The traditional point of view assumed that dominating part of
degrees of freedom in our World are attributed to physical fields.
However it became clear soon that such concept complicates
the construction of Quantum Gravity: it is necessary to introduce small
distance cutoffs for all integrals in the theory in order to make it
sensible. As a consequence, our World should be described on a
three-dimensional discrete lattice with the period of the order of Planck
length. Lately some physicists share an even more radical
point of view: instead of the three-dimensional lattice, complete
description of Nature requires only a two-dimensional one, situated on
the space boundary of our World. This approach is based on the
so-called holographic principle. The name is related to the optical
hologram, which is essentially a two-dimensional record of a
three-dimensional object. The holographic principle consists of two
main statements:
1) All information contained in some region of space can be
recorded (represented) on the boundary of that region.
2) The theory, formulated on the boundaries of the considered
region of space, must have no more than one degree of freedom per
Planck area:
\begin{equation}
\label{Hol_f:1}
N\le \frac{A}{A_{pl}},\quad A_{pl}=\frac{G\hbar}{c^3}.
\end{equation}
Thus, the key piece in the holographic principle is
the assumption that all the information about the Universe can be
encoded on some two-dimensional surface --- the holographic
screen. Such approach leads to a new interpretation of cosmological
acceleration and to an absolutely unusual understanding of Gravity. The
Gravity is understood as an entropy force, caused by variation of
information connected to positions of material bodies. More
precisely, the quantity of information related to matter and its
position is measured in terms of entropy. Relation between the
entropy and the information states that the information change is
exactly the negative entropy change $\Delta I=-\Delta S$. Entropy change
due to matter displacement leads to the so-called entropy force, which,
as will be proven below, has the form of gravity. Its origin
therefore lies in the universal tendency of any macroscopic theory
to maximize the entropy. The dynamics can be constructed in terms
of entropy variation and it does not depend on the details of
microscopic theory. In particular, there is no fundamental field
associated with the entropy force. The entropy forces are typical
for macroscopic systems like colloids and biophysical systems. Big
colloid molecules, placed in thermal environment of smaller
particles, feel the entropy forces. Osmose is another phenomenon
governed by the entropy forces.
Probably the best known example of the entropy force is the elasticity of a polymer
molecule. A single polymer molecule can be modeled as a
composition of many monomers of fixed length. Each monomer can
freely rotate around the fixation point and choose any spacial
direction. Each of such configurations has the same energy. When the
polymer molecule is placed into a thermal bath, it prefers to form a
ring as the entropically most preferable configuration: there are
many more such configurations when the polymer molecule is short,
than those when it is stretched. The statistical tendency to transit
into the maximum entropy state transforms into the macroscopic
force, in the considered case---into the elastic force.
Let us consider a small piece of holographic screen and a particle
of mass $m$ approaching it. According to the holographic principle,
the particle affects the amount of the information (and therefore of the
entropy) stored on the screen. It is natural to assume that
entropy variation near the screen is linear on the displacement
$\Delta x$:
\begin{equation}
\Delta S = 2\pi k_B \frac{mc}{\hbar} \Delta x. \label{delta_s}
\end{equation}
The factor $2\pi$ is introduced for convenience, which the reader
will appreciate solving the problems of this section. In order to
understand why this quantity should be proportional to mass,
let us imagine that the particle has split into two or more particles
of smaller mass. Each of those particles produces its own entropy
change when displaced by $\Delta x$. As entropy and mass are both
additive, then it is natural that the former is proportional to the
latter. According to the first law of thermodynamics, the entropy
force related to information variation satisfies the equation
\begin{equation}
F\Delta x = T\Delta S. \label{delta_x}
\end{equation}
If we know the entropy gradient, which can be found from (\ref{delta_s}),
and the screen temperature, we can calculate the entropy
force.
An observer moving with acceleration $a$, feels the
temperature (the Unruh temperature)
\begin{equation}
\label{Hol_f_Unruh:4}
k_B T_U=\frac{1}{2\pi}\frac\hbar c a.
\end{equation}
Let us assume that the total energy of the system equals $E$. Let us
make a simple assumption that the energy is uniformly distributed
over all $N$ bits of information on the holographic screen. The
temperature is then defined as the average energy per bit:
\begin{equation}
E =\frac12 N k_B T. \label{average_e}
\end{equation}
Equations (\ref{delta_s})--(\ref{average_e}) allow one to describe the holographic
dynamics, and as a particular case---the dynamics of the Universe, and
all that without the notion of Gravity.
11.10.1
Problem 11.10.1
For the interacting holographic dark energy $Q = 3\alpha H \rho_L,$ with the Hubble radius as the IR cutoff, find the depending on the time for the scale factor, the Hubble parameter and the deceleration parameter.
solution
Setting $L = H^{-1}$ in $\rho_{L}=3c^2M_p^2L^{-2}$ and working with the equality, it becomes
\begin{equation}\label{W_m_IHDE}
\rho_{H} = 3\, c^{2}M^{2}_pH^{2}.
\end{equation}
The effective equation of state parameter takes on the form
\begin{equation}
w_H =-\frac23\frac{\dot{H}}{H^2}-\alpha-1, \label{w_H}
\end{equation}
As we see from \eqref{w_H}, the parameter responsible for interaction, $\alpha$,contributes to accelerated expansion when it is positive.
In this case, the Friedmann equation has an exact solution, and so for the Hubble parameter, we obtain
\begin{equation}
H= \frac{2(1-\alpha c^2)}{3-2\alpha c^2} \frac{A}{t},\label{H_H}
\end{equation}
where $A$ is an integration constant, and $t$ is the cosmic time. The time dependence of the scale factor $a(t)$ has the form
\begin{equation}
a(t)=a_0 t^{\frac{2(1-\alpha c^2)}{3-2\alpha c^2} },\label{a_H}
\end{equation}
where $a_0$ is an integration constant.
In conclusion, we find that for this model the deceleration parameter is $q = -1 - \frac{\dot{H}}{H^2}:$
\begin{equation}
q(t)=\frac{1}{2(1-2\alpha c^2)},\label{q_H}
\end{equation}
Clearly, if we assume in equations \eqref{H_H}-\eqref{q_H} that $ \alpha = 0,$ we obtain expressions for a Universe filled with non-relativistic matter.
As seen in this model, the deceleration parameter is constant throughout the evolution of the Universe and,
therefore, cannot explain the change of phases from a slow (matter dominated) expansion to an accelerated
(dark energy dominated) expansion of the Universe.
11.10.2
Problem 11.10.2
Show that for the choice $\rho_{hde}\propto H^2$ ($\rho_{hde}=\beta H^2$, $\beta=const$)an interaction is the only way to have an equation of state different from that of the dust.
solution
If we assume $\rho_{hde}=\beta H^2$,this definition implies
\[\dot\rho_{hde}=-3\beta H^3\left(1+\frac{w_{de}}{1+r}\right)=-3 H\left(1+\frac{w_{de}}{1+r}\right).\]
Without loss of generality we can choose the interaction in the form $Q=\Gamma\rho_{hde}$, $\Gamma$where is an arbitrary (generally variable) function. Inserting $\dot\rho_{hde}$ in the conservation equation for DE we obtain relation between parameter $w_{de}$ and interaction $\Gamma$,
\[w_{de}=-\left(1+\frac1r\right)\frac\Gamma{3H}.\]
The interaction parameter $\Gamma/(3H)$ together with the ratio $r$ determines the equation of state parameter $w_{de}$. For the choice $\rho_{hde}=\beta H^2$ an interaction is the only way to have an equation of state different from that of the dust ($w=0$).
11.10.3
Problem 11.10.3
Calculate the derivative \[\frac{d\rho_{de}}{d\ln a}\] for the holographic dark energy model, where IR cut-off $L$ is chosen to be equal to the future event horizon. (after [1])
solution
The holographic dark energy density is
\[\rho_{de}=3c^2M_{Pl}^2L^{-2},\]
where $c$ is a numerical constant, and $M_{Pl}=1/\sqrt{8\pi G}$ is the reduced Planck mass. In the considered case
\[L(a)=l_e(a)=a\int_t^\infty\frac{dt'}{a(t')}=a\int_a^\infty\frac{da'}{a'^2H}.\]
Taking the derivative one finds
\[\frac{d\rho_{de}}{d\ln a}= a\frac{d\rho_{de}}{da}=-6M_{Pl}^2H^2\Omega_{de}\left(1-\frac{\sqrt{\Omega_{de}}}{c}\right).\]
11.10.4
Problem 11.10.4
Find the effective state parameter value $w_{eff}$, such that \[\rho'_{de}+3(1+w_{eff})\rho_{de}=0\] for the holographic dark energy model, considered in the previous problem, with the interaction of the form $Q=3\alpha H\rho_{de}$.
solution
For the chosen type of interaction
\[\rho'_{de}+3(1+w_{de})\rho_{de}=-3\alpha H\rho_{de}.\]
Using
\[\frac{d\rho_{de}}{d\ln a}= a\frac{d\rho_{de}}{da}=-6m_{Pl}^2H^2\Omega_{de}\left(1-\frac{\sqrt{\Omega_{de}}}{c}\right),\]
one obtains
\[w_{eff}=w_{de}+\alpha=-\frac13-\frac23\frac{\sqrt{\Omega_{de}}}{c}.\]
11.10.5
Problem 11.10.5
Analyze how fate of the Universe depends on the parameter $c$ in the holographic dark energy model, where IR cut-off $L$ is chosen to be equal to the future event horizon.
solution
For the flat Universe with dark energy, the dark energy would eventually dominate the density, $\Omega_{de}\to1$, hence the effective equation of state of the holographic dark energy evolves dynamically, and
\[w_{eff}\to-\frac13-\frac2{3c}.\]
For $c>1$, the parameter $w_{eff}>-1$, and the holographic dark energy behaves as quintessence. For $c<1$, $w_{eff}<-1$ could be realized. For such models, the Universe would end in a Big Rip.
11.10.6
Problem 11.10.6
In the case of interacting holographic Ricci dark energy with interaction is given by
\begin{eqnarray}
\label{intrate}
Q=\gamma H \rho_{_{\cal R}},
\end{eqnarray}
where $\gamma$ is a dimensionless parameter,
find the dependence of the density of dark energy and dark matter on the scale factor.
solution
The energy density of dark energy in the IRDE model is defined as
\begin{equation}\label{RDE}
\rho_{_{\cal R}} = 3\alpha M_{p}^{2} \left( \dot{H}+2H^{2} + \frac{k}{a^{2}}\right),
\end{equation}
where $\alpha$ is a dimensionless parameter.
Note that $\rho_{_{\cal R}}$ is proportional to the Ricci scalar curvature
\begin{equation}
{\cal{R}} = -6\left(\dot{H}+2H^{2}+\frac{k}{a^{2}}\right).
\end{equation}
The time evolution of the scale factor $a(t)$ is described by the Friedmann equation
\begin{eqnarray}
H^{2} = \frac{1}{3M_{p}^{2}}
(\rho_{_{\cal R}}+\rho_{m}+\rho_{\gamma}+\rho_{k}),
\label{Friedmann}
\end{eqnarray}
where $\rho_{_{\cal R}}, \rho_{m}, \rho_{\gamma}$ and $\rho_{k}$ represent the energy densities of dark energy,
matter, radiation and curvature, respectively.
The energy density of radiation is given by
$\rho_{\gamma}=\rho_{\gamma 0}a^{-4}$, where $\rho_{\gamma 0}$
is the present value of radiation density.
We adopt the convention that $a(t_{0})=1$ for the present age of the Universe
$t_{0} \approx 14$ Gyr.
The Friedmann equation (\ref{Friedmann})
is take the form
\begin{eqnarray}
\nonumber
\frac{ \alpha}{2} \frac{d^{2}H^{2}}{dx^{2}}
- \left(1-\frac{7\alpha}{2} -\frac{\alpha \gamma}{2} \right)\frac{d H^{2}}{dx}
-(3-6\alpha-2\alpha \gamma) H^{2}
&&\\
\label{EOH}
- \frac{\rho_{\gamma 0}}{3M_{p}^{2}} e^{-4x}
-\{1- \alpha(1+\gamma ) \} k e^{-2x}
&=&0,
\end{eqnarray}
where $x=\ln{a}$.
The solution to eq. (\ref{EOH}) is obtained as
\begin{eqnarray}\label{solution}
\frac{H^{2}}{H_{0}^{2}} &=& A_{+} e^{\sigma_{+} x} + A_{-} e^{\sigma_{-} x}
+ A_{\gamma} e^{-4 x} + A_{k} e^{-2 x} ,
\end{eqnarray}
where
\begin{eqnarray}
\label{sigmapm}
\sigma_{\pm}
&=& \frac{2-7\alpha-\alpha \gamma \pm\sqrt{(2-\alpha)^{2}
-2\alpha (\alpha +2) \gamma+\alpha^{2}\gamma^{2}}}{2\alpha},
\end{eqnarray}
$\Omega_{\gamma 0}=\rho_{\gamma 0}/\rho_{c0}$,
$\Omega_{k 0}= - k/H_{0}^{2}$ and
$\rho_{c0}=3M_{p}^{2}H_{0}^{2}$.
Note that $\sigma_{\pm}$ can be imaginary for
sufficiently large $\alpha$ and $\gamma$.
This implies that there is a parameter region where $H^{2}$
has oscillatory behavior.
However, this region is not phenomenologically viable.
The constants $\Omega_{\gamma 0}$ and $\Omega_{k 0}$ are the present
value of $\Omega_{\gamma}$ and $\Omega_{k}$, respectively.
The constants $A_{\gamma}$, $A_{k}$ and $A_{\pm}$ are given by
\begin{eqnarray}
A_{\gamma} &=& \Omega_{\gamma 0} ,
\end{eqnarray}
\begin{eqnarray}
A_{k} &=& \Omega_{k 0} ,
\end{eqnarray}
\begin{eqnarray}
\label{rholambda}
A_{\pm} &=& \pm \frac{
\alpha(\sigma_{\mp}+3) \Omega_{k 0}
+ 2 \Omega_{\Lambda 0}
-\alpha (1-\Omega_{\gamma 0})(\sigma_{\mp}+4) }
{\alpha(\sigma_{+}-\sigma_{-})}.
\end{eqnarray}
Replacing eq. (\ref{solution}) to eq. (\ref{RDE}),
the Ricci dark energy density is given by
\begin{eqnarray}
\label{rdes}
\rho_{_{\cal R}}
&=& \rho_{c0}
\sum_{i=+,-} \alpha
\left(\frac{ \sigma_{i}}{2}+2
\right)
A_{i} e^{\sigma_{i}x} .
\end{eqnarray}
Moreover , the matter density is
\begin{eqnarray}
\label{rms}
\rho_{m}
&=& \rho_{c0}
\sum_{i=+,-}
\left\{1-
\alpha \left(\frac{ \sigma_{i}}{2} +2 \right)
\right\}
A_{i} e^{\sigma_{i}x}.
\end{eqnarray}
The equation of state of dark energy can be found by
substituting eq. (\ref{rdes}) into the following expression:
\begin{eqnarray}
w_{_{\cal R}} = -1-\frac{1}{3}
\left(
\gamma + \frac{1}{\rho_{_{\cal R}}}\frac{d\rho_{_{\cal R}}}{dx}
\right).
\end{eqnarray}
In eqs.~(\ref{rdes}) and (\ref{rms}), the term proportional to $e^{\sigma_{-}x}$ is dominant in the past( $a$ $\ll$ 1), while
the term proportional to $e^{\sigma_{+}x}$ is dominant in the future ($a$ $\gg$ 1).
As an illustration, let us consider the case $\alpha$ $=$ 0.45 and $\gamma$
$=$ 0.15, which corresponds to $\sigma_{+}$ $\approx$ 0.25 and
$\sigma_{-}$ $\approx$ $-3.0$.
In the past ($a$ $\ll$ 1), the ratio of eq.~(\ref{rms}) to eq.~(\ref{rdes}) is
$\rho_m/\rho_{_{\cal R}}$ $\approx$ $\alpha^{-1}(2+\sigma_{-}/2)^{-1}-1$ $\approx$ 3.4,
while $\rho_m/\rho_{_{\cal R}}$ $\approx$ $\alpha^{-1}(2+\sigma_{+}/2)^{-1}-1$
$\approx$ 0.045 in the future ($a$ $\gg$ 1).
11.10.7
Problem 11.10.7
Find the exact solutions for linear interactions between Ricci DE and DM, if the energy density of Ricci DE is given by $ \rho_x =\left(2\dot H + 3\alpha H^2\right)/\Delta,$ where $\Delta=\alpha -\beta$ and $\alpha,\,\beta$ are constants.
solution
The Friedmann equation can be written as
\begin{equation}
\label{1}
3H^2=\rho = \rho_m + \rho_x,
\end{equation}
In terms of the variable $\eta = 3\ln(a/a_0)$, the compatibility between the global conservation equation
\begin{equation}
\label{2}
\rho ' = d\rho/d\eta= -\rho_c-(1+\omega_x)\rho_x,
\end{equation}
and the equation deduced from the expression of the modified holographic Ricci dark energy
\begin{equation}
\label{3}
\rho ' = -\alpha\rho_m-\beta\rho_x,
\end{equation}
namely, $(\rho_m + \gamma_x\rho_x)=(\alpha\rho_m + \beta\rho_x)$, gives a relation between the EoS parameter of the dark energy component $\omega_x = \gamma_x - 1$ and the ratio $r= \rho_m/\rho_x$
\begin{equation}
\label{4}
\omega_x=(\alpha - 1)r+\beta-1.
\end{equation}
Solving the system of equations (\ref{1}) and (\ref{3}), we get $\rho_m$ and $\rho_x$ in terms of $\rho$ and $\rho'$:
\begin{equation}
\label{5}
\rho_m=-(\beta\rho+\rho')/\Delta, \qquad \rho_x=(\alpha\rho+\rho')/\Delta.
\end{equation}
The interaction between the dark components is introduced through the term $Q$ by means of splitting the Eq.(\ref{3}) into $\rho'_m+ \alpha \rho_m = - Q $ and $ \rho'_x+ \beta \rho_x = Q $. Then, differentiating $\rho_m$ or $\rho_x$ in (\ref{5}) and using the expression of $Q$, we obtain a second order differential equation for the total energy density $\rho$ Chimento (2010)
\begin{equation}
\label{6}
\rho''+(\alpha+\beta)\rho'+\alpha\beta\rho = Q \Delta.
\end{equation}
For a given interaction $Q$, solving Eq. (\ref{6}) gives us the total energy density $\rho$ and the energy densities $\rho_{m}$ and $\rho_{x}$ after using Eq. (\ref{5}).
The general linear interaction $ Q$ Chimento (2010), linear in $\rho_{m}$, $\rho_{x}$, $\rho$, and $\rho'$, can be written as
\begin{eqnarray}
\label{7}
Q= c_1 \frac{(\gamma_s - \alpha)(\gamma_s-\beta)}{\Delta}\,\rho + c_2 (\gamma_s-\alpha)\rho_m \\
\nonumber
- c_3 (\gamma_s -\beta)\rho_x -c_4 \frac{(\gamma_s - \alpha)(\gamma_s-\beta)}{\gamma_s\Delta}\, \rho',
\end{eqnarray}
where $\gamma_s$ is constant and the coefficients $c_{i}$ fulfill the condition $c_{1}+c_{2}+c_{3}+c_{4}=1$ Chimento (2010).
Now, using Eqs. (\ref{5}), we rewrite the interaction (\ref{7}) as a linear combination of $\rho$ and $\rho'$,
\begin{equation}
\label{8}
Q=\frac{u\rho+\gamma^{-1}_{s}[u-(\gamma_s-\alpha)(\gamma_s-\beta)]\rho'}{\Delta},
\end{equation}
where $u=c_1(\gamma_s -\alpha)(\gamma_s-\beta)-c_{2}\beta(\gamma_s-\alpha)-c_{3}\alpha(\gamma_s-\beta)$.
Placing the interaction (\ref{8}) into the source equation (\ref{6}), we obtain
\begin{equation}
\label{9}
\rho'' + (\gamma_s+ \gamma^{+})\rho'+ \gamma_{s}\gamma^{+}\rho=0.
\end{equation}
where the roots of the characteristic polynomial associated with the second order linear differential equation (\ref{9}) are $\gamma_{s}$ and $\gamma^{+}=(\beta\alpha -u)/{\gamma_s}$.
In what follows, we adopt $\gamma^{+}=1$ in order to mimic the dust-like behavior of the Universe at early times. In that case, the general solution of (\ref{9}) is $\rho=b_1a^{-3\gamma_s}+b_2a^{-3}$, from which we obtain
\begin{equation}
\label{10a}
\rho_m=\frac{(\gamma_s-\beta)b_1a^{-3\gamma_s}+(1-\beta)b_2a^{-3}}{\Delta},
\end{equation}
\begin{equation}
\label{10b}
\rho_x=\frac{(\alpha-\gamma_s)b_1a^{-3\gamma_s}+(\alpha-1)b_2a^{-3}}{\Delta}.
\end{equation}
Interestingly, Eqs. (\ref{10a}) and (\ref{10b}) tell us that the interaction (\ref{8}) seems to be a good candidate for alleviating the cosmic coincidence problem, since the ratio $\Omega_{c}/\Omega_{x}$ becomes bounded for all times.
11.10.8
Problem 11.10.8
Find the equation of motion for the relative density
\begin{equation}\label{eq7}
\Omega_q=\frac{n^2}{H^2T^2},
\end{equation}
were
\begin{equation}\label{AGE_U}
T=\int_{0}^{a}{\frac{d{a}'}{H{a}'}}.
\end{equation}
of interacting agegraphic dark energy and the deceleration parameter, for sets of interaction term $Q=3\alpha H\rho_q;\; 3\beta H\rho_m;\; 3\gamma H\rho_{tot}.$
solution
From Eq.~(\ref{eq7}),
we get
\begin{equation}\label{eq13}
\Omega_q^\prime=\Omega_q\left(-2\frac{\dot{H}}{H^2}
-\frac{2}{n}\sqrt{\Omega_q}\right).
\end{equation}
Differentiating Eq.~\eqref{eq7} it is easy to find that
\begin{equation}\label{eq14}
-\frac{\dot{H}}{H^2}=\frac{3}{2}\left(1-\Omega_q\right)
+\frac{\Omega_q^{3/2}}{n}-\frac{Q}{6M_p^2 H^3}.
\end{equation}
Therefore, we obtain the equation of motion for $\Omega_q$,
\begin{equation}\label{eq15}
\Omega_q^\prime=\Omega_q\left[\left(1-\Omega_q\right)
\left(3-\frac{2}{n}\sqrt{\Omega_q}\right)
-\frac{Q}{3M_p^2 H^3}\right],
\end{equation}
where
\begin{equation}\label{eq16}
\frac{Q}{3M_p^2 H^3}=\left\{
\begin{array}{ll}
3\alpha\Omega_q & {\rm ~for~~} Q=3\alpha H\rho_q \\
3\beta\left(1-\Omega_q\right) & {\rm ~for~~} Q=3\beta H\rho_m \\
3\gamma & {\rm ~for~~} Q=3\gamma H\rho_{tot}
\end{array}
\right..
\end{equation}
From
(\ref{AGE_U}) and~(\ref{eq7}), we get the
EoS of the ADE, namely
\begin{equation}\label{eq17}
w_q=-1+\frac{2}{3n}\sqrt{\Omega_q}-\frac{Q}{3H\rho_q},
\end{equation}
where
\begin{equation}\label{eq18}
\frac{Q}{3H\rho_q}=\left\{
\begin{array}{ll}
\alpha & {\rm ~for~~} Q=3\alpha H\rho_q \\
\beta\left(\Omega_q^{-1}-1\right) & {\rm ~for~~} Q=3\beta H\rho_m\\
\gamma\,\Omega_q^{-1} & {\rm ~for~~} Q=3\gamma H\rho_{tot}
\end{array}
\right..
\end{equation}
Using Eq.~(\ref{eq14}), the deceleration parameter is given by
\begin{equation}\label{eq19}
q\equiv -\frac{\ddot{a}a}{\dot{a}^2}=-1-\frac{\dot{H}}{H^2} =\frac{1}{2}-\frac{3}{2}\Omega_q
+\frac{\Omega_q^{3/2}}{n}-\frac{Q}{6M_p^2 H^3}.
\end{equation}