- SM of particle physics
- describes elementary particles and their interactions.
- is well test with experiments.

IMG Source: https://en.wikipedia.org/wiki/File:Standard_Model_of_Elementary_Particles.svg

The orange numbers at the right bottom of each particle is the degrees of freedom it has. Here are some comments.

- Photons have only two DoF because it is mass 0. Same reason can apply to gluon. But according to symmetry, there are 8 kinds of gluons.
- W bosons carry charges. This is where the 2 come from.
- Electrons and quarks have antiparticles. So there DoF will be doubled after counting the spin.
- Each quark have 3 different colors and this gives us the 3 when calculating there DoF.

Finally, we can make this table.

Partilces | Higgs | Messengers | Quarks | Leptons |
---|---|---|---|---|

DoF | 1 | 27 | 72 | 18 |

We can see that the heaviest particle is top quark with a mass of \(m_t = 170 \mathrm{GeV}\).

If temperature of the universe \(T \gg m_t\), all particles should be in relativistic regime and the decay (annihilation) and inverse decay (inverse annihilation) are in equilibrium so all particles contribute to the thermal quantities in a relativistic way.

\[ \begin{align}\begin{aligned}g_B = 28\\g_F = 90\end{aligned}\end{align} \]

Then

\[g _ * = g_B + \frac{7}{8} g _ F = 106.75\]

For convinience, define the following reduced Planck mass

\[8\pi G = \frac{1}{M _ p ^2}\]

And it’s good to know its value, which is \(2.4\times 10^{18} \mathrm{GeV}\).

We would like to know the relation between expansion and temperature. We already know that the energy density is

\[\rho = g _ * \frac{\pi^2}{30} T^4\]

So the expansion is

\[H^2 = \frac{8\pi G}{3}\rho = 106.75 \times \frac{\pi^2}{30} \frac{T^4}{3 M_p^2}\]

So Hubble function is

\[H \approx 3 \frac{T^2}{M_p}\]

As temperature drops down, particle dacay (annihilation) will be greater than its inverse which is suppressed by Boltzmann factor \(\exp (-m/T)\). The decay rate is so quick that the particle will almost dispear before the universe expand a lot.

So when the temperature drops below the mass of a particle, it won’t contribute to the energy density. Their DoF will just dispear.

For example, if \(T~\mathrm{MeV}\), Higgs and W and Z will decay and quarks are combined with gluons. So we only have **photons, electrons, neutrinos** as elementary particles, that is \(g_* = 10.75\).

The Hubble function,

\[H \approx \frac{T^2}{M _ p ^2}\]

We can generally prove that decay rate is much faster than the expansion rate. ............... To be added.

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