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Sep 13, 2020 at 12:37 comment added Independent Physics But that Einstein's approximation which Schwarzschild refers to is the 1915 $g_{tt}= 1-(α/r)$ , which is Hilbert's r-solution.
Sep 13, 2020 at 10:31 comment added TimRias @Manuel That comment was made about an approximation. In this case a large radius approximation. And while the exact answers may be equal, the expressions truncated at finite order may differ (by a higher order quantity). It may even occur that the exact expression in one coordinate truncates at finite order, while it does not in the other.
Sep 12, 2020 at 23:51 comment added Independent Physics Why does Schwarzschild state that both R and r solution differ for the Mercury's orbit? I quote: "Since α/r is nearly equal to twice the square of the velocity of the planet (with the velocity of light as unit), for Mercury the parenthesis differs from 1 only for quantities of the order 10^−12. Therefore r is virtually identical to R and Mr. Einstein’s approximation is adequate to the strongest requirements of the practice."
Sep 12, 2020 at 21:34 history answered benrg CC BY-SA 4.0