- #1
- 61
- 9
Do black holes accumulate neutrinos ? Do the neutrinos that fall in, decay in some way , maybe interact with the condensed matter, or just stay "parked" in perpetual loops ?
Any unpowered object crossing the photon sphere (1.5 times the Schwarzschild radius) will fall in, yes. I don't think that means that the photon sphere collects neutrinos. They certainly don't stay there - they continue down through the event horizon to the singularity.I guess the "photon sphere" would collect neutrinos ?
It's the inner boundary of the unstable orbits region for massive particles, where they would require infinite angular momentum to orbit. So if a neutrino (or any massive inertial object) has a trajectory touches the photon sphere, it must end up in the hole.I would say the photon sphere has no relevance at all to neutrinos.
Supreme noob question, but if all timelike worldlines terminate, does that mean time stops for these things?Any unpowered object crossing the photon sphere (1.5 times the Schwarzschild radius) will fall in, yes. I don't think that means that the photon sphere collects neutrinos. They certainly don't stay there - they continue down through the event horizon to the singularity.
As @phyzguy says, we strongly suspect that GR stops being an accurate description of the world somewhere along the way, but we don't know where. Do note that the singularity is not a zero-volume infinite-density point. It's not part of the manifold, so defining a volume doesn't make sense. It's also more like a time than a place - once you are through the event horizon all timelike worldlines terminate in the singularity, similar to the way that all timelike worldlines go through next Monday in our normal experience.
GR models spacetime as a manifold. Singularities aren't part of the manifold, so worldlines that reach it go out of scope of GR, and the worldlines just end. It isn't that the clock stops ticking, it's that there is no clock after a certain tick. Note that any real clock will be destroyed before it reaches the singularity. But a point-like ideal clock follows an ideal time-like worldline to the singularity and then just isn't anymore.Supreme noob question, but if all timelike worldlines terminate, does that mean time stops for these things?
Not the expert here, but gravity affects light, which has no mass, so why wouldn’t it affect neutrinos? But I’m guessing it depends upon the escape velocity of the massive body and how close the neutrinos get.I fess up - I was thinking of neutrinos as massless, that is not the case.Maybe a better thought would be, neutrinos around an intense gravity source, like a neutron star - would they go through the star easiy ? There must be some interaction with mass I guess, as that is how they are detected - but the idea of a neutron star having an "atmosphere" of neutrinos could be interesting
It takes very, very little energy to give a neutrino escape velocity. So I very much doubt that there's a neutrino soup around anything. Neutrinos don't interact with much else; they'll pass through almost anything, so mostly just whizz through matter without slowing down.I fess up - I was thinking of neutrinos as massless, that is not the case.Maybe a better thought would be, neutrinos around an intense gravity source, like a neutron star - would they go through the star easiy ? There must be some interaction with mass I guess, as that is how they are detected - but the idea of a neutron star having an "atmosphere" of neutrinos could be interesting
Actually, neutron stars are so dense that most neutrinos would not get through on a direct hit, and would be absorbed
Are you sure about this? There's no more material than in the progenitor star, and we have no problem seeing neutrinos from the sun's core, which pass through almost as much material.
Fair enough. I still don't think you're going to find a neutrino atmosphere around a neutron star, because mechanisms to brake them in free space are few and far between. I'd make the same objection to your primordial neutrinos: anything entering a system at even a modest fraction of the speed of light is leaving again unless there's some mechanism to bleed off its energy.Actually, neutron stars are so dense that most neutrinos would not get through on a direct hit,
The mean free path scales as λ∝(σn)−1∝V/N\lambda\propto (\sigma n)^{-1} \propto V/N.
I said neutrinos from any source other than relic could even graze the neutron star with minimal deflection, let alone orbit. In dispute is head on collisions, which I’ll address separately. As for relic neutrinos, if one passes close to a neutron star, and it’s rleative speed is .1c, it will be orbitally captured. The escape velocity at the surface of a neutron star is typically .3c. Whether such neutrinos occur is not known, but it would happen for some plausible neutrino mass, and some plausible relic neutrino models.Fair enough. I still don't think you're going to find a neutrino atmosphere around a neutron star, because mechanisms to brake them in free space are few and far between. I'd make the same objection to your primordial neutrinos: anything entering a system at even a modest fraction of the speed of light is leaving again unless there's some mechanism to bleed off its energy.
I don't think there's serious friction between neutrinos and normal matter, and happening to come in on a braking trajectory has to be rare. The only easily available mechanism is colliding with the star, as you note. Or am I misunderstanding something?
Yes I’m sure. In addition to back of the envelope calculations I’ve done similar to what matterwave posted, see the followingAre you sure about this? There's no more material than in the progenitor star, and we have no problem seeing neutrinos from the sun's core, which pass through almost as much material.
If the escape velocity is .3c then anything free-falling in and grazing the star would have to have picked up .3c from gravitational acceleration on top of its velocity at infinity, surely? Unless it was already bound, which I don't think relic neutrinos would be.As for relic neutrinos, if one passes close to a neutron star, and it’s rleative speed is .1c, it will be orbitally captured. The escape velocity at the surface of a neutron star is typically .3c.
I don't see how this can be true. Imagine a slightly lumpy sphere, and a detector pointing at some spot on the sphere. If I put more matter in the lumps where I am not measuring, are you telling me the probability of interaction goes up?
If you go with the "light year of lead" story, are you telling me if I made the chunk of lead twice as wide it could be four times shorter?
Oops, you are right.If the escape velocity is .3c then anything free-falling in and grazing the star would have to have picked up .3c from gravitational acceleration on top of its velocity at infinity, surely? Unless it was already bound, which I don't think relic neutrinos would be.
I was wrong. The thing that matters is the amount of material subtended by the same angle. So the probability of interaction goes as 1/r^2. (Not cubed)
Surely the thing that matters is the amount of material in a tube with a small cross section along the path of a sample particle whose interaction is in question. i.e. what @PAllen said in #22.I was wrong. The thing that matters is the amount of material subtended by the same angle. So the probability of interaction goes as 1/r^2. (Not cubed)
That’s a completely different question that the probability of absorption or scattering of a single neutrino by some body, or equivalently, the expected fraction of neutrino beam that is scattered away or absorbed. It is these that are relevant to a random neutrino from some source in a galaxy making its way through a neutron star.Imagine there was a neutrino source behind the sun. Your neutrino detector detects some source, and it has some solid angle of acceptance. The probability of interaction depends on the number of solar nuclei in that acceptance. Halve the radius of the sun, and that number of nuiclei quadruples.