# Ideal gas laws in cosmology usage

1. Feb 17, 2014

### Mordred

The ideal gas has the following requirements.

1) there are no intermolecular forces between the molecules.
2) the volume of the gas is negligible compared to the volume of the container they occupy.
3) the interactions between the particles and the container is perfectly elastic (total kinetic energy is conserved).

equation of state for an ideal gas is given by pV=nRT

given that the compression factor is the ratio of the molar volume of the real gas Vm to the molar volume of the ideal gas Vmo at the same pressure and temperature.
Z=compression factor

Z=Vm\Vmo

where Z=1, ideal gas behavior
Z<1, attractive forces dominate volume of gas is less than an ideal gas
Z>1, repulsive forces dominate volume of gas is greater than an ideal gas.

Thus far I understand the above, however the above define the gas in a container, I know that the ideal gas laws have been applied in Cosmology applications but I am unclear of how they apply the container portion described above.
Does anyone have any good articles that cover ideal gas law applications specifically in Cosmology usage? Also any material covering rate of diffusion of an ideal gas in an open system would also be handy.

edit:also articles detailing Gibb's law would also help, as I already understand how the FLRW equations of state can be used to describe the thermodynamics of the universe. However I would like to correlate that to Gibbs law described below

Tds=d(pV)+PdV

Last edited: Feb 17, 2014
2. Feb 17, 2014

### Mordred

found what I was after, turns out its already been done lol.

http://arxiv.org/abs/0708.2962

however I came across this reference

Neugebauer-Meinel dust, thus far I've only found articles that use the Neugebauer-Meinel dust metrics. I would like to find articles covering how the metrics was developed and its implications

Last edited: Feb 17, 2014
3. Apr 6, 2014

### Mordred

I could still use some advice on the energy density to pressure relation defined by the equations of state in cosmology. I would like to eventually write an article on the ideal gas laws in cosmology. However the part I am stumped on is how the volume is defined.

As near as I can determine the volume is defined as the region in thermal equilibrium. With the boundary (container) being the seperation point where the same metrics can no longer apply.

Would that be accurate or is there a better definement for rules 2 and 3 of the ideal gas laws posted in the OP.

4. Apr 6, 2014

### Ken G

I'm not clear why you need to define a volume at all. It seems to me that no "container" is needed for the ideal gas law, or for any equation of state. Is not the purpose of an equation of state to determine the momentum flux tensor at any point, given the temperature and density? The momentum flux tensor just tells you the momentum flux through any imaginary sector of a surface, and that's a tensor because the momentum is a vector and its flux through a surface relates to the vector normal to the surface. You want to know that tensor because for relativistic gas, it affects the EFE and the expansion dynamics, and although for nonrelativistic gas it doesn't affect the dynamics, it does affect the temperature evolution via the first law of thermodynamics. So why do you need to define a volume? Laws of thermodynamics that relate to changes in volume can just replace dV/V = 3da/a, where a is the scale factor that couples the thermodynamics to the dynamics. Any actual volumes should be irrelevant, done away with by the cosmological principle, is that not so?

5. Apr 7, 2014

### timmdeeg

Mordred, to my understanding the ideal FRW fluid should not be interpreted as to consist of an ideal gas, rather as a perfectly homogenous matter density. The fluid doesn't know about atoms and molecules.
But perhaps I am misinterpreting your question.

EDIT: In contrast to dust an ideal gas exerts pressure, which then contributes to the Friedmann equation accordingly. Perphaps that's what you are attempting to calculate?

Last edited: Apr 7, 2014
6. Apr 7, 2014

### Mordred

An example is in the paper I posted in post 2.

Theorem 1
For the materials with energy-momentum tensor (5), the cosmological principle or the FRW metric implies the isentropic process dS≡0.
This can be checked as follows. Let V= Ωa3, where Ω is any given constant comoving volume independent of a,

I think Ken may have hit upon it though, as the paper correlates the scale factor to the energy-momentum conservation law.

let V=$\Omega$a3.

This works in a Overall FRW metric, however I've seen examples where ideal gas laws are applied in localized applications such as blackhole dynamics, supernova regions etc. The paper above uses gibb's law of ideal gases to define the temperature history rather than the inverse of the scale factor.

Keep in mind I am a self taught wannabe cosmologist, so how the ideal gas laws apply in cosmology is something I'm still working on. I have a few text books such as thermodynamics of the early universe. Physics of the interstellar medium. Which has ideal gas formulas being used however glossed over how volume is defined. They make statements such as the volume increases and the temperature and pressure drops. without going into detail on the volume or pressure vs energy density relation. Hence my trouble understanding with my gas laws understanding from my industrial gas laws understanding

7. Apr 7, 2014

### George Jones

Staff Emeritus
For a normal ideal gas, pressure makes a negligible to the stress-energy tensor and can be safely ignored, i.e., as a source of gravity in GR, an ideal gas is well-modeled as a perfect fluid that is dust. If I get some time today (unlikely) or tomorrow, I will post quantitative demonstration of that $p << \rho c^2$.

At the other extreme, the pressure of a relativistic gas cannot be ignored. If the gas is relativistic enough, it will have the same equation of state as a photon gas, $p = \rho c^2/3$, even for massive particles.

8. Apr 7, 2014

### Ken G

Note that relativistic gases can be viewed as ideal gases too, such as the CMB.

9. Apr 7, 2014

### Ken G

Mordred, perhaps another way to think about "the volume of the container" is to just take a chunk of comoving matter, and think of its volume as the volume in question. That is an arbitrary volume, but that's OK, the answer won't depend on what volume you choose, so any volume works as long as it's large. Even if the universe is empty, you can keep track of what volume you are talking about by imagining there is some highly nonrelativistic "trace dust" in there, you'd need that anyway to know what your own coordinates mean. Since highly nonrelativistic dust does not do much "trading places" with other nonrelativistic dust, the volumes are fairly well defined by such "comoving matter," and indeed identifying volumes like that is called using "comoving coordinates."

10. Apr 7, 2014

### Mordred

I look forward with enthusiasm to any feedback you offer.

That is similar to how I've been interpreting volume. Essentially the volume of a gas as being definable with the same metrics and values. The boundary being where those same metrics and values no longer apply. However that is my interpretation, and all the research I've done on cosmology thermodynamic applications haven't led me to a better descriptive. ie for non relativistic gases the volume being defined to the region prior to redshift occurrence.

$P\propto$a-3=V-1 where V is the volume.

Globally scale factor is easy to determine volume, however localized in such examples as differing layers of say for example a supernova made me wonder if that definition is sufficient.

Last edited: Apr 7, 2014
11. Apr 7, 2014

### Ken G

The definition in terms of the scale parameter is the best one. I'm not sure why you want the "edge of the container" to be where the metric or values no longer apply, cosmology uses the "cosmological principle" which stipulates that the metric is the same everywhere at any given age of the comoving dust in the universe. So there is never anywhere the metric changes, and the volumes we talk about are completely arbitrary, they don't have any edges to point to, but the volumes chosen don't matter anyway, they're just a visualization aid.

12. Apr 8, 2014

### timmdeeg

That would be interesting.

The pressure of an ideal gas is given by $p = \rho RT/M$, with M the molar mass. So, as long as nothing heats (e.g. thermal equilibrium with radiation) the gas, indeed it's pressure is negligible.

Last edited: Apr 8, 2014
13. Apr 8, 2014

### Mordred

yeah I gotcha on that aspect, looking closer at some examples from accretion disk fluid calculations. They involve stress energy and energy momentum tensors to define the volume aspect.

dU=TdS-PdV

section 3.1 of this black hole accretion disk manual covering the fluid equations of an accretion disk

http://arxiv.org/pdf/1104.5499v3.pdf

14. Apr 8, 2014

### Ken G

Exactly, fluids don't need walls, but what is bothering you has to do with certain imprecise notions in fluid mechanics like "coarse graining" and so forth. I think it can be made formal, but everything involves some level of idealization. The bottom line is, if you want to take a lot of particles and say they are a "fluid" that is moving in some "continuous spacetime", you are doing idealization and I'm not sure you ever get a better justification than that it seems to be working.

15. Apr 8, 2014

### Mordred

Yeah it was bothering me, now that I understand the mechanism, some of the articles/texts books I have are making more sense. Thanks for the help

16. Apr 11, 2014

### timmdeeg

Provided the universe contains nothing than matter, which is modelled by a perfect fluid, the isotropic pressure is negligible as mentioned by George Jones.

However, if the matter is at rest (its thought particles have zero proper velocity), why then is the pressure not exactly zero? Could someone kindly explain?

17. Apr 11, 2014

### Ken G

That actually isn't necessarily true-- a universe with a significant amount of relativistic "hot dark matter" would have a gravitationally important pressure. What seems to rule that out is that we need it to clump up more than relativistic matter would.
If the proper velocity is zero, then the pressure is exactly zero. But everything has some temperature, so some random proper motions. But it's negligible, as you say.

18. Apr 11, 2014

### timmdeeg

Ok, thanks for confirming. I was referring to the ideal FRW case, perfect fluid, not to a realistic universe.

19. Apr 11, 2014

### Ken G

The ideal FRW case still allows proper motions of the particles, so they can have pressure, but the "fluid approximation" is in play, and the fluid has no proper motion. It's like, you can have a gas in a balloon, and the balloon can have zero proper motion, but the gas has a pressure because it is only the whole fluid that has no proper motion.

20. Apr 11, 2014

### timmdeeg

Ah, so $p = 0$ isn't compelling, unless one has a certain reason, e.g. Einstein's static universe. Or do I have the freedom to allow some pressure and cancel it with the cosmological constant even in that case?

21. Apr 11, 2014

### Ken G

If it's negligible, it doesn't much matter what you do with it, you can cancel it somehow easily.

22. Apr 11, 2014

### timmdeeg

Ok, thanks for clarifying.

23. Apr 13, 2014

### George Jones

Staff Emeritus
The euqation of state for a cosmological fluid is taken to relate pressure and energy density by $P=w\rho c^2$, with $w$ a constant. Consequently, for pressureless dust $w=0$. Below, I approximate $w$ for a non-relativistic ideal gas.

Consider a volume $V$ that contains $N$ particles (particles, atoms, molecules, whatever) of $m_p$, so that the total mass in $V$ is $m = N m_p$. Thus,

$$\frac{N}{V} = \frac{m}{Vm_p} = \frac{\rho_m}{m_p}$$

and the ideal gal law $PV=NkT$ becomes

$$P=\frac{NkT}{V} = \frac{\rho_m}{m_p}kT$$

where $\rho_m$ is the mass-density of stuff in $V$. The internal energy of the stuff in $V$ is one-half $kT$ per particle per degree of freedom $f$.

Hence, the energy density is

\begin{align} \rho c^2 &= \rho _{m}c^2 + \frac{f}{2} \frac{NkT}{V}\\ &= \rho _m c^2 + \frac{f}{2} \frac{\rho_m}{m_p}kT \end{align}

and

\begin{align} w &= \frac{P}{\rho c^2}\\ &= \frac{\frac{\rho _{m}}{m_{p}}kT}{\rho _{m}c^{2}+\frac{f}{2}\frac{\rho _{m}}{m_{p}}kT}\\ &= \frac{kT}{m_{p}c^{2}+\frac{f}{2}kT}. \end{align}

For hydrogen H2 gas at 1000 Kelvin,

$$w = \frac{\left( 1.38\times 10^{-23}\right) \left( 1000\right) }{2\left( 1.67\times 10^{-27}\right) \left( 3.00\times 10^{8}\right) ^{2}+\frac{5}{2}\left( 1.38\times 10^{-23}\right) \left( 1000\right) }=4.6\times 10^{-11}.$$

24. Apr 13, 2014

### Mordred

nice and clear example, thanks a bunch for that