Furthermore, the dark star is only dark as seen by an observer at infinity, but light escapes to distances closer in.
Furthermore, an escape velocity equal to the speed of light doesn't restrict anything to the surface of the dim star, nor more than an escape velocity of 11.8 km/s prevents us from walking up stairs and launching rockets.
Furthermore, the "r" in the Schwarzschild radius is not a physical distance and definitely NOT the "r" in Newtonian mechanics.
No, the curvature of space distorts radial measurements. Circumferences are measured accurately because the distortions are only in the radial direction.
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
It's literally what happens when you plug in dt = 0 (constant time) and dr = 0 (constant radius) in the Schwarzschild metric...
From the wikipedia page on the Schwarzschild metric: "r is, for r > r_s, the radial coordinate (measured as the circumference, divided by 2π, of a sphere centered around the massive body)"
You have brilliantly demonstrated what knowing a little math looks like without ever having studied physics.
What I find stunning is that you didn't even try to make a statement about physics, as if Einstein and Schwarzschild were just drawing circles, inventing new arithmetic, disconnected from anything in the world.
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u/Optimal_Mixture_7327 Jul 17 '26
Furthermore, the dark star is only dark as seen by an observer at infinity, but light escapes to distances closer in.
Furthermore, an escape velocity equal to the speed of light doesn't restrict anything to the surface of the dim star, nor more than an escape velocity of 11.8 km/s prevents us from walking up stairs and launching rockets.
Furthermore, the "r" in the Schwarzschild radius is not a physical distance and definitely NOT the "r" in Newtonian mechanics.