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Following up on the questions raise here and trying to get a little more clarity on what it means to be 'local' and 'flat'. In the first chapter of Gravitation, the author(s) state:

The geometry of spacetime is locally Lorentzian everywhere

I can visualize this. If I'm 8 light minutes from a large star, I can obviously see curvature in the path of a planet around that star, but if I take smaller and smaller measurements of shorter duration, so that for all intents and purposes I'm measuring a point on a geodesic around this star, then any experiment I do with test particles will meet the predictions of a Lorentzian geometry.

But I don't follow this logic if I do the experiment at the center of this mass. If I do my experiment at the center of this star, then no matter how small I make my laboratory, even making it small enough to be considered a point, I'm still going to see curvature all around me.

How is the geometry of spacetime is locally flat at the exact center of a exceedingly large mass?

Edit: In trying to understand this statement from Gravitation, I'm trying to understand how a local geometry might not be flat. An image of the geodesics around a singularity came to mind as a place on the manifold that wasn't differentiable. This is the context for the question.

Quark Soup
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    Related: How is spacetime locally Lorentzian? – G. Smith Feb 11 '21 at 18:18
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    The phrase “center of curvature” does not have any meaning in GR. In general, spacetime is locally flat at every point and has curvature at every point. There is no contradiction because, as has been explained multiple times in response to your past questions, “locally flat” is very different from “flat”. – G. Smith Feb 11 '21 at 18:21
  • I hope you can step back and appreciate how contradictory that statement appears to someone trying to understand this subject. – Quark Soup Feb 11 '21 at 18:24
  • Spacetime as a manifold is locally flat at the centre of a large mass because also there exists a tangent space with a Lorentzian metric. – Thomas Wening Feb 11 '21 at 18:24
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    It should be intuitively obvious that the surface of the Earth is “locally flat” but not “flat”. Spacetime works in the same way, but in four dimensions instead of two. – G. Smith Feb 11 '21 at 18:31
  • It seems somewhat strange to me that the place where local spacetime flatness troubles you the most — the center of a massive body — is the place where in Newtonian physics its gravitational force is zero. – G. Smith Feb 11 '21 at 18:37
  • @G.Smith - "The center of a massive body is the place where in Newtonian physics its gravitational force is zero". That's a good point. The acceleration is greatest at the surface of the star and returns to zero at the center. It's differentiable at that point. I was thinking of a black hole where the center is a singularity that is not differentiable. – Quark Soup Feb 11 '21 at 18:44
  • Note that while the force is zero at the center, the curvature is not. The right side of the Einstein field equations (the energy-momentum tensor) is nonzero there, so the left side (the Einstein curvature tensor) is nonzero. – G. Smith Feb 11 '21 at 18:47
  • @G.Smith - Does the same logic apply to a singularity? – Quark Soup Feb 11 '21 at 18:53
  • I think of various curvature invariants as becoming infinite as you approach a singularity (which means that at least some components of the curvature are becoming infinite). I don’t know how to think about the energy-momentum tensor at the singularity. I imagine that it doesn’t make mathematical sense to do so, but I’ll let someone else chime in on that. – G. Smith Feb 11 '21 at 18:59
  • @G.Smith - Thank you. It's helpful for me (possibly others) to understand what "locally flat" means by understanding how something might not be "locally flat". Continuing this discussion below. – Quark Soup Feb 11 '21 at 19:15
  • From https://physics.stackexchange.com/a/144458/123208 "A singularity in GR is like a piece that has been cut out of the manifold. It's not a point or point-set at all. Because of this, formal treatments of singularities have to do a lot of nontrivial things to define stuff that would be trivial to define for a point set. For example, the formal definition of a timelike singularity is complicated, because it has to be written in terms of light-cones of nearby points." (emphasis mine) – PM 2Ring Feb 11 '21 at 19:49

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The phrase "then no matter how small I make my laboratory, even making it small enough to be considered a point, I'm still going to see curvature all around me" is always true, no matter how far you are away from any star.

Spacetime is locally Lorentzian in a way similar to that a parabola can be locally approximated by a line. In no finite region will you find that the parabola is congruent to the line (unless it is a degenerate parabola of course). But you are guaranteed that you can find a line that concides with the parabola in one point and with the same slope.

Similarly you are guaranteed to find a Lorentzian metric and a connection that conincide in one point with the gravitational manifold.

oliver
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  • Let me use this space to continue the thread of thought in the original post here. If the mass was so great so as to cause a singularity, would you have a point on this manifold that was not differentiable? Would the geometry at this point be locally Lorentzian? – Quark Soup Feb 11 '21 at 19:05
  • In an essential singularity the metric and connection are, by definition singular, and hence, have no value which could be considered locally Lorentzian at all. But in a conventional star there is no essential singularity. Probably a coordinate singularity, but then you can always choose suitable coordinates to make it disappear. – oliver Feb 11 '21 at 23:01
  • Thank you. Between you and G. Smith I've figured out that 1.) I was getting a coordinate singularity mixed up with a physical singularity. This can be fixed by choosing another coordinate set. 2.) The geodesics around a normal mass are differentiable, they don't approach a physical singularity at the center of the mass. 3.) The local geometry of spacetime is not Lorentzian at a singularity. Have I got that right? – Quark Soup Feb 11 '21 at 23:05
  • @GluonSoup: as to 3): take the simple hyperpola $f(x)=1/x$ as a basic metaphor: you can't approximate it by a line at $x_0=0$ because that would require $f(x_0)$ and $f^\prime(x_0)$ to be be defined. By contrast, $f(x)=x^2/x$ is also singular at $x_0=0$, but this singularity is removable (similar to a coordinate singularity) and then you can approximate it by $x$ at $x_0=0$ even if it is not defined there. Locality in $locally Lorentzian$ is defined by the metric and it's derivatives/connection. If they are not defined in the point of interest, you're in trouble with applying that definition. – oliver Feb 14 '21 at 16:52
  • as to 2) if you mean by "normal mass" an extended mass and not a point mass, then the metric, the connection, and the geodesics are differentiable and so it is meaningful (and true) to talk about them being locally Lorentzian. – oliver Feb 14 '21 at 16:56