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Or rather, does mass-less particles orbit the matter in the universe due to the infinite range of gravity? And if yes, in what shape are they orbiting? A ring as on Saturn? Maybe a whirlpool due to the expansion of space? A full shell?

Qmechanic
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BlackCap
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  • Just a few points: Gravity is not a force, so it does not have an infinite range, rather it is a manifestation of Spacetime curvature. Second, something orbiting something else in GR is understood as geodesic motion. So you can have null geodesic motion within a universe / a solution of Einstein's equations, but I don't know what light orbiting the universe would mean. – Dr. Ikjyot Singh Kohli Sep 18 '16 at 19:00
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    I think the question would be interpreted to ask whether in a cosmological solution there is a lightlike horizon at large or infinite a(t), the scale factor? – Bob Bee Sep 18 '16 at 19:25
  • @BobBee Exactly. – BlackCap Sep 18 '16 at 20:54

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I illustrate in this post how a cosmology with a constant density of energy in the vacuum will in fact accelerate outwards. This will not permit the orbiting of particles or light on a large scale.

There is the Godel solution that has closed timelike curves. This is different than an orbit, but a case where geodesics have a time parameter that is topologically a circle. The anti-de Sitter spacetime is similar in that $AdS_n$ has topology $\mathbb R^{n-1}\times\mathbb S^1$. Here the spatial surface is the $\mathbb R^{n-1}$ and time is $S^1$. This spacetime is different from the spacetime that approximates what we observe that is $\mathbb S^{n-1}\times\mathbb R$. However, it is possible that our observable cosmology emerges from and anti-de Sitter spacetime. Maybe the inflationary spacetime that spins off cosmology at a lower vacuum energy is an anti-de Sitter spacetime.

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Lawrence B. Crowell
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    The late time cosmological spacetime with dark energy dominating and exponentially expanding the universe does indeed have a Killing, or lightlike horizon. The lightlike orbits at that value of r are indeed geodesics. The Killing vector goes from timelike to space like across it. If the question is opened up I can expand. – Bob Bee Sep 18 '16 at 23:29