Derivative of a rotating unit vector

[Sergey S]

I think this is a textbook-style question, if I am wrong, please redirect me to the forum section where I should have posted this. This is my first time here, so I am sorry if I am messing it up.

Homework Statement

We have an n-dimensional vector \vec{r} with a constant length \|\vec{r}\|=1. The vector rotates with an angular speed \vec{ω}, which is also an n-dimensional vector and is a given function of time. I want to find first derivative of \vec{r} with respect to time: \frac{d\vec{r}}{dt}-?

2. Homework Equations

The Attempt at a Solution

I don't really know how to tackle the problem. I think I've solved this for the 2-dimensional case, where \vec{r}\in\mathbb R^2 rotates with angular speed \dot{Θ} (so Θ would denote angular displacement from the original orientation). I doubt that it will be of any help with the general n-dimensional case, but here it is:

We can rewrite \vec{r} as a product of a constant unit vector \vec{a}=const and a rotational matrix T_Θ:

\vec{r}=T_Θ\vec{a}, where T_Θ=\begin{pmatrix}<br /> \cosΘ &amp; -\sinΘ\\<br /> \sinΘ &amp; \cosΘ\\<br /> \end{pmatrix} (1)

So time derivative of \vec{r} can be written as:

\frac{d\vec{r}}{dt}=\frac{d(T_Θ)}{dt}\vec{a} (2)

Then we do math:

\frac{d(T_Θ)}{dt}=\dot{Θ}\begin{pmatrix}<br /> -\sinΘ&amp; -\cosΘ\\<br /> \cosΘ&amp; -\sinΘ\\<br /> \end{pmatrix}=<br /> (-1)\dot{Θ}\begin{pmatrix}<br /> \cos(\frac{\pi}{2}-Θ)&amp; \sin(\frac{\pi}{2}-Θ)\\<br /> -\sin(\frac{\pi}{2}-Θ)&amp; \cos(\frac{\pi}{2}-Θ)\\<br /> \end{pmatrix}=<br /> (-1)\dot{Θ}\begin{pmatrix}<br /> \cos(Θ-\frac{\pi}{2})&amp; -\sin(Θ-\frac{\pi}{2})\\<br /> \sin(Θ-\frac{\pi}{2})&amp; \cos(Θ-\frac{\pi}{2})\\<br /> \end{pmatrix}<br />

Matrix \begin{pmatrix}<br /> \cos(Θ-\frac{\pi}{2})&amp; -\sin(Θ-\frac{\pi}{2})\\<br /> \sin(Θ-\frac{\pi}{2})&amp; \cos(Θ-\frac{\pi}{2})\\<br /> \end{pmatrix}<br /> is a rotation matrix that rotates a vector through the angle Θ-\frac{\pi}{2}. You can do the same rotation with two matrices, first of which would rotate the vector through the angle Θ (it is T_Θ), and second through the angle -\frac{\pi}{2} (we'll name the later matrixT_0).

T_0=\begin{pmatrix}<br /> 0&amp; 1\\<br /> -1&amp; 0\\<br /> \end{pmatrix}

So we rewrite equation (2) as:

\frac{d\vec{r}}{dt}=(-1)\dot{Θ}T_ΘT_0\vec{a}

Rotating a vector through the angle (-\frac{\pi}{2}) and then multiplying it with (-1) is the same as rotating through the angle \frac{\pi}{2}, so:

\frac{d\vec{r}}{dt}=\dot{Θ}T_ΘT_1\vec{a}, where T_1=\begin{pmatrix}<br /> 0&amp; -1\\<br /> 1&amp; 0\\<br /> \end{pmatrix}

So using (1) we can write:

\frac{d\vec{r}}{dt}=\dot{Θ}\begin{pmatrix}<br /> 0&amp; -1\\<br /> 1&amp; 0\\<br /> \end{pmatrix}\vec{r}

Sorry for the long post. I guess what I've shown above might be the clumsiest way to deal with the problem. And it is hardly of any help with the n-dimensional case.

 

[Sunfire]

This problem is solved here
http://en.wikipedia.org/wiki/Rotating_reference_frame (scroll down to "Time derivatives in the two frames")

for the case $\omega$=const.

Perhaps you can start here and assume angular velocity that is a function of time.

 

[Sergey S]

I have spend some time thinking about it. First, for 3-dimensional case we can use well known result from classical mechanics and say that \frac{d\vec{r}}{dt}=\vec{ω}\times\vec{r}. I don't know if the result holds for the general case.

Well, at least now I know how to prove that result I've just mentioned, going from what I've written in the opening post to the 3d case is a matter of simple geometry.

 

[D H]

There is a generalization to N dimensions. Look at your result for the 2D case:
\frac{d\vec{r}}{dt}=\dot{Θ}\begin{pmatrix}<br /> 0&amp; -1\\<br /> 1&amp; 0\\<br /> \end{pmatrix}\vec{r}
Note that this can be rewritten as
\frac{d\vec{r}}{dt}<br /> \begin{pmatrix}<br /> 0&amp; -\omega_{x,y} \\<br /> \omega_{x,y} &amp; 0<br /> \end{pmatrix}<br /> \, \vec{r}
where I'm denoting your scalar \dot \Theta as \omega_{x,y}. I'm simply using \omega to denote the rotation rate. The subscript x,y denotes the fact that this is a rotation in the x-y plane. I'll have more to say on why I'm using that strange notation a bit later.

Now let's look at your 3D result, \frac {d\vec r}{dt} = \vec \omega \times \vec r. The cross product can be written in matrix form, resulting in
\frac {d\vec r}{dt} =<br /> \begin{pmatrix}<br /> 0&amp; -\omega_z &amp; \omega_y \\<br /> \omega_z &amp; 0 &amp; -\omega_x \\<br /> -\omega_y &amp; \omega_x &amp; 0<br /> \end{pmatrix}<br /> \, \vec r<br />
Here the subscript denotes rotation about an axis. Rotation about an axis is a concept that is unique to three dimensions. The 2D concept of rotation in a plane does generalize to higher dimensions. For example, in three dimensions, instead of thinking about a rotation about the z axis, think about it as a rotation in the x-y plane. With this notation, the above becomes
\frac {d\vec r}{dt} =<br /> \begin{pmatrix}<br /> 0&amp; -\omega_{x,y} &amp; \omega_{x,z} \\<br /> \omega_{y,x} &amp; 0 &amp; -\omega_{y,z} \\<br /> -\omega_{x,z} &amp; \omega_{y,z} &amp; 0<br /> \end{pmatrix}<br /> \, \vec r<br />

At this point you might see a pattern emerging. The pattern is that in both cases, the matrix is skew-symmetric and the off-diagonal elements are the rotation rate in the plane indicated by the indices. Here's the four dimensional analog:
\frac {d\vec r}{dt} =<br /> \begin{pmatrix}<br /> 0&amp; -\omega_{x,y} &amp; \omega_{x,z} &amp; -\omega_{x,w} \\<br /> \omega_{y,x} &amp; 0 &amp; -\omega_{y,z} &amp; \omega_{y,w} \\<br /> -\omega_{x,z} &amp; \omega_{y,z} &amp; 0 &amp; -\omega_{z,w} \\<br /> \omega_{x,w} &amp; -\omega_{y,w} &amp; \omega_{z,w} &amp; 0<br /> \end{pmatrix}<br /> \, \vec r<br />

So where does this skew-symmetric matrix come from? The answer is the time derivative of the rotation matrix T that describes the rotation. Regardless of dimension, the time derivative of this matrix can be written in the form \dot T = S T where S is a skew-symmetric matrix.

If you want to learn more, I suggest you start reading up on Lie groups and Lie algebras.

 

[Sergey S]

Thank you D H! The idea of planar rotations in four dimensions is kind of hard to digest (well, even with 3-d for me it takes effort to visualize some operations). I will probably have to do the differentiation for the 4-d case to get the feel of it, but now that I know the answer it should be easier =)

Also the way you written the 4x4 matrix suggests that there are six components of rotation in 4 dimensions, that is yet another thing where I'll probably need some time to get my head around it.

I've just got a textbook on Lie groups, going to read it soon.

Thanks again!