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 P: 71 Actually I'm really having problems with the concept of Galilean relativity and I think it is because I don't understand Galilean spacetime properly. Consider a world line and its underlying spatial trajectory $$\bar{w}\colon \mathbb{R}\to \mathbb{R}^{4}\colon t \mapsto (t,\bar{x}(t))$$ $$\bar{x}\colon \mathbb{R}\to \mathbb{R}^{3}\colon t \mapsto (x,y,z)$$ where $\mathbb{R}^{3}$ with the Euclidean structure and $\mathbb{R}^{4}$ with the Galilean structure. The acceleration of the world line is given by $$\bar{a}\colon \mathbb{R}\to \mathbb{R}^{4}\colon t\mapsto \frac{d^{2}\bar{w}}{dt^{2}}=(0,\frac{d^{2}\bar{x}}{dt^{2}})\equiv(0,\ti lde{a}(t))$$ A force field is given by $\bar{F}\colon \mathbb{R}^{4}\to \mathbb{R}^{3}$ and it can be evaluated along a world line by using $\bar{F}(\bar{w}(t))=m \tilde{a}(t)$ When change frame using a Galilean transformation $$t=t'+t_{t}\quad\quad \bar{x}=t'\bar{u}+R\cdot\bar{x}'+\bar{t}_{\bar{x}}$$ we find that $$\tilde{a}(t)=R\cdot \tilde{a}'(t')\Leftrightarrow \bar{F}(\bar{w}(t))=m\tilde{a}(t)=m R\cdot \tilde{a}'(t')=R\cdot \bar{F}(\bar{w}'(t'))$$ So for inertial frames ($R=id$) we find that $\bar{F}(\bar{w}(t))=\bar{F}(\bar{w}'(t'))$. Is this then the second aspect of Galilean relativity? And what about the other aspect that states that laws have the same form after a Galilean transformation? Both aspects of invariance are for example discussed here.