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The Schwarzschild radius from general relativity is given to be $r = \frac{2GM}{c^2}$.

One can obtain the same answer using classical calculations. That is, the escape velocity of a particle is given by $v = \sqrt\frac{2GM}{r}$, which can be arranged to give $r = \frac{2GM}{v^2}$, which can be interpreted as the maximum radius for which a particle travelling at velocity $v$ cannot escape. By treating light as simply a particle travelling at velocity $c$ and substituting in the above equation, one arrives at the Schwarzschild radius.

Is it just a coincidence that the classical approach gives the same result as the general relativity result, or is there some merit to the classical approach?

Kenshin
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1 Answers1

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It's just a coincidence that the factor of 2 comes out right, as the kinetic energy of the photon doesn't have a denominator. If you use isotropic coordinates, the "radius" of the black hole becomes M/2 instead of 2M (in natural units), where the "radius" in isotropic coordinates is the value of $\sqrt{x^2+y^2+z^2}$ on the horizon. The coincidence is dependent on Schwarzschild coordinates, and has no deeper significance.

But the dependence on GM and r is demanded by dimensional analysis, and is not coincidental.

Ron Maimon
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