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Layman here, but EE and BS physics. I know that light is affected by gravity. But are neutrinos? During the collapse of a star into a neutron star, as the electrons join protons to form neutrons (e.g., or the collapse of a star to a black hole?), I read that the only things that can get "out" instantaneously are the neutrinos. (this even applies to a normal star I presume, as the photons take "forever" to exit). But I know gravity is extreme in these instances, to say the least, so does gravity NOT affect the neutrinos? This would seem to be contradictory.

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    $\begingroup$ Related: physics.stackexchange.com/q/15263/2451 $\endgroup$
    – Qmechanic
    Commented Apr 17, 2012 at 11:29
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    $\begingroup$ There are good answers below (particular @LeosOndra); but I wanted to add that everything is effected by gravity, according to general relativity. Because gravity is a distortion in space-time itself, it doesn't matter what the properties of the particle/object is. $\endgroup$ Commented Apr 17, 2012 at 13:00

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Neutrinos are certainly affected by gravity. However extreme gravity may be around the collapsing core of a massive star, the real problem is great density of matter. Neutrinos interact with the stellar matter much less than other particles so they escape much easier, though the very center of the collapsing core is opaque even to them.

In fact, the main problem of the supernova-to-be massive star is not where to get energy (it is at hand in the form of potential gravitational energy) but how to get rid of it! Energy must be carried away from the core to enable collapse - and it is carried away by the neutrinos.

However, even neutrinos don't escape from the collapsing star instantaneously. About 1 per cent of their energy is absorbed in outer layers, reversing their collapse to explosion - the visible firework of the supernova. The rest (99 per cent !) of the original gravitational energy is quietly carried away by neutrinos.

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    $\begingroup$ Only the first sentence addresses the question and only then as a statement rather than a quantitative argument. $\endgroup$
    – ProfRob
    Commented Jul 16, 2016 at 15:47
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All particles, even massless ones, are affected by gravity - it is just a question of degree.

The (kinetic) energy of the neutrinos produced in a supernova are of the order of 10 MeV.

If the neutrinos have a rest mass energy of say an eV (though it might be much less than this), then their gravitational potential energy were they situated on the surface of a proto-neutron star (10 km radius and about $1 M_{\odot}$), would only be of order 0.1 eV.

The neutrinos are therefore almost unaffected by the gravitational potential of the supernova remnant and escape to infinity with their kinetic energy hardly lowered.

That is not to say that all neutrinos are not strongly affected by gravity. The neutrinos from the big bang have kinetic energies less than an meV. If cosmic neutrino background neutrinos have mass of an eV or even tenths of an eV then they will be strongly affected by the gravitational potentials of large galaxies or clusters of galaxies and will "clump" as a result. More details of the calculation of gravitational clumping of neutrinos in the potentials of galaxies and clusters via the Vlasov equation can be found in Ringwald & Wong (2004).

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  • $\begingroup$ are cosmic background neutrinos expected to be 50-50 split among left-handed and right-handed? $\endgroup$
    – lurscher
    Commented Oct 23, 2021 at 3:50
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The main question is best answered by the linked question, but the neutron star part of this question is another matter.

Common particles that try to escape a neutron star find themselves hindered not by just gravity, but also by the electroweak force. And with the density of neutron stars, the latter is very strong too. Neutrinos aren't so affected by electroweak forces, which is why they "instantaneously" escape neutron stars.

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  • $\begingroup$ Neutrinos are of course affected by the weak force. That is how they interact. The reason they escape from neutron stars is that the cross-section is too small to stop them. The question however was about gravity. $\endgroup$
    – ProfRob
    Commented May 19, 2016 at 22:47
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It's thought that neutrino oscillations are, in some way, related to quantum gravity, or that quantum gravity is involved in the process by which neutrinos oscillate between different flavours. As for whether or not their trajectories are impacted by gravitational fields, one has to be reminded of the supernova observations and the fact that neutrinos seem to escape supernovae faster than light. In light of these observations, it stands to reason that either something is slowing down the light (a collapsing gravitational field, perhaps) or that the neutrinos are outpacing light via some quantum effect.

My suspicion is that the trajectory of a neutrino may be altered by quantum gravity effects so long as very large numbers are taken into consideration when estimating the probability of that occurring. What's more interesting to me, however, are the flavour oscillations. These may occur, in my opinion, if a background particle (a gravity particle) were to exchange a Z boson with the neutrino. The upshot of thinking this was is that we may infer, from this, that this gravity particle is probably the electron. Furthermore, in collapsing gravitational fields the electron would be the perfect candidate to explain the 'slowing down' of light from supernovae.

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  • $\begingroup$ The mechanism by which light is slowed down is opacity. Even in the sun, the photon diffusion times are in average several thousand years before finally being emitted in the vacuum. In a supernova light transmission is stopped by the same mechanism, but the violent explosion releases it all after a while. Neutrinos suffer no opacity effects $\endgroup$
    – lurscher
    Commented Oct 22, 2021 at 20:05
  • $\begingroup$ Fine, but that opacity is ultimately caused by the fact that photons interact with subatomic particles (such as electrons). We know that the electron absorbs and emits photons. We know stars contain electrons. Hence, it stands to reason that electrons are to some extent responsible for slowing down the light escaping supernovae. No? $\endgroup$
    – Sam Cottle
    Commented Oct 24, 2021 at 6:57
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I'm a layman also, but can answer your first question by saying that the general theory and definition of gravity involves anything with mass. Because neutrinos are particles and have mass then yes, they are affected by gravity. Photons are subatomic particles also. Since we can see that photons bend their stream while passing planets and other large gravitational masses it lends credence to the idea that invisible neutrinos will bend in gravitational fields as well. As for collapsing stars I agree with the above assertion that the density of that immense reaction would cause certain behaviors that could cause the destruction of everything but the neutrinos. That's relying on faith in our current understanding of neutrinos as nearly indestructible. A neutrino could pass through hundreds of thousands of miles of steel without being harmed - while an atom traveling at the same speed would be completely disintegrated on impact. That is a good analogy to the extreme gravitational and destructive forces at play in a collapsing star.

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  • $\begingroup$ I am not sure indestructible is the right word. Do you mean that these particles are not made of smaller constituent particles? $\endgroup$ Commented Apr 7, 2016 at 12:43
  • $\begingroup$ You're right about that. I think quasi-indestructible might have been a better word. Any physical thing that can pass through 300,000 miles of a solid steel rod without stopping or significantly slowing down is fairly indestructible in my mind, but perhaps not completely. As for theories on neutrinos being comprised of different particles - I entertain it but it's not something I tend to accept as unconditionally as others have. I apologize if my wording offended any theories. But yes, I do have my own. $\endgroup$ Commented Apr 14, 2016 at 16:54
  • $\begingroup$ Well, stopping because you hit a light year of lead does not necessarily equate to a destroyed object. One could imagine hitting a wall of lead/steel, whatever, and not falling apart but that would not mean said object was indestructible. Neutrinos can pass through huge amounts of mass because they almost never interact with anything, not because of an inherent indestructible nature. A proton hitting the same wall may stop, but it will not automatically be destroyed either... $\endgroup$ Commented Apr 14, 2016 at 19:29
  • $\begingroup$ I agree. "Almost" never is the thing that makes them destructible. $\endgroup$ Commented Apr 15, 2016 at 12:27
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    $\begingroup$ We do observe these "other" particles though. That is how we know a neutrino was present, by observing its predicted decay/daughter products. We also used neutrino observations before, during, and after the famous SN1987A to help resolve whether neutrinos do indeed have mass (which we believe they do). When a star collapses, not everything gets destroyed. Most of the released energy goes into neutrinos and a few percent goes into the resulting shock wave. In all cases, a massive core remains in some form with $M \geq M_{\odot}$... $\endgroup$ Commented Apr 15, 2016 at 20:01

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