The Newtonian gravitational potential is given by:


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?

  • $\begingroup$ Have you seen how the Schwarzschild metric is derived by solving the vacuum Einstein field equations? $\endgroup$
    – Ghoster
    Jul 15 at 0:40
  • 2
    $\begingroup$ 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. $\endgroup$
    – Ghoster
    Jul 15 at 0:57
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/18981/2451 , physics.stackexchange.com/q/24934/2451 and links therein. $\endgroup$
    – Qmechanic
    Jul 15 at 10:21
  • $\begingroup$ The singularity has zero radius, so the equation you are trying to use does not apply. $\endgroup$ Jul 15 at 15:51

2 Answers 2


The question that you are asking, is what Penrose and Hawking spent their early careers establishing by doing a tonne of maths to prove the results.

Instead of just assuming that a black hole would have the singularity, they considered realistic scenarios. Even in the most optimistic case of starting with a constant density big ball and then slowly letting that collapse, once the Event Horizon forms outside the complete ball, then they found that there does not exist a force, known to mankind, that is strong enough to prevent the collapse into a singularity. Even if you have electric charge so as to make the ball want to fly apart, or you consider nuclear strong forces, there is just nothing strong enough to prevent the collapse.

That is the true reason we trust that there would be a singularity, not the handwavy answers that popular science gives an impression of.


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.

At present there is no known physical mechanism that balances this gravitationally driven collapse (given enough mass), which even manages to overcome quantum degeneracy pressure so reaching the singularity.

It must also be emphasised that the definition of singularity is much more subtle in General Relativity than in Newtonian mechanics. This is because in Newtonian mechanics we know it to be an artefact of description (such as divergences when considering point charges) and spacetime remains fixed. In General Relativity, however, since spacetime itself is dynamic, the definition of singularity requires the use of geodesic completeness. Once incomplete geodesics have been identified, we then proceed to characterise them (in this case we speak of curvature singularities).


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