Difference between revisions of "Exactly Integrable n-dimensional Universes"

Consider an $(n+1)$-dimensional homogeneous and isotropic Lorentzian spacetime with the metric \begin{equation} \label{1} ds^2=g_{\mu\nu} dx^\mu dx^\nu=- dt^2+a^2(t)g_{ij} d x^i d x^j,\quad i,j=1,\dots,n, \end{equation} where $t$ is the cosmological (or cosmic) time and $g_{ij}$ is the metric of the $n$-dimensional Riemannian manifold $M$ of constant curvature characterized by an indicator, $k=-1,0,1$; $M$ is an $n$-hyperboloid, the flat space $\Bbb R^n$, or an $n$-sphere, with the respective metric \begin{equation} \label{2} g_{ij} d x^i d x^j=\frac1{1-kr^2}\, d r^2+r^2\, d\Omega^2_{n-1}, \end{equation} where $r>0$ is the radial variable and $d\Omega_{n-1}^2$ denotes the canonical metric of the unit sphere $S^{n-1}$. The Einstein equations: \begin{equation} \label{3} G_{\mu\nu}+\Lambda g_{\mu\nu}=8\pi G T_{\mu\nu}, \end{equation} where $G_{\mu\nu}$ is the Einstein tensor, $G$ the universal gravitational constant, and $\Lambda$ the cosmological constant, the speed of light is set to unity, and $T_{\mu\nu}$ is the energy-momentum tensor of an ideal cosmological fluid given by \begin{equation} \label{4} T^{\mu\nu}=\mbox{diag}\{\rho_m,p_m,\dots,p_m\}, \end{equation} with $\rho_m$ and $p_m$ the $t$-dependent matter energy density and pressure. Inserting the metric (\ref{1})--(\ref{2}) into (\ref{3}) we arrive at the Friedman equations \begin{equation}\label{5} H^2=\frac{16\pi G}{n(n-1)}\rho-\frac k{a^2}, \end{equation} \begin{equation} \dot{H}=-\frac{8\pi G}{n-1}(\rho+p)+\frac k{a^2},\label{6} \end{equation} in which \begin{equation} \label{7} H=\frac{\dot{a}}a, \end{equation} denotes the usual Hubble "constant", $\dot{f}=df/dt$, and $\rho,p$ are the effective energy density and pressure, related to $\rho_m,p_m$ through: \begin{equation} \label{8} \rho=\rho_m+\frac{\Lambda}{8\pi G},\quad p=p_m-\frac{\Lambda}{8\pi G}. \end{equation}

With (\ref{1}) and (\ref{4}) and (\ref{8}), the energy-conservation law, $\nabla_\nu T^{\mu\nu}=0$, takes the form \begin{equation} \label{9} \dot{\rho}_m+n(\rho_m+p_m)H=0. \end{equation} It is readily seen that (\ref{5}) and (\ref{9}) imply (\ref{6}). In other words, the full cosmological governing equations consist of (\ref{5}) and (\ref{9}) only.

Recall that the perfect-fluid cosmological model spells out a relation between the energy density $\rho_m$ and pressure $p_m$ of the matter source expressed by the so-called barotropic equation of state, \begin{equation} \label{10} p_m=w \rho_m, \end{equation} where $w$ is a constant so that $w=0$ leads to a vanishing pressure, $p_m=0$, corresponding to the dust model; $w=-1$ the vacuum model, and $w=1/n$ the radiation-dominated model. Inserting (\ref{10}) into (\ref{9}), we have \begin{equation}\label{11} \dot{\rho}_m+n(1+w)\rho_m \frac{\dot{a}}a=0, \end{equation} which can be integrated to yield \begin{equation}\label{12} \rho_m=\rho_0 a^{-n(1+w)}, \end{equation} where $\rho_0>0$ is an integration constant. Using (\ref{12}) in (\ref{8}), we arrive at the relation \begin{equation}\label{13} \rho=\rho_0 a^{-n(1+w)}+\frac{\Lambda}{8\pi G}. \end{equation}

From (\ref{5}) and (\ref{13}), we get the following equation of motion for the scale factor $a$: \begin{equation} \label{14} \dot{a}^2=\frac{16\pi G\rho_0}{n(n-1)}a^{-n(1+w)+2}+\frac{2\Lambda}{n(n-1)} a^2-k. \end{equation}

Rewrite equation (\ref{14}) as \begin{equation} \label{15} \dot{a}=\pm\sqrt{c_0 a^{-n(1+w)+2}+\Lambda_0 a^2},\quad c_0=\frac{16\pi G\rho_0}{n(n-1)},\quad\Lambda_0=\frac{2\Lambda}{n(n-1)}. \end{equation} Integration of (\ref{15}) gives \begin{equation} \label{15a} \pm\int a^{-1}\left(c_0 a^{-n(1+w)}+\Lambda_0 \right)^{-\frac12}d a=t+C. \end{equation} It is clear that the integral on the left-hand side of (\ref{15a}) satisfies the integrability condition stated in the Chebyshev theorem for any $n$ and any rational $w$. We have just seen that (\ref{15}) can be integrated directly for any rational $w$. Apply $a>0$ and get from (\ref{15}) the equation \begin{equation} \label{16} \frac{d}{d t}\ln a=\pm\sqrt{c_0 a^{-n(1+w)}+\Lambda_0}, \end{equation} or equivalently, \begin{equation}\label{17} \dot{u}=\pm\sqrt{c_0 e^{-n(1+w)u}+\Lambda_0},\quad u=\ln a. \end{equation} Set \begin{equation}\label{18} \sqrt{c_0 e^{-n(1+w)u}+\Lambda_0}=v. \end{equation} Then \begin{equation}\label{19} u=\frac{\ln c_0}{n(1+w)}-\frac 1{n(1+w)} \ln(v^2-\Lambda_0). \end{equation} Inserting (\ref{19}) into (\ref{17}), we find \begin{equation}\label{20} \dot{v}=\mp\frac12 n(1+w)(v^2-\Lambda_0), \end{equation} whose integration gives rise to the expressions \begin{equation} \label{21} v(t)=\left\{\begin{array}{cc} \frac{v_0}{1\pm \frac12 n(1+w)v_0t}, & \Lambda_0=0;\\ &\\ \sqrt{\Lambda_0}\frac{1+C_0 e^{\mp n(1+w)\sqrt{\Lambda_0} t}}{1-C_0 e^{\mp n(1+w)\sqrt{\Lambda_0} t}}, \quad C_0=\frac{v_0-\sqrt{\Lambda_0}}{v_0+\sqrt{\Lambda_0}}, & \Lambda_0>0;\\ &\\ \sqrt{-\Lambda_0}\tan\left(\mp\frac12 n(1+w)\sqrt{-\Lambda_0} t +\arctan\frac{v_0}{\sqrt{-\Lambda_0}}\right), & \Lambda_0<0, \end{array} \right. \end{equation} where $v_0=v(0)$. Hence, in terms of $v$, we obtain the time-dependence of the scale factor $a$: \begin{equation} \label{22} a^{n(1+w)}(t)=\frac{8\pi G \rho_0}{\frac12 n(n-1)v^2(t)-\Lambda}. \end{equation}

When $\Lambda=0$, we combine (\ref{21}) and (\ref{22}) to get \begin{equation} \label{x1} a^{n(1+w)}(t)=4\pi G\rho_0\left(\frac n{n-1}\right)(1+w)^2 t^2. \end{equation} When $\Lambda>0$, we similarly obtain \begin{equation} \label{x2} a^{n(1+w)}(t)=\frac{8\pi G\rho_0}{\Lambda}\sinh^2\left(\sqrt{\frac{n\Lambda}{2(n-1)}}(1+w) t\right). \end{equation} Both results (\ref{x1}) and (\ref{x2}) lead to an expanding Universe. We now consider the case when $\Lambda<0$ and rewrite (\ref{22}) as \begin{equation} \label{23} a^{n(1+w)}(t)=\frac{8\pi G \rho_0}{(-\Lambda)}\cos^2\left(\sqrt{\frac{n(-\Lambda)}{2(n-1)} }(1+w)t \mp\arctan\sqrt{\frac{n(n-1)}{-2\Lambda}}\, v_0\right). \end{equation} If we require $a(0)=0$, then (\ref{23}) leads to \begin{equation} \label{24} a^{n(1+w)}(t)=\frac{8\pi G \rho_0}{(-\Lambda)}\sin^2\sqrt{\frac{n(-\Lambda)}{2(n-1)} }(1+w)t, \end{equation} which gives rise to a periodic Universe so that the scale factor $a$ reaches its maximum $a_m$, \begin{equation} \label{25} a^{n(1+w)}_m=\frac{8\pi G \rho_0}{(-\Lambda)}, \end{equation} at the times \begin{equation}\label{26} t=t_{m,k}=\left(\frac\pi2+k\pi\right)\frac1{(1+w)}\sqrt{\frac{2(n-1)}{n(-\Lambda)}},\quad k\in\Bbb Z, \end{equation} and shrinks to zero at the times \begin{equation}\label{27} t=t_{0,k}=\frac{k\pi}{(1+w)}\sqrt{\frac{2(n-1)}{n(-\Lambda)}},\quad k\in\Bbb Z. \end{equation}

\begin{equation} %\label{21} q(t)=\left\{\begin{array}{cc} \frac n2 (1+w)-1, & \Lambda_0=0;\\ &\\ \frac{n(1+w)}{2\cosh^2\left(\sqrt{\frac{n\Lambda}{2(n-1)}}(1+w)t\right)}-1, & \Lambda_0>0;\\ &\\ \frac{n(1+w)}{2\cos^2\left(\sqrt{\frac{n\Lambda}{2(n-1)}}(1+w)t\right)}-1, & \Lambda_0<0. \end{array} \right. \end{equation}

The equation (\ref{14}) now reads \begin{equation} \label{32} \dot{a}^2=\frac{16\pi G\rho_0}{n(n-1)}a^{-n(1+w)+2}-k. \end{equation} In order to apply Chebyshev's theorem, we now assume that $w$ is rational. Thus we see that the question whether (\ref{32}) may be integrated in cosmological time is equivalent to whether \begin{eqnarray} I&=&\int a^{\frac12 n(1+w)-1}\left(-k a^{n(1+w)-2}+\sigma\right)^{-\frac12}\, d a\nonumber\\ &=&\frac2{n(1+w)}\int \left(-k u^{\gamma}+\sigma\right)^{-\frac12}\, d u,\quad u=a^{\frac12 n(1+w)},\quad\sigma=\frac{16\pi G\rho_0}{n(n-1)},\label{33} \end{eqnarray} is an elementary function of $u$, where \begin{equation} \label{34} \gamma=2\left(1-\frac2{n(1+w)}\right). \end{equation} By (\ref{34}), we see that (\ref{33}) is not elementary unless $1/\gamma$ or $(2-\gamma)/(2\gamma)$ is an integer. The case when $\gamma=0$ or $w=(2-n)/n$ is trivial since it renders $a(t)$ a linear function through (\ref{32}). That is, (\ref{32}) may only be integrated directly in cosmological time when $w$ satisfies one of the following: \begin{eqnarray} w&=&\frac{4N}{n(2N-1)} -1,\quad N=0,\pm1,\pm2,\dots;\\ w&=&\frac2n+\frac1{nN}-1,\quad N=\pm1,\pm2,\dots. \end{eqnarray} In particular, in the special situations when $n=3$, we have \begin{equation} w=-1,\dots,-\frac23,-\frac59,-\frac12,-\frac7{15},-\frac49,-\frac37,\dots,-\frac29,-\frac15,-\frac16,-\frac19,0,\frac13, \end{equation} so that $-1$ and $-1/3$ are the only limiting points.

The equation (\ref{33}) now becomes \begin{equation} I=\frac32\int(-k u^{-1}+\sigma)^{-\frac12}\,d u,\quad u=a^{\frac23}. \end{equation} When $k=1$ (closed Universe), we may use the substitutions $U=\sqrt{\sigma u-1}$ to carry out the integration, which gives us the explicit solution \begin{eqnarray} \label{41} &&a^{\frac13}\sqrt{\sigma a^{\frac23}-1}-a_0^{\frac13}\sqrt{\sigma a_0^{\frac23}-1} +\frac1{\sqrt{\sigma}}\ln\left(\frac{\sqrt{\sigma a^{\frac23}-1}+\sqrt{\sigma} a^{\frac13}}{\sqrt{\sigma a_0^{\frac23}-1}+\sqrt{\sigma} a_0^{\frac13}}\right)=\frac23\sigma t,\\ &&\quad t\geq0, \quad a_0=a(0),\nonumber \end{eqnarray} where $a_0$ satisfies the consistency condition $\sigma a_0^{\frac23}\geq1$ or \begin{equation} a_0\geq\left(\frac3{8\pi G \rho_0}\right)^{\frac32}, \end{equation} which spells out minimum size of the Universe in terms of $\rho_0$ whose initial energy density in view of (\ref{12}) is given by \begin{equation} \rho_m(0)=\frac{64}9 \pi^2 G^2\rho^3_0. \end{equation} When $k=-1$ (open Universe), we may likewise use the substitutions $U=\sqrt{\sigma u+1}$ to obtain the solution \begin{eqnarray} \label{44} &&a^{\frac13}\sqrt{\sigma a^{\frac23}+1}-a_0^{\frac13}\sqrt{\sigma a_0^{\frac23}+1} -\frac1{\sqrt{\sigma}}\ln\left(\frac{\sqrt{\sigma a^{\frac23}+1}+\sqrt{\sigma} a^{\frac13}}{\sqrt{\sigma a_0^{\frac23}+1}+\sqrt{\sigma} a_0^{\frac13}}\right)=\frac23\sigma t,\\ &&\quad t\geq0, \quad a_0=a(0),\nonumber \end{eqnarray} where no restriction is imposed on the initial value of the scale factor $a=a(t)$. In particular, if we adopt the big bang scenario, we can set $a_0=0$ to write down the special solution \begin{equation} a^{\frac13}\sqrt{\sigma a^{\frac23}+1} -\frac1{\sqrt{\sigma}}\ln\left({\sqrt{\sigma a^{\frac23}+1}+\sqrt{\sigma} a^{\frac13}}\right)=\frac23\sigma t,\quad t\geq0. \end{equation} The solutions (\ref{41}) and (\ref{44}) may collectively and explicitly be recast in the form of an elegant single formula: \begin{eqnarray} \label{44a} &&a^{\frac13}\sqrt{\sigma a^{\frac23}-k}-a_0^{\frac13}\sqrt{\sigma a_0^{\frac23}-k} +\frac{k}{\sqrt{\sigma}}\ln\left(\frac{\sqrt{\sigma a^{\frac23}-k}+\sqrt{\sigma} a^{\frac13}}{\sqrt{\sigma a_0^{\frac23}-k}+\sqrt{\sigma} a_0^{\frac13}}\right)=\frac23\sigma t,\\ &&\quad t\geq0, \quad a_0=a(0),\quad k=\pm1.\nonumber \end{eqnarray} We see that, in both closed and open situations, $k=\pm1$, respectively, the Universe grows following a power law of the type $a(t)=\mbox{\footnotesize O}(t^{\frac32})$ for all large time so that a greater Newton's constant or initial energy density gives rise to a greater growth rate.

\begin{equation} \label{48} ({a}')^2=\frac{16\pi G\rho_0}{n(n-1)}a^{-n(1+w)+4}+\frac{2\Lambda}{n(n-1)} a^4-ka^2. \end{equation}

When $k=0$, the conformal time version of (\ref{15}) reads \begin{equation} \label{49} a'=\pm a^2\sqrt{c_0 a^{-n(1+w)}+\Lambda_0 }, \end{equation} whose integration is \begin{equation} \label{50} \pm\int a^{-2}\left(c_0 a^{-n(1+w)}+\Lambda_0 \right)^{-\frac12}d a=\eta+C. \end{equation} Consequently Chebyshev's theorem indicates that, when $\Lambda\neq0$, the left-hand side of (\ref{50}) is elementary if and only if $\frac1{n(1+w)}$ or $\frac1{n(1+w)}-\frac12$ is an integer (again we exclude the trivial case $w=-1$), or more explicitly, $w$ satisfies one of the following conditions: \begin{eqnarray} w&=&-1+\frac1{nN},\quad N=\pm1,\pm2,\dots;\label{x4}\\ w&=&-1+\frac1{n\left(N+\frac12\right)},\quad N=0,\pm1,\pm2,\dots. \end{eqnarray}

The equation (\ref{49}) now becomes \begin{equation} a'=\pm a^2\sqrt{c_0 a^{-1}+\Lambda_0}, \end{equation} which can be integrated to yield the solution \begin{equation} \frac{c_0}a+\Lambda_0=\frac{c_0^2}4(\eta+C)^2, \end{equation} where $C$ is an integrating constant.

The Friedmann equation now becomes \begin{equation} \label{53} a'=\pm a\sqrt{c_0 a^{-n(1+w)+2}-k}, \end{equation} which in view of Chebyshev's theorem can be integrated in terms of elementary functions when $w$ is any rational number. In fact, as before, (\ref{53}) may actually be integrated to yield its exact solution expressed in elementary functions for any $w$: \begin{equation} \label{57} \pm\left(1-\frac12 n(1+w)\right)\eta+C=\left\{\begin{array}{rll} &-\frac1v,&\quad k=0;\\ &&\\ &\arctan v,&\quad k=1;\\ &&\\ &\frac12\ln\left|\frac{v-1}{v+1}\right|,&\quad k=-1,\end{array}\right. \end{equation} where $C$ is an integration constant and \begin{equation} v=\sqrt{c_0 a^{-n(1+w)+2}-k}\quad \mbox{ or }\quad a^{-n(1+w)+2}=\frac1{c_0} (v^2+k). \end{equation}

To obtain inflationary solutions, we assume \begin{equation} n(1+w)>2, \end{equation} which includes the dust and radiation matter situations since $n\geq3$. We assume the initial condition $a(0)=0$. When $k=0$, (\ref{57}) gives us the solution \begin{equation}\label{xx1} a^{n(1+w)-2}(\eta)=\frac{4\pi G\rho_0}{n(n-1)}\left(n(1+w)-2\right)^2\eta^2,\quad \eta\geq0. \end{equation} When $k=1$, (\ref{57}) renders the solution \begin{equation} a^{n(1+w)-2}(\eta)=\frac{16\pi G \rho_0}{n(n-1)}\sin^2\left(\frac12(n[1+w]-2)\eta\right),\quad \eta\geq0, \end{equation} which gives rise to a periodic Universe. When $k=-1$, (\ref{57}) yields \begin{equation} a^{n(1+w)-2}(\eta)=\frac{16\pi G \rho_0}{n(n-1)}\sinh^2\left(\frac12(n[1+w]-2)\eta\right),\quad \eta\geq0, \end{equation} which leads to an inflationary Universe. It is interesting to notice that the closed Universe situation here ($k=1,\Lambda=0$), in conformal time, is comparable to the flat Universe with a negative cosmological constant ($k=0,\Lambda<0$), in cosmological time, and the open Universe situation here ($k=-1,\Lambda=0$), in conformal time, is comparable to the flat Universe with a positive cosmological constant ($k=0, \Lambda>0$), also in cosmological time.