Why do strings in black holes not collapse to a singularity?

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String theory suggests that black holes may be represented as stringy fuzzballs, which do not collapse into a singularity but instead extend to the event horizon. The fuzzball conjecture posits that strings accumulate at the event horizon, countering the gravitational forces that would typically lead to a singularity. Unlike traditional black holes, where compressive forces dominate, the dynamics of string configurations allow for non-spherical, extended states that maintain their structure. This prevents the collapse into a point, as the strings exhibit a complex geometry that does not yield a traditional horizon. The discussion highlights the differences in behavior between stringy and non-stringy matter in the context of black hole physics.
brooknorton
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I've read that string theory suggests that black holes may be stringy fuzzballs without a singularity at the center. Instead, strings pile up all the way to the event horizon. My understanding of non-stringy black holes is that the gravitational compressive force overcomes all known repulsive forces (such as electrostatic repulsion between particles) and so the compressed matter collapses with nothing stopping an indefinitely denser collapse, resulting in a sigularity at the black hole center. Now the fuzzball conjecture posits a stringy ball that extends to the event horizon. My question is, why don't the strings collapse to a singularity, just like non-stringy matter does? Don't the strings have the same mass and gravitational effect as their non-stringy counterpart-particles (eg - a neutron represented by a string, or as a particle/wave, still has the same mass and gravitational attraction)? Don't stringy particles have the same repulsive forces as their non-stringy counterparts? What allows a fuzzball to build matter to the event horizon without collapse?
 
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Apparently, there is no traditional horizon in the fuzzball picture.

Mathur, The information paradox: A pedagogical introduction: "Why doesn’t the energy of the Schwarzschild hole all fall into the origin at r = 0? Consider the simpler case of a single string in flat space. It would seem that a string in the shape of a circle must shrink to a point under its tension so there should be no extended string states. But of course we can get extended states: the string profile is not circularly symmetric, and while each segment of the string is indeed trying to shrink, the whole exited string maintains a nonzero size in its evolution. Similarly, the nonextremal microstates are not spherically symmetric, and they cannot be sliced in a time independent manner. Each part of the geometry is dynamical, and the whole structure maintains a nontrivial structure without generating a traditional horizon."
 
"Supernovae evidence for foundational change to cosmological models" https://arxiv.org/pdf/2412.15143 The paper claims: We compare the standard homogeneous cosmological model, i.e., spatially flat ΛCDM, and the timescape cosmology which invokes backreaction of inhomogeneities. Timescape, while statistically homogeneous and isotropic, departs from average Friedmann-Lemaître-Robertson-Walker evolution, and replaces dark energy by kinetic gravitational energy and its gradients, in explaining...

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