Ricci scalar
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Observational cosmology problems
Problem 1
problem id: 10_1
Show that the Ricci scalar can be written in the following form \[R=6H^2[q(t)-\Omega(t)].\]
\[R=g^{\mu\nu}R_{\mu\nu}=-6\left(\frac{\ddot a}{a}+H^2+\frac k{a^2}\right)=6H^2\left(-\frac{\ddot a}{aH^2}-1-\frac k{a^2H^2}\right),\] \[H^2=\frac{8\pi G}3\rho-\frac k{a^2}\Rightarrow1+\frac k{a^2H^2}=\Omega,\quad\Omega\equiv\frac\rho{\rho_{cr}},\quad \rho_{cr}\equiv\frac{3H^2}{8\pi G},\]\[\frac{\ddot a}{aH^2}\equiv q\Rightarrow R=6H^2[q(t)-\Omega(t)].\]
Problem 2
problem id: 10_2
Show that for a one-component flat Universe filled with ideal fluid with the EoS $p=w\rho$ the Ricci scalar can be written in the following form \[R=8\pi G\rho(3w-1).\]
Use result of the previous problem \[R=6H^2[q(t)-\Omega(t)].\] For the one-component flat Universe \[q=\Omega(1+3w)/2,\] and taking into account that \[\Omega=\frac{8\pi G}{3H^2}\rho\] one finds \[R=8\pi G\rho(3w-1).\]
Problem 3
problem id: 10_3
Find time derivative of the scalar curvature for the spatially flat Universe filled with a barotropic fluid.
Using the conservation and Friedmann equations we obtain \begin{align} \nonumber \dot\rho & =-\sqrt{3\rho}(\rho+p),\\ \nonumber \dot p & = \frac{dp}{d\rho}\dot\rho=-\sqrt{3\rho}(\rho+p)\frac{dp}{d\rho}, \end{align} where we have used $p=p(\rho)$. According to the result of the problem (10_1) $R=-\rho+3p$. Consequently \[\dot R=-\dot\rho+3\dot p=-\dot\rho\left(1-3\frac{dp}{d\rho}\right)=\sqrt{3\rho}(\rho+p)\left(1-3\frac{dp}{d\rho}\right).\]
Problem 4
problem id: 10_4
Compare entropy of the Sun in the present state and right after its transformation to a black hole due to compression.
Let us treat the Sun as an ideal gas composed of nucleons, then \[S_\odot=k_BN\left\{\ln\left[\left(\frac{mk_BT}{2\pi\hbar^2}\right)^{3/2}\frac V N\right]+\frac52\right\}.\] here $N\approx10^{57}$, $V\approx1.4\times10^{33}cm^3$, $T\approx10^7K$, so \[S_\odot\approx1.4\times10^{53}k_B\approx2\times10^{35}\frac J K.\] On the other hand, for a black hole with the solar mass one obtains \[S_{BH,\odot}=\frac{k_B}4\frac{A_{BH,\odot}}{l_{Pl}^2}\approx10^{77}k_B\approx1.4\times10^{54}\frac J K.\] Therefore \[\frac{S_{BH,\odot}}{S_\odot}\approx0.7\times10^{19}.\] As $\Delta S=-\Delta I$ ($I$ is amount of information), the transformation of the Sun into the black hole is followed by huge loss of information (the information paradox). It relates to the fact that the simplest (uncharged and non-rotating) black hole is characterized by just single parameter --- its mass, while the Sun has quite complicated structure. However we derived entropy of the Sun $S_\odot$ in assumption that the Sun can be described solely by number of nucleons forming an ideal gas.
Problem 5
problem id: 10_5
Show that black holes have negative thermal capacity.
Temperature $T$ of a black hole is inversely proportional to its mass $M$: \[T\propto M^{-1}.\] Therefore when mass increases the black hole does not heat and rather cools down and therefore its thermal capacity is negative. The same property holds for any Newtonnian self-gravitating system. Therefore for such systems one cannot use the canonical distribution and only the micro-canonical one is allowed.
Problem 6
problem id: 10_6
What are the physical reasons for the negative thermal capacity in Newtonnian self-gravitating systems?
For a stable system of $N$ particles coupled by the potential forces $\vec{F}_k$ the virial theorem holds: \[2\langle T\rangle=-\sum\limits_{k=1}^N\langle\vec{r}_k\vec{F}_k\rangle,\] where $\langle T\rangle$ is average total kinetic energy and $\vec{F}_k$ is the force acting on the $k$-th particle. In particular, when the force has corresponding potential energy of interaction $V(r)$ proportional to $n$-th power of the inter-particle distance $r$, the virial theorem takes on the form \[2\langle T\rangle =n\langle U\rangle,\] i.e. doubled avarage total kinetic energy $T$ equals to $n$ times the average average total potential energy $U$. For the Newtonnian interaction ($n=-1$): \[2\langle T\rangle =-\langle U\rangle.\] Let us apply this theorem to evolution of massive stars. Black holes can form on final stages of evolution of such objects. Stationary state of the star corresponds to hydrostatic and thermal equilibrium. In the equilibrium state the negative potential energy of the star (the energy of the gravity compression) by absolute value is twice greater than the thermal energy (the kinetic one). It is this peculiarity that leads to the negative thermal capacity of the considered system. During the compression the potential energy transforms into the kinetic energy os the falling stellar layers, and its depth heats up, so that the thermal energy acquired by the star during the compression is exactly twice greater of the energy lost due to the radiation.
Problem 7
problem id: 10_7
How the negative thermal capacity of the stars (in particular the Sun) affects the course of the nuclear reactions inside them?
Due to the negative thermal capacity the temperature in central part of the star grows (see the previous problem) and thus the continuous thermonuclear synthesis of chemical elements takes pace. For instance, the reaction of transformation of hydrogen into helium in the present Sun occurs at temperature of $15$ millions degrees. When, $4$ millions years later, all the hydrogen in the center of the Sun will be transformed into helium, further synthesis of carbon atoms from the helium ones will require considerably higher temperature, about $100$ millions degrees (because the electric charge of the helium nuclei is twice greater than that of hydrogen, so one needs much higher temperature to close them in to distance of $10^{-13}cm$). Exactly this temperature will be provided by the negative thermal capacity of the Sun to the moment of burning up in its depth the thermonuclear reaction of transformation of helium into carbon.
Problem 8
problem id: 10_8
From the three parameters (mass, charge and angular momentum) of the Kerr-Newman black hole form all possible length scales.
\[R_1=\frac{MG}{c^2},\quad R_2=\frac{\sqrt{G}Q}{c^2},\quad R_3=\frac{J}{Mc}.\]
Problem 9
problem id: 10_9
Following [O. Luongo and H. Quevedo, Self-accelerated universe induced by repulsive e?ects as an alternative to dark energy and modi?ed gravities, arXiv: (1507.06446)] let us introduce the parameter $\lambda=-\frac{\ddot a}{a}=qH^2$, so that $\lambda<0$ when the Universe is accelerating, whereas for $\lambda>0$ the Universe decelerates. Luongo and H. Quevedo showed, that the parameter $\lambda$ can be considered as an eigenvalue of the curvature tensor defined in special way. In particular, for FLRW metric the curvature tensor $R$ can be expressed as a ($6\times6$)-matrix \[R=diag(\lambda,\lambda,\lambda,r,r,r),\quad r\equiv\frac13\rho.\] The curvature eigenvalues reflect the behavior of the gravitational interaction and if gravity becomes repulsive in some regions, the eigenvalues must change accordingly; for instance, if repulsive gravity becomes dominant at a particular point, one would expect at that point a change in the sign of at least one eigenvalue. Moreover, if the gravitational field does not diverge at infinity, the eigenvalue must have an extremal at some point before it changes its sign. This means that the extremal of the eigenvalue can be interpreted as the onset of repulsion. Obtain the onset of repulsion condition in terms of cosmographic parameters.
As mentioned above, the onset of repulsion is determined by an extremal of the eigenvalue, i.e. $\dot\lambda=0$, \[\lambda=qH^2\rightarrow\dot\lambda=\dot qH^2+2qH\dot H=0.\] Using the result of the previous problem for $\dot q$ and $\dot H$ we find that repulsion onset condition $\dot\lambda$ reduces to \[j=-q.\]
Problem 10
problem id: 10_10
Represent result of the previous problem in terms of the Hubble parameter and its time derivatives.
Solving the system of equations \begin{align} \nonumber \dot H &=-H^2(1+q),\\ \nonumber \ddot H &=H^3(j+3q+2), \end{align} with respect to the variables $q$ and $j$, one finds that the condition $j=-q$ transforms into \[\frac{\ddot H}{H}=-2\dot H.\]