For a typical spiral galaxy $M_{gal}\approx10^{12} M_\odot$, $R\approx30kpc$, for globular clusters, $M\approx10^{6} M_\odot$, $R\approx100pc$, for galaxy clusters, $M_C\approx10^{16} M_\odot$, $R_C\approx3Mpc$, so for all these structures the equation holds.

\begin{tabular}{|c|c|} \hline UNIQ-MathJax12-QINU (UNIQ-MathJax13-QINU) & Object \\ \hline UNIQ-MathJax14-QINU & Galaxy \\ UNIQ-MathJax15-QINU & Universe \\ UNIQ-MathJax16-QINU & globular cluster, GMC, etc. \\ UNIQ-MathJax17-QINU & supercluster \\ UNIQ-MathJax18-QINU & nuclei \\ UNIQ-MathJax19-QINU & electron \\ UNIQ-MathJax20-QINU & Solar system, planetary nebula\\ \hline \end{tabular}

Let the scalar field $\varphi(t)$ is a single-valued function of time, then it is possible to reparametrize the Hubble parameter in terms of $\varphi$: \[ H(t) = H[\varphi(t)]. \] Taking time derivatives in the Friedman equation \[ \frac{1}{2}\, \dot{\varphi}^2 + V = 3 H^2, \] one arrives at the results: \[ \dot{\varphi} ( \ddot{\varphi} + V' ) = 6 H \dot{H},\quad \dot{H} \equiv H' \dot{\varphi}. \] Taking into account the Klein Gordon equation \[ \ddot{\varphi} + 3 H \dot{\varphi} + V' = 0, \] it follows, that for $\dot{\varphi} \neq 0$ and $H \neq 0$ one gets \[ \dot{\varphi} = - 2 H', \quad \dot{H} = - \frac{1}{2}\, \dot{\varphi}^2 \leq 0. \] Thus the Hubble parameter is a semi-monotonically decreasing function of time.

\[\dot H=-\frac12(\rho+p).\] If $\rho+p\ge0$ (null energy condition), $H$ is a semi-monotonically decreasing function of time.

For single-field models in which the scalar field is a single-valued function of time in some interval, it is possible to reparametrize the Hubble parameter in terms of $\varphi$: \[H(t)=H[\varphi(t)]\] Replacing the time derivatives $\dot{\varphi} = - 2 H'$ (see the problem \ref{SSC_0}) in the Friedmann equation we find \[V=3H^2-2H'^2.\] The latter expression can be used to reconstruct the potential if the evolution history $H[\varphi(t)]$ is known, or for given $V(\varphi)$ this is a first-order differential equation for $H[\varphi(t)]$.

Replacing the time derivatives in the Friedmann equation using the results of the previous problem, one finds \[ 2 H^{\prime\, 2} - 3 H^2 + V(\varphi) = 0. \] There are two kinds of stationary points; a point where $\dot{\varphi} = H' = 0$ is an end point of the evolution if \[ \ddot{\varphi} = 4 H' H'' = 0, \] which happens if $H''$ is finite. In contrast, if \[ \ddot{\varphi} = 4 H' H'' \neq 0, \] $H''$ necessarily diverges in such a way as to make $\ddot{\varphi}$ finite: $H'' \propto 1/H'$.

For such a scalar field to exist it is required that \[ H' = - \frac{1}{2}\, \dot{\varphi} = \frac{\omega \varphi_0}{2} \sin \omega t = \frac{\omega}{2} \sqrt{\varphi_0^2 - \varphi^2}. \] There are infinitely many stationary points \[ \omega t_n = n \pi, \quad \varphi(t_n) = (-1)^n \varphi_0, \] where $H' = 0$. Now \[ H'' = - \frac{1}{2} \frac{\omega \varphi}{\sqrt{\varphi_0^2 - \varphi^2}}, \] and therefore $H''$ diverges at all stationary points $t_n$, but in such a way that \[ 4 H' H'' = - \omega^2 \varphi = \ddot{\varphi}. \] Then all stationary points in the considered model are turning points.

\begin{align} H & = H_0 - \frac{1}{4} \omega \varphi_0^2 \arccos \left( \frac{\varphi}{\varphi_0} \right) +\frac{1}{4} \omega \varphi \sqrt{\varphi_0^2 - \varphi^2} \\ & = H_0 - \frac{1}{4} \omega^2 \varphi_0^2 t + \frac{1}{8} \omega \varphi_0^2 \sin 2 \omega t. \end{align}

The corresponding solution for the scale factor is \[ a(t) = a(0) \exp\left\{H_0 t - \frac{1}{8} \omega^2 \varphi_0^2 t^2 + \frac{1}{16} \left( 1 - \cos 2 \omega t \right)\right\}. \] which is a gaussian, slightly modulated by an oscillating function of time (see figure).

Scalefactor $a(t)$ for an eternally oscillating scalar field.

The potential giving rise to this behavior reads \begin{align} V & = 3 H^2 - 2 H^{\prime\, 2} \\ & = 3 \left( H_0 - \frac{1}{4}\, \omega \varphi_0^2 \arccos \left( \frac{\varphi}{\varphi_0} \right) + \frac{1}{4} \omega \varphi \sqrt{ \varphi_0^2 - \varphi^2} \right)^2 - \frac{\omega^2}{2} \left( \varphi_0^2 - \varphi^2 \right). \end{align} Observe, that this potential keeps track of the number of oscillations the scalar field has performed through the arccos-function, so ultimately $V$ increases indefinitely as a function of time, whilst the volume of a representative domain of space decreases rapidly.

If $H$ becomes negative then the Universe inevitably collapses.If $H$ never becomes negative, it must tend to a vanishing or positive final minimum, which can be reached either in finite or infinite time. The universe then ends up in a Minkowski or in a de Sitter state. These conclusions are a consequence of the non-positivity of $\dot{H}$ (see problem #SSC_0, which implies that a negative $H$ can never return to larger values at later times.

In order to establish the existence of end points or asymptotic end points of evolution at non-negative values of $H$, we first consider the locus of all stationary points, defined by \[ \dot{\varphi} = - 2 H' = 0 \quad \Rightarrow \quad V = 3 H^2 \geq 0 . \] It follows that stationary points can occur only in the region of positive or vanishing potential. In particular this holds for end points, which therefore do not occur in a region of negative potential. Moreover, it is clear that a Minkowski final state occurs only at a stationary point where $V = 0$, whereas all stationary points with $V > 0$ correspond to de Sitter states. To correspond to an end point of the evolution, $H''$ must be finite at these stationary points to guarantee that $\ddot{\varphi} = 0$ as well. From the results of the problem \ref{SSC_1} it follows that \[ V' = 2 H' (3 H - 2 H''), \] and therefore $V' = 0$ if $H' = 0$ and $H$ and $H''$ are finite. As a result end points of the evolution necessarily occur at an extremum of $V$, but only if $V \geq 0$ there.

We distinguish the cases $v_0 > 0$, $v_0 = 0$ and $v_0 < 0$. The stationary points are represented graphically by the curves in the $\varphi$-$H$-plane in figure.

• Critical curves $H'[\varphi] = 0$ for quadratic potentials with $v_0
• 

  a)

• 

  b)

• 

  c)

Critical curves $H'[\varphi] = 0$ for quadratic potentials with $v_0 > 0$ (a)), $v_0 = 0$ (b)) and $v_0 < 0$ (c)).

For $v_0 > 0$ there exists a stationary point for any value of $\varphi$, but the potential has a unique minimum at $\varphi = 0$, which is the only stationary point where $V' = 0$, and therefore the only end point. Indeed, once this point is reached $H$ can not decrease anymore and we have final state of de Sitter type.

For $v_0 = 0$ the critical curves become straight lines, crossing at the origin where $H = 0$ at $V = 0$. This is still a stationary point with $\ddot{\varphi} = 0$ representing a Minkoswki state, but as $V'$ is not defined there it is really to be interpreted as a limit of the previous case. There are no evolution curves flowing from the domain $H > 0$ to the domain $H < 0$.

Substitution into equation $2 H^{\prime\, 2} - 3 H^2 + V(\varphi) = 0 $ then leads to the equalities \[ 3 h_0^2 - 2 h_1^2 = v_0, \quad h_1(3h_0 - 4 h_2) = 0, \quad 4h_1(h_1 - 4 h_3) + \frac{8}{3}\, h_2 ( 3 h_0 - 4 h_2 ) = m^2, \quad ..., \] from which the solutions $H[\varphi]$ can be reconstructed (see figure below).

Solutions $H[\varphi]$ for quadratic potentials (problem #SSC_6_1) with $v_0 > 0$.

Solutions $H[\varphi]$ for quadratic potentials (problem #SSC_6_1) with $v_0 < 0$.

Using the results of previous problem, one can define the number of $e$-folds in some interval of time: \[ N = \int_{t_1}^{t_2} dt H = - \int_{\varphi_1}^{\varphi_2} d\varphi\, \frac{H}{2H'} = - \frac{1}{2}\, \int_{\varphi_1}^{\varphi_2} d\varphi\, \frac{h_0 + h_1 \varphi + h_2 \varphi^2 + ...}{h_1 + 2 h_2 \varphi + 3 h_3 \varphi^2 + ...}. \] This number can get sizeable contributions only in regions where the slow-roll condition is satisfied: \[ \varepsilon = - \frac{\dot{H}}{H^2} = \frac{2H^{\prime\, 2}}{H^2} < 1 \quad \Rightarrow \quad 3H^2 - V < H^2. \] Thus we simultaneously have \[ V < 3 H^2 \quad \mbox{and} \quad V > 2H^2 \quad \Leftrightarrow \quad 0 \leq \frac{V}{3} < H^2 < \frac{V}{2}. \] In most cases this holds only for a relatively narrow range of field values.

In both cases $\dot{\varphi} = 0$, but at end points in addition $\ddot{\varphi} = 0$, which can happen only at extrema of the potential $V[\varphi]$. However, if the end point occurs at a relative maximum or saddle point of the potential, this end point will be classically unstable. Indeed, the field can remain there for an indefinite period of time, but any slight change in the initial conditions will cause it to move on to lower values of the Hubble parameter. Nevertheless, such a period of temporary slow roll of the field creates the right conditions for a period of inflation.

The Hubble parameter and potential giving rise to this solution can be constructed following the same procedure as for the eternally oscillating field (see problem #SSC_2), with the result \[ H = h + \frac{1}{4}\, \omega \varphi^2, \quad V[\varphi] = v_0 - \frac{\mu^2}{2}\, \varphi^2 + \frac{\lambda}{4}\, \varphi^4, \] where \[ v_0 = 3 h^2, \quad \mu^2 = \omega^2 - 3 \omega h, \quad \lambda = \frac{3 \omega^2}{4}. \] Thus we obtain a quartic potential; for $\mu^2 > 0$ it has minima in which reflection symmetry is spontaneously broken. The exponential solution ends asymptotically at the unstable maximum of the potential where $\dot{\varphi} = \ddot{\varphi} = 0$. As such it represents an end point of the evolution, but a minimal change in the initial conditions for the scalar field will turn the end point into a reflection point (if it starts at lower $H$), or it will overshoot the maximum (if it starts at higher $H$). Thus the end point is unstable, but the exponential decay will still provide a good approximation to first part of the evolution of the universe for solutions $H[\varphi]$ coming close to the maximum of the potential (see figure below).

Critical curves of stationary points and solutions $H[\varphi]$ for a quartic potential with spontaneous symmetry breaking.

The exponential scalar field leads to a behavior of the scale factor given by \[ a(t) = a_0\, e^{ht + \frac{1}{8} \varphi^2_0\, (1 - e^{-2\omega t})}. \] Thus for $h > 0$ this epoch in the evolution of the Universe ends in an asymptotic de Sitter state with Hubble constant $h$. Afterwards, the scalar field will roll further down the potential; provided $3h \leq \omega \leq 6h$ it will oscillate around the minimum until it comes to rest in another de Sitter or a Minkowski state, again depending on the value of $h$. In particular, for $\omega \geq 3h$ the model has a final de Sitter or Minkowski state in which $\dot{\varphi} = 0$ and \[ \langle \varphi^2 \rangle = \frac{\mu^2}{\lambda} = \frac{4}{3} \left(1 - \frac{3h}{\omega} \right). \] In this final state the energy density is \[ \langle V \rangle = v_0 - \frac{\mu^4}{4\lambda} = \frac{\omega}{3} ( 6h - \omega ). \]

The energy density for the solution of problem #SSC_9 is \[ \rho_s(t) = \frac{1}{2}\, \dot{\varphi}^2 + V = 3 H^2 = 3 \left( h + \frac{1}{4}\, \omega\, \varphi_0^2\, e^{-2 \omega t} \right)^2. \] Now the solution for the scale factor \[ a(t) = a_0\, e^{ht + \frac{1}{8} \varphi^2_0\, (1 - e^{-2\omega t})}. \] shows, that before reaching the first turning point at $\varphi = 0$ the scale factor increases by an additional number of $e$-folds given by \[ N = \frac{1}{8}\, \varphi_0^2. \] Therefore the initial energy density at $t = 0$ can be written as \[ \rho_s(0) = 3 ( h + 2N \omega )^2. \] If we take this initial energy density to equal the Planck density: $\rho_s(0) = 1$, this establishes a simple relation between $h$, $\omega$ and $N$.

Taking the final energy density $\langle V \rangle$ (see problem #SSC_10) equal to the observed energy density of the Universe today: \[ \langle V \rangle = 3 H_0^2 = 1.04 \times 10^{-120} \] in Planck units, being so close to zero, one can set to an extremely good approximation $\omega = 6h$, and \[ 3 h^2 ( 1 + 12 N )^2 = 1, \quad \mu^2 = 18 h^2, \quad \lambda = 27 h^2. \] The lower limit on $N$ for inflation as derived from the CMB observations is $N \geq 60$, which requires \begin{equation} h \leq 0.8 \times 10^{-3}.\label{h_estimate} \end{equation} Now expanding $\varphi$ around its vacuum expectation value \[ \varphi = \frac{\mu}{\sqrt{\lambda}} + \chi, \] the potential becomes \[ V = \frac{1}{2}\, m_{\chi}^2 \chi^2 + \frac{\alpha}{3}\, m_{\chi} \chi^3 + \frac{\lambda}{4}\, \chi^4, \] where \[ m_{\chi} = 6h, \quad \alpha = 9h , \quad \lambda = 27 h^2. \] According to the estimate (\ref{h_estimate}) the upper limits on these parameters are \[ m_{\chi} = 0.48 \times 10^{-2}, \quad \alpha = 0.72 \times 10^{-2}\approx1/137, \quad \lambda = 0.17 \times 10^{-4}. \] Converting to particle physics units, the upper limit on the mass is $m_{\chi} \leq 1.2 \times 10^{-16}$ GeV. This suggests that the inflaton could be associated with a GUT scalar of Brout-Englert-Higgs type.

There are several ways to obtain the result:
1) \[q=-\frac{\ddot a }{aH^2};\quad \frac{\ddot a}{a}=-\frac16(\rho+3p);\quad H^2=\frac13\rho;\] \[q=\frac{\dot\varphi^2-V}{\frac{\dot\varphi^2}2+V};\] 2) \[q=\frac12\Omega_{tot}+\frac32\sum\limits_iw_i\Omega_i.\] For a flat single-component Universe one obtains \[q=\frac12+\frac32w=\frac12+\frac32\frac{\dot\varphi^2-2V}{\dot\varphi^2+2V}=\frac{\dot\varphi^2-V}{\frac{\dot\varphi^2}2+V};\] 3) \[q=\frac{d}{dt}\frac 1 H-1=-\frac{\dot H}{H^2}-1=\frac{3H^2-V}{H^2}-1=\frac{2H^2-V}{H^2}=\frac{\dot\varphi^2-V}{\frac{\dot\varphi^2}2+V}\]

By substituting the Hubble function \[H^2=\frac13\left(\frac12\dot\varphi^2+V(\varphi)\right)\] into the Klein-Gordon equation we obtain \[\ddot\varphi+\sqrt3\sqrt{\frac12\dot\varphi^2+V(\varphi)}\dot\varphi + \frac{dV}{d\varphi}=0\] Introducing $\dot\varphi=\sqrt{f(\varphi)}$ and changing the independent variable from $t$ to $\varphi$, transform last equation into \[\frac12\frac{df(\varphi)}{d\varphi}+\sqrt3\sqrt{\frac12f(\varphi+V(\varphi)}\sqrt{f(\varphi)}+\frac{dV}{d\varphi}=0.\]

Slow-roll inflation corresponds to \[\varepsilon\equiv-\frac{\dot H}{H}\ll1.\] For power law expansion $H=n/t$ so that $\varepsilon=n^{-1}$. Consequently, slow-roll inflation occurs when $n\gg1$.

\[\bar H=\frac q t,\quad \bar H'=-\frac q{t^2}.\] Using \[\bar H'=-\frac{1+3w}{2}\bar H^2,\] one obtains \[w=\frac{2-q}{3q}.\]

\[H=\frac{\dot a}{a}=\frac\alpha t+\frac\beta{t_0},\] \[q=-\frac{\ddot a}{aH^2}=\frac{\alpha t_0^2}{(\beta t +\alpha t_0)^2}-1,\] \[j=\frac{\ddot a}{aH^3}=1+\frac{(2t_0-3\beta t-3\alpha t_0)\alpha t_0^2}{(\beta t+\alpha t_0)^3}.\]

\[t\to0:\] \[a\to a_0\left(\frac{t}{t_0}\right)^\alpha,\quad H\to\frac\alpha t,\quad q\to-1+\frac1\alpha,\quad j\to 1-\frac3\alpha + \frac2{\alpha^2};\] \[t\to\infty:\] \[a\to a_0\exp\left[\beta\left(\frac{t}{t_0}-1\right)\right],\quad H\to\frac\beta{t_0},\quad q\to-1,\quad j\to1.\]

\[\rho = 3H^2,\] \[p=-2\frac{\ddot a}a -H^2=-2\dot H-3H^2,\] \[w=\frac p\rho=-\frac23\frac{\dot H}{H^2}-1,\] \[H=\frac\alpha{t}+\frac\beta{t_0},\quad \dot H=-\frac\alpha{t^2},\] \[w=\frac23\frac\alpha{t^2}\left(\frac\alpha{t}+\frac\beta{t_0}\right)^{-2}-1.\]

The energy density and pressure of the quintessence minimally coupled to gravity can be given by \[\rho=\frac12\dot\varphi^2+V(\varphi),\quad p=\frac12\dot\varphi^2-V(\varphi),\] Using the hybrid expansion law and relation \[p+\rho=-2\dot H=\frac{2\alpha}{t^2}\] we find \[\varphi(t)=\sqrt{2\alpha}\ln(t)+\varphi_1\] \[V(t)=3\left(\frac\alpha{t}+\frac\beta{t_0}\right)^{2}-\frac\alpha{t^2},\] where $\varphi_1$ is the integration constant. The potential as a function of the scalar field $\varphi$ is then given by the following expression: \[V(\varphi)=3\beta^2e^{-\sqrt{\frac2\alpha}(\varphi_0-\varphi_1)}+\alpha(3\alpha-1)e^{-\sqrt{\frac2\alpha}(\varphi-\varphi_1)} + 6\alpha\beta e^{-\frac12\sqrt{\frac2\alpha}(\varphi+\varphi_0-2\varphi_1)}\] where $\varphi_0=\varphi_1+\sqrt{2\alpha}\ln(t_0)$.

For the tachyon field \[\frac p\rho=w=-1+\dot\varphi^2.\] Any realization of the hybrid expansion law gives \[\frac23\frac\alpha{t^2}\left(\frac\alpha{t}+\frac\beta{t_0}\right)^{-2}-1.\] Consequently, \[\dot\varphi= \sqrt{\frac{2\alpha}{3}}\left(\alpha+\beta\frac{t}{t_0}\right)^{-1}\] Integration of the latter results in the following \[\varphi(t)=\sqrt{\frac{2\alpha t_0^2}{3\beta}}\ln{(\beta t+\alpha t_0)}+\varphi_2\] and \[V(t)=3\left(\frac\alpha{t}+\frac\beta{t_0}\right)^{2}\sqrt{1-\frac{2\alpha t_0^2}{3(\beta t+\alpha t_0)^2}}.\] where $\varphi_2$ is an integration constant. The corresponding tachyon potential is given by \[V(\varphi)=\frac{3 \beta ^2}{t_{0}^2}e^{\sqrt{\frac{6\beta^2}{\alpha t_{0}^2}}(\varphi-\varphi_{2})}\sqrt{1-\frac{2}{3}\alpha t_{0}^2 e^{\sqrt{\frac{6\beta^2}{\alpha t_{0}^2}}(\varphi-\varphi_{2})}}\left(\alpha t_{0}-e^{\frac{1}{2}\sqrt{\frac{6\beta^2}{\alpha t_{0}^2}}(\varphi-\varphi_{2})}\right)^{-2}.\]

In case of the phantom scenario, the hybrid expansion law ansatz must be slightly modified in order to acquire self consistency. In particular, we rescale time as $t\rightarrow t_{s}-t$, where $t_{s}$ is a sufficiently positive reference time. Thus, the hybrid expansion law ansatz becomes \[a(t)=a_{0}\left(\frac{t_{s}-t}{t_{s}-t_{0}}\right)^{\alpha}e^{\beta \left(\frac{t_{s}-t}{t_{s}-t_{0}}-1\right)},\] where $\alpha<0$. Then \[ H=-\frac{\alpha}{t_{s}-t}-\frac{\beta}{t_{s}-t_{0}}, \] \[ \dot{H}=-\frac{\alpha}{(t_{s}-t)^2}, \] \[ q=\frac{\alpha (t_{s}-t_{0})^2}{[\beta (t_{s}-t)+\alpha (t_{s}-t_{0})]^{2}}-1. \]

\[\varphi(t)=\sqrt{-2\alpha }\ln(t_{s}-t)+\varphi_{3},\] \[V{(t)}=3\left(\frac{\alpha}{t_{s}-t}+\frac{\beta}{t_{s}-t_{0}}\right)^2-\frac{\alpha}{(t_{s}-t)^2},\] \[V(\varphi) = 3\beta^{2}e^{-\sqrt{-\frac{2}{\alpha }}(\varphi_{0}-\varphi_{3})}+\alpha(3\alpha-1)e^{-\sqrt{-\frac{2}{\alpha }}(\varphi-\varphi_{3})} +6\alpha\beta e^{\frac{1}{2}\sqrt{-\frac{2}{\alpha }}(\varphi+\varphi_{0}-2\varphi_{3})}, \] where $\varphi_{0}=\varphi_{3}+\sqrt{-2\alpha }\ln(t_{s}-t_{0})$. We observe that $\alpha<0$ leads to $q<0$ (acceleration) and \[\dot{H}=-\frac{\alpha}{(t_{s}-t)^2}>0\] (super acceleration)

If we admit the energy-momentum tensor of a perfect fluid, then the field equations of the BI universe are found as, \begin{eqnarray} \label{feforgm}\frac{{\dot{a}_{1}}{\dot{a}_{2}}}{a_{1} a_{2}}+\frac{{\dot{a}_{1}}{\dot{a}_{3}}}{a_{1} a_{3}}+\frac{{\dot{a}_{2}}{\dot{a}_{3}}}{a_{2} a_{3}}&=&\rho,\\ \frac{{\ddot{a}_{1}}}{a_{1}}+\frac{{\ddot{a}_{3}}}{a_{3}}+\frac{{\dot{a}_{1}}{\dot{a}_{3}}}{a_{1} a_{3}}&=& -p,\\ \frac{{\ddot{a}_{2}}}{a_{2}}+\frac{{\ddot{a}_{1}}}{a_{1}}+\frac{{\dot{a}_{2}}{\dot{a}_{1}}}{a_{2} a_{1}}&=&-p,\\ \frac{{\ddot{a}_{3}}}{a_{3}}+\frac{{\ddot{a}_{2}}}{a_{2}}+\frac{{\dot{a}_{3}}{\dot{a}_{2}}}{a_{3} a_{2}}&=&-p. \end{eqnarray}

Inserting the directional Hubble parameters and their time derivatives \[\dot H_1=\frac{\ddot a_1}{a_1}-\left(\frac{\dot a_1}{a_1}\right)^2,\ \dot H_2=\frac{\ddot a_2}{a_2}-\left(\frac{\dot a_2}{a_2}\right)^2,\ \dot H_3=\frac{\ddot a_3}{a_3}-\left(\frac{\dot a_3}{a_3}\right)^2\] into the modified Friedmann equations we obtain \begin{align} \nonumber H_1H_2+H_1H_3+H_2H_3 & =\rho,\\ \nonumber \dot H_1+ H_1^2 +\dot H_3+ H_3^2 +H_1H_3& =-p,\\ \nonumber \dot H_1+ H_1^2 +\dot H_2+ H_2^2 +H_1H_2& =-p,\\ \nonumber \dot H_2+ H_2^2 +\dot H_3+ H_3^2 +H_2H_3& =-p. \end{align}

\[\rho_{cr}=H_1H_2+H_1H_3+H_2H_3.\]

The energy conservation equation $T^\mu_{\nu;\mu}=0$ yields \[\dot\rho+3\bar H(\rho+p)=0,\quad \bar H\equiv \frac13(H_1+H_2+H_3)=\frac13\left(\frac{\dot a_1}{a_1} +\frac{\dot a_2}{a_2} +\frac{\dot a_3}{a_3}\right),\] where $\bar H$ represents the mean of the three directional Hubble parameters in the BI Universe.

Adding the three latter Friedmann equations (see Problem \ref{bi_2}) one obtains \begin{equation}\label{bi_5_1} 2\frac{d}{dt}\sum\limits_{i=1}^3 H_i+2(H_1^2+H_2^2+H_3^2)+H_1H_2+H_1H_3+H_2H_3=-3p. \end{equation} where $\bar H$ represents the mean of the three directional Hubble parameters in the BI Universe. Substituting \[\sum\limits_{i=1}^3 H_i^2=\left(\sum\limits_{i=1}^3 H_i\right)^2-2(H_1H_2+H_1H_3+H_2H_3)\] and \[H_1H_2+H_1H_3+H_2H_3=\rho\] into equation (\ref{bi_5_1}), we then obtain \[\frac{d}{dt}\sum\limits_{i=1}^3 H_i+\left(\sum\limits_{i=1}^3 H_i\right)^2=\frac32(\rho-p).\] Using the mean of the three directional Hubble parameters $\bar H$ we obtain a nonlinear first order differential equation \[\dot{\bar H}+3\bar H^2=\frac12(\rho-p).\]

\[\bar H=\frac13\frac{d}{dt}\ln(a_1a_2a_3)=\frac13\frac{d}{dt}\ln V=\frac13\frac{\dot V}{V}.\]

Using the relation between volume $V$ and the mean Hubble parameter $\bar H$, obtained in the previous problem, one finds \[\dot{\bar H}=\frac13\frac{\ddot V}{V}-3\bar H^2.\] As \[\dot{\bar H}+3\bar H^2=\frac12(\rho-p),\] we obtain \[\ddot V-\frac32(\rho-p)V=0.\]

The equations \begin{align} \nonumber \dot H_1+ 3H_1\bar H & =\frac12(\rho-p),\\ \nonumber \dot H_2+ 3H_2\bar H & =\frac12(\rho-p),\\ \nonumber \dot H_3+ 3H_3\bar H & =\frac12(\rho-p), \end{align} allow us to write the generic solution of the directional Hubble parameters, \[H_i(t)=\frac1{\mu(t)}\left[K_i+\frac12\int\mu(t)(\rho(t)-p(t))dt\right],\quad i=1,2,3,\] where $K_i$s are the integration constants. The integration factor $\mu$ is defined as, \[\mu(t)=\exp\left(3\int\bar H(t)dt\right).\] As can be seen, the initial values (integration constants) determine the solution of each directional Hubble parameter. These values are the origin of the anisotropy. Note that the generic solution of the directional Hubble parameters is incomplete. To obtain exact solutions of the Hubble parameters and therefore the Einstein equations, one has to know the state equation for the component which fills the Universe.

By using the energy conservation equation \[\dot\rho+3\bar H(\rho+p)=0\to\dot\rho+4\bar H\rho=0,\] and the volume representation of the mean Hubble parameter \[\bar H=\frac13\frac{d}{dt}\ln(a_1a_2a_3)=\frac13\frac{d}{dt}\ln V=\frac13\frac{\dot V}{V}.\] we obtain (with $\rho\to\rho_r$, $V\to V_r$): \[\rho_r=\rho_{r0}\left(\frac{V_{r0}}{V_r}\right)^{4/3}.\] Here the density and the volume element is normalized to the present time $t_0$. The parameters $\rho_{r0}$ and $V_{r0}$ are the normalized density and normalized volume elements.

For the radiation dominated case \[\ddot V_r-\frac32(\rho-p)V_r=0\to\ddot V_r-V_r\rho_r=0.\] (see problem 8). Using \[\rho_r=\rho_{r0}\left(\frac{V_{r0}}{V_r}\right)^{4/3}\] we obtain (for $V_{r0}=1$) \[\ddot V_r-\rho_{r0}V_r^{-1/3}=0.\] Multiplying this equation with the $\dot V_r$ and integrating it, yields, \[\dot V_r^2-3\rho_{r0}V_r^{2/3}=0.\] Hence, the exact solution of the volume evolution equation is \[V_r=(2H_0t)^{3/2}.\] The mean Hubble parameter of the radiation dominated case is \[\bar H=\frac13\frac{\dot V_r}{V}=\frac1{2t}.\]

The generic solution of the directional Hubble parameters (see problem 9) is \[H_i(t)=\frac1{\mu(t)}\left[K_i+\frac12\int\mu(t)(\rho(t)-p(t))dt\right],\quad i=1,2,3,\] Using the expression for the mean Hubble parameter obtained in the previous problem, one finds \[\mu_r(t)=\exp(3\int\bar H(t)dt)\] By direct substitution of the integration factor $\mu_r$ and the equation of state $p_r=\rho_r/3$ of the radiation dominated case we obtain for the directional Hubble parameters that are normalized to the present-day time $t_0$ the following results \[H_{r,i}t_0=\alpha_{r,i}\left(\frac{t_0}{t}\right)^{3/2}+\frac12\frac{t_0}{t};\quad \alpha_{r,i}\equiv\frac{K_{r,i}}{t_0}.\]

The normalized scale factors $a_i$ can be obtained from the directional Hubble parameters \[H_{r,i}=\alpha_{r,i}\frac1{t_0}\left(\frac{t_0}{t}\right)^{3/2}+\frac12\frac{1}{t},\] with a direct integration in terms of cosmic time, \[a_{r,i}=\exp\left[-2\alpha_{r,i}\left(\sqrt{\frac{t_0}{t}}-1\right)\right]\left(\frac{t_0}{t}\right)^{1/2}.\] The scale factors of the BI radiation dominated model has the contribution from anisotropic expansion/contraction \[\exp\left[-2\alpha_{r,i}\left(\sqrt{\frac{t_0}{t}}-1\right)\right]\] as well as the standard matter dominated FLRW contribution $(t/t_0)^{1/2}$. These two different dynamical behaviors in three directional scale factors of the BI universe indicate that the FLRW part of the scale factor becomes dominant when time starts reaching the present-day. On the other hand, in the early times of the BI model, the expansion is completely dominated by the anisotropic part.

Equations of state for the considered components read: \begin{eqnarray}\label{eosrm} {p}_{m}=0,\phantom{a}{p}_{r}=\frac{1}{3}{\rho}_{r}. \end{eqnarray} \noindent As a result, the energy conservation equations in the radiation-matter period are \begin{eqnarray} \dot{\rho}_{r}=-4 \bar H_{rm} {\rho}_{r},\phantom{a} \dot{\rho}_{m}=-3 \bar H_{rm}{\rho}_{m}, \label{energyconsermatradzero} \end{eqnarray} \noindent Using the definition \[\bar H=\frac13\frac{\dot V}{V}\] one obtains \begin{eqnarray} {\rho}_{r}=\rho_{r,0}\left(\frac{V_{rm,0}}{V_{rm}}\right)^{4/3},\phantom{a} {\rho}_{m}=\rho_{m,0}\frac{V_{rm,0}}{V_{rm}}. \label{energyconsermatrad} \end{eqnarray}

In the considered case of radiation+matter dominated BI Universe \[\bar H +3H^2=\frac12(\rho-p) \to \dot{\bar H}_{rm}+3{\bar H}^2_{rm}=\frac12\left(\frac{2}{3}\rho_{r}+\rho_{m}\right).\] Substitution of \[\bar H=\frac13\frac{\dot V_{rm}}{V_{rm}}\] gives \[{\ddot{V}_{rm}}-\frac32\left(\rho_{m,0}+\frac{2}{3}\frac{\rho_{r,0}}{V_{rm}^{1/3}}\right)=0.\] Multiplying this equation with $\dot{V}_{rm}$, integrate it in terms of time, and substitute the normalized densities \[{\rho}_{r,0}=3\bar H^2_{0}\Omega_{r,0},\quad{\rho}_{m,0}=3\bar H^2_{0}\Omega_{m,0},\] we then obtain \[{{\dot V}_{rm}}^2-9\bar H^2_{0}\Omega_{m,0} V_{rm} - 9\bar H^2_{0}\Omega_{r,0} V_{rm}^{2/3}=0.\]

\[\left(\frac{\bar H_{rm}}{\bar H_0}\right)^2=\frac{\Omega_{m,0}}{V_{rm}}+\frac{\Omega_{r,0}}{V_{rm}^{4/3}}.\]

From $F(\rho,p)=0$ it follows that \[\frac{\partial F}{\partial\rho}d\rho + \frac{\partial F}{\partial p}dp=0\] which leads to \[c_s^2=\frac{dp}{d\rho}=-\frac{\frac{\partial F}{\partial\rho}}{\frac{\partial F}{\partial p}}.\]

Inserting $p=w\rho$ in \[\frac{\partial F}{\partial\rho}d\rho + \frac{\partial F}{\partial p}dp=0\] and using the definition of the speed of sound we obtain \[\frac{d\rho}{\rho}=\frac{dw}{c_s^2-w}.\] Combining this expression with the continuity equation for the fluid results in equation \[\frac{dw}{(c_s^2-w)(1+w)}=-3\frac{da}a=3\frac{dz}{1+z}.\] For $c_s^2=const$ parameter of EOS evolves as \[w=\frac{c_s^2\frac{1+w_0}{c_s^2-w_0}(1+z)^{3(1+c_s^2)}-1}{\frac{1+w_0}{c_s^2-w_0}(1+z)^{3(1+c_s^2)}+1}.\] From this relation it immediately follows that \[\rho=\rho_0\frac{c_s^2-w_0}{c_s^2-w}= \rho_0\frac{c_s^2-w_0}{1+c_s^2}\left[\frac{1+w_0}{c_s^2-w_0}(1+z)^{3(1+c_s^2)}+1\right]\] and \[p=c_s^2\rho-\rho_0(c_s^2-w_0)= \rho_0\frac{c_s^2-w_0}{1+c_s^2}\left[\frac{1+w_0}{c_s^2-w_0}(1+z)^{3(1+c_s^2)}-1\right].\]

Dimensionality of the fundamental constants $c,\hbar,G_D$ in $D=4+n$ dimensions can be determined as \[[G_d]=L^{D-1}T^{-2}M^{-1},\quad \hbar=L^2T^{-1}M,\quad c=LT^{-1}.\] Note that the dimension of the space affects only the dimensionality of the Newton's constant $G_D$, because the universal gravitation law transforms with changes of dimensionality of the space as the following \[F=G_D\frac{M_1M_2}{R^{D-2}}.\] Use the combination \[[G_D^\alpha\hbar^\beta c^\gamma]= L^{\alpha(D-1)+2\beta+\gamma} T^{-2\alpha-\beta-\gamma} M^{-\alpha+\beta-\gamma}\] to find that \[L_{P(D)}=\left(\frac{G_D\hbar}{c^3}\right)^{\frac{1}{D-2}}\quad T_{P(D)}=\left(\frac{G_D\hbar}{c^{D+1}}\right)^{\frac{1}{D-2}}\quad M_{P(D)}=\left(\frac{c^{5-D}\hbar^{D-3}}{G_D}\right)^{\frac{1}{D-2}}.\]

Slow roll inflation corresponds to \[\varepsilon\equiv-\frac{\dot H}{H}\ll1.\] For power law expansion $H=n/t$ so that $\varepsilon=n^{-1}$. Consequently, slow roll inflation occurs when $n\gg1$.

Differentiating the first Friedmann equation with respect to time and substituting $\dot\rho_{dm}$ and $\dot\rho_{de}$ from the corresponding conservation equations, one obtains the equation \[2\dot H=-(1+w_{dm})\rho_{dm}-(1+w_{de})\rho_{de}.\] The acceleration is given by the relation $\ddot a=a(\dot H+H^2)$. Using $3H^2=\rho_{dm}+\rho_{de}$ we obtain \[\ddot a=-\frac a6 [(1+3w_{dm})\rho_{dm}+(1+3w_{de})\rho_{de}].\] The condition $\ddot a>0$ leads to the inequality \[\rho_{de}>-\frac{1+3w_{dm}}{1+3w_{de}}\rho_{dm}.\]

\[c_s^2=\frac13\left(\frac34\frac{\rho_m}{\rho_r}+1\right)^{-1},\quad \frac{\rho_m}{\rho_r}=a\frac{\rho_{m0}}{\rho_{r0}},\quad \frac{\rho_{m0}}{\rho_{r0}}=\frac1{a_{eq}},\] \[c_s^2=\frac13\frac1{\left(\frac34\frac{a}{a_{eq}}+1\right)}=\frac43\frac{1}{(4+3y)}.\]

\[\bar H^2=a^2\rho,\quad (8\pi G/3=1),\] \[\rho'+3\bar H\rho(1+w)=0,\] \[\bar{H'}=a^2\rho-\frac32a^2\rho-\frac32a^2\rho w\to\bar{H'}=-\frac{1+3w}2\bar H^2.\]

\[r=1+3\frac{\dot H}{H^2}+\frac{\ddot H}{H^3},\quad s=-\frac{2}{3H}\frac{3H\dot H+\ddot H}{3H^2+2\dot H}.\]

One can estimate the current observation time using other well known parameters. For example the period when when matter density was equal to that of radiation $z=3410\pm40$ (this value would be $1.69$ time greater if one takes under radiation only photons). It means that all length scales in that epoch were $3400$ times less than today. The CMB temperature was $9300$Ê. Age of that epoch was $51100\pm1200$ years. In this epoch the Universe expanded much faster: $H=(10.6\pm0.2)\ km\ sec^{-1}\ pc^{-1}$. We can also give our cosmic observational time by quoting the value of some parameters at Universe, and the CMB temperature was then 9300K, as hot as an A-type star. The age at that epoch was $t_{eq} = (51100 \pm 1200)$ years. And at that epoch the Universe was expanding much faster than today, actually $H_{eq} = (10.6 ± 0.2)\ km\ s^{-1}$ (note this is per 'pc', not 'Mpc').

For the Sun: \[p\approx\frac{GM_\odot^2}{R_\odot^4}\approx10^{16}J/m63;\] \[\rho\ge\frac{M_\odot c^2}{\frac43\pi R_\odot^3}\sim10^{21}J/m^3.\]

For a perfect fluid, \[T_{\alpha\beta}= (\rho+p)u_\alpha u_\beta-pg_{\alpha\beta}.\] The conservation equation, $\nabla^\alpha T_{\alpha\beta}=0$, thus gives \[\nabla^\alpha T_{\alpha\beta}=(\rho+p)u^\alpha\nabla^\alpha u_\beta+u^\beta\nabla^\alpha[(\rho+p)u_\alpha]-\nabla_\beta p=0.\] Contracting with $u^\beta$, we find that \[\nabla^\alpha[(\rho+p)u_\alpha]-u^\gamma\nabla_\gamma p=0.\] Substituting this back into $\nabla^\alpha T_{\alpha\beta}$, we get \[(\rho+p)u^\alpha\nabla^\alpha u_\beta+u^\gamma\nabla_\gamma pu_\beta-\nabla_\beta p=0,\] or, equivalently, \[(\rho+p)u^\alpha\nabla^\alpha u_\beta=\left(g^{\alpha\beta}-u^\beta u^\gamma\right)\nabla_\gamma p=0.\]

The conservation equations are \begin{align}\label{335_1} \nonumber\dot\rho_m=3H\rho_m & =0\\ \dot\rho_X+3H(1+w_X)\rho_X=0. \end{align} Using (\ref{335_1}) and \[\rho_m=\rho_X=H^2,\quad 8\pi G=1/M_p^2=1,\] and introducing $\Omega_i=\rho_i/(3H^2)$ $i=m,X$ we obtain \[w_X=-1-\frac1{3H}\frac{\dot\rho_X}{\rho_X}=-1-\frac1{3H\Omega_X}\left(\frac{2\Omega_X}H\frac{dH}{dt} +\frac{d\Omega_X}{dt}\right)=-1-\frac23\left(\frac{d\ln H}{d\ln a}+\frac12\frac{d\ln\Omega_X}{d\ln a}\right).\] Substituting this $w_X$ into the Friedman equation \[2\dot H+3H^2=-p,\] one finally finds \[\frac{d\ln H}{d\ln a}-\frac12\frac{\Omega_X}{1-\Omega_X}\frac{d\ln\Omega_X}{d\ln a}+\frac32=0.\]

i) The event horizon cut-off is given by \[R_E=a\int\limits_t^\infty\frac{dt'}{a(t')}=\int\limits_a^\infty\frac{da'}{a'2H}.\] In this case, the event horizon $R_E$ is considered as the spatial scale. Consequently, with the dark energy density $\rho_{hde}=3c^2R_E^2$ and $\Omega_{hde}=\rho_{hde}/(3H^2)$, we obtain \[\int\limits_a^\infty\frac{d\ln a'}{Ha'}=\frac{c}{Ha}\Omega_{hde}^{-1/2}.\] Taking the derivative with respect to $\ln a$, we get \[\frac{d\ln H}{d\ln a}+\frac12\frac{d\ln\Omega_{hde}}{d\ln a}=\frac{\sqrt{\Omega_{hde}}} c-1.\] Because (see the previous problem) \[w_{hde}=-1-\frac23\left(\frac{d\ln H}{d\ln a}+\frac12\frac{d\ln\Omega_{hde}}{d\ln a}\right)\] one finally finds \[w_{hde}=-\frac13\left(\frac23\sqrt{\Omega_{hde}}+1\right).\] The acceleration condition $w<-1/3$ is satisfied for $c>0$.
ii) Conformal time cut-off is given by \[\eta_{hde}=\int\limits_0^a\frac{dt'}{a(t')}=\int\limits_0^a\frac{da'}{a'^2H}.\] In this case, the conformal time is considered as a temporal scale, and we can again convert it to a spatial scale after multiplication by the speed of light. Proceeding the same way as in the previous case one obtains \[\frac{d\ln H}{d\ln a}+\frac12\frac{d\ln\Omega_{hde}}{d\ln a}+\frac{\sqrt{\Omega_{hde}}}{ac}=0.\] and \[w_{hde}=\frac23\frac{\sqrt{\Omega_{hde}}}{c}(1+z)-1,\] which corresponds to an acceleration when $c>\sqrt{\Omega_{hde}}(1+z).$
iii) The cosmic age cut-off is defined as \[t_{hde}=\int\limits_0^tdt'=\int\limits_0^a\frac{da'}{a'H}.\] In this case, the age of Universe is considered as a time scale. The corresponding spatial scale is again obtained after multiplication by the speed of light. Proceeding the same way as in the two previous cases one finds \[\int\limits_0^\infty\frac{d\ln a'}{H}=\frac c H \Omega_{hde}^{-1/2}.\] Equation of state for holographic dark energy \[w_{hde}=\frac{2}{3c}\sqrt{\Omega_{hde}}-1.\] Accelerated expansion requires $c>\sqrt{\Omega_{hde}}$.

A simple way to derive the Jeans scale is to compare the sound crossing time $t_{SC}\approx R/v_S$ to the free-fall time of a sphere of radius $R$, $t_{ff}\approx1/\sqrt{G\rho}$. The physical meaning of this criterion is that in order to make the system stable the sound waves must cross the overdense region to communicate pressure changes before collapse occurs. The maximum space scale (Jeans scale) can be found from the condition \[R_J\approx t_{ff}v_S.\] It then follows that \[R_J\approx\frac{v_S}{\sqrt{G\rho}}.\]

For a star with $N$ baryons, the gravitational energy per baryon is $E_G\sim-GNm_p^2/R$, and the kinetic energy of relativistic degenerate gas is $E_K\sim p_F c\sim\hbar cN^{1/3}/R$ where $p_F$ is the Fermi momentum. Consequently, the total energy is: \[E=\hbar cN^{1/3}/R-GNm_p^2/R.\] For the system to be stable, the maximal number of baryons $N$ is obtained by setting $E=0$ in the expression above. The result is the Chandrasekhar mass: \[M_{Chandra}=m_p\left(\frac{\hbar c}{Gm_p^2}\right)^{3/2}=m_p(\alpha\alpha_g)^{-3/2}\approx1.8M_\odot.\] where $\alpha=e^2/(\hbar c)$ is the fine structure constant. This simple derivation result is close to the more precise value, derived via the equations of stellar structure for degenerate matter, $1.4M_\odot$.

Using the ansatz for $H$, the equation of motion \(\ddot\varphi3H\dot\varphi+V'(\varphi)=0\) is integrated to give \[\frac12\dot\varphi^2=3G^2(\varphi)-V(\varphi),\] where an integration constant is absorbed into the definition of . Using this result, the first Friedmann equation becomes the "generating equation" \[V(\varphi)=3G^2(\varphi)-2\left[G'(\varphi)\right]^2.\] The evolution of the scalar field and the Hubble parameter are given by \[\dot\varphi=-2G'(\varphi),\quad H=G(\varphi).\] We need $G(\varphi)>0$ if the Universe is expanding. If we solve generation equation for a given potential $V(\varphi)$ and obtain the generating function $G(\varphi)$, the whole solution spectra can be found.

The equation of state parameter of the scalar field is \[w=\frac p\rho=-1+\frac43\frac{G'^2}{G^2}.\] At the point satisfying $V(\varphi)=3G^2(\varphi)$ the equation of state becomes $w=-1$ and the scalar field will behaves as if it were a cosmological constant.

\[a(t)=a_0\left(\frac{\Omega_{m0}}{\Omega_{\Lambda0}}\right)^{1/3} \left[sh\left(\frac32\sqrt{\Omega_{\Lambda0}}H_0t\right)\right]^{2/3};\] \[t_0=\frac23H_0^{-1}\Omega_{\Lambda0}^{-1/2}arsh\sqrt{\frac{\Omega_{\Lambda0}}{\Omega_{m0}}} =t_{\Lambda}arsh\sqrt{\frac{\Omega_{\Lambda0}}{\Omega_{m0}}} =t_{\Lambda}arcth\sqrt{\Omega_{\Lambda0}}.\] According to the WMAP data \[t_{\Lambda}\equiv\frac23H_0^{-1}(\Omega_{\Lambda0})^{-1/2}\simeq10.768\times10^9\ years\] and \[t_0=13.7\ Gyr.\] According to the "PLANCK" data \[t_{\Lambda}=11.74\ Gyr\] and \[t_0=13.8\ Gyr.\]

\[\rho_m=\frac{a_0^3}{a^3}\rho_{m0}=\rho_r;\] \[\frac a{a_0}=\frac{\rho_{r0}}{\rho_{m0}}=\frac{\Omega_{r0}}{\Omega_{m0}}=2.9088\times10^{-4};\] \[1+z=\frac{a_0}a\to z^*=3436.9;\] \[t(z^*)=\frac1{H_0}\int\limits_0^{a/a_0}\frac{dx}x \frac1{\sqrt{\Omega_{\Lambda0}+\Omega_{curv}x^{-2}+\Omega_{m0}x^{-3}+\Omega_{r0}x^{-4}}};\] \[t(z^*=3436.9)=50.152\ years.\]

\[\rho_m=\frac{a_0^3}{a^3}\rho_{m0}=\rho_\Lambda;\] \[\frac a{a_0}=\left(\frac{\rho_{m0}}{\rho_{\Lambda}}\right)^{1/3}= \left(\frac{\Omega_{m0}}{\Omega_{\Lambda}}\right)^{1/3}=0.7719\times10^{-4};\] \[1+z=\frac{a_0}a\to z^*=0.2956;\] \[t(z^*)=\frac1{H_0}\int\limits_0^{a/a_0}\frac{dx}x \frac1{\sqrt{\Omega_{\Lambda0}+\Omega_{curv}x^{-2}+\Omega_{m0}x^{-3}+\Omega_{r0}x^{-4}}};\] \[t(z^*=0.2956)=10.309\ Gyr\to 3.508\ Gyr \ ago.\]

\[V(\varphi)=(3\alpha^2-\alpha)\exp[-2\Delta\varphi/\sqrt{2\alpha}];\] \[\Delta\varphi(t)=\sqrt{2\alpha}\ln t.\]

In this case \[H^2=\frac13\lambda\varphi^2,\quad \frac{dV_a}{d\varphi}=2\lambda\varphi.\] Therefore, from \[-3H(\varphi)\left(\frac{dV_a}{d\varphi}\right)^{-1}d\varphi=dt.\] one finds that \[\Delta\varphi(t)=\pm2\sqrt{\frac\lambda3}\Delta t.\] Hence, $\dot\varphi$ is constant. Letting $\varphi(t_0=0)=0$, and using eqs. (\ref{es_1}) and (\ref{es_4}) we get \begin{align} \nonumber V(\varphi) & = \lambda\varphi^2-\frac23\lambda,\\ \nonumber a(t) & =a_0e^{-\frac\lambda3t^2}. \end{align} Obviously, one would be tempted to pick $\lambda<0$ in order to make $a(t)$ a growing function of $t$, but that would make $\varphi(t)$ an imaginary function of $t$.

Proceeding the same way as in the previous problem one finds \begin{align} \nonumber \varphi(t) & =\varphi_0e^{\pm4\sqrt{\frac\lambda3}t},\\ \nonumber V(\varphi) & = \lambda\varphi^4-\frac83\lambda\varphi^2,\\ \nonumber a(t) & =a_0\exp\left[-\frac{\varphi_0^2}8e^{\pm8\sqrt{\frac\lambda3}(t-t_0)}\right]. \end{align}

Proceeding the same way as above one finds \begin{align} \nonumber \varphi(t) & =\left[\varphi_0^{2-n}\pm2n(n-2)\sqrt{\frac\lambda3}(t-t_0)\right]^{-\frac1{n-2}};\\ \nonumber V(\varphi) & = \lambda\varphi^{2n}-\frac23\lambda n^2\varphi^{2(n-1)};\\ \nonumber a(t) & =a_0\exp\left\{-\frac1{4n}\left[\varphi_0^{2-n}\pm2n(n-2)\sqrt{\frac\lambda3}(t-t_0)\right]^{-\frac2{n-2}}\right\}. \end{align}

Simply break up the Hubble parameter under the integral into a series, integrate, and use the numerous formulas already in the Cosmography section. (see Expanding Universe: slowdown or speedup?, or the Visser convergence article.)

\[\frac1{1+z}=\frac{a(t)}{a_0}=1+H_0(t-t_0)-\frac{q_0 H_0^2}{2!}(t-t_0)^2+\frac{j_0 H_0^3}{3!}(t-t_0)^3+O(|t-t_0|^4).\] This is the power series for $a(t)$, and it serves as the basis of all of our power series. However, it has a pole at $z=-1$. By complex variable theory, this means that the radius of convergence is at most $1$, so data with $z>1$ is outside of the convergence radius. (NOTE: I'm a bit confused about the fine points of this - I don't fully understand why we look at the convergence radius of this particular relation, to be honest. This is something you should discuss)

$z=-1$ corresponds to $a=\infty$ - we can't "look past" infinity.

The pole is now located at \[\Delta z=-(1+z_{pivot}),\] which again physically corresponds to a Universe that had undergone infinite expansion, $a=\infty$. The radius of convergence is now \[|\Delta z|\le(1+z_{pivot}),\] and we expect the pivoted version of the Hubble law to fail for \[z>1+2z_{pivot}.\]

a) What is the "new $y$-redshift" in terms of the old $z$? \[y=\frac z{1+z},\quad z=\frac y{1-y}.\] b) The past and the future correspond, in terms of $z$-redshift to $z\in(0,\infty)$ and $z\in(-1,0)$ respectively. What are these intervals in terms of $y$-redshift? Past: $y\in(0,1)$, future: $y\in(-\infty,0)$.
c) get the correlation between luminosity distance and $y$-redshift up to the second power: \[d_L(y)=d_H y\left\{1-\frac12(-3+q_0)y+O(y^2)\right\}.\]

We can't extrapolate through the big bang. Thus, we assume that the convergence radius of $y$-redshift-based formulas should be $1$ (see the picture)

\begin{align} \nonumber d_F(z) & =d_H z\left\{1-\frac12q_0z+O(z^2)\right\}.\\ \nonumber d_P(z) & =d_H z\left\{1-\frac12(1+q_0)z+O(z^2)\right\}.\\ \nonumber d_Q(z) & =d_H z\left\{1-\frac12(2+q_0)z+O(z^2)\right\}.\\ \nonumber d_A(z) & =d_H z\left\{1-\frac12(3+q_0)z+O(z^2)\right\}. \end{align}

\[\mu_D(z)=25+\frac5{\ln(10)}\left\{\ln(d_H/Mpc)+\ln z+\frac12(1-q_0)z+O(z^2)\right\}.\]

\begin{align} \nonumber d_F(y) & =d_H y\left\{1-\frac12(-2+q_0)y+O(y^2)\right\}.\\ \nonumber d_P(y) & =d_H y\left\{1-\frac12(-1+q_0)y+O(y^2)\right\}.\\ \nonumber d_Q(y) & =d_H y\left\{1-\frac{q_0}2y+O(y^2)\right\}.\\ \nonumber d_A(y) & =d_H y\left\{1-\frac12(1+q_0)y+O(y^2)\right\}. \end{align}

\[d_L(z) =d_H z\left\{1-\frac12(-1+q_0)z +\frac16[q_0+3q_0^2-(j_0+\Omega_0)]z^2+O(z^3)\right\},\] \[\Omega_0=1+\frac{kc^2}{H_0^2a_0^2}=1+\frac{kd_H^2}{a_0^2}.\]

\[d_L=\frac1{H_0}\cdot\left[y_4+y_4^2\cdot\left(\frac12-\frac{q_0}2\right)\right].\]

Recalling problem #cg_10, we see that the term that includes $K$ is proportional to $(H_0a_0/c\gg1)^{-2}$, so for all practical (cosmographic) purposes, our Universe is indistinguishable from a flat Universe. If the Universe is really flat, then we will never be able to strictly prove its flatness from cosmographic methods - we will only be able to put ever-stricter bounds on $H_0a_0/c\gg1$. If the Universe is not flat, then with good enough data, we will be able to determine the non-flatness <!-(see Visser3(Simple good))-->.

The more parameters you have, the larger the region in parameter space, the less strict the limitations on the parameters and the higher the possibility of degeneracies. So the complexity of the formula should be in correspondence with the amount and accuracy of available experimental data!

From the formula in the formulation of the problem, take $\Delta t/t\ll1$. This gives $\Delta t_s=[a(t_s)/a(t_o)]\Delta t_o$. Then, \[\Delta z\equiv\frac{a(t_o+\Delta t_o)}{a(t_s+\Delta t_s)}-\frac{a(t_o)}{a(t_s)}\approx\left[\frac{\dot a(t_o)-\dot a(t_s)}{a(t_s)}\right]\Delta t_o,\] which automatically gives \[\frac{\Delta z}{\Delta t_{obs}}=(1+z)H_{obs}-H(z).\]

\[z(D)=\frac{H_0D}c+\frac{1+q_0}2\frac{H_0^2D^2}{c^2}+\frac{6(1+q_0)+j_0}6\frac{H_0^3D^3}{c^3} +O\left(\left[\frac{H_0^4D^4}{c^4}\right]\right).\]

\[D(z)=\frac{cz}{H_0}\left[1-\left(1+\frac{q_0}2\right)z +\left(1+q_0+\frac{q_0^2}2-\frac{q_0}6\right)z^2 O(z^3)\right].\]

\[\frac{\Delta z}{\Delta t_0} = -H_0q_0z+\frac12H_0(q_0^2-j_0)z^2+ \frac12H_0\left[\frac13(s_0+4q_0j_0)+j_0-q_0^2-q_0^3\right]z^3 +O(z^4).\]

\[C_P=0,\quad C_V<0.\]

A partial solution reads: \[\dot a(t)=c_0\eta_0(t-t_\odot)^{\eta_0-1}+c_1\eta_1(t-t_\odot)^{\eta_1-1}+\ldots\] Also note: the most general lesson to be learned is this: If in the vicinity of any cosmological milestone, the input scale factor $a(t)$ is a generalized power series, then all physical observables ($H$, $q$, the Riemann tensor, etc.) will likewise be generalized power series, with related indicial exponents that can be calculated from the indicial exponents of the scale factor. Whether or not the particular physical observable then diverges at the cosmological milestone is "simply" a matter of calculating its dominant indicial exponent in terms of those occurring in the scale factor.

a) ($0<\eta_0<\eta_1\ldots$).
b) ($\eta_0<\eta_1\ldots$), $\eta_0<0$ and $c_0\ne0$.
c) $\eta_0=0$ and $\eta_1>0$ with $c_0>0$ and $\eta_1$ non-integer.
d) $\eta_0=0$ and $\eta_i\in Z^+$. \end{description}

Taking the exact form of the Hubble parameter as a ratio of series, and keeping only the most dominant terms, we have (assuming that the first power coefficient isn't $0$) \[H=\frac{\dot a}a\sim\frac{c_0\eta_0(t-t_\odot)^{\eta_0-1}}{c_0(t-t_\odot)^{\eta_0}}=\frac{\eta_0}{t-t_\odot};\quad (\eta_0\ne0).\] When the first power coefficient is zero (which corresponds to either a sudden singularity or extremality event), since the next power coefficient must be greater (by definition), we have the following: \[H\sim\frac{c_1\eta_1(t-t_\odot)^{\eta_1-1}}{c_0}=\eta_1\frac{c_1}{c_0}(t-t_\odot)^{\eta_1-1};\quad (\eta_0=0;\ \eta-1>0).\] Combining all of this, we get the following: \[\lim_{t\to t_\odot}H=\left\{ \begin{array}{lcl} +\infty & \eta_0>0; & {}\\ sign(c_1)\infty & \eta_0=0; & \eta_1\in(0,1);\\ c_1/c_0 & \eta_0=0; & \eta_1=1;\\ 0 & \eta_0=0; & \eta_1>1;\\ -\infty & \eta_0<0. & {}\\ \end{array} \right.\] Other things (cosmographic parameters, components of the Ricci tensor, validity of the Energy conditions) are analyzed analogously

\[w9t)=\frac p\rho=-\frac{H^2(1-2q)+kc^2/a^2}{3(H^2+kc^2/a^2)}= -\frac{(1-2q)+kc^2/(H^2a^2)}{3(1+kc^2/(H^2a^2))}.\]

\[8\pi G_N=\frac{d\rho}{dt}=-6c^2H\left[(1+q)H^2+\frac{kc^2}{a^2}\right],\] \[8\pi G_N=\frac{dp}{dt}=2c^2H\left[(1-j)H^2+\frac{kc^2}{a^2}\right].\] Then \[\kappa_0=-\frac13\left[\frac{1-j_0+kc^2/(H_0^2a_0^2)}{1+q_0+kc^2/(H_0^2a_0^2)}\right]\] which approximates (using $H_0a_0/c\gg1$) to \[\kappa_0=-\frac13\left[\frac{1-j_0}{1+q_0}\right]\] We therefore see an important note: to get the FIRST term in the series expansion of the EOS contains the both the deceleration parameter and the jerk (third order observational term), the latter of which is very weakly bounded by observations. This is a significant problem in testing models (even with crude linear tests, as they require the slope parameter - see the previous problem)

Starting off, it's easy to see that \[\frac{d^2p}{d\rho^2}=\frac{\ddot p-\kappa\ddot\rho}{(\dot\rho)^2}.\] Then \begin{align} \nonumber\left.\frac{d^2p}{d\rho^2}\right|_0 & = -\frac{(1+kc^2/[H_0^2a_0^2])}{6\rho_0(1+q_0+kc^2/[H_0^2a_0^2])^3} \left\{s_0(1+q_0)+j_0(1+j_0+4q_0+q_0^2)+q_0(1+2q_0)\right.\\ {} & \left.+(s_0+j_0+q_0+q_0j_0)\frac{kc^2}{H_0^2a_0^2}\right\}.\ \end{align} In the approximation $H_0a_0/c\gg1$ this reduces to \[\left.\frac{d^2p}{d\rho^2}\right|_0 = -\frac{s_0(1+q_0)+j_0(1+j_0+4q_0+q_0^2)+q_0(1+2q_0)}{6\rho_0(1+q_0)^3}.\]