Deceleration Parameter

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

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

Using expression for $\dddot a$ obtained in the problem 40, one finds \[\ddddot a=\frac M3\frac{\ddot a}{a^3}-M\frac{\dot a^2}{a^4}+\frac\Lambda3\ddot a.\] For the snap parameter \[s\equiv a^3\frac{\ddddot a}{\dot a^4}\] one obtains \[s=-3(A+B)q-3A.\] Substitution of the parameters \begin{align} \nonumber A &=\frac23(j+q);\\ \nonumber B &=\frac23(\frac12j-q). \end{align} introduced in the problem 40, one finally finds \[s+2(q+j)+qj=0.\] This fourth order ODE is equivalent to the Friedmann equation and has an advantage that it appears as a constraint on directly measurable quantities.

For small values of $a(t)$ density and pressure of the Chaplygin gas reduces to that of dust $\rho\propto a^{-3}$, and for large $a$ one gets the de Sitter Universe: $\rho=const$, $p=-\rho$. In between these two regimes one can use the approximation \begin{equation}\label{scalar_5}\rho=\sqrt A+\frac{B}{\sqrt{2A}}a^{-6}.\end{equation} Thus $\sqrt A$ plays the role of a cosmological constant. We insert this to the Friedmann equation with $\Lambda$ and follow the procedure of eliminating the constants by differentiation. This leads to an approximate constraint \begin{equation}\label{scalar_6}s+5(q+j)+qj=0.\end{equation}

\begin{equation}\label{scalar_7}s+(3\alpha+2)(q+j)+qj=0.\end{equation} Note that for $\alpha=1$ we reproduce the above obtained result for the Chaplygin gas, while $\alpha=0$ we return to the results obtained in the problem (43). If we want to exclude the parameter $\alpha$ from the latter equation we must take one more derivative of the Friedmann equation and introduce an additional cosmological parameter \[l=\frac1a\frac{d^5a}{dt^5}\left(\frac1a\frac{da}{dt}\right)^{-5}.\] As a result one obtains \begin{equation}\label{scalar_8} -2qs-2jq^2-lq-2sj-3sq^2-j^2q-lj+s^2-3q^2j-qsj+j^3-2j^2q^2=0. \end{equation} This constraint is again approximate and is valid only in the regime where the higher order terms in the expansion of $\rho$ can be dropped.

Making use of the definition of the deceleration parameter \[q(t)=-\frac{\ddot a a}{\dot a^2}=\frac{d}{dt}\left(\frac1H\right)-1,\] it is easy to obtain \[\bar q(t_0)=-1+\frac1{t_0H_0}.\]

Using results of the previous problem one finds \begin{equation}\label{background_5_3}t_0=\frac{H_0^{-1}}{1+\bar q}.\end{equation} Naturally current age of the Universe is proportional to $H_0^{-1}$, but the proportionality coefficient is solely determined by the average value of the deceleration parameter. It is worth noting that this purely kinematic result depends on curvature of the Universe, nor on number of components filling it, nor on the type of gravity theory used.

As the average value of the deceleration parameter $\bar q$ is of order of unity then from (see result of the previous problem) \[t_0=\frac{H_0^{-1}}{1+\bar q}\] it immediately follows that the Hubble time $H_0)^{-1}$ represents a characteristic time scale at any stage of the evolution of the Universe.

\begin{align} \nonumber NEC&\Rightarrow & -\frac{\ddot a}{a}+\frac{\dot a^2}{a^2}+\frac{k}{a^2}\ge&0,\\ \nonumber WEC&\Rightarrow & \frac{\dot a^2}{a^2}+\frac{k}{a^2}\ge&0,\\ \nonumber SEC&\Rightarrow & \frac{\ddot a}{a}\le&0,\\ \nonumber DEC&\Rightarrow & \frac{\ddot a}{a}+2\left[\frac{\dot a^2}{a^2}+\frac{k}{a^2}\right]\ge&0. \end{align}

\begin{align} \label{background_6_5} NEC&\Rightarrow q\ge1,\\ \nonumber SEC&\Rightarrow q\ge0,\\ \nonumber DEC&\Rightarrow q\le2. \end{align} There is no WEC among above conditions because it is always satisfied for arbitrary real $a(t)$.

The conditions (\ref{background_6_5}) considered separately in principle allow a possibility for both decelerated ($q>0$) and accelerated ($q<0$) expansion of the Universe. The constraints by NEC in (\ref{background_6_5}) have clear sense: as follows from the second Friedmann equation, the inequality $\rho+3p\le0$ gives necessary condition for accelerated expansion of the Universe, i.e. the accelerated expansion of the Universe is possible only in presence of components with high negative pressure $p<-\rho/3$. The SEC excludes existence of such components. As a result, $q\ge0$ in this case. At the same time, NEC and DEC are compatible with the condition $p<-\rho/3$ and therefore they allow the regimes with $q<0$. It worth noting that even before the discovery of the accelerated expansion of the Universe in 1997, Visser [M. Visser, Science 276, 88 (1997), M. Visser, Phys. Rev. D 56, 7578(1997)] already concluded, basing on analysis of the energy conditions, that current observations suggest that SEC was violated sometime between the epoch of galaxy formation and the present. This implies that no possible combination of "normal" matter is capable of fitting the observational data.

\[d_L(z)=(1+z)\int\limits_0^z\frac{dz'}{H(z')}=(1+z)H_0\int\limits_0^z du\exp\left\{-\int\limits_0^u [1+q(v)]d[\ln(1+v)]\right\}.\]

\begin{equation}\label{distance_7_3}d_L=\frac{z}{H_0}\left[1+\left(\frac{1-q_0}{2}\right)z+O(z^2)\right],\end{equation} where in the spatially flat case \[q_0=\frac12\sum\limits_i\Omega_{i0}(1+3w_i).\]

\begin{align} \nonumber d_L(z)= & \frac{cz}{H_0}\left[ 1+\frac12(1-q_0)z-\frac16(1-q_0-3q_0^2+j_0)z^2\right.\\ +&\left.\frac1{24}(2-2q_0-15q_0^2-15q_0^3+5j_0+10j_0q_0+s_0)z^3+O(z^4)\right]. \end{align} As was expected, the latter decomposition contains the cosmographic parameters defined in terms of the higher order time derivatives of the scale factor ($j\propto d^3a/dt^3$).

\begin{equation}\label{distance_7_5}d_L=a_0r_1(1+z),\quad r_1=\int\limits_{t_1}^{t_0}\frac{dt}{a(t)}.\end{equation} In the considered case \[r_1=\frac{3t_0}{a_0}\left[1-\left(\frac{t_1}{t_0}\right)^{1/3}\right],\quad 1+z=\frac{a_0}{a_1}=\left(\frac{t_0}{t_1}\right)^{2/3},\] thus \[r_1=\frac{3t_0}{a_0}\left[1-\left(1+z\right)^{-1/2}\right]=\frac{2}{a_0H_0} \left[1-\left(1+z\right)^{-1/2}\right].\] We took into account here that in the Einstein-de Sitter model $h_0t_0=2/3$. For the luminosity distance one ultimately obtains \[d_L=a_0r_1(1+z)=\frac2{H_0}\left(1+z-\sqrt{1+z}\right).\]

In the flat case, differentiating the expression \[d_L(z)=(1+z)H_0^{-1}\int\limits_0^z\frac{dz'}{E(z')},\] one obtains \[\frac{d(d_l)}{dz}=\frac{1+z}{H(z)}+\frac{d_L(z)}{1+z}\Rightarrow H(z)=\frac{(1+z)^2}{d'_L(1+z)-d_L},\] where prime denotes derivative with respect to redshift $z$.

Using result of the previous problem, one finds \[q(z)=-1\frac{\dot H}{H^2}=-1+(1+z)\frac{H'}{H}=1-\frac{(1+z)^2d''_L(z)}{d'_L(1+z)-d_L},\] where prime denotes derivative with respect to redshift $z$. The latter formula can be represented in the form \[q(N)=-1-\frac{H'(N)}{H(N)}=1+\frac{d''_L(N)+d'_L(N)}{d'_L(N)+d_L(N)}.\] Here primes denote derivatives with respect to $N\equiv\ln a=-\ln(1+z)$. For arbitrary geometry \[q(z)=\frac{1+\omega_K d_L(z)d'_L(z)/(1+z)}{1+\omega_K d_L^2(z)/(1+z)^2}-\frac{(1+z)^2d''_L(z)}{d'_L(z)(1+z)-d_L}.\]

The component with $q=1$ represents the ultra-relativistic matter (radiation) for which $a\propto t^{1/2}$ and therefore $H=1/(2t)$. In this case the particle horizon is finite and equals to \[L_p=a(t)\int\limits_0^t\frac{dt'}{a(t')}=2t=\frac1{H(t)}=R_H.\] The component with $q=-1$ represents the cosmological constant for which $a\propto e^{Ht}$, $H=const$. In this case the particle horizon is absent because the origin of the Universe is moved to $t=-\infty$. The event horizon has the following size \[L_e=a(t)\int\limits_t^\infty\frac{dt'}{a(t')}=\frac1H=R_H.\]

\[\frac{d}{dt}(R_H)=c\frac{d}{dt}\left(\frac1H\right)= -\frac{c}{H^2}\left(\frac{\ddot a}{a}-\frac{\dot a^2}{a^2}\right)=c(1+q).\] As one can see, the Hubble sphere contracts when $q<-1$, remains stationary when $q=-1$ and expands when $q>-1$.

In the Universe with decelerated expansion ($q>0$) the Hubble's sphere has velocity exceeding the light speed by the quantity $cq$ and thus it overtakes the galaxies situated on its surface. Therefore the galaxies initially situated outside the Hubble's sphere will initially enter inside. Galaxies at distance $R>R_H$ are later $R<R_H$, and their superluminal recession in the course of time becomes subluminal. The light emitted toward the observer by a galaxy outside the Hubble sphere recedes until it enters inside the sphere. Therefore it starts approaching us and becomes available to observations. Therefore, all decelerating Universes lack event horizons unless they terminate at some future time. In uniformly expanding Universes $q=0$ the Hubble surface and the galaxies situated on it have equal velocities. Thus number of galaxies in the observable Universe remains constant. Then both particle horizon and event horizon are absent in such Universes. In case of the accelerating expansion $q<0$ the Hubble sphere has velocity which is less than the light speed by the quantity $q$ and thus it falls behind the galaxies, and therefore number of them decreases inside the Hubble sphere. All accelerating Universes have the property that galaxies at distance $R<R_H$ are later $R>R_H$, and their subluminal recession in the course of time becomes superluminal. Light emitted outside the Hubble sphere recedes from the observer and can never approach the observer. There are events that can never be observed, and such Universe have event horizons.

Using the definitions of the particle horizon \[L_p=a(t)\int\limits_0^t\frac{dt'}{a(t')}\] and the event horizon \[L_e=a(t)\int\limits_t^\infty\frac{dt'}{a(t')},\] one finds \begin{align} \nonumber \frac{dL_p}{dt}=&\frac{d}{dt}\left[a(t)\int\limits_0^t\frac{dt'}{a(t')}\right]=L_p(z)H(z)+1;\\ \label{distance_11_3} \frac{dL_e}{dt}=&\frac{d}{dt}\left[a(t)\int\limits_t^\infty\frac{dt'}{a(t')}\right]=L_e(z)H(z)-1. \end{align}

Differentiating the relations obtained in the previous problem w.r.t. time, one obtains \begin{align} \nonumber \frac{d^2L_p}{dt^2}=&H(1-qHL_p);\\ \label{distance_11_5} \frac{dL_e}{dt}=&-H(1+qHL_e). \end{align}

Current value of the proper distance $L_p$ is \[L_p=a_0\int\limits_0^{t_0}\frac{dt'}{a(t')}=a_0\eta|_0^{\eta_0}=a_0\eta_0,\] where $\eta$ is the conformal time. Using the expressions for $a_0$ and $\eta_0$ obtained in problems 40 and 41, one finds \[L_p=a_0\int\limits_0^{r_0}\frac{dr}{\sqrt{1-kr^2}} =a_0\int\limits_0^{t_0}\frac{dt}{a(t)}=\frac1{H_0} \left\{\begin{array}{lcr} 2, & k=0, & q_0=1/2;\\ \frac{\arcsin{\frac{\sqrt{2q_0-1}}{q_0}}}{\sqrt{2q_0-1}}, & k=1, & q_0>1/2;\\ \frac{\mathrm{arcsinh}{\frac{\sqrt{1-2q_0}}{q_0}}}{\sqrt{1-2q_0}}, & k=-1, & q_0<1/2. \end{array}\right.\]

In these power-law Universes we have \[H\equiv\frac{\dot a}{a}=\frac n t, \quad q\equiv-\frac{\ddot a}{aH^2}=\frac{1-n}t.\] Hence $n<1$ in a decelerating ($q>0$) Universe.

In the considered case \[L_p=a(t)\int\limits_0^t\frac{dt'}{a(t')}=\frac{t}{1-n}=\frac t n \frac1q=\frac{R_h}{q}\]

We cite below an extremely bright discussion of this question in [E.Harrison, Hubble's spheres and particle horizons, The Astrophisical Journal, 383, 63, 1991] " In the study of causal connections, the Hubble sphere bounded by the Hubble surface is as important as the observable universe bounded by the particle horizon. Let two comoving bodies be separated by a distance $L$ sufficiently small that each lies in the observable Universe of the other. Each body remains thereafter permanently in the other's observable Universe, and the ratio $L/(ct)$ during expansion depands on the behavior of the Hubble sphere. In a decelerating Universe the Hubble sphere expands faster than Universe, and a body at distance $L$ either is inside or will soon be inside the Hubble sphere. Hence, any two bodies must eventually recede from each other at subluminal velocity, and the ratio $L/(ct)$ will then decrease in time. In an accelerating Universe the Hubble sphere expands slower then universe, and a body at distance $L$ either is outside or will soon be outside the Hubble sphere. Hence, any two bodies must eventually recede from each other at superluminal velocity and the ratio $L/(ct)$ will then increase in time. How can causally connected distances of $L\gg ct$ exist? The answeris that the universe passes through a period of accelerated expansion, and causal connections of $L<ct$, established before acceleration, expand superluminally outside the Hubble sphere... A period of accelerated expansion distend all previously established causal connections and increases the distance to the particle horizon."

In order to describe the cosmological expansion one commonly uses two sets of coordinates: the physical (or Euler) coordinates ($R,\theta,\varphi$) and comoving (or fixed, Lagrangian) coordinates ($r,\theta,\varphi$). (The angular coordinates are the same for both sets as the cosmological expansion is assumed to be radial.) The two sets are related by the formula $R(t)=a(t)r$. Therefore a point which is fixed w.r.t. cosmological expansion, i.e. with constant coordinates ($r,\theta,\varphi$), has additional radial acceleration \begin{equation}\nonumber \left.\frac{d^2R}{dt^2}\right|_{expansion}=R\frac{\ddot a}{a}=-qH^2R.\end{equation} Consequently, in the Newtonian limit the radial force $F$ per unit mass at a distance $R$ from a point mass $m$ is given by \begin{equation}\nonumber F=-\frac{m}{R^2}-q(t)H^2(t)R.\end{equation} The force consists of the usual $1/R^2$ inwards component due to the central (point) mass $m$ and a cosmological component proportional to $R$ that is directed outwards (inwards) when the expansion of the Universe is accelerating (decelerating).

In this case, the Hubble parameter and, hence, the deceleration parameter become time-independent and are given by $H=\sqrt{\Lambda/3}$ and $q=-1$. Thus, the force, obtained in the previous problem, also becomes time-independent, \begin{equation}\nonumber F=-\frac{m}{R^2}+\frac13\Lambda R.\end{equation}

\begin{equation}\nonumber F=-\frac{M(R)}{R^2}+\frac13\Lambda R.\end{equation} where $M(R)$ is the total mass of the object contained within the radius $R$. If the object has the radial density $\rho(R)$ then \[M(R)=\int\limits_0^R4\pi R'^2\rho(R')dR'.\]

Although the de Sitter background is not an accurate representation of our universe, the SCM is dominated by dark-energy in a form consistent with a simple cosmological constant. Even in the simple Newtonian case (previous problems), we see immediately that there is an obvious, but profound, difference between the cases $\Lambda=0$ and $\Lambda\ne0$. In the former, the force on a constituent particle of a galaxy or cluster (say) is attractive for all values of $R$ and tends gradually to zero as $R\to\infty$ (for any sensible radial density profile). In the latter case, however, the force on a constituent particle (or equivalently its radial acceleration) vanishes at the finite radius $R_F$ which satisfies $R_F=[3M(R_F)/\Lambda]^{1/3}$, beyond which the net force becomes repulsive. This suggests that a non-zero $\Lambda$ should set a maximum size, dependent on mass, for galaxies and clusters. In the Newtonian limit, the speed of a particle in a circular orbit of radius $r$ is given by \[V(R)=\sqrt{\frac{M(R)}{R}-\Lambda R^2},\quad \left(F=-\frac{M(R)}{R^2}+\frac13\Lambda R\right).\] from which it is clear that no circular orbit can exist beyond the radius $R_F$.

The radius $R_S$ may be determined as the minimum of the (time-dependent) effective potential for a test particle in orbit about the central mass [R. Nandra, A. N. Lasenby and M. P. Hobson, The effect of an expanding universe on massive objects, arXiv:1104.4458] . Remaining within the frames of the Newtonian approximation (a weak gravitational field and low velocities), the equation of motion for the test particle is \begin{equation}\nonumber \ddot R\approx-\frac m{R^2}-q(t)H^2(t)R+\frac{L^2}{R^3}.\end{equation} Here $L$ is angular momentum per unit mass. This is simply the Newtonian radial force expression with the inclusion of a centrifugal term. The test particle moves in the effective one-dimensional potential $V(R)$ \begin{equation}\label{1301_69_e3} V(R)=+\frac{L^2}{2R^2}-\frac m{R}+\frac12q(t)H^2(t)R^2.\end{equation} Extrema of the effective potential in which the test particle moves occur at the $R$-values for which $d^2R/dt^2=0$, namely the solutions of \begin{equation}\nonumber -\frac m{R^2}-q(t)H^2(t)R+\frac{L^2}{R^3}=0.\end{equation} Consider the function \begin{equation}\nonumber y = -mR-q(t)H^2(t)R^4+L^2.\end{equation} The polynomial has single extremum --- a minimum in the point \[R^*=R_S=\left(-\frac m{4q(t)H^2(t)}\right)^{1/3}\] Condition for existence of real roots of the equation (\ref{1301_69_e3}) reads $y(R^*)\le0$. Critical value of the angular momentum for which the minimum of the effective potential disappears can be found from the condition $y(R^*)=0$ and it equals \[L_{crit}=\frac{\sqrt3}{2}m^{1/2}\left(-\frac m{4q(t)H^2(t)}\right)^{1/6}= \left(-\frac{27m^4}{256q(t)H^2(t)}\right)^{1/6}\]

\[\rho_{min}(t)=\frac{3m}{4\pi R_S^3(t)}=-\frac{3q(t)H^2(t)}{\pi}.\] The latter relation is valid only for $q(t)<0$ (accelerating expansion). For $q(t)>0$, $\rho_{min}(t)=0$.

Using result of the previous problem one obtains \[\Omega_{min}\equiv\frac{\rho_{min}}{\rho_{crit}}=-8q(t).\]

Since $\rho_m(t)=\Omega_m(t)\rho_{crit}(t)$, then \[\delta_{min}(t)=-\left[1+\frac{8q(t)}{\Omega_m(t)}\right].\] For the present time ($\Omega_{m0}\approx0.3$, $q_0\approx-0.55$) one finds that $\delta_{min0}\approx14$.