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 Wox Feb29-12 05:04 AM

Galilean principle of relativity

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

$$\left[\begin{matrix}t'\\x'\\y'\\z'\\1\end{matrix}\right]= \left[\begin{matrix} 1&0&0&0&t_{t}\\ u_{x}&R_{11}&R_{12}&R_{13}&t_{x}\\ u_{y}&R_{21}&R_{22}&R_{23}&t_{y}\\ u_{z}&R_{31}&R_{32}&R_{33}&t_{z}\\ 0&0&0&0&1 \end{matrix}\right] \cdot\left[\begin{matrix}t\\x\\y\\z\\1\end{matrix}\right]$$

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 space-time have the same form in frames related by a Galilean transformation. Newton's second law of motion for example given by $F=m.a$ in one frame becomes $F'=m.a'$ in the second frame, while $F$ and $F'$ 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 $R=id$ (i.e. inertial frames of reference). Again Newton's second law of motion: $F=F'$ and $a=a'$.

Is this a correct understanding of Galilean relativity?

 Wox Feb29-12 10:00 AM

Re: Galilean principle of relativity

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.

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