# Neutron degeneracy pressure and electron degeneracy pressure

1. Jan 19, 2008

### CJames

The words neutron degeneracy pressure and electron degeneracy pressure are thrown around a lot when talk of white dwarfs, neutron stars, and black holes comes up. Despite this, I can't find a quantitative description of these critical pressures ANYWHERE. It is almost always described in terms of solar masses or density, never pressure.

How much pressure is required to keep matter degenerate? I wouldn't think it would be this hard to find the answer.

I know you could find the answer using the density and gravitational pull, but I doubt newton's law of gravitation is still valid under these conditions, so I don't know how to calculate the gravitational pull either.

Can anybody help?

2. Jan 19, 2008

### jonmtkisco

Have you looked at the Wikipedia articles relating to this topic, such as "Degenerate Matter", "Compact Star", etc? The articles references other sources of information on this subject. I believe that all the math has been worked out for various categories of stellar collapse and supernovae.

Jon

3. Jan 19, 2008

### stevebd1

According to a document I found regarding white dwarfs and neutron stars, in respect of the neutron star-

"If no phase transitions occur the EoS will approach $\text{p} = \rho$/3" (which is the equation of state for ultrarelativistic matter)

Based on this, if a neutron star has a density of $$10^{9}\,\text{tonnes/cm}^{3}$$, this would provide an energy density of $$9\,\text{x}\,10^{34}\,\text{joules/m}^{3}$$, and because the units for energy density and pressure are synonymous, the pressure would therefore approach $$3\,\text{x}\,10^{34}\,\text{N/m}^{2}$$

Though this isn't conclusive, it provides a marker, or a ball park figure.

Link to the document dated July 2007 (statement on page 20)-
http://www.tat.physik.uni-tuebingen.de/~kokkotas/NSBHGW_files/GTR_course4.pdf [Broken]

regards
Steve

Last edited by a moderator: May 3, 2017
4. Jan 19, 2008

### CJames

Thank you jonmtkisco and stevebd1.

For anybody interested, the reason I asked was because I was wondering what theoretical limits would need to be reached to produce color superconducting quark matter in the lab. From wikipedia (http://en.wikipedia.org/wiki/Neutron_star) I have the maximum density of a neutron star at 10^18 Kg/m^3, which I would conjecture to be the density at which it collapses into quark matter. That comes to 10^9 tonnes/cm^3.

Thanks to jonmtkisco I found this link (http://epubl.luth.se/1402-1757/2005/25/LTU-LIC-0525-SE.pdf). In the first paper of the appendix, figure 7, color superconducting quark (CFL) matter appears to have a minimum pressure of 100 MeV/fm^3, which, if the energy is converted to mass, works out to 1.78*10^8 tonnes/cm^3.

If steve is right, then this pressure is equal to the density.

If not, if we take quark matter to have a density of 10^9 tonnes/cm^3 as given above, then theoretically you could reach the required pressure by converting (1.78/10=) 18 percent of the mass into energy and perfectly applying pressure using that energy.

5. Jan 19, 2008

### stevebd1

I noticed I made a typing error in my original post, the density for the neutron star should have read 10^9 tonnes/cm^3 (not m^3) which I have now corrected, the other figures stay as they are. Intrestingly, when I express the pressure, 3x10^34 N/m^2, from my original post as a density I get 3.33x10^8 tonnes/cm^3, which is within the same order as the figure you came up with. For a neutron star of 10^9 tonnes/cm^3, your figure of 1.78x10^8 tonnes/cm^3 for pressure implies an equation of state of around $\text{p} = \rho$/6 compared to the $\text{p} = \rho$/3 I worked with.

regards
Steve

Last edited: Jan 19, 2008
6. Jan 19, 2008

### CJames

It's always a good sign when you find the same order of magnitude! I haven't looked at your link but the quote does say "If no phase transitions occur." Since we're actually talking about the border between degenerate neutrons and degenerate quarks, that would be a phase change, which may explain the difference in equation of state?

7. Jan 22, 2008

### stevebd1

I did a bit of research and found the info below which seems to imply that the equation of state for neutron matter changes as the density increases. (I'd be interested to know how you converted the units from MeV/fm^3 to tonnes/cm^3, CJames).

Variation in equation of state (EoS) and relevant pressures in a typical 2.5 solar mass neutron star.
(Note- This is based on a pure neutron star and doesn’t include for the number of protons and electrons in the neutron liquid or the potential collapse of neutron matter into quark matter at about 10^9 tonnes/cm^3, what it provides is a simple model of how the equation of state (and pressure) varies through the various layers of the neutron star towards the core).

Neutron star approx. 24 km in diameter-

Density
(source- http://var.astro.cz/brno/perseus4_2002_clanek2.pdf pages 2-3)

Atmosphere $$\rightarrow$$ outer crust (12.00 km to 11.90 km approx.)
> 10^6 g/cm^3 (1 tonne/cm^3)
Plasma of H, He, C and Fe with a large density gradient.

Outer crust $$\rightarrow$$ inner crust (11.90 km to 11.70 km approx.)
10^6 g/cm^3 to 4x10^11 g/cm^3 (1 tonne/cm^3 to 4x10^4 tonnes/cm^3)
Solid region of heavy nuclei (Fe56>Ni62>Kr118) and electrons.

Inner crust $$\rightarrow$$ neutron liquid (11.70 km to 10.30 km approx.)
4x10^11 g/cm^3 to 2x10^14 g/cm^3 (4x10^4 tonnes/cm^3 to 0.2x10^9 tonnes/cm^3)
Neutron ‘drip’ takes place at about 3x10^5 tonnes/cm^3, a phase where free neutrons begin to leak out of the nuclei
Neutron-rich nuclei with a superfluid of neutrons and ‘normal’ electrons.

Neutron liquid $$\rightarrow$$ core (10.30 km to 8.00 km approx.)
2x10^14 g/cm^3 to 8x10^14 g/cm^3 (0.2x10^9 tonnes/cm^3 to 0.8x10^9 tonnes/cm^3)
As a comparision, atomic nuclei has density of about 0.3x10^9 tonnes/cm^3
Mainly a superfluid of neutrons with a small concentration of superfluid protons and ‘normal’ electrons.

Core (8.00 km to zero)
8x10^14 g/cm^3 to a possible 20x10^14 g/cm^3* (0.8x10^9 tonnes/cm^3 to 2.0x10^9 tonnes/cm^3*)
Behaviour of nuclear interactions not known
Meson (kaon or pion) condensate? Transition to a neutron solid or quark matter? Hyperons? quark-gluon plasma? Bose-Einstein condensates? Higher mass baryons?

*A critical density of 20x10^14 g/cm^3 (2.0x10^9 tonnes/cm^3) appears to be the norm for a neutron star with a 12 km radius and a mass of 2.48 solar masses which can increase to 40x10^14 g/cm^3 (4 billion tonnes per cm^3) for neutron stars with a radius of 9 km and a mass of 1.46 solar masses.

Equation of State and pressure
Equation of state for pure neutron matter (the dependence of pressure versus energy density.
(source- http://www.rpi.edu/dept/phys/Courses/Astronomy/NeutStarsAJP.pdf Fig. 11, page 903)

100 Mev/fm^3 = 0.178x10^9 tonnes/cm^3

Pressure / energy density (MeV/fm^3) = pressure / density (10^9 tonnes/cm^3) = EoS $$\rightarrow$$ pressure (10^34 N/m^2)

20 / 200 = 0.035 / 0.356 = 1 / 10 $$\rightarrow$$ 0.315x10^34 N/m^2

100 / 400 = 0.178 / 0.712 = 1 / 4 $$\rightarrow$$ 1.602

250 / 600 = 0.445 / 1.068 = 1 / 2.4 $$\rightarrow$$ 4.005

450 / 800 = 0.801 / 1.424 = 1 / 1.77 $$\rightarrow$$ 7.209

650 / 900 = 1.157 / 1.602 = 1 / 1.54 $$\rightarrow$$ 10.413

1050 / 1200 = 1.869 / 2.136 = 1 /1.14 $$\rightarrow$$ 16.821

Putting these figures into the cross section at the top of the page, an EoS of 1/10 (pressure- 0.315x10^34 N/m^2) occurs just below the inner crust within the neutron liquid (about 2 km down). This increases to 1/4 (1.602x10^34 N/m^2) just outside the core (about 4 km down) then increases to 1/2.4 (4.005 x10^34 N/m^2) at about 5 km down where the density is estimated to be 1 billion tonnes per cm^3, eventually reaching 1/1.14 (16.821x10^34 N/m^2) somewhere within the core where the critical density is estimated to be 2 billion tonnes/cm^3 approx.

regards
Steve

Last edited by a moderator: May 3, 2017
8. Jan 25, 2008

### CJames

Thanks steve, I couldn't for the life of me find info like that anywhere. Sorry I didn't see this sooner.

It's not really just a straightforward conversion, since we are also converting mass into energy the units aren't the same but...

1 fm = 10^-15 m
1 cm = 10^-2 m

so

100 MeV/fm^3 = 100 * (10^13)^3 MeV/cm^3 = 10^41 MeV/cm^3

and from http://en.wikipedia.org/wiki/MeV

1 MeV/c^2 = 1.783 * 10^-30 Kg

so to express 10^41 MeV in Kg we multiply by 1.783 * 10^-30 and get

1.783 * 10^11 Kg/cm^3
= 1.783 * 10^8 tonnes/cm^3