Editing Appendix/Mathematics/EulerAngles (section)

===Using Matrix Notation===

<table border="1" align="right" cellpadding="1">
<tr><td align="center">Astrophysics Example</td></tr>
<tr><td align="center">[[File:EulerAnglesExample02.jpeg|250px|Euler Angles Example02]]</td></tr>
</table>
With more finesse, we can write the general rotation matrix that links the ''body frame'', <math>(x_1, x_2, x_3)</math>, to the ''inertial/laboratory'' frame, <math>(X, Y, Z)</math>, as the product of the three rotations about the corresponding axes:
<table border="0" align="center" cellpadding="3">
<tr>
  <td align="right">
<math>\hat{R}(\phi, \theta, \psi) = \hat{R}_3(\psi) \cdot \hat{R}_1(\theta) \cdot \hat{R}_3(\phi)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
{\begin{bmatrix}
\cos\psi & \sin\psi & 0  \\
-\sin\psi & \cos\psi & 0 \\
0 & 0 & 1 
\end{bmatrix}} \cdot
{\begin{bmatrix}
1 & 0 & 0  \\
0 & \cos\theta & \sin\theta \\
0 & -\sin\theta & \cos\theta 
\end{bmatrix}} \cdot
{\begin{bmatrix}
\cos\phi & \sin\phi & 0  \\
-\sin\phi & \cos\phi & 0 \\
0 & 0 & 1 
\end{bmatrix}} \, .
</math>
  </td>
</tr>

<tr>
<td align="center" colspan="3">
[https://phas.ubc.ca/~berciu/TEACHING/PHYS206/LECTURES/FILES/euler.pdf Berciu's online class notes], bottom of p. 3
</td>
</tr>
</table>

<span id="RotationMatrix">Carrying out the matrix multiplications</span>, starting from the right, gives,
<table border="0" align="center" cellpadding="3">
<tr>
  <td align="right">
<math>\hat{R}(\phi, \theta, \psi)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
{\begin{bmatrix}
\cos\psi & \sin\psi & 0  \\
-\sin\psi & \cos\psi & 0 \\
0 & 0 & 1 
\end{bmatrix}} \cdot
{\begin{bmatrix}
\cos\phi & \sin\phi & 0  \\
-\sin\phi \cos\theta & \cos\theta \cos\phi & \sin\theta \\
\sin\theta\sin\phi & -\sin\theta \cos\phi & \cos\theta 
\end{bmatrix}} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
{\begin{bmatrix}
(\cos\psi \cos\phi - \sin\psi \sin\phi \cos\theta)  
& (\sin\phi \cos\psi + \sin\psi \cos\theta \cos\phi)
& (\sin\psi \sin\theta) \\
(-\sin\psi \cos\phi - \sin\phi \cos\theta \cos\psi) 
& ( - \sin\psi \sin\phi + \cos\psi \cos\theta\cos\phi )
& \sin\theta \cos\psi \\
\sin\theta\sin\phi & -\sin\theta \cos\phi & \cos\theta
\end{bmatrix}} \, .
</math>
  </td>
</tr>
</table>

This precisely matches the Euler-angle expression for the rotation matrix, <math>\hat{R}</math>, that we derived above using a more brute force approach.

----

For the [[#Simple_Numerical_Example|'''simple numerical example''' established above]], these three derivation steps give, respectively,
<table border="0" align="center" cellpadding="2">
<tr>
  <td align="right">
<math>\hat{R}(\phi, \theta, \psi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
{\begin{bmatrix}
0.9659 & 0.2588 & 0  \\
-0.2588 & 0.9659 & 0 \\
0 & 0 & 1 
\end{bmatrix}}
\cdot
{\begin{bmatrix}
1 & 0 & 0  \\
0 & 0.9063 & 0.4226 \\
0 & -0.4226 & 0.9063 
\end{bmatrix}}
\cdot
{\begin{bmatrix}
0.8660 & 0.5000 & 0  \\
-0.5000 & 0.8660 & 0 \\
0 & 0 & 1 
\end{bmatrix}}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
{\begin{bmatrix}
0.9659 & 0.2588 & 0  \\
-0.2588 & 0.9659 & 0 \\
0 & 0 & 1 
\end{bmatrix}}
\cdot
{\begin{bmatrix}
0.8660 & 0.500 & 0  \\
-0.4532 & 0.7849 & 0.4226 \\
0.2113 & -0.3660 & 0.9063 
\end{bmatrix}}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
{\begin{bmatrix}
0.7192 & 0.6861 & 0.1094  \\
-0.6619 & 0.6287 & 0.4082 \\
0.2113 & -0.3660 & 0.9063 
\end{bmatrix}} \, .
</math>
  </td>
</tr>
</table>

----


<span id="Rinverse">Following</span> the [[#Transpose|steps provided above]], we recognize that the inverse &#8212; or transpose &#8212; of this rotation matrix is,
<table border="0" align="center" cellpadding="3">
<tr>
  <td align="right">
<math>\hat{R}^{-1} = \hat{R}^T</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
{\begin{bmatrix}
(\cos\psi \cos\phi - \sin\psi \sin\phi \cos\theta)  
&(-\sin\psi \cos\phi - \sin\phi \cos\theta \cos\psi)
& (\sin\theta\sin\phi) \\

 (\sin\phi \cos\psi + \sin\psi \cos\theta \cos\phi) 
& ( - \sin\psi \sin\phi + \cos\psi \cos\theta\cos\phi )
& (-\sin\theta \cos\phi) \\

(\sin\psi \sin\theta) & (\sin\theta \cos\psi) & (\cos\theta)
\end{bmatrix}} \, .
</math>
  </td>
</tr>
</table>

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

From the last row of the column labeled "Proper Euler angles" in [https://en.wikipedia.org/wiki/Euler_angles#Rotation_matrix Wikipedia's discussion of the ''rotation matrix''], we find,
<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right">
<math>~Z_1 X_2 Z_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\begin{bmatrix}
(c_1c_3 - c_2s_1s_3) & (-c_1s_3 - c_2 c_3s_1) & (s_1s_2) \\
(c_3s_1 + c_1c_2s_3) & (c_1c_2c_3 - s_1s_3)  & (-c_1s_2) \\
(s_2s_3) & (c_3s_2) & (c_2)
\end{bmatrix}
</math>
  </td>
</tr>
</table>
The equivalent expression can be found at the top of p. 4 of [https://phas.ubc.ca/~berciu/TEACHING/PHYS206/LECTURES/FILES/euler.pdf Professor Berciu's online class notes]; it reads,
<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right">
<math>~\hat{R}(\phi, \theta, \psi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\begin{bmatrix}
(\cos\phi \cos\psi - \cos\theta \sin\phi \sin\psi) & (-\cos\phi \sin\psi - \cos\theta \cos\psi \sin\phi) & (\sin\phi \sin\theta) \\
(\cos\psi \sin\phi + \cos\phi \cos\theta \sin\psi) & (\cos\phi \cos\theta \cos\psi - \sin\phi \sin\psi)  & (-\cos\phi \sin\theta) \\
(\sin\theta \sin\psi) & (\cos\psi \sin\theta) & (\cos\theta)
\end{bmatrix}
</math>
  </td>
</tr>
</table>

These two rotation-matrix expressions are equivalent to one another, but they do not match our derived expression for <math>\hat{R}</math>.  Instead, then match our expression for <math>\hat{R}^{-1}</math>.  <font color="red">It is not (yet) clear to us why this is the case.</font>

----

For the [[#Simple_Numerical_Example|'''simple numerical example''' established above]], the rotation matrix on the right-hand-side of both of these expressions gives,
<table border="0" align="center" cellpadding="2">
<tr>
  <td align="left">
<math>
{\begin{bmatrix}
0.7192 & -0.6619 & 0.2113 \\
0.6861 & 0.6287 & -0.3660 \\
0.1094 & 0.4082 & 0.9063 
\end{bmatrix}} \, .
</math>
  </td>
</tr>
</table>

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