# Calculating energy density in the early Universe

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1. Nov 23, 2014

### Denver Dang

1. The problem statement, all variables and given/known data
I've been told to calculate the energy density in the early Universe. It states that it is completely dominated by neutrinos (3 species), photons, electrons, and positrons.

2. Relevant equations

3. The attempt at a solution
Now, I've found an equation to calculate the energy density for bosons and fermions, with different $g$-factors (degree of freedom) for each type of particle. So that doesn't seem to complicated. But, in order to get the energy density of the early Universe, I need to incorporate all particles in this density - is my guess.
My thought is just to calculate and energy density for each particle, and then sum up, and divide by the number of different particles. But that just seems to trivial, so is that the way to go, or am I missing something in order to do this correctly ?

2. Nov 24, 2014

### Staff: Mentor

Right.
There is no need to do that.

The non-trivial part is the calculation of the individual energy densities.

3. Nov 24, 2014

### Denver Dang

So actually, when I have:
$$\rho = \frac{\pi^{2}}{30}gT^{4} \text{for bosons}$$
$$\rho = \frac{7}{8} \frac{\pi^{2}}{30}gT^{4} \text{for fermions} ,$$
where
$g_{\gamma} = 2, g_{\nu} = g_{\bar{\nu}} = 1, g_{e^-} = g_{e^+} = 2$
I actually just end up with 6 different densities (Some of them the same of course due to the degree of freedom), that I sum up, and that's pretty much it ? Almost seems to easy :)

4. Nov 24, 2014

### Staff: Mentor

You have to find the temperatures, and those formulas don't take rest mass of electrons and positrons into account (if early enough, that should be fine).

5. Nov 24, 2014

### Denver Dang

Ah, well, the problem states, that I have to find it at $T = 1 MeV$, so that shouldn't be a problem.

6. Nov 24, 2014

### Orodruin

Staff Emeritus
Well, the electron mass is 0.5 MeV so it does not seem safe to neglect it if T = 1 MeV.

7. Nov 24, 2014

### Denver Dang

@Orodruin
Well, can't I just not neglect it ? It's only 6 equations, so it's not that big of a problem. Or am I missing something ?

8. Nov 24, 2014

### Orodruin

Staff Emeritus
Not neglecting it would involve using a more general expression for the energy density than provided in the OP. The mass implies a different equilibrium distribution and hence a different energy density. Other than that, no problem.

9. Nov 24, 2014

### Staff: Mentor

Also note that the neutrino temperature can be different at that point.
Things get interesting.

10. Nov 24, 2014

### Denver Dang

But why exactly do I need to neglect electrons/positrons ? If they have masses of about 0.5 MeV, the the neutrinos have above and below that, with a lot. Why are they not neglected ?

11. Nov 24, 2014

### Orodruin

Staff Emeritus
You are not supposed to neglect electrons and positrons. What we said is that you have to take their masses into account.

Neutrinos can be considered massless for the purposes of this problem.

12. Nov 24, 2014

### Denver Dang

Hmmm, I see why I could ignore the electron neutrino, but doesn't the tau neutrino have large masses compared to electrons ?

13. Nov 24, 2014

### Orodruin

Staff Emeritus
No.

Being a picky neutrino physicist: Flavor neutrino states do not have definite masses but are linear superpositions of mass eigenstates. The square of the mass eigenstate masses differ by less than 1 eV^2.

Thus, all neutrino masses are negligible in this setting.

14. Nov 24, 2014

### Denver Dang

But in this setting, I have the equation above, which gives me the density for different particles, if I choose the right degree of freedom. And the only difference between photons/electrons/positrons and neutrinos are the degree of freedom 1 instead of 2. That shouldn't give a so small value, that it could be ignored, compared to the other particles ?

And T is the same in all equations, i.e. T = 1 MeV, isn't it ?

15. Nov 24, 2014

### Orodruin

Staff Emeritus
As pointed out by mfb in post #4, your expressions for the energy densities hold only for massless particles and do not hold if the particle masses cannot be neglected.

16. Nov 24, 2014

### Staff: Mentor

The temperature as function of time was different for neutrinos and the other particles. This became more notable for colder temperatures, I'm not sure how important it is at 1 MeV. Today photons have 3 K while neutrinos have 2 K.

17. Nov 25, 2014

### Orodruin

Staff Emeritus
Neutrino decoupling happens at about 1.5 MeV. However, they do not really become colder until electron decoupling when the photons are heated. The question is how non-relativistic the electrons have become and thus how much the photons have been heated already and how many electrons are left. At 1 MeV I would not dare to make an off-the-top-of-my-head estimate without making some checks.