3
$\begingroup$

I'm trying to make a point that there is curvature of spacetime from the metric expansion that contributes to the dynamics of a galaxy. This curvature would be in addition to the curvature caused by the visible mass/energy content of the galaxy. I got back a note from an editor saying

"In the language of relativity physicists, “locally flat”, also called “locally Lorentz”, means flat at first order in separation from any chosen point. Of course, at second order one sees the influence of the Riemann curvature tensor, i.e. of the curvature."

Can someone interpret this for me? When a book says that the local geometry of spacetime is flat, how local is that? Microscopically, the size of a football field, a solar system, a galaxy? What's the cutoff for a 'local' geometry?

$\endgroup$
7
  • 3
    $\begingroup$ A simple analogy is that the parabola $y=x^2$ is locally flat at $x=0$. There is no region around $x=0$ where it is completely flat. In GR local flatness just means that the first derivatives of the metric vanish in Riemann normal coordinates. $\endgroup$
    – G. Smith
    Jan 22 at 23:14
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/595179/2451 and links therein. $\endgroup$
    – Qmechanic
    Jan 23 at 0:00
  • $\begingroup$ I figure it to be curvature that's not perceptible using all the magnification energy that's accessible in the observable region during at least some of the time, but your question's good enough that I'm wondering how the spatio-temporal curvature revealed by the 1919 eclipse (that 1st confirmed GR) might fit into that picture. (Maybe it would've equalled the negative energy involved in whatever gyrations of the astronomical bodies involved would've separated the last previous one of those eclipses capable of revealing GR from the one that actually did.) $\endgroup$
    – Edouard
    Jan 23 at 2:34
  • $\begingroup$ In the glossary to the 1997 ed. of his book titled "The Inflationary Universe", Guth describes "gravitation" (which the Encyclopedia Britannica considers to be synonymous with "gravity") variously as "negative energy", "potential energy", an "attraction", or an "interaction", but refrains from referring to its effectiveness as "energy" per se, because it has "infinite range", and can only be considered as existing in a field (in comparison to Newtonian gravity, which, although it can be considered either as a field or as a force acting at a distance, explained the 1919 effects inadequately). $\endgroup$
    – Edouard
    Jan 23 at 14:22
  • $\begingroup$ Unlike the gravitational field as described by Guth, fields in general are not necessarily infinite, even if they can (as described by Vilenkin) have infinite effects: The hypothetical scalar field whose effects would include an infinite expansion of space does not have numerically negative values. $\endgroup$
    – Edouard
    Jan 23 at 14:36
7
$\begingroup$

In coordinates that have the dimension of length, the dimensions of the Riemann curvature tensor are inverse length squared. Therefore at each point the components of this tensor establish length scales which you can loosely think of as radii of curvature. “Local” refers to a region whose length scales are small compared with any of these curvature-based length scales.

For example, near the horizon of a stellar-mass black hole, the radii of curvature are on the scale of kilometers. Therefore dynamics in a local region on the scale of, say, meters is barely affected by the curvature. The differences from the Minkowski metric within this local region are on the order of one part in a million.

For an even smaller region, the differences from Minkowskian are even more negligible. For a larger region, they are less negligible. There is no “cutoff”, but just “smaller and more flat” or “bigger and less flat”. Over a scale of kilometers the spacetime is not flat at all. Over no region is it perfectly flat.

To understand mathematically how spacetime is “flat at first order”, but not second order, at every point, look at the metric tensor in Riemann normal coordinates:

$$g_{\mu\nu}=\eta_{\mu\nu}-\frac13R_{\mu\sigma\nu\tau}x^\sigma x^\tau+O(|x|^3).$$

If we write

$$R\sim\left(\frac{1}{L_\text{curvature}}\right)^2$$

then the deviations from the Minkowski metric over a region of linear scale $L_\text{region}$ are of order $$\left(\frac{L_\text{region}}{L_\text{curvature}}\right)^2.$$

$\endgroup$
1
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Jan 23 at 18:59
1
$\begingroup$

What G. Smith said is correct, but I want to also point out that

there is curvature of spacetime from the metric expansion that contributes to the dynamics of a galaxy

is incorrect. There's no such effect.

There are various ways to see this. Here's one. FLRW spacetimes have no Weyl curvature, only Ricci curvature. Ricci curvature is nonpropagating: the field equation directly equates it to the stress-energy at the exact location of the curvature. Therefore all of the curvature of FLRW spacetimes comes from locally present stress-energy. The only significant contributors to the stress-energy tensor are ordinary matter, dark matter, and dark energy. Once you've accounted for those, you're done. There is no mechanism in GR by which any extra "metric expansion" effect could enter into the dynamics.

Note that, within galaxies, none of the major contributors to the stress-energy tensor has a net outward motion, nor a tendency to dilute with the scale factor. The only reason that ordinary and dark matter dilute by $1/a^3$ at larger scales is that the voids between superclusters increase in size and the average density including those voids therefore decreases. This has no effect on the dynamics of individual galaxies. There's no mechanism in GR by which it could have an effect.

The reason for this very common misconception seems to be that people think of FLRW spacetime as a background, i.e., as what you start with before adding the galaxies. It's actually the gravitational field of a uniform matter distribution, and that matter is the same matter that makes up the galaxies. If you put a galaxy on a FLRW background, you're double-counting the matter. To get the correct dynamics, you need to put the galaxy on a de Sitter background, because that's what's actually left behind when the ordinary and dark matter clump.

$\endgroup$
11
  • $\begingroup$ I'm amazed that anyone would use a theory that's failed so many tests as a proof. Your theory doesn't even agree with itself anymore, it's disproven itself. There is no such thing as 'Dark Matter'. How many tests does it need to fail before you'll see reason? You no longer have any theoretical foundation for your belief, either. SUSY is dead. You have no idea what what causes the flattening of galaxy rotation curves, but you're absolutely sure it's not metric expansion. It's like talking to a cult member. $\endgroup$
    – Gluon Soup
    Jan 23 at 13:35
  • 1
    $\begingroup$ @GluonSoup Dark matter may or may not exist in real life, but it definitely exists in the standard cosmological model, which is what your question appears to be about. If you don't want answers about standard cosmology then say so in the question, and tag it appropriately. $\endgroup$
    – benrg
    Jan 23 at 16:45
  • $\begingroup$ 1.) I did not add that tag 2.) The tag is 'Cosmology', not 'Quackery'. $\endgroup$
    – Gluon Soup
    Jan 23 at 16:51
  • $\begingroup$ @benrg: There is no mechanism in GR by which any extra "metric expansion" effect could enter into the dynamics How about this: dark energy is a dynamical field, its current values of energy and pressure depend on the history of cosmological expansion, and these values are (potentially) observable in local dynamics. $\endgroup$
    – A.V.S.
    Jan 23 at 16:52
  • 1
    $\begingroup$ @A.V.S. (and Gluon Soup), I'm saying there are no other effects in standard cosmology, the theory. Many people believe that there is an extra effect in standard cosmology and that's demonstrably not the case. I don't know what's going on in the real world. $\endgroup$
    – benrg
    Jan 23 at 19:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.