For simplicity we suppose the particle moves in a direction perpendicular to two opposite walls of an expanding box. Normally, of course, the particle rebounds in different directions, but the final result is just the same. The walls are perfect reflectors and therefore, relative to the wall, the particle rebounds with the same speed as when it strikes the wall. During the collision, the direction of motion is reversed, but the speed relative to the wall remains unchanged. Because the wall is receding, the particle returns to the center of the box with slightly reduced speed. Each time the particle strikes a receding wall it returns with reduced speed. Using the Hubble law it can be shown that a particle of mass m and speed U, moving within an expanding box, obeys the law that mU is proportional to $1/L$. The product mU is the momentum. As the box gets larger the momentum gets smaller. The length L expands in the same way as the scaling factor R, and the momentum therefore obeys the important law: \[mUR=const\] This law holds not only for particles in an expanding box but also for particles moving freely in an expanding Universe. Remarkably, the general relativity equation of motion of a freely moving particle in the uniformly curved space of an expanding Universe gives exactly the same result. This illustrates how the cosmic box not only helps us to understand what happens but also allows us to employ very simple methods to derive important results.

Using $p\propto a^{-1} $ find for non-relativistic particle \[E_{kin} \propto a^{-2} \] In terms of the redshift this gives \[E_{kin} =E_{kin0} (1+z)^{2} \] Consider now a relativistic particle. In this case, $E\propto p$ and in terms of the redshift, \[E=E_{0} (1+z)\] Consequently, at redshift $z=1$ , when the Universe is half its present size, the kinetic energy of a freely moving nonrelativistic particle is four times its present value, and the energy of a relativistic particle is twice its present value.

We have seen before that individual particles, moving freely, lose their energy when enclosed in an expanding box. Exactly the same thing happens to a gas consisting of many particles. Particles composing a gas continually collide with one another; between collisions they move freely and lose energy in the way described for free particles; during their encounters they exchange energy, but collisions do not change the total energy. The temperature of a gas therefore varies with expansion in the same manner as the energy of a single (nonrelativistic) particle: gas temperature is proportional to $1/a^{2} $ . If $T$ denotes temperature, and $T_{0} $ the present temperature, then \[T=T_{0} \left(1+z\right)^{2} \]

The number of photons in our cosmic box (and in the Universe) is a measure of its entropy. The total number of photons in the cosmic box is $N_{\lambda } =n_{\lambda } V$ . The photon density $n_{\lambda } $ varies as $T^{3} $ , and therefore varies as $1/a^{3} $. But $V$ varies as $a^{3} $ , hence also $VT^{3} =const$ . Thus the entropy of the thermal radiation in the cosmic box is constant during expansion. This is just another way of saying that the total number of photons $N_{\lambda } $ (and, consequently, entropy) in the box is constant. Actually, their number is slowly increased by the light emitted by stars and other sources, but this contribution is so small that for most purposes it can be ignored.

This unexpected increase in mass occurs because the radiation exerts pressure on the walls of the box and the walls contain stresses. These stresses in the walls are a form of energy that equals 3PV, where P is the pressure of the radiation. The pressure equals $\frac{1}{3} \rho $, and the energy in the walls is therefore $\rho V$and has a mass equivalent of $M=\rho V$ . The mass of the box is therefore increased by the mass $M$ of the radiation and the mass M of the stresses in the walls, giving a total increase of 2M. In the Universe there are no walls: nonetheless, the radiation still behaves as if it had a gravitational mass twice what is normally expected. Instead of using $\rho $ , we must use $\rho +3P$ as in the second Friedmann equation. This feature of general relativity explains why in a collapsing star, where all particles are squeezed to high energy, increasing the pressure, contrary to expectation, hastens the collapse of the star.

\[j=\frac{1}{H^{3} } \frac{\dddot{a}}{a} =\frac{1}{aH^{3} } \frac{d}{dt} \left(\frac{\ddot{a}}{aH^{2} } aH^{2} \right)=-\frac{1}{aH^{3} } \frac{d}{dt} \left(qaH^{2} \right)=q+2q^{2} -\frac{\dot{q}}{H} \]

Using \[\frac{\ddot{a}}{a} =\dot{H}+H^{2} \] we find \[\dot{\rho }+3H\left(\rho +p\right)=6H\left(-f(t)+\frac{\dot{f}(t)}{2H} +g(t)\right)\]

In this case$f(t)=g(t)=\Lambda /3,\; \; \dot{f}=0$ and \[\dot{\rho }+3H\left(\rho +p\right)=6H\left(-f(t)+\frac{\dot{f}(t)}{2H} +g(t)\right)\to \dot{\rho }+3H\left(\rho +p\right)=0\]

For \textbf{$f(t)=g(t)=\Lambda /3$} \[\dot{\rho }+3H\left(\rho +p\right)=6H\left(-f(t)+\frac{\dot{f}(t)}{2H} +g(t)\right)\to \dot{\rho }+3H\left(\rho +p\right)=\dot{\Lambda }(t)\] which corresponds to \textbf{$\Lambda (t)CDM$}model.</p> </div> </div></div> The de Sitter spacetime is the solution of the vacuum Einstein equations with a positive cosmological constant$\Lambda $ . To describe the geometry of this spacetime one usually takes the spatially flat metric \[ds^{2} =dt^{2} -a^{2} (t)d\vec{x}^{2} \] with the scale factor \[a(t)=a_{0} e^{Ht} \] The Hubble parameter is thus a fixed constant. <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 10''' <p style="color: #999;font-size: 11px">problem id: </p> Show that the de Sitter spacetime has a constant four-dimensional curvature. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[R=-6\left(\frac{\ddot{a}}{a} +H^{2} \right)=-6\left(\dot{H}+2H^{2} \right)=-12H^{2} \] As \[H^{2} =\frac{8\pi G}{3} \rho _{\Lambda } =\frac{\Lambda }{3} \] then \[R=-4\Lambda \]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 11''' <p style="color: #999;font-size: 11px">problem id: </p> In the de Sitter spacetime transform the FRLW metric into the explicitly conformally flat metric. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Introduce $dt\equiv a\left(\eta \right)d\eta $ to obtain \[ds^{2} =dt^{2} -a^{2} (t)d\vec{x}^{2} =a^{2} \left(\eta \right)\left(d\eta ^{2} -d\vec{x}^{2} \right)\] where the conformal time $\eta $ and the scale factor $a\left(\eta \right)$ are \[\eta =-\frac{1}{H} e^{-Ht} ,\quad a\left(\eta \right)=-\frac{1}{H\eta } \] The conformal time $\eta $ changes from $-\infty $ to $0$ when the proper time $t$ goes from $-\infty $ to $+\infty $.</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 12''' <p style="color: #999;font-size: 11px">problem id: </p> (Problems 12-13, A.Vilenkin, Many worlds in one, Hill and Wang, New York, 2006)} In thirties of XX-th century a cyclic Universe model was popular. This model predicted alternating stages of expansion and contraction. Show that such model contradicts the second law of thermodynamics. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Second law requires that entropy, which is a measure of disorder, should grow in each cycle of cosmic evolution. If the Universe had already gone through an infinite number of cycles, it would have reached the maximum-entropy state of thermal equilibrium ("heat death"). We certainly do not find ourselves in such a state.</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 13''' <p style="color: #999;font-size: 11px">problem id: </p> P. Steinhardt and N. Turok proposed a model of cyclic Universe where the expansion rate in each cycle is greater than the contraction one so that volume of the Universe grows from one cycle to the other. Show that this model does not contradict the second law of thermodynamics and is free of the heat death problem. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">The contradiction to the second principle of thermodynamics and therefore the heat death problem are absent in the considered model, because the amount of expansion in a cycle is greater than the amount of contraction. So the volume of the Universe is increased after each cycle. The entropy of our observable region is now the same as the entropy of a similar region in the preceding cycle, but the entropy of the entire Universe has increased, simply because the volume of the Universe is now greater. As time goes on, both the entropy and the total volume grow without bound. The state of maximum of entropy is never reached.</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 14''' <p style="color: #999;font-size: 11px">problem id: </p> If a closed Universe appeared as a quantum fluctuation, so what is the upper limit of its existence? (see A.Vilenkin, Many worlds in one, Hill and Wang, New York, 2006) <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">The energy of a closed Universe is always equal to zero. The energy of matter is positive, the gravitational energy is negative, and it turns out that in a closed Universe the two contributions exactly cancel each other. Thus if a closed Universe were to arise as a quantum fluctuation, there would be no need to borrow energy from the vacuum $\left(\Delta E=0\right)$ and because $\Delta E\Delta t\ge \hbar $, the lifetime of the fluctuation could be arbitrary long.</p> </div> </div></div> ---- ---- ---- =UNIQ--h-4--QINU NEW Problems in Dark Energy Category = ---- ---- ---- ==UNIQ--h-5--QINUBianchi I Model== [[Bianchi I Model|'''Bianchi I Model''']] (after Esra Russell, Can Battal Kılınç, Oktay K. Pashaev, Bianchi I Models: An Alternative Way To Model The Present-day Universe, arXiv:1312.3502) Theoretical arguments and indications from recent observational data support the existence of an anisotropic phase that approaches an isotropic one. Therefore, it makes sense to consider models of a Universe with an initially anisotropic background. The anisotropic and homogeneous Bianchi models may provide adequate description of anisotropic phase in history of Universe. One particular type of such models is Bianchi type I (BI) homogeneous models whose spatial sections are flat, but the expansion rates are direction dependent, \[ds^2={c^2}dt^2-a^{2}_{1}(t)dx^2-a^{2}_{2}(t)dy^2-a^{2}_{3}(t)dz^2\] where $a_{1}$, $a_{2}$ and $a_{3}$ represent three different scale factors which are a function of time $t$. <div id="bianchi_01"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 1''' <p style="color: #999;font-size: 11px">problem id: bianchi_01</p> Find the field equations of the BI Universe. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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}</p> </div> </div></div> <div id="bi_2"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 2''' <p style="color: #999;font-size: 11px">problem id: bi_2</p> Reformulate the field equations of the BI Universe in terms of the directional Hubble parameters. \[H_1\equiv\frac{\dot{a_1}}{a_1},\ H_2\equiv\frac{\dot{a_2}}{a_2},\ H_3\equiv\frac{\dot{a_3}}{a_3}.\] <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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}</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 3''' <p style="color: #999;font-size: 11px">problem id: </p> The BI Universe has a flat metric, which implies that its total density is equal to the critical density. Find the critical density. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[\rho_{cr}=H_1H_2+H_1H_3+H_2H_3.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 4''' <p style="color: #999;font-size: 11px">problem id: </p> Obtain an analogue of the conservation equation $\dot\rho+3H(\rho+p)=0$ for the case of the BI Universe. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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.</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 5''' <p style="color: #999;font-size: 11px">problem id: </p> Obtain the evolution equation for the mean of the three directional Hubble parameters $\bar H$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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 \[\dot{\bar H}+3\bar H^2=\frac12(\rho-p).\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 6''' <p style="color: #999;font-size: 11px">problem id: </p> Show that the system of equations for the BI Universe \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} can be transformed to the following \begin{align} \nonumber H_1H_2+H_1H_3+H_2H_3 & =\rho,\\ \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} </div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 7''' <p style="color: #999;font-size: 11px">problem id: </p> Show that the mean of the three directional Hubble parameters $\bar H$ is related to the elementary volume of the BI Universe $V\equiv a_1a_2a_3$ as \[\bar H=\frac13\frac{\dot V}{V}.\] <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[\bar H=\frac13\frac{d}{dt}\ln(a_1a_2a_3)=\frac13\frac{d}{dt}\ln V=\frac13\frac{\dot V}{V}.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 8''' <p style="color: #999;font-size: 11px">problem id: </p> Obtain the volume evolution equation of the BI model. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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 \[\ddot V-\frac32(\rho-p)V=0.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 9''' <p style="color: #999;font-size: 11px">problem id: </p> Find the generic solution of the directional Hubble parameters. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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.</p> </div> </div></div> ==UNIQ--h-6--QINURadiation dominated BI model == <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 10''' <p style="color: #999;font-size: 11px">problem id: </p> Find the energy density of the radiation dominated BI Universe in terms of volume element $V_r$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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.</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 11''' <p style="color: #999;font-size: 11px">problem id: </p> Find the mean Hubble parameter of the radiation dominated case. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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 \[\bar H=\frac13\frac{\dot V_r}{V}=\frac1{2t}.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 12''' <p style="color: #999;font-size: 11px">problem id: </p> Find the directional expansion rates of the radiation dominated model. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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 \[H_{r,i}t_0=\alpha_{r,i}\left(\frac{t_0}{t}\right)^{3/2}+\frac12\frac{t_0}{t};\quad \alpha_{r,i}\equiv\frac{K_{r,i}}{t_0}.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 13''' <p style="color: #999;font-size: 11px">problem id: </p> Find time dependence for the scale factors $a_i$ in the radiation dominated BI Universe. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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.</p> </div> </div></div> <div id="bianchi_02"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 14''' <p style="color: #999;font-size: 11px">problem id: bianchi_02</p> Find the partial energy densities for the two components of the BI Universe dominated by radiation and matter in terms of volume element $V_{rm}$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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}</p> </div> </div></div> <div id="bianchi_03"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 15''' <p style="color: #999;font-size: 11px">problem id: bianchi_03</p> Obtain time evolution equation for the total volume $V_{rm}$ in the BI Universe dominated by radiation and matter. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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 \[{{\dot V}_{rm}}^2-9\bar H^2_{0}\Omega_{m,0} V_{rm} - 9\bar H^2_{0}\Omega_{r,0} V_{rm}^{2/3}=0.\]</p> </div> </div></div> <div id="bianchi_04"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 16''' <p style="color: #999;font-size: 11px">problem id: bianchi_04</p> Using result of the previous problem, obtain a relation between the mean Hubble parameter and the volume element. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[\left(\frac{\bar H_{rm}}{\bar H_0}\right)^2=\frac{\Omega_{m,0}}{V_{rm}}+\frac{\Omega_{r,0}}{V_{rm}^{4/3}}.\]</p> </div> </div></div> ---- =UNIQ--h-7--QINU NEW problems in Dark Matter Category = ==UNIQ--h-8--QINUGeneralized models of unification of dark matter and dark energy== [[Generalized models of unification of dark matter and dark energy|'''Generalized models of unification of dark matter and dark energy''']] (see N. Caplar, H. Stefancic, Generalized models of unification of dark matter and dark energy (arXiv: 1208.0449)) <div id="gmudedm_1"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 1''' <p style="color: #999;font-size: 11px">problem id: gmudedm_1</p> The equation of state of a barotropic cosmic fluid can in general be written as an implicitly defined relation between the fluid pressure $p$ and its energy density $\rho$, \[F(\rho,p)=0.\] Find the speed of sound in such fluid. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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 \[c_s^2=\frac{dp}{d\rho}=-\frac{\frac{\partial F}{\partial\rho}}{\frac{\partial F}{\partial p}}.\]</p> </div> </div></div> <div id="gmudedm_2"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 2''' <p style="color: #999;font-size: 11px">problem id: gmudedm_2</p> For the barotropic fluid with a constant speed of sound $c_s^2=const$ find evolution of the parameter of EOS, density and pressure with the redshift. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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 \[p=c_s^2\rho-\rho_0(c_s^2-w_0)= \rho_0\frac{c_s^2-w_0}{1+c_s^2}\left[\frac{1+w_0}{c_s^2-w_0}(1+z)^{3(1+c_s^2)}-1\right].\] </p> </div> </div></div> ---- =UNIQ--h-9--QINU Tutti Frutti = [[Planck_scales_and_fundamental_constants#Problem_15|'''New problem in Cosmo warm-up Category:''']] <div id="TF_1"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 1''' <p style="color: #999;font-size: 11px">problem id: TF_1</p> Construct planck units in a space of arbitrary dimension. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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 \[L_{P(D)}=\left(\frac{G_D\hbar}{c^3}\right)^{\frac{1}{D-2}}\quad T_{P(D)}=\left(\frac{G_D\hbar}{c^{D+1}}\right)^{\frac{1}{D-2}}\quad M_{P(D)}=\left(\frac{c^{5-D}\hbar^{D-3}}{G_D}\right)^{\frac{1}{D-2}}.\]</p> </div> </div></div> '''New problem in Inflation Category:''' <div id="TF_2"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 2''' <p style="color: #999;font-size: 11px">problem id: TF_2</p> Show that for power law $a(t)\propto t^n$ expansion slow roll inflation occurs when $n\gg1$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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 $n\gg1$.</p> </div> </div></div> <div id="TF_3"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 3''' <p style="color: #999;font-size: 11px">problem id: TF_3</p> Find the general condition to have accelerated expansion in terms of the energy densities of the darks components and their EoS parameters <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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 \[\rho_{de}>-\frac{1+3w_{dm}}{1+3w_{de}}\rho_{dm}.\]</p> </div> </div></div> <div id="dec_5"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 4''' <p style="color: #999;font-size: 11px">problem id: dec_5</p> Complementing the assumption of isotropy with the additional assumption of homogeneity predicts the space-time metric to become of the Robertson-Walker type, predicts the redshift of light $z$, and predicts the Hubble expansion of the Universe. Then the cosmic luminosity distance-redshift relation for comoving observers and sources becomes \[d_L(z)=\frac{cz}{H_0}\left[1-(1-q_0)\frac z2\right]+O(z^3)\] with $H_0$ and $q_0$ denoting the Hubble and deceleration parameters, respectively. Show that this prediction holds for arbitrary spatial curvature, any theory of gravity (as long as space-time is described by a single metric) and arbitrary matter content of the Universe.(see 1212.3691) </div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 5''' <p style="color: #999;font-size: 11px">problem id: </p> Show that in the Universe filled by radiation and matter the sound speed equals to \[c_s^2=\frac13\left(\frac34\frac{\rho_m}{\rho_r}+1\right)^{-1}.\] </div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 6''' <p style="color: #999;font-size: 11px">problem id: </p> Show that result of the previous problem can be presented in the following form \[c_s^2=\frac43\frac{1}{(4+3y)},\quad y\equiv\frac a{a_{eq}},\] where $a_{eq}$ is the scale factor value in the moment when matter density equals to that of radiation. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[c_s^2=\frac13\left(\frac34\frac{\rho_m}{\rho_r}+1\right)^{-1},\quad \frac{\rho_m}{\rho_r}=a\frac{\rho_{m0}}{\rho_{r0}},\quad \frac{\rho_{m0}}{\rho_{r0}}=\frac1{a_{eq}},\] \[c_s^2=\frac13\frac1{\left(\frac34\frac{a}{a_{eq}}+1\right)}=\frac43\frac{1}{(4+3y)}.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 7''' <p style="color: #999;font-size: 11px">problem id: </p> Show that in the flat Universe filled by non-relativistic matter and radiation the effective radiation parameter $w_{tot}=p_{tot}/\rho_{tot}$ equals \[w_{tot}=\frac1{3(1+y)},\quad y\equiv\frac a{a_{eq}},\] where $a_{eq}$ is the scale factor value in the moment when matter density equals to that of radiation. </div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 8''' <p style="color: #999;font-size: 11px">problem id: </p> Show that in spatially flat one-component Universe the following hold \[\bar{H'}=-\frac{1+3w}2\bar H^2.\] <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[\bar H^2=a^2\rho,\quad (8\pi G/3=1),\] \[\rho'+3\bar H\rho(1+w)=0,\] \[\bar{H'}=a^2\rho-\frac32a^2\rho-\frac32a^2\rho w\to\bar{H'}=-\frac{1+3w}2\bar H^2.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 9''' <p style="color: #999;font-size: 11px">problem id: </p> Express statefinder parameters in terms Hubble parameter and its derivatives with respect to cosmic times. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[r=1+3\frac{\dot H}{H^2}+\frac{\ddot H}{H^3},\quad s=-\frac{2}{3H}\frac{3H\dot H+\ddot H}{3H^2+2\dot H}.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 10''' <p style="color: #999;font-size: 11px">problem id: </p> Find temperature of radiation and Hubble parameter in the epoch when matter density was equal to that of radiation (Note that it was well before the last scattering epoch). <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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').</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 11''' <p style="color: #999;font-size: 11px">problem id: </p> Estimate the mass-energy density $\rho$ and pressure $p$ at the center of the Sun and show that $\rho\gg p$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">For the Sun: \[p\approx\frac{GM_\odot^2}{R_\odot^4}\approx10^{16}J/m63;\] \[\rho\ge\frac{M_\odot c^2}{\frac43\pi R_\odot^3}\sim10^{21}J/m^3.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 12''' <p style="color: #999;font-size: 11px">problem id: </p> For a perfect fluid show that ${T^{\alpha\beta}}_{;\alpha}=0$ implies \[(\rho+p)u^\alpha\nabla_\alpha u^\beta=h^{\beta\gamma}\nabla_\gamma p,\] where $h_{\alpha\beta}\equiv g_{\alpha\beta}-u_\alpha u_\beta$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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, \[(\rho+p)u^\alpha\nabla^\alpha u_\beta=\left(g^{\alpha\beta}-u^\beta u^\gamma\right)\nabla_\gamma p=0.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 13''' <p style="color: #999;font-size: 11px">problem id: </p> Show that in flat Universe filled by non-relativistic matter and a substance with the state equation $p_X=w_X\rho_X$ the following holds \[\frac{d\ln H}{d\ln a}-\frac12\frac{\Omega_X}{1-\Omega_X}\frac{d\ln\Omega_X}{d\ln a}+\frac32=0.\] <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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 \[\frac{d\ln H}{d\ln a}-\frac12\frac{\Omega_X}{1-\Omega_X}\frac{d\ln\Omega_X}{d\ln a}+\frac32=0.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 14''' <p style="color: #999;font-size: 11px">problem id: </p> (after Ming-Jian Zhang, Cong Ma, Zhi-Song Zhang, Zhong-Xu Zhai, Tong-Jie Zhang, Cosmological constraints on holographic dark energy models under the energy conditions) Using result of the previous problem, find EoS parameter $w_{hde}$ for holographic dark energy, taking the IR cut-off scale equal to the following: <br /> i) event horizon; <br />ii) conformal time; <br />iii) Cosmic age. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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 $c>0$. <br /> 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 $c>\sqrt{\Omega_{hde}}(1+z).$ <br /> 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 $c>\sqrt{\Omega_{hde}}$.</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 15''' <p style="color: #999;font-size: 11px">problem id: </p> According to so-called Jeans criterion exponential growth of the perturbation, and hence instability, will occur for wavelengths that satisfy: \[k<\frac{\sqrt{4\pi G\rho}}{v_S}\equiv k_J.\] In other words, perturbations on scales larger than the Jeans scale, defined as follows: \[R_J=\frac\pi {k_J}\] will become unstable and collapse. Give a physical interpretation of this criterion. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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 \[R_J\approx\frac{v_S}{\sqrt{G\rho}}.\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 16''' <p style="color: #999;font-size: 11px">problem id: </p> According to the Jeans criterion, initial collapse occurs whenever gravity overcomes pressure. Put differently, the important scales in star formation are those on which gravity operates against electromagnetic forces, and thus the natural dimensionless constant that quantifies star formation processes is given by: \[\alpha_g=\frac{Gm_p^2}{e^2}\approx8\times10^{-37}.\] Estimate the maximal mass of a white dwarf star in terms of $\alpha_g$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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, $1.4M_\odot$.</p> </div> </div></div> ---- The formation of a star, or indeed a star cluster, begins with the collapse of an overdense region whose mass is larger than the Jeans mass, defined in terms of the Jeans mass $R_J$(???), \[M_J=\frac43\pi\rho\left(\frac{R_J}2\right)^3\propto\frac{T^{3/2}}{\rho^{1/2}}.\] (why $T^{3/2}$, if in gases it is $T^{1/2}$???) Overdensities can arise as a result of turbulent motions in the cloud. At the first stage of the collapse, the gas is optically thin and isothermal, whereas the density increases and $M_J\propto\rho^{-1/2}$. As a result, the Jeans mass decreases and smaller clumps inside the originally collapsing region begin to collapse separately. Fragmentation is halted when the gas becomes optically thick and adiabatic, so that $M_J\propto\rho^{1/2}$, as illustrated in fig. 1. This process determines the opacity-limited minimum fragmentation scale for low mass stars, and is given by: \[M_{min}\approx m_p\alpha_g^{-3/2}\alpha^{-1}\left(\frac{m_e}{m_p}\right)^{1/4}\approx0.01M_\odot.\] Of course, this number, which is a robust scale and confirmed in simulations, is far smaller than the observed current epoch stellar mass range, for which the characteristic stellar mass is $\sim0.5M_\odot$. Fragmentation also leads to the formation of star clusters, where many stars with different masses form through the initial collapse of a large cloud. In reality, however, the process of star formation is more complex, and the initial collapse of an overdense clump is followed by accretion of cold gas at a typical rate of $v_S^3/G$, where $v_S$ is the speed of sound. This assumes spherical symmetry, but accretion along filaments, which is closer to what is actually observed, yields similar rates. The gas surrounding the protostellar object typically has too much angular momentum to fall directly onto the protostar, and as a result an accretion disk forms around the central object. The final mass of the star is fixed only when accretion is halted by some feedback process. ---- <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 17''' <p style="color: #999;font-size: 11px">problem id: </p> Using the "generating function" $G(\varphi)$, \[H(\varphi,\dot\varphi)=-\frac1{\dot\varphi}\frac{dG^2(\varphi)}{d\varphi},\] make transition from the two coupled differential equations with respect to time \[3H^2=\frac12\dot\varphi^2+V(\varphi);\] \[\ddot\varphi+3H\dot\varphi+V'(\varphi)=0.\] to one non-linear first order differential equation with respect to the scalar field. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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.</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 18''' <p style="color: #999;font-size: 11px">problem id: </p> (Hyeong-Chan Kim, Inflation as an attractor in scalar cosmology, arXiv:12110604) Express the EoS parameter of the scalar field in terms of the generating function and find the condition under which the scalar field behaves as the cosmological constant. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">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.