# How can we be sure that black hole's singularity is not a missunderstanding? [duplicate]

The Newtonian gravitational potential is given by:

$$\phi=-\dfrac{GM}{r}$$

Which appears in the Schwarzschild metric tensor with a so-called singularity at $$r=0$$. Nonetheless, I can't get why is it necessarily the case that a singularity should form.

From Newtonian formulas, we would get that there is an infinite potential at the center of the Earth. But that is not the case because $$M$$ decreases along with $$r$$. When you get to the center of our planet, all the matter (approx.) is bordering you, so discounting the insane pressure in that place you are actually pulled towards the exterior in all directions (so that pull ends up cancelling).

My question is; doesn't it happen the same with black holes? In the same way the gravitational force is not infinite at $$r=0$$ because the mass $$M$$ becomes zero too, then why is this logic dismissed when studying the black holes case?

• Have you seen how the Schwarzschild metric is derived by solving the vacuum Einstein field equations? Commented Jul 15, 2023 at 0:40
• Newtonian gravitational potential is a concept that makes sense only in the weak gravity away from the hole. When Schwarzschild solved the Einstein field equations, he used the large-$r$, not small-$r$, behavior of the metric to identify a constant of integration as the mass of the hole. The singularity comes from looking at curvature invariants, not at the Newtonian potential. Commented Jul 15, 2023 at 0:57
• Possible duplicates: physics.stackexchange.com/q/18981/2451 , physics.stackexchange.com/q/24934/2451 and links therein. Commented Jul 15, 2023 at 10:21
• The singularity has zero radius, so the equation you are trying to use does not apply. Commented Jul 15, 2023 at 15:51

The situation you are considering is the Oppenheimer-Snyder collapse. It should be made clear that the Schwarzschild's solution is valid $$\textbf{outside}$$ the collapsing spherical body. The metric inside is not a solution of the $$\textbf{vacuum}$$ Einstein's equations. However, if the initial mass is sufficiently large, it is possible for the collapsing body, and thus the surface of the collapsing object, to go below its own Schwarzschild radius forming a Black Hole.