1
Problem 1
problem id: gnd_1
Derive Friedmann equations for the spatially n-dimensional Universe.
solution
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}
2
Problem 2
problem id: gnd_2
Obtain the energy conservation law for the case of n-dimensional Universe.
solution
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.
3
Problem 3
problem id: gnd_3
Obtain relation between the energy density and scale factor in the case of a two-component n-dimensional Universe dominated by the cosmological constant and a barotropic fluid.
solution
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}
4
Problem 4
problem id: gnd_4
Obtain equation of motion for the scale factor for the previous problem.
solution
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}
To integrate (\ref{14}), we recall Chebyshev's theorem:
For rational numbers $p,q,r$ ($r\neq0$) and nonzero real numbers $\alpha,\beta$, the integral $\int x^p(\alpha+\beta x^r)^q\,d x$ is elementary if and only if at least one of the quantities
\begin{equation}\label{cd}
\frac{p+1}r,\quad q,\quad \frac{p+1}r+q,
\end{equation}
is an integer.
Another way to see the validity of the Chebyshev theorem is to represent the integral of concern by a hypergeometric function such that when a quantity in (\ref{cd}) is an integer the hypergeometric function is reduced into an elementary function. Consequently, when $k=0$ or $\Lambda=0$, and $w$ is rational, the Chebyshev theorem enables us to know that, for exactly what values of $n$ and $w$, the equation (\ref{14}) may be integrated.
5
Problem 5
problem id: gnd_4_0
Obtain analytic solutions for the equation of motion for the scale factor of the previous problem for spatially flat ($k=0$) Universe.
solution
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}
6
Problem 6
problem id: gnd_5
Use the analytic solutions obtained in the previous problem to study cosmology with $w>-1$ and $a(0)=0$.
solution
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}
7
Problem 7
problem id: gnd_6
Use results of the previous problem to calculate the deceleration $q=-a\ddot a/\dot a^2$ parameter.
solution
\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}
8
Problem 8
problem id: gnd_4_1
Obtain the exact solvability conditions for the case $\Lambda=0$ in equation (\ref{14}) (see problem #gnd_4 ).
solution
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.
9
Problem 9
problem id: gnd_4_2
Obtain the explicit solutions for the case $n=3$ and $w=-5/9$ in the previous problem.
solution
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 $a(t)=\mbox{O}(t^{\frac32})$
for all large time so that
a greater Newton's constant or initial energy density gives rise to a greater growth rate.
10
Problem 10
problem id: gnd_10
Rewrite equation of motion for the scale factor (\ref{14}) in terms of the conformal time.
solution
\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}
11
Problem 11
problem id: gnd_11
Find the rational values of the equation of state parameter $w$ which provide exact integrability of the equation (\ref{48}) with $k=0$.
solution
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}
12
Problem 12
problem id: gnd_12
Obtain explicit solutions for the case $w=\frac1n-1$ in the previous problem.
solution
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.
13
Problem 13
problem id: gnd_13
Obtain exact solutions for the equation (\ref{48}) with $\Lambda=0$.
solution
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}
14
Problem 14
problem id: gnd_14
Obtain inflationary solutions using the results of the previous problem.
solution
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.