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I'm having some trouble understanding the concept for this more than likely due to my lacking mathematical background.

I am currently reading Roger Penrose's The Road to Reality page 394 specifically if anyone has a direct reference.

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Spacetime ${\cal N}$ is described as being bundled with base space $\mathbb{E}^1$, time, and fibre $\mathbb{E}^3$. Now unless I've completely missed something, points in space are not equivalent at different times as described earlier in the text, instead each point in space for some particle under motion is on a different fibre $\mathbb{E}^3$.

That's all well and good, now here's where my question lies. The structure of ${\cal N}$ is equivalent to the Galilean case ${\cal G}$ by a "sliding" of the $\mathbb{E}^3$ fibres. What exactly is meant by a sliding of these fibres, both mathematically and conceptually.

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I think you have misunderstood the text slightly. In figure 17.7, figure (a) shows a general Newton-Cartan spacetime with random gravitational fields. The trajectories of the freely moving particle worldlines are curves, and there is no global transformation that can simultaneously make them all straight.

Figure (b) shows the special case where the gravitational field is uniform throughout all of space (though it can vary in time). That means at any instant in time every object in the associated bundle has the same acceleration ${\bf g}(t)$ (for some value of ${\bf g}(t)$). So if we can use a coordinate transformation and switch to coordinates that are accelerating with a proper acceleration of $-{\bf g}(t)$. In these coordinates objects are not accelerating, so the space is equivalent to the Galilean space shown in figure (c).

So when Penrose talks about sliding the bundles he means we use a coordinate transformation (that is time dependant so it's different for each bundle) to cancel out the acceleration. But note again that it is only for the special case where the gravitational acceleration is independant of position in the space $E^3$. This is obviously physically unreasonable.

You might be interested to read up about Rindler coordinates. In the relativistic world this is the coordinate transformation that eliminates the acceleration when the acceleration is independant of time.

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  • $\begingroup$ That makes much more sense now I did misunderstand the text. Also thank you for the link to rindler coordinates. $\endgroup$
    – Skeletor
    May 15, 2015 at 6:56

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