Difference between revisions of "New from June"
(Created page with "<div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem ''' <p style= "color: #999;font-size: 11px">problem id: 150_0</p> Comprehensive explanations o...") |
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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.] | 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.] | ||
<div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;"> | <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;"> | ||
− | + | # 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. | |
− | + | # The balloon is expanded by the pressure difference between the inside and the outside, but the Universe is not being expanded by pressure. | |
− | + | # 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. | |
− | + | # 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. | |
− | + | ||
− | + | ||
</p> </div></div></div> | </p> </div></div></div> | ||
Revision as of 23:11, 18 June 2015
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.]
- 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.
- The balloon is expanded by the pressure difference between the inside and the outside, but the Universe is not being expanded by pressure.
- 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.
- 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.
\paragraph{To chapter 3, section 15, if absent}
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 $t_0=H_0^{-1}$.
\paragraph{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 $T=\hbar c^3/8\pi kGM$.
\paragraph{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, $V/(dV/d\varphi)$.
paragraph{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} \end{enumerate}
paragraph{Cardassian Model} [K. Freese and M. Lewis, Cardassian Expansion: a Model in which the Universe is Flat, Matter Dominated, and Accelerating, arXiv: 0201229] 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} \end{enumerate}
\paragraph{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} {\it Problems \ref{150_8}-\ref{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} \begin{enumerate} <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem ''' <p style="color: #999;font-size: 11px">problem id: 150_8</p> Consider a cosmological model in a flat Universe where the only component is a pressureless fluid with constant bulk viscosity (UNIQ-MathJax149-QINU). The pressureless fluid represent both the baryon and dark matter components. Find the dependence UNIQ-MathJax150-QINU for the considered model. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;"> The conservation equation in terms of the scale factor and the first Friedmann equation are UNIQ-MathJax324-QINU UNIQ-MathJax325-QINU Here UNIQ-MathJax151-QINU is total density of the baryon and dark matter components. Having excluded the Hubble parameter and changed the independent variable from the scale factor UNIQ-MathJax152-QINU to the redshift UNIQ-MathJax153-QINU, one finds UNIQ-MathJax326-QINU The solution of this equation is: UNIQ-MathJax327-QINU </p> </div></div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem ''' <p style="color: #999;font-size: 11px">problem id: 150_9</p> Find UNIQ-MathJax154-QINU and UNIQ-MathJax155-QINU for the model of Universe considered in the previous problem. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;"> Substitute the solution UNIQ-MathJax156-QINU obtained in the previous problem into the first Friedmann equation to obtain UNIQ-MathJax328-QINU In the considered model the bulk viscous matter is the only component of the flat Universe. Consequently, UNIQ-MathJax157-QINU and one finally obtains UNIQ-MathJax329-QINU The obtained expression allows to write the scale factor in terms of the cosmic time. Let us transform the expression for the Hubble parameter UNIQ-MathJax330-QINU to the following form UNIQ-MathJax331-QINU For UNIQ-MathJax158-QINU and UNIQ-MathJax159-QINU (UNIQ-MathJax160-QINU implies UNIQ-MathJax161-QINU and contradicts the observations) one finds UNIQ-MathJax332-QINU </p> </div></div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem ''' <p style="color: #999;font-size: 11px">problem id: 150_10</p> Analyze the expression for the scale factor UNIQ-MathJax162-QINU obtained in the previous problem for different types of the bulk viscosity. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;"> \begin{enumerate} \item UNIQ-MathJax163-QINU When UNIQ-MathJax164-QINU then the obtained solution reproduces the de Sitter-like Universe, UNIQ-MathJax333-QINU \item UNIQ-MathJax165-QINU In this case the considered model exactly reproduces the de Sitter-like Universe, UNIQ-MathJax334-QINU The model predicts an Universe in an eternal accelerated expansion. \item UNIQ-MathJax166-QINU In this case the Universe expands forever (decelerated expansion epoch is absent). \end{enumerate}