# Packing within a neutron star

1. May 18, 2012

### lntz

Hey,

So i just read 2 articles, the first talked about the mathematical problem of 'packing', IE how many smaller objects of the same size can fit inside a shape without overlapping or covering the boundaries of the shape. and the second article was talking about a recent neutron star that was discovered to be 2.04 times the mass of the sun. It was suggested that a neutron star shouldn't be able to get this heavy without becoming ablack hole.

So my knowledge on this stuff is fairly limited, so please bear with me. (i'm currently doing my A-Levels). my ideas will probably come across as very 'classical' since that is what i am most familiar with.

So if i have a regular solid object, say a lump of iron, i might say that it can not be compressed. The atoms are packed together tightly, and don't 'want' to get any closer to one another. at a guess i'd attribute this to the strong nuclear force not being able to overcome electro-static repulsion. I understand that within an object such as a star, gravity is a massive force, but does it simply push these atoms closer together? or is there something else going on? what i read about packing also said that it may be possible for atoms to 'overlap'... also i've read the term 'collapse of an atom'. what does that mean?

Sorry for the long winded question, but i would really appreciate any help you can give me on this.

2. May 18, 2012

### Dickfore

3. May 18, 2012

### ImaLooser

The pressure due to gravity in a neutron star is extremely high. In the core there are no atoms. All of the space of an atom is squeezed out. It is like a giant atomic nucleus, but the center of a neutron star is ten times denser than that.

4. May 18, 2012

### Whovian

No, there's nothing really trying to overcome electrostatic repulsion at all, the strong nuclear force really has pretty much nothing to do with anything at this point. Simply that most forces in the everyday world won't be able to overcome the electrostatic repulsion.

Overlap I'm not sure about, collapse of an atom is sort of what makes a Neutron Star a Neutron Star.

So, first, you should know that the radius of a nucleus is tiny compared to the atom, right? (A Hydrogen Atom's nucleus, if I remember correctly, is something like 1/1000 of the radius of a Hydrogen Atom.) In a Neutron Star, what happens is the gravitational force gets even stronger than the repulsion between the electrons in the shell, and all the atoms collapse into one nucleus. According to theory, the electrons should fall onto the protons and they decay into neutrons, which is why we call it a Neutron Star. So, to answer your question, it's simply that one would require a relatively strong force to compress matter to this stage.

5. May 21, 2012

### Drakkith

Staff Emeritus
The distance between atomic nuclei in iron is much MUCH larger than the range of the strong nuclear force. The metallic bonds between the atoms are a certain length that minimizes the potential energy of those bonds. If you increased the pressure the bonds would shorten a little bit, but that would increase the potential energy of the bonds. (As it takes energy to make them shorter) Upon release of this pressure the whole lump would expand slightly, releasing the energy.

In an atom you have a very small nucleus surrounded by a "cloud" of electrons. The electron clouds of each atom are what interact and form bonds between atoms in a material. In a metal such as iron, each atoms electron cloud is overlapping with multiple other atoms forming bonds between each one and making the material very strong as a result.

Now, when gravity crushes these atoms in a star, it causes them to get closer together, but only to a point. Because of Quantum effects as the atoms and their electron clouds are crushed closer together the electrons have a greater and greater certainty in their positions. The uncertainty principle tells us that as this certainty in position gets more and more accurate the uncertainty in their momentum becomes greater and greater. At a certain point the electrons are moving near the speed of light and a further reduction in size would require the addition so much energy that it is actually more favorable for the electrons to simply combine with protons to form neutrons. (Neutrons are more massive than protons and electrons combined. The extra mass comes from the energy the electrons had due to very high momentum) At this point the atom "collapses" as all of the electrons combine with the protons in the nucleus. Without an electron cloud there is nothing to resist the reduction in size, as neutrons don't repel each other like electrons do. So the material collapses in a process we call a supernova.

6. May 21, 2012

### zonde

There are no atoms in the star - there are ions and electrons i.e. plasma.

Quantum mechanics does not say that you need more energy to make sharper position measurement. But the way you say it it sounds like that.

Why wouldn't it be more favorable for electrons to be squeezed out of star's core (plasma is perfect)?

Besides protons turn into neutrons when there is too big proton density. So it does not seem so straight forward that electrons have very significant role in neutron production.

So basically the question is - why should one believe that this is fair description of reality?

7. May 22, 2012

### Drakkith

Staff Emeritus
You are correct, the star is composed of plasma. Forgive my rookie mistake, the iron lump in the example threw me off.

No, this doesn't have anything to do with measurement, but with the electrons themselves needing energy to increase their velocity as the momentum becomes more uncertain. At least that is what has been explained to me.

Not sure but I would guess that the attraction to the positive ions keeps them there.

Protons cannot turn into neutrons without an electron present, otherwise charge isn't conserved.

Not sure what you are asking.

8. May 22, 2012

### ImaLooser

The core of the star is composed of neutron superfluid, proton superconductive superfluid, and relativistic electrons. I would not expect it to behave like a plasma.

The crust is iron nuclei that are extremely dense and polymerized by the magnetic field. One expert says that below a billion K they are not ionized. Hard to believe, but it could be.

9. May 22, 2012

### zonde

And I don't see how this has anything to do with quantum mechanics.

Well maybe if we start with Pauli exclusion principle. We know that there are only so many electrons that can occupy potential wells of atom orbitals. So large number of free electrons around positively charged core should compete for potential well just like in atom. But what are the arguments why this has anything to do with uncertainty principle? And why we can use it to make quantitative assessment of effect from Pauli exclusion principle?

And even more important - Pauli exclusion principle does not say that you can sqeeze more electrons into potential well by adding energy. It is rather the other way around - more energy keeps electrons out of potential well.

Let's say we have sphere of neutral plasma. Then divide it into two overlapping spheres of negative electrons and positive ions. Now let's say we squeeze poisitively charged ion sphere into the smaller sphere. It overlaps with the part of electron sphere of the same radius but smaller charge and in the sum will have positive charge. The shell of electrons that remains outside this combined positively charged sphere will be attracted toward it and will fall down in it until positive charge is reduced to the level where electrostatic attraction is balanced by degeneracy pressure of electrons.
If electrostatic repulsion of ions at this stage is such that it is balanced by gravitational attraction then we have equilibrium state with slightly negatively charged surface and positively charged core.

And in particular:
"Nuclei which decay by positron emission may also decay by electron capture. For low-energy decays, electron capture is energetically favored by 2mec2 = 1.022 MeV, since the final state has an electron removed rather than a positron added. As the energy of the decay goes up, so does the branching ratio towards positron emission."

10. May 23, 2012

### Staff: Mentor

In potential wells, $\frac{\hbar^2}{L^2}$ with the length scale L determines the difference between energy levels. If you want to squeeze more electrons in the same volume, you have to occupy higher energy levels. If you want to reduce the volume where the electrons are in, you have to increase the energy of the electrons (as the energy levels rise). This can be viewed as a result from the uncertainty principle, where the smaller size is similar to a "measurement" of the position (you know that it is inside).

11. May 23, 2012

### zonde

Let's say that if you want to squeeze more electrons in the same potential well, you have to occupy higher energy levels.

You will have to expand on this.
Single potential well where there are some electrons and then we squeeze additional (initially free) electrons in the same well? If that is so then certainly we do not increase energy of electrons, instead we lower energy of (free) electrons but by smaller amount than that for the first electrons in the same potential well.

Or we start with many potential wells that are occupied by electrons at the base energy level? And then we squeeze potential wells together so that we actually get one big potential well that just can't be occupied with all electrons at base energy level. But in that case the major process is squeezing of potential wells and there is no reason to believe that electrons can "push" this squeezing of potential wells by being more energetic. Electrons can only fall in pre-existing potential well or they can be kicked out of potential well by adding energy.

I do not see similarity there. Probably you will have to explain a bit more about this "smaller size". Is it smaller size of potential well, smaller size of particular energy level in potential well, smaller size of electron, shorter De Broglie wavelength of electron or something else.

12. May 24, 2012

### Drakkith

Staff Emeritus
13. May 25, 2012

### Staff: Mentor

I start with a single potential well and some electrons in it and try to reduce the size of the potential well. This requires energy, which can be expressed via the degeneracy pressure.

Smaller size of the potential well.

14. May 25, 2012

### zonde

Wanted to ask what I should reread but it turned out that the page was updated few days ago.

Now there is reference to Fowler from which it is clear that degeneracy pressure quantitatively is estimated from Fermi-Dirac statistics and uncertainty principle is just red herring.

15. May 26, 2012

### zonde

I would like to bring Fermi-Dirac statistics into our discussion - see my replay to Drakkith.

I believe that Fermi-Dirac statistics describes what you get when you make phase spaces of many electrons overlap at certain place. And that momentum of electrons at that particular place should be different. But it more like describes what happens when you increase count of electrons (electron phase spaces) at that place rather than when you reduce that place (in whatever sense). But most likely Fermi-Dirac statistics can be used to analyse problem from that perspective too.

Anyway I would say that you can't confine electron phase spaces to the same place completely. Electrons has to have different momentum where they phase spaces overlap so their phase spaces are bound to extend in different directions and to different extents, I mean - they can't overlap everywhere.

16. Jun 6, 2012

### Darwin123

The neutrons in a neutron star form a Fermi gas. Because of the Pauli exclusion
principle, no two neutrons in the neutron star can have identical quantum levels.
Therefore, the bigger the density of neutrons the larger the energy of the
ground state. Therefore, the pressure of the neutron star goes up.
This is basically a wave effect more than a particle effect. The functional
dependence of pressure on density in a neutron gas has to be calculated
with quantum mechanics, on the assumption that the neutron waves satisfy
the Pauli exclusion principle. Basically, the repulsion takes place because
waves associated with neighboring particles cancel each other out. There
is no classical analog that I can give you to explain the pressure of a Fermi
gas.
There are other Fermi gases in nature that exist. The free electrons
in a metal form a Fermi gas. They also have a pressure associated with the
Pauli exclusion principle. The electrons in white dwarf matter form a Fermi gas.
Although different in degree than the neutrons in a neutron star, they obey
similar principles.
The behavior of free carriers in metals and semiconductors satisfies the laws of
a Fermi gas. The behavior of the Fermi gas has been investigated thoroughly because
of metals and semiconductors. The modern electronics industry depends on knowing
the behavior of Fermi gases. The behavior of neutrons in a neutron star isn't
qualitatively different from these gases.
The neutrons do not repel each other due to a classical force. Neutrons are
neutral, so electrostatic repulsion doesn't affect them. There is nothing to line
up their magnetic dipoles, so the magnetic fields don't directly affect the
pressure. However, one can use quantum mechanics to determine what is called
an exchange potential of the neutrons. The exchange potential acts on large
distance scales like a Newtonian force. So the short answer to what causes
the pressure in a neutron star is the exchange potential force.
The pressure of a Fermi gas is caused by the way the waves interact,
not the forces that act on it. In a neutron star, the force due to the Pauli
exclusion principle has to balance the force of gravity.
The exchange potential force on a neutron has to equal the force of gravity on
that neutron. This is a very rough, approximate way to visualize what goes on in a neutron star. In my opinion, it is better to go back to the original quantum mechanics
to estimate what is going on. If you absolutely have to use a classical analog in your
thinking, then think about the exchange potential force.
The behavior of neutrons in a neutron star satisfy Fermi statistics.
The diameter of the neutron star is determined by a balance between the
pressure of the Fermi gas of neutrons and the gravity. The more mass in the star,
the larger the star, the greater the gravitational field and the greater greater
the escape velocity at the surface of the neutron star. If the escape velocity at
the surface of the star is greater than the speed of light, "c", then the neutron
star has to become a black hole. This presents an approximate limit to the mass
of a neutron star. Supposedly, you can't have a neutron star with a mass far above
this limit.