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Derivative of a rotating unit vector

  1. Aug 9, 2014 #1
    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.

    1. The problem statement, all variables and given/known data

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

    2. Relevant equations
    3. 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 [itex]\vec{r}\in\mathbb R^2[/itex] rotates with angular speed [itex]\dot{Θ}[/itex] (so [itex]Θ[/itex] 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 [itex]\vec{r}[/itex] as a product of a constant unit vector [itex]\vec{a}=const[/itex] and a rotational matrix [itex]T_Θ[/itex]:

    [itex]\vec{r}=T_Θ\vec{a}[/itex], where [itex]T_Θ=\begin{pmatrix}
    \cosΘ & -\sinΘ\\
    \sinΘ & \cosΘ\\
    \end{pmatrix}[/itex] (1)

    So time derivative of [itex]\vec{r}[/itex] can be written as:

    [itex]\frac{d\vec{r}}{dt}=\frac{d(T_Θ)}{dt}\vec{a}[/itex] (2)

    Then we do math:

    [itex]\frac{d(T_Θ)}{dt}=\dot{Θ}\begin{pmatrix}
    -\sinΘ& -\cosΘ\\
    \cosΘ& -\sinΘ\\
    \end{pmatrix}=
    (-1)\dot{Θ}\begin{pmatrix}
    \cos(\frac{\pi}{2}-Θ)& \sin(\frac{\pi}{2}-Θ)\\
    -\sin(\frac{\pi}{2}-Θ)& \cos(\frac{\pi}{2}-Θ)\\
    \end{pmatrix}=
    (-1)\dot{Θ}\begin{pmatrix}
    \cos(Θ-\frac{\pi}{2})& -\sin(Θ-\frac{\pi}{2})\\
    \sin(Θ-\frac{\pi}{2})& \cos(Θ-\frac{\pi}{2})\\
    \end{pmatrix}
    [/itex]

    Matrix [itex]\begin{pmatrix}
    \cos(Θ-\frac{\pi}{2})& -\sin(Θ-\frac{\pi}{2})\\
    \sin(Θ-\frac{\pi}{2})& \cos(Θ-\frac{\pi}{2})\\
    \end{pmatrix}
    [/itex] is a rotation matrix that rotates a vector through the angle [itex]Θ-\frac{\pi}{2}[/itex]. You can do the same rotation with two matrices, first of which would rotate the vector through the angle Θ (it is [itex]T_Θ[/itex]), and second through the angle [itex]-\frac{\pi}{2}[/itex] (we'll name the later matrix[itex]T_0[/itex]).

    [itex]T_0=\begin{pmatrix}
    0& 1\\
    -1& 0\\
    \end{pmatrix}[/itex]

    So we rewrite equation (2) as:

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

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

    [itex]\frac{d\vec{r}}{dt}=\dot{Θ}T_ΘT_1\vec{a}[/itex], where [itex]T_1=\begin{pmatrix}
    0& -1\\
    1& 0\\
    \end{pmatrix}[/itex]

    So using (1) we can write:

    [itex]\frac{d\vec{r}}{dt}=\dot{Θ}\begin{pmatrix}
    0& -1\\
    1& 0\\
    \end{pmatrix}\vec{r}[/itex]

    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.
     
    Last edited: Aug 9, 2014
  2. jcsd
  3. Aug 10, 2014 #2
  4. Aug 11, 2014 #3
    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 [itex]\frac{d\vec{r}}{dt}=\vec{ω}\times\vec{r}[/itex]. 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.
     
  5. Aug 11, 2014 #4

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    There is a generalization to N dimensions. Look at your result for the 2D case:
    [tex]\frac{d\vec{r}}{dt}=\dot{Θ}\begin{pmatrix}
    0& -1\\
    1& 0\\
    \end{pmatrix}\vec{r}[/tex]
    Note that this can be rewritten as
    [tex]\frac{d\vec{r}}{dt}
    \begin{pmatrix}
    0& -\omega_{x,y} \\
    \omega_{x,y} & 0
    \end{pmatrix}
    \, \vec{r}[/tex]
    where I'm denoting your scalar [itex]\dot \Theta[/itex] as [itex]\omega_{x,y}[/itex]. I'm simply using [itex]\omega[/itex] to denote the rotation rate. The subscript [itex]x,y[/itex] 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, [itex]\frac {d\vec r}{dt} = \vec \omega \times \vec r[/itex]. The cross product can be written in matrix form, resulting in
    [tex]\frac {d\vec r}{dt} =
    \begin{pmatrix}
    0& -\omega_z & \omega_y \\
    \omega_z & 0 & -\omega_x \\
    -\omega_y & \omega_x & 0
    \end{pmatrix}
    \, \vec r
    [/tex]
    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
    [tex]\frac {d\vec r}{dt} =
    \begin{pmatrix}
    0& -\omega_{x,y} & \omega_{x,z} \\
    \omega_{y,x} & 0 & -\omega_{y,z} \\
    -\omega_{x,z} & \omega_{y,z} & 0
    \end{pmatrix}
    \, \vec r
    [/tex]

    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:
    [tex]\frac {d\vec r}{dt} =
    \begin{pmatrix}
    0& -\omega_{x,y} & \omega_{x,z} & -\omega_{x,w} \\
    \omega_{y,x} & 0 & -\omega_{y,z} & \omega_{y,w} \\
    -\omega_{x,z} & \omega_{y,z} & 0 & -\omega_{z,w} \\
    \omega_{x,w} & -\omega_{y,w} & \omega_{z,w} & 0
    \end{pmatrix}
    \, \vec r
    [/tex]

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

    If you want to learn more, I suggest you start reading up on Lie groups and Lie algebras.
     
  6. Aug 15, 2014 #5
    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!
     
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