6
$\begingroup$

One property of spatial infinity is that all spacelike geodesics end at it. Since spacelike geodesics can have different directions, I do not understand why spatial infinity is a point. It looks more like a 2 sphere instead of a point.

I will provide more information. Let us pick a point other than the spatial infinity in the conformal diagram. Usually, people draw the conformal diagram in a plane or represent it on the surface of a cylinder. So this point on a plane represents a 2 sphere. But spatial infinity is literally a point. Why?

Improve this question
$\endgroup$
8
  • $\begingroup$ You have to give more context. Sometimes, infinity is a point, and sometimes, physicists talk about a "sphere at infinity". Those are two distinct concepts, and only with more information can we tell which one you are talking about. $\endgroup$
    – ACuriousMind ♦
    Commented Nov 5, 2015 at 17:28
  • $\begingroup$ Hi Drake. I've linked a question that I'm fairly sure covers the same material as yours. If you don't agree then shout and I'll reopen this question. $\endgroup$
    – John Rennie
    Commented Nov 5, 2015 at 17:36
  • $\begingroup$ Hi @JohnRennie. I don't think these 2 questions are the same. In fact, in that link, I asked a different question about whether all spacelike curves end at spatial infinity. Although in that link, I stated that spatial infinity is at $\tau=0, \rho=\pm\pi$, which looks like a point, I did not mean that a spatial infinity is a point because I suppressed 2 dimensions. Therefore, I think I asked a new question in this post. Please open this question. Thanks! $\endgroup$
    – Drake Marquis
    Commented Nov 5, 2015 at 19:46
  • $\begingroup$ Hi Drake. Reopened as requested! $\endgroup$
    – John Rennie
    Commented Nov 5, 2015 at 20:27
  • $\begingroup$ Alexandroff extension. $\endgroup$
    – AccidentalFourierTransform
    Commented Sep 8, 2018 at 21:03

3 Answers 3

Reset to default
5
$\begingroup$

Indeed the plane is conformal to the punctured sphere (by stereographic projection), rather than the open disc. This means that its conformal boundary is the single point at infinity on the sphere. This is an aspect of the uniformization theorem in 2-dimensions, but it's true in all dimensions.

To see why the plane is not conformal to the open disc, consider that a conformal map from the plane to the disc would be a bounded holomorphic function, and hence constant by Liouville's theorem in complex analysis.

In higher dimensions it follows another theorem of Liouville. Those hidden spheres of angles in the Penrose diagram you were asking about get squashed to zero size at infinity. Note that the situation is different for Minkowski space, whose conformal compactification has topology $S^1 \times S^d$. In general signature $(p,q)$ the compactification has topology $S^p \times S^q$. See this question for instance.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Notice though that in QFT the compactification $S^1\times S^d$ is not acceptable (since it contains closed timelike curves), and CFTs live on the universal cover $\mathbb{R}\times S^d$. $\endgroup$
    – Peter Kravchuk
    Commented Sep 10, 2018 at 6:15
3
$\begingroup$

It may seem awkward or inconsistent that on a Penrose diagram, $\mathscr{I}^+$ and $\mathscr{I}^-$ are shown as lines (representing 3-dimensional things), while $i^0$, $i^+$, and $i^-$ are points (representing 2-spheres). The figure shows why this actually makes sense.

enter image description here

Given a finite region of spacetime S, we can find a point like P that is spacelike with respect to the whole region, and a point like Q that is timelike with respect to the whole region. It is not possible to find a point that is lightlike in relation to every point is S.

This is obviously not a rigorous argument like Ryan Thorngren's, but hopefully it is helpful as a supplement to that answer, in order to build intuition. Please don't upvote this answer without upvoting his, since his is more rigorous and deserves the bounty.

Improve this answer
$\endgroup$
-1
$\begingroup$

Let me answer my question.

By the definition of conformal flatness, $\nabla_a\Omega|_{i^0}=0$, where $\Omega$ is the conformal factor, and $i^0$ is the spatial infinity. So the spatial infinity is singular, and I think that is the reason people think spatial infinity is a point.

Improve this answer
$\endgroup$
5
  • $\begingroup$ But having spatial infinity be a point is just a choice. If you are trying to motivate that choice you should expand your answer. $\endgroup$
    – Timaeus
    Commented Nov 8, 2015 at 4:16
  • $\begingroup$ @Timaeus Why do people make the choice that spatial infinity is a point? To conformally compactify Minkowski spacetime, we use this conformal factor $\Omega=2\cos\frac{T+R}{2}\cos\frac{T-R}{2}$ and the metric becomes $d\tilde s^2=-dT^2+dR^2+\sin^2R(d\theta^2+\sin^2\theta d\phi^2)$. The spatial infinity is at $T=0,R=\pi$ which is a pole on $S^3$ and a coordinate singularity. So it seems to me that this spatial infinity is still a $S^2$. Then people want the spatial infinity to be a point probably because $\nabla_a\Omega|_{i^0}=0$. $\endgroup$
    – Drake Marquis
    Commented Nov 9, 2015 at 4:37
  • $\begingroup$ @Timaeus I really have no idea... $\endgroup$
    – Drake Marquis
    Commented Nov 9, 2015 at 4:38
  • $\begingroup$ @DrakeMarquis: This is a great question, thanks for asking it. But I don't think your answer is right. The metric also becomes singular in this sense at null infinity, but null infinity is represented on a Penrose diagram by a surface, not a point. $\endgroup$
    – user4552
    Commented Sep 8, 2018 at 20:57
  • $\begingroup$ @Timaeus: But having spatial infinity be a point is just a choice. I don't think it is just a choice, for the reasons explained in Ryan Thorngren's answer. $\endgroup$
    – user4552
    Commented Sep 8, 2018 at 20:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.