Editing Appendix/Mathematics/EulerAngles (section)

==Basic Relations==

In terms of the (black) unit vectors of the <math>(X, Y, Z)</math> Cartesian coordinate system shown in the left panel of Figure 1, we can specify the (red) vector, <math>\vec{A}</math>, by the expression,
<div align="center">
<math>\vec{A} = \vec{e}_X (A_X) + \vec{e}_Y (A_Y) + \vec{e}_Z (A_Z)</math>
</div>
where the three coefficients, <math>(A_X, A_Y, A_Z)</math>, give the length (and <math>\pm</math> direction) of each vector component.  Alternatively, in terms of the (black) unit vectors of the <math>(x_1, x_2, x_3)</math> Cartesian coordinate system shown in the right panel of Figure 1, the same (red) vector is specified by the expression,
<div align="center">
<math>~\vec{A} = \vec{e}_1 (A_1) + \vec{e}_2 (A_2) + \vec{e}_3 (A_3) \, .</math>
</div>

<table border="1" align="center" cellpadding="3">
<tr><td align="center" colspan="2">'''Figure 1'''</td></tr>
<tr>
  <td align="center">
[[File:BerciuFig1a.png|250px|Berciu Figure 1a]]
  </td>
  <td align="center">
[[File:BerciuFig1bAgain.png|250px|Berciu Figure 1b]]
  </td>
</tr>
</table>

It is clear from the nature of unit vectors and vector dot-products that the value of the coefficient, <math>A_1</math> &#8212; which explicitly appears in the second of these two expressions &#8212; may be obtained from the dot product, <math>\vec{e}_1 \cdot \vec{A}</math>.  The same must be true if we insert, for <math>\vec{A}</math>, the first of the two expressions; that is to say,
<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right">
<math>A_1 = \vec{e}_1 \cdot \vec{A}</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\vec{e}_1 \cdot\vec{e}_X (A_X) + \vec{e}_1 \cdot\vec{e}_Y (A_Y) + \vec{e}_1 \cdot\vec{e}_Z (A_Z) \, .</math>
  </td>
</tr>
</table>
Analogously, we can write,
<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right">
<math>A_2 = \vec{e}_2 \cdot \vec{A}</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\vec{e}_2 \cdot\vec{e}_X (A_X) + \vec{e}_2 \cdot\vec{e}_Y (A_Y) + \vec{e}_2 \cdot\vec{e}_Z (A_Z) \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>A_3 = \vec{e}_3 \cdot \vec{A}</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\vec{e}_3 \cdot\vec{e}_X (A_X) + \vec{e}_3 \cdot\vec{e}_Y (A_Y) + \vec{e}_3 \cdot\vec{e}_Z (A_Z) \, .</math>
  </td>
</tr>
</table>
<span id="FormMatrix">This set of three relations</span> therefore provides a generic way to express one set of Cartesian vector projections in terms of the other.  When written in ''matrix form'', the set becomes,
<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right">
<math>\begin{bmatrix}A_1 \\ A_2 \\ A_3 \end{bmatrix}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\begin{bmatrix}
\vec{e}_1 \cdot\vec{e}_X & \vec{e}_1 \cdot\vec{e}_Y & \vec{e}_1 \cdot\vec{e}_Z  \\
\vec{e}_2 \cdot\vec{e}_X & \vec{e}_2 \cdot\vec{e}_Y & \vec{e}_2 \cdot\vec{e}_Z \\
\vec{e}_3 \cdot\vec{e}_X & \vec{e}_3 \cdot\vec{e}_Y & \vec{e}_3 \cdot\vec{e}_Z 
\end{bmatrix}
\cdot
\begin{bmatrix}A_X \\ A_Y \\ A_Z \end{bmatrix}
= \hat{R} \cdot
\begin{bmatrix}A_X \\ A_Y \\ A_Z \end{bmatrix} \, .
</math>
  </td>
</tr>
</table>

Alternatively, we could "dot" the unit-vector triplet <math>(\vec{e}_X, \vec{e}_Y, \vec{e}_Z)</math> into the vector, <math>\vec{A}</math>, in which case it would be easy to demonstrate that mapping the other direction is accomplished via the relation,
<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right">
<math>\begin{bmatrix}A_X \\ A_Y \\ A_Z \end{bmatrix}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\hat{R}^{-1} \cdot
\begin{bmatrix}A_1 \\ A_2 \\ A_3 \end{bmatrix} \, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right">
<math>\hat{R}^{-1}</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\begin{bmatrix}
\vec{e}_X \cdot\vec{e}_1 & \vec{e}_X \cdot\vec{e}_2 & \vec{e}_X \cdot\vec{e}_3  \\
\vec{e}_Y \cdot\vec{e}_1 & \vec{e}_Y \cdot\vec{e}_2 & \vec{e}_Y \cdot\vec{e}_3 \\
\vec{e}_Z \cdot\vec{e}_1 & \vec{e}_Z \cdot\vec{e}_2 & \vec{e}_Z \cdot\vec{e}_3 
\end{bmatrix} \, .
</math>
  </td>
</tr>
</table>

<table border="1" align="center" width="80%" cellpadding="8">
<tr><td align="left">

Adopting the shorthand notation used in [https://phas.ubc.ca/~berciu/TEACHING/PHYS206/LECTURES/FILES/euler.pdf Berciu's class notes], we will define,
<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right">
<math>\vec{A}_\mathrm{body} \equiv \begin{bmatrix}A_1 \\ A_2 \\ A_3 \end{bmatrix}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>\vec{A}_\mathrm{XYZ} \equiv \begin{bmatrix}A_X \\ A_Y \\ A_Z \end{bmatrix} \, .</math>
  </td>
</tr>
</table>
To be clear, these are not different vectors.  They are, rather, two different coordinate representations of the same vector as illustrated in Figure 1.  Using this notation, mapping from one representation to the other is accomplished via the compact expressions,
<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right">
<math>\vec{A}_\mathrm{body} = \hat{R} \cdot \vec{A}_{XYZ}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; or, &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>\vec{A}_\mathrm{XYZ} = \hat{R}^{-1} \cdot \vec{A}_\mathrm{body} \, .</math>
  </td>
</tr>
</table>

</td></tr>
</table>


<span id="Transpose">'''Matrix Transpose:'''</span>&nbsp; By definition, a matrix <math>\hat{M}</math> is the ''transpose'' of the matrix <math>\hat{N}</math> &#8212; that is, <math>\hat{M} = \hat{N}^T</math> &#8212; if <math>m_{ij} = n_{ji}</math> for all matrix-element indices, <math>i, j</math>.  We recognize, therefore, that
<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right">
<math>\hat{R}^{-1} = \hat{R}^T </math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>\hat{R} = [\hat{R}^{-1}]^T \, .</math>
  </td>
</tr>
</table>
Specifically, this means that all three diagonal elements (for which, <math>i=j</math>) are the same in <math>\hat{R}</math> and <math>\hat{R}^{-1}</math>; but the locations of paired off-diagonal elements are swapped.