Curvilinear Coordinates in Geometric Algebra |
| Advanced Science |
| 29 Sep 2022 |
Example:
$$ x = r \cos { \varphi } $$ $$ y = r \sin { \varphi } $$
$$ r = \sqrt{ { x }^{ 2} + { y }^{ 2}}$$ $$ \tan { \varphi } = \frac{ y}{ x}$$
We only allow coordinates $ \left\{ { w }_{ k}\right\}$ where the coordinate change $ \left\{ { w }_{ k}\right\} \rightarrow \left\{ { r }_{ i}\right\}$ is continuously differentiable and has an invertible diffential at any point in $ U$. Therefore the partial derivatives $ { \, \partial r }_{ i} / { \, \partial w }_{ k} \,$ and $ { \, \partial w }_{ i} / { \, \partial r }_{ k} \,$ exist and are continuous.
Let's check the existence of $ { \, \partial r }_{ i} / { \, \partial w }_{ k} \,$ for polar coordinates.
$$ \frac{\partial \left( { x \left( {r,\varphi } \right)} \right)}{\, \partial r} = \cos { \varphi } $$ $$ \frac{\partial \left( { x \left( {r,\varphi } \right)} \right)}{\, \partial \varphi } = - r \sin { \varphi } $$
$$ \frac{\partial \left( { y \left( {r,\varphi } \right)} \right)}{\, \partial r} = \sin { \varphi } $$ $$ \frac{\partial \left( { y \left( {r,\varphi } \right)} \right)}{\, \partial \varphi } = r \cos { \varphi } $$
Let's check the existence of $ { \, \partial w }_{ i} / { \, \partial r }_{ k} \,$ for polar coordinates.
$$ \frac{\partial \left( { r \left( {x,y} \right)} \right)}{\, \partial x} = \frac{ 1}{ 2} \frac{ 2 x}{ \sqrt{ { x }^{ 2} + { y }^{ 2}}}$$ $$ \frac{\partial \left( { r \left( {x,y} \right)} \right)}{\, \partial y} = \frac{ 1}{ 2} \frac{ 2 y}{ \sqrt{ { x }^{ 2} + { y }^{ 2}}}$$
$$ \frac{ \partial \varphi \left( {x,y} \right)}{\, \partial x} = - \frac{ y}{ { x }^{ 2} + { y }^{ 2}}$$ $$ \frac{ \partial \varphi \left( {x,y} \right)}{\, \partial y} = \frac{ x}{ { x }^{ 2} + { y }^{ 2}}$$
A curvilinear coordinate system $ \left\{ { w }_{ k}\right\}$ determines two bases at each point in $ U$, $ \left\{ { \vec{w} }_{ k}\right\}$ and $ \left\{ { \vec{w} }^{ j}\right\}$ in which $ j$ is a superscript, not an exponent. They are defined by
$$ { \vec{w} }_{ k} = \frac{\, \vec{\partial r}}{ { \, \partial w }_{ k}} = \frac{ \partial \left( { { r }_{ i} { \vec{e} }_{ i}} \right)}{ { \, \partial w }_{ k}} = \frac{ \partial { r }_{ i}}{ { \, \partial w }_{ k}} { \vec{e} }_{ i}$$
and
$$ { \vec{w} }^{ j} = \nabla { w }_{ j} = { \vec{e} }_{ i} { \partial }_{ i} { w }_{ j} = \frac{ \partial { w }_{ j}}{ { \, \partial r }_{ i}} { \vec{e} }_{ i}$$
Example: We define
$$ \vec{r} = r \cos { \varphi } { \vec{e} }_{ x} + r \sin { \varphi } { \vec{e} }_{ y}$$
What are the basis vectors $ { \vec{w} }_{ r}$ and $ { \vec{w} }_{ \varphi }$ for the polar coordinates definition? Generally we have
$$\fbox{$ \displaystyle { \vec{w} }_{ k} = \frac{ \partial { r }_{ i}}{ { \, \partial w }_{ k}} { \vec{e} }_{ i} $} \qquad \mbox{Curvilinear base vectors} \tag{1}$$
and thus in our case
$$ { \vec{w} }_{ r} = \frac{ \partial x \left( {r,\varphi } \right)}{\, \partial r} { \vec{e} }_{ x} + \frac{ \partial y \left( {r,\varphi } \right)}{\, \partial r} { \vec{e} }_{ y} = \cos { \varphi } { \vec{e} }_{ x} + \sin { \varphi } { \vec{e} }_{ y}$$
$$ { \vec{w} }_{ \varphi } = \frac{ \partial x \left( {r,\varphi } \right)}{\, \partial \varphi } { \vec{e} }_{ x} + \frac{ \partial y \left( {r,\varphi } \right)}{\, \partial \varphi } { \vec{e} }_{ y} = - r \sin { \varphi } { \vec{e} }_{ x} + r \cos { \varphi } { \vec{e} }_{ y}$$
Let's check whether these curvilinear base vectors are orthogonal. They are orthogonal if the scalar product turns out to be zero.
$$\begin{eqnarray} { \vec{w} }_{ r} \cdot { \vec{w} }_{ \varphi } &=& \left( { \cos { \varphi } { \vec{e} }_{ x} + \sin { \varphi } { \vec{e} }_{ y}} \right) \cdot \left( { - r \sin { \varphi } { \vec{e} }_{ x} + r \cos { \varphi } { \vec{e} }_{ y}} \right) \\ { \vec{w} }_{ r} \cdot { \vec{w} }_{ \varphi } &=& \left( { \cos { \varphi } { \vec{e} }_{ x}} \right) \cdot \left( { - r \sin { \varphi } { \vec{e} }_{ x} + r \cos { \varphi } { \vec{e} }_{ y}} \right) + \left( { \sin { \varphi } { \vec{e} }_{ y}} \right) \cdot \left( { - r \sin { \varphi } { \vec{e} }_{ x} + r \cos { \varphi } { \vec{e} }_{ y}} \right) \\ { \vec{w} }_{ r} \cdot { \vec{w} }_{ \varphi } &=& \left( { \cos { \varphi } { \vec{e} }_{ x}} \right) \cdot \left( { - r \sin { \varphi } { \vec{e} }_{ x}} \right) + \left( { \sin { \varphi } { \vec{e} }_{ y}} \right) \cdot \left( { r \cos { \varphi } { \vec{e} }_{ y}} \right) \\ { \vec{w} }_{ r} \cdot { \vec{w} }_{ \varphi } &=& - r \sin { \varphi } \cos { \varphi } { \vec{e} }_{ x} + r \sin { \varphi } \cos { \varphi } \\ { \vec{w} }_{ r} \cdot { \vec{w} }_{ \varphi } &=& 0\end{eqnarray}$$
What about the magnitude of these base vectors?
$$\begin{eqnarray} \left| { { \vec{w} }_{ r}} \right| &=& 1 \\ \left| { { \vec{w} }_{ \varphi }} \right| &=& r\end{eqnarray}$$
Obviosuly curvilinear base vectors are nort necessarily orthonormal.
Let's now examine the reciprocal base vectors $ { \vec{w} }^{ r}$ and $ { \vec{w} }^{ \varphi }$. Generally we have
$$\fbox{$ \displaystyle { \vec{w} }^{ j} = \frac{ \partial { w }_{ j}}{ { \, \partial r }_{ i}} { \vec{e} }_{ i} $} \qquad \mbox{Reciprocal base vectors} \tag{2}$$
and thus in our case
$$\begin{eqnarray} { \vec{w} }^{ r} &=& \frac{ \partial r \left( {x,y} \right)}{\, \partial x} { \vec{e} }_{ x} + \frac{ \partial r \left( {x,y} \right)}{\, \partial y} y \\ { \vec{w} }^{ r} &=& \frac{ x}{ \sqrt{ { x }^{ 2} + { y }^{ 2}}} { \vec{e} }_{ x} + \frac{ y}{ \sqrt{ { x }^{ 2} + { y }^{ 2}}} { \vec{e} }_{ y} \\ { \vec{w} }^{ r} &=& \frac{ x}{ r} { \vec{e} }_{ x} + \frac{ y}{ r} { \vec{e} }_{ y}\end{eqnarray}$$
$$\begin{eqnarray} { \vec{w} }^{ \varphi } &=& \frac{ \partial \varphi \left( {x,y} \right)}{\, \partial x} { \vec{e} }_{ x} + \frac{ \partial \varphi \left( {x,y} \right)}{\, \partial y} y \\ { \vec{w} }^{ \varphi } &=& \frac{\partial \left( { \arctan \left( { \frac{ y}{ x}} \right)} \right)}{\, \partial x} { \vec{e} }_{ x} + \frac{\partial \left( { \arctan \left( { \frac{ y}{ x}} \right)} \right)}{\, \partial y} { \vec{e} }_{ y}\end{eqnarray}$$
$$ { \vec{w} }^{ \varphi } = - \frac{ y}{ { x }^{ 2} + { y }^{ 2}} { \vec{e} }_{ x} + \frac{ x}{ { x }^{ 2} + { y }^{ 2}} { \vec{e} }_{ y} = - \frac{ y}{ { r }^{ 2}} { \vec{e} }_{ x} + \frac{ x}{ { r }^{ 2}} { \vec{e} }_{ y}$$
What about the magnitude of these reciprocal base vectors?
$$\begin{eqnarray} \left| { { \vec{w} }^{ r}} \right| &=& \sqrt{ {\left( \frac{ x}{ r} \right)}^{ 2} + {\left( \frac{ y}{ r} \right)}^{ 2}} &=& 1 \\ \left| { { \vec{w} }^{ \varphi }} \right| &=& \sqrt{ {\left( \frac{ y}{ { r }^{ 2}} \right)}^{ 2} + {\left( \frac{ x}{ { r }^{ 2}} \right)}^{ 2}} &=& \frac{ 1}{ r}\end{eqnarray}$$
We obviously have
$$\begin{eqnarray} \left| { { \vec{w} }_{ r}} \right| \left| { { \vec{w} }^{ r}} \right| &=& 1 \\ \left| { { \vec{w} }_{ \varphi }} \right| \left| { { \vec{w} }^{ \varphi }} \right| &=& 1\end{eqnarray}$$
In general neither $ \left\{ { \vec{w} }_{ k}\right\}$ nor $ \left\{ { \vec{w} }^{ j}\right\}$ is an orthonormal basis but we have
$$ { \vec{w} }_{ k} \cdot { \vec{w} }^{ j} = \left( { \frac{ \partial { r }_{ l}}{ { \, \partial w }_{ k}} { \vec{e} }_{ l}} \right) \cdot \left( { \frac{ \partial { w }_{ j}}{ { \, \partial r }_{ i}} { \vec{e} }_{ i}} \right) = \left( { \frac{ \partial { r }_{ i}}{ { \, \partial w }_{ k}} { \vec{e} }_{ i}} \right) \cdot \left( { \frac{ \partial { w }_{ j}}{ { \, \partial r }_{ i}} { \vec{e} }_{ i}} \right) = \frac{ \partial { r }_{ i}}{ { \, \partial w }_{ k}} \frac{ \partial { w }_{ j}}{ { \, \partial r }_{ i}} = \frac{ \partial { w }_{ j}}{ { \, \partial w }_{ k}} = \left\{ \begin{array}{r@{\quad:\quad}l}
1 & for j = k\\ 0 & for j \ne k\\
\end{array}\right. \tag{3}$$
The reciprocal base vector $ { \vec{w} }^{ j}$ is obviously parallel to $ { \vec{w} }_{ j}$. We thus have $ { \vec{w} }^{ j} \wedge { \vec{w} }_{ j} = 0$ and therefore $ { \vec{w} }^{ j} \cdot { \vec{w} }_{ j} = { \vec{w} }^{ j} { \vec{w} }_{ j} = 1$. This implies that $ { \vec{w} }^{ j}$ is the inverse of $ { \vec{w} }_{ j}$.
$$\fbox{$ \displaystyle { \vec{w} }^{ j} = { { \vec{w} }_{ j} }^{ - 1} = \frac{ 1}{ { \left| { { \vec{w} }_{ j}} \right| }^{ 2}} { \vec{w} }_{ j} $} \tag{4}$$
This is super useful since $ { \vec{w} }_{ j}$ is often easier to calculate than $ { \vec{w} }^{ j}$. Unlike $ { \vec{e} }_{ i}$ the bases $ { \vec{w} }_{ k}$ and $ { \vec{w} }^{ j}$ can vary from point to point. Thus they cannot be treated as constant when differentiating.
Let's consider
$$ \nabla \wedge { \vec{w} }^{ j} = \left( { { \vec{e} }_{ k} { \partial }_{ k}} \right) \wedge \left( { \frac{ \partial { w }_{ j}}{ { \, \partial r }_{ i}} { \vec{e} }_{ i}} \right) = \left( { { \vec{e} }_{ k} \frac{ \partial }{ { \partial }_{ { r }_{ k}}}} \right) \wedge \left( { \frac{ \partial { w }_{ j}}{ { \, \partial r }_{ i}} { \vec{e} }_{ i}} \right)$$
For $ k = i$ we have a wedge prooduct of parallel vectors and thus null. For $ i \ne k$ we get pairs like
$$ \left( { { \vec{e} }_{ k} \frac{ \partial }{ { \partial }_{ { r }_{ k}}}} \right) \wedge \left( { \frac{ \partial { w }_{ j}}{ { \, \partial r }_{ i}} { \vec{e} }_{ i}} \right) + \left( { { \vec{e} }_{ i} \frac{ \partial }{ { \partial }_{ { r }_{ i}}}} \right) \wedge \left( { \frac{ \partial { w }_{ j}}{ { \, \partial r }_{ k}} { \vec{e} }_{ k}} \right) = \left( { \frac{ \partial }{ { \partial }_{ { r }_{ k}}} \frac{ \partial { w }_{ j}}{ { \, \partial r }_{ i}} { \vec{e} }_{ k}} \right) \wedge { \vec{e} }_{ i} + \left( { \frac{ \partial }{ { \partial }_{ { r }_{ i}}} \frac{ \partial { w }_{ j}}{ { \, \partial r }_{ k}} { \vec{e} }_{ i}} \right) \wedge { \vec{e} }_{ k} = \left( { \left( { \frac{ { \partial }^{ 2} { w }_{ j}}{ { \, \partial r }_{ i} { \, \partial r }_{ k}} - \frac{ { \partial }^{ 2} { w }_{ j}}{ { \, \partial r }_{ i} { \, \partial r }_{ k}}} \right) { \vec{e} }_{ k}} \right) \wedge { \vec{e} }_{ i} = 0$$
and thus
$$\fbox{$ \displaystyle \nabla \wedge { \vec{w} }^{ j} = 0 $}$$
Theorem: Let $ \left( \begin{array}{ccc}
{ w }_{ 1} & { w }_{ 2} & { w }_{ n}\\
\end{array}\right)$ be a curvilinear coordinate system in $ U$ then the gradient in curvilinear coordinates is given by
$$\fbox{$ \displaystyle \nabla = { \vec{w} }^{ j} { \partial }_{ { w }_{ j}} $} \tag{5}$$
Substituting Eq. 2 into this equation gives
$$\begin{eqnarray} \nabla &=& { \vec{w} }^{ j} { \partial }_{ { w }_{ j}} \\ \nabla &=& \frac{ \partial { w }_{ j}}{ { \, \partial r }_{ i}} { \vec{e} }_{ i} \frac{ \partial }{ { \partial }_{ { w }_{ j}}} \\ \nabla &=& { \vec{e} }_{ i} \frac{ \partial }{ { \, \partial r }_{ i}} \\ \nabla &=& { \vec{e} }_{ i} { \partial }_{ i}\end{eqnarray}$$
and thus the gradient in cartesian coordinates.
$$ \nabla = { \vec{w} }^{ j} { \partial }_{ { w }_{ j}}$$
With Eq. 4 and Eq. 1 and Eq. 5
$$ { \vec{w} }^{ j} = \frac{ 1}{ { \left| { { \vec{w} }_{ j}} \right| }^{ 2}} { \vec{w} }_{ j}$$ $$ { \vec{w} }_{ k} = \frac{ \partial { r }_{ i}}{ { \, \partial w }_{ k}} { \vec{e} }_{ i}$$ $$ \nabla = { \vec{w} }^{ j} { \partial }_{ { w }_{ j}}$$
we get
$$\fbox{$ \displaystyle \nabla = \frac{ 1}{ { \left| { { \vec{w} }_{ j}} \right| }^{ 2}} { \vec{w} }_{ j} { \partial }_{ { w }_{ j}} $} \qquad \mbox{Gradient in curvilinear coordinates} \tag{6}$$
We also have
$$ { \vec{w} }_{ j} = \left| { { \vec{w} }_{ j}} \right| { \vec{e} }_{ { w }_{ j}}$$ $$ { \vec{e} }_{ { w }_{ j}} = \frac{ 1}{ \left| { { \vec{w} }_{ j}} \right|} { \vec{w} }_{ j}$$
where $ { \vec{e} }_{ { w }_{ j}}$ is the curvilinear unit vectors. This lets us write Eq. 6 like this
$$\fbox{$ \displaystyle \nabla = \frac{ 1}{ \left| { { \vec{w} }_{ j}} \right|} { \vec{e} }_{ { w }_{ j}} { \partial }_{ { w }_{ j}} $} \qquad \mbox{Gradient in curvilinear coordinates} \tag{7}$$
$$ \vec{r} \left( {\alpha ,\beta ,r} \right) = \left( \begin{array}{c} r \cos { \alpha } \sin { \beta } \\ r \sin { \alpha } \sin { \beta } \\ r \cos { \beta } \end{array}\right)$$
Our aim is to calculate Eq. 6 for this coordinate system ($ { w }_{ 1} = \alpha $, $ { w }_{ 2} = \beta $, $ { w }_{ 3} = r$). We first determine the three
$$ { \vec{w} }_{ k} = \frac{ \partial { r }_{ i}}{ { \, \partial w }_{ k}} { \vec{e} }_{ i}$$ $$ { \vec{e} }_{ { w }_{ j}} = \frac{ 1}{ \left| { { \vec{w} }_{ j}} \right|} { \vec{w} }_{ j}$$
and their magnitudes.
$$\begin{eqnarray} { \vec{w} }_{ \alpha } &=& \frac{ \partial x}{ { \partial }_{ \alpha }} { \vec{e} }_{ x} + \frac{ \partial y}{ { \partial }_{ \alpha }} { \vec{e} }_{ y} + \frac{ \partial z}{ { \partial }_{ \alpha }} { \vec{e} }_{ z} \\ { \vec{w} }_{ \alpha } &=& - r \sin { \alpha } \sin { \beta } { \vec{e} }_{ x} + r \cos { \alpha } \sin { \beta } { \vec{e} }_{ y} \\ \left| { { \vec{w} }_{ \alpha }} \right| &=& \sqrt{ { r }^{ 2} { \sin { \alpha } }^{ 2} { \sin { \beta } }^{ 2} + { r }^{ 2} { \cos { \alpha } }^{ 2} { \sin { \beta } }^{ 2}} \\ \left| { { \vec{w} }_{ \alpha }} \right| &=& r \sin { \beta } \\ { \vec{e} }_{ \alpha } &=& - \sin { \alpha } { \vec{e} }_{ x} + \cos { \alpha } { \vec{e} }_{ y}\end{eqnarray}$$
$$\begin{eqnarray} { \vec{w} }_{ \beta } &=& \frac{ \partial x}{ { \partial }_{ \beta }} { \vec{e} }_{ x} + \frac{ \partial y}{ { \partial }_{ \beta }} { \vec{e} }_{ y} + \frac{ \partial z}{ { \partial }_{ \beta }} { \vec{e} }_{ z} \\ { \vec{w} }_{ \beta } &=& r \cos { \alpha } \cos { \beta } { \vec{e} }_{ x} + r \sin { \alpha } \cos { \beta } { \vec{e} }_{ y} - r \sin { \beta } { \vec{e} }_{ z} \\ \left| { { \vec{w} }_{ \beta }} \right| &=& \sqrt{ { r }^{ 2} { \cos { \alpha } }^{ 2} { \cos { \beta } }^{ 2} + { r }^{ 2} { \sin { \alpha } }^{ 2} { \cos { \beta } }^{ 2} + { r }^{ 2} { \sin { \beta } }^{ 2}} \\ \left| { { \vec{w} }_{ \beta }} \right| &=& r \\ { \vec{e} }_{ \beta } &=& \cos { \alpha } \cos { \beta } { \vec{e} }_{ x} + \sin { \alpha } \cos { \beta } { \vec{e} }_{ y} - \sin { \beta } { \vec{e} }_{ z}\end{eqnarray}$$
$$\begin{eqnarray} { \vec{w} }_{ r} &=& \frac{ \partial x}{ { \partial }_{ r}} { \vec{e} }_{ x} + \frac{ \partial y}{ { \partial }_{ r}} { \vec{e} }_{ y} + \frac{ \partial z}{ { \partial }_{ r}} { \vec{e} }_{ z} \\ { \vec{w} }_{ r} &=& \cos { \alpha } \sin { \beta } { \vec{e} }_{ x} + \sin { \alpha } \sin { \beta } { \vec{e} }_{ y} + \cos { \beta } { \vec{e} }_{ z} \\ \left| { { \vec{w} }_{ r}} \right| &=& \sqrt{ { \cos { \alpha } }^{ 2} { \sin { \beta } }^{ 2} + { \sin { \alpha } }^{ 2} { \sin { \beta } }^{ 2} + { \cos { \beta } }^{ 2}} \\ \left| { { \vec{w} }_{ r}} \right| &=& 1 \\ { \vec{e} }_{ r} &=& \cos { \alpha } \sin { \beta } { \vec{e} }_{ x} + \sin { \alpha } \sin { \beta } { \vec{e} }_{ y} + \cos { \beta } { \vec{e} }_{ z}\end{eqnarray}$$
Assembling these results into Eq. 6 gives
$$\begin{eqnarray} \nabla &=& \frac{ - r \sin { \alpha } \sin { \beta } { \vec{e} }_{ x} + r \cos { \alpha } \sin { \beta } { \vec{e} }_{ y}}{ { r }^{ 2} { \sin { \beta } }^{ 2}} { \partial }_{ \alpha } + \frac{ r \cos { \alpha } \cos { \beta } { \vec{e} }_{ x} + r \sin { \alpha } \cos { \beta } { \vec{e} }_{ y} - r \sin { \beta } { \vec{e} }_{ z}}{ { r }^{ 2}} { \partial }_{ \beta } + \left( { \cos { \alpha } \sin { \beta } { \vec{e} }_{ x} + \sin { \alpha } \sin { \beta } { \vec{e} }_{ y} + \cos { \beta } { \vec{e} }_{ z}} \right) { \partial }_{ { w }_{ j}} \\ \nabla &=& \frac{ - \sin { \alpha } { \vec{e} }_{ x} + \cos { \alpha } { \vec{e} }_{ y}}{ r \sin { \beta } } { \partial }_{ \alpha } + \frac{ \cos { \alpha } \cos { \beta } { \vec{e} }_{ x} + \sin { \alpha } \cos { \beta } { \vec{e} }_{ y} - \sin { \beta } { \vec{e} }_{ z}}{ r} { \partial }_{ \beta } + \left( { \cos { \alpha } \sin { \beta } { \vec{e} }_{ x} + \sin { \alpha } \sin { \beta } { \vec{e} }_{ y} + \cos { \beta } { \vec{e} }_{ z}} \right) { \partial }_{ { w }_{ j}}\end{eqnarray}$$
$$\fbox{$ \displaystyle \nabla = \frac{ 1}{ r \sin { \beta } } { \vec{e} }_{ \alpha } { \partial }_{ \alpha } + \frac{ 1}{ r} { \vec{e} }_{ \beta } { \partial }_{ \beta } + { \vec{e} }_{ r} { \partial }_{ r} $} \qquad \mbox{Gradient in spherical coordinates}$$
This corresponds to Eq. 1 in Gradient in Kugelkoordinaten if we apply this operator to a scalar field. What happens if we apply this operator to a vector field expressed in spherical coordinates?
$$ \vec{A} = { A }_{ \alpha } { \vec{e} }_{ \alpha } + { A }_{ \beta } { \vec{e} }_{ \beta } + { A }_{ r} { \vec{e} }_{ r}$$
$$\begin{eqnarray} \nabla \vec{A} &=& \left( { \frac{ 1}{ r \sin { \beta } } { \vec{e} }_{ \alpha } { \partial }_{ \alpha } + \frac{ 1}{ r} { \vec{e} }_{ \beta } { \partial }_{ \beta } + { \vec{e} }_{ r} { \partial }_{ r}} \right) \left( { { A }_{ \alpha } { \vec{e} }_{ \alpha } + { A }_{ \beta } { \vec{e} }_{ \beta } + { A }_{ r} { \vec{e} }_{ r}} \right) \\ \nabla \vec{A} &=& \frac{ 1}{ r \sin { \beta } } { \vec{e} }_{ \alpha } { \partial }_{ \alpha } \left( { { A }_{ \alpha } { \vec{e} }_{ \alpha } + { A }_{ \beta } { \vec{e} }_{ \beta } + { A }_{ r} { \vec{e} }_{ r}} \right) + \frac{ 1}{ r} { \vec{e} }_{ \beta } { \partial }_{ \beta } \left( { { A }_{ \alpha } { \vec{e} }_{ \alpha } + { A }_{ \beta } { \vec{e} }_{ \beta } + { A }_{ r} { \vec{e} }_{ r}} \right) + { \vec{e} }_{ r} { \partial }_{ r} \left( { { A }_{ \alpha } { \vec{e} }_{ \alpha } + { A }_{ \beta } { \vec{e} }_{ \beta } + { A }_{ r} { \vec{e} }_{ r}} \right) \\ \nabla \vec{A} &=& \frac{ 1}{ r \sin { \beta } } { \vec{e} }_{ \alpha } { \partial }_{ \alpha } \left( { { A }_{ \alpha } { \vec{e} }_{ \alpha } + { A }_{ \beta } { \vec{e} }_{ \beta } + { A }_{ r} { \vec{e} }_{ r}} \right) + \frac{ 1}{ r} { \vec{e} }_{ \beta } { \partial }_{ \beta } \left( { { A }_{ \alpha } { \vec{e} }_{ \alpha } + { A }_{ \beta } { \vec{e} }_{ \beta } + { A }_{ r} { \vec{e} }_{ r}} \right) + { \vec{e} }_{ r} { \partial }_{ r} \left( { { A }_{ \alpha } { \vec{e} }_{ \alpha } + { A }_{ \beta } { \vec{e} }_{ \beta } + { A }_{ r} { \vec{e} }_{ r}} \right)\end{eqnarray}$$
Note that $ { \vec{e} }_{ \alpha } , { \vec{e} }_{ \beta } , { \vec{e} }_{ r}$ have to be differentiated as well.
$$\begin{eqnarray} \nabla \vec{A} &=& \frac{ 1}{ r \sin { \beta } } { \vec{e} }_{ \alpha } { \partial }_{ \alpha } \left( { { A }_{ \alpha } { \vec{e} }_{ \alpha } + { A }_{ \beta } { \vec{e} }_{ \beta } + { A }_{ r} { \vec{e} }_{ r}} \right) \\ \frac{ 1}{ r} { \vec{e} }_{ \beta } { \partial }_{ \beta } \left( { { A }_{ \alpha } { \vec{e} }_{ \alpha } + { A }_{ \beta } { \vec{e} }_{ \beta } + { A }_{ r} { \vec{e} }_{ r}} \right) \\ { \vec{e} }_{ r} { \partial }_{ r} \left( { { A }_{ \alpha } { \vec{e} }_{ \alpha } + { A }_{ \beta } { \vec{e} }_{ \beta } + { A }_{ r} { \vec{e} }_{ r}} \right)\end{eqnarray}$$
$$\begin{eqnarray} \nabla \vec{A} &=& \frac{ 1}{ r \sin { \beta } } { \vec{e} }_{ \alpha } \left( { \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \alpha } { \vec{e} }_{ \alpha } + { A }_{ \alpha } \frac{\partial \left( { { \vec{e} }_{ \alpha }} \right)}{\, \partial \alpha } + \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial \alpha } { \vec{e} }_{ \beta } + { A }_{ \beta } \frac{\partial \left( { { \vec{e} }_{ \beta }} \right)}{\, \partial \alpha } + \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial \alpha } { \vec{e} }_{ r} + { A }_{ r} \frac{\partial \left( { { \vec{e} }_{ r}} \right)}{\, \partial \alpha }} \right) \\ + \frac{ 1}{ r} { \vec{e} }_{ \beta } \left( { \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \beta } { \vec{e} }_{ \alpha } + { A }_{ \alpha } \frac{\partial \left( { { \vec{e} }_{ \alpha }} \right)}{\, \partial \beta } + \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial \beta } { \vec{e} }_{ \beta } + { A }_{ \beta } \frac{\partial \left( { { \vec{e} }_{ \beta }} \right)}{\, \partial \beta } + \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial \beta } { \vec{e} }_{ r} + { A }_{ r} \frac{\partial \left( { { \vec{e} }_{ r}} \right)}{\, \partial \beta }} \right) \\ + { \vec{e} }_{ r} \left( { \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial r} { \vec{e} }_{ \alpha } + { A }_{ \alpha } \frac{\partial \left( { { \vec{e} }_{ \alpha }} \right)}{\, \partial r} + \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial r} { \vec{e} }_{ \beta } + { A }_{ \beta } \frac{\partial \left( { { \vec{e} }_{ \beta }} \right)}{\, \partial r} + \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial r} { \vec{e} }_{ r} + { A }_{ r} \frac{\partial \left( { { \vec{e} }_{ r}} \right)}{\, \partial r}} \right)\end{eqnarray}$$
$$\begin{eqnarray} \nabla \vec{A} &=& \frac{ 1}{ r \sin { \beta } } \left( { \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \alpha } + { A }_{ \alpha } { \vec{e} }_{ \alpha } \frac{\partial \left( { { \vec{e} }_{ \alpha }} \right)}{\, \partial \alpha } + \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ \beta } + { A }_{ \beta } { \vec{e} }_{ \alpha } \frac{\partial \left( { { \vec{e} }_{ \beta }} \right)}{\, \partial \alpha } + \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ r} + { A }_{ r} { \vec{e} }_{ \alpha } \frac{\partial \left( { { \vec{e} }_{ r}} \right)}{\, \partial \alpha }} \right) \\ + \frac{ 1}{ r} \left( { \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \beta } { \vec{e} }_{ \beta } { \vec{e} }_{ \alpha } + { A }_{ \alpha } { \vec{e} }_{ \beta } \frac{\partial \left( { { \vec{e} }_{ \alpha }} \right)}{\, \partial \beta } + \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial \beta } + { A }_{ \beta } { \vec{e} }_{ \beta } \frac{\partial \left( { { \vec{e} }_{ \beta }} \right)}{\, \partial \beta } + \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial \beta } { \vec{e} }_{ \beta } { \vec{e} }_{ r} + { A }_{ r} { \vec{e} }_{ \beta } \frac{\partial \left( { { \vec{e} }_{ r}} \right)}{\, \partial \beta }} \right) \\ + \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ \alpha } + { A }_{ \alpha } { \vec{e} }_{ r} \frac{\partial \left( { { \vec{e} }_{ \alpha }} \right)}{\, \partial r} + \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ \beta } + { A }_{ \beta } { \vec{e} }_{ r} \frac{\partial \left( { { \vec{e} }_{ \beta }} \right)}{\, \partial r} + \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial r} + { A }_{ r} { \vec{e} }_{ r} \frac{\partial \left( { { \vec{e} }_{ r}} \right)}{\, \partial r}\end{eqnarray}$$
We have
$$\begin{eqnarray} { \vec{e} }_{ \alpha } { \vec{e} }_{ \alpha } &=& { \vec{e} }_{ \alpha } \cdot { \vec{e} }_{ \alpha } &=& 1 \\ \frac{\partial \left( { { { \vec{e} }_{ \alpha } }^{ 2}} \right)}{\, \partial \alpha } &=& 2 { \vec{e} }_{ \alpha } \frac{\partial \left( { { \vec{e} }_{ \alpha }} \right)}{\, \partial \alpha } &=& 0\end{eqnarray}$$and thus
$$\begin{eqnarray} \nabla \vec{A} &=& \frac{ 1}{ r \sin { \beta } } \left( { \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \alpha } + \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ \beta } + { A }_{ \beta } { \vec{e} }_{ \alpha } \frac{\partial \left( { { \vec{e} }_{ \beta }} \right)}{\, \partial \alpha } + \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ r} + { A }_{ r} { \vec{e} }_{ \alpha } \frac{\partial \left( { { \vec{e} }_{ r}} \right)}{\, \partial \alpha }} \right) \\ + \frac{ 1}{ r} \left( { \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \beta } { \vec{e} }_{ \beta } { \vec{e} }_{ \alpha } + { A }_{ \alpha } { \vec{e} }_{ \beta } \frac{\partial \left( { { \vec{e} }_{ \alpha }} \right)}{\, \partial \beta } + \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial \beta } + \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial \beta } { \vec{e} }_{ \beta } { \vec{e} }_{ r} + { A }_{ r} { \vec{e} }_{ \beta } \frac{\partial \left( { { \vec{e} }_{ r}} \right)}{\, \partial \beta }} \right) \\ + \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ \alpha } + { A }_{ \alpha } { \vec{e} }_{ r} \frac{\partial \left( { { \vec{e} }_{ \alpha }} \right)}{\, \partial r} + \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ \beta } + { A }_{ \beta } { \vec{e} }_{ r} \frac{\partial \left( { { \vec{e} }_{ \beta }} \right)}{\, \partial r} + \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial r}\end{eqnarray}$$
Let's consider the last term in the first row.
$$\begin{eqnarray} { \vec{e} }_{ \alpha } \frac{\partial \left( { { \vec{e} }_{ r}} \right)}{\, \partial \alpha } &=& { \vec{e} }_{ \alpha } \frac{\partial \left( { \cos { \alpha } \sin { \beta } { \vec{e} }_{ x} + \sin { \alpha } \sin { \beta } { \vec{e} }_{ y} + \cos { \beta } { \vec{e} }_{ z}} \right)}{\, \partial \alpha } \\ { \vec{e} }_{ \alpha } \frac{\partial \left( { { \vec{e} }_{ r}} \right)}{\, \partial \alpha } &=& { \vec{e} }_{ \alpha } \left( { - \sin { \alpha } { \vec{e} }_{ x} + \cos { \alpha } { \vec{e} }_{ y}} \right) \sin { \beta } \\ { \vec{e} }_{ \alpha } \frac{\partial \left( { { \vec{e} }_{ r}} \right)}{\, \partial \alpha } &=& { { \vec{e} }_{ \alpha } }^{ 2} \sin { \beta } \\ { \vec{e} }_{ \alpha } \frac{\partial \left( { { \vec{e} }_{ r}} \right)}{\, \partial \alpha } &=& \sin { \beta } \end{eqnarray}$$
Let's consider the last term in the second row.
$$\begin{eqnarray} { \vec{e} }_{ \beta } \frac{\partial \left( { { \vec{e} }_{ r}} \right)}{\, \partial \beta } &=& { \vec{e} }_{ \beta } \frac{\partial \left( { \cos { \alpha } \sin { \beta } { \vec{e} }_{ x} + \sin { \alpha } \sin { \beta } { \vec{e} }_{ y} + \cos { \beta } { \vec{e} }_{ z}} \right)}{\, \partial \beta } \\ { \vec{e} }_{ \beta } \frac{\partial \left( { { \vec{e} }_{ r}} \right)}{\, \partial \beta } &=& { \vec{e} }_{ \beta } \left( { \cos { \alpha } \cos { \beta } { \vec{e} }_{ x} + \sin { \alpha } \cos { \beta } { \vec{e} }_{ y} - \sin { \beta } { \vec{e} }_{ z}} \right) \\ { \vec{e} }_{ \beta } \frac{\partial \left( { { \vec{e} }_{ r}} \right)}{\, \partial \beta } &=& { { \vec{e} }_{ \beta } }^{ 2} \\ { \vec{e} }_{ \beta } \frac{\partial \left( { { \vec{e} }_{ r}} \right)}{\, \partial \beta } &=& 1\end{eqnarray}$$
We look at the fourth term in the last row
$$\begin{eqnarray} { \vec{e} }_{ r} \frac{\partial \left( { { \vec{e} }_{ \beta }} \right)}{\, \partial r} &=& { \vec{e} }_{ r} \frac{\partial \left( { \cos { \alpha } \cos { \beta } { \vec{e} }_{ x} + \sin { \alpha } \cos { \beta } { \vec{e} }_{ y} - \sin { \beta } { \vec{e} }_{ z}} \right)}{\, \partial r} \\ { \vec{e} }_{ r} \frac{\partial \left( { { \vec{e} }_{ \beta }} \right)}{\, \partial r} &=& 0\end{eqnarray}$$
and the second term in the last row.
$$\begin{eqnarray} { \vec{e} }_{ r} \frac{\partial \left( { { \vec{e} }_{ \alpha }} \right)}{\, \partial r} &=& { \vec{e} }_{ r} \frac{\partial \left( { { \vec{e} }_{ \alpha }} \right)}{\, \partial r} \\ { \vec{e} }_{ r} \frac{\partial \left( { { \vec{e} }_{ \alpha }} \right)}{\, \partial r} &=& 0\end{eqnarray}$$This gives
$$\begin{eqnarray} \nabla \vec{A} &=& \frac{ 1}{ r \sin { \beta } } \left( { \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \alpha } + \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ \beta } + { A }_{ \beta } { \vec{e} }_{ \alpha } \frac{\partial \left( { { \vec{e} }_{ \beta }} \right)}{\, \partial \alpha } + \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ r} + { A }_{ r} \sin { \beta } } \right) \\ + \frac{ 1}{ r} \left( { \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \beta } { \vec{e} }_{ \beta } { \vec{e} }_{ \alpha } + { A }_{ \alpha } { \vec{e} }_{ \beta } \frac{\partial \left( { { \vec{e} }_{ \alpha }} \right)}{\, \partial \beta } + \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial \beta } + \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial \beta } { \vec{e} }_{ \beta } { \vec{e} }_{ r} + { A }_{ r}} \right) \\ + \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ \alpha } + \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ \beta } + \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial r}\end{eqnarray}$$
We check out the third term in the first row.
$$\begin{eqnarray} { \vec{e} }_{ \alpha } \frac{\partial \left( { { \vec{e} }_{ \beta }} \right)}{\, \partial \alpha } &=& { \vec{e} }_{ \alpha } \frac{\partial \left( { \cos { \alpha } \cos { \beta } { \vec{e} }_{ x} + \sin { \alpha } \cos { \beta } { \vec{e} }_{ y} - \sin { \beta } { \vec{e} }_{ z}} \right)}{\, \partial \alpha } \\ { \vec{e} }_{ \alpha } \frac{\partial \left( { { \vec{e} }_{ \beta }} \right)}{\, \partial \alpha } &=& { \vec{e} }_{ \alpha } \left( { - \sin { \alpha } \cos { \beta } { \vec{e} }_{ x} + \cos { \alpha } \cos { \beta } { \vec{e} }_{ y}} \right) \\ { \vec{e} }_{ \alpha } \frac{\partial \left( { { \vec{e} }_{ \beta }} \right)}{\, \partial \alpha } &=& { \vec{e} }_{ \alpha } \left( { - \sin { \alpha } { \vec{e} }_{ x} + \cos { \alpha } { \vec{e} }_{ y}} \right) \cos { \beta } \\ { \vec{e} }_{ \alpha } \frac{\partial \left( { { \vec{e} }_{ \beta }} \right)}{\, \partial \alpha } &=& \cos { \beta } \end{eqnarray}$$
We check out the second term in the second row.
$$\begin{eqnarray} { \vec{e} }_{ \beta } \frac{\partial \left( { { \vec{e} }_{ \alpha }} \right)}{\, \partial \beta } &=& { \vec{e} }_{ \beta } \frac{\partial \left( { - \sin { \alpha } { \vec{e} }_{ x} + \cos { \alpha } { \vec{e} }_{ y}} \right)}{\, \partial \beta } \\ { \vec{e} }_{ \beta } \frac{\partial \left( { { \vec{e} }_{ \alpha }} \right)}{\, \partial \beta } &=& 0\end{eqnarray}$$
This gives
$$ \nabla \vec{A} = \frac{ 1}{ r \sin { \beta } } \left( { \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \alpha } + \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ \beta } + { A }_{ \beta } \cos { \beta } + \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ r} + { A }_{ r} \sin { \beta } } \right) + \frac{ 1}{ r} \left( { \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \beta } { \vec{e} }_{ \beta } { \vec{e} }_{ \alpha } + \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial \beta } + \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial \beta } { \vec{e} }_{ \beta } { \vec{e} }_{ r} + { A }_{ r}} \right) + \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ \alpha } + \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ \beta } + \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial r}$$
and after rearraging the terms
$$\begin{eqnarray} \nabla \vec{A} &=& \frac{ 1}{ r \sin { \beta } } \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \alpha } + \frac{ 1}{ r \sin { \beta } } \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ \beta } + \frac{ 1}{ r \sin { \beta } } { A }_{ \beta } \cos { \beta } + \frac{ 1}{ r \sin { \beta } } \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ r} + \frac{ 1}{ r \sin { \beta } } { A }_{ r} \sin { \beta } + \frac{ 1}{ r} \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \beta } { \vec{e} }_{ \beta } { \vec{e} }_{ \alpha } + \frac{ 1}{ r} \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial \beta } + \frac{ 1}{ r} \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial \beta } { \vec{e} }_{ \beta } { \vec{e} }_{ r} + \frac{ 1}{ r} { A }_{ r} + \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ \alpha } + \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ \beta } + \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial r} \\ \nabla \vec{A} &=& \frac{ 1}{ r \sin { \beta } } \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \alpha } + \frac{ 1}{ r \sin { \beta } } { A }_{ \beta } \cos { \beta } + \frac{ 1}{ r \sin { \beta } } { A }_{ r} \sin { \beta } + \frac{ 1}{ r} \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial \beta } + \frac{ 1}{ r} { A }_{ r} + \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial r} + \frac{ 1}{ r \sin { \beta } } \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ \beta } + \frac{ 1}{ r \sin { \beta } } \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ r} + \frac{ 1}{ r} \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \beta } { \vec{e} }_{ \beta } { \vec{e} }_{ \alpha } + \frac{ 1}{ r} \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial \beta } { \vec{e} }_{ \beta } { \vec{e} }_{ r} + \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ \alpha } + \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ \beta }\end{eqnarray}$$
This result has a scalar component and a bi-vector component.
$$ \nabla \vec{A} = \nabla \cdot \vec{A} + \nabla \wedge \vec{A}$$
We consider the scalar component first.
$$\begin{eqnarray} \nabla \cdot \vec{A} &=& \frac{ 1}{ r \sin { \beta } } \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \alpha } + \frac{ 1}{ r \sin { \beta } } { A }_{ \beta } \cos { \beta } + \frac{ 2}{ r} { A }_{ r} + \frac{ 1}{ r} \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial \beta } + \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial r} \\ \nabla \cdot \vec{A} &=& \frac{ 1}{ r \sin { \beta } } \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \alpha } + \frac{ 1}{ r \sin { \beta } } { A }_{ \beta } \cos { \beta } + \frac{ 2}{ r} { A }_{ r} + \frac{ 1}{ r} \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial \beta } + \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial r} \\ \nabla \cdot \vec{A} &=& \left( { \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial r} + 2 \frac{ 1}{ r} { A }_{ r}} \right) + \left( { \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial \beta } \frac{ 1}{ r \sin { \beta } } \sin { \beta } + { A }_{ \beta } \frac{ 1}{ r \sin { \beta } } \cos { \beta } } \right) + \frac{ 1}{ r \sin { \beta } } \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \alpha } \\ \nabla \cdot \vec{A} &=& \frac{ 1}{ { r }^{ 2}} \left( { \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial r} { r }^{ 2} + 2 { A }_{ r} r} \right) + \frac{ 1}{ r \sin { \beta } } \left( { \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial \beta } \sin { \beta } + { A }_{ \beta } \cos { \beta } } \right) + \frac{ 1}{ r \sin { \beta } } \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \alpha }\end{eqnarray}$$
$$\fbox{$ \displaystyle \nabla \cdot \vec{A} = \frac{ 1}{ { r }^{ 2}} \frac{\partial \left( { { A }_{ r} { r }^{ 2}} \right)}{\, \partial r} + \frac{ 1}{ r \sin { \beta } } \frac{\partial \left( { { A }_{ \beta } \sin { \beta } } \right)}{\, \partial \beta } + \frac{ 1}{ r \sin { \beta } } \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \alpha } $} \qquad \mbox{Divergence of a vector field}$$
We also have a bi-vecor component.
$$ \nabla \wedge \vec{A} = \frac{ 1}{ r \sin { \beta } } \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ \beta } + \frac{ 1}{ r \sin { \beta } } \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ r} + \frac{ 1}{ r} \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \beta } { \vec{e} }_{ \beta } { \vec{e} }_{ \alpha } + \frac{ 1}{ r} \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial \beta } { \vec{e} }_{ \beta } { \vec{e} }_{ r} + \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ \alpha } + \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ \beta }$$
$$\fbox{$ \displaystyle \nabla \wedge \vec{A} = \left( { \frac{ 1}{ r \sin { \beta } } \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial \alpha } - \frac{ 1}{ r} \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \beta }} \right) { \vec{e} }_{ \alpha } { \vec{e} }_{ \beta } + \left( { \frac{ 1}{ r} \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial \beta } - \frac{\partial \left( { { A }_{ \beta }} \right)}{\, \partial r}} \right) { \vec{e} }_{ \beta } { \vec{e} }_{ r} + \left( { \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial r} - \frac{ 1}{ r \sin { \beta } } \frac{\partial \left( { { A }_{ r}} \right)}{\, \partial \alpha }} \right) { \vec{e} }_{ r} { \vec{e} }_{ \alpha } $} \qquad \mbox{Curl of a vector field}$$
$$ \vec{r} \left( {\alpha ,\beta ,r} \right) = \left( \begin{array}{c} r \cos { \alpha } \\ r \sin { \alpha } \\ z \end{array}\right)$$
Our aim is to calculate Eq. 6 for this coordinate system ($ { w }_{ 1} = \alpha $, $ { w }_{ 2} = \beta $, $ { w }_{ 3} = r$). We first determine the three
$$ { \vec{w} }_{ k} = \frac{ \partial { r }_{ i}}{ { \, \partial w }_{ k}} { \vec{e} }_{ i}$$ $$ { \vec{e} }_{ { w }_{ j}} = \frac{ 1}{ \left| { { \vec{w} }_{ j}} \right|} { \vec{w} }_{ j}$$
and their magnitudes.
$$\begin{eqnarray} { \vec{w} }_{ \alpha } &=& \frac{ \partial x}{\, \partial \alpha } { \vec{e} }_{ x} + \frac{ \partial y}{\, \partial \alpha } { \vec{e} }_{ y} + \frac{ \partial z}{\, \partial \alpha } { \vec{e} }_{ z} \\ { \vec{w} }_{ \alpha } &=& - r \sin { \alpha } { \vec{e} }_{ x} + r \cos { \alpha } { \vec{e} }_{ y} \\ \left| { { \vec{w} }_{ \alpha }} \right| &=& r \\ { \vec{e} }_{ \alpha } &=& - \sin { \alpha } { \vec{e} }_{ x} + \cos { \alpha } { \vec{e} }_{ y}\end{eqnarray}$$
$$\begin{eqnarray} { \vec{w} }_{ z} &=& \frac{ \partial x}{\, \partial z} { \vec{e} }_{ x} + \frac{ \partial y}{\, \partial z} { \vec{e} }_{ y} + \frac{ \partial z}{\, \partial z} { \vec{e} }_{ z} \\ { \vec{w} }_{ z} &=& { \vec{e} }_{ z} \\ \left| { { \vec{w} }_{ z}} \right| &=& 1 \\ { \vec{e} }_{ z} &=& { \vec{e} }_{ z}\end{eqnarray}$$
$$\begin{eqnarray} { \vec{w} }_{ r} &=& \frac{ \partial x}{\, \partial r} { \vec{e} }_{ x} + \frac{ \partial y}{\, \partial r} { \vec{e} }_{ y} + \frac{ \partial z}{\, \partial r} { \vec{e} }_{ z} \\ { \vec{w} }_{ r} &=& \cos { \alpha } { \vec{e} }_{ x} + \sin { \alpha } { \vec{e} }_{ y} \\ \left| { { \vec{w} }_{ r}} \right| &=& 1 \\ { \vec{e} }_{ r} &=& \cos { \alpha } { \vec{e} }_{ x} + \sin { \alpha } { \vec{e} }_{ y}\end{eqnarray}$$
Assembling these results into Eq. 6 gives
$$\begin{eqnarray} \nabla &=& \frac{ 1}{ { \left| { { \vec{w} }_{ j}} \right| }^{ 2}} { \vec{w} }_{ j} { \partial }_{ { w }_{ j}} \\ \nabla &=& \frac{ 1}{ { r }^{ 2}} \left( { - r \sin { \alpha } { \vec{e} }_{ x} + r \cos { \alpha } { \vec{e} }_{ y}} \right) { \partial }_{ \alpha } + { \vec{e} }_{ z} { \partial }_{ z} + \left( { \cos { \alpha } { \vec{e} }_{ x} + \sin { \alpha } { \vec{e} }_{ y}} \right) { \partial }_{ r}\end{eqnarray}$$
$$\fbox{$ \displaystyle \nabla = \frac{ 1}{ r} { \vec{e} }_{ \alpha } { \partial }_{ \alpha } + { \vec{e} }_{ z} { \partial }_{ z} + { \vec{e} }_{ r} { \partial }_{ r} $} \qquad \mbox{Gradient in cylindrical coordinates}$$
This corresponds to LINK in LINK if we apply this operator to a scalar field. What happens if we apply this operator to a vector field expressed in cylindrical coordinates?
$$ \vec{A} = { A }_{ \alpha } { \vec{e} }_{ \alpha } + { A }_{ z} { \vec{e} }_{ z} + { A }_{ r} { \vec{e} }_{ r}$$
$$\begin{eqnarray} \nabla \vec{A} &=& \left( { \frac{ 1}{ r} { \vec{e} }_{ \alpha } { \partial }_{ \alpha } + { \vec{e} }_{ z} { \partial }_{ z} + { \vec{e} }_{ r} { \partial }_{ r}} \right) \left( { { A }_{ \alpha } { \vec{e} }_{ \alpha } + { A }_{ z} { \vec{e} }_{ z} + { A }_{ r} { \vec{e} }_{ r}} \right) \\ \nabla \vec{A} &=& \frac{ 1}{ r} { \vec{e} }_{ \alpha } { \partial }_{ \alpha } \left( { { A }_{ \alpha } { \vec{e} }_{ \alpha } + { A }_{ z} { \vec{e} }_{ z} + { A }_{ r} { \vec{e} }_{ r}} \right) + { \vec{e} }_{ z} { \partial }_{ z} \left( { { A }_{ \alpha } { \vec{e} }_{ \alpha } + { A }_{ z} { \vec{e} }_{ z} + { A }_{ r} { \vec{e} }_{ r}} \right) + { \vec{e} }_{ r} { \partial }_{ r} \left( { { A }_{ \alpha } { \vec{e} }_{ \alpha } + { A }_{ z} { \vec{e} }_{ z} + { A }_{ r} { \vec{e} }_{ r}} \right)\end{eqnarray}$$
$$\begin{eqnarray} \nabla \vec{A} &=& \frac{ 1}{ r} { \vec{e} }_{ \alpha } \left( { \frac{ \partial }{\, \partial \alpha } \left( { { A }_{ \alpha } { \vec{e} }_{ \alpha }} \right) + \frac{ \partial }{\, \partial \alpha } \left( { { A }_{ z} { \vec{e} }_{ z}} \right) + \frac{ \partial }{\, \partial \alpha } \left( { { A }_{ r} { \vec{e} }_{ r}} \right)} \right) + { \vec{e} }_{ z} \left( { \frac{ \partial }{\, \partial z} \left( { { A }_{ \alpha } { \vec{e} }_{ \alpha }} \right) + \frac{ \partial }{\, \partial z} \left( { { A }_{ z} { \vec{e} }_{ z}} \right) + \frac{ \partial }{\, \partial z} \left( { { A }_{ r} { \vec{e} }_{ r}} \right)} \right) \\ + { \vec{e} }_{ r} \left( { \frac{ \partial }{\, \partial r} \left( { { A }_{ \alpha } { \vec{e} }_{ \alpha }} \right) + \frac{ \partial }{\, \partial r} \left( { { A }_{ z} { \vec{e} }_{ z}} \right) + \frac{ \partial }{\, \partial r} \left( { { A }_{ r} { \vec{e} }_{ r}} \right)} \right)\end{eqnarray}$$
$$\begin{eqnarray} \nabla \vec{A} &=& \frac{ 1}{ r} { \vec{e} }_{ \alpha } \left( { \frac{ \partial { A }_{ \alpha }}{\, \partial \alpha } { \vec{e} }_{ \alpha } + { A }_{ \alpha } \frac{ \partial { \vec{e} }_{ \alpha }}{\, \partial \alpha } + \frac{ \partial { A }_{ z}}{\, \partial \alpha } { \vec{e} }_{ z} + { A }_{ z} \frac{ \partial { \vec{e} }_{ z}}{\, \partial \alpha } + \frac{ \partial { A }_{ r}}{\, \partial \alpha } { \vec{e} }_{ r} + { A }_{ r} \frac{ \partial { \vec{e} }_{ r}}{\, \partial \alpha }} \right) \\ + { \vec{e} }_{ z} \left( { \frac{ \partial { A }_{ \alpha }}{\, \partial z} { \vec{e} }_{ \alpha } + { A }_{ \alpha } \frac{ \partial { \vec{e} }_{ \alpha }}{\, \partial z} + \frac{ \partial { A }_{ z}}{\, \partial z} { \vec{e} }_{ z} + { A }_{ z} \frac{ \partial { \vec{e} }_{ z}}{\, \partial z} + \frac{ \partial { A }_{ r}}{\, \partial z} { \vec{e} }_{ r} + { A }_{ r} \frac{ \partial { \vec{e} }_{ r}}{\, \partial z}} \right) \\ + { \vec{e} }_{ r} \left( { \frac{ \partial { A }_{ \alpha }}{\, \partial r} { \vec{e} }_{ \alpha } + { A }_{ \alpha } \frac{ \partial { \vec{e} }_{ \alpha }}{\, \partial r} + \frac{ \partial { A }_{ z}}{\, \partial r} { \vec{e} }_{ z} + { A }_{ z} \frac{ \partial { \vec{e} }_{ z}}{\, \partial r} + \frac{ \partial { A }_{ r}}{\, \partial r} { \vec{e} }_{ r} + { A }_{ r} \frac{ \partial { \vec{e} }_{ r}}{\, \partial r}} \right) \\ + { \vec{e} }_{ r} \left( { \frac{ \partial { A }_{ \alpha }}{\, \partial r} { \vec{e} }_{ \alpha } + { A }_{ \alpha } \frac{ \partial { \vec{e} }_{ \alpha }}{\, \partial r} + \frac{ \partial { A }_{ z}}{\, \partial r} { \vec{e} }_{ z} + { A }_{ z} \frac{ \partial { \vec{e} }_{ z}}{\, \partial r} + \frac{ \partial { A }_{ r}}{\, \partial r} { \vec{e} }_{ r} + { A }_{ r} \frac{ \partial { \vec{e} }_{ r}}{\, \partial r}} \right)\end{eqnarray}$$
$$\begin{eqnarray} \nabla \vec{A} &=& \frac{ 1}{ r} \left( { \frac{ \partial { A }_{ \alpha }}{\, \partial \alpha } + { A }_{ \alpha } { \vec{e} }_{ \alpha } \frac{ \partial { \vec{e} }_{ \alpha }}{\, \partial \alpha } + \frac{ \partial { A }_{ z}}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ z} + { A }_{ z} { \vec{e} }_{ \alpha } \frac{ \partial { \vec{e} }_{ z}}{\, \partial \alpha } + \frac{ \partial { A }_{ r}}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ r} + { A }_{ r} { \vec{e} }_{ \alpha } \frac{ \partial { \vec{e} }_{ r}}{\, \partial \alpha }} \right) \\ + \frac{ \partial { A }_{ \alpha }}{\, \partial z} { \vec{e} }_{ z} { \vec{e} }_{ \alpha } + { A }_{ \alpha } { \vec{e} }_{ z} \frac{ \partial { \vec{e} }_{ \alpha }}{\, \partial z} + \frac{ \partial { A }_{ z}}{\, \partial z} + { A }_{ z} { \vec{e} }_{ z} \frac{ \partial { \vec{e} }_{ z}}{\, \partial z} + \frac{ \partial { A }_{ r}}{\, \partial z} { \vec{e} }_{ z} { \vec{e} }_{ r} + { A }_{ r} { \vec{e} }_{ z} \frac{ \partial { \vec{e} }_{ r}}{\, \partial z} \\ + \frac{ \partial { A }_{ \alpha }}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ \alpha } + { A }_{ \alpha } { \vec{e} }_{ r} \frac{ \partial { \vec{e} }_{ \alpha }}{\, \partial r} + \frac{ \partial { A }_{ z}}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ z} + { A }_{ z} { \vec{e} }_{ r} \frac{ \partial { \vec{e} }_{ z}}{\, \partial r} + \frac{ \partial { A }_{ r}}{\, \partial r} + { A }_{ r} { \vec{e} }_{ r} \frac{ \partial { \vec{e} }_{ r}}{\, \partial r}\end{eqnarray}$$
We have
$$\begin{eqnarray} { \vec{e} }_{ \alpha } { \vec{e} }_{ \alpha } &=& { \vec{e} }_{ \alpha } \cdot { \vec{e} }_{ \alpha } &=& 1 \\ \frac{\partial \left( { { { \vec{e} }_{ \alpha } }^{ 2}} \right)}{\, \partial \alpha } &=& 2 { \vec{e} }_{ \alpha } \frac{\partial \left( { { \vec{e} }_{ \alpha }} \right)}{\, \partial \alpha } &=& 0\end{eqnarray}$$
and thus
$$\begin{eqnarray} \nabla \vec{A} &=& \frac{ 1}{ r} \left( { \frac{ \partial { A }_{ \alpha }}{\, \partial \alpha } + \frac{ \partial { A }_{ z}}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ z} + { A }_{ z} { \vec{e} }_{ \alpha } \frac{ \partial { \vec{e} }_{ z}}{\, \partial \alpha } + \frac{ \partial { A }_{ r}}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ r} + { A }_{ r} { \vec{e} }_{ \alpha } \frac{ \partial { \vec{e} }_{ r}}{\, \partial \alpha }} \right) \\ + \frac{ \partial { A }_{ \alpha }}{\, \partial z} { \vec{e} }_{ z} { \vec{e} }_{ \alpha } + { A }_{ \alpha } { \vec{e} }_{ z} \frac{ \partial { \vec{e} }_{ \alpha }}{\, \partial z} + \frac{ \partial { A }_{ z}}{\, \partial z} + \frac{ \partial { A }_{ r}}{\, \partial z} { \vec{e} }_{ z} { \vec{e} }_{ r} + { A }_{ r} { \vec{e} }_{ z} \frac{ \partial { \vec{e} }_{ r}}{\, \partial z} \\ + \frac{ \partial { A }_{ \alpha }}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ \alpha } + { A }_{ \alpha } { \vec{e} }_{ r} \frac{ \partial { \vec{e} }_{ \alpha }}{\, \partial r} + \frac{ \partial { A }_{ z}}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ z} + { A }_{ z} { \vec{e} }_{ r} \frac{ \partial { \vec{e} }_{ z}}{\, \partial r} + \frac{ \partial { A }_{ r}}{\, \partial r}\end{eqnarray}$$
Let's consider the last term in the first row.
$$\begin{eqnarray} { \vec{e} }_{ \alpha } \frac{ \partial { \vec{e} }_{ r}}{\, \partial \alpha } &=& { \vec{e} }_{ \alpha } \frac{ \partial \left( { \cos { \alpha } { \vec{e} }_{ x} + \sin { \alpha } { \vec{e} }_{ y}} \right)}{\, \partial \alpha } \\ { \vec{e} }_{ \alpha } \frac{ \partial { \vec{e} }_{ r}}{\, \partial \alpha } &=& { \vec{e} }_{ \alpha } \left( { - \sin { \alpha } { \vec{e} }_{ x} + \cos { \alpha } { \vec{e} }_{ y}} \right) \\ { \vec{e} }_{ \alpha } \frac{ \partial { \vec{e} }_{ r}}{\, \partial \alpha } &=& 1\end{eqnarray}$$
This gives
$$\begin{eqnarray} \nabla \vec{A} &=& \frac{ 1}{ r} \left( { \frac{ \partial { A }_{ \alpha }}{\, \partial \alpha } + \frac{ \partial { A }_{ z}}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ z} + \frac{ \partial { A }_{ r}}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ r} + { A }_{ r}} \right) \\ \frac{ \partial { A }_{ \alpha }}{\, \partial z} { \vec{e} }_{ z} { \vec{e} }_{ \alpha } + \frac{ \partial { A }_{ z}}{\, \partial z} + \frac{ \partial { A }_{ r}}{\, \partial z} { \vec{e} }_{ z} { \vec{e} }_{ r} \\ \frac{ \partial { A }_{ \alpha }}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ \alpha } + \frac{ \partial { A }_{ z}}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ z} + \frac{ \partial { A }_{ r}}{\, \partial r}\end{eqnarray}$$
and thus
$$\begin{eqnarray} \nabla \vec{A} &=& \frac{ 1}{ r} \frac{ \partial { A }_{ \alpha }}{\, \partial \alpha } + \frac{ 1}{ r} \frac{ \partial { A }_{ z}}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ z} + \frac{ 1}{ r} \frac{ \partial { A }_{ r}}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ r} + \frac{ 1}{ r} { A }_{ r} + \frac{ \partial { A }_{ \alpha }}{\, \partial z} { \vec{e} }_{ z} { \vec{e} }_{ \alpha } + \frac{ \partial { A }_{ z}}{\, \partial z} + \frac{ \partial { A }_{ r}}{\, \partial z} { \vec{e} }_{ z} { \vec{e} }_{ r} + \frac{ \partial { A }_{ \alpha }}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ \alpha } + \frac{ \partial { A }_{ z}}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ z} + \frac{ \partial { A }_{ r}}{\, \partial r} \\ \nabla \vec{A} &=& \frac{ 1}{ r} \frac{ \partial { A }_{ \alpha }}{\, \partial \alpha } + \frac{ 1}{ r} { A }_{ r} + \frac{ \partial { A }_{ z}}{\, \partial z} + \frac{ \partial { A }_{ r}}{\, \partial r} + \frac{ 1}{ r} \frac{ \partial { A }_{ z}}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ z} + \frac{ \partial { A }_{ \alpha }}{\, \partial z} { \vec{e} }_{ z} { \vec{e} }_{ \alpha } + \frac{ 1}{ r} \frac{ \partial { A }_{ r}}{\, \partial \alpha } { \vec{e} }_{ \alpha } { \vec{e} }_{ r} + \frac{ \partial { A }_{ \alpha }}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ \alpha } + \frac{ \partial { A }_{ r}}{\, \partial z} { \vec{e} }_{ z} { \vec{e} }_{ r} + \frac{ \partial { A }_{ z}}{\, \partial r} { \vec{e} }_{ r} { \vec{e} }_{ z} \\ \nabla \vec{A} &=& \frac{ 1}{ r} \frac{ \partial { A }_{ \alpha }}{\, \partial \alpha } + \frac{ 1}{ r} { A }_{ r} + \frac{ \partial { A }_{ z}}{\, \partial z} + \frac{ \partial { A }_{ r}}{\, \partial r} + \left( { \frac{ 1}{ r} \frac{ \partial { A }_{ z}}{\, \partial \alpha } - \frac{ \partial { A }_{ \alpha }}{\, \partial z}} \right) { \vec{e} }_{ \alpha } { \vec{e} }_{ z} + \left( { \frac{ 1}{ r} \frac{ \partial { A }_{ r}}{\, \partial \alpha } - \frac{ \partial { A }_{ \alpha }}{\, \partial r}} \right) { \vec{e} }_{ \alpha } { \vec{e} }_{ r} + \left( { \frac{ \partial { A }_{ r}}{\, \partial z} - \frac{ \partial { A }_{ z}}{\, \partial r}} \right) { \vec{e} }_{ z} { \vec{e} }_{ r}\end{eqnarray}$$
This result has a scalar component and a bi-vector component.
$$ \nabla \vec{A} = \nabla \cdot \vec{A} + \nabla \wedge \vec{A}$$
We consider the scalar component first.
$$\fbox{$ \displaystyle \nabla \cdot \vec{A} = \frac{ 1}{ r} \frac{ \partial { A }_{ \alpha }}{\, \partial \alpha } + \frac{ 1}{ r} { A }_{ r} + \frac{ \partial { A }_{ z}}{\, \partial z} + \frac{ \partial { A }_{ r}}{\, \partial r} $} \qquad \mbox{Divergence of a vector field}$$
This corresponds to what we have found in Eq. 2:
$$\begin{eqnarray} \mbox{div}\, { \vec{A}} &=& \frac{ 1}{ r} \left( { \frac{ \partial }{\, \partial r} \left( { { A }_{ r} r} \right) + \frac{ \partial }{\, \partial \varphi } { A }_{ \alpha } + \frac{ \partial }{\, \partial z} \left( { { A }_{ z} r} \right)} \right) \\ \mbox{div}\, { \vec{A}} &=& \frac{ 1}{ r} \left( { \frac{ \partial { A }_{ r}}{\, \partial r} r + { A }_{ r} + \frac{ \partial { A }_{ \alpha }}{\, \partial \alpha } + \frac{ \partial { A }_{ z}}{\, \partial z} r + { A }_{ z} \frac{ \partial r}{\, \partial z}} \right) \\ \mbox{div}\, { \vec{A}} &=& \frac{ \partial { A }_{ r}}{\, \partial r} + \frac{ 1}{ r} { A }_{ r} + \frac{ 1}{ r} \frac{ \partial { A }_{ \alpha }}{\, \partial \alpha } + \frac{ \partial { A }_{ z}}{\, \partial z} + \frac{ 1}{ r} { A }_{ z} \frac{ \partial r}{\, \partial z} \\ \mbox{div}\, { \vec{A}} &=& \frac{ 1}{ r} \frac{ \partial { A }_{ \alpha }}{\, \partial \alpha } + \frac{ 1}{ r} { A }_{ r} + \frac{ \partial { A }_{ z}}{\, \partial z} + \frac{ \partial { A }_{ r}}{\, \partial r} + \frac{ 1}{ r} { A }_{ z} \frac{ \partial r}{\, \partial z} \\ \mbox{div}\, { \vec{A}} &=& \frac{ 1}{ r} \frac{ \partial { A }_{ \alpha }}{\, \partial \alpha } + \frac{ 1}{ r} { A }_{ r} + \frac{ \partial { A }_{ z}}{\, \partial z} + \frac{ \partial { A }_{ r}}{\, \partial r}\end{eqnarray}$$
We also have a bi-vecor component.
$$\fbox{$ \displaystyle \nabla \wedge \vec{A} = \left( { \frac{ 1}{ r} \frac{ \partial { A }_{ z}}{\, \partial \alpha } - \frac{ \partial { A }_{ \alpha }}{\, \partial z}} \right) { \vec{e} }_{ \alpha } { \vec{e} }_{ z} + \left( { \frac{ 1}{ r} \frac{ \partial { A }_{ r}}{\, \partial \alpha } - \frac{ \partial { A }_{ \alpha }}{\, \partial r}} \right) { \vec{e} }_{ \alpha } { \vec{e} }_{ r} + \left( { \frac{ \partial { A }_{ r}}{\, \partial z} - \frac{ \partial { A }_{ z}}{\, \partial r}} \right) { \vec{e} }_{ z} { \vec{e} }_{ r} $} \qquad \mbox{Curl of a vector field} \tag{8}$$
This corresponds to what we have found in Eq. 3.
$$ \mbox{rot}\, { \vec{A}} = \left( { \frac{ 1}{ r} \frac{ \partial { A }_{ z}}{\, \partial \varphi } - \frac{ \partial { A }_{ \varphi }}{\, \partial z}} \right) { \vec{e} }_{ r} + \left( { \frac{ \partial { A }_{ r}}{\, \partial z} - \frac{ \partial { A }_{ z}}{\, \partial r}} \right) { \vec{e} }_{ \varphi } + \frac{ 1}{ r} \left( { \frac{ \partial \left( { r { A }_{ \varphi }} \right)}{\, \partial r} - \frac{ \partial { A }_{ r}}{\, \partial \varphi }} \right) { \vec{e} }_{ z}$$
$$\begin{eqnarray} \nabla \wedge \vec{A} &=& { \mbox{rot}\, { \vec{A}} }^{ *} \\ \nabla \wedge \vec{A} &=& \left( { \left( { \frac{ 1}{ r} \frac{ \partial { A }_{ z}}{\, \partial \varphi } - \frac{ \partial { A }_{ \varphi }}{\, \partial z}} \right) { \vec{e} }_{ r} + \left( { \frac{ \partial { A }_{ r}}{\, \partial z} - \frac{ \partial { A }_{ z}}{\, \partial r}} \right) { \vec{e} }_{ \varphi } + \frac{ 1}{ r} \left( { \frac{ \partial \left( { r { A }_{ \varphi }} \right)}{\, \partial r} - \frac{ \partial { A }_{ r}}{\, \partial \varphi }} \right) { \vec{e} }_{ z}} \right) { \vec{e} }_{ r} { \vec{e} }_{ \varphi } { \vec{e} }_{ z} \\ \nabla \wedge \vec{A} &=& \left( { \frac{ 1}{ r} \frac{ \partial { A }_{ z}}{\, \partial \varphi } - \frac{ \partial { A }_{ \varphi }}{\, \partial z}} \right) { \vec{e} }_{ r} { \vec{e} }_{ r} { \vec{e} }_{ \varphi } { \vec{e} }_{ z} + \left( { \frac{ \partial { A }_{ r}}{\, \partial z} - \frac{ \partial { A }_{ z}}{\, \partial r}} \right) { \vec{e} }_{ \varphi } { \vec{e} }_{ r} { \vec{e} }_{ \varphi } { \vec{e} }_{ z} + \frac{ 1}{ r} \left( { \frac{ \partial \left( { r { A }_{ \varphi }} \right)}{\, \partial r} - \frac{ \partial { A }_{ r}}{\, \partial \varphi }} \right) { \vec{e} }_{ z} { \vec{e} }_{ r} { \vec{e} }_{ \varphi } { \vec{e} }_{ z} \\ \nabla \wedge \vec{A} &=& \left( { \frac{ 1}{ r} \frac{ \partial { A }_{ z}}{\, \partial \varphi } - \frac{ \partial { A }_{ \varphi }}{\, \partial z}} \right) { \vec{e} }_{ \varphi } { \vec{e} }_{ z} + \left( { \frac{ \partial { A }_{ z}}{\, \partial r} - \frac{ \partial { A }_{ r}}{\, \partial z}} \right) { \vec{e} }_{ r} { \vec{e} }_{ z} + \frac{ 1}{ r} \left( { \frac{ \partial \left( { r { A }_{ \varphi }} \right)}{\, \partial r} - \frac{ \partial { A }_{ r}}{\, \partial \varphi }} \right) { \vec{e} }_{ r} { \vec{e} }_{ \varphi } \\ \nabla \wedge \vec{A} &=& \left( { \frac{ 1}{ r} \frac{ \partial { A }_{ z}}{\, \partial \alpha } - \frac{ \partial { A }_{ \alpha }}{\, \partial z}} \right) { \vec{e} }_{ \alpha } { \vec{e} }_{ z} + \left( { \frac{ \partial { A }_{ z}}{\, \partial r} - \frac{ \partial { A }_{ r}}{\, \partial z}} \right) { \vec{e} }_{ r} { \vec{e} }_{ z} + \frac{ 1}{ r} \left( { \frac{ \partial \left( { r { A }_{ \alpha }} \right)}{\, \partial r} - \frac{ \partial { A }_{ r}}{\, \partial \alpha }} \right) { \vec{e} }_{ r} { \vec{e} }_{ \alpha } \\ \nabla \wedge \vec{A} &=& \left( { \frac{ 1}{ r} \frac{ \partial { A }_{ z}}{\, \partial \alpha } - \frac{ \partial { A }_{ \alpha }}{\, \partial z}} \right) { \vec{e} }_{ \alpha } { \vec{e} }_{ z} + \frac{ 1}{ r} \left( { \frac{ \partial \left( { r { A }_{ \alpha }} \right)}{\, \partial r} - \frac{ \partial { A }_{ r}}{\, \partial \alpha }} \right) { \vec{e} }_{ r} { \vec{e} }_{ \alpha } + \left( { \frac{ \partial { A }_{ z}}{\, \partial r} - \frac{ \partial { A }_{ r}}{\, \partial z}} \right) { \vec{e} }_{ r} { \vec{e} }_{ z} \\ \nabla \wedge \vec{A} &=& \left( { \frac{ 1}{ r} \frac{ \partial { A }_{ z}}{\, \partial \alpha } - \frac{ \partial { A }_{ \alpha }}{\, \partial z}} \right) { \vec{e} }_{ \alpha } { \vec{e} }_{ z} + \frac{ 1}{ r} \left( { \frac{ \partial { A }_{ r}}{\, \partial \alpha } - \frac{ \partial \left( { r { A }_{ \alpha }} \right)}{\, \partial r}} \right) { \vec{e} }_{ \alpha } { \vec{e} }_{ r} + \left( { \frac{ \partial { A }_{ r}}{\, \partial z} - \frac{ \partial { A }_{ z}}{\, \partial r}} \right) { \vec{e} }_{ z} { \vec{e} }_{ r} \\ \nabla \wedge \vec{A} &=& \left( { \frac{ 1}{ r} \frac{ \partial { A }_{ z}}{\, \partial \alpha } - \frac{ \partial { A }_{ \alpha }}{\, \partial z}} \right) { \vec{e} }_{ \alpha } { \vec{e} }_{ z} + \frac{ 1}{ r} \left( { \frac{ \partial { A }_{ r}}{\, \partial \alpha } - \left( { { A }_{ \alpha } + r \frac{ \partial { A }_{ \alpha }}{\, \partial r}} \right)} \right) { \vec{e} }_{ \alpha } { \vec{e} }_{ r} + \left( { \frac{ \partial { A }_{ r}}{\, \partial z} - \frac{ \partial { A }_{ z}}{\, \partial r}} \right) { \vec{e} }_{ z} { \vec{e} }_{ r} \\ \nabla \wedge \vec{A} &=& \left( { \frac{ 1}{ r} \frac{ \partial { A }_{ z}}{\, \partial \alpha } - \frac{ \partial { A }_{ \alpha }}{\, \partial z}} \right) { \vec{e} }_{ \alpha } { \vec{e} }_{ z} + \left( { \frac{ 1}{ r} \frac{ \partial { A }_{ r}}{\, \partial \alpha } - \frac{ 1}{ r} { A }_{ \alpha } - \frac{ \partial { A }_{ \alpha }}{\, \partial r}} \right) { \vec{e} }_{ \alpha } { \vec{e} }_{ r} + \left( { \frac{ \partial { A }_{ r}}{\, \partial z} - \frac{ \partial { A }_{ z}}{\, \partial r}} \right) { \vec{e} }_{ z} { \vec{e} }_{ r}\end{eqnarray}$$
This is not 100% compatible with Eq. 8. We have to revisit this.
$$ \nabla = { \vec{w} }^{ j} { \partial }_{ { w }_{ j}} = \frac{ 1}{ { \left| { { \vec{w} }_{ j}} \right| }^{ 2}} { \vec{w} }_{ j} { \partial }_{ { w }_{ j}} = \frac{ 1}{ \left| { { \vec{w} }_{ j}} \right|} { \vec{e} }_{ { w }_{ j}} { \partial }_{ { w }_{ j}}$$
Let's consinder the general case of non-orthogonal basis vectors now. From definitions Eq. 1 and Eq. 2
$$ { \vec{w} }_{ k} = \frac{ \partial { r }_{ i}}{ { \, \partial w }_{ k}} { \vec{e} }_{ i}$$ $$ { \vec{w} }^{ j} = \frac{ \partial { w }_{ j}}{ { \, \partial r }_{ i}} { \vec{e} }_{ i}$$
we got (see Eq. 3)
$$ { \vec{w} }_{ k} \cdot { \vec{w} }^{ j} = \frac{ \partial { w }_{ j}}{ { \, \partial w }_{ k}} = \left\{ \begin{array}{r@{\quad:\quad}l}
1 & for j = k\\ 0 & for j \ne k\\
\end{array}\right.$$
so for a given index $ i$ we have
$$ { \vec{w} }_{ i} \cdot { \vec{w} }^{ i} = 1 \qquad \mbox{(no sum)}$$
The vectors $ { \vec{w} }_{ i}$ and $ { \vec{w} }^{ i}$ do not necessarily have to be parallel for the projection $ { \vec{w} }_{ i} \cdot { \vec{w} }^{ i}$ to be 1.
Let $ G$ be an $ n$-dimensional geometric algebra such that $ { I }^{ 2} \ne 0$ (-1 for $ n = 3$, 1 for $ n = 4$,...). Let $ { \vec{e} }_{ i}$ be a basis for the vectors in this algebra. Then the reciprocal bases $ { \vec{e} }^{ i}$ is defined as
$$\fbox{$ \displaystyle { \vec{e} }^{ i} = \left( { {\left( - 1 \right)}^{ i - 1} \left( { { \vec{e} }_{ 1} \wedge \wedge { \tilde{\vec{e}} }_{ i} \wedge \wedge { \vec{e} }_{ n}} \right)} \right) \cdot { I }^{ - 1} $} \tag{9}$$
in which $ { \tilde{\vec{e}} }_{ i}$ means that this index is omitted from the outer product and in which the dot represents the generalized inner product of two blades, that is the lowest grade part of their geometric product. In Geometric product of ... we have produced that geometric product of a bi-bvector with a vector and got a grade-1 component and a grade-3 component. In Geometric product of ... we multiplied two bi-vectors and got a grade-0 component and a grade-2 component. The generalized inner product is of grade $ n - k$ where $ n$ is the grade of the factor with the higher grade and $ k$ the grade of the factor with lower grade. In Eq. 9 we produce the inner product of a $ n - 1$ vector with an $ n$ - vector. This yields a result of grade-1.
Theorem: If $ { \vec{e} }_{ i}$ is an orthonormal basis then $ { \vec{e} }_{ i} = { \vec{e} }^{ i}$.
Proof: We have
$$ { \vec{e} }^{ i} = \left( { {\left( - 1 \right)}^{ i - 1} \left( { { \vec{e} }_{ 1} \wedge \wedge { \tilde{\vec{e}} }_{ i} \wedge \wedge { \vec{e} }_{ n}} \right)} \right) \cdot { I }^{ - 1}$$
Since the base vectors are orthogonal we can convert the outer product into a geometric product. The inner product extracts the lowest grade component.
$$\begin{eqnarray} { \vec{e} }^{ i} &=& {\left( - 1 \right)}^{ i - 1} \langle \left( { { \vec{e} }_{ 1} { \tilde{\vec{e}} }_{ i} { \vec{e} }_{ n}} \right) \left( { { \vec{e} }_{ n} { \vec{e} }_{ 1}} \right)\rangle_{ lo w e s t} \\ { \vec{e} }^{ i} &=& {\left( - 1 \right)}^{ i - 1} \langle { \vec{e} }_{ 1} { \tilde{\vec{e}} }_{ i} { \vec{e} }_{ n} { \vec{e} }_{ n} { \vec{e} }_{ 1}\rangle_{ lo w e s t} \\ { \vec{e} }^{ i} &=& {\left( - 1 \right)}^{ i - 1} \langle { \vec{e} }_{ 1} { \tilde{\vec{e}} }_{ i} { \vec{e} }_{ n-1} { \vec{e} }_{ n-1} { \vec{e} }_{ 1}\rangle_{ lo w e s t} \\ { \vec{e} }^{ i} &=& {\left( - 1 \right)}^{ i - 1} {\left( - 1 \right)}^{ i - 1} \langle { \vec{e} }_{ i}\rangle_{ lo w e s t} \\ { \vec{e} }^{ i} &=& {\left( - 1 \right)}^{ i - 1} {\left( - 1 \right)}^{ i - 1} { \vec{e} }_{ i} \\ { \vec{e} }^{ i} &=& { \vec{e} }_{ i}\end{eqnarray}$$
One of the main purposes of the reciprocal frame is to quickly compute the coordinates of a vector with respect to a non-orthogonal basis. In fact, the reciprocal basis is so constructed that if $ \vec{b} = { b }_{ i} { \vec{e} }_{ i}$ defines a given vector then
$$\fbox{$ \displaystyle { b }_{ i} = { \vec{e} }^{ i} \cdot \vec{b} $}$$
Let $ G$ be an $ n$-dimensional geometric algebra such that $ { I }^{ 2} \ne 0$ (-1 for $ n = 3$, 1 for $ n = 4$,...). Let $ { \vec{w} }_{ i}$ be a basis for the vectors in this algebra. Then the reciprocal bases $ { \vec{w} }^{ i}$ is defined as
$$\fbox{$ \displaystyle { \vec{w} }^{ i} = \left( { {\left( - 1 \right)}^{ i - 1} \left( { { \vec{w} }_{ 1} \wedge \wedge { \tilde{\vec{w}} }_{ i} \wedge \wedge { \vec{w} }_{ n}} \right)} \right) \cdot { I }^{ - 1} $} \tag{10}$$
in which $ { \tilde{\vec{e}} }_{ i}$ means that this index is omitted from the outer product and in which the dot represents the generalized inner product of two blades, that is the lowest grade part of their geometric product. In Geometric product of ... we have produced that geometric product of a bi-bvector with a vector and got a grade-1 component and a grade-3 component. In Geometric product of ... we multiplied two bi-vectors and got a grade-0 component and a grade-2 component. The generalized inner product is of grade $ n - k$ where $ n$ is the grade of the factor with the higher grade and $ k$ the grade of the factor with lower grade. In Eq. 9 we produce the inner product of a $ n - 1$ vector with an $ n$ - vector. This yields a result of grade-1.
Theorem: If $ { \vec{e} }_{ i}$ is an orthonormal basis then $ { \vec{e} }_{ i} = { \vec{e} }^{ i}$.
Proof: We have
$$ { \vec{e} }^{ i} = \left( { {\left( - 1 \right)}^{ i - 1} \left( { { \vec{e} }_{ 1} \wedge \wedge { \tilde{\vec{e}} }_{ i} \wedge \wedge { \vec{e} }_{ n}} \right)} \right) \cdot { I }^{ - 1}$$
Since the base vectors are orthogonal we can convert the outer product into a geometric product. The inner product extracts the lowest grade component.
$$\begin{eqnarray} { \vec{e} }^{ i} &=& {\left( - 1 \right)}^{ i - 1} \langle \left( { { \vec{e} }_{ 1} { \tilde{\vec{e}} }_{ i} { \vec{e} }_{ n}} \right) \left( { { \vec{e} }_{ n} { \vec{e} }_{ 1}} \right)\rangle_{ lowest} \\ { \vec{e} }^{ i} &=& {\left( - 1 \right)}^{ i - 1} \langle { \vec{e} }_{ 1} { \tilde{\vec{e}} }_{ i} { \vec{e} }_{ n} { \vec{e} }_{ n} { \vec{e} }_{ 1}\rangle_{ lowest} \\ { \vec{e} }^{ i} &=& {\left( - 1 \right)}^{ i - 1} \langle { \vec{e} }_{ 1} { \tilde{\vec{e}} }_{ i} { \vec{e} }_{ n-1} { \vec{e} }_{ n-1} { \vec{e} }_{ 1}\rangle_{ lowest} \\ { \vec{e} }^{ i} &=& {\left( - 1 \right)}^{ i - 1} {\left( - 1 \right)}^{ i - 1} \langle { \vec{e} }_{ i}\rangle_{ lo w e s t} \\ { \vec{e} }^{ i} &=& {\left( - 1 \right)}^{ i - 1} {\left( - 1 \right)}^{ i - 1} { \vec{e} }_{ i} \\ { \vec{e} }^{ i} &=& { \vec{e} }_{ i}\end{eqnarray}$$
One of the main purposes of the reciprocal frame is to quickly compute the coordinates of a vector with respect to a non-orthogonal basis. In fact, the reciprocal basis is so constructed that if $ \vec{b} = { b }_{ i} { \vec{e} }_{ i}$ defines a given vector then
$$\fbox{$ \displaystyle { b }_{ i} = { \vec{e} }^{ i} \cdot \vec{b} $}$$
Let
$$ { w }_{ 1} = \left( \begin{array}{c} 2\\ 1\\ 0 \end{array}\right)$$ $$ { \vec{w} }_{ 2} = \left( \begin{array}{c} 1\\ 2\\ 0 \end{array}\right)$$ $$ { \vec{w} }_{ 3} = \left( \begin{array}{c} 0\\ 0\\ 1 \end{array}\right)$$
be a non-orthonogal basis of $ { R }^{ 3}$. We show that the basis in non-orthonogla by computing
$$ { \vec{w} }_{ 1} \cdot { \vec{w} }_{ 2} = 4$$
Let's now consider the vector
$$ \vec{b} = 2 { \vec{e} }_{ x} + 3 { \vec{e} }_{ y} + 4 { \vec{e} }_{ z} \tag{11}$$
What will be the components of this vector with respect to the basis $ { \vec{w} }_{ i}$? Let's calculate Eq. 10.
$$\begin{eqnarray} { \vec{w} }^{ 1} &=& \left( { {\left( - 1 \right)}^{ 1 - 1} \left( { { \vec{w} }_{ 2} \wedge { \vec{w} }_{ 3}} \right)} \right) \cdot { I }^{ - 1} \\ { \vec{w} }^{ 2} &=& \left( { {\left( - 1 \right)}^{ 2 - 1} \left( { { \vec{w} }_{ 1} \wedge { \vec{w} }_{ 3}} \right)} \right) \cdot { I }^{ - 1} \\ { \vec{w} }^{ 3} &=& \left( { {\left( - 1 \right)}^{ 3 - 1} \left( { { \vec{w} }_{ 1} \wedge { \vec{w} }_{ 2}} \right)} \right) \cdot { I }^{ - 1}\end{eqnarray}$$
We rewrite the wedge products as geometric products.
$$\begin{eqnarray} { \vec{w} }^{ 1} &=& \left( { \frac{ 1}{ 2} \left( { { \vec{w} }_{ 2} { \vec{w} }_{ 3} - { \vec{w} }_{ 3} { \vec{w} }_{ 2}} \right)} \right) \cdot { I }^{ - 1} \\ { \vec{w} }^{ 2} &=& - \left( { \frac{ 1}{ 2} \left( { { \vec{w} }_{ 1} { \vec{w} }_{ 3} - { \vec{w} }_{ 3} { \vec{w} }_{ 1}} \right)} \right) \cdot { I }^{ - 1} \\ { \vec{w} }^{ 3} &=& \left( { \frac{ 1}{ 2} \left( { { \vec{w} }_{ 1} { \vec{w} }_{ 2} - { \vec{w} }_{ 2} { \vec{w} }_{ 1}} \right)} \right) \cdot { I }^{ - 1}\end{eqnarray}$$
and do the generalized inner product trick shown above.
$$\begin{eqnarray} { \vec{w} }^{ 1} &=& \frac{ 1}{ 2} \langle \left( { { \vec{w} }_{ 2} { \vec{w} }_{ 3} - { \vec{w} }_{ 3} { \vec{w} }_{ 2}} \right) { \vec{e} }_{ 3} { \vec{e} }_{ 2} { \vec{e} }_{ 1}\rangle_{ lowest} \\ { \vec{w} }^{ 2} &=& - \frac{ 1}{ 2} \langle \left( { { \vec{w} }_{ 1} { \vec{w} }_{ 3} - { \vec{w} }_{ 3} { \vec{w} }_{ 1}} \right) { \vec{e} }_{ 3} { \vec{e} }_{ 2} { \vec{e} }_{ 1}\rangle_{ lowest} \\ { \vec{w} }^{ 3} &=& \frac{ 1}{ 2} \langle \left( { { \vec{w} }_{ 1} { \vec{w} }_{ 2} - { \vec{w} }_{ 2} { \vec{w} }_{ 1}} \right) { \vec{e} }_{ 3} { \vec{e} }_{ 2} { \vec{e} }_{ 1}\rangle_{ lowest}\end{eqnarray}$$We need to substitute $ { \vec{w} }_{ i}$ now.
$$ { \vec{w} }_{ 1} = 2 { \vec{e} }_{ x} + { \vec{e} }_{ y}$$ $$ { \vec{w} }_{ 2} = { \vec{e} }_{ x} + 2 { \vec{e} }_{ y}$$ $$ { \vec{w} }_{ 3} = { \vec{e} }_{ z}$$
This gives
$$\begin{eqnarray} { \vec{w} }^{ 1} &=& \frac{ 1}{ 2} \langle \left( { \left( { { \vec{e} }_{ x} + 2 { \vec{e} }_{ y}} \right) { \vec{e} }_{ z} - { \vec{e} }_{ z} \left( { { \vec{e} }_{ x} + 2 { \vec{e} }_{ y}} \right)} \right) { \vec{e} }_{ z} { \vec{e} }_{ y} { \vec{e} }_{ x}\rangle_{ lowest} \\ { \vec{w} }^{ 2} &=& - \frac{ 1}{ 2} \langle \left( { \left( { 2 { \vec{e} }_{ x} + { \vec{e} }_{ y}} \right) { \vec{e} }_{ z} - { \vec{e} }_{ z} \left( { 2 { \vec{e} }_{ x} + { \vec{e} }_{ y}} \right)} \right) { \vec{e} }_{ z} { \vec{e} }_{ y} { \vec{e} }_{ x}\rangle_{ lowest} \\ { \vec{w} }^{ 3} &=& \frac{ 1}{ 2} \langle \left( { \left( { 2 { \vec{e} }_{ x} + { \vec{e} }_{ y}} \right) \left( { { \vec{e} }_{ x} + 2 { \vec{e} }_{ y}} \right) - \left( { { \vec{e} }_{ x} + 2 { \vec{e} }_{ y}} \right) \left( { 2 { \vec{e} }_{ x} + { \vec{e} }_{ y}} \right)} \right) { \vec{e} }_{ z} { \vec{e} }_{ y} { \vec{e} }_{ x}\rangle_{ lowest}\end{eqnarray}$$and then
$$\begin{eqnarray} { \vec{w} }^{ 1} &=& \frac{ 1}{ 2} \langle \left( { { \vec{e} }_{ x} { \vec{e} }_{ z} + 2 { \vec{e} }_{ y} { \vec{e} }_{ z} - \left( { { \vec{e} }_{ z} { \vec{e} }_{ x} + 2 { \vec{e} }_{ z} { \vec{e} }_{ y}} \right)} \right) { \vec{e} }_{ z} { \vec{e} }_{ y} { \vec{e} }_{ x}\rangle_{ lowest} \\ { \vec{w} }^{ 2} &=& - \frac{ 1}{ 2} \langle \left( { 2 { \vec{e} }_{ x} { \vec{e} }_{ z} + { \vec{e} }_{ y} { \vec{e} }_{ z} - \left( { 2 { \vec{e} }_{ z} { \vec{e} }_{ x} + { \vec{e} }_{ z} { \vec{e} }_{ y}} \right)} \right) { \vec{e} }_{ z} { \vec{e} }_{ y} { \vec{e} }_{ x}\rangle_{ lowest} \\ { \vec{w} }^{ 3} &=& \frac{ 1}{ 2} \langle \left( { 2 \left( { { \vec{e} }_{ x} { \vec{e} }_{ x} + 2 { \vec{e} }_{ x} { \vec{e} }_{ y}} \right) + \left( { { \vec{e} }_{ y} { \vec{e} }_{ x} + 2 { \vec{e} }_{ y} { \vec{e} }_{ y}} \right) - \left( { \left( { 2 { \vec{e} }_{ x} { \vec{e} }_{ x} + { \vec{e} }_{ x} { \vec{e} }_{ y}} \right) + 2 \left( { 2 { \vec{e} }_{ y} { \vec{e} }_{ x} + { \vec{e} }_{ y} { \vec{e} }_{ y}} \right)} \right)} \right) { \vec{e} }_{ z} { \vec{e} }_{ y} { \vec{e} }_{ x}\rangle_{ lowest}\end{eqnarray}$$
$$\begin{eqnarray} { \vec{w} }^{ 1} &=& \frac{ 1}{ 2} \langle \left( { { \vec{e} }_{ x} { \vec{e} }_{ z} + 2 { \vec{e} }_{ y} { \vec{e} }_{ z} - { \vec{e} }_{ z} { \vec{e} }_{ x} - 2 { \vec{e} }_{ z} { \vec{e} }_{ y}} \right) { \vec{e} }_{ z} { \vec{e} }_{ y} { \vec{e} }_{ x}\rangle_{ lowest} \\ { \vec{w} }^{ 2} &=& - \frac{ 1}{ 2} \langle \left( { 2 { \vec{e} }_{ x} { \vec{e} }_{ z} + { \vec{e} }_{ y} { \vec{e} }_{ z} - 2 { \vec{e} }_{ z} { \vec{e} }_{ x} - { \vec{e} }_{ z} { \vec{e} }_{ y}} \right) { \vec{e} }_{ z} { \vec{e} }_{ y} { \vec{e} }_{ x}\rangle_{ lowest} \\ { \vec{w} }^{ 3} &=& \frac{ 1}{ 2} \langle \left( { \left( { 2 + 4 { \vec{e} }_{ x} { \vec{e} }_{ y}} \right) + { \vec{e} }_{ y} { \vec{e} }_{ x} + 2 - \left( { 2 + { \vec{e} }_{ x} { \vec{e} }_{ y}} \right) - \left( { 4 { \vec{e} }_{ y} { \vec{e} }_{ x} + 2} \right)} \right) { \vec{e} }_{ z} { \vec{e} }_{ y} { \vec{e} }_{ x}\rangle_{ lowest}\end{eqnarray}$$
$$\begin{eqnarray} { \vec{w} }^{ 1} &=& \frac{ 1}{ 2} \langle \left( { 2 { \vec{e} }_{ x} { \vec{e} }_{ z} + 4 { \vec{e} }_{ y} { \vec{e} }_{ z}} \right) { \vec{e} }_{ z} { \vec{e} }_{ y} { \vec{e} }_{ x}\rangle_{ lowest} \\ { \vec{w} }^{ 2} &=& - \frac{ 1}{ 2} \langle \left( { 4 { \vec{e} }_{ x} { \vec{e} }_{ z} + 2 { \vec{e} }_{ y} { \vec{e} }_{ z}} \right) { \vec{e} }_{ z} { \vec{e} }_{ y} { \vec{e} }_{ x}\rangle_{ lowest} \\ { \vec{w} }^{ 3} &=& \frac{ 1}{ 2} \langle \left( { 2 + 4 { \vec{e} }_{ x} { \vec{e} }_{ y} + { \vec{e} }_{ y} { \vec{e} }_{ x} + 2 - 2 - { \vec{e} }_{ x} { \vec{e} }_{ y} - 4 { \vec{e} }_{ y} { \vec{e} }_{ x} - 2} \right) { \vec{e} }_{ z} { \vec{e} }_{ y} { \vec{e} }_{ x}\rangle_{ lowest}\end{eqnarray}$$
$$\begin{eqnarray} { \vec{w} }^{ 1} &=& \frac{ 1}{ 2} \langle \left( { 2 { \vec{e} }_{ x} { \vec{e} }_{ z} + 4 { \vec{e} }_{ y} { \vec{e} }_{ z}} \right) { \vec{e} }_{ z} { \vec{e} }_{ y} { \vec{e} }_{ x}\rangle_{ lowest} \\ { \vec{w} }^{ 2} &=& - \frac{ 1}{ 2} \langle \left( { 4 { \vec{e} }_{ x} { \vec{e} }_{ z} + 2 { \vec{e} }_{ y} { \vec{e} }_{ z}} \right) { \vec{e} }_{ z} { \vec{e} }_{ y} { \vec{e} }_{ x}\rangle_{ lowest} \\ { \vec{w} }^{ 3} &=& \frac{ 1}{ 2} \langle \left( { 6 { \vec{e} }_{ x} { \vec{e} }_{ y}} \right) { \vec{e} }_{ z} { \vec{e} }_{ y} { \vec{e} }_{ x}\rangle_{ lowest}\end{eqnarray}$$
$$\begin{eqnarray} { \vec{w} }^{ 1} &=& \frac{ 1}{ 2} \langle 2 { \vec{e} }_{ x} { \vec{e} }_{ z} { \vec{e} }_{ z} { \vec{e} }_{ y} { \vec{e} }_{ x} + 4 { \vec{e} }_{ y} { \vec{e} }_{ z} { \vec{e} }_{ z} { \vec{e} }_{ y} { \vec{e} }_{ x}\rangle_{ lowest} \\ { \vec{w} }^{ 2} &=& - \frac{ 1}{ 2} \langle 4 { \vec{e} }_{ x} { \vec{e} }_{ z} { \vec{e} }_{ z} { \vec{e} }_{ y} { \vec{e} }_{ x} + 2 { \vec{e} }_{ y} { \vec{e} }_{ z} { \vec{e} }_{ z} { \vec{e} }_{ y} { \vec{e} }_{ x}\rangle_{ lowest} \\ { \vec{w} }^{ 3} &=& \frac{ 1}{ 2} \langle 6 { \vec{e} }_{ x} { \vec{e} }_{ y} { \vec{e} }_{ z} { \vec{e} }_{ y} { \vec{e} }_{ x}\rangle_{ lowest}\end{eqnarray}$$
$$\begin{eqnarray} { \vec{w} }^{ 1} &=& \frac{ 1}{ 2} \langle 2 { \vec{e} }_{ z} { \vec{e} }_{ y} + 4 { \vec{e} }_{ x}\rangle_{ lowest} \\ { \vec{w} }^{ 2} &=& - \frac{ 1}{ 2} \langle - 4 { \vec{e} }_{ y} + 2 { \vec{e} }_{ x}\rangle_{ lowest} \\ { \vec{w} }^{ 3} &=& \frac{ 1}{ 2} \langle 6 { \vec{e} }_{ z}\rangle_{ lowest}\end{eqnarray}$$
$$\begin{eqnarray} { \vec{w} }^{ 1} &=& 2 { \vec{e} }_{ x} \\ { \vec{w} }^{ 2} &=& 2 { \vec{e} }_{ y} - { \vec{e} }_{ x} \\ { \vec{w} }^{ 3} &=& 3 { \vec{e} }_{ z}\end{eqnarray}$$
Now that we have calculated the reciprocal basis $ { \vec{w} }^{ i}$ we can calculate the coefficients in $ \vec{b} = { b }_{ i} { \vec{w} }_{ i}$ like this
$$ { b }_{ i} = { \vec{w} }^{ i} \cdot \vec{b}$$
for our vector Eq. 11. This gives
$$\begin{eqnarray} { b }_{ 1} &=& \left( { 2 { \vec{e} }_{ x}} \right) \cdot \left( { 2 { \vec{e} }_{ x} + 3 { \vec{e} }_{ y} + 4 { \vec{e} }_{ z}} \right) \\ { b }_{ 2} &=& \left( { 2 { \vec{e} }_{ y} - { \vec{e} }_{ x}} \right) \cdot \left( { 2 { \vec{e} }_{ x} + 3 { \vec{e} }_{ y} + 4 { \vec{e} }_{ z}} \right) \\ { b }_{ 3} &=& \left( { 3 { \vec{e} }_{ z}} \right) \cdot \left( { 2 { \vec{e} }_{ x} + 3 { \vec{e} }_{ y} + 4 { \vec{e} }_{ z}} \right)\end{eqnarray}$$
$$\begin{eqnarray} { b }_{ 1} &=& 4 \\ { b }_{ 2} &=& 8 \\ { b }_{ 3} &=& 12\end{eqnarray}$$
We check this result by calculating $ \vec{b} = { b }_{ i} { \vec{w} }_{ i}$ with these coefficients.
$$ \vec{b} = { b }_{ i} { \vec{w} }_{ i}$$
$$\begin{eqnarray} \vec{b} &=& 4 \left( \begin{array}{c} 2\\ 1\\ 0 \end{array}\right) + 8 \left( \begin{array}{c} 1\\ 2\\ 0 \end{array}\right) + 12 \left( \begin{array}{c} 0\\ 0\\ 1 \end{array}\right) \\ \vec{b} &=& \left( \begin{array}{c} 8\\ 4\\ 0 \end{array}\right) + \left( \begin{array}{c} 8\\ 16\\ 0 \end{array}\right) + \left( \begin{array}{c} 0\\ 0\\ 1 \cdot 2 \end{array}\right)\end{eqnarray}$$
This result is wrong. We should revisit this chapter.