Neutron star question

twofish-quant

Our theories make some very strong statements on some things. For example, lepton interactions have a very firm theory, and that theory just says that "nothing weird will happen with electrons, muons, and tau leptons" in neutron stars. You ask electroweak theory what happens to stuff at neutron star densities, you get firm numbers.

The theory might be wrong, but there is no reason to think that the theory is wrong. One thing is that we can get electrons up to 90 GeV on earth, whereas the energies for neutron stars is 100 MeV.

For stuff to do with quarks, we can't get numbers that are quite as firm. When you increase things to neutron star densities, you get all sorts of effects that we don't know how to calculate, so we don't get firm numbers.

Some things we have strong knowledge of, some things we don't. Lepton processes are something that we have strong knowledge of. Baryon processes we don't, and then magnetic fields and convection are total mysteries.

And sometimes it doesn't matter. If you build a dog house using Newtonian physics, then it turns out that GR effects don't matter. Same with neutron stars. In a lot of situations, you can ask "how wrong do we have to be before it makes a difference?" and it turns out that there is a huge room for error.

zonde

This paper does not have Appendix C - it ends with references.

And we know about particle physics in "open" environment. We have theoretical ideas about particle physics in a volume of dense plasma. These ideas might work just fine if we have reasonably correct model for degeneracy pressure. And I doubt that.

Electrons do not turn into muons directly with or without neutrinos. Just like muons do not turn into electrons directly with or without neutrinos. They can do that only through intermediate state of W^- boson. Are we on the same line so far?

zonde

Give some reference that works. Or alternatively we can try to go trough the details ourselves.

twofish-quant

It's on page 822. Appendix C - Derivation of the Weak Interaction Rates

Any particular reason? Degeneracy pressure is just Fermi-Dirac statistics. You count the number of states, count the number of particles, put it into a canonical ensemble, and boom, you have the equation of state. It's pretty well covered in any graduate level textbook on solid state physics or thermodynamics (i.e. Kerson Huang's or Kittel/Kromer).

If there is a particular solid state/HEP phenomenon that you think isn't being modeled that will make a big difference, then it would be useful to state what that is.

I just care about the cross sections, and conservation of electron/muon number makes that cross-section zero for the energies that are relevant here.

I'd be *very* interested if you can dig up a reference for someone that has done the calculation that Bruenn did which shows a significant generation of muons in neutron stars.

twofish-quant

I just checked

http://articles.adsabs.harvard.edu//...pJS...58..771B

It's on pages 822 to 834.

There is one important process that Bruenn doesn't include which is the effects of Bremsstrahlung and that's calculated in gory detail at

http://arxiv.org/abs/astro-ph/9711132

twofish-quant

Also here is a paper that describes what happens if you put in "new physics" (i.e. flavor violation lepton processes)

http://arxiv.org/abs/1010.0883

IMHO, it doesn't make much of a difference.....

zonde

Thanks, I will look what I can make out of it. And thanks for other papers too.

Yes there is particular reason.
When degeneracy pressure is calculated for astronomical bodies it is assumed that distribution of charged particles is homogeneous. But electrons might (should) be partially squeezed out of the body due to larger degeneracy pressure.
Another thing is that degeneracy pressure should be higher in the middle of the body.

Look, I am trying to get us on the same page. If I read about muon decay in wikipedia the things you say do not make much sense.

Can you read that page (the part about muon decay)? It does not speak about any cross sections. It speaks about "decay width" and "Fermi's golden rule" and these do not seem to be related to any cross sections.

So what I am missing?

twofish-quant

Also, if you are interested in the nuclear equation of state.....

http://www.astro.sunysb.edu/dswesty/lseos.html

And several other EOS that are mentioned in this paper

http://arxiv.org/abs/1108.0848
New Equations of State in Simulations of Core-Collapse Supernovae

And this one

http://arxiv.org/abs/1202.5791
Equation of State for Proto-Neutron Star

And if you want to take a walk on the wild side (hyperions!!!! strange quark matter!!!!)

http://arxiv.org/abs/1205.3621
Hadron-Quark Crossover and Massive Hybrid Stars with Strangeness

I'm a lot less familiar with that physics, since I was able to get numbers out by using Lattimer-Swesty as a "black box". For the radiation hydro part, I had to tinker with the particle cross sections and implement some of these equations, so I know a bit more about what goes into that part of the code.

twofish-quant

Nope. It's assumed that there is no charge at scales that are larger than atomic scales. You can show that any sort of non-zero charge above atomic scales is very, very quickly going to get canceled out.

Astrophysical plasma are conductive, but they have zero net charge.

Nope. The pressure is going to work on everything in the same way. Pressure forces don't cause charge separation. Once you have any sort of charge separation, the particles are going to correct that at near the speed of light, whereas pressure forces act at the speed of sound.

Pressure is going to be higher. Whether degeneracy pressure is going to be higher depends on a lot of things, but it usually is. One reason why we think that the middle of neutron stars are "quark soup" is that if you have more particles you have less degeneracy pressure and that reduces the mass you need before everything goes into a black hole.

But this is something that goes into the calculations.

That's because wikipedia is missing information. It would be nice if that article were linked to a intro textbook on quantum field theory.

OK. Here's some intro quantum field theory.....

What you end up with when you do a QFT calculation is to answer the equation, if particles X1, X2, X3 with energies E1, E2, E3, go into a region with angles theta1, theta2, theta3 etc. What is the probability that you will come up with particles X4, X5, X6. You calculate these by writing down the Feyman diagrams, and going through a lot of nasty math.

Now if you want to calculate decay rates, you are asking if I go in with a muon with any energy at any angle, what is the probability that I will come out with an electron, neutrino pair with any energy. To do the calculation, you use a math trick called Fermi's golden rule, and then you also, end up with a probability distribution that contains a number called a decay width.

So the article is talking about cross sections.....

Once you have cross sections, you can use symmetry arguments to figure out stuff. For example, every muon that has ever been seen to decay to an electron has emitted a muon neutrino, that means that in order to produce a muon from an electron, you need to reverse the reaction and add some extremely highly energetic muon neutrinos, and you don't have these present in neutron stars. The only mechanisms you can use to create muon neutrinos are bremstrahlung and pair production. Both of these are thermal which means that the muons you make are going to be at the temperature of the surroundings, and you don't get enough high energy muon neutrinos to make muons.

It's not that people are being stupid when they ignore muons in neutron stars. It's that people have thought about the problem and concluded that you aren't going to get them (except at the core where everything goes wild). At the core things are weird enough so that you can create heavy baryons and let those decay into muons. But the cross sections are so high, that the muons are trapped and only get out with difficulty.

zonde

Well, if we would speak about thermal pressure then yes, you would be right. But we are not speaking about thermal pressure, we are speaking about degeneracy pressure instead. And so you are wrong. Fermi-Dirac statistics describes population of identical fermions. So electron degeneracy pressure is acting inside population of electrons and proton degeneracy pressure is acting inside population of protons.

And if because of degeneracy pressure outcomes with low energy electron are forbidden the probabilities should come out different.

Hmm, no.
I have different symmetry argument. In order for electron to decay into muon we must have very high energy electron and we produce muon, electron neutrino and muon antineutrino.

twofish-quant

Pressure is pressure. More precisely pressure is - d(internal energy) / d (volume).

Thermal pressure, degeneracy pressure. It's all the same thing. There is a standard way of calculating pressure of bulk matter. There are some assumptions, and it's known to break in some situations, but high density/high pressure situations make the method work even better.

Yup and neutrino degeneracy pressure acts on neutrinos. You end up dividing up different species of particles. Count the number of states. If the lower states are filled up, you have to put the particle into the lowest available state. That gives you the internal energy. You then take the derivative with respect to volume, and that gives you the pressure.

Yup, and that's in there.

If you look on page 822, equation (C2), there is a expression for the occupation probability for a given state. Equations (C4) (C5) and (C6) gives you the emissivity, absorption and the scattering cross sections in terms of the raw reaction rates, and you'll see expressions (F(E)) and (1-F(E)) scattered about those terms.

Won't work.

One of the constraints of quantum field theory is that everything has to work the same in any reference frame, so if I do my calculations in the rest frame of the electron, everything should end up the same.

So I transform everything into the rest frame of the electron where it has zero momentum. At that point, all you are left with is the rest mass of the electron, and since that's less than the muon, you don't have the energy to generate a muon, electron neutrino and muon antineutrino. In the absence of another particle, you can't get a particle to decay by accelerating it. Special relativity won't allow it.

(Now if you accelerate an electron, and then put it into an electric field, *then* you can get particle creation through Bremstrahlung, it's not enough to generate muons unless you are very deep in the core at which point all heck breaks loose anyway.)

One of my points here is that physicists aren't idiots. A lot of times you get posts about how stupid scientists are that they haven't thought of this ridiculously simple idea, when it fact lots of people have thought very deeply about that topic.

If you can come up with a mechanism by which you can get a non-trivial muon flux in the outer layers of a neutron star, *that* would be worth a paper, but people have thought about it, and the conclusion is that you can't. Also, you can invoke the tooth fairy. If invoking muons, would resolve some sort of observational difficulty, that *also* would be worth a paper. (i.e. I have no clue how to make muons in neutron stars, but if we wave our magic wand, and produce muons, then something cool happens.)

As it is, the theory says that you aren't going to get muons. The observations provide no reasons why we think muons are important. If you want to work on figuring out ways of getting muons in neutron stars, you are welcome to do so, but there are probably more promising areas of research else where.

zonde

Then let me rephrase my statement. Charges will get separated in electron degenerate plasma because electrons will move faster.



Yes, you are right of course. It would have to be at least two particle collision to speak about electron with very high energy.

But I would like to drop this line about muons. It occurred to me that I am not sure if quantum states of electrons and muons are completely independent. And without that assumption the problem becomes too vague.

Doubt is very important thing in science, wouldn't you say? And it doesn't mean that physicists are idiots.

twofish-quant

I just did a crash course in quantum field theory. Here is another crash course in computational astrophysical modelling.

One thing about astrophysics is that a lot of astrophysical phenomenon can be modelling through what I call "time scale decoupling." What happens is that you look at the time scales over which two different processes happen and if they are very different, you can "decouple" them by assuming that process one is in local equilibrium and then process two interacts with process two by changing the equilibrium.

You squeeze a tube of toothpaste. You don't have to calculate how your actions affect the protons and electrons because the time scale of the atomic process is very different than that of the squeezing process.

Now in neutron stars, electron processes happen over the course of nanoseconds whereas pressure processes happen over the course of milliseconds. That means that the two processes decouple, and if the time scale you are interested in is the pressure timescale, the atomic processes are in equilibrium.

Put another way any charge imbalance is going to resolve itself in a few nanoseconds, which means that if you look at processes over the course of milliseconds, the material is going to be in charge equilibrium.

Now, if you can come up with an argument in which a charge imbalance can remain for several milliseconds so you can't assume equilibrium, that gets interesting. This happens a lot in the interstellar medium. As density goes down, so does the sound speed so you end up in situations you can't "separate" pressure processes and atomic processes.

The reason I got sucked into neutrino physics is that I'm interested in radiation hydrodynamics. You can show that neutrino processes will happen over the course of milliseconds which means that the neutrinos are not going to be in equilibrium, which means that I have to go over the details of those processes, whereas because nuclear processes are in equilibrium, I can just load in a file and I don't have to think about them,

This is because neutrinos interact through weak nuclear forces which are much weaker than EM. Electrons are going to be in equilibrium.

Quantum states of electrons and muons are independent as far as we can tell. Now it turns out that because neutrinos have mass that quantum states of electron neutrinos and muon neutrinos aren't independent.

Also, it's not a vague problem. Given particles X1, X2, X3, with momentum vectors p1, p2, and p2 into a system, what pops out?

Now that I think about it. Brehstrahlung won't work. It's an electromagnetic process so you aren't going to get muons from that. What might work is the reaction

electron + electron antineutrino -> muon + muon antineutrino

The trouble with this is that you need a source of 100+ MeV antineutrinos. This is difficult because the main source of antineutrinos is pair production, which means that you are going to get 20-30 MeV ones.

Trying to work out a particle/nuclear reaction chain is a lot like doing a crossword puzzle.

There's "I don't know"-doubt and "this doesn't exist"-doubt. If someone just asks "so how do muons work in neutron stars" that's one thing. For someone to assert "physicists have not considered the role of muons in neutron stars therefore the whole concept is suspect" is another.

twofish-quant

One other thing, I thought of a quick test to see if the assumption of "timescale decoupling" works.

Is it round?

If you have a round object it means that the atomic processes are in local equilibrium so that you can do pressure calculations assuming local equilibrium. If it's not round, that means that the atomic processes aren't allowing the object to go into pressure equilibrium.

zonde

So the question is at what timescales electrons will arrive at distribution described by Fermi-Dirac statistics and if that process will happen at speed of sound or speed of light. Now if we look at atomic orbitals I suppose there is no question that electrons can't occupy lower already occupied quantum state despite electrostatic attraction. If it wouldn't be so all electrons would fall into 1s orbital before they could be "bounced" out of it (at speed of sound).

Another point. Pressure arise from elastic collisions between particles. Now we can ask what energy determines outcome of elastic collision - is it all the energy of the particle or is it only the portion of energy that it can give up by falling into another available quantum state. And I say it's the later. So it seems that fully degenerate particles will not exert any pressure on other kind of particles. So it is rather confusing to call "degeneracy pressure" a "pressure".


Just to be clear about my hesitation.
Let's say we have ion without any electrons. Now we let muon occupy some (muon) orbital of that ion. And you say that electron orbitals will remain exactly the same around that ion. Say we add some electron to that ion and then let muon decay and we won't observe any anomalous radiation as electron hypothetically will fall into "proper" electron orbital. Right? And I am not sure about that.

But if we assume that you are right then we can discuss it further.
We can look at it from different side. Let's say that there appears some muon inside electron degenerate core of the star. And it is at rest meaning that all available electron quantum states have equally high energy in any direction. In order for it to decay it should emit electron into very high energy quantum state and in order to conserve momentum in it's rest frame it should emit very high energy neutrinos in other directions. So theoretically we can reach level where muon rest energy is simply not enough to do that i.e. it will be stable.
I am speaking now only on qualitative level. I do not say that it should work on quantitative level.


Okay, point taken.