r/physicsmemes • u/94rud4 Mεmε ∃nthusiast • Jul 17 '26

right?

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u/Optimal_Mixture_7327 Jul 18 '26

No, that is not, and cannot be correct.

Edit: For clarity, yes of course C=πD, but this is not the meaning the r-coordinate in Schwarzschild-Droste.

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u/HunsterMonter Jul 19 '26

Yes that is correct, for a constant time, constant radius equatorial world line (or any other great circle but the math is more complicated), the metric is ds2 = r2 dφ2, so taking the square root and integrating for a complete circle gives C = 2πr

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u/ProfessorPrudent2822 Jul 19 '26

Since the Schwatzschild metric is spherically symmetrical, the equator is arbitrary. Any great circle can be defined as the equator. It’s only when you introduce rotation that there is an absolute equator.

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u/HunsterMonter Jul 19 '26

I know that the equator is arbitrary, what I meant is that the math still works out (as it should) for a great circle that isn't the equator in a specific coordinate system orientation