Theoretically, what is the difference between a black hole and a point particle of certain nonzero mass? Of course, the former exists while it's not clear whether the latter exists or not, but both have infinite density.
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1$\begingroup$ Possibly related: physics.stackexchange.com/q/9529 $\endgroup$– qftmeCommented Jul 18, 2011 at 15:54
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$\begingroup$ Could micro black hoe evaporate into fundamental particles $\endgroup$– John PrymeCommented Oct 9, 2021 at 16:21
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4 Answers
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We should probably distinguish between a particle being "point-like" and a particle being "structure-less". In classical mechanics we talk of "point-like" particles, objects with no extension. It is the case that in general relativity any "point-like" mass would be inside of its event horizon and so would be a black hole.
In quantum-mechanics even a "structure-less" particle - a particle with no consitituent parts - is wave-like and has extension, though not a fixed size, and it can never be come exactly point-like since that would take an infinite amount of energy. I do not believe it to be the case, therefore, that quantum-mechanically all particles are black holes in any sense.
answered Jul 18, 2011 at 17:24
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1$\begingroup$ I am tempted to infer from your answer that with sufficient energy the wave function for a particle could become compact enough to become a black hole. I imagine this is wrong, but I would like a sense of why. $\endgroup$ Commented Jul 19, 2011 at 16:33
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$\begingroup$ @Zassounotsukushi, you've phrased this in terms of QM, but the answer is really classical. Say we have a thoroughly classical object like a rock. In its own rest frame, it has some mass-energy density. Transform into another frame, and its mass-energy goes up by a factor of gamma, while its volume goes down by gamma, increasing its density of mass-energy by gamma^2. But this doesn't make it a black hole. The definition of a black hole can be stated in language that is independent of the frame of reference, so an object that isn't a b.h. in one frame can't be a b.h. in another. $\endgroup$– user4552Commented Jul 20, 2011 at 1:19
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One big difference is that all electrons, for example, are identical, but all black holes are not. In particular, a black hole can have any mass at all, whereas a particle like an electron has a fixed value for its mass. This property of fundamental particles like electrons is ultimately what allows us to define fixed scales of length and time in the laws of physics. In a universe that didn't have massive fundamental particles, the laws of physics would have a certain kind of symmetry called conformal invariance, which would make it impossible to construct clocks or rulers according to universally standardized rules.
Another difference is that there are fundamental particles such as electrons and neutrinos that are stable (don't spontaneously undergo radioactive decay), whereas it is believed that black holes will ultimately evaporate into fundamental particles.
You say that both have infinite density, but this is probably not actually true. The mass of a particle like an electron is probably attributable to the soup of virtual particles that surrounds it, whereas in general relativity a black hole's mass really is localized at a mathematical point. (Of course this is kind of an unfair comparison, since we know that GR is wrong below the Planck scale. It's possible that GR's singularities aren't really singularities. I'm just trying to give an answer in terms of established physical theories.)
It's tempting to imagine that fundamental particles are black holes, but this is not possible. Classically, a spinning, charged black hole has constraints on its angular momentum and its charge in relation to its mass. Otherwise, there is no event horizon, and we have a naked singularity rather than a black hole. An electron violates both of these limits, but we don't observe that electrons have the properties predicted for these naked singularities. For example, naked singularities have closed timelike curves in the spacetime surrounding them, which would violate causality, but there is no evidence that electrons cause causality violation.
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A point particle is a mathematical simplification, not a real thing. In reality, every particle has size, but for most considerations it's ok to neglect that size and treat it as a point particle.
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$\begingroup$ Wrong. Thanks to finite size $\pm \vec{\delta}$ one can observe a body, introduce a mass or geometrical center of it with the coordinate $\vec{R}$ and never confuse the position $\vec{R}$ with the position of the body $\vec{R} \pm \vec{\delta}$. $\endgroup$ Commented Jul 18, 2011 at 15:55
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$\begingroup$ @Vladimir You're right, you can deduce the size and shape of some particles with various methods. However, as I say above, "for most considerations it's ok to neglect that size...". $\endgroup$– AndySCommented Jul 18, 2011 at 16:03
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$\begingroup$ As soon as you neglect the size, you tend to fall in error that $R$ is sufficient to describe a body, especially "elementary" particle. All difficulties in QFT origin from this error - postulating point-like elementary particles and trying to construct a theory with this wrong idea. $\endgroup$ Commented Jul 18, 2011 at 16:08
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5$\begingroup$ "every particle has size" Actually this is a dicey statement. If the particle is structureless, then is can not deform under an impulse, and a force applied to one side would have to be transmitted instantaneously to the other side allowing a local violation of causality. Ouch. Which is one of the reasons why we expect a more complicated underling theory to explain how GR and QFTs get along at the real small scale. $\endgroup$ Commented Jul 18, 2011 at 16:38
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$\begingroup$ @dmckee: an external (long range) force can act on all parts of a body nearly equally. $\endgroup$ Commented Jul 18, 2011 at 19:45
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The notion of a point-like particle (electron, for example) is a very bad notion dwelling in Physics despite experiments say contrary. But if an electron tends to explode, the gravitating mass tends to collapse. Both features are manifestations of our blunders in describing physical phenomena.
answered Jul 18, 2011 at 15:41
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$\begingroup$ the links in your answer are funny though it is not understanble their relevance to the question. Why do you make such stuff up....to attract downvotes !! ?? $\endgroup$– Rajesh DCommented Jul 18, 2011 at 15:53
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$\begingroup$ Because some geeks speak quite seriously of self-action in Physics. $\endgroup$ Commented Jul 18, 2011 at 15:58