Ambiguity in sense of rotation given a rotation matrix A

[Kashmir]

Goldstein 3rd Ed pg 161.

Im not able to understand this paragraph about the ambiguity in the sense of rotation axis given the rotation matrix A, and how we ameliorate it.
Any help please.

"The prescriptions for the direction of the rotation axis and for the rotation angle are not unambiguous. Clearly if $\mathbf{R}$ is an eigenvector, so is $-\mathbf{R}$; hence the sense of the direction of the rotation axis is not specified. Further, $-\Phi$ satisfies Eq. (4.61) if $\Phi$ does. Indeed, it is clear that the eigenvalue solution does not uniquely fix the orthogonal transformation matrix A. From the determinantal secular equation (4.52), it follows that the inverse matrix $\mathrm{A}^{-1}=\tilde{\mathrm{A}}$ has the same eigenvalues and eigenvectors as A. However, the ambiguities can at least be ameliorated by assigning $\Phi$ to $A$ and $-\Phi$ to $A^{-1}$, and fixing the sense of the axes of rotation by the right-hand screw rule"

 

[PeroK]

Isn't this simply establishing a clockwise/anticlockwise convention for the rotation angle in relation to the axis?

 

[Kashmir]

Isn't this simply establishing a clockwise/anticlockwise convention for the rotation angle in relation to the axis?

I'm not sure. This is what I know: We have been given a matrix $A$ which represents a rotation of vectors. It's eigenvector $R$ corresponding to +1 eigenvalue specify the line around which rotation happens. Trace(A) =$1+2cos\Phi$. Both $\Phi$ and $-\Phi$ satisfy it.

 

[PeroK]

I'm not sure. This is what I know: We have been given a matrix $A$ which represents a rotation of vectors. It's eigenvectors $R$ specify the line of rotation. Trace(A) =$1+2cos\Phi$. Both $\Phi$ and $-\Phi$ satisfy it.

Technically, a matrix doesn't have specific eigenvectors, but eigenspaces. If $\vec v$ is an eigenvector, then so is $-\vec v$ and, in general, $\alpha \vec v$ for any scalar $\alpha$.

In other words, you have a line, but a choice of two directions. Compare the positive and negative z-axes.

 

[Kashmir]

Technically, a matrix doesn't have specific eigenvectors, but eigenspaces. If $\vec v$ is an eigenvector, then so is $-\vec v$ and, in general, $\alpha \vec v$ for any scalar $\alpha$.

In other words, you have a line, but a choice of two directions. Compare the positive and negative z-axes.

Yes. I agree. But how to relate it to what the author is trying to say? Given A isn't there an ambiguity in the sense of rotation?

 

[PeroK]

Yes. I agree. But how to relate it to what the author is trying to say?

Goldstsein is describing, in slighty elaborate old-fashioned language, precisely what I've said. And is proposing a "right-hand rule" to remove the ambiguity.

 

[Kashmir]

Goldstsein is describing, in slighty elaborate old-fashioned language, precisely what I've said. And is proposing a "right-hand rule" to remove the ambiguity.

Yes the wording was tough for me. Thank you again for helping me. :)

 

[PeroK]

For example, take a rotation of $\theta$ about the z-axis. You can describe that in four ways, using the conventional right-hand rule:

1) Anticlockwise rotation of $\theta$ about the positive z-axis.

2) Clockwise rotation of $-\theta$ (or $2\pi - \theta$) about the positive z-axis.

3) Clockwise rotation of $\theta$ about the negative z-axis.

4) Anticlockwise rotation of $-\theta$ (or $2\pi - \theta$) about the negative z-axis.

 

[Kashmir]

For example, take a rotation of $\theta$ about the z-axis. You can describe that in four ways, using the conventional right-hand rule:

1) Anticlockwise rotation of $\theta$ about the positive z-axis.

2) Clockwise rotation of $-\theta$ (or $2\pi - \theta$) about the positive z-axis.

3) Clockwise rotation of $\theta$ about the negative z-axis.

4) Anticlockwise rotation of $-\theta$ (or $2\pi - \theta$) about the negative z-axis.

The matrix actually doesn't specify which one actually happens out of these 4 ?

 

[PeroK]

The matrix actually doesn't specify which one actually happens out of these 4 ?

In general, two matrices are equal iff all entries are equal. There can only be one matrix for this rotation. But, as above, that matrix will have a one-dimensional eigenspace corresponding to the axis of rotation and that defines two unit vectors with opposite directions. If we choose a right-hand rule, then that gets rid of two of the descriptions (the clockwise ones can go and we always describe rotations relative to the anticlockwise direction). That leaves us with:

1) Anticlockwise rotation of $\theta$ about the positive z-axis.

4) Anticlockwise rotation of $-\theta$ (or $2\pi - \theta$) about the negative z-axis.

We have a single rotational matrix, $R$, which can be described by:

1) The unit vector $\hat n$ and angle $\theta$

4) The unit vector $-\hat n$ and angle $-\theta$.

It's the same matrix, but two mappings onto the set of unit vectors and angle of rotation. Two descriptions of what that matrix does.

Generally, we do not try to remove that ambiguity. That's something we accept.

 

[vanhees71]

There is no ambiguity given the right-hand rule and the axis of rotation. So we have a well-defined function $(\vec{n},\varphi) \mapsto \hat{R} \in \mathrm{SO}(3)$, where $\vec{n} \in \text{S}_1$ (the unit sphere in 3D Euclidean space) and $\varphi \in [0,2 \pi)$. That's one way to parametrize uniquely SO(3).

It's, however, not an injective map, i.e., to $\hat{R} \in \mathrm{SO}(3)$ there are two orientations of the axis of rotation $\pm \vec{n}$. If $(\vec{n},\vec{\varphi})$ maps to $\hat{R}$, so also $(-\vec{n},2 \pi-\varphi)$ maps to the same $\hat{R}$.

 

[wrobel]

Actually the angular velocity is a principle object for kinematics of a rigid body. The angular velocity admits no ambiguity and does not require any angles of rotations for its definition. .

 

[vanhees71]

Indeed, the angular velocity defined via the unique rotation of the body-fixed frame's Cartesian basis wrt. the space-fixed frame's Cartesian basis.