# Are there null geodesics inside null infinity?

Looking at a Penrose diagram for Minkowski space, you would think that you could draw a null geodesic running from $$i^0$$, along $$\mathscr{I}^+$$, and ending up on $$i^+$$. In fact there would be many such geodesics, due to the rotational symmetry that is hidden on the two-dimensional diagram. Similarly we could draw null geodesics running along $$\mathscr{I}^-$$.

These seem like nonphysical artifacts that we don't want around. It would be weird if there were some dynamics going on "off stage" like this that we could never detect. We don't have a metric on the boundary, so it seems like there is no real definition of a geodesic. Also, we can't apply the Einstein field equations at the boundary, since they are written as differential equations that only make sense if you have an open neighborhood surrounding any given point.

On the other hand, we do have null geodesics that intersect null infinity. So it seems a little delicate to set things up mathematically so that null geodesics can get to null infinity but can't have any motion within null infinity.

Is there some way that it's encoded into the formalism that we don't have such things going on? Is it in one of the complicated technical conditions written into the definition of asymptotic flatness? How do we make sure that it is allowed for geodesics to intersect null infinity, but not allowed for them to dwell there?

• I don't think that it is a problem that null infinity is ruled by null geodesics. That reflects only the geometry and doesn't imply that there are massless particles there. Just like in a vacume solutions you have plenty of null curves, yet space-time is empty.
– MBN
Jan 23, 2019 at 8:09

A Penrose diagram is a convenient way of representing the causal structure of a given spacetime. Of course, different spacetimes may have equivalent causal structures, as long as their metrics are conformally equivalent. When we work with the Penrose diagram for Minkowski spacetime, we are typically juggling two different spacetimes: Minkowski spacetime $$M$$ on the one hand, and the "Einstein static universe" $$E$$ on the other hand. Both have the same causal structure, but their metrics are different: conformally equivalent, but not isometric. Spacetime $$M$$ is an open subset of spacetime $$E$$. Let $$\overline{M}$$ denote the closed subset obtained from $$M$$ by adjoining the boundary (including $$\mathscr{I}^\pm$$), so $$M\subset\overline{M}\subset E.$$ The central messages of this answer are:

• The Minkowski metric $$g$$ on $$M$$ cannot be extended to $$\overline{M}$$. This follows from the fact that null geodesics which are affinely parameterized with respect to the Minkowski metric on $$M$$ never reach the boundary for any finite value of the affine parameter, and a coordinate transform cannot change this. (The quality of being affinely parameterized is coordinate-independent.)

• On $$E$$ we have a different metric $$\hat g$$ that is well-defined everywhere on the closed subset $$\overline{M}$$, including on the boundary. The metric $$\hat g$$ is conformally equivalent to $$g$$ within the subset $$M$$, so null geodesics in $$M$$ are also (segments of) null geodesics in $$E$$. Furthermore, there are (finite segments of) null geodesics within the boundary of $$\overline{M}$$, as long as "geodesic" is defined with respect to $$\hat g$$.

• Even though $$g$$ and $$\hat g$$ are in the same conformal equivalence class and therefore have the same null geodesics wherever both metrics are defined (namely within $$M$$), they don't give the same affine parameterizations of those null geodesics. This is why we can have a well-defined metric $$\hat g$$ on $$\overline{M}$$ which is in the same conformal class as the Minkowski metric $$g$$ on $$M$$, even though $$g$$ itself cannot itself be extended to $$\overline{M}$$.

• When we talk about a geodesic "reaching" a point (on $$\mathscr{I}^+$$, for example), we are implicitly talking about reaching that point at a finite value of the affine parameter. For example, a spacetime is called "geodesically complete" if every affinely parameterized geodesic can be continued to arbitrarily large values of the affine parameter. The "affine parameter" qualifier is essential, so that Minkowski spacetime can be regarded as geodesically complete.

...it seems a little delicate to set things up mathematically so that null geodesics can get to null infinity but can't have any motion within null infinity. Is there some way that it's encoded into the formalism that we don't have such things going on?

Yes. We simply need to keep track of which metric ($$g$$ or $$\hat g$$) we are using to define the affine parameterization. If we are using the Minkowski metric $$g$$, then null geodesics in $$M$$ never reach $$\mathscr{I}^\pm$$; and if we are using the conformally-equivalent metric $$\hat g$$, then null geodesics in $$M$$ pass through $$\mathscr{I}^\pm$$ (which also contains [finite segments of] null geodesics), so $$\mathscr{I}^\pm$$ isn't really a "boundary" at all unless we arbitrarily choose to ignore everything in $$E$$ outside of $$\overline{M}$$.

The rest of this answer makes this mathematically explicit. Instead of starting with Minkowski spacetime $$(M,g)$$ and constructing its conformal compactification (which is what the Penrose diagram represents), I'll start with the Einstein static universe $$(E,\hat g)$$, because it's bigger, and then I'll construct $$M$$ as a topologically open subset of $$E$$ and review how the metrics $$g$$ and $$\hat g$$ are related to each other.

To define the Einstein static universe $$(E,\hat g)$$, start with the smooth manifold $$E=\mathbb{R}\times S^{D-1} \hskip2cm \text{(topologically)}, \tag{1}$$ where $$D$$ is the number of dimensions. Use a coordinate $$T$$ to cover the $$\mathbb{R}$$ factor. For the factor $$S^{D-1}$$, use the standard metric on the unit sphere, written as $$d\Omega_{D-1}^2=dR^2+\sin^2(R)d\Omega_{D-2}^2 \\ 0\leq R\leq \pi. \tag{2}$$ Intuitively, the coordinate $$R$$ is analogous to latitude measured from one of the poles, so that the equator is at $$R=\pi/2$$, and the second term on the right-hand side of (2) is the metric of a constant-latitude slice of $$S^{D-1}$$, which has topology $$S^{D-2}$$ and radius $$\sin(R)$$. The Einstein static universe $$E$$ is defined to be the manifold (1) equipped with the metric $$\hat g$$ defined by $$\hat g=dT^2-\left(dR^2+\sin^2(R)d\Omega_{D-2}^2\right) \\ -\infty The spacetime $$(E,\hat g)$$ is geodesically complete, geometrically curved, and topologically non-trivial (not homeomorphic to $$\mathbb{R}^D$$). This is the Einstein static universe.

Now we can construct $$D$$-dimensional Minkowski spacetime $$(M,g)$$ by taking a particular open subset $$M\subset E$$ and giving it a new metric $$g$$. Within $$M$$, the new metric $$g$$ will be conformally equivalent to $$\hat g$$, so null geodesics in $$M$$ still coincide with (segments of) null geodesics in $$E$$. This important fact is stated with exceptional clarity in [$$1$$]:

Two metrics related by an overall scalar multiple function are said to be "conformally related," or related by a "Weyl rescaling" or "Weyl transformation." The light cones of two such metrics $$g_{\mu\nu}$$ and $$A^2(x)g_{\mu\nu}$$ are obviously the same, and hence so are the null curves. ... in fact the null geodesic curves are also the same, but ... the affine parameters on those curves are not the same.

(I italicized the part that is central to this answer.) To construct $$M$$, consider the topologically-open subset of $$E$$ defined by $$0\leq |T|+R<\pi \tag{4}.$$ This subset is nicely illustrated in the following figure from [$$2$$]:

In this figure, $$\mathbb{R}\times S^{D-1}$$ is depicted as $$\mathbb{R}\times S^1$$, so each horizontal circle really represents a sphere $$S^{D-1}$$. The subset of $$E$$ defined by (4) corresponds to the shaded region in the figure. This subset will be denoted $$M$$. It is topologically trivial (homeomorphic to $$\mathbb{R}^D$$), because a neighborhood of the point $$R=\pi$$ is omitted from each of the spheres $$S^{D-1}$$ that it intersects. (At $$T=0$$, only the point $$R=\pi$$ itself is omitted.) The subset $$M$$ by itself is represented by the following Penrose diagram, which is from [$$3$$]:

This is half of the diamond-shaped region from the preceding figure, the two halves differing only in the values of the angular coordinates $$\Omega_{D-2}$$. Note that the symbols $$\mathcal{I}^\pm$$ in these pictures mean the same thing as the symbols $$\mathscr{I}^\pm$$ in this answer (and in the OP).

We could continue using the curved metric (3) on $$M$$, but then $$M$$ would not be geodesically complete: some null geodesics in $$(E,\hat g)$$ pass through $$(M,\hat g)$$ but are not completely contained in $$M$$. We can make $$M$$ geodesically complete by replacing the metric (3) with the new metric $$g$$ defined by $$g=\frac{dT^2-\left(dR^2+\sin^2(R)d\Omega_{D-2}^2\right)}{ \cos^2 U\cos^2 V} \hskip1cm 0\leq |T|+R<\pi \tag{5}$$ with $$U,V$$ being abbreviations for $$(T\pm R)/2$$. This metric is well-defined everywhere on $$M$$, where the denominator is non-zero, but it is not well-defined everywhere on $$E$$. In particular, the denominator goes to zero on the boundary of $$\overline{M}$$. In particular, it goes to zero on $$\mathscr{I}^\pm$$. That's okay, because the manifold $$M$$ does not include $$\mathscr{I}^\pm$$ due to the restriction $$|T|+R<\pi$$.

Here is the central point: With respect to the metric $$\hat g$$ defined by (3), we have null geodesics inside $$\mathscr{I}^\pm$$, namely any worldline for which the angular coordinates $$\Omega_{D-2}$$ are constant and $$|T|+R=\pi$$. However, we cannot say whether or not these worldlines are "geodesics" with respect to the metric $$g$$ defined by (5) (or (9), below), because $$g$$ is not defined on $$\mathscr{I}^\pm$$.

Despite appearances, the metric $$g$$ defined by (5) is isometric to (not just conformally equivalent to) to the usual Minkowski metric. In particular, the metric (5) describes flat spacetime. To make this explicit, define new coordinates $$t,r$$ by $$t=u+v \hskip2cm r=u-v \tag{6}$$ with $$u=\tan U \hskip2cm v=\tan V. \tag{7}$$ These equations are wel-defined for all $$0\leq U,V < \pi/2$$, so the coordinates $$t,r$$ cover all of $$M$$, along with the angular coordinates that are collectively denoted $$\Omega_{D-2}$$ in (5). The corresponding ranges are $$-\infty By taking the differential of (6), we can relate $$dt,dr$$ to $$dT,dR$$. Using those relationships in (5) gives the Minkowski metric $$g=dt^2-(dr^2+r^2 d\Omega_{D-2}^2). \tag{9}$$ Within the subset $$M$$, the new metric (9) is conformally equivalent to the original metric (3), even though it is geometrically different. According to the excerpt that was highlighted above, this means that each null geodesic in $$(M,g)$$ is a (segment of a) null geodesic in $$(E,\hat g)$$. However, given a null geodesic, the quality of being affinely parameterized depends on which metric we use. A null geodesic that is affinely parameterized according to $$\hat g$$ is not necessarily affinely parameterized according to $$g$$, and conversely. A null geodesic that is affinely parameterized with respect to the Minkowski metric (9) never reaches the boundary $$\mathscr{I}^+$$ for any finite value of the affine parameter, whereas the same null geodesic affinely parameterized with respect to the metric (3) does reach the boundary at a finite value of the affine parameter, and it continues right through the boundary to the other side, which doesn't even exist in $$M$$. Furthermore, the boundary itself contains (finite segments of) null geodesics in $$E$$, whereas $$M$$ does not include the boundary at all.

In summary, the situation described in the OP never occurs. It does not occur in $$(M,g)$$ because $$(M,g)$$ is a geodesically-complete spacetime that excludes $$\mathscr{I}^+$$, and affinely-parameterized null geodesics in $$(M,g)$$ don't reach $$\mathscr{I}^+$$. It does not occur in $$(\overline{M},\hat g)$$ either, because null geodesics in $$(\overline{M},\hat g)$$ pass right through $$\mathscr{I}^+$$ to the other side in $$(E,\hat g)$$.

References:

[$$1$$] Jacobson (2007), "HW#3—Phys 675—Fall 2007," http://www.physics.umd.edu/grt/taj/675d/675dhw3.pdf

[$$2$$] Felinska (2010?), "Carter-Penrose diagrams and black holes ," http://www.ift.uni.wroc.pl/~blaschke/master/Felinska.pdf

[$$3$$] Aretakis (2013), "Lecture Notes on General Relativity," https://web.math.princeton.edu/~aretakis/columbiaGR.pdf

• Thanks very much for putting so much effort into this answer. I hate to say it, but I think you spend the whole answer addressing what I would consider to be more of a side issue in my question. My main question was whether, when we consider Minkowski space as a manifold-with-boundary that includes null infinity, we can pick out things that can be considered null geodesics lying completely inside null infinity. Of course standard Minkowski space does not include the boundary, but that is sort of side-stepping the question.
– user4552
Feb 6, 2019 at 23:52
• @BenCrowell In order to define what "geodesic" means inside null infinity, I was thinking we need a metric that's defined there, and in order to get a metric that's defined there, we need to consider a different-but-conformally-equivalent metric, hence the spacetime I called $E$. If we stick with the usual Minkowski metric on $M$, then I guess I don't know what "geodesic" would mean on the boundary, because that metric is undefined there. Is there another way to make "geodesic" well-defined without changing the metric? (This is a sincere question; I still have lots to learn!) Feb 7, 2019 at 1:03
• @BenCrowell I updated the answer to use a notation that explicitly distinguishes between $M$ and its closure $\overline{M}$, which includes $\mathscr{I}^\pm$, and also to clearly specify which metric is being used (the Minkowski metric $g$, which is defined on $M$ and cannot be extended to $\overline{M}$, versus the conformally-equivalent metric $\hat g$ which is defined on all of $E\supset\overline{M}\supset M$). I also re-worded the beginning to help bring out the key points: we have null geodesics inside $\mathscr{I}^\pm$ wrt $\hat g$, but $g$ isn't defined there (and can't be extended). Feb 8, 2019 at 15:34

This feels like something that is a definitional thing to me that is however you make it. Infinitesimally in from conformal null infinity, the spacetime is infinitesimally close to Minkowski space, and the normal to your outbound null vector is definitely a null 3-surface, so it makes sense to me to say that conformal infinity is a null 3-surface too. but yeah, it's not a point on the underlying manifold, so it's a bit dishonest to extend the metric there.

… you would think that you could draw a null geodesic running from $$i^0$$, along $$\mathscr{I}^+$$, and ending up on $$i^+$$. In fact there would be many such geodesics, due to the rotational symmetry that is hidden on the two-dimensional diagram.

This could be illustrated by a following image of a 3D version of a “diagram” for Minkowski space:

One could see, that through each point of $$\mathscr{I}^+$$ only one null geodesics of the auxiliary “diagram” space (conformally isometric to the original spacetime) passes, as one would expect on a null hypersurface. We could contrast this with behavior with boundary of AdS covering space which would be timelike in a auxiliary space. Moreover, any two embedding of AdS boundary would only differ by a conformal factor of the induced metric. So one could argue that AdS boundary (unlike $$\mathscr{I}^\pm$$) does have naturally defined null geodesics, inducing “normal” causal structure at the boundary.

In order to understand what is going on with $$\mathscr{I}^+$$ of asymptotically flat spacetimes we could turn to holography.

It would be weird if there were some dynamics going on "off stage" like this that we could never detect. We don't have a metric on the boundary, so it seems like there is no real definition of a geodesic. Also, we can't apply the Einstein field equations at the boundary, since they are written as differential equations that only make sense if you have an open neighborhood surrounding any given point.

Holography is not about having some external dynamics going on the boundary. Boundary is already present in the observable dynamics in the bulk via the boundary conditions. Objects, such as S-matrix are defined using entities from the boundary. Holography just gives us another viewpoint to look at the same physical reality in a language fully intrinsic to the boundary. And yes, there would be no Einstein equations there, that is why holographic dual of a gravity theory in the bulk would not have gravity at the boundary.

Flat space holography is a much less studied subject than AdS/CFT but over the last few years there has been some successes in 3D and 4D cases. But the general idea is the same as with other holographic theories: quantum gravity theory in an asymptotically flat spacetime (“the bulk”) has an equivalent description in terms of a nongravitational quantum theory living on the boundary. See [1 ] for a review of older results and expository chapters of [2] for recent insights.

In a gravitational theory the symmetries at the boundary of a spacetime are described by the Asymptotic Symmetry Group (ASG). For asymptotically flat spacetimes this is the Bondi–Metzner–Sachs (BMS) group. The BMS group is acting on null infinity and so in a holographic viewpoint this group would be the group of symmetries of a boundary theory.

Since we do not (yet) know what kind of theory would be living at the boundary, the obvious approach would be to constrain possible theories by using the symmetries, implementing a BMS version of conformal bootstrap programme. The important recent insight is realization that the BMS group is a conformal extension of Carroll group [3].

The term “Carroll group” has been coined by Lévy–Leblond as a reference to a phrase by the Red Queen:

Now, here, you see, it takes all the running you can do, to keep in the same place.

Carroll group is a Inönü–Wigner contraction of Poincaré group in the limit $$c\to 0$$, just like Galilei group corresponds to a limit $$c\to \infty$$. Another way of one can look at this group as a local group of manifold transformations preserving a special kind of structure: the Carroll manifold structure [4]. Carroll manifolds are a dual of Newton–Cartan manifolds, a geometrization of Newtonian gravity.

So we have come to a natural conclusion (a conjecture, hinging on the validity of holographic principle): $$\mathscr{I}^+$$ is a Carroll manifold with the topology $$\mathbb{R}\times S_2$$, with $$\mathbb{R}$$ serving as Carroll time, and this Carroll time evolution of a given space point provides us with parametrization of null geodesic in a embedding “diagram” space. At the same time this manifold does not have the notion of null geodesics internally, but it does have nontrivial conformal Carrollian field theory residing on it and providing a holographically dual description of dynamics in the asymptotically flat spacetime.

Is there some way that it's encoded into the formalism that we don't have such things going on? Is it in one of the complicated technical conditions written into the definition of asymptotic flatness? How do we make sure that it is allowed for geodesics to intersect null infinity, but not allowed for them to dwell there?

As @MBN already noted, geodesics by themselves are simply curves. Their existence does not necessarily mean there is some nontrivial dynamics. And, as we have seen, the “stuff” that actually dwells on the boundary is not independent of the bulk dynamics, and though the geometry there is non-Lorentzian it does have meaning.

1. Melas, E. (2011). Open problems and results in the group theoretic approach to quantum gravity via the BMS group and its generalizations. In Journal of Physics: Conference Series (Vol. 283, No. 1, p. 012023), doi:10.1088/1742-6596/283/1/012023.

2. Bagchi, A., Basu, R., Kakkar, A., & Mehra, A. (2016). Flat Holography: Aspects of the dual field theory. Journal of High Energy Physics, 2016(12), 147, doi:10.1007/JHEP12(2016)147, arXiv:1609.06203.

3. Duval, C., Gibbons, G. W., & Horvathy, P. A. (2014). Conformal Carroll groups and BMS symmetry. Classical and Quantum Gravity, 31(9), 092001, doi:10.1088/0264-9381/31/9/092001, arXiv:1402.5894.

4. Duval, C., Gibbons, G. W., Horvathy, P. A., & Zhang, P. M. (2014). Carroll versus Newton and Galilei: two dual non-Einsteinian concepts of time. Classical and Quantum Gravity, 31(8), 085016, doi:10.1088/0264-9381/31/8/085016, arXiv:1402.0657.

It sounds like you're thinking about using null infinity as a Cauchy surface, and worried about the null geodesics within infinity representing a particle propagating "through" the initial condition?

Are there null geodesics running through $$\mathscr{J}^-$$? Yes.
Is that a problem? No. The issue I think you're worried about has nothing to do with $$\mathscr{J}$$ being at infinity, and everything to do with it being null. And it has a solution.
Long story short: For a massless KG equation, or something else with null-propagating solutions, a null Cauchy surface is something called a characteristic surface of the problem. Instead of providing the usual $$\phi, \dot{\phi}$$ initial data on the null surface, you only need to provide $$\phi$$. Because the initial data is specified differently on a surface like this, there's not an issue with modes propagating along the boundary surface.
If you want me to elaborate on that leave a comment, I do have an example in mind. (Or work it out yourself... the massless 2d KG equation has a general solution $$f(t-x)+g(t+x)$$. What happens if you specify initial data on a pair of null surfaces?)