This page supplements the previous coordinate transformation page
by focusing on the many ways to generate and interpret the transformation matrix, \({\bf Q}\).
Remember, one more time, that the transform matrix rotates the coordinate system, not the object.
Quick Review
The transformation matrix, \({\bf Q}\), is used in coordinate transformations of vectors
and tensors as follows.
\[
{\bf Q} =
\left[ \matrix {\;\;\;\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta} \right]
\]
where \(\theta\) is the angle between the old and new axes. This is simply
a special case of the general 3-D case discussed below.
Transformation Matrix Properties
Transformation matrices have several special properties that, while easily seen in this
discussion of 2-D vectors, are equally applicable to 3-D applications as well.
This list is useful for checking the accuracy of a transformation matrix if questions
arise. While a matrix still could be wrong even if it passes all these checks,
it is definitely wrong if it fails even one!
The determinant of \({\bf Q}\) equals one.
The transpose of \({\bf Q}\) is its inverse.
The dot product of any row or column with itself equals one.
Ex: \( (\cos \theta \; {\bf i} + \sin \theta \; {\bf j}) \cdot
(\cos \theta \; {\bf i} + \sin \theta \; {\bf j}) = 1 \)
The dot product of any row with any other row equals zero.
Ex: \( (\cos \theta \; {\bf i} + \sin \theta \; {\bf j}) \cdot
(-\sin \theta \; {\bf i} + \cos \theta \; {\bf j}) = 0 \)
The dot product of any column with any other column equals zero.
Ex: \( (\cos \theta \; {\bf i} - \sin \theta \; {\bf j}) \cdot
(\sin \theta \; {\bf i} + \cos \theta \; {\bf j}) = 0 \)
Recall from above that the dot product of any two different rows or columns of
a transformation matrix is zero, while the dot product of any row or column
with itself is one. This can be written in matrix and tensor notation as
\[
{\bf Q} \cdot {\bf Q}^T = {\bf I} \qquad \qquad \text{and} \qquad \qquad
\lambda_{ik} \lambda_{jk} = \delta_{ij}
\]
This shows that the transpose of a transformation matrix is also
its inverse.
The general definition of \({\bf Q}\), in 3-D, is
\[
{\bf Q} = \left[
\matrix { \cos(x',x) & \cos(x',y) & \cos(x',z) \\
\cos(y',x) & \cos(y',y) & \cos(y',z) \\
\cos(z',x) & \cos(z',y) & \cos(z',z) } \right]
\]
where \((x',x)\) represents the angle between the \(x'\) and \(x\) axes,
\((x',y)\) is the angle between the \(x'\) and \(y\) axes, etc.
3-D Reduction to 2-D
So a rotation about the \(z\)-axis means that \(\cos(z',z) = 1\) because
the angle between \(z'\) and \(z\) remains 0°. Meanwhile,
\(\cos(x',z) = \cos(y',z) = \cos(z',x) = \cos(z',y) = 0\) because the
angles between these axes remains 90°.
The angle between \(x'\) and \(y\) is \((90^\circ - \theta)\), and
\(\cos(x',y) = \cos(90^\circ - \theta) = \sin \theta\).
Likewise, the angle between \(y'\) and \(x\) is \((90^\circ + \theta)\), and
\(\cos(y',x) = \cos(90^\circ + \theta) = -\sin \theta\).
An alternative way of interpreting and generating the transformation matrix is as follows.
\[
{\bf Q} = \left[
\matrix { \left( \matrix{\text{x-comp} \\ \text{of } {\bf i'}} \right) &
\left( \matrix{\text{y-comp} \\ \text{of } {\bf i'}} \right) &
\left( \matrix{\text{z-comp} \\ \text{of } {\bf i'}} \right) \\
\left( \matrix{\text{x-comp} \\ \text{of } {\bf j'}} \right) &
\left( \matrix{\text{y-comp} \\ \text{of } {\bf j'}} \right) &
\left( \matrix{\text{z-comp} \\ \text{of } {\bf j'}} \right) \\
\left( \matrix{\text{x-comp} \\ \text{of } {\bf k'}} \right) &
\left( \matrix{\text{y-comp} \\ \text{of } {\bf k'}} \right) &
\left( \matrix{\text{z-comp} \\ \text{of } {\bf k'}} \right) }
\right]
\]
where "x-comp of \({\bf i'}\)" means the x-component of the \({\bf i'}\) unit vector.
Pay close attention here to which terms have primes on them and which don't.
Don't confuse this with "the x'-component" because "the x'-comp of \({\bf i'}\)"
is simply 1. Yet another way to say this is, "the first component of the \({\bf i'}\)
unit vector in the non-primed reference \(x, y, z\) coordinate system."
Transformation Webpage
This webpage performs coordinate transforms on 3-D tensors.
It also reports the transformation matrix. Try it out. Note that rotations on the webpage are
always about global axes, e.g., a rotation in the 1-2 plane will be about the 3-axis.
Successive Rotations - Roe Convention
The 3-D transformation matrix can be viewed as a series of three successive
rotations about coordinate axes. There must be dozens of variations of this
since any combination of axes can be chosen in any order to rotate about.
One popular choice is the so-called Roe convention.
As shown in the figure,
it consists of (i) a rotation through angle \(\psi\) about the z-axis, then
(ii) a rotation of angle \(\theta\)
about the new y-axis (which has already rotated itself due to the first rotation
about z), and finally, (iii) a second rotation of \(\phi\) about the now-tilted
z-axis.
The Roe convention is very popular despite one key challenge it contains. That
is... if \(\theta = 0\), then \(\psi\) and \(\phi\) become indistinguishable
and only the sum of the two is what matters.
In the figure above, \(\psi = 60^\circ, \theta = 30^\circ,\) and \(\phi = 45^\circ\).
These Roe angles give a coordinate transformation matrix equal to
Going In Reverse - Determining Roe Angles from a Matrix
Suppose you have a transformation matrix populated with values as shown here
and would like to know what Roe transformation angles were responsible
for producing the matrix.
\[
{\bf Q} = \left[ \matrix { q_{11} & q_{12} & q_{13} \\
q_{21} & q_{22} & q_{23} \\ q_{31} & q_{32} & q_{33} } \right]
\]
The first and easiest step is to look at the \(q_{33}\) component. It is simply the cosine of \(\theta\),
so \(\theta = \text{Cos}^{-1}(q_{33})\).
The second step is to determine \(\psi\) as follows
\[
{q_{32} \over q_{31}} = { \sin \psi \sin \theta \over \cos \psi \sin \theta } =
{ \sin \psi \over \cos \psi } = \tan \psi
\]
So
\[
\psi = \text{Tan}^{-1} \left( {q_{32} \over q_{31}} \right)
\]
\(\phi\) is determined in a very similar manner.
\[
{q_{23} \over -q_{13}} = { \sin \theta \sin \phi \over \sin \theta \cos \phi } =
{ \sin \phi \over \cos \phi } = \tan \phi
\]
So
\[
\phi = \text{Tan}^{-1} \left({q_{23} \over -q_{13}} \right)
\]
In summary
\[
\psi = \text{Tan}^{-1} \left( {q_{32} \over q_{31}} \right) \qquad \qquad
\theta = \text{Cos}^{-1} \left( q_{33} \right) \qquad \qquad
\phi = \text{Tan}^{-1} \left({q_{23} \over -q_{13}} \right)
\]
UNLESS!!!....... \(\theta\) proves to be very small!
In this case, \(\sin \theta\) will be very small and therefore, so will
\(q_{13}, q_{23}, q_{31}\), and \(q_{32}\) because all these terms contain \(\sin \theta\).
This can lead to major round-off errors in any such calculations.
Fortunately, the solution for this (\(\theta \rightarrow 0\)) is to recall that \(\psi\) and
\(\phi\) become indistinguishable from each other. This permits \(\phi\) to be set to zero, and
\(\psi\) to be computed according to \(\psi = \text{Sin}^{-1}(q_{12})\).
Example: Roe Angles from Matrix
Let's start with the matrix in the previous example and work backwards.
\[
{\bf Q} =
\left[ \matrix {
-0.3062 & \;\;\; 0.8839 & -0.3536 \\
-0.9186 & -0.1768 & \;\;\; 0.3536 \\
\;\;\; 0.2500 & \;\;\; 0.4330 & \;\;\; 0.8660
} \right]
\]
Start with the \(q_{33}\) term to determine \(\theta\).
\[
\theta = \text{Cos}^{-1}(q_{33}) = \text{Cos}^{-1}(0.8660) = 30^\circ
\]
Since \(\theta\) is not near zero, \(\psi\) and \(\phi\) can be computed as follows.
\[
\psi = \text{Tan}^{-1}\left({q_{32} \over q_{31}}\right) = \text{Tan}^{-1}\left({0.4330 \over 0.2500}\right) = 60^\circ
\]
\[
\phi = \text{Tan}^{-1}\left({q_{23} \over -q_{13}}\right) = \text{Tan}^{-1}\left({0.3536 \over 0.3536}\right) = 45^\circ
\]
So as expected, the original values are recovered: \( \psi = 60^\circ, \; \theta = 30^\circ,\)
and \(\phi = 45^\circ\).
Rotation About An Axis
Yet another way of specifying the orientation of a transformed coordinate system is through
a rotation axis vector, \({\bf p}\), and a rotation angle, \(\alpha\), about
the \({\bf p}\) axis.
In this case, the transformation matrix is written as
It is very important to recognize that \({\bf p}\) is a unit vector. Using any
other length will lead to incorrect results.
Also, it's actually more common to use
\(\theta\) or \(\phi\) as the rotation angle instead of \(\alpha\). But
\(\alpha\) is being used here to minimize confusion with
\(\psi, \theta,\) and \(\phi\) used as the Roe convention angles.
It is easy to visualize the coordinate rotations in this method. For example,
the 2-D case can be reproduced by noting that the rotation is about the z-axis,
so the vector is \({\bf p} = (0, 0, 1)\). This leads to
As with the Roe angles, it is possible to back-out \({\bf p}\) and \(\alpha\) given
a transformation matrix, \({\bf Q}\). The first step is to determine \(\alpha\).
This is done by taking the trace of the matrix.
\[
\alpha = \text{Cos}^{-1} \left\{ {1 \over 2} \Big[ \text{tr}({\bf Q}) - 1 \Big] \right\}
\]
Once \(\alpha\) is determined, the components of \({\bf p}\) are computed by
\[
p_1 = { q_{23} - q_{32} \over 2 \sin \alpha } \qquad \qquad
p_2 = { q_{31} - q_{13} \over 2 \sin \alpha } \qquad \qquad
p_3 = { q_{12} - q_{21} \over 2 \sin \alpha }
\]
The three equations can be conveniently summarized in tensor notation as
\[
p_i = { \epsilon_{ijk} \, q_{jk} \over 2 \sin \alpha}
\]
Note that \({\bf p}\) becomes undefined when \(\alpha = 0\). This means, "the axis
you rotate about doesn't matter if you don't rotate in the first place."
Backing Out P and α
Determine the single rotation, \({\bf p}\) and \(\alpha\), that is equivalent to
that produced by the Roe angles used earlier: \(\psi = 60^\circ, \theta = 30^\circ,\)
and \(\phi = 45^\circ\).
Recall that the Roe angles \(\psi = 60^\circ, \theta = 30^\circ,\)
and \(\phi = 45^\circ\) lead to
\[
\begin{eqnarray}
p_1 &=& \;\;\; { 0.3536 - 0.4330 \over 2 \; \sin 108^\circ} &=& -0.0417\\
\\
p_2 &=& { 0.2500 - (-0.3536) \over 2 \; \sin 108^\circ} &=& \;\;\; 0.3173\\
\\
p_3 &=& { 0.8839 - (-0.9186) \over 2 \; \sin 108^\circ} &=& \;\;\; 0.9475
\end{eqnarray}
\]
So the Roe angles \(\psi = 60^\circ, \theta = 30^\circ,\)
and \(\phi = 45^\circ\) are equivalent to a single rotation of
\(108^\circ\) about the axis given by
\({\bf p} = (-0.0417\), \(0.3173\), \(0.9475)\).
Both sets lead to the new \(x', y', z'\) axes pointing in the same
directions.
The above example closes the loop on converting rotations from one system, Roe,
to a second system, about an axis. In this case, and any other, the
transformation matrix \({\bf Q}\) is the go-between permitting the conversion.
The procedure is always to calculate the \({\bf Q}\) matrix in the first system
and then back-out the angles in the second system from \({\bf Q}\).
Inverses and Transposes
All coordinate transformation matrices possess a rather remarkable
property - their transpose is their inverse. So it becomes trivial
to compute the inverse of such a matrix.
\[
{\bf Q} =
\left[ \matrix {
-0.3062 & \;\;\; 0.8839 & -0.3536 \\
-0.9186 & -0.1768 & \;\;\; 0.3536 \\
\;\;\; 0.2500 & \;\;\; 0.4330 & \;\;\; 0.8660
} \right]
\]
Multiply it by its transpose to demonstrate that the result is the identity matrix
and therefore, its transpose must also be its inverse.
As hard as this and the previous page have been, there is still much more to tackle.
For starters, there is a very similar line of development for rotating
objects around in a fixed coordinate system... just the opposite of
what we have done here, which was rotating the coordinate system around
while the object itself remained fixed. This will be addressed in later sections.
We will see that the rotation matrix (no longer a transformation matrix)
is just the transpose/inverse of the transformation matrix.
Thank You
Thank you for visiting this webpage. Feel free to
email me if you have questions.
And on a different note, if you plan on buying a
Tesla, please consider
using my referral code!
https://ts.la/robert67717