I How Much Pressure To Cause Matter To Degenerate?

[Hornbein]

TL;DR Summary

How much pressure does there have to be to cause degeneration of matter into white dwarf and neutron star material?

How much pressure does there have to be to cause degeneration of matter into white dwarf or neutron star material? I tried to find this online and all I could get were the Chandrashekar limits on mass of heavenly bodies. Related, but not the same thing. Presumably once this pressure is reached at the center of a heavenly body then the whole thing collapses. That seems right but it's hard to be sure about such things.

A very rough estimate is good enough. This is just a casual question for the sake of curiosity.

 

[snorkack]

How much pressure does there have to be to cause degeneration of matter into white dwarf or neutron star material?

I don´t quite get a distinction between "degenerate" and "condensed" matter.

I tried to find this online and all I could get were the Chandrashekar limits on mass of heavenly bodies. Related, but not the same thing. Presumably once this pressure is reached at the center of a heavenly body then the whole thing collapses. That seems right but it's hard to be sure about such things.

I believe the opposite. The whole thing becomes stable to collapse - and unstable to runaway heating or cooling.

 

[PeterDonis]

How much pressure does there have to be to cause degeneration of matter into white dwarf and neutron star material?

The criterion is not based on pressure. It's usually stated in terms of density; see this post:

https://www.physicsforums.com/threa...n-shapiro-teukolsky-1983.1048744/post-6859842

As that post notes, the specific density criterion will depend on chemical composition. What pressure is required to compress the material to that density will also depend on chemical composition.

Presumably once this pressure is reached at the center of a heavenly body then the whole thing collapses.

No, degeneracy by itself is not enough to make an object collapse.

 

[Hornbein]

The criterion is not based on pressure. It's usually stated in terms of density; see this post:

https://www.physicsforums.com/threa...n-shapiro-teukolsky-1983.1048744/post-6859842

As that post notes, the specific density criterion will depend on chemical composition. What pressure is required to compress the material to that density will also depend on chemical composition.


No, degeneracy by itself is not enough to make an object collapse.

Aha, so the pressure isn't directly relevant. Well, as long as I've done the work here's my estimate of the pressure at the center of a white dwarf at the Chandrashekar limit under the (false) assumption of uniform density. I hope I haven't made a mistake with the units. The pressure is 10^19 Pascals.

This could be good enough for the crude purpose I'm pursuing. That is, if the pressure is that high then collapse into neutron matter can't be far away, eh?

Chandrashekar white dwarf radius : 10^6 kilometers

Mass : 1.44 solar masses

Density : 1 x 10^9 kilograms per cubic meter

G = 6.67 x 10^-11 N m^2/kg^2

ρ = 10^9 kg/m^3

R = 10^6km

P = (2/3) * π * G * ρ² * R²

P = (2/3) * π * 6.67 x 10^-11 N m^2/kg^2 * 10^9kg²/m^6 * 10^9m ²

P = (2/3) * π * 6.67 x 10^-2 N 10^9/m ² * 10^9m ²

P = (2/3) * π * 6.67 x 10^-2 N 10^18/m ²

Estimate

P = 10^19N / m^2

That's ten quintillion newtons per square meter or the same number of Pascals.

 

[PeterDonis]

if the pressure is that high then collapse into neutron matter can't be far away, eh?

Pressure isn't directly relevant for that either; density is. The onset of neutronization, which is what triggers the collapse from a white dwarf to a neutron star, happens, roughly speaking, when the density gets high enough for the electron and proton wave functions to overlap significantly, so that the weak interactions that drive neutronization get driven towards the neutron endpoint.

The definitive reference is Shapiro & Teukolsky. Some discussion of the relevant criteria can be found in this Insights article:

https://www.physicsforums.com/insights/why-there-are-maximum-mass-limits-for-compact-objects/

 

[Hornbein]

At any rate, in the weird project I am pursuing if my math is correct then to my great surprise the pressure in the center would be at least ten quadrillion times less than in a white dwarf, a mere thousand Pascals. Collapse would not be immanent. I must have done something wrong, but the margin for error is quite large.

 

[snorkack]

The definitive reference is Shapiro & Teukolsky. Some discussion of the relevant criteria can be found in this Insights article:

https://www.physicsforums.com/insights/why-there-are-maximum-mass-limits-for-compact-objects/

But while the Insights article takes some properties of "degenerate" matter for granted, it does not say what defines a matter as "degenerate" - as in, what is sufficient set of properties to make matter degenerate?

 

[PeterDonis]

while the Insights article takes some properties of "degenerate" matter for granted, it does not say what defines a matter as "degenerate" - as in, what is sufficient set of properties to make matter degenerate?

Roughly speaking, "degenerate" means that you have a lump of fermions in which the temperature is much less than the Fermi energy.

Note that this is not a binary property; there is a continuum between one idealized extreme in which the Fermi energy is zero (fermions at infinite separation) and the temperature is finite, and the other idealized extreme in which the temperature is zero (no thermal motion at all) and the Fermi energy is finite (fermions at finite separation). The latter idealization is often used in models of "degenerate" matter, but of course no real object will have exactly zero temperature, so there is always some thermal contribution. But as long as the thermal contribution is negligible compared to the Fermi energy, "degenerate" is a reasonable description. This is true of white dwarfs and neutron stars, for example.

Shapiro & Teukolsky go into much more detail.

 

[snorkack]

Roughly speaking, "degenerate" means that you have a lump of fermions in which the temperature is much less than the Fermi energy.

Yes, and I see that the definition of Fermi energy is grossly problematic.

 

[PeterDonis]

the definition of Fermi energy is grossly problematic

How so?

 

[snorkack]

https://en.wikipedia.org/wiki/Fermi_energy

The Fermi energy is a concept in quantum mechanics usually referring to the energy difference between the highest and lowest occupied single-particle states in a quantum system of non-interacting fermions at absolute zero temperature. In a Fermi gas, the lowest occupied state is taken to have zero kinetic energy, whereas in a metal, the lowest occupied state is typically taken to mean the bottom of the conduction band.

Note how the second sentence backtracks on the first, because all metals (certainly at any observable pressure) possess occupied states far below the conduction band!

 

[PeterDonis]

https://en.wikipedia.org/wiki/Fermi_energy

Note how the second sentence backtracks on the first, because all metals (certainly at any observable pressure) possess occupied states far below the conduction band!

Note how you're treating Wikipedia as thought it were a good reference that explains everything properly. It isn't.

The reason the occupied states below the conduction band are ignored in models of metals is that there are no transitions to or from those states. So there is never any change in their occupation numbers. So the model can just ignore them, and only has to include the states that can have changes in their occupation numbers.

But that does not mean that the states at the bottom of the conduction band are treated as having zero kinetic energy. They're not. The Fermi energy in the metal is still the total kinetic energy of the highest occupied state; it's not the difference between that energy and the energy of the bottom of the conduction band.

So there's no problem at all.