# Null infinity reachable by timelike worldlines?

Usually, Penrose diagrams are marked with points and segments being named past/future timelike infinity $$i^{-,+}$$, past/future null infinity $$\mathscr{I}^{-,+}$$ and spacelike infinity $$i^0$$ -- see for example the diagram of german wikipedia's Penrose diagram page (figure below).

(Source: de.wikipedia.org/wiki/Penrose-Diagramm, 23.11.2023)

I was wondering how those notions are justified. I understand, that $$i^0$$ is not reachable by any timelike worldline. However, consider e.g. a timelike wordline, starting in the center and proceeding at an angle, say $$40°$$ to the x-axis. Such a timelike wordline reaches future null infinity $$\mathscr{I}^+$$. So why is it, that timelike wordlines are supposed to always end in $$i^+$$, when I just provided a counter example. Did I misunderstand something?

• Of course it takes an infinite time to travel to spatial infinity. Nov 23, 2023 at 16:27
• But $i^+$ lies on $t=\infty$ too, doesn't it? Nov 23, 2023 at 17:48
• Yes, but your assumption that a worldline starting at the center with 40° stays 40° is incorrect, it will curve inwards like the orange worldline in these coordinates. The orange worldline looks like constant velocity, you'd need acceleration to keep a fixed angle below 45° on such a spacetimediagram. Only 45° will stay 45°. On the top and bottom of the diagram where the worldline looks as if it were moving to the left it is in fact moving to the right, as you can see on the curves of constant x. Nov 23, 2023 at 18:37
• @Yukterez They defined the worldline to be everywhere inclined at 40°, so it is. It's not a geodesic. The question isn't about geodesics. Nov 24, 2023 at 5:09
• @benrg - the statement that you have to end up in i+ can only be valid for timelike geodesics, if you accelerate all the way up to infinity so that your worldline stays at constant angle that's a different story. Nov 24, 2023 at 11:30

So why is it, that timelike wordlines are supposed to always end in $$i^{+}$$, when I just provided a counter example. Did I misunderstand something?

You provided an example of a timelike curve that reaches null infinity. Arguably, such curve could not be called a realistic worldline. Note, that if we assume that a massive object (such as particle) travels along such curve then in order to “outrun” (some of) photons travelling in the same direction it must be approaching the speed of light and hence would gain infinite energy from the viewpoint of an observer at origin.

So if we consider e.g. a scattering process in Minkowski spacetime with finite total energy, then worldlines of incoming massive particles originate at past timelike infinity $$i^{-}$$ and outgoing massive particles end up at at future timelike infinity $$i^{+}$$. While null infinities are for massless particles/waves.

It's not necessarily true that the end points should be at $$i^{\pm}$$. You can have time-like curves whose end points are on $$\mathscr{I}^{\pm}$$. It is helpful to look at the future or past set of a given trajectory. Note that, for any point $$p\in\mathbb{M}$$, the future set $$I^+[p]$$ is same as $$I^+[\gamma]$$ for any time-like curve $$\gamma$$ with past end point at $$p$$. Then clearly, for any two timelike curves $$\alpha$$ and $$\beta$$ with the same past end point $$p$$, we have $$I^+[\alpha]=I^+[\beta]=I^+[p]$$. Similar argument can be said for past set. So take $$p=i^-$$. Then note that $$I^+[i^-]=\mathbb{M}=I^-[i^+]$$. This essentially implies that any time-like curve which starts from $$i^-$$ will end at $$i^+$$

Let me add the definition for future/past set: $$I^{\pm}[p]$$ is defined as collection of all such points (subspace of $$\mathbb{M}$$) which can be reached from the point $$p$$ by a future (for $$I^+$$) / past (for $$I^-$$) directed time-like curve

• Not all timelike curves, but all timelike geodesics. To avoid i+ you'd need proper acceleration all the way, which gets your v to the limit of c. Nov 24, 2023 at 1:20
• Why just timelike geodesics and not timelike curves? The definition of chronological sets $I^{\pm}$ includes all timelike curves. Even if there is proper acceleration, the curve will reach arbitrarily close to $\mathscr{I}$ , but will have end points on $i$ only. The argument using $I^{\pm}$ is very generic as pointed out by Penrose (see section on TIPs and TIFs (pg 295 onwards) from Spinors and Spacetime Vol-II)
– KP99
Nov 24, 2023 at 5:27
• Could you elaborate on the implication you state (before the definition)? Doesn't the equation $I^+[i^-] = M = I^-[i^+]$ only imply that a time-like curve that starts at $i^-$ CAN have end-point $i^+$ and vice versa? Because $\mathscr{I}^+ \subset M$ and so $\mathscr{I}^+ \subset I^+[i^-]$, right? Or is the argument, that because $\mathscr{I}^+ \subset I^-[i^+]$, then one can always "extend" a curve from $\mathscr{I}^+$ to continue to $i^+$? Nov 27, 2023 at 21:51
• You might be confusing physical spacetime 𝕄 with unphysical spacetime M. 𝓘 is boundary of M while 𝕄 is the interior of M. So 𝓘 is not a part of $I^{\pm}$. Moreover $I[\alpha]=I[\beta]$ <-> p is past/end point : this statement is an if and if statement. So any time like curve which starts at i^- will end at i^+ and vice versa
– KP99
Nov 28, 2023 at 6:16