Editing Appendix/Mathematics/EulerAngles (section)

==Extra Illustrations==

For the most part, we have focused our discussion on the case where three simple rotations are carried out '''in the order''' depicted in Figure 2.  This sequence  &#8212; first <math>\phi</math> (about the original <math>Z</math> axis), second <math>\theta</math> (about the line of nodes), third <math>\psi</math> (about the new, tilted <math>\boldsymbol{x_3}</math> axis) &#8212; has been denoted by the mathematical expression,
<div align="center">
<math>\hat{R}(\phi, \theta, \psi) = \hat{R}_3(\psi) \cdot \hat{R}_1(\theta) \cdot \hat{R}_3(\phi) \, .</math><br />
[https://phas.ubc.ca/~berciu/TEACHING/PHYS206/LECTURES/FILES/euler.pdf Berciu's online class notes], bottom of p. 3
</div>
It is extremely important to appreciate that the '''order''' in which the rotations are carried out matters!  On the right-hand-side of our adopted expression, the matrix that specifies the first rotation <math>(\phi)</math> is placed farthest to the right, while the matrix that specifies the third/last rotation <math>(\psi)</math> is placed farthest to the left.  In the paragraph that follows, we explicitly demonstrate that a different mapping arises if the first and third rotation matrices are swapped.  But, as we shall illustrate, once the '''order''' of rotations has been specified, it doesn't matter whether the steps that are taken to combine matrix elements begins with multiplication of the first and second rotation matrices, or begins with multiplication of the second and third matrices.

===The Order of Rotations Matters===

Let's swap the order of the rotations that are identified in Figure 2 as #1 <math>(\phi)</math> and #3 <math>(\psi)</math>.  The resulting combined matrix expression is,

<table border="0" align="center" cellpadding="3">

<tr>
  <td align="right">
<math>\hat{R}_3(\phi) \cdot \hat{R}_1(\theta) \cdot \hat{R}_3(\psi)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
{\begin{bmatrix}
\cos\phi & \sin\phi & 0  \\
-\sin\phi & \cos\phi & 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\psi & \sin\psi & 0  \\
-\sin\psi & \cos\psi & 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}
\cos\phi & \sin\phi & 0  \\
-\sin\phi & \cos\phi & 0 \\
0 & 0 & 1 
\end{bmatrix}} \cdot
{\begin{bmatrix}
\cos\psi & \sin\psi & 0  \\
-\sin\psi \cos\theta & \cos\theta \cos\psi & \sin\theta \\
\sin\theta \sin\psi & -\sin\theta \cos\psi & \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\phi \cos\psi - \sin\phi \sin\psi \cos\theta) & (\cos\phi \sin\psi + \sin\phi \cos\theta \cos\psi) & (\sin\phi \sin\theta)  \\
(-\sin\phi \cos\psi - \cos\phi \sin\psi \cos\theta) & (-\sin\phi \sin\psi + \cos\phi \cos\theta \cos\psi) & (\sin\theta \cos\phi) \\
(\sin\theta \sin\psi) & (-\sin\theta \cos\psi) & (\cos\theta) 
\end{bmatrix}} \, .
</math>
  </td>
</tr>
</table>
This does '''not''' match the [[#Using_Matrix_Notation|expression that was derived earlier]] for <math>\hat{R}(\phi,\theta,\psi)</math>; nor does it match [[#Rinverse|the inverse]], <math>\hat{R}^{-1}</math>.  This illustrates that the '''order of specified rotations''' does matter.  It is worth pointing out that the ''diagonal elements'' of all of these rotation-matrix expressions are identical, but the off-diagonal terms differ.

===Steps Taken to Combine Matrix Elements===

In our [[#Using_Matrix_Notation|above discussion]], we carried out the multiplication of three rotation matrices &#8212; strategically combining the elements of various columns and rows &#8212; by starting from the right, that is, by first multiplying the 2<sup>nd</sup> and 3<sup>rd</sup> matrices.  Here, beginning with the same expression for <math>\hat{R}(\phi, \theta,\psi)</math>, we multiply the three matrices by starting from the left, that is, by first multiplying the 1<sup>st</sup> and 2<sup>nd</sup> matrices.


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


This is precisely the same result that we obtained previously, illustrating that once the '''order''' of the multiplications has been established, it doesn't matter how pairs of matrices are grouped in order to carry out the row-column multiplications.