# Minkowski spacetime conformal infinity: why not allow the full range of $R$?

Let $(\mathbb{R}^4,\eta)$ be Minkowski spacetime. We want to describe infinity as a place in some bigger manifold containing Minkowski spacetime. The idea is to work with null incoming/outgoing geodesics and compactify along these, to preserve the causal structure.

So define the advanced/retarded null coordinates $u=t-r$ and $v=t+r$. These have ranges $(-\infty,\infty)$ both with restriction $u\leq v$.

Then we define $U = \arctan u$ and $V=\arctan v$ both with ranges $(-\pi/2,\pi/2)$ and $U\leq V$.

We finaly define $$T=V+U,\quad R=V-U.$$

These have ranges $0\leq R<\pi$ and $|T|+R<\pi$. These coordinates with these ranges still just describe Minkowski spacetime, albeit in a rather awkward coordinate system.

Then we want to add the missing points describing infinity. So the obvious idea would be to extend the ranges of the coordinates allowing them to describe a bigger manifold and picture Minkowski spacetime as the submanifold defined by the above constraints inide of it.

Now the canonical way to do it is to extend $T$ to cover $(-\infty,\infty)$ and to extend $R$ to cover $0\leq R\leq \pi$. In this way the bigger manifold is $\mathbb{R}\times S^3$.

Now, why is that? If we further allow $R$ to run from $(0,\infty)$ or even $(-\infty,\infty)$ we still get a bigger manifold, properly containing the previous one, which in turn properly contains Minkowski spacetime.

There is certainly a reason behind this choice, but I can't see it.

• This is a good question, but I want to nitpick some aspects of its statement. It focuses a lot on the details of the four-dimensional construction of a Penrose diagram in spherical coordinates, but I don't think there are actually any issues here that couldn't be addressed in the simpler 1+1 context. It doesn't distinguish manifolds from manifolds-with-boundary, or extensions from maximal analytic extensions. – Ben Crowell Aug 11 '18 at 2:28
• The question also never mentions the metric, which makes the topic vacuous. In the absence of a metric, Minkowski space can be embedded in any 4-manifold. For instance, the Penrose diagram of the Einstein static universe does sort of look like the bigger extension the OP seems to have in mind. – Ben Crowell Aug 11 '18 at 2:33
• @BenCrowell thanks for the comments. After thinking quite a lot about then I think I've understood the point. I've added one answer with what I found to be the answer to my question. Could you point me out if I understood it correctly? Thanks for the help! – user1620696 Sep 4 '18 at 16:06

## 3 Answers

Basically, you want to have a handle to speak of the limiting points "at infinity" and not add some disconnected copies or foliations of Minkowski space. In other words, the extra space you are adding by going from $R<\pi$ to $R\leq \pi$ can be properly thought of as limiting points of the original space, while going to $R<\infty$ adds a lot of extra space that cannot easily be mapped back to the original coordinates.

In terms of paths, an outgoing light ray in Minkowski space would asymptotically approach $R=\pi$, but will not reach any value $R>\pi$.

Update: You can intuitively think of the limiting points of Minkowski space as those points where $t$ or $r$ "become infinite". (Note that the angular coordinates do not play much of a rôle here, so there could be basically any number of those.) Given that the original metric is $\text{d} s^2=-\text{d} t^2+\text{d} r^2+\dots$, these boundary points can be infinitely far away in space, $\int\text{d} s^2\to\infty$, or time, $\int\text{d} s^2\to-\infty$, or infinitely far out in affine distance along null geodesics. The conformal diagram map the infinite ranges of $t$ and $r$ to finite open intervals (e.g. of $T$ and $R$), and by adding the endpoints we have a well-defined way to talk about the "points at infinity" and how they relate to the causal structure of spacetime.

• This is a good answer. It might be helpful to point out how it implicitly deals with the metric, which was a missing ingredient in the question. In the first paragraph, the reason we would think of the boundary points $i^0$, $i^+$, and $i^-$ as the limiting points of the original space is that the metric distance to such points is infinite. In the second paragraph, what defines an outgoing light ray (as opposed to any other path) is the metric, and points at $\mathscr{I}^+$ lie at infinite affine distance due to the metric. – Ben Crowell Aug 11 '18 at 2:37

For starters, the metric tensor would become infinite for $|U|,|V|=\frac{\pi}{2}$, so OP's coordinate extension is not allowed.

• I've noticed that, but this seems to be a result of the conformal factor vanishing on null infinity. I don't see how to deal with it though. How does one correctly extend spacetime to add conformal infinity ? – user1620696 Aug 10 '18 at 19:04

After the comments of @BenCrowell I've thought better about the question and I believe I've found the answer myself. I'm posting my conclusions. If it is wrong in some aspect I'd like to be warned in comments.

The "motivation" for all of this would be to realize infinity as a "place". The idea of infinity depends then on how we approach it: through timelike lines, lightlike lines or spacelike lines. Which one we choose obviously would depend on the problem at hand.

To analyze for instance radiation carried to infinity by a massless Klein-Gordon field we would need to approach infinity through light-like lines. To state assymptotic in/out states with massless particles we would also need to approach infinity through lightlike lines and so forth.

In the case of approaching through light-like lines, heuristically speaking it becomes clear that this "place" would be reached by following null geodesics until we reach their endpoints. These geodesics fall into two categories: the ingoing and outgoing. To describe then we introduce coordinates $u = t-r$ and $v=t+r$ which respectively have the meanings:

1. $u$ at an event is the coordinate time at which an observer at the origin emits light that reaches that event.
2. $v$ at an event is the coordinate time at which an observer at the origin perceives light emitted far away that is detected at that event.

The ingoing lightlike geodesics are then $v$ constant lines and the outgoing lightlike geodesics are then $u$ constant lines.

The "place" we would like to define as the infinity for lightlike directions is the endpoints of these geodesics.

We face one issue however. Minkowski spacetime is geodesically complete. This means that the inextendible geodesics already contain all possible endpoints. So even if we "pull infinity closer" by transforming the coordinates to $U=\arctan u$ and $V = \arctan v$ the endpoints corresponding to $U,V=\pm \pi/2$ cannot be added to Minkowski spacetime.

This reflects on the metric. As pointed out in comments, the metric is essential for this discussion. In $U,V$ coordinates the metric becomes

$$ds^2=\dfrac{1}{\cos^2 U\cos^2 V}\left(dUdV-\frac{1}{4}\sin^2(U-V)(d\theta^2-\sin^2\theta d\phi^2)\right)$$

It is clear that at the so sought endpoints of these geodesics $ds^2$ blows up as a reflection that Minkowski spacetime is already geodesically complete and hence inextendible as a pseudo-Riemannian manifold.

Physically this would reflect the fact that our transformation performed a change in scale to bring infinity closer, but at infinity there should be one infinite amount of stretching, since after all, it was originally infinitely far away.

We can however, drop the divergent term and consider a totally new metric which is $$d\tilde{s}^2=dUdV-\frac{1}{4}\sin^2(U-V)(d\theta^2-\sin^2\theta d\phi^2).$$

Now the same ranges of coordinates $U,V$ for Minkowski spacetime with this metric is another spacetime and this one is not geodesically complete. This one can actually be extended by adding in the endpoints of the geodesics.

The extension can of course be performed in the coordinates $U,V$. Still, we see from the metric that it is periodic with respect to $V-U$. Hence the desired extension will be periodic with respect to $V-U$.

Introducing $V-U = R$ as one new coordinate and $T = U+V$ as a companion one, we see that even if we extend $R$ to run throught the whole real line, we will just be introducing "double labels" to same events since the extension we are performing must by consistency be $\pi$-periodic in $R$.

In that setting we can take one of the possible ranges, say $0\leq R\leq \pi$ while there is no such periodicity requirement on top of $T$ which may run freely from $-\infty$ to $\infty$.

Finally with this we obtain a new unphysical spacetime $(M,\tilde{g})$ whose region described by $U,V\in (-\pi/2,\pi/2)$ is not Minkowski spacetime which was inextendible, but rather is conformal to Minkowski spacetime, reflecting the original change of scale introduced to bring infinity closer.

The so-desired infinity $\mathscr{I}$ is then defined on this unphysical manifold with the unphysical metric which admitted the desired extension.