
\section{Torsion}
In this section,  we define the \textbf{Frenet Frame}, an orthonormal basis for $\mathbb{R}^3$ that is intrinsic to a smooth curve at a given point and changes as we move along that curve. We also introduce the notion of \textbf{torsion}: a measurement of how quickly the ``direction'' of the osculating plane changes along a curve $\mathcal{C} \subset \mathbb{R}^3$.  

Although the above gives a heuristic explanation of torsion, it is somewhat wrong---after all, a plane does not have a ``direction''.  However, we bypass this by instead measuring the rate of change of the direction of the unit normal vector to the osculating plane. 

Torsion is a \underline{scalar} quantity which is \underline{signed}; the $\pm$ sign depends upon whether the curve is right-handed or left-handed.



\subsection{Binormal Vector and Normal Plane}
The \textbf{binormal vector} is defined as
\begin{equation}
\boxed{
\mathbf{B} : = \mathbf{T} \times \mathbf{N}  \quad  \textrm{(Binormal Vector)}.
}
\end{equation}
This tells us that $\mathbf{B}$ is perpendicular to both $ \mathbf{T}$ and $\mathbf{N}$. Since the tangent and normal vectors span the osculating plane, this means that $\mathbf{B}$ is perpendicular to the osculating plane. This will be useful in finding equations of osculating planes.

The \textbf{normal plane} is defined as the plane spanned by $\mathbf{N}$ and $\mathbf{B}$. This plane is always perpendicular to $\mathbf{T}$ (intuitively, perpendicular to the curve at the given point). 

The vectors $\mathbf{T}, \mathbf{N}, \mathbf{B}$ together form a basis of orthogonal vectors for $\mathbb{R}^3$. This collection of vectors---defined relative to some curve $\mathbf{r}$, and changing as we move along the curve---is called the \textbf{Frenet frame} for $\mathbf{r}$.





\subsection{Torsion: Deriving a Formula} 
Consider the vector-valued function $\frac{d \mathbf{B}}{d s}$, with the derivative taken with respect to arclength parametrization.  We claim that this vector is perpendicular to both $\mathbf{B}$ and $\mathbf{T}$. 

Since $\vert \mathbf{B} \vert = 1$ for all $s$, it follows that $\frac{d \mathbf{B}}{ds} \perp \mathbf{B}$ as discussed previously (see Worked Problems in Section 1.1) . For $\mathbf{T}$, we compute the derivative of ${\bf B}$:
\begin{eqnarray*}
\frac{d \mathbf{B}}{ds} & = & \frac{d}{ds} \big(\mathbf{T} \times \mathbf{N} \big) = \frac{d \mathbf{T}}{ds} \times \mathbf{N} + \mathbf{T} \times \frac{d \mathbf{N}}{ds} \\[1ex]
\quad & = & \kappa \underbrace{(\mathbf{N} \times \mathbf{N})}_{= \mathbf{0}} + \underbrace{\mathbf{T} \times \frac{d \mathbf{N}}{ds}}_{\perp \mathbf{T}}
\end{eqnarray*} 
Since $\frac{d \mathbf{B}}{ds} \perp \mathbf{T}$ and also $\frac{d \mathbf{B}}{ds} \perp \mathbf{B}$, we know that $\frac{d \mathbf{B}}{ds}$ is parallel to the unit normal vector of the plane spanned by $\mathbf{T}$ and $\mathbf{B}$---which is the vector $\mathbf{N}$!

In particular,  for each $s$, there exists some scalar quantity $\tau = \tau (s)$ such that
\begin{equation}\label{1.5eq-torsion}
\boxed{
\frac{d \mathbf{B}}{ds} = - \tau \mathbf{N}.
}
\end{equation}
This scalar quantity $\tau = \tau (s)$ is called the \textbf{torsion} of $\mathbf{r}$ at $s$. (The letter $\tau$ (i.e. ``tau'') comes from the Greek alphabet,  and rhymes with ``cow''.)


\subsection{Torsion: A Worked Example}
Let us consider again the helix
$$
\mathbf{r} (t) = a \cos (t) \mathbf{i} + a \sin (t) \mathbf{j} + b t \mathbf{k}, \, a > 0.
$$
From earlier work, we know that
$$
\mathbf{T} (t) = - \frac{a}{\sqrt{a^2 + b^2}} \sin (t) \mathbf{i} + \frac{a}{\sqrt{a^2 + b^2}} \cos (t) \mathbf{j} + \frac{b}{\sqrt{a^2 + b^2}}\, \mathbf{k},
$$
$$
\mathbf{N} (t) = - \cos (t) \mathbf{i} - \sin (t) \mathbf{j},
$$
It follows that
\begin{align*}
\mathbf{B} (t) & = \mathbf{T} \times \mathbf{N} =
\frac{1}{\sqrt{a^2 + b^2}}
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
- a \sin (t) & a \cos (t) & b \\
- \cos (t) & - \sin (t) & 0
\end{vmatrix}
\\
&=
\frac{1}{\sqrt{a^2 + b^2}} (b \sin (t) \mathbf{i} - b \cos (t) \mathbf{j} + a \mathbf{k})
\end{align*}
We also know from previous examples that the curvature is given by
\begin{equation}\label{1.5eq-helixcurvature}
\kappa = \frac{a}{a^2 + b^2}
\end{equation}
We now calculate the torsion by comparing the left-hand and right-hand sides of \eqref{1.5eq-torsion}.  From the left-hand side, we rewrite
\begin{eqnarray*}
\frac{d \mathbf{B}}{ds} & = & \frac{d \mathbf{B} / dt}{ds /dt} = \frac{\mathbf{B}' (t)}{\sqrt{a^2 + b^2}} \\[1ex]
\quad & = & \frac{1}{\sqrt{a^2 + b^2}} \cdot \underbrace{\frac{1}{\sqrt{a^2 + b^2}} \big(b \cos (t) \mathbf{i} + b \sin (t) \mathbf{j} \big)}_{\mathbf{B}' (t)}
\end{eqnarray*}
Simplifying leads to the expression
$$
\frac{d \mathbf{B}}{ds} = \frac{-b}{a^2 + b^2} (- \cos (t) \mathbf{i} - \sin (t) \mathbf{j} ) = \frac{-b}{a^2 + b^2} \mathbf{N},
$$
and our torsion equation \eqref{1.5eq-torsion} then gives
\begin{equation}\label{1.5eq-helixtorsion}
\tau = \frac{b}{a^2 + b^2}.
\end{equation}

\begin{remark}
From equations \eqref{1.5eq-helixcurvature} and \eqref{1.5eq-helixtorsion}, we see that both the curvature and the torsion of the helix are constant and non-zero for all parameters $t$.  This is one reason why we use the helix for so many examples---because these quantities ($\kappa$ and $\tau$) are well-behaved.
\end{remark}

\subsection{The Frenet-Serret Formulas.} Notice that the vectors $\mathbf{T}' = \frac{d \mathbf{T}}{ds}$ and $\mathbf{B}' = \frac{d\mathbf{B}}{ds}$ both satisfy nice formulas in terms of the Frenet frame:
$$
\frac{d \mathbf{T}}{ds} = \kappa \mathbf{N}, \ \   \frac{d \mathbf{B}}{ds} = - \tau \mathbf{N}.
$$
This leads one to ask whether there is a similar formula for $\frac{d \mathbf{N} }{ds}$? We derive such a formula here.

\smallskip
Recall: the Frenet frame $\mathbf{T}, \mathbf{N}, \mathbf{B}$ consists of mutually perpendicular unit vectors.  By definition, we let $\mathbf{B} = \mathbf{T} \times \mathbf{N}$; however,  by mutual orthogonality,  we must also have
$$
\mathbf{N} = \mathbf{B} \times \mathbf{T}, \, \mathbf{T} = \mathbf{N} \times \mathbf{B}.
$$
Hence,  similar to our calculations for torsion, we have
\begin{eqnarray*}
\frac{d \mathbf{N}}{ds} & = & \frac{d}{ds} (\mathbf{B} \times \mathbf{T}) = \frac{d \mathbf{B}}{ds} \times \mathbf{T} + \mathbf{B} \times \frac{d \mathbf{T}}{ds} \\
\\
(\textrm{Definition of } \tau \textrm{ and } \kappa) & = & (- \tau \mathbf{N}) \times \mathbf{T} + \mathbf{B}\times (\kappa \mathbf{N}) \\[1ex]
\quad & = & \tau (\underbrace{\mathbf{T} \times \mathbf{N}}_{= \mathbf{B}}) - \kappa (\underbrace{\mathbf{N} \times \mathbf{B}}_{=\mathbf{T}}).
\end{eqnarray*}
Therefore, we have the formula
\begin{equation}\label{1.5eq-frenetnormal}
\frac{d \mathbf{N}}{ds} = \tau \mathbf{B} - \kappa \mathbf{T}.
\end{equation}
This leads to the \textbf{Frenet-Serret formulas},  which in matrix form are:
\begin{equation}\label{1.5eq-frenetmatrix}
\frac{d}{ds}
\begin{pmatrix}
\mathbf{T}\\
\mathbf{N}\\
\mathbf{B}
\end{pmatrix}
=
\begin{pmatrix}
0 & \kappa & 0 \\
- \kappa & 0 & \tau \\
0 & - \tau & 0 
\end{pmatrix}
\begin{pmatrix}
\mathbf{T} \\
\mathbf{N} \\
\mathbf{B}
\end{pmatrix}
\end{equation}

\subsection{Congruence of Curves in $\mathbb{R}^3$} We conclude with a  deeper result concerning the curvature and torsion of space curves. This is the \textit{Fundamental Theorem of Space Curves}. The proof is omitted.

\begin{theorem}
Suppose $\mathcal{C}_1$ and $\mathcal{C}_2$ are two curves with the same curvature $\kappa (t)$ and torsion $\tau (t)$,  and $\kappa (t) \neq 0$. Then, the curves $\mathcal{C}_1$ and $\mathcal{C}_2$ are \textbf{congruent}---that is, $\mathcal{C}_2$ can be translated and rotated so as to coincide with $\mathcal{C}_1$.
\end{theorem}

This allows us to immediately classify when two curves are, in the geometric sense of \textit{congruence}, ``copies'' of one another. As a nice and simple application, we have the following result.

\begin{corollary}
If $\mathcal{C}$ is a curve with $\kappa$ and $\tau$ both nonzero and constant (independent of $t$), then the curve is congruent to some helix $\mathcal{H}$. If $\kappa$ is nonzero and constant, but $\tau=0$ for all $t$, then the curve is a circle.


\end{corollary}


\section*{Worked Problems}
\begin{enumerate}[1.]




\item{\it Find  the equations of the normal and osculating plane for the helix
$$
\mathbf{r} (t) = \cos (t) \mathbf{i} + \sin (t) \mathbf{j} + t \mathbf{k},
$$
at $t = 0$. }

\medskip

To find the normal and osculating planes at $t = 0$, we need to determine their normal vectors $\mathbf{T}$ and $\mathbf{B}$ at $t = 0$. From the Worked Example above, with $a=b=1$, we have
\begin{align*}
\mathbf{T} (t) & = \frac{1}{\sqrt{2}} \sin (t) \mathbf{i} + \frac{1}{\sqrt{2}} \cos (t) \mathbf{j} + \frac{1}{\sqrt{2}} \mathbf{k},
\\
\mathbf{B} (t) & = \frac{1}{\sqrt{2}} (\sin (t) \mathbf{i} - \cos (t) \mathbf{j} + \mathbf{k}).
\end{align*}
At $t = 0$,  we have $\mathbf{r} = \langle 1, 0, 0 \rangle$, and
$$
\mathbf{T} = \frac{1}{\sqrt{2}} \langle 0,1,1 \rangle \quad \mathbf{N} = \langle -1, 0,0 \rangle, \quad 
\mathbf{B} = \frac{1}{\sqrt{2}}\langle 0, - 1,1\rangle
$$
The \underline{normal plane} must be perpendicular to $\jj+\kk$ (we took a multiple of $\mathbf{T}$ to simplify the calculations), and passes through $(1,0,0)$.  The equation of the plane is
$$
0 \cdot (x - 1) +  (y - 0) +  (z-0) = 0 \ \Leftrightarrow\  y + z = 0.
$$
Similarly, the \underline{osculating plane} is perpendicular to $-\jj+\kk$ and passes through $(1,0,0)$, so that its equation is
$$
0 \cdot (x - 1) -(y - 0) +(z - 0) = 0 \ \Leftrightarrow \ -y + z = 0.
$$
Hence,  our equations for the planes are:
\begin{eqnarray*}
\textrm{(Normal Plane)} & \quad & y+z = 0 \\[1ex]
\textrm{(Osculating Plane)} & \quad & -y +z = 0
\end{eqnarray*}








\item 
{\it Recall the Frenet-Serret formulas:
\begin{align*}
&{\bf T}'(s)=\kappa {\bf N},\\
&{\bf N}'(s)=-\kappa {\bf T}+\tau {\bf B},\\
&{\bf B}'(s)=-\tau {\bf N},
\end{align*}
where $\kappa$ is the curvature and $\tau$ is the torsion. Find a vector-valued function ${\bf W}$ such that these formulas can be written as 
${\bf T}'(s)={\bf W}\times {\bf T}$,  ${\bf N}'(s)={\bf W}\times {\bf N},$
${\bf B}'(s)={\bf W}\times {\bf B}$.}

\bigskip

Let ${\bf W}=a{\bf T}+b{\bf N}+c{\bf B}$, where $a,b,c$ are parameters. We are looking for $a,b,c$ such that
\begin{align*}
\kappa {\bf N}&= {\bf W}\times {\bf T}=b{\bf N}\times {\bf T}+c{\bf B}\times {\bf T}=-b{\bf B}+c{\bf N},
\\
-\kappa {\bf T}+\tau{\bf B}
&= {\bf W}\times {\bf N}=a{\bf T}\times {\bf N}+c{\bf B}\times {\bf N}=a{\bf B}-c{\bf T}
,\\
-\tau{\bf N}
&={\bf W}\times {\bf B}=a{\bf T}\times {\bf B}+c{\bf N}\times {\bf B}=-a{\bf N}+b{\bf T}.
\end{align*}
This means $a=\tau$, $b=0$, $c=\kappa$, ${\bf W}=\tau{\bf T}+\kappa{\bf B}$.

\end{enumerate}




\section*{Practice Problems}
\begin{enumerate}[1.]


\item
{
Let $C$ be the curve $x=e^{2t}$, $y=2e^t $, $z=t$ for $t \in\mathbb{R}$. 
Find all points 
on the curve where the 
osculating plane is parallel to the plane $x-6y+18z=0$. (If there are no such points, prove it.)}

\bigskip


\item
{
Prove that if ${\bf r}(t)$ is a smooth curve such that $\kappa\neq 0$ and ${\bf r}+\kappa^{-1}{\bf N}$ (the center of the osculating circle) is constant, then
${\bf r}(t)$ is a circle. 
(Hint: By the uniqueness theorem for curves (see above), it suffices to prove that $\kappa$ is constant and the torsion $\tau$ is 0.) }


\bigskip
\item Let $C$ be a smooth curve parametrized by $\bfr(t):\mathbb{R}\to\mathbb{R}^3$. Assume that (0,0,0) belongs to the osculating plane at some point $\bfr(t_0)$. Prove that at $t=t_0$,
$$
(\bfr\times \bfr')\cdot \bfr''=0.
$$




\end{enumerate}



