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All timelike geodesics that pass through the coordinate origin of AdS (under the standard parameterization) execute simple harmonic motion about the origin with a period proportional to the AdS radius, as discussed at AdS Space Boundary and Geodesics and on Carroll pg. 327. This statement seems to pick out a special point in space about which all orbits oscillate.

But AdS is maximally symmetric and therefore completely homogeneous, so all points must be physically equivalent and anything that appears special about the origin must end up being just an artifact of the choice of coordinates.

How can we reconcile these two facts? The center of a particle's oscillation seems physical and coordinate-independent. How can there be periodic oscillation without the oscillation being centered at any special point? (After all, given a particle trajectory, there is no coordinate-independent sense of an "initial position at $t = 0$" about which it should oscillate.) Would an observer in AdS ever simultaneously observe two test particles oscillating about different points?

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  • $\begingroup$ I don't think that's what maximally symmetric means - all points related by the action of the symmetry group should physically equivalent. Surely not every point has to be physically equivalent with every other point. $\endgroup$
    – zzz
    Commented Jan 9, 2017 at 8:27
  • $\begingroup$ @bianchira Maximally symmetric definitely does imply that every point is physically equivalent. As Carroll says, it implies homogeneity and isotropy about every point, and homogeneity is the equivalent concept to translational invariance. $\endgroup$
    – tparker
    Commented Jan 9, 2017 at 13:12
  • $\begingroup$ Not sure what the word "translation" means to you but to me in this context it means action of a global symmetry. $\endgroup$
    – zzz
    Commented Jan 9, 2017 at 19:37
  • $\begingroup$ @bianchira That's not what "translation" means. It refers specifically to spacetime translation. Of course "all points related by the action of the symmetry group should physically equivalent," that's what a symmetry group means. The work "maximally" refers to the size of the symmetry group, and implies that it is a large as possible. $\endgroup$
    – tparker
    Commented Jan 9, 2017 at 20:43
  • $\begingroup$ I guess I'm asking which "translations" are you thinking of that might act non-trivially on the poincare disc origin? $\endgroup$
    – zzz
    Commented Jan 9, 2017 at 21:19

3 Answers 3

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In short, the point is that your statement "The center of a particle's oscillation seems physical and coordinate-independent" is incorrect. The answer to the question "How can there be periodic oscillation without the oscillation being centered at any special point?" is the following.

A single timelike geodesic in pure AdS can't be sensibly said to do anything; it just sits there. In order to claim it's executing simple harmonic motion, you must introduce a reference location with respect to which it's executing this motion. In your question, do you this by introducing an origin of some set of global AdS coordinates. But since AdS is maximally symmetric, there are an infinite set of global coordinate charts, each one of which we can think as being labeled by the geodesic at the origin of that chart (plus some extra rotational freedom, but that doesn't matter for the current purpose). In other words: given a particular timelike geodesic in AdS, it is always possible to construct a global coordinate chart in which that geodesic lies at the origin. (I should therefore note that there is no such thing as the origin, as you wrote; there can only be an origin.)

Thus your statement that "All timelike geodesics that pass through the coordinate origin of AdS...execute simple harmonic motion about the origin..." can be rephrased in a coordinate-independent way as "The relative proper separation between any two intersecting timelike geodesics in AdS undergoes SHM as a function of the proper time along those geodesics." It is now clear how the issue is resolved: you are not considering just a single geodesic, but the relative separation of two geodesics.

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For posterity, I want to explain exactly what I was confused about. Imagine that Alice observes a massive particle oscillating periodically about her origin:

enter image description here

So far, so good. But now Bob comes over and sets up another coordinate system whose origin $x = 0$ is just shifted over from Alice's by a small bit:

enter image description here

The particle also passes through this coordinate system's origin, but it doesn't oscillate periodically about it. But massive particles whose trajectories pass through the origin (of Bob's coordinates) are supposed to oscillate about the origin. Apparent contradiction.

As Sebastian pointed out, the resolution is that if the black sine curve is in fact a geodesic, then Bob's blue coordinate system is not a geodesic. It's a valid coordinate system, but if Bob were attached to that coordinate system then he wouldn't be inertial, but would feel acceleration. Therefore Bob's coordinate system isn't the usual polar-type coordinates which we we usually use for AdS, and it isn't true that all timelike geodesics that pass through its origin must oscillate around that origin.

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  • $\begingroup$ If I’m not mistaken, the geodesic version of the blue line would itself look sinusoidal, right? $\endgroup$
    – user76284
    Commented Jun 13, 2020 at 18:55
  • $\begingroup$ @user76284 Yes, I think so. $\endgroup$
    – tparker
    Commented Jun 13, 2020 at 19:34
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Indeed, you are right in your statement that timelike geodesics oscillate through the origin, and together with the fact that AdS is homogeneous and isotropic this presents a bit of a puzzle, because clearly the origin of AdS is not a special point.

The resolution is that the origin is coordinate dependent in the same way that the North Pole of a sphere is coordinate dependent. If you took a general geodesic oscillating through the origin and transformed to a new set of coordinates, generically the new trajectory would not be oscillating through the origin of the new coordinate system.

The idea in discussing the geodesics in the way Carroll does is that, due to the homogeneity, if we wish to consider all possible geodesics it suffices to simply consider all geodesics passing through a given point, say the origin. The symmetries can then be used to transform those geodesics into ones that pass through different points. A key point is that the timelike geodesics passing through the origin only represent a subset of all possible geodesics.

The flat space analogy would be to consider the set of all straight lines through the origin if one were interested in flat space geodesics, omitting ones that did not pass through the origin because these can be trivially obtained via translations. In the current case, because AdS is curved it does not have "translations" per se, with $x^{\mu} \rightarrow x^{\mu} + C^{\mu}$ if $x^{\mu}$ are the coordinates and $C^{\mu}$ are a collection of constants, but it does have isometries which move the origin.

In fact, for an even more compact description of timelike geodesics, one could just provide the one of a particle sitting at the origin, and all others could be obtained via the symmetries. In flat space this would be like supplying a single straight line, and claiming that all other straight lines could be obtained via rotations and translations.

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  • $\begingroup$ Yes, but this was my confusion: take a timelike trajectory oscillating about the origin and translate your system of coordinates so that the new origin is at one tip of the sine curve. I figured in the new coordinate system you'd get a timelike trajectory like ($1 - \cos(t)$) that oscillates about a point that isn't the origin, contradicting the statement that all timelike trajectories oscillate about the origin. But Sebastian's answer clears thing up. $\endgroup$
    – tparker
    Commented Jan 10, 2017 at 17:04
  • $\begingroup$ Ok. I'm not sure at this point what has been made clear and what has not, so it might be worth emphasizing that in a given coordinate system, not all timelike geodesics oscillate around the origin (which is itself dependent on the coordinate system). $\endgroup$ Commented Jan 10, 2017 at 17:31
  • $\begingroup$ Yep, I'm clear now. My confusion was whether a timelike geodesic can pass through the origin without oscillating about it, e.g. $r(t) = 1 - \cos t$ $\endgroup$
    – tparker
    Commented Jan 10, 2017 at 17:34

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