\part{Frames}

\chapter{Frames of coordinate systems}

Readers will be familiar with coordinate systems.

Coordinate frame is a Basis that formed based on some arbitrary 'origin' coordinate. It encapsulates the idea 
For instance, at some coordinate $p$, one may want a frame representing the directions of the increase of each coordinate (cartesian, cylindrical, spherical frames), or perhaps one wants a frame with directions along a curve, orthogonal to the bend of a curve, and another direction orthogonal to both of those (Frenet-Serret frame).


We first study the former; frames of coordinate systems.

Note that although many frames we use are orthonormal basis', but this is not a requirement; we'll see examples of these much later on.

coordinate frame
\begin{definition}
Let $x$ be a coordinate system and $\mathbf{r}$ map the coordinate system to $S$, then the \emph{orthonormal coordinate frame with respect to $p$} is the orthonormal basis of vectors defined by the set of $\mathbf{e}_{x}(p) = \frac{ \mathbf{r}_{x}(p) }{ \| \mathbf{r}_{x}(p) \| }$ for each coordinate.
\end{definition}


scale factor

divide by scale factor to make normalized coordinate frame




Cartesian frame

\[(x,y,z)\]

\[\mathbf{r}(x,y,z) = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\]

	\[ \mathcal{F} =  \{  \mathbf{e}_{x} , \mathbf{e}_{y} ,  \mathbf{e}_{z} \}   = \{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} ,  \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} ,  \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \} = \{ \mathbf{i} \mathbf{j}, \mathbf{k}\} \]

This is a \emph{fixed frame}; the frame is the same relative to any coordinate point. Often students are confused when learning other coordinate frames because the cartesian coordinate frame is fixes, and they expect all others to be fixed too, but the fact that we have nonfixed frames is precisely why we like to deal with them.


Cylindrical frame
\[(r,\theta,z)\]
\[\mathbf{r}(r,\theta,z) = \begin{bmatrix} r \cos (\theta)  \\ r \sin (\theta) \\ z \end{bmatrix}\]

\[ \mathcal{F} =  \{  \mathbf{e}_{r} , \mathbf{e}_{\theta} ,  \mathbf{e}_{z} \}   = \{ \begin{bmatrix} \cos (\theta) \\ \sin (\theta) \\ 0 \end{bmatrix} ,  \begin{bmatrix} -\sin (\theta) \\ \cos (\theta) \\ 0 \end{bmatrix} ,  \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \} \]


Spherical frame
\[(\rho,\theta,\phi)\]
\[ \mathbf{r}(\rho,\theta,\phi) = \begin{bmatrix} \rho \sin (\phi) \cos (\theta)  \\ \rho \sin (\phi) \sin (\theta) \\ \rho \cos (\phi) \end{bmatrix}\]
	\[ \mathcal{F} =  \{  \mathbf{e}_{\rho} , \mathbf{e}_{\theta} ,  \mathbf{e}_{\phi} \}   = \{ \begin{bmatrix} \sin (\phi) \cos (\theta) \\ \sin (\phi) \sin (\theta) \\ \cos (\phi) \end{bmatrix} ,  \begin{bmatrix} \cos (\phi) \cos (\theta) \\ \cos (\phi)  \sin (\theta) \\ -\sin (\phi) \end{bmatrix} ,  \begin{bmatrix} \cos (\phi) \\ -\sin (\phi) \\ 0 \end{bmatrix} \} \]



\section{Applications for line integrals}
\section{Applications for surface and flux integrals}


\chapter{Frenet-Serret frame}


Say we want to make an orthonormal frame not based on coordinates of points, but based on position along a differentiable curve with the direction of its tangent, its direction of acceleration and then direction orthomormal to both of those.

For some differentiable curve $\mathbf{r}$, $\mathbf{r}'$ represents it's 'tangent'; the direction that the curve is currently 'facing'. We'd love a normalized version of this to be in this 'differentiable curve frame' of ours. 

We'd also like its 'acceleration' $\mathbf{r}''$; this tells us in which direction the curve is 'curving' towards. However is taking the derivative of a vector function sufficient to give us a new orthogonal vector functions?

\begin{proposition}
\[\mathbf{f} \perp \mathbf{f}'\]
\end{proposition}

So far so good, we normalize this 'acceleration' vector and add it to our frame. The last member of the frame can be done by taking the cross product of the 2 vectors; since they are both normalized, the result of this is also a normal vector.

The frame we've just described is called the \emph{Frenet-Serret frame}.

\section{Frenet-Serret frame}

\begin{definition}
Let $\mathbf{r}$ be a differentiable curve, the \emph{Frenet-Serret frame of $\mathbf{r}$} is an orthonormal basis spanning \(\mathbb{R}^3\) relative to a a point on the differentiable curve \(\mathbf{r}(s)\) that is parametrized by arclength \(s\).

	\[ \mathbf{T}(s) := \mathbf{r}'(s)\]
	\[ \mathbf{N}(s) := \frac{\mathbf{T}'(s)}{\| \mathbf{T}'(s) \|}\]
	\[ \mathbf{B}(s) := \mathbf{T}(s) \times \mathbf{N}(s)\]
	$\mathbf{T}$ (tangent) direction tangent to the differentiable curve
	$\mathbf{N}$ (normal) direction the differentiable curve is turning into against its tangent
	$\mathbf{B}$ (binormal) direction orthogonal to $\mathbf{T}, \mathbf{N}$

\end{definition}

By the chain rule, one can express the Frenet-Serret frame with respect to time rather than arclength.

\begin{proposition}
	\[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\mathbf{r}'(t)}\]
	\[ \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\| \mathbf{T}'(t) \|}\]
	\[ \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t)\]
\end{proposition}

For developing mathematical theory, it is more natural to deal with arclength parametrizations, however calculations may be easier with respect to time instead. We'll therefore continue using arclength parametrizations for theory and use time parametrizations in concrete examples.







\begin{definition}
The \emph{curvature of a differentiable curve} is the measure of directional change on the tangential plane.
\[ \kappa (s) = \| \mathbf{T}'(s) \| \]
\[ \kappa (t) = \frac{\| \mathbf{T}'(t) \|}{\| \mathbf{r}'(t)\|} \]
\end{definition}


\begin{definition}
The \emph{torsion of a differentiable curve} is the measure of directional change out of the tangential plane.
\[ \tau(s) = \| \mathbf{B}'(s) \| \]
\[ \tau (t) = \frac{\| \mathbf{B}'(t) \|}{\| \mathbf{r}'(t)\|} \]
\end{definition}

\begin{theorem}[Frenet-Serret formulae]
\[ \mathbf{T}'(s) = \kappa(s) \mathbf{N}(s) \]
\[ \mathbf{N}'(s) = -\kappa(s) \mathbf{T}(s) +\tau(s) \mathbf{B}(s) \]
\[ \mathbf{B}'(s) = -\tau(s) \mathbf{N}(s) \]


\[ \begin{bmatrix} \mathbf{T}'(s) \\  \mathbf{N}'(s) \\ \mathbf{B}'(s) \end{bmatrix} = \begin{bmatrix} 0 & \kappa(s) & 0 \\ -\kappa(s) & 0 \tau (s) \\ 0 & -\tau(s) & 0 \end{bmatrix} \begin{bmatrix}  \mathbf{T}(s)  \\ \mathbf{N}(s) \\ \mathbf{B}(s) \end{bmatrix}\]
\[  = -\kappa(s) \mathbf{T}(s) +\tau(s) \mathbf{B}(s) \]
\[ \mathbf{B}'(s) = -\tau(s) \mathbf{N}(s) \]

\end{theorem}



\section{Curvature}

\[ \kappa(t) = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t)\|^{3}} \]
\section{Torsion}
\[ \tau(t) = \frac{ ( \mathbf{r}'(t) \times  \mathbf{r}''(t) )\cdot \mathbf{r}'''(t)  }{\| \mathbf{r}'(t) \times \mathbf{r}''(t)\|^{2}}\]


\begin{itemize}
\item Osculating plane $\mathrm{span}\{ \mathbf{T} , \mathbf{N} \}$
\item Rectifying plane $\mathrm{span}\{ \mathbf{T} , \mathbf{B} \}$
\item Normal plane $\mathrm{span}\{ \mathbf{N} , \mathbf{B} \} $
\end{itemize}



