What happens when a neutron star collapses into a black hole?

[chill_factor]

Nope. What happens is that there are more energy states available, and so the effect of having the limited number of energy states disappears.



Great!

If you can have fermions at relativistic states, then degeneracy pressure should disappear. Now figuring out how to set up that sort of experiment in the lab is something I'll leave for other people to do.



Not quite. What happens is that the energy levels change so that fermi and boltzman distributions converge to something that is different than non-relativistic gases.

http://en.wikipedia.org/wiki/Chandrasekhar_limit

ok i think i understand what you're saying. in the Boltzmann limit the energy states are far apart and degeneracy doesn't matter. in the "relativistic fermi distribution", if there's enough total system energy, the particles will be forced to be far apart which is the end result. ordinarily that results in degeneracy pressure, which is the resistance of the particles to be in such high energy states, but if there's enough gravitational potential it will happen anyways.

i don't think that process will ever be obervable in the lab except with probably diamond anvils creating metallic hydrogen, which is sort of degenerate.

 

[Bernie G]

"If I create a floor of photons, then I can't walk on it."

Then why is radiation pressure dominant in the largest stars, but radiation or relativistic pressure not at work in a stellar collapse? According to http://en.wikipedia.org/wiki/Radiation_pressure#In_stellar_interiors "In the Sun, radiation pressure is still quite small when compared to the gas pressure. In the heaviest stars, radiation pressure is the dominant pressure component.[6]"

Also see: http://hyperphysics.phy-astr.gsu.edu/hbase/starlog/staradpre.html which states "Extremely massive stars (more than approximately 40 solar masses), which are very luminous and thus have very rapid stellar winds, lose mass so rapidly due to radiation pressure that they tend to strip off their own envelopes before they can expand to become red supergiants, and thus retain extremely high surface temperatures (and blue-white color) from their main sequence time onwards. Stars cannot be more than about 120 solar masses because the outer layers would be expelled by the extreme radiation."

 

[twofish-quant]

"In the Sun, radiation pressure is still quite small when compared to the gas pressure. In the heaviest stars, radiation pressure is the dominant pressure component.[6]"

It's not so much the amount of pressure rather than how the pressure changes with density. Photons can be viewed as a gas, and gas is "soft" whereas when you have degeneracy pressure it starts acting like a solid, and solids are "hard".

Also, in the case of neutron stars, there's nothing generating radiation pressure so that all you have is degeneracy pressure.

 

[Bernie G]

This is about the amount of pressure after collapse, the point being that after collapse (rho)(c^2)/3 would exert more pressure than the degeneracy pressure of neutrons was capable of. Of course there's nothing generating radiation pressure in a neutron star; the radiation comes after collapse, and this thread is about what happens if a neutron star collapses. After there are no longer neutrons there is no neutron degeneracy pressure. But whatever the neutrons collapse to is capable of exerting pressure. Its correct for you to call it a "soft photon gas", but this soft neutron gas would crush your "hard" neutrons.

A minor detail about neutron hardness or neutrons acting like a solid (although it doesn't matter after neutron collapse): as neutrons near collapse their shape is apparently no longer round as the space between them fills up. Neutrons just ain't hard enough at collapse. But I am always amazed at the strength of neutrons up to collapse; the pressure numbers are staggering almost beyond belief.

 

[Bernie G]

My mistake, a typo. The above should read: "Its correct for you to call it a "soft photon gas", but this soft photon gas would crush your "hard" neutrons."

 

[twofish-quant]

This is about the amount of pressure after collapse, the point being that after collapse (rho)(c^2)/3 would exert more pressure than the degeneracy pressure of neutrons was capable of.

It turns out that the total amount of pressure doesn't matter. What matters are pressure differences. If you inflate a balloon to 100 psi, but it's in a 100 psi environment, nothing happens. Now if you inflate the balloon to 0.5 psi but put it into a vacuum, it blows up.

So pressure is irrelevant. What matter is the difference in pressure, and how pressure changes when you change the physical situation. Take that balloon. If you blow it up to 1000 psi and squeeze it. If the pressure in that balloon stays 1000 psi, then it will not react when you squeeze it.

But whatever the neutrons collapse to is capable of exerting pressure.

No. If the neutrons break up into smaller particles, that the number of states and that reduces pressure. Now if the neutrons combined and formed *bigger* particles, then you'd reduce the number of states and that would stop the collapse until you add even more mass. However, we haven't seen any particles that are larger than the one's we know, and there are good reasons for thinking that even if those particles did exist they would decay into smaller particles.

A minor detail about neutron hardness or neutrons acting like a solid (although it doesn't matter after neutron collapse): as neutrons near collapse their shape is apparently no longer round as the space between them fills up. Neutrons just ain't hard enough at collapse. But I am always amazed at the strength of neutrons up to collapse; the pressure numbers are staggering almost beyond belief.

One thing about astrophysics is that you stop being impressed by large numbers. What's important is the order of magnitude, and when I think about neutron stars, I think of the number 15. The density at the core of the neutron star is 10^15 g/cm^3. So when I think about astrophysical objects, I just look at the order of magnitude. My brain can't handle thinking about 10^51 or 10^54, but it can handle 51 and 54.

 

[chill_factor]

It turns out that the total amount of pressure doesn't matter. What matters are pressure differences. If you inflate a balloon to 100 psi, but it's in a 100 psi environment, nothing happens. Now if you inflate the balloon to 0.5 psi but put it into a vacuum, it blows up.

So pressure is irrelevant. What matter is the difference in pressure, and how pressure changes when you change the physical situation. Take that balloon. If you blow it up to 1000 psi and squeeze it. If the pressure in that balloon stays 1000 psi, then it will not react when you squeeze it.
No. If the neutrons break up into smaller particles, that the number of states and that reduces pressure. Now if the neutrons combined and formed *bigger* particles, then you'd reduce the number of states and that would stop the collapse until you add even more mass. However, we haven't seen any particles that are larger than the one's we know, and there are good reasons for thinking that even if those particles did exist they would decay into smaller particles.
One thing about astrophysics is that you stop being impressed by large numbers. What's important is the order of magnitude, and when I think about neutron stars, I think of the number 15. The density at the core of the neutron star is 10^15 g/cm^3. So when I think about astrophysical objects, I just look at the order of magnitude. My brain can't handle thinking about 10^51 or 10^54, but it can handle 51 and 54.

i hate to be one of those annoying nitpickers but whether the balloon blows up or not depends on structural strength of the plastic.

also, do neutrons actually repel each other? not through degeneracy pressure, but through other means?

 

[Chronos]

Neutrons are electrically neutral so aside from Pauli exclusion there is no known repulsive force between neutrons.

 

[Bernie G]

This thread is about what happens if a neutron star collapses. Are you saying there is no relativistic pressure of about (rho)(c^2)/3 after neutron collapse?

 

[Bernie G]

Neutron star core pressure is about equivalent to supporting the entire weight of the sun on one square inch of the Earth's surface. Or setting off over a billion H-bombs and containing it in 1 cc. That impresses me. A neutron is a mighty strong thing.

When a neutron collapses it does not turn into nothing, it turns into something and there is conservation of mass-energy. Logically that something is quark type matter and radiation.

 

[twofish-quant]

This thread is about what happens if a neutron star collapses. Are you saying there is no relativistic pressure of about (rho)(c^2)/3 after neutron collapse?

It's actually P = \rho ^(4/3). What you'll find is that when P = \rho ^(4/3) there are no stable solutions so that the object must collapse. What matters is not whether or not there is pressure but how the pressure changes with density.

 

[twofish-quant]

Neutron star core pressure is about equivalent to supporting the entire weight of the sun on one square inch of the Earth's surface. Or setting off over a billion H-bombs and containing it in 1 cc.

So what?

The problem is that you can't do arithmetic with it. Imagine a billion H-bombs going off at the same time. Now imagine a million. Now imagine a trillion. The pictures look the same, but a trillion H-bombs is a very different situation than a million H-bombs.

It helps a lot *not* to get impressed by large numbers, and then think of 6, 9, and 12.

When a neutron collapses it does not turn into nothing, it turns into something and there is conservation of mass-energy. Logically that something is quark type matter and radiation.

Except that because quarks are lighter than neutrons they are even more unstable to collapse than raw neutrons.

The problem is that until relativity breaks down, there is nothing that can stop the collapse.

 

[Bernie G]

we're making progress. A billion or a trillion H-bombs contained in 1 cc is impressive to me, so is the weight of the sun on 1 square inch of the earth.

When neutrons collapse they should convert to mostly radiation and only a small amount of quark matter, so the net pressure should be pretty close to (rho)(c^2)/3. If the quarks ultimately collapse to radiation that's OK too. Where do you get the coefficient (4/3) from? My guess it is from a pressure formula that is not applicable here.

What do you mean by until relativity breaks down? Effects at a neutron stars surface or core, or effects at a black holes surface or core?

 

[twofish-quant]

When neutrons collapse they should convert to mostly radiation and only a small amount of quark matter, so the net pressure should be pretty close to (rho)(c^2)/3.

That's not possible without breaking the standard model. It doesn't conserve baryon number. Neutrons have to break down to quarks.

Also when the neutron star starts to collapse into a black hole, the densities aren't extraordinarily high.

What do you mean by until relativity breaks down? Effects at a neutron stars surface or core, or effects at a black holes surface or core?

At black hole singularity, relativity has to break done.

 

[Bernie G]

Well, super collider experiments show a smashed nucleus breaks down to mostly radiation plus 3 quarks and a little bit of other small exotic particles. I always considered densities rather large in a neutron star. At a black hole singularity, relativity and everything breaks down; that's why I don't believe in a point singularity. On the other hand if a 2 solar mass neutron star of 12 km radius was to hypothetically entirely collapse to a radiation/quark star of 4 km radius , the average density would go up by a factor of 27. Thats large density too. Larger radiation/quark stars would have less density and core pressure than smaller radiation/quark stars.

 

[Drakkith]

Well, super collider experiments show a smashed nucleus breaks down to mostly radiation plus 3 quarks and a little bit of other small exotic particles.

Particles collide at enormous energies inside of colliders, and it is that extra energy that is converted to radiation and other particles. I'm not sure if you would say that a nucleus "breaks down" under enormous pressures or not. It seems to me that as the pressure increases the atom is converted into a higher energy state as the electrons and protons are transformed into neutrons. Beyond that, if the neutrons form quark matter, then wouldn't they be in an even higher energy state, and not release particles or radiation? Looks like the nucleons "merge" together in a quark soup or something.

 

[Bernie G]

"Beyond that, if the neutrons form quark matter, then wouldn't they be in an even higher energy state, and not release particles or radiation? Looks like the nucleons "merge" together in a quark soup or something."

I could go along with that. I want to calculate the radius of a star of relativistic material using the viral theorem. Do you have a suggested pressure formula as a function of density for this quark soup?

 

[Bernie G]

I think the pressure of this quark soup would also be (rho)(c^2)/3. See:
http://www.strw.leidenuniv.nl/~nefs/FermionGas.pdf

 

[Bernie G]

Pressure of (rho)(c^2)/3 is enormous and I think this would prevent collapse of the quark star, although the radius would be smaller than the Schwarzschild radius.

 

[Bernie G]

I think this star gravitationally is Newtonian, except for an atmosphere where effective gravity shifts to relativistic.

(rho)(c^2)/3 indicates a supporting energy of M(c^2)/3 , which indicates gravitational potential energy of 2M(c^2)/3 .

 

[Orion1]

Neutron stars if not anything else, are composed mostly of fundamental particles called quarks and gluons in the form of degenerate neutrons. And according to Loop Quantum Gravity (LQG), when a neutron star collapses, the matter inside a black hole becomes a relativistic quark-gluon plasma, which gravitationally collapses and rebounds in an oscillation on the order of a Planck singularity. In this sense, black hole cores are the ultimate atom smashers in the Universe.

The pressures and densities cited by this thread are used to describe the pressure and densities inside neutron stars, also known as a 'Fermion gas', which depending on the mathematical model of their Equation of State, can be relativistic or non-relativistic, and inside black holes, with the consideration of relativistic quark-gluon plasma pressures and densities, curve into relativistic Planck pressure and Planck density.

The pressures and densities inside the core of a black hole are on the order of Planck pressure and Planck density.

Reference:
Loop quantum gravity - Wikipedia

 

[Bernie G]

The Wiki article on Loop quantum gravity is vague. Can you give an equation for the Plank pressure or density in a black hole?

I think a gravitationally contained quark matter star would experience a pressure of about (rho)(c^2)/3 and have a radius between R = (1.2GM)/(c^2) and R = (1.5GM)/(c^2).

 

[Bernie G]

I did the calculation for gravitational potential energy using a density profile of
[1 - (r^2)/(R^2)] and came up with 1.1 G (M^2)/R. Not sure if this is correct. Does anybody have the equation for gravitational potential energy using a density profile of
[1 - (r^2)/(R^2)] ?

 

[Bernie G]

1.1 G (M^2)/R seems too high for the Newtonian gravitational energy based on [1 - (r^2)/(R^2)] ... I think at most it should be 1.0 G (M^2)/R ... I'll go over it again and will post it here, but won't get to it for several days.