I want to rotate a $(X,Y,Z)$ coordinate-system around it $Z$-axis. For the coordinates this can be done with the rotation matrix:

$$ R_Z(\theta)= \begin{pmatrix} cos \theta & -\sin(\theta) & 0\\ \sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 1 \end{pmatrix} $$

($R_Z(\theta)$ because I want to rotate around the $Z$-axis). To get the final coordinates, take the dot product of this with the position vector $\vec{u}$:

$$ \vec{u_{rot}} = R_Z \cdot \vec{u} $$

Does this also work too for the velocity vector $\vec{v}$?


Yes, but only because the rotation is about the origin.

The velocity is the time derivative of the position, so if the position at time $t$ is $\vec u(t)$, then the velocity is $$\vec v(t) = \frac{d}{dt} \vec u(t) = \lim_{h\to 0} \frac1h ( \vec u(t+h) - \vec u(t) ) $$

If we compute this in the rotated coordinate system, we get

$$ \begin{align}\lim_{h\to 0} \frac1h ( R\vec u(t+h) - R\vec u(t)) &= \lim_{h\to 0} \frac1h R(\vec u(t+h) - \vec u(t)) \\&= \lim_{h\to 0} R\frac1h (\vec u(t+h) - \vec u(t)) \\&= R\lim_{h\to 0} \frac1h (\vec u(t+h) - \vec u(t)) = R \vec v(t) \end{align}$$

the first two equals signs are because $R$ is a linear transformation; the third is because $R$ is continuous and so preserves limits.

Note that this is not true for a rotation $T$ about a different point than the origin, because then $T\vec u(t+h)-T\vec u(t) \neq T(\vec u(t+h)-\vec u(t))$ in general.

  • $\begingroup$ Okay, that explains it. So the same $R_Z$ is used, that was something else I wasn't sure of $\endgroup$ – Mathias711 Apr 5 '14 at 13:50

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