New problems
Contents
New to Observational Cosmology
New from June 2015
To Chapter 2
problem id: 150_0
Comprehensive explanations of the expanding Universe often use the balloon analogy. Although the balloon analogy is useful, one must guard against misconceptions that it can generate. Point out the misconceptions that appear when using this analogy [see M.O. Farooq, Observational constraints on dark energy cosmological model parameters, arXiv: 1309.3710.]
(a) The surface that is expanding is two dimensional; the "center" of the balloon is in the third dimension and is not part of the surface, which has no center.
(b) The balloon is expanded by the pressure difference between the inside and the outside, but the Universe is not being expanded by pressure.
(c) Pressure couples to gravity in the Einstein equation, so the addition of (positive) pressure to the Universe would slow, not increase, the expansion rate. It is the negative pressure which is only capable to generate the accelerated expansion of the Universe.
(d) If the dots on the balloon represent galaxies, they too will expand. But real galaxies do not expand due to general Hubble expansion because they are gravitationally bound objects. We can make a better analogy by gluing solid objects (like 10 cent coins) to the surface of the balloon to represent galaxies, so that they do not expand when the balloon expands.
problem id: 150_1
(into the cosmography and extended deceleration parameter) Show that \[\frac{d\dot a}{da}=-Hq.\]
\[\frac{d\dot a}{da}=\frac{d\dot a}{dt}\frac{dt}{da}=\frac{\ddot a}{\dot a}=\frac{\ddot a}{aH}\equiv-Hq.\]
problem id: 150_2
Give a physical interpretation of the conservation equation.
The energy density has a time dependence determined by the conservation equation, \[\dot\rho=-3H\rho-3Hp.\] The two terms define the behavior of the uniform fluid which contains the energy in a dynamic Universe. The $H$ term provides the "friction", while the density term tracks the reduction in density due to the volume increase during expansion and the pressure term tracks the reduction in pressure energy during expansion.
problem id: 150_04
Find evolution equation for the density parameter $\Omega$ of the single-fluid FLRW models with the linear equation of state $p=w\rho$.
\begin{align} \nonumber \Omega&=\frac\rho{3H^2},\quad (8\pi G=1),\\ \nonumber N&=\log\frac{a}{a_0},\\ \nonumber \frac d{dN}&=\frac1H\frac{d}{dt},\\ \nonumber \frac{d\Omega}{dN}&=\frac13\frac{\frac{d\rho}{dN}H^2-2\rho H\frac{dH}{dN}}{H^4},\\ \nonumber \frac{d\rho}{dN}&=\frac1H\frac{d\rho}{dt}=-3\rho(1+w),\\ \nonumber \frac{dH}{dN}&=-(1+q)H,\\ \nonumber \frac{d\Omega}{dN}&=-[2q-3w-1]\Omega. \end{align} For a single-component fluid \[q=\frac12(1+3w)\Omega\] and one finally obtains \[\frac{d\Omega}{dN}=-(1+3w)(1-\Omega)\Omega.\]
problem id: 150_05
Solve the previous problem for the multi-component case.
The evolution equations for an $n$-fluid model use the density parameters $\Omega_1$, $\Omega_2\ldots\Omega_n$, as dynamical variables. Relation \[\frac{d\Omega}{dN}=-[2q-3w-1]\Omega\] can easily be generalized or an $n$-fluid model to give \[\frac{d\Omega_i}{dN}=-[2q-(1+3w_i)]\Omega_i,\quad i=1\ldots n,\] where \[q=\frac12\sum\limits_i(1+3w_i)\Omega_i.\] For the total density parameter \[\Omega_{tot}=\sum\limits_{i=1}^n\Omega_i\] one obtains the evolution equation \[\frac{d\Omega_{tot}}{dN}=-2q(1-\Omega_{tot}).\]
problem id: 150_06
Use the conformal time to prove existence of smooth transition from the radiation-dominated era to the matter dominated one.
Rewrite the first Friedmann equation in the form \[H^2=\frac{\rho_{eq}}{3}\left[\left(\frac{a_{eq}}{a}\right)^3+\left(\frac{a_{eq}}{a}\right)^4\right],\] where $a_{eq}$ is the scale factor when the densities of matter and radiation are the same, and $\rho_{eq}$ is the density in each component at that time. This equation can be solved exactly using conformal time \[\frac{da}{dt}=\frac1a\frac{da}{d\eta},\quad \left(\frac{da}{d\eta}\right)^2=\frac13\rho_{tot}a^4,\] \[\frac{a(\eta)}{a_{eq}}=2(\sqrt2-1)\left(\frac\eta{\eta_{eq}}\right)+[1-2(\sqrt2-1)]\left(\frac\eta{\eta_{eq}}\right)^2,\quad \eta_{eq}\equiv\frac{2(\sqrt2-1)}{a_{eq}}\sqrt{\frac{3}{\rho_{eq}}}.\] We see a smooth transition between the two characteristic behaviors of radiation domination $\eta\ll\eta_{eq}$, $a\propto\eta$ and matter domination $\eta\gg\eta_{eq}$, $a\propto\eta^2$. There is no good way to rewrite this relation in terms of the cosmic times.
problem id: 150_07
Consider a set of the cosmographic parameters built from the Hubble parameter and its time derivatives [see S. Carloni, A new approach to the analysis of the phase space of f(R)-gravity, arXiv:1505.06015) ] \[Q\equiv\frac{\dot H}{H^2},\quad J\equiv\frac{\ddot H H}{\dot H^2},\quad S\equiv\frac{\dddot H H^2}{\dot H^3},\ldots \] Express them in terms of the canonic cosmographic parameters $q,j,s\dots$.
\[Q=\frac{\dot H}{H^2}=\left(\frac{\ddot a}a-H^2\right)\frac1{H^2}=-(q+1);\] \[J=\ddot H \frac{H}{\dot H^2}=\left(\frac{\dddot a}a-\frac{\dot a\ddot a}{a^2}-2H\dot H\right)\frac{H}{\dot H^2}=\frac j{Q^2}+\frac q{Q^2}-\frac2Q=\frac{j+3q+2}{(1+q)^2}.\]
problem id: 150_08
Consider another set of the cosmographic parameters [see S. Carloni, A new approach to the analysis of the phase space of f(R)-gravity, arXiv:1505.06015) ] \[\bar Q\equiv\frac{H_{,N}}{H},\quad \bar J\equiv\frac{H_{,NN}}{H},\quad \bar S\equiv\frac{H_{,NNN}}{H},\ldots,\] where \[H_{,N}\equiv \frac{dH}{d\ln a}.\] Express them in terms of the Hubble parameter and its time derivatives.
\[\frac{d}{d\ln a}=\frac1H\frac{d}{dt};\] \[\bar Q\equiv\frac{\dot H}{H^2},\quad \bar J\equiv\frac{\ddot H}{H^3}-\frac{\dot H^2}{H^4},\quad \bar S\equiv\frac{\dddot H}{H^4}+3\frac{\dot H^3}{H^6}-4\frac{\dot H\ddot H}{H^5}.\]
problem id: 150_09
Express the Ricci scalar and its time derivatives in terms of the $\bar Q$, $\bar J$ and $\bar S$.
\begin{align} \nonumber R&= -6\left[(\bar Q+2)H^2+\frac k{a^2}\right];\\ \nonumber \dot R&= -6H\left\{\left[\bar J+\bar Q(\bar Q+4)\right]H^2-\frac{2k}{a^2}\right\};\\ \nonumber \ddot R&= -6H^2\left\{\left[\bar S+4\bar J(\bar Q+1)+(\bar Q+8)\bar Q^2\right]H^2-2(2-\bar Q)\frac{k}{a^2}\right\}. \end{align}
problem id: 150_3
Show that for a perfect fluid with EoS $p=w(a)\rho$ the adiabatic sound speed can be represented in the form \[c_S^2=w(a)-\frac13\frac{d\ln(1+w)}{d\ln a}.\]
\[c_S^2\equiv\frac{d p}{d\rho}=w(a)+\rho\frac{dw(a)}{d\rho};\] \[\frac{dw}{d\rho}=\frac{dw}{da}\frac{da}{d\rho};\] \[\dot\rho+3H(\rho+p)\Rightarrow\frac{da}{d\rho}=-\frac13\frac a{\rho(1+w)},\] \[c_S^2=w(a)+\rho\frac{dw}{d a}\left(-\frac13\frac a{\rho(1+w)}\right)=w(a)-\frac13\frac{d\ln(1+w)}{d\ln a}.\]
problem id: 150_4
Obtain equation for $\ddot\rho(t)$, where $\rho(t)$ is energy density of an ideal fluid participating in the cosmological expansion.
Take time derivative of the conservation equation to obtain \begin{align}\nonumber \frac{d}{dt}[\dot\rho+3H(\rho+p)]&=\ddot\rho+3\dot H(\rho+p)+3H(\dot\rho+\dot p)=\\ \nonumber &=\ddot\rho+3\dot H(\rho+p)+3H\dot\rho(1+c_S^2)=\\ \nonumber & =\ddot\rho+H(\rho+p)-9H^2(\rho+p)(1+c_S^2)=0. \end{align} It then follows that \[\ddot\rho=-3[\dot H+3H^2(1+c_S^2)](\rho+p).\]
problem id: 150_5
Show that in the case of the flat Friedmann metric, the third power of the scale factor $\varphi\equiv a^3$ satisfies the equation \[\frac{d^2\varphi}{dt^2}=\frac32(\rho-p)\varphi,\quad 8\pi G=1.\] Check validity of this equation for different cosmological components: non-relativistic matter, cosmological constant and a component with EoS $p=w\rho$.
\[\frac{d^2 a^3}{dt^2}=3\left(2H^2+\frac{\ddot a}a\right)a^3=3\left[\frac23\rho-\frac16(\rho+3p)\right]a^3=\frac32(\rho-p)\varphi.\]
problem id: 150_6
The lookback time is defined as the difference between the present day age of the Universe and its age at redshift $z$, i.e. the difference between the age of the Universe at observation $t_0$ and the age of the Universe, $t$, when the photons were emitted. Find the lookback time for the Universe filled with non-relativistic matter, radiation and a component with the EoS $p=w(z)\rho$.
From the definitions of redshift $1+z=1/a$ we have \[\frac{dz}{dt}=-\frac{\dot a}{a^2}=-H(z)(1+z)\] or \[dt=-\frac{dz}{H(z)(1+z)}.\] The lookback time is defined as \[t_0-t=\int\limits_t^{t_0}dt=\int\limits_0^z \frac{dz'}{H(z')(1+z')}=\frac1{H_0}\int\limits_0^z \frac{dz'}{E(z')(1+z')}\] where \[E(z)=\sqrt{\Omega_r(1+z)^4+\Omega_m(1+z)^3+\Omega_k(1+z)^2+\Omega_w\exp\left(3\int\limits_0^z dz'\frac{1+w(z')}{1+z'}\right)}.\] From the definition of lookback time it is clear that the cosmological time or the time back to the Big Bang, is given by \[t(z)=\int\limits_z^\infty\frac{dz'}{H(z')(1+z')}.\]
problem id: 150_7
Show that the Hubble radius grows faster than the expanding Universe in the case of power law expansion $a(t)\propto t^\alpha$ with $\alpha<1$ (the decelerated expansion).
In the case of the power law expansion with $\alpha<1$ (the decelerated expansion) the Hubble radius indeed grows faster than the expanding Universe: $R_H=H^{-1}\propto t$, while $a(t)\propto t^\alpha$. In power law situations the Hubble radius has an expansion velocity \[\frac{d}{dt}\left(\frac1H\right)=\frac1\alpha\] greater than light speed. This behavior is true only for a decelerating Universe composed of matter and radiation. This is not a physical velocity, violating special relativity, but the velocity of expansion of the metric itself.
To chapter 3
problem id: 150_015
Show, that in the Milne Universe the age of the Universe is equal to the Hubble time.
In an empty (Milne) Universe since $H=\dot a/a=t^{-1}$, the age of the Universe $t_0$ is equal to the Hubble time $t_0=H_0^{-1}$.
To chapter 4 The black holes
problem id: new2015_1
see E. Berti, A Black-Hole Primer: Particles, Waves, Critical Phenomena and Superradiant instabilities (arXiv:1410.4481[gr-qc])
A Newtonian analog of the black hole concept is a so-called "dark star". If we consider light as a corpuscle traveling at speed $c$, light cannot escape to infinity whenever $V_{esc}>c$, where \[V_{esc}^2=\frac{2GN}R.\] Therefore the condition for existence of "dark stars" in Newtonian mechanics is \[\frac{2GN}{c^2R}\ge1.\]
Can this condition be satisfied in the Newtonian mechanics?
A naive argument tells us that as we pile up more and more material of constant density $\rho_0$, the ratio $M/R$ increases: \begin{equation}\label{new2015_1_e1} \frac M R=\frac43\pi R^2\rho_0. \end{equation} This equation would seem to suggest that dark stars could indeed form. However, we must include the binding energy $U$, \begin{equation}\label{new2015_1_e2} U=-\int\frac{GMdM}{r} =-\int\limits_0^R\frac{G}{r}\left(\frac43\pi r^3\rho_0\right)\frac4\pi r^2\rho_0 dr=-\frac{16G\pi^2}{15}\rho_0^2R^5. \end{equation} The total mass $M_T$ of the hypothetical dark star is given by the rest mass $M$ plus the binding energy \begin{equation}\label{new2015_1_e3} \frac{M_T}R=\frac43\pi R^2\rho_0-\frac{16G\pi^2}{15}\rho_0^2R^4=\frac M R\left[1-\frac35\frac G{c^2}\frac M R\right] \le\frac5{12}\frac{c^2}G, \end{equation} where the upper limit is obtained by maximizing the function in the range (\ref{new2015_1_e1}). Thus, the dark star criterion (\ref{new2015_1_e1}) is never satisfied, even for the unrealistic case of constant-density matter. In fact, the endpoint of Newtonian gravitational collapse depends very sensitively on the equation of state, even in spherical symmetry.
problem id: 150_017
Derive the relation $T\propto M^{-1}$ from the Heisenberg uncertainty principle and the fact that the size of a black hole is given by the Schwarzschild radius.
The position of quanta inside a black hole can only be known within $\Delta x=2r_{Sh}=4GM/c^2$. Thus $\Delta t =2r_{Sh}/c=4GM/c^3$. According to the uncertainty principle $\Delta E\Delta t\ge\hbar$. Thus \[\Delta T=\frac{\Delta E}k\approx\frac{\bar c^3}{4kGM}\] Up to a numeric factor this relation coincides with the black hole Hawking temperature $T=\hbar c^3/8\pi kGM$.
To chapter 8
problem id: 2501_06
Why the cosmological constant cannot be used as a source for inflation in the inflation model?
The cosmological constant do provide the accelerated expansion of Universe needed to realize the inflation: $a(t)\propto e^{Ht}$, $q=-1$. However, in approximation of the cosmological constant, $H$ is constant for all time. Therefore a dynamical mechanism for the limited time of inflation is needed. The physical mechanism for the existence of an approximately constant value of $H$ which lasts for a limited time is given by a scalar field. For a large initial potential energy of scalar field the state equation parameter $w\approx-1$ and the scalar field in process of the "slow roll" imitates the cosmological constant for sufficiently long period of time to solve the flatness and causal problems. Then due to shape of the potential the scalar field exits from the slow-roll regime, oscillates about its' potential minimum decaying into less massive particles insuring that inflation time is finite.
problem id: 2501_09
Show that inflation ends when the parameter \[\varepsilon\equiv\frac{M_P^2}{16\pi}\left(\frac{dV}{d\varphi}\frac1V\right)^2=1.\]
Using definition of the parameter $\varepsilon$ one finds \[\varepsilon=-\frac{\dot H}{H^2}.\] In the slow-roll approximation \[H^2=\frac{8\pi}{3M_P^2V},\] \[3H\dot\varphi=-\frac{dV}{d\varphi}.\] Therefore \[\frac{\ddot a}a=H^2(1-\varepsilon).\] The condition $\varepsilon=1$ is equivalent to $\ddot a=0$. When the value $\varepsilon=1$ is reached due to variation of the potential shape the Universe exits the regime of the accelerated expansion (inflation). Around tend, the inflaton field(s) typically begin oscillating around the minimum of the potential. (see Inflation and the Higgs Scalar, DANIEL GREEN (1412.2107).)
problem id: 2501_10
How does the number of e-folds $N$ depend on the slow-roll parameter $\varepsilon$?
\[N\propto\int\limits_{\varphi_{end}}^{\varphi_{initial}}d\varphi\frac{V}{dV/d\varphi} \propto\int\limits_{\varphi_{end}}^{\varphi_{initial}}d\varphi/\sqrt\varepsilon.\] The number of $e$-folds depends on how fast the field is $f$-decreasing. The number of e-folds, $N$ , is inversely proportional to the square root of the slow roll parameter $\varepsilon$ or proportional to the inverse fractional change of the potential with the field, $V/(dV/d\varphi)$.
To chapter 9
problem id: 150_021
Using the by dimensional analysis for cosmological constant $\Lambda > 0$, define the set of fundamental "de Sitter units" of length, time and mass.
\begin{align} \nonumber L_{dS}&=\sqrt{1/\Lambda}\\ \nonumber T_{dS}&=\frac1c\sqrt{1/\Lambda}\\ \nonumber M_{dS}&=\frac\hbar c\sqrt{\Lambda} \end{align}
problem id: 150_022
In the Newtonian approximation, find the force acting on the point unit mass in the Universe filled by non-relativistic matter and cosmological constant. (see Chiu Man Ho and Stephen D. H. Hsu, The Dark Force: Astrophysical Repulsion from Dark Energy, arXiv: 1501.05592)
The Einstein equation with cosmological constant $\Lambda$ is \[R_{\mu\nu}-\frac12g_{\mu\nu}R=8\pi GT_{\mu\nu}+g_{\mu\nu}\Lambda.\] Contracting both sides with $g^{\mu\nu}$, one gets \[R=-8\pi GT-4\Lambda,\] where $T=T_\mu^\mu$ is the trace of the matter (including dark matter) energy-momentum tensor. This can be substituted in the original equation to obtain \[R_{\mu\nu}8\pi G\left(T_{\mu\nu}-\frac12T\right)-g_{\mu\nu}\Lambda.\] In the Newtonian limit, one can decompose the metric tensor as \[g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu},\quad |h_{\mu\nu}|\ll1.\] Specifically we are interested in the $00$-component of the Einstein equation. We parameterize the $00$th-component of the metric tensor as \[g_{00}=1+2\Phi,\] where $\Phi$ is the Newtonian gravitational potential. To leading order, one can show that [S. Weinberg, Gravitation and Cosmology, John Wiley \& Sons, Inc., 1972.] \[R_{00}\approx\frac12\vec\nabla^2g_{00}=\vec\nabla^2\Phi.\] In the inertial frame of a perfect fluid, its $4$-velocity is given by $u_\mu=(1,0)$ and we have \[T_{\mu\nu}=(\rho+p)u_{\mu\nu}-pg_{\mu\nu}=diag(\rho,p).\] where $\rho$ is the energy density and $p$ is the pressure. For a Newtonian (non-relativistic) fluid, the pressure is negligible compared to the energy density, and hence $T\approx T_{00}=\rho$. As a result, in the Newtonian limit, the 00-component of the Einstein equation reduces to \[\vec\nabla^2\Phi=4\pi G\rho-\Lambda,\] which is just the modified Poisson equation for Newtonian gravity, including cosmological constant. This equation can also be derived from the Poisson equation of Newtonian gravity, $\vec\nabla^2\Phi=4\pi G(\rho+3p)$, with source terms from matter and dark energy; $p\approx 0$ for non-relativistic matter, and $p=-\rho$ for a cosmological constant. Assuming spherical symmetry, we have \[\vec\nabla^2\Phi=\frac1{r^2}\frac\partial{\partial r}\left(\frac{\partial\Phi}{\partial r}\right)\] and the Poisson equation is easily solved to obtain \[\Phi=-\frac{GM}{r}-\frac\Lambda6r^2,\] where $M$ is the total mass enclosed by the volume $4/3\pi r^3$. The corresponding gravitational field strength is given by \[\vec{g}=-\vec\nabla\Phi=\left(-\frac{GM}{r^2}+\frac\Lambda3r\right)\frac{\vec{r}}r.\]
problem id: 150_023
Consider a spatially flat FLRW Universe, which consists of two components: the non-relativistic matter and the scalar field $\varphi$ in the potential $V(\varphi)$. Find relation between the scalar field potential and the deceleration parameter.
From the definition of the energy density and the pressure for the scalar field it follows that \begin{equation}\label{150_023_e1} V(\varphi)=\frac12(\rho_\varphi-p_\varphi). \end{equation} The deceleration parameter by definition is \[q=-\frac{\ddot a}{aH^2}=\frac1{6M_{Pl}^2H^2}(\rho_m+\rho_\varphi+3p_\varphi).\] Substituting $\rho_m+\rho_\varphi=3M_{Pl}^2H^2$, one finds \[p_\varphi=(2q-1)M_{Pl}^2H^2.\] Substituting $\rho_m=3M_{Pl}^2H^2-\rho_\varphi$ and $p_\varphi=(2q-1)M_{Pl}^2H^2$ into (\ref{150_023_e1}), one finds \[V[\varphi(z)]=\rho_{m0}\left[\frac{(2-q)}{3\Omega_{m0}}\frac{H^2}{H_0^2}-\frac12(1+z)^3\right].\]
problem id: 150_024
Find relation between the deceleration parameter and the derivative $d\varphi/dz$ for the Universe considered in the previous problem.
From the definition of the energy density and the pressure for the scalar field it follows that \begin{equation}\label{150_024_e1} \dot\varphi^2=\rho_\varphi+p_\varphi. \end{equation} Substitute the relations $\rho_m=3M_{Pl}^2H^2-\rho_\varphi$ and $p_\varphi=(2q-1)M_{Pl}^2H^2$ (see the previous problem) into (\ref{150_024_e1}) to obtain \[\left(\frac{d\varphi}{dz}\right)^2=M_{Pl}^2\left[\frac{2(1+q)}{(1+z)^2-3\Omega_{m0}}(1+z)\frac{H_0^2}{H^2}\right].\]
problem id: new_30
Find the sound speed for the modified Chaplygin gas with the state equation \[p=B\rho-\frac A{\rho^\alpha}.\]
\[c_s^2=\frac{\delta p}{\delta\rho}=\frac{\dot p}{\dot\rho}=-\alpha w+(1+\alpha)B,\quad w=\frac p\rho.\] (place after the problem 81, chapter 9 of the book.)
A couple of problems for the SCM
problem id: 150_026
Let $N=\ln(a/a_0)$, where $a_0=a(t_0)$ and $t_0$ is some chosen reference time. Usually the reference time is the present time and in that case $\tau=-\ln(1+z)$. Find $\Omega_m(N)$ and $\Omega_\Lambda(N)$ for the SCM.
\begin{align} \nonumber \Omega_m&=\frac{\rho_m}{3H^2}=\frac{\Omega_{m0}\exp(-3N)}{\Omega_{m0}\exp(-3N)+\Omega_{\Lambda0}};\\ \nonumber \Omega_\Lambda&=\frac{\rho_\Lambda}{3H^2}=\frac{\Omega_{\Lambda0}}{\Omega_{m0}\exp(-3N)+\Omega_{\Lambda0}}. \end{align}
problem id: 150_027
Express the cosmographic parameters $H,q,j$ as functions of $N=\ln a/a_0$ for the SCM.
\begin{align} \nonumber H&=H_0\sqrt{\Omega_{m0}\exp(-3N)+\Omega_{\Lambda0}},\\ \nonumber q&=-1+\frac32\Omega_m=-1+\frac32\frac{\Omega_{m0}\exp(-3N)}{\Omega_{m0}\exp(-3N)+\Omega_{\Lambda0}},\\ \nonumber j&=1. \end{align}
Cardassian Model
[K. Freese and M. Lewis, Cardassian Expansion: a Model in which the Universe is Flat, Matter Dominated, and Accelerating, arXiv: 0201229] Cardassian Model is a modification to the Friedmann equation in which the Universe is flat, matter dominated, and accelerating. An additional term, which contains only matter or radiation (no vacuum contribution), becomes the dominant driver of expansion at a late epoch of the universe. During the epoch when the new term dominates, the universe accelerates. The authors named this period of acceleration by the Cardassian era. (The name Cardassian refers to a humanoid race in Star Trek whose goal is to take over the universe, i.e., accelerated expansion. This race looks foreign to us and yet is made entirely of matter.) Pure matter (or radiation) alone can drive an accelerated expansion if the first Friedmann equation is modified by the addition of a new term on the right hand side as follows: \[H^2=A\rho+B\rho^n,\] where the energy density $\rho$ contains only ordinary matter and radiation, and $n<2/3$. In the usual Friedmann equation $B=0$. To be consistent with the usual result, we take \[A=\frac{8\pi}{3M_{Pl}^2},\] where $M_{Pl}^2\equiv1/G$.
problem id: 150_cardas1
Show that once the new term dominates the right hand side of the Friedmann equation, we have accelerated expansion.
When the new term is so large that the ordinary first term can be neglected, we find \[a\propto t^{\frac2{3n}}.\] Indeed, assuming that the Universe is filled solely by the non-relativistic matter, one finds \[H\propto\rho^{n/2}\propto a^{-3n/2},\quad a\propto a^{-3n/2+1},\quad a^{-3n/2-1}da\propto dt\Rightarrow a\propto t^{2/(3n)},\] so the expansion is superluminal (accelerated) for $n<2/3$. For example, for $n=2/3$ we have $a\propto t$; for $n=1/3$ we have $a\propto t^2$; and for $n=1/6$ we have $a\propto t^4$. The case of $n=2/3$ produces a term $H^2\propto a^{-2}$ in the FLRW equation; such a term looks similar to a curvature term but it is generated here by matter in a universe with a flat geometry. Note that for $n=1/3$ the acceleration is constant, for $n>1/3$ the acceleration is diminishing in time, while for $n<1/3$ the acceleration is increasing (the cosmic jerk).
problem id: 150_cardas2
Let us represent the Cardassian model in the form \[H^2\propto \rho+\rho_X,\quad\rho_X=\rho^n.\] Find the parameter $w_X$ of the EoS $p_X=w_X\rho_X$, assuming that the Universe is filled exclusively by the non-relativistic matter.
Use the result of the previous problem \[a\propto t^{2/(3n)}.\] As \[a\propto t^{\frac{2}{3(w+1)}},\] one finds that \[w=n-1\]in the considered case.
problem id: 150_cardas3
Show that the result obtained in the previous problem takes place for arbitrary one-component fluid with $w_X=const$.
The relation \[\rho_X(z)=\rho_{X0}\exp\left[3\int\limits_0^zdz'\frac{1+w_X(z')}{1+z'}\right]\] holds for arbitrary component with the EoS $p_X=w_X\rho_X$. Consequently \[\frac{d\rho_X}{dz}=3\rho_X\frac{1+w_X(z)}{1+z}.\] The same result can be obtained immediately from the conservation equation using the relation \[\frac d{dt}=\frac{dz}{dt}\frac d{dz}=-(1+z)H.\] Taking into account that $\rho_x=\rho^n$ and \[\rho=(1+z)^{3(w_X+1)}\] for $w_X=const$, one finds \[\rho_X=(1+z)^{3(1+w)n}.\] Substitution of the latter expression into the equation for $\rho_X$ gives \[w_X=n-1.\]
problem id: 150_cardas4
Generalize the previous problem for the case of two-component ideal liquid (non-relativistic matter $+$ radiation) with density $\rho=\rho_m+\rho_r$.
Use the result of the previous problem \[\frac{d\rho_X}{dz}=3\rho_X\frac{1+w_X}{1+z}.\] In the considered case \[\rho_X=\rho^n=\left(\rho_m+\rho_r\right)^n=\left(\rho_{m0}(1+z)^3+\rho_{r0}(1+z)^4\right)^n.\] Substitution of the latter expression into the equation for $\rho_X$ gives \[n\rho^{n-1}\left(\rho_{m0}(1+z)^2+\rho_{r0}(1+z)^3\right)=3\rho^n\frac{1+w_X}{1+z};\] \[w_X=\frac n{3\rho}\left(\rho_{m0}(1+z)^3+\rho_{r0}(1+z)^4\right)-1=\frac{3n\rho+n\rho_{r0}(1+z)^4-3\rho}{3\rho},\] and one finally obtains \[w_X=n-1+\frac n{3\rho}\rho_{r0}(1+z)^4,\] which is very slowly varying function with the redshift. If $\rho_{r0}\ll\rho_{m0}$, at late times, parameter $w_X$ is almost constant and it is identical to a dark energy component with a constant equation of state. But in early times, as one can not ignore the radiation component, one has to take the general EoS parameter $w_X$ which is not constant.
problem id: 150_cardas5
Show that we can interpret the Cardassian empirical term in the modified Friedmann equation as the superposition of a quintessential fluid with $w=n-1$ and a background of dust.
Equivalence of the equations \[H^2=A\rho_m+B(\rho_m)^n\] and \[H^2=A\rho_{m0}a^{-3}+Ba^{-3n},\] taking into account that $\rho_i\propto a^{-3(1+w_i)}$ allows to interpret the Cardassian empirical term in the modified Friedmann equation as the superposition of a quintessential fluid with $w=n-1$ and a background of dust with $w=0$.
problem id: 150_cardas6
We have two parameters in the original Cardassian model: $B$ and $n$. Make the transition $\{B,n\}\to\{z_{eq},n\}$, where $z_{eq}$ is the redshift value at which the second term $B\rho^n$ starts to dominate.
The second term starts to dominate at the redshift $z_{eq}$ when $A\rho(z_{eq})=B\rho^n(z_{eq})$, i.e., when \begin{equation}\label{150_cardas6_e1} \frac B A=\rho_0^{1-n}(1+z_{eq})^{3(1-n)}. \end{equation} In the Cardassian model today, we have \begin{equation}\label{150_cardas6_e2} H_0^2=A\rho_0+B\rho_0^n, \end{equation} so that \begin{equation}\label{150_cardas6_e3} A=\frac{H_0^2}{\rho_0}-B\rho_0^{n-1}. \end{equation} From Eqs.(\ref{150_cardas6_e1}) and (\ref{150_cardas6_e3}), we have \[B=\frac{H_0^2(1+z_{eq})^{3(1-n)}}{\rho_0^n\left[1+(1+z_{eq})^{3(1-n)}\right]}.\]
problem id: 150_cardas7
What is the current energy density of the Universe in the Cardassian model? Show that the corresponding energy density is much smaller than in the standard Friedmann cosmology, so that only matter can be sufficient to provide a flat geometry.
From \[H^2=A\rho+B\rho^n\] and (see the previous problem) \[\frac B A=\rho_0^{1-n}(1+z_{eq})^{3(1-n)}\] we have \[H^2=A\left[\rho+\rho_0^{1-n}(1+z_{eq})^{3(1-n)}\rho^n\right].\] Evaluating this equation today with $A=8\pi/3M_{Pl}^2$, we obtain \[H_0^2=\frac{8\pi}{3M_{Pl}^2}\rho_0\left[1+(1+z_{eq})^{3(1-n)}\right].\] The energy density $\rho_0$ is, by definition, the critical density. Consequently, \[\rho_0=\rho_{cr}=\frac{3H_0^2M_{Pl}^2}{8\pi\left[1+(1+z_{eq})^{3(1-n)}\right]}.\] This result can be presented in the form \[\rho_{cr}=\rho_{crFLRW}\times F(z_{eq},n),\quad \rho_{crFLRW}=\frac{3H_0^2M_{Pl}^2}{8\pi},\quad F(z_{eq},n)\equiv\left[1+(1+z_{eq})^{3(1-n)}\right]^{-1}.\] For example, if $z_{eq}=1$, then $F=\{1/3,1/5,3/20\}$ for $n=\{1/3,2/3,1/6\}$. We see that the value of the critical density can be much lower than in the standard Friedmann cosmology.
problem id: 150_cardas8
Let us represent the basic relation of Cardassian model in the following way \[H^2=A\rho\left[1+\left(\frac\rho{\rho_{car}}\right)^{n-1}\right],\] where $\rho_{car}=\rho(z_{eq})$ is the energy density at which the two terms are equal. Find the function $\rho(z_{eq})$ under assumption that the Universe is filled with non-relativistic matter and radiation.
\[\rho(z_{eq})=\rho_m(z_{eq})+\rho_r(z_{eq})=\rho_{m0}(1+z_{eq})^3\left[1+\frac{\Omega_{r0}}{\Omega_{m0}}(1+z_{eq})\right].\]
problem id: 150_cardas9
Let Friedmann equation is modified to be \[H^2=\frac{8\pi G}{3}g(\rho),\] where $\rho$ consists only of non-relativistic matter. Find the effective total pressure.
In the case of adiabatic expansion one has \begin{align} \nonumber dE+pdV&=0,\\ \nonumber d(g(\rho)a^3)+p_{tot}d(a^3)&=0,\\ \nonumber a^3dg+3a^2gda+3p_{tot}a^2da&=0,\\ \nonumber \dot g+3Hg&=-3p_{tot}H,\\ \nonumber \frac{dg}{dt}&=\frac{dg}{d\rho}\frac{d\rho}{dt},\\ \nonumber \frac{d\rho}{dt}&=-3H\rho,\\ \nonumber -\frac{dg}{d\rho}3H\rho+3Hg&=-3p_{tot}H,\\ \nonumber p_{tot}&=\rho\frac{dg}{d\rho}-g. \end{align} The same relation is commonly used also in presence of the radiation. This is incorrect, as the relation \[\frac{d\rho}{dt}=-3H\rho\] does not hold with radiation.
problem id: 150_cardas10
Find the speed of sound in the Cardassian model. [P.Gandolo, K. Freese, Fluid Interpretation of Cardassian Expansion, 0209322 ]
The Cardassian model \[H^2=\frac{8\pi}{M_{Pl}^2}\rho_m+B\rho_m^n\] can equivalently be written as \[H^2=\frac{8\pi}{M_{Pl}^2}\rho_m\left[1+\left(\frac{\rho_m}{\rho_{card}}\right)^{n-1}\right]\equiv\frac{8\pi}{M_{Pl}^2}\rho.\] In the previous problem we have shown that \[p_{tot}=\rho\frac{dg}{d\rho}-\rho.\] It allows to represent the speed of sound as \[c^2=\frac{\partial p_{tot}}{\partial\rho}=\frac{\partial p_{tot}/\partial\rho_m}{\partial\rho/\partial\rho_m}=\rho_m\frac{\left(\partial^2\rho/\partial\rho_m^2\right)}{\partial\rho/\partial\rho_m}.\] All the derivatives are calculated at constant entropy. Finally for the sound speed in the model we obtain \[c^2=-\frac{n(1-n)}{n+\left(\frac{\rho_{card}}{\rho_m}\right)^{1-n}}.\] This value is not guaranteed to be positive. So this model should be considered as an effective description at scales where $c^2>0$.
problem id: 150_cardas11
Find the deceleration parameter for the canonic Cardassian model.
\begin{align} \nonumber q(z)&=\frac12\frac{(1+z)}{E^2(z)}\frac{dE^2(z)}{dz}-1,\\ \nonumber E^2&\equiv\frac{H^2}{H_0^2}=\Omega_{m0}(1+z)^3+(1-\Omega_{m0})(1+z)^{3n},\\ \nonumber q(z)&=\frac12-\frac32(1-n)\frac{\kappa(z)}{1+\kappa(z)},\\ \nonumber \kappa(z)&\equiv\left(\frac{1-\Omega_{m0}}{\Omega_{m0}}\right)(1+z)^{-3(1-n)}. \end{align}
Models with Cosmic Viscosity
A Universe filled with a perfect fluid represents quite a simple which seems to be in good agreement with cosmological observations. But, on a more physical and realistic basis we can replace the energy-momentum tensor for the simplest perfect fluid by introducing cosmic viscosity. The energy momentum tensor with bulk viscosity is given by \[T_{\mu\nu}=(\rho=p-\xi\theta)u_\mu u_\nu+(p-\xi\theta)g_{\mu\nu},\] where $\xi$ is bulk viscosity, and $\theta\equiv3H$ is the expansion scalar. This modifies the equation of state of the cosmic fluid. The Friedmann equations with inclusion of the bulk viscosity, i.e. using the energy-momentum tensor $T_{\mu\nu}$, read \begin{align} \nonumber \frac{\dot a^2}{a^2}&=\frac13\rho,\quad \rho=\rho_m+\rho_\Lambda,\quad 8\pi G=1;\\ \nonumber \frac{\ddot a^2}{a}&=-\frac16(\rho+3p-9\xi H). \end{align} Problems #150_8-#150_14 are inspired by A. Avelino and U. Nucamendi, Can a matter-dominated model with constant bulk viscosity drive the accelerated expansion of the universe? arXiv:0811.3253
problem id: 150_8
Consider a cosmological model in a flat Universe where the only component is a pressureless fluid with constant bulk viscosity ($\xi=const$). The pressureless fluid represent both the baryon and dark matter components. Find the dependence $\rho_m(z)$ for the considered model.
The conservation equation in terms of the scale factor and the first Friedmann equation are \[a\frac{d\rho_m}{da}-3(3H\xi-\rho_m)=0,\] \[H^2=\frac{8\pi G}3\rho_m.\] Here $\rho_m$ is total density of the baryon and dark matter components. Having excluded the Hubble parameter and changed the independent variable from the scale factor $a$ to the redshift $z$, one finds \[(1+z)\frac{d\rho_m}{dz}-3\rho_m+\gamma\rho_m^{1/2}=0,\quad \gamma\equiv9\sqrt{8\pi G/3}\xi.\] The solution of this equation is: \[\rho_m(z)=\left[\frac\gamma3+\left(\rho_{m0}^{1/2}-\frac\gamma3\right)(1+z)^{3/2}\right]^2.\]
problem id: 150_9
Find $H(z)$ and $a(t)$ for the model of Universe considered in the previous problem.
Substitute the solution $\rho_m(z)$ obtained in the previous problem into the first Friedmann equation to obtain \[H(z)=H_0\left[\frac{\bar\gamma}3+\left(\Omega_{m0}^{1/2}-\frac{\bar\gamma}3\right)(1+z)^{3/2}\right],\quad \bar\xi\equiv\frac{24\pi G}{H_0}\xi.\] In the considered model the bulk viscous matter is the only component of the flat Universe. Consequently, $\Omega_{m0}=1$ and one finally obtains \[H(z)=\frac13H_0\left[\bar\gamma+\left(3-\bar\gamma\right)(1+z)^{3/2}\right].\] The obtained expression allows to write the scale factor in terms of the cosmic time. Let us transform the expression for the Hubble parameter \[H(a)=\frac13H_0\frac{\bar\xi a^{3/2}+3-\bar\xi}{a^{3/2}}\] to the following form \[H(t-t_0)=3\int\limits_1^a\frac{{a'}^{1/2}}{\bar\xi {a'}^{3/2}+3-\bar\xi}\ da'.\] For $\xi\ne0$ and $\bar\xi a^{3/2}+3-\bar\xi>0$ ($\bar\xi a^{3/2}+3-\bar\xi<0$ implies $H<0$ and contradicts the observations) one finds \[a(t)=\left[\frac{3\exp\left(\frac12\bar\xi H(t-t_0)-3+\bar\xi\right)}{\bar\xi}\right]^{2/3}.\]
problem id: 150_10
Analyze the expression for the scale factor $a(t)$ obtained in the previous problem for different types of the bulk viscosity.
(a) $0<\bar\xi<3$
When $t\to\infty$ then the obtained solution reproduces the de Sitter-like Universe,
\[a(t)\propto\exp\left[\frac{\bar\xi}3H(t-t_0)\right]\]
(b) $\bar\xi=3$
In this case the considered model exactly reproduces the de Sitter-like Universe,
\[a(t)=\exp\left[H_0(t-t_0)\right]\]
The model predicts an Universe in an eternal accelerated expansion.
(c) $\bar\xi>3$
In this case the Universe expands forever (decelerated expansion epoch is absent).
problem id: 150_11
Show that the Universe in the considered model with $\xi=const$ had the Big Bang in the past for all values of the bulk viscosity in the interval $0<\bar\xi<3$ and determine how far in the past (in terms of the cosmic time) it happened.
Using the result of Problem #150_9 \[a(t)=\left[\frac{3\exp\left(\frac12\bar\xi H(t-t_0)-3+\bar\xi\right)}{\bar\xi}\right]^{2/3},\] one finds the time of the Big Bang as $a(t_{BB})=0$: \[t_{BB}=t_0+\frac2{\bar\xi H_0}\ln\left(1-\frac{\bar\xi}3\right).\]
problem id: 150_12
Show that the result of the previous Problem for zero bulk viscosity ($\xi=0$) correctly reproduces the lifetime of the matter-dominated Universe.
For $\xi\to0$ \[H_0(t_0-t_{BB})=\frac23.\]
problem id: 150_13
As we have seen in the previous problems, in the interval $0<\bar\xi<3$ the Universe begins with a Big-Bang followed by an eternal expansion and this expansion begins with a decelerated epoch followed by an eternal accelerated one. The transition between the decelerated-accelerated expansion epochs depends on the value of $\bar\xi$. Find the value of the scale factor where the transition happens.
Using \[H(a)=\frac13H_0\frac{\bar\xi a^{3/2}+3-\bar\xi}{a^{3/2}},\] one obtains \[\frac{d\dot a}{da}=H+a\frac{dH}{da}=\frac13H_0\left(\bar\xi+\frac{\bar\xi-3}{2a^{3/2}}\right).\] In order to find the scale factor value $a_t$, corresponding to the transition from the decelerated expansion to the accelerated one, we use the fact that $d\dot a/da\equiv -Hq$. Thus the condition $q=0$ is equivalent to the requirement $d\dot a/da=0$, therefore \[a_t=\left(\frac{3-\bar\xi}{2\bar\xi}\right)^{2/3},\quad z_t=\left(\frac{2\bar\xi}{3-\bar\xi}\right)^{2/3}-1.\]
problem id: 150_14
Analyze the dependence \[a_t=\left(\frac{3-\bar\xi}{2\bar\xi}\right)^{2/3},\] obtained in the previous problem.
1. For $0<\bar\xi<1$ the transition between the decelerated epoch to the accelerated one takes place in the future $a_t>1$.
2. For $\bar\xi\to0$ the value $a_t\to\infty$ (the distant future).
3. For $\bar\xi=1$ the transition takes place today ($a_t=1$).
4. For $1<\bar\xi<3$ the transition takes place in the past of the Universe ($0<a_t<1$).
5. For $\bar\xi\to3$ the transition is close to the Big-Bang time ($a_t\to0$).
problem id: 150_15
Find the deceleration parameter $q(a,\bar\xi)$ for the cosmological model presented in the Problem #150_8.
\[q(a)=-\frac{\ddot a}{aH^2}.\] The term $\ddot a/a$ can be calculated from the second Friedmann equation, that for a matter-dominated universe with bulk viscosity reads: \[\frac{\ddot a}a=-\frac{4\pi G}{3}(\rho_m-9\xi H).\] From the definition of \[\bar\xi\equiv\frac{24\pi G}{H_0}\xi,\] using \[\rho_m=\frac3{8\pi G}H^2\] one obtains that \[\frac{\ddot a}a=\frac12(\bar\xi H_0-H)H.\] Using \[H(a)=\frac13H_0\frac{\bar\xi a^{3/2}+3-\bar\xi}{a^{3/2}},\] we find \[q(a,\bar\xi)=\frac12\left[\frac{3-\bar\xi(1+2a^{3/2})}{3-\bar\xi(1-a^{3/2})}\right].\]
problem id: 150_16
Analyze behavior of the deceleration parameter $q(a,\bar\xi)$ obtained in the previous problem for different values of the bulk viscosity $\bar\xi(\xi)$.
(a) For the case $\bar\xi=0$ we have $q=1/2$ that corresponds to a matter-dominated universe with null bulk viscosity.
(b) For the case $\bar\xi=3$ we have $q=-1$ that corresponds to the de Sitter Universe.
(c) For the case $0<\bar\xi<3$ it is always a decreasing function from $q(0)=1/2$ to $q(\infty)=-1$ with a transition from positive to negative values in the value of the scale factor \[a_t=\left[\frac{3-\bar\xi}{2\bar\xi}\right]^{3/2}.\]
(d) For the case $\bar\xi>3$ the Universe expands forever, and this expansion is always accelerated (there is no decelerating epoch or acceleration-deceleration transition).
When $a\to a_*\equiv(1-3/\bar\xi)^{2/3}$ $q\to-\infty$ and when $a\to\infty$ then $q\to-1$. It is a negative and increasing function but it never becomes positive, its maximum value is $-1$.
(e) The case $\bar\xi<3$ predicts an eternal decelerated expanding Universe. The universe begins with a Big-Bang and expands forever until to reach its maximum value $a_*=(1-3/\bar\xi)^{2/3}$ when $t\to\infty$. Deceleration parameter is a positive and increasing monotonic function where its minimum value is $1/2$.
problem id: 150_17
Use result of the problem \ref{150_15} to find the current value of the deceleration parameter and make sure that for $\bar\xi=1$ the transition from the decelerated to accelerated epochs of the Universe takes place today.
From the expression \[a_t=\left[\frac{3-\bar\xi}{2\bar\xi}\right]^{3/2}.\] we obtain the deceleration parameter evaluated today as \[q(a=1,\bar\xi)=\frac{1-\bar\xi}{2}.\] When $\bar\xi=1$ the transition from the decelerated to accelerated epochs of the Universe takes place today ($a=1$). For $\bar\xi<1$ we have a decelerated Universe in the present and for $\bar\xi>1$ we have an accelerated one today.
problem id: 150_18
Find the curvature scalar $R(a,\xi)$ for the cosmological model presented in the Problem #150_8.
For a flat Universe \[R=-6\left(\frac{\ddot a}a +H^2\right).\] Using \[\frac{\ddot a}a=\frac12H(\bar\xi H_0-H), \quad H(a)=\frac13H_0\frac{\bar\xi a^{3/2}+3-\bar\xi}{a^{3/2}},\] we find \[R=-3H(a)[\bar\xi+H(a)]\Rightarrow R(a,\bar\xi)=-\frac13H_0^2\left[\frac{(3-\bar\xi)^2}{a^3}+\frac{5\bar\xi(3-\bar\xi)}{a^{3/2}}+4\bar\xi^2\right].\]
problem id: 150_19
Let us consider a flat homogeneous and isotropic Universe filled by a fluid with bulk viscosity. We shall assume that the EoS for the fluid is $p=w\rho$, $w=const$ and that the viscosity coefficient $\xi(\rho)=\xi_0\rho^\nu$. Find the dependence $\rho(a)$ for the considered model.
The Friedmann equations for the considered model read \[H^2=\frac{8\pi G}{3}\rho;\] \[\dot\rho+3H[(1+w)\rho-3\xi_0\rho^\nu H]=0.\] Substitution of $H$ into the conservation equation gives \[\dot\rho+3H[(1+w)\rho-\xi^*_0\rho^{\nu+1/2}H]=0,\quad \xi_0^*\equiv\sqrt{24\pi G}\xi_0.\] Solution of the latter equation reads \[\rho(a)=\left[\frac{\xi_0^*}{1+w}+\frac C{1+w}a^r\right]\frac{1}{\frac12-\nu},\quad r\equiv 3(\nu-1/2)(1+w).\] Here $C$ is an integration constant.
problem id: 150_0013
Usually the inflationary models of the early Universe contain two distinct phases. During the first phase entropy of the Universe remains constant. The second phase is essentially non-adiabatic, particles are produced through the damping of the coherent oscillations of the inflaton field by coupling to other fields and by its subsequent decay. Find relation between the bulk viscosity and the entropy production [J. A. S. Lima, R. Portugal, I. Waga, Bulk viscosity and deflationary universes, arXiv:0708.3280].
Use the first law of thermodynamics \[d(\rho a^3)+pd(a^3)=TdS\] and the conservation equation in presence of viscosity \[\dot\rho+3H(\rho+p+\Pi)=0,\quad \Pi=-3H\xi,\] one finds \[\frac{T\dot S}{a^3}=\dot\rho+3H(\rho+p)=-3H\Pi.\] Consequently, \[\Pi=-\frac{T\dot S}{3Ha^3}.\] As $\dot S>0$ in an expanding Universe $(H>0)$ then $\Pi$ is negative. It agrees with the definition $\Pi=-3H\xi$, where $\xi$ is the positive bulk viscosity coefficient.
New from March 2015
New from December 2014
Exactly Integrable n-dimensional Universes
Exactly Integrable n-dimensional Universes
UNSORTED NEW Problems
The history of what happens in any chosen sample region is the same as the history of what happens everywhere. Therefore it seems very tempting to limit ourselves with the formulation of Cosmology for the single sample region. But any region is influenced by other regions near and far. If we are to pay undivided attention to a single region, ignoring all other regions, we must in some way allow for their influence. E. Harrison in his book Cosmology, Cambridge University Press, 1981 suggests a simple model to realize this idea. The model has acquired the name of "Cosmic box" and it consists in the following.
Imaginary partitions, comoving and perfectly reflecting, are used to divide the Universe into numerous separate cells. Each cell encloses a representative sample and is sufficiently large to contain galaxies and clusters of galaxies. Each cell is larger than the largest scale of irregularity in the Universe, and the contents of all cells are in identical states. A partitioned Universe behaves exactly as a Universe without partitions. We assume that the partitions have no mass and hence their insertion cannot alter the dynamical behavior of the Universe. The contents of all cells are in similar states, and in the same state as when there were no partitions. Light rays that normally come from very distant galaxies come instead from local galaxies of long ago and travel similar distances by multiple re?ections. What normally passes out of a region is reflected back and copies what normally enters a region.
Let us assume further that the comoving walls of the cosmic box move apart at a velocity given by the Hubble law. If the box is a cube with sides of length $L$, then opposite walls move apart at relative velocity $HL$.Let us assume that the size of the box $L$ is small compared to the Hubble radius $L_{H} $ , the walls have a recession velocity that is small compared to the velocity of light. Inside a relatively small cosmic box we use ordinary everyday physics and are thus able to determine easily the consequences of expansion. We can even use Newtonian mechanics to determine the expansion if we embed a spherical cosmic box in Euclidean space.
Problem 1
problem id:
As we have shown before (see Chapter 3): $p(t)\propto a(t)^{-1} $ , so all freely moving particles, including galaxies (when not bound in clusters), slowly lose their peculiar motion and ultimately become stationary in expanding space. Try to understand what happens by considering a moving particle inside an expanding cosmic box
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.
Problem 2
problem id:
Show that 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.
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.
Problem 3
problem id:
Let the cosmic box is filled with non-relativistic gas. Find out how the gas temperature varies in the expanding cosmic box.
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 \[T=T_{0} \left(1+z\right)^{2} \]
Problem 4
problem id:
Show that entropy of the cosmic box is conserved during its expansion.
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.
Problem 5
problem id:
Consider a (cosmic) box of volume V, having perfectly reflecting walls and containing radiation of mass density $\rho $. The mass of the radiation in the box is $M=\rho V$ . We now weigh the box and find that its mass, because of the enclosed radiation, has increased not by M but by an amount 2M. Why?
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.
Problem 6
problem id:
Show that the jerk parameter is \[j(t)=q+2q^{2} -\frac{\dot{q}}{H} \]
\[j=\frac{1}{H^{3} } \frac{\dddot{a}}{a} =\frac{1}{aH^{3} } \frac{d}{dt} \left(\frac{\ddot{a}}{aH^{2} } aH^{2} \right)=-\frac{1}{aH^{3} } \frac{d}{dt} \left(qaH^{2} \right)=q+2q^{2} -\frac{\dot{q}}{H} \]
Problem 7
problem id:
We consider FLRW spatially flat Universe with the general Friedmann equations \[\begin{array}{l} {H^{2} =\frac{1}{3} \rho +f(t),} \\ {\frac{\ddot{a}}{a} =-\frac{1}{6} \left(\rho +3p\right)+g(t)} \end{array}\] Obtain the general conservation equation.
Using \[\frac{\ddot{a}}{a} =\dot{H}+H^{2} \] we find \[\dot{\rho }+3H\left(\rho +p\right)=6H\left(-f(t)+\frac{\dot{f}(t)}{2H} +g(t)\right)\]
Problem 8
problem id:
Show that for extra driving terms in the form of the cosmological constant the general conservation equation (see previous problem) transforms in the standard conservation equation.
In this case$f(t)=g(t)=\Lambda /3,\; \; \dot{f}=0$ and \[\dot{\rho }+3H\left(\rho +p\right)=6H\left(-f(t)+\frac{\dot{f}(t)}{2H} +g(t)\right)\to \dot{\rho }+3H\left(\rho +p\right)=0\]
Problem 9
problem id:
Show that case $f(t)=g(t)=\Lambda /3$ corresponds to $\Lambda (t)CDM$ model.
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-14--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.