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 $T_Θ=\begin{pmatrix}<br /> \cosΘ & -\sinΘ\\<br /> \sinΘ & \cosΘ\\<br /> \end{pmatrix}$ (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:

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

Matrix $\begin{pmatrix}<br /> \cos(Θ-\frac{\pi}{2})& -\sin(Θ-\frac{\pi}{2})\\<br /> \sin(Θ-\frac{\pi}{2})& \cos(Θ-\frac{\pi}{2})\\<br /> \end{pmatrix}$ 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 matrix$T_0$).

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

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 $T_1=\begin{pmatrix}<br /> 0& -1\\<br /> 1& 0\\<br /> \end{pmatrix}$

So using (1) we can write:

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

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:
$$\frac{d\vec{r}}{dt}=\dot{Θ}\begin{pmatrix}<br /> 0& -1\\<br /> 1& 0\\<br /> \end{pmatrix}\vec{r}$$
Note that this can be rewritten as
$$\frac{d\vec{r}}{dt}<br /> \begin{pmatrix}<br /> 0& -\omega_{x,y} \\<br /> \omega_{x,y} & 0<br /> \end{pmatrix}<br /> \, \vec{r}$$
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
$$\frac {d\vec r}{dt} =<br /> \begin{pmatrix}<br /> 0& -\omega_z & \omega_y \\<br /> \omega_z & 0 & -\omega_x \\<br /> -\omega_y & \omega_x & 0<br /> \end{pmatrix}<br /> \, \vec r$$
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
$$\frac {d\vec r}{dt} =<br /> \begin{pmatrix}<br /> 0& -\omega_{x,y} & \omega_{x,z} \\<br /> \omega_{y,x} & 0 & -\omega_{y,z} \\<br /> -\omega_{x,z} & \omega_{y,z} & 0<br /> \end{pmatrix}<br /> \, \vec r$$

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:
$$\frac {d\vec r}{dt} =<br /> \begin{pmatrix}<br /> 0& -\omega_{x,y} & \omega_{x,z} & -\omega_{x,w} \\<br /> \omega_{y,x} & 0 & -\omega_{y,z} & \omega_{y,w} \\<br /> -\omega_{x,z} & \omega_{y,z} & 0 & -\omega_{z,w} \\<br /> \omega_{x,w} & -\omega_{y,w} & \omega_{z,w} & 0<br /> \end{pmatrix}<br /> \, \vec r$$

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!