Do black holes accumulate neutrinos?

  • #1
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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 ?
 

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  • #2
phyzguy
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Neutrinos that fall into a black hole are trapped inside, just like any other particle that falls in. Exactly what happens to the matter inside a black hole is not known. Classical GR says that all of the mass-energy is compressed into a zero volume, infinite density "singularity" at the center of the black hole. No one believes this is really what happens, but we have no theory that works at these extreme conditions.
 
  • #3
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I guess the "photon sphere" would collect neutrinos ?
 
  • #4
Ibix
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I guess the "photon sphere" would collect neutrinos ?
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.
 
  • #5
PAllen
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The photon sphere is where massless particles can have unstable orbits around the BH. Any closer, and they spiral into the BH, without a single orbit. A neutrino has a small mass, so it cannot orbit tat all at the photon sphere. I would say the photon sphere has no relevance at all to neutrinos.
 
  • #6
Ibix
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I would say the photon sphere has no relevance at all to 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.
 
  • #7
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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.
Supreme noob question, but if all timelike worldlines terminate, does that mean time stops for these things?
 
  • #8
Ibix
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Supreme noob question, but if all timelike worldlines terminate, does that mean time stops for these things?
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.

Here be dragons. Quantum gravitic dragons, we presume.
 
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  • #9
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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
 
  • #10
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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
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.
 
  • #11
Ibix
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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.

It's the not slowing down bit that means that they don't end up hanging around anywhere.
 
  • #12
PAllen
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Actually, neutron stars are so dense that most neutrinos would not get through on a direct hit, and would be absorbed. On the other hand, any neutrino from decay would have enough energy to graze a neutron star without being captured or defepjected much. On the other hand, depending on exact neutrino mass, so called relic neutrinos (the neutrino analog of CMB), might be low enough energy to be moving at a modest fraction of c relative to the neutron star. These could be gravitationally captured, and perhaps orbit.
 
  • #13
Vanadium 50
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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.
 
  • #14
Matterwave
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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.
The matter density matters a lot in this calculation. The mean free path scales as ##\lambda\propto (\sigma n)^{-1} \propto V/N##. So given a constant ##N##, the mean free path scales as the volume not radius. The length of the path that the neutrino has to take to "escape" obviously scales as the radius. So if the radius is crunched by a factor of 10,000, the mean free path is crunched by a factor of ##10^{12}##. If memory serves, the mean free path of neutrinos inside neutrons stars is on the order of centimeters, so they actually do take a significant time to diffuse out. This neutrino diffusion is one way (iirc actually the primary way above some temperature like a Trillion Kelvin) in which neutron stars cool. My memory on the last point is quite hazy so do take it with a grain of salt.
 
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  • #15
Ibix
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Actually, neutron stars are so dense that most neutrinos would not get through on a direct hit,
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?
 
  • #16
Vanadium 50
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The mean free path scales as λ∝(σn)−1∝V/N\lambda\propto (\sigma n)^{-1} \propto V/N.
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?

There is an effect that I have neglected, and that's coherence. But that's a factor of up to a few hundred for few MeV neutrinos.
 
  • #17
PAllen
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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?
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.
 
  • #18
PAllen
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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.
Yes I’m sure. In addition to back of the envelope calculations I’ve done similar to what matterwave posted, see the following

https://arxiv.org/abs/nucl-th/0307101

This work finds MFP in meters, up to hundreds of meters for modest energy neutrinos. The only neutrinos likely to pass through a neutron star are very low energy neutrinos because the cross section becomes unbelievably tiny for low energy neutrinos.
 
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  • #19
Ibix
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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.
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.
 
  • #20
Matterwave
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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?
A uniform body is assumed in the calculation I provided. Adding in complications with regards to non-uniformly dense bodies means one has to care about that non-uniformity and incorporate it into the MFP calculations. I don't think such a calculation is necessary for this case. The point is that the density affects the mean free path - which should make intuitive sense.

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?
The mean free path of a neutrino in normal lead is roughly 1 Light year. If you make that lead twice as dense by making it 4 times shorter and twice as wide - the mean free path of the neutrino would be roughly 1/2 light years (which is now longer than your modified block). I'm not really sure where you're going with this, but I don't think this example conflicts with your intuition either. If you make the block 100 times shorter without modifying it in any other way, you change the mean free path by a factor of 100 as well. Where this differs from the star example is that in the example of the star, the star gets smaller in all 3 dimensions but the distance the neutrino has to travel to "get out" only changes as R.
 
  • #21
PAllen
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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.
Oops, you are right.
 
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  • #22
PAllen
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Another trivial model of scattering is to imagine a tube of area of the cross section, times a given length. Then, the probability of interaction within that length scales as the density of the material because the number of targets within the tube scales this way. A neutron star density is 1013 times lead. Note that a light year is about 1013 km.
 
  • #23
Vanadium 50
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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)
 
  • #24
Matterwave
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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)
I'm not sure how to respond to this statement...are you disputing the relationship ##\lambda_{mfp}\propto (\sigma n)^{-1}##?
 
  • #25
jbriggs444
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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.
If you scale size down by a factor of two in each dimension then you have eight times the density and 1/2 the distance to traverse. So the probability of interaction is indeed 1/r^2. But the amount of material subtended by a solid angle is up by a factor of eight for density and down by a factor of eight for volume. It's a wash and r does not enter in.
 

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