# Light’s temperature and particle’s temperature from 3000K

1. Aug 23, 2012

### nonspace

Light’s temperature and particle’s temperature from 3000K

The universe is estimated to have been in the thermal equilibrium state(3000K) around 380,000 years after its birth.
From this, we induce the current temperature of CMB - 2.7K.

In case of particles existing outside(example. at Void area) of the cluster of galaxies structure (or, if not affected by local heat source), what is the temperature currently predicted? How can we get the real answer?

Have a nice day!

2. Aug 23, 2012

### Kraflyn

Hi.

$\delta s^2 =\delta t^2 - a^2(t) \delta S^2$
and solve Einstein equations. These give You 2 equations regarding energy density $\rho$ and pressure $p$ as functions of scale parameter $a(t)$. One of them is
$\dot \rho = -3 \frac{\dot a (t)}{a(t)} \left( \rho +p\right)$
For photon gas, $p_R=\rho_R /3$ for statistical reasons. So when calculated, it turns out that density $\rho$ of photon gas behaves as
$\rho_R = 1/a^4 (t)$
Density $\rho_R$ is not $1/a^3 (t)$ as one might expect, because photon wavelengths are elongated because of space-time expanding. Expansion scale factor being $a(t)$, calculated density seems right.

So all photons emitted back then lost their energy by now by factor $1+z=a_0 / a(t)$ due to expansion of space.
When everything is calculated through, factor $1+z$ turns out to be about 1100. So temperature dropped from around 3000K to 3000K/1100=2,73K.

It is not so trivial to show that age of universe back then was 380 000y at the time of creation of atoms when photons stopped scattering off free particles and became free. You can check any book on Cosmology to find more details on it. Calculations involve cross sections for electron to be captured in hydrogen and so on. I can elaborate it if You want to - if You can't download any recombination/decoupling reading from net. Explanation requires knowledge of quantum mechanics II and thermodynamics of fermions.

However, You can calculate $1+z$ easily once You know the age of emission of photons. We use the second equation coming from solving Einstein equations:
$\frac{\ddot a (t)}{a(t)}=-\frac{4 \pi}{3} \left( \rho + 3p \right)$
After decoupling of matter and radiation, universe became matter dominated. So $\rho$ for previous equation is $\rho_M$. It decreases as $\rho_M=1/a^3 (t)$, with $p_M =0$ due to matter becoming non-relativistic. So equation to solve is
$\frac{\ddot a (t)}{a(t)}=-\frac{4 \pi}{3} \frac{1}{a^3 (t)}$
Solution is $a(t)=t^{2/3}$ up to a constant factor. Hence $t_0$ being 13,4Gy and $t$ being 380 000y at decoupling, one finds $1+z=1100$.

Bianchi relations for FLRW metric are first law of thermodynamics. Yes, expansion is adiabatic: there is no energy transfer, because there is no communication possible with whatever lies outside of our particle horizon.

I hope this helps.

Cheers.

Last edited: Aug 23, 2012
3. Aug 24, 2012

### nonspace

Dear Kraflyn!
Thank you very much!
I’m sorry. I apologize for my poor English.

My question is not radiation’s temperature but matter’s(intergalactic medium’s) temperature.

How do derive it from thermal equilibrium state(3000K)?

4. Aug 24, 2012

### Staff: Mentor

It would be an adiabatic expansion, too. However, matter interacts with other matter and with radiation (e.g. from stars), its distribution is not so uniform and so on. The result is that there are very hot areas, and even a few areas which are colder than the CMB. See Wikipedia for more information.

5. Aug 24, 2012

### Kraflyn

Hello.

You're welcome.

Ah, I see Your point now. Yes just apply adiabatic expansion law together with the usual equations of classical thermodynamics. Matter became non-relativistic at recombination and universe entered matter dominated era. So the usual classical thermodynamics holds. You may take matter to consist purely of hydrogen. This is just a theoretical assumption: the hydrogen assumption. You probably know there is some exotic kind of ordinary matter hidden in space, called Dark Matter. The weakly interacting matter. So maybe there is more than just hydrogen. So if You suppose it's just hydrogen, then

$pV^{\kappa} = const.$
$pV=NRT$
$\kappa = 1,4$

should bring You there. The result is lower than 2,7K. So this approach has some wrong assumption in it. The capital wrong assumption is that matter forms a gas. Furthermore, this approach does not take nuclear reactions in stars in account. Heat is being produced in nuclear reactions, so adiabatic assumption for classical matter falls badly. Matter is clustered, so no gas. Plus Dark matter might behave in some unexpected way too. So You might wish to start with adding heat from nuclear reactions. I'd say one may take all hydrogen to be active in stars, so one should: A) calculate power radiated in nuclear reactions or B) read some data on net on it, and then make an average over entire horizon. We know mass of matter in visible universe, so we indirectly know number of hydrogen atoms. All that is required is mean temperature of a star. I don't have such data right here= in front of me, of course. However, one way to find out about the temperature of mass is like described. Interestingly: You can do it without knowing about 3000K or 380 000y. It might be interesting to relate recombination to energy released through nuclear reactions. This would relate number of hydrogen atoms to 3000K and early universe. I believe we don't have such relation yet so far.

This means, You can safely discard any information You might get from classical thermodynamics. The interesting part comes from stars. Temperature comes from nuclei.

Keep us informed.

Cheers.

6. Aug 25, 2012

### nonspace

Dear Kraflyn,

Thank you very much! I really appreciate your explains.

7. Aug 25, 2012

### Lino

Kraflyn, would you recommend any particular book / source specifically in relation to dating the age of the universe at this point (i.e. at 380 000y)?

Regards,

Noel.