# Are neutrinos affected by gravity?

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.

• – Qmechanic Apr 17 '12 at 11:29
• 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. – DilithiumMatrix Apr 17 '12 at 13:00

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.

• Only the first sentence addresses the question and only then as a statement rather than a quantitative argument. – Rob Jeffries Jul 16 '16 at 15:47

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).

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.

• 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. – Rob Jeffries May 19 '16 at 22:47

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.

• I am not sure indestructible is the right word. Do you mean that these particles are not made of smaller constituent particles? – honeste_vivere Apr 7 '16 at 12:43
• 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. – John Muggins Apr 14 '16 at 16:54
• 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... – honeste_vivere Apr 14 '16 at 19:29
• I agree. "Almost" never is the thing that makes them destructible. – John Muggins Apr 15 '16 at 12:27
• 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 $M \geq M_{\odot}$... – honeste_vivere Apr 15 '16 at 20:01