What is metric (invariant) in Newton mechanics (equivalent to spacetime interval of Minkowski space)?

Question

The answer might be obvious for those with much experience, but I could not get it via web search.

https://en.wikipedia.org/wiki/Minkowski_space

From the second postulate of special relativity, together with homogeneity of spacetime and isotropy of space, it follows that the spacetime interval between two arbitrary events called 1 and 2 is:

${\displaystyle c^{2}\left(t_{1}-t_{2}\right)^{2}-\left(x_{1}-x_{2}\right)^{2}-\left(y_{1}-y_{2}\right)^{2}-\left(z_{1}-z_{2}\right)^{2}.}$

The invariance of the interval under coordinate transformations between inertial frames follows from the invariance of

${\displaystyle c^{2}t^{2}-x^{2}-y^{2}-z^{2}}$

If we go back from second postulate (principle of the constancy of light speed) back to Newtonian, what is equivalent formula for space-time interval (metric) will be?

For context: I'm trying to assess correctness of physics in sci-fi https://en.wikipedia.org/wiki/Orthogonal_(series) where author tries to describe universe with (as I understood it) space-time interval of (changed Metric signature of "our" Universe of (1,3) to (4,0), also see http://www.gregegan.net/ORTHOGONAL/00/PM.html):

${\displaystyle c^{2}\left(t_{1}-t_{2}\right)^{2}+\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}+y_{2}\right)^{2}+\left(z_{1}-z_{2}\right)^{2}.}$

Thinking about it I realized I could not even "guess" what the metric is in Newtonian mechanics.

Answer 1

In relativistic mechanics we have 4-dimensional space-time and the Minkowski metric $$(\Delta s)^2 = c^2(t_1-t_2)^2-(x_1-x_2)^2-(y_1-y_2)^2-(z_1-z_2)^2$$ which is invariant to Lorentz tranformations (containing boosts and space-rotations), or more generally to Poincaré transformations (containing boosts, space-rotations and space/time-translations).

On the other hand, in Newtonian mechanics we have 3-dimensional space and 1-dimensional time, which are two completeley separate concepts. So here we have two separate metrics for space and time. The space-metric is $$(\Delta\vec{r})^2 = (x_1-x_2)^2 + (y_1-y_2)^2+(z_1-z_2)^2$$ and the time-metric simply is $$(\Delta t)^2 = (t_1-t_2)^2.$$ These two metrics are invariant to Galilean transformations (including boosts, space-rotations, space-translations and time-translations).

Answer 2

A useful way to parametrize the cases is by writing $$\Delta \vec S \cdot \Delta \vec S= \Delta t^2 - E (\Delta x^2 +\Delta y^2+\Delta z^2),$$ where $E=-1$ for Euclidean space with signature $(+,+,+,+)$ and $E=+1$ for Minkowski spacetime $(+,-,-,-)$, and $E=0$ for a Galilean-Newtonian spacetime $(+,0,0,0)$.

For constant squared-interval, the $E=-1$ case is a 3-D hypersphere, the $E=+1$ case is a double-sheeted 3-D hyperboloid, and the $E=0$ case is a 3-D hyperplane.

• As noted in the comments to your question, the Galilean-Newtonian case has a degenerate metric (since the determinant is zero). It has two degenerate metrics with signatures $(+,0,0,0)$ [for the temporal metric] and $(0,+,+,+)$ [for the spatial metric]... and these are 4-dimensional metrics for a 4-dimensional vector space.

In the spacetime-cases, $E$ can be associated with a maximum signal speed, which is finite for special relativity and infinite for Galilean relativity.

• To see this on a diagram for a 2-D space[time], try my Desmos visualization:
  
  robphy v8e spacetime diagrammer for relativity (v8e-2021) - t-UP https://www.desmos.com/calculator/emqe6uyzha and play with the E-slider.

(related: Euclidean space to Minkowski spacetime )

• While "infinite maximum-signal speed" is implicit in Galilean relativity,
  historically, it was not featured like the "speed of light" postulate for special relativity. If it were, I would think that we would have found special relativity sooner than 1905.
  
  (In https://www.physicsforums.com/threads/why-is-minkowski-spacetime-non-euclidean.1016402/post-6647528 , I argue that Felix Klein had the mathematical structure in place to propose Minkowski spacetime geometry... but not the physical intuition or motivation to do so.)

• Of course, the metric does more than assigning square-intervals to displacements. It also defines the notion of "orthogonality" between two vectors, via the dot-product (gotten from the polarization identities).