Difference between revisions of "Causal Structure"

Suppose $\tilde{\eta},\tilde{\chi}$ span infinite or semi-infinite values. Then we can always make the following sequence of coordinate transformations:

The scale factor in the closed dS Universe is \begin{equation} a(t)=H_\Lambda^{-1}\cosh (H_\Lambda t),\qquad t\in (-\infty, +\infty), \end{equation} so on integration, for conformal time we obtain \begin{equation} \eta (t)=\int\limits_{-\infty}^{t}\frac{dt}{a(t)}=\arcsin\big[\tanh (H_\Lambda t)\big]+\frac{\pi}{2} \in (0,\pi). \end{equation} We choose the integration constant here so that $\eta=0$ corresponds to $t=-\infty$ and $\eta=\pi$ to $t=+\infty$. The full metric then can be written as \begin{equation} ds^{2}_{dS}=\frac{H_\Lambda^2}{\sin^2 \eta}\big[d\eta^2 -d\chi^2 -\sin^2 \chi d\Omega^2 \big]. \label{dSclosed} \end{equation} As the values of $\eta$ span a finite interval, $(\eta,\chi)$ are already conformal coordinates. The conformal diagram is again a square \begin{equation} \eta\in [0,\pi],\qquad \chi \in[0,\pi], \end{equation} with the difference from the radiation dominated Universe that the edges $\eta=0,\pi$ do not represent a singularity anymore, but instead correspond to infinite (and regular) past and future respectively. Both horizons are given again by \begin{equation} \eta_{e}=\pi-\chi,\qquad \eta_{p}=\chi \end{equation} and exist at all times. The spacelike boundaries of the conformal diagram correspond to $t\to \pm\infty$, and therefore to infinite values of affine parameter. This can be shown if one remembers the general formula for the cosmological redshift: \begin{equation} \text{const}=\omega a =\frac{dt}{d\lambda}a,\quad \Rightarrow\quad \lambda=\text{const}\cdot\int\limits^{t}dt \cosh(H_\Lambda t) \underset{t\to\pm\infty}{\longrightarrow}\infty . \end{equation} The timelike boundary of the diagram corresponds to the opposite pole, there is no real boundary there, the same as on a sphere: as particles propagate across the pole, their radial coordinate begins to decrease again, while the worldline on the conformal diagram is reflected from $\chi=\pi$. Thus by definition the spacetime is (null) geodesically complete.

The de Sitter Universe. The closed sections coordinates cover the whole space, which is geodesically complete. There are no singularities: the horizontal boundaries of the diagram correspond to infinite past and future.

Let us introduce dimensionless coordinates $t=H_\Lambda T$ and $r=H_\Lambda R$. The inverse relations then are \begin{equation} \sin^2 \eta =\frac{1}{\cosh^2 t-r^2 \sinh^2 t}, \quad \sin^2 \chi =\frac{r^2}{\cosh^2 t-r^2 \sinh^2 t}. \end{equation} Using them, after some algebra from (\ref{dSclosed}) we get \begin{equation} ds_{dS}^2 =\big[1-H_\Lambda^2 R^2\big]dT^2 -\frac{dR^2}{1-H_\Lambda^2 R^2}-R^2 d\Omega^2 , \label{dSstatic} \end{equation} which resembles the Schwarzschild line element (and this is not a coincidence).

The de Sitter Universe with contour lines of the "static" coordinates $(T,R)$. The solid lines are $T=const$, and the dashed ones $R=const$. The coordinates become singular on both horizons, so the whole spacetime is divided by them into four sectors, each separately covered by the regular coordinate chart $(R,T)$. Spacetime is actually static only in the left and right sectors (I and III), where $T$ is a timelike coordinate and $R$ spacelike: the static patches are bounded by the horizons.

• As $|\tanh (H_\Lambda T)|<1$, we have $|\cos\eta| \leq |\cos\chi|$. This condition cuts out two out of four sectors from the square conformal diagram, I and III: one is $\eta\in [\chi, \pi-\chi]$ and the other is $\eta\in [\pi-\chi, \chi]$. In both $|\sin\eta|\geq |\sin\chi|$, so $R \leq H_{\Lambda}^{-1}$;
• The particle horizon $\eta=\chi$ corresponds to $R=H_\Lambda^{-1}$ and $T=-\infty$. It is one part of the boundary of the region (in two parts) covered by coordinates $(T,R)$. The event horizon $\eta=\pi-\chi$ corresponds to $R=H_\Lambda^{-1}$ and $T=+\infty$ and is the other part of the boundary of this region.
• $T=const$ is $\cos\eta =t\cos\chi $ and $R=const$ is $\sin\eta =r^{-1}\sin\chi$.
• In regions II and IV the needed relation is obtained if we simply replace $\tanh$ with $\coth$ in the first relation:

\begin{equation} \coth (H_\Lambda T)=-\frac{\cos\eta}{\cos\chi},\qquad H_\Lambda R=\frac{\sin\chi}{\sin\eta}, \end{equation} as can be checked explicitly by substitution into (\ref{dSstatic}), which again gives (\ref{dSclosed}). In these regions $R$ is a timelike coordinate, and $T$ is spacelike. The geodesics of comoving massive particles are $\chi=const$, one of them $\chi=\pi/2$ corresponds to $T=0$. Thus in the lower part of the diagram, where $R\in (+\infty,H_\Lambda^{-1})$, the spacetime is contracting; in the upper part, where $R\in (H_\Lambda^{-1},\infty)$, it is expanding. The coordinate frame is not static. The relations between $(T,R)$ and $(\eta,\chi)$ in static and non-static regions mirros those in Schwarzshild black hole solution between the static and the global (Kruskal-Szekeres) coordinates. Here $\eta$ and $\chi$ are the global coordinates.

As $t\in(-\infty,+\infty)$, \begin{equation} \tilde{\eta} =\int\frac{dt}{a(t)}=H_\Lambda \int\limits_{+\infty}^{t}dt\; e^{-H_\Lambda t} =-e^{-H_\Lambda t} \in (-\infty,0). \end{equation} Here we choose $+\infty$ as the lower limit, because at $-\infty$ the integral diverges.

The de Sitter Universe in flat sections' coordinates, which cover only half of it. The boundary -- the particle horizon -- consists of three different infinities.

After getting \begin{equation} \sinh \tilde{\eta}=-\frac{1}{\sinh(H_\Lambda t)}, \end{equation} the first part is checked straightforwardly; in the open de Sitter $\tilde{\chi}\in [0,+\infty)$, and $\tilde{\eta}\in(-\infty,0)$. The region covered by coordinates $(\tilde{\eta},\tilde{\chi})$ is $\{\eta>\chi+\pi/2\}$, only one eighth part of the full diagram.

The de Sitter Universe in open sections' coordinates, which cover only $1/8^{\text{th}}$ of the full diagram.

Coordinate transformation gives \begin{equation} ds^2 =\frac{1}{\cos^2 \chi -\sin^2 \eta}\big[d\eta^2 -d\chi^2 -\Psi^{2}(\eta,\chi)d\Omega^2 \big]. \end{equation} Here $r\in [0,+\infty)$ and $t\in (-\infty,+\infty)$. Comparing with the relation between $(\tilde{\eta},\tilde{\chi})$ with conformal coordinates $(\eta,\chi)$ in the open de Sitter universe, where $\tilde{\eta}\in (-\infty,0)$, we see that the difference is that $\tilde{\eta}$ spans $(-\infty,0)$, while now $t$ spans twice the range, $(-\infty,\infty)$. Therefore the conformal diagram is composed of two triangles, one the same as for open de Sitter and one for its time-reversed copy. Accordingly there now appear future timelike infinity $i^+$ and future lightlike infinity $J^+$.

The Minkowski spacetime in two different pairs of conformal coordinates and the full set of infinities. Thin dashed and solid lines show the images of coordinate grid $(t,r)$. The one on the right corresponds to another choice of conformal coordinates, discussed in the next problem,

We start from spherical coordinates \begin{equation} ds^2 =dt^2 -dr^2 -r^2 d\Omega^2 . \end{equation}

Minkowski metric $ds^2 =dT^2 -dR^2$, rewritten in terms of $(\tau, r)$ such that \begin{equation} T=\tau \cosh r ,\quad R=\tau \sinh r , \end{equation} is the metric of the Milne Universe. As Minkowski space is complete, the Milne Universe then is a \emph{part} of Minkowski in different variables. This part is where $T>R$. The boundary $T=R$ is the future light cone, so the whole Milne Universe is represented by the triangle homothetic to the whole space but 4 times smaller in area, with the common node of future infinity.

The Milne Universe on Minkowski's conformal diagram. It has the same shape and shares the same future infinity, but covers one eighth of the area. The boundary is the past horizon.

For flat Universe $\tilde{\chi}=r$, for the open $\tilde{\chi}=\sinh r$, so in both cases $\tilde{\chi}\in (0,+\infty)$. Strong energy condition implies that $w>-1/3$, so \begin{equation} \rho \sim a^{-3(1+w)}=a^{-n}, \end{equation} where $n>2$. From the first Friedman equation then after simple manipulations we obtain that \begin{equation} \frac{da}{dt}\sim a^{-\theta}, \end{equation} where $\theta$ is some positive number. Therefore both \begin{equation} t\sim \int da\; a^{\theta}, \quad\text{and}\quad \eta =\int \frac{da}{a}a^{\theta} \end{equation} converge at $a\to 0$ and diverge at $a\to \infty$. Consequently, the integration constant can be chosen so that $\eta\in (0,+\infty)$. The conformal structure is the same as that of the \emph{upper} half of Minkowski spacetime. The Big Bang singularity at $\eta=0$ is at the cut, and there are spacelike infinity, future infinity and future null infinity.

Conformal diagram for an open or flat Universe filled with matter which satisfies the strong energy condition.

As seen in the previous problem, if strong energy condition were satisfied, we would have $\dot{a}\sim a^{-\theta}$ with some positive $\theta$; this is not the case, so the condition is violated. As $t\in (0,+\infty)$, \begin{equation} \eta\sim \int \frac{dt}{t^n} \end{equation} diverges at small $a$ (thus also small $t$) and converges at $a\to \infty$ (thus as large $t$). So integration constant can be chosen so that $\eta\in (-\infty,0)$. The conformal structure is the same as \emph{lower half} of Minkowski spacetime. The cut is regular future infinity, and from Minkowski there are spacelike infinity, past null infinity, and past infinity. The point of past infinity corresponds to Big Bang and is singular.

Conformal diagram for an open or flat Universe filled with matter which violates the strong energy condition (SEC), such as in the case of power-law inflation.