### Solution

A1  ?? Find the total energy $E$ of the universe as a function of its radius $R$ in the form
$E = -\frac{\alpha}{R} + \frac{\beta}{R^2} + E_K,$
where $E_K$ is the kinetic energy of the universe.

Let $M=M_s + M_q$ and calculate the gravitational energy $E_G$ of the universe as a function of its radius $R$. It's important that $M>0$, so gravity compresses the universe in TCM just as it compresses ordinary matter with a total mass $M$.

Gauss's law for the gravitational field has the form $\int_\Gamma \vec{g} \cdot d\vec{S} = -4\pi Gm,$ where $m$ is the mass enclosed within the closed surface $\Gamma$. Let's choose a sphere with radius $r$ and center coinciding with the center of the universe. So
$- 4\pi r^2 g = -4\pi G m\frac{r^3}{R^3} \quad \Rightarrow \quad g = GM \frac{r}{R^3}.$
Gravitational potential at the surface of the universe $\varphi(R) = -GM/R$, so
$\varphi(r) = -\frac{GM}{R} - \frac{GM}{R^3}\int\limits_r^R r \, dr = - \frac{3GM}{2R} + \frac{GMr^2}{2R^3}.$

Now we can calculate the change $\Delta E_G$ in the gravitational potential energy as $M \to M + \Delta M$.
$\Delta E_G = \int\limits_0^{R} \left( - \frac{3GM}{2R} + \frac{GMr^2}{2R^3} \right) \frac{4\pi r^2 dr}{\frac{4}{3}\pi R^3} \Delta M = -\frac{3GM\Delta M}{2R}+\frac{3GM\Delta M}{10R} = -\frac{6 GM \Delta M}{5R}.$
Thus
$E_G = -\frac{6G}{5R} \int\limits_0^{M} m \, dm = -\frac{3GM^2}{5R}.$

The internal energy $E_q$ of the quintessence is
$E_q = \frac{4}{3}\pi R^3 \cdot A \left( \frac{-M_q}{\frac{4}{3}\pi R^3} \right)^{5/3} = A \left( \frac{3 \pi}{4} \right)^{2/3} \frac{\left(-M_q \right)^{5/3}}{R^2},$
so the total energy $E$ of the universe is
$E = -\frac{3G(M_s + M_q)^2}{5R} + A \left( \frac{3 \pi}{4} \right)^{2/3} \frac{\left(-M_q \right)^{5/3} }{R^2} +E _K$

Answer: $E = -\frac{3G(M_s + M_q)^2}{5R} + A \left( \frac{3 \pi}{4} \right)^{2/3} \frac{\left(-M_q \right)^{5/3}}{R^2} +E _K$
$\alpha = \frac{3G(M_s+M_q)^2}{5}, \quad \beta = -M_q^{5/3}A \left( \frac{3 \pi}{4} \right)^{2/3}$

A2  ?? For given $M_s$, $M_q$, $A$ and $E$ of the universe find a range of radii which it can have. Write the expression in terms of $\alpha$, $\beta$ and $E$.

The kinetic energy $E_K$ of the universe is not less than zero, so
$E_K = E + \frac{\alpha}{R}- \frac{\beta}{R^2} \geq 0$
A part of thits equation that depends on the $R$ is infinitely negative when $R \to 0$ and goes to zero from the positive side when $R \to +\infty$, so it definitely has a maximum.

Let's multiply the equation for $E_K$ by $R^2$:
$ER^2 + \alpha R - \beta \geq 0, \quad \mathcal{D} = \alpha^2 + 4 \beta E.$

For $E \neq 0$ we have:

• If $\mathcal{D} < 0$, there are no roots for $R$, so the existence of such a universe is impossible.
• If $\mathcal{D} = 0$, there is only a one root $R=\frac{-\alpha}{2E}$ which is grater then zero, so this universe is a solution.
• If $\mathcal{D} > 0$ there are two roots $R = \frac{-\alpha \pm \sqrt{\alpha^2 + 4 \beta E}}{2E}$As we can see from the picture above if $E<0$ the universe could exists if its $R$ is in a range between soultions and if $E \geq 0$ the universe could exists for every $R$ grater than positive root.

For $E=0$ we have the root $R=\beta/\alpha$ and any $R$ that is greater than or equal to this value is possible.

Answer: $\begin{split} E < -\frac{\alpha^2}{4 \beta} \quad \Rightarrow \quad & R \in \emptyset\\ E \in \left[ -\frac{\alpha^2}{4 \beta}, 0 \right) \quad \Rightarrow \quad & R \in \left[ \frac{-\alpha + \sqrt{\alpha^2 + 4 \beta E}}{2E}, \frac{-\alpha - \sqrt{\alpha^2 + 4 \beta E}}{2E} \right] \\ E=0 \quad \Rightarrow \quad & R \geq \frac{\beta}{\alpha}\\ E > 0 \quad \Rightarrow \quad & R \geq \frac{-\alpha + \sqrt{\alpha^2 + 4 \beta E}}{2E} \end{split}$

A3  ?? Could the universe described by TCM be stable, i.e., have a constant radius? If so, describe it quantitatively.

Since the potential energy has a minimum the equilibrium point exists. If $E=-\alpha^2/4\beta$, then the radius $R$ of the universe is constant and
$R = \frac{2\beta}{\alpha}.$

Answer: If $E=-\alpha^2/4\beta$, then the radius $R$ of the universe is constant and
$R = \frac{2\beta}{\alpha}.$

A4  ?? How long will the universe continue to expand at a positive acceleration?

Since $v(R,t) = \dot{R}$, let's find the kinetic energy of the universe:
$E_K = \int\limits_0^R \frac{M_s r^2 \dot{R}^2}{2R^2} \frac{4\pi r^2 \, dr}{\frac{4}{3}\pi R^3} = \frac{3M_s\dot{R}^2}{2R^5} \int\limits_0^R r^4 \, dr = \frac{3 M_s \dot{R}^2}{10}.$
Thus
$\frac{3 M_s \dot{R}^2}{10} = \gamma \dot{R}^2 = \frac{\alpha}{R}- \frac{\beta}{R^2}$
and a minimum radius of the universe is $R_0=\beta/\alpha$. Let the velocity $\dot{R}$ reach its maximum when $R=R_1$, so
$\left( \frac{\alpha}{R} - \frac{\beta}{R^2} \right)' \bigg|_{R=R_1} = 0 \quad \Rightarrow \quad -\frac{\alpha}{R_1^2}+\frac{2\beta}{R_1^3} = 0 \quad \Rightarrow \quad R_1 = \frac{2 \beta}{\alpha} = 2 R_0.$

The equation of motion is
$\frac{dR}{\sqrt{\frac{\alpha}{R} - \frac{\beta}{R^2}}} = \frac{dt}{\sqrt{\gamma}}.$
Let's use the substitution $R = R_0 x$, so
$\int\limits_{R_0}^{2R_0}\frac{dR}{\sqrt{\frac{\alpha}{R} - \frac{\beta}{R^2}}} = \frac{\sqrt{\beta^3}}{\alpha^2}\int\limits_1^2\frac{x \, dx}{ \sqrt{ x -1}} = \frac{\sqrt{\beta^3}}{\alpha^2} \left( \int\limits_1^2 \frac{(x-1) \, d(x-1)}{\sqrt{x-1}} +\int\limits_1^2 \frac{d(x-1)}{\sqrt{x-1}} \right) = \frac{\sqrt{\beta^3}}{\alpha^2} \left( \frac{2}{3} + 2\right) = \frac{8 \sqrt{\beta^3}}{3 \alpha^2}.$

Thus
$T = \frac{8 \sqrt{\beta^3 \gamma}}{3 \alpha^2} = \sqrt{\frac{10}{3}} \frac{5}{3 \pi} \frac{\sqrt{M_s} \left(-M_q^{5/3}A \right)^{3/2}}{G^2(M_s+M_q)^4}$

Answer: $T = \sqrt{\frac{10}{3}} \frac{5}{3 \pi} \frac{\sqrt{M_s} \left(-M_q^{5/3}A \right)^{3/2}}{G^2(M_s+M_q)^4}$