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Furthermore, would a black hole be able to tear apart a quark pair? Would there be a difference in tidal forces on a scale that small?

qtConstellation
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  • Related: https://physics.stackexchange.com/q/312722/123208 & https://physics.stackexchange.com/q/4243/123208 and the links therein. – PM 2Ring Dec 06 '20 at 00:58
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    Note that that huge factor is comparing the nuclear force between 2 nucleons to the gravitational force between them. It doesn't tell you about the gravitational force of a black hole which is far more massive than a single nucleon. – PM 2Ring Dec 06 '20 at 01:00
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    In classical GR, tidal forces become arbitrarily large as you get arbitrarily close to the singularity. – G. Smith Dec 06 '20 at 01:25
  • black holes are heavy bodies concentrated in a very low space. So the weak gravitational force becomes a strong dominant force. Even the light could not escape black hole – Anonymous Dec 06 '20 at 02:39
  • @safesphere In GR, "tidal force" means geodesic deviation, which depends on the Riemann curvature tensor. For a Schwarzschild black hole, every component of the Riemann tensor becomes infinite as $r\to 0$ so geodesic deviation becomes infinite as one approaches the singularity. – G. Smith Dec 06 '20 at 07:04
  • @safesphere Is Wikipedia’s statement that “Mathematically, the tidal force in general relativity is described by the Riemann curvature tensor” not sufficient? Is it your point of view that the phrase “except inside a black hole” accidentally got left out? – G. Smith Dec 07 '20 at 04:36
  • @G.Smith Everything in GR is described by the Riemann curvature tensor. It means nothing specific. The inner Schwarzschild spacetime is a $4\text{D}$ cylinder, which is flat in the direction of the radial fall. So unless you can rigiriously show how the Riemannian curvature creates tidal forces inside a black hole, the only thing this quote implies is a false confidence. – safesphere Dec 07 '20 at 05:19
  • @G.Smith "For a Schwarzschild black hole, every component of the Riemann tensor becomes infinite as $r\to 0$" - This certainly is not true. Based on your link, most components are zero, including the diagonal $R_{3434}$ component at $r=0$. In a radial fall, the gravitational acceleration is a spatial gradient of the time dilation, by $r$ outside, but by $t$ inside. However, Schwarzschild is static in $t$, so the latter gradient is zero. There is no gravitational acceleration inside the horizon in a radial fall. And without the gravitational acceleration, there are no tidal forces either. – safesphere Dec 10 '20 at 15:29

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It is possible to "saturate" the strong force by adding enough protons and neutrons to a nucleus so that it breaks apart all on its own, despite the strong force. But you cannot "saturate" the gravitational force by adding more mass to a massive object- its gravity just keeps getting stronger and stronger.

And for a black hole that is small, the tidal forces it produces get arbitrarily large the closer you get to it.

This suggests that a black hole can shred nuclei.

niels nielsen
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  • A remarkable journey Sir - approaching a 50k reputation while playing a base guitar! +1 – safesphere Dec 07 '20 at 04:36
  • @safesphere yes it's been quite the journey. 2 months ago I got a cancer diagnosis that looked fatal; now the docs say they can cure it with surgery and I go under the knife on the 15th so you might not see much of me here for a while. I'll try playing the bass on my back in bed! – niels nielsen Dec 07 '20 at 16:48
  • My prayers are with you Sir! – safesphere Dec 07 '20 at 22:30