Is this derivation for schwarzschild radius for a black hole of mass $M$ correct?

Consider a body of mass $M$. We know that light can’t escape a black hole. Speed of light being the highest possible could be set as the escape velocity.(??) Then $$\text{Escape velocity}^2=(2GM/r)$$ Solving for $r$ we get $$r=2GM/v^2$$ Since $v=c$; $$r=2GM/c^2$$ My only problem with this derivation is that shouldn’t we be using relativistic mechanics instead of newtonian? If we do use Relativistic mechanics,is there any proof similar to this one?

• – Qmechanic May 2 '18 at 12:29
• – Kyle Kanos May 2 '18 at 12:39
• The coincidence between your derivation of the Schwarzschild radius and the General Relativistic result is just that. The notion of an escape velocity is not applicable. – Rob Jeffries May 2 '18 at 13:05

The problem with your derivation is that the Newtonian notion of escape velocity is not the same as the General Relativity notion of light not being able to escape.

In Newtonian gravity, if you shine a beam of light outwards from the radius $2GM/c^2$, the light will actually leave the radius and go all the way out to infinity. Of course, the light will slow down to arbitrarily slow speeds. If you shine the light from a radius of less than $2GM/c^2$, the light will still make it outside the "event horizon" and just orbit in an ellipse. Therefore, in purely Newtonian gravity, we do not have a sharp event horizon.

In general relavity, once behind the event horizon, light cannot escape at all. Furthermore, the light will always be traveling at the speed of $c$, and can not speed up or slow down. So the two physical pictures are quite different.

Notice that in Newtonian Gravity, a black hole isn't black. You can still see stuff come out of it. It's just that that stuff can't escape out to infinity, but it can still make it pretty far.

• "Of course, the light will slow down to arbitrarily slow speeds." Just to avoid a misunderstanding, you mean the coordinate speed of light slows down. Locally measured it's c, outside and inside the event horizon as well. – timm May 2 '18 at 15:30
• I suppose my answer is confusingly written. In my second paragraph, I am explaining what will happen, according to Newtonian physics, if you were to release a particle at velocity $c$ from the "event horizon." The particle would go out to infinity, getting slower and slower. This is of course not what happens in relativistic physics. – user1379857 May 2 '18 at 15:40

If you assume that a sufficiently compact object does have an event horizon, then the relationship has to be of the form

$$r=\alpha GM/c^2,$$

where $\alpha$ is a unitless constant. This is because there is no other way to combine the mass with the universal constants $G$ and $c$ in order to get units of distance. The fact that the Newtonian derivation gets $\alpha=2$ right is just luck.