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.

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    $\begingroup$ Related: How is spacetime locally Lorentzian? $\endgroup$
    – G. Smith
    Feb 11, 2021 at 18:18
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    $\begingroup$ 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”. $\endgroup$
    – G. Smith
    Feb 11, 2021 at 18:21
  • $\begingroup$ I hope you can step back and appreciate how contradictory that statement appears to someone trying to understand this subject. $\endgroup$
    – Quark Soup
    Feb 11, 2021 at 18:24
  • $\begingroup$ 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. $\endgroup$ Feb 11, 2021 at 18:24
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    $\begingroup$ 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. $\endgroup$
    – G. Smith
    Feb 11, 2021 at 18:31

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – Quark Soup
    Feb 11, 2021 at 19:05
  • $\begingroup$ 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. $\endgroup$
    – oliver
    Feb 11, 2021 at 23:01
  • $\begingroup$ 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? $\endgroup$
    – Quark Soup
    Feb 11, 2021 at 23:05
  • $\begingroup$ @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. $\endgroup$
    – oliver
    Feb 14, 2021 at 16:52
  • $\begingroup$ 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. $\endgroup$
    – oliver
    Feb 14, 2021 at 16:56

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