormal vector also does not change. In other words, the derivative of the binormal vector, ${\displaystyle {\frac {db}{ds}}}$, is 0.

Thus, let us consider the derivative of the binormal vector.

Consider the two equations ${\displaystyle b\cdot b=1}$ and ${\displaystyle b\cdot t=0}$ and differentiate both of them in respect to the arc length parameter, to obtain ${\displaystyle b\cdot {\frac {db}{ds}}=0}$ and ${\displaystyle {\frac {db}{ds}}\cdot t+b\cdot {\frac {dt}{ds}}}$ and consequently ${\displaystyle {\frac {db}{ds}}\cdot t=-b\cdot {\frac {dt}{ds}}=-\kappa b\cdot p=0}$ indicating that ${\displaystyle {\frac {db}{ds}}}$ is orthogonal to both b and p and thus lies on the principal normal line. Thus, we define the torsion ${\displaystyle \tau (s)}$ to be the proportionality between them:

${\displaystyle {\frac {db}{ds}}(s)=-\tau (s)p(s)}$

The negative sign is there so that it turns out that the torsion is positive for a right-handed helix.

We can take the dot product on both sides to get ${\displaystyle \tau (s)=-p(s)\cdot {\frac {db}{ds}}(s)}$

Its reciprocal, ${\displaystyle {\frac {1}{\tau }}}$ is called the radius of torsion.

## Exercises

1. Prove that a curve is a plane curve if and only if its torsion is the null vector at every point.