# Are orbits possible in de sitter space?

Since the de sitter space has constant positive curvature does that mean that objects can't orbit around other objects?

## 2 Answers

Couchyam's answer is for undeformed de Sitter space - empty if you will. So no orbits in that.

What you are asking is if one put an object the mass of the earth in a de Sitter space, could objects the mass of the moon orbit the earth thing?

It would seem that the earth like mass would follow a prescribed path, but since its a rule in General Relativity that any small patch of space is flat, the moon like thing can orbit the earth like thing.

In fact our universe is likely well modelled by a de sitter space: https://en.wikipedia.org/wiki/De_Sitter_universe so its obvious that orbits can exist.

You have to be careful with definitions here. Saying that our universe is a de Sitter space is only an approximation. That's where Couchyam's answer comes in - for a true exact de Sitter space, which is necessarily smooth and featureless with only vanishing small test particles allowed.

You can also look at the limiting case in a lumpy de Sitter like universe that we are in. If you take a completely flat spacetime, then anything can orbit anything else at any distance, but with a de Sitter universe at some point the expansion would rip extremely weakly bound orbits apart.

• Nice answer, but I thought I should comment that I accounted for deformed de Sitter at the beginning of my answer in the parenthetical remark. – TotallyRhombus Sep 4 '15 at 14:24
• Thanks - I saw that but I think its a bit hidden, so I wanted to expand a bit for the lumpy de Sitter like case. – Tom Andersen Sep 7 '15 at 21:49

I'll assume you are asking about geodesics in de Sitter (If the objects contribute to the energy-momentum tensor they perturb the metric and perform orbits).

De Sitter space can be thought of as the set of points in 4+1-dimensional Minkowski space that satisfy $$X_1^2+X_2^2+X_3^2+X_4^2=1+X_0^2.$$ Cross sections of constant $X_0$ are 3-spheres, so you might think that time-like or null geodesics can go in orbits around the spacetime. However, if you start at a certain value of comoving coordinates (say on $S^3$), it turns out that there is no time-like or null geodesic that returns to your starting point. The easiest way to see this is through a Penrose diagram for de Sitter.

The Penrose diagram for de Sitter is shown below: (from Les Houches 2001)

Penrose diagrams encode the causal structure of spacetime. Every point in de Sitter corresponds to a point in the above square, and diagonal lines at $45^\circ$ angles correspond to the world-lines of null geodesics or light rays. Two light rays are shown above, going from the line $I^-$ to $I^+$. The 'North Pole' and 'South Pole' are antipodal points on $S^3$, while $I^+$ and $I^-$ correspond to points at the infinite future and infinite past (if you want, you can think of these as equivalence classes of time-like geodesics). Hence, a ray of light has 'just enough time' in the lifetime of a de Sitter universe to travel from $\theta=0$ to $\theta=\pi$, and can't get back to its original comoving position at $\theta=0$. In fact, it's clear that any two time-like geodesics that attempt to go around $S^3$ will intersect at most once.