Sunday 9 August 2020

general relativity - Classical approach to Schwarzschild radius


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?



Answer



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.


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...