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From my understanding of general relativity, energy curves spacetime to produce an effect that we call gravity. The greater the concentration of energy, the greater the curvature and hence the greater the gravitational force we feel. However, I have repeatedly seen online that general relativity predicts that black holes have a 1D singularity of infinite density, which is an issue since the concept of infinitely small doesn't really exist physically.

What part of general relativity says that the singularity in a black hole must be infinitely dense? Why can't the explanation just be that the object has a sufficiently large mass to deform spacetime enough to produce an event horizon?

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  • $\begingroup$ The singularity is a point at which general relativity fails to be predictive since, as you say correctly, infinities are non-physical. $\endgroup$
    – Charlie
    May 19, 2021 at 22:04
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    $\begingroup$ Related: physics.stackexchange.com/q/18981/2451 and links therein. $\endgroup$
    – Qmechanic
    May 19, 2021 at 22:11
  • $\begingroup$ The links there were helpful - pointed me to the Penrose–Hawking singularity theorems and No-hair theorem which was what I was looking for $\endgroup$ May 19, 2021 at 22:39
  • $\begingroup$ I believe part of the notion of the singularity comes from the fact that there currently is no widely-accepted theory that describes a mechanism that would explain any sort of force or pressure which could resist the gravitational collapse at the point of black hole formation, so there is no expected reason that the collapse should ever stop. Which means there's no limit to how small a region the matter ultimately occupies, which means there's no limit to how dense it will become. And this is confounded by the fact that one can never verify such a theory, because of the event horizon. $\endgroup$
    – geshel
    May 20, 2021 at 0:26

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I think at least part of your question could be rephrased as

Why can't a black hole consist of a dense object with a nonzero radius less than its Schwarzschild radius?

If that's the case, the answer is that it's not possible to have a stable distribution of matter inside the Schwarzschild radius. One way to understand this is to note that inside a Swarzschild black hole, the "radial" coordinate (which is the surface area of the corresponding $2$-sphere divided by $4\pi$) is timelike. There is no force which could prevent matter from falling toward $r=0$ for precisely the same reason that outside the black hole, there is no force which could prevent matter from progressing towards next Wednesday.

The worldlines of massive particles are always timelike. Loosely speaking, outside of a black hole's event horizon this means that the worldlines can accelerate one way or another but they are always oriented in the direction of increasing $t$; inside of a black hole's event horizon, they can accelerate one way or another but are always oriented in the direction of decreasing $r$.

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None of what you're saying is true. A black hole singularity isn't a one-dimensional point. Its dimensionality is undefined. It doesn't have infinite density according to GR, because it isn't part of the spacetime manifold, so GR doesn't define whether there is any matter there.

There is a notion of a strong curvature singularity. When you have this type of singularity, infalling matter's density approaches infinity as it approaches the singularity. This is the closest thing that GR has to what you've been thinking of as a singularity of infinite density.

However, the singularity of a Schwarzschild black hole is not a strong curvature singularity. Infalling matter gets spaghettified, but not infinitely compressed. Actually its density stays constant.

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  • $\begingroup$ Do you have a source saying the density of infalling stuff to a swarchild metric black hole remains at constant density? $\endgroup$ May 20, 2021 at 3:08
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    $\begingroup$ @shai horowitz, sorry but I did not notice that your comment refers to my answer. In Schwarzschild interior solution the density is constant. As compactness parameter $\alpha$ reaches the critical value $8/9$ an initial event horizon arises at the center. On it the density is still constant and equal $3\alpha$. You may want to look on the mathematical formulation for this solution under physics.stackexchange.com/a/679431/281096 . $\endgroup$
    – JanG
    Jan 3, 2022 at 12:00
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In my understanding, the black hole singularity is not a point in space but the description of physical quantities behavior for compactness parameter $\alpha~ (\equiv r_S/R)~$ approaching some critical value $\alpha_{c}~$. In the case of Schwarzschild interior solution $\alpha_{c}=8/9~$. The energy density at the singularity has finite value $3\alpha_{c}~$, while the pressure diverges. This behavior can be found in all static spherically perfect fluid solutions. The respective values of the critical parameter $\alpha_{c}~(\le 8/9)~$and the finite energy density at the singularity depend on the solution. Opposite to it, the pressure behavior is in all solutions the same. For $r \rightarrow 0~~$it diverges like $p \sim r^{-2}~$, generating the universal force equal $2~c^{4}/G$, as conjectured by Gibbons (arXiv: hep-th/0210109v1 [qr-qc]).

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