(app_pseudo_vector)=
# Polar and axial vectors

```{admonition} Reference
{cite:t}`Ando2019`
```

```{figure} ../fig/parity-transformation.png
:name: parity-transformation
Parity transformation
```

Polar and axial vectors are vectors characterized by their behaviors for the following *parity transformation*: 
```{math}
:label: eq:app_PseudoVector_nonref_0
\begin{split}
	&\bar{x}_{i} = - x_{i} \\
	&\bar{\mathbf{e}}_{i} = - \mathbf{e}_{i}
\end{split}
```
Importantly the parity transformation changes a system from right-handed ($x$) to left-handed ($\bar{x}$), and *vice versa*. In the right-handed system, the cross products between the base vectors are summarized in the following matrix form: 
```{math}
:label: eq:app_PseudoVector_nonref_1
\left(
\begin{array}{lll}
\mathbf{e}_{x} \times \mathbf{e}_{x} &\mathbf{e}_{x} \times \mathbf{e}_{y} &\mathbf{e}_{x} \times \mathbf{e}_{z} \\
\mathbf{e}_{y} \times \mathbf{e}_{x} &\mathbf{e}_{y} \times \mathbf{e}_{y} &\mathbf{e}_{y} \times \mathbf{e}_{z} \\
\mathbf{e}_{z} \times \mathbf{e}_{x} &\mathbf{e}_{z} \times \mathbf{e}_{y} &\mathbf{e}_{z} \times \mathbf{e}_{z}
\end{array}
\right)
=
\left(
\begin{array}{rrr}
0 &\mathbf{e}_{z} &-\mathbf{e}_{y} \\
-\mathbf{e}_{z} &0 &\mathbf{e}_{x} \\
\mathbf{e}_{y} &-\mathbf{e}_{x} &0
\end{array}
\right)
```
In the left-handed system, we use the *left-screw law*, i.e., a screw moves forward when it is rotated counterclockwise. Therefore, the cross-product rule conserves: 
```{math}
:label: eq:app_PseudoVector_nonref_2
\left(
\begin{array}{lll}
\bar{\mathbf{e}}_{x} \times \bar{\mathbf{e}}_{x} &\bar{\mathbf{e}}_{x} \times \bar{\mathbf{e}}_{y} &\bar{\mathbf{e}}_{x} \times \bar{\mathbf{e}}_{z} \\
\bar{\mathbf{e}}_{y} \times \bar{\mathbf{e}}_{x} &\bar{\mathbf{e}}_{y} \times \bar{\mathbf{e}}_{y} &\bar{\mathbf{e}}_{y} \times \bar{\mathbf{e}}_{z} \\
\bar{\mathbf{e}}_{z} \times \bar{\mathbf{e}}_{x} &\bar{\mathbf{e}}_{z} \times \bar{\mathbf{e}}_{y} &\bar{\mathbf{e}}_{z} \times \bar{\mathbf{e}}_{z}
\end{array}
\right)
=
\left(
\begin{array}{rrr}
0 &\bar{\mathbf{e}}_{z} &-\bar{\mathbf{e}}_{y} \\
-\bar{\mathbf{e}}_{z} &0 &\bar{\mathbf{e}}_{x} \\
\bar{\mathbf{e}}_{y} &-\bar{\mathbf{e}}_{x} &0
\end{array}
\right)
```
By making use of the permutation symbol, these relations may be written as 
```{math}
:label: eq:app_PseudoVector_nonref_3
\begin{split}
	&\mathbf{e}_{i} = \epsilon_{ijk} \mathbf{e}_{j} \times \mathbf{e}_{k}~~~~\text{no sum on}~j, k \\
	&\bar{\mathbf{e}}_{i} = \epsilon_{ijk} \bar{\mathbf{e}}_{j} \times \bar{\mathbf{e}}_{k}~~~~\text{no sum on}~j, k
\end{split}
```
or
```{math}
:label: eq:app_PseudoVector_nonref_4
\begin{split}
	&\mathbf{e}_{i} \times \mathbf{e}_{j} = \epsilon_{kij} \mathbf{e}_{k} \\
	&\bar{\mathbf{e}}_{i} \times \bar{\mathbf{e}}_{j} = \epsilon_{kij} \bar{\mathbf{e}}_{k}
\end{split}
```
Also 
```{math}
:label: eq:app_PseudoVector_nonref_5
	\epsilon_{ijk} = \mathbf{e}_{i} \cdot ( \mathbf{e}_{j} \times \mathbf{e}_{k} ) = \bar{\mathbf{e}}_{i} \cdot ( \bar{\mathbf{e}}_{j} \times \bar{\mathbf{e}}_{k} ) = \bar{\epsilon}_{ijk}
```
Then, Kronecker's delta is the dot products between the base vectors:
```{math}
:label: eq:app_PseudoVector_nonref_6
	\delta_{ij} = \mathbf{e}_{i} \cdot \mathbf{e}_{j} = \bar{\mathbf{e}}_{i} \cdot \bar{\mathbf{e}}_{j}
```

Let us begin by investigating the behavior of the position vector $\mathbf{r}$ under the parity transformation. 
```{math}
:label: eq:app_PseudoVector_nonref_7
	\bar{\mathbf{r}}
	= \bar{x}_{i} \bar{\mathbf{e}}_{i}
	= ( - x_{i} ) ( - \mathbf{e}_{i} )
	= x_{i} \mathbf{e}_{i}
	= \mathbf{r}
```
Obviously, the position vector does not change under the parity transformation. The velocity vector $\mathbf{v}$ is the temporal derivative of $\mathbf{r}$, i.e., $\mathbf{v} = d \mathbf{r} / dt$. Therefore, 
```{math}
:label: eq:app_PseudoVector_nonref_8
	\bar{\mathbf{v}}
	= \frac{d \bar{\mathbf{r}}}{dt}
	= \frac{d \bar{x _{i}} \bar{\mathbf{e}}_{i}}{dt}
	= \frac{d \bar{x _{i}}}{dt} \bar{\mathbf{e}}_{i}
	= \left( - \frac{d x _{i}}{dt} \right) \left( - \mathbf{e}_{i} \right)
	= \frac{d x _{i}}{dt} \mathbf{e}_{i}
	= \mathbf{v}
```
The velocity vector is also unchangeable under the parity transformation. This is of course also true for the momentum vector $\mathbf{p} = m \mathbf{v}$. Vectors unchangeable under the parity transformation are called *polar vectors*. 

The angular momentum is given by the cross product $\mathbf{L} = \mathbf{r} \times \mathbf{p}$. Its behavior under the parity transformation is as follows: 
```{math}
:label: eq:app_PseudoVector_nonref_9
	\bar{\mathbf{L}}
	= \bar{\mathbf{r}} \times \bar{\mathbf{p}}
	= \bar{x}_{i} \bar{\mathbf{e}}_{i} \times \bar{p}_{j} \bar{\mathbf{e}}_{j}
	= \bar{x}_{i} \bar{p}_{j} \bar{\mathbf{e}}_{i} \times \bar{\mathbf{e}}_{j}
	= (- x_{i} ) (- p_{j}) \bar{\mathbf{e}}_{i} \times \bar{\mathbf{e}}_{j}
	= x_{i} p_{j} \bar{\mathbf{e}}_{i} \times \bar{\mathbf{e}}_{j}
```
where the components were transformed using the polar character. Then, 
```{math}
:label: eq:app_PseudoVector_nonref_10
	x_{i} p_{j} \bar{\mathbf{e}}_{i} \times \bar{\mathbf{e}}_{j}
	= \epsilon_{kij} x_{i} p_{j} \bar{\mathbf{e}}_{k}
	= - \epsilon_{kij} x_{i} p_{j} \mathbf{e}_{k}
	= - \mathbf{L}
```
Thus, $\bar{\mathbf{L}} = - \mathbf{L}$. This result clearly shows that the parity transformation changes the direction of the angular momentum. It should however be noted that the direction of rotation represented by $\mathbf{L}$ does not change. For example, for a point mass rotating counterclockwise about the $z$ axis on the $xy$ plane, $\mathbf{L}$ directs the positive $z$. In the left-handed system $\bar{x}$ obtained by the parity transformation, $\bar{\mathbf{L}} = - \mathbf{L}$, so that the direction of $\bar{\mathbf{L}}$ is the negative $z$. This vector also represents the counterclockwise rotation on the $xy$ plane since the left-screw law is applied in the $\bar{x}$ system. Therefore, the reversal of $\mathbf{L}$ is required to maintain the physical state of the rotation. Vectors possessing this character are *axial vectors* (or pseudo vectors). 

The differential operator $\nabla$ is a vector-like operator written by 
```{math}
:label: eq:app_PseudoVector_nonref_11
	\nabla = \frac{\partial}{\partial x_{i}} \mathbf{e}_{i}
```
This can be regarded as polar vector as shown below: 
```{math}
:label: eq:app_PseudoVector_nonref_12
	\bar{\nabla} 
	= \frac{\partial}{\partial \bar{x}_{i}} \bar{\mathbf{e}}_{i}
	= \left( - \frac{\partial}{\partial x_{i}} \right) \left( - \mathbf{e}_{i} \right)
	= \frac{\partial}{\partial x_{i}} \mathbf{e}_{i}
	= \nabla
```
The vorticity defined by 
```{math}
:label: eq:app_PseudoVector_nonref_13
	\boldsymbol{\omega} = \nabla \times \mathbf{v}
```
is an axial vector as demonstrated below 
```{math}
:label: eq:app_PseudoVector_nonref_14
	\bar{\boldsymbol{\omega}}
	= \bar{\nabla} \times \bar{ \mathbf{v} }
	= \epsilon_{ijk} \frac{\partial \bar{v}_{k}}{\partial \bar{x}_{j}}  \bar{\mathbf{e}}_{i}
	= - \epsilon_{ijk} \frac{\partial v_{k}}{\partial x_{j}} \mathbf{e}_{i}
	= - \nabla \times \mathbf{v}
	= - \boldsymbol{\omega}
```

It is obvious that polar vectors cannot be added to axial vectors since their behaviors under the transformation are different. In the equation of motion of fluids, we see 
```{math}
:label: eq:app_PseudoVector_nonref_15
	\frac{\partial \mathbf{v}}{\partial t} + \nabla \left( \frac{v^{2}}{2} \right) + \boldsymbol{\omega} \times \mathbf{v} = - \frac{\nabla p}{\rho}
```
The first term is polar vector. The second term is also polar due to the polar nature of $\nabla$. Therefore, the third term must also be polar vector, showing that the cross product between an axial vector and a polar vector generates a polar vector. Let us develop lookup tables for the results of cross and dot products between vectors. In the following the superscripts $(p)$ and $(a)$ denote polar and axial vectors, respectively. First, the cross product $\mathbf{A}$ between polar vectors $\mathbf{a}^{(p)}$ and $\mathbf{b}^{(p)}$ is 
```{math}
:label: eq:app_PseudoVector_nonref_16
	\bar{ \mathbf{A} }
	= \bar{ \mathbf{a} }^{(p)} \times \bar{ \mathbf{b} }^{(p)}
	= \epsilon_{ijk} \bar{a}_{j} \bar{b}_{k} \bar{\mathbf{e}}_{i}
	= - \epsilon_{ijk} a_{j} b_{k} \mathbf{e}_{i}
	= - \mathbf{a}^{(p)} \times \mathbf{b}^{(p)}
	= - \mathbf{A}~~~~(\text{axial vector})
```
```{math}
:label: eq:app_PseudoVector_nonref_17
\begin{split}	
	\bar{ \mathbf{A} }
	&= \bar{ \mathbf{a} }^{(a)} \times \bar{ \mathbf{b} }^{(a)}
	= ( \bar{a}^{(a)}_{i} \bar{\mathbf{e}}_{i}) \times ( \bar{b}^{(a)}_{j} \bar{\mathbf{e}}_{j} )
	= ( a^{(a)}_{i} \bar{\mathbf{e}}_{i}) \times ( b^{(a)}_{j} \bar{\mathbf{e}}_{j} )
	= a^{(a)}_{i} b^{(a)}_{j} \bar{\mathbf{e}}_{i} \times \bar{\mathbf{e}}_{j} \\
	&= \epsilon_{kij} a^{(a)}_{i} b^{(a)}_{j} \bar{\mathbf{e}}_{k}
	= - \epsilon_{kij} a^{(a)}_{i} b^{(a)}_{j} \mathbf{e}_{k}
	= - \mathbf{a}^{(a)} \times \mathbf{b}^{(a)}
	= - \mathbf{A}~~~~(\text{axial vector})
\end{split}
```
where we used the inversion of the components of the axial vector: $\bar{a}^{(a)}_{i} \bar{\mathbf{e}}_{i} = a^{(a)}_{i} \bar{\mathbf{e}}_{i}$. Then, 
```{math}
:label: eq:app_PseudoVector_nonref_18
\begin{split}	
	\bar{ \mathbf{A} }
	&= \bar{ \mathbf{a} }^{(p)} \times \bar{ \mathbf{b} }^{(a)}
	= ( \bar{a}^{(p)}_{i} \bar{\mathbf{e}}_{i}) \times ( \bar{b}^{(a)}_{j} \bar{\mathbf{e}}_{j} )
	= ( - a^{(p)}_{i} \bar{\mathbf{e}}_{i}) \times ( b^{(a)}_{j} \bar{\mathbf{e}}_{j} )
	= - a^{(p)}_{i} b^{(a)}_{j} \bar{\mathbf{e}}_{i} \times \bar{\mathbf{e}}_{j} \\
	&= - \epsilon_{kij} a^{(p)}_{i} b^{(a)}_{j} \bar{\mathbf{e}}_{k}
	= \epsilon_{kij} a^{(p)}_{i} b^{(a)}_{j} \mathbf{e}_{k}
	= \mathbf{a}^{(p)} \times \mathbf{b}^{(a)}
	= \mathbf{A}~~~~(\text{polar vector})
\end{split}
```
```{math}
:label: eq:app_PseudoVector_nonref_19
\begin{split}	
	\bar{ \mathbf{A} }
	&= \bar{ \mathbf{a} }^{(a)} \times \bar{ \mathbf{b} }^{(p)}
	= ( \bar{a}^{(a)}_{i} \bar{\mathbf{e}}_{i}) \times ( \bar{b}^{(p)}_{j} \bar{\mathbf{e}}_{j} )
	= ( a^{(a)}_{i} \bar{\mathbf{e}}_{i}) \times ( - b^{(p)}_{j} \bar{\mathbf{e}}_{j} )
	= - a^{(a)}_{i} b^{(p)}_{j} \bar{\mathbf{e}}_{i} \times \bar{\mathbf{e}}_{j} \\
	&= - \epsilon_{kij} a^{(a)}_{i} b^{(p)}_{j} \bar{\mathbf{e}}_{k}
	= \epsilon_{kij} a^{(a)}_{i} b^{(p)}_{j} \mathbf{e}_{k}
	= \mathbf{a}^{(a)} \times \mathbf{b}^{(p)}
	= \mathbf{A}~~~~(\text{polar vector})
\end{split}
```
Types of scalars generated by dot products are as follows. 
```{math}
:label: eq:app_PseudoVector_nonref_20
	\bar{\alpha} 
	= \bar{\mathbf{a}}^{(p)} \cdot \bar{\mathbf{b}}^{(p)}
	= \bar{a}^{(p)}_{i} \bar{\mathbf{e}}_{i} \cdot \bar{b}^{(p)}_{j} \bar{\mathbf{e}}_{j} 
	= \bar{a}^{(p)}_{i} \bar{b}^{(p)}_{j} \delta_{ij} 
	= \bar{a}^{(p)}_{i} \bar{b}^{(p)}_{i}
	= (-a^{(p)}_{i}) (-b^{(p)}_{i})
	= a^{(p)}_{i} b^{(p)}_{i}
	= \alpha~~~~(\text{scalar})
```
```{math}
:label: eq:app_PseudoVector_nonref_21
	\bar{\alpha} 
	= \bar{\mathbf{a}}^{(a)} \cdot \bar{\mathbf{b}}^{(a)}
	= \bar{a}^{(a)}_{i} \bar{\mathbf{e}}_{i} \cdot \bar{b}^{(a)}_{j} \bar{\mathbf{e}}_{j} 
	= a^{(a)}_{i} \bar{\mathbf{e}}_{i} \cdot b^{(a)}_{j} \bar{\mathbf{e}}_{j} 
	= a^{(a)}_{i} b^{(a)}_{j} \delta_{ij}
	= a^{(a)}_{i} b^{(a)}_{i}
	= \alpha~~~~(\text{scalar})
```
```{math}
:label: eq:app_PseudoVector_nonref_22
	\bar{\alpha} 
	= \bar{\mathbf{a}}^{(p)} \cdot \bar{\mathbf{b}}^{(a)}
	= \bar{a}^{(p)}_{i} \bar{\mathbf{e}}_{i} \cdot \bar{b}^{(a)}_{j} \bar{\mathbf{e}}_{j} 
	= -a^{(p)}_{i} b^{(a)}_{j} \delta_{ij} 
	= -a^{(p)}_{i} b^{(a)}_{i}
	= -\alpha~~~~(\text{pseudo scalar})
```
```{math}
:label: eq:app_PseudoVector_nonref_23
	\bar{\alpha} 
	= \bar{\mathbf{a}}^{(a)} \cdot \bar{\mathbf{b}}^{(p)}
	= \bar{a}^{(a)}_{i} \bar{\mathbf{e}}_{i} \cdot \bar{b}^{(p)}_{j} \bar{\mathbf{e}}_{j} 
	= -a^{(a)}_{i} b^{(p)}_{j} \delta_{ij} 
	= -a^{(a)}_{i} b^{(p)}_{i}
	= -\alpha~~~~(\text{pseudo scalar})
```

For the cross product: 
| | polar | axial |
| :--- | :---: | :---: |
| **polar** | axial | polar |
| **axial** | polar | axial |

For the dot product: 
| | polar | axial |
| :--- | :---: | :---: |
| **polar** | scalar | pseudo scalar |
| **axial** | pseudo scalar | scalar |


In the discussion on the Stokes drag, we set 
```{math}
:label: eq:app_PseudoVector_nonref_24
	\mathbf{v} = \nabla \times \mathbf{A} + \mathbf{u}
```
where $\mathbf{v}$ and $\mathbf{u}$ are polar vectors. Therefore, $\nabla \times \mathbf{A}$ should also be polar to add them each other. As we discussed $\nabla$ behaves like polar vector, so that, $\mathbf{A}$ should be an axial vector. The vector potential $\mathbf{A}$ obviously depends on $\mathbf{u}$ and should be proportional to $\mathbf{u}$ since $\nabla \times \mathbf{A}$ needs to produce $- \mathbf{u}$ at the sphere surface to satisfy the boundary condition $\mathbf{v} = 0$ at $r = a$. On the other hand, the vector potential should also be a function of $\mathbf{r}$. The velocity and position vectors are polar vectors. Therefore, $\mathbf{r} \times \mathbf{u}$ is only the form satisfying all the requirements for $\mathbf{A}$. Thus, 
```{math}
:label: eq:app_PseudoVector_nonref_25
	\mathbf{A} = g(r) \mathbf{e}_{r} \times \mathbf{u}
```