Did the Big Bang happen at a point?

Question

TV documentaries invariably show the Big Bang as an exploding ball of fire expanding outwards. Did the Big Bang really explode outwards from a point like this? If not, what did happen?

Answer 1

The simple answer is that no, the Big Bang did not happen at a point. Instead, it happened everywhere in the universe at the same time. Consequences of this include:

• The universe doesn't have a centre: the Big Bang didn't happen at a point so there is no central point in the universe that it is expanding from.

• The universe isn't expanding into anything: because the universe isn't expanding like a ball of fire, there is no space outside the universe that it is expanding into.

In the next section, I'll sketch out a rough description of how this can be, followed by a more detailed description for the more determined readers.

A simplified description of the Big Bang

Imagine measuring our current universe by drawing out a grid with a spacing of 1 light year. Although obviously, we can't do this, you can easily imagine putting the Earth at (0, 0), Alpha Centauri at (4.37, 0), and plotting out all the stars on this grid. The key thing is that this grid is infinite$^1$ i.e. there is no point where you can't extend the grid any further.

Now wind time back to 7 billion years after the big bang, i.e. about halfway back. Our grid now has a spacing of half a light year, but it's still infinite - there is still no edge to it. The average spacing between objects in the universe has reduced by half and the average density has gone up by a factor of $2^3$.

Now wind back to 0.0000000001 seconds after the big bang. There's no special significance to that number; it's just meant to be extremely small. Our grid now has a very small spacing, but it's still infinite. No matter how close we get to the Big Bang we still have an infinite grid filling all of space. You may have heard pop science programs describing the Big Bang as happening everywhere and this is what they mean. The universe didn't shrink down to a point at the Big Bang, it's just that the spacing between any two randomly selected spacetime points shrank down to zero.

So at the Big Bang, we have a very odd situation where the spacing between every point in the universe is zero, but the universe is still infinite. The total size of the universe is then $0 \times \infty$, which is undefined. You probably think this doesn't make sense, and actually, most physicists agree with you. The Big Bang is a singularity, and most of us don't think singularities occur in the real universe. We expect that some quantum gravity effect will become important as we approach the Big Bang. However, at the moment we have no working theory of quantum gravity to explain exactly what happens.

$^1$ we assume the universe is infinite - more on this in the next section

For determined readers only

To find out how the universe evolved in the past, and what will happen to it in the future, we have to solve Einstein's equations of general relativity for the whole universe. The solution we get is an object called the metric tensor that describes spacetime for the universe.

But Einstein's equations are partial differential equations, and as a result, have a whole family of solutions. To get the solution corresponding to our universe we need to specify some initial conditions. The question is then what initial conditions to use. Well, if we look at the universe around us we note two things:

1. if we average over large scales the universe looks the same in all directions, that is it is isotropic

2. if we average over large scales the universe is the same everywhere, that is it is homogeneous

You might reasonably point out that the universe doesn't look very homogeneous since it has galaxies with a high density randomly scattered around in space with a very low density. However, if we average on scales larger than the size of galaxy superclusters we do get a constant average density. Also, if we look back to the time the cosmic microwave background was emitted (380,000 years after the Big Bang and well before galaxies started to form) we find that the universe is homogeneous to about $1$ part in $10^5$, which is pretty homogeneous.

So as the initial conditions let's specify that the universe is homogeneous and isotropic, and with these assumptions, Einstein's equation has a (relatively!) simple solution. Indeed this solution was found soon after Einstein formulated general relativity and has been independently discovered by several different people. As a result the solution glories in the name Friedmann–Lemaître–Robertson–Walker metric, though you'll usually see this shortened to FLRW metric or sometimes FRW metric (why Lemaître misses out I'm not sure).

Recall the grid I described to measure out the universe in the first section of this answer, and how I described the grid shrinking as we went back in time towards the Big Bang? Well the FLRW metric makes this quantitative. If $(x, y, z)$ is some point on our grid then the current distance to that point is just given by Pythagoras' theorem:

$$ d^2 = x^2 + y^2 + z^2 $$

What the FLRW metric tells us is that the distance changes with time according to the equation:

$$ d^2(t) = a^2(t)(x^2 + y^2 + z^2) $$

where $a(t)$ is a function called the [scale factor]. We get the function for the scale factor when we solve Einstein's equations. Sadly it doesn't have a simple analytical form, but it's been calculated in answers to the previous questions What was the density of the universe when it was only the size of our solar system? and How does the Hubble parameter change with the age of the universe?. The result is:

The value of the scale factor is conventionally taken to be unity at the current time, so if we go back in time and the universe shrinks we have $a(t) < 1$, and conversely in the future as the universe expands we have $a(t) > 1$. The Big bang happens because if we go back to time to $t = 0$ the scale factor $a(0)$ is zero. This gives us the remarkable result that the distance to any point in the universe $(x, y, z)$ is:

$$ d^2(t) = 0(x^2 + y^2 + z^2) = 0 $$

so the distance between every point in the universe is zero. The density of matter (the density of radiation behaves differently but let's gloss over that) is given by:

$$ \rho(t) = \frac{\rho_0}{a^3(t)} $$

where $\rho_0$ is the density at the current time, so the density at time zero is infinitely large. At the time $t = 0$ the FLRW metric becomes singular.

No one I know thinks the universe did become singular at the Big Bang. This isn't a modern opinion: the first person I know to have objected publically was Fred Hoyle, and he suggested Steady State Theory to avoid the singularity. These days it's commonly believed that some quantum gravity effect will prevent the geometry from becoming singular, though since we have no working theory of quantum gravity no one knows how this might work.

So to conclude: the Big Bang is the zero time limit of the FLRW metric, and it's a time when the spacing between every point in the universe becomes zero and the density goes to infinity. It should be clear that we can't associate the Big Bang with a single spatial point because the distance between all points was zero so the Big Bang happened at all points in space. This is why it's commonly said that the Big Bang happened everywhere.

In the discussion above I've several times casually referred to the universe as infinite, but what I really mean is that it can't have an edge. Remember that our going-in assumption is that the universe is homogeneous i.e. it's the same everywhere. If this is true the universe can't have an edge because points at the edge would be different from points away from the edge. A homogenous universe must either be infinite, or it must be closed i.e. have the spatial topology of a 3-sphere. The recent Planck results show the curvature is zero to within experimental error, so if the universe is closed the scale must be far larger than the observable universe.

Answer 2

My view is simpler and observational.

Observations say that the current state of the observable universe is expanding: i.e. clusters of galaxies are all receding from our galaxy and from each other.

The simplest function to fit this observation is a function that describes an explosion in four-dimensional space, which is how the Big Bang came into our world.

There are experts on explosive debris that can reconstruct the point where the explosion happened in a three-dimensional explosion. In four dimensions the function that describes the expansion of space also leads to the conclusion that there is a beginning of the universe from which we count the time after the Big Bang.

The BB model has survived, modified to fit the observation of homogeneity (quantum fluctuations before 10^-32 seconds) and the observation that the expansion we measure seems to be accelerating (the opening of the cone in the picture)

^Source

Note that in the picture the "Big Bang" zero points is "fuzzy". That is because, before 10^-32 seconds where it is expected that quantum mechanical effects dominate, there is no definitive theory joining both general relativity and quantum mechanics. There exists an effective quantization of gravity but the theory has not come up with a solid model.

Thus extrapolating with a mathematical model -- derived from completely classical equations -- to the region where the "origin" of the universe was where we know a quantum mechanical solution is necessary, is not warranted.

Take the example of the potential around a point charge. The classical electrodynamic potential goes as $\frac{1}{r}$, which means that at $r=0$ the potential is infinite. We know though that, at distances smaller than a Fermi, quantum mechanical effects take over: even though the electron is a point charge, no infinities exist. Similarly, one expects that a definitive quantized gravity unified with the other forces model will avoid infinities, justifying the fuzziness at the origin shown in the picture of the BB.

In conclusion, in the classical relativistic mechanics' solution of the Big Bang, there was a "beginning point singularity" which as the universe expanded from the four-dimensional explosion, is the ancestor in the timeline of each point in our present-day universe. The surface of a balloon analogy is useful: the points of the two-dimensional surface can be extrapolated to an original "point" when the blowing expansion starts, but all points were there at the beginning.

The need for a quantum mechanical solution for distances smaller than 10^-32 demanded from the extreme homogeneity of the Cosmic Microwave Background radiation confirms that quantum mechanical effects are needed for the beginning, which will make the beginning fuzzy. Physicists are still working on the quantization of gravity to extrapolate to what "really happened".

Addendum by Gerold Broser

There are two further illustrations:

• Nature Research: Box: Timeline of the inflationary Universe, nature.com, April 15, 2009

_{(source: nature.com)}

• Kavli Institute for Particle Astrophysics and Cosmology (KIPAC): Inflation, Stanford University, July 31, 2012

Edit since a question has been made a duplicate of the above:

Was the singularity at the big bang a black hole? [duplicate]

Black hole singularities come from the solutions of general relativity, and in general describe very large masses which distort spacetime , and have a horizon, after which nothing comes out and everything ends up on the singularity, the details depending on the metric used . You see above in the history of the universe image that the description in the previous sentence does not fit the universe. Galaxies and clusters of galaxies are receding from each other which led to the Big Bang model, and what more, the expansion is accelerating as seen in the image.

So the Big Bang mathematics do not follow the black hole mathematics.

Answer 3

The answer is that we don't know. Why? Because the theory of gravity which we have and use, GR, has a singularity. Things which should be finite in a physical theory, like the density, become infinite. And theories with a singularity are simply wrong, they need a modification, and this modification is necessary not only at the singularity itself, but already in some environment of this singularity.

Moreover, we already know for independent reasons that a modification is necessary: Because if one looks at times $10^{-44}$ s after the singularity, quantum gravity becomes important, which is an unknown theory.

And we have also empirical evidence that the most trivial model based on well-established theories (GR with SM for matter) fails: It is the so-called horizon problem. It requires, for its solution, some accelerated expansion in the very early universe. One can propose models which lead to such an expansion based on particle theory, theories usually named "inflation" (imho very misleading, as I explain here), but they usually use speculative extensions of the SM like GUTs, supersymmetry, strings and so on. So, even the details of a particle theory which would give inflation are unknown.

So, while big bang theory is well established, if one thinks about everything having been as dense as inside the Sun, and I would say reliable if as dense as inside a neutron star, we have not much reason to believe that the theories remain applicable for much higher densities, and certainly not for the density becoming infinite.

From a purely mathematical point of view one cannot tell anything about the singularity itself too. If one considers, for example, the metric in the most usual FLRW coordinates $ds^2=d\tau^2-a^2(\tau)(dx^2+dy^2+dz^2)$, then the singularity would be a whole $\mathbb{R}^3$. The limit of the distance between the points would be zero (which is the reason why one usually prefers the picture with a point singularity). On the other hand, the limit of what one point which moves toward the singularity can causally influence in its future remains (without inflation) a small region, which in no way tends to cover the whole universe, which corresponds much better with a whole $\mathbb{R}^3$ space singularity.

Answer 4

In addition to what the others have said, let me explain a simple analogy for the expansion of the universe.

Consider a balloon, the surface of which is considered as the universe. Let's draw dots on the balloon which symbolize galaxies. Now, blow the balloon. All the galaxies will start to separate from each other. Now suppose you are on one of the galaxies. You will observe all the galaxies moving away from you, and you would be led to the conclusion that you are on the center of the universe. This is what every galaxy would observe. Thats why there is no center for the universe's expansion.

I hope you enjoyed my analogy.

Answer 5

The explosion that you have seen is actually 4 dimensional representation of the universe. If we are representing universe in 4D then big bang had happened at a point and is expanding as a hollow sphere. But in 3D the big bang should have happened in every point of the universe and is expanding into every direction. This interpretation is using Friedman model of universe.

Answer 6

[Editorial note: This answer was meant to be a comment to @good_ole_ray's comment to John Rennie's answer but the 600 characters comment limit...you know.]

Re "galaxies seem to be moving away from a common center"

"common center" is more appropriate than one may think at first sight.

Sure, it's not that kind of center 99 % of the people understand as such: a single point surrounded by other points with the outermost points in ideally equal distance to the center, i.e. things known as sphere, ball, orb, globe or bowl, hollow or not doesn't matter.

The center I'm talking about here is so 'common', in the meaning of 'joint', because all existing points in our universe are this center.

It's easier to understand if one imagines the young universe, rather tiny at the beginning. It then looked more like a point as we know it from our day-to-day life.

But it evolved, it expanded and it expanded in a way that between any two points (or units of space) another point (or unit of space) arose. Such "pushing" the former two points (or units of space) apart from each other.

And this happens since 13.7 bn years, at any point of the universe so that the points that was one once are many now. Or in other words: any of the points is now far, far away from each of the other points that were at the same position once. But they're still the center because they once were the center. This, their, property hasn't changed because they didn't move because of a proper motion but because new space arose between them.

And why is this? Because the Big Bang wasn't an explosion in the common sense. Since there was no space into which something could have exploded into. Space, and time, for that matter, only started to exist with the Big Bang.

It also happens slowly on a small scale. The latest value of the Hubble parameter is $71_{-3.0}^{+2.4} \frac {km}{Mpc \cdot s}$ which is rather small on a small scale (if one considers an AU [~150m km] to be small – but compared to astronomical dimensions that's even tiny anyway):

$$ 1 \space Mpc = 3.09 \cdot 10^{22} \space m $$ $$ 1 \space AU = 1.5 \cdot 10^{11} \space m $$

So the (theoretical) increase of the average distance between sun and earth due to the universe's expansion can be calculated to

$$ v_{\Delta{AU}} = 3.44 \cdot 10^{-7} \space \frac {m}{s} = 10.86 \space \frac {m}{yr}. $$

But since that happened for such a long time the former small scale became large scale everywhere but in the vicinity of our galaxis (or, to be precise: in the vicinity of any [subjective] observation point in the universe). And, be aware of that this applies only to space itself. It does not mean that the earth drifts actually away from the sun, or that you move away from your beloved ones constantly, and vice versa. Remember, there's gravity, the weakest of the four fundamental interactions according to its factors

$$ m_1 \cdot m_2 \cdot \frac {1}{r^2} $$

but the most relentless when it comes to masses.

"galaxies seem to be moving away from a common center" also isn't true for all galaxies observed from any observation point. The spectral lines of the Andromeda galaxy, for instance, are blueshifted. That means it is close enough to us that it's proper motion towards us is greater than the drift away from us caused by the universe's expansion:

                 Andromeda ($ 300 ± 4 \frac {km}{s} $)
                ←----------------------------  
     ⊙
                                                  ␣  ---→
                 Expansion speed at 2.5m ly, the distance of Andromeda (~$ 54.42 \frac {km}{s} $)

Legend:   - ≙ $ 10 \frac {km}{s} $

[Final editorial note: Well, this was a little bit more than 600 characters.]

P.S.: @good_ole_ray I hope you have the opportunity to read this before it is flagged as not appropriate, or even worse, because it doesn't really address the original question.

Answer 7

Our best theory for modelling cosmology is GR. Now, the equations of GR support either a bounded or unbounded universe. To decide between them would mean setting certain boundary conditions.

Einstein himself originally chose an unbounded, static universe because that he felt reflected the cosmological assumptions at the time: space is infinite and barely changes. To achieve this, in 1917 barely two years after duscovering GR, he introduced a new term into GR, the cosmological constant. This produced a cosmological pressure that counteracted gravity resulting in a static universe.

Friedmann, however, in 1922, assuming the homogeneity and isotropy of space, showed that then GR implied that the spatial metric must have constant curvature, and so was either a sphere (the 3d surface of a 4d ball), a hyperbolic space or flat. The latter two spacetimes are unbounded but the first is bounded. He also showed that these spacetimes were dynamic and so either contracting or expanding in time, or some combination of the two and derived an equation for the scale factor. Einstein, however, was unwilling to accept Friedmann's vision of an evolving universe and dismissed his work.

Now, in 1912 Vesto Slipher had discovered that the light from galaxies were red-shifted implying that they were all receding from the stand-point of earth and at varying speeds. At that time, they were not known to be galaxies and in fact, the entire universe was thought to just consist of the Milky Way. There had been earlier suggestions that the universe may be much larger than supposed, primarily by Kant who published such a supposition in 1755 in his General History of Nature and Theory of the Heavens.

It was Hubble, by calibrating distances using Cepheid variables, who showed a decade later that these astronomical bodies were much too far away to be part of the Milky Way and were galaxies in their own right. Suddenly, the universe had grown a great deal larger. And then in 1929, combining his observations with that of Slipher derived what was once called the Hubble law, but is now called the Hubble-Lemaitre law, linking together the distance of a star from earth and the amount of red-shift its light had shifted by.

It turned out that Hubble's discovery had already been anticipated by a Belgian priest and theoretical physicist, Lemaitre, two years earlier in 1927 in his paper, A Homogeneous Universe of Constant Mass and Increasing Radius Accounting for the Radial Velocity of Extragalactic Nebulae. In this work, Lemaitre expanded upon Friedmann's cosmology, though his work was done independently, in essence by choosing the expanding spherical Friedmann metric. Einstein, still holding on to his vision of a static universe, also dismissed this work too, saying "your calculations are correct but your physics is atrocious". It was to Lemaitres theory, especially after he also theorised 'a primeval atom' from which the universe sprang from, that Fred Hoyle dismissively called 'the Big Bang Theory', a name that stuck.

Now, at the time of the Big Bang, all distances shrink to zero and Lemaitre's spherical universe shrinks to a point, a point of infinite density and temperature. Its easy to see in this picture that tge Big Bang happened everywhere all, all at once simply because everywhere is just a point. Moreover, this also is suggestive of spacetime itself being 'created'. Whilst Lemaitre himself chose a closed universe - the surface of a sphere, Friedmann showed that an open universe was possible, either flat or a hyperbolic hyperboloid. Is a Big Bang here also possible, a time when distances between points approached zero and where density and temperature approached infinity? Well, yes: take an infinite expanse of space with a certain fixed density of mass and halve the distances, then the density will cube. By iteration, we see the density rapidly increase towards infinity. Thus, even in an open universe, where spacetime extends out infinitely, it is possible to have a Big Bang. In this case, it began everywhere, and all at once.

But what does this mean for the topology of spacetime? Have we somehow squeezed an infinite spacetime into a point? No. There is a topological property called compactness that doesn't rely on a metric (sometimes called a geometry, because to do geometry requires measuring distances and angles and it is precisely a metric that enables this). A sphere is compact but both hyperbolic hyperboloid and flat space are non-compact. However, at the time of the Big Bang, or to be more precise, as we approach it, the distances between all points approach zero. So geometrically, it looks as though this spacetime approaches a point, but in fact, it does not. No matter how close the points are, if we go out far enough, which we can in an open spacetime, we will find the distances between points become appreciable.

It's only at the time of the Big Bang does the metric go to zero and states that all points have zero distance between them. And so is geometrically a point, whilst at the same time being non-compact. This is bizarre. And what it really points to is the likelihood of new physics here. Moreover, we should recall GR does not deal with non-degenerate metrics. In fact, metrics by definition are non-degenerate.

Answer 8

The short and simple answer is: no, absolutely not. It's a 3D hypersurface.

The reason is simple: the past light cones of different space-time points have different intersections with the initial hypersurface, thereby precluding any possibility of identifying all of the points of the initial hypersurface as being one and the same point.

Instead, the initial hypersurface is a three-dimensional surface on which the space-time metric reduces to time-like rank 1, and the contravariant metric reduces - up to conformal equivalence - to space-like rank 3. Another way of describing this is:

The initial hypersurface is a 3-D slice of Newton-Cartan Geometry.

Or, to put it a little bit differently, and more directly, taking $t = 0$ as the time of the initial hypersurface:

The $t → 0$ limit is a physical realization of the Galilean limit.

The hypersurface is an envelope of light cones - namely of all the past light cones. They are all tangent to the hypersurface. So, it is a null surface. The reason the in-surface metric is zero is because distances are being measured, with this metric, in light speed units and light speed on it is infinity.

So, you have an infinite case of the King's Thumb Problem. That doesn't mean that points are all "next to each other". It just means that it's a null surface and that its internal geometry is no longer connected to the geometry given by the space-time metric.

It's the very same situation, by that way, that happens with space-like surfaces in Newton-Cartan geometry.

For simplicity, assume that the underlying metric is an FRW metric and that its spatial part is Euclidean - which is true to a high degree of accuracy. It is actually still an unresolved problem whether the spatial sections of FRW have non-zero curvature and, if non-zero, whether it is positive or negative. But it doesn't matter, because the observations to follow apply to all three examples of the universe (positive, zero, or negative curvature). It's just easier to describe in the Euclidean case.

The metric then has the form: $$ds^2 = dt^2 - A(t) \left(dx^2 + dy^2 + dz^2\right),$$ where it is presented as a metric for proper time, denoted $s$; such that $A(t) > 0$ for $t > 0$ and $A(0) = 0$. The term $A(t)$ is called the growth factor.

This is the metric in the co-moving frame. The light cone of any space-time point $\left(x_0,y_0,z_0,t_0\right)$ is the locus of all curves $\left(x(t),y(t),z(t),t\right)$, such that $$|𝐯(t)| = c(t) ≡ \frac{1}{\sqrt{A(t)}},\quad 𝐯(t) = \frac{d𝐫(t)}{dt},\quad 𝐫(t) = \left(x(t),y(t),z(t)\right),$$ such that $$𝐫\left(t_0\right) = \left(x_0, y_0, z_0\right).$$ That is: for all intents and purposes, the growth factor is just $A(t) = 1/c(t)^2$ and $c(t)$ is a time-variable light speed.

This works out consistently with the fields, such as the electromagnetic field. So, you end up getting a constitutive law $𝐃 = ε(t) 𝐄$ and $𝐁 = μ(t) 𝐇$ for the Einstein-Maxwell-Lorentz Lagrangian density $𝔏 = -¼ ε_0 c g^{μρ}g^{νσ} \sqrt{|g|} F_{μν} F_{ρσ}$, with $ε(t) = ε_0 \sqrt{A(t)}$, $μ(t) = μ_0 \sqrt{A(t)}$ and $ε(t)μ(t) = A(t) = 1/c(t)^2$, provided we normalize $A(t) = 1$ at the current time $t = \text{now}$.

So, it really is $A(t) = 1/c(t)^2$ ... and $c(0) = ∞$.

The question of what happens to the past light cones as they approach $t = 0$ depends on how $A(t)$ scales in the vicinity of $t = 0$. If it goes as $A(t) ≃ t^γ$, then for $γ > 2$ the weak-energy principle is violated, and for $γ = 2$ the light cones splay out to infinity.

Only for $γ = 2$ - which is borderline weak-energy-violating - can you consistently assert that all of the $t = 0$ surface is one point.

For $γ < 2$, the light cones land on the boundary compact subset of the initial hypersurface - a strict subset. In fact, each space-time point in the $t > 0$ sector will become associated with a unique such subset. So, you could effectively treat the entire $t > 0$ side of the universe as just a family of spheres on the initial hypersurface and the universe - itself - as a holographic projection of the hypersurface: one $t > 0$ space-time point per hypersurface sphere.

Edit: The sphere, on the initial hypersurface, for the past lightcone of the space-time point where you are at, now, is none other than the outermost, $t = 0$, sphere of the sky at your location and time. The sky is, itself, your past light cone. What lies beyond that - based on a literal reading of the FRW metric - is what I'm about to describe.

The case of greatest interest is where $A(t)$ becomes radiation dominant near $t = 0$. For that case, $γ = 1$ and the metric becomes: $$ds^2 = dt^2 - t \left(dx^2 + dy^2 + dz^2\right),$$ near $t = 0$.

This geometry - and by extension: any instance of the FRW metric that has this geometry asymptotically near $t = 0$ - has the following properties:

• It is signature changing at $t = 0$, the $t < 0$ sector being a 4D Euclidean time-less space. Thus, it is not a pseudo-Riemannian geometry at all. Only the $t > 0$ sector is.
• It is geodesically complete! However, the property of [pseudo-]Riemannian geometries that a geodesic be uniquely determined by a tangent breaks down at the $t = 0$ hypersurface.
• For curves in the $t = 0$ hypersurface, "geodesic" has to be defined in the limit, i.e. the envelope time-like and/or null geodesics on the $t > 0$ side; since the in-surface part of the metric is zero at $t = 0$.
• For $t > 0$, the null-geodesics reflect parabolically off the $t = 0$ surface and reverse direction in time.
• Past light cones reverse direction and splay out to infinity at $t = 0$, in a shape like a pointed Mexican hat.
• The future light cones of each space-time point at $t > 0$ is contained inside its past light cone.
• Timelike geodesics in the co-moving frame pass through the $t = 0$ surface into the Euclidean $t < 0$ sector.
• Timelike geodesics not in the co-moving frame reflect as catenary curves off the $t = 0$ hypersurface and reverse direction in time.
• Each time-like worldline has a past-directed extension to a time-like worldline that intersects itself in the future of any given point on the timeline.
• The space-like geodesic worldlines near $t = 0$, on both the $t > 0$ and $t < 0$ side have sinusoidal variation in $t$, but do not intersect the $t = 0$ hypersurface.