# New problems

## Contents

- 1 New on the matter creation model
- 2 New on the Ricci scalar etc
- 3 New on the deceleration parameter
- 4 New on cosmology with power and hybrid expansion laws
- 5 New to Quantum cosmology
- 6 New to Observational Cosmology
- 7 New from June 2015
- 8 New from March 2015
- 9 New from December 2014
- 10 Exactly Integrable n-dimensional Universes
- 11 UNSORTED NEW Problems

# New on the matter creation model

* As we have seen above, negative pressure is the key ingredient required to achieve the accelerated expansion. Such pressure naturally appears when the system deviates from the thermodynamic equilibrium. In particular, as was first pointed out by Zel'dovich [Ya. B. Zel'dovich, JETP Lett. 12, 307 (1970)], the negative pressure is generated in process of particle creation due to the gravitational field. Physically, one may think that the (classical) time varying gravitational field works like a 'pump' supplying energy to the quantum fields. Construction of the models including the essentially quantum process of particle creation is problematic because of the difficulties connected with its inclusion into the classical field equations of Einstein. One can however avoid those difficulties on the phenomenological level [J.Lima, arXiv 0807.3379].*

*We shall now consider an open thermodynamical system where the number of fluid particles $N$ is not preserved. So the particle conservation equations*
\[\dot N_{;\mu}=0\Rightarrow \dot n + \theta n=0\]
*is now modified as*
\[\dot n + \theta n=n\Gamma,\]
*where $N_{\mu}=nu_\mu$ is the particle flow vector, $u_\mu$ is the particle four velocity, $\theta=u^\mu_{;\mu}$ is the fluid expansion, (for the FLRW Universe, $\theta=u^\mu_{;\mu}=3H$), $\dot n = n_{,\mu}u^{mu}$ and $\Gamma$ stands for the rate of change of the particle number in a comoving volume $V$, $\Gamma>0$ indicates particle creation while $\Gamma<0$ means particle annihilation. Any non-zero $\Gamma$ will effectively behave as a bulk viscous pressure of the thermodynamical fluid and nonequilibrium thermodynamics should come into the picture.*

**Problem 1**

problem id: 150_dp0

Express the first law of thermodynamics in terms of the specific quantities: the energy density $\rho$ and the particle number density $n$.

For a closed thermodynamical system the first law of thermodynamics states the conservation of internal energy $E$ as \[dE=dQ-pdV,\] where $p$ is the thermodynamic pressure, $V$ is any comoving volume and $dQ$ represents the heat received by the system in time $dt$. Now defining $\rho=E/V$ as the energy density, $n=N/V$ as the particle number density and $dq=dQ/N$ as the heat per unit particle, the above conservation law takes the form \[d\left(\frac\rho n\right)=dq-pd\left(\frac1n\right).\] Note that this equation (known as Gibb's equation) is also true when the thermodynamical system is not closed, i.e., $N$ is not constant $[N=N(t)]$.

**Problem 2**

problem id: 150_dp0

In the context of particle creation, let us consider spatially flat FRLW model of the Universe as an open thermodynamical system. Obtain the Friedmann equations for the cosmic fluid having energy-momentum tensor (in this section we will use the metric $(-,+,+,+)$ \[T_{\mu\nu}=(\rho+p+\Pi)u_\mu u_\nu+(p+\Pi)g_{\mu\nu}.\] For cosmological models with particle creation the cosmic fluid may be considered as a perfect fluid and the dissipative term $\Pi$ as the effective bulk viscous pressure due to particle production. In other words, the cosmic substratum is not a conventional dissipative fluid, rather a perfect fluid with varying particle number.

Using the standard procedure for transition from the Einstein's field equations to the Friedmann equations (see Chapter 2), one obtains \begin{align} \nonumber \rho & = 3H^2;\\ \nonumber \dot H & = -\frac12(\rho+p+\Pi), \quad 8\pi G=1. \end{align} And the energy conservation relation $T^{\mu\nu}_{;\nu}$ gives \[\dot\rho+\theta(\rho+p+\Pi)=0.\]

**Problem 3**

problem id: 150_dp0

Show that for the adiabatic process the effective bulk viscous pressure due to particle production is entirely determined by the particle production rate.

Represent the Gibb's relation \[dq=d\left(\frac\rho n\right)+pd\left(\frac1n\right)\] in the form \[Tds=d\left(\frac\rho n\right)+pd\left(\frac1n\right).\] Remember that here $s$ is the entropy per particle and $T$ is the temperature of the fluid. Using the conservation equations $\dot n + \theta n=n\Gamma$ and $\dot\rho+\theta(\rho+p+\Pi)=0$, we obtain \[nT\dot s=-\Pi\theta-\Gamma(\rho+p).\] For isentropic process, the entropy per particle remains constant, i.e. $\dot s=0$ and hence we have \[\Pi=-\frac\Gamma\theta(\rho+p).\] Thus, the effective bulk viscous pressure due to particle production is entirely determined by the particle production rate. So we may say that at least for the adiabatic process, a dissipative fluid is equivalent to a perfect fluid with varying particle number.

**Problem 4**

problem id: 150_dp0

An equivalent way to see the derivation of the field equations (7)-(8) (???) is to consider the energy-momentum tensor in the Einstein field equations as a total energy momentum tensor \[T^{(eff)}_{\mu\nu}=T^{(\gamma)}_{\mu\nu}+T^{(c)}_{\mu\nu},\] where $T^{(\gamma)}_{\mu\nu}$ is the energy-momentum tensor for the fluid with the EoS $p = (\gamma- 1) \rho$, i.e. \[T^{(\gamma)}_{\mu\nu}=(\rho+p)u_\mu u_\nu+pg_{\mu\nu}.\] and \[T^{(c)}_{\mu\nu}=p_c\left(u_\mu u_\nu+g_{\mu\nu}\right)\] is the energy-momentum tensor which corresponds to the matter creation term. Show that energy-momentum tensor $T^{(eff)}_{\mu\nu}$ provides us with the matter creation pressure \[\Pi=-\frac\Gamma\theta(\rho+p),\] obtained in the previous problem.

For the considered case the Einstein field equations are \[G_{\mu\nu}=T^{(eff)}_{\mu\nu}=T^{(\gamma)}_{\mu\nu}+T^{(c)}_{\mu\nu}.\] The Bianchi identity gives \[g^{\nu\sigma}\left(T^{(\gamma)}_{\mu\nu}+T^{(c)}_{\mu\nu}\right)_{;\sigma}=0.\] Using the explicit form of tensors $T^{(\gamma)}_{\mu\nu}$ and $T^{(c)}_{\mu\nu}$, one finds that the latter equality is equivalent to the conservation equation \begin{equation}\label{2016_07_p04_eq1} \dot\rho+\theta(\rho+p+\Pi)=0 \end{equation} From the other hand, Gibbs equation \[Tds=d\left(\frac\rho n\right)+pd\left(\frac1n\right).\] for the adiabatic process, taking into account the relation $\dot n +\theta n=n\Gamma$, directly leads to the conservation equation \begin{equation}\label{2016_07_p04_eq2} \dot\rho+\theta\left(1-\frac\Gamma\theta\right)(\rho+p)=0. \end{equation} Comparing the equations (\ref{2016_07_p04_eq1}) and (\ref{2016_07_p04_eq1}), one can reproduce the result, obtained in the previous problem \[\Pi=-\frac\Gamma\theta(\rho+p).\]

**Problem 5**

problem id: 150_dp0

Find the effective parameter $w_{eff}$ of the EoS for the cosmological model with particle creation.

\begin{align} \nonumber w_{eff} &=\frac{p_{tot}}{\rho_{tot}}=\frac{\rho+\Pi}\rho,\\ \nonumber \Pi &=-\frac\Gamma\theta(p+\rho)=-\gamma\rho\frac\Gamma\theta,\quad w+1\equiv\gamma,\\ \nonumber w_{eff} &=-1+\gamma\left(1-\frac\Gamma\theta\right). \end{align} We see that $w_{eff}$ represents quintessence for $\Gamma<\theta$ ($\Gamma<3H$), phantom energy for $\Gamma>\theta$ ($\Gamma>3H$), and $\Gamma=\theta$ ($\Gamma=3H$) represents cosmological constant.

**Problem 6**

problem id: 150_dp0

Find the deceleration parameter $q$ for the cosmological model with particle creation.

The deceleration parameter $q$ is a measure of state of acceleration/deceleration of the Universe, defined as \begin{align} \nonumber q &=-1-\frac{\dot H}{H^2},\\ \nonumber \dot H &=-\frac32\gamma H^2\left(1-\frac\Gamma\theta\right). \end{align} Consequently, \[q=-1+\frac32\gamma\left(1-\frac\Gamma\theta\right)=-1+\frac32\gamma\left(1-\frac\Gamma{3H}\right).\]

**Problem 7**

problem id: 150_dp0

Obtain the equation for the scale factor in cosmological model with particle creation.

Combining the modified Friedmann equations with the EoS $p=w\rho$, one obtains the equation for the scale factor \[\frac{\ddot a}a +\frac{H^2}2\left[1+3w-\frac{1+w}H\Gamma\right]=0.\] Setting $\Gamma=0$, we reproduce the second Friedmann equation \[\frac{\ddot a}a=-\frac\rho6(1+3w).\]

**Problem 8**

problem id: 150_dp0

Calculate the derivative $dS/dt$ and find $S(t)$ for the Creation Matter Model (CMM).

In an open thermodynamical system, the entropy change $dS$ can be decomposed into an entropy flow $d_fS$ and the entropy creation matter $d_cS$ \[dS = d_fS+d_cS,\] with $d_cS>0$. As in a homogeneous system $d_fS=0$, but there is entropy production due to matter creation, so we have \begin{align} \nonumber \frac d{dt}\left(nsV\right)&=\dot nsV+ns\frac{dV}{dt},\quad \dot s=0,\\ \nonumber ns\frac{dV}{dt}&=ns\frac{da^3}{dt}=ns3HV=nsV\theta,\\ \nonumber \frac{dS}{dt} &=\frac d{dt}\left(nsV\right)=\dot nsV+nsV\theta=sV(\dot n +n\theta)=sVn\Gamma=S\Gamma. \end{align} Here $s$ is the entropy per particle. After integration we obtain \[S(t)=S_0\exp\left[3\int\limits_{a_0}^a\frac\Gamma\theta\frac{da}a\right]\] with $S_0=S(t_0)$ and $a_0=a(t_0)$.

**Problem 9**

problem id: 150_dp0

Calculate the scale factor dependence of the Universe's temperature for the Creation Matter Model (CMM).

For a substance with zero chemical potential \[nsT=\rho+p.\] For isentropic process, the entropy per particle remains constant, i.e. $\dot s=0$ and consequently \[\dot nsT+ns\dot T=\dot\rho(1+w)=\dot\rho\gamma.\] Using \[\dot\rho=-\theta(\rho+p+\Pi),\quad \Pi=-\frac\Gamma\theta(\rho+p),\] we obtain \[\frac{\dot T}T=(1-\gamma)\theta\left(1-\frac\Gamma\theta\right).\] Using \[\frac{dT}{dt}=\frac{dT}{da}Ha\] and $\theta=3H$, we find \[T=T_0a^{-3(\gamma-1)}\exp\left[3(\gamma-1)\int\limits_{a_0}^a\frac\Gamma\theta\frac{da}a\right],\] or using result of the previous problem \[T=T_1\left(\frac S{a^3}\right)^{\gamma-1},\quad T_1\equiv\frac{T_0}{S_0^{\gamma-1}}.\]

**Problem 10**

problem id: 150_dp0

Solve the previous problem using the thermodynamic identity \[T\left(\frac{\partial p}{\partial T}\right)_n+n\left(\frac{\partial\rho}{\partial n}_T\right)=\rho+p.\]

Combining the thermodynamic identity with the conservation equation, we obtain \[\frac{\dot T}T=\left(\frac{\partial p}{\partial\rho}\right)_n\frac{\dot n}n.\] In particular, for $\Gamma=0$, $\dot n +3nH=0$ and radiation or ultra relativistic particles ($p=\rho/3$) the temperature evolution law becomes \[\frac{\dot T}T=-\frac{\dot a}a\Rightarrow aT=const.\] If gravitational particles source is present, then $\dot n +3nH=n\Gamma$ and \[\frac{\dot T}T=\left(\frac{\partial p}{\partial\rho}\right)_n\frac{\dot n}n=\left(\frac{\partial p}{\partial\rho}\right)_n\left(\Gamma-3H\right).\]

**Problem 11**

problem id: 150_dp0

Solve the previous problem for cosmological substance with $\partial p/\partial\rho=w=const$.

Using \[\dot H+\frac32(1+w)H^2\left(1-\frac\Gamma{3H}\right)=0,\] we find \[\Gamma-3H=\frac2{1+w}\frac{\dot H}H\] and \[\frac{\dot T}T=\frac{2w}{1+w}\frac{\dot H}H.\] Consequently, the temperature of the fluid is governed by the following law \[T=T_0\left(\frac H{H_0}\right)^{\frac{2w}{1+w}}.\]

**Problem 12**

problem id: 150_dp0

Show that the matter creation process can be effectively described by the dimensionless ratio \[\beta=\frac\Gamma\theta=\frac\Gamma{3H}.\]

If $\Gamma\ll3H$, then $\beta\ll1$, and the matter creation process is qualitatively small. It leads to $n\propto R^{-3}$ and $H=2/(3t)$, as it should be for the Einstein-de Sitter model. The opposite case $\Gamma\gg3H$ corresponds to the extreme physical situation when the matter creation process is so intense that it compensates with excess the matter density decrease due to the expansion. behavior of such type could probably be realized only in very early Universe, as it took place, for example, during the inflation stage, which corresponded to the reheating. The intermediate and physically more realistic situation appears if this ratio is less or of order of unity: $\Gamma\le3H$. In particular, if $\Gamma=3H$, then density decrease due to the expansion is exactly compensated by its increase due to the matter creation, so the particle number density remains constant.

**Problem 13**

problem id: 150_dp0

How is the evolution of $a(t)$ affected by rate of change of the particle number $\Gamma$?

In the considered model, the Hubble parameter for the Universe, filled with the cosmic fluid with EoS $p=w\rho$, satisfies the dynamical equation \[\dot H+\frac32(1+w)H^2\left(1-\frac\Gamma{3H}\right)=0.\] The de Sitter solution $\dot H=0$ for $\Gamma=3H=constant$ is possible regardless of the EoS defining the cosmic fluid. Since the Universe is evolving, such a solution is unstable, and, as long as $\Gamma\ll3H$, conventional solutions without particle production are recovered. From the above equation, one may conclude that the main effect of $\Gamma$ is to provoke a dynamic instability in the space-time thereby allowing a transition from a de Sitter regime ($\Gamma\approx3H$) to a conventional solution, and vice versa.

**Problem 14**

problem id: 150_dp0

Let us assume a radiation dominated Universe ($w=1/3$, $\Gamma=\Gamma_r$). The dynamics is determined by the ratio $\Gamma_r/(3H)$ (see previous problem). The particle production must be strongly suppressed, $\Gamma_r/(3H)\ll1$, when the Universe enters the radiation phase. The simplest formula satisfying such a criterion is linear, namely: $\Gamma_r/(3H)=H/H_I$, where $H_I$ is the inflationary expansion rate associated with the initial de Sitter expansion ($H\le H_I$). Find the Hubble parameter dependence on the scale factor $H(a)$ for the radiation phase.

For $w=1/3$, $\Gamma=\Gamma_r$, $\Gamma_r/(3H)=H/H_I$, \[\dot H+\frac32(1+w)H^2\left(1-\frac\Gamma{3H}\right)=0\Rightarrow\dot H+2H^2\left(1-\frac H{H_I}\right)=0,\] or \[a\frac{dH}{da}+2H^2\left(1-\frac H{H_I}\right)=0\Rightarrow\frac{dH}{H-H_I}-\frac{dH}H=2\frac{da}a.\] The solution of this equation is \[H(a)=\frac{H_I}{1+Da^2}.\] Here $D\ge0$ is an integration constant.

**Problem 15**

problem id: 150_dp0

Using the result obtained in the previous problem, show that in the considering model the Universe evolves continuously from de Sitter stage to the standard radiation phase.

Integrating the equation \[H(a)=\frac{H_I}{1+Da^2},\] we obtain \[H_It=\ln\left(\frac a{a_*}\right)+\frac12Da_*^2\left(\frac a{a_*}\right)^2,\] where the integration constant$a_*$ defines the transition from the de Sitter stage to the beginning of the standard radiation epoch. At early times ($a\ll a_*$), when the logarithmic term dominates, one finds $a\approx a_*\exp(H_It)$ (de Sitter stage), while at late times, with $a\gg a_*$ and $H\ll H_I$ the obtained result reduces to \[a\approx\left(\frac2D\right)^{1/2}(H_It)^{1/2}\] and the standard radiation phase is reached.

**Problem 16**

problem id: 150_dp0

Find the radiation energy density $\rho_r$ as function of scale factor for the rate of change of the particle number $\Gamma=\Gamma_r$, $\Gamma_r/(3H)=H/H_I$ ($H_I$ is the inflationary expansion rate).

The radiation energy density is $\rho_r(a)=3H^2(a)$ ($8\pi G=1$) and using \[H(a)=\frac{H_I}{1+Da^2},\] we obtain \[\rho_r(a)=\frac{3H_I^2}{(1+Da^2)^2}\] or \[\rho_r(a)=\frac{\rho_I}{\left[1+Da_*^2\left(\frac a{a_*}\right)^2\right]^2}\] Here $\rho_I=3H_I^2$ is the greatest value of the energy density, characterizing the initial de Sitter stage and the integration constant $a_*$ defines the transition from the de Sitter stage to the beginning of the standard radiation epoch. We see again that the conventional radiation phase, $\rho_r\propto a^{-4}$, is attained when $a\gg a_*$.

**Problem 17**

problem id: 150_dp0

For cosmological model considered in the problems \ref{2016_07_p14}-\ref{2016_07_p16} find how did the cosmic temperature evolve at the very early stages?

For adiabatic production of relativistic particles the energy density scales as $\rho_r\propto T^4$ and the equation obtained in previous problem \[\rho_r(a)=\frac{\rho_I}{\left[1+Da_*^2\left(\frac a{a_*}\right)^2\right]^2}\] implies that \[T=\frac{T_I}{\sqrt{1+Da_*^2\left(\frac a{a_*}\right)^2}}\] where $T_I$ is the temperature of the initial de Sitter phase which must be uniquely determined by the scale $H_I$ - the inflationary expansion rate associated to the initial de Sitter phase. We see that the expansion proceeds isothermally: $T=const$ during the de Sitter phase ($a\ll a_*$) . After the de Sitter stage, the temperature decreases continuously in the course of the expansion. For $a\gg a_*$ and $H\ll H_I$, we obtain $T\propto a^{-1}$. Accordingly, the comoving number of photons becomes constant since $n\propto a^{-3}$, as expected for the standard radiation stage.

**Problem 18**

problem id: 150_dp0

Solve Problem \ref{2016_07_p14} for a substance with EoS $p=w\rho=(\gamma-1)\rho$ and $\Gamma=\Gamma_0H^2$ with $\Gamma_0=3\beta/H_e$.

For this choice of $\Gamma$, $H$ can be found from \[\dot H+\frac32(1+w)H^2\left(1-\frac\Gamma{3H}\right)=0,\] as \[H(a)=\frac{H_e}{\beta-(1-\beta)\left(\frac a{a_e}\right)^{\frac{3\gamma}2}},\] where $H_e$ and $a_e$ are chosen to be the values of the Hubble parameter and the scale factor respectively at some instant. If we identify $a_e$ with some intermediate value of the scale factor $a$ corresponding to $\ddot a=0$, i.e. to the transition from de Sitter accelerating phase to the standard decelerating matter phase, then we have \[\dot H+H^2=0\Rightarrow\dot H_e=-H^2_e,\] and using \[\frac\Gamma{3H}=1+\frac2{3\gamma}\frac{\dot H}{H^2},\] we find \[\beta=1-\frac2{3\gamma}.\]

**Problem 19**

problem id: 150_dp0

Find the deceleration parameter for the model considered in previous problem.

Using \[q=-1+\frac32\gamma\left(1-\frac\Gamma{3H}\right)\] for \[H(a)=\frac{H_e}{\beta-(1-\beta)\left(\frac a{a_e}\right)^{\frac{3\gamma}2}}\] we obtain \[q(z)=-1+\frac{3\gamma}2\left[1-\frac{\beta}{\beta+(1-\beta)(1+z)^{-3\gamma/2}}\right],\quad \frac{a_e}a=1+z.\]

**Problem 20**

problem id: 150_dp0

Let us now consider how the CMM (Creation matter model) works at later stages of the Universe's evolution in the transition from Einstein-de Sitter to a late time de Sitter stage. Due to the conservation of baryon number one have to take into account only the production rate of cold dark matter particles $\Gamma_{dm}$. What is the exact form of $\Gamma_{dm}$?

The evolution equation for the Hubble parameter in the flat $\Lambda$CDM reads \begin{equation}\label{2016_07_p20_e1} \dot H+\frac32H^2\left[1-\left(\frac{H_f}H\right)^2\right]=0, \end{equation} where $H_f$ is Hubble parameter of the final de Sitter stage $H\ge H_f$. All available observations are in accordance with the $\Lambda$CDM evolution both at the background and perturbative levels. Thus the form of $\Gamma_{dm}$ can be found from the requirement that the evolution equation for the model with the production of cold dark matter particles ($w=0$, $\Gamma=\Gamma_{dm}$) \begin{equation}\label{2016_07_p20_e2} \dot H+\frac32H^2\left(1-\frac{\Gamma_{dm}}{3H}\right)=0, \end{equation} leads to the same background evolution. Comparing equations (\ref{2016_07_p20_e1}) and (\ref{2016_07_p20_e2}) we find that for the fulfillment of this condition it is necessary that \[\frac{\Gamma_{dm}}{3H}=\left(\frac{H_f}H\right)^2,\] or $\Gamma_{dm}\propto H^{-1}$. The limiting value of the creation rate, $\Gamma_{dm}=3H_f$, leads to a late time de Sitter phase: $\dot H=0$, $H=H_f$ thereby showing that the de Sitter solution now becomes an attractor at late times.

**Problem 21**

problem id: 150_dp0

Let us consider the dynamical system in the context of bulk viscosity. In that case, we need to replace the pressure as $p\to p-3\zeta H$, where $\zeta$ is the coefficient of bulk viscosity. At what choice of the particle production rate $\Gamma$ the creation matter model would be equivalent to the bulk viscosity model?

In presence of the bulk viscosity the Friedmann equations are modified as \begin{align} \nonumber \rho &=3H^2, \nonumber \dot H &=-\frac12(\rho+p)+\frac32\zeta H. \end{align} Using the barotropic EoS: $p=(\gamma-1)\rho$, the second Friedmann equation can be written as \[\dot H=-\frac32\gamma H^2\left(1-\frac\zeta{\gamma H}\right),\] which coincides with \[\dot H=-\frac32\gamma H^2\left(1-\frac\Gamma{3H}\right),\] if one takes \[\zeta = \frac\Gamma3\gamma.\]

## Dynamical analysis of the Creation Matter Model

**Problem 22**

problem id: 150_dp0

Since the equation \[\dot H=-\frac32\gamma H^2\left(1-\frac\Gamma{3H}\right)\] is a one-dimensional first order differential equation, hence, the dynamics is obtained from the study of its critical points (or, fixed points) in which $\dot H=0$. Show that the fixed points $H_*$ in which $H_*\ne0$ represent the de Sitter Universe.

If $H=H_*$ is the fixed point, then $H_*=0$ or $\Gamma(H_*)=3H_*$. The fixed points $H_*\ne0$ correspond to $H=const$ and these solutions describe the de Sitter Universe.

**Problem 23**

problem id: 150_dp0

Find the fixed points and determine their nature for the case with the particle creation rate $\Gamma=n/H$, $n>0$.

Let us consider the general form of the equation $\dot h=f(H)$. If \[\frac{df(H_*)}{dH}<0\] at the fixed point $H_*$, then the fixed point is asymptotically stable (attractor), and on the other hand, if we have \[\frac{df(H_*)}{dH}<0,\] then the fixed point has unstable nature (repeller). The repeller point is suitable for early Universe, since it can describe the inflationary epoch, whereas the attractor point can describe the stable accelerating phase for the later time. To find the fixed points for the considered case with the particle creation rate $\Gamma=n/H$, $n>0$, one should solve the equation \[-\frac32\gamma H^2\left(1-\frac n{3H^2}\right)=0.\] The non-zero solutions are $H_*=\pm\sqrt{n/3}$. For the above given form for $\Gamma$, one has \[f(H)=-\frac32\gamma\left(H^2-\frac n3\right)\] and thus \[\frac{df(\pm\sqrt{n/3})}{dH}=\mp\sqrt{3n},\] which means that $H_*=\sqrt{n/3}$ is an attractor and $H_*=-\sqrt{n/3}$ is a repeller.

**Problem 24**

problem id: 150_dp0

Find the fixed points for the case with the particle creation rate $\Gamma=-\Gamma_0+mH+n/H$, $n>0$.

If $\Gamma(H)$ is a polynomial function of $H$, then the fixed point condition $\Gamma(H_*)=3H_*$ (for $H_*\ne0$), is a polynomial equation which has as many solutions (not necessary real solutions) as the highest power of the polynomial equation $\Gamma(H_*)=3H_*$. Hence, with the given particle creation rate we have to consider the following second-order polynomial equation \[(m-3)H_*^2-\Gamma_0H_*+n=0.\] In order to have the two critical points, as many as the inflationary phases of the Universe, we are interested in the case when $m\ne3$, and \[n\ne\frac{\Gamma_0^2}{4(m-3)}.\] Solving this equation, we find the critical points \[H_\pm=\frac{\Gamma_0}{2(m-3)}\left(1\pm\sqrt{1+\frac{4n(3-m)}{\Gamma_0^2}}\right).\] The case $m=3$ is special in a sense that there is only one critical point $H_-=n/\Gamma_0$, which is always an attractor for $\Gamma_0>0$, and a repeller for $\Gamma_0<0$.

**Problem 25**

problem id: 150_dp0

Perform dynamic analysis of the solutions, obtained in previous problem.

To perform the dynamical analysis, we have to divide the parameter space ($m,n\Gamma_0$) to different regions by the following conditions: \[\left(\Gamma_0>0,\ \Gamma_0<0;\quad m>3,\ m<3;\quad n>0,\ n<0;\quad\frac{4(3-m)}{\Gamma_0^2}>-1,\ \frac{4(3-m)}{\Gamma_0^2}<-1.\right)\] To have a non-singular Universe (without the Big Bang singularity) with the accelerated phases both at early and late times, one possibility is to have two critical points $H_+>H_->0$, where $H_+$ is a repeller and $H_-$ must be an attractor. If so, in principle, when the Universe leaves the point $H_+$, realizing the inflationary phase, and when it comes asymptotically to $H_-$, it enters into the current accelerated phase. Of course, the viability of the background has to be checked dealing with cosmological perturbations and comparing the theoretical predictions with the observations. For the considered model, the above described scenario can only happen in the region of the parameter space given by \[(m,n,\Gamma_0)=\left\{\Gamma_0>0,\ m>3,\ n\ge0,\ \frac{4(3-m)}{\Gamma_0^2}>-1\right\}.\]

**Problem 26**

problem id: 150_dp0

Prove that the dynamics generated by the creation matter model is equivalent to that driven by the single scalar field.

We use the energy density $\rho_\varphi$ and pressure $p_\varphi$ of the scalar field $\varphi$ given by \begin{align} \nonumber \rho_\varphi &=\frac12\dot\varphi^2+V(\varphi),\\ \nonumber p_\varphi &=\frac12\dot\varphi^2-V(\varphi). \end{align} Recall that the creation matter model is described by Friedmann equations \begin{align} \nonumber 3H^2&=\rho,\\ \nonumber \dot H &=-\frac32\gamma H^2\left(1-\frac\Gamma{3H}\right). \end{align} To show the equivalence, we perform the replacement \[\rho\to\rho_\varphi;\quad p-\frac\Gamma{3H}\gamma\rho\to p_\varphi.\] The Friedmann equations will become \begin{align} \label{2016_07_p26_e1} 3H^2&=\rho_\varphi,\\ \nonumber 2\dot H &=-\dot\varphi^2. \end{align} Note that Eqs. (\ref{2016_07_p26_e1}) uses the equations of General Relativity for a single scalar field. It means that we are dealing with the equivalence between an open system (the creation matter model) and the one driven by a single scalar field in the context of General Relativity. Further, the effective EoS parameter is given by \[w_{eff}=-1-\frac23\frac{\dot H}{H^2}=-1+\gamma\left(1-\frac\Gamma{3H}\right).\] Of course, \[w_{eff}=w_\varphi=\frac{\dot\varphi^2-2V}{\dot\varphi^2+2V}.\] Note that $2\dot H=-\dot\varphi^2<0$ implies that $w_{eff}>-1$, and thus one has $\Gamma<3H$ (the quintessence era).

**Problem 27**

problem id: 150_dp0

Find the scalar field $\varphi(H)$ and the potential $V(\varphi)$, satisfying the condition that the dynamics generated by the creation matter model is equivalent to the dynamics driven by a single scalar field.

From the Friedmann equations (see the previous problem) \begin{align} \nonumber 3H^2&=\rho_\varphi,\\ \nonumber 2\dot H &=-\dot\varphi^2; \end{align} with \begin{align} \nonumber \rho_\varphi&=\frac12\dot\varphi^2+V(\varphi),\\ \nonumber \dot H &=-\frac32\gamma H^2\left(1-\frac\Gamma{3H}\right); \end{align} we obtain \[V(\varphi)=\frac32H^2\left(2-\gamma+\frac{\gamma\Gamma}{3H}\right),\] \[\dot\varphi=\sqrt{-2\dot H}=\sqrt{3\gamma H^2\left(1-\frac\Gamma{3H}\right)}.\] Performing the change of variables \[dt=\frac{dH}{\dot H},\] we find \[\varphi=\int\sqrt{-\frac2{\dot H}}dH=-\frac2{\sqrt{\gamma}}\int\frac{dH}{\sqrt{3H^2-\Gamma H}}.\]

**Problem 28**

problem id: 150_dp0

Solve the previous problem for case \[\Gamma=-\Gamma_0+mH+n/H.\]

In the particular case $\Gamma=-\Gamma_0+mH+n/H$ one has \[\varphi=-\frac2{\sqrt{\gamma}}\int\frac{dH}{\sqrt{(3-m)H^2+\Gamma_0H-n}}.\] In the domain of interest (see Problem \ref{2016_07_p25}) \[(m,n,\Gamma_0)=\left\{\Gamma_0>0,\ m>3,\ n\ge0,\ \frac{4(3-m)}{\Gamma_0^2}>-1\right\},\] the integral could be solved analytically, giving as a result \[\varphi=\frac2{\sqrt{(m-3)\gamma}}\arcsin\left[\frac{m-3}\omega\left(\frac{\Gamma_0}{m-3}-2H\right)\right],\quad \omega\equiv\sqrt{\Gamma_0^2+4(3-m)n}.\] Inverting this expression, we find \[H=\frac1{2(m-3)}\left[\Gamma_0-\omega\sin\left(\frac{\sqrt{(m-3)\gamma}}2\varphi\right)\right].\] Using \[V(\varphi)=\frac32H^2\left(2-\gamma+\frac{\gamma\Gamma}{3H}\right)=\frac12\left[\left(6+(m-3)\gamma\right)H^2 -\gamma\Gamma_0H+\gamma n\right],\] we obtain \[V(\varphi)=\frac3{4(m-3)^2}\left[\Gamma_0-\omega\sin\left(\frac{\sqrt{(m-3)\gamma}}2\varphi\right)\right]^2 - \frac{\gamma\omega^2}{8(m-3)}\cos^2\left(\frac{\sqrt{(m-3)\gamma}}2\varphi\right).\]

**Problem 29**

problem id: 150_dp0

Prove equivalence of the dynamics generated by the creation matter model with the dynamics driven by a decaying vacuum energy density model.

For a generic decaying vacuum model Friedmann equations reduce to \begin{align} \label{2016_07_p29_e1} \rho+\Lambda(t)&=3\left(H^2+\frac k{a^2}\right),\\ \nonumber p-\Lambda(t)&=-2\frac{\ddot a}a-\left(H^2+\frac k{a^2}\right), \end{align} where $\rho$ and $p$ are the energy density and the pressure of the dominant fluid component (dark matter) with EoS $p=w\rho$ ($p=(\gamma-1)\rho$). The decaying vacuum causes a change in the number of particles of dark matter, so the equation describing particle concentration has a source term, i.e., \[\dot n+3Hn=n\Gamma^*.\] Here, $\Gamma^*$ is the rate of change of the number of particles. By combining Friedmann equations (\ref{2016_07_p29_e1}), or more directly, from the total energy conservation law one finds \begin{equation}\label{2016_07_p29_e2} \dot\rho+3H(\rho+p)=-\dot\Lambda(t). \end{equation} Since in the considered model the vacuum decay is the unique source of particle creation, we can write \[\dot\Lambda(t)=-\zeta n\Gamma,\] where $\zeta$ is a positive phenomenological parameter. At last, we can combine Friedmann equations in order to obtain the dynamics of decaying vacuum model, \begin{equation}\label{2016_07_p29_e3} \frac{\ddot a}a + \Delta(H^2+k^2)-\frac{1+w}2\Lambda(t)=0,\quad \Delta\equiv\frac{3w+1}2. \end{equation} A similar equation for the creation matter model has the form \begin{equation}\label{2016_07_p29_e4} \frac{\ddot a}a + \Delta(H^2+k^2)+\frac12\Pi=0, \end{equation} where the term $\Pi$ is the effective bulk pressure due to particle production (see the previous Problems). Comparing (\ref{2016_07_p29_e3}) and (\ref{2016_07_p29_e4}) we find \[\Pi=-(1+w)\Lambda(t)=-\gamma\Lambda(t).\] We have seen above that \[\Pi=-\gamma\rho\frac\Gamma{3H}.\] Consequently, both the theories are equivalent, if we set \[\Gamma=3H\frac\Lambda\rho.\] Note that above identification holds regardless of the curvature of the Universe.

**Problem 30**

problem id: 150_dp0

Show that for spatially flat Universe the inequalities $\Gamma\ll H$ and $\Lambda\ll H^2$ should be satisfied simultaneously.

For a spatially flat geometry, we have $\rho=3H^2$ and consequently \[\Gamma=3H\frac\Lambda\rho\Rightarrow\frac\Gamma H=\frac\Lambda{H^2}.\] In particular, if $\Gamma\ll H$, we find that $\Lambda\ll H^2$.

**Problem 31**

problem id: 150_dp0

Let us consider the particular case $\Lambda(t)=const$. Show that for such choice the creation matter model corresponds to the standard LCDM.

The conservation equation for the creation matter model is \[\dot\rho+3H(\rho+p+\Pi)=0.\] For cold dark matter with $w=0$ ($\gamma=1$) and a spatially flat geometry $\Pi=-\Gamma H$ and the conservation equation transforms into \begin{equation}\label{2016_07_p31_e1} \dot\rho+3H(\rho-\Gamma H)=0. \end{equation} In the considered case \[\Gamma=3H\frac\Lambda\rho=\frac\Lambda H\] and (\ref{2016_07_p31_e1}) takes on the form \begin{equation}\label{2016_07_p31_e2} \dot\rho+3H(\rho-\Lambda)=0. \end{equation} Performing the integration, one finds \begin{equation}\label{2016_07_p31_e3} \rho(a)=\Lambda+\rho_{dm0}a^{-3}, \end{equation} where $\rho_{dm0}$ is an integration constant that must quantify the current amount of matter that is clustering to the present time. The equation (\ref{2016_07_p31_e3}) describes the dynamics of a creation matter model that behaves like the LCDM model.

**Problem 32**

problem id: 150_dp0

Calculate the cosmographic parameter $j(H)$ for the cosmological model with particle creation.

\[j=\frac{\ddot H}{H^3}-3q-2;\] \[\dot H=-\frac32\gamma H^2\left(1-\frac\Gamma{3H}\right);\] \[q=-1+\frac32\gamma\left(1-\frac\Gamma{3H}\right).\] Let us introduce the function \[F\equiv\frac32\gamma\left(1-\frac\Gamma{3H}\right).\] In terms of this function \[\dot H=-H^2 F.\] As \[\frac d{dt}=\dot H\frac d{dH},\] we obtain \[\ddot H=\frac d{dt}\dot H=\dot H\frac{d\dot H}{dH}=-\dot H\frac d{dH}(H^2F)=H^3\left(2F+H\frac{dF}{dH}\right),\] \[\frac{dF}{dH}=-\frac12\frac\gamma H\left(\frac{d\Gamma}{dH}-\frac\Gamma H\right).\] Consequently, \[j=1+F\left(2F+H\frac{dF}{dH}\right)-3F,\] \[\frac{dF}{dH}=-\frac12\frac\gamma H\left(\frac{d\Gamma}{dH}-\frac\Gamma H\right),\] \[j=1+F\left[2F-3-\frac\gamma2\left(\frac{d\Gamma}{dH}-\frac\Gamma H\right)\right].\] The result can be checked by passing to the LCDM limit ($\Gamma\to0$, $\gamma=1$ or $\gamma=-1$ ???). In this case $F\to3/2$ or $F\to0$. In both cases $j=1$ in accordance with the LCDM.

**Problem 33**

problem id: 150_dp0

Solve the previous problem using the cosmographical relation \[j=-\frac1H\frac{dq}{dt}+q(1+2q).\]

Using $q=-1+F$ and \[\dot H=-H^2 F,\quad F\equiv\frac32\gamma\left(1-\frac\Gamma{3H}\right),\] we find \[\frac{dq}{dt}=\dot H\frac{dF}{dH}=-H^2 F\frac{dF}{dH}\] and \[j=1+F\left(2F+H\frac{dF}{dH}\right)-3F,\] \[\frac{dF}{dH}=-\frac12\frac\gamma H\left(\frac{d\Gamma}{dH}-\frac\Gamma H\right),\] \[j=1+F\left[2F-3-\frac\gamma2\left(\frac{d\Gamma}{dH}-\frac\Gamma H\right)\right],\] which coincides with the result of the previous problem.

* However, in connection with the particle production at the expense of gravitational field of the expanding Universe, we recall that long back ago, Zeldovich [Ya. B. Zeldovich, JETP Lett. 12, 307 (1970)] introduced some bulk viscosity mechanism which is responsible for particle production. However, later on, Lima and Germano [J. A. S. Lima and A. S. M. Germano, Phys. Lett. A 170, 373 (1992); For a more general macroscopic approach see also R. Silva, J. A. S. Lima and M. O. Calv~ao, Gen. Rel. Grav. 34, 865 (2002).] showed that although both the processes, namely the bulk viscosity mechanism by Zeldovich [Ya. B. Zeldovich, JETP Lett. 12, 307 (1970).] and the gravitational particle production produce the same dynamics of the Universe, but in principle they are completely different from a thermodynamical point of view. Since we describe the particle production, we would like also to note an analogy which exists between the models driven by the particle production and the models of Steady State Cosmology developed in [F. Hoyle and J. V. Narlikar, Proc. Roy. Soc. A 282, 191 (1964); ibdem, 290, 143 (1966); see also Action at a Distance in Physics and Cosmology, 1974 (New York, Freeman).]. *

**Problem 34**

problem id: 150_dp0

In the model with $\Gamma=-\Gamma_0+mH$, where $0\le m\le3$ and $-3H_0\le\gamma\le0$, find the time dependence of the scale factor.

For $\Gamma=-\Gamma_0+mH$ the evolution equation for the hubble parameter takes on the form \[\dot H+\frac32H^2\left(1-\frac m3+\frac{\Gamma_0}{3H}\right).\] Its solution reads \begin{equation}\label{2016_07_p34_e1} H(t)=\frac{\Gamma_0/(m-3)}{1-\exp(\Gamma_0t/2)}. \end{equation} Then for the scale factor one finds \begin{equation}\label{2016_07_p34_e2} a(t)=a_0\left[H_0\left(\frac{m-3}{\Gamma_0}-1\right)\left(e^{-\Gamma_0t/2}-1\right)\right]^{\frac2{3-m}}. \end{equation} In the limit $\Gamma_0\to0$ \[a(t)=a_0\left[\frac{3-m}2H_0t\right]^{\frac2{3-m}}.\] Naturally at $m=0$ this solution is reduced to that of the Einstein-de Sitter model (Universe filled with non-relativistic matter).

**Problem 35**

problem id: 150_dp0

In the model considered in the previous problem, express age of the Universe in terms of the parameters $\Gamma_0$ and $m$.

Setting $a=a_0$ in the expression for the scale factor obtained in the previous problem, one can find age of the Universe: \begin{equation}\label{2016_07_p35_e1} t_0=2\Gamma_0\ln\left(1-\frac{\Gamma_0/H_0}{m-3}\right). \end{equation}

**Problem 36**

problem id: 150_dp0

Find deceleration parameter in the model with matter creation rate equal to $\Gamma$.

As we have seen above, in the considered model the evolution equation for the scale factor takes on the form \[a\ddot a+\frac12\left(1-\frac\Gamma H\right)\dot a^2=0.\] Using the definition of the deceleration parameter \[q\equiv-\frac{\ddot aa}{\dot a^2},\] one finds \[q=\frac12\left(1-\frac\Gamma H\right).\]

**Problem 37**

problem id: 150_dp0

Show that in the creation matter model with $\Gamma=-\Gamma_0+mH$ in the case $\Gamma_0=0$ the expansion of Universe is always accelerated if $m>1$ and always decelerated if $m<1$.

Using the result of the previous problem, one obtains \[q=\frac12\left[1-m+\Gamma_0/H\right].\] It immediately follows then that at $\Gamma_0=0$ the deceleration parameter does not depend on time. It easy to see that at $m>1$ the expansion of Universe is always accelerated, and it is always decelerated for $m<1$. As the transition from the decelerated expansion to the accelerated one is required by the observation, the expression for the matter creation rate $\Gamma$ must contain the constant term $-\Gamma_0\ne0$.

**Problem 38**

problem id: 150_dp0

Find the deceleration parameter as a function of the redshift in the creation matter model.

Exclude time from the above obtained expressions for the scale factor (\ref{2016_07_p34_e2}) and Hubble parameter (\ref{2016_07_p34_e1}), and use the relation \[1+z=\frac{a_0}a\] to find the hubble parameter as function of the redshift \[H(z)=\frac{\Gamma_0}{m-3}+\left(H_0-\frac{\Gamma_0}{m-3}\right)(1+z)^{\frac{3-m}2}.\] Then \begin{equation}\label{2016_07_p38_e1} q(z)=\frac12\left[1-m+\frac{\Gamma_0}{H(z)}\right] =\frac12\left[1-m+\frac1{\frac1{m-3}+\left(\frac{H_0}{\Gamma_0}-\frac1{m-3}\right)(1+z)^{\frac{3-m}2}}\right]. \end{equation} In the limit $\Gamma_0\to0$ \[q=\frac{1-m}2\] and for $m\to0$ \[q(z)=\frac12\left[1+\frac{\Gamma_0/H_0}{(1+\frac{\Gamma_0}{3H_0})(1+z)^{\frac32}-\frac{\Gamma_0}{3H_0}}\right].\]

**Problem 39**

problem id: 150_dp0

Find the redshift value $z_t$ corresponding to the transition from the decelerated expansion to the accelerated one in the creation matter model.

Using the expression (\ref{2016_07_p38_e1}) for $q(z)$ obtained in the previous Problem with the condition $q(z_t)=0$, one finds that the transition from the decelerated expansion to the accelerated one in the cobsidered model takes place at the redshift value \[z_t=\left[\frac{2\Gamma_0/H_0}{(1-m)(m-3-\Gamma_0/H_0)}\right]^{\frac2{3-m}}-1.\]

**Problem 40**

problem id: 150_dp0

Express age of the Universe in terms of the $z_t$ (see the previous Problem) in the creation matter model with $m=0$.

For the case $m=0$ \[\frac{\Gamma_0}{3H_0}=-\frac{1}{1+2(1+z_t)^{-\frac32}}.\] Substituting the latter expression into the result (\ref{2016_07_p35_e1}) of the Problem \ref{2016_07_p35} \[t_0=2\Gamma_0\ln\left(1-\frac{\Gamma_0/H_0}{m-3}\right),\] one obtains \[H_0t_0=\frac{4+2(1+z_t)^{\frac32}}{3(1+z_t)^{\frac32}}\ln\left[1+\frac{(1+z_t)^{\frac32}}2\right].\]

# New on the Ricci scalar etc

**Problem 1**

problem id: 10_1

Show that the Ricci scalar can be written in the following form \[R=6H^2[q(t)-\Omega(t)].\]

\[R=g^{\mu\nu}R_{\mu\nu}=-6\left(\frac{\ddot a}{a}+H^2+\frac k{a^2}\right)=6H^2\left(-\frac{\ddot a}{aH^2}-1-\frac k{a^2H^2}\right),\] \[H^2=\frac{8\pi G}3\rho-\frac k{a^2}\Rightarrow1+\frac k{a^2H^2}=\Omega,\quad\Omega\equiv\frac\rho{\rho_{cr}},\quad \rho_{cr}\equiv\frac{3H^2}{8\pi G},\]\[\frac{\ddot a}{aH^2}\equiv q\Rightarrow R=6H^2[q(t)-\Omega(t)].\]

**Problem 2**

problem id: 10_2

Show that for a one-component flat Universe filled with ideal fluid with the EoS $p=w\rho$ the Ricci scalar can be written in the following form \[R=8\pi G\rho(3w-1).\]

Use result of the previous problem \[R=6H^2[q(t)-\Omega(t)].\] For the one-component flat Universe \[q=\Omega(1+3w)/2,\] and taking into account that \[\Omega=\frac{8\pi G}{3H^2}\rho\] one finds \[R=8\pi G\rho(3w-1).\]

**Problem 3**

problem id: 10_3

Find time derivative of the scalar curvature for the spatially flat Universe filled with a barotropic fluid.

Using the conservation and Friedmann equations we obtain \begin{align} \nonumber \dot\rho & =-\sqrt{3\rho}(\rho+p),\\ \nonumber \dot p & = \frac{dp}{d\rho}\dot\rho=-\sqrt{3\rho}(\rho+p)\frac{dp}{d\rho}, \end{align} where we have used $p=p(\rho)$. According to the result of the problem (#10_1) $R=-\rho+3p$. Consequently \[\dot R=-\dot\rho+3\dot p=-\dot\rho\left(1-3\frac{dp}{d\rho}\right)=\sqrt{3\rho}(\rho+p)\left(1-3\frac{dp}{d\rho}\right).\]

**Problem 4**

problem id: 10_4

Compare entropy of the Sun in the present state and right after its transformation to a black hole due to compression.

Let us treat the Sun as an ideal gas composed of nucleons, then \[S_\odot=k_BN\left\{\ln\left[\left(\frac{mk_BT}{2\pi\hbar^2}\right)^{3/2}\frac V N\right]+\frac52\right\}.\] here $N\approx10^{57}$, $V\approx1.4\times10^{33}cm^3$, $T\approx10^7K$, so \[S_\odot\approx1.4\times10^{53}k_B\approx2\times10^{35}\frac J K.\] On the other hand, for a black hole with the solar mass one obtains \[S_{BH,\odot}=\frac{k_B}4\frac{A_{BH,\odot}}{l_{Pl}^2}\approx10^{77}k_B\approx1.4\times10^{54}\frac J K.\] Therefore \[\frac{S_{BH,\odot}}{S_\odot}\approx0.7\times10^{19}.\] As $\Delta S=-\Delta I$ ($I$ is amount of information), the transformation of the Sun into the black hole is followed by huge loss of information (the information paradox). It relates to the fact that the simplest (uncharged and non-rotating) black hole is characterized by just single parameter --- its mass, while the Sun has quite complicated structure. However we derived entropy of the Sun $S_\odot$ in assumption that the Sun can be described solely by number of nucleons forming an ideal gas.

**Problem 5**

problem id: 10_5

Show that black holes have negative thermal capacity.

Temperature $T$ of a black hole is inversely proportional to its mass $M$: \[T\propto M^{-1}.\] Therefore when mass increases the black hole does not heat and rather cools down and therefore its thermal capacity is negative. The same property holds for any Newtonnian self-gravitating system. Therefore for such systems one cannot use the canonical distribution and only the micro-canonical one is allowed.

**Problem 6**

problem id: 10_6

What are the physical reasons for the negative thermal capacity in Newtonnian self-gravitating systems?

For a stable system of $N$ particles coupled by the potential forces $\vec{F}_k$ the virial theorem holds: \[2\langle T\rangle=-\sum\limits_{k=1}^N\langle\vec{r}_k\vec{F}_k\rangle,\] where $\langle T\rangle$ is average total kinetic energy and $\vec{F}_k$ is the force acting on the $k$-th particle. In particular, when the force has corresponding potential energy of interaction $V(r)$ proportional to $n$-th power of the inter-particle distance $r$, the virial theorem takes on the form \[2\langle T\rangle =n\langle U\rangle,\] i.e. doubled avarage total kinetic energy $T$ equals to $n$ times the average average total potential energy $U$. For the Newtonnian interaction ($n=-1$): \[2\langle T\rangle =-\langle U\rangle.\] Let us apply this theorem to evolution of massive stars. Black holes can form on final stages of evolution of such objects. Stationary state of the star corresponds to hydrostatic and thermal equilibrium. In the equilibrium state the negative potential energy of the star (the energy of the gravity compression) by absolute value is twice greater than the thermal energy (the kinetic one). It is this peculiarity that leads to the negative thermal capacity of the considered system. During the compression the potential energy transforms into the kinetic energy os the falling stellar layers, and its depth heats up, so that the thermal energy acquired by the star during the compression is exactly twice greater of the energy lost due to the radiation.

**Problem 7**

problem id: 10_7

How the negative thermal capacity of the stars (in particular the Sun) affects the course of the nuclear reactions inside them?

Due to the negative thermal capacity the temperature in central part of the star grows (see the previous problem) and thus the continuous thermonuclear synthesis of chemical elements takes pace. For instance, the reaction of transformation of hydrogen into helium in the present Sun occurs at temperature of $15$ millions degrees. When, $4$ millions years later, all the hydrogen in the center of the Sun will be transformed into helium, further synthesis of carbon atoms from the helium ones will require considerably higher temperature, about $100$ millions degrees (because the electric charge of the helium nuclei is twice greater than that of hydrogen, so one needs much higher temperature to close them in to distance of $10^{-13}cm$). Exactly this temperature will be provided by the negative thermal capacity of the Sun to the moment of burning up in its depth the thermonuclear reaction of transformation of helium into carbon.

**Problem 8**

problem id: 10_8

From the three parameters (mass, charge and angular momentum) of the Kerr-Newman black hole form all possible length scales.

\[R_1=\frac{MG}{c^2},\quad R_2=\frac{\sqrt{G}Q}{c^2},\quad R_3=\frac{J}{Mc}.\]

**Problem 9**

problem id: 10_9

Following [O. Luongo and H. Quevedo, Self-accelerated universe induced by repulsive e?ects as an alternative to dark energy and modi?ed gravities, arXiv: (1507.06446)] let us introduce the parameter $\lambda=-\frac{\ddot a}{a}=qH^2$, so that $\lambda<0$ when the Universe is accelerating, whereas for $\lambda>0$ the Universe decelerates. Luongo and H. Quevedo showed, that the parameter $\lambda$ can be considered as an eigenvalue of the curvature tensor defined in special way. In particular, for FLRW metric the curvature tensor $R$ can be expressed as a ($6\times6$)-matrix \[R=diag(\lambda,\lambda,\lambda,r,r,r),\quad r\equiv\frac13\rho.\] The curvature eigenvalues reflect the behavior of the gravitational interaction and if gravity becomes repulsive in some regions, the eigenvalues must change accordingly; for instance, if repulsive gravity becomes dominant at a particular point, one would expect at that point a change in the sign of at least one eigenvalue. Moreover, if the gravitational field does not diverge at infinity, the eigenvalue must have an extremal at some point before it changes its sign. This means that the extremal of the eigenvalue can be interpreted as the onset of repulsion. Obtain the onset of repulsion condition in terms of cosmographic parameters.

As mentioned above, the onset of repulsion is determined by an extremal of the eigenvalue, i.e. $\dot\lambda=0$, \[\lambda=qH^2\rightarrow\dot\lambda=\dot qH^2+2qH\dot H=0.\] Using the result of the previous problem for $\dot q$ and $\dot H$ we find that repulsion onset condition $\dot\lambda$ reduces to \[j=-q.\]

**Problem 10**

problem id: 10_10

Represent result of the previous problem in terms of the Hubble parameter and its time derivatives.

Solving the system of equations \begin{align} \nonumber \dot H &=-H^2(1+q),\\ \nonumber \ddot H &=H^3(j+3q+2), \end{align} with respect to the variables $q$ and $j$, one finds that the condition $j=-q$ transforms into \[\frac{\ddot H}{H}=-2\dot H.\]

# New on the deceleration parameter

**Problem 1**

problem id: 150_dp0

Show that the fact that in the present time $t_0$ the relation $H_0t_0\approx1$ holds unavoidably follows existance of the accelerated expansion stage of the Universe evolution.

The combination of the definitions \[H=\frac{\dot a}{a}=\frac{d\ln a}{dt},\quad q(t)=-\frac{\ddot a a}{\dot a^2}\] give the equation \[\dot H+(1+q)H^2=0.\] Integrating this equation between the limits $x=0$ and $x=1$, $x\equiv t/t_0$, and taking into account that the present value $H_0t_0\approx1$ we obtain \[0\approx\int\limits_0^1q(x)dx.\] At a very high redshift $q(x)>0$. There must be periods (or a period) of accelerated expansion in the history of the universe in order to turn the integral $\int_0^1q(x)dx$ to zero.

**Problem 2**

problem id: 150_dp2

Suppose that $dq/dt=f(q)$. Find the Hubble parameter in terms of $q$.

\[1+q=\frac d{dt}\left(\frac1H\right),\quad \frac{dq}{dt}=f(q)\Rightarrow dt=\frac{dq}{f(q)},\] \[\frac1{H(q)}=\int\frac{1+q}{f(q)}dq.\]

**Problem 3**

problem id: 150_dp2

Find connection between the scalar curvature and deceleration parameter for the flat Universe.

Using the relation \[R=-6\left(\frac{\ddot a}a+H^2\right)\] and definition of the deceleration parameter \[q\equiv-\frac{\ddot a}{aH^2},\] one finds \[R=-6H^2(1-q).\]

**Problem 4**

problem id: 150_dp3

Show that the deceleration parameter $q$ relates the density of the Universe $\rho$ to the critical density $\rho_{cr}$ through \[q=\frac12(1+3w)\frac\rho{\rho_{cr}}.\]

\[H^2=\frac{8\pi G}3\rho-\frac k{a^2}\Rightarrow1+\frac k{a^2H^2}=\frac{\rho}{\rho_{cr}}.\] Using \[q=\frac12(1+3w)\left(1+\frac k{a^2H^2}\right),\] we obtain \[q=\frac12(1+3w)\frac\rho{\rho_{cr}}.\]

**Problem 5**

problem id: 150_dp4

Solve the previous problem using the second Friedmann's equation.

\[-\frac{\ddot a}a=\frac{4\pi G}3(\rho+3p)\Rightarrow-\frac{\ddot a}{aH^2}=\frac{4\pi G}{3H^2}(1+w)\rho,\] \[q=\frac12(1+3w)\frac\rho{\rho_{cr}}.\]

# New on cosmology with power and hybrid expansion laws

Let us consider a general class of power-law cosmology described by the scale factor \[a(t)=a_0\left(\frac t{t_0}\right)^\alpha\] where $\alpha$ is a dimensionless positive parameter.

**Problem 6**

problem id: 150_dp5

Obtain deceleration parameter in the power-law cosmology.

\[q\equiv=-\frac{\ddot a}{aH^2}=\frac1\alpha-1.\] The positivity of $\alpha$ leads to $q>-1$.

**Problem 7**

problem id: 150_dp6

Express the scale factor $a(t)$ and Hubble parameter $H(z)$ in terms of the deceleration parameter.

\[a(t)=a_0\left(\frac t{t_0}^{\frac1{1+q}}\right),\] \[H(z)=H_0(1+z)^{1+q}.\]

**Problem 8**

problem id: 150_dp7

Obtain the comoving distance $r(z)$ in the power-law cosmology.

\[r(z)=\frac1{\sqrt{|\Omega_c|}}F\left(|\Omega_c|\int\limits_0^zdz'\frac{H_0}{H(z')}\right),\quad \Omega_c\equiv-\frac k{a_0^2H_0^2}.\] \[F(x)\equiv\left\{\begin{array}{ll} \sinh(x), & k=-1;\\ x, & k=0;\\ \sin(x), & k=+1. \end{array}\right.\] Using result of the previous problem, one finally obtains \[r(z)=\frac1{\sqrt{|\Omega_c|}} F\left(|\Omega_c|q^{-1}\left[1-(1+z)^{-q}\right]\right).\]

**Problem 9**

problem id: 150_dp8

Find the comoving distance in the Milne model.

The Milne model (empty Universe with $k=-1$) can be viewed as a power-law cosmology with $\Omega_c=1$, $\alpha=1$ or $q=0$. For this model therefore \[r(z)=\sinh\left[\ln(1=z)\right].\]

**Problem 10**

problem id: 150_dp9

In the power-law cosmology, find time dependence of the CMB temperature in terms of the deceleration parameter.

\[\frac{T_0}T=\frac a{a_0}=\left(\frac t{t_0}\right)^\alpha =\left(\frac t{t_0}\right)^{\frac1{1+q}},\] \[T(t)=T_0\left(\frac t{t_0}\right)^{-\alpha} =T_0\left(\frac t{t_0}\right)^{-\frac1{1+q}}.\]

Let us consider now a simple generalization of power-law cosmology, called the hybrid expansion law [O. Akarsu et. al. Cosmology with hybrid expansion law: scalar field reconstruction of cosmic history and observational constraints, arXiv:gr-qc/1307.4911] \[a(t)=a_0\left(\frac t{t_0}\right)^\alpha e^{\beta\left(\frac t{t_0}-1\right)}\] where $\alpha$ and $\beta$ are non-negative constants.

**Problem 11**

problem id: 150_dp10

Find Hubble parameter, deceleration parameter and jerk parameter for the hybrid expansion law.

\begin{align} \nonumber H & = \frac\alpha t + \frac\beta{t_0},\\ \nonumber q & = \frac{\alpha t_0^2}{(\beta t+\alpha t_0)^2} -1,\\ \nonumber j & = \frac{\alpha t_0^2}{(\beta t+\alpha t_0)^2}(2t_0-3\beta t-3\alpha t_0)+1. \end{align}

**Problem 12**

problem id: 150_dp11

Find asymptotes of the scale factor and of the cosmographic parameters for the hybrid expansion law at $t\to0$ and $t\to\infty$.

For $t\to0$ the cosmological parameters approximate to the following: \[a\to a_0\left(\frac t{t_0}\right)^\alpha,\quad H\to\frac\alpha t,\quad q\to-1+\frac1\alpha,\quad j\to1-\frac3\alpha+\frac2{\alpha^2}.\] In the limit $t\to\infty$ we have \[a\to a_0 e^{\beta\left(\frac t{t_0}-1\right)},\quad H\to\frac\beta{t_0},\quad q\to-1,\quad j\to1.\]

**Problem 13**

problem id: 150_dp12

Find the moment of time $t_{tr}$ when the transition from deceleration to acceleration takes place.

\[q=0\to\frac{t_{tr}}{t_0}=\frac{\sqrt\alpha-\alpha}{\beta}.\]

**Problem 14**

problem id: 150_dp13

Find the range of the parameter $\alpha$ variation for the hybrid expansion law.

Use results of the previous problem. From the condition $t_{tr}>0$ it follows that \[\sqrt\alpha-\alpha>0\Rightarrow0<\alpha<1.\]

**Problem 15**

problem id: 150_dp14

In 1983, Berman [M. Berman, A special law of variation for Hubble's parameter. Nuovo Cimento B 74, 182 (1983)] proposed a special law of variation of Hubble parameter in FLRW space-time, which yields a constant value of DP, \[H=Da^{-n}.\] Find time dependence of the scale factor providing the constant deceleration parameter.

Integration of the relation \(H=Da^{-n}\) leads to \[a(t)=(nDt+c_1)^{\frac1n}\] where $c_1$ is a constant of integration.

**Problem 16**

problem id: 150_dp15

For Berman's law of variation for Hubble's parameter \(H=Da^{-n}\) find the deceleration parameter and analyze what values of the parameter $n$ correspond to accelerated expansion, and which --- to decelerated one.

The value of deceleration parameter in this case is \[q=n-1,\] which is a constant. As usual, sign of $q$ determines the expansion regime. A positive sign of $q$, i.e., $n>1$ corresponds to the standard decelerating model whereas the negative sign of $q$, i.e. $0<n<1$, corresponds to accelerating model. The expansion of the Universe at a constant rate corresponds to $n=1$, that is, $q=0$.

Linearly varying deceleration parameter

Inspired by O. Akarsu et al. Probing kinematics and fate of the Universe with linearlytime-varying deceleration parameter, arXiv:gr-qc/1305.5190 A general approach is to expand the deceleration parameter in Taylor series is \[q(x)=q_0+q_1\left(1-\frac x{x_0}\right)+q_2\left(1-\frac x{x_0}\right)^2+\ldots\] where $x$ is a some cosmological parameter as cosmic scale factor $a$, cosmic redshift $z$, cosmic time $t$ etc. As the first step one can take the following linear approximation \[q(x)=q_0+q_1\left(1-\frac x{x_0}\right).\]

**Problem 17**

problem id: 150_dp16

Consider the linearly varying deceleration parameter in terms of cosmic redshift $z$. Analyze advantages and defeats of such interpretation.

In this case \[\frac x{x_0}=\frac{a_0}a=1+z\] and \[q=q_0+q_1z.\] We note that the deceleration parameter grows monotonically with no limits as we go back to earlier times of the Universe. On the other hand, the model is well behaved in the future such that the Universe either expands forever (provided that $q_0\ge q_1-1$ or ends with a Big Rip in finite future (provided that $q_0<q_1-1$). However, the predicted future of the Universe using such parametrization cannot be considered so reliable since the observational constraints obtained using it with high redshift data are not reliable.

**Problem 18**

problem id: 150_dp17

Unlimited growth of the deceleration parameter for large $z$ in the parametrization, used in the previous problem, forces us to consider the linearly varying deceleration parameter in terms of scale factor. Make transition to such parametrization.

Substituting \[\frac x{x_0}=\frac{a_0}a\] we obtain \[q=q_0+q_1\left(1-\frac a{a_0}\right).\] Such parametrization can be written in terms of redshift as follows: \[q=q_0+q_1\frac z{1+z}.\]

**Problem 19**

problem id: 150_dp19

Treating the Universe as a dynamical system it is useful to consider the parametrization of the deceleration parameter directly in terms of cosmic time $t$. Make transition to such parametrization.

Such parametrization can be obtained by substitution \[\frac x{x_0}=\frac t{t_0},\quad q=q_0+q_1\left(1-\frac t{t_0}\right).\] In contrast to the two previous problems, this model cannot be used for observational analysis directly. One should first solve it for the scale factor explicitly, and then obtain the time red-shift relation. This procedure can be facilitated if one first solves the equation with separating variables \[\frac{dH}{dt}=-H^2(1+q).\] For the chosen parametrization \[H(t)=2\frac{t_0}t\left[2(1+q_0+q_1)t_0-q_1t\right]^{-1}.\] Using the obtained expression for $H(t)$ to integrate the equation \[\frac{dz}{dt}=-(1+z)H(t),\] we obtain \[q=q_0+q_1\left(1-\frac{2(1+q_0+q_1)}{q_1+(2+2q_0+q_1)(1+z)^{1+q_0+q_1}}\right).\]

**Problem 20**

problem id: 150_dp19

Consider a single-component flat Universe with a fluid described by an EoS parameter expressed as a first order Taylor expansion in cosmic time: \[w=w_0+w_1(1-t),\] where $w_0$ and $w_1$ are real constants and $t$ is the normalized time. Find the corresponding parametrization for the deceleration parameter.

\[q=\frac\Omega2+\frac32\sum\limits_iw_i\Omega_i=\frac12+\frac32\left[w_0+w_1(1-t)\right].\] Such an approach enables us to use the cosmographic (model-free) constraints to test the EoS for different components.

**Problem 21**

problem id: 150_dp20

In a Universe filled with a fluid characterized by the EoS parameter $w=w_0+w_1(1-t/t_0)$, find time dependence of the scale factor. [S. Kumar, Probing the matter and dark energy sources in a viable Big Rip model of the Universe, arXiv:1404.1910]

Using \[3H^2=\rho,\quad (8\pi G=1),\] \[2\dot H+3H^2=-p,\] find the following solution for the scale factor: \[a(t)=a_0\left[\frac{w_1t}{2(1+w_0+w_1)t_0-w_1t}\right]^{\frac2{3(1+w_0+w_1)}},\] where $a_1$ is a constant of integration, the first constant of integration being assumed zero.

**Problem 22**

problem id: 150_dp21

In a Universe filled with a fluid characterized by the EoS parameter $w=w_0+w_1(1-t/t_0)$, find the Hubble parameter, deceleration parameter and the jerk parameter.

Using expression for the scale factor obtained in the previous problem, one finds \begin{align} \nonumber H(t) & =\frac{4t_0}{3t[2(1+w_0+w_1)t_0-w_1t]},\\ \nonumber q(t) & =\frac12 +\frac32\left[w_0+w_1\left(1-\frac t{t_0}\right)\right],\\ \nonumber j(t) & =\frac{27w_1^2t^2}{8t_0^2} -\frac{9w_1Wt}{4t_0} +\frac{W(1+W)}2,\quad W\equiv1=3w_0+3w_1. \end{align}

**Problem 23**

problem id: 150_dp22

In a Universe filled with a fluid characterized by the EoS parameter $w=w_0+w_1(1-t/t_0)$, find the pressure and energy density.

\begin{align} \nonumber \rho(t) & =\frac13H^2= \frac{16t_0^2}{3t^2[2(1+w_0+w_1)t_0-w_1t]^2},\\ \nonumber \rho(t) & =w(t)\rho(t)= \frac{16t_0[-w_1t+(w_0+w_1)t_0]}{3t^2[2(1+w_0+w_1)t_0-w_1t]^2}. \end{align}

**Problem 24**

problem id: 150_dp23

Show that the Universe filled with a fluid characterized by the EoS parameter $w=w_0+w_1(1-t/t_0)$ achieves de Sitter phase ($q=-1$) at the end of its half life.

As we have seen in the previous problems, that the parameters $a,H,p,\rho$ diverge to infinity at two distinct epochs $t=0$ and $t=2(1+w_0+w_1)t_0/w_1$. On the other hand, the parameters $w,q$ and $j$ are well behaved from former to the later epoch. It implies that in this model, the Universe begins with Big Bang at $t=0$ and ends in Big Rip at $t_{BR}=2(1+w_0+w_1)t_0/w_1$. The de Sitter time $t_{dS}$, which is the solution of $q(t=t_{dS})=-1$, is $t_{dS}=(1+w_0+w_1)t_0/w_1$. Thus, the Universe achieves de Sitter phase at the end of its half life.

# New to Quantum cosmology

**Problem 1**

problem id: 150_dp24

Find the solutions corrected by LQC Friedmann equations for a matter dominated Universe.

Solving the corrected first Friedmann equation and the conservation equation for a matter dominated Universe \[H^2=\rho3\left(1-\frac\rho{\rho_c}\right),\quad \dot\rho+3H\rho=0,\] one obtains the following quantities \[a(t)=\left(\frac34\rho_ct^2+1\right)^{1/3},\quad\rho(t)=\frac{\rho_c}{\frac34\rho_ct^2+1},\quad H(t)=\frac{\frac12\rho_ct}{\frac34\rho_ct^2+1}\] For small values of the energy density $\rho\ll\rho_c$ the solutions of standard Friedmann equations are recovered.

# New to Observational Cosmology

**Problem 1**

problem id: 150_o1

Find ratio between the illuminance of the Earth surface by the Sun and other light sources in the Universe (it is a quantitative resolution of the Olbers paradox).

As we mentioned above (in the Warm up chapter), finiteness of the Universe's lifetime enables to qualitatively resolve the Olbers paradox: the night sky is dark because light from the distant galaxies do not have time to reach the Earth during the lifetime of the Universe equal to $t_0\sim H_0^{-1}$. Let us now make some qualitative estimates. The observations show that the luminosity density of the galaxies in the observed Universe equals to \[nL\approx2\times10^8L_\odot Mpc^{-3}.\] Compared to the terrestrial standards, the Universe is not a well illuminated place: the above cited luminosity density is equivalent to one $40W$ lamp per a sphere of $1$ à.u. radius. Illuminance created by all the galaxies situated inside the horizon $R_{hor}\sim cH_0$ is \[E_g\approx nL\int\limits_0^{c/H_0}dr=nL(c/H_0)\sim2\times10^{-11}L_\odot AU^{-2}.\] Comparing this quantity with the illuminance on the Earth due to the Sun $E_{sun}=\frac{L_\odot}{4\pi AU^2}\approx0.08L_\odot AU^{-2},$ one finds that \[E_g/E_{sun}\sim3\times10^{-10}.\] Due to the cosmological principle the Universe is weakly illuminated as the Earth. That is why darkness of the night sky should not amaze us.

**Problem 2**

problem id: 150_o2

Consider light emitted by surface of the Sun and observed on the Earth. Determine the shift of the observed frequency compared to the analogues light frequency emitted by atoms on the Earth.

For a given atom transition ratio of the light frequencies (observed in the point $1$) between that emitted from the points $2$ and $1$, equals to \[\frac{\nu_2}{\nu_1}=\left(\frac{g_{00}(x_2)}{g_{00}(x_1)}\right)^{1/2}.\] In the weak field limit \[g_{00}\approx -1-2\Phi,\] $\Phi\ll1$, so that \[\frac{\Delta\nu}\nu=\frac{\nu_2}{\nu_1}-1=\Phi(x_2)-\Phi(x_1).\] Let us apply this relation to the case when light is emitted by surface of the Sun and is observed on the Earth. Gravity potential of the Sun is \[\Phi_\odot=-G\frac{M_\odot}{R_\odot},\] where solar mass and radius respectively are \begin{align} \nonumber M_\odot&=1.97\times10^{33}g;\\ \nonumber R_\odot&=0.695\times10^{6}km. \end{align} Finally one obtains \[\Phi_\odot=-2.12\times10^{-6}.\] The terrestrial gravity potential can be neglected compared to the solar one. Therefore one finally obtains that the relative redshift of the light arrived from the Sun approximately equals to $-2.12\times10^{-6}$ (compared to the light frequency emitted by atoms on the Earth).

**Problem 3**

problem id: 150_o3

Find the distance to a distant galaxy using recessional velocity as measured by the Doppler redshift.

The Hubble distance is given by \[R=\frac{V}{H_0}=\left[\frac{(z+1)^2-1}{(z+1)^2+1}\right]\frac{c}{H_0}\] and can be calculated from the wavelength shift of any spectral line.

**Problem 4**

problem id: 150_o4

Show that galaxies situated outside the Hubble sphere have imaginary redshift.

\[z=\sqrt{c+V}{c-V}-1=\sqrt{c+HR}{c-HR}-1.\] The Hubble radius is \[R_H=\frac c H,\] so for $R>R_H$ we obtain imaginary values of the redshift.

**Problem 5**

problem id: 150_o5

Find the exact relativistic Doppler velocity-redshift relation.

\[(1+z)_{Dop}=\sqrt{c+V}{c-V}.\] The exact relativistic Doppler velocity-redshift relation is \[V_{Dop}(z)=c\frac{2z+z^2}{2+2z+z^2}.\] For $z\to\infty$, the velocity $V_{Dop}(z)\to c$ corresponding to the limit $V_{Dop}\le c$.

**Problem 6**

problem id: 150_o7

Find $V_{exp}(z)$ for three cosmological models: Einstein-de Sitter, Milne and de Sitter.

For $\Omega_m=1$, $\Omega_\Lambda=0$ \[V_{exp}=H_0R(z)=H_0\int\limits_0^z\frac{cdz'}{H(z')}=\frac{2\left(\sqrt{1+z}-1\right)}{\sqrt{1+z}}c.\] For $\Omega_m=0$, $\Omega_\Lambda=0$ \[V_{exp}=\frac{z\left(1+z/2\right)}{1+z}c.\] For $\Omega_m=0$, $\Omega_\Lambda=1$ \[V_{exp}=zc.\]

**Problem 7**

problem id: 150_o8

Some luminous object has apparent stellar magnitude $m=20$, and the absolute one is $M=-15$. Determine distance to it.

Use the relation \[m-M=5\log_{10}\left(\frac{d_L}{Mpc}\right)+25.\]

**Problem 8**

problem id: 150_o9

Due to a random coincidence the Balmer series of singly ionized helium atom in a distant star overlap with the Balmer series of hydrogen in the Sun. How fast this star recedes from us?

Energy levels of any system, composed of two particles with opposite charges ($-e,Ze$) coupled solely by the electrostatic attractive forces, are \[E_n\sim Z^2.\] Thus the photon wavelength generated by the electron transitions is \[\lambda_n\sim\frac1{Z^2}.\] Therefore for stationary atoms \[\lambda_{He}=\frac14\lambda_H.\] For a distant star moving with velocity $V$, \[\lambda_{He}'=\lambda_{He}\sqrt{\frac{1+V/c}{1-V/c}}.\] The condition of the spectral lines overlapping reads \[\lambda_{He}'=\lambda_H.\] Then \[V\approx0.88c.\]

**Problem 9**

problem id: 150_o10

A gas cloud rotates around a super-massive black hole with mass equal to $M=3.6\times10^6M_\odot$ (it is a possible interpretation of recent observations). Assuming that distance between these objects is of order of $60$ lightyears, determine the expected Doppler shift.

\[G\frac{m_{cl}M}{R^2}=\frac{m_{cl}V^2}{R}.\] \[V\approx 2.9\times10^4m/s.\] In assumption of small Doppler shift one obtains \[\frac{\Delta\lambda}{\lambda}\approx\frac V c\approx 9.6\times10^{-5}.\]

**Problem 10**

problem id: 150_o11

Explain why transition from the cosmological time to the redshift can be considered as a quantitative characteristic describing evolution of the Universe.

During evolution of the Universe the wavelength $\lambda$ of freely propagating light increases as \[\lambda(t_0)=\lambda(t)\frac{a(t_0)}{a(t)}.\] The redshift is \[z=\frac{\lambda(t_0)-\lambda(t)}{\lambda(t_0)}.\] Using the first equation, one finds \[a=\frac a{1+z}.\] Measuring redshift of the observed object we can know how large the Universe was in the moment of the emission of the light. relative size of the Universe provides us with a measurable characteristic of the time in cosmology. The cosmologists use the redshift as a measure of time: $z=0$ means today, $z=\infty$ at the moment of the Big Bang. The advantage of the redshift before time is that the former is a directly measurable quantity.

**Problem 11**

problem id: 150_o12

Show, that for small separations the "Hubble law" $dz=Hdr$ holds. In other words, we have replaced the velocity by the redshift.

The photon trajectory is given by $ds^2=0$. Introducing the conformal time $\eta$, one obtains \[ds^2=0\rightarrow d\eta^2-dr^2=0.\] The solution for the radial comoving distance between source and observer is \[r=\pm\eta+const.\] Having made the transformation \[d\eta=\frac{d\eta}{dt}\frac{dt}{da}\frac{da}{dz}dz=-\frac a{\dot a}dz=-\frac{dz}{H(z)},\] we find the "Hubble law" for the solution $d\eta=-dr$ \[dz=Hdr.\]

**Problem 12**

problem id: 150_o13

Solve the previous problem using the Taylor series of the scale factor in terms of time.

Small redshifts correspond to small time difference between the moments of light emission $t_e$ and its observation $t_0$. It allows to limit the decomposition of $a(t_e)$ by the first terms \[a(t_e)\approx a(t_0)-\dot a(t_0-t_e)\approx a_0(1-H_0 r).\] Using the relation between the red shift and the scale factor \[1+z=\frac{a(t_0)}{a(t_e)}\approx\frac1{1-H_0 r}\approx1+H_0r.\] Therefore for $z\ll1$ the Hubble law can be represented in the form \[z=H_0r.\]

**Problem 13**

problem id: 2501_02o

Why the Linear Distance-Redshift Law in Near Space?

Consider a nearby galaxy and the change of the scale factor during the small time interval $dt$ when the light travels from this galaxy to the observer: \[a(t_{obs})=a(t_{em})+\left.\frac{da}{dt}\right|_{t=t_{em}}dt.\] When the distance is short and the expansion speed much less than the speed of light, then the distance $R$ measured by the astronomer using any of the available ways is practically equal to the metric distance $a(t)r$ and the time interval can be approximated as $dt\approx a(t)r/c=R/c$. From this it follows that \[a(t_{obs})/a(t_{em})=1+z=1+\frac{\dot a}a \frac R c.\] and finally \[cz=HR.\]

**Problem 14**

problem id: 150_o1

Determine physical distance to an object emitted light with redshift $z$ in the flat expanding Universe.

The physical distance reads \[R(t)=a(t)r(t),\] where $r(t)$ is the comoving distance \[r(t)=\int\limits_t^{t_0}\frac{dt'}{a(t')}.\] Therefore \[R(t)=\frac1{1+z}\int\limits_t^{t_0}\frac{dt'}{a(t')}=\frac1{1+z}\int\limits_t^{t_0}d\eta.\] The differentials of the conformal time $d\eta$ and redshift $dz$ are related by \[d\eta=\frac{d\eta}{dt}\frac{dt}{da}\frac{da}{dz}dz=-\frac a{\dot a}dz=-\frac{dz}{H(z)}.\] Thus the physical distance to an object is related to the redshift $z$ of the emitted light by \[R(t)=\frac1{1+z}\int\limits_0^z\frac{dz'}{H(z')}.\]

**Problem 15**

problem id: 150_o16

Find comoving distance to a presently observed galaxy as a function of the redshift.

The photon equation of motion reads $ds^2=0$. Consider a radial trajectory with the observer in the origin. In this case one obtains for the spatially flat metrics \[ds^2=a^2(t)(d\eta^2-dr^2)=0,\] where $r$ is the radial comoving coordinate. Taking into account that \[d\eta=-\frac{dz}{H(z)},\] one obtains \[r(z)=\int\limits_0^z\frac{dz'}{H(z')}.\] The same result can be obviously obtained using result of the previous problem \[r(z)=\frac{R(z)}{a(z)}=\int\limits_0^z\frac{dz'}{H(z')}.\]

**Problem 16**

problem id: 150_o17

Solve the previous problem for the flat Universe dominated by non-relativistic matter.

In this case \[H(z)=H_0(1+z)^{3/2}\] and \[r(z)=\frac2{H_0}\left(1-\frac1{\sqrt{1+z}}\right).\]

**Problem 17**

problem id: 150_o18

Determine the recession velocity due to the cosmological expansion for an object emitted light with the redshift $z$ in the flat expanding Universe.

\[V=\dot R(t)=\dot a(t)=H(z)R(z)=\frac{H(z)}{1+z}\int\limits_0^z\frac{dz'}{H(z')}.\]

**Problem 18**

problem id: 150_o19

Obtain relations between velocity of cosmological expansion and redshift.

The exact velocity-proper distance relation (Hubble law) reads \[V_{exp}=HR.\] For the observer on the Earth living at the time $t=t_0$ the observed redshift serves as a measure of distance at this epoch. For photons moving along the comoving coordinate $r$ one obtains $cdt=\pm adr$. Consequently \[a_0 dr=\frac{cdz}{H(z)}.\] Integration over the redshift gives the proper distance-redshift relation \[R(t_0,z)\equiv R(z)=\int\limits_0^z\frac{cdz'}{H(z')}.\] Using the $R(z)$ relation one gets the exact velocity-redshift relation \[V_{exp}(z)=H_0R(z)=H_0\int\limits_0^z\frac{cdz'}{H(z')},\] where the expansion velocity is for an object with the redshift $z$ observed at the time $t=t_0$.

**Problem 19**

problem id: 150_o20

Find dependence of the Hubble parameter $H$ on the redshift $z$ for the case of one-component Universe composed of the non-relativistic matter.

For the non-relativistic matter dominated Universe one has \[\rho=\rho_0/a^3=\rho_0(1+z)^3.\] By the definition of the relative density \[\Omega\equiv\frac{\rho}{\rho_{cr}};\quad \rho_{cr}\equiv\frac{3H^2}{8\pi G},\] then \[\frac\Omega{\Omega_0}=\frac\rho{\rho_0}\frac{H_0^2}{H^2}=(1+z)^3\frac{H_0^2}{H^2};\] \[\Omega H^2=(1+z)^3\Omega_0H_0^2.\] Rewrite the first Friedmann equation in the form \begin{align} \nonumber a^2 H^2(1-\Omega) & =-k;\\ \nonumber a^2 H^2(1-\Omega) & =a_0^2 H_0^2(1-\Omega_0), \end{align} as $k=const$. Taking into account that $a_0/a=1+z$, one obtains \[H^2(1-\Omega) =(1+z)^2 H_0^2(1-\Omega_0).\] Substituting the above obtained expression for the $\Omega H^2$, one finally obtains \[H=H_0(1+z)(1+\Omega_0 z)^{1/2}.\]

**Problem 20**

problem id: 150_o21

Find dependence of relative density $\Omega$ on the redshift $z$ for the case of one-component Universe composed of the non-relativistic matter.

Substituting the expression for the relative density obtained in the previous problem \[\Omega = (1+z)^3\Omega_0\frac{H_0^2}{H^2}\] into the obtained $ibid$ expression for the Hubble parameter, one finds \[\Omega=\Omega_0\frac{1+z}{1+\Omega_0z}.\]

**Problem 21**

problem id: 150_o22

Construct a scheme to determine sign of acceleration of the scale factor, based on measurement of the supernovae bursts characteristics.

The Hubble law tells nothing about the magnitude, sign and the very possibility of non-uniform expansion of the Universe. It is valid under the approximation which is insensitive to the acceleration. In order to investigate the non-linear effects one needs data for high redshifts. If the observations detect deviations from the linear law, then it enables us to determine the acceleration sign basing on the magnitude and sign of the deviation. If the detected deviation lies towards increasing distance at fixed redshift then the acceleration is positive. The distance is estimated by luminosity of the source under assumption that the considered set of the sources represents the standard candles, i.e. has fixed luminosity with sufficiently high accuracy.

**Problem 22**

problem id: 150_o23

Show that the time derivative of the redshift for the light emitted at time $t$ and registered at time $t_0$ can be determined as \[\dot z\equiv\frac{dz}{dt_0}=H(t_0)(1+z)-H(t).\]

Make use of the definition \[1+z=\frac{a(t_0)}{a(t)},\] then \[\frac{dz}{dt_0}=\frac{\dot a(t_0)a(t)-a(t_0)\dot a(t)\frac{dt}{dt_0}}{a^2(t)}.\] Taking into account that \[\frac{dt}{dt_0}=\frac1{1+z},\] one obtains \begin{align} \nonumber \dot z & = \frac{\dot a(t_0)}{a(t)}-\frac{a(t_0)\dot a(t)}{a^2(t)}\frac1{1+z}=\\ \nonumber & = \frac{\dot a(t_0)}{a(t_0)}\frac{a(t_0)}{a(t)} - \frac{a(t_0)}{a(t)}\frac{\dot a(t)}{a(t)}\frac1{1+z}=\\ \nonumber & = H(t_0)(1+z)-H(t)=H_0(1+z)-H(z). \end{align}

**Problem 23**

problem id: 150_o24

In a flat one-component Universe with the state equation $p=w\rho$ at time $t_0$ one registers a light signal with redshift $z$. What values of the EoS parameter $w$ lead to \[\frac{dz}{dt_0}>0?\] Explain physical sense of the obtained result.

In the flat one-component Universe one has \[H(z)=H_0(1+z)^{\frac32(1+w)}.\] Using result of the previous problem one obtains \[\frac{dz}{dt_0}=H_0(1+z)-H_0(1+z)^{\frac32(1+w)}=H_0(1+z)\left[1-(1+z)^{\frac32(1+w)-1}\right]\] Thus $dz/dt_0>0$ if \[w<-1/3.\] In the flat one-component case this condition provides the accelerated expansion of Universe because in this case \[\frac{\ddot a}{a}=-\frac{4\pi G}{3}(\rho+3p).\]

**Problem 24**

problem id: 150_o26

Show that the luminosity distance can be generally presented as \[d_L=\frac{1+z}{H_0\sqrt{\Omega_{k0}}}\sinh\left(H_0\sqrt{\Omega_{k0}}\int\limits_0^z\frac{dz}{H(z)}\right),\] where $\Omega_{k0}$ is the relative contribution of the spatial curvature.

Using that \[\rho_{curv}=-\frac{3}{8\pi G}\frac{k}{R^2},\] where $R$ is the (comoving) radius of curvature of the open or closed Universe, one finds \[\Omega_{k0}=-\frac{k}{RH_0}.\] It is easy to see that the expression \[d_L=\frac{1+z}{H_0\sqrt{\Omega_{k0}}}\sinh\left(H_0\sqrt{\Omega_{k0}}\int\limits_0^z\frac{dz}{H(z)}\right)\] reproduces the well known one for the luminosity distance \[d_L=(1+z)\left\{ \begin{array}{ll} R\sinh\left[\frac{c}{H_0R}\int\limits_0^z\frac{dz'}{E(z')}\right], & open;\\ \frac{c}{H_0}\int\limits_0^z\frac{dz'}{E(z')}, & open;\\ R\sin\left[\frac{c}{H_0R}\int\limits_0^z\frac{dz'}{E(z')}\right], & closed; \end{array}\right.\] where $E(z)\equiv H(z)/H_0$.

**Problem 25**

problem id: 150_o27

\it
*When galaxy formation started in the history of the Universe remains unclear. Studies of the cosmic microwave background indicate that the Universe, after initial cooling (following the Big Bang), was reheated and reionized by hot stars in newborn galaxies at a redshift in the range $6<z<14$. Here we report a spectroscopic redshift of $z=6.96$*. [M. Iue et al., A galaxy at a redshift z=6.96 , arXiv:0609393]

\bf Estimate age of the Universe when the first galaxies were created.

According to the SCM, age of the Universe corresponding to the redshift $z$ can be determined by the formula \[t(z)=\frac1{H_0}\int_0^{1/(1+z)}\frac{dx}{x\sqrt{\Omega_{0\Lambda}+\Omega_{0M}x^{-3}}}.\] Redshift $z\simeq7$ corresponds to just $750$ Myr after the Big Bang or only $6\%$ of its present age.

**Problem 26**

problem id: 150_o30

Express probability to find two galaxies in the infinitesimally small volumes $dV_1$ and $dV_2$ in terms of the correlation function $\xi(\vec{r_1},\vec{r_2})$, if average density of the galaxies in the considered volume equals $\bar n$ and total number of the galaxies is $N$. Point out main properties of the correlation function.

For absolutely random homogeneous distribution of galaxies the probability $dP_1$ to find a galaxy in a infinitesimally small volume $dV_1$ is proportional to the volume $dV_1$ and to the average galaxy number $\bar n$:
\[dP_1 = \frac{\bar n}{N}dV_1,\]
where $N$ is total number of galaxies in the *sample*. If there were no galaxy clusterization effect, then the probability to find galaxies in the volumes $dV_1$ and $dV_2$ would be equal to the product $dP_1dP_2$. Any clusterization will lead to deviation from this simple formula. It allows to define the two-point correlation function $\xi(\vec{r_1}-\vec{r_2})$:
\[dP_{12}=\frac{\bar n^2}{N^2}[1+\xi(\vec{r_1}-\vec{r_2})]dV_1dV_2.\]
The homogeneity implies that $\xi$ depends only on $|\vec{r_1}-\vec{r_2}|$. Assumption that the distribution is random on sufficiently large scales leads to the condition:
\[\xi(\vec{r_1}-\vec{r_2})\to0\]
for sufficiently large $r=|\vec{r_1}-\vec{r_2}|$. The function $xi$ must have values in the range $-1<\xi<\infty$. Positive value of $xi$ means that the galaxies tend to the cluster, and negative value --- that they repel. If $xi$ is positive on short distances then it must become negative with increasing distance in order to keep constant total density.

**Problem 27**

problem id: 150_o31

\bf Find relation between the spatial $\xi(r)$ and angular $w(\theta)$ correlation functions.

In order to obtain the spatial correlation function $\xi(r)$ one needs three-dimensional information on the galaxy distribution. Such information requires intensive observation of the redshifts and therefore long time of the telescope processing time. Until the recent time all large catalogues of galaxies recorded their position on the sky regardless their depth (redshift or distance). Therefore we can immediately obtain only the angular correlation function $w(\theta)$. The spatial correlation function $\xi(r)$ can be reconstructed from $w(\theta)$ under assumption of homogeneity and isotropy. The angular correlation function $w(\theta)$ is defined as the probability to observe a pair of galaxies in the solid angles $d\Omega_1$ and $d\Omega_2$ separated by angle $\theta$ on the celestial sphere. If average surface density in the unit solid angle equals to $\bar\sigma$, then \begin{equation}\label{150_o31_e1} dP_{12}\propto \bar\sigma^2[1+w(\theta)]d\Omega_1d\Omega_2. \end{equation} Let us assume for simplicity that all galaxies have the same absolute magnitude (luminosity). It means that all the galaxy have the same limiting distance up to which they are still detectable. Let this limiting distance equals to $D$. Then the angular distribution is obtained from the three-dimensional spatial distribution by integration up to the depth $D$: \begin{equation}\label{150_o31_e2} dP_{12}\propto \left\{\int\limits_0^Dr_1^2dr_1\int\limits_0^Dr_2^2dr_2[1+\xi(r)]\right\}d\Omega_1d\Omega_2. \end{equation} For the angular distance $\theta$ the corresponding spatial separation $r$ is \[r^2=r_1^2+r_2^2-2r_1r_2\cos\theta.\] Comparing (\ref{150_o31_e1}) and (\ref{150_o31_e2}) one finds \[w(\theta)\propto\int\limits_0^Ddr_1\int\limits_0^Ddr_2\,r_1^2r_2^2\ \xi\left[(r_1^2+r_2^2-2r_1r_2\cos\theta)^{1/2}\right].\]

**Problem 28**

problem id: 150_o32

Show that power decay law of the spatial correlations leads to power decay law for the angular correlations.

Firstly, take into account that the correlation is essential only for small angular distances. It allows to set \[\cos\theta\simeq1-\frac12\theta^2.\] Secondly, $\xi$ essentially differs from zero only on small $r$. Therefore $r_1-r_2\ll r_1+r_2.$ Then \[4r_1r_2=(r_1+r_2)^2-(r_1-r_2)^2\approx(r_1+r_2)^2.\] Due to fast decay of the correlation functions we can extend the upper integration bound to infinity without much loss of accuracy. Let us introduce new variables \begin{align} \nonumber u & = \frac1{2D}(r_1+r_2);\\ \nonumber v & = \frac1{D\theta}(r_1-r_2). \end{align} Then \[w(\theta)\propto \int\limits_0^\infty du\int\limits_0^\infty dv\ u^4\xi\left[(u^2+v^2)^{1/2}D\theta\right]\theta.\] Let \[\xi(r)=Ar^{-\alpha}.\] Then \[w(\theta)\propto\left\{\int\int\ du\ dv\ u^4(u^2+v^2)^{-\alpha/2}D^{-\alpha}\right\}\theta^{-\alpha+1}.\] The conclusion is that power law decay of the spatial correlation function results in the power law decay of the angular correlation.

**Problem 29**

problem id: 150_o33

Find two-point correlation function for $N$ galaxies distributed on a straight line with average density $\bar n$ in non-overlapping clusters of length $a$. Density of galaxies inside the cluster is constant and equals to $n_c$. The clusters are randomly distributed.

In this case we can construct the correlation function $\xi(r)$ in the following way: Take a random galaxy. If $r>a$ then the second galaxy will be situated in another cluster randomly distributed with respect to the first, so that $\xi=0$. If $r<a$ then the second galaxy will be in the same cluster so that the probability to find it in the infinitesimal volume $dV_2$ equals to $n_cdV_2/N$. Thus \[dP_{12}=\frac{\bar nn_c}{N^2}=\frac{\bar n^2}{N^2}\left(1+\frac{n_c-\bar n}{\bar n}\right)\] and therefore $\xi(r)=(n_c-\bar n)/\bar n$, so \[\xi(r)=\left\{\begin{array}{ll}(n_c-\bar n)/\bar n, & r<a;\\ 0, & r>a.\end{array}\right.\] If the clusterization takes place on scales of order $a$ then on the same scales we must observe a sharp bend in behavior of the correlation function. And vise versa, presence of such a sharp bend in the observational data should point on the clusterization effect present.

**Problem 30**

problem id: 150_o34

In a flat matter-dominated Universe of age $t_0$ light from a certain galaxy exhibits a redshift $z=0.95$. How long has it taken the light signal to reach us from this galaxy?

Let the light signal starts off at $t=t_1$ to reach us at $t=t_0$. Consider the time interval $dt'$ where $t_1<t'<t_0$. In this time interval the light signal covers a distance $cdt'$, but by the time $t=t_0$, this will have expanded to $cdt'\times a(t=t_0)/a(t')=cdt'a(0)/a(t')$. In a matter-dominated Universe $a(0)/a(t')=(t_0/t')^{2/3}$. Hence, the total distance travelled by the light signal will be \[D=a(0)\int\limits_{t_1}^{t_0}\frac{cdt'}{a(t')}=ct_0^{2/3} \int\limits_{t_1}^{t_0}\frac{dt'}{t'^{2/3}}=3ct_0\left[1-\left(\frac{t_1}{t_0}\right)^{1/3}\right].\] The redshift is given by $1+z=a(0)/a(t_1)=(t_0/t_1)^{2/3}$. Hence, the time elapsed \[t_{el}=3t_0\left[1-\frac1{(1+z)^{1/2}}\right]\approx0.85t_0.\]

# New from June 2015

## To Chapter 2

**Problem 1**

problem id: 150_0

Comprehensive explanations of the expanding Universe often use the balloon analogy. Although the balloon analogy is useful, one must guard against misconceptions that it can generate. Point out the misconceptions that appear when using this analogy [see M.O. Farooq, Observational constraints on dark energy cosmological model parameters, arXiv: 1309.3710.]

(a) The surface that is expanding is two dimensional; the "center" of the balloon is in the third dimension and is not part of the surface, which has no center.

(b) The balloon is expanded by the pressure difference between the inside and the outside, but the Universe is not being expanded by pressure.

(c) Pressure couples to gravity in the Einstein equation, so the addition of (positive) pressure to the Universe would slow, not increase, the expansion rate. It is the negative pressure which is only capable to generate the accelerated expansion of the Universe.

(d) If the dots on the balloon represent galaxies, they too will expand. But real galaxies do not expand due to general Hubble expansion because they are gravitationally bound objects. We can make a better analogy by gluing solid objects (like 10 cent coins) to the surface of the balloon to represent galaxies, so that they do not expand when the balloon expands.

**Problem 2**

problem id: 150_1

(into the cosmography and extended deceleration parameter) Show that \[\frac{d\dot a}{da}=-Hq.\]

\[\frac{d\dot a}{da}=\frac{d\dot a}{dt}\frac{dt}{da}=\frac{\ddot a}{\dot a}=\frac{\ddot a}{aH}\equiv-Hq.\]

**Problem 3**

problem id: 150_2

Give a physical interpretation of the conservation equation.

The energy density has a time dependence determined by the conservation equation, \[\dot\rho=-3H\rho-3Hp.\] The two terms define the behavior of the uniform fluid which contains the energy in a dynamic Universe. The $H$ term provides the "friction", while the density term tracks the reduction in density due to the volume increase during expansion and the pressure term tracks the reduction in pressure energy during expansion.

**Problem 4**

problem id: 150_04

Find evolution equation for the density parameter $\Omega$ of the single-fluid FLRW models with the linear equation of state $p=w\rho$.

\begin{align} \nonumber \Omega&=\frac\rho{3H^2},\quad (8\pi G=1),\\ \nonumber N&=\log\frac{a}{a_0},\\ \nonumber \frac d{dN}&=\frac1H\frac{d}{dt},\\ \nonumber \frac{d\Omega}{dN}&=\frac13\frac{\frac{d\rho}{dN}H^2-2\rho H\frac{dH}{dN}}{H^4},\\ \nonumber \frac{d\rho}{dN}&=\frac1H\frac{d\rho}{dt}=-3\rho(1+w),\\ \nonumber \frac{dH}{dN}&=-(1+q)H,\\ \nonumber \frac{d\Omega}{dN}&=-[2q-3w-1]\Omega. \end{align} For a single-component fluid \[q=\frac12(1+3w)\Omega\] and one finally obtains \[\frac{d\Omega}{dN}=-(1+3w)(1-\Omega)\Omega.\]

**Problem 5**

problem id: 150_05

Solve the previous problem for the multi-component case.

The evolution equations for an $n$-fluid model use the density parameters $\Omega_1$, $\Omega_2\ldots\Omega_n$, as dynamical variables. Relation \[\frac{d\Omega}{dN}=-[2q-3w-1]\Omega\] can easily be generalized or an $n$-fluid model to give \[\frac{d\Omega_i}{dN}=-[2q-(1+3w_i)]\Omega_i,\quad i=1\ldots n,\] where \[q=\frac12\sum\limits_i(1+3w_i)\Omega_i.\] For the total density parameter \[\Omega_{tot}=\sum\limits_{i=1}^n\Omega_i\] one obtains the evolution equation \[\frac{d\Omega_{tot}}{dN}=-2q(1-\Omega_{tot}).\]

**Problem 6**

problem id: 150_06

Use the conformal time to prove existence of smooth transition from the radiation-dominated era to the matter dominated one.

Rewrite the first Friedmann equation in the form \[H^2=\frac{\rho_{eq}}{3}\left[\left(\frac{a_{eq}}{a}\right)^3+\left(\frac{a_{eq}}{a}\right)^4\right],\] where $a_{eq}$ is the scale factor when the densities of matter and radiation are the same, and $\rho_{eq}$ is the density in each component at that time. This equation can be solved exactly using conformal time \[\frac{da}{dt}=\frac1a\frac{da}{d\eta},\quad \left(\frac{da}{d\eta}\right)^2=\frac13\rho_{tot}a^4,\] \[\frac{a(\eta)}{a_{eq}}=2(\sqrt2-1)\left(\frac\eta{\eta_{eq}}\right)+[1-2(\sqrt2-1)]\left(\frac\eta{\eta_{eq}}\right)^2,\quad \eta_{eq}\equiv\frac{2(\sqrt2-1)}{a_{eq}}\sqrt{\frac{3}{\rho_{eq}}}.\] We see a smooth transition between the two characteristic behaviors of radiation domination $\eta\ll\eta_{eq}$, $a\propto\eta$ and matter domination $\eta\gg\eta_{eq}$, $a\propto\eta^2$. There is no good way to rewrite this relation in terms of the cosmic times.

**Problem 7**

problem id: 150_07

Consider a set of the cosmographic parameters built from the Hubble parameter and its time derivatives [see S. Carloni, A new approach to the analysis of the phase space of f(R)-gravity, arXiv:1505.06015) ] \[Q\equiv\frac{\dot H}{H^2},\quad J\equiv\frac{\ddot H H}{\dot H^2},\quad S\equiv\frac{\dddot H H^2}{\dot H^3},\ldots \] Express them in terms of the *canonic* cosmographic parameters $q,j,s\dots$.

\[Q=\frac{\dot H}{H^2}=\left(\frac{\ddot a}a-H^2\right)\frac1{H^2}=-(q+1);\] \[J=\ddot H \frac{H}{\dot H^2}=\left(\frac{\dddot a}a-\frac{\dot a\ddot a}{a^2}-2H\dot H\right)\frac{H}{\dot H^2}=\frac j{Q^2}+\frac q{Q^2}-\frac2Q=\frac{j+3q+2}{(1+q)^2}.\]

**Problem 8**

problem id: 150_08

Consider another set of the cosmographic parameters [see S. Carloni, A new approach to the analysis of the phase space of f(R)-gravity, arXiv:1505.06015) ] \[\bar Q\equiv\frac{H_{,N}}{H},\quad \bar J\equiv\frac{H_{,NN}}{H},\quad \bar S\equiv\frac{H_{,NNN}}{H},\ldots,\] where \[H_{,N}\equiv \frac{dH}{d\ln a}.\] Express them in terms of the Hubble parameter and its time derivatives.

\[\frac{d}{d\ln a}=\frac1H\frac{d}{dt};\] \[\bar Q\equiv\frac{\dot H}{H^2},\quad \bar J\equiv\frac{\ddot H}{H^3}-\frac{\dot H^2}{H^4},\quad \bar S\equiv\frac{\dddot H}{H^4}+3\frac{\dot H^3}{H^6}-4\frac{\dot H\ddot H}{H^5}.\]

**Problem 9**

problem id: 150_09

Express the Ricci scalar and its time derivatives in terms of the $\bar Q$, $\bar J$ and $\bar S$.

\begin{align} \nonumber R&= -6\left[(\bar Q+2)H^2+\frac k{a^2}\right];\\ \nonumber \dot R&= -6H\left\{\left[\bar J+\bar Q(\bar Q+4)\right]H^2-\frac{2k}{a^2}\right\};\\ \nonumber \ddot R&= -6H^2\left\{\left[\bar S+4\bar J(\bar Q+1)+(\bar Q+8)\bar Q^2\right]H^2-2(2-\bar Q)\frac{k}{a^2}\right\}. \end{align}

**Problem 10**

problem id: 150_3

Show that for a perfect fluid with EoS $p=w(a)\rho$ the adiabatic sound speed can be represented in the form \[c_S^2=w(a)-\frac13\frac{d\ln(1+w)}{d\ln a}.\]

\[c_S^2\equiv\frac{d p}{d\rho}=w(a)+\rho\frac{dw(a)}{d\rho};\] \[\frac{dw}{d\rho}=\frac{dw}{da}\frac{da}{d\rho};\] \[\dot\rho+3H(\rho+p)\Rightarrow\frac{da}{d\rho}=-\frac13\frac a{\rho(1+w)},\] \[c_S^2=w(a)+\rho\frac{dw}{d a}\left(-\frac13\frac a{\rho(1+w)}\right)=w(a)-\frac13\frac{d\ln(1+w)}{d\ln a}.\]

**Problem 11**

problem id: 150_4

Obtain equation for $\ddot\rho(t)$, where $\rho(t)$ is energy density of an ideal fluid participating in the cosmological expansion.

Take time derivative of the conservation equation to obtain \begin{align}\nonumber \frac{d}{dt}[\dot\rho+3H(\rho+p)]&=\ddot\rho+3\dot H(\rho+p)+3H(\dot\rho+\dot p)=\\ \nonumber &=\ddot\rho+3\dot H(\rho+p)+3H\dot\rho(1+c_S^2)=\\ \nonumber & =\ddot\rho+H(\rho+p)-9H^2(\rho+p)(1+c_S^2)=0. \end{align} It then follows that \[\ddot\rho=-3[\dot H+3H^2(1+c_S^2)](\rho+p).\]

**Problem 12**

problem id: 150_5

Show that in the case of the flat Friedmann metric, the third power of the scale factor $\varphi\equiv a^3$ satisfies the equation \[\frac{d^2\varphi}{dt^2}=\frac32(\rho-p)\varphi,\quad 8\pi G=1.\] Check validity of this equation for different cosmological components: non-relativistic matter, cosmological constant and a component with EoS $p=w\rho$.

\[\frac{d^2 a^3}{dt^2}=3\left(2H^2+\frac{\ddot a}a\right)a^3=3\left[\frac23\rho-\frac16(\rho+3p)\right]a^3=\frac32(\rho-p)\varphi.\]

**Problem 13**

problem id: 150_6

The lookback time is defined as the difference between the present day age of the Universe and its age at redshift $z$, i.e. the difference between the age of the Universe at observation $t_0$ and the age of the Universe, $t$, when the photons were emitted. Find the lookback time for the Universe filled with non-relativistic matter, radiation and a component with the EoS $p=w(z)\rho$.

From the definitions of redshift $1+z=1/a$ we have \[\frac{dz}{dt}=-\frac{\dot a}{a^2}=-H(z)(1+z)\] or \[dt=-\frac{dz}{H(z)(1+z)}.\] The lookback time is defined as \[t_0-t=\int\limits_t^{t_0}dt=\int\limits_0^z \frac{dz'}{H(z')(1+z')}=\frac1{H_0}\int\limits_0^z \frac{dz'}{E(z')(1+z')}\] where \[E(z)=\sqrt{\Omega_r(1+z)^4+\Omega_m(1+z)^3+\Omega_k(1+z)^2+\Omega_w\exp\left(3\int\limits_0^z dz'\frac{1+w(z')}{1+z'}\right)}.\] From the definition of lookback time it is clear that the cosmological time or the time back to the Big Bang, is given by \[t(z)=\int\limits_z^\infty\frac{dz'}{H(z')(1+z')}.\]

**Problem 14**

problem id: 150_7

Show that the Hubble radius grows faster than the expanding Universe in the case of power law expansion $a(t)\propto t^\alpha$ with $\alpha<1$ (the decelerated expansion).

In the case of the power law expansion with $\alpha<1$ (the decelerated expansion) the Hubble radius indeed grows faster than the expanding Universe: $R_H=H^{-1}\propto t$, while $a(t)\propto t^\alpha$. In power law situations the Hubble radius has an expansion velocity \[\frac{d}{dt}\left(\frac1H\right)=\frac1\alpha\] greater than light speed. This behavior is true only for a decelerating Universe composed of matter and radiation. This is not a physical velocity, violating special relativity, but the velocity of expansion of the metric itself.

## To chapter 3

**Problem 15**

problem id: 150_015

Show, that in the Milne Universe the age of the Universe is equal to the Hubble time.

In an empty (Milne) Universe since $H=\dot a/a=t^{-1}$, the age of the Universe $t_0$ is equal to the Hubble time $t_0=H_0^{-1}$.

## To chapter 4 The black holes

**Problem 16**

problem id: new2015_1

see E. Berti, A Black-Hole Primer: Particles, Waves, Critical Phenomena and Superradiant instabilities (arXiv:1410.4481[gr-qc])

A Newtonian analog of the black hole concept is a so-called "dark star". If we consider light as a corpuscle traveling at speed $c$, light cannot escape to infinity whenever $V_{esc}>c$, where \[V_{esc}^2=\frac{2GN}R.\] Therefore the condition for existence of "dark stars" in Newtonian mechanics is \[\frac{2GN}{c^2R}\ge1.\]

Can this condition be satisfied in the Newtonian mechanics?

A naive argument tells us that as we pile up more and more material of constant density $\rho_0$, the ratio $M/R$ increases: \begin{equation}\label{new2015_1_e1} \frac M R=\frac43\pi R^2\rho_0. \end{equation} This equation would seem to suggest that dark stars could indeed form. However, we must include the binding energy $U$, \begin{equation}\label{new2015_1_e2} U=-\int\frac{GMdM}{r} =-\int\limits_0^R\frac{G}{r}\left(\frac43\pi r^3\rho_0\right)\frac4\pi r^2\rho_0 dr=-\frac{16G\pi^2}{15}\rho_0^2R^5. \end{equation} The total mass $M_T$ of the hypothetical dark star is given by the rest mass $M$ plus the binding energy \begin{equation}\label{new2015_1_e3} \frac{M_T}R=\frac43\pi R^2\rho_0-\frac{16G\pi^2}{15}\rho_0^2R^4=\frac M R\left[1-\frac35\frac G{c^2}\frac M R\right] \le\frac5{12}\frac{c^2}G, \end{equation} where the upper limit is obtained by maximizing the function in the range (\ref{new2015_1_e1}). Thus, the dark star criterion (\ref{new2015_1_e1}) is never satisfied, even for the unrealistic case of constant-density matter. In fact, the endpoint of Newtonian gravitational collapse depends very sensitively on the equation of state, even in spherical symmetry.

**Problem 17**

problem id: 150_017

Derive the relation $T\propto M^{-1}$ from the Heisenberg uncertainty principle and the fact that the size of a black hole is given by the Schwarzschild radius.

The position of quanta inside a black hole can only be known within $\Delta x=2r_{Sh}=4GM/c^2$. Thus $\Delta t =2r_{Sh}/c=4GM/c^3$. According to the uncertainty principle $\Delta E\Delta t\ge\hbar$. Thus \[\Delta T=\frac{\Delta E}k\approx\frac{\bar c^3}{4kGM}\] Up to a numeric factor this relation coincides with the black hole Hawking temperature $T=\hbar c^3/8\pi kGM$.

## To chapter 8

**Problem 18**

problem id: 2501_06

Why the cosmological constant cannot be used as a source for inflation in the inflation model?

The cosmological constant do provide the accelerated expansion of Universe needed to realize the inflation: $a(t)\propto e^{Ht}$, $q=-1$. However, in approximation of the cosmological constant, $H$ is constant for all time. Therefore a dynamical mechanism for the limited time of inflation is needed. The physical mechanism for the existence of an approximately constant value of $H$ which lasts for a limited time is given by a scalar field. For a large initial potential energy of scalar field the state equation parameter $w\approx-1$ and the scalar field in process of the "slow roll" imitates the cosmological constant for sufficiently long period of time to solve the flatness and causal problems. Then due to shape of the potential the scalar field exits from the slow-roll regime, oscillates about its' potential minimum decaying into less massive particles insuring that inflation time is finite.

**Problem 19**

problem id: 2501_09

Show that inflation ends when the parameter \[\varepsilon\equiv\frac{M_P^2}{16\pi}\left(\frac{dV}{d\varphi}\frac1V\right)^2=1.\]

Using definition of the parameter $\varepsilon$ one finds \[\varepsilon=-\frac{\dot H}{H^2}.\] In the slow-roll approximation \[H^2=\frac{8\pi}{3M_P^2V},\] \[3H\dot\varphi=-\frac{dV}{d\varphi}.\] Therefore \[\frac{\ddot a}a=H^2(1-\varepsilon).\] The condition $\varepsilon=1$ is equivalent to $\ddot a=0$. When the value $\varepsilon=1$ is reached due to variation of the potential shape the Universe exits the regime of the accelerated expansion (inflation). Around tend, the inflaton field(s) typically begin oscillating around the minimum of the potential. (see Inflation and the Higgs Scalar, DANIEL GREEN (1412.2107).)

**Problem 20**

problem id: 2501_10

How does the number of e-folds $N$ depend on the slow-roll parameter $\varepsilon$?

\[N\propto\int\limits_{\varphi_{end}}^{\varphi_{initial}}d\varphi\frac{V}{dV/d\varphi} \propto\int\limits_{\varphi_{end}}^{\varphi_{initial}}d\varphi/\sqrt\varepsilon.\] The number of $e$-folds depends on how fast the field is $f$-decreasing. The number of e-folds, $N$ , is inversely proportional to the square root of the slow roll parameter $\varepsilon$ or proportional to the inverse fractional change of the potential with the field, $V/(dV/d\varphi)$.

## To chapter 9

**Problem 21**

problem id: 150_021

Using the by dimensional analysis for cosmological constant $\Lambda > 0$, define the set of fundamental "de Sitter units" of length, time and mass.

\begin{align} \nonumber L_{dS}&=\sqrt{1/\Lambda}\\ \nonumber T_{dS}&=\frac1c\sqrt{1/\Lambda}\\ \nonumber M_{dS}&=\frac\hbar c\sqrt{\Lambda} \end{align}

**Problem 22**

problem id: 150_022

In the Newtonian approximation, find the force acting on the point unit mass in the Universe filled by non-relativistic matter and cosmological constant. (see Chiu Man Ho and Stephen D. H. Hsu, The Dark Force: Astrophysical Repulsion from Dark Energy, arXiv: 1501.05592)

The Einstein equation with cosmological constant $\Lambda$ is \[R_{\mu\nu}-\frac12g_{\mu\nu}R=8\pi GT_{\mu\nu}+g_{\mu\nu}\Lambda.\] Contracting both sides with $g^{\mu\nu}$, one gets \[R=-8\pi GT-4\Lambda,\] where $T=T_\mu^\mu$ is the trace of the matter (including dark matter) energy-momentum tensor. This can be substituted in the original equation to obtain \[R_{\mu\nu}8\pi G\left(T_{\mu\nu}-\frac12T\right)-g_{\mu\nu}\Lambda.\] In the Newtonian limit, one can decompose the metric tensor as \[g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu},\quad |h_{\mu\nu}|\ll1.\] Specifically we are interested in the $00$-component of the Einstein equation. We parameterize the $00$th-component of the metric tensor as \[g_{00}=1+2\Phi,\] where $\Phi$ is the Newtonian gravitational potential. To leading order, one can show that [S. Weinberg, Gravitation and Cosmology, John Wiley \& Sons, Inc., 1972.] \[R_{00}\approx\frac12\vec\nabla^2g_{00}=\vec\nabla^2\Phi.\] In the inertial frame of a perfect fluid, its $4$-velocity is given by $u_\mu=(1,0)$ and we have \[T_{\mu\nu}=(\rho+p)u_{\mu\nu}-pg_{\mu\nu}=diag(\rho,p).\] where $\rho$ is the energy density and $p$ is the pressure. For a Newtonian (non-relativistic) fluid, the pressure is negligible compared to the energy density, and hence $T\approx T_{00}=\rho$. As a result, in the Newtonian limit, the 00-component of the Einstein equation reduces to \[\vec\nabla^2\Phi=4\pi G\rho-\Lambda,\] which is just the modified Poisson equation for Newtonian gravity, including cosmological constant. This equation can also be derived from the Poisson equation of Newtonian gravity, $\vec\nabla^2\Phi=4\pi G(\rho+3p)$, with source terms from matter and dark energy; $p\approx 0$ for non-relativistic matter, and $p=-\rho$ for a cosmological constant. Assuming spherical symmetry, we have \[\vec\nabla^2\Phi=\frac1{r^2}\frac\partial{\partial r}\left(\frac{\partial\Phi}{\partial r}\right)\] and the Poisson equation is easily solved to obtain \[\Phi=-\frac{GM}{r}-\frac\Lambda6r^2,\] where $M$ is the total mass enclosed by the volume $4/3\pi r^3$. The corresponding gravitational field strength is given by \[\vec{g}=-\vec\nabla\Phi=\left(-\frac{GM}{r^2}+\frac\Lambda3r\right)\frac{\vec{r}}r.\]

**Problem 23**

problem id: 150_023

Consider a spatially flat FLRW Universe, which consists of two components: the non-relativistic matter and the scalar field $\varphi$ in the potential $V(\varphi)$. Find relation between the scalar field potential and the deceleration parameter.

From the definition of the energy density and the pressure for the scalar field it follows that \begin{equation}\label{150_023_e1} V(\varphi)=\frac12(\rho_\varphi-p_\varphi). \end{equation} The deceleration parameter by definition is \[q=-\frac{\ddot a}{aH^2}=\frac1{6M_{Pl}^2H^2}(\rho_m+\rho_\varphi+3p_\varphi).\] Substituting $\rho_m+\rho_\varphi=3M_{Pl}^2H^2$, one finds \[p_\varphi=(2q-1)M_{Pl}^2H^2.\] Substituting $\rho_m=3M_{Pl}^2H^2-\rho_\varphi$ and $p_\varphi=(2q-1)M_{Pl}^2H^2$ into (\ref{150_023_e1}), one finds \[V[\varphi(z)]=\rho_{m0}\left[\frac{(2-q)}{3\Omega_{m0}}\frac{H^2}{H_0^2}-\frac12(1+z)^3\right].\]

**Problem 24**

problem id: 150_024

Find relation between the deceleration parameter and the derivative $d\varphi/dz$ for the Universe considered in the previous problem.

From the definition of the energy density and the pressure for the scalar field it follows that \begin{equation}\label{150_024_e1} \dot\varphi^2=\rho_\varphi+p_\varphi. \end{equation} Substitute the relations $\rho_m=3M_{Pl}^2H^2-\rho_\varphi$ and $p_\varphi=(2q-1)M_{Pl}^2H^2$ (see the previous problem) into (\ref{150_024_e1}) to obtain \[\left(\frac{d\varphi}{dz}\right)^2=M_{Pl}^2\left[\frac{2(1+q)}{(1+z)^2-3\Omega_{m0}}(1+z)\frac{H_0^2}{H^2}\right].\]

**Problem 25**

problem id: new_30

Find the sound speed for the modified Chaplygin gas with the state equation \[p=B\rho-\frac A{\rho^\alpha}.\]

\[c_s^2=\frac{\delta p}{\delta\rho}=\frac{\dot p}{\dot\rho}=-\alpha w+(1+\alpha)B,\quad w=\frac p\rho.\] (place after the problem 81, chapter 9 of the book.)

## A couple of problems for the SCM

**Problem 26**

problem id: 150_026

Let $N=\ln(a/a_0)$, where $a_0=a(t_0)$ and $t_0$ is some chosen reference time. Usually the reference time is the present time and in that case $\tau=-\ln(1+z)$. Find $\Omega_m(N)$ and $\Omega_\Lambda(N)$ for the SCM.

\begin{align} \nonumber \Omega_m&=\frac{\rho_m}{3H^2}=\frac{\Omega_{m0}\exp(-3N)}{\Omega_{m0}\exp(-3N)+\Omega_{\Lambda0}};\\ \nonumber \Omega_\Lambda&=\frac{\rho_\Lambda}{3H^2}=\frac{\Omega_{\Lambda0}}{\Omega_{m0}\exp(-3N)+\Omega_{\Lambda0}}. \end{align}

**Problem 27**

problem id: 150_027

Express the cosmographic parameters $H,q,j$ as functions of $N=\ln a/a_0$ for the SCM.

\begin{align} \nonumber H&=H_0\sqrt{\Omega_{m0}\exp(-3N)+\Omega_{\Lambda0}},\\ \nonumber q&=-1+\frac32\Omega_m=-1+\frac32\frac{\Omega_{m0}\exp(-3N)}{\Omega_{m0}\exp(-3N)+\Omega_{\Lambda0}},\\ \nonumber j&=1. \end{align}

## Cardassian Model

[K. Freese and M. Lewis, Cardassian Expansion: a Model in which the Universe is Flat, Matter Dominated, and Accelerating, arXiv: 0201229] Cardassian Model is a modification to the Friedmann equation in which the Universe is flat, matter dominated, and accelerating. An additional term, which contains only matter or radiation (no vacuum contribution), becomes the dominant driver of expansion at a late epoch of the universe. During the epoch when the new term dominates, the universe accelerates. The authors named this period of acceleration by the Cardassian era. (The name Cardassian refers to a humanoid race in Star Trek whose goal is to take over the universe, i.e., accelerated expansion. This race looks foreign to us and yet is made entirely of matter.) Pure matter (or radiation) alone can drive an accelerated expansion if the first Friedmann equation is modified by the addition of a new term on the right hand side as follows: \[H^2=A\rho+B\rho^n,\] where the energy density $\rho$ contains only ordinary matter and radiation, and $n<2/3$. In the usual Friedmann equation $B=0$. To be consistent with the usual result, we take \[A=\frac{8\pi}{3M_{Pl}^2},\] where $M_{Pl}^2\equiv1/G$.

**Problem 28**

problem id: 150_cardas1

Show that once the new term dominates the right hand side of the Friedmann equation, we have accelerated expansion.

When the new term is so large that the ordinary first term can be neglected, we find \[a\propto t^{\frac2{3n}}.\] Indeed, assuming that the Universe is filled solely by the non-relativistic matter, one finds \[H\propto\rho^{n/2}\propto a^{-3n/2},\quad a\propto a^{-3n/2+1},\quad a^{-3n/2-1}da\propto dt\Rightarrow a\propto t^{2/(3n)},\] so the expansion is superluminal (accelerated) for $n<2/3$. For example, for $n=2/3$ we have $a\propto t$; for $n=1/3$ we have $a\propto t^2$; and for $n=1/6$ we have $a\propto t^4$. The case of $n=2/3$ produces a term $H^2\propto a^{-2}$ in the FLRW equation; such a term looks similar to a curvature term but it is generated here by matter in a universe with a flat geometry. Note that for $n=1/3$ the acceleration is constant, for $n>1/3$ the acceleration is diminishing in time, while for $n<1/3$ the acceleration is increasing (the cosmic jerk).

**Problem 29**

problem id: 150_cardas2

Let us represent the Cardassian model in the form \[H^2\propto \rho+\rho_X,\quad\rho_X=\rho^n.\] Find the parameter $w_X$ of the EoS $p_X=w_X\rho_X$, assuming that the Universe is filled exclusively by the non-relativistic matter.

Use the result of the previous problem \[a\propto t^{2/(3n)}.\] As \[a\propto t^{\frac{2}{3(w+1)}},\] one finds that \[w=n-1\]in the considered case.

**Problem 30**

problem id: 150_cardas3

Show that the result obtained in the previous problem takes place for arbitrary one-component fluid with $w_X=const$.

The relation \[\rho_X(z)=\rho_{X0}\exp\left[3\int\limits_0^zdz'\frac{1+w_X(z')}{1+z'}\right]\] holds for arbitrary component with the EoS $p_X=w_X\rho_X$. Consequently \[\frac{d\rho_X}{dz}=3\rho_X\frac{1+w_X(z)}{1+z}.\] The same result can be obtained immediately from the conservation equation using the relation \[\frac d{dt}=\frac{dz}{dt}\frac d{dz}=-(1+z)H.\] Taking into account that $\rho_x=\rho^n$ and \[\rho=(1+z)^{3(w_X+1)}\] for $w_X=const$, one finds \[\rho_X=(1+z)^{3(1+w)n}.\] Substitution of the latter expression into the equation for $\rho_X$ gives \[w_X=n-1.\]

**Problem 31**

problem id: 150_cardas4

Generalize the previous problem for the case of two-component ideal liquid (non-relativistic matter $+$ radiation) with density $\rho=\rho_m+\rho_r$.

Use the result of the previous problem \[\frac{d\rho_X}{dz}=3\rho_X\frac{1+w_X}{1+z}.\] In the considered case \[\rho_X=\rho^n=\left(\rho_m+\rho_r\right)^n=\left(\rho_{m0}(1+z)^3+\rho_{r0}(1+z)^4\right)^n.\] Substitution of the latter expression into the equation for $\rho_X$ gives \[n\rho^{n-1}\left(\rho_{m0}(1+z)^2+\rho_{r0}(1+z)^3\right)=3\rho^n\frac{1+w_X}{1+z};\] \[w_X=\frac n{3\rho}\left(\rho_{m0}(1+z)^3+\rho_{r0}(1+z)^4\right)-1=\frac{3n\rho+n\rho_{r0}(1+z)^4-3\rho}{3\rho},\] and one finally obtains \[w_X=n-1+\frac n{3\rho}\rho_{r0}(1+z)^4,\] which is very slowly varying function with the redshift. If $\rho_{r0}\ll\rho_{m0}$, at late times, parameter $w_X$ is almost constant and it is identical to a dark energy component with a constant equation of state. But in early times, as one can not ignore the radiation component, one has to take the general EoS parameter $w_X$ which is not constant.

**Problem 32**

problem id: 150_cardas5

Show that we can interpret the Cardassian empirical term in the modified Friedmann equation as the superposition of a quintessential fluid with $w=n-1$ and a background of dust.

Equivalence of the equations \[H^2=A\rho_m+B(\rho_m)^n\] and \[H^2=A\rho_{m0}a^{-3}+Ba^{-3n},\] taking into account that $\rho_i\propto a^{-3(1+w_i)}$ allows to interpret the Cardassian empirical term in the modified Friedmann equation as the superposition of a quintessential fluid with $w=n-1$ and a background of dust with $w=0$.

**Problem 33**

problem id: 150_cardas6

We have two parameters in the original Cardassian model: $B$ and $n$. Make the transition $\{B,n\}\to\{z_{eq},n\}$, where $z_{eq}$ is the redshift value at which the second term $B\rho^n$ starts to dominate.

The second term starts to dominate at the redshift $z_{eq}$ when $A\rho(z_{eq})=B\rho^n(z_{eq})$, i.e., when \begin{equation}\label{150_cardas6_e1} \frac B A=\rho_0^{1-n}(1+z_{eq})^{3(1-n)}. \end{equation} In the Cardassian model today, we have \begin{equation}\label{150_cardas6_e2} H_0^2=A\rho_0+B\rho_0^n, \end{equation} so that \begin{equation}\label{150_cardas6_e3} A=\frac{H_0^2}{\rho_0}-B\rho_0^{n-1}. \end{equation} From Eqs.(\ref{150_cardas6_e1}) and (\ref{150_cardas6_e3}), we have \[B=\frac{H_0^2(1+z_{eq})^{3(1-n)}}{\rho_0^n\left[1+(1+z_{eq})^{3(1-n)}\right]}.\]

**Problem 34**

problem id: 150_cardas7

What is the current energy density of the Universe in the Cardassian model? Show that the corresponding energy density is much smaller than in the standard Friedmann cosmology, so that only matter can be sufficient to provide a flat geometry.

From \[H^2=A\rho+B\rho^n\] and (see the previous problem) \[\frac B A=\rho_0^{1-n}(1+z_{eq})^{3(1-n)}\] we have \[H^2=A\left[\rho+\rho_0^{1-n}(1+z_{eq})^{3(1-n)}\rho^n\right].\] Evaluating this equation today with $A=8\pi/3M_{Pl}^2$, we obtain \[H_0^2=\frac{8\pi}{3M_{Pl}^2}\rho_0\left[1+(1+z_{eq})^{3(1-n)}\right].\] The energy density $\rho_0$ is, by definition, the critical density. Consequently, \[\rho_0=\rho_{cr}=\frac{3H_0^2M_{Pl}^2}{8\pi\left[1+(1+z_{eq})^{3(1-n)}\right]}.\] This result can be presented in the form \[\rho_{cr}=\rho_{crFLRW}\times F(z_{eq},n),\quad \rho_{crFLRW}=\frac{3H_0^2M_{Pl}^2}{8\pi},\quad F(z_{eq},n)\equiv\left[1+(1+z_{eq})^{3(1-n)}\right]^{-1}.\] For example, if $z_{eq}=1$, then $F=\{1/3,1/5,3/20\}$ for $n=\{1/3,2/3,1/6\}$. We see that the value of the critical density can be much lower than in the standard Friedmann cosmology.

**Problem 35**

problem id: 150_cardas8

Let us represent the basic relation of Cardassian model in the following way \[H^2=A\rho\left[1+\left(\frac\rho{\rho_{car}}\right)^{n-1}\right],\] where $\rho_{car}=\rho(z_{eq})$ is the energy density at which the two terms are equal. Find the function $\rho(z_{eq})$ under assumption that the Universe is filled with non-relativistic matter and radiation.

\[\rho(z_{eq})=\rho_m(z_{eq})+\rho_r(z_{eq})=\rho_{m0}(1+z_{eq})^3\left[1+\frac{\Omega_{r0}}{\Omega_{m0}}(1+z_{eq})\right].\]

**Problem 36**

problem id: 150_cardas9

Let Friedmann equation is modified to be \[H^2=\frac{8\pi G}{3}g(\rho),\] where $\rho$ consists only of non-relativistic matter. Find the effective total pressure.

In the case of adiabatic expansion one has \begin{align} \nonumber dE+pdV&=0,\\ \nonumber d(g(\rho)a^3)+p_{tot}d(a^3)&=0,\\ \nonumber a^3dg+3a^2gda+3p_{tot}a^2da&=0,\\ \nonumber \dot g+3Hg&=-3p_{tot}H,\\ \nonumber \frac{dg}{dt}&=\frac{dg}{d\rho}\frac{d\rho}{dt},\\ \nonumber \frac{d\rho}{dt}&=-3H\rho,\\ \nonumber -\frac{dg}{d\rho}3H\rho+3Hg&=-3p_{tot}H,\\ \nonumber p_{tot}&=\rho\frac{dg}{d\rho}-g. \end{align} The same relation is commonly used also in presence of the radiation. This is incorrect, as the relation \[\frac{d\rho}{dt}=-3H\rho\] does not hold with radiation.

**Problem 37**

problem id: 150_cardas10

Find the speed of sound in the Cardassian model. [P.Gandolo, K. Freese, Fluid Interpretation of Cardassian Expansion, 0209322 ]

The Cardassian model \[H^2=\frac{8\pi}{M_{Pl}^2}\rho_m+B\rho_m^n\] can equivalently be written as \[H^2=\frac{8\pi}{M_{Pl}^2}\rho_m\left[1+\left(\frac{\rho_m}{\rho_{card}}\right)^{n-1}\right]\equiv\frac{8\pi}{M_{Pl}^2}\rho.\] In the previous problem we have shown that \[p_{tot}=\rho\frac{dg}{d\rho}-\rho.\] It allows to represent the speed of sound as \[c^2=\frac{\partial p_{tot}}{\partial\rho}=\frac{\partial p_{tot}/\partial\rho_m}{\partial\rho/\partial\rho_m}=\rho_m\frac{\left(\partial^2\rho/\partial\rho_m^2\right)}{\partial\rho/\partial\rho_m}.\] All the derivatives are calculated at constant entropy. Finally for the sound speed in the model we obtain \[c^2=-\frac{n(1-n)}{n+\left(\frac{\rho_{card}}{\rho_m}\right)^{1-n}}.\] This value is not guaranteed to be positive. So this model should be considered as an effective description at scales where $c^2>0$.

**Problem 38**

problem id: 150_cardas11

Find the deceleration parameter for the canonic Cardassian model.

\begin{align} \nonumber q(z)&=\frac12\frac{(1+z)}{E^2(z)}\frac{dE^2(z)}{dz}-1,\\ \nonumber E^2&\equiv\frac{H^2}{H_0^2}=\Omega_{m0}(1+z)^3+(1-\Omega_{m0})(1+z)^{3n},\\ \nonumber q(z)&=\frac12-\frac32(1-n)\frac{\kappa(z)}{1+\kappa(z)},\\ \nonumber \kappa(z)&\equiv\left(\frac{1-\Omega_{m0}}{\Omega_{m0}}\right)(1+z)^{-3(1-n)}. \end{align}

## Models with Cosmic Viscosity

A Universe filled with a perfect fluid represents quite a simple which seems to be in good agreement with cosmological observations. But, on a more physical and realistic basis we can replace the energy-momentum tensor for the simplest perfect fluid by introducing cosmic viscosity. The energy momentum tensor with bulk viscosity is given by
\[T_{\mu\nu}=(\rho=p-\xi\theta)u_\mu u_\nu+(p-\xi\theta)g_{\mu\nu},\]
where $\xi$ is bulk viscosity, and $\theta\equiv3H$ is the expansion scalar. This modifies the equation of state of the cosmic fluid. The Friedmann equations with inclusion of the bulk viscosity, i.e. using the energy-momentum tensor $T_{\mu\nu}$, read
\begin{align}
\nonumber \frac{\dot a^2}{a^2}&=\frac13\rho,\quad \rho=\rho_m+\rho_\Lambda,\quad 8\pi G=1;\\
\nonumber \frac{\ddot a^2}{a}&=-\frac16(\rho+3p-9\xi H).
\end{align}
*Problems #150_8-#150_14 are inspired by A. Avelino and U. Nucamendi, Can a matter-dominated model with constant bulk viscosity drive the accelerated expansion of the universe? arXiv:0811.3253*

**Problem 39**

problem id: 150_8

Consider a cosmological model in a flat Universe where the only component is a pressureless fluid with constant bulk viscosity ($\xi=const$). The pressureless fluid represent both the baryon and dark matter components. Find the dependence $\rho_m(z)$ for the considered model.

The conservation equation in terms of the scale factor and the first Friedmann equation are \[a\frac{d\rho_m}{da}-3(3H\xi-\rho_m)=0,\] \[H^2=\frac{8\pi G}3\rho_m.\] Here $\rho_m$ is total density of the baryon and dark matter components. Having excluded the Hubble parameter and changed the independent variable from the scale factor $a$ to the redshift $z$, one finds \[(1+z)\frac{d\rho_m}{dz}-3\rho_m+\gamma\rho_m^{1/2}=0,\quad \gamma\equiv9\sqrt{8\pi G/3}\xi.\] The solution of this equation is: \[\rho_m(z)=\left[\frac\gamma3+\left(\rho_{m0}^{1/2}-\frac\gamma3\right)(1+z)^{3/2}\right]^2.\]

**Problem 40**

problem id: 150_9

Find $H(z)$ and $a(t)$ for the model of Universe considered in the previous problem.

Substitute the solution $\rho_m(z)$ obtained in the previous problem into the first Friedmann equation to obtain \[H(z)=H_0\left[\frac{\bar\gamma}3+\left(\Omega_{m0}^{1/2}-\frac{\bar\gamma}3\right)(1+z)^{3/2}\right],\quad \bar\xi\equiv\frac{24\pi G}{H_0}\xi.\] In the considered model the bulk viscous matter is the only component of the flat Universe. Consequently, $\Omega_{m0}=1$ and one finally obtains \[H(z)=\frac13H_0\left[\bar\gamma+\left(3-\bar\gamma\right)(1+z)^{3/2}\right].\] The obtained expression allows to write the scale factor in terms of the cosmic time. Let us transform the expression for the Hubble parameter \[H(a)=\frac13H_0\frac{\bar\xi a^{3/2}+3-\bar\xi}{a^{3/2}}\] to the following form \[H(t-t_0)=3\int\limits_1^a\frac{{a'}^{1/2}}{\bar\xi {a'}^{3/2}+3-\bar\xi}\ da'.\] For $\xi\ne0$ and $\bar\xi a^{3/2}+3-\bar\xi>0$ ($\bar\xi a^{3/2}+3-\bar\xi<0$ implies $H<0$ and contradicts the observations) one finds \[a(t)=\left[\frac{3\exp\left(\frac12\bar\xi H(t-t_0)-3+\bar\xi\right)}{\bar\xi}\right]^{2/3}.\]

**Problem 41**

problem id: 150_10

Analyze the expression for the scale factor $a(t)$ obtained in the previous problem for different types of the bulk viscosity.

(a) $0<\bar\xi<3$

When $t\to\infty$ then the obtained solution reproduces the de Sitter-like Universe,
\[a(t)\propto\exp\left[\frac{\bar\xi}3H(t-t_0)\right]\]
(b) $\bar\xi=3$

In this case the considered model exactly reproduces the de Sitter-like Universe,
\[a(t)=\exp\left[H_0(t-t_0)\right]\]
The model predicts an Universe in an eternal accelerated expansion.

(c) $\bar\xi>3$

In this case the Universe expands forever (decelerated expansion epoch is absent).

**Problem 42**

problem id: 150_11

Show that the Universe in the considered model with $\xi=const$ had the Big Bang in the past for all values of the bulk viscosity in the interval $0<\bar\xi<3$ and determine how far in the past (in terms of the cosmic time) it happened.

Using the result of Problem #150_9 \[a(t)=\left[\frac{3\exp\left(\frac12\bar\xi H(t-t_0)-3+\bar\xi\right)}{\bar\xi}\right]^{2/3},\] one finds the time of the Big Bang as $a(t_{BB})=0$: \[t_{BB}=t_0+\frac2{\bar\xi H_0}\ln\left(1-\frac{\bar\xi}3\right).\]

**Problem 43**

problem id: 150_12

Show that the result of the previous Problem for zero bulk viscosity ($\xi=0$) correctly reproduces the lifetime of the matter-dominated Universe.

For $\xi\to0$ \[H_0(t_0-t_{BB})=\frac23.\]

**Problem 44**

problem id: 150_13

As we have seen in the previous problems, in the interval $0<\bar\xi<3$ the Universe begins with a Big-Bang followed by an eternal expansion and this expansion begins with a decelerated epoch followed by an eternal accelerated one. The transition between the decelerated-accelerated expansion epochs depends on the value of $\bar\xi$. Find the value of the scale factor where the transition happens.

Using \[H(a)=\frac13H_0\frac{\bar\xi a^{3/2}+3-\bar\xi}{a^{3/2}},\] one obtains \[\frac{d\dot a}{da}=H+a\frac{dH}{da}=\frac13H_0\left(\bar\xi+\frac{\bar\xi-3}{2a^{3/2}}\right).\] In order to find the scale factor value $a_t$, corresponding to the transition from the decelerated expansion to the accelerated one, we use the fact that $d\dot a/da\equiv -Hq$. Thus the condition $q=0$ is equivalent to the requirement $d\dot a/da=0$, therefore \[a_t=\left(\frac{3-\bar\xi}{2\bar\xi}\right)^{2/3},\quad z_t=\left(\frac{2\bar\xi}{3-\bar\xi}\right)^{2/3}-1.\]

**Problem 45**

problem id: 150_14

Analyze the dependence \[a_t=\left(\frac{3-\bar\xi}{2\bar\xi}\right)^{2/3},\] obtained in the previous problem.

1. For $0<\bar\xi<1$ the transition between the decelerated epoch to the accelerated one takes place in the future $a_t>1$.

2. For $\bar\xi\to0$ the value $a_t\to\infty$ (the distant future).

3. For $\bar\xi=1$ the transition takes place today ($a_t=1$).

4. For $1<\bar\xi<3$ the transition takes place in the past of the Universe ($0<a_t<1$).

5. For $\bar\xi\to3$ the transition is close to the Big-Bang time ($a_t\to0$).

**Problem 46**

problem id: 150_15

Find the deceleration parameter $q(a,\bar\xi)$ for the cosmological model presented in the Problem #150_8.

\[q(a)=-\frac{\ddot a}{aH^2}.\] The term $\ddot a/a$ can be calculated from the second Friedmann equation, that for a matter-dominated universe with bulk viscosity reads: \[\frac{\ddot a}a=-\frac{4\pi G}{3}(\rho_m-9\xi H).\] From the definition of \[\bar\xi\equiv\frac{24\pi G}{H_0}\xi,\] using \[\rho_m=\frac3{8\pi G}H^2\] one obtains that \[\frac{\ddot a}a=\frac12(\bar\xi H_0-H)H.\] Using \[H(a)=\frac13H_0\frac{\bar\xi a^{3/2}+3-\bar\xi}{a^{3/2}},\] we find \[q(a,\bar\xi)=\frac12\left[\frac{3-\bar\xi(1+2a^{3/2})}{3-\bar\xi(1-a^{3/2})}\right].\]

**Problem 47**

problem id: 150_16

Analyze behavior of the deceleration parameter $q(a,\bar\xi)$ obtained in the previous problem for different values of the bulk viscosity $\bar\xi(\xi)$.

(a) For the case $\bar\xi=0$ we have $q=1/2$ that corresponds to a matter-dominated universe with null bulk viscosity.

(b) For the case $\bar\xi=3$ we have $q=-1$ that corresponds to the de Sitter Universe.

(c) For the case $0<\bar\xi<3$ it is always a decreasing function from $q(0)=1/2$ to $q(\infty)=-1$ with a transition from positive to negative values in the value of the scale factor \[a_t=\left[\frac{3-\bar\xi}{2\bar\xi}\right]^{3/2}.\]
(d) For the case $\bar\xi>3$ the Universe expands forever, and this expansion is always accelerated (there is no decelerating epoch or acceleration-deceleration transition).

When $a\to a_*\equiv(1-3/\bar\xi)^{2/3}$ $q\to-\infty$ and when $a\to\infty$ then $q\to-1$. It is a negative and increasing function but it never becomes positive, its maximum value is $-1$.

(e) The case $\bar\xi<3$ predicts an eternal decelerated expanding Universe. The universe begins with a Big-Bang and expands forever until to reach its maximum value $a_*=(1-3/\bar\xi)^{2/3}$ when $t\to\infty$. Deceleration parameter is a positive and increasing monotonic function where its minimum value is $1/2$.

**Problem 48**

problem id: 150_17

Use result of the problem \ref{150_15} to find the current value of the deceleration parameter and make sure that for $\bar\xi=1$ the transition from the decelerated to accelerated epochs of the Universe takes place today.

From the expression \[a_t=\left[\frac{3-\bar\xi}{2\bar\xi}\right]^{3/2}.\] we obtain the deceleration parameter evaluated today as \[q(a=1,\bar\xi)=\frac{1-\bar\xi}{2}.\] When $\bar\xi=1$ the transition from the decelerated to accelerated epochs of the Universe takes place today ($a=1$). For $\bar\xi<1$ we have a decelerated Universe in the present and for $\bar\xi>1$ we have an accelerated one today.

**Problem 49**

problem id: 150_18

Find the curvature scalar $R(a,\xi)$ for the cosmological model presented in the Problem #150_8.

For a flat Universe \[R=-6\left(\frac{\ddot a}a +H^2\right).\] Using \[\frac{\ddot a}a=\frac12H(\bar\xi H_0-H), \quad H(a)=\frac13H_0\frac{\bar\xi a^{3/2}+3-\bar\xi}{a^{3/2}},\] we find \[R=-3H(a)[\bar\xi+H(a)]\Rightarrow R(a,\bar\xi)=-\frac13H_0^2\left[\frac{(3-\bar\xi)^2}{a^3}+\frac{5\bar\xi(3-\bar\xi)}{a^{3/2}}+4\bar\xi^2\right].\]

**Problem 50**

problem id: 150_19

Let us consider a flat homogeneous and isotropic Universe filled by a fluid with bulk viscosity. We shall assume that the EoS for the fluid is $p=w\rho$, $w=const$ and that the viscosity coefficient $\xi(\rho)=\xi_0\rho^\nu$. Find the dependence $\rho(a)$ for the considered model.

The Friedmann equations for the considered model read \[H^2=\frac{8\pi G}{3}\rho;\] \[\dot\rho+3H[(1+w)\rho-3\xi_0\rho^\nu H]=0.\] Substitution of $H$ into the conservation equation gives \[\dot\rho+3H[(1+w)\rho-\xi^*_0\rho^{\nu+1/2}H]=0,\quad \xi_0^*\equiv\sqrt{24\pi G}\xi_0.\] Solution of the latter equation reads \[\rho(a)=\left[\frac{\xi_0^*}{1+w}+\frac C{1+w}a^r\right]\frac{1}{\frac12-\nu},\quad r\equiv 3(\nu-1/2)(1+w).\] Here $C$ is an integration constant.

**Problem 51**

problem id: 150_0013

Usually the inflationary models of the early Universe contain two distinct phases. During the first phase entropy of the Universe remains constant. The second phase is essentially non-adiabatic, particles are produced through the damping of the coherent oscillations of the inflaton field by coupling to other fields and by its subsequent decay. Find relation between the bulk viscosity and the entropy production [J. A. S. Lima, R. Portugal, I. Waga, Bulk viscosity and deflationary universes, arXiv:0708.3280].

Use the first law of thermodynamics \[d(\rho a^3)+pd(a^3)=TdS\] and the conservation equation in presence of viscosity \[\dot\rho+3H(\rho+p+\Pi)=0,\quad \Pi=-3H\xi,\] one finds \[\frac{T\dot S}{a^3}=\dot\rho+3H(\rho+p)=-3H\Pi.\] Consequently, \[\Pi=-\frac{T\dot S}{3Ha^3}.\] As $\dot S>0$ in an expanding Universe $(H>0)$ then $\Pi$ is negative. It agrees with the definition $\Pi=-3H\xi$, where $\xi$ is the positive bulk viscosity coefficient.

# New from March 2015

# New from December 2014

# Exactly Integrable n-dimensional Universes

Exactly Integrable n-dimensional Universes

# UNSORTED NEW Problems

*The history of what happens in any chosen sample region is the same as the history of what happens everywhere. Therefore it seems very tempting to limit ourselves with the formulation of Cosmology for the single sample region. But any region is influenced by other regions near and far. If we are to pay undivided attention to a single region, ignoring all other regions, we must in some way allow for their influence. E. Harrison in his book Cosmology, Cambridge University Press, 1981 suggests a simple model to realize this idea. The model has acquired the name of "Cosmic box" and it consists in the following.*

*Imaginary partitions, comoving and perfectly reflecting, are used to divide the Universe into numerous separate cells. Each cell encloses a representative sample and is sufficiently large to contain galaxies and clusters of galaxies. Each cell is larger than the largest scale of irregularity in the Universe, and the contents of all cells are in identical states. A partitioned Universe behaves exactly as a Universe without partitions. We assume that the partitions have no mass and hence their insertion cannot alter the dynamical behavior of the Universe. The contents of all cells are in similar states, and in the same state as when there were no partitions. Light rays that normally come from very distant galaxies come instead from local galaxies of long ago and travel similar distances by multiple re?ections. What normally passes out of a region is reflected back and copies what normally enters a region.*

*Let us assume further that the comoving walls of the cosmic box move apart at a velocity given by the Hubble law. If the box is a cube with sides of length $L$, then opposite walls move apart at relative velocity $HL$.Let us assume that the size of the box $L$ is small compared to the Hubble radius $L_{H} $ , the walls have a recession velocity that is small compared to the velocity of light. Inside a relatively small cosmic box we use ordinary everyday physics and are thus able to determine easily the consequences of expansion. We can even use Newtonian mechanics to determine the expansion if we embed a spherical cosmic box in Euclidean space.*

**Problem 1**

problem id:

As we have shown before (see Chapter 3): $p(t)\propto a(t)^{-1} $ , so all freely moving particles, including galaxies (when not bound in clusters), slowly lose their peculiar motion and ultimately become stationary in expanding space. Try to understand what happens by considering a moving particle inside an expanding cosmic box

For simplicity we suppose the particle moves in a direction perpendicular to two opposite walls of an expanding box. Normally, of course, the particle rebounds in different directions, but the final result is just the same. The walls are perfect reflectors and therefore, relative to the wall, the particle rebounds with the same speed as when it strikes the wall. During the collision, the direction of motion is reversed, but the speed relative to the wall remains unchanged. Because the wall is receding, the particle returns to the center of the box with slightly reduced speed. Each time the particle strikes a receding wall it returns with reduced speed. Using the Hubble law it can be shown that a particle of mass m and speed U, moving within an expanding box, obeys the law that mU is proportional to $1/L$. The product mU is the momentum. As the box gets larger the momentum gets smaller. The length L expands in the same way as the scaling factor R, and the momentum therefore obeys the important law: \[mUR=const\] This law holds not only for particles in an expanding box but also for particles moving freely in an expanding Universe. Remarkably, the general relativity equation of motion of a freely moving particle in the uniformly curved space of an expanding Universe gives exactly the same result. This illustrates how the cosmic box not only helps us to understand what happens but also allows us to employ very simple methods to derive important results.

**Problem 2**

problem id:

Show that at redshift $z=1$ , when the Universe is half its present size, the kinetic energy of a freely moving nonrelativistic particle is four times its present value, and the energy of a relativistic particle is twice its present value.

Using $p\propto a^{-1} $ find for non-relativistic particle \[E_{kin} \propto a^{-2} \] In terms of the redshift this gives \[E_{kin} =E_{kin0} (1+z)^{2} \] Consider now a relativistic particle. In this case, $E\propto p$ and in terms of the redshift, \[E=E_{0} (1+z)\] Consequently, at redshift $z=1$ , when the Universe is half its present size, the kinetic energy of a freely moving nonrelativistic particle is four times its present value, and the energy of a relativistic particle is twice its present value.

**Problem 3**

problem id:

Let the cosmic box is filled with non-relativistic gas. Find out how the gas temperature varies in the expanding cosmic box.

We have seen before that individual particles, moving freely, lose their energy when enclosed in an expanding box. Exactly the same thing happens to a gas consisting of many particles. Particles composing a gas continually collide with one another; between collisions they move freely and lose energy in the way described for free particles; during their encounters they exchange energy, but collisions do not change the total energy. The temperature of a gas therefore varies with expansion in the same manner as the energy of a single (nonrelativistic) particle: gas temperature is proportional to $1/a^{2} $ . If $T$ denotes temperature, and $T_{0} $ the present temperature, then \[T=T_{0} \left(1+z\right)^{2} \]

**Problem 4**

problem id:

Show that entropy of the cosmic box is conserved during its expansion.

The number of photons in our cosmic box (and in the Universe) is a measure of its entropy. The total number of photons in the cosmic box is $N_{\lambda } =n_{\lambda } V$ . The photon density $n_{\lambda } $ varies as $T^{3} $ , and therefore varies as $1/a^{3} $. But $V$ varies as $a^{3} $ , hence also $VT^{3} =const$ . Thus the entropy of the thermal radiation in the cosmic box is constant during expansion. This is just another way of saying that the total number of photons $N_{\lambda } $ (and, consequently, entropy) in the box is constant. Actually, their number is slowly increased by the light emitted by stars and other sources, but this contribution is so small that for most purposes it can be ignored.

**Problem 5**

problem id:

Consider a (cosmic) box of volume V, having perfectly reflecting walls and containing radiation of mass density $\rho $. The mass of the radiation in the box is $M=\rho V$ . We now weigh the box and find that its mass, because of the enclosed radiation, has increased not by M but by an amount 2M. Why?

This unexpected increase in mass occurs because the radiation exerts pressure on the walls of the box and the walls contain stresses. These stresses in the walls are a form of energy that equals 3PV, where P is the pressure of the radiation. The pressure equals $\frac{1}{3} \rho $, and the energy in the walls is therefore $\rho V$and has a mass equivalent of $M=\rho V$ . The mass of the box is therefore increased by the mass $M$ of the radiation and the mass M of the stresses in the walls, giving a total increase of 2M. In the Universe there are no walls: nonetheless, the radiation still behaves as if it had a gravitational mass twice what is normally expected. Instead of using $\rho $ , we must use $\rho +3P$ as in the second Friedmann equation. This feature of general relativity explains why in a collapsing star, where all particles are squeezed to high energy, increasing the pressure, contrary to expectation, hastens the collapse of the star.

**Problem 6**

problem id:

Show that the jerk parameter is \[j(t)=q+2q^{2} -\frac{\dot{q}}{H} \]

\[j=\frac{1}{H^{3} } \frac{\dddot{a}}{a} =\frac{1}{aH^{3} } \frac{d}{dt} \left(\frac{\ddot{a}}{aH^{2} } aH^{2} \right)=-\frac{1}{aH^{3} } \frac{d}{dt} \left(qaH^{2} \right)=q+2q^{2} -\frac{\dot{q}}{H} \]

**Problem 7**

problem id:

We consider FLRW spatially flat Universe with the general Friedmann equations \[\begin{array}{l} {H^{2} =\frac{1}{3} \rho +f(t),} \\ {\frac{\ddot{a}}{a} =-\frac{1}{6} \left(\rho +3p\right)+g(t)} \end{array}\] Obtain the general conservation equation.

Using \[\frac{\ddot{a}}{a} =\dot{H}+H^{2} \] we find \[\dot{\rho }+3H\left(\rho +p\right)=6H\left(-f(t)+\frac{\dot{f}(t)}{2H} +g(t)\right)\]

**Problem 8**

problem id:

Show that for extra driving terms in the form of the cosmological constant the general conservation equation (see previous problem) transforms in the standard conservation equation.

In this case$f(t)=g(t)=\Lambda /3,\; \; \dot{f}=0$ and \[\dot{\rho }+3H\left(\rho +p\right)=6H\left(-f(t)+\frac{\dot{f}(t)}{2H} +g(t)\right)\to \dot{\rho }+3H\left(\rho +p\right)=0\]

**Problem 9**

problem id:

Show that case $f(t)=g(t)=\Lambda /3$ corresponds to $\Lambda (t)CDM$ model.

For \textbf{$f(t)=g(t)=\Lambda /3$} \[\dot{\rho }+3H\left(\rho +p\right)=6H\left(-f(t)+\frac{\dot{f}(t)}{2H} +g(t)\right)\to \dot{\rho }+3H\left(\rho +p\right)=\dot{\Lambda }(t)\] which corresponds to \textbf{$\Lambda (t)CDM$}model.</p> </div> </div></div> The de Sitter spacetime is the solution of the vacuum Einstein equations with a positive cosmological constant$\Lambda $ . To describe the geometry of this spacetime one usually takes the spatially flat metric \[ds^{2} =dt^{2} -a^{2} (t)d\vec{x}^{2} \] with the scale factor \[a(t)=a_{0} e^{Ht} \] The Hubble parameter is thus a fixed constant. <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 10''' <p style="color: #999;font-size: 11px">problem id: </p> Show that the de Sitter spacetime has a constant four-dimensional curvature. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[R=-6\left(\frac{\ddot{a}}{a} +H^{2} \right)=-6\left(\dot{H}+2H^{2} \right)=-12H^{2} \] As \[H^{2} =\frac{8\pi G}{3} \rho _{\Lambda } =\frac{\Lambda }{3} \] then \[R=-4\Lambda \]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 11''' <p style="color: #999;font-size: 11px">problem id: </p> In the de Sitter spacetime transform the FRLW metric into the explicitly conformally flat metric. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Introduce $dt\equiv a\left(\eta \right)d\eta $ to obtain \[ds^{2} =dt^{2} -a^{2} (t)d\vec{x}^{2} =a^{2} \left(\eta \right)\left(d\eta ^{2} -d\vec{x}^{2} \right)\] where the conformal time $\eta $ and the scale factor $a\left(\eta \right)$ are \[\eta =-\frac{1}{H} e^{-Ht} ,\quad a\left(\eta \right)=-\frac{1}{H\eta } \] The conformal time $\eta $ changes from $-\infty $ to $0$ when the proper time $t$ goes from $-\infty $ to $+\infty $.</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 12''' <p style="color: #999;font-size: 11px">problem id: </p> (Problems 12-13, A.Vilenkin, Many worlds in one, Hill and Wang, New York, 2006)} In thirties of XX-th century a cyclic Universe model was popular. This model predicted alternating stages of expansion and contraction. Show that such model contradicts the second law of thermodynamics. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Second law requires that entropy, which is a measure of disorder, should grow in each cycle of cosmic evolution. If the Universe had already gone through an infinite number of cycles, it would have reached the maximum-entropy state of thermal equilibrium ("heat death"). We certainly do not find ourselves in such a state.</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 13''' <p style="color: #999;font-size: 11px">problem id: </p> P. Steinhardt and N. Turok proposed a model of cyclic Universe where the expansion rate in each cycle is greater than the contraction one so that volume of the Universe grows from one cycle to the other. Show that this model does not contradict the second law of thermodynamics and is free of the heat death problem. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">The contradiction to the second principle of thermodynamics and therefore the heat death problem are absent in the considered model, because the amount of expansion in a cycle is greater than the amount of contraction. So the volume of the Universe is increased after each cycle. The entropy of our observable region is now the same as the entropy of a similar region in the preceding cycle, but the entropy of the entire Universe has increased, simply because the volume of the Universe is now greater. As time goes on, both the entropy and the total volume grow without bound. The state of maximum of entropy is never reached.</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 14''' <p style="color: #999;font-size: 11px">problem id: </p> If a closed Universe appeared as a quantum fluctuation, so what is the upper limit of its existence? (see A.Vilenkin, Many worlds in one, Hill and Wang, New York, 2006) <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">The energy of a closed Universe is always equal to zero. The energy of matter is positive, the gravitational energy is negative, and it turns out that in a closed Universe the two contributions exactly cancel each other. Thus if a closed Universe were to arise as a quantum fluctuation, there would be no need to borrow energy from the vacuum $\left(\Delta E=0\right)$ and because $\Delta E\Delta t\ge \hbar $, the lifetime of the fluctuation could be arbitrary long.</p> </div> </div></div> ---- =UNIQ--h-20--QINU Tutti Frutti = [[Planck_scales_and_fundamental_constants#Problem_15|'''New problem in Cosmo warm-up Category:''']] <div id="TF_1"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 1''' <p style="color: #999;font-size: 11px">problem id: TF_1</p> Construct planck units in a space of arbitrary dimension. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Dimensionality of the fundamental constants $c,\hbar,G_D$ in $D=4+n$ dimensions can be determined as \[[G_d]=L^{D-1}T^{-2}M^{-1},\quad \hbar=L^2T^{-1}M,\quad c=LT^{-1}.\] Note that the dimension of the space affects only the dimensionality of the Newton's constant $G_D$, because the universal gravitation law transforms with changes of dimensionality of the space as the following \[F=G_D\frac{M_1M_2}{R^{D-2}}.\] Use the combination \[[G_D^\alpha\hbar^\beta c^\gamma]= L^{\alpha(D-1)+2\beta+\gamma} T^{-2\alpha-\beta-\gamma} M^{-\alpha+\beta-\gamma}\] to find that \[L_{P(D)}=\left(\frac{G_D\hbar}{c^3}\right)^{\frac{1}{D-2}}\quad T_{P(D)}=\left(\frac{G_D\hbar}{c^{D+1}}\right)^{\frac{1}{D-2}}\quad M_{P(D)}=\left(\frac{c^{5-D}\hbar^{D-3}}{G_D}\right)^{\frac{1}{D-2}}.\]</p> </div> </div></div> '''New problem in Inflation Category:''' <div id="TF_2"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 2''' <p style="color: #999;font-size: 11px">problem id: TF_2</p> Show that for power law $a(t)\propto t^n$ expansion slow roll inflation occurs when $n\gg1$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Slow roll inflation corresponds to \[\varepsilon\equiv-\frac{\dot H}{H}\ll1.\] For power law expansion $H=n/t$ so that $\varepsilon=n^{-1}$. Consequently, slow roll inflation occurs when $n\gg1$.</p> </div> </div></div> <div id="TF_3"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 3''' <p style="color: #999;font-size: 11px">problem id: TF_3</p> Find the general condition to have accelerated expansion in terms of the energy densities of the darks components and their EoS parameters <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Differentiating the first Friedmann equation with respect to time and substituting $\dot\rho_{dm}$ and $\dot\rho_{de}$ from the corresponding conservation equations, one obtains the equation \[2\dot H=-(1+w_{dm})\rho_{dm}-(1+w_{de})\rho_{de}.\] The acceleration is given by the relation $\ddot a=a(\dot H+H^2)$. Using $3H^2=\rho_{dm}+\rho_{de}$ we obtain \[\ddot a=-\frac a6 [(1+3w_{dm})\rho_{dm}+(1+3w_{de})\rho_{de}].\] The condition $\ddot a>0$ leads to the inequality \[\rho_{de}>-\frac{1+3w_{dm}}{1+3w_{de}}\rho_{dm}.\]</p> </div> </div></div> <div id="dec_5"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 4''' <p style="color: #999;font-size: 11px">problem id: dec_5</p> Complementing the assumption of isotropy with the additional assumption of homogeneity predicts the space-time metric to become of the Robertson-Walker type, predicts the redshift of light $z$, and predicts the Hubble expansion of the Universe. Then the cosmic luminosity distance-redshift relation for comoving observers and sources becomes \[d_L(z)=\frac{cz}{H_0}\left[1-(1-q_0)\frac z2\right]+O(z^3)\] with $H_0$ and $q_0$ denoting the Hubble and deceleration parameters, respectively. Show that this prediction holds for arbitrary spatial curvature, any theory of gravity (as long as space-time is described by a single metric) and arbitrary matter content of the Universe.(see 1212.3691) </div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 5''' <p style="color: #999;font-size: 11px">problem id: </p> Show that in the Universe filled by radiation and matter the sound speed equals to \[c_s^2=\frac13\left(\frac34\frac{\rho_m}{\rho_r}+1\right)^{-1}.\] </div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 6''' <p style="color: #999;font-size: 11px">problem id: </p> Show that result of the previous problem can be presented in the following form \[c_s^2=\frac43\frac{1}{(4+3y)},\quad y\equiv\frac a{a_{eq}},\] where $a_{eq}$ is the scale factor value in the moment when matter density equals to that of radiation. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[c_s^2=\frac13\left(\frac34\frac{\rho_m}{\rho_r}+1\right)^{-1},\quad \frac{\rho_m}{\rho_r}=a\frac{\rho_{m0}}{\rho_{r0}},\quad \frac{\rho_{m0}}{\rho_{r0}}=\frac1{a_{eq}},\] \[c_s^2=\frac13\frac1{\left(\frac34\frac{a}{a_{eq}}+1\right)}=\frac43\frac{1}{(4+3y)}.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 7''' <p style="color: #999;font-size: 11px">problem id: </p> Show that in the flat Universe filled by non-relativistic matter and radiation the effective radiation parameter $w_{tot}=p_{tot}/\rho_{tot}$ equals \[w_{tot}=\frac1{3(1+y)},\quad y\equiv\frac a{a_{eq}},\] where $a_{eq}$ is the scale factor value in the moment when matter density equals to that of radiation. </div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 8''' <p style="color: #999;font-size: 11px">problem id: </p> Show that in spatially flat one-component Universe the following hold \[\bar{H'}=-\frac{1+3w}2\bar H^2.\] <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[\bar H^2=a^2\rho,\quad (8\pi G/3=1),\] \[\rho'+3\bar H\rho(1+w)=0,\] \[\bar{H'}=a^2\rho-\frac32a^2\rho-\frac32a^2\rho w\to\bar{H'}=-\frac{1+3w}2\bar H^2.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 9''' <p style="color: #999;font-size: 11px">problem id: </p> Express statefinder parameters in terms Hubble parameter and its derivatives with respect to cosmic times. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[r=1+3\frac{\dot H}{H^2}+\frac{\ddot H}{H^3},\quad s=-\frac{2}{3H}\frac{3H\dot H+\ddot H}{3H^2+2\dot H}.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 10''' <p style="color: #999;font-size: 11px">problem id: </p> Find temperature of radiation and Hubble parameter in the epoch when matter density was equal to that of radiation (Note that it was well before the last scattering epoch). <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">One can estimate the current observation time using other well known parameters. For example the period when when matter density was equal to that of radiation $z=3410\pm40$ (this value would be $1.69$ time greater if one takes under radiation only photons). It means that all length scales in that epoch were $3400$ times less than today. The CMB temperature was $9300$Ê. Age of that epoch was $51100\pm1200$ years. In this epoch the Universe expanded much faster: $H=(10.6\pm0.2)\ km\ sec^{-1}\ pc^{-1}$. We can also give our cosmic observational time by quoting the value of some parameters at Universe, and the CMB temperature was then 9300K, as hot as an A-type star. The age at that epoch was $t_{eq} = (51100 \pm 1200)$ years. And at that epoch the Universe was expanding much faster than today, actually $H_{eq} = (10.6 ± 0.2)\ km\ s^{-1}$ (note this is per 'pc', not 'Mpc').</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 11''' <p style="color: #999;font-size: 11px">problem id: </p> Estimate the mass-energy density $\rho$ and pressure $p$ at the center of the Sun and show that $\rho\gg p$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">For the Sun: \[p\approx\frac{GM_\odot^2}{R_\odot^4}\approx10^{16}J/m63;\] \[\rho\ge\frac{M_\odot c^2}{\frac43\pi R_\odot^3}\sim10^{21}J/m^3.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 12''' <p style="color: #999;font-size: 11px">problem id: </p> For a perfect fluid show that ${T^{\alpha\beta}}_{;\alpha}=0$ implies \[(\rho+p)u^\alpha\nabla_\alpha u^\beta=h^{\beta\gamma}\nabla_\gamma p,\] where $h_{\alpha\beta}\equiv g_{\alpha\beta}-u_\alpha u_\beta$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">For a perfect fluid, \[T_{\alpha\beta}= (\rho+p)u_\alpha u_\beta-pg_{\alpha\beta}.\] The conservation equation, $\nabla^\alpha T_{\alpha\beta}=0$, thus gives \[\nabla^\alpha T_{\alpha\beta}=(\rho+p)u^\alpha\nabla^\alpha u_\beta+u^\beta\nabla^\alpha[(\rho+p)u_\alpha]-\nabla_\beta p=0.\] Contracting with $u^\beta$, we find that \[\nabla^\alpha[(\rho+p)u_\alpha]-u^\gamma\nabla_\gamma p=0.\] Substituting this back into $\nabla^\alpha T_{\alpha\beta}$, we get \[(\rho+p)u^\alpha\nabla^\alpha u_\beta+u^\gamma\nabla_\gamma pu_\beta-\nabla_\beta p=0,\] or, equivalently, \[(\rho+p)u^\alpha\nabla^\alpha u_\beta=\left(g^{\alpha\beta}-u^\beta u^\gamma\right)\nabla_\gamma p=0.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 13''' <p style="color: #999;font-size: 11px">problem id: </p> Show that in flat Universe filled by non-relativistic matter and a substance with the state equation $p_X=w_X\rho_X$ the following holds \[\frac{d\ln H}{d\ln a}-\frac12\frac{\Omega_X}{1-\Omega_X}\frac{d\ln\Omega_X}{d\ln a}+\frac32=0.\] <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">The conservation equations are \begin{align}\label{335_1} \nonumber\dot\rho_m=3H\rho_m & =0\\ \dot\rho_X+3H(1+w_X)\rho_X=0. \end{align} Using (\ref{335_1}) and \[\rho_m=\rho_X=H^2,\quad 8\pi G=1/M_p^2=1,\] and introducing $\Omega_i=\rho_i/(3H^2)$ $i=m,X$ we obtain \[w_X=-1-\frac1{3H}\frac{\dot\rho_X}{\rho_X}=-1-\frac1{3H\Omega_X}\left(\frac{2\Omega_X}H\frac{dH}{dt} +\frac{d\Omega_X}{dt}\right)=-1-\frac23\left(\frac{d\ln H}{d\ln a}+\frac12\frac{d\ln\Omega_X}{d\ln a}\right).\] Substituting this $w_X$ into the Friedman equation \[2\dot H+3H^2=-p,\] one finally finds \[\frac{d\ln H}{d\ln a}-\frac12\frac{\Omega_X}{1-\Omega_X}\frac{d\ln\Omega_X}{d\ln a}+\frac32=0.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 14''' <p style="color: #999;font-size: 11px">problem id: </p> (after Ming-Jian Zhang, Cong Ma, Zhi-Song Zhang, Zhong-Xu Zhai, Tong-Jie Zhang, Cosmological constraints on holographic dark energy models under the energy conditions) Using result of the previous problem, find EoS parameter $w_{hde}$ for holographic dark energy, taking the IR cut-off scale equal to the following: <br /> i) event horizon; <br />ii) conformal time; <br />iii) Cosmic age. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">i) The event horizon cut-off is given by \[R_E=a\int\limits_t^\infty\frac{dt'}{a(t')}=\int\limits_a^\infty\frac{da'}{a'2H}.\] In this case, the event horizon $R_E$ is considered as the spatial scale. Consequently, with the dark energy density $\rho_{hde}=3c^2R_E^2$ and $\Omega_{hde}=\rho_{hde}/(3H^2)$, we obtain \[\int\limits_a^\infty\frac{d\ln a'}{Ha'}=\frac{c}{Ha}\Omega_{hde}^{-1/2}.\] Taking the derivative with respect to $\ln a$, we get \[\frac{d\ln H}{d\ln a}+\frac12\frac{d\ln\Omega_{hde}}{d\ln a}=\frac{\sqrt{\Omega_{hde}}} c-1.\] Because (see the previous problem) \[w_{hde}=-1-\frac23\left(\frac{d\ln H}{d\ln a}+\frac12\frac{d\ln\Omega_{hde}}{d\ln a}\right)\] one finally finds \[w_{hde}=-\frac13\left(\frac23\sqrt{\Omega_{hde}}+1\right).\] The acceleration condition $w<-1/3$ is satisfied for $c>0$. <br /> ii) Conformal time cut-off is given by \[\eta_{hde}=\int\limits_0^a\frac{dt'}{a(t')}=\int\limits_0^a\frac{da'}{a'^2H}.\] In this case, the conformal time is considered as a temporal scale, and we can again convert it to a spatial scale after multiplication by the speed of light. Proceeding the same way as in the previous case one obtains \[\frac{d\ln H}{d\ln a}+\frac12\frac{d\ln\Omega_{hde}}{d\ln a}+\frac{\sqrt{\Omega_{hde}}}{ac}=0.\] and \[w_{hde}=\frac23\frac{\sqrt{\Omega_{hde}}}{c}(1+z)-1,\] which corresponds to an acceleration when $c>\sqrt{\Omega_{hde}}(1+z).$ <br /> iii) The cosmic age cut-off is defined as \[t_{hde}=\int\limits_0^tdt'=\int\limits_0^a\frac{da'}{a'H}.\] In this case, the age of Universe is considered as a time scale. The corresponding spatial scale is again obtained after multiplication by the speed of light. Proceeding the same way as in the two previous cases one finds \[\int\limits_0^\infty\frac{d\ln a'}{H}=\frac c H \Omega_{hde}^{-1/2}.\] Equation of state for holographic dark energy \[w_{hde}=\frac{2}{3c}\sqrt{\Omega_{hde}}-1.\] Accelerated expansion requires $c>\sqrt{\Omega_{hde}}$.</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 15''' <p style="color: #999;font-size: 11px">problem id: </p> According to so-called Jeans criterion exponential growth of the perturbation, and hence instability, will occur for wavelengths that satisfy: \[k<\frac{\sqrt{4\pi G\rho}}{v_S}\equiv k_J.\] In other words, perturbations on scales larger than the Jeans scale, defined as follows: \[R_J=\frac\pi {k_J}\] will become unstable and collapse. Give a physical interpretation of this criterion. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">A simple way to derive the Jeans scale is to compare the sound crossing time $t_{SC}\approx R/v_S$ to the free-fall time of a sphere of radius $R$, $t_{ff}\approx1/\sqrt{G\rho}$. The physical meaning of this criterion is that in order to make the system stable the sound waves must cross the overdense region to communicate pressure changes before collapse occurs. The maximum space scale (Jeans scale) can be found from the condition \[R_J\approx t_{ff}v_S.\] It then follows that \[R_J\approx\frac{v_S}{\sqrt{G\rho}}.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 16''' <p style="color: #999;font-size: 11px">problem id: </p> According to the Jeans criterion, initial collapse occurs whenever gravity overcomes pressure. Put differently, the important scales in star formation are those on which gravity operates against electromagnetic forces, and thus the natural dimensionless constant that quantifies star formation processes is given by: \[\alpha_g=\frac{Gm_p^2}{e^2}\approx8\times10^{-37}.\] Estimate the maximal mass of a white dwarf star in terms of $\alpha_g$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">For a star with $N$ baryons, the gravitational energy per baryon is $E_G\sim-GNm_p^2/R$, and the kinetic energy of relativistic degenerate gas is $E_K\sim p_F c\sim\hbar cN^{1/3}/R$ where $p_F$ is the Fermi momentum. Consequently, the total energy is: \[E=\hbar cN^{1/3}/R-GNm_p^2/R.\] For the system to be stable, the maximal number of baryons $N$ is obtained by setting $E=0$ in the expression above. The result is the Chandrasekhar mass: \[M_{Chandra}=m_p\left(\frac{\hbar c}{Gm_p^2}\right)^{3/2}=m_p(\alpha\alpha_g)^{-3/2}\approx1.8M_\odot.\] where $\alpha=e^2/(\hbar c)$ is the fine structure constant. This simple derivation result is close to the more precise value, derived via the equations of stellar structure for degenerate matter, $1.4M_\odot$.</p> </div> </div></div> ---- The formation of a star, or indeed a star cluster, begins with the collapse of an overdense region whose mass is larger than the Jeans mass, defined in terms of the Jeans mass $R_J$(???), \[M_J=\frac43\pi\rho\left(\frac{R_J}2\right)^3\propto\frac{T^{3/2}}{\rho^{1/2}}.\] (why $T^{3/2}$, if in gases it is $T^{1/2}$???) Overdensities can arise as a result of turbulent motions in the cloud. At the first stage of the collapse, the gas is optically thin and isothermal, whereas the density increases and $M_J\propto\rho^{-1/2}$. As a result, the Jeans mass decreases and smaller clumps inside the originally collapsing region begin to collapse separately. Fragmentation is halted when the gas becomes optically thick and adiabatic, so that $M_J\propto\rho^{1/2}$, as illustrated in fig. 1. This process determines the opacity-limited minimum fragmentation scale for low mass stars, and is given by: \[M_{min}\approx m_p\alpha_g^{-3/2}\alpha^{-1}\left(\frac{m_e}{m_p}\right)^{1/4}\approx0.01M_\odot.\] Of course, this number, which is a robust scale and confirmed in simulations, is far smaller than the observed current epoch stellar mass range, for which the characteristic stellar mass is $\sim0.5M_\odot$. Fragmentation also leads to the formation of star clusters, where many stars with different masses form through the initial collapse of a large cloud. In reality, however, the process of star formation is more complex, and the initial collapse of an overdense clump is followed by accretion of cold gas at a typical rate of $v_S^3/G$, where $v_S$ is the speed of sound. This assumes spherical symmetry, but accretion along filaments, which is closer to what is actually observed, yields similar rates. The gas surrounding the protostellar object typically has too much angular momentum to fall directly onto the protostar, and as a result an accretion disk forms around the central object. The final mass of the star is fixed only when accretion is halted by some feedback process. ---- <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 17''' <p style="color: #999;font-size: 11px">problem id: </p> Using the "generating function" $G(\varphi)$, \[H(\varphi,\dot\varphi)=-\frac1{\dot\varphi}\frac{dG^2(\varphi)}{d\varphi},\] make transition from the two coupled differential equations with respect to time \[3H^2=\frac12\dot\varphi^2+V(\varphi);\] \[\ddot\varphi+3H\dot\varphi+V'(\varphi)=0.\] to one non-linear first order differential equation with respect to the scalar field. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Using the ansatz for $H$, the equation of motion \(\ddot\varphi3H\dot\varphi+V'(\varphi)=0\) is integrated to give \[\frac12\dot\varphi^2=3G^2(\varphi)-V(\varphi),\] where an integration constant is absorbed into the definition of . Using this result, the first Friedmann equation becomes the "generating equation" \[V(\varphi)=3G^2(\varphi)-2\left[G'(\varphi)\right]^2.\] The evolution of the scalar field and the Hubble parameter are given by \[\dot\varphi=-2G'(\varphi),\quad H=G(\varphi).\] We need $G(\varphi)>0$ if the Universe is expanding. If we solve generation equation for a given potential $V(\varphi)$ and obtain the generating function $G(\varphi)$, the whole solution spectra can be found.</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 18''' <p style="color: #999;font-size: 11px">problem id: </p> (Hyeong-Chan Kim, Inflation as an attractor in scalar cosmology, arXiv:12110604) Express the EoS parameter of the scalar field in terms of the generating function and find the condition under which the scalar field behaves as the cosmological constant. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">The equation of state parameter of the scalar field is \[w=\frac p\rho=-1+\frac43\frac{G'^2}{G^2}.\] At the point satisfying $V(\varphi)=3G^2(\varphi)$ the equation of state becomes $w=-1$ and the scalar field will behaves as if it were a cosmological constant.