# In <Cosmology> written by Weinberg, in page 109

1. Jul 26, 2011

In <Cosmology> written by Weinberg, in page 109, it says that

$\epsilon=\alpha_{B}T^4+\frac{3}{2}n_{B}Nk_{B}T$
$p=\frac{1}{3}\alpha_{B}T^4+n_{B}Nk_{B}T$

$\alpha_{B}T^4$is the average energy density of black-body radition, then what does $\frac{3}{2}n_{B}Nk_{B}T$means? the energy density of non-relativistic particles? If so, then should the press of non-relativistic particles be zero in the second equation?

2. Jul 26, 2011

### bcrowell

Staff Emeritus
Doesn't Weinberg define any of the notation in the surrounding text? Could you give us as much context as possible? You can notate w=p/ϵ for various pure forms of matter or radiation. Radiation has w=1/3, dust w=0, and the cosmological constant w=-1. This seems to be something with w=2/3.

3. Jul 26, 2011

### BillSaltLake

I think it's just the (non-relitavistic) energy of the free hydrogen & He bouncing around with the photons. These both have kinetic energy of 3/2 kT per particle.

4. Jul 26, 2011

### bcrowell

Staff Emeritus
If it's nonrelativistic, shouldn't it have p=0, like any form of dust?

5. Jul 26, 2011

$\alpha_B$comes from $\int_0^{\infty} h \nu n(\nu)d\nu=\alpha_B T^4$
so it means the average energy density of radiation. $n_B$means the number density of bayons. $N$means the non-relativistic particles per baryon.

6. Jul 26, 2011

I refered to the Statistical books and found that for non-relativistic gas the press is$\frac{2}{3}n \bar{\epsilon}$, $n$is the number density and $\bar{\epsilon}$is the energy density. I think the baryons should be treated as gas rather than as dust.

7. Jul 26, 2011

### bcrowell

Staff Emeritus
If they're non-relativistic, then it doesn't make sense to me that they contribute to P.

8. Jul 27, 2011

### Chalnoth

Not before recombination. Before recombination the baryons are a plasma and do experience pressure.

9. Jul 27, 2011

I agree with you, but why after recombination, the baryons do not experience pressure.

10. Jul 27, 2011

### bcrowell

Staff Emeritus
It's not that the pressure is exactly zero, it's that the pressure here is being measured in units where c=1. In those units, any nonrelativistic particles contibute negligibly to the pressure of a gas. To convert to these units from SI units, you have to divide all the pressures by c2, which makes them negligible if the particles are nonrelativistic.

I'll move this to the relativity forum, where we may be able to attract the attention of someone who is knowledgeable and also has a copy of the book handy.

11. Jul 27, 2011

### bcrowell

Staff Emeritus
Bump. Is there anyone in the relativity forum who has this book and can help to clear up what's going on?

12. Jul 27, 2011

### pervect

Staff Emeritus
Amazon.com let me read the pages in question by searching for "the equilibrium era"
(first I looked for pg 109, saw that that was in the chapte equilibrium era, then I searched for equilibrium era).

Weinberg is considering a non-relativistic gas of baryons and radiation. He's asking the question : "When we have only photons, temperature falls as 1/a(t). But if we had only baryons, he argues that temperature would fall as 1/a(t)^2. So he wants to know which dominates when we have thermal equilibrium, and he proceeds to derive it from thermodynamics.

The sorts of non-relativistic particles in this gas are protons, helium nuclei, and electrons. N is a number of order unity, which gives the number of particles / number of baryons.

So n_B is the number density of baryons, N n_b is the number density of particles

(3/2) (N n_b) k T is just the total energy, which by the equipartition theorem for the non-relativistic gas in equilibrium is (1/2) kT per particle.

Weinberg uses k_B where I write k - it's just Boltman's constant. And E isn't quite right for the symbol he uses , but it's close enough I hope to not be confusing if I use it rather than his symbol.

Weinberg wants to calculate the change in entropy, delta-S, which he writes as k $\sigma$, where sigma is the entropy per unit baryon.

Thermo isn't my forte, so I'll consult another textbook, MTW, to see if it tells us the same thing. (Which it does)

MTW argues that in curves space-time we wite

d(energy in a volume containing a constant number of baryons ) =
- p d(volume) + T * d(entropy)

This should be equivalent to Weinberg's 2.21, i.e.

d(k sigma) = [d (E/n_b) + p d(1/n_b)] / T

which is equivalent to MTW's formulation ( a few pages more in the textbook)

d(rho/n) + P d(1/n) - T ds = 0, MTW uses the more familar rho for the energy density rather than the E.

Therefore, we apparently do need to include the pressure-volume terms in computing the entropy change, the P d(1/n) terms. In flat space-time we interpret them as "work done by the expanding gas", in curved space-time the interpretation is a bit hazier, but we still need them. Maybe someone better than I am with thermo can give a more convincing explanation of why they're needed, other than "two textbooks say so".

Last edited: Jul 27, 2011
13. Jul 28, 2011

The particles in the air, as we all know, are non-relativistic, but they can express press.

However, I still wonder what is the criterion to determine whether they can press or not, the press of dust is 0 and why. You can refer to #8.

14. Jul 28, 2011

What is the criterion to determine whether they can express press or not?

15. Jul 28, 2011

the press should be reflected in the equation, the question is what does press include. Of course it includes press from radiation and anything else? In my opinion, baryons or dusts, whatever, should have press, as far as they constitutes the gas, baryon gas or dust gas. However, the press of non-relativistic particles if $p=\frac{2}{3} \bar{\rho}$, $\bar{\rho}$is the energy density, and it is proportional to the temperature T, it is very small compared to the energy density from radiation, and can be neglected.

In Weinberg's book, the press from the baryons is very small, too. It is included for strict consideration and does not have any effect in the later calculation.

16. Jul 28, 2011

### Chalnoth

Well, in this case, being ionized means that they experience the pressure of photons. I'm not sure, however, if this makes any difference for the expansion rate before recombination. But it does make a difference in terms of how matter behaves when falling into potential wells (baryons tend to "bounce" before recombination due to this pressure, dark matter does not).

Now that I think about it, I think that the equation presented isn't talking about this, but instead the normal ideal gas pressure, which is very, very low but nonzero for most of the history of the universe.

17. Jul 28, 2011

### bcrowell

Staff Emeritus
Thanks very much, pervect, for taking the time to write up the summary for us!

I think I understand now what the issue is.

First let me see if I can clear up for nadia why we usually take a nonrelativistic gas to have p=0.

Let's do a numerical example.

Say we have a photon gas with a mass-energy density of 1 J/m3. The pressure of this gas is 0.3 Pa.

Now let's take a 1 m3 volume containing one mole of helium gas at room temperature. It basically doesn't matter whether the temperature is room temperature or the temperature of the sun; all that matters is that it's nonrelativistic. The mass-energy is made up almost entirely of the rest mass of the helium -- not its kinetic energy. Its density is 4x1014 J/m3. The pressure at room temperature is 2x103 Pa.

So for the photon gas, the ratio of pressure to mass-energy density is higher by 11 orders of magnitude. This is why we normally ignore the pressure of nonrelativistic matter when we're computing a stress-energy tensor.

So now let's look at Weinberg's equations:

$\epsilon=\alpha_{B}T^4+\frac{3}{2}n_{B}Nk_{B}T$
$p=\frac{1}{3}\alpha_{B}T^4+n_{B}Nk_{B}T$

The key here is that his $\epsilon$ is not the mass-energy density, it's the energy density. In other words, Weinberg isn't just being idiosyncratic in his notation; he really means something different by $\epsilon$ than the mass-energy density that everyone normally notates as $\rho$. Now that pervect has supplied us with the context, it makes sense that Weinberg can get away with this, because he's only writing down $\epsilon$ so that he can differentiate it. Therefore the constant contributed by the baryons' density of rest mass doesn't matter, even though it's actually huge compared to the energy density.

18. Jul 28, 2011

### BillSaltLake

A solid dust particle will have some thermal energy, but only on the order of a few kT. If the particle contains, say, a trillion baryons, its energy per baryon (and pressure per baryon) is a trillionth of that of free hydrogen, so dust pressure is relatively negligible.

19. Jul 28, 2011

### bcrowell

Staff Emeritus
"Dust" is standard terminology in GR for any perfect fluid that has $p/\rho\approx 0$. See, e.g., http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll8.html . It doesn't refer to literal "dust." It's common, for example, to refer to galaxies as dust in cosmology.