Physics near null infinity

Question

The concept of null infinity $\mathscr{I}$ is standard in general relativity, and more recently in the analysis of infrared structure of gravity (see e.g. the article by Strominger). I am curious about explicit examples in which the physics at or near null infinity is important, not just as a boundary condition.

For example, since observers can never reach null infinity, how relevant is any analysis (e.g. gravitational wave memory) near $\mathscr{I}^+$? Unlike $\mathscr{I}$, to me other parts of "infinity" (in the sense of Penrose diagram) has a more transparent meaning: for example, in the case of spatial infinity $i^0$ we can think about physics "far away" from an isolated object we are interested in. Similarly, the past/future timelike infinity $i^\pm$ has clear meaning in terms of observers at "early/late times" relative to some processes (e.g. scattering).

I have trouble thinking about physics near $\mathscr{I}^\pm$: certainly, massless perturbations (e.g. gravitational or electromagnetic waves) propagate at the speed of light and hence they are perturbations which will reach $\mathscr{I}^+$ for asymptotically flat spacetimes. What difference does different observers near different parts of $\mathscr{I}^+$ have? The only physical thing I know is really that massless perturbations reach $\mathscr{I}^+$ and that $\mathscr{I}^+$ can sometimes be used as initial data.

Remark: it would be best if the (potential) answer(s) are phrased in terms of observers or some experimental considerations. For example, I used scattering experiment to make sense of $i^\pm$ (since in those cases, one typically assumes e.g. in QFT that the field is asymptotically free/non-interacting), though not the only way.

Update 1: Note that, for example, in Strominger's work a lot of effort has been put in understanding gravitational memory near $\mathscr{I}$. So unless I am missing something, I don't see how that is important observationally (as he seems to claim) if we cannot come close to $\mathscr{I}$.

Answer 1

Future null infinity $\mathscr{I}^+$, is the natural location to analyse the observables for distant observers (eventhough as Ben Crowell points out, no observer would ever reach or come close to $\mathscr{I}^+$.) It is where radiation that has travelled a long time ``lives''. It is after all the limit of the (closure of the) spacetime as you a follow a null ray outwards. Hence, it serves as a first approximation to what a distant oberserver would see. (In fact, one could define a "distant observer" as one for which the finite distance effects are negligable.) Note that neither $i^0$ nor $i^+$ would serve for this purpose because radiation does not come near to either.

For example, when numerical relativists calculate the waveform of a black hole merger, this waveform is extracted at $\mathscr{I}^+$. Similarly, by studying the gravitational wave memory at $\mathscr{I}^+$, one obtains an approximation to the memory effects for distant observers. The effects of memory are more easily understood and formulate at $\mathscr{I}^+$ because the asymptotic symmetries of the spacetime are exact rather than approximate at a finite distance.

Answer 2

What difference does different observers near different parts of $\mathcal{I}^+$ have?

I'm not sure it makes sense to talk about an observer near a point on $\mathscr{I}^+$, for a couple of reasons.

First off, the elephants' graveyard where massive particles and observers go to die isn't null infinity, it's timelike infinity. Null infinity is where photons go to die, not observers.

Also, no point or region of space is close to these idealized points. The whole system is defined very carefully so that every point is infinitely far from these idealized points.

Spacetime doesn't change its behavior as you head off toward null infinity. The equivalence principle says that every neighborhood of spacetime looks the same.

By the way, I think the standard way to produce the right symbol in latex or mathjax is \mathscr{I}, which makes $ \mathscr{I}$, not \mathcal{I}, which makes $\mathcal{I}$.