Matter creation
← All additional problem collections
Observational cosmology problems
As we have seen above, negative pressure is the key ingredient required to achieve the accelerated expansion. Such pressure naturally appears when the system deviates from the thermodynamic equilibrium. In particular, as was first pointed out by Zel'dovich [Ya. B. Zel'dovich, JETP Lett. 12, 307 (1970)], the negative pressure is generated in process of particle creation due to the gravitational field. Physically, one may think that the (classical) time varying gravitational field works like a 'pump' supplying energy to the quantum fields. Construction of the models including the essentially quantum process of particle creation is problematic because of the difficulties connected with its inclusion into the classical field equations of Einstein. One can however avoid those difficulties on the phenomenological level [J.Lima, arXiv 0807.3379].
We shall now consider an open thermodynamical system where the number of fluid particles $N$ is not preserved. So the particle conservation equations \[\dot N_{;\mu}=0\Rightarrow \dot n + \theta n=0\] is now modified as \[\dot n + \theta n=n\Gamma,\] where $N_{\mu}=nu_\mu$ is the particle flow vector, $u_\mu$ is the particle four velocity, $\theta=u^\mu_{;\mu}$ is the fluid expansion, (for the FLRW Universe, $\theta=u^\mu_{;\mu}=3H$), $\dot n = n_{,\mu}u^{mu}$ and $\Gamma$ stands for the rate of change of the particle number in a comoving volume $V$, $\Gamma>0$ indicates particle creation while $\Gamma<0$ means particle annihilation. Any non-zero $\Gamma$ will effectively behave as a bulk viscous pressure of the thermodynamical fluid and nonequilibrium thermodynamics should come into the picture.
Problem 1
problem id: 150_dp0-problem-1
Express the first law of thermodynamics in terms of the specific quantities: the energy density $\rho$ and the particle number density $n$.
For a closed thermodynamical system the first law of thermodynamics states the conservation of internal energy $E$ as \[dE=dQ-pdV,\] where $p$ is the thermodynamic pressure, $V$ is any comoving volume and $dQ$ represents the heat received by the system in time $dt$. Now defining $\rho=E/V$ as the energy density, $n=N/V$ as the particle number density and $dq=dQ/N$ as the heat per unit particle, the above conservation law takes the form \[d\left(\frac\rho n\right)=dq-pd\left(\frac1n\right).\] Note that this equation (known as Gibb's equation) is also true when the thermodynamical system is not closed, i.e., $N$ is not constant $[N=N(t)]$.
Problem 2
problem id: 150_dp0-problem-2
In the context of particle creation, let us consider spatially flat FRLW model of the Universe as an open thermodynamical system. Obtain the Friedmann equations for the cosmic fluid having energy-momentum tensor (in this section we will use the metric $(-,+,+,+)$ \[T_{\mu\nu}=(\rho+p+\Pi)u_\mu u_\nu+(p+\Pi)g_{\mu\nu}.\] For cosmological models with particle creation the cosmic fluid may be considered as a perfect fluid and the dissipative term $\Pi$ as the effective bulk viscous pressure due to particle production. In other words, the cosmic substratum is not a conventional dissipative fluid, rather a perfect fluid with varying particle number.
Using the standard procedure for transition from the Einstein's field equations to the Friedmann equations (see Chapter 2), one obtains \begin{align} \nonumber \rho & = 3H^2;\\ \nonumber \dot H & = -\frac12(\rho+p+\Pi), \quad 8\pi G=1. \end{align} And the energy conservation relation $T^{\mu\nu}_{;\nu}$ gives \[\dot\rho+\theta(\rho+p+\Pi)=0.\]
Problem 3
problem id: 150_dp0-problem-3
Show that for the adiabatic process the effective bulk viscous pressure due to particle production is entirely determined by the particle production rate.
Represent the Gibb's relation \[dq=d\left(\frac\rho n\right)+pd\left(\frac1n\right)\] in the form \[Tds=d\left(\frac\rho n\right)+pd\left(\frac1n\right).\] Remember that here $s$ is the entropy per particle and $T$ is the temperature of the fluid. Using the conservation equations $\dot n + \theta n=n\Gamma$ and $\dot\rho+\theta(\rho+p+\Pi)=0$, we obtain \[nT\dot s=-\Pi\theta-\Gamma(\rho+p).\] For isentropic process, the entropy per particle remains constant, i.e. $\dot s=0$ and hence we have \[\Pi=-\frac\Gamma\theta(\rho+p).\] Thus, the effective bulk viscous pressure due to particle production is entirely determined by the particle production rate. So we may say that at least for the adiabatic process, a dissipative fluid is equivalent to a perfect fluid with varying particle number.
Problem 4
problem id: 150_dp0-problem-4
An equivalent way to see the derivation of the field equations (7)-(8) (???) is to consider the energy-momentum tensor in the Einstein field equations as a total energy momentum tensor \[T^{(eff)}_{\mu\nu}=T^{(\gamma)}_{\mu\nu}+T^{(c)}_{\mu\nu},\] where $T^{(\gamma)}_{\mu\nu}$ is the energy-momentum tensor for the fluid with the EoS $p = (\gamma- 1) \rho$, i.e. \[T^{(\gamma)}_{\mu\nu}=(\rho+p)u_\mu u_\nu+pg_{\mu\nu}.\] and \[T^{(c)}_{\mu\nu}=p_c\left(u_\mu u_\nu+g_{\mu\nu}\right)\] is the energy-momentum tensor which corresponds to the matter creation term. Show that energy-momentum tensor $T^{(eff)}_{\mu\nu}$ provides us with the matter creation pressure \[\Pi=-\frac\Gamma\theta(\rho+p),\] obtained in the previous problem.
For the considered case the Einstein field equations are \[G_{\mu\nu}=T^{(eff)}_{\mu\nu}=T^{(\gamma)}_{\mu\nu}+T^{(c)}_{\mu\nu}.\] The Bianchi identity gives \[g^{\nu\sigma}\left(T^{(\gamma)}_{\mu\nu}+T^{(c)}_{\mu\nu}\right)_{;\sigma}=0.\] Using the explicit form of tensors $T^{(\gamma)}_{\mu\nu}$ and $T^{(c)}_{\mu\nu}$, one finds that the latter equality is equivalent to the conservation equation \begin{equation}\label{2016_07_p04_eq1} \dot\rho+\theta(\rho+p+\Pi)=0 \end{equation} From the other hand, Gibbs equation \[Tds=d\left(\frac\rho n\right)+pd\left(\frac1n\right).\] for the adiabatic process, taking into account the relation $\dot n +\theta n=n\Gamma$, directly leads to the conservation equation \begin{equation}\label{2016_07_p04_eq2} \dot\rho+\theta\left(1-\frac\Gamma\theta\right)(\rho+p)=0. \end{equation} Comparing the equations (\ref{2016_07_p04_eq1}) and (\ref{2016_07_p04_eq1}), one can reproduce the result, obtained in the previous problem \[\Pi=-\frac\Gamma\theta(\rho+p).\]
Problem 5
problem id: 150_dp0-problem-5
Find the effective parameter $w_{eff}$ of the EoS for the cosmological model with particle creation.
\begin{align} \nonumber w_{eff} &=\frac{p_{tot}}{\rho_{tot}}=\frac{\rho+\Pi}\rho,\\ \nonumber \Pi &=-\frac\Gamma\theta(p+\rho)=-\gamma\rho\frac\Gamma\theta,\quad w+1\equiv\gamma,\\ \nonumber w_{eff} &=-1+\gamma\left(1-\frac\Gamma\theta\right). \end{align} We see that $w_{eff}$ represents quintessence for $\Gamma<\theta$ ($\Gamma<3H$), phantom energy for $\Gamma>\theta$ ($\Gamma>3H$), and $\Gamma=\theta$ ($\Gamma=3H$) represents cosmological constant.
Problem 6
problem id: 150_dp0-problem-6
Find the deceleration parameter $q$ for the cosmological model with particle creation.
The deceleration parameter $q$ is a measure of state of acceleration/deceleration of the Universe, defined as \begin{align} \nonumber q &=-1-\frac{\dot H}{H^2},\\ \nonumber \dot H &=-\frac32\gamma H^2\left(1-\frac\Gamma\theta\right). \end{align} Consequently, \[q=-1+\frac32\gamma\left(1-\frac\Gamma\theta\right)=-1+\frac32\gamma\left(1-\frac\Gamma{3H}\right).\]
Problem 7
problem id: 150_dp0-problem-7
Obtain the equation for the scale factor in cosmological model with particle creation.
Combining the modified Friedmann equations with the EoS $p=w\rho$, one obtains the equation for the scale factor \[\frac{\ddot a}a +\frac{H^2}2\left[1+3w-\frac{1+w}H\Gamma\right]=0.\] Setting $\Gamma=0$, we reproduce the second Friedmann equation \[\frac{\ddot a}a=-\frac\rho6(1+3w).\]
Problem 8
problem id: 150_dp0-problem-8
Calculate the derivative $dS/dt$ and find $S(t)$ for the Creation Matter Model (CMM).
In an open thermodynamical system, the entropy change $dS$ can be decomposed into an entropy flow $d_fS$ and the entropy creation matter $d_cS$ \[dS = d_fS+d_cS,\] with $d_cS>0$. As in a homogeneous system $d_fS=0$, but there is entropy production due to matter creation, so we have \begin{align} \nonumber \frac d{dt}\left(nsV\right)&=\dot nsV+ns\frac{dV}{dt},\quad \dot s=0,\\ \nonumber ns\frac{dV}{dt}&=ns\frac{da^3}{dt}=ns3HV=nsV\theta,\\ \nonumber \frac{dS}{dt} &=\frac d{dt}\left(nsV\right)=\dot nsV+nsV\theta=sV(\dot n +n\theta)=sVn\Gamma=S\Gamma. \end{align} Here $s$ is the entropy per particle. After integration we obtain \[S(t)=S_0\exp\left[3\int\limits_{a_0}^a\frac\Gamma\theta\frac{da}a\right]\] with $S_0=S(t_0)$ and $a_0=a(t_0)$.
Problem 9
problem id: 150_dp0-problem-9
Calculate the scale factor dependence of the Universe's temperature for the Creation Matter Model (CMM).
For a substance with zero chemical potential \[nsT=\rho+p.\] For isentropic process, the entropy per particle remains constant, i.e. $\dot s=0$ and consequently \[\dot nsT+ns\dot T=\dot\rho(1+w)=\dot\rho\gamma.\] Using \[\dot\rho=-\theta(\rho+p+\Pi),\quad \Pi=-\frac\Gamma\theta(\rho+p),\] we obtain \[\frac{\dot T}T=(1-\gamma)\theta\left(1-\frac\Gamma\theta\right).\] Using \[\frac{dT}{dt}=\frac{dT}{da}Ha\] and $\theta=3H$, we find \[T=T_0a^{-3(\gamma-1)}\exp\left[3(\gamma-1)\int\limits_{a_0}^a\frac\Gamma\theta\frac{da}a\right],\] or using result of the previous problem \[T=T_1\left(\frac S{a^3}\right)^{\gamma-1},\quad T_1\equiv\frac{T_0}{S_0^{\gamma-1}}.\]
Problem 10
problem id: 150_dp0-problem-10
Solve the previous problem using the thermodynamic identity \[T\left(\frac{\partial p}{\partial T}\right)_n+n\left(\frac{\partial\rho}{\partial n}_T\right)=\rho+p.\]
Combining the thermodynamic identity with the conservation equation, we obtain \[\frac{\dot T}T=\left(\frac{\partial p}{\partial\rho}\right)_n\frac{\dot n}n.\] In particular, for $\Gamma=0$, $\dot n +3nH=0$ and radiation or ultra relativistic particles ($p=\rho/3$) the temperature evolution law becomes \[\frac{\dot T}T=-\frac{\dot a}a\Rightarrow aT=const.\] If gravitational particles source is present, then $\dot n +3nH=n\Gamma$ and \[\frac{\dot T}T=\left(\frac{\partial p}{\partial\rho}\right)_n\frac{\dot n}n=\left(\frac{\partial p}{\partial\rho}\right)_n\left(\Gamma-3H\right).\]
Problem 11
problem id: 150_dp0-problem-11
Solve the previous problem for cosmological substance with $\partial p/\partial\rho=w=const$.
Using \[\dot H+\frac32(1+w)H^2\left(1-\frac\Gamma{3H}\right)=0,\] we find \[\Gamma-3H=\frac2{1+w}\frac{\dot H}H\] and \[\frac{\dot T}T=\frac{2w}{1+w}\frac{\dot H}H.\] Consequently, the temperature of the fluid is governed by the following law \[T=T_0\left(\frac H{H_0}\right)^{\frac{2w}{1+w}}.\]
Problem 12
problem id: 150_dp0-problem-12
Show that the matter creation process can be effectively described by the dimensionless ratio \[\beta=\frac\Gamma\theta=\frac\Gamma{3H}.\]
If $\Gamma\ll3H$, then $\beta\ll1$, and the matter creation process is qualitatively small. It leads to $n\propto R^{-3}$ and $H=2/(3t)$, as it should be for the Einstein-de Sitter model. The opposite case $\Gamma\gg3H$ corresponds to the extreme physical situation when the matter creation process is so intense that it compensates with excess the matter density decrease due to the expansion. behavior of such type could probably be realized only in very early Universe, as it took place, for example, during the inflation stage, which corresponded to the reheating. The intermediate and physically more realistic situation appears if this ratio is less or of order of unity: $\Gamma\le3H$. In particular, if $\Gamma=3H$, then density decrease due to the expansion is exactly compensated by its increase due to the matter creation, so the particle number density remains constant.
Problem 13
problem id: 150_dp0-problem-13
How is the evolution of $a(t)$ affected by rate of change of the particle number $\Gamma$?
In the considered model, the Hubble parameter for the Universe, filled with the cosmic fluid with EoS $p=w\rho$, satisfies the dynamical equation \[\dot H+\frac32(1+w)H^2\left(1-\frac\Gamma{3H}\right)=0.\] The de Sitter solution $\dot H=0$ for $\Gamma=3H=constant$ is possible regardless of the EoS defining the cosmic fluid. Since the Universe is evolving, such a solution is unstable, and, as long as $\Gamma\ll3H$, conventional solutions without particle production are recovered. From the above equation, one may conclude that the main effect of $\Gamma$ is to provoke a dynamic instability in the space-time thereby allowing a transition from a de Sitter regime ($\Gamma\approx3H$) to a conventional solution, and vice versa.
Problem 14
problem id: 150_dp0-problem-14
Let us assume a radiation dominated Universe ($w=1/3$, $\Gamma=\Gamma_r$). The dynamics is determined by the ratio $\Gamma_r/(3H)$ (see previous problem). The particle production must be strongly suppressed, $\Gamma_r/(3H)\ll1$, when the Universe enters the radiation phase. The simplest formula satisfying such a criterion is linear, namely: $\Gamma_r/(3H)=H/H_I$, where $H_I$ is the inflationary expansion rate associated with the initial de Sitter expansion ($H\le H_I$). Find the Hubble parameter dependence on the scale factor $H(a)$ for the radiation phase.
For $w=1/3$, $\Gamma=\Gamma_r$, $\Gamma_r/(3H)=H/H_I$, \[\dot H+\frac32(1+w)H^2\left(1-\frac\Gamma{3H}\right)=0\Rightarrow\dot H+2H^2\left(1-\frac H{H_I}\right)=0,\] or \[a\frac{dH}{da}+2H^2\left(1-\frac H{H_I}\right)=0\Rightarrow\frac{dH}{H-H_I}-\frac{dH}H=2\frac{da}a.\] The solution of this equation is \[H(a)=\frac{H_I}{1+Da^2}.\] Here $D\ge0$ is an integration constant.
Problem 15
problem id: 150_dp0-problem-15
Using the result obtained in the previous problem, show that in the considering model the Universe evolves continuously from de Sitter stage to the standard radiation phase.
Integrating the equation \[H(a)=\frac{H_I}{1+Da^2},\] we obtain \[H_It=\ln\left(\frac a{a_*}\right)+\frac12Da_*^2\left(\frac a{a_*}\right)^2,\] where the integration constant$a_*$ defines the transition from the de Sitter stage to the beginning of the standard radiation epoch. At early times ($a\ll a_*$), when the logarithmic term dominates, one finds $a\approx a_*\exp(H_It)$ (de Sitter stage), while at late times, with $a\gg a_*$ and $H\ll H_I$ the obtained result reduces to \[a\approx\left(\frac2D\right)^{1/2}(H_It)^{1/2}\] and the standard radiation phase is reached.
Problem 16
problem id: 150_dp0-problem-16
Find the radiation energy density $\rho_r$ as function of scale factor for the rate of change of the particle number $\Gamma=\Gamma_r$, $\Gamma_r/(3H)=H/H_I$ ($H_I$ is the inflationary expansion rate).
The radiation energy density is $\rho_r(a)=3H^2(a)$ ($8\pi G=1$) and using \[H(a)=\frac{H_I}{1+Da^2},\] we obtain \[\rho_r(a)=\frac{3H_I^2}{(1+Da^2)^2}\] or \[\rho_r(a)=\frac{\rho_I}{\left[1+Da_*^2\left(\frac a{a_*}\right)^2\right]^2}\] Here $\rho_I=3H_I^2$ is the greatest value of the energy density, characterizing the initial de Sitter stage and the integration constant $a_*$ defines the transition from the de Sitter stage to the beginning of the standard radiation epoch. We see again that the conventional radiation phase, $\rho_r\propto a^{-4}$, is attained when $a\gg a_*$.
Problem 17
problem id: 150_dp0-problem-17
For cosmological model considered in the problems \ref{2016_07_p14}-\ref{2016_07_p16} find how did the cosmic temperature evolve at the very early stages?
For adiabatic production of relativistic particles the energy density scales as $\rho_r\propto T^4$ and the equation obtained in previous problem \[\rho_r(a)=\frac{\rho_I}{\left[1+Da_*^2\left(\frac a{a_*}\right)^2\right]^2}\] implies that \[T=\frac{T_I}{\sqrt{1+Da_*^2\left(\frac a{a_*}\right)^2}}\] where $T_I$ is the temperature of the initial de Sitter phase which must be uniquely determined by the scale $H_I$ - the inflationary expansion rate associated to the initial de Sitter phase. We see that the expansion proceeds isothermally: $T=const$ during the de Sitter phase ($a\ll a_*$) . After the de Sitter stage, the temperature decreases continuously in the course of the expansion. For $a\gg a_*$ and $H\ll H_I$, we obtain $T\propto a^{-1}$. Accordingly, the comoving number of photons becomes constant since $n\propto a^{-3}$, as expected for the standard radiation stage.
Problem 18
problem id: 150_dp0-problem-18
Solve Problem \ref{2016_07_p14} for a substance with EoS $p=w\rho=(\gamma-1)\rho$ and $\Gamma=\Gamma_0H^2$ with $\Gamma_0=3\beta/H_e$.
For this choice of $\Gamma$, $H$ can be found from \[\dot H+\frac32(1+w)H^2\left(1-\frac\Gamma{3H}\right)=0,\] as \[H(a)=\frac{H_e}{\beta-(1-\beta)\left(\frac a{a_e}\right)^{\frac{3\gamma}2}},\] where $H_e$ and $a_e$ are chosen to be the values of the Hubble parameter and the scale factor respectively at some instant. If we identify $a_e$ with some intermediate value of the scale factor $a$ corresponding to $\ddot a=0$, i.e. to the transition from de Sitter accelerating phase to the standard decelerating matter phase, then we have \[\dot H+H^2=0\Rightarrow\dot H_e=-H^2_e,\] and using \[\frac\Gamma{3H}=1+\frac2{3\gamma}\frac{\dot H}{H^2},\] we find \[\beta=1-\frac2{3\gamma}.\]
Problem 19
problem id: 150_dp0-problem-19
Find the deceleration parameter for the model considered in previous problem.
Using \[q=-1+\frac32\gamma\left(1-\frac\Gamma{3H}\right)\] for \[H(a)=\frac{H_e}{\beta-(1-\beta)\left(\frac a{a_e}\right)^{\frac{3\gamma}2}}\] we obtain \[q(z)=-1+\frac{3\gamma}2\left[1-\frac{\beta}{\beta+(1-\beta)(1+z)^{-3\gamma/2}}\right],\quad \frac{a_e}a=1+z.\]
Problem 20
problem id: 150_dp0-problem-20
Let us now consider how the CMM (Creation matter model) works at later stages of the Universe's evolution in the transition from Einstein-de Sitter to a late time de Sitter stage. Due to the conservation of baryon number one have to take into account only the production rate of cold dark matter particles $\Gamma_{dm}$. What is the exact form of $\Gamma_{dm}$?
The evolution equation for the Hubble parameter in the flat $\Lambda$CDM reads \begin{equation}\label{2016_07_p20_e1} \dot H+\frac32H^2\left[1-\left(\frac{H_f}H\right)^2\right]=0, \end{equation} where $H_f$ is Hubble parameter of the final de Sitter stage $H\ge H_f$. All available observations are in accordance with the $\Lambda$CDM evolution both at the background and perturbative levels. Thus the form of $\Gamma_{dm}$ can be found from the requirement that the evolution equation for the model with the production of cold dark matter particles ($w=0$, $\Gamma=\Gamma_{dm}$) \begin{equation}\label{2016_07_p20_e2} \dot H+\frac32H^2\left(1-\frac{\Gamma_{dm}}{3H}\right)=0, \end{equation} leads to the same background evolution. Comparing equations (\ref{2016_07_p20_e1}) and (\ref{2016_07_p20_e2}) we find that for the fulfillment of this condition it is necessary that \[\frac{\Gamma_{dm}}{3H}=\left(\frac{H_f}H\right)^2,\] or $\Gamma_{dm}\propto H^{-1}$. The limiting value of the creation rate, $\Gamma_{dm}=3H_f$, leads to a late time de Sitter phase: $\dot H=0$, $H=H_f$ thereby showing that the de Sitter solution now becomes an attractor at late times.
Problem 21
problem id: 150_dp0-problem-21
Let us consider the dynamical system in the context of bulk viscosity. In that case, we need to replace the pressure as $p\to p-3\zeta H$, where $\zeta$ is the coefficient of bulk viscosity. At what choice of the particle production rate $\Gamma$ the creation matter model would be equivalent to the bulk viscosity model?
In presence of the bulk viscosity the Friedmann equations are modified as \begin{align} \nonumber \rho &=3H^2, \nonumber \dot H &=-\frac12(\rho+p)+\frac32\zeta H. \end{align} Using the barotropic EoS: $p=(\gamma-1)\rho$, the second Friedmann equation can be written as \[\dot H=-\frac32\gamma H^2\left(1-\frac\zeta{\gamma H}\right),\] which coincides with \[\dot H=-\frac32\gamma H^2\left(1-\frac\Gamma{3H}\right),\] if one takes \[\zeta = \frac\Gamma3\gamma.\]
Dynamical analysis of the Creation Matter Model
Problem 22
problem id: 150_dp0-problem-22
Since the equation \[\dot H=-\frac32\gamma H^2\left(1-\frac\Gamma{3H}\right)\] is a one-dimensional first order differential equation, hence, the dynamics is obtained from the study of its critical points (or, fixed points) in which $\dot H=0$. Show that the fixed points $H_*$ in which $H_*\ne0$ represent the de Sitter Universe.
If $H=H_*$ is the fixed point, then $H_*=0$ or $\Gamma(H_*)=3H_*$. The fixed points $H_*\ne0$ correspond to $H=const$ and these solutions describe the de Sitter Universe.
Problem 23
problem id: 150_dp0-problem-23
Find the fixed points and determine their nature for the case with the particle creation rate $\Gamma=n/H$, $n>0$.
Let us consider the general form of the equation $\dot h=f(H)$. If \[\frac{df(H_*)}{dH}<0\] at the fixed point $H_*$, then the fixed point is asymptotically stable (attractor), and on the other hand, if we have \[\frac{df(H_*)}{dH}<0,\] then the fixed point has unstable nature (repeller). The repeller point is suitable for early Universe, since it can describe the inflationary epoch, whereas the attractor point can describe the stable accelerating phase for the later time. To find the fixed points for the considered case with the particle creation rate $\Gamma=n/H$, $n>0$, one should solve the equation \[-\frac32\gamma H^2\left(1-\frac n{3H^2}\right)=0.\] The non-zero solutions are $H_*=\pm\sqrt{n/3}$. For the above given form for $\Gamma$, one has \[f(H)=-\frac32\gamma\left(H^2-\frac n3\right)\] and thus \[\frac{df(\pm\sqrt{n/3})}{dH}=\mp\sqrt{3n},\] which means that $H_*=\sqrt{n/3}$ is an attractor and $H_*=-\sqrt{n/3}$ is a repeller.
Problem 24
problem id: 150_dp0-problem-24
Find the fixed points for the case with the particle creation rate $\Gamma=-\Gamma_0+mH+n/H$, $n>0$.
If $\Gamma(H)$ is a polynomial function of $H$, then the fixed point condition $\Gamma(H_*)=3H_*$ (for $H_*\ne0$), is a polynomial equation which has as many solutions (not necessary real solutions) as the highest power of the polynomial equation $\Gamma(H_*)=3H_*$. Hence, with the given particle creation rate we have to consider the following second-order polynomial equation \[(m-3)H_*^2-\Gamma_0H_*+n=0.\] In order to have the two critical points, as many as the inflationary phases of the Universe, we are interested in the case when $m\ne3$, and \[n\ne\frac{\Gamma_0^2}{4(m-3)}.\] Solving this equation, we find the critical points \[H_\pm=\frac{\Gamma_0}{2(m-3)}\left(1\pm\sqrt{1+\frac{4n(3-m)}{\Gamma_0^2}}\right).\] The case $m=3$ is special in a sense that there is only one critical point $H_-=n/\Gamma_0$, which is always an attractor for $\Gamma_0>0$, and a repeller for $\Gamma_0<0$.
Problem 25
problem id: 150_dp0-problem-25
Perform dynamic analysis of the solutions, obtained in previous problem.
To perform the dynamical analysis, we have to divide the parameter space ($m,n\Gamma_0$) to different regions by the following conditions: \[\left(\Gamma_0>0,\ \Gamma_0<0;\quad m>3,\ m<3;\quad n>0,\ n<0;\quad\frac{4(3-m)}{\Gamma_0^2}>-1,\ \frac{4(3-m)}{\Gamma_0^2}<-1.\right)\] To have a non-singular Universe (without the Big Bang singularity) with the accelerated phases both at early and late times, one possibility is to have two critical points $H_+>H_->0$, where $H_+$ is a repeller and $H_-$ must be an attractor. If so, in principle, when the Universe leaves the point $H_+$, realizing the inflationary phase, and when it comes asymptotically to $H_-$, it enters into the current accelerated phase. Of course, the viability of the background has to be checked dealing with cosmological perturbations and comparing the theoretical predictions with the observations. For the considered model, the above described scenario can only happen in the region of the parameter space given by \[(m,n,\Gamma_0)=\left\{\Gamma_0>0,\ m>3,\ n\ge0,\ \frac{4(3-m)}{\Gamma_0^2}>-1\right\}.\]
Problem 26
problem id: 150_dp0-problem-26
Prove that the dynamics generated by the creation matter model is equivalent to that driven by the single scalar field.
We use the energy density $\rho_\varphi$ and pressure $p_\varphi$ of the scalar field $\varphi$ given by \begin{align} \nonumber \rho_\varphi &=\frac12\dot\varphi^2+V(\varphi),\\ \nonumber p_\varphi &=\frac12\dot\varphi^2-V(\varphi). \end{align} Recall that the creation matter model is described by Friedmann equations \begin{align} \nonumber 3H^2&=\rho,\\ \nonumber \dot H &=-\frac32\gamma H^2\left(1-\frac\Gamma{3H}\right). \end{align} To show the equivalence, we perform the replacement \[\rho\to\rho_\varphi;\quad p-\frac\Gamma{3H}\gamma\rho\to p_\varphi.\] The Friedmann equations will become \begin{align} \label{2016_07_p26_e1} 3H^2&=\rho_\varphi,\\ \nonumber 2\dot H &=-\dot\varphi^2. \end{align} Note that Eqs. (\ref{2016_07_p26_e1}) uses the equations of General Relativity for a single scalar field. It means that we are dealing with the equivalence between an open system (the creation matter model) and the one driven by a single scalar field in the context of General Relativity. Further, the effective EoS parameter is given by \[w_{eff}=-1-\frac23\frac{\dot H}{H^2}=-1+\gamma\left(1-\frac\Gamma{3H}\right).\] Of course, \[w_{eff}=w_\varphi=\frac{\dot\varphi^2-2V}{\dot\varphi^2+2V}.\] Note that $2\dot H=-\dot\varphi^2<0$ implies that $w_{eff}>-1$, and thus one has $\Gamma<3H$ (the quintessence era).
Problem 27
problem id: 150_dp0-problem-27
Find the scalar field $\varphi(H)$ and the potential $V(\varphi)$, satisfying the condition that the dynamics generated by the creation matter model is equivalent to the dynamics driven by a single scalar field.
From the Friedmann equations (see the previous problem) \begin{align} \nonumber 3H^2&=\rho_\varphi,\\ \nonumber 2\dot H &=-\dot\varphi^2; \end{align} with \begin{align} \nonumber \rho_\varphi&=\frac12\dot\varphi^2+V(\varphi),\\ \nonumber \dot H &=-\frac32\gamma H^2\left(1-\frac\Gamma{3H}\right); \end{align} we obtain \[V(\varphi)=\frac32H^2\left(2-\gamma+\frac{\gamma\Gamma}{3H}\right),\] \[\dot\varphi=\sqrt{-2\dot H}=\sqrt{3\gamma H^2\left(1-\frac\Gamma{3H}\right)}.\] Performing the change of variables \[dt=\frac{dH}{\dot H},\] we find \[\varphi=\int\sqrt{-\frac2{\dot H}}dH=-\frac2{\sqrt{\gamma}}\int\frac{dH}{\sqrt{3H^2-\Gamma H}}.\]
Problem 28
problem id: 150_dp0-problem-28
Solve the previous problem for case \[\Gamma=-\Gamma_0+mH+n/H.\]
In the particular case $\Gamma=-\Gamma_0+mH+n/H$ one has \[\varphi=-\frac2{\sqrt{\gamma}}\int\frac{dH}{\sqrt{(3-m)H^2+\Gamma_0H-n}}.\] In the domain of interest (see Problem \ref{2016_07_p25}) \[(m,n,\Gamma_0)=\left\{\Gamma_0>0,\ m>3,\ n\ge0,\ \frac{4(3-m)}{\Gamma_0^2}>-1\right\},\] the integral could be solved analytically, giving as a result \[\varphi=\frac2{\sqrt{(m-3)\gamma}}\arcsin\left[\frac{m-3}\omega\left(\frac{\Gamma_0}{m-3}-2H\right)\right],\quad \omega\equiv\sqrt{\Gamma_0^2+4(3-m)n}.\] Inverting this expression, we find \[H=\frac1{2(m-3)}\left[\Gamma_0-\omega\sin\left(\frac{\sqrt{(m-3)\gamma}}2\varphi\right)\right].\] Using \[V(\varphi)=\frac32H^2\left(2-\gamma+\frac{\gamma\Gamma}{3H}\right)=\frac12\left[\left(6+(m-3)\gamma\right)H^2 -\gamma\Gamma_0H+\gamma n\right],\] we obtain \[V(\varphi)=\frac3{4(m-3)^2}\left[\Gamma_0-\omega\sin\left(\frac{\sqrt{(m-3)\gamma}}2\varphi\right)\right]^2 - \frac{\gamma\omega^2}{8(m-3)}\cos^2\left(\frac{\sqrt{(m-3)\gamma}}2\varphi\right).\]
Problem 29
problem id: 150_dp0-problem-29
Prove equivalence of the dynamics generated by the creation matter model with the dynamics driven by a decaying vacuum energy density model.
For a generic decaying vacuum model Friedmann equations reduce to \begin{align} \label{2016_07_p29_e1} \rho+\Lambda(t)&=3\left(H^2+\frac k{a^2}\right),\\ \nonumber p-\Lambda(t)&=-2\frac{\ddot a}a-\left(H^2+\frac k{a^2}\right), \end{align} where $\rho$ and $p$ are the energy density and the pressure of the dominant fluid component (dark matter) with EoS $p=w\rho$ ($p=(\gamma-1)\rho$). The decaying vacuum causes a change in the number of particles of dark matter, so the equation describing particle concentration has a source term, i.e., \[\dot n+3Hn=n\Gamma^*.\] Here, $\Gamma^*$ is the rate of change of the number of particles. By combining Friedmann equations (\ref{2016_07_p29_e1}), or more directly, from the total energy conservation law one finds \begin{equation}\label{2016_07_p29_e2} \dot\rho+3H(\rho+p)=-\dot\Lambda(t). \end{equation} Since in the considered model the vacuum decay is the unique source of particle creation, we can write \[\dot\Lambda(t)=-\zeta n\Gamma,\] where $\zeta$ is a positive phenomenological parameter. At last, we can combine Friedmann equations in order to obtain the dynamics of decaying vacuum model, \begin{equation}\label{2016_07_p29_e3} \frac{\ddot a}a + \Delta(H^2+k^2)-\frac{1+w}2\Lambda(t)=0,\quad \Delta\equiv\frac{3w+1}2. \end{equation} A similar equation for the creation matter model has the form \begin{equation}\label{2016_07_p29_e4} \frac{\ddot a}a + \Delta(H^2+k^2)+\frac12\Pi=0, \end{equation} where the term $\Pi$ is the effective bulk pressure due to particle production (see the previous Problems). Comparing (\ref{2016_07_p29_e3}) and (\ref{2016_07_p29_e4}) we find \[\Pi=-(1+w)\Lambda(t)=-\gamma\Lambda(t).\] We have seen above that \[\Pi=-\gamma\rho\frac\Gamma{3H}.\] Consequently, both the theories are equivalent, if we set \[\Gamma=3H\frac\Lambda\rho.\] Note that above identification holds regardless of the curvature of the Universe.
Problem 30
problem id: 150_dp0-problem-30
Show that for spatially flat Universe the inequalities $\Gamma\ll H$ and $\Lambda\ll H^2$ should be satisfied simultaneously.
For a spatially flat geometry, we have $\rho=3H^2$ and consequently \[\Gamma=3H\frac\Lambda\rho\Rightarrow\frac\Gamma H=\frac\Lambda{H^2}.\] In particular, if $\Gamma\ll H$, we find that $\Lambda\ll H^2$.
Problem 31
problem id: 150_dp0-problem-31
Let us consider the particular case $\Lambda(t)=const$. Show that for such choice the creation matter model corresponds to the standard LCDM.
The conservation equation for the creation matter model is \[\dot\rho+3H(\rho+p+\Pi)=0.\] For cold dark matter with $w=0$ ($\gamma=1$) and a spatially flat geometry $\Pi=-\Gamma H$ and the conservation equation transforms into \begin{equation}\label{2016_07_p31_e1} \dot\rho+3H(\rho-\Gamma H)=0. \end{equation} In the considered case \[\Gamma=3H\frac\Lambda\rho=\frac\Lambda H\] and (\ref{2016_07_p31_e1}) takes on the form \begin{equation}\label{2016_07_p31_e2} \dot\rho+3H(\rho-\Lambda)=0. \end{equation} Performing the integration, one finds \begin{equation}\label{2016_07_p31_e3} \rho(a)=\Lambda+\rho_{dm0}a^{-3}, \end{equation} where $\rho_{dm0}$ is an integration constant that must quantify the current amount of matter that is clustering to the present time. The equation (\ref{2016_07_p31_e3}) describes the dynamics of a creation matter model that behaves like the LCDM model.
Problem 32
problem id: 150_dp0-problem-32
Calculate the cosmographic parameter $j(H)$ for the cosmological model with particle creation.
\[j=\frac{\ddot H}{H^3}-3q-2;\] \[\dot H=-\frac32\gamma H^2\left(1-\frac\Gamma{3H}\right);\] \[q=-1+\frac32\gamma\left(1-\frac\Gamma{3H}\right).\] Let us introduce the function \[F\equiv\frac32\gamma\left(1-\frac\Gamma{3H}\right).\] In terms of this function \[\dot H=-H^2 F.\] As \[\frac d{dt}=\dot H\frac d{dH},\] we obtain \[\ddot H=\frac d{dt}\dot H=\dot H\frac{d\dot H}{dH}=-\dot H\frac d{dH}(H^2F)=H^3\left(2F+H\frac{dF}{dH}\right),\] \[\frac{dF}{dH}=-\frac12\frac\gamma H\left(\frac{d\Gamma}{dH}-\frac\Gamma H\right).\] Consequently, \[j=1+F\left(2F+H\frac{dF}{dH}\right)-3F,\] \[\frac{dF}{dH}=-\frac12\frac\gamma H\left(\frac{d\Gamma}{dH}-\frac\Gamma H\right),\] \[j=1+F\left[2F-3-\frac\gamma2\left(\frac{d\Gamma}{dH}-\frac\Gamma H\right)\right].\] The result can be checked by passing to the LCDM limit ($\Gamma\to0$, $\gamma=1$ or $\gamma=-1$ ???). In this case $F\to3/2$ or $F\to0$. In both cases $j=1$ in accordance with the LCDM.
Problem 33
problem id: 150_dp0-problem-33
Solve the previous problem using the cosmographical relation \[j=-\frac1H\frac{dq}{dt}+q(1+2q).\]
Using $q=-1+F$ and \[\dot H=-H^2 F,\quad F\equiv\frac32\gamma\left(1-\frac\Gamma{3H}\right),\] we find \[\frac{dq}{dt}=\dot H\frac{dF}{dH}=-H^2 F\frac{dF}{dH}\] and \[j=1+F\left(2F+H\frac{dF}{dH}\right)-3F,\] \[\frac{dF}{dH}=-\frac12\frac\gamma H\left(\frac{d\Gamma}{dH}-\frac\Gamma H\right),\] \[j=1+F\left[2F-3-\frac\gamma2\left(\frac{d\Gamma}{dH}-\frac\Gamma H\right)\right],\] which coincides with the result of the previous problem.
However, in connection with the particle production at the expense of gravitational field of the expanding Universe, we recall that long back ago, Zeldovich [Ya. B. Zeldovich, JETP Lett. 12, 307 (1970)] introduced some bulk viscosity mechanism which is responsible for particle production. However, later on, Lima and Germano [J. A. S. Lima and A. S. M. Germano, Phys. Lett. A 170, 373 (1992); For a more general macroscopic approach see also R. Silva, J. A. S. Lima and M. O. Calv~ao, Gen. Rel. Grav. 34, 865 (2002).] showed that although both the processes, namely the bulk viscosity mechanism by Zeldovich [Ya. B. Zeldovich, JETP Lett. 12, 307 (1970).] and the gravitational particle production produce the same dynamics of the Universe, but in principle they are completely different from a thermodynamical point of view. Since we describe the particle production, we would like also to note an analogy which exists between the models driven by the particle production and the models of Steady State Cosmology developed in [F. Hoyle and J. V. Narlikar, Proc. Roy. Soc. A 282, 191 (1964); ibdem, 290, 143 (1966); see also Action at a Distance in Physics and Cosmology, 1974 (New York, Freeman).].
Problem 34
problem id: 150_dp0-problem-34
In the model with $\Gamma=-\Gamma_0+mH$, where $0\le m\le3$ and $-3H_0\le\gamma\le0$, find the time dependence of the scale factor.
For $\Gamma=-\Gamma_0+mH$ the evolution equation for the hubble parameter takes on the form \[\dot H+\frac32H^2\left(1-\frac m3+\frac{\Gamma_0}{3H}\right).\] Its solution reads \begin{equation}\label{2016_07_p34_e1} H(t)=\frac{\Gamma_0/(m-3)}{1-\exp(\Gamma_0t/2)}. \end{equation} Then for the scale factor one finds \begin{equation}\label{2016_07_p34_e2} a(t)=a_0\left[H_0\left(\frac{m-3}{\Gamma_0}-1\right)\left(e^{-\Gamma_0t/2}-1\right)\right]^{\frac2{3-m}}. \end{equation} In the limit $\Gamma_0\to0$ \[a(t)=a_0\left[\frac{3-m}2H_0t\right]^{\frac2{3-m}}.\] Naturally at $m=0$ this solution is reduced to that of the Einstein-de Sitter model (Universe filled with non-relativistic matter).
Problem 35
problem id: 150_dp0-problem-35
In the model considered in the previous problem, express age of the Universe in terms of the parameters $\Gamma_0$ and $m$.
Setting $a=a_0$ in the expression for the scale factor obtained in the previous problem, one can find age of the Universe: \begin{equation}\label{2016_07_p35_e1} t_0=2\Gamma_0\ln\left(1-\frac{\Gamma_0/H_0}{m-3}\right). \end{equation}
Problem 36
problem id: 150_dp0-problem-36
Find deceleration parameter in the model with matter creation rate equal to $\Gamma$.
As we have seen above, in the considered model the evolution equation for the scale factor takes on the form \[a\ddot a+\frac12\left(1-\frac\Gamma H\right)\dot a^2=0.\] Using the definition of the deceleration parameter \[q\equiv-\frac{\ddot aa}{\dot a^2},\] one finds \[q=\frac12\left(1-\frac\Gamma H\right).\]
Problem 37
problem id: 150_dp0-problem-37
Show that in the creation matter model with $\Gamma=-\Gamma_0+mH$ in the case $\Gamma_0=0$ the expansion of Universe is always accelerated if $m>1$ and always decelerated if $m<1$.
Using the result of the previous problem, one obtains \[q=\frac12\left[1-m+\Gamma_0/H\right].\] It immediately follows then that at $\Gamma_0=0$ the deceleration parameter does not depend on time. It easy to see that at $m>1$ the expansion of Universe is always accelerated, and it is always decelerated for $m<1$. As the transition from the decelerated expansion to the accelerated one is required by the observation, the expression for the matter creation rate $\Gamma$ must contain the constant term $-\Gamma_0\ne0$.
Problem 38
problem id: 150_dp0-problem-38
Find the deceleration parameter as a function of the redshift in the creation matter model.
Exclude time from the above obtained expressions for the scale factor (\ref{2016_07_p34_e2}) and Hubble parameter (\ref{2016_07_p34_e1}), and use the relation \[1+z=\frac{a_0}a\] to find the hubble parameter as function of the redshift \[H(z)=\frac{\Gamma_0}{m-3}+\left(H_0-\frac{\Gamma_0}{m-3}\right)(1+z)^{\frac{3-m}2}.\] Then \begin{equation}\label{2016_07_p38_e1} q(z)=\frac12\left[1-m+\frac{\Gamma_0}{H(z)}\right] =\frac12\left[1-m+\frac1{\frac1{m-3}+\left(\frac{H_0}{\Gamma_0}-\frac1{m-3}\right)(1+z)^{\frac{3-m}2}}\right]. \end{equation} In the limit $\Gamma_0\to0$ \[q=\frac{1-m}2\] and for $m\to0$ \[q(z)=\frac12\left[1+\frac{\Gamma_0/H_0}{(1+\frac{\Gamma_0}{3H_0})(1+z)^{\frac32}-\frac{\Gamma_0}{3H_0}}\right].\]
Problem 39
problem id: 150_dp0-problem-39
Find the redshift value $z_t$ corresponding to the transition from the decelerated expansion to the accelerated one in the creation matter model.
Using the expression (\ref{2016_07_p38_e1}) for $q(z)$ obtained in the previous Problem with the condition $q(z_t)=0$, one finds that the transition from the decelerated expansion to the accelerated one in the cobsidered model takes place at the redshift value \[z_t=\left[\frac{2\Gamma_0/H_0}{(1-m)(m-3-\Gamma_0/H_0)}\right]^{\frac2{3-m}}-1.\]
Problem 40
problem id: 150_dp0-problem-40
Express age of the Universe in terms of the $z_t$ (see the previous Problem) in the creation matter model with $m=0$.
For the case $m=0$ \[\frac{\Gamma_0}{3H_0}=-\frac{1}{1+2(1+z_t)^{-\frac32}}.\] Substituting the latter expression into the result (\ref{2016_07_p35_e1}) of the Problem \ref{2016_07_p35} \[t_0=2\Gamma_0\ln\left(1-\frac{\Gamma_0/H_0}{m-3}\right),\] one obtains \[H_0t_0=\frac{4+2(1+z_t)^{\frac32}}{3(1+z_t)^{\frac32}}\ln\left[1+\frac{(1+z_t)^{\frac32}}2\right].\]