# Derivation of Schwarzschild radius from escape velocity

1. Dec 3, 2013

### greypilgrim

Hi,

Is it pure coincidence that if you put $c=v_e=\sqrt{2GM/R}$ in the escape velocity, you end up with the Schwarzschild radius $R=2GM/c^2$?

The derivation of the escape velocity is purely classical mechanics. It involves $E_{kin}=mv^2/2$ which is incorrect in special relativity even for massive particles and is entirely useless for massless photons.
One could argue that with the constants $G,M,c$ at hand the form of the Schwarzschild radius follows from dimensional analysis, but that doesn't explain why the factor $2$ is present in both derivations.

2. Dec 3, 2013

### PAllen

Newtonian escape velocity can be derived without use energy.

That Newtonian escape velocity = c matches SC radius is generally considered a coincidence. Note that in Newtonian mechanics, there is nothing special about c, and nothing preventing a body having escape velocity > c, nor preventing projectiles with v>c that can escape.

3. Dec 4, 2013