# Category:Deceleration Parameter

## Contents

- 1 Classification of models of Universe based on the deceleration parameter
- 2 Deceleration as a cosmographic parameter
- 3 Cosmological scalars and the Friedmann equation
- 4 Averaging deceleration parameter
- 5 Energy conditions in terms of the deceleration parameter
- 6 Distance-Deceleration Parameter Relations
- 7 Horizons
- 8 Power-Law Universes
- 9 The Effects of a Local Expansion of the Universe

**Problem 1**

problem id: 1301_1

Show that for a spatially flat Universe consisting of one component with equation of state $p=w\rho$ the deceleration parameter is equal to $q=(1+3w)/2$.

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

**Problem 2**

problem id: 1301_2

Find generalization of the relation $q=(1+3w)/2$ for the non-flat case.

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

**Problem 3**

problem id: 1301_3

Find deceleration parameter for multi-component non-flat Universe.

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

**Problem 4**

problem id: 1301_4

Show that the expression for deceleration parameter obtained in the previous problem can be presented in the following form \[q=\frac{\Omega_{total}}2+\frac32\sum\limits_i w_i \Omega_i.\]

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

**Problem 5**

problem id:

Let us consider the model of two-component Universe [ J. Ponce de Leon, cosmological model with variable equations of state for matter and dark energy, arXiv:1204.0589]. Such approximation is sufficient to achieve good accuracy on each stage of its evolution. At the present time the two dark components - dark matter and dark energy - are considered to be dominating. We neglect meanwhile the interaction between the components and as a result they separately satisfy the conservation equation. Let us assume that the state equation parameter for each component depends on the scale factor \begin{align} \nonumber p_{de} & = W(a) \rho_{de};\\ \nonumber p_m & = w(a)\rho_{m}. \end{align} Express the relative densities $\Omega_m$ and $\Omega_{de}$ in terms of the deceleration parameter and the state equation parameters $w(a)$ and $W(a)$.

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}

**Problem 6**

problem id: 1301_6

Find the upper and lower limits on the deceleration parameter using results of the previous problem.

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 $q$, \[\frac{1+3W}{2}\left(1+\frac{k}{a^2H^2}\right)\le q\le \frac{1+3w}{2}\left(1+\frac{k}{a^2H^2}\right).\]

**Problem 7**

problem id: 1301_7

Find the relation between the total pressure and the deceleration parameter for the flat one-component Universe

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 \[p=\frac{H^2}{8\pi G}(2q-1).\]

**Problem 8**

problem id: 1301_8

The expansion of pressure by the cosmic time is given by \[p(t)=\left.\sum\limits_{k=0}^\infty\frac1{k!}\frac{d^kp}{dt^k}\right|_{t=t_0}(t-t_0)^k.\] Using cosmography parameters , evaluate the derivatives up to the fourth order.

**Problem 9**

problem id: 1301_9

Show that for one-component flat Universe filled with ideal fluid of density $\rho$.

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

**Problem 10**

problem id: 1301_10

For what values of the state parameter $w$ the rate of expansion of a one-component flat Universe increases with time?

**Problem 11**

problem id: 1301_11

Express the age of the spatially flat Universe filled with a single component with equation of state $p=w\rho$ through the deceleration parameter.

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 \[t_0=\frac{H_0^{-1}}{1+q}.\]

**Problem 12**

problem id: 1301_12

Suppose we know the current values of the Hubble constant $H_0$ and the deceleration parameter $q_0$ for a closed Universe filled with dust only. How many times larger will it ever become? Find lifetime of such a Universe.

**Problem 13**

problem id: 1301_13

In a closed Universe filled with non-relativistic matter the current values of the Hubble constant is $H_0$, the deceleration parameter is $q_0$. Find the current age of this Universe.

**Problem 14**

problem id: 1301_14

In a closed Universe filled with dust the current value of the Hubble constant is $H_0$ and of the deceleration parameter $q_0$.

a) What is the total proper volume of the Universe at present time?

b) What is the total current proper volume of space occupied by matter which we are presently observing?

c) What is the total proper volume of space which we are directly observing?

**Problem 15**

problem id: 1301_15

For the closed ($k=+1$) model of Universe, filled with non-relativistic matter, show that solutions of the Friedmann equations can be represented in terms of the two parameters $H_0$ and $q_0$. [Y.Shtanov, Lecture Notes on theoretical cosmology, 2010 ]

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 $q_0$.

**Problem 16**

problem id: 1301_16

Do the same as in the previous problem for the case of open ($k=-1$) model of Universe.

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 $q_0$.

# Classification of models of Universe based on the deceleration parameter

**Problem 17**

problem id: 1301_17

When the rate of expansion never changes, and $\dot a$ is constant, the scaling factor is proportional to time $t$, and the deceleration term is zero. When the Hubble parameter is constant, the deceleration parameter $q$ is also constant and equal to $-1$, as in the de Sitter model. All models can be characterized by whether they expand or contract, and accelerate or decelerate. Build a classification of such type using the signature of the Hubble parameter and the 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).

**Problem 18**

problem id: 1301_18

Can the Universe variate its type of evolutions in frames of the classification given in the previous problem?

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.

**Problem 19**

problem id: 1301_19

Point out possible regime of expansion of Universe in the case of constant deceleration parameter.

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 $q<-1$.

**Problem 20**

problem id: 1301_20

Having fixed the material content we can classify the model of Universe using the connection between the deceleration parameter and the spatial geometry. Perform this procedure in the case of Universe filled with non-relativistic matter.

\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}

**Problem 21**

problem id: 1301_21

Models of the Universe can be classified basing on the relation between the deceleration parameter and age of the Universe. Build such a classification in the case of Universe filled with non-relativistic matter (the Einstein-de Sitter Universe)

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}

**Problem 22**

problem id: 1301_22

Show that sign of the deceleration parameter determines the difference between the actual age of the Universe and the Hubble time.

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.

**Problem 23**

problem id: 1301_23

Suppose the flat Universe is filled with non-relativistic matter with density $\rho_m$ and some substance with equation of state $p_X=w\rho_X$. Express the deceleration parameter through the ratio $r\equiv\rho_m/\rho_X$.

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

**Problem 24**

problem id: 1301_23_1

Obtain Friedmann equations for the case of spatially flat $n$-dimensional Universe. (see Shouxin Chen, Gary W. Gibbons, Friedmann's Equations in All Dimensions and Chebyshev's Theorem, arXiv: 1409.3352)

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}

**Problem 25**

problem id: 1301_23_2

Analyze exact solutions of the Friedmann equations obtained in the previous problem in the case of flat ($k=0$) $n$-dimensional Universe filled with a barotropic liquid with the state equation \begin{equation} \label{10} p_m=w \rho_m. \end{equation} and obtain corresponding explicit expressions for the deceleration parameter.

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}

# Deceleration as a cosmographic parameter

**Problem 26**

problem id: 1301_24

Make transition from the derivatives w.r.t. cosmological time to that w.r.t. conformal time in definitions of the Hubble parameter and the deceleration parameter.

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

**Problem 27**

problem id: 1301_25

Make Taylor expansion of the scale factor in time using the cosmographic parameters.

\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}

**Problem 28**

problem id: 1301_26

Make Taylor expansion of the redshift in time using the cosmographic parameters.

\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}

**Problem 29**

problem id: 1301_27

Show that \[q(t)=\frac{d}{dt}\left(\frac1H\right)-1.\]

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

**Problem 30**

problem id: 1301_28

Show that the deceleration parameter as a function of the red shift satisfies the following relations \begin{align} \nonumber q(z)=&\frac{1+z}{H}\frac{d H}{dz}-1;\\ \nonumber q(z)=&\frac12(1+z)\frac1{H^2}\frac{d H^2}{dz}-1;\\ \nonumber q(z)=&\frac12\frac{d\ln H^2}{d\ln(1+z)}-1;\\ \nonumber q(z)=&\frac{d\ln H}{dz}(1+z)-1. \end{align}

**Problem 31**

problem id: 1301_29

Show that the deceleration parameter as a function of the scale factor satisfies the following relations \begin{align} \nonumber q(a)=&-\left(1+\frac{a\frac{dH}{da}}{H}\right);\\ \nonumber q(a)=&\frac{d\ln(aH)}{d\ln a}. \end{align}

**Problem 32**

problem id: 1301_30

Show that the derivatives $dH/dz$, $d^2H/dz^2$, $d^3H/dz^3$ and $d^4H/dz^4$ can be expressed through the deceleration parameter $q$ and other cosmological parameters.

\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}

**Problem 33**

problem id: 1301_31

Use results of the previous problem to make Taylor expansion of the Hubble parameter in redshift.

\[H(z)=H_0+\left.\frac{dH}{dz}\right|_{z=0}z +\frac12\left.\frac{d^2H}{dz^2}\right|_{z=0}z^2 +\frac16\left.\frac{d^3H}{dz^3}\right|_{z=0}z^3+\dots=\] \[=H_0(1+(1+q_0)z+\frac12(j_0-q_0^2)z^2+\frac16(3q_0^2+3q_0^3-4q_0j_0-3j_0-s)z^3).\] 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}

**Problem 34**

problem id: 1301_32

Express derivatives of the Hubble parameter squared w.r.t. the redshift $d^iH^2/dz^i$, $i=1,2,3,4$ in terms of the cosmographic parameters.

\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}

**Problem 35**

problem id: 1301_33

Express the current values of deceleration and jerk parameters in terms of $N=-\ln(1+z)$.

\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}

**Problem 36**

problem id: 1301_34

Express time derivatives of the Hubble parameter in terms of the cosmographic parameters.

\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}

**Problem 37**

problem id: 1301_35

Express the deceleration parameter as power series in redshift $z$ or $y$-redshift $z/(1+z)$.

\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}

**Problem 38**

problem id: 1301_36

Show that the Hubble parameter is connected to the deceleration parameter by the integral relation \[H=H_0\exp\left[\int\limits_0^z[q(z')+1]d\ln (1+z')\right].\]

**Problem 39**

problem id: 1301_37

Show that derivatives of the lower order parameters can be expressed through the higher ones, for instance \[\frac{dq}{d\ln(1+z)}=j-q(2q+1).\]

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 \[\frac{dq}{d\ln(1+z)}=j-q(2q+1).\]

# Cosmological scalars and the Friedmann equation

*Dunajski and Gibbons [M. Dunajski, Gary Gibbons, Cosmic Jerk, Snap and Beyond, arXiv:0807.0207] proposed an original way to test the General Relativity and the cosmological models based on it. The procedure implies expressing the Friedmann equation in terms of directly measurable cosmological scalars constructed out of higher derivatives of the scale factor, i.e. cosmographic parameters $H,q,j,s,l$. In other words, the key idea is to treat the Friedmann equations as one algebraic constraint between the scalars. This links the measurement of the cosmological parameters to a test of General Relativity, or any of its modifications.*

**Problem 40**

problem id: 1301_38

Express the curvature parameter $k$ terms of the cosmographic parameters for the case of Universe filled with non-interacting cosmological constant and non-relativistic matter.

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

**Problem 41**

problem id: 1301_39

Do the same as in the previous problem for the case of Universe filled with non-interacting non-relativistic matter $\rho_m=M_m/a^3$ and radiation $\rho_r=M_r/a^4$.

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

**Problem 42**

problem id: 1301_40

Check the expressions for the curvature $k$ obtained in the previous problem for two cases: a) a flat Universe solely filled with non-relativistic matter; b) a flat Universe solely filled with radiation.

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 $j=3$.

**Problem 43**

problem id: 1301_41

Find relation between the cosmographic parameters free of any cosmological parameter for the case of Universe considered in the problem #1301_38.

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

**Problem 44**

problem id: 1301_42

Perform the same procedure for the Chaplygin gas with the equation of state $p=-A/\rho$.

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}

**Problem 45**

problem id: 1301_43

Perform the same procedure for the generalized Chaplygin gas with the equation of state $\rho=-A/\rho^\alpha$.

\begin{equation}\label{scalar_7}s+(3\alpha+2)(q+j)+qj=0.\end{equation} Note that for $\alpha=1$ we reproduce the above obtained result for the Chaplygin gas, while $\alpha=0$ we return to the results obtained in the problem (\ref{1301_41}). If we want to exclude the parameter $\alpha$ from the latter equation we must take one more derivative of the Friedmann equation and introduce an additional cosmological parameter \[l=\frac1a\frac{d^5a}{dt^5}\left(\frac1a\frac{da}{dt}\right)^{-5}.\] As a result one obtains \begin{equation}\label{scalar_8} -2qs-2jq^2-lq-2sj-3sq^2-j^2q-lj+s^2-3q^2j-qsj+j^3-2j^2q^2=0. \end{equation} This constraint is again approximate and is valid only in the regime where the higher order terms in the expansion of $\rho$ can be dropped.

# Averaging deceleration parameter

*Since the deceleration parameter $q$ is a slowly varying quantity (e.g. $q = 1/2$ for matter-dominated case and $q = -1$ in the Universe dominated by dark energy in form of cosmological constant), then the useful information is contained in its time average value, which is very interesting to obtain without integration of the equations of motions for the scale factor. Let us see how it is possible [J.Lima, Age of the Universe, Average Deceleration Parameter and Possible Implications for the End of Cosmology, arXiv:0708.3414]. For that purpose let us define average value $\bar q$ of this parameter on time interval $[0,t_0]$ with the expression \[\bar q(t_0)=\frac1{t_0}\int\limits_0^{t_0}q(t)dt.\]*

**Problem 46**

problem id: 1301_44

Express current value of the average deceleration parameter in terms of the Hubble parameter.

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 \[\bar q(t_0)=-1+\frac1{t_0H_0}.\]

**Problem 47**

problem id: 1301_45

Show that current age of the Universe depends solely on average value of the deceleration parameter.

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.

**Problem 48**

problem id: 1301_46

Show that the Hubble time $H_0)^{-1}$ represents a characteristic time scale for age of the Universe at any stage of the evolution.

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.

# Energy conditions in terms of the deceleration parameter

Dynamic model-independent constraints on the kinematics of the Universe can further be obtained from the so-called energy conditions. These conditions, based on quite general physical principles, impose restrictions on the components of the energy-momentum tensor $T_{\mu\nu}$. In choosing a model for the medium (a model, but not the equation of state!), these conditions can be transformed into inequalities restricting the possible values of pressure and density of the medium. In terms of density and pressure the energy conditions take on the form \begin{align} \nonumber NEC&\Rightarrow & \rho+p\ge&0, & &\\ \nonumber WEC&\Rightarrow & \rho\ge&0,&\rho+p\ge&0,\\ \nonumber SEC&\Rightarrow & \rho+3p\ge&0,&\rho+p\ge&0,\\ \nonumber DEC&\Rightarrow & \rho\ge&0,&-\rho\le p\le&\rho. \end{align} Here, NEC, WEC, SEC, and DEC correspond to the zero, weak, strong, and dominant energy conditions. Because these conditions do not require any definite equation of state for the substance filling the Universe, they impose very simple and model-independent constraints on the behavior of the energy density and pressure. Hence, the energy conditions provide one of the possibilities for explaining the evolution of the Universe on the basis of quite general principles.

**Problem 49**

problem id: 1301_47

Express the energy conditions in terms of the scale factor and its derivatives.

\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}

**Problem 50**

problem id: 1301_48

Transform the energy conditions to constraints on the deceleration parameter for the flat Universe.

\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 $a(t)$.

**Problem 51**

problem id: 1301_49

Analyze the constraints on the regimes of accelerated and decelerated expansion of the Universe following from the energy conditions.

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.

# Distance-Deceleration Parameter Relations

In cosmology there are many different and equally natural definitions of the notion of *distance* between two objects or events. In particular, the luminosity distance $d_L$ of an object at redshift $z$ is $d_L=(L/2\pi F)^{1/2}$, where $L$ is the bolometric luminosity for a given object and $F$ is the bolometric energy flux received from that object. The expression for the luminosity distance in a FLRW Universe is
\begin{equation}\label{distance_7_1}
d_L(z)=(1+z)\left\{
\begin{array}{lr}
R\sinh\left[\frac1{H_0R}\int\limits_0^z\frac{dz'}{E(z')}\right], & open\\
H_0^{-1}\int\limits_0^z\frac{dz'}{E(z')}, & flat\\
R\sin\left[\frac1{H_0R}\int\limits_0^z\frac{dz'}{E(z')}\right], & closed
\end{array}
\right.\end{equation}
(Here and below we set $c=1$). Here $R$ is the (comoving) radius of curvature of the open or closed Universe, $E=H/H_0$.

**Problem 52**

problem id: 1301_50

Represent the luminosity distance in the flat Universe in terms of the deceleration parameter.

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

**Problem 53**

problem id: 1301_51

Find expression for the luminosity distance up to terms of order of $z^2$.

\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 \[q_0=\frac12\sum\limits_i\Omega_{i0}(1+3w_i).\]

**Problem 54**

problem id: 1301_52

Find expression for the luminosity distance in the next order $O(z^3)$ in the redshift.

\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 ($j\propto d^3a/dt^3$).

**Problem 55**

problem id: 1301_53

Calculate the luminosity distance in the Universe filled with non-relativistic matter (Einstein-de Sitter model ($k=0$)).

\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 \[d_L=a_0r_1(1+z)=\frac2{H_0}\left(1+z-\sqrt{1+z}\right).\]

**Problem 56**

problem id: 1301_54

The expression (\ref{distance_7_1}) allows to find the luminosity distance given the function $H(z)$. Solve the inverse problem for the flat case: find the Hubble parameter as a function of the luminosity distance [S. Nesseris and J. Garcia-Bellido, Comparative analysis of model-independent methods for exploring the nature of dark energy, arXiv:1306.4885].

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 $z$.

**Problem 57**

problem id: 1301_55

Use result of the previous problem to express deceleration parameter a s a function of luminosity distance and its derivatives.

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 \[q(z)=\frac{1+\omega_K d_L(z)d'_L(z)/(1+z)}{1+\omega_K d_L^2(z)/(1+z)^2}-\frac{(1+z)^2d''_L(z)}{d'_L(z)(1+z)-d_L}.\]

# Horizons

**Problem 58**

problem id: 1301_56

The Hubble sphere in general does not coincide with the horizons, except when it becomes degenerate with the particle horizon at $q=1$ and with event horizon at $q=-1$. Show that.

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
\[L_e=a(t)\int\limits_t^\infty\frac{dt'}{a(t')}=\frac1H=R_H.\]

**Problem 59**

problem id: 1301_57

Show that the Hubble sphere contracts when $q<-1$, remains stationary when $q=-1$ and expands when $q>-1$.

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

**Problem 60**

problem id: 1301_58

Analyze correspondence between kinematics of the Hubble sphere and boundaries of the observable Universe for different expansion regimes.

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.

**Problem 61**

problem id: 1301_59

Calculate the derivatives $dL_p/dt$ and $dL_e/dt$.

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}

**Problem 62**

problem id: 1301_60

Calculate the derivatives $d^2L_p/dt^2$ and $d^2L_e/dt^2$.

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}

**Problem 63**

problem id: 1301_61

Find the expressions for the current particle horizon in the single-component Universe filled with non-relativistic matter.

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 \ref{1301_38} and \ref{1301_39}, one finds \[L_p=a_0\int\limits_0^{r_0}\frac{dr}{\sqrt{1-kr^2}} =a_0\int\limits_0^{t_0}\frac{dt}{a(t)}=\frac1{H_0} \left\{\begin{array}{lcr} 2, & k=0, & q_0=1/2;\\ \frac{\arcsin{\frac{\sqrt{2q_0-1}}{q_0}}}{\sqrt{2q_0-1}}, & k=1, & q_0>1/2;\\ \frac{\mathrm{arcsinh}{\frac{\sqrt{1-2q_0}}{q_0}}}{\sqrt{1-2q_0}}, & k=-1, & q_0<1/2. \end{array}\right.\]

# Power-Law Universes

**Problem 64**

problem id: 1301_62

Assume that the scale factor $a(t)$ varies as $t^n$, where $t$ is the age of Universe and $n=const$. Show that in decelerated expanding Universe $n<1$.

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.

**Problem 65**

problem id: 1301_63

Show that if the scale factor $a(t)$ varies as $t^n$, \[L_p=\frac{R_h}{q}.\]

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

**Problem 66**

problem id: 1301_64

How, in a Universe of age $t$ can causally connected distances of $L\gg ct$ exist?

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."

# The Effects of a Local Expansion of the Universe

**Problem 67**

problem id: 1301_65

Considering the radial motion of a test particle in a spatially-flat expanding Universe find the Newtonian limit the radial force $F$ per unit mass at a distance $R$ from a point mass $m$.

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).

**Problem 68**

problem id: 1301_66

Find the Newtonian limit of the radial force $F$ per unit mass at a distance $R$ from a point mass $m$ in a Universe which contains no matter (or radiation), but only dark energy in the form of a non-zero cosmological constant $\Lambda$.

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

**Problem 69**

problem id: 1301_67

Solve the previous problem for the case finite (i.e. non-pointlike) spherically-symmetric massive objects.

\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 \[M(R)=\int\limits_0^R4\pi R'^2\rho(R')dR'.\]

**Problem 70**

problem id: 1301_68

Show that a non-zero $\Lambda$ should set a maximum size, dependent on mass, for galaxies and clusters.

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 $R_F$.

**Problem 71**

problem id: 1301_69

The radius $R=R_F$ (see previous problem) does not necessarily correspond to the maximum possible size of the galaxy or cluster: many of the gravitationally-bound particles inside $R_F$ may be in unstable circular orbits. Therefore the so-called "outer" radius becomes of great importance, which one may interpret as the maximum size of the object, that is corresponding to the largest stable circular orbit $R_S$. Find in the Newtonian limit radius of the largest stable circular orbit $R_S$ in the case of gravity field created by a point mass $m$.

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 \[L_{crit}=\frac{\sqrt3}{2}m^{1/2}\left(-\frac m{4q(t)H^2(t)}\right)^{1/6}= \left(-\frac{27m^4}{256q(t)H^2(t)}\right)^{1/6}\]

**Problem 72**

problem id: 1301_70

If we consider $R_S(t)$ (see previous problem) as the maximum possible size of a massive object at cosmic time $t$, and assume that the object is spherically-symmetric and has constant density, then it follows that there exists a time-dependent minimum density (due to the maximum size). Determine the corresponding density.

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

**Problem 73**

problem id: 1301_71

Show that minimum relative density $\Omega_{min}\equiv\rho_{min}/\rho_{crit}$ is determined only by the deceleration parameter.

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

**Problem 74**

problem id: 1301_72

An important characteristics -- the minimum fractional density contrast $\delta_{min}(t)\equiv[\rho_{min}(t)-\rho_m(t)]/\rho_m(t)$ is immediately related to the deceleration parameter. Estimate this quantity in the SCM.

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 $\delta_{min0}\approx14$.

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