IOAA 2018, Теория — Задача 11. Thermal History of the Universe

• (a) 14.46 Gyr

  The Hubble parameter has dimensions of inverse time [T⁻¹]; t_H = 1/H.

• (b) $8.59 \times 10^{-27}$ kg/m³

  ρ_c = 3H²/(8πG).

• (c) $\Omega_\Lambda = \frac{\Lambda c^2}{3H^2},\quad \Omega_k = -\frac{kc^2}{a^2H^2}$

• (d) ρ_r ∝ a⁻⁴ and a = 1/(1+z), hence ρ_r ∝ (1+z)⁴.

• (e) -1

• (f) $5.4 \times 10^{-44}$ s

  t_P ≈ √(ħG/c⁵).

• (g) $1.22 \times 10^{19}$ GeV

  M_P = √(ħc/G).

• (h) ε_r ∝ T⁴ and ε_r = ρ_r c² ∝ (1+z)⁴, so T ∝ 1+z.

• (i) $1 + z_{eq} = 2.4 \times 10^4\,\Omega_{m0}h^2$

• (j) 1.3 s

  The official solution gives t = 1.3 s ~ 1 s.

(a) One can argue this in different ways. One can note that the Hubble parameter is often called Hubble constant, but actually it’s not a constant. The dimensions of Hubble parameter is the inverse of time $[T^{-1}]$. Alternatively, one can simply look at unit of H$_0$ in the table of constants and conclude the same. (1 point)

It’s natural to define a timescale as the reciprocal of Hubble parameter:$t_H = \frac{1}{H}$, and this is the Hubble time which is a characteristic timescale of Universe expansion. (2 points)

Present-day Hubble time t$_{\mathrm{H0}}$ =14.46 Gyr. (2 points)

(b) From Friedmann Equation, the critical density is defined in the way of $\left(\frac{\dot a}{a}\right)^2 = \frac{8\pi G}{3}\rho_c$, thus $\rho_c = \frac{3H^2}{8\pi G}$. (2 points)

Substitute H$_0$ = 2.19 x 10$^{-18}$ s$^{-1}$ into above, we have $\rho_{c0} = 8.59 \times 10^{-27}\ kg/m^3$ (3 points)

(c) Ω$_{\mathrm{m}}$ + Ω$_{\mathrm{r}}$ + Ω$_{Λ}$ + Ω$_{\mathrm{k}}$ = 1

(2 points)

Comparing this with the Friedmann equation $H^2 = \frac{8\pi G}{3}(\rho_m + \rho_r) + \frac{\Lambda c^2}{3} - \frac{kc^2}{a^2}$.

$\Omega_\Lambda = \frac{\Lambda c^2}{3H^2}$ and $\Omega_k = -\frac{kc^2}{a^2H^2}$ (4 points)

(d) Pressure exerted by the radiation is $p = \frac{1}{3}\rho_rc^2$

Thus, the Fluid Equation becomes, $\dot{\rho_r} + 3\frac{\dot a}{a}\left(\rho_r + \frac{1}{3}\rho_r\right) = 0$ (2 points)

$\frac{\dot{\rho_r}}{\rho_r} = -4\frac{\dot a}{a}$ implying

Thus $\rho_r \propto a^{-4}$. (3 points)

And we know $a = \frac{1}{1+z}$, hence $\rho_r \propto (1+z)^4$. (2 points)

(e) In the fluid equation, $\dot\rho + 3\frac{\dot a}{a}(\rho + q\rho) = 0$.

But the cosmological constant does not evolve. So $\dot{\rho_\Lambda} = 0$

$\frac{3\dot a\rho_\Lambda}{a}(1+q) = 0$ which means q = -1. (4 points)

(f) From the units, one can figure out that $\hbar$ has the dimensions of $[ML^2T^{-1}]$.

G has the dimensions of $[M^{-1}L^3T^{-2}]$.

And speed of light c has dimensions of $[LT^{-1}]$. (5 points)

We assume that Planck time is given by $\hbar^xc^yG^z$. In order to make it a time dimensional quantity, the following equation must hold:$T = (ML^2T^{-2})^x(LT^{-1})^y(M^{-1}L^3T^{-2})^z. T = M^{x-z}L^{2x+y+3z}T^{-x-y-2z}$.

(3 points)

From $x - z = 0, 2x + y + 3z = 0, -x - y - 2z = 1$, we get $x = \frac{1}{2}, y = \frac{-5}{2}, z = \frac{1}{2}$. Thus, the Planck time is approximately $t_p \approx \sqrt{\hbar G/c^5}$. (3 points)

Substitute the numerical values into it and get:$t_P = 5.4 \times 10^{-44}s$. (2 points)

(g) Schwarzschild radius of a black hole equals to $2l_P$: $r_s = \frac{2GM_P}{c^2} = 2ct_P = 2c\sqrt{\hbar G/c^5}$, (3 points)

$M_P = \sqrt{\hbar c/G}$. (2 points)

$M_Pc^2 = \sqrt{\hbar c^5/G} = 1.22 \times 10^{19}GeV$. (2 points)

(h) Given the energy density, and the result of problem d, we find that $\varepsilon_r \propto T^4, \varepsilon_r = \rho_rc^2, \rho_r \propto (1+z)^4$.

Thus $T \propto 1 + z, \frac{T}{1+z} = const.$ (4 points)

(i) The matter-radiation equality means $\Omega_r(z_{eq}) = \Omega_m(z_{eq})$. We need to derive the behavior of this two density parameters. Obviously $\Omega_m(z) = \Omega_{m0}(1+z)^3$ and $\Omega_r(z) = \Omega_{r0}(1+z)^4$. (4 points)

Thus $1 + z_{eq} = \frac{\Omega_{m0}}{\Omega_{r0}}$. $\Omega_{r0}$ is the density parameter of CMB radiation and can be calculated. (1 point)

By the definition of density parameter of photon radiation (footnote gamma: $\gamma$, $\Omega_{\gamma0} = \frac{\rho_{\gamma0}}{\rho_{c0}} = \frac{\varepsilon_{\gamma0}}{c^2}\frac{8\pi G}{3H^2} = \frac{\varepsilon_{\gamma0}}{c^2}\frac{8\pi G}{3H_{100}^2h^2}$, where $H_{100} = 100\ km/s/Mpc$. (4 points)

The energy density of blackbody photon radiation is $\varepsilon_\gamma = \frac{\pi^2}{15\hbar^3c^3}(k_BT)^4$, thus

$\Omega_{\gamma0}h^2 = \frac{\pi^2}{15\hbar^3c^5}k_B^4\frac{8\pi G}{3H_{100}^2}T_0^4 = 2.47 \times 10^{-5}$. (4 points)

But there is also an additional contribution from neutrinos of about 68%, so the total density parameter of radiation is$\Omega_{r0}h^2 = \Omega_{\gamma0}h^2 \times 1.68 = 4.15 \times 10^{-5}$. (2 points)

Hence $1 + z_{eq} = \frac{\Omega_{m0}}{\Omega_{r0}} = 2.4 \times 10^4\Omega_{m0}h^2$. (1 point)

(j) At that time, the redshift is

Thus the scale factor $a = \frac{1}{1+z} = 2.35 \times 10^{-10}$. (2 points)

Only considering radiation dominated Universe, Friedmann Equation can be written as

$\left(\frac{\acute a}{a}\right)^2 = \frac{8\pi G}{3}\rho_r = \frac{8\pi G\rho_{r_0}}{3a^4},$ (2 points)

$\frac{\acute a}{a} = \frac{1}{a^2}\sqrt{\frac{8\pi G\rho_{r0}}{3}}$

$a^2H = \sqrt{\frac{8\pi G\Omega_{r0}}{3}\rho_{c0}} = \sqrt{H_0^2\Omega_{r0}}$ (1 points)

Thus, we have, $t = \frac{1}{2H} = \frac{a^2}{2\sqrt{H_0^2\Omega_{r0}.}} = 1.3s \sim 1s$.

Thus neutrino decoupled at about 1 s after big bang. (3 points)