5
$\begingroup$

Given a spacetime displacement $\textbf{x}$, we can define the interval $I(\textbf{x})$ as the square of the time measured by a clock that moves inertially along $\textbf{x}$. If we assume that the interval can be derived from a bilinear function $f$ as $I(\textbf{x})=f(\textbf{x},\textbf{x})$, then the ability to measure $I$ implies the ability to measure $f$ as well. That is, if you have a norm, and you assume it comes from a bilinear inner product, you automatically get an inner product for free. Or in more physical terms, if you have a clock and a way to tell whether a world-line is inertial, you have a way to measure the metric.

But is there some nice physical or mathematical way to see that the interval should be derivable from a bilinear function? If we rule out the degenerate Galilean case, then $I$ must be compatible with Euclidean geometry for spacelike displacements, and the Euclidean metric is bilinear. So this makes it kind of plausible that the spacetime metric should be bilinear as well. But is there any really nice way to show that it has to be bilinear?

The bilinear form of the Euclidean metric is basically the Pythagorean theorem, which is a statement about parallelism. Is the bilinearity of the spacetime metric interpretable in some nice way as a statement about parallelism?

$\endgroup$
10
  • $\begingroup$ What do you mean by "spacetime displacement"? Spacetime does not naturally come with a linear structure. Are you referring to infinitesimal displacements? As it stands the question is unclear. $\endgroup$
    – TimRias
    Nov 1, 2019 at 8:10
  • $\begingroup$ @mmeent: The question is about special relativity. I originally tagged it as SR, but someone else changed it to GR. $\endgroup$
    – user4552
    Nov 1, 2019 at 12:21
  • $\begingroup$ For a physical argument about the derivation of the bilinear form, this is a good start arxiv.org/abs/physics/0302045 . You can start with this and then extend to show that the metric is indeed bilinear. You can also look at the Introduction and Chapter 1 of "The Geometry of Minkowski Spacetime" by Naber. $\endgroup$ Nov 1, 2019 at 12:30
  • $\begingroup$ @YuzurihaInori: I'm familiar with the Pal paper, but don't immediately see how to apply it here. It assumes that we have a preferred system of coordinates for a given observer, and it doesn't discuss the metric. $\endgroup$
    – user4552
    Nov 1, 2019 at 12:38
  • $\begingroup$ @BenCrowell Ah, that makes much more sense. $\endgroup$
    – TimRias
    Nov 1, 2019 at 12:38

1 Answer 1

3
$\begingroup$

This isn't anywhere near an answer... but just a possible guide into the literature.

On my list of things to read (now pushed further back) are papers on the foundations of spacetime geometry along the lines of the Ehlers-Pirani-Schild (EPS) approach, which tries to motivate the Lorentzian structure of spacetime.

Along those lines are approaches to consider a Finslerian geometry

Some links to the literature:
http://www.phy.olemiss.edu/~luca/Topics/geom/finsler.html

Happy hunting!

$\endgroup$
1
  • $\begingroup$ Nice, thanks! Some interesting material from Jaconson: "In the EPS paper, the quadratic nature of the light cone is derived from their axioms. This is not very satifying however, since one of the axioms is not particularly physically natural. [...]the axiom states that the function f (B) = λ(C)λ(C 0) is a twice differentiable function on spacetime." "It seems that in order to generalize known physical theories to non-quadratic Finsler metrics one must introduce further structure (e.g. a spacetime volume element) and the result is not nearly as simple or “natural”..." $\endgroup$
    – user4552
    Nov 1, 2019 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.