fannyfanfanz said:
Thank you for your help! I am still a little confused, and I have several more questions.
For the example that you gave above, you said that "this represents a single axis rotation by arccos(1/4)." Where does the 1/4 come from?
From the trace of the matrix.
Let T be the matrix that transforms a vector as represented in the (x,y,z) system to the vector's representation in the (x',y',z') system. One way to look at this transformation matrix is the direction cosine matrix:
<br />
T =<br />
\begin{pmatrix}<br />
\hat x \cdot \hat x' & \hat y \cdot \hat x' & \hat z \cdot \hat x' \\<br />
\hat x \cdot \hat y' & \hat y \cdot \hat y' & \hat z \cdot \hat y' \\<br />
\hat x \cdot \hat z' & \hat y \cdot \hat z' & \hat z \cdot \hat z'<br />
\end{pmatrix}<br />
There are no cosines here, so why is this called the "direction cosine matrix"? The answer is simple. The inner product of two vectors ##\vec a## and ##\vec b## is ##\vec a \cdot \vec b = ||\vec a|| \, ||\vec b|| \, \cos\theta##, where ##\theta## is the angle between the two vectors. If the two vectors are unit vectors, the magnitudes are just one and this simplifies to ##\hat a \cdot \hat b = \cos \theta##.
There are a number of other ways of looking at rotations in three dimensional space. Euler angles, for example. Another is the single axis rotation (aka eigen rotation). Suppose frame B is generated by frame A by rotating frame A by an angle ##\phi## about some axis ##\hat u = (u_x, u_y, u_z) ##. Here, ##\hat u## is a unit vector and rotation is positive/negative per the right hand thumb convention. The transformation matrix given this single axis rotation is:
<br />
T =<br />
\begin{pmatrix}<br />
\cos\phi + u_x^{\,2} (1-\cos\phi) &<br />
u_x u_y (1-\cos\phi) + u_z \sin\phi &<br />
u_x u_z (1-\cos\phi) - u_y \sin\phi \\<br />
u_y u_x (1-\cos\phi) - u_z \sin\phi &<br />
\cos\phi + u_y^{\,2} (1-\cos\phi) &<br />
u_y u_z (1-\cos\phi) + u_x \sin\phi \\<br />
u_z u_x (1-\cos\phi) + u_y \sin\phi &<br />
u_z u_y (1-\cos\phi) - u_x \sin\phi &<br />
\cos\phi + u_z^{\,2} (1-\cos\phi)<br />
\end{pmatrix}<br />
A much more compact representation is ##T_{ij} = u_i u_j (1-\cos\phi) + \delta_{ij}\cos \phi + \epsilon_{ijk}u_k\sin\phi##. If you don't know what that means, don't worry about it.
So where did that 1/4 come from? The three angles you have specified, between x and x', y and y', and z and z', are the diagonal elements of the transformation matrix. I'll denote these angles ##\alpha##, ##\beta##, and ##\gamma##. From the direction cosine matrix formulation, your information let's us fill in part of the transformation matrix:
<br />
T =<br />
\begin{pmatrix}<br />
\cos\alpha & ? & ? \\<br />
? & \cos \beta & ? \\<br />
? & ? & \cos \gamma<br />
\end{pmatrix}<br />
There's not much information here, but there is enough to compute the trace of the matrix. That's just ##\operatorname{tr}(T) = \cos\alpha + \cos\beta + \cos\gamma##. Now look at the single axis rotation formulation. From that, the trace of the matrix is ##\operatorname{tr}(T) = 3\cos\phi + (u_x^{\,2}+u_y^{\,2}+u_z^{\,2})(1-\cos\phi)##, and since ##\hat u## is a unit vector, this reduces to ##2\cos\phi + 1##.
Equating the two results, we get ##2\cos\phi + 1 = \cos\alpha + \cos\beta + \cos\gamma##, so ##\cos\phi = \frac {\cos\alpha + \cos\beta + \cos\gamma - 1} 2##. I picked simple values for those angles to make the numbers come out nice and clean. Choosing ##\alpha=\beta=\gamma=60^{\circ}## immediately leads to ##\cos\phi = \frac 1 4##.
How did you get [+/-1+/-sqrt(5)] / 4 in the transformation matrix?
The upper left element of the transformation matrix is ##T_{1,1} = \cos\alpha = \cos\phi + u_x^{\,2} (1-\cos\phi)##. Solving for ##u_x^{\,2}## yields ##u_x^{\,2} = \frac {\cos \alpha - \cos \phi} {1 - \cos\phi}##. This gives the magnitude of ##u_x## but not the sign. We don't know the sign. With my nice-values example, we know that ##u_x^{\,2} = \frac 1 3## but we don't know if ##u_x = \frac {\sqrt 3} 3## or ##-\,\frac {\sqrt 3} 3##. The same goes for the other two elements of the rotation axis.
This problem is not specific to my example. It is pervasive. You chose a poor representation.
What other information do I need to define that specific angle?
You need to choose a better way of relating your two frames. There are a number of ways to represent rotations in three dimensional space. I've focused on direction cosine matrices and single axis rotations, but there are also Euler angles (actually a family of representations; there are 24 in total), quaternions, Rodrigues parameters, modified Rodrigues parameters, and others. You need to pick a good scheme. You picked a bad one.One final point: You asked in the opening post "What is angle between the projection of z' on the xy plane and the x axis?" There's a potential problem here. You can calculate that angle, but that calculated value will be meaningless if the two frames are very close to being aligned with one another. You have experimental data "related to my research project on light polarization." That projection onto the xy plane will remove all signal from your data, leaving just random noise if the two frames aren't separated by much.
