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I studied physics many years ago and now 're-doing' my college degree by myself for fun.

I used to think of the path of a particle in Newtonian mechanics as a a curve in $\mathbb{R}^3$, in other words a mapping $t \in \mathbb{R} \rightarrow x(t) \in \mathbb{R}^3$.

However, while reading Schutz's book "Geometrical methods of mathematical physics", I came across an example in chapter 2 section 10 of his book where he says that the view of spacetime taken by Newtonian physics has a natural fiber bundle structure, with $\mathbb{R}$ representing the base manifold (time), and $\mathbb{R}^3$ representing the fiber (particle's position). In this interpretation, which I can see makes sense as time has an absolute meaning in Newtonian mechanics and hence can serve as base manifold since every particle in the fiber would agree on the value of $t$ through the projection onto the base, the path of a particle would be a section of the fiber $\sigma: \mathbb{R} \rightarrow \mathbb{R} \times \mathbb{R}^3$.

But I am not sure the two interpretations are equivalent, I don't think they are? So I am a bit confused now and my question is which interpretation is correct?

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  • $\begingroup$ Why aren't they equivalent? This is a trivial bundle, so the projection is simply onto $\mathbb R$, and from the definition of sections, this implies that $\sigma(t)=(t, \vec x(t))$. $\endgroup$
    – Nihar Karve
    Mar 6, 2021 at 8:39
  • $\begingroup$ I suppose what I am struggling with is that under the interpretation of a path there is only one `copy' of $\mathbb{R}^3$ needed, whereas under the section interpretation there are infinite number of copies of $\mathbb{R}^3$, one for each value of $t$. But perhaps as you say this distinction is trivial for a trivial bundle. $\endgroup$
    – user228424
    Mar 6, 2021 at 8:56

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If a fibre bundle $\pi:E\to B$ has $E=B\times X$ (i.e. a trivial fibre bundle), then the map $\pi$ is simply the projection onto the first factor. However, sections of fibre bundles $\sigma: B\to E$ satisfy $\pi(\sigma(x))=x \ \forall x\in B$. This means that a section of a trivial bundle is diagonal on the base space, i.e. $\sigma$ sends $t\to(t,\vec x(t))$ - this seemingly extra parameter in the codomain is redundant. Thus the fibre bundle picture coincides with the "natural" picture: $t\in\mathbb R\to \vec x(t)\in\mathbb R^3$.

In English, the natural mapping $\mathbb R\to\mathbb R^3$ says "give me a time, I'll give you the position corresponding to that time", while the section of the trivial bundle says "give me a time, I'll give you the same time back, as well as the position corresponding to that time".

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