Two points with zero velocity in some inertial frame move in a straight line I've been trying to solve a problem in Arnold's Mathematical Methods of Classical Mechanics in which I'm supposed to show that given a mechanical system of two points such that they have zero velocity in some inertial frame then the motion of the two points will stay on the line connecting them in the initial moment. I think I should use the invariance of the solution to Newton's equation under uniform translation in the direction of this line, but I don't see how.
 A: Try using invariance under rotations instead.
Here is an informal outline. You know from examples 1 and 2 (in the textbook) that invariance under time translations and spacial translations implies
$$\ddot{\mathbf x}_i = \mathbf f_i(\mathbf x_1 - \mathbf x_2, \dot{\mathbf x}_1 - \dot{\mathbf x}_2), \quad i \in \{1, 2\}.$$
If at any moment the vectors $\mathbf x_1 - \mathbf x_2$ and $\dot{\mathbf x}_1 - \dot{\mathbf x}_2$ are parallel, then a rotation $G$ about $\mathbf x_1 - \mathbf x_2$ leaves both invariant, and, by example 3,
\begin{equation}
    \mathbf f_i(\mathbf x_1 - \mathbf x_2, \dot{\mathbf x}_1 - \dot{\mathbf x}_2) = \mathbf f_i(G(\mathbf x_1 - \mathbf x_2), G(\dot{\mathbf x}_1 - \dot{\mathbf x}_2)) = G \mathbf f_i(\mathbf x_1 - \mathbf x_2, \dot{\mathbf x}_1 - \dot{\mathbf x}_2),
\end{equation}
implying that the accelerations of both points will also be parallel to $\mathbf x_1 - \mathbf x_2$ at that moment. Now consider the configuration at the initial moment and what that implies.
(Actually I think something like this is rigorous enough, since spending too much effort on rigor would be against the spirit of the book.)
