### Solution

A1  ?? Does the Hubble law imply that the solar system is in the region of the Universe where the Big Bang occurred? (Because Hubble made his observations from the Earth!) Explain your answer with a drawing and formula.

Let's consider that $\vec{r}_0$ is a position of the Earth and $\vec{x}$ is a position of a certain point in the universe. So $\vec{r} = \vec{x} - \vec{r}_0$ is a position relative to the Earth.

According to the Hubble law a time derivative $\dot{\vec{R}}$ of any vector $\vec{R}$ is $H\vec{R}$. Thus
$\dot{\vec{r}}= - \dot{\vec{r}}_0 + \dot{\vec{x}} = H \vec{r}$
and the Hubble constant $H$ is independent of the observer's position.

Answer: The Hubble's law doesn't imply that the solar system is in the region of the Big Bang.

A2  ?? Calculate the gravitational potential energy $E_G$ of the NU at a time $t$ when its radius is equal to $R$. Give your answer as a formula expressing $E_G$ in terms of $M$ and $R$.

Remark: if the mass of the system increases from $m$ to $m + \Delta m$ without any changing the relative mass distribution, the gravitational potential energy increases by

$\Delta E_G = \int\limits_0^{\Delta m} \varphi \, dm,$

where $\varphi$ is a gravitational potential at the point of mass $dm$. Also $\varphi$ is chosen so that $\varphi=0$ on infinity.

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
$E_G = -\frac{6G}{5R} \int\limits_0^{M} m \, dm = -\frac{3GM^2}{5R}.$

Answer: $E_G = -\frac{6G}{5R} \int\limits_0^{M} m \, dm = -\frac{3GM^2}{5R}.$

A3  ?? Calculate the kinetic energy $E_K$ of the NU at the same time $t$. Give your answer as a formula expressing $E_K$ in terms of $M$, $H$ and $R$.

We can calculate $E_K$ by integrating:
$E_K = \int\limits_0^{R} \frac{(Hr)^2}{2} \cdot M \frac{4\pi r^2 \, dr}{\frac{4}{3} \pi R^3} = \frac{3H^2 M}{2R^3} \int\limits_0^R r^4 \, dr = \frac{3H^2MR^2}{10}$

Answer: $E_K = \frac{3H^2MR^2}{10}$

A4  ?? At what relation between $\rho(t)$ and $H(t)$ will the expansion of the NU stop at a finit size and be replaced by compression? Give the answer in the form of an inequality.

Total energy of the universe
$E = -\frac{3GM^2}{5R} + \frac{3H^2MR^2}{10},$
where $M=4\pi \rho R^3/3,$ so
$E = -\frac{16 \pi^2}{15}\rho^2 GR^5 + \frac{2\pi}{5} H^2 \rho R^5.$
If $E<0$ the universe has maximum size when $E = E_G$, so
$E = \frac{2\pi}{5} \rho R^5 \left[ - \frac{8 \pi}{3} \rho G + H^2 \right] < 0 \quad \Rightarrow \quad H^2 < \frac{8 \pi}{3} \rho G$

Answer: $H^2 < \frac{8 \pi}{3} \rho G$

A5  ?? Let the total energy of matter in the NU, i.e. the sum of kinetic energy and potential energy of gravitational interaction, be equal to
$E = -\frac{2}{15} Mc^2,$
where $c$ is the speed of light in vacuum. Find the maximum radius of the NU in the process of expansion. Write down the formula and get a numerical answer in parsecs ($1 ~\text{parsec}$ is approximately equal to $3.2~\text{light years}$, or $3\cdot10^{16}~\text{m}$).

Maximum radius is reached when $E=E_G$, so
$-\frac{2}{15}Mc^2 = -\frac{3GM^2}{5R_\text{max}} \quad \Rightarrow \quad R_\text{max} = \frac{9GM}{2c^2}$

Answer: $R_\text{max} = \frac{9GM}{2c^2} = 3 \cdot 10^{28}~\text{m}= 10^{12}~\text{pc}$

A6  ?? For the conditions described in question A5, find the total lifetime of the NU from the Big Bang to the Big Implosion. Write down the formula and get a numerical answer in years.

Remark: $$\int\limits_{0}^{1} \frac{\sqrt{x}}{\sqrt{1-x}} d x=(y=\sqrt{1-x})=2 \int\limits_{0}^{1} \sqrt{1-y^{2}} d y$$

As we discussed earlier, the total energy of the universe
$E = -\frac{3GM^2}{5R} + \frac{3H^2MR^2}{10},$
where $\dot{R} = HR$. Thus,
$\dot{R} = \pm \sqrt{\frac{2GM}{R} + \frac{10E}{3M}} = \pm\sqrt{2GM \left(\frac{1}{R} - \frac{1}{R_\text{max}} \right) }.$
The lifetime consists of the expansion time and the compression time, which are equal. Let's find the expansion time:
$T_e = \int\limits_0^{R_\text{max}} \frac{dR}{\sqrt{2GM\left(\frac{1}{R} - \frac{1}{R_\text{max}} \right)}} = \sqrt{\frac{R_\text{max}^3}{2GM}} \int\limits_0^1 \frac{\sqrt{\frac{R}{R_\text{max}}} \, d\left(\frac{R}{R_\text{max}} \right)}{\sqrt{1 - \frac{R}{R_\text{max}}}}.$

To calculate the integral mentioned in the remark, let's use the substitution $y = \sin \phi$, $dy = \cos \phi \, d \phi$:
$2 \int\limits_0^1 \sqrt{1-y^2}dy = 2 \int\limits_0^{\pi/2}\cos^2 \phi \, d\phi = \int\limits_0^{\pi/2} \left( \cos 2\phi +1 \right) d \phi = 0 + \frac{\pi}{2}.$

Finally,
$T = 2T_e = \pi \sqrt{\frac{R_\text{max}^3}{2GM}} = \pi \sqrt{\frac{9R_\text{max}^3}{2c^2 R_\text{max}}} = \frac{3\pi R_\text{max}}{c\sqrt{2}}$

Answer: $T = \frac{3\pi R_\text{max}}{c\sqrt{2}} = 2 \cdot 10^{13}~\text{yr}$

A7  ?? What are the options for further motion of matter in the Universe at different ratios between the density and the Hubble constant? Describe the qualitative behavior of the NU radius for each of the possible cases. Draw graphs: three different pairs of graphs showing the NU radius $R$ and the rate of its expansion $\dot{R}$ versus time $t$ (keep in mind that at the time of the Big Bang $t = 0$ and the radius of the NU is considered to be almost zero). Draw the graphs qualitatively, showing all the important details and features.