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Is there a 3D coordinate transform which turns rotation in cartesian coordinates into translation in the transformed coordinate system? It would be sufficient if the transformation has the desired property only for small rotation angles.

In 2D, the polar transform turns a rotation of $(x,y)$ by the angle $\alpha$ into a translation in polar coordinates $(r,\varphi+\alpha)$. In spherical coordinates, the desired property only holds for rotations about the z-axis and not for more complex rotations, e.g., about the z- and y-axis.

Under a different question a user hinted in the comments that "stereographic coordinates" work. However, I do not see why (and I cannot ask there because I lack reputation).

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    $\begingroup$ the group of rotations is non-abelian, while the group of translations is abelian, so it cannot be done (in 2D both groups are abelian) $\endgroup$ Commented Aug 27 at 12:58
  • $\begingroup$ You have to define what you mean in your question: under the usual interpretation of the terms, the answer is negative even in 2d. $\endgroup$ Commented Aug 27 at 14:13

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Your question is somewhat self-answering.

A translation in spherical coordinates maps $(\rho,\theta,\phi)$ to $(\rho,\theta+\alpha,\phi+\beta)$. In Cartesian coordinates, $$\begin{cases}x&=r\sin(\theta)\,\cos(\phi),\\y&=r\sin(\theta)\,\sin(\phi),\\z&=r\cos(\theta).\end{cases}\to\begin{cases}x&=r\sin(\theta+\alpha)\,\cos(\phi+\beta),\\y&=r\sin(\theta+\alpha)\,\sin(\phi+\beta),\\z&=r\cos(\theta+\alpha).\end{cases}$$

After expansion using the sum-of-angles formulas, you will get terms combining sines and cosines that do not express linearly with respect to the original coordinates. Hence the transformation is non-linear, which rules out rotations, among others.

You can get the Cartesian expression of the transformation by transforming the Cartesian coordinates to spherical, and plugging them in the spherical to Cartesian equations, with spherical translation.

You can also use (WLOG $r=1$):

$$\cos(\theta)=z, \sin(\theta)=\sqrt{x^2+y^2},\\\cos(\phi)=\frac x{\sqrt{x^2+y^2+z^2}},\sin(\phi)=\frac y{\sqrt{x^2+y^2+z^2}}$$ (plus a sign discussion).

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