#### PeterDonis

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Yes, but that's not why the metric is non-Euclidean. The relationship between ##R## and ##r## is irrelevant here. See below.within that metric a surface of constant [itex]R[/itex] would have surface area [itex]4\pi R^2>4\pi r^2[/itex] would it not?

No. What indicates that a surface of constant ##t## has negative curvature is, as I said, that the coefficient of ##dR^2## in the metric is not 1. The radial coordinate in this metric is ##R##, not ##r##; you don't need to know anything about ##r## to determine that the spatial metric is non-Euclidean.That would indicate [itex]R^+ \times S^2[/itex] has negative curvature and not be Euclidean

Put another way, what makes a surface of constant ##t## non-Euclidean in g(1916) is that a 2-sphere of constant ##R## has surface area ##4 \pi R^2##, but the proper distance between a 2-sphere at ##R## and a 2-sphere at ##R + dR## is *greater* than ##dR##, whereas if the space were Euclidean, that proper distance would be equal to ##dR##. For small ##dR##, the proper distance between the 2-spheres is ##dR / \sqrt{1 - r_s / R}##. That is, the ratio of the proper distance between two nearby 2-spheres to the difference in their areas is larger than it would be if the space were Euclidean.

No; g(current) is also non-Euclidean, and for the same basic reason. The only difference is that in g(current), ##r## plays the role that ##R## plays in g(1916): in g(current), the proper distance between two nearby 2-spheres at ##r## and ##r + dr## is larger than ##dr##, but the surface area of a 2-sphere at ##r## is ##4 \pi r^2##.as in g(current).