## 物理代写|宇宙学代写cosmology代考|Neutrinos

The next constituent we need to consider are neutrinos. Unlike photons and baryons, cosmic neutrinos have not been observed directly, so arguments about their contribution to the energy density are necessarily theoretical. However, these theoretical arguments are quite strong, based on very well-understood physics. Moreover, the CMB anisotropies constrain the total density in relativistic particles $\Omega_{\mathrm{r}} h^2$ in the early universe. Experiments such as the Planck satellite have found clear evidence for an amount of relativistic particles (in addition to the known photons) that is consistent with the expected neutrino contribution.
Let us sum up what we know about these particles:

• There are three generations of neutrinos ${ }^7$.
• There is one spin degree of freedom each for the neutrino and antineutrino of each generation.
• Neutrinos are fermions and follow the Fermi-Dirac distribution function when in equilibrium.

We can use this information to evaluate the energy density of neutrinos in the universe, relating it to the photon energy density $\rho_\gamma$ for convenience. The first two items on the list imply that the degeneracy factor of neutrinos is equal to 6 . The third means we need to change the denominator in the integrand in Eq. (2.73) to $e^{p / T}+1$. The resulting FermiDirac energy integral is smaller by a factor of $7 / 8$ compared to the corresponding BoseEinstein integral. Finally, since the energy density of a massless particle scales as $T^4$, we can write
$$\rho_v=3 \times \frac{7}{8} \times\left(\frac{T_v}{T}\right)^4 \rho_\gamma .$$
We then only need to determine $T_v$, which, as you might have guessed from how we wrote this equation, is different from the photon temperature $T$.

For this, let us first consider the production of neutrinos in the early universe. A basic understanding of the interaction rates of neutrinos (Fig. 1.4) enables us to argue that neutrinos were once kept in equilibrium with the rest of the cosmic plasma. At later times, they lost contact with the plasma because their interactions are weak. The tricky part in determining the neutrino temperature is the annihilation of electrons and positrons when the cosmic temperature was of order the electron mass. Neutrinos lost contact with the cosmic plasma slightly before this annihilation, so they inherited almost none of the associated energy. The photons, which acquired the vast majority of it, are therefore hotter than the neutrinos.

The epoch at which the energy density in matter equals that in radiation is called matterradiation equality. It has a special significance for the generation of large-scale structure and for the development of CMB anisotropies, because perturbations grow at different rates in the two different eras (note that for large-scale structure, there is a third era: that of dark energy domination today; see Exercise 2.14). It is therefore a useful exercise to calculate the epoch of matter-radiation equality. To do this, we need to compute the energy density of both matter and radiation, and then find the value of the scale factor at which they were equal.

Using Eq. (2.76) and Eq. (2.82), we see that, as long as $T_v$ is much larger than all neutrino masses, the total energy density in radiation is
$$\frac{\rho_{\mathrm{r}}}{\rho_{\mathrm{cr}}}=\frac{4.15 \times 10^{-5}}{h^2 a^4} \equiv \frac{\Omega_{\mathrm{r}}}{a^4} .$$
To calculate the epoch of matter-radiation equality, we equate Eqs. (2.85) and (2.72) to find
$$a_{\mathrm{eq}}=\frac{4.15 \times 10^{-5}}{\Omega_{\mathrm{m}} h^2} .$$
A different way to express this epoch is in terms of redshift $z$; the redshift of equality is
$$1+z_{\text {eq }}=2.38 \times 10^4 \Omega_{\mathrm{m}} h^2 .$$
Note that, as the amount of matter in the universe today, $\Omega_{\mathrm{m}} h^2$, goes up, the redshift of equality also goes up.

## 物理代写|宇宙学代写cosmology代考|中微子

• 有三代中微子${ }^7$ .
• 每一代的中微子和反中微子都有一个自旋自由度。

$$\rho_v=3 \times \frac{7}{8} \times\left(\frac{T_v}{T}\right)^4 \rho_\gamma .$$

## 物理代写|宇宙学代写cosmology代考|物质-辐射相等的纪元

$$\frac{\rho_{\mathrm{r}}}{\rho_{\mathrm{cr}}}=\frac{4.15 \times 10^{-5}}{h^2 a^4} \equiv \frac{\Omega_{\mathrm{r}}}{a^4} .$$为了计算物质-辐射相等的时间，我们用等号。(2.85)和(2.72)查找
$$a_{\mathrm{eq}}=\frac{4.15 \times 10^{-5}}{\Omega_{\mathrm{m}} h^2} .$$表示这个纪元的另一种方式是红移 $z$;等式的红移是
$$1+z_{\text {eq }}=2.38 \times 10^4 \Omega_{\mathrm{m}} h^2 .$$请注意，作为当今宇宙中的物质总量， $\Omega_{\mathrm{m}} h^2$，上升，相等的红移也上升。

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