Galilean invariance

Remarks

Example: let the equations be $m_i \ddot x_i = \sum_{j \ne i} f_{ij}(\abs{x_i - x_j})\,(x_i - x_j)$ with $f_{ij} = f_{ji}$, on the set where all $x_i$ are distinct. For an image motion, $\ddot{\tilde x}_i(t) = R\, \ddot x_i(t - s)$ and $\tilde x_i - \tilde x_j = R(x_i - x_j)$, with the same length, since $R$ is orthogonal. So both sides of each equation are multiplied by $R$, and the image is again a motion. Non-example: a particle tied to the origin by a spring, $m \ddot x = -k x$ with $k > 0$, is not Galilean invariant. For the shift $\tilde x(t) = x(t) + a$ with $a \ne 0$, one gets $m \ddot{\tilde x} = -k x = -k \tilde x + k a \ne -k \tilde x$. The fixed origin breaks the symmetry; the system is not isolated.