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Galilean principle of relativityby Wox
Tags: galilean relativity 
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#1
Feb2912, 05:04 AM

P: 71

A Galilean transformation consists of a rotation (in space), a boost (in space) and a translation (in space and time). This can be represented for homogeneous coordinates as
[tex] \left[\begin{matrix}t'\\x'\\y'\\z'\\1\end{matrix}\right]= \left[\begin{matrix} 1&0&0&0&t_{t}\\ u_{x}&R_{11}&R_{12}&R_{13}&t_{x}\\ u_{y}&R_{21}&R_{22}&R_{23}&t_{y}\\ u_{z}&R_{31}&R_{32}&R_{33}&t_{z}\\ 0&0&0&0&1 \end{matrix}\right] \cdot\left[\begin{matrix}t\\x\\y\\z\\1\end{matrix}\right] [/tex] To me there seem to be two principles of relativity in frames that are related by a Galilean transformation. The first says that all physical laws described in Galilean spacetime have the same form in frames related by a Galilean transformation. Newton's second law of motion for example given by [itex]F=m.a[/itex] in one frame becomes [itex]F'=m.a'[/itex] in the second frame, while [itex]F[/itex] and [itex]F'[/itex] transform under a Galilean transformation. The second says that all physical laws are the same in frames that are related by a Galilean transformation with [itex]R=id[/itex] (i.e. inertial frames of reference). Again Newton's second law of motion: [itex]F=F'[/itex] and [itex]a=a'[/itex]. Is this a correct understanding of Galilean relativity? 


#2
Feb2912, 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. 


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