# Electron degeneracy pressure in QM

1. Feb 9, 2010

### Naty1

In BLACK HOLES AND TIME WARPS by Kip Thorne there is a fascinating discussion about how electron degeneracy pressure fights gravity in dense stars, beginning on page 146. He says the following (abbreviated excerpts) :

"..Quantum mechanics insists that when already dense matter is compressed a bit....the energy of the degeneracy motion must increase....If the electon is already moving at close to the speed of light...the additional energy goes into inertia...these two different fates of energy (added speed vs added inertia) produce different increases in the electron's pressure and thus different resistances to compression."

In effect this means that a star's resistance to gravitational collapse decreases with increasing density, not what intuition might suggest.

[Without this difference, if I understood the discussion, we'd not have any black holes.]

My question is why this difference in pressure happens...do we have a physical interpretation....why electron velocity has a different effect on pressure than inertial change??

Apparently the general relationship of white dwarf matter density is governed by the Stoner-Anderson equation of state which shows that..."matters resistance to compression decreases smoothly from 5/3 to 4/3...as electrons speed into the relativistic domain.."

and within the equation might be the answer to my question above....I did not find any Wikipedia entry for Stoner-Anderson...

Thanks

Last edited: Feb 10, 2010
2. Feb 9, 2010

### Matterwave

Relativistic electrons cannot exert the same degeneracy pressure as non-relativistic electrons. There are several derivations of this using either E=p^2/2m+mc^2 or E^2=p^2c^2+m^2c^4.
The decreased ability to keep up degeneracy pressure comes out in the math, but I've never really heard of a conceptual answer...

3. Feb 9, 2010

### bcrowell

Staff Emeritus
Here is my own explanation of what Thorne is talking about (see subsection 4.4.3): http://www.lightandmatter.com/html_books/genrel/ch04/ch04.html#Section4.4 [Broken]

Essentially the idea is that one factor in determining the pressure exerted by a gas is the time it takes a particle to rattle back and forth in the space it's confined in. If this time is short, then there are more impacts, and more pressure. For a degenerate gas, energy gets higher as you increase the density. (This is just the Heisenberg uncertainty principle.) Once the speeds become relativistic, you can no longer decrease this time between impacts by adding energy. The dependence of pressure on density therefore becomes less drastic for a highly relativistic degenerate gas ($P \propto \rho^{4/3}$) than for a nonrelativistic one ($P \propto \rho^{5/3}$).

Last edited by a moderator: May 4, 2017
4. Feb 10, 2010

### Ich

bcrowell,

I did a quick web search yesterday to refresh my memory, but your explanation is far better than everything else I found.

5. Feb 10, 2010

### Altabeh

But to my own knowledge, the more energy you add to any material system, the faster its particles oscillate so the more impacts will happen. This has nothing to do with relativistic speeds as far as the total energy of the system amounts to a certain binding energy.

Maybe relativity here makes the particles go beyond this energy and then impacts happening in the system decrease because particles are no longer bound to that system due to its collapse!! Can you discuss this?

AB

6. Feb 10, 2010

### Ich

This is not about oscillations, and not even about impacts. It's about (symmetric) momentum transport across a surface, wich is proportional to the number of particles crossing per time unit, which in turn is proportional to the particle velocity, which is bounded.

7. Feb 10, 2010

### Altabeh

I just shared my own knowledge of how the number of impacts in a mateial system varies by adding energy into it which was seemed to be converse from bcrowell's viewpoint! Even if we assume that you've not misunderstood me, the number of particles crossing through the surface (of material system) per time unit does strongly depend on the energy stored in the system. The velocity of particle is bounded but if the energy injected into the system gets higher, this velocity is no longer limited unless we say that conservation law of energy and momentum do not hold!

As stars generates energy by nuclear fusion, there must be heat produced from the reactions of fusion which in turn prevents gravitational collapse of the star. But as the heat carries energy, so the energy of system increases, giving rise to the increase of the number of particles crossing through the surface of star. With this happening, the radius of star gets smaller untill gravitational force can take the control of particles going out of star which in turn creates a black hole or a neutron star if the initial mass of star is below the Chandrasekhar limit (CL). Otherwise the star becomes a white dwarf with a final mass again below CL. But I'm asking, assuming that the mass of star is less than CL, why the storage of energy in a very huge amount produced through nuclear fusion does not lead to the collapse of star, say, like a car blast, and doesn't this mean that energy is not conserved inside a star which loses mass while the particles velocity remains bounded? Is this even meaningful?

AB

8. Feb 10, 2010

### Ich

ok, either I completely misunderstand what you're saying, or you're starting from a number of major misconceptions about the topic of this thread (degeneracy pressure) and energy conservation. Maybe somebody else, who understands better what you're saying, wants to answer.

9. Feb 10, 2010

### Naty1

bcrowell:
Can we also attribute this to the Pauli exclusion principle ?? In your write up you discuss Pauli at norelativistic speeds...and I assume it continues at relativistic speeds.

Thats a real nice way to think about this. Also, I am guessing that all this is where mostly quantum mechanics applies and einstein's GR gets real fuzzy...His pressure,momentum,energy relationships would likely be different at this scale,right??
Do

Also, I read a bit further and came across electron capture!!! Holy cow it's been along, long time since I studied a bit of that in the dark ages and had forgotten it entirely...
And I see you have some tensor discussion prior to 4.4.3..I'll check that out because at other locations when attempting to learn tensors on my own, my attention span proved shorter than the concepts!!!
Thanks ...

10. Feb 10, 2010

### Naty1

Post #7 makes virtually no sense to me..so if it does to others, an interpretation would be helpful....

edit: I just saw Ichs comments..I agree...post #7 seems riddled with misconceptions.

separately,
Does relativistic quantum mechanics (or any other kind) tell us anything heat production at these ultra compact densities?

Seems like even absent thermonuclear reactions there would be substantial heat produced yet I have not seen it mentioned. So I'm guessing that as the electrons begin to act as a degenerate gas and their collapse slows or halts, whatever heat is generated just dissipates. If the star is small enough it just sits cold and dark; otherwise it collapses into a black hole...and as far as I know they are so cold its theorized all of them are still absorbing radiation from "our" space...they are colder the the CMBR.

11. Feb 10, 2010

### Altabeh

Could someone specify these misconceptions so I can be able to correct them? I'm not basically an astrophysicist, and I just asked for more information upon the main issue and the one brought up by bcrowell through this sentence:

This comes right after If this time is short, then there are more impacts meaning that if time doesn't get shorter, then there are less impacts happening in the system which sounds logical but is in contrast to the above quote. Because adding more energy into system would lead to more impacts, I think, if the system is not going to collapse otherwise the energy would not be conserved!

AB

12. Feb 10, 2010

### Ich

Whats hardest not to interpret as a misconception is this:
The energy of the system doesn't increase. Nuclear reactions change mass into heat, the sum is the same - except that radiation and solar wind carry some energy away.
Then, pressure is not the number of particles passing through a star's surface. You pick any surface within the star, count how much momentum flows from one side to the other and back, and that's the definition of pressure at said surface.
With a given particle density, an increase in momentum p changes two things:
1) the momentum per particle increases ~p (obviously)
2) the velocity increases ~p for a slow particle, but stays ~=c for very fast particles

So when all particles travel at nearly the speed of light, you can't make them faster. Thus, the number of particles crossing through the surface does no longer increase with momentum.

Then, we're talking about a degenerate gas. That's where a few million K count as "cold".
Heat is totally neglected, the particle momentum is dictated only by the Heisenberg uncertainty principle, as ~number density n³.
So you have pressure = n*(momentum per particle)*(particle speed), which changes as
dP~ dn^(1+1/3+1/3) for slow particles, and
dP~dn^(1+1/3+0) for relativistic particles.

13. Feb 10, 2010

### Altabeh

Now I got it! You know, as this first was brought up in General Relativity forum, as someone looking strictly from the standpoint of conservation law of energy, I thought by "adding" bcrowell meant some sort of external energy (for instance, infalling particles/bodies hitting the star) would be injected into the system and this energy happens to expose itself as heat so if this does not affect system then...!

Anyways, thanks for clarification and one more question: How does one really get dP~ dn^(1+1/3+1/3) or dP~dn^(1+1/3+0)? Is this based upon experiments or it just has some mathematical reasoning behind it?

AB

14. Feb 10, 2010

### Ich

You assume that the particles' momentum is defined by the uncertainty principle.
With V=Volume per particle = 1/n, n=1/V=number density=particles per Volume, p=momentum per particle you get
hbar/2 = const. = x*p = V^(1/3)*p ~ n^(-1/3)*p, so
p~n^(1/3),
n~p^3
Then, as explained before, pressure
P ~ n*(momentum per particle)*(particle velocity) =
= p^3*p*p = n^(5/3) if velocity ~ p
= p^3*p*1 = n^(4/3) if velocity = const. = c

15. Feb 10, 2010

### twofish-quant

Just to amplify this a bit, the reason you can't decrease the time between impacts is that no matter how much energy you add to the system, the particles are not going to move back and forth at faster than the speed of light.

This has a very important implication. You can change the particles that are rattling back and forth, but no matter what the particles are, they can't move back and forth at faster than the speed of light, and so the time between impacts is always going to be bounded. Which means that no matter what the particles are, eventually if you pile on enough matter, it's going to collapse.

16. Feb 10, 2010

### twofish-quant

There is a limit to how fast particles can oscillate. They can't move faster than light, which is where relativity comes in.

17. Feb 10, 2010

### Ich

1000 posts in 125 days and science advisor. Congratulations.

18. Feb 10, 2010

### Altabeh

This was my point, too!

And thank you Ich for your explanation. But one bothering thing is that why would one make use of the Uncertainty inequality as an equality? I think such usages of UP are extensively approximate and won't give much reliable estimations of real numbers!

AB

19. Feb 10, 2010

### Naty1

Originally Posted by Altabeh
.

Did you read the reference given in post #3??.....that gives some understanding of why this effect decreases as any material is compressed to such high densities that relativistic electrons can no longer follow their macroscopic behavior...

But this quantum behavior is by no means at all obvious...

20. Feb 10, 2010

### Altabeh

I also send my congrats to twofish-quant for his useful notes! I don't know what field he is active in, but hope for him to stay with us here for as long as he can!

AB