# Geodesically complete incompleteness and the most awful spacetimes

The notion of a singularity took quite a while to define in the 1960's and 1970's. Defining such a notion in general relativity has many pitfalls, as it was fairly easy to make it too narrow or too broad.

One popular definition was geodesic completeness : a spacetime is said to be geodesically incomplete if there exists a maximally-extended geodesic $\gamma$ of finite half-length, that is, for an inextendible curve defined on $I = (a,b)$, with the starting point $a' \in (a,b)$, we have

$$\int_{a'}^b \|\gamma'(\lambda)\| d\lambda \leq \infty $$As this notion is equivalent to completeness in the case of Riemannian spaces, this was thought to be a satisfying notion of a spacetime having a singularity. It can be however shown that one can construct a spacetime which is geodesically complete, but admits a timelike curve of finite half-length. Physically, this means that on such a spacetime, an observer can travel along a realistic path and just disappear in a singularity.

## 1. Bounded acceleration

First, one needs to define what a reasonable timelike curve is, in those circumstances. Even in Minkowski space, it is possible to have an inextendible incomplete curve if it is allowed to be arbitrary. To construct it, we need a timelike curve with unbounded acceleration. Pick some curve with an acceleration depending on its parameter

$$\|\ddot\gamma(\tau)\| = \alpha(\tau)^2$$Along with the usual normalization condition for timelike curves

$$\|\dot\gamma(\tau)\| = -1$$This gives us, in two dimensional Minkowski space,

\begin{eqnarray} \ddot{x}^2(\tau) - \ddot{t}^2(\tau) &=& \alpha(\tau)^2\\ \dot{x}^2(\tau) - \dot{t}^2(\tau) &=& -1 \end{eqnarray}Or, using the lightcone coordinates $v = (t+x)/2$, $u = (t - x)/2$

\begin{eqnarray} \ddot{u}(\tau) \ddot{v}(\tau) &=& \alpha(\tau)^2\\ \dot{u}(\tau)\dot{v}(\tau) &=& -1 \end{eqnarray}Meaning that $\dot{u} = -\dot{v}^{-1}$, so that

$$\ddot{u} = \frac{\ddot{v}}{\dot{v}^2}$$ $$ (\frac{\ddot{v}}{\dot{v}})^2 = \alpha(\tau)^2$$If we concentrate on the positive class of solutions (the other will simply be the mirror solutions with either curves going in the other direction or past-directed curves), the equation is just

$$\ddot{v} - \alpha(\tau)\dot{v} = 0$$with the simple solution

\begin{eqnarray} \dot v(\tau) &=& A e^{\int_1^{\tau} \alpha(\xi) d\xi}\\ \dot u(\tau) &=& A^{-1} e^{-\int_1^{\tau} \alpha(\xi) d\xi} \end{eqnarray}Switching back to canonical coordinates

\begin{eqnarray} \dot t(\tau) &=& \sinh(\int_1^{\tau} \alpha(\xi) d\xi)\\ \dot x(\tau) &=& \cosh(\int_1^{\tau} \alpha(\xi) d\xi) \end{eqnarray}As the curve accelerates more and more, it will get closer and closer to a null curve, and due to this, its proper time will shrink. If we pick $\alpha$ to be unbounded and growing fast enough, this can mean that the total proper time can be finite even at infinity. Since we picked $|\dot \gamma(\tau)| = 1$, the only way we can do this is if $t(\tau) = \infty$ for some finite $\tau$. A rather simple way to do this is to pick the acceleration to be $\tan(\tau)$, which diverges for $\tau = \pm \pi / 2$.

$$\dot t(\tau) = \sinh(-\ln(\cos(\tau)))$$ $$\dot x(\tau) = \cosh(-\ln(\cos(\tau)))$$The time coordinate obviously diverges for $\tau = \pi/2$, it can be seen from the actual expression

\begin{eqnarray} t(\tau) = \int \sinh(-\ln(\cos(\tau))) d\tau = \frac 12 (- \sin (\tau) - \ln( \cos(t/2) - \sin(t/2) )) + \ln( \cos(t/2) + \sin(t/2) ) \end{eqnarray}which simply becomes $-\ln(0)/2$ at $\tau = \pm \pi / 2$.

We can prove this in a more rigorous manner by proving that the curve defined on $(-\pi / 2, \pi / 2)$ is inextendible, by showing that it has no endpoint. If the curve had an endpoint $p$, for every neighbourhood of this point there would be a proper time $\tau'$ for which every further point $\gamma(t)$, $t > t'$ belongs to that neighbourhood. Consider the neighbourhood formed by $(t_a, t_b) \times (x_a, x_b)$. Since the function $t(\tau)$ is strictly increasing and divergent, there will always be points for $\tau > \tau'$ do not belong to this interval.

The proper time is then simply of $\pi$, which is indeed finite. This observer would then only live for a finite time in an inextendible curve. As this case isn't particularly physical (it's in fact very reminescent of the space invader scenario), we will ask that all timelike curves considered be of bounded acceleration.

Even with this condition, it is still possible to find geodesically complete spacetimes with observers of finite proper time. There are two constructions of such a spacetime, due to Geroch[2] and Beem[3]. Both of them involve some sort of "geodesic trap" in their construction, regions of space that attract all geodesics but leave accelerated observers free to leave. Due to that infinite stacking of geodesic traps, themselves fairly elaborate, they are perhaps the worst spacetimes I've ever seen describes.

## 2. The Geroch construction

The Geroch example, a 2-dimensional spacetime, is made by stacking together an infinite amount of copies of the geodesic trap mentionned. This trap is composed of 4 layers, called the region $A$, $B$, $C$ and $D$. Each stack of the trap is progressively shorter in some sense, so that an accelerated observer, after avoiding all geodesic traps, will finally crash on a singularity at the end of the stack in finite time.

### Region $A$

Region $A$ is the 2-dimensional covering space of anti-de Sitter space, up to some factor depending on the stack we consider.

$$ds^2 = (\frac{1}{2})^{2n} [-(1 + x^2) dt^2 + \frac{1}{(1+x^2)} dx^2]$$The point of region $A$ is that anti de Sitter space has the property of focusing timelike geodesics. Any timelike geodesic crossing the point $x$ at the time $t_0$ will go through it again at the time $t_0 + \pi$. Because of this, region $A$ will last for a time interval of exactly $\pi$. To show this, the non-zero Christoffel symbols in those coordinates are

$${\Gamma^x}_{xx} = - \frac{x}{1+x^2},\ {\Gamma^x}_{tt} = x + x^3, {\Gamma^t}_{xt} = {\Gamma^t}_{tx} = \frac{x}{1+x^2}$$So that the geodesic equation is

\begin{eqnarray} \ddot{x}(\tau) &=& \frac{x}{1+x^2} \dot{x}^2 - \dot{t}^2 (x + x^3)\\ \ddot{t}(\tau) &=& -2 \frac{x}{1+x^2} \dot{x} \dot{t}\\ \end{eqnarray}For timelike geodesics, $g(u,u) = -1$, we have the equality

$$(\frac{1}{2})^{2n} ( \frac{1}{(1+x^2)} \dot{x}^2 -(1 + x^2) \dot{t}^2) = -1$$And as the metric tensor is independant of the $t$ coordinate, there exists a Killing vector $\xi = \partial_t$, such that $g(\xi, u)$ is a constant,

$$(\frac{1}{2})^{2n} (1 + x^2) \dot{t} = E $$or

$$ \dot{t} = (\frac{1}{2})^{-2n} \frac{E}{(1 + x^2)} $$This gives us

$$\dot{x}^2 = -(\frac{1}{2})^{-2n} (1+x^2) + (\frac{1}{2})^{-4n} E^2 $$ \begin{equation} \ddot{x}(\tau) = -(\frac{1}{2})^{-2n} x \end{equation}The solution is just

$$x(\tau) = A \sin(2^n \tau - \tau_0)$$ The solution is periodic in $\tau$ of period $2\pi / 2^n$ $$\ddot{t}(\tau) &=& -2 \frac{x}{1+x^2} \dot{x} \dot{t}$$### Region $B$

Region $B$ is a transition between region $A$ and region $C$, region $C$ being simply Minkowski space at its interface with region $B$. We will then need a smooth transition function $f$ so that $f(x) = 0$ for $x < 0$ and $f(x) = 1$ for $x > 1$, so that if we want region $B$ to last for a time interval of some constant $b$, our metric becomes

\begin{eqnarray} ds^2 &=& f(\frac{t}{b} - 3) (\frac{1}{2})^{2n} [-(1 + x^2) dt^2 + \frac{1}{(1+x^2)} dx^2]\\ &+& [1 - f(\frac{t}{b} - 3)] (\frac{1}{2})^{2n} [-dt^2 + dx^2] \end{eqnarray}### Region $C$

Region $C$ is conformally related to Minkowski space, with two removed regions, for instance we can remove the closed sets defined by $x \geq b$, $t \in [-2b, 2b]$ and $x \leq b$, $t \in [-2b, 2b]$, leaving a narrow channel between them. The conformal factor $\Omega(x,t)$ has the following properties :

- It is invariant under both space and time reversal : $\Omega(x,t) = \Omega(-x,t) = \Omega(x,-t)$
- It reduces to the metric of regions $B$ and $D$ on its boundaries : $\Omega(x, \pm 3b) = 2^{-2n}$, and all derivatives of the conformal factor also vanish on those boundaries.
- The center is guaranteed to be close to Minkowski space : $\Omega(0, t) = 2^{-2n}$
- To avoid any obvious singularity, $\Omega \to \infty$ as it approaches the boundary of the removed regions, so that any geodesic approaching them will be of infinite length

### Region $D$

Region $D$ is a transition between region $C$ and another version of region $A$, except slightly translated by an amount $c$ and with a different factor. \begin{eqnarray} ds^2 &=& f(- \frac{t}{b} - 3) (\frac{1}{2})^{2(n-1)} [-(1 + (x - c)^2) dt^2 + \frac{1}{(1+(x - c)^2)} dx^2] \\ &+& [1 - f(- \frac{t}{b} - 3)] (\frac{1}{2})^{2n} [-dt^2 + dx^2] \end{eqnarray}### The stack

For a given value of $n$, the spacetime made by stacking together $D$, $C$, $B$ and $A$ will be called $M_n$. The exact structure of $M_n$ is the following : $M_n$ is defined by three values, $n$, $b_n$ and $c_n$. The stack starts with region $D$, between $t = -4b_n$ and $t = -3b_n$, followed by the region $C$ between $t = -3b_n$ and $t = 3b_n$, then rebion $B$ between $t = 3b_n$ and $t = 4b_n$, and finally region $A$ between $t = 4b_n$ and $t = 4b_n + \pi$.

The behaviour of geodesics in the various layers of $M_n$

The final spacetime we want to make is composed by the lower half ($t < 0$) of anti-de Sitter space, which we'll call $M_0$, and then we'll stack every $M_i$, $i \in \mathbb N$, so that the coordinates $(x, t = 4 b_n)$ of $M_n$ are identified with the coordinates $(x + c_n, t = 4 b_{n-1} + \pi)$ of $M_{n-1}$.## 3. The Beem construction

The Beem spacetime has the topology of a subset of $\mathbb R^2$ with the metric of the form

$$ds^2 = -H(x,t) dxdt$$### Blocking sets

## Bibliography

Last updated :

*2018-11-14 10:07:16*