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Consider a flat Newtonian or a flat Galilean 2+1 spacetime. So, mainly a flat 2D Euclidean space, evolving over time, where each time-slice is connected with the next one by a world line. Like in this picture:

enter image description here

I am confused about how exactly to understand the metric structure of world lines in these contexts. Do the world lines in Newtonian or Galilean spacetimes follow the Euclidean metric, just like the individual time-slices in the global structure? Can we even speak of distances between the time-slices in a meaningful way?

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  • $\begingroup$ What do you mean by the world lines “following” a Euclidean metric? $\endgroup$
    – J. Murray
    Jan 26, 2022 at 21:00
  • $\begingroup$ Just whether the Euclidean metric applies to them, whether it is possible to calculate their length via the Euclidean metric. $\endgroup$
    – Maverick
    Jan 26, 2022 at 21:43
  • $\begingroup$ Related: Why Galilean spacetime is not E4? $\endgroup$
    – Quillo
    May 23, 2023 at 12:01

1 Answer 1

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A Newtonian spacetime is not a metric manifold, and so the length of a worldline is not a well-defined notion. Instead, it is a smooth manifold $\mathcal M$ equipped with a torsion-free connection $\nabla$ and a privileged function $t:\mathcal M \rightarrow \mathbb R$ which has the properties that $(i)$ $\mathrm dt$ is a nowhere-vanishing covector field and $(ii)$ $\nabla (\mathrm dt)=0$.

The interpretation of $t$ is that it maps an event $p\in\mathcal M$ to the absolute time of that event $t(p)$. The condition that $\mathrm dt$ is nowhere vanishing essentially tells us that $\mathcal M$ can be foliated into time slices $\Sigma_t$ of constant $t$, while the condition that $\nabla(\mathrm dt)=0$ tells us that the "flow of time" is uniform. Given a tangent vector $X\in T\mathcal M$, we may categorize it as future-directed, spatial, or past-directed based on whether $\mathrm dt(X)>0, \mathrm dt(X) = 0,$ or $\mathrm dt(X)<0$, respectively.

The worldline of a particle is therefore a future-directed curve through $\mathcal M$ (that is, a curve whose tangent vector is everywhere future-directed). If $x^\mu$ are the coordinates along the worldline, then Newton's 2nd law can be written

$$m\big(\ddot x^\mu + \Gamma^\mu_{\alpha\beta}\dot x^\alpha \dot x^\beta\big) = F^\mu$$

where $F^\mu$ is a spatial vector field, i.e. $\mathrm dt(F) = 0$.

Given a worldline which begins at some event $p$ and ends at another event $q$, the time elapsed for an observer following that worldline is simply $t(q)-t(p)$ and is therefore independent of the worldline itself; this is reflected in the fact that Newtonian spacetime does not feature kinematic time dilation.

Can we even speak of distances between the time-slices in a meaningful way?

Given two time slices $\Sigma_t$ and $\Sigma_{t'}$, one can speak of the difference in absolute time $t'-t$, which could be interpreted as a kind of "distance." However, because Newtonian spacetime is not a metric manifold, one cannot define the length of a generic curve, and so this is a very different notion of distance than the one present on a manifold equipped with a metric.


It's not necessary to refer to a metric - or indeed, any notion of distance - in order to formulate Newton's laws in this way. That being said, a typical Newtonian model takes the spacetime to be $\mathcal M := \mathbb R\times Q$ for some manifold $Q$, and $t(a,b,c,d)=a$. We also assume the existence of a privileged vector field $V$ and a symmetric $(0,2)$-tensor $\gamma$ which has the following properties:

  1. $\mathrm dt(V) = 1$
  2. $\forall Y, \gamma(X,Y)=0 \iff X=k V$ for some $k\in \mathbb R$
  3. $\forall X, \gamma(X,X)\geq 0$ with equality if and only if $X=kV$ for some $k\in \mathbb R$

The vector field $V$ is intuitively the vector field which points in the direction of increasing $t$. Because $\gamma$ is degenerate along $V$, it is not a metric on $\mathcal M$; however, the restriction of $\gamma$ to each time slice $\Sigma_t\simeq Q$ constitutes a Riemannian metric on the spatial vectors. In coordinates $(t,x^1,x^2,x^3)$ - in which the absolute time is taken to be the zeroth coordinate $x^0$ - $\gamma$ takes the form

$$\gamma_{ij} = \pmatrix{0&\matrix{0&0&0}\\ \matrix{0\\0\\0} & g_{ij}}$$ where $g_{ij}$ are the components of a Riemannian metric on $Q$.

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  • $\begingroup$ This is a very insightful answer! Can you give some references of books/articles, which follows this approach? I only saw it in Frederic Schuller's lectures. $\endgroup$
    – ProphetX
    Feb 4, 2022 at 16:25
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    $\begingroup$ @ProphetX You may be interested in this very detailed review. Note that it takes a slightly different approach to Schuller and requires a bit more structure, but there are a number of different ways to rigorously formalize Newtonian spacetime. $\endgroup$
    – J. Murray
    Feb 4, 2022 at 16:34
  • $\begingroup$ Thanks! So Schuller's approach is 'not really found' in the literature? Here for example the axioms of a Newtonian spacetime assume different mathematical structures, similar to the Newton-Cartan formalism, where we have two metrics with a compatibility condition. $\endgroup$
    – ProphetX
    Feb 4, 2022 at 16:51
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    $\begingroup$ @ProphetX I'm afraid I don't have an answer for that - I don't know enough of the literature to say. Note that Schuller's approach is essentially based on axioms 1-3 of the above review, which is basically the bare minimum requirement for stating Newton's laws. Axioms 4-5 (which Schuller leaves out) essentially just endow the spatial time slices with a Euclidean metric, which is done in every model I can think of but is not strictly necessary to write down the laws. $\endgroup$
    – J. Murray
    Feb 4, 2022 at 22:18
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    $\begingroup$ @ProphetX That being said, in Schuller's german-language mechanics class, he does refer to this additional structure (in a somewhat different form - he assumes the existence of a tensor field which looks like my $\gamma_{ij}$ in a suitable chart). If you speak German or are willing to sift through the auto-translated subtitles, you can find that lecture here $\endgroup$
    – J. Murray
    Feb 4, 2022 at 22:19

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