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

\[L_S=\left(\frac{Ge^2}{c^4}\right)^{1/2};\] \[M_S=\left(\frac{e^2}{G}\right)^{1/2};\] \[T_S=\left(\frac{Ge^2}{c^6}\right)^{1/2}.\]

\[L_S=L_{Pl}\alpha^{1/2}\] \[M_S=M_{Pl}\alpha^{1/2}\] \[T_S=T_{Pl}\alpha^{1/2}\]

\[F_{max}=\frac{Mc^2}{2R_S}=\frac{Mc^2}{4MG/c^2}=\frac{c^4}{4G}\approx3\times10^{43}N;\] \[P_{max}=\frac{Mc^2}{2R_S/c}=F_{max}c=\frac{c^5}{4G}\approx9\times10^{51}W.\]

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{2612_5_1} \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{2612_5_2} U=-\int\frac{GMdM}{r}=-\int\frac G r \left(\frac43\pi r^3\rho_0\right)4\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 $U$ \begin{equation}\label{2612_5_3} \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}, \end{equation} where the upper limit is obtained by maximizing the function in the range (\ref{2612_5_1}). Thus, the dark star criterion (\ref{2612_5_1}) 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.

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 \[E(z)=\sqrt{\Omega_r(1+z)^4+\Omega_m(1+z)^3+\Omega_k(1+z)^2+\Omega_w\exp\left(\int\limits_0^zdz'\frac{1+w(z')}{1+z'}\right)}.\] 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 \[a(t)=\left(\frac34\rho_ct^2+1\right)^{1/3},\quad \rho(t)=\frac{\rho_c}{\frac34\rho_ct^2+1},\quad H(t)=\frac{\frac12\rho_c t}{\frac34\rho_ct^2+1}\] For small values of the energy density ($\rho\ll\rho_c$) we recover the solutions of standard Friedmann equations.

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

The main argument of the opponents to the cyclic cosmology model of the 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.

We can rewrite the Ricci scalar $R$ as function of redshift $z$, \[R=6H\left[(1+z)\frac{dH}{dz}-2H\right].\] For $z=0$ we have \begin{align} \nonumber R_0 & = 6H\left(\left.\frac{dH}{dz}\right|_{z=0}-2H_0\right),\\ \nonumber \left.\frac{dR}{dz}\right|_{z=0} & =6\left[\left(\left.\frac{dH}{dz}\right|_{z=0}\right)^2-H_0 \left(3\left.\frac{dH}{dz}\right|_{z=0}-\left.\frac{d^2H}{dz^2}\right|_{z=0}\right)\right]. \end{align}

\[\frac{\ddot a}{a}=-\frac{4\pi G}{3}\rho(1+3w).\] In the flat case \[\rho=\frac{3}{8\pi G}H^2\] and \[q=-\frac{\dot a}{aH^2}=\frac12(1+3w).\]

\[\frac{\ddot a}{a}=-\frac{4\pi G}{3}\rho(1+3w).\] Substitution \[\rho=\frac{3}{8\pi G}\left(H^2+\frac k{a^2}\right)\] gives \[q=-\frac{\dot a}{aH^2}=\frac12(1+3w)\left(1+\frac k{a^2H^2}\right).\]

In this case \[q=-\frac{\dot a}{aH^2}=\frac12(1+3\frac p\rho)\left(1+\frac k{a^2H^2}\right) =\] \[=\frac12\left[1+3\frac p\rho\left(1+\frac k{a^2H^2}\right)\right]+\frac k{2a^2H^2},\] \[1+\frac k{a^2H^2}=\sum\limits_i\Omega_i,\quad\frac p\rho=\frac{\sum\limits_iw_i\Omega_i}{\sum\limits_i\Omega_i},\] \[q=\frac12\left(1+3\sum\limits_i\Omega_i\right)+\frac k{2a^2H^2}.\]

Using results of the previous problem one obtains \[q=\frac12\left(1+3\sum\limits_iw_i\Omega_i\right)+\frac k{2a^2H^2}=\frac12\left(1+\frac k{a^2H^2}\right)+\frac32\sum\limits_iw_i\Omega_i=\] \[=\frac12\sum\limits_i\Omega_i+\frac32\sum\limits_iw_i\Omega_i=\frac{\Omega_{total}}2+\frac32\sum\limits_i w_i \Omega_i.\]

In the considered case \[\Omega_m +\Omega_{de}=1+\frac{k}{a^2H^2},\] \[q=\frac12+\frac32\left[W\Omega_{de}+w\Omega_m\right]+\frac{k}{2a^2H^2}.\] These can, formally, be regarded as two equations for $\Omega_{de}$ and $\Omega_{m}$. Solving them we get \begin{align} \nonumber \Omega_{m} & = \frac{2q-1-3W}{3(w-W)}-\frac{k(1+3W)}{3a^2H^2(w-W)};\\ \nonumber \Omega_{m} & = \frac{2q-1-3w}{3(w-W)}-\frac{k(1+3w)}{3a^2H^2(w-W)}. \end{align}

We note that the denominator in these expressions is always positive because $W<0$ for dark energy. Thus, the fact that $\Omega_{m(de)}>0$ imposes an upper and lower limit on $q$, \[\frac{1+3W}{2}\left(1+\frac{k}{a^2H^2}\right)\le q\le \frac{1+3w}{2}\left(1+\frac{k}{a^2H^2}\right).\]

It follows from the conservation equation that \[p=-\frac{\dot\rho}{3H}\frac{w}{1+w}.\] Using \[w = \frac{2q-1}3;\quad \dot\rho=\frac{3}{4\pi G}H\dot H,\quad \dot H=-H^2(1+q)\] one finds \[p=\frac{H^2}{8\pi G}(2q-1).\]

\[q=-1-\frac12\frac{d\ln\rho}{d\ln a}.\]

For spatially flat one-component Universe with state equation $p=w\rho$ the scale factor is \[a\propto t^\frac{2}{3(1+w)}\] and therefore \[H=\frac{2}{3(1+w)}\frac1t.\] Using $q=(1+3w)/2$ one can find a simple relation between the current age of the Universe and the DP \[t_0=\frac{H_0^{-1}}{1+q}.\]

The Friedman equations can then be represented in the form \begin{align} \label{background_2_29} H^2 & +\frac1{a^2} =\frac{8\pi G}3\rho,\\ \nonumber 2\frac{\ddot a}{a} & +H^2+\frac1{a^2} = 0. \end{align} Using this equations one can find the relations between the current values of the Universe's parameters \begin{align} \label{background_2_30} H_0^2 & =\frac1{a_0^2(2q_0-1)},\\ \nonumber q_0 & = \frac{4\pi G}{3H_0^2}\rho_0. \end{align} Note that in general \[q_0=\frac{4\pi G}3\frac{\rho_0+3p_0}{H_0^2}=\frac{\rho_0+3p_0}{2\rho_{0,crit}},\quad \rho_{0,crit}=\frac{3H_0^2}{8\pi G}.\] For Universe filled only with non-relativistic matter one has $\Omega_m = 2q_0$. It is easy to see that $q_0>1/2$ and $\Omega_m>1$ as was expected in the closed model. Using (\ref{background_2_30}) one can rewrite the equation for scale factor in the form \begin{equation}\label{background_2_31}\dot a^2=\frac\alpha a-1,\quad \alpha\equiv\frac{2q_0}{H_0(2q_0-1)^{3/2}}.\end{equation} It is easy to see that the considered model includes both parameters $H_0$ and $q_0$. integration of (\ref{background_2_31}) gives \[t=\int \sqrt{\frac{a}{\alpha-a}}\,da.\] Substitution \[a=\frac\alpha2(1-\cos\tau)\] leads to \begin{equation}\label{background_2_33}t=\frac\alpha2(\tau-\sin\tau).\end{equation} Because of the relation $dt=ad\tau$ it is evident that the variable $\tau$ is the conformal time. Taking the constants of integration so that $a=0$ as $t=0$ (and $\tau=0$) we can see that $a=a_0$ at $\tau=\tau_0$. Consequently, \[\cos\tau_0=\frac{1-q_0}{q_0},\quad \sin\tau_0=\frac{\sqrt{2q_0-1}}{q_0}.\] From (\ref{background_2_33}) it follows that the age of Universe in close model is \[t_0=\frac\alpha2(\tau_0-\sin\tau_0)=\frac{q_0}{H_0(2q_0-1)^{3/2}}\left(\arccos{\frac{1-q_0}{q_0}} - \frac{\sqrt{2q_0-1}}{q_0}\right).\] The maximum of scale factor reaches at $\tau=\pi$, \[a_{\rm max}=\alpha=\frac{2q_0}{H_0(2q_0-1)^{3/2}}.\] As one can see, all parameters of the model can be expressed in terms of parameters $H_0$ and $q_0$.

In the case of open Universe the formulae (\ref{background_2_29})-(\ref{background_2_31}) transform into \begin{align} \label{background_2_37} H^2 & -\frac1{a^2} =\frac{8\pi G}3\rho,\\ \nonumber 2\frac{\ddot a}{a} & +H^2-\frac1{a^2} = 0,\\ \label{background_2_38} H_0^2 & =\frac1{a_0^2(1-2q_0)},\\ \nonumber q_0 & = \frac{4\pi G}{3H_0^2}\rho_0, \end{align} \begin{equation}\label{background_2_39}\dot a^2=\frac\beta a+1,\quad \beta\equiv\frac{2q_0}{H_0(1-2q_0)^{3/2}}.\end{equation} Now $q_0<1/2$, $0\le\Omega_m< 1$. The solution of (\ref{background_2_39}) in the conformal time parametrization is \[a=\frac\beta2(\cosh\tau-1),\quad t=\frac\beta2(\sinh\tau-\tau).\] Current value $\tau_0$ of the conformal time is defined by the relation \[\cosh\tau_0=\frac{1-q_0}{q_0}.\] The age of the Universe is \[t_0=\frac\beta2(\sinh\tau_0-\tau_0) = \frac{q_0}{H_0(1-q_0)^{3/2}}\left(\frac{\sqrt{1-2q_0}}{q_0} - \ln\frac{1-q_0+\sqrt{1-2q_0}}{q_0}\right).\] In this case again all characteristics of the model are expressed in terms of the parameters $H_0$ and $q_0$.

a) $H>0$, $q>0$: expanding and decelerating
b) $H>0$, $q<0$: expanding and accelerating
c) $H<0$, $q>0$: contracting and decelerating
d) $H<0$, $q<0$: contracting and accelerating
e) $H>0$, $q=0$: expanding, zero deceleration
f) $H<0$, $q=0$: contracting, zero deceleration
g) $H=0$, $q=0$: static
There is little doubt that we live in an expanding Universe, and hence only (a), (b), and (e) are possible. Evidences in favor of the fact that the expansion is presently accelerating continuously grows in number and therefore the current dynamics belongs to type (b).

Of course, generally speaking, both the Hubble parameter and deceleration parameter can change their sign during the evolution. Therefore the evolving Universe can transit from one type to another. It is one of the basic tasks of cosmology to follow this evolution and clarify its causes.

In the case of constant deceleration parameter the Universe would exhibit decelerating expansion if $q>0$, an expansion with constant rate if $q=0$, accelerating power-law expansion if $-1<q<0$, exponential expansion (also known as de Sitter expansion) if $q=-1$ and super-exponential expansion if $q<-1$.

\begin{align} \nonumber q>\frac12 & :\quad closed\ spherical\ space & k&=+1;\\ \nonumber q=\frac12 & :\quad open\ flat\ space & k&=0;\\ \nonumber q<\frac12 & :\quad open\ hyperbolic\ space & k&=-1. \end{align}

For the Einstein-de Sitter Universe where the age is \[t^*=\frac23H^{-1}\] we have \begin{align} \nonumber q>\frac12 & :\quad age<t^*;\\ \nonumber q=\frac12 & :\quad age=t^*;\\ \nonumber q<\frac12 & :\quad age>t^*. \end{align}

In a decelerating Universe with $q>0$, the age of the Universe will be less than the Hubble time, because at earlier times it was expanding at a faster rate, whereas a Universe that has always been accelerating, that is, $q<0$ for all time, will have an age that is greater than the Hubble time. A Universe that expands at a constant rate, $q=0$, has an age equal to the Hubble time.

\[q=-\frac{\ddot a/a}{H^2}=\frac{\frac{4\pi G}{3}(\rho+3p)}{\frac{8\pi G}{3}\rho}=\frac12+\frac32\frac p\rho= \frac12+\frac32\frac{w\rho_X}{\rho_m+\rho_X} =\frac12+\frac32\frac w{(1+r)}.\]

Consider an $(n+1)$-dimensional homogeneous and isotropic Lorentzian spatially flat spacetime with the metric \begin{equation} \label{1} ds^2=g_{\mu\nu} dx^\mu dx^\nu=-dt^2+a^2(t)g_{ij} dx^i dx^j,\quad i,j=1,\dots,n, \end{equation} where $t$ is the cosmological (or cosmic) time and $g_{ij}$ is the metric of an $n$-dimensional Riemannian manifold $M$ of constant scalar curvature characterized by an indicator, $k=-1,0,1$, so that $M$ is an $n$-hyperboloid, the flat space $\Bbb R^n$, or an $n$-sphere, with the respective metric \begin{equation} \label{2} g_{ij}dx^i dx^j=\frac1{1-kr^2}\,dr^2+r^2\,d\Omega^2_{n-1}, \end{equation} where $r>0$ is the radial variable and $d\Omega_{n-1}^2$ denotes the canonical metric of the unit sphere $S^{n-1}$ in $\Bbb R^n$. Inserting the metric (\ref{1})--(\ref{2}) into the Einstein equations \begin{equation} G_{\mu\nu}+\Lambda g_{\mu\nu}=8\pi G T_{\mu\nu}, \end{equation} where $G_{\mu\nu}$ is the Einstein tensor, $G$ the universal gravitational constant, and $\Lambda$ the cosmological constant, the speed of light is set to unity, and $T_{\mu\nu}$ is the energy-momentum tensor of an ideal cosmological fluid given by \begin{equation} \label{4} T^{\mu\nu}=\mbox{\footnotesize diag}\{\rho_m,p_m,\dots,p_m\}, \end{equation} with $\rho_m$ and $p_m$ the $t$-dependent matter energy density and pressure, we arrive at the Friedmann equations \begin{align} H^2=&\frac{16\pi G}{n(n-1)}\rho-\frac k{a^2},\label{5}\\ \dot{H}=&-\frac{8\pi G}{n-1}(\rho+p)+\frac k{a^2},\label{6} \end{align} in which $\rho,p$ are the effective energy density and pressure related to $\rho_m,p_m$ through \begin{equation} \label{8} \rho=\rho_m+\frac{\Lambda}{8\pi G},\quad p=p_m-\frac{\Lambda}{8\pi G}. \end{equation} On the other hand, recall that, with (\ref{1}) and (\ref{4}) and (\ref{8}), the energy-conservation law, $\nabla_\nu T^{\mu\nu}=0$, takes the form \begin{equation} \label{9} \dot{\rho}_m+n(\rho_m+p_m)H=0. \end{equation}

Inserting (\ref{10}) into (\ref{9}), we have \begin{equation}\label{11} \dot{\rho}_m+n(1+w)\rho_m \frac{\dot{a}}a=0, \end{equation} which can be integrated to yield \begin{equation}\label{12} \rho_m=\rho_0 a^{-n(1+w)}, \end{equation} where $\rho_0>0$ is an integration constant. Using (\ref{12}) in (\ref{8}), we arrive at the relation \begin{equation}\label{13} \rho=\rho_0 a^{-n(1+w)}+\frac{\Lambda}{8\pi G}. \end{equation} From (\ref{5}) and (\ref{13}), we get the following equation of motion for the scale factor $a$: \begin{equation} \label{14} \dot{a}^2=\frac{16\pi G\rho_0}{n(n-1)}a^{-n(1+w)+2}+\frac{2\Lambda}{n(n-1)} a^2-k. \end{equation} To integrate (\ref{14}), we recall Chebyshev's theorem: For rational numbers $p,q,r$ ($r\neq0$) and nonzero real numbers $\alpha,\beta$, the integral \[\int x^p(\alpha+\beta x^r)^q\,dx\] is elementary if and only if at least one of the quantities \begin{equation}\label{cd} \frac{p+1}r,\quad q,\quad \frac{p+1}r+q, \end{equation} is an integer. Another way to see the validity of the Chebyshev theorem is to represent the integral of concern by a hypergeometric function such that when a quantity in (\ref{cd}) is an integer the hypergeometric function is reduced into an elementary function. Consequently, when $k=0$ or $\Lambda=0$, and $w$ is rational, the Chebyshev theorem enables us to know that, for exactly what values of $n$ and $w$, the equation (\ref{14}) may be integrated. For the spatially flat situation $k=0$ we rewrite equation (\ref{14}) as \begin{equation} \label{15} \dot{a}=\pm\sqrt{c_0 a^{-n(1+w)+2}+\Lambda_0 a^2},\quad c_0=\frac{16\pi G\rho_0}{n(n-1)},\quad\Lambda_0=\frac{2\Lambda}{n(n-1)}. \end{equation} Then (\ref{15}) reads \begin{equation} \label{15a} \pm\int a^{-1}\left(c_0 a^{-n(1+w)}+\Lambda_0 \right)^{-\frac12}da=t+C. \end{equation} It is clear that the integral on the left-hand side of (\ref{15a}) satisfies the integrability condition stated in the Chebyshev theorem for any $n$ and any rational $w$. It turns out that (\ref{15}) might be integrated for any real $w$ as well, not necessarily rational. To do so we apply $a>0$ and get from (\ref{15}) the equation \begin{equation} \label{16} \frac{d}{dt}\ln a=\pm\sqrt{c_0 a^{-n(1+w)}+\Lambda_0}, \end{equation} or equivalently, \begin{equation}\label{17} \dot{u}=\pm\sqrt{c_0 e^{-n(1+w)u}+\Lambda_0},\quad u=\ln a. \end{equation} Set \begin{equation}\label{18} \sqrt{c_0 e^{-n(1+w)u}+\Lambda_0}=v. \end{equation} Then \begin{equation}\label{19} u=\frac{\ln c_0}{n(1+w)}-\frac 1{n(1+w)} \ln(v^2-\Lambda_0). \end{equation} Inserting (\ref{19}) into (\ref{17}), we find \begin{equation}\label{20} \dot{v}=\mp\frac12 n(1+w)(v^2-\Lambda_0), \end{equation} whose integration gives rise to the expressions \begin{equation} \label{21} v(t)=\left\{\begin{array}{rl} &v_0\left(1\pm \frac12 n(1+w)v_0t\right)^{-1},\quad \Lambda_0=0;\\ &\\ & \sqrt{\Lambda_0}(1+C_0 e^{\mp n(1+w)\sqrt{\Lambda_0} t})(1-C_0 e^{\mp n(1+w)\sqrt{\Lambda_0} t})^{-1},\\& C_0=(v_0-\sqrt{\Lambda_0})(v_0+\sqrt{\Lambda_0})^{-1},\quad \Lambda_0>0;\\ &\\ &\sqrt{-\Lambda_0}\tan\left(\mp\frac12 n(1+w)\sqrt{-\Lambda_0} t +\arctan\frac{v_0}{\sqrt{-\Lambda_0}}\right),\quad \Lambda_0<0, \end{array} \right. \end{equation} where $v_0=v(0)$. Hence, in terms of $v$, we obtain the time-dependence of the scale factor $a$: \begin{equation} \label{22} a^{n(1+w)}(t)=\frac{8\pi G \rho_0}{\frac12 n(n-1)v^2(t)-\Lambda}. \end{equation} We now assume \(w>-1\) in the equation of state in our subsequent discussion. We are interested in solutions satisfying \(a(0)=0.\) When $\Lambda=0$, we combine (\ref{21}) and (\ref{22}) to get \begin{equation} \label{x1} a^{n(1+w)}(t)=4\pi G\rho_0\left(\frac n{n-1}\right)(1+w)^2 t^2. \end{equation} When $\Lambda>0$, we similarly obtain \begin{equation}\label{x2} a^{n(1+w)}(t)=\frac{8\pi G\rho_0}{\Lambda}\sinh^2\left(\sqrt{\frac{n\Lambda}{2(n-1)}}(1+w) t\right). \end{equation} When $\Lambda<0$ we rewrite (\ref{22}) as \begin{equation} \label{23} a^{n(1+w)}(t)=\frac{8\pi G \rho_0}{(-\Lambda)}\cos^2\left(\sqrt{\frac{n(-\Lambda)}{2(n-1)} }(1+w)t \mp\arctan\sqrt{\frac{n(n-1)}{-2\Lambda}}\, v_0\right). \end{equation} If we require $a(0)=0$, then (\ref{23}) leads to the conclusion \begin{equation} \label{24} a^{n(1+w)}(t)=\frac{8\pi G \rho_0}{(-\Lambda)}\sin^2\sqrt{\frac{n(-\Lambda)}{2(n-1)} }(1+w)t, \end{equation} which gives rise to a periodic Universe so that the scale factor $a$ reaches its maximum $a_m$, \begin{equation} \label{25} a^{n(1+w)}_m=\frac{8\pi G \rho_0}{(-\Lambda)}, \end{equation} at the times \begin{equation}\label{26} t=t_{m,k}=\left(\frac\pi2+k\pi\right)\frac1{(1+w)}\sqrt{\frac{2(n-1)}{n(-\Lambda)}},\quad k\in\Bbb Z, \end{equation} and shrinks to zero at the times \begin{equation}\label{27} t=t_{0,k}=\frac{k\pi}{(1+w)}\sqrt{\frac{2(n-1)}{n(-\Lambda)}},\quad k\in\Bbb Z. \end{equation} Using the equations (\ref{x1},\ref{x2},\ref{24}), one easily obtains the expressions for the DP \begin{equation} q(t)=\left\{\begin{array}{cc} \frac n2 (1+w)-1, & \Lambda_0=0;\\ &\\ \frac{n(1+w)}{2\cosh^2\left(\sqrt{\frac{n\Lambda}{2(n-1)}}(1+w)t\right)}-1, & \Lambda_0>0;\\ &\\ \frac{n(1+w)}{2\cos^2\left(\sqrt{\frac{n\Lambda}{2(n-1)}}(1+w)t\right)}-1, & \Lambda_0<0. \end{array} \right. \end{equation}

\[H=\frac{a'}{a^2},\quad q=-\left(\frac{aa''}{a'^2}-1\right),\] where prime denotes derivative w.r.t. the conformal time.

\begin{align} \nonumber a(t)=&a_0\left[1+H_0(t-t_0)-\frac12q_0H_0^2(t-t_0)^2+ \frac1{3!}j_0H_0^3(t-t_0)^3\right.\\ \label{background_4_3} &\left.+\frac1{4!}s_0H_0^4(t-t_0)^4+\frac1{5!}l_0H_0^5(t-t_0)^5 + O\left((t-t_0)^6\right)\right]. \end{align}

\begin{align} \nonumber 1+z=&\left[1+H_0(t-t_0)-\frac12q_0H_0^2(t-t_0)^2+ \frac1{3!}j_0H_0^3(t-t_0)^3\right.\\ \nonumber &\left.+\frac1{4!}s_0H_0^4(t-t_0)^4+\frac1{5!}l_0H_0^5(t-t_0)^5 + O\left((t-t_0)^6\right)\right]^{-1}.\\ \nonumber z=&H_0(t-t_0)+\left(1+\frac{q_0}2\right)H_0^2(t-t_0)^2+\cdots. \end{align}

\[q(t)=\frac{d}{dt}\left(\frac1H\right)-1=-\frac{\dot H}{H^2}-1=-\frac{\frac{\ddot a}{a}-H^2}{H^2}-1=q.\]

\begin{align} \frac{dH}{dz}=&\frac{1+q}{1+z}H;\\ \nonumber \frac{d^2H}{dz^2}=&\frac{j-q^2}{(1+z)^2}H;\\ \nonumber \frac{d^3H}{dz^3}=&\frac{H}{(1+z)^3}\left(3q^2+3q^3-4qj-3j-s\right);\\ \nonumber \frac{d^4H}{dz^4}=&\frac{H}{(1+z)^4}\left(-12q^2-24q^3-15q^4+32qj+25q^2j+7qs+12j-4j^2+8s+l\right). \end{align}

\[H(z)=H_0+\left.\frac{dH}{dz}\right|_{z=0}z +\frac12\left.\frac{d^2H}{dz^2}\right|_{z=0}z^2 +\frac16\left.\frac{d^3H}{dz^3}\right|_{z=0}z^3+\dots=\] \[=H_0(1+(1+q_0)z+\frac12(j_0-q_0^2)z^2+\frac16(3q_0^2+3q_0^3-4q_0j_0-3j_0-s)z^3).\] There is a decomposition for the inverse Hubble parameter \begin{align} \nonumber \frac{d}{dz}\left(\frac1H\right)=&-\frac1{H^2}\frac{dH}{dz}=-\frac{1+q}{1+z}\frac1H;\\ \nonumber \frac{d^2}{dz^2}\left(\frac1H\right)=& 2\left(\frac{1+q}{1+z}\right)^2\frac1H =\left(\frac{2+4q+3q^2-j}{(1+z)^2}\right)\frac1H ;\\ \nonumber \frac1{H(z)}=& \frac1{H_0}\left[1-(1+q_0)z+\frac{2+4q_0+3q_0^2-j_0}{6}z^2+\dots\right]. \end{align}

\begin{align} \nonumber \frac{dH^2}{dz}=&\frac{2H^2}{1+z}(1+q),\\ \nonumber \frac{d^2H^2}{dz^2}=&\frac{2H^2}{(1+z)^2}(1+2q+j),\\ \nonumber \frac{d^3H^2}{dz^3}=&\frac{2H^2}{(1+z)^3}(-qj-s),\\ \nonumber \frac{d^4H^2}{dz^4}=&\frac{2H^2}{(1+z)^4}(4qj+3qs+3q^2j-j^2+4s+l). \end{align}

\begin{align} \nonumber q_0=& \left.-\frac1{H^2}\left\{\frac12\frac{d\left(H^2\right)}{dN}+H^2\right\}\right|_{N=0},\\ \nonumber j_0=&\left.\left\{\frac1{2H^2}\frac{d^2\left(H^2\right)}{dN^2} +\frac3{2H^2}\frac{d\left(H^2\right)}{dN}+1\right\}\right|_{N=0}. \end{align}

\begin{align} \dot H=& -H^2(1+q),\\ \nonumber \ddot H=& H^3(j+3q+2),\\ \nonumber \dddot H=& H^4[s-4j-3q(q+4)-6],\\ \nonumber \ddddot H=& H^5[l-5s+10(q+2)j+30(q+2)q+24]. \end{align}

\begin{align} \nonumber q(z)=&q_0 + \left(-q_0-2q_0^2+j_0\right)z+\frac12\left(2q_0+8q_0^2+8q_0^3-7q_0j_0-4j_0-s_0\right)z^2+O(z^3),\\ \nonumber q(y)=&q_0 +\left(-q_0-2q_0^2+j_0\right)y+\frac12\left(4q_0+8q_0^3-7q_0j_0-2j_0-s_0\right)y^2+O(y^3). \end{align}

Differentiate the relation \[q(z)=\frac{1+z}{H}\frac{d H}{dz}-1\] to find \[\frac{dq}{dz}=\frac1H\frac{dH}{dz}-\frac{1+z}{H^2}\left(\frac{dH}{dz}\right)^2+\frac{1+z}{H}\frac{d^2H}{dz^2}.\] Using \[\frac{dH}{dz}=\frac{1+q}{1+z}H,\quad \frac{d^2H}{dz^2}=\frac{j-q^2}{(1+z)^2}H,\] one obtains \[\frac{dq}{d\ln(1+z)}=j-q(2q+1).\]

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 \[k=a^2H^2(4q-j-1).\]

Consider an $(n+1)$-dimensional homogeneous and isotropic Lorentzian spacetime with the metric \begin{equation} \label{1} ds^2=g_{\mu\nu} dx^\mu dx^\nu=- dt^2+a^2(t)g_{ij} d x^i d x^j,\quad i,j=1,\dots,n, \end{equation} where $t$ is the cosmological (or cosmic) time and $g_{ij}$ is the metric of the $n$-dimensional Riemannian manifold $M$ of constant curvature characterized by an indicator, $k=-1,0,1$; $M$ is an $n$-hyperboloid, the flat space $\Bbb R^n$, or an $n$-sphere, with the respective metric \begin{equation} \label{2} g_{ij} d x^i d x^j=\frac1{1-kr^2}\, d r^2+r^2\, d\Omega^2_{n-1}, \end{equation} where $r>0$ is the radial variable and $d\Omega_{n-1}^2$ denotes the canonical metric of the unit sphere $S^{n-1}$. The Einstein equations: \begin{equation} \label{3} G_{\mu\nu}+\Lambda g_{\mu\nu}=8\pi G T_{\mu\nu}, \end{equation} where $G_{\mu\nu}$ is the Einstein tensor, $G$ the universal gravitational constant, and $\Lambda$ the cosmological constant, the speed of light is set to unity, and $T_{\mu\nu}$ is the energy-momentum tensor of an ideal cosmological fluid given by \begin{equation} \label{4} T^{\mu\nu}=\mbox{diag}\{\rho_m,p_m,\dots,p_m\}, \end{equation} with $\rho_m$ and $p_m$ the $t$-dependent matter energy density and pressure. Inserting the metric (\ref{1})--(\ref{2}) into (\ref{3}) we arrive at the Friedman equations \begin{equation}\label{5} H^2=\frac{16\pi G}{n(n-1)}\rho-\frac k{a^2}, \end{equation} \begin{equation} \dot{H}=-\frac{8\pi G}{n-1}(\rho+p)+\frac k{a^2},\label{6} \end{equation} in which \begin{equation} \label{7} H=\frac{\dot{a}}a, \end{equation} denotes the usual Hubble "constant", $\dot{f}=df/dt$, and $\rho,p$ are the effective energy density and pressure, related to $\rho_m,p_m$ through: \begin{equation} \label{8} \rho=\rho_m+\frac{\Lambda}{8\pi G},\quad p=p_m-\frac{\Lambda}{8\pi G}. \end{equation}

With (\ref{1}) and (\ref{4}) and (\ref{8}), the energy-conservation law, $\nabla_\nu T^{\mu\nu}=0$, takes the form \begin{equation} \label{9} \dot{\rho}_m+n(\rho_m+p_m)H=0. \end{equation} It is readily seen that (\ref{5}) and (\ref{9}) imply (\ref{6}). In other words, the full cosmological governing equations consist of (\ref{5}) and (\ref{9}) only.

Recall that the perfect-fluid cosmological model spells out a relation between the energy density $\rho_m$ and pressure $p_m$ of the matter source expressed by the so-called barotropic equation of state, \begin{equation} \label{10} p_m=w \rho_m, \end{equation} where $w$ is a constant so that $w=0$ leads to a vanishing pressure, $p_m=0$, corresponding to the dust model; $w=-1$ the vacuum model, and $w=1/n$ the radiation-dominated model. Inserting (\ref{10}) into (\ref{9}), we have \begin{equation}\label{11} \dot{\rho}_m+n(1+w)\rho_m \frac{\dot{a}}a=0, \end{equation} which can be integrated to yield \begin{equation}\label{12} \rho_m=\rho_0 a^{-n(1+w)}, \end{equation} where $\rho_0>0$ is an integration constant. Using (\ref{12}) in (\ref{8}), we arrive at the relation \begin{equation}\label{13} \rho=\rho_0 a^{-n(1+w)}+\frac{\Lambda}{8\pi G}. \end{equation}

From (\ref{5}) and (\ref{13}), we get the following equation of motion for the scale factor $a$: \begin{equation} \label{14} \dot{a}^2=\frac{16\pi G\rho_0}{n(n-1)}a^{-n(1+w)+2}+\frac{2\Lambda}{n(n-1)} a^2-k. \end{equation}

Rewrite equation (\ref{14}) as \begin{equation} \label{15} \dot{a}=\pm\sqrt{c_0 a^{-n(1+w)+2}+\Lambda_0 a^2},\quad c_0=\frac{16\pi G\rho_0}{n(n-1)},\quad\Lambda_0=\frac{2\Lambda}{n(n-1)}. \end{equation} Integration of (\ref{15}) gives \begin{equation} \label{15a} \pm\int a^{-1}\left(c_0 a^{-n(1+w)}+\Lambda_0 \right)^{-\frac12}d a=t+C. \end{equation} It is clear that the integral on the left-hand side of (\ref{15a}) satisfies the integrability condition stated in the Chebyshev theorem for any $n$ and any rational $w$. We have just seen that (\ref{15}) can be integrated directly for any rational $w$. Apply $a>0$ and get from (\ref{15}) the equation \begin{equation} \label{16} \frac{d}{d t}\ln a=\pm\sqrt{c_0 a^{-n(1+w)}+\Lambda_0}, \end{equation} or equivalently, \begin{equation}\label{17} \dot{u}=\pm\sqrt{c_0 e^{-n(1+w)u}+\Lambda_0},\quad u=\ln a. \end{equation} Set \begin{equation}\label{18} \sqrt{c_0 e^{-n(1+w)u}+\Lambda_0}=v. \end{equation} Then \begin{equation}\label{19} u=\frac{\ln c_0}{n(1+w)}-\frac 1{n(1+w)} \ln(v^2-\Lambda_0). \end{equation} Inserting (\ref{19}) into (\ref{17}), we find \begin{equation}\label{20} \dot{v}=\mp\frac12 n(1+w)(v^2-\Lambda_0), \end{equation} whose integration gives rise to the expressions \begin{equation} \label{21} v(t)=\left\{\begin{array}{cc} \frac{v_0}{1\pm \frac12 n(1+w)v_0t}, & \Lambda_0=0;\\ &\\ \sqrt{\Lambda_0}\frac{1+C_0 e^{\mp n(1+w)\sqrt{\Lambda_0} t}}{1-C_0 e^{\mp n(1+w)\sqrt{\Lambda_0} t}}, \quad C_0=\frac{v_0-\sqrt{\Lambda_0}}{v_0+\sqrt{\Lambda_0}}, & \Lambda_0>0;\\ &\\ \sqrt{-\Lambda_0}\tan\left(\mp\frac12 n(1+w)\sqrt{-\Lambda_0} t +\arctan\frac{v_0}{\sqrt{-\Lambda_0}}\right), & \Lambda_0<0, \end{array} \right. \end{equation} where $v_0=v(0)$. Hence, in terms of $v$, we obtain the time-dependence of the scale factor $a$: \begin{equation} \label{22} a^{n(1+w)}(t)=\frac{8\pi G \rho_0}{\frac12 n(n-1)v^2(t)-\Lambda}. \end{equation}

When $\Lambda=0$, we combine (\ref{21}) and (\ref{22}) to get \begin{equation} \label{x1} a^{n(1+w)}(t)=4\pi G\rho_0\left(\frac n{n-1}\right)(1+w)^2 t^2. \end{equation} When $\Lambda>0$, we similarly obtain \begin{equation} \label{x2} a^{n(1+w)}(t)=\frac{8\pi G\rho_0}{\Lambda}\sinh^2\left(\sqrt{\frac{n\Lambda}{2(n-1)}}(1+w) t\right). \end{equation} Both results (\ref{x1}) and (\ref{x2}) lead to an expanding Universe. We now consider the case when $\Lambda<0$ and rewrite (\ref{22}) as \begin{equation} \label{23} a^{n(1+w)}(t)=\frac{8\pi G \rho_0}{(-\Lambda)}\cos^2\left(\sqrt{\frac{n(-\Lambda)}{2(n-1)} }(1+w)t \mp\arctan\sqrt{\frac{n(n-1)}{-2\Lambda}}\, v_0\right). \end{equation} If we require $a(0)=0$, then (\ref{23}) leads to \begin{equation} \label{24} a^{n(1+w)}(t)=\frac{8\pi G \rho_0}{(-\Lambda)}\sin^2\sqrt{\frac{n(-\Lambda)}{2(n-1)} }(1+w)t, \end{equation} which gives rise to a periodic Universe so that the scale factor $a$ reaches its maximum $a_m$, \begin{equation} \label{25} a^{n(1+w)}_m=\frac{8\pi G \rho_0}{(-\Lambda)}, \end{equation} at the times \begin{equation}\label{26} t=t_{m,k}=\left(\frac\pi2+k\pi\right)\frac1{(1+w)}\sqrt{\frac{2(n-1)}{n(-\Lambda)}},\quad k\in\Bbb Z, \end{equation} and shrinks to zero at the times \begin{equation}\label{27} t=t_{0,k}=\frac{k\pi}{(1+w)}\sqrt{\frac{2(n-1)}{n(-\Lambda)}},\quad k\in\Bbb Z. \end{equation}

\begin{equation} %\label{21} q(t)=\left\{\begin{array}{cc} \frac n2 (1+w)-1, & \Lambda_0=0;\\ &\\ \frac{n(1+w)}{2\cosh^2\left(\sqrt{\frac{n\Lambda}{2(n-1)}}(1+w)t\right)}-1, & \Lambda_0>0;\\ &\\ \frac{n(1+w)}{2\cos^2\left(\sqrt{\frac{n\Lambda}{2(n-1)}}(1+w)t\right)}-1, & \Lambda_0<0. \end{array} \right. \end{equation}

The equation (\ref{14}) now reads \begin{equation} \label{32} \dot{a}^2=\frac{16\pi G\rho_0}{n(n-1)}a^{-n(1+w)+2}-k. \end{equation} In order to apply Chebyshev's theorem, we now assume that $w$ is rational. Thus we see that the question whether (\ref{32}) may be integrated in cosmological time is equivalent to whether \begin{eqnarray} I&=&\int a^{\frac12 n(1+w)-1}\left(-k a^{n(1+w)-2}+\sigma\right)^{-\frac12}\, d a\nonumber\\ &=&\frac2{n(1+w)}\int \left(-k u^{\gamma}+\sigma\right)^{-\frac12}\, d u,\quad u=a^{\frac12 n(1+w)},\quad\sigma=\frac{16\pi G\rho_0}{n(n-1)},\label{33} \end{eqnarray} is an elementary function of $u$, where \begin{equation} \label{34} \gamma=2\left(1-\frac2{n(1+w)}\right). \end{equation} By (\ref{34}), we see that (\ref{33}) is not elementary unless $1/\gamma$ or $(2-\gamma)/(2\gamma)$ is an integer. The case when $\gamma=0$ or $w=(2-n)/n$ is trivial since it renders $a(t)$ a linear function through (\ref{32}). That is, (\ref{32}) may only be integrated directly in cosmological time when $w$ satisfies one of the following: \begin{eqnarray} w&=&\frac{4N}{n(2N-1)} -1,\quad N=0,\pm1,\pm2,\dots;\\ w&=&\frac2n+\frac1{nN}-1,\quad N=\pm1,\pm2,\dots. \end{eqnarray} In particular, in the special situations when $n=3$, we have \begin{equation} w=-1,\dots,-\frac23,-\frac59,-\frac12,-\frac7{15},-\frac49,-\frac37,\dots,-\frac29,-\frac15,-\frac16,-\frac19,0,\frac13, \end{equation} so that $-1$ and $-1/3$ are the only limiting points.

The equation (\ref{33}) now becomes \begin{equation} I=\frac32\int(-k u^{-1}+\sigma)^{-\frac12}\,d u,\quad u=a^{\frac23}. \end{equation} When $k=1$ (closed Universe), we may use the substitutions $U=\sqrt{\sigma u-1}$ to carry out the integration, which gives us the explicit solution \begin{eqnarray} \label{41} &&a^{\frac13}\sqrt{\sigma a^{\frac23}-1}-a_0^{\frac13}\sqrt{\sigma a_0^{\frac23}-1} +\frac1{\sqrt{\sigma}}\ln\left(\frac{\sqrt{\sigma a^{\frac23}-1}+\sqrt{\sigma} a^{\frac13}}{\sqrt{\sigma a_0^{\frac23}-1}+\sqrt{\sigma} a_0^{\frac13}}\right)=\frac23\sigma t,\\ &&\quad t\geq0, \quad a_0=a(0),\nonumber \end{eqnarray} where $a_0$ satisfies the consistency condition $\sigma a_0^{\frac23}\geq1$ or \begin{equation} a_0\geq\left(\frac3{8\pi G \rho_0}\right)^{\frac32}, \end{equation} which spells out minimum size of the Universe in terms of $\rho_0$ whose initial energy density in view of (\ref{12}) is given by \begin{equation} \rho_m(0)=\frac{64}9 \pi^2 G^2\rho^3_0. \end{equation} When $k=-1$ (open Universe), we may likewise use the substitutions $U=\sqrt{\sigma u+1}$ to obtain the solution \begin{eqnarray} \label{44} &&a^{\frac13}\sqrt{\sigma a^{\frac23}+1}-a_0^{\frac13}\sqrt{\sigma a_0^{\frac23}+1} -\frac1{\sqrt{\sigma}}\ln\left(\frac{\sqrt{\sigma a^{\frac23}+1}+\sqrt{\sigma} a^{\frac13}}{\sqrt{\sigma a_0^{\frac23}+1}+\sqrt{\sigma} a_0^{\frac13}}\right)=\frac23\sigma t,\\ &&\quad t\geq0, \quad a_0=a(0),\nonumber \end{eqnarray} where no restriction is imposed on the initial value of the scale factor $a=a(t)$. In particular, if we adopt the big bang scenario, we can set $a_0=0$ to write down the special solution \begin{equation} a^{\frac13}\sqrt{\sigma a^{\frac23}+1} -\frac1{\sqrt{\sigma}}\ln\left({\sqrt{\sigma a^{\frac23}+1}+\sqrt{\sigma} a^{\frac13}}\right)=\frac23\sigma t,\quad t\geq0. \end{equation} The solutions (\ref{41}) and (\ref{44}) may collectively and explicitly be recast in the form of an elegant single formula: \begin{eqnarray} \label{44a} &&a^{\frac13}\sqrt{\sigma a^{\frac23}-k}-a_0^{\frac13}\sqrt{\sigma a_0^{\frac23}-k} +\frac{k}{\sqrt{\sigma}}\ln\left(\frac{\sqrt{\sigma a^{\frac23}-k}+\sqrt{\sigma} a^{\frac13}}{\sqrt{\sigma a_0^{\frac23}-k}+\sqrt{\sigma} a_0^{\frac13}}\right)=\frac23\sigma t,\\ &&\quad t\geq0, \quad a_0=a(0),\quad k=\pm1.\nonumber \end{eqnarray} We see that, in both closed and open situations, $k=\pm1$, respectively, the Universe grows following a power law of the type $a(t)=\mbox{O}(t^{\frac32})$ for all large time so that a greater Newton's constant or initial energy density gives rise to a greater growth rate.

\begin{equation} \label{48} ({a}')^2=\frac{16\pi G\rho_0}{n(n-1)}a^{-n(1+w)+4}+\frac{2\Lambda}{n(n-1)} a^4-ka^2. \end{equation}

When $k=0$, the conformal time version of (\ref{15}) reads \begin{equation} \label{49} a'=\pm a^2\sqrt{c_0 a^{-n(1+w)}+\Lambda_0 }, \end{equation} whose integration is \begin{equation} \label{50} \pm\int a^{-2}\left(c_0 a^{-n(1+w)}+\Lambda_0 \right)^{-\frac12}d a=\eta+C. \end{equation} Consequently Chebyshev's theorem indicates that, when $\Lambda\neq0$, the left-hand side of (\ref{50}) is elementary if and only if $\frac1{n(1+w)}$ or $\frac1{n(1+w)}-\frac12$ is an integer (again we exclude the trivial case $w=-1$), or more explicitly, $w$ satisfies one of the following conditions: \begin{eqnarray} w&=&-1+\frac1{nN},\quad N=\pm1,\pm2,\dots;\label{x4}\\ w&=&-1+\frac1{n\left(N+\frac12\right)},\quad N=0,\pm1,\pm2,\dots. \end{eqnarray}

The equation (\ref{49}) now becomes \begin{equation} a'=\pm a^2\sqrt{c_0 a^{-1}+\Lambda_0}, \end{equation} which can be integrated to yield the solution \begin{equation} \frac{c_0}a+\Lambda_0=\frac{c_0^2}4(\eta+C)^2, \end{equation} where $C$ is an integrating constant.

The Friedmann equation now becomes \begin{equation} \label{53} a'=\pm a\sqrt{c_0 a^{-n(1+w)+2}-k}, \end{equation} which in view of Chebyshev's theorem can be integrated in terms of elementary functions when $w$ is any rational number. In fact, as before, (\ref{53}) may actually be integrated to yield its exact solution expressed in elementary functions for any $w$: \begin{equation} \label{57} \pm\left(1-\frac12 n(1+w)\right)\eta+C=\left\{\begin{array}{rll} &-\frac1v,&\quad k=0;\\ &&\\ &\arctan v,&\quad k=1;\\ &&\\ &\frac12\ln\left|\frac{v-1}{v+1}\right|,&\quad k=-1,\end{array}\right. \end{equation} where $C$ is an integration constant and \begin{equation} v=\sqrt{c_0 a^{-n(1+w)+2}-k}\quad \mbox{ or }\quad a^{-n(1+w)+2}=\frac1{c_0} (v^2+k). \end{equation}

To obtain inflationary solutions, we assume \begin{equation} n(1+w)>2, \end{equation} which includes the dust and radiation matter situations since $n\geq3$. We assume the initial condition $a(0)=0$. When $k=0$, (\ref{57}) gives us the solution \begin{equation}\label{xx1} a^{n(1+w)-2}(\eta)=\frac{4\pi G\rho_0}{n(n-1)}\left(n(1+w)-2\right)^2\eta^2,\quad \eta\geq0. \end{equation} When $k=1$, (\ref{57}) renders the solution \begin{equation} a^{n(1+w)-2}(\eta)=\frac{16\pi G \rho_0}{n(n-1)}\sin^2\left(\frac12(n[1+w]-2)\eta\right),\quad \eta\geq0, \end{equation} which gives rise to a periodic Universe. When $k=-1$, (\ref{57}) yields \begin{equation} a^{n(1+w)-2}(\eta)=\frac{16\pi G \rho_0}{n(n-1)}\sinh^2\left(\frac12(n[1+w]-2)\eta\right),\quad \eta\geq0, \end{equation} which leads to an inflationary Universe. It is interesting to notice that the closed Universe situation here ($k=1,\Lambda=0$), in conformal time, is comparable to the flat Universe with a negative cosmological constant ($k=0,\Lambda<0$), in cosmological time, and the open Universe situation here ($k=-1,\Lambda=0$), in conformal time, is comparable to the flat Universe with a positive cosmological constant ($k=0, \Lambda>0$), also in cosmological time.

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.

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.

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 \[T=T_{0} \left(1+z\right)^{2} \]

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.

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.

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

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

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

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-5--QINU NEW Problems in Cosmo warm-up Category = [[Play with Numbers after Sivaram|'''Play with Numbers after Sivaram''']] <div id="Siv_1"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 1''' <p style="color: #999;font-size: 11px">problem id: Siv_1</p> (after C.Sivaram, Dark Energy may link the numbers of Rees, arXiv: 0710.4993) Given $\Lambda$-dominated Universe, the requirement that for various large scale structures (held together by self gravity) to form a variety of length scales, their gravitational self energy density should at least match the ambient vacuum energy repulsion, as was shown to imply [16. C Sivaram, Astr. Spc. Sci, 219, 135; IJTP, 33, 2407, 1994, 17. C Sivaram, Mod. Phys. Lett., 34, 2463, 1999] a scale invariant mass-radius relationship to the form (for the various structures): \[\frac M{R^2}\approx\sqrt\Lambda\frac{c^2} G.\] This equation predicts a universality of $M/R^2$ for a large variety of structures. Check this statement for such structures as a galaxy, a globular cluster, a galaxy cluster. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">For a typical spiral galaxy $M_{gal}\approx10^{12} M_\odot$, $R\approx30kpc$, for globular clusters, $M\approx10^{6} M_\odot$, $R\approx100pc$, for galaxy clusters, $M_C\approx10^{16} M_\odot$, $R_C\approx3Mpc$, so for all these structures the equation holds.</p> </div> </div></div> <div id="Siv_2"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 2''' <p style="color: #999;font-size: 11px">problem id: Siv_2</p> (after C.Sivaram, Scaling Relations for self-Similar Structures and the Cosmological Constant, arXiv: 0801.1218) In recent papers [13. Sivaram, C.: 1993a, Mod. Phys. Lett. 8,321.; 14. Sivaram, C.: 1993b, Astrophys. Spc. Sci. 207, 317.; 15. Sivaram, C.: 1993c, Astron. Astrophys. 275, 37.; 16. Sivaram, C.: 1994a, Astrophysics. Spc. Sci., 215, 185.; 17. Sivaram, C.: 1994b. Astrphysics .Spc .Sci., 215,191.; 18. Sivaram, C.: 1994c. Int. J. Theor. Phys. 33, 2407.], it was pointed out that the surface gravities of a whole hierarchy of astronomical objects (i.e. globular clusters, galaxies, clusters, super clusters, GMC's etc.) are more or less given by a universal value $a_0\approx cH_0\approx 10^{-8} cm\ s^{-2}$ a o ƒ° cHo ƒ° 10-8 cms-2. Thus \[a=\frac{GM}{R^2}\approx a_0\] for all these objects, $M$ being their typical mass and $R$ their typical radius. Also interestingly enough it was also pointed out [4. Sivaram, C.: 1982, Astrophysics. Spc. Sci. 88,507.; 5. Sivaram, C.: 1982, Amer. J. Phy. 50, 279.; 6. Sivaram, C.: 1983, Amer. J. Phys. 51, 277.; 7. Sivaram, C.: 1983, Phys. Lett. 60B, 181.] that the gravitational self energy of a typical elementary particle (hadron) was shown to be \[E_G\approx\frac{Gm^3 c}{\hbar}\approx\hbar H_0\] implying the same surface gravity value for the particle \[a_h=\frac{GM}{r^2}\approx \frac{Gm^3 c}{\hbar}\times\frac c\hbar\approx cH_0\approx a_0.\] Calculate actual value of the ratio \[\frac M{R^2}\approx\sqrt\Lambda\frac{c^2} G\sim1\] for such examples as a galaxy, whole Universe, globular cluster, a GMC, a supercluster, nuclei, an electron, Solar system, planetary nebula. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\begin{tabular}{|c|c|} \hline $M/R^2$ ($g/cm^2$) & Object \\ \hline $10^{45}/(3\times10^{22})^2\sim1$ & Galaxy \\ $10^{56}/(10^{28})^2\sim1$ & Universe \\ $10^{38}/(10^{19})^2\sim1$ & globular cluster, GMC, etc. \\ $few\times10^{48}/(few\times10^{24})^2\sim1$ & supercluster \\ $10^{-23}/(10^{-12})^2\sim1$ & nuclei \\ $10^{-27}/(10^{-13})^2\sim1$ & electron \\ $10^{33}/(10^{16})^2\sim1$ & Solar system, planetary nebula\\ \hline \end{tabular}</p> </div> </div></div> ---- =UNIQ--h-6--QINU NEW Problems in Dark Energy Category = ==UNIQ--h-7--QINUSingle Scalar Cosmology== [[Single Scalar Cosmology|'''Single Scalar Cosmology''']] The discovery of the Higgs particle has confirmed that scalar fields play a fundamental role in subatomic physics. Therefore they must also have been present in the early Universe and played a part in its development. About scalar fields on present cosmological scales nothing is known, but in view of the observational evidence for accelerated expansion it is quite well possible that they take part in shaping our Universe now and in the future. In this section we consider the evolution of a flat, isotropic and homogeneous Universe in the presence of a single cosmic scalar field. Neglecting ordinary matter and radiation, the evolution of such a Universe is described by two degrees of freedom, the homogeneous scalar field $\varphi(t)$ and the scale factor of the Universe $a(t)$. The relevant evolution equations are the Friedmann and Klein-Gordon equations, reading (in the units in which $c = \hbar = 8 \pi G = 1$) \[ \frac{1}{2}\, \dot{\varphi}^2 + V = 3 H^2, \quad \ddot{\varphi} + 3 H \dot{\varphi} + V' = 0, \] where $V[\varphi]$ is the potential of the scalar fields, and $H = \dot{a}/a$ is the Hubble parameter. Furthermore, an overdot denotes a derivative w.r.t.\ time, whilst a prime denotes a derivative w.r.t.\ the scalar field $\varphi$. ---- New subsection in this section [[Single_Scalar_Cosmology#Exact_Solutions_for_the_Single_Scalar_Cosmology| '''Exact Solutions for the Single Scalar Cosmology''']] ---- <div id="SSC_0"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 1''' <p style="color: #999;font-size: 11px">problem id: SSC_0</p> Show that the Hubble parameter cannot increase with time in the single scalar cosmology. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Let the scalar field $\varphi(t)$ is a single-valued function of time, then it is possible to reparametrize the Hubble parameter in terms of $\varphi$: \[ H(t) = H[\varphi(t)]. \] Taking time derivatives in the Friedman equation \[ \frac{1}{2}\, \dot{\varphi}^2 + V = 3 H^2, \] one arrives at the results: \[ \dot{\varphi} ( \ddot{\varphi} + V' ) = 6 H \dot{H},\quad \dot{H} \equiv H' \dot{\varphi}. \] Taking into account the Klein Gordon equation \[ \ddot{\varphi} + 3 H \dot{\varphi} + V' = 0, \] it follows, that for $\dot{\varphi} \neq 0$ and $H \neq 0$ one gets \[ \dot{\varphi} = - 2 H', \quad \dot{H} = - \frac{1}{2}\, \dot{\varphi}^2 \leq 0. \] Thus the Hubble parameter is a semi-monotonically decreasing function of time.</p> </div> </div></div> <div id="SSC_00"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 2''' <p style="color: #999;font-size: 11px">problem id: SSC_00</p> Show that if the Universe is filled by a substance which satisfies the null energy condition then the Hubble parameter is a semi-monotonically decreasing function of time. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[\dot H=-\frac12(\rho+p).\] If $\rho+p\ge0$ (null energy condition), $H$ is a semi-monotonically decreasing function of time.</p> </div> </div></div> <div id="SSC_0_1"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 3''' <p style="color: #999;font-size: 11px">problem id: SSC_0_1</p> For single-field scalar models express the scalar field potential in terms of the Hubble parameter and its derivative 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;">For single-field models in which the scalar field is a single-valued function of time in some interval, it is possible to reparametrize the Hubble parameter in terms of $\varphi$: \[H(t)=H[\varphi(t)]\] Replacing the time derivatives $\dot{\varphi} = - 2 H'$ (see the problem \ref{SSC_0}) in the Friedmann equation we find \[V=3H^2-2H'^2.\] The latter expression can be used to reconstruct the potential if the evolution history $H[\varphi(t)]$ is known, or for given $V(\varphi)$ this is a first-order differential equation for $H[\varphi(t)]$.</p> </div> </div></div> <div id="SSC_1"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 4''' <p style="color: #999;font-size: 11px">problem id: SSC_1</p> Obtain first-order differential equation for the Hubble parameter $H$ as function of $\varphi$ and find its stationary points. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Replacing the time derivatives in the Friedmann equation using the results of the previous problem, one finds \[ 2 H^{\prime\, 2} - 3 H^2 + V(\varphi) = 0. \] There are two kinds of stationary points; a point where $\dot{\varphi} = H' = 0$ is an end point of the evolution if \[ \ddot{\varphi} = 4 H' H'' = 0, \] which happens if $H$ is finite. In contrast, if \[ \ddot{\varphi} = 4 H' H'' \neq 0, \] $H$ necessarily diverges in such a way as to make $\ddot{\varphi}$ finite: $H \propto 1/H'$.</p> </div> </div></div> <div id="SSC_2"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 5''' <p style="color: #999;font-size: 11px">problem id: SSC_2</p> Consider eternally oscillating scalar field of the form $\varphi(t) = \varphi_0 \cos \omega t$ and analyze stationary points in such a model. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">For such a scalar field to exist it is required that \[ H' = - \frac{1}{2}\, \dot{\varphi} = \frac{\omega \varphi_0}{2} \sin \omega t = \frac{\omega}{2} \sqrt{\varphi_0^2 - \varphi^2}. \] There are infinitely many stationary points \[ \omega t_n = n \pi, \quad \varphi(t_n) = (-1)^n \varphi_0, \] where $H' = 0$. Now \[ H'' = - \frac{1}{2} \frac{\omega \varphi}{\sqrt{\varphi_0^2 - \varphi^2}}, \] and therefore $H$ diverges at all stationary points $t_n$, but in such a way that \[ 4 H' H'' = - \omega^2 \varphi = \ddot{\varphi}. \] Then all stationary points in the considered model are turning points.</p> </div> </div></div> <div id="SSC_3"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 6''' <p style="color: #999;font-size: 11px">problem id: SSC_3</p> Obtain explicit solution for the Hubble parameter in the model considered in the previous problem. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\begin{align} H & = H_0 - \frac{1}{4} \omega \varphi_0^2 \arccos \left( \frac{\varphi}{\varphi_0} \right) +\frac{1}{4} \omega \varphi \sqrt{\varphi_0^2 - \varphi^2} \\ & = H_0 - \frac{1}{4} \omega^2 \varphi_0^2 t + \frac{1}{8} \omega \varphi_0^2 \sin 2 \omega t. \end{align}</p> </div> </div></div> <div id="SSC_4"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 7''' <p style="color: #999;font-size: 11px">problem id: SSC_4</p> Obtain explicit time dependence for the scale factor in the model of problem [[#SSC_2]]. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">The corresponding solution for the scale factor is \[ a(t) = a(0) \exp\left\{H_0 t - \frac{1}{8} \omega^2 \varphi_0^2 t^2 + \frac{1}{16} \left( 1 - \cos 2 \omega t \right)\right\}. \] which is a gaussian, slightly modulated by an oscillating function of time (see figure). [[File:Osc phi.png|center|thumb|400px|Scalefactor $a(t)$ for an eternally oscillating scalar field.]]</p> </div> </div></div> <div id="SSC_5"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 8''' <p style="color: #999;font-size: 11px">problem id: SSC_5</p> Reconstruct the scalar field potential $V(\varphi)$ needed to generate the model of problem [[#SSC_2]]. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">The potential giving rise to this behavior reads \begin{align} V & = 3 H^2 - 2 H^{\prime\, 2} \\ & = 3 \left( H_0 - \frac{1}{4}\, \omega \varphi_0^2 \arccos \left( \frac{\varphi}{\varphi_0} \right) + \frac{1}{4} \omega \varphi \sqrt{ \varphi_0^2 - \varphi^2} \right)^2 - \frac{\omega^2}{2} \left( \varphi_0^2 - \varphi^2 \right). \end{align} Observe, that this potential keeps track of the number of oscillations the scalar field has performed through the arccos-function, so ultimately $V$ increases indefinitely as a function of time, whilst the volume of a representative domain of space decreases rapidly.</p> </div> </div></div> <div id="SSC_6_00"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 9''' <p style="color: #999;font-size: 11px">problem id: SSC_6_00</p> Describe possible final states for the Universe governed by a single scalar field at large times. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">If $H$ becomes negative then the Universe inevitably collapses.If $H$ never becomes negative, it must tend to a vanishing or positive final minimum, which can be reached either in finite or infinite time. The universe then ends up in a Minkowski or in a de Sitter state. These conclusions are a consequence of the non-positivity of $\dot{H}$ (see problem [[#SSC_0]], which implies that a negative $H$ can never return to larger values at later times.</p> </div> </div></div> <div id="SSC_6_0"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 10''' <p style="color: #999;font-size: 11px">problem id: SSC_6_0</p> Formulate conditions for existence of end points of evolution in terms of the potential $V(\varphi)$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">In order to establish the existence of end points or asymptotic end points of evolution at non-negative values of $H$, we first consider the locus of all stationary points, defined by \[ \dot{\varphi} = - 2 H' = 0 \quad \Rightarrow \quad V = 3 H^2 \geq 0 . \] It follows that stationary points can occur only in the region of positive or vanishing potential. In particular this holds for end points, which therefore do not occur in a region of negative potential. Moreover, it is clear that a Minkowski final state occurs only at a stationary point where $V = 0$, whereas all stationary points with $V > 0$ correspond to de Sitter states. To correspond to an end point of the evolution, $H$ must be finite at these stationary points to guarantee that $\ddot{\varphi} = 0$ as well. From the results of the problem \ref{SSC_1} it follows that \[ V' = 2 H' (3 H - 2 H''), \] and therefore $V' = 0$ if $H' = 0$ and $H$ and $H$ are finite. As a result end points of the evolution necessarily occur at an extremum of $V$, but only if $V \geq 0$ there.</p> </div> </div></div> <div id="SSC_6_1"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 11''' <p style="color: #999;font-size: 11px">problem id: SSC_6_1</p> Consider a single scalar cosmology described by the quadratic potential \[ V = v_0 + \frac{m^2}{2}\, \varphi^2. \] Describe all possible stationary points and final states of the Universe in this model. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">We distinguish the cases $v_0 > 0$, $v_0 = 0$ and $v_0 < 0$. The stationary points are represented graphically by the curves in the $\varphi$-$H$-plane in figure. UNIQ--gallery-00000000-QINU '''Critical curves $H'[\varphi] = 0$ for quadratic potentials with $v_0 > 0$ (a)), $v_0 = 0$ (b)) and $v_0 < 0$ (c)). ''' <br /> For $v_0 > 0$ there exists a stationary point for any value of $\varphi$, but the potential has a unique minimum at $\varphi = 0$, which is the only stationary point where $V' = 0$, and therefore the only end point. Indeed, once this point is reached $H$ can not decrease anymore and we have final state of de Sitter type. For $v_0 = 0$ the critical curves become straight lines, crossing at the origin where $H = 0$ at $V = 0$. This is still a stationary point with $\ddot{\varphi} = 0$ representing a Minkoswki state, but as $V'$ is not defined there it is really to be interpreted as a limit of the previous case. There are no evolution curves flowing from the domain $H > 0$ to the domain $H < 0$. For $v_0 < 0$ there are no stationary points in the region $\varphi^2 < 2 |v_0| /m^2$, and the solutions can cross into the domain of negative $H$ there.</p> </div> </div></div> <div id="SSC_7"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 12''' <p style="color: #999;font-size: 11px">problem id: SSC_7</p> Obtain actual solutions for the model of previous problem using the power series expansion \[ H[\varphi] = h_0 + h_1 \varphi + h_2 \varphi^2 + h_3 \varphi^3 + ... \] Consider the cases of $v_0 > 0$ and $v_0 < 0$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Substitution into equation $2 H^{\prime\, 2} - 3 H^2 + V(\varphi) = 0 $ then leads to the equalities \[ 3 h_0^2 - 2 h_1^2 = v_0, \quad h_1(3h_0 - 4 h_2) = 0, \quad 4h_1(h_1 - 4 h_3) + \frac{8}{3}\, h_2 ( 3 h_0 - 4 h_2 ) = m^2, \quad ..., \] from which the solutions $H[\varphi]$ can be reconstructed (see figure below). [[File:Epsgt0v2.png|center|thumb|400px|Solutions $H[\varphi]$ for quadratic potentials (problem [[#SSC_6_1]]) with $v_0 > 0$.]] [[File:Epslt0v2.png|center|thumb|400px|Solutions $H[\varphi]$ for quadratic potentials (problem [[#SSC_6_1]]) with $v_0 < 0$.]] </p> </div> </div></div> <div id="SSC_8"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 13''' <p style="color: #999;font-size: 11px">problem id: SSC_8</p> Estimate main contribution to total expansion factor of the Universe. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Using the results of previous problem, one can define the number of $e$-folds in some interval of time: \[ N = \int_{t_1}^{t_2} dt H = - \int_{\varphi_1}^{\varphi_2} d\varphi\, \frac{H}{2H'} = - \frac{1}{2}\, \int_{\varphi_1}^{\varphi_2} d\varphi\, \frac{h_0 + h_1 \varphi + h_2 \varphi^2 + ...}{h_1 + 2 h_2 \varphi + 3 h_3 \varphi^2 + ...}. \] This number can get sizeable contributions only in regions where the slow-roll condition is satisfied: \[ \varepsilon = - \frac{\dot{H}}{H^2} = \frac{2H^{\prime\, 2}}{H^2} < 1 \quad \Rightarrow \quad 3H^2 - V < H^2. \] Thus we simultaneously have \[ V < 3 H^2 \quad \mbox{and} \quad V > 2H^2 \quad \Leftrightarrow \quad 0 \leq \frac{V}{3} < H^2 < \frac{V}{2}. \] In most cases this holds only for a relatively narrow range of field values.</p> </div> </div></div> <div id="SSC_9_0"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 14''' <p style="color: #999;font-size: 11px">problem id: SSC_9_0</p> Explain difference between end points and turning points of the scalar field evolution. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">In both cases $\dot{\varphi} = 0$, but at end points in addition $\ddot{\varphi} = 0$, which can happen only at extrema of the potential $V[\varphi]$. However, if the end point occurs at a relative maximum or saddle point of the potential, this end point will be classically unstable. Indeed, the field can remain there for an indefinite period of time, but any slight change in the initial conditions will cause it to move on to lower values of the Hubble parameter. Nevertheless, such a period of temporary slow roll of the field creates the right conditions for a period of inflation.</p> </div> </div></div> <div id="SSC_9"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 15''' <p style="color: #999;font-size: 11px">problem id: SSC_9</p> Show that the exponentially decaying scalar field \[ \varphi(t) = \varphi_0 e^{-\omega t} \] can give rise to unstable end points of the evolution. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">The Hubble parameter and potential giving rise to this solution can be constructed following the same procedure as for the eternally oscillating field (see problem [[#SSC_2]]), with the result \[ H = h + \frac{1}{4}\, \omega \varphi^2, \quad V[\varphi] = v_0 - \frac{\mu^2}{2}\, \varphi^2 + \frac{\lambda}{4}\, \varphi^4, \] where \[ v_0 = 3 h^2, \quad \mu^2 = \omega^2 - 3 \omega h, \quad \lambda = \frac{3 \omega^2}{4}. \] Thus we obtain a quartic potential; for $\mu^2 > 0$ it has minima in which reflection symmetry is spontaneously broken. The exponential solution ends asymptotically at the unstable maximum of the potential where $\dot{\varphi} = \ddot{\varphi} = 0$. As such it represents an end point of the evolution, but a minimal change in the initial conditions for the scalar field will turn the end point into a reflection point (if it starts at lower $H$), or it will overshoot the maximum (if it starts at higher $H$). Thus the end point is unstable, but the exponential decay will still provide a good approximation to first part of the evolution of the universe for solutions $H[\varphi]$ coming close to the maximum of the potential (see figure below). [[File:Quarticv2.png|center|thumb|400px|Critical curves of stationary points and solutions $H[\varphi]$ for a quartic potential with spontaneous symmetry breaking.]] </p> </div> </div></div> <div id="SSC_10"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 16''' <p style="color: #999;font-size: 11px">problem id: SSC_10</p> Analyze all possible final states in the model of previous problem. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">The exponential scalar field leads to a behavior of the scale factor given by \[ a(t) = a_0\, e^{ht + \frac{1}{8} \varphi^2_0\, (1 - e^{-2\omega t})}. \] Thus for $h > 0$ this epoch in the evolution of the Universe ends in an asymptotic de Sitter state with Hubble constant $h$. Afterwards, the scalar field will roll further down the potential; provided $3h \leq \omega \leq 6h$ it will oscillate around the minimum until it comes to rest in another de Sitter or a Minkowski state, again depending on the value of $h$. In particular, for $\omega \geq 3h$ the model has a final de Sitter or Minkowski state in which $\dot{\varphi} = 0$ and \[ \langle \varphi^2 \rangle = \frac{\mu^2}{\lambda} = \frac{4}{3} \left(1 - \frac{3h}{\omega} \right). \] In this final state the energy density is \[ \langle V \rangle = v_0 - \frac{\mu^4}{4\lambda} = \frac{\omega}{3} ( 6h - \omega ). \]</p> </div> </div></div> <div id="SSC_11"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 17''' <p style="color: #999;font-size: 11px">problem id: SSC_11</p> Express initial energy density of the model of problem [[#SSC_9]] in terms of the $e$-folding number $N$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">The energy density for the solution of problem [[#SSC_9]] is \[ \rho_s(t) = \frac{1}{2}\, \dot{\varphi}^2 + V = 3 H^2 = 3 \left( h + \frac{1}{4}\, \omega\, \varphi_0^2\, e^{-2 \omega t} \right)^2. \] Now the solution for the scale factor \[ a(t) = a_0\, e^{ht + \frac{1}{8} \varphi^2_0\, (1 - e^{-2\omega t})}. \] shows, that before reaching the first turning point at $\varphi = 0$ the scale factor increases by an additional number of $e$-folds given by \[ N = \frac{1}{8}\, \varphi_0^2. \] Therefore the initial energy density at $t = 0$ can be written as \[ \rho_s(0) = 3 ( h + 2N \omega )^2. \] If we take this initial energy density to equal the Planck density: $\rho_s(0) = 1$, this establishes a simple relation between $h$, $\omega$ and $N$.</p> </div> </div></div> <div id="SSC_12"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 18''' <p style="color: #999;font-size: 11px">problem id: SSC_12</p> Estimate mass of the particles corresponding to the exponential scalar field considered in problem [[#SSC_9]]. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Taking the final energy density $\langle V \rangle$ (see problem [[#SSC_10]]) equal to the observed energy density of the Universe today: \[ \langle V \rangle = 3 H_0^2 = 1.04 \times 10^{-120} \] in Planck units, being so close to zero, one can set to an extremely good approximation $\omega = 6h$, and \[ 3 h^2 ( 1 + 12 N )^2 = 1, \quad \mu^2 = 18 h^2, \quad \lambda = 27 h^2. \] The lower limit on $N$ for inflation as derived from the CMB observations is $N \geq 60$, which requires \begin{equation} h \leq 0.8 \times 10^{-3}.\label{h_estimate} \end{equation} Now expanding $\varphi$ around its vacuum expectation value \[ \varphi = \frac{\mu}{\sqrt{\lambda}} + \chi, \] the potential becomes \[ V = \frac{1}{2}\, m_{\chi}^2 \chi^2 + \frac{\alpha}{3}\, m_{\chi} \chi^3 + \frac{\lambda}{4}\, \chi^4, \] where \[ m_{\chi} = 6h, \quad \alpha = 9h , \quad \lambda = 27 h^2. \] According to the estimate (\ref{h_estimate}) the upper limits on these parameters are \[ m_{\chi} = 0.48 \times 10^{-2}, \quad \alpha = 0.72 \times 10^{-2}\approx1/137, \quad \lambda = 0.17 \times 10^{-4}. \] Converting to particle physics units, the upper limit on the mass is $m_{\chi} \leq 1.2 \times 10^{-16}$ GeV. This suggests that the inflaton could be associated with a GUT scalar of Brout-Englert-Higgs type.</p> </div> </div></div> <div id="SSC_13"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 19''' <p style="color: #999;font-size: 11px">problem id: SSC_13</p> Calculate the deceleration parameter for flat Universe filled with the scalar field in form of quintessence. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">There are several ways to obtain the result: <br /> '''1)''' \[q=-\frac{\ddot a }{aH^2};\quad \frac{\ddot a}{a}=-\frac16(\rho+3p);\quad H^2=\frac13\rho;\] \[q=\frac{\dot\varphi^2-V}{\frac{\dot\varphi^2}2+V};\] '''2)''' \[q=\frac12\Omega_{tot}+\frac32\sum\limits_iw_i\Omega_i.\] For a flat single-component Universe one obtains \[q=\frac12+\frac32w=\frac12+\frac32\frac{\dot\varphi^2-2V}{\dot\varphi^2+2V}=\frac{\dot\varphi^2-V}{\frac{\dot\varphi^2}2+V};\] '''3)''' \[q=\frac{d}{dt}\frac 1 H-1=-\frac{\dot H}{H^2}-1=\frac{3H^2-V}{H^2}-1=\frac{2H^2-V}{H^2}=\frac{\dot\varphi^2-V}{\frac{\dot\varphi^2}2+V}\] </p> </div> </div></div> <div id="SSC_14_"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 20''' <p style="color: #999;font-size: 11px">problem id: SSC_14_</p> When considering dynamics of scalar field $\varphi$ in flat Universe, let us define a function $f(\varphi)$ so that $\dot\varphi=\sqrt{f(\varphi)}$. Obtain the equation describing evolution of the function $f(\varphi)$. (T. Harko, F. Lobo and M. K. Mak, Arbitrary scalar field and quintessence cosmological models, arXiv: 1310.7167) <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">By substituting the Hubble function \[H^2=\frac13\left(\frac12\dot\varphi^2+V(\varphi)\right)\] into the Klein-Gordon equation we obtain \[\ddot\varphi+\sqrt3\sqrt{\frac12\dot\varphi^2+V(\varphi)}\dot\varphi + \frac{dV}{d\varphi}=0\] Introducing $\dot\varphi=\sqrt{f(\varphi)}$ and changing the independent variable from $t$ to $\varphi$, transform last equation into \[\frac12\frac{df(\varphi)}{d\varphi}+\sqrt3\sqrt{\frac12f(\varphi+V(\varphi)}\sqrt{f(\varphi)}+\frac{dV}{d\varphi}=0.\]</p> </div> </div></div> ---- ==UNIQ--h-8--QINUThe Power-Law Cosmology== [[The Power-Law Cosmology|'''The Power-Law Cosmology''']] <div id="PWL_1"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 1''' <p style="color: #999;font-size: 11px">problem id: PWL_1</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="PWL_2"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 2''' <p style="color: #999;font-size: 11px">problem id: PWL_2</p> Show that in the power-law cosmology the scale factor evolution $a\propto\eta^q$ in conformal time transforms into $a\propto t^p$ in physical (cosmic) time with \[p=\frac{q}{1+q}.\] </div> <div id="PWL_3"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 3''' <p style="color: #999;font-size: 11px">problem id: PWL_3</p> Show that if $a\propto\eta^q$ then the state parameter $w$ is related to the index $q$ by the following \[w=\frac{2-q}{3q}=const.\] <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[\bar H=\frac q t,\quad \bar H'=-\frac q{t^2}.\] Using \[\bar H'=-\frac{1+3w}{2}\bar H^2,\] one obtains \[w=\frac{2-q}{3q}.\]</p> </div> </div></div> ---- ==UNIQ--h-9--QINUHybrid Expansion Law== [[Hybrid Expansion Law|'''Hybrid Expansion Law''']] In problems [[#SSC_18]] - [[#SSC_19_0]] we follow the paper of Ozgur Akarsu, Suresh Kumar, R. Myrzakulov, M. Sami, and Lixin Xu4, Cosmology with hybrid expansion law: scalar field reconstruction of cosmic history and observational constraints (arXiv:1307.4911) to study expansion history of Universe, using the hybrid expansion law---a product of power-law and exponential type of functions \[a(t)=a_0\left(\frac{t}{t_0}\right)^\alpha\exp\left[\beta\left(\frac{t}{t_0}-1\right)\right],\] where $\alpha$ and $\beta$ are non-negative constants. Further $a_0$ and $t_0$ respectively denote the scale factor and age of the Universe today. <div id="SSC_18"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 1''' <p style="color: #999;font-size: 11px">problem id: SSC_18</p> Calculate Hubble parameter, deceleration parameter and jerk parameter for hybrid expansion law. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[H=\frac{\dot a}{a}=\frac\alpha t+\frac\beta{t_0},\] \[q=-\frac{\ddot a}{aH^2}=\frac{\alpha t_0^2}{(\beta t +\alpha t_0)^2}-1,\] \[j=\frac{\ddot a}{aH^3}=1+\frac{(2t_0-3\beta t-3\alpha t_0)\alpha t_0^2}{(\beta t+\alpha t_0)^3}.\]</p> </div> </div></div> <div id="SSC_18_2"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 2''' <p style="color: #999;font-size: 11px">problem id: SSC_18_2</p> For hybrid expansion law find $a, H, q$ and $j$ in the cases of very early Universe $(t\to0)$ and for the late times $(t\to\infty)$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[t\to0:\] \[a\to a_0\left(\frac{t}{t_0}\right)^\alpha,\quad H\to\frac\alpha t,\quad q\to-1+\frac1\alpha,\quad j\to 1-\frac3\alpha + \frac2{\alpha^2};\] \[t\to\infty:\] \[a\to a_0\exp\left[\beta\left(\frac{t}{t_0}-1\right)\right],\quad H\to\frac\beta{t_0},\quad q\to-1,\quad j\to1.\] </p> </div> </div></div> <div id="SSC_18_3"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 3''' <p style="color: #999;font-size: 11px">problem id: SSC_18_3</p> In general relativity, one can always introduce an effective source that gives rise to a given expansion law. Using the ansatz of hybrid expansion law obtain the EoS parameter of the effective fluid. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[\rho = 3H^2,\] \[p=-2\frac{\ddot a}a -H^2=-2\dot H-3H^2,\] \[w=\frac p\rho=-\frac23\frac{\dot H}{H^2}-1,\] \[H=\frac\alpha{t}+\frac\beta{t_0},\quad \dot H=-\frac\alpha{t^2},\] \[w=\frac23\frac\alpha{t^2}\left(\frac\alpha{t}+\frac\beta{t_0}\right)^{-2}-1.\]</p> </div> </div></div> <div id="SSC_19"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 4''' <p style="color: #999;font-size: 11px">problem id: SSC_19</p> We can always construct a scalar field Lagrangian which can mimic a given cosmic history. Consequently, we can consider the quintessence realization of the hybrid expansion law. Find time dependence for the the quintessence field $\varphi(t)$ and potential $V(t)$, realizing the hybrid expansion law. Obtain the dependence $V(\varphi)$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">The energy density and pressure of the quintessence minimally coupled to gravity can be given by \[\rho=\frac12\dot\varphi^2+V(\varphi),\quad p=\frac12\dot\varphi^2-V(\varphi),\] Using the hybrid expansion law and relation \[p+\rho=-2\dot H=\frac{2\alpha}{t^2}\] we find \[\varphi(t)=\sqrt{2\alpha}\ln(t)+\varphi_1\] \[V(t)=3\left(\frac\alpha{t}+\frac\beta{t_0}\right)^{2}-\frac\alpha{t^2},\] where $\varphi_1$ is the integration constant. The potential as a function of the scalar field $\varphi$ is then given by the following expression: \[V(\varphi)=3\beta^2e^{-\sqrt{\frac2\alpha}(\varphi_0-\varphi_1)}+\alpha(3\alpha-1)e^{-\sqrt{\frac2\alpha}(\varphi-\varphi_1)} + 6\alpha\beta e^{-\frac12\sqrt{\frac2\alpha}(\varphi+\varphi_0-2\varphi_1)}\] where $\varphi_0=\varphi_1+\sqrt{2\alpha}\ln(t_0)$.</p> </div> </div></div> <div id="SSC_19_1"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 5''' <p style="color: #999;font-size: 11px">problem id: SSC_19_1</p> Quintessence paradigm relies on the potential energy of scalar fields to drive the late time acceleration of the Universe. On the other hand, it is also possible to relate the late time acceleration of the Universe with the kinetic term of the scalar field by relaxing its canonical kinetic term. In particular this idea can be realized with the help of so-called tachyon fields, for which \[\rho=\frac{V(\varphi)}{\sqrt{1-\dot\varphi^2}},\quad p=-V(\varphi)\sqrt{1-\dot\varphi^2}.\] Find time dependence of the tachyon field $\varphi(t)$ and potential $V(t)$, realizing the hybrid expansion law. Construct the potential $V(\varphi)$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">For the tachyon field \[\frac p\rho=w=-1+\dot\varphi^2.\] Any realization of the hybrid expansion law gives \[\frac23\frac\alpha{t^2}\left(\frac\alpha{t}+\frac\beta{t_0}\right)^{-2}-1.\] Consequently, \[\dot\varphi= \sqrt{\frac{2\alpha}{3}}\left(\alpha+\beta\frac{t}{t_0}\right)^{-1}\] Integration of the latter results in the following \[\varphi(t)=\sqrt{\frac{2\alpha t_0^2}{3\beta}}\ln{(\beta t+\alpha t_0)}+\varphi_2\] and \[V(t)=3\left(\frac\alpha{t}+\frac\beta{t_0}\right)^{2}\sqrt{1-\frac{2\alpha t_0^2}{3(\beta t+\alpha t_0)^2}}.\] where $\varphi_2$ is an integration constant. The corresponding tachyon potential is given by \[V(\varphi)=\frac{3 \beta ^2}{t_{0}^2}e^{\sqrt{\frac{6\beta^2}{\alpha t_{0}^2}}(\varphi-\varphi_{2})}\sqrt{1-\frac{2}{3}\alpha t_{0}^2 e^{\sqrt{\frac{6\beta^2}{\alpha t_{0}^2}}(\varphi-\varphi_{2})}}\left(\alpha t_{0}-e^{\frac{1}{2}\sqrt{\frac{6\beta^2}{\alpha t_{0}^2}}(\varphi-\varphi_{2})}\right)^{-2}.\] </p> </div> </div></div> <div id="SSC_19_2"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 6''' <p style="color: #999;font-size: 11px">problem id: SSC_19_2</p> Calculate Hubble parameter and deceleration parameter for the case of phantom field in which the energy density and pressure are respectively given by \[\rho =-\frac{1}{2}\dot{\varphi}^2+V(\varphi),\quad p =-\frac{1}{2}\dot{\varphi}^2-V(\varphi).\] <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">In case of the phantom scenario, the hybrid expansion law ansatz must be slightly modified in order to acquire self consistency. In particular, we rescale time as $t\rightarrow t_{s}-t$, where $t_{s}$ is a sufficiently positive reference time. Thus, the hybrid expansion law ansatz becomes \[a(t)=a_{0}\left(\frac{t_{s}-t}{t_{s}-t_{0}}\right)^{\alpha}e^{\beta \left(\frac{t_{s}-t}{t_{s}-t_{0}}-1\right)},\] where $\alpha<0$. Then \[ H=-\frac{\alpha}{t_{s}-t}-\frac{\beta}{t_{s}-t_{0}}, \] \[ \dot{H}=-\frac{\alpha}{(t_{s}-t)^2}, \] \[ q=\frac{\alpha (t_{s}-t_{0})^2}{[\beta (t_{s}-t)+\alpha (t_{s}-t_{0})]^{2}}-1. \]</p> </div> </div></div> <div id="SSC_19_0"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 7''' <p style="color: #999;font-size: 11px">problem id: SSC_19_0</p> Solve the problem [[#SSC_19]] for the case of phantom field. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[\varphi(t)=\sqrt{-2\alpha }\ln(t_{s}-t)+\varphi_{3},\] \[V{(t)}=3\left(\frac{\alpha}{t_{s}-t}+\frac{\beta}{t_{s}-t_{0}}\right)^2-\frac{\alpha}{(t_{s}-t)^2},\] \[V(\varphi) = 3\beta^{2}e^{-\sqrt{-\frac{2}{\alpha }}(\varphi_{0}-\varphi_{3})}+\alpha(3\alpha-1)e^{-\sqrt{-\frac{2}{\alpha }}(\varphi-\varphi_{3})} +6\alpha\beta e^{\frac{1}{2}\sqrt{-\frac{2}{\alpha }}(\varphi+\varphi_{0}-2\varphi_{3})}, \] where $\varphi_{0}=\varphi_{3}+\sqrt{-2\alpha }\ln(t_{s}-t_{0})$. We observe that $\alpha<0$ leads to $q<0$ (acceleration) and \[\dot{H}=-\frac{\alpha}{(t_{s}-t)^2}>0\] (super acceleration)</p> </div> </div></div> <div id="SSC_19_12"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 8''' <p style="color: #999;font-size: 11px">problem id: SSC_19_12</p> Find EoS parameter for the case of phantom field. </div> ---- ==UNIQ--h-10--QINUBianchi I Model== [[Bianchi I Model|'''Bianchi I Model''']] (after Esra Russell, Can Battal Kılınç, Oktay K. Pashaev, Bianchi I Models: An Alternative Way To Model The Present-day Universe, arXiv:1312.3502) Theoretical arguments and indications from recent observational data support the existence of an anisotropic phase that approaches an isotropic one. Therefore, it makes sense to consider models of a Universe with an initially anisotropic background. The anisotropic and homogeneous Bianchi models may provide adequate description of anisotropic phase in history of Universe. One particular type of such models is Bianchi type I (BI) homogeneous models whose spatial sections are flat, but the expansion rates are direction dependent, \[ds^2={c^2}dt^2-a^{2}_{1}(t)dx^2-a^{2}_{2}(t)dy^2-a^{2}_{3}(t)dz^2\] where $a_{1}$, $a_{2}$ and $a_{3}$ represent three different scale factors which are a function of time $t$. <div id="bianchi_01"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 1''' <p style="color: #999;font-size: 11px">problem id: bianchi_01</p> Find the field equations of the BI Universe. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">If we admit the energy-momentum tensor of a perfect fluid, then the field equations of the BI universe are found as, \begin{eqnarray} \label{feforgm}\frac{{\dot{a}_{1}}{\dot{a}_{2}}}{a_{1} a_{2}}+\frac{{\dot{a}_{1}}{\dot{a}_{3}}}{a_{1} a_{3}}+\frac{{\dot{a}_{2}}{\dot{a}_{3}}}{a_{2} a_{3}}&=&\rho,\\ \frac{{\ddot{a}_{1}}}{a_{1}}+\frac{{\ddot{a}_{3}}}{a_{3}}+\frac{{\dot{a}_{1}}{\dot{a}_{3}}}{a_{1} a_{3}}&=& -p,\\ \frac{{\ddot{a}_{2}}}{a_{2}}+\frac{{\ddot{a}_{1}}}{a_{1}}+\frac{{\dot{a}_{2}}{\dot{a}_{1}}}{a_{2} a_{1}}&=&-p,\\ \frac{{\ddot{a}_{3}}}{a_{3}}+\frac{{\ddot{a}_{2}}}{a_{2}}+\frac{{\dot{a}_{3}}{\dot{a}_{2}}}{a_{3} a_{2}}&=&-p. \end{eqnarray}</p> </div> </div></div> <div id="bi_2"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 2''' <p style="color: #999;font-size: 11px">problem id: bi_2</p> Reformulate the field equations of the BI Universe in terms of the directional Hubble parameters. \[H_1\equiv\frac{\dot{a_1}}{a_1},\ H_2\equiv\frac{\dot{a_2}}{a_2},\ H_3\equiv\frac{\dot{a_3}}{a_3}.\] <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Inserting the directional Hubble parameters and their time derivatives \[\dot H_1=\frac{\ddot a_1}{a_1}-\left(\frac{\dot a_1}{a_1}\right)^2,\ \dot H_2=\frac{\ddot a_2}{a_2}-\left(\frac{\dot a_2}{a_2}\right)^2,\ \dot H_3=\frac{\ddot a_3}{a_3}-\left(\frac{\dot a_3}{a_3}\right)^2\] into the modified Friedmann equations we obtain \begin{align} \nonumber H_1H_2+H_1H_3+H_2H_3 & =\rho,\\ \nonumber \dot H_1+ H_1^2 +\dot H_3+ H_3^2 +H_1H_3& =-p,\\ \nonumber \dot H_1+ H_1^2 +\dot H_2+ H_2^2 +H_1H_2& =-p,\\ \nonumber \dot H_2+ H_2^2 +\dot H_3+ H_3^2 +H_2H_3& =-p. \end{align}</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 3''' <p style="color: #999;font-size: 11px">problem id: </p> The BI Universe has a flat metric, which implies that its total density is equal to the critical density. Find the critical density. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[\rho_{cr}=H_1H_2+H_1H_3+H_2H_3.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 4''' <p style="color: #999;font-size: 11px">problem id: </p> Obtain an analogue of the conservation equation $\dot\rho+3H(\rho+p)=0$ for the case of the BI Universe. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">The energy conservation equation $T^\mu_{\nu;\mu}=0$ yields \[\dot\rho+3\bar H(\rho+p)=0,\quad \bar H\equiv \frac13(H_1+H_2+H_3)=\frac13\left(\frac{\dot a_1}{a_1} +\frac{\dot a_2}{a_2} +\frac{\dot a_3}{a_3}\right),\] where $\bar H$ represents the mean of the three directional Hubble parameters in the BI Universe.</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 5''' <p style="color: #999;font-size: 11px">problem id: </p> Obtain the evolution equation for the mean of the three directional Hubble parameters $\bar H$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Adding the three latter Friedmann equations (see Problem \ref{bi_2}) one obtains \begin{equation}\label{bi_5_1} 2\frac{d}{dt}\sum\limits_{i=1}^3 H_i+2(H_1^2+H_2^2+H_3^2)+H_1H_2+H_1H_3+H_2H_3=-3p. \end{equation} where $\bar H$ represents the mean of the three directional Hubble parameters in the BI Universe. Substituting \[\sum\limits_{i=1}^3 H_i^2=\left(\sum\limits_{i=1}^3 H_i\right)^2-2(H_1H_2+H_1H_3+H_2H_3)\] and \[H_1H_2+H_1H_3+H_2H_3=\rho\] into equation (\ref{bi_5_1}), we then obtain \[\frac{d}{dt}\sum\limits_{i=1}^3 H_i+\left(\sum\limits_{i=1}^3 H_i\right)^2=\frac32(\rho-p).\] Using the mean of the three directional Hubble parameters $\bar H$ we obtain a nonlinear first order differential equation \[\dot{\bar H}+3\bar H^2=\frac12(\rho-p).\]</p> </div> </div></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 the system of equations for the BI Universe \begin{align} \nonumber H_1H_2+H_1H_3+H_2H_3 & =\rho,\\ \nonumber \dot H_1+ H_1^2 +\dot H_3+ H_3^2 +H_1H_3& =-p,\\ \nonumber \dot H_1+ H_1^2 +\dot H_2+ H_2^2 +H_1H_2& =-p,\\ \nonumber \dot H_2+ H_2^2 +\dot H_3+ H_3^2 +H_2H_3& =-p, \end{align} can be transformed to the following \begin{align} \nonumber H_1H_2+H_1H_3+H_2H_3 & =\rho,\\ \nonumber \dot H_1+ 3H_1\bar H & =\frac12(\rho-p),\\ \nonumber \dot H_2+ 3H_2\bar H & =\frac12(\rho-p),\\ \nonumber \dot H_3+ 3H_3\bar H & =\frac12(\rho-p). \end{align} </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 the mean of the three directional Hubble parameters $\bar H$ is related to the elementary volume of the BI Universe $V\equiv a_1a_2a_3$ as \[\bar H=\frac13\frac{\dot V}{V}.\] <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[\bar H=\frac13\frac{d}{dt}\ln(a_1a_2a_3)=\frac13\frac{d}{dt}\ln V=\frac13\frac{\dot V}{V}.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 8''' <p style="color: #999;font-size: 11px">problem id: </p> Obtain the volume evolution equation of the BI model. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Using the relation between volume $V$ and the mean Hubble parameter $\bar H$, obtained in the previous problem, one finds \[\dot{\bar H}=\frac13\frac{\ddot V}{V}-3\bar H^2.\] As \[\dot{\bar H}+3\bar H^2=\frac12(\rho-p),\] we obtain \[\ddot V-\frac32(\rho-p)V=0.\]</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> Find the generic solution of the directional Hubble parameters. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">The equations \begin{align} \nonumber \dot H_1+ 3H_1\bar H & =\frac12(\rho-p),\\ \nonumber \dot H_2+ 3H_2\bar H & =\frac12(\rho-p),\\ \nonumber \dot H_3+ 3H_3\bar H & =\frac12(\rho-p), \end{align} allow us to write the generic solution of the directional Hubble parameters, \[H_i(t)=\frac1{\mu(t)}\left[K_i+\frac12\int\mu(t)(\rho(t)-p(t))dt\right],\quad i=1,2,3,\] where $K_i$s are the integration constants. The integration factor $\mu$ is defined as, \[\mu(t)=\exp\left(3\int\bar H(t)dt\right).\] As can be seen, the initial values (integration constants) determine the solution of each directional Hubble parameter. These values are the origin of the anisotropy. Note that the generic solution of the directional Hubble parameters is incomplete. To obtain exact solutions of the Hubble parameters and therefore the Einstein equations, one has to know the state equation for the component which fills the Universe.</p> </div> </div></div> ==UNIQ--h-11--QINURadiation dominated BI model == <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 10''' <p style="color: #999;font-size: 11px">problem id: </p> Find the energy density of the radiation dominated BI Universe in terms of volume element $V_r$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">By using the energy conservation equation \[\dot\rho+3\bar H(\rho+p)=0\to\dot\rho+4\bar H\rho=0,\] and the volume representation of the mean Hubble parameter \[\bar H=\frac13\frac{d}{dt}\ln(a_1a_2a_3)=\frac13\frac{d}{dt}\ln V=\frac13\frac{\dot V}{V}.\] we obtain (with $\rho\to\rho_r$, $V\to V_r$): \[\rho_r=\rho_{r0}\left(\frac{V_{r0}}{V_r}\right)^{4/3}.\] Here the density and the volume element is normalized to the present time $t_0$. The parameters $\rho_{r0}$ and $V_{r0}$ are the normalized density and normalized volume elements.</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> Find the mean Hubble parameter of the radiation dominated case. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">For the radiation dominated case \[\ddot V_r-\frac32(\rho-p)V_r=0\to\ddot V_r-V_r\rho_r=0.\] (see problem 8). Using \[\rho_r=\rho_{r0}\left(\frac{V_{r0}}{V_r}\right)^{4/3}\] we obtain (for $V_{r0}=1$) \[\ddot V_r-\rho_{r0}V_r^{-1/3}=0.\] Multiplying this equation with the $\dot V_r$ and integrating it, yields, \[\dot V_r^2-3\rho_{r0}V_r^{2/3}=0.\] Hence, the exact solution of the volume evolution equation is \[V_r=(2H_0t)^{3/2}.\] The mean Hubble parameter of the radiation dominated case is \[\bar H=\frac13\frac{\dot V_r}{V}=\frac1{2t}.\]</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> Find the directional expansion rates of the radiation dominated model. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">The generic solution of the directional Hubble parameters (see problem 9) is \[H_i(t)=\frac1{\mu(t)}\left[K_i+\frac12\int\mu(t)(\rho(t)-p(t))dt\right],\quad i=1,2,3,\] Using the expression for the mean Hubble parameter obtained in the previous problem, one finds \[\mu_r(t)=\exp(3\int\bar H(t)dt)\] By direct substitution of the integration factor $\mu_r$ and the equation of state $p_r=\rho_r/3$ of the radiation dominated case we obtain for the directional Hubble parameters that are normalized to the present-day time $t_0$ the following results \[H_{r,i}t_0=\alpha_{r,i}\left(\frac{t_0}{t}\right)^{3/2}+\frac12\frac{t_0}{t};\quad \alpha_{r,i}\equiv\frac{K_{r,i}}{t_0}.\]</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> Find time dependence for the scale factors $a_i$ in the radiation dominated BI Universe. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">The normalized scale factors $a_i$ can be obtained from the directional Hubble parameters \[H_{r,i}=\alpha_{r,i}\frac1{t_0}\left(\frac{t_0}{t}\right)^{3/2}+\frac12\frac{1}{t},\] with a direct integration in terms of cosmic time, \[a_{r,i}=\exp\left[-2\alpha_{r,i}\left(\sqrt{\frac{t_0}{t}}-1\right)\right]\left(\frac{t_0}{t}\right)^{1/2}.\] The scale factors of the BI radiation dominated model has the contribution from anisotropic expansion/contraction \[\exp\left[-2\alpha_{r,i}\left(\sqrt{\frac{t_0}{t}}-1\right)\right]\] as well as the standard matter dominated FLRW contribution $(t/t_0)^{1/2}$. These two different dynamical behaviors in three directional scale factors of the BI universe indicate that the FLRW part of the scale factor becomes dominant when time starts reaching the present-day. On the other hand, in the early times of the BI model, the expansion is completely dominated by the anisotropic part.</p> </div> </div></div> <div id="bianchi_02"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 14''' <p style="color: #999;font-size: 11px">problem id: bianchi_02</p> Find the partial energy densities for the two components of the BI Universe dominated by radiation and matter in terms of volume element $V_{rm}$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Equations of state for the considered components read: \begin{eqnarray}\label{eosrm} {p}_{m}=0,\phantom{a}{p}_{r}=\frac{1}{3}{\rho}_{r}. \end{eqnarray} \noindent As a result, the energy conservation equations in the radiation-matter period are \begin{eqnarray} \dot{\rho}_{r}=-4 \bar H_{rm} {\rho}_{r},\phantom{a} \dot{\rho}_{m}=-3 \bar H_{rm}{\rho}_{m}, \label{energyconsermatradzero} \end{eqnarray} \noindent Using the definition \[\bar H=\frac13\frac{\dot V}{V}\] one obtains \begin{eqnarray} {\rho}_{r}=\rho_{r,0}\left(\frac{V_{rm,0}}{V_{rm}}\right)^{4/3},\phantom{a} {\rho}_{m}=\rho_{m,0}\frac{V_{rm,0}}{V_{rm}}. \label{energyconsermatrad} \end{eqnarray}</p> </div> </div></div> <div id="bianchi_03"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 15''' <p style="color: #999;font-size: 11px">problem id: bianchi_03</p> Obtain time evolution equation for the total volume $V_{rm}$ in the BI Universe dominated by radiation and matter. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">In the considered case of radiation+matter dominated BI Universe \[\bar H +3H^2=\frac12(\rho-p) \to \dot{\bar H}_{rm}+3{\bar H}^2_{rm}=\frac12\left(\frac{2}{3}\rho_{r}+\rho_{m}\right).\] Substitution of \[\bar H=\frac13\frac{\dot V_{rm}}{V_{rm}}\] gives \[{\ddot{V}_{rm}}-\frac32\left(\rho_{m,0}+\frac{2}{3}\frac{\rho_{r,0}}{V_{rm}^{1/3}}\right)=0.\] Multiplying this equation with $\dot{V}_{rm}$, integrate it in terms of time, and substitute the normalized densities \[{\rho}_{r,0}=3\bar H^2_{0}\Omega_{r,0},\quad{\rho}_{m,0}=3\bar H^2_{0}\Omega_{m,0},\] we then obtain \[{{\dot V}_{rm}}^2-9\bar H^2_{0}\Omega_{m,0} V_{rm} - 9\bar H^2_{0}\Omega_{r,0} V_{rm}^{2/3}=0.\]</p> </div> </div></div> <div id="bianchi_04"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 16''' <p style="color: #999;font-size: 11px">problem id: bianchi_04</p> Using result of the previous problem, obtain a relation between the mean Hubble parameter and the volume element. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[\left(\frac{\bar H_{rm}}{\bar H_0}\right)^2=\frac{\Omega_{m,0}}{V_{rm}}+\frac{\Omega_{r,0}}{V_{rm}^{4/3}}.\]</p> </div> </div></div> ---- =UNIQ--h-12--QINU NEW problems in Dark Matter Category = ==UNIQ--h-13--QINUGeneralized models of unification of dark matter and dark energy== [[Generalized models of unification of dark matter and dark energy|'''Generalized models of unification of dark matter and dark energy''']] (see N. Caplar, H. Stefancic, Generalized models of unification of dark matter and dark energy (arXiv: 1208.0449)) <div id="gmudedm_1"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 1''' <p style="color: #999;font-size: 11px">problem id: gmudedm_1</p> The equation of state of a barotropic cosmic fluid can in general be written as an implicitly defined relation between the fluid pressure $p$ and its energy density $\rho$, \[F(\rho,p)=0.\] Find the speed of sound in such fluid. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">From $F(\rho,p)=0$ it follows that \[\frac{\partial F}{\partial\rho}d\rho + \frac{\partial F}{\partial p}dp=0\] which leads to \[c_s^2=\frac{dp}{d\rho}=-\frac{\frac{\partial F}{\partial\rho}}{\frac{\partial F}{\partial p}}.\]</p> </div> </div></div> <div id="gmudedm_2"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 2''' <p style="color: #999;font-size: 11px">problem id: gmudedm_2</p> For the barotropic fluid with a constant speed of sound $c_s^2=const$ find evolution of the parameter of EOS, density and pressure with the redshift. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Inserting $p=w\rho$ in \[\frac{\partial F}{\partial\rho}d\rho + \frac{\partial F}{\partial p}dp=0\] and using the definition of the speed of sound we obtain \[\frac{d\rho}{\rho}=\frac{dw}{c_s^2-w}.\] Combining this expression with the continuity equation for the fluid results in equation \[\frac{dw}{(c_s^2-w)(1+w)}=-3\frac{da}a=3\frac{dz}{1+z}.\] For $c_s^2=const$ parameter of EOS evolves as \[w=\frac{c_s^2\frac{1+w_0}{c_s^2-w_0}(1+z)^{3(1+c_s^2)}-1}{\frac{1+w_0}{c_s^2-w_0}(1+z)^{3(1+c_s^2)}+1}.\] From this relation it immediately follows that \[\rho=\rho_0\frac{c_s^2-w_0}{c_s^2-w}= \rho_0\frac{c_s^2-w_0}{1+c_s^2}\left[\frac{1+w_0}{c_s^2-w_0}(1+z)^{3(1+c_s^2)}+1\right]\] and \[p=c_s^2\rho-\rho_0(c_s^2-w_0)= \rho_0\frac{c_s^2-w_0}{1+c_s^2}\left[\frac{1+w_0}{c_s^2-w_0}(1+z)^{3(1+c_s^2)}-1\right].\] </p> </div> </div></div> ---- =UNIQ--h-14--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.