# The Milne Universe

### Problem 1: the solution

Find the solutions of the Friedman equations for the Milne Universe: the expanding Universe with $\rho\to 0$ and $k=-1$. Why is it necessarily spatially open? What is the scalar curvature of this spacetime?

The Friedman equations with $\rho\to 0$ reduce to \begin{align*} (\dot{a})^{2}=-k,\\ \ddot{a}=0, \end{align*} and thus we have that $k=-1$, and the solution for the scale factor is $a=t$ (or $a(t)=ct$ if we preserve $c$). The metric of the Milne Universe therefore has the form \begin{equation}\label{Milne-metric} ds^{2}=dt^{2} -t^{2}\big[dr^{2}+\sinh^{2}r d\Omega^{2}\big]. \end{equation} Using the expression for the scalar curvature in the FLRW metric, we get \[R = - 6\left( \frac{\ddot a}{a} + \frac{\dot a^2}{a^2} + \frac{k}{a^2} \right)=0,\] as it should be in a spacetime with no matter whatsoever.

### Problem 2: alternative derivation

Let us consider the Minkowski spacetime in spherical spatial coordinates $(T,R,\theta,\phi)$. Let at some moment of initial explosion a cloud of particles emerge from the origin with all possible velocities $v<c$ in all directions, which stay constant. Their mass is considered negligible, so that they do not interact and do not affect the underlying spacetime. The larger is the velocity of a particle, the further away from the origin it is at a given moment of time, so the velocity of a particle $v$, or alternatively its "rapidity"
\[r=\text{artanh}\, v
\equiv\tfrac{1}{2}\ln\frac{1+v}{1-v}\]
serve as radial coordinates in the region $R<T$. Let $\tau$ be the proper time of the particle. Show that the region $R<T$ in coordinates $(\tau,r,\theta,\phi)$ *is* the Milne Universe$^*$.

$^*$In fact, this is the way Milne in his papers of 1935 and 1948 introduced this spacetime, trying to show that Big Bang can be described by pure kinematics and in the frame of Special Theory of Relativity only. This is in general not possible, but his renowned example is very instructive.

The velocity and Lorentz-factor of a particle expressed through its rapidity are
\[v=\tanh r,\qquad \gamma=\cosh r.\]
Then taking into account the time dilation
\[\tau=\gamma^{-1}T=T/\cosh r,\]
and the obvious $R=vT$, we obtain
\begin{align}
T&=\tau \cosh r,\nonumber\\
\label{Milne-Mink}
R&=\tau \sinh r.
\end{align}
Now it is not hard to see that
\[dT^{2}-dR^{2}=d\tau^{2}-\tau^{2}dr^{2},\]
and the metric of the original Minkowski spacetime turns out to describe the Milne Universe:
\[ds^{2}=dT^{2}-dR^{2}-R^{2}d\Omega^{2}
=d\tau^{2}
-\tau^{2}\big[dr^{2}+\sinh^2 r d\Omega^{2}\big].\]
As the Milne Universe is just a re-parametrized Minkowski, not only its scalar curvature is zero, but the curvature tensor as well (as should be in the absence of any matter). Its spatial slices $T=const$ are, obviously, flat, while the slices $\tau=const$, which are *hyperboloids* $T^{2}-R^{2}=\tau^{2}$, have constant *negative* (scalar) curvature $k=-1$.
It should be noted however, that though the (part of) spacetime is the same, the Milne and Minkowski metrics describe quite different frames of reference, and the observables in them (redshifts for example) will be different too.

### Problem 3: deeper relation

Let the density of matter in the Milne Universe (in the comoving frame) be small but finite. Find the dependence of (number) density on the distance to the horizon $R=T$ in the Minkowski spacetime (the laboratory frame with regard to the experiment of the Big Bang), if the distribution in the Milne Universe is homogeneous. What is the total number of particles (galaxies) in each of the frames of reference?

a) In the Milne frame the spatial metric of the slice $\tau=const$ is
\[dl^{2}_{\tau}
=\tau^2 dr^2 +\tau^2 \sinh^{2}r d\Omega^2,\]
so the volume element is (integrating over $d\Omega$)
\[dV_{\tau}=4\pi \tau^{3}\sinh^{2}r\, dr.\]
If $n_{\tau}=const$ is the particles' number density, the total number of particles within a sphere of radius (radial coordinate, to be precise) $r_{\star}$ is
\[N_{\tau}(r_{\star})=4\pi\;
\underbrace{n_{\tau}(\tau)\tau^{3}}_{const}
\int\limits_{0}^{r_{\star}}dr\,\sinh^{2}r
\xrightarrow[r_{\star}\to\infty]{}\infty,\]
which is natural, as we assume the number density is constant at a given time on an unbounded hyperboloid of infinite volume.
b) In the Minkowski frame the spatial metric of the slice $T=const$ is
\[dl^{2}_{T}=dR^{2}+R^{2}d\Omega,\]
so the volume element is $dV_{T}=4\pi R^2 dR$. We can rewrite it in terms of $r$, while keeping in mind that we are talking of the slice $T=const$, and thus from (\ref{Milne-Mink}) $R=T\tanh r$ and $dR=Tdr/\cosh^{2}r$ we get
\[dV_{T}=4\pi R^{2}dR
=4\pi T^{3}\frac{\sinh^{2}r\,dr}{\cosh^{4}r}
=4\pi \tau^{3}\frac{\sinh^{2}r\,dr}{\cosh r}.\]
As $r$ is the *comoving* coordinate, and the number of particles is conserved, it stays the same in the layer between $r$ and $r+dr$ regardless of the chosen spatial slice, so
\[dN[r,dr]=n_{\tau}dV_{\tau}=n_{T}dV_{T}\]
and
\[n_{T}=n_{\tau}\cosh r =n_{\tau}\gamma
=\frac{n_{\tau}}{\sqrt{1-R^2 /T^2}}.\]
We could have obtained this result by simple observation that the number (or mass) density in the Minkowski frame changes due to the contraction of the radial distance between the particles by the Lorentz-factor $\gamma$.
However, when considering the slice $T=const$, we now also have to take into account that $\tau$ is a function of $R$:
\[\tau^{2}=T^{2}-R^{2},\]
so due to $n_{\tau}\tau^{3}=const$, we get
\begin{equation}\label{Milne-density}
n_{T}=n_{0}\frac{(T_{0}/T)^{3}}
{\big[1-R^2 /T^2\big]^{2}},
\end{equation}
which leads to the correct divergence of the integral for the total number of particles on the horizon $R=T$:
\begin{align*}
N_{T}&=\int n_{T}dV_{T}
=4\pi\;n_{0}T_{0}^{3} \int\limits_{0}^{T}
\frac{R^2\,dR /T^3}{(1-R^2 /T^2)^{2}}=\\
&= \Big\| \xi=R/T\;\;\Big\|
=4\pi\;n_{\tau}\tau^{3} \int\limits_{0}^{1}
\frac{\xi^{2}\,d\xi}{(1-\xi^2)^{2}}=\\
&=\Big\| \xi=\tanh r\Big\|
=4\pi\;n_{\tau}\tau^{3} \int\limits_{0}^{\infty}
\sinh^{2}r\,dr\to\infty.
\end{align*}
In terms of mass density the divergence of $\rho_{T}$ at the horizon means that even for arbitrarily small density $\rho_{0}$ in the comoving frame the density in the laboratory frame becomes arbitrarily large close enough to the horizon. Thus in the neighborhood of the horizon we cannot actually consistently neglect the mass distribution and the curvature of spacetime it induces, and even for small $\rho_0$ we cannot define the world time $T$ in all of the originally Minkowski spacetime.