Shape of universe from the metric's scalar factor

Question

I am trying to understand FRW model and how one interpreted it. I missed a lecture and I am now trying to go through a friends notes. The teacher presented this metric as an example $$ds^2=-dt^2+\frac{t}{c}(dx^2+dy^2+dz^2) $$ where $c$ is a constant. He then states what type of geometry the universe has: flat, closed or opened. I don't understand how I can determine that.

I know that $a=\sqrt{\frac{t}{c}}$ and then the Hubble constant is $H=\frac{1}{2t}$. One way to see the criteria for the different cases are $k\geq 0$ closed, $k= 0$ flat or $k\leq0 $ for open. Where $k$ is form the Friedmann equation $$H^2=\frac{8\pi G}{3}\rho + \frac{k}{a^2} $$ and $\rho$ is the matter density.

I don't know how to prove it mathematically but my reasoning is that it cannot be closed. Because $t$ will continue to grow so the universe must be expanding. How to narrow it done further escapes me now.

Answer 1

Another way to determine the shape then the one you provided is to look at the density $\rho$. The different cases would then be:

for closed $$\rho \geq \frac{3H^2}{8\pi G} $$ for open $$\rho \leq \frac{3H^2}{8\pi G} $$ and flat

$$\rho=\frac{3H^2}{8\pi G} $$

Answer 2

$\newcommand{\dd}{{\rm d}}$ $\newcommand{\vect}[1]{{\bf #1}}$

There are many things going on there

Cosmological principle

In general a metric of the form

$$ \dd s^2 = c^2\dd t^2 - a^2(t) \dd \vect{x}^2 \tag{1} $$

Satisfies that the cosmological principle: universe is homogeneous an isotropic. In fact, imagine that $t$ is fixed, the effect of $a(t)$ is just "stretching" the coordinates, so if the universe satisfies the cosmological principle at a given time, it will follow it at any time.

Now, if the universe is to be homogeneous, its curvature $\kappa$ must be the same everywhere. There are only a handful of geometries that satisfy this constraint, one is the plane ($\kappa = 0$), other is the sphere ($\kappa = +1$) and finally an hyperbolic universe ($\kappa = -1$). Note that I deliberately used $\kappa = \pm 1$, because the metric can be scaled through the factor $a$ in Eq. (1), so the actual value of the curvature is unimportant.

The most general metric for an universe that follows the cosmological principle is

$$ \dd s^2 = c^2 \dd t^2 -a^2 \left(\frac{1}{1 - \kappa r^2} \dd r^2 + r^2\dd \Omega^2\right) \tag{2} $$

Einstein's field equations

The equations that describe the coupled evolution of the spacetime geometry and its content are

$$ R_{\mu\nu} -\frac{1}{2}Rg_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4}T_{\mu\nu} \tag{3} $$

where $R_{\mu\nu}$ is the Ricci curvature tensor, $R$ the scalar curvature $R = R^\mu_\mu$, $g_{\mu\nu}$ the metric tensor, which can be directly extracted from Eq. (2) $\dd s^2 = g_{\mu\nu}\dd x^\mu \dd x^\nu$, $\Lambda$ the cosmological constant and $T_{\mu\nu}$ the stress-energy tensor which contains information about the physical content of the universe, for example $T^{00} = \rho$

This equation can be then explicitly written in terms of the metric $g_{\mu\nu}$ which results in a relation between the geometry $\kappa$ and the energy content $\rho c^2$ of the universe. The result is what we know as Friedmann equations

$$ \frac{\dot{a}^2 + \kappa c^2}{a^2} = \frac{8 \pi G \rho + \Lambda c^2}{3} \tag{4} $$

which is the result of calculating the 00 component of Eq. (3). If you take the trace instead, you get

$$ \frac{\ddot{a}}{a} = -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3} \tag{5} $$

Set $\Lambda = 0$ in Eq. (4) and you will conclude about the dynamics of $a$ in terms of the geometry $\kappa$ and the density $\rho$

The fate of the universe

Let us consider for instance a universe for which $\Lambda =0$ and define

$$ \rho_c = \frac{3H^2}{8\pi G} ~~~\mbox{and}~~~ H = \frac{\dot{a}}{a} $$

You can see that Eq. (4) becomes

$$ 1 = \frac{\rho}{\rho_c} - \frac{\kappa c^2}{a^2H^2} \tag{5} $$

which can be solved for $\kappa$

$$ \kappa = \frac{a^2 H^2}{c^2}\left(\frac{\rho}{\rho_c} - 1\right) $$

This establishes a clear connection between density and geometry

1. $\rho > \rho_c$, then $\kappa > 0$ and the universe has spherical geometry (closed)

2. $\rho < \rho_c$ then $\kappa> 0$ and the universe has hyperbolical geometry (open)

3. $\rho = \rho_c$ in this case $k = 0$, which corresponds to the Euclidean flat space

Answer 3

Meaning of Closed, Open and Flat Universe

Let us start with the simple definitions of a closed, open and flat universe. A closed universe is by definition one with $k=-1$, a flat universe is one with $k=0$ and an open universe is one $k=1$ regardless of what is happening to that universe.

The Different Types of Density

Their are three main types of density in the universe: $$ \newcommand{\p}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\f}[2]{\frac{ #1}{ #2}} \newcommand{\l}[0]{\left(} \newcommand{\r}[0]{\right)} \newcommand{\mean}[1]{\langle #1 \rangle}\newcommand{\e}[0]{\varepsilon} \newcommand{\ket}[1]{\left|#1\right>} \begin{align} \text{Matter (non-relativistic):}~~ & \rho_M=\f{\rho_{M,0}}{a^3} \\ \text{Radiation:}~~ & \rho_R=\f{\rho_{R,0}}{a^4} \\ \text{Vacuum:}~~ & \rho_V=\rho_{V,0}\end{align}$$ where the last of these is assuming that the cosmological constant is infact constant.

With $\rho_V \ne 0$

With the vacuum density things get complicated, you can have a closed universe ($k=-1$) which is accelerating, a flat universe $k=0$ that recollapses etc. There is a nice picture of the different regions in 'An introduction to modern cosmology' by A.Liddle pg54 if you can get your hands on it.

With $\rho_V=0$

Things are a little nice a flat universe $k=0$ has $H\rightarrow 0$ as $t\rightarrow \infty$, a closed universe recollapses and an open universe expands forever. The dynamical evolution equation states that: $$\dot a^2-\f{8\pi G \rho a^2}{3}=-k\tag{1}$$ now if $\dot a=0$ we have: $$a^2=\f{3k}{8\pi G \rho}$$ if $k=-1$ this has no solution and thus $\dot a$ will never pass through zero, and the universe must keep expanding forever. If $k=1$ this does has a solution and their will be a point where $\dot a=0$ and thus we can except the universe to collapse*. If $k=0$ then going back to equation (1) we can see that since $\rho$ goes like $a^{-3}$ (at least) the only place $\dot a=0$ is at $a=\infty$ thus in this case we get a universe which only stops expanding at $a=\infty$ (which must correspond to $t=\infty$).