# Difference between revisions of "Dynamical Forms of Dark Energy"

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\delta S = \int {d^4 x\sqrt { - g} } \left[ { \left( {\nabla ^\mu \nabla _\mu \varphi } \right) - V_{,\varphi } (\varphi )} \right]\delta \varphi - \int {d^4 x\sqrt { - g} } \nabla _\mu \left( {\delta \varphi \nabla ^\mu \varphi } \right) | \delta S = \int {d^4 x\sqrt { - g} } \left[ { \left( {\nabla ^\mu \nabla _\mu \varphi } \right) - V_{,\varphi } (\varphi )} \right]\delta \varphi - \int {d^4 x\sqrt { - g} } \nabla _\mu \left( {\delta \varphi \nabla ^\mu \varphi } \right) | ||

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=== Problem 62 === | === Problem 62 === | ||

Find the energy density and pressure of the phantom field. | Find the energy density and pressure of the phantom field. |

## Latest revision as of 03:51, 3 December 2012

## The Quintessence

*The cosmological constant represents nothing but the simplest realization of the dark energy - the hypothetical substance introduced to explain the accelerated expansion of the Universe. There is a dynamical alternative to the cosmological constant - the scalar fields, formed in the post-inflation epoch. The most popular version is the scalar field $\varphi$ evolving in a properly designed potential $V(\varphi)$. Numerous models of such type differ by choice of the scalar field Lagrangian. The simplest model is the so-called quintessence. In antique and medieval philosophy this term (literally "the fifth essence", after the earth, water, air and fire) meant the concentrated extract, the creative force, penetrating all the material world. We shall understand the quintessence as the scalar field in a potential, minimally coupled to gravity, i.e. feeling only the influence of space-time curvature. Besides that we restrict ourselves to the canonic form of the kinetic energy. The action for fields of such type takes the form*
\[S=\int d^4x \sqrt{-g}\; L=\int d^4x \sqrt{-g}\left[\frac12g^{\mu\nu}\frac{\partial\varphi}{\partial x^\mu} \frac{\partial\varphi}{\partial x^\nu}-V(\varphi)\right].\]
*The equations of motion for the scalar field are obtained as usual, by variation of the action with respect to the field (see Chapter "Inflation").*

### Problem 1

Obtain the Friedman equations for the case of flat Universe filled with quintessence.

\begin{align*} H^2 & =\frac{8\pi G}{3}\left[\frac12\dot\varphi^2+V(\varphi)\right]\\ \frac{\ddot a}{a} & =-\frac{8\pi G}{3}\left[\dot\varphi^2-V(\varphi)\right] \end{align*}

### Problem 2

Obtain the general solution of the Friedman equations for the Universe filled with free scalar field, $V(\varphi)=0$.

The equations with the free scalar field read \[\dot H=\frac{k}{a^2}-\dot\varphi^2,\] \[\ddot\varphi+3H\dot\varphi=0,\] \[X\equiv H^2-\frac13\dot\varphi^2+\frac{k}{a^2}=0.\] (Here $4\pi G=1$). The second equation can be represented in the form \[\frac{d\ln(\dot\varphi a^3)}{dt}=0.\] Consequently, \[\dot\varphi=c_0a^{-3}.\] The initial set of equations then reduces to two \begin{align*} \dot H & =\frac{k}{a^2}-\frac{c_0^2}{a^6},\\ H^2 & =\frac{c_1}{a^6}-\frac{k}{a^2},\quad c_1\equiv\frac{c_0^2}{3}. \end{align*} Equating $\dot H/\dot a=dH/da$, one obtains from the first equation (the second Friedman equation) \[\frac{\dot H}{\dot a}=\frac{dH}{da}=\left(\frac{k}{a^2}-\frac{c_0^2}{a^6}\right)\frac{1}{Ha}.\] Integration gives \begin{equation}\label{Hpm} H=\pm\sqrt{\frac{c_1}{a^6}-\frac{k}{a^2}+c_2}, \end{equation} which can be further integrated to yield \[t=\pm\int\frac{da}{\sqrt{c_1 a^{-4}+c_2 a^2 - k}}.\] Here $c_2$ is a new arbitrary constant. The $\pm$ signs reflect time reversal invariance of the original equations. Choosing the positive sign one obtains \[a(t)\underset{t\to\infty}{\sim} a_0 \exp(\sqrt{c_2}t).\] Here $c_2$ is effective cosmological constant. For $c_2\ne0$ this solution contradicts the one obtained in the previous Chapter: $w_{DE}\to-1$ for $\dot\varphi^2\gg V(\varphi)$, and we consider the free scalar field with $\dot\varphi^2\gg V(\varphi)$. The contradiction can be eliminated if one takes into account (\ref{Hpm}), according to which \[\frac{\dot a}{a}=\pm\sqrt{c_1 a^{-6}-ka^{-2}}\] and \[dt=\pm\frac{da}{\sqrt{c_1a^{-4}-k}}.\] Consequently $c_2=0$. Thus the above obtained solution with $c_2\ne0$ is incorrect. The correct solution at $t\to\infty$ reads \[a(t)\left\{ \begin{array}{ccl} = & c_1^{-1/6}t^{1/3} & k=0\\ \underset{t\to\infty}{\sim} & c_1^{-1/4} & k=+1\\ \underset{t\to\infty}{\sim} & t & k=-1 \end{array} \right. .\] The last solutions correspond to power-law \[H\underset{t\to\infty}{\sim}\frac{const}{t}.\]

### Problem 3

Show that in the case of Universe filled with non-relativistic matter and quintessence the following relation holds: \[\dot H=-4\pi G(\rho_m+\dot\varphi^2).\]

\[\dot H=\frac{\ddot a}{a}-H^2=-4\pi G(\rho_{tot}+p_{tot})=-4\pi G (\rho_m+\dot\varphi^2).\]

### Problem 4

Show that in the case of Universe filled with non-relativistic matter and quintessence the Friedman equations \[H^2=\frac{8\pi G}{3}\left[\rho_m+\frac12\dot\varphi^2+V(\varphi)\right],\] \[\dot H =-4\pi G(\rho_m+\dot\varphi^2)\] can be transformed to the form \[\frac{8\pi G}{3H_0^2}\left(\frac{d\varphi}{dx}\right)^2=\frac{2}{3H_0^2x}\frac{d\ln H}{dx}-\frac{\Omega_{m0}x}{H^2};\] \[\frac{8\pi G}{3H_0^2}V(x)=\frac{H^2}{H_0^2}-\frac{x}{6H_0^2}\frac{d H^2}{dx}-\frac12\Omega_{m0}x^3;\] \[x\equiv1+z.\]

Take time derivative of the first Friedman equation to obtain \[\frac{dH^2}{dt}=\frac{8\pi G}{3}\left[\frac{d}{dt}(\rho_m+\rho_{DE})\right].\] Using the conservation equation, one finds \[\frac{d}{dt}(\rho_m+\rho_{DE})=-3H(\rho_m+\dot\varphi^2).\] Substitution \[\frac{d}{dt}=-H(1+z)\frac{d}{dz}\] finally yields \[\frac{8\pi G}{3H_0^2}\left(\frac{d\varphi}{dx}\right)^2=\frac{2}{3H_0^2x}\frac{d\ln H}{dx}-\frac{\Omega_{m0}x}{H^2},\quad x\equiv1+z.\] Then transform the equation $\dot H=-4\pi G(\rho_m+\dot\varphi^2)$ to the form \[\frac12(1+z)\frac{dH^2}{dz}=8\pi G\left(\frac12\rho_m+\rho_{DE}-V\right)\] and switching to relative densities we obtain \[\frac16(1+z)\frac{1}{H_0^2}\frac{dH^2}{dz}=\frac12\Omega_m+\Omega_{DE}-\frac{8\pi G}{3H_0^2}V.\] Using \[\frac{H^2}{H_0^2}=\sum\limits_i\Omega_i,\] one finally obtains \[\frac{8\pi G}{3H_0^2}V(x)=\frac{H^2}{H_0^2}-\frac{x}{6H_0^2}\frac{d H^2}{dx}-\frac12\Omega_{m0}x^3.\]

### Problem 5

Show that the conservation equation for quintessence can be obtained from the Klein-Gordon equation \[\ddot\varphi+3H\dot\varphi+\frac{dV}{d\varphi}=0.\]

Multiply the Klein-Gordon equation by $\dot\varphi$ to obtain \[\frac{d}{dt}\left(\frac{\dot\varphi^2}{2}\right)+6H\frac{\dot\varphi^2}{2}=-\frac{dV}{dt}.\] Then using expressions for density and pressure of the scalar field, we get \[\frac{d\rho_\varphi}{dt}-\frac{dV}{dt}+3H(\rho_\varphi+p_\varphi)=-\frac{dV}{dt},\] \[\frac{d\rho_\varphi}{dt}+3H\rho_\varphi(1+w)=0,\quad w\equiv\frac{p_\varphi}{\rho_\varphi}.\] The last expression can be rewritten as \[\frac{d\rho_\varphi}{d\ln a}+3\rho_\varphi(1+w)=0.\] The last two expressions are different forms of the conservation equation.

### Problem 6

Find the explicit form of Lagrangian describing the dynamics of the Universe filled with the scalar field in potential $V(\varphi)$. Use it to obtain the equations of motion for the scale factor and the scalar field.

\[S=S_g+S_\varphi;\quad S_g=-\frac12\int d^4x\sqrt{-g}R;\quad R=g^{\mu\nu}(x)R_{\mu\nu}(x);\] \[S_\varphi=\int d^4x \sqrt{-g}\left[\frac12g^{\mu\nu}\frac{\partial\varphi}{\partial x_\mu} \frac{\partial\varphi}{\partial x_\nu}-V(\varphi)\right];\quad 8\pi G=1;\] \[L=-\frac12\sqrt{-g}R+\sqrt{-g}\left[\frac12g^{\mu\nu}\frac{\partial\varphi}{\partial x_\mu} \frac{\partial\varphi}{\partial x_\nu}-V(\varphi)\right]\] \[R=-6\left(\frac{\ddot a}{a}+\frac{\dot a^2}{a^2}+\frac{k}{a^2}\right);\quad \sqrt{-g}\propto a^3;\] \[L=3(a^2\ddot a+a\dot a^2+ak)+a^3\left(\frac12\dot\varphi^2 - V(\varphi)\right);\] \[\ddot\varphi+3\frac{\dot a}{a}\varphi+\frac{dV}{d\varphi}=0.\]

### Problem 7

In the flat Universe filled with scalar field $\varphi$ obtain the isolated equation for $\varphi$ only. (See S.Downes, B.Dutta, K.Sinha, arXiv:1203.6892)

The Klein-Gordon equation for the scalar field in the FLRW-background is (see the previous problem) \[\ddot\varphi+3H\dot\varphi+\frac{dV}{d\varphi}=0\] The dynamics of scale factor is governed by the evolution equation \[\dot H=-\frac12\dot\varphi^2,\quad 8\pi G=1.\] The evolution and Klein-Gordon equations are a pair of coupled nonlinear differential equations for scale factor $a$ and scalar field $\varphi$, subject to the relation \[3H^2=\frac12\dot\varphi^2+V.\] Define a new variable $N$: \[N=\ln\frac{a(t)}{a_0},\quad \frac{dN}{dt}=H.\] Since $\dot\varphi=H\varphi'$ ($\varphi'\equiv d\varphi/dN$), the Friedman equation becomes \[3H^2=\frac12\varphi'^2H^2+V.\] So long as $|\varphi'|<\sqrt6$, for Hubble parameter one has \[H^2=\frac13\frac{V}{1-\frac16\varphi'^2}.\] Computing $\ddot\varphi$, one finds \[\ddot\varphi=\dot H\varphi'+H^2\varphi''=H^2\left(\varphi''-\frac12\varphi'^3\right).\] Therefore \[H^2\left(\varphi''-\frac12\varphi'^3+3\varphi'\right)+\frac{dV}{d\varphi}=0.\] Using the relation for $H^2$ one finally obtains \[\varphi''+3\left(1-\frac16(\varphi')^6\right)\left(\varphi'+\frac{d\ln V}{d\varphi}\right)=0.\]

### Problem 8

What is the reason for the requirement that the scalar field's evolution in the quintessence model is slow enough?

The scalar field in form of the quintessence was introduced to explain the accelerated expansion of the Universe. The latter is possible under condition $w < - 1/3$. On the other hand in the case of quintessence we have $$ w = {\frac{\frac{1 }{ 2}\dot \varphi ^2 - V} {\frac{1}{ 2}\dot \varphi ^2 + V}}. $$ Therefore the condition $w < - 1/3$ can be satisfied only by slowly evolving scalar field ($\dot\varphi ^2 \ll V$).

### Problem 9

Find the potential and kinetic energies for quintessence with the given state parameter $w$.

\[V=K\frac{1-w}{1+w};\quad K\equiv\frac12\dot\varphi^2.\]

### Problem 10

Find the dependence of state equation parameter $w$ for scalar field on the quantity \[x=\frac{\dot\varphi^2}{2V(\varphi)}\] and determine the ranges of $x$ corresponding to inflation in the slow-roll regime, matter-dominated epoch and the rigid state equation ($p\sim\rho$) limit correspondingly.

The function \[w(x) = {{x - 1} \over {x + 1}}\] monotonically increases from the minimum value $w_{\min } = - 1$ at $x=0$ to the asymptotic one $w=+1$ at $x \to \infty $ , corresponding to $V=0$. The slow-roll inflation regime corresponds to the region $x \ll 1,\;p \sim - \rho $, the case of non-relativistic matter---to $x \sim 1,\;p \sim 0$, and finally $x \gg 1,\;p \sim \rho $ corresponds to the rigid state equation.

### Problem 11

Show that if kinetic energy $K=\dot\varphi^2/2$ of a scalar field is initially much greater than its potential energy $V(\varphi)$, then it will decrease as $a^{-6}$.

If $V(\varphi)\ll\dot\varphi^2/2$, then $w\approx1$ and $\rho\propto a^{-3(1+w)}\propto a^{-6}$.

### Problem 12

Show that the energy density of a scalar field $\varphi$ behaves as $\rho_\varphi\propto a^{-n}$, $0\le n\le6$.

$$ \rho _\varphi = \rho _{\varphi0} \exp {\left( - \int 3[1 + w( \varphi)]\frac{da}{a}\right)} $$ where $$w( \varphi ) = \frac{p_\varphi}{\rho _\varphi} = \frac{\frac12\dot\varphi^2 - V(\varphi)}{\frac12\dot\varphi^2 + V(\varphi)}.$$ For a steep potential $\dot\varphi \gg V(\varphi )$ the equation of state parameter $w(\varphi )$ approaches the rigid matter limit $w(\varphi ) \to 1$, , whereas in the case of a flat potential $\dot\varphi \ll V(\varphi)$ one has $w(\varphi)\to-1$. Hence the energy density scales as $a^{-n}$ with $0\le n\le 6$.

### Problem 13

Show that dark energy density with the state equation $p=w(a)\rho(a)$ can be presented as a function of scale factor in the form \[\rho=\rho_0 a^{-3[1+\bar w(a)]},\] where $\bar w(a)$ is the parameter $w$ averaged in the logarithmic scale $$ \bar w(a) \equiv \frac{\int w(a)d\ln a}{\int {d\ln a} }. $$

Transform the initial expression for the dark energy density to obtain $$ \frac{\rho}{\rho_0} = e^{ - 3\int (1 + w(a))d\ln a } = \left( e^{\ln a} \right)^{\frac{- 3\int (1 + w(a))d\ln a}{\int d\ln a} } = a^{ - 3\left(1 + \frac{\int w(a)d\ln a}{\int d\ln a}\right) }. $$ The former expression directly follows from the conservation equation for the dark energy.

### Problem 14

Consider the case of Universe filled with non-relativistic matter and quintessence with the state equation $p=w\rho$ and show that the first Friedman equation can be presented in the form \[H^2(z)=H_0^2\left[\Omega_{m0}(1+z)^3+(1-\Omega_{m0})e^{3\int_0^z\frac{dz'}{1+z'}(1+w(z'))}\right].\]

### Problem 15

Show that for the model of the Universe considered in the previous problem the state equation parameter $w(z)$ can be presented in the form \[w(z)=\frac{\frac23(1+z)\frac{d\ln H}{dz}-1}{1-\frac{H_0^2}{H^2}\Omega_{m0}(1+z)^3}.\]

Using the result of the previous problem, one finds \[\frac{\frac{H^2}{H_0^2}-\Omega_{m0}(1+z)^3}{1-\Omega_{m0}}=e^{3\int_0^z\frac{dz'}{1+z'}(1+w(z'))}.\] Taking logarithm and derivative with respect to $z$, one obtains \[\frac13\frac{d}{dz}\left[\ln\left(\frac{\frac{H^2}{H_0^2}-\Omega_{m0}(1+z)^3}{1-\Omega_{m0}}\right)\right]=\frac{1+w(z)}{1+z},\] and finally \[w(z)=\frac{\frac23(1+z)\frac{d\ln H}{dz}-1}{1-\frac{H_0^2}{H^2}\Omega_{m0}(1+z)^3}.\]

### Problem 16

Show that the result of the previous problem can be presented in the form \[w(z)=-1+(1+z)\frac{2/3E(z)E'(z)-\Omega_{m0}(1+z)^2}{E^2(z)-\Omega_{m0}(1+z)^3},\quad E(z)\equiv\frac{H(z)}{H_0}.\]

The standard Friedman equation for a flat FLRW model \[E^2(z)=\Omega_{m0}(1+z)^3+(1-\Omega_{m0})\exp\left(3\int\limits_0^z\frac{1+w(z')}{1+z'}dz'\right)\] can be represented in the form \[\int\limits_0^z\frac{1+w(z')}{1+z'}dz'=\frac13\ln\left(\frac{E^2(z)-\Omega_{m0}(1+z)^3}{1-\Omega_{m0}}\right).\] On differentiating both sides with respect to $z$ one obtains the required relation.

### Problem 17

Show that decreasing of the scalar field's energy density with increasing of the scale factor slows down as the scalar field's potential energy $V(\varphi)$ starts to dominate over the kinetic energy density $\dot\varphi^2/2$.

$$ \rho _\varphi \sim a^{ - 3(1 + w_\varphi )} ;\quad w_\varphi = {{p_\varphi } \over {\rho _\varphi }} = {{{1 \over 2}\dot \varphi ^2 - V\left( \varphi \right)} \over {{1 \over 2}\dot \varphi ^2 + V\left( \varphi \right)}}. $$ When $V(\varphi)\gg\dot\varphi^2$ one has $w_\varphi\to-1$, which leads to small expression in the exponent for $\rho_\varphi$.

### Problem 18

Express the time derivative $\dot\varphi$ through the quintessence' density $\rho_\varphi$ and the state equation parameter $w_\varphi$.

Use the expressions for density and pressure of the scalar field $$ \rho_\varphi = {1 \over 2}\dot \varphi ^2 + V\left( \varphi \right);\ p_\varphi = {1 \over 2}\dot \varphi ^2 - V\left( \varphi \right);\ p_\varphi =w\rho_\varphi $$ to obtain $$ \dot \varphi = \sqrt {\left( {1 + w_\varphi } \right)\rho _\varphi }. $$

### Problem 19

Estimate the magnitude of the scalar field variation $\Delta\varphi$ during time $\Delta t$.

The variation of the scalar field at the time interval $\Delta t$ can be estimated as $$ \Delta \varphi \approx \dot \varphi \Delta t. $$ Using the results of the previous problem one finds $$ \Delta \varphi \approx \sqrt {\rho _\varphi (1 + w)} \Delta t. $$ For the entire time of the evolution of the Universe it equals to $$ \Delta \varphi \approx \sqrt {\rho _\varphi (1 + w)} H_0^{-1}. $$

### Problem 20

Show that in the radiation-dominated or matter-dominated epoch the variation of the scalar field is small, and the measure of its smallness is given by the relative density of the scalar field.

Use the result of the previous problem: $$ \Delta \varphi \approx \sqrt {\rho _\varphi (1 + w)} \Delta t. $$ If the 'Hubble friction' $3H\dot\varphi$ is small (which is true in the radiation-dominated or matter-dominated epoch), then one can take the Hubble time $H^{ - 1} $ as the characteristic time scale, so that $$ \Delta \varphi \approx \sqrt {(1 + w)\rho _\varphi /H^2 } \approx \sqrt {(1 + w)\Omega _\varphi }, $$ which proves the required statement.

### Problem 21

Show that in the quintessence $(w>-1)$ dominated Universe the condition $\dot{H}<0$ always holds.

$$w = - 1 - {2 \over 3}{{\dot H} \over {H^2 }},$$ so $\dot{H}<0$ always holds for $w>-1$.

### Problem 22

Consider simple bouncing solution of Friedman equations that avoid singularity. This solution requires positive spatial curvature $k=+1$, negative cosmological constant $\Lambda<0$ and a "matter" source with equation of state $p=w\rho$ with $w$ in the range \[-1<w<-\frac13.\]
In the special case $w=-2/3$ Friedman equations describe a constrained harmonic oscillator (a simple harmonic Universe). Find the corresponding solutions.

(Inspired by P.Graham et al. arXiv:1109.0282)

Let $8\pi G=1$. The energy density is \[\rho=\Lambda+\rho_0a^{-3(1+w)}=-|\Lambda|+\frac{\rho_0}{a}.\] The second Friedman equation in this case is \[\ddot a+\frac13|\Lambda|a=\frac{\rho_0}{6}.\] This equation describes a constrained simple harmonic oscillator and the solution is \[a=\frac{\rho_0}{2\Lambda}+a_0\cos(\omega t+\psi),\] where $\psi$ is an arbitrary phase and \[\omega=\sqrt{\frac{|\Lambda|}{3}}.\] The amplitude $a_0$ can be found from the first Friedman equation \[a_0=\frac{1}{|\Lambda|}\sqrt{\frac{\rho_0^2}{4}-3|\Lambda|}.\] This requires $\rho_0^2\ge12|\Lambda|$ for positivity of the radicand.

### Problem 23

Derive the equation for the simple harmonic Universe (see previous problem), using the results of problem #DE04.

### Problem 24

Barotropic liquid is a substance for which pressure is a single--valued function of density. Is quintessence generally barotropic?

No, it is not, as there are such potentials $V(\varphi)$, for which $p_\varphi$ is not single-valued function of $\rho_\varphi$.

### Problem 25

Show that a scalar field oscillating near the minimum of potential is not a barotropic substance.

In the case of a scalar field oscillating near the minimum of a potential, pressure is not a single-valued function of the density, so it is not a barotropic substance. Indeed, such field has zero value of the kinetic term in each extremal point of the oscillations and zero value of the potential energy at the minimum point, which means that in terms of the state equation the ratio $w=p_\varphi/\rho_\varphi$ equals $-1$ and $+1$ respectively. So $p_\varphi$ passes through zero twice per each oscillation period, each time with lesser value of $\rho_\varphi$, because $\rho_\varphi$ decreases with expansion of the Universe.

### Problem 26

For a scalar field $\varphi$ with state equation $p=w\rho$ and relative energy density $\Omega_\varphi$ calculate the derivative \[w'=\frac{dw}{d\ln a}.\]

First, one can use the definitions of density and pressure for the scalar field $$ \rho_\varphi = \frac12\dot\varphi^2 + V(\varphi);\quad p_\varphi = \frac12\dot\varphi^2 - V(\varphi);\quad p_\varphi =w\rho_\varphi $$ to express the potential and kinetic energy of the field through the density and the state parameter $w$ in the form $$ V = \frac12(1-w)\rho ;\quad \frac12\dot\varphi^2 = \frac12(1 + w)\rho. $$ Then \[w=1-\frac{2V}{\rho}\quad \Rightarrow \quad w' \equiv {{dw} \over {d\ln a}} \equiv \frac{\dot w}{H}=\frac1H\left(\frac{2V\dot\rho}{\rho^2} - \frac{2\dot V}{\rho}\right).\] Taking into account that $\rho=2V/(1-w)$ and $\dot\rho=-3H(1+w)\rho,$ one obtains \[\frac{2\dot V}{\rho}=\frac{dV}{d\varphi}\frac{\dot\varphi}{V}(1-w),\] \[\frac{2V\dot\rho}{\rho^2}=-3H(1+w)(1-w).\] Substitution of $\dot\varphi=\sqrt{(1+w)\rho}$ and \[\rho=\frac{3H^2 M_{Pl}^2}{8\pi}\Omega_\varphi,\] finally yields $$ w' = - 3\left( {1 - w^2 } \right) - M_{Pl}\sqrt {\frac{3\Omega _\varphi(1 + w)}{8\pi} }( 1 - w ) \frac{dV/d\varphi }{V}. $$

### Problem 27

Calculate the sound speed in the quintessence field $\varphi(t)$ with potential $V(\varphi)$.

$$ c_{s\varphi }^2 = {{dp_\varphi } \over {d\rho _\varphi }} = {{\dot p_\varphi } \over {\dot \rho _\varphi }};\quad p_\varphi = {1 \over 2}\dot \varphi ^2 - V(\varphi );\quad \rho _\varphi = {1 \over 2}\dot \varphi ^2 + V(\varphi ); $$ $$ c_{s\varphi }^2 = 1 + {2 \over 3}{{V'(\varphi )} \over {H\dot \varphi }} $$

### Problem 28

Find the dependence of quintessence energy density on redshift for the state equation $p_{DE}=w(z)\rho_{DE}$.

\[\rho_{DE}(z)=\rho_{DE}(z=0)\exp \left[ 3\int\limits_0^z dz'\frac{1+w(z')}{1+z'}\right].\]

### Problem 29

The equation of state $p=w(a)\rho$ for quintessence is often parameterized as $w(a)=w_0 + w_1(1-a)$. Show that in this parametrization energy density and pressure of the scalar field take the form: $$ \rho(a) \propto a^{-3[1+w_{\it eff}(a)]},\quad p(a) \propto (1+w_{\it eff}(a))\rho(a), $$ where $$ w_{\it eff}(a)=(w_0+w_1)+(1-a)w_1/\ln a. $$

Make use of the relation: \[\rho(a)\propto\exp\left\{-3\int\limits_0^a\frac{da'}{a'}(1+w(a'))\right\}.\]

### Problem 30

Find the dependence of Hubble parameter on redshift in a flat Universe filled with non-relativistic matter with current relative density $\Omega_{m0}$ and dark energy with the state equation $p_{DE}=w(z)\rho_{DE}$.

\[ \frac{H^2(z)}{H_0^2} = \Omega_{m0}(1 + z)^3 + (1 - \Omega_{m0})\exp \left[ 3\int\limits_0^z dz'\frac{1+w(z')}{1+z'}\right]. \]

### Problem 31

Show that in a flat Universe filled with non--relativistic matter and arbitrary component with the state equation $p=w(z)\rho$ the first Friedman equation can be presented in the form: \[w(z)=-1+\frac13\frac{d\ln(\delta H^2/H_0^2)}{d\ln(1+z)},\] where \[\delta H^2 = H^2 - \frac{8\pi G}{3}\rho_m\] describes the contribution into the Universe's expansion rate of all components other than matter.

Use the result of the previous problem to obtain \[ \ln\left(\frac{\delta H^2}{H_0^2}\right) = \ln(1 - \Omega _{m0}) + 3\int\limits_0^z [1+w(z')]d[\ln(1+z')]. \] The required expression for $w(z)$ can be obtained by differentiation of the latter with respect to $\ln(1+z)$.

### Problem 32

Express the time derivative of a scalar field through its derivative with respect to redshift $d\varphi/dz.$

\[\frac{d\varphi}{dt}=\frac{d\varphi}{da}\frac{da}{dt}=\frac{d\varphi}{dz}\frac{dz}{da}\frac{da}{dt}=\frac{d\varphi}{dz} \left(-\frac{1}{a^2}\right)aH=- (1 + z)H\frac{d\varphi } {dz};\] $$ \frac{d\varphi } {dt} = - (1 + z)H\frac{d\varphi } {dz}. $$

### Problem 33

Show that the particle horizon does not exist for the case of quintessence because the corresponding integral diverges (see Chapter 2(3)).

### Problem 34

Show that in a Universe filled with quintessence the number of observed galaxies decreases with time.

The Hubble sphere's surface recedes with velocity equal to $V_H=(1+q)c$. For quintessence, with $-1<w<-1/3$, the deceleration parameter is \[q=\frac12(1+3w)<0,\] so $V_H<c$. Thus the galaxies on the Hubble sphere, which are receding with the velocity $V_g=c$, 'overtake' the surface itself.

### Problem 35

Let $t$ be some time in the distant past $t\ll t_0$. Show that in a Universe dominated by a substance with state parameter $w>-1$ the current cosmic horizon (see Chapter 3) is \[R_h(t_0)\approx\frac32(1+\langle w\rangle)t_0,\] where $\langle w\rangle$ is the time-averaged value of $w$ from $t$ to the present time \[\langle w\rangle\equiv\frac{1}{t_0}\int\limits_t^{t_0} w(t)dt.\]

From definition \[\dot R_h=\frac32(1+w).\] Then, integrating from $t$ to $t_0$, we find that \[R_h(t_0)-R_h(t)=\frac32(1+\langle w\rangle)t_0-\frac32t.\] Now, for any $w>-1$, $\rho$ drops as the universe expands and since $R_h\propto\rho^{-1/2}$, clearly $R_h(t)\ll R_h(t_0)$. Therefore, \[R_h(t_0)\approx\frac32(1+\langle w\rangle)t_0.\]

### Problem 36

From WMAP$^*$ observations we infer that the age of the Universe is $t_0\approx13.7\cdot10^9$ years and cosmic horizon equals to $R_h(t_0)=H_0^{-1}\approx13.5\cdot10^9$ light-years. Show that these data imply existence of some substance with equation of state $w<-1/3$, - "dark energy".

$^*$ Wilkinson Microwave Anisotropy Probe is a spacecraft which measures differences in the temperature of the Big Bang's remnant radiant heat - the cosmic microwave background radiation - across the full sky.

Use the result of previous problem \[R_h(t_0)\approx\frac32(1+\langle w\rangle)t_0.\] For $R_h(t_0)=H_0^{-1}\approx13.5\cdot10^9$ light-years and $t_0\approx13.7\cdot10^9$ years one obtains $\langle w\rangle<-1/3$. This result was obtained independently of the Type Ia supernova data. But an analysis of the latter reveals that values $\langle w\rangle\ge-1/3$ are apparently ruled out, so in fact $w<-1/3$.

### Problem 37

The age of the Universe today depends upon the equation-of-state of the dark energy. Show that the more negative parameter $w$ is, the older Universe is today.

The age of the Universe is \[t_0=\int\limits_0^{t_0}dt\int\limits_0^\infty\frac{dz}{(1+z)H(z)},\] where $H(z)$ depends upon the equation-of-state parameter of the dark energy $w$, so that \[ H_0t_0=\int\limits_0^\infty\frac{dz}{(1+z)^{5/2}\sqrt{\Omega_m +\Omega_{DE}(1+z)^{3w}}}.\] The more negative $w$ is, the more accelerated the expansion is and the older Universe is today for the fixed $H_0$.

### Problem 38

Consider a Universe filled with dark energy with state equation depending on the Hubble parameter and its derivatives, \[p=w\rho+g(H,\dot H, \ddot H,\ldots,;t).\] What equation does the Hubble parameter satisfy in this case?

Use the Friedman equation in the form $$H^2 = \frac{\kappa ^2 } 3\rho ,\quad \dot H = - \frac{\kappa^2 }2\left( {\rho + p} \right);\quad \kappa ^2 = 8\pi G, $$ to obtain the following equation for the Hubble parameter: $$ \dot H + \frac 3 2 (1 + w)H^2 + \frac{\kappa ^2 } {2} g(H,\dot H,\ddot H, \cdots ;t) = 0. $$

### Problem 39

Show that taking function $g$ (see the previous problem) in the form \[g(H,\dot H, \ddot H)=-\frac{2}{\kappa^2}\left(\ddot H + \dot H + \omega_0^2 H + \frac32(1+w)H^2-H_0\right),\ \kappa^2=8\pi G\] leads to the equation for the Hubble parameter identical to the one for the harmonic oscillator, and find its solution.

Direct substitution of $g$ into the equation for the Hubble parameter, obtained in the previous problem, gives the equation $$ \ddot H + \omega _0^2 H = H_0. $$ Its solution is $$ H(t) = \frac{H_0 } {\omega _0^2 } + H_1 \sin (\omega _0 t + \delta _0 ) $$ where $H_1$ and $\delta _0 $ are integration constants. (see D.Saez-Gomez, arXiv:0804.4586, hep-th).

### Problem 40

Find the time dependence of the Hubble parameter in the case of function $g$ (see problem #DE64) in the form \[g(H;t)= -\frac{2\dot f(t)}{\kappa^2f(t)}H,\ \kappa^2=8\pi G\] where $f(t)=-\ln(H_1+H_0\sin\omega_0t)$, $H_1>H_0$ is arbitrary function of time.

Such choice of the function $g$ reduces the equation for the Hubble parameter to the form $$ \dot H - \frac{\dot f(t)}{f(t)}H + \frac{3} {2}(1 + w)H^2 = 0 ,$$ which is nothing but the well-known Bernoulli equation $\dot y +f(t) y=d(t)y^\alpha$. For the case $f(t)=-\ln(H_1+H_0\sin\omega_0t)$, $H_1>H_0$ its solution reads $$ H(t) = \frac{H_1+H_0\sin\omega_0t} {\frac 32(1 + w) + \kappa}, $$ where $\kappa$ is the integration constant.

### Problem 41

Show that in an open Universe the scalar field potential $V[\varphi(\eta)]$ depends monotonically on the conformal time $\eta$.

\[V(\varphi)=\frac12(\rho_\varphi-p_\varphi)=\frac{1-w_\varphi}{2}\rho-\varphi;\] From the conservation equation in terms of conformal time $$ \rho '_\varphi (\eta ) + 3{{a'} \over a}(1 + w_\varphi )\rho _\varphi (\eta ) = 0 $$ one obtains \[ \rho _\varphi (\eta ) = \rho _\varphi (\eta _0 )\left( {{{a_0 } \over {a(\eta )}}} \right)^{3\left( {1 + w_\varphi } \right)}, \] and finally $$ V\left[ {\varphi (\eta )} \right] = {{1 - w_\varphi } \over 2}\rho _\varphi (\eta _0 )\left( {{{a_0 } \over {a(\eta )}}} \right)^{3\left( {1 + w_\varphi } \right)}. $$ It follows that $ V\left[ {\varphi (\eta )} \right] $ decreases monotonically with growing $\eta$, because $a(\eta)$ increases monotonically in the open Universe.

### Problem 42

Reconstruct the dependence of the scalar field potential $V(a)$ on the scale factor basing on given dependencies for the field's energy density $\rho_\varphi(a)$ and state equation parameter $w(a)$.

\begin{align*} \rho _\varphi & = \frac12\dot\varphi^2 + V(\varphi ); & w & = \frac{p_\varphi}{\rho_\varphi} = \frac{ \frac12\dot\varphi^2 - V(\varphi)}{\frac12\dot\varphi^2 + V(\varphi )};\\ V & = \frac{\dot\varphi^2}{2}\frac{1 - w}{1 + w}; & V & = \left( \rho_\varphi - V \right)\frac{1 - w}{1 + w}; \end{align*} $$ V(a) = [1 - w(a)]\rho_\varphi (a). $$

### Problem 43

Find the quintessence potential providing the power law growth of the scale factor $a\propto t^p$, where the accelerated expansion requires $p>1$.

Use Friedman equations for the Universe filled with quintessence to express $V(\varphi)$ and $\dot\varphi$ in terms of $H$ and $\dot H$. It allows one to obtain the system of equations to describe the parametric dependence of $V$ on $\varphi$: \begin{align*} V & =\frac{3H^2}{8\pi G}\left(1+\frac{\dot H}{3H^2}\right);\\ \varphi & =\int dt\left(-\frac{\dot H}{4\pi G}\right)^{1/2}. \end{align*} For the case $a\propto t^p$ the second equation leads to \[\frac{\varphi}{M_{Pl}}=\sqrt{\frac{p}{4\pi}}\ln t,\quad M_{Pl}^2\equiv\frac1G.\] Excluding the time variable from the expression for the potential one obtains \[V(\varphi)=V_0\exp\left(-\sqrt{\frac{16\pi}{p}}\frac{\varphi}{M_{Pl}}\right).\] Taking into account that accelerated expansion requires $p>1$, the obtained result can be treated in the following way: the scalar field in the potential obtained above can be considered dark energy, i.e. a substance that provides the accelerated expansion.

### Problem 44

Let $a(t)$, $\rho(t)$, $p(t)$ be solutions of Friedman equations. Show that for the case $k=0$ the function $\psi_n\equiv a^n$ is the solution of "Schr\"odinger equation" $\ddot\psi_n=U_n\psi_n$ with potential [see A.V.Yurov, arXiv:0905.1393] \[U_n=n^2\rho-\frac{3n}{2}(\rho+p).\]

\[\frac{\ddot\psi_n}{\psi_n}=n(n-1)\left(\frac{\dot a}{a}\right)^2+n\frac{\ddot a}{a}.\] Using the relations \[\left(\frac{\dot a}{a}\right)^2=\rho,\quad \frac{\ddot a}{a}=-\frac12(\rho+3p),\quad \left(\frac{8\pi G}{3}=1\right),\] one finds \[U_n=n^2\rho-\frac{3n}{2}(\rho+p).\]

### Problem 45

Consider flat FLRW Universe filled with a scalar field $\varphi$. Show that in the case when $\varphi=\varphi(t)$, the Einstein equations with the cosmological term are reduced to the "Schrödinger equation" \[\ddot\psi=3(V+\Lambda)\psi\] with $\psi=a^3$. Derive the equation for $\varphi(t)$ (see A.V.Yurov, arXiv:0305019).

\[\frac{\ddot\psi}{\psi}=6\left(\frac{\dot a}{a}\right)^2+3\frac{\ddot a}{a}=6\left(\frac13\rho +\frac\Lambda3\right) +3\left(-\frac16(\rho+3p)\frac\Lambda3\right)=3\left(\frac12\rho-\frac12 p+\Lambda\right).\] Here \(8\pi G=1\). Using the expressions for the scalar field \[\rho=\frac12\dot\varphi^2+V,\quad p=\frac12\dot\varphi^2-V,\] one obtains \[\ddot\psi=3(V+\Lambda)\psi.\] Then \[\dot\varphi^2=-2\frac{d}{dt}\frac{\dot a}{a}=-\frac23\frac{d^2}{dt^2}\ln a^3=-\frac23\frac{d^2}{dt^2}\ln\psi.\]

### Problem 46

Consider FLRW space-time filled with non-interacting matter and dark energy components. Assume the following forms for the equation of state parameters of matter and dark energy \[w_m=\frac{1}{3(x^\alpha+1)},\quad w_{DE}=\frac{\bar{w}x^\alpha}{x^\alpha+1},\] where $x=a/a_*$ with $a_*$ being some reference value of $a$, $\alpha$ is some positive constant and $\bar{w}$ is a negative constant. Analyze the dynamics of the Universe in this model. (see S.Kumar,L.Xu, arXiv:1207.5582)

Using conservation equations for matter and dark energy
\[\dot\rho_m+3H\rho_m(1+w_m)=0,\quad \dot\rho_{DE}+3H\rho_{DE}(1+w_{DE})=0,\]
one finds
\begin{align*}
\rho_m & =\frac{C_1(x^\alpha+1)^{1/\alpha}}{x^4}, & \rho_{DE} & = \frac{C_2(x^\alpha+1)^{-3\bar{w}/\alpha}}{x^3},\\
p_m & =\frac{C_1(x^\alpha+1)^{(1-\alpha)/\alpha}}{3x^4}, & p_{DE} & =\frac{C_2\bar{w}(x^\alpha+1)^{-(3\bar{w}+\alpha)/\alpha}}{x^3},
\end{align*}
where $C_1$ and $C_2$ are positive integration constants.

For $x\ll1$
\[\rho_m\approx\frac{C_1}{x^4},\quad p_m=\frac13\rho_m,\quad \rho_{DE}\approx\frac{C_2}{x^3}.\]
In this case matter (relativistic) dominates over dark energy, as expected.

For $x\gg1$
\[\rho_m\approx\frac{C_1}{x^3},\quad p_m=\frac{C_1}{3x^{(3+\alpha)}}\]
Matter density decreases exactly as in the cosmological dust model. Since $\alpha>0$, $\rho_m\gg p_m$ as in the (non-relativistic) matter dominated Universe. Furthermore, in this limit
\[\rho_{DE}\gg\rho_m,\quad |p_{DE}|\gg p_m.\]
Thus the considered model describes evolution of the Universe from the early radiation-dominated phase to the present dark energy-dominated phase.

## Tracker Fields

*A special type of scalar fields - the so-called tracker fields - was discovered at the end of the nineties. The term reflects the fact that a wide range of initial values for the fields of such type rapidly converges to the common evolutionary track. The initial values of energy density for such fields may vary by many orders of magnitude without considerable effect on the long-time asymptote. The peculiar property of tracker solutions is the fact that the state equation parameter for such a field is determined by the dominant component of the cosmological background.*

*It should be stressed that, unlike the standard attractor, the tracker solution is not a fixed point (in the sense of a solution corresponding to the fixed point in a system of autonomous differential equations ): the ratio of the scalar field energy density to that of background component (matter or radiation) continuously changes as the quantity $\varphi$ descends along the potential. It is well desirable feature because we want the energy density $\varphi$ to exceed ultimately the background density and to transfer the Universe into the observed phase of the accelerated expansion.*

*Below we consider a number of concrete realizations of the tracker fields.*

### Problem 47

Show that initial value of the tracker field should obey the condition $\varphi_0=M_{Pl}$.

### Problem 48

Show that densities of kinetic and potential energy of the scalar field $\varphi$ in the potential of the form \[V(\varphi)=M^4\exp(-\alpha\varphi M),\quad M\equiv\frac{M_{PL}^2}{16\pi}\] are proportional to the density of the concomitant component (matter or radiation) and therefore it realizes the tracker solution.

The Klein-Gordon equation in the considered case takes the form $$\ddot \varphi + {2 \over {(1 + w)t}}\dot \varphi - \alpha M^3 \exp ( - {{\alpha \varphi } {M) = 0}}.$$ A particular solution can be sought in the form $\varphi = A\ln t$, and substitution of this into the first equation leads to conditions: $$\left\{ {\begin{array}{rcl} A\frac{\alpha }{M} & = & 2 \\ \\ A\frac{1 - w}{1 + w} & = & \alpha M^3 \end{array}} \right.,\quad A^2 = 2M^4 \frac{1 + w}{1 - w}.$$ Then one finds $$\rho = \frac{2M^4 }{1 - w}t^{ - 2}, \quad p = \frac{2M^4 w}{1 - w}t^{ - 2} $$ and therefore $$w_\varphi = w.$$

### Problem 49

Consider a scalar field potential \[V(\varphi)=\frac A n\varphi^{-n},\] where $A$ is a dimensional parameter and $n>2$. Show that the solution $\varphi(t)\propto t^{2/(n+2)}$ is a tracker field under condition $a(t)\propto t^m$, $m=1/2$ or $2/3$ (either radiation or non-relativistic matter dominates).

### Problem 50

Show that the scalar field energy density corresponding to the tracker solution in the potential \[V(\varphi)=\frac A n\varphi^{-n}\] (see the previous problem #DE73) decreases slower than the energy density of radiation or non-relativistic matter.

\begin{align*} V & \propto \varphi ^{ - n} ;\\ \varphi & \propto t^\alpha = t^{\frac{2}{n + 2}} ;\\ \dot \varphi & \propto t^{\frac{2}{n + 2} - 1} = t^{ - \left( {\frac{n}{n + 2}} \right)} ;\\ \dot \varphi ^2 & \propto t^{ - \frac{2n}{n + 2}} = \varphi ^{ - n}; \end{align*} $$\rho _\varphi = \frac{1}{2}\dot \varphi ^2 + V(\varphi ) \propto \varphi ^{ - n} = t^{ - 2(1 - \alpha )}. $$ Recall that the energy density decreases as $t^{ - 2} $ both for matter and radiation. Then the condition $\alpha > 0$ proves the required statement.

### Problem 51

Find the equation of state parameter $w_\varphi\equiv p_\varphi/\rho_\varphi$ for the scalar field of problem #DE73.

As is known, \[ \rho \propto a^{ - 3(1 + w)} = t^{ - 3m(1 + w)}. \] Then \[ 2(1 - \alpha ) = 3m(1 + w),\, \alpha = \frac{2}{n + 2}, \] and \[ w_\varphi = - \frac{2}{3}\frac{(1 - \alpha )}{m} - 1. \] Recall that \[a \propto t^{\frac{2}{3(1 + w)}} \] and therefore \begin{align*} m & = \frac{2}{3(1 + w)},\\ w_\varphi & = w\frac{n}{n + 2} - \frac{2}{n + 2}. \end{align*}

### Problem 52

Use explicit form of the tracker field in the potential of problem #DE73 to verify the value of $w_\varphi$ obtained in the previous problem.

The expression \[w_\varphi = w\frac{n}{n + 2} - \frac{2}{n + 2}\] can be checked using the explicit form of the solution for the scalar field and the following definition \[ w_\varphi = \frac{p_\varphi }{\rho _\varphi } = \frac{\frac{1}{2}\dot \varphi ^2 - V(\varphi )}{\frac{1}{2}\dot \varphi ^2 + V(\varphi )}. \] Substitute the explicit expressions for corresponding quantities to obtain $$ w_\varphi = \frac{\frac{1}{2}\left( {\alpha Ct^{\alpha - 1} } \right)^2 - A\frac{{C^{ - n} }}{n}t^{ - \frac{2n}{n + 2}}} {\frac{1}{2}\left( {\alpha Ct^{\alpha - 1} } \right)^2 + A\frac{{C^{ - n} }}{n}t^{ - \frac{2n}{n + 2}}} = \frac{\frac{1}{2}\alpha ^2 - A\frac{C^{ - (n + 2)} }{n}}{\frac{1}{2}\alpha ^2 + A\frac{C^{ - (n + 2)}}{n}}. $$ Then recall that \[ C = \left( {\frac{A(1 + w)}{(1 + w)\alpha ^2 + \alpha (1 - w)}} \right)^{1/n + 2} \] and \[\alpha = \frac{2}{n + 2}\] and rewrite it in the form $$AC^{ - (n + 2)} = \frac{2(4 - nw + n)}{(1 + w)(n + 2)^2},$$ to finally obtain \[ w_\varphi = w\frac{n}{n + 2} - \frac{2}{n + 2}. \]

## The K-essence

Let us introduce the quantity $$X\equiv \frac{1}{2}{{g}^{\mu \nu }}\frac{\partial \varphi }{\partial {{x}^{\mu }}}\frac{\partial \varphi }{\partial {{x}^{\nu }}}$$ and consider action for the scalar field in the form $$ S=\int{{{d}^{4}}x\sqrt{-g}}\; L\left( \varphi ,X \right), $$ where Lagrangian $L$ is generally speaking an arbitrary function of variables $\varphi$ and $X.$ The dark energy model realized due to modification of the kinetic term with the scalar field, is called the $k$-essence. The traditional action for the scalar field corresponds to $$ L\left( \varphi ,X \right)=X-V(\varphi ). $$ In the problems proposed below we restrict ourselves to the subset of Lagrangians of the form $$ L\left( \varphi ,X \right)=K(X)-V(\varphi ), $$ where $K(X)$ is a positively defined function of kinetic energy $X$. In order to describe a homogeneous Universe we should choose $$ X=\frac{1}{2}{\dot{\varphi}^{2}}. $$

### Problem 53

Find the density and pressure of the $k$-essence.

Using the standard definitions \[T_{\mu\nu}=\frac{2}{\sqrt{-g}}\frac{\delta S}{\delta g^{\mu\nu}},\quad \rho_\varphi=T_{00},\quad p_\varphi=T_{ii};\] one finds \[p_\varphi=K(X)-V(\varphi);\] \[\rho_\varphi=2X\frac{\partial K(X)}{\partial X}-K(X) +V(\varphi).\]

### Problem 54

Construct the equation of state for the $k$-essence.

\[{{w}_{\varphi }}=\frac{K(X)-V(\varphi )}{2X\frac{\partial K(X)}{\partial X}-K(X)+V(\varphi )}.\]

### Problem 55

Find the sound speed in the $k$-essence.

$$ c_{s}^{2}=\frac{\frac{\partial {{p}_{\varphi }}}{\partial X}}{\frac{\partial {{\rho }_{\varphi }}}{\partial X}}=\frac{K(X)}{2X\frac{{{\partial }^{2}}K}{\partial {{X}^{2}}}+\frac{\partial K}{\partial X}}. $$

### Problem 56

The sound speed $c_s$ in any medium must satisfy two fundamental requirements: first, the sound waves must be stable and second, its velocity value should be low enough to preserve the causality condition. Therefore \[0\le c_s^2\le1.\] Reformulate the latter condition in terms of scale factor dependence for the equation of state parameter $w(a)$ for the case of the $k$-essence.

Present the expression for the sound speed in the form: $$ c_{s}^{2}=\frac{dp/da}{d\rho/da} $$ and use relations \[\frac{d\rho}{da}=-3\frac{3(1+w)}{a}\rho;\] \[\frac{dp}{da}=\frac{dw}{da}\rho+w\frac{d\rho}{da}=\frac{a\frac{dw}{da}-3w(1+w)}{a}\rho;\] to obtain \[c_s^2=\frac{3w(1+w)-w'}{3(1+w)}>0,\] where \[w'\equiv\frac{dw}{d\ln a}.\] The quintessence range is limited by the condition \[-1\le w\le-\frac13.\] The stability condition for the sound waves should be satisfied in the range, leading to relation \[w'\le3w(1+w).\] The requirement $c_s^2\le1$ provides validity of the causality principle and leads to the additional bound \[w'\ge-3(1-w^2).\]

### Problem 57

Find the state equation for the simplified $k$-essence model with Lagrangian $L=F(X)$ (the so-called pure kinetic $k$-essence).

\[\begin{aligned} & {{\rho }_{\varphi }}=2X{{F}_{X}}-F;\quad {{F}_{X}}\equiv \frac{\partial F}{\partial X} \\ & p=F; \\ & {{w}_{\varphi }}=\frac{F}{2X{{F}_{X}}-F} .\\ \end{aligned}\]

### Problem 58

Find the equation of motion for the scalar field in the pure kinetic $k$-essence.

The equation of motion can be obtained either from the Euler-Lagrange equations for the scalar field action or by substitution of the explicit expressions for density and pressure into the conservation equation for the $k$-essence. The result is $$ F_X\ddot{\varphi }+{{F}_{XX}}\dot{\varphi}^2\ddot{\varphi }+3H{F}_{X}\dot{\varphi}=0, $$ or in terms of the kinetic energy $X$ $$ \left( {{F}_{X}}+2{{F}_{XX}}X \right)\dot{X}+6H{{F}_{X}}X=0. $$

### Problem 59

Show that the scalar field equation of motion for the pure kinetic $k$-essence model gives the tracker solution.

The equation of motion for the scalar field in the pure kinetic $k$-essence model (see the previous problem) can be integrated to give $$ XF_{X}^{2}=k{{a}^{-6}}, $$ where the constant $k>0.$ The solution $X(a)$ has one important property: behavior of all $k$-essence characteristics (${{\rho }_{\varphi }},{{p}_{\varphi }},{{w}_{\varphi }}$) as functions of the scale factor is completely determined solely by the function $F(X)$ and they do not depend on evolution of energy densities of other components. All the dependence of the $k$-essence on other components comes from $a(t)$ only. But the latter reads $a(t)\propto {{\rho }_{tot}},$ so it is determined by the energy density of the dominant component. Therefore the solution ${{\rho }_{\varphi }}$ belongs to the type of the tracker solutions, which were presented before for the quintessence model in a specially designed potential.

## Phantom Energy

*The full amount of available cosmological observational data shows that the state equation parameter $w$ for dark energy lies in a narrow range near the value $w=-1$. In the previous subsections we considered the region $-1\le w\le-1/3$. The lower bound $w=-1$ of the interval corresponds to the cosmological constant, and all the remainder can be covered by the scalar fields with canonic Lagrangians. Recall that the upper bound $w=-1/3$ appears due to the necessity to provide the observed accelerated expansion of Universe. What other values of parameter $w$ can be used? The question is very hard to answer for the energy component we know so little about. General Relativity restricts possible values of the energy - momentum tensor by the so-called "energy conditions" (see Chapter 2). One of the simplest among them is the so-called Null Dominant Energy Condition (NDEC) $\rho+p\ge0$. The physical motivation of the latter is to avoid the vacuum instability. Applied to the dynamics of Universe, the NDEC requires that density of any allowed energy component cannot grow with the expansion of the Universe. The cosmological constant with $\dot\rho_\Lambda=0$, $\rho_\Lambda=const$ represents the limiting case. Because of our ignorance concerning the nature of dark energy it is reasonable to question whether this mysterious substance can differ from the already known "good" sources of energy and if it can violate the NDEC. Taking into account that dark energy must have positive density (it is necessary to make the Universe flat) and negative pressure (to provide the accelerated expansion of Universe), the violation of the NDEC must lead to $w<-1$. Such substance is called the phantom energy. The phantom field $\varphi$ minimally coupled to gravity has the following action:*
\[S=\int d^4x \sqrt{-g}L=-\int d^4x \sqrt{-g}\left[\frac12g^{\mu\nu}\frac{\partial\varphi}{\partial x_\mu} \frac{\partial\varphi}{\partial x_\nu}+V(\varphi)\right],\]
*which differs from the canonic action for the scalar field only by the sign of the kinetic term.*

### Problem 60

Show that the action of a scalar field minimally coupled to gravitation \[S=\int d^4x\sqrt{-g}\left[\frac12(\nabla\varphi)^2-V(\varphi)\right]\] leads, under the condition $\dot\varphi^2/2<V(\varphi)$, to $w_\varphi<-1$, i.e. the field is phantom.

### Problem 61

Obtain the equation of motion for the phantom scalar field described by the action of the previous problem.

Proceeding like in problem \ref{inf5} of Chapter Inflation, one obtains $$ \delta S = \int {d^4 x\sqrt { - g} } \left[ { \left( {\nabla ^\mu \nabla _\mu \varphi } \right) - V_{,\varphi } (\varphi )} \right]\delta \varphi - \int {d^4 x\sqrt { - g} } \nabla _\mu \left( {\delta \varphi \nabla ^\mu \varphi } \right) $$ and therefore $$ \left( {\nabla ^\mu \nabla _\mu \varphi } \right) - V'(\varphi ) = 0. $$ The resulting evolution equation for the homogeneous phantom scalar field in the expanding Universe is $$ \ddot \varphi + 3H\dot \varphi - V'\left( \varphi \right) = 0. $$

### Problem 62

Find the energy density and pressure of the phantom field.

### Problem 63

Show that the phantom energy density grows with time. Find the dependence $\rho(a)$ for $w=-4/3$.

\[ \rho = \rho _0 a^{ - 3(1 + w)} = \rho _0 a. \]

### Problem 64

Show that the phantom scalar field violates all four energety conditions.

The corresponding energy conditions in terms of energy density $\rho$ and pressure $p$ read \[\rho+3p\ge0,\ \rho+p\ge0\ (SEC)\] \[\rho+p\ge0\ (NEC)\] \[\rho+p\ge0,\ \rho>0\ (WEC)\] \[|p|\le\rho,\ \rho\ge0\ (DEC)\] In terms of the state equation parameter the energy conditions take the following form: \[1+3w\ge0,\ 1+w\ge0\ (SEC)\] \[1+w\ge0\ (NEC)\] \[1+w>0,\ \rho\ge0\ (WEC)\] \[|w|\le1,\ \rho\ge0\ (DEC)\] It is easy to see that all the above mentioned conditions are violated in the case of phantom energy with $w<-1$.

### Problem 65

Show that in the phantom scalar field $(w<-1)$ dominated Universe the condition $\dot{H}>0$ always holds.

\[w = - 1 - {\frac 2 3}{\frac{\dot H}{H^2 }},\] therefore $\dot{H}>0$ always holds for $w<-1$.

### Problem 66

As we have seen in Chapter 3, the Friedman equations, describing spatially flat Universe, possess the duality, which connects the expanding and contracting Universe by appropriate transformation of the state equation. Consider the Universe where the weak energetic condition $\rho\ge0,\ \rho+p\ge0$ holds and show that the ideal liquid associated with the dual Universe is a phantom liquid or the cosmological constant.

It follows from the weak energy condition that $w\ge-1$, then for the dual Universe one finds $w'=-(w+2)\le-1$.

### Problem 67

Show that the Friedman equations for the Universe filled with dark energy in the form of cosmological constant and a substance with the state equation $p=w\rho$ can be presented in the form of nonlinear oscillator (see M.Dabrowski arXiv:0307128 ) \[\ddot X-\frac{D^2}{3}\Lambda X+D(D-1)kX^{1-2/D}=0\] where \[X=a^{D(w)},\quad D(w)=\frac32(1+w).\]

### Problem 68

Show that the Universe dual to the one filled with a free scalar field, is described by the state equation $p=-3\rho$.

In the considered case one obtains by definition: \[\rho=\frac{\dot\varphi^2}{2};\quad p=\frac{\dot\varphi^2}{2},\] thus $w=1$. For the dual Universe $w'=-(w+2)=-3$.

### Problem 69

Show that in the phantom component of dark energy the sound speed exceeds the light speed.

In the units with dimensional lightspeed the state equation takes the form: \[p = w\rho c^2.\] The speed of sound is then \[c_s^2 = \left| {\left( {{\frac{\partial p} {\partial \rho }}} \right)_S } \right| = \left| w \right|c^2 >c^2.\]

### Problem 70

Construct the phantom energy model with negative kinetic term in the potential satisfying the slow-roll conditions \[\frac 1 V \frac{dV}{d\varphi}\ll1\] and \[\frac 1 V \frac{d^2V}{d\varphi^2}\ll1.\]

## Disintegration of Bound Structures

*Historically the first criterion for decay of gravitationally bound systems due to the phantom dark energy was proposed by Caldwell, Kamionkowski and Weinberg (CKW) (see arXiv:astro-ph/0302506v1). The authors argue that a satellite orbiting around a heavy attracting body becomes unbound when total repulsive action of the dark energy inside the orbit exceeds the attraction of the gravity center. Potential energy of gravitational attraction is determined by the mass $M$ of the attracting center, while the analogous quantity for repulsive potential equals to $\rho+3p$ integrated over the volume inside the orbit. It results in the following rough estimate for the disintegration condition*
\begin{equation}\label{disintegration}
-\frac{4\pi}{3}(\rho+3p)R^3\simeq M.
\end{equation}

### Problem 71

Show that for $w\ge-1$ a system gravitationally bound at some moment of time (Milky Way for example) remains bound forever.

If $w \ge - 1$ then the quantity $\rho + 3p$ does not increase with time, so if the condition \[ - {\frac{4\pi } 3}(\rho + 3p)R^3 \le M\] is satisfied at some moment of time, then it will be satisfied forever, and the system will remain gravitationally bound.

### Problem 72

Show that in the phantom energy dominated Universe any gravitationally bound system will dissociate with time.

If $w < - 1$ (the case of phantom energy), then the quantity $ - (\rho + 3p)$ increases with time and the condition \[ - {\frac{4\pi } 3}(\rho + 3p)R^3 \le M\] will be violated sooner or later.

### Problem 73

Show that in a Universe filled with non-relativistic matter a hydrogen atom will remain a bound system forever.

Consider a spherical coordinate system $r,\theta ,\varphi$ with origin in the nucleus of the hydrogen atom. Position of the electron of mass $m$, orbiting in the equatorial plane $\theta = \pi /2$, is described by the functions $r(t),\varphi (t)$. As the electron is subject to radial forces only, its angular momentum $L = mr^2 \dot \varphi $ is conserved and we define the integral of motion for the unit mass electron as $$ L \equiv r^2 \dot \varphi. $$ In the absence of the cosmological expansion the equation of motion for $r(t)$ takes the well-known form $$ \ddot r - {\frac{L^2 }{r^3 }} = - {\frac C {r^2 }}. $$ Now let us take into account the expansion effect. The point, which takes part in the cosmological expansion, has the radial acceleration equal to $$ \ddot r = r{\frac{\ddot a} a}. $$ The latter term can be considered as a radial force, acting on the unit mass, so it should be inserted into the original equation of motion to obtain the following $$ \ddot r - {\frac{L^2 } {r^3 }} = - {\frac C {r^2 }} + r{\frac{\ddot a} a}. $$ For the case of non-relativistic matter $a \propto t^{2/3} $ and the term $\ddot a/a$ is negative, i.e. it has the same sign as the Coulomb force. It means that the electron will stay bound forever.

### Problem 74

Demonstrate, that any gravitationally bound system with mass $M$ and radius (linear scale) $R$, immersed in the phantom background $\left( {w < - 1} \right)$ will decay in time \[t \simeq P\frac{|1+3w|}{|1+w|}\frac29\sqrt{\frac{3}{2\pi}}\] before Big Rip. Here \[P=2\pi\sqrt{\frac{R^3}{GM}}\] is the period on the circular orbit of radius $R$ around the considered system.

The disintegration condition (\ref{disintegration}) for a fixed value of the state equation parameter $w<-1$ takes the form \[\frac{4\pi}{3}\rho_{DE}|1+3w|R^3= M,\] where the dark energy density $\rho_{DE}$ can be related to the scale factor by \[\rho_{DE}=\frac{3}{8\pi G}H_0^2(1-\Omega_m)a^{3|1+w|},\] with $\Omega_m$ being the relative density of matter. Time dependence for the scale factor can be found from the Friedman equation \[\left(\frac{\dot a}{a}\right)=H_0^2\left(\Omega_m a^{-3}+(1-\Omega_m)a^{3|1+w|}\right),\] where one can neglect the matter contribution to obtain \[t-t_0=\frac23\frac{1-a^{-3/2|1+w|}}{|1+w|H_0\sqrt{1-\Omega_m}}.\] The Big Rip here corresponds to the limit $a\to\infty$, so one obtains for the Big Rip time: \[t_{rip}-t_0=\frac23|1+w|^{-1}H_0^{-1}(1-\Omega_m)^{-1/2}.\] Substitution of this explicit solution into the disintegration condition gives the following relation for the disintegration time: \[t_{rip}-t_{dis}=\frac23\frac{\sqrt{|1+3w|}}{|1+w|}\frac{4\pi}{3}\sqrt{\frac{R^3}{M}}\sqrt{\frac{3}{8\pi G}}= P\frac{\sqrt{|1+3w|}}{|1+w|}\frac29\sqrt{\frac{3}{2\pi}}.\] Note that both $H_0$ and $\Omega_m$ cancel out in the final expression, in which only $P$ and $w$ remain.

### Problem 75

Use the result of the previous problem to determine the time of disintegration for the following systems: galaxy clusters, Milky Way, Solar System, Earth, hydrogen atom. Consider the case $w=-3/2$.

According to the results of the previous problem, the disintegration time depends only on the characteristic orbiting period $P$ of the system with the prefactor depending on the state equation parameter. For $w=-3/2$ the corresponding relation reads: \[t_{rip}-t_{dis}\simeq0.6P.\] Results of substitution of the corresponding periods are summarized in Table.

Time | System disintegrated |

$t_{rip} - 2$ Gyr | Galaxy Clusters |

$t_{rip} - 120$ Myr | Milky Way |

$t_{rip} - 6$ months | Solar System |

$t_{rip} - 60$ minutes | Earth |

$t_{rip} - 10^{-19}$ s | Atoms |

## Big Rip, Pseudo Rip, Little Rip

The future finite-time singularity is an essential element of phantom cosmology (see S.Nojiri, S. Odintsov, arXiv:hep-th/0505215). One may classify the future singularities as in the following way (see S.Nojiri, S. Odintsov and S.Tsajikava, arXiv:hep-th/0501025):

1. For $t\to t_s$, $a\to\infty$, $\rho\to\infty$, $|p|\to\infty$ ("Big Rip").

The density of phantom dark energy and scale factor become infinite at some finite time $t_s$.

2. For $t\to t_s$, $a\to a_s$, $\rho\to\rho_s$ or $\rho\to0$, $|p|\to\infty$ ("sudden singularity").

The condition $w<-1$ is necessary for future singularities, but it is not sufficient. If $w$ approaches to $-1$ sufficiently rapidly, then it is possible to have a model in which there are no future singularities. Models without future singularities in which $\rho_{DE}$ increases with time will eventually lead to dissolution of bound systems. This process received the name "Little Rip" (see P.Frampton, K.Ludwick and R.Scherrer, arXiv:1106.4996). In the Big Rip the scale factor and energy density diverge at finite future time. As opposed to Big Rip in the $\Lambda$CDM, there is no such divergence. Little Rip represents an interpolation between these two limit cases.

3. For $t\to t_s$, $a\to a_s\ne0$, $\rho\to\infty$, $|p|\to\infty$.

4. For $t\to t_s$, $a\to a_s\ne0$, $\rho\to\rho_s$ (including $\rho_s=0$), while derivatives of $H$ diverge.

Here $t_s$, $a_s\ne0$ and $\rho_s$ are constants.

### Problem 76

For the flat Universe composed of matter $(\Omega_m\simeq0.3)$ and phantom energy $(w=-1.5)$ find the time interval left to the Big Rip.

* Immediate consequence of approaching the Big Rip is the dissociation of bound systems due to negative pressure inside them.*

### Problem 77

Show that all little-rip models can be described by condition $\ddot f>0$ where $f(t)$ is a nonsingular function such that $a(t)=\exp[f(t)]$.

In order to realize the Little Rip it is sufficient to require that $a(t)=\exp[f(t)]$, where $f(t)$ is a nonsingular function. Using first Friedman equation one finds $\rho=3H^2=3\dot f^2$, ($8\pi G=1$). The necessary condition to make $\rho$ an increasing function of the scale factor is the following \[\frac{d\rho}{dt}=6\frac{\dot f}{\dot a}\ddot f>0\] which is satisfied as long as \[\ddot f>0.\]

### Problem 78

Consider the approach of the following authors (see S. Nojiri, S.D. Odintsov, and S. Tsujikawa, Phys. Rev. D 71, 063004 (2005); S. Nojiri and S.D. Odintsov, Phys. Rev. D 72, 023003 (2005); H. Stefancic, Phys. Rev. D 71, 084024 (2005)), who expressed the pressure as a function of the density in the form \[p=-\rho-f(\rho).\] Show that condition $f(\rho>0)$ ensures that the density increases with the scale factor.

Use the conservation equation $\dot\rho+3H(\rho+p)=0$.

### Problem 79

Find the dependencies $a(\rho)$ and $t(\rho)$ for the case of flat Universe filled by a substance with the following state equation \[p=-\rho-f(\rho).\]

Integrate the equation $\dot\rho+3H(\rho+p)=0$ to obtain the following \[a=\exp\left(\int\frac{d\rho}{3f(\rho)}\right),\ t=\int\frac{d\rho}{\sqrt{3\rho}f(\rho)}.\]

### Problem 80

Solve the previous problem in the case of \(f(\rho)A\rho^\alpha,\ \alpha=const.\)

\begin{align*} a & =\exp\left[\frac{\left(\frac{\rho}{\rho_0}\right)^{1-\alpha}}{3A(1-\alpha)}\right], & \alpha & \ne1;\\ a & =\left(\frac{\rho}{\rho_0}\right)^{1/(3A)}, & \alpha & =1;\\ t & =t_0+\frac{2}{\sqrt3}\frac{1}{\kappa A}\frac{\rho^{-\alpha+1/2}}{1-2\alpha}, & \alpha & \ne\frac12;\\ t & =t_0+\frac{1}{\sqrt3\kappa A}\ln\left(\frac{\rho}{\rho_0}\right), & \alpha & =\frac12.\\ \end{align*}

### Problem 81

Find the condition for big-rip singularity in the case $p=-\rho-f(\rho).$

The condition for a big-rip singularity implies that \[t=\int\frac{d\rho}{\sqrt{3\rho}f(\rho)}\] converges.

### Problem 82

Show that taking a power law for $f(\rho)$, namely $f(\rho)=A\rho^\alpha$ a future singularity can be avoided for $\alpha\le1/2$.

Use solution of the previous problem.

### Problem 83

Solve the previous problem using the condition for absence of future singularities obtained in Problem #RIPS_1.

In this case we have \[\frac{\rho}{\rho_1}=\left[\frac{3A}{2\sqrt\rho_1}\ln\left(\frac{a}{a_c}\right)+1\right]^2\] where $\rho_1$ is a constant and $\rho=\rho_1$ and $a=a_c$ at $t=t_1$. Time dependence of the density is given by \[\frac{\rho}{\rho_1}=e^{\sqrt3A(t-t_1)}.\] Eliminating $\rho$ one finds \[\frac{a}{a_c}=\exp\left\{\frac{2\sqrt\rho_1}{3A}\left[e^{\frac{\sqrt3A}{2}(t-t_1)}-1\right]\right\}.\] By comparing $a=e^f$ with this relation, it is easy to see that \[f=\frac{2\sqrt\rho_1}{3A}\left[e^{\frac{\sqrt3A}{2}(t-t_1)}-1\right]+\ln a_c\] and \[\ddot f=\frac{A\sqrt\rho_1}{2}\exp\left[\frac{\sqrt3A}{2}(t-t_1)\right]>0\] because $A>0$ (phantom phase). Hence, the future singularities are absent.

### Problem 84

Formulate the condition for the absence of a finite-time future (Big Rip) singularity in terms of function $\rho(a)$ .

With the definition $N\equiv\ln a$, the first Friedman equation can be rewritten to the following form \[t=\int\sqrt{\frac{3}{\rho(N)}}dN.\] The condition to avoid the finite-time future singularity implies that it takes infinite time for the Big Rig singularity to appear. This means that \[\int\limits_{N_0}^\infty\frac{dN}{\sqrt{\rho(N)}}\to\infty.\]

*The problems below develop an alternative approach to investigate the singularities in the phantom Universe (see P-H. Chavanis, arXiv:1208.1195)*

### Problem 85

Consider the polytropic equation of state \[p=\alpha\rho+k\rho^{1+1/n}\equiv-\rho+\rho\left(1+\alpha+k\rho^{1/n}\right)\] under assumption $-1<\alpha\le1$. The case $\alpha=-1$ is treated separately in Problem #RIPS_7_4. The additional assumption $1+\alpha+k\rho^{1/n}\le0$ (and necessary condition $k<0$) guarantees that the density increases with the scale factor. This corresponds to phantom Universe. Find explicit dependence $\rho(a)$ and analyze limits $a\to0$ and $a\to\infty$.

The conservation equation for polytropic equation of state reads \[\frac{d\rho}{dt}+3H\rho\left(1+\alpha+k\rho^{1/n}\right)=0.\] Assuming $1+\alpha+k\rho^{1/n}\le0$, this equation can be integrated to yield \[\rho=\frac{\rho_*}{\left[1-(a/a_*)^{3(1+\alpha)/n}\right]^n}.\] Here $\rho_*=[(\alpha+1)/|k|]^n$ and $a_*$ is a constant of integration. For $n>0$, the density is defined only when $a<a_*$. Then $a\to0$ results in $\rho\to\rho_*$ and $p\to-\rho_8$. If $a\to a_*$, then \[\rho\sim\rho_*\left[\frac{n}{3(1+\alpha)}\right]^n\frac{1}{(1-a/a_*)^n}\to\infty\] and $p\to\infty$. For $n<0$ the density is defined only when $a>a_*$. Then $a\to a_*$ leads to \[\rho\sim\rho_*\left[\frac{3(1+\alpha)}{|n|}\right]^{|n|}(a/a_*-1)\to0\] In the same limit, $p\to-\infty$ for $n>-1$, $p$ tends to a finite value for $n=-1$ and $p\to0$ for $n<-1$. If $a\to\infty$ then $\rho\to\rho_*$ and $p\to-\rho_*$.

### Problem 86

Consider the previous problem with $\alpha=-1$ and $k<0$. This equation of state was introduced by Nojiri and Odintsov (see problem #RIPS_7_3). Chavanis re-derives their results in a more transparent form.

For the case $n>0$, the conservation equation can be integrated to give \[\rho=\frac{\rho_*}{[\ln(a_*/a)]^n},\] where $\rho_*=(n/3|k|)^n$ and $a_*$ is a constant of integration. The density is defined for $a\le a_*$. If $a\to0$ then $\rho\to0$ and $p\to0$. If $a\to a_*$ then $\rho\to\infty$ and $p\to-\infty$. For the case $n<0$ one gets \[\rho=\frac{\rho_*}{[\ln(a/a_*)]^n},\] where $\rho_*=(|n|/3|k|)^n$ and $a_*$ is a constant of integration. The density is defined for $a\ge a_*$. If $a\to a_*$ then $\rho\to0$. In the same limit , $p\to0$ for $n<-1$, $p$ tends to a finite value for $n=-1$, and $p\to-\infty$ for $n>-1$. If $a\to\infty$ then $\rho\to\infty$ and $p\to-\infty$.

The little-rip dissociates all bound structures, but

the strength of the dark energy is not enough to rip

apart space-time as there is no finite-time singularity

P. Frampton, K. Ludwick1, and R. Scherrer

(see A. Astashenok, S. Nojiri, S. Odintsov, and R. Scherrer, arXiv:1203.1976)

### Problem 87

Show that for any bound system the rip always occurs when either $H$ diverges or $\dot H$ diverges (assuming $\dot H>0$ ( expansion of Universe is accelerating)).

As the universe expands, the relative acceleration between two points separated by a distance $l$ is given by $l\ddot a/a$. If there is a particle with mass $m$ at every point, an observer at one of the masses will measure an inertial force on the other mass equal to the following \[F_{iner}=ml\frac{\ddot a}{a}=ml(\dot H +H^2).\] It follows that for any bound system the rip always occurs when either $H$ diverges or $\dot H$ diverges.

### Problem 88

Solve the previous problem in terms of function $f(\rho)$.

If $p=-\rho-f(\rho)$, then \[H^2=\frac\rho3,\ \dot H=-\frac12(\rho+p)=\frac12f(\rho).\] Consequently, \[F_{iner}=ml\frac{\ddot a}{a}=ml(\dot H +H^2)=ml\left(\frac13\rho+\frac12f(\rho)\right).\] The case $f(\rho)\to\infty$ at $\rho\to\rho_f$ describes the sudden singularity and the rip of any system, while $f(\rho)\to-\infty$ at $\rho\to\rho_f<\infty$ corresponds to the crush.

### Problem 89

Perform analysis of possible singularities in terms of characteristics of the scalar field $\varphi$ with the potential $V(\varphi)$.

For the scalar field and its potential, one can derive the following relations: \[\varphi(x)=\varphi_0\pm\frac{2}{\sqrt3}\int\limits_{x_0}^x\frac{dx}{\sqrt{|f(x)|}},\ x\equiv\sqrt\rho;\] \[V(x)=x^2+\frac12f(x).\] For the crush and sudden future singularity the potential of scalar field has a pole. For the sudden future singularity potential $V(\varphi)\to+\infty$, and for the crush $V(\varphi)\to-\infty$.

*All the Big Rip, Little Rip and Pseudo-Rip arise from the assumption that the dark energy density $\rho(a)$ is monotonically increasing. Let us investigate what will happen if this assumption is broken and then propose a so-called "Quasi-Rip" scenario, which is driven by a type of quintom dark energy. In this work, we consider an explicit model of Quasi-Rip in details. We show that Quasi-Rip has an unique feature different from Big Rip, Little Rip and Pseudo-Rip. Our universe has a chance to be rebuilt in the ash after the terrible rip. This might be the last hope in the "hopeless" rip.*

### Problem 90

So-called soft singularities are characterized by a diverging $\ddot a$ whereas both the scale factor $a$ and $\dot a$ are finite. Analyze features of intersections between the soft singularities and geodesics.

The geodesic equations in a flat Friedman space-time are \begin{align*} \frac{d^2 x^\alpha}{d\lambda^2}+2\frac{\dot a}{a}\frac{dt}{d\lambda}\frac{dx^\alpha}{d\lambda} & =0,\\ \frac{d^2 t}{d\lambda^2}+a\dot a\sum\limits_\alpha\left(\frac{dx^\alpha}{d\lambda}\right)^2 & =0. \end{align*} where $\lambda$ is an affine parameter. These equations are singular only for vanishing scale factor. (Therefore, the existence of solutions $t(\lambda)$ and $x^\alpha(\lambda)$ is assured by the Cauchy-Peano theorem for any nonzero $a$ (including the soft singularity). Thus the functions $t(\lambda)$ and $x^\alpha(\lambda)$, i.e. the geodesics, can be continued through the singularity occurring at finite scale factor. In other words, as the Christoffel symbols depend only on the first derivative of the scale factor, they are regular at these singularities. Hence, the geodesics are well behaved and they can cross the singularity. One can argue that the particles crossing the singularity will generate the geometry of the spacetime, providing in such a way a "soft rebirth" of the universe after the singularity crossing (see Z. Keresztes et al., arXiv:1204.1199).

### Problem 91

(see F.Cannata, A. Kamenshchik, D.Regoli, arXiv:0801.2348) The power law cosmological evolution $a(t)\propto t^\beta$ leads to the Hubble parameter $H(t)\propto 1/t$. Consider a "softer" version of the cosmological evolution given by the law \[H(t)=\frac{S}{t^\alpha},\] where $S$ is a positive constant and $0<\alpha<1$. Analyze the dynamics of such model at $t\to 0$.

Integrating \[\frac{d}{dt}\left(\ln a\right)=\frac{S}{t^\alpha}\] one obtains \[\ln\left(\frac{a(t)}{a(0)}\right)=\frac{S}{1-\alpha}t^{1-\alpha}.\] At $t=0$ singularity is present, but it is different from the traditional Big Bang singularity. If $t\ne0$ ($t>0$) the righthand side is finite and hence one cannot have $a(0)=0$ in the lefthand side. Hence $a(0)>0$, while \[a(t)=a(0)\frac{S}{t^\alpha}\exp\left(\frac{S}{1-\alpha}t^{1-\alpha}\right)\underset{t\to0}{\to}\infty.\] This type of singularity received the name "soft Bing Bang" singularity because the cosmological scale factor is finite (and non-zero) while its time derivative, the Hubble variable and the scalar curvature are singular. It is interesting to note that in the limit $t\to\infty$ both $a(t)$ and $\dot a(t)$ tend to infinity, but they do not encounter any cosmological singularity because the Hubble variable and its derivatives tend to zero.

### Problem 92

Reconstruct the potential of the scalar field model, producing the given cosmological evolution $H(t)$.

Combine Friedman equation \[H^2=\rho\] ($8\pi G/3=1$) and conservation equation \[\dot\rho+3H(\rho + p)\] to find \[\dot H=-\frac32(\rho+p).\] Using \begin{align*} \rho & =\frac12\dot\varphi^2+V(\varphi)\\ p & =\frac12\dot\varphi^2-V(\varphi) \end{align*} one obtains \begin{align} V & =\frac13\dot H+H^2 \label{RIPS_67_1}\\ \dot\varphi^2 & =- \frac23\dot H \label{RIPS_67_2} \end{align} Equation (\ref{RIPS_67_1}) represents the potential as a function of time $t$. Integrating equation (\ref{RIPS_67_2}) one can find the scalar field as a function of time. Inverting this dependence we can obtain the time parameter as a function of $\varphi$ and substituting the corresponding formula into equation (\ref{RIPS_67_1}) one arrives to the uniquely reconstructed potential $V(\varphi)$. It is necessary to stress that this potential reproduces a given cosmological evolution only for some special choice of initial conditions on the scalar field and its time derivative.

### Problem 93

Reconstruct the potential of the scalar field model, producing the cosmological evolution \begin{equation}\label{RIPS_68} H(t)=\frac{S}{t^\alpha}, \end{equation} using the technique described in the previous problem.

\begin{equation}\dot\varphi = \pm \sqrt{-\frac23\dot H}=\pm\sqrt{\frac23\alpha S}t^{-\frac{\alpha+1}{2}}.\label{RIPS_68_1}\end{equation} We shall choose the positive sign, without loosing generality. Integration gives \begin{equation}\varphi(t) = \sqrt{\frac23\alpha S}\frac{2t^{-\frac{\alpha+1}{2}}}{1-\alpha}.\label{RIPS_68_2}\end{equation} up to an arbitrary constant. Inverting the last relation we find \[t(\varphi)=\left[\left(\frac{3}{2\alpha S}\right)^{1/2}\frac{1-\alpha}{2}\varphi\right]^{\frac{2}{1-\alpha}}.\] The ultimate result is \[V(\varphi)=\frac{S^2}{\left[\sqrt{\frac{3}{2\alpha S}}\frac{1-\alpha}{2}\varphi\right]^{\frac{4}{1-\alpha}}} - \frac{\alpha S}{\left[\sqrt{\frac{3}{2\alpha S}}\frac{1-\alpha}{2}\varphi\right]^{\frac{2(\alpha+1)}{1-\alpha}}}\] This potential provides the cosmological evolution (\ref{RIPS_68}) if one chooses initial conditions compatible with Eqs. (\ref{RIPS_68_1}) and (\ref{RIPS_68_1}). Naturally, there are also other cosmological evolutions, generated by other initial conditions.

## The Statefinder

In the models including dark energy in different forms it is useful to introduce a pair of cosmological parameters $\{r,s\}$, which is called the statefinder (see V.Sahni, T.Saini, A.Starobinsky, U.Alam astro-ph/0201498): \[r\equiv\frac{\dddot a}{aH^3},\ s\equiv\frac{r-1}{3(q-1/2)}.\] These dimensionless parameters are constructed from the scale factor and its derivatives. Parameter $r$ is the next member in the sequence of the kinematic characteristics describing the Universe's expansion after the Hubble parameter $H$ and the deceleration parameter $q$ (see Chapter "Cosmography"). Parameter $s$ is the combination of $q$ and $r$ chosen in such a way that it is independent of the dark energy density. The values of these parameters can be reconstructed with high precision basing on the available cosmological data. After that the statefinder can be successfully used to identify different dark energy models.

### Problem 94

Explain the advantages for the description of the current Universe's dynamics brought by the introduction of the statefinder.

The fundamental characteristics can be either of geometric nature, if they are derived immediately from the space-time metric, or of physical one, if they depend on properties of the fields carrying the dark energy. The physical variables are clearly model-dependent, while the geometrical ones are more universal. Besides that, the latter are free from uncertainties which appear in measurements of physical quantities, such as e.g. the energy density. That is why geometrical variables are more convenient to describe the current expansion of the Universe and the properties of dark energy.

### Problem 95

Express the statefinder $\{r,s\}$ in terms of the total density, pressure and their time derivatives for a spatially flat Universe.

$$ r = 1 + {9 \over 2}{{(\rho + p)\dot p} \over {\rho \dot \rho }},\quad s = {{\rho + p} \over p}{{\dot p} \over {\dot \rho }} $$

### Problem 96

Show that for a flat Universe filled with a two-component liquid composed of non--relativistic matter (dark matter + baryons) and dark energy with relative density $\Omega _{DE} = \rho _{DE} /\rho _{cr} $ the statefinder takes the form $$ r = 1 + {\frac92}\Omega _{DE} w(1 + w) - {\frac32}\Omega _{DE} {\frac{\dot w}{H}}; $$ $$ s = 1 + w - {\frac13}{\frac{\dot w}{wH}};\quad w \equiv {\frac{p_{DE} } {\rho _{DE} }}. $$

Make use of the conservation equation for the total density $$ \dot \rho + 3H(\rho + p) = 0 $$ and that for the dark energy density $$ \dot \rho _{_{DE}} + 3H(\rho _{_{DE}} + p_{_{DE}} ) = 0 $$ Take into account also that pressure of non-relativistic matter can be neglected.

### Problem 97

Express the statefinder in terms of Hubble parameter $H(z)$ and its derivatives.

\begin{align*} r(x) & = 1 - 2{{H'} \over H}x + \left\{ {{{H''} \over H} + \left( {{{H'} \over H}} \right)^2 } \right\}x^2 ;\\ s(x) & = {{r(x) - 1} \over {3\left( {q(x) - 1/2} \right)}};\\ q(x) & = {{H'(x)} \over H}x - 1;\quad x \equiv 1 + z. \end{align*}

### Problem 98

Find the statefinders

**a)** for dark energy in the form of cosmological constant;
b)

**for the case of time--independent state equation parameter $w$;**

**c)**for dark energy in the form of quintessence.

\begin{align*} a) \left\{ {r,s} \right\} & = \left\{ {1,0} \right\};\\ b) \left\{ {r,s} \right\} & = \left\{ {1 + {9 \over 2}\Omega _{_{DE}} w(1 + w),1 + w} \right\};\\ c) \left\{ {r,s} \right\} & = \left\{ {1 + {{12\pi G\dot \varphi ^2 } \over {H^2 }} + {{8\pi G\dot V} \over {H^3 }},{{2\left( {\dot \varphi ^2 + {{2\dot V} \over {3H}}} \right)} \over {\dot \varphi ^2 - 2V}}} \right\} \end{align*}

### Problem 99

Express the photometric distance $d_L(z)$ through the current values of parameters $q$ and $s$.

\[ H\left( z \right) = H_0 + \left( {{{dH} \over {dz}}} \right)_{z = 0} z + {1 \over 2}\left( {{{d^2 H} \over {dz^2 }}} \right)_{z = 0} z^2 + \cdots ; \] It was shown in Chapter 3 that $$ {{dH} \over {dz}} = {{q + 1} \over {z + 1}}H;\quad {{d^2 H} \over {dz^2 }} = {{r - 1 + 2(1 + q) - (1 + q)^2 } \over {(1 + z)^2 }}, $$ thus $$ H(z) = H_0 \left\{ {1 + \left( {q_0 + 1} \right)z + {1 \over 2}\left[ {r_0 - 1 + 2\left( {q_0 + 1} \right) - \left( {q_0 + 1} \right)^2 } \right]z^2 + \cdots } \right\}, $$ and the photometric distance equals to $$ d_L = (1 + z)\int_0^z {{{dz'} \over {H(z')}}} $$ Finally one obtains: $$ d_L (z) = H_0^{ - 1} z\left\{ {1 + {1 \over 2}\left( {1 - q_0 } \right)z + {1 \over 6}\left[ {3\left( {1 + q_0 } \right)^2 - 5\left( {1 + q_0 } \right) + 1 - r_0 } \right]z^2 + \cdots } \right\}. $$

## Crossing the Phantom Divide

*In the quintessence model of dark energy $-1<w<-1/3$. In the phantom model with negative kinetic energy $w<-1$. Recent cosmological data seem to indicate that there occurred the crossing of the phantom divide line in the near past. This means that equation of state parameter $w_{DE}$ crosses the phantom divide line $w_{DE}=-1$. This crossing to the phantom region is possible neither for an ordinary minimally coupled scalar field nor for a phantom field. There are at least three ways to solve this problem. If dark energy behaves as quintessence at early stage, and evolves as phantom at the later stage, a natural suggestion would be to consider a 2-field model (quintom model): a quintessence and a phantom. The next possibility, discussed in the next Chapter, is to consider an interacting model, in which dark energy interacts with dark matter. Yet another possibility would be that General Relativity fails at cosmological scales. In this case quintessence or phantom energy can cross the phantom divide line in a modified gravity theory. We investigate this approach in Chapter 12.*

### Problem 100

Show that at the point of transition between the quintessence and the phantom phases $\dot H$ vanishes.

The parameter $w=p/\rho$ can be expressed in the form \[w=-\left(1+\frac{2\dot H}{3H^2}\right).\] Then in the quintessence phase ($-1<w<-1/3$) $\dot H<0$ . By contrast, in the phantom phase ($w<-1$) $\dot H>0$. Then on the point of transition $\dot H=0$.

### Problem 101

Show that the sound speed of a single perfect barotropic fluid is diverges when $w$ crosses the phantom divide line.

If the fluid is barotropic, the adiabatic sound speed is determined by \[c_s^2=\frac{p'}{\rho'}=\frac{(w\rho)'}{\rho'}=w+w'\frac{\rho}{\rho'}=w-\frac{w'}{3\bar{H}(1+w)}.\] Here the prime denotes derivative with respect to conformal time, $\bar{H}\equiv\bar{a}/a$ is the conformal Hubble parameter. The sound speed is apparently divergent when $w$ crosses $-1$, which leads to instability of dark energy perturbations.

### Problem 102

Find a dynamical law for the equation of state parameter $w=p/\rho$ in the barotropic cosmic fluid (see N.Caplar, H.Stefancic, arXiv:1208.0449).

The equation of state of a barotropic cosmic fluid can in general be written as \[F(\rho, p(\rho))=0.\] Here $p(\rho)=w\rho$. This relation implies that $\rho$ and $p$ can be considered functions of $w$, i.e. $\rho=\rho(w)$ and $p=p(w)=w\rho(w)$. The speed of sound in the barotropic fluid is \[c_s^2=\frac{dp}{d\rho}=-\frac{\frac{\partial F}{\partial \rho}}{\frac{\partial F}{\partial p}}.\] Using the identity \[\frac{\partial F}{\partial \rho}d\rho+\frac{\partial F}{\partial p}dp= \frac{\partial F}{\partial \rho}d\rho+\frac{\partial F}{\partial p}(\rho dw+wd\rho)=0,\] one obtains \[\frac{d\rho}{\rho}=\frac{dw}{c_s^2-w}.\] Combining this expression with the conservation equation \[\frac{d\rho}{\rho}+3(1+w)\frac{da}{a}=0,\] one finally finds \[\frac{dw}{(c_s^2-w)(1+w)}=-3\frac{da}{a}=3\frac{dz}{1+z}.\] As $p$ and $\rho$ are functions of $w$, one has $c_s=c_s(w)$. Consequently, the obtained expression is a dynamical law for the parameter $w$.

### Problem 103

Using the results of previous problem, find the functions $w(z)$, $\rho(z)$ and $p(z)$ for the simplest possibility $c_S=const$.

By integration of the equation \[\frac{dw}{(c_s^2-w)(1+w)}=3\frac{dz}{1+z}\] one obtains \[w=\frac{c_s^2\frac{1+w_0}{c_s^2-w_0}(1+z)^{3(1+c_s^2)}-1}{\frac{1+w_0}{c_s^2-w_0}(1+z)^{3(1+c_s^2)}+1}\] with $w\equiv w(z=0)$. Integrating \[\frac{d\rho}{\rho}=\frac{dw}{c_s^2-w}\] one finds \[\rho=\rho_0\frac{c_s^2-w_0}{c_s^2-w}=\rho_0\frac{c_s^2-w_0}{1+c_s^2} \left[\frac{1+w_0}{c_s^2-w_0}(1+z)^{3(1+c_s^2)}+1\right]\] and \[\rho=c_s^2\rho-\rho_0(c_s^2-w_0)=\rho_0\frac{c_s^2-w_0}{1+c_s^2} \left[c_s^2\frac{1+w_0}{c_s^2-w_0}(1+z)^{3(1+c_s^2)}-1\right].\]

### Problem 104

Realize the procedure described in the problem #DE117_11 for the case of a minimally coupled scalar field $\varphi$ with potential $V(\varphi)$ in a spatially flat Universe.

The time derivative of the scalar field can be expressed as (see problem #DE117_11) \[\dot\varphi=\frac{d\varphi}{dw}\frac{dw}{dt}=-3H(c_s^2-w)(1+w)\frac{d\varphi}{dw}.\] On the other hand, \[\dot\varphi^2=\rho+p=(1+w)\rho(w).\] Combining these two expressions and using \[H^2=\frac13\rho\ (8\pi G=1)\] one obtains \[d\varphi=\mp\frac{dw}{\sqrt{3(1+w)}(c_s^2-w)}.\] Integration of this equation leads to $\varphi=\varphi(w)$. In a similar way, \[V(\varphi)=\frac12(1-w)\rho(w)=\rho_0\frac{c_s^2-w_0}{c_s^2-w}(1-w).\]

### Problem 105

Consider the case of Universe filled with non-relativistic matter and quintessence and show that the condition to cross the phantom divide line $w=-1$ is equivalent to sign change in the following expression \[\frac{dH^2(z)}{dz}-3\Omega_{m0}H_0^2(1+z)^2.\]

For the Universe model considered the following holds (see problem #DE55_2): \[w(z)=\frac{\frac23(1+z)\frac{d\ln H}{dz}-1}{1-\frac{H_0^2}{H^2}\Omega_{m0}(1+z)^3}.\] Then the condition $w(z)=-1$ can be rewritten as \[\frac{dH^2(z)}{dz}-3\Omega_{m0}H_0^2(1+z)^2=0.\]

### Problem 106

Consider a model with the scale factor of the form \[a=a_c\left(\frac{t}{t-t_s}\right)^n,\] where $a_c$ is a constant, $n>0$, $t_s$ is the time of a Big Rip singularity. Show that on the interval $0<t<t_s$ there is crossing of the phantom divide line $w=-1$.

For $t\ll t_s$ we have $a\sim t^n$ and thus \[w=-1-\frac{2\dot H^2}{3H^2}=-1+\frac{2}{3n}>-1,\] whereas for $t\to t_s$, \[w=-1-\frac{2}{3n}<-1.\]

### Problem 107

Show, that for the model considered in the previous problem the parameter $H(t)$ and density $\rho(t)$ achieve their minimal values at the phantom divide point. (see K.Bamba, S.Capozziello, S.Nojiri, S.Odintsov, arXiv:1205.3421)

For \[a=a_c\left(\frac{t}{t-t_s}\right)^n,\] the parameters $H(t)$ and $\rho(t)$ are given by \[H=n\left(\frac1t+\frac{1}{t_s-t}\right),\] \[\rho(t)=3n^2\left(\frac1t+\frac{1}{t_s-t}\right)^2.\] At $t=t_s/2$ the parameters $H$ and $\rho$ take their extremal values \[H_{min}=\frac{4n}{t_s},\quad \rho_{min}=\frac{48n^2}{t_s^2}.\] Time derivative $\dot\rho$ is equal to \[\dot\rho= \pm2\rho\sqrt{\frac{\rho}{3n^2}-\frac{4}{nt_s}\sqrt{\frac\rho3}}.\] Here, the plus (minus) denotes the expression of $\dot\rho$ for $t>t_s/2$ ($0<t<t_s/2$), i.e. the phantom (non-phantom phase), because density of dark energy must increase in the phantom phase. Using the conservation equation in the form $\dot\rho=3Hf(\rho)$, one obtains \[f(\rho)=\pm\frac{2\rho}{3n}\sqrt{1-\frac{4n}{t_s}\sqrt{\frac3\rho}}.\] We see that $f(\rho_{min})=0$. This means that at the phantom crossing point, both $H$ and $\rho$ take their minimum values.

### Problem 108

Find condition of intersection with the line $w=-1$ for the quintom Lagrangian \[L=\frac12g^{\mu\nu}\left(\frac{\partial\varphi}{\partial x^\mu}\frac{\partial\varphi}{\partial x^\nu} - \frac{\partial\psi}{\partial x^\mu}\frac{\partial\psi}{\partial x^\nu} \right)-W(\varphi,\psi).\]

In a FLRW Universe the density and pressure of quintom are \[\rho=\frac12\dot\varphi^2 - \frac12\dot\psi^2 +W;\] \[p=\frac12\dot\varphi^2 - \frac12\dot\psi^2 -W,\] and \[w=\frac{\frac12\dot\varphi^2 - \frac12\dot\psi^2 -W}{\frac12\dot\varphi^2 - \frac12\dot\psi^2 +W}.\] Condition $w=-1$ requires \[\dot\varphi^2=\dot\psi^2.\] The quintom model does not require a static field (a field with zero kinetic term) to get a cosmological constant. We only need that $\varphi$ and $\psi$ evolve in the same way.