I have read this:


The weak force has a very short range, gravity is extremely weak on the subatomic scale, and neutrinos, as leptons, do not participate in the strong interaction. Thus, neutrinos typically pass through normal matter unimpeded and undetected.

Now we know that from black holes even light cannot escape.

And neutrinos should have rest mass. But neutrinos are not affected by gravity on the subatomic scale, because gravity is very weak on that scale, and neutrinos interact very weakly.

The only reason even light cannot escape a black hole is gravity (stress energy) that bends spacetime. But since neutrinos are not affected by it on the subatomic scale, neutrinos should pass through a black hole.


  1. Do neutrinos pass through black holes like through normal matter? Has there been any experiment to measure whether we can detect neutrinos that passed through black holes?
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    $\begingroup$ How can neutrinos not be "affected" by spacetime when spacetime is where and when they exist? $\endgroup$ – Peter - Reinstate Monica Jun 29 '18 at 10:19
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    $\begingroup$ Gravity is weak on a subatomic scale compared to electromagnetism. It's still just as strong for neutrinos as it is for anything else. $\endgroup$ – Dmitry Grigoryev Jun 29 '18 at 10:23
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    $\begingroup$ This is a little like asking whether a bullet aimed straight down a bottomless hole will ever come out the other side. There is no other side. $\endgroup$ – Beanluc Jun 29 '18 at 17:34
  • $\begingroup$ The Wikipedia statement is a simplification which is true for "normal" situations, i.e. those found on Earth. Gravity is weak most places in the universe, but not in a black hole, whether or not you're looking at a subatomic scale. So black holes do strongly interact with neutrinos. $\endgroup$ – Yly Jun 30 '18 at 4:53

neutrinos are not affected by gravity on the subatomic scale

Where do you get this idea? It's not a part of mainstream physics.

One of the main ideas behind general relativity is that gravity isn't just another force battling it out in the arena of physics; it is the arena. That is, gravity is really just the geometry of spacetime, and anything (including a neutrino field) in spacetime is experiencing that geometry.

So no, standard physics does not predict that neutrinos will be able to escape a black hole. Instead, standard physics predicts that everything that enters a black hole's horizon will be unable to exit the horizon, including neutrinos. No one has done an experiment that can test this theory because we don't have very good neutrino telescopes or very good access to black holes. And it's possible that standard physics is wrong. But that's all we can say for now.

EDIT: In the comments below, the OP points out the underlying confusion that led to this question, which is the role of gravitons in black-hole physics. First of all, gravitons are (as @probably_someone points out) not really a part of mainstream physics. In particular, we don't know how to formulate a complete working theory of gravity using quantum field theory. We can quantize linearized gravity, which is the basic reason anyone really bothers to talk about gravitons, but that doesn't extend to nonlinear gravitational systems. And that's the key point: black holes are very nonlinear (unless you're very far away).

One consequence of this is that you can't model a black hole as a particle that interacts with other particles via exchange of gravitons. That's just not how our current theory of physics works. There's a related question with a very nice answer here, where Jerry Schirmer points out that the graviton is an excitation of the gravitational field, and not the field itself — but it's the field that makes a black hole, not its excitations. You might want to appeal to quantum field theory in curved spacetime, but even then, you basically assume a background curvature to spacetime. And it's that background curvature that affects the motion of the neutrino and traps it inside the horizon.

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  • $\begingroup$ thank you. Where I am confused is whether gravitons cause the bending of spacetime, or if not, then what do gravitons do? If gravity is just the bending of spacetime, and not like an EM interaction, where photons are mediating, then why do we need anything to mediate? Or do gravitons mediate the bending? $\endgroup$ – Árpád Szendrei Jun 28 '18 at 18:20
  • $\begingroup$ @ÁrpádSzendrei Gravitons are not really part of mainstream physics yet, either. $\endgroup$ – probably_someone Jun 28 '18 at 18:25
  • $\begingroup$ @Arpad You can ask the same question about gravitational waves. They are mainstream these days . $\endgroup$ – my2cts Jun 28 '18 at 18:53
  • $\begingroup$ @ÁrpádSzendrei I've updated my answer to fill in some of the details here. The upshot is that you shouldn't rely on the idea of gravitons, especially when black holes are involved. $\endgroup$ – Mike Jun 28 '18 at 20:44
  • $\begingroup$ @Mike, thank you. I understand that gravitons are only hypothetical, but I would like to know whether in the theory, gravitons mediate the bending of spacetime or do they mediate something else? $\endgroup$ – Árpád Szendrei Jun 29 '18 at 0:00

No, you have a wrong understanding of the Wikipedia statement. Wikipedia says only "Gravity is extremely weak" on a subatomic scale. It does not say "neutrinos are not affected by gravity".

They cannot pass through a black hole just as light cannot pass through a black hole. Photons are even lighter (no mass is as light as can be!) than neutrinos, and photons are certainly at "subatomic scales" (they are fundamental particles!) and so if photons cannot escape black holes, neutrinos can't either. (In fact nothing can - that's why they are black holes.)

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    $\begingroup$ To be precise: Black holes cannot prevent quantum tunneling, so even the biggest holes let escape particles and light. It is extremely rare, but still possible. See also Hawking radiation. $\endgroup$ – Thorsten S. Jun 28 '18 at 23:11
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    $\begingroup$ The dual meaning of 'Light' when referring to photons really confused me for a few seconds... $\endgroup$ – bendl Jun 29 '18 at 17:19
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    $\begingroup$ @bendl Photons are light, everyone knows that ;) $\endgroup$ – wedstrom Jun 29 '18 at 17:50

The black hole will gobble up the neutrinos for lunch and think "Mmm, what a yummy little snack!" :)

Seriously. Neutrinos are just tiny bits of massive matter, and thus they will be consumed by the hole in the same fashion that any other matter will be. The non-interactivity of neutrinos is that they are not interactive with the electromagnetic and strong force, which removes most of the interaction with ordinary matter because these two forces are very strong, and thus comprise the bulk of what makes ordinary matter highly interactive, while the other two forces, which neutrinos do interact with - that is, the weak force and gravity, are much weaker. Thus the first two forces account for the vast majority of the interactivity of ordinary matter and so particles ignoring them will have greatly reduced interactivity. But in extreme conditions, these "weak" interactions can become much stronger, and the black hole is an example of an extreme condition.

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Gravity being weak on the subatomic scale means that on the subatomic scale, the gravitational effect between two particles is weak compared to other effects. When it comes to gravity, black holes are pretty much the opposite of "subatomic scale" and "weak". So a system with a black hole and a neutrino isn't in a subatomic scale anymore.

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  • $\begingroup$ It's also true that "normal" gravities, like those on Earth are effectively weak because A) subatomic particles are much closer to each other. That's why electron orbitals in stationary crystals on Earth are not noticeably distorted by the Earth's gravity; and B) Many are moving so much faster than matter we're used to. Neutrinos are only barely affected by gravity in "gravitational lensing" and don't orbit the Earth or anything because their velocities are too high. Black holes just have enough gravity that the "gravitational lensing" can actually cause orbits or prevent things from escaping. $\endgroup$ – H. H. Jul 27 at 20:49

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