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Consider two points in the empty (isotropic and homogeneous) space: since the only vector that "makes sense" (the only vector that we can define) is the vector given by the difference of the two positions $\vec{r}$, the force $\vec{f}$ between these two point particles must be "central", $\vec{f} = F(r) \hat{r}$.

How to build, using a purely geometric reasoning, the most general force between two points, one of which has a formal "spin" that defines a preferred direction $\hat{s}$?

Now we have 2 objects: $\vec{s}$ and $\vec{r}$, so one may guess that $$ \vec{f} = ( \hat{r}-\hat{s} \hat{r}\cdot\hat{s}) Q(|\hat{r}-\hat{s} \hat{r}\cdot\hat{s}|, \vec{r}\cdot \vec{s}) + \hat{s} P(|\hat{r}-\hat{s} \hat{r}\cdot\hat{s}|, \vec{r}\cdot \vec{s}) +\hat{r} \times \hat{s} W(|\hat{r}-\hat{s} \hat{r}\cdot\hat{s}|, \vec{r}\cdot \vec{s}) $$

Is this correct or other terms are possible? is there a clever way to write such a force? The point is that the preferred direction reduces the symmetry of the problem from $O(3)$ to $O(2)$, the rotations in the plane orthogonal to $\hat{s}$.

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