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Wox
Wox is offline
#2
Feb29-12, 10:00 AM
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
[tex]
\bar{w}\colon \mathbb{R}\to \mathbb{R}^{4}\colon t \mapsto (t,\bar{x}(t))
[/tex]
[tex]
\bar{x}\colon \mathbb{R}\to \mathbb{R}^{3}\colon t \mapsto (x,y,z)
[/tex]
where [itex]\mathbb{R}^{3}[/itex] with the Euclidean structure and [itex]\mathbb{R}^{4}[/itex] with the Galilean structure. The acceleration of the world line is given by
[tex]
\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))
[/tex]
A force field is given by [itex]\bar{F}\colon \mathbb{R}^{4}\to \mathbb{R}^{3}[/itex] and it can be evaluated along a world line by using [itex]\bar{F}(\bar{w}(t))=m \tilde{a}(t)[/itex] When change frame using a Galilean transformation
[tex]
t=t'+t_{t}\quad\quad \bar{x}=t'\bar{u}+R\cdot\bar{x}'+\bar{t}_{\bar{x}}
[/tex]
we find that
[tex]
\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'))
[/tex]

So for inertial frames ([itex]R=id[/itex]) we find that [itex]\bar{F}(\bar{w}(t))=\bar{F}(\bar{w}'(t'))[/itex]. 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.