Would a Neutron Star Stop a Neutrino?

[Islam Hassan]

Like the question says, would a neutrino be stopped by the very high density matter in a neutron star?

IH

 

[bcrowell]

Googling shows that there are various people trying to calculate this, e.g.,

http://arxiv.org/abs/astro-ph/9806285
http://prola.aps.org/abstract/PR/v133/i4B/pB1046_1

Just as a first guess, I would have expected by default that a neutrino entering a neutron star would have about the same probability of being absorbed as a neutrino entering a main-sequence star, since they encounter about the same amount of mass. The abstract of the second paper talks about a mechanism that produces a deviation from this expectation, causing a neutron star to be completely transparent to neutrinos with low energies.

 

[Vanadium 50]

The first guess is pretty close. And since we know that probability is low, we know that neutron stars are largely transparent to neutrinos.

 

[Islam Hassan]

Googling shows that there are various people trying to calculate this, e.g.,

http://arxiv.org/abs/astro-ph/9806285
http://prola.aps.org/abstract/PR/v133/i4B/pB1046_1

Just as a first guess, I would have expected by default that a neutrino entering a neutron star would have about the same probability of being absorbed as a neutrino entering a main-sequence star, since they encounter about the same amount of mass. The abstract of the second paper talks about a mechanism that produces a deviation from this expectation, causing a neutron star to be completely transparent to neutrinos with low energies.

Funny, isn't density a factor at all? I would have thought that the sheer density of matter in a neutron star would significantly raise the probability of impact. After all, if you can precisely align a neutrino's trajectory with a nucleon, wouldn't impact be guaranteed? How would impact depend only on the magnitude of mass encountered and not on its spatial 'packing'?

IH

 

[zhermes]

Like the question says, would a neutrino be stopped by the very high density matter in a neutron star?

It is a fairly simple problem to estimate (in an order of magnitude sense), which is much better than guessing. The cross section for interaction between a neutrino and a nucleon (e.g. neutron) is about $$\sigma_{n \nu} \approx 10^{-42} \textrm{ cm}^2$$. If you assume a neutron star is ~1.4 solar masses, with a radius of about 10km, composed entirely of nucleons -- you can calculate the mean free path as $$l = (n \sigma)^{-1}$$, where n is the number density of nucleons.

The mean free path (l) is the average distance a neutrino will go before colliding. If the mean free path is smaller than 10km, you can expect the neutrino to collide; otherwise you would expect it to escape freely.

~~~~~

The exact answer depends on the details of neutron star structure which is still uncertain.

 

[PAllen]

Funny, isn't density a factor at all? I would have thought that the sheer density of matter in a neutron star would significantly raise the probability of impact. After all, if you can precisely align a neutrino's trajectory with a nucleon, wouldn't impact be guaranteed? How would impact depend only on the magnitude of mass encountered and not on its spatial 'packing'?

IH

I'm not sure this part has been answered. A nucleon is complex object (an acquaintance who assisted programming lattice QCD likened it to weather forecasting, only worse), consisting of quarks and gluons (mostly). From the perspective of a neutrino, it is mostly empty space. Thus, first order, squeezing a given mass into a smaller volume does not change the interaction cross section.

Given that it takes light years of lead to reach 50% chance of absorbing a neutrino, that is enough to conclude that neutrons stars have a small chance of stopping a neutrino.

 

[Islam Hassan]

I'm not sure this part has been answered. A nucleon is complex object (an acquaintance who assisted programming lattice QCD likened it to weather forecasting, only worse), consisting of quarks and gluons (mostly). From the perspective of a neutrino, it is mostly empty space. Thus, first order, squeezing a given mass into a smaller volume does not change the interaction cross section.

Given that it takes light years of lead to reach 50% chance of absorbing a neutrino, that is enough to conclude that neutrons stars have a small chance of stopping a neutrino.

What if we assume that a good deal of the center of a neutron star is quark gluon plasma; would that enhance the chances of impact?

IH

 

[PAllen]

What if we assume that a good deal of the center of a neutron star is quark gluon plasma; would that enhance the chances of impact?

IH

Not unless it were much denser than a neutron. In fact it is less dense.

 

[PAllen]

It is a fairly simple problem to estimate (in an order of magnitude sense), which is much better than guessing. The cross section for interaction between a neutrino and a nucleon (e.g. neutron) is about $$\sigma_{n \nu} \approx 10^{-42} \textrm{ cm}^2$$. If you assume a neutron star is ~1.4 solar masses, with a radius of about 10km, composed entirely of nucleons -- you can calculate the mean free path as $$l = (n \sigma)^{-1}$$, where n is the number density of nucleons.

The mean free path (l) is the average distance a neutrino will go before colliding. If the mean free path is smaller than 10km, you can expect the neutrino to collide; otherwise you would expect it to escape freely.

~~~~~

The exact answer depends on the details of neutron star structure which is still uncertain.

I verified the given cross section as ballpark (depends sensitively on neutrino energy). I also calculate that the nucleon density in the core of a neutron star is 5*10^38 nucleon/cm^3 (assuming I didn't make a mistake combining various published figures). This gives a rather short mean free path. This suggests both a neutron star and a normal star are likely to absorb a neutrino.

So, unless there is a flaw in the above, my previous answers were incorrect.

 

[zhermes]

This gives a rather short mean free path.

From that calculation, I get l ~ 1000 cm. Extremely short indeed; neutron stars are definitely opaque to neutrinos.

This suggests both a neutron star and a normal star are likely to absorb a neutrino.

A normal star is very unlikely to absorb a neutrino. The interaction probability is not dependent on only the total mass---as the above calculation indicates. The number density of nucleons scales like $$R^{-3}$$ while the path-length of interest only scales like R---thus a larger (less dense) distribution is far less effective at absorbing a neutrino.

 

[PAllen]

From that calculation, I get l ~ 1000 cm. Extremely short indeed; neutron stars are definitely opaque to neutrinos.


A normal star is very unlikely to absorb a neutrino. The interaction probability is not dependent on only the total mass---as the above calculation indicates. The number density of nucleons scales like $$R^{-3}$$ while the path-length of interest only scales like R---thus a larger (less dense) distribution is far less effective at absorbing a neutrino.

Right. Of course, since the cross section is proportional to neutrino energy squared, and the given figure is ballpark for 1 Mev neurtrinos, if you consider 1 ev neutrinos, you have mean free path of 10^15 cm, even through neutron star core material. So very low energy neutrinos are the ultimate ghost particles.

 

[quasi44]

Right. Of course, since the cross section is proportional to neutrino energy squared, and the given figure is ballpark for 1 Mev neurtrinos, if you consider 1 ev neutrinos, you have mean free path of 10^15 cm, even through neutron star core material. So very low energy neutrinos are the ultimate ghost particles.

Based on what we think we knew last week. So when the recent studies in Japan are confirmed, it's either as we thought last week or we have to consider our model flawed in that we had thus far only considered neutrinos of a given type, but now with the possibility that some are linking to a higher state to become a charged particle, we would have to assume an impact; or something else that changes the environmental conditions.

 

[PAllen]

Based on what we think we knew last week. So when the recent studies in Japan are confirmed, it's either as we thought last week or we have to consider our model flawed in that we had thus far only considered neutrinos of a given type, but now with the possibility that some are linking to a higher state to become a charged particle, we would have to assume an impact; or something else that changes the environmental conditions.

I could be wrong, but the derivation and formula I reviewed purported to be independent of neutrino flavor, thus not affected by oscillation. Perhaps the extremely small magnetic moment that follows if the Japan results are confirmed would be relevant at low energies. Please explain what you mean by 'linking to a higher state to become a charged particle'. I don't know what that means or how it relates to recent findings.

 

[quasi44]

I could be wrong, but the derivation and formula I reviewed purported to be independent of neutrino flavor, thus not affected by oscillation. Perhaps the extremely small magnetic moment that follows if the Japan results are confirmed would be relevant at low energies. Please explain what you mean by 'linking to a higher state to become a charged particle'. I don't know what that means or how it relates to recent findings.

Well, their finding suggested that they are registering hits by neutrinos with an electron, though that is not what they fired. To pick up an electron, it would have to hit something, or have some requirement satisfied as to allow it to take one from something else.

 

[quasi44]

I could be wrong, but the derivation and formula I reviewed purported to be independent of neutrino flavor, thus not affected by oscillation. Perhaps the extremely small magnetic moment that follows if the Japan results are confirmed would be relevant at low energies. Please explain what you mean by 'linking to a higher state to become a charged particle'. I don't know what that means or how it relates to recent findings.

Oh, yeah, sorry. Not a charged particle, but a particle pair with charge.

 

[PAllen]

Well, their finding suggested that they are registering hits by neutrinos with an electron, though that is not what they fired. To pick up an electron, it would have to hit something, or have some requirement satisfied as to allow it to take one from something else.

This still makes no sense to me. Can you point to a reference?

 

[PAllen]

From what I see of the recent Japanese results, there is no real surprise at all - just verification of muon->electron neutrino oscillation, and possible quantification of the last previously unknown neutrino mixing angle. There may be relevance for matter-antimatter symmetry, but I just don't see anything that would imply anomalously high cross sections for low energy neutrino interactions.

 

[quasi44]

From what I see of the recent Japanese results, there is no real surprise at all - just verification of muon->electron neutrino oscillation, and possible quantification of the last previously unknown neutrino mixing angle. There may be relevance for matter-antimatter symmetry, but I just don't see anything that would imply anomalously high cross sections for low energy neutrino interactions.

So then we fire muon neutrinos, get back electron neutrinos, and still expect that NOTHING acted on the neutrino, and that nothing other than a matter/antimatter symmetry function has taken place, and that this oscillation conversion is something we should have expected? If so, then what's the point? We aren't talking about a pointless money soak in Japan by the University and the international research partners, or are we?

 

[PAllen]

So then we fire muon neutrinos, get back electron neutrinos, and still expect that NOTHING acted on the neutrino, and that nothing other than a matter/antimatter symmetry function has taken place, and that this oscillation conversion is something we should have expected? If so, then what's the point? We aren't talking about a pointless money soak in Japan by the University and the international research partners, or are we?

Neutrino oscillation involves no interaction at all. It happens, for example, in empty space as solar fusion neutrinos make their way to earrth. There is nothing wasteful about the Japanese experiment. Very few quantitative details about neutrinos are known to high precision:

1) mass
2) mixing angles

Japanese data can provide information about these and other things, as well as verifying expectations.

There is just nothing I see reading the reports on the results that *remotely* suggests revision of nucleon interaction cross section.

 

[Vanadium 50]

The T2K results are irrelevant for this question.

I still don't quite follow the scaling argument. Consider an ordinary star: as a neutrino exits, it passes near N nuclei, with a probability p of interacting with any, for a total probability Np. From solar neutrinos, we know Np is small.

Now, increase the star's density. p is unchanged. Consider a line from the center along the neutrino's path: it intersects N nuclei. If I just move the nuclei radially when compressing the star, doesn't that hold N constant?

What am I missing?

 

[PAllen]

The T2K results are irrelevant for this question.

I still don't quite follow the scaling argument. Consider an ordinary star: as a neutrino exits, it passes near N nuclei, with a probability p of interacting with any, for a total probability Np. From solar neutrinos, we know Np is small.

Now, increase the star's density. p is unchanged. Consider a line from the center along the neutrino's path: it intersects N nuclei. If I just move the nuclei radially when compressing the star, doesn't that hold N constant?

What am I missing?

The same mistake I made at first. Consider billiard balls. Consider trying to hit a ball through the packed configuration versus a dispersed configuration. Not the same, are they?

A bit more precisely, you model the particle as effectively tracing out a volume equal to cross-section * R(adius). The number 'targets' per unit volume is N/R^3, N being the total number of targets (e.g. total nucleons in the star). Then, the number of targets in the 'interaction tube' of the particle is: cross-section * N / R^2. For a star, this is essentially zero, for the neutron star, especially the core region (based on data I looked up), it is rather large.

Of course, better yet is to deal with mean free path as Zhermes suggested.