Use the relation \[\rho_{curv}=-\frac3{8\pi G}\frac{k}{R^2},\] where $R$ is the (comoving) radius of curvature of the open or closed Universe, to find \[\Omega_{k0}=-\frac{k}{RH_0}.\] 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 commonly known expression for the luminosity distance \[ d_L(z)=(1+z)\left\{ \begin{array}{lr} R\sinh\left[\frac1{H_0R}\int\limits_0^z\frac{dz'}{E(z')}\right], & open\\ H_0^{-1}\int\limits_0^z\frac{dz'}{E(z')}, & flat\\ R\sin\left[\frac1{H_0R}\int\limits_0^z\frac{dz'}{E(z')}\right], & closed \end{array} \right.\]

The energy density has time dependence determined by the conservation equation \[\dot{\rho}=-3H\rho-3Hp.\] The two terms determine behavior of the homogeneous 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.

From the second Friedmann equation \[\frac{\ddot a}{a}=-\frac{4\pi G}{3}(\rho+3p)\] clearly shows that both for matter and for radiation domination, the acceleration is negative; an expanding Universe dominated by the energy density of matter or radiation will decelerate. The acceleration in a matter dominated phase goes as $a^{-2}$ while in a radiation dominated phase it goes as $a^{-3}$. In either case the deceleration is large when $a(t)$ is small and then slows as the scale factor grows.

\begin{tabular}{|l|l|l|l|l|} \hline & UNIQ-MathJax8-QINU & UNIQ-MathJax9-QINU & UNIQ-MathJax10-QINU & UNIQ-MathJax11-QINU\\ \hline matter & UNIQ-MathJax12-QINU & UNIQ-MathJax13-QINU & UNIQ-MathJax14-QINU & UNIQ-MathJax15-QINU\\ \hline radiation & UNIQ-MathJax16-QINU & UNIQ-MathJax17-QINU & UNIQ-MathJax18-QINU & UNIQ-MathJax19-QINU\\ \hline UNIQ-MathJax20-QINU & constant & UNIQ-MathJax21-QINU & UNIQ-MathJax22-QINU & constant\\ \hline \end{tabular}

Indeed, in the case of the power-paw expansion with $\alpha<1$ (decelerated expansion) the Hubble radius grows faster than the Universe expands: $R_H=H^{-1}\propto t$ while $a(t)\propto t^\alpha$. In power law situations the Hubble radius has expansion velocity greater than the light speed \[\frac{d}{dt}\left(\frac1H\right)=\frac1\alpha.\] 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.

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.

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).)

\[N\propto\int\limits_{\varphi_{end}}^{\varphi_{initial}}d\varphi\frac{V}{dV/d\varphi} \propto\int\limits_{\varphi_{end}}^{\varphi_{initial}}d\varphi/\sqrt\varepsilon.\] 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, $V/(dV/d\varphi)$.

\[1+q=\frac{d}{dt}\left(\frac1H\right),\quad \frac{dq}{dt}=f(q)\to dt=\frac{dq}{f(q)},\] \[\frac1{H(q)}=\int\frac{1+q}{f(q)}dq.\] (see S. Carloni et al., A new approach to reconstruction methods in f(R) gravity, arXiv:1005.1840)

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 $r$, $cdt=\pm adr$. Consequently \[a_0dr=\frac{cdz}{H(z)}.\] 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 \[V_{exp}(z)=H_0 R(z)=H_0\int\limits_0^z\frac{cdz'}{H(z')}.\] where the expansion velocity is taken for an object with the redshift $z$ observed at the time $t=t_0$.

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 \[cz=HR.\]

\[(1+z)_{Dop}=\sqrt{\frac{c+V}{c-V}}.\] 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 $V_{Dop}le c$. \item \label{2501_04o} {\bf Show, that $V_{Dop}$ and $V_{exp}$ give the same results only in the first order of $V/c$.} \item \label{2501_05o} {\bf Find $V_{exp}(z)$ for the three cosmological models: Einstein-de Sitter, Milne and de Sitter.} For $\Omega_m=1$, $\Omega_\Lambda=0$: \[V_{exp}(z)=H_0 R(z)=H_0\int\limits_0^z\frac{cdz'}{H(z')}=\frac{2(\sqrt{1+z}-1)}{\sqrt{1+z}}c.\] For $\Omega_m=0$, $\Omega_\Lambda=0$: \[V_{exp}(z)=\frac{z(1+z/2)}{(1+z)}c.\] For $\Omega_m=0$, $\Omega_\Lambda=1$: \[V_{exp}(z)=zc.\]

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

Solving the corrected first Friedmann equation and the conservation equation for a matter dominated Universe \[H^2=\frac\rho3\left(1-\frac\rho{\rho_c}\right),\quad \dot\rho+3H\rho=0;\] 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 solution of standard Friedmann equations is recovered.

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

\[f'(R)R_{\mu\nu}-\frac12f(R)g_{\mu\nu}-\left(\nabla_\mu\nabla_\nu-g_{\mu\nu}\square\right)f'(R)=\kappa T_{\mu\nu}.\] Here \[T_{\mu\nu}=-\frac{2}{\sqrt{-g}}\frac{\delta S_m}{\delta g^{\mu\nu}},\] $\nabla_\mu$ is covariant derivative associated with the Levi-Civita connection of the metric, and $\square=\nabla_\mu\nabla^\mu$.

\begin{equation}\label{f_r_2_e_1} f'(R)R-2f(R)+3\square f'(R)=\kappa T, \end{equation} where $T=g^{\mu\nu}T_{\mu\nu}$. $R$ and $T$ are related in differential rather in algebraic way like $R=-kT$ in general relativity. Note that equality $T=0$ does not imply that $R$ equals to zero or a constant.

For $R=const$ and $T_{\mu\nu}=0$ the equation \ref{f_r_2_e_1} obtained in the previous problem takes on the form \[f'(R)R-2f(R)=0.\] For given function $f(R)$ this equation represents an algebraic equation for $R$. If $R=0$ is a root of this equation then it immediately follows from the field equations that $R_{\mu\nu}=0$ and the maximally symmetric solution is the Minkowski space-time. If the equation has a root $R=const=C$ then \[R_{\mu\nu}=\frac C4 g_{\mu\nu}\] and the maximally symmetric solution corresponds to the de- Sitter or anti-de Sitter (or to the cosmological constant in usual General Relativity) depending on sign of $C$.

\[H^2=\frac\kappa{3f'}\left[\rho+\frac12(Rf'-f)-3H\dot Rf''\right];\] \[2\dot H+H^2=-\frac\kappa{f'}\left[p+\dot R^2f'''+2H\dot Rf''+\ddot Rf''+\frac12(f-Rf')\right].\]

To reproduce (imitate) the de Sitter equation of state (i.e. the cosmological constant) $w_{eff}=-1$ the following condition is required \[\frac{f'''}{f''}=\frac{\dot RH-\ddot R}{\dot R^2}.\]

The result for $n\ne1$ is \[w_{eff}=-\frac{6n^2-7n+1}{6n^2-9n+3},\] and $\alpha$ in terms of $n$ reads \[\alpha=\frac{-2n^2+3n-1}{n-2}.\] Appropriate choice of $n$ leads to desired value of $w_{eff}$. for example $n=2$ leads to $w_{eff}=-1$, $\alpha=\infty$. Arbitrary dependence $a(t)$ would result in time dependence of $w_{eff}$.

\[w_{eff}=-1+\frac{2(n+2)}{3(2n+1)(n+1).}\] In particular, for $n=1$ one finds that $w_{eff}=-2/3$ (accelerated expansion). Note that in this case positive values of $n$ imply presence of terms inversely proportional to $R$, in contrast to the case considered in the previous problem.

Let us show an important consequence of the formula \begin{equation}\label{12_1} F=-\frac m{R^2}-q(t)H^2(t)R. \end{equation} Consider a Universe, which contains no matter (or radiation), but only dark energy in the form of a non-zero cosmological constant $\Lambda$. In this case, the Hubble parameter and, hence, the deceleration parameter become time-independent and are given by $H=\sqrt{\Lambda/3}$ and $q=-1$. Thus, the force (\ref{12_1}) also becomes time-independent, \begin{equation}\label{12_3} F=-\frac m{r^2}+\frac13\Lambda r. \end{equation} For the case of spatially finite (i.e. non-pointlike) spherically-symmetric massive objects (12.3) is replaced by \begin{equation}\label{12_4} F=-\frac{M(r)}{r^2}+\frac13\Lambda r, \end{equation} where $M(r)$ is the total mass of the object contained within the radius $r$. If the object has the radial density $\rho(r)$ then \[M(r)=\int\limits_0^r4\pi r'^2\rho(r')dr'.\] Although the de Sitter background is not an accurate representation of our Universe, the SCM is dominated by dark-energy in a form consistent with a simple cosmological constant. Even in the simple Newtonian case (\ref{12_4}), we see immediately that there is an obvious, but profound, difference between the cases $\Lambda=0$ and $\Lambda\ne0$. In the former, the force on a constituent particle of a galaxy or cluster (say) is attractive for all values of $r$ and tends gradually to zero as $r\to\infty$ (for any sensible radial density profile). In the latter case, however, the force on a constituent particle (or equivalently its radial acceleration) vanishes at the finite radius $r_F$ which satisfies $r_F=[3M(R_F)/\Lambda]^{1/3}$, beyond which the net force becomes repulsive. This suggests that a non-zero $\Lambda$ should set a maximum size, dependent on mass, for galaxies and clusters.

Using the relation $\rho_m=M/a^3$ ($M=const$) we rewrite the first Friedmann equation in the form \begin{equation}\label{1301_38_e1} \dot a^2+k=\frac13\frac M a+\frac13\Lambda a,\quad 8\pi G=1.\end{equation} Then we differentiate the latter equation two times to find \begin{align} \label{1301_38_e2} \ddot a=&-\frac16\frac M{a^2}=\frac13\Lambda a,\\ \nonumber \dddot a=&-\frac13\frac{M\dot a}{a^3}=\frac13\Lambda\dot a. \end{align} Using the definitions of the cosmographic parameters \[H=\frac{\dot a}{a},\quad q=-a\frac{\ddot a}{\dot a^2},\quad j=a^2\frac{\dddot a}{\dot a^3},\] we represent (\ref{1301_38_e1}) in the form \begin{align} \nonumber q &= \frac12A-B;\\ \nonumber j &= A+B;\\ \nonumber A &\equiv\frac13 \frac{M}{a^3H^2};\\ \nonumber B &\equiv\frac13 \frac{\Lambda}{H^2}. \end{align} Then we find \begin{align} \nonumber A &=\frac23(j+q);\\ \nonumber B &=\frac23(\frac12j-q). \end{align} The Friedmann equation (\ref{1301_38_e1}) in terms of the above introduced variables $A$ and $B$ takes on the form \[\frac k{a^2}=(A+B-1)H^2\] or \[k=a^2H^2(j-1).\]

We represent the first Friedmann equation in the form \begin{equation}\label{1301_39_e1} \frac{\dot a^2}{a^2}+\frac k{a^2}=\frac{M_m}{a^3}+\frac{M_r}{a^4},\quad \frac{8\pi G}{3}=1.\end{equation} We twice differentiate the latter equation to find \begin{align} \label{1301_39_e2} \ddot a=&-\frac12\frac{M_m}{a^2}-\frac{M_r}{a^3},\\ \nonumber \dddot a=&\frac{M_m}{a^3}\dot a+3\frac{M_r}{a^4}\dot a. \end{align} Using definitions of the cosmographic parameters $q$ and $j$, one obtains \begin{align} \nonumber q &= \frac12A+B;\\ \nonumber j &= A+3B;\\ \nonumber A &\equiv\frac{M_m}{a^3H^2};\\ \nonumber B &\equiv\frac{M_r}{a^4H^2}. \end{align} Then we find \begin{align} \nonumber A &=-2j+6q;\\ \nonumber B &=j-2q. \end{align} The Friedmann equation (\ref{1301_39_e1}) in terms of the above introduced variables $A$ and $B$ takes on the form \[\frac k{a^2}=(A+B-1)H^2\] or in terms of the cosmographic parameters \[k=a^2H^2(4q-j-1).\]

In the first case $B=0$, $q=1/2$. Then $A=1$ and $k=0$. Note that in this case $j=1$. In the second case $A=0$, $q=1$. Then $B=1$, $k=0$. In that case $j=3$.

Using expression for $\dddot a$ obtained in the problem \ref{1301_38}, one finds \[\ddddot a=\frac M3\frac{\ddot a}{a^3}-M\frac{\dot a^2}{a^4}+\frac\Lambda3\ddot a.\] For the snap parameter \[s\equiv a^3\frac{\ddddot a}{\dot a^4}\] one obtains \[s=-3(A+B)q-3A.\] Substitution of the parameters \begin{align} \nonumber A &=\frac23(j+q);\\ \nonumber B &=\frac23(\frac12j-q). \end{align} introduced in the problem \ref{1301_38}, one finally finds \[s+2(q+j)+qj=0.\] This fourth order ODE is equivalent to the Friedmann equation and has an advantage that it appears as a constraint on directly measurable quantities.

For small values of $a(t)$ density and pressure of the Chaplygin gas reduces to that of dust $\rho\propto a^{-3}$, and for large $a$ one gets the de Sitter Universe: $\rho=const$, $p=-\rho$. In between these two regimes one can use the approximation \begin{equation}\label{scalar_5}\rho=\sqrt A+\frac{B}{\sqrt{2A}}a^{-6}.\end{equation} Thus $\sqrt A$ plays the role of a cosmological constant. We insert this to the Friedmann equation with $\Lambda$ and follow the procedure of eliminating the constants by differentiation. This leads to an approximate constraint \begin{equation}\label{scalar_6}s+5(q+j)+qj=0.\end{equation} Analogous procedure for the generalized Chaplygin gas leads to the equation \begin{equation}\label{scalar_7}s+(3\alpha+2)(q+j)+qj=0.\end{equation} Note that for $\alpha=1$ we reproduce the above obtained result for the Chaplygin gas, while $\alpha=0$ we return to the results obtained in the problem (\ref{1301_41}). If we want to exclude the parameter $\alpha$ from the latter equation we must take one more derivative of the Friedmann equation and introduce an additional cosmological parameter \[l=\frac1a\frac{d^5a}{dt^5}\left(\frac1a\frac{da}{dt}\right)^{-5}.\] As a result one obtains \begin{equation}\label{scalar_8} -2qs-2jq^2-lq-2sj-3sq^2-j^2q-lj+s^2-3q^2j-qsj+j^3-2j^2q^2=0. \end{equation} This constraint is again approximate and is valid only in the regime where the higher order terms in the expansion of $\rho$ can be dropped.

The main argument of the opponents to the cyclic model of Universe was based on the so-called Tolman Entropy Conundrum (R.C. Tolman, Relativity, Thermodynamics and Cosmology. Oxford University Press (1934)): the entropy of the Universe necessarily increases, due to the second law of thermodynamics, and therefore cycles become larger and longer in the future, smaller and shorter in the past, implying that a Big Bang must have occurred at A FINITE time in the past.

\begin{align} \nonumber L_S&=L_{Pl}\alpha^{1/2};\\ \nonumber M_S&=M_{Pl}\alpha^{1/2};\\ \nonumber T_S&=T_{Pl}\alpha^{1/2}; \end{align}

By Gauss's theorem we have $[G]=L^NM^{-1}T^{-2}$; $[e^2]=L^NMT^{-2}$.

Evolution of the considered universe is described by the equations \[H^2=\frac2{n(n-1)}\rho-\frac k{a^2},\quad 8\pi G=1,\] \[\rho=\rho_m+\Lambda,\] \[\dot\rho_m+nH(\rho_m+p_m)=0.\] Using the conservation equation one finds \[\rho=\rho_0a^{-n(1+w)}+\Lambda.\] The first Friedmann equation then takes on the form \[\dot a^2=\frac2{n(n-1)}\rho_0a^{-n(1+w)+2}+\frac{2\Lambda}{n(n-1)}a^2-k.\] In the notions \[-n(1+w)\equiv N,\quad \frac2{n(n-1)}=M\] the Friedmann equation reads \begin{equation}\label{new_28_e_1} \dot a^2=M\rho_0a^{N+2}+M\Lambda a^2-k. \end{equation} Differentiation of the latter equation gives \begin{align}\label{new_28_e_2} \ddot a&=\frac12M\rho_0(N+2)a^{N+1}+M\Lambda a;\\ \nonumber\dddot a&=\frac{(N+1)(N+2)}2M\rho_0a^N\dot a+M\Lambda\dot a. \end{align} Transforming to the cosmographic parameters, one gets \begin{align}\label{new_28_e_3} q&\equiv-\frac{\ddot a a}{\dot a^2}=-\frac12M\rho_0(N+2)\frac{a^N}{H^2}-\frac{M\Lambda}{H^2},\\ \nonumber j&\equiv\frac{\dddot a a^2}{\dot a^3}=\frac{(N+1)(N+2)}2M\rho_0(N+2)\frac{a^N}{H^2}+\frac{M\Lambda}{H^2}. \end{align} Using the notions \[A\equiv-M\rho_0(N+2)\frac{a^N}{H^2},\quad B\equiv\frac{M\Lambda}{H^2},\] the equation (\ref{new_28_e_3}) can be presented in the form \begin{align}\label{new_28_e_4} q&=\frac12A-B,\\ \nonumber j&=-\frac{N+1}2A+B. \end{align} It then follows that \begin{equation}\label{new_28_e_5} A=-\frac2N(q+j),\quad B=-\frac1N[(N+1)q+j]. \end{equation} The first Friedmann equation in terms of the parameters $A$ and $B$ becomes \begin{equation}\label{new_28_e_6} \frac{k}{a^2H^2}=-\frac1{N+2}A+B-1, \end{equation} or transforming to the cosmographic parameters $q$ and $j$ \begin{equation}\label{new_28_e_7} k=a^2H^2\left\{\frac2{N(N+2)}(q+j)-\frac1N\left[(N+1)q+j\right]-1\right\}, \end{equation} For $n=3$, $w=0$ ($N=-3$) the obtained relation for spatial curvature reduces to \[k=a^2H^2(j-1),\] which reproduces result of the problem #1301_38.

Using the relation for $\dddot a$ obtained in the previous problem, one finds \[\ddddot a=\frac{N(N+1)(N+2)}2M\rho_0a^{N-1}\dot a^2+\frac{(N+1)(N+2)}2M\rho_0a^{N}\ddot a+M\Lambda\ddot a.\] Using the definition \[s=\frac{a^3}{\dot a^4}\ddddot a\] we obtain \[s=\left(\frac{N+1}2A-B\right)q-\frac{N(N+1)}{2}A,\] where $A$ and $B$ are defined in the previous problem. Transforming to the cosmographic parameters, the latter relation can be presented in the form \[s-\left\{\frac2N(q+j)+\frac1N[(N+1)q+j]q-\frac6N(q+j)\right\}=0.\] For $N=-3$ the obtained result reproduces that of the problem #1301_41.

\[c_s^2=\frac{\delta p}{\delta\rho}=\frac{\dot p}{\dot\rho}=-\alpha w+(1+\alpha)B,\quad w=\frac p\rho.\]