# Galilean Transformation

The transformation between inertial systems are given by affine transformations of $\mathbb{R}^{1+3}$. These are given by $t'=\lambda t+\vec{c}^\top \vec{x} +a$ and $\vec{x}'=\vec{v}t+M\vec{x}+\vec{b}$ where M is a matrix. But the Galilean transformations do not include the $\vec{c}^\top\vec{x}$ term in the time part.This means $\vec{c}=0$. Why is it, that this holdes ? I can't find an argument for that.

The argument is that in Galilean transformation you assume time to be absolute. Translating this requirement in an operative statement, you want intervals of time between to events to be invariant under the transformation from a reference system to another. In formulas $$\Delta t = \Delta t' \quad\to\quad t_2-t_1 = t_2'-t_1' = \lambda\Delta t + \vec{c}^T\Delta\vec{x}$$ where $\Delta\vec{x}$ is the spatial distance between two events and the $t$s are the corresponding times. Given the hypothesis, if the equality must hold for whatever passage between frames, than it must be $\vec{c}=0$ and $\lambda=1$. The only transformation of time allowed is the addition of a constant parameter $a$ appearing in the general transformation you wrote, which is physically just a time shift in the synchronization procedure of the clocks of the different references.