Characteristic Parameters and Scales

12.1.1

Problem 12.1.1

Calculate the dark energy density and the cosmological constant value.

solution

For quark one obtains the following: SCM assumes $\Omega _\Lambda = 0.73,$ and the Hubble constant value is generally accepted to be $H_0 = 70\, km\cdot s^{-1}\cdot Mpc^{- 1} = 2.26 \cdot 10^{ - 18}s^{-1}.$ Then the critical density corresponds to the following value \[ \rho_{cr} = \frac{3H_0^2}{8\pi G} = 0.92 \cdot 10^{-29} g/cm^3 \] and \[\rho _\Lambda = \Omega _\Lambda \rho_{cr} = 0.67 \cdot 10^{-29} g/cm^3.\]

12.1.2

Problem 12.1.2

Estimate total number of baryons in the Universe.

solution

$$ N = \frac{M}{m_n}=\frac{\frac{4}{3}\pi \left( \frac{c}{H_0} \right)^3\rho }{m_n}=\frac{\frac{4}{3}\pi \alpha \left( \frac{c}{H_0}\right)^3\frac{3H_0^2}{8\pi G}}{m_n}= \frac{1}{2}\alpha \frac{H_0^2}{Gm_n}\left( \frac{c}{H_0} \right)^3=\frac{1}{2}\alpha \frac{M_{Pl}}{m_n}\frac{M_{Pl}c^2}{\hbar H_0}\simeq 10^{80}\quad (\alpha \simeq 0.04) $$

12.1.3

Problem 12.1.3

Find dependence of relative density of dark energy $\Omega_\Lambda$ on the redshift. Plot $\Omega_\Lambda(z)$.

solution

Neglecting the contribution of radiation ($\rho _{r0} \ll \rho _{m0} < \rho _\Lambda$) one obtains $$ \Omega _\Lambda (z) = \frac{\rho _\Lambda }{\rho _{cr}(z)} = \frac{\rho _\Lambda}{\rho _{m0}(1 + z)^3 + \rho _\Lambda } = \frac{\Omega _{\Lambda 0}} {\Omega _{m0}(1 + z)^3 + \Omega _{\Lambda 0}}; $$ $$ \Omega _\Lambda (z) \simeq \frac{1}{\frac{1}{3}(1 + z)^3+ 1} $$

• 

12.1.4

Problem 12.1.4

Estimate total number of stars in the Universe.

solution

$$ \begin{gathered} \rho_{cr} \equiv \frac{3H_0^2}{8\pi G} \simeq 10^{-29}\mbox{g/cm}^3; \\ \rho_{bar}\simeq 0.04\rho _{cr}; \\ R_H \approx 10^{28}\mbox{cm}; \\ M_{bar} \approx 0.2 \times 10^{55}\,cm; \\ M_ \odot\simeq 10^{33}\,cm; \\ N_ \odot \approx 10^{21}. \\ \end{gathered} $$

12.1.5

Problem 12.1.5

Find the ratio of dark energy density to the energy density of electric field of intensity $1\,V/m$. Compare the dark energy density with gravitational field energy density on the Earth surface.

solution

Transit from the relative density $\Omega_{DE}$ to th usual one $\rho_{DE}=\Omega_{DE}\rho_{cr}.$ The present value of the critical density of Universe is $$\rho_{cr}\simeq 9\times 10^{-10} \;\mbox{J/m}^3,$$ therefore $$ \rho_{DE}\simeq 6.3\times 10^{-10} \;J/m^3 $$ which is equivalent to the mass density $$ \rho _{DE}^{(m)}\simeq 0.7\times 10^{-26}\;kg/m^3\simeq 3.9\;GeV/s^2\,m^3\simeq 4 \;proton mass/m^3 $$ Compare $\rho_{DE}$ with the energy density of the electric field $\rho_{E}$ with intensity $E=1\;V/m$ (such electric fields are easily available for laboratory measurements). For this field one gets $$ \rho_E = \frac{\varepsilon_0E^2}{2} = 4.4 \times 10^{- 12}\mbox{J/m}^3. $$ As one can see, it is possible to measure fields with energy density which is orders of magnitude less than the dark energy density. However the measurement of dark energy is far from being trivial. Unlike the the electric field, the dark energy cannot be switched on and off. The dark energy, at least in the form of cosmological constant, is homogeneously distributed in space, and we presently have no means to create local regions of increased dark energy density, similar to a capacitor in the case of the electric field. At last let us compare the dark energy density under interest to that of gravity field on the Earth surface where the laboratory measurements are performed. The latter equals to $$ \rho _G = \frac{g^2}{8\pi G} = 5.7 \times 10^{+10} \mbox{J/m}^3. $$ It is clear that if the background exceeds the measured quantity in twenty orders of magnitude, then reliable identification of this quantity requires a super-precision experiment.

12.1.6

Problem 12.1.6

Estimate the distance between two neutral hydrogen atoms at which the gravitational force of their attraction is balanced by the repulsion force generated by dark energy in the form of cosmological constant. Make the same estimates for the Sun-Earth system.

solution

$$ \ddot R = \frac{\ddot a}{a}R. $$ The acceleration generated by the cosmological constant is $$\begin{gathered} \ddot R = \frac{\ddot a}{a}R = - \frac{4\pi G}{3}(\rho _\Lambda + 3p_\Lambda )R = \frac{8\pi G}{3}\rho _\Lambda R; \\ \frac{\rho _{\Lambda 0}}{\rho _{cr0}} = \Omega _{\Lambda 0};\quad \rho _{cr0} = \frac{3H_0^2}{8\pi G} \\ \end{gathered} $$ Then the contribution of the cosmological constant into the current acceleration is $\ddot R = \Omega _{\Lambda 0}H_0^2R.$ This quantity is to be compared with the usual gravity acceleration generated by the mass $M,$ $$ g = G\frac{M}{R^2} $$ Then for the case of hydrogen atom one obtains \[R = \left(\frac{Gm_p}{H_0^2\Omega _{\Lambda 0}} \right)^{1/3}\simeq 0.3 \, m.\] For the Sun-Earth system the analogues estimate gives $$ R=\frac{2\pi R_{a.u.}}{H_0\sqrt{\Omega_{\Lambda 0}}}\simeq 0.5 \,Mpc, $$ where $R_{a.u.}-$ is the mean Sun-Earth distance.

12.1.7

Problem 12.1.7

Calculate magnitude of physical acceleration.

solution

$$ \begin{gathered} \dot V = (\dot H + H^2)R; \\ H \equiv \frac{\dot a}{a};\;\dot H = \frac{\ddot aa - \dot a^2}{a^2} =\frac{\ddot a}{a} - H^2; \\ \dot H + H^2 = - \frac{4\pi G}{3}(\rho + 3p) = - \frac{4\pi G}{3}( - 2\rho _\Lambda + \rho _m) = \\ = \frac{8\pi G}{3}\left( \rho _\Lambda - \frac{1}{2}\rho _m \right) = H_0^2\left( \Omega _{\Lambda 0} - \frac{1}{2}\Omega _{m 0} \right); \\ \dot V = H_0^2R\left( \Omega _{\Lambda 0 } - \frac{1}{2}\Omega _{m 0} \right) \\ \end{gathered} $$ $$ \begin{gathered} R = 1\, Mpc;\\ \dot V \simeq 10^{ - 11}\, sm/s^2 \\ \end{gathered} $$

12.1.8

Problem 12.1.8

How far can one see in the Universe?

solution

The particle horizon limits the causally connected region of the Universe, which is in principle observable at given time moment. by definition it is $$ L_{p} = a\int\limits_{0}^t\frac{dt}{a\left( t \right)}. $$ where $a(t) = A^{1/3}\sinh ^{2/3}\left( t/{t_\Lambda }\right)$ and $t_\Lambda \equiv \frac{2}{3}\left(H_0 \right)^{-1}\left( \Omega _{\Lambda 0}\right)^{ - 1/2}\simeq 10.768 \cdot 10^{9}$ years. $t_0 = t_\Lambda Arth\sqrt {\Omega _{\Lambda 0}}\simeq 13.7\cdot10^9 $ years. \[A \equiv \frac{\Omega_{m0}}{\Omega _{\Lambda 0}} = \frac{1 - \Omega _{\Lambda 0}}{\Omega _{\Lambda 0}} = 0.37.\] After integration one finally obtains $L_{p} = 47.563\cdot 10^9$ light years.

12.1.9

Problem 12.1.9

Find age of the Universe.

solution

Using the result of the previous problem, one obtains the age of the Universe from the condition $a(t_0) = a_0 = 1$ $$ t_0 = \frac{2}{3}\left( H_0\right)^{-1}\left(\Omega _{\Lambda 0}\right)^{-1/2}Arsh \sqrt{\frac{\Omega _{\Lambda 0}}{\Omega _{m0}}} $$ or $$ t_0 = t_\Lambda Arsh\sqrt {\frac{\Omega _{\Lambda 0}}{\Omega _{m0}}} $$ Using the relation $Arth(x) = Arsh\left(x/\sqrt {1 - x^2}\right)$ one gets \[ t_0 = t_\Lambda Arth\sqrt {\Omega _{\Lambda 0}} \simeq 13.7\cdot 10^9 years.\]

12.1.10

Problem 12.1.10

Give a qualitative explanation why the age of the Universe in SCM is considerably larger than the age of matter dominated Universe (the Einstein-de Sitter model).

solution

In the Einstein-de Sitter model the expansion is always decelerating, while in the SCM the decelerated expansion turns to the accelerated one. Therefore for a given Hubble constant $H_0$ the preceding value in the Einstein-de Sitter model was greater than that in the SCM, and the average expansion rate was also greater, so the observed size of the Universe was reached in a smaller period of time.