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Effect of cosmological expansion on bound systems in realistic cosmologies

  1. Sep 5, 2012 #1

    bcrowell

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    This paper dates to 1998:
    Cooperstock, Faraoni, and Vollick, "The influence of the cosmological expansion on local systems," http://arxiv.org/abs/astro-ph/9803097v1

    They show that systems such as the solar system, galaxies, and clusters of galaxies experience nonzero effects from cosmological expansion. They estimate these effects numerically, and show that they're undetectably small. They are, however, fun to think about.

    At the time the paper was written, we didn't know about dark energy. It turns out to be a little tricky to adapt the results of the paper to modern cosmological models. Well, tricky for me, at least -- I messed it up repeatedly in this thread: https://www.physicsforums.com/showthread.php?p=4061068#post4061068

    The effect of cosmological expansion is felt, in a circularly orbiting system, as an anomalous acceleration [itex](\ddot{a}/a)\mathbf{r}[/itex], where a is the cosmological scale factor. The direction of this effect depends on the sign of the cosmological acceleration [itex]\ddot{a}[/itex], not the sign of the cosmological expansion [itex]\dot{a}[/itex]. In an expanding matter-dominated universe, it's inward rather than outward!

    However, the effect of this perturbation on an orbit is a whole different issue. If I'm understanding correctly how to retrofit the Cooperstock results to an a(t) that isn't what they anticipated, it looks like the sign of the secular trend [itex]\dot{r}[/itex] in the orbital radius r depends on the sign of [itex](d/dt)(\ddot{a}/a)[/itex]. This does not relate directly to the direction of the anomalous acceleration. In the matter-dominated cosmology that Cooperstock assumed, the inward anomalous acceleration produces, very counterintuitively, an expansion of the orbital radius. If you were free to pick any function a(t), you could devise examples in which the orbit would contract in response to cosmological expansion, which would be pretty crazy!

    However, the FRW equations tell us this:

    [tex]\frac{\ddot{a}}{a} = \frac{1}{3}\Lambda-\frac{4\pi}{3}(\rho+3P)[/tex]

    That means that [itex]\dot{r}[/itex] is proportional to [itex]-(d/dt)(\rho+3P)[/itex], with a positive constant of proportionality. The cosmological constant term vanishes when you differentiate, so although a nonzero [itex]\Lambda[/itex] can affect the sign of the anomalous acceleration, it has no effect on the secular trend in r.

    For simplicity, let's assume a universe in which dust and dark energy are dominant, so that we can take P=0. Then it looks to me like the secular trend in the radius of a circular orbit is proportional to [itex]-d\rho/dt[/itex].

    So:

    (1) In a matter-dominated closed universe, this tells us that r expands slightly while the universe expands, then contracts slightly while the universe contracts. Not too surprising. (And remember, the effect is much too small to measure, much tinier than the variation in a(t).)

    (2) In a flat or negative-curvature matter-dominated universe, r expands.

    (3) In a vacuum-dominated universe, the effect vanishes.

    Does this all seem right? Am I still messing anything up?

    It seems kind of funny that 1 and 2 are so much like what we would have naively expected from some kind of struggle between binding and cosmological expansion, even though the effect really comes from quantities like the third derivative of the scale factor, and not on the rate of expansion itself. Seems like some kind of voodoo that must have a deeper explanation. Meanwhile 3 suggests that all of this is just some silly coincidence that arises out of various grotty, complicated dynamical effects that almost vanish, and that only work in this particular way for circular orbits. (It really does all change if the system isn't a circularly orbiting one. E.g., if you scoop out a spherical vacuum from a cosmological spacetime, Birkhoff's theorem guarantees that local effects vanish exactly.)
     
    Last edited: Sep 5, 2012
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  3. Sep 5, 2012 #2

    pervect

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    There's a simple way to think of all of this which I find intuitively convenient. I'm not sure how well it can be formally justified because the space-time geometry isn't quite static (but it varies very slowly, so we're using a quasi-static approximation).

    This simple (oversimple?) way to think of the the anomalous accelerations as being caused by the mass inside a sphere of the radius of the desired orbit - in particular, the Komar mass, the integral of rho+3P.

    In the purely static case, we can say that the surface integral of the force-at-infinity multipled by the surface area of the enclosing sphere is equal to the mass enclose by the sphere.

    So one would think of the change in the size of the orbit you mention as being due to the "cosmological fluid" leaving the enclosed orbit as the universe expands.

    Ned Wright seems to have a similar view, though he doesn't spell it out quite as explicitly. I've bolded the section from his FAQ which makes me think he shares a similar view.

    http://www.astro.ucla.edu/~wright/cosmology_faq.html#SS

    So as the universe expands, comsological fluid leaves the solar system. This causes the orbits to expand.

    If we break the cosmological fluid into dark matter and dark energy, the results would depend on what exactly the dark energy was due to. If it was purely a cosmological constant, then the dark energy inside a radius R would stay constant. There are various proposals for dark energy / quintessence, so the details of what happens to it might vary with the model.
     
  4. Sep 5, 2012 #3

    mfb

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    To estimate the order of magnitude (I don't care about factors of 2 here): This paper gives q0 = −1.0 ± 0.4 where [itex]q_0 = \frac{\ddot{a}(t_0)a(t_0)}{\dot{a}^2(t_0)}[/itex]

    By definition, a(t0)=1 and [itex]\frac{\dot{a}(t_0)}{a(t_0)} = H = 2\cdot 10^{-18}/s[/itex].

    This allows to calculate [itex]\frac{\ddot{a}(t_0)}{a(t_0)} = q_0 H^2 = -5 \cdot 10^{-36} \frac{1}{s^2}[/itex]

    Earth has an orbital radius of ~150*10^6 km and an orbital acceleration of 6*10^(-3) m/s^2. The accelerated expansion of the universe gives a force of about -7.5*10^(-25) m/s^2. The relative modification of the orbit has a similar ratio, and is about 20 pm, less than the diameter of a hydrogen atom.

    Sun has an orbital radius of ~25000 ly and an orbital acceleration of 2*10^(-10) m/s^2. The accelerated expansion of the universe gives a force of about 1*10^(-15)m/s^2.
    Still too small to be relevant, but it can be seen that at the scale of galaxy clusters, the accelerated expansion becomes important.
     
  5. Sep 5, 2012 #4

    bcrowell

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    Very cool, pervect -- thanks for the link. I'd had the same idea of relating it to the mass inside the sphere of the earth's orbit, and just worked this out over breakfast and verified that it does come out exactly right, including the factors of 2 and pi.

    Outlining the math:

    [tex] \omega^2 r = GM/r^2 [/tex]
    [tex] L/m=\omega r^2 = const [/tex]
    [tex] r/r_o = (M+(4\pi/3) \rho_o r_o^3)/(M+(4\pi/3) \rho r^3) [/tex]
    [tex] \dot{r}/r_o \approx -(4\pi/3) G \omega_o^{-2}\dot{\rho} [/tex]
    [tex] \ddot{a}/a = G[\Lambda/3-4\pi/4(\rho+3P)] [/tex]
    [tex] P\approx 0 [/tex]
    [tex] \dot{r}/r_o = \omega_o^{-2} (d/dt)(\ddot{a}/a) [/tex]

    This is exactly equivalent to Cooperstock's result.

    Ned Wright seems to take this seriously as physics...!? I'm not sure I buy it as more than a heuristic -- or I'd certainly need some convincing. A homogeneous dust in a homogeneous cosmology can't make a nonvanishing force on a test particle. It's true that you can get such an effect in Newtonian physics if the solar system is embedded in a spherically symmetric cloud centered on the sun, and there can be a secular trend if the cloud gradually thins out by being transported away radially. The effect is independent of the radius of the cloud. In the Newtonian case, an infinite radius for the cloud seems to lead to a paradox, since by symmetry there can't be any force. This was how Newton proposed to get a static cosmology. I think the paradox is probably resolved by careful consideration of perturbations on the result of taking the limit of infinite radius. Conceivably you could get something similar in some kind of GR-based theory, but it's not obvious to me how that would arise.
     
    Last edited: Sep 5, 2012
  6. Sep 5, 2012 #5

    bcrowell

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    Hi, mfb--

    Since the result ends up being proportional only to [itex]\dot{\rho}[/itex] (the rate at which the density of "dust" drops), we can avoid the large error bars on [itex]\ddot{a}/a[/itex]. I think your estimate of the anomalous acceleration of the earth matches Cooperstock's, but I don't think your estimate of the force on the sun can be right...? Cooperstock estimates the effects on galaxy clusters and finds that they're insignificant, but I haven't checked his math. To get a scale where the effects are significant, I think you have to go to the largest scales at which the universe has structure (about 20% of 1/Ho) -- and I don't think structures at this scale are gravitationally bound (and certainly not in circular orbits, which is a crucial assumption of the Cooperstock result).
     
  7. Sep 5, 2012 #6

    bcrowell

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    A few more random thoughts:

    I think my analysis of the Newtonian paradox (vanishing force from a uniform cosmological fluid, despite nonvanishing force from a sphere of fluid with any finite radius) does work. Suppose the fluid has any finite r.m.s. deviation [itex]\epsilon[/itex] from uniformity. We have [itex]n[/itex] concentric shells. Each exterior shell makes a force of order [itex]\epsilon[/itex], and summing n many random terms, each of some finite size, gives a result with an infinitely large variance as n approaches infinity.

    One way in which GR-based models differ from Newtonian ones is that there is a scale ~1/Ho beyond which gravitational effects haven't had time to propagate to us. This provides kind of a natural cut-off to the summing of gravitational forces over concentric shells. However, I still don't see how we can get a nonvanishing force in a homogeneous cosmological fluid.

    Birkhoff's theorem say that we can't get any nonvanishing effect if there is no cosmological fluid in our vicinity. I.e., if you take an FRW model and scoop out a spherical vacuum region, the metric inside that sphere is Minkowski. So maybe in this sense it makes some sense to associate the effect with the local cosmological fluid...?

    I'm not sure it makes sense to interpret [itex]\rho+3P[/itex] as the Komar mass, since the Komar mass only makes sense in a stationary spacetime, and cosmological spacetimes aren't stationary. But in realistic cosmologies, P is negligible for the present epoch.

    I wonder if this analysis applies in any interesting way to Big Rip models.
     
  8. Sep 6, 2012 #7

    pervect

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    I was thinking about this a bit, and I think the best argument is the homogeneity one - basically if the cosmology is homogeneous and isotropic, you don't expect the rest of the universe (outside the orbit) to contribute anything to the motion. You seem to have thought of the same argument and a bit more to it.

    If you did have a timelike Killing vector (which we don't, quite), one would expect that the surface integral of the force-at-infinity was equal to the enclosed mass, however. I'm not sure if this is a familiar result for you, or one that would take "a lot of convincing". (But it's from Wald, about pg 288-289). If your reservations aren't about that part, but are about the part where we don't actually have a time-like Killing vector, I'd have to agree that it's not clear how seriously to take it
     
  9. Sep 6, 2012 #8

    bcrowell

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    Actually I just convinced myself this morning that my homogeneity argument doesn't work.

    Say you have an FRW cosmology, which is perfectly homogeneous. You release two test particles, A and B, close to one another, with initially parallel velocities. In an expanding universe, their velocities will start to diverge. If A is an observer, that observer will perceive B to have accelerated. If he wants to talk about "forces," he can say that there was a fictitious force acting on it (or an anomalous acceleration, which amounts to the same thing). Of course we really shouldn't conceptualize GR in terms of forces at all. And GR is also nonlinear and based on a field equation, not linear and based on instantaneous action at a distance, so you really can't say that such a "force" is a superposition of forces from everywhere in the universe.

    Homogeneity doesn't prevent this "force" from existing. The symmetry is broken by the fact that we singled out A and B. If we'd swapped the roles, making B the observer, then B would have inferred a "force" on A that was in the opposite direction.

    So maybe Wright's interpretation really is more than a heuristic. If we have a spherical region of an FRW spacetime from which the matter has been scooped out, then Birkhoff's theorem says that the metric inside is Minkowski, and test particles inside that region, released with initially parallel velocities, will never diverge. We can now add matter back into that region, restoring the missing cosmological fluid and also putting the sun and earth in. Because Newtonian gravity is locally a good approximation, the failure of parallelism of world-lines of test particles has to be the same as predicted by Newtonian gravity, based only on the matter that we put back in. In the frame of the sun, we can choose to call this a Newtonian force on the earth. In the frame of the earth, we can say the sun is accelerating the same amount, but in the opposite direction. Although it's not correct to think of the result as the global sum of all forces in the universe, I think it works as a way of finding the difference between the configuration with the hole and the configuration without it.

    One way of "test-driving" this argument would be simply to apply it to the original test particles A and B. If they're close enough so that Newtonian gravity is a meaningful approximation in some region that includes them both, then we ought to be able to calculate correctly the failure of parallelism of their world-lines by attributing it to a sphere of cosmological fluid, centered on one particle, with its radius being equal to the distance between them.
     
  10. Sep 6, 2012 #9

    Mentz114

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    I don't know if this helps, but we can compare circular orbits in the Schwarzschild vacuum with the same thing in the Kottler spacetime, which is a BH in a lambda vac solution.

    In the first case the angular velocity is
    [tex]
    \omega_{S}=\frac{1}{r}\sqrt{\frac{M}{r - 3\,M}}
    [/tex]
    and for the Kottler
    [tex]
    \omega_{K}=\frac{1}{r}\sqrt{\frac{M-{r}^{3}\,{H}^{2}}{r - 3\,M}}
    [/tex]
    I think H is actually [itex]\Lambda/3[/itex]. It looks as if the effect of the CC is to make ω smaller for the same r. But the r in the Schwrazschild coords may not be the same as the r in the Kottler metric.
    If the second equation is correct, it predicts that no circular orbit exists for M-r3H2<0 .
     
    Last edited: Sep 6, 2012
  11. Sep 6, 2012 #10

    bcrowell

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    @Mentz114: Interesting. Is this the spacetime that I would call Schwarzschild-de Sitter space, with the following metric?

    [tex] d s^2 = \left(1-\frac{2m}{r}-\frac{1}{3}\Lambda r^2\right) d t^2 - \frac{ d r^2}{1-\frac{2m}{r}-\frac{1}{3}\Lambda r^2} - r^2 d \theta^2-r^2\sin^2\theta d \phi^2 [/tex]

    Cf. http://en.wikipedia.org/wiki/De_Sitter–Schwarzschild

    The size of the horizon does seem to be different than it would be with [itex]\Lambda\ne 0[/itex]. For small [itex]\Lambda[/itex], I think the coordinate singularity happens at [itex]r\approx 2m+(8/3)\Lambda m^3[/itex], so for [itex]\Lambda>0[/itex], its r coordinate is slightly increased. I think this is where the horizon is as well. Whether this is a "real" increase depends on whether the Schwarzschild r is comparable between the [itex]\Lambda= 0[/itex] and [itex]\Lambda\ne 0[/itex] cases...? If it is real, then this matches up with what we would theoretically see for material objects in a spacetime whose expansion is accelerating: a slight increase in the object's physical size due to the apparent acceleration that goes like [itex](\ddot{a}/a)\mathbf{r}[/itex]. (The Davis and Lineweaver article in Scientific American discusses this briefly, although I've never seen it addressed in a scientific paper.)

    There is clearly no secular trend in the size of the event horizon, since the metric has no t-dependent stuff in it (and therefore it's static in the exterior region, where Schwarzschild t is timelike). The lack of a secular trend is consistent with what we see for circular orbits, which is that [itex]\Lambda[/itex] causes no secular trend, because [itex](d/dt)(\ddot{a}/a)[/itex] vanishes.

    This is nice because it shows that although there is no secular trend in orbital radii, there is a static effect, and unlike my discussion of the radius of the event horizon, this effect is clearly stated in coordinate-independent language.

    A further thought re something I said in #8:

    It's tempting to try to test this in a vacuum-dominated cosmology. However, I think you can't do that, because scooping a hole out doesn't work. You could make a spacetime in which [itex]\Lambda[/itex] was equal to a constant except in some spherical region where it was set to zero. But I think that leads to a violation of the Einstein field equations.
     
    Last edited: Sep 6, 2012
  12. Sep 6, 2012 #11

    Mentz114

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    Yes, that's it.

    I missed that, searching for 'Kottler'.

    My first guess is that the two r coordinates are not the same but we should be able to find a factor that gives a valid comparison.

    I need to read the paper(s) cited above before I can understand this bit.

    I've been playing with a perturbation of the FLRW that satisfies the EFE to first order and gives some attraction between comoving observers. It's not going very well yet.

    [Edit]The proper acceleration for the stationary frame in the Kottler case is greater than that for the Schwarzschild spacetime, being,
    [tex]
    \frac{M-{r}^{3}\,{H}^{2}}{{r}^{2}\,\sqrt{\frac{-2\,M-{r}^{3}\,{H}^{2}+r}{r}}}
    [/tex]
     
    Last edited: Sep 6, 2012
  13. Sep 6, 2012 #12

    bcrowell

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    This is it: http://www.mso.anu.edu.au/~charley/papers/LineweaverDavisSciAm.pdf

    I had an "oh, duh" moment just now, finally realized that the origin of the [itex](\ddot{a}/a)\mathbf{r}[/itex] expression for the anomalous acceleration is really trivial. For two test particles initially at rest relative to the Hubble flow, the distance between them simply grows in proportion to a. So this expression just follows trivially.

    So I think the idea in #8 does pass the test I described, except in cases where [itex]\Lambda\ne 0[/itex], where you can't scoop a hole out of the cosmological constant without violating the field equations.
     
  14. Sep 6, 2012 #13

    Mentz114

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    Ben, number #8 makes sense.
    I still have some imagination problem with the effect of [itex]\Lambda>0[/itex] on the kinematics of the Schwarzschild spacetime. My last claculation was the tidal tensor for the stationary observer frame which has components [itex]T_{rr}=(2\,M+{r}^{3}\,{H}^{2})/r^3,\ \ T_{\theta\theta}=T_{\phi\phi}=({r}^{3}\,{H}^{2}-M)/r^3[/itex] and the trace is [itex]3H^2[/itex]. So the CC is causing tidal effects. The falling cloud of dust is differently shaped, and changes volume. This looks right to me.
     
  15. Sep 6, 2012 #14

    pervect

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    I think the homogeneity argument still works if you apply it correctly - which is to say that you can ignore all the matter outside a sphere containing the two particles. As you do later on in the post.

    If you look at the effect of the matter contained within said sphere, both particles will be attracted towards the center of a sphere (assuming positive rho and no cosmological constant) and hence you predict the expected deacceleration (again given positive rho).

    Also, in the force view, and for general physical insight (but NOT for ease of calculation, sadly), I'd suggest using local fermi coodinates rather than cosmological coordinates. The picture one gets in fermi coordinates is one where the particles that were initially moving away from each other slowing down due to attractive "forces", rather than one of particles that are stationary within an expanding spacetime.

    It's really more in the spirit of GR to think of the particles as following geodesic paths but if you apply the "force" view correctly you get the same predictions. I still find the "force" view more intuitive, I suppose I'm still at heart used to the old Newtons law picture somewhere deep in my mind.

    Yes, exactly.


    To actually perform the calculation, I'd suggest using standard FRW coordinates.

    IF we take the metric for a flat FRW space time, -dt^2 + a(t)^2*(dx^2+dy^2+dz^2), an orthonormal basis of one forms is just dt, a(t)dx, a(t)dy, a(t)dz.

    We can then compute [itex]R_{\hat{x}\hat{t}\hat{x}\hat{t}}[/itex] in the orthonormal basis to be
    [tex] - \frac{d^2 a(t)}{dt^2} / a(t) [/tex]

    By the geodesic equation, this will give the relative acceleration between geodesics per unit distance, i.e.

    [tex]R_{\hat{a}\hat{b}\hat{c}\hat{d}} u^\hat{b} u^\hat{d}[/tex]

    where u is a unit vector in the t direction, i.e [itex]\hat{t}[/itex]

    We want to compare this to the quasi-Newtonian prediction. If the distance between particles is d, the enclosed mass is

    M = 4/3 pi rho (d/2)^3 = (1/6) pi rho d^3

    a = 2 * M / (d/2)^2 = 8M / d^2 = 4/3 pi rho d

    With zero cosmological constant and pressure, the Friedmann equations give us
    a'' / a = (4/3) pi rho

    so the two results tally,.
     
  16. Sep 6, 2012 #15

    pervect

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    By the time I get around to composing a post, you've already figured it out :-0
     
  17. Sep 8, 2012 #16

    bcrowell

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    I think the result of this discussion is that I now pretty much understand this to my own satisfaction. I've written up a discussion in my GR book that describes my current state of understanding. For anyone who's interested or who may have comments, the material is here http://www.lightandmatter.com/html_books/genrel/ch08/ch08.html#Section8.2 [Broken] (subsection 8.2.10), with some preliminary material developed earlier in the example titled "Scooping out a hole."

    There are a few loose ends about which I'm not completely satisfied.

    Cooperstock makes a very long and complicated calculation in an appendix. The point seems to be that they get two equations, A.15 and A.17. As far as I can tell, A.15, an expression for the acceleration of test particles, is painfully trivial, and I don't understand why anything this fancy is needed in order to derive it. A.17 is a big, ugly expression for the metric in Fermi normal coordinates. It's not trivial, but I don't see why they even need it anywhere else in the paper.

    Presumably the reason they felt compelled to do all this was so that they could address the contrary results derived by various other people, as described in the first page of section 4. Since these other papers are all old and paywalled, I don't have convenient access to them. It seems trivially obvious to me that, as Cooperstock says, you need to consider a local inertial frame if you want to see what really happens according to a local observer -- you clearly can't just pick some random coordinate system and expect the coordinates to have clear interpretations. It doesn't seem difficult to me to verify that their A.15, is what you get in a local inertial frame. Therefore it makes me uncomfortable trying to imagine how really smart people like Dicke could have gotten the wrong result before.
     
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  18. Sep 10, 2012 #17

    pervect

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    The idea behind using Fermi normal coordinates is that they do represent, locally, as best as possible, the coordinates that a local inertial frame would assign.

    While mathematically involved to calculate with, introducing Fermi normal coordinates make sure one hasn't overlooked something.

    For instance, the local Fermi observer's idea of "now" is different from the cosmological "now", though the two hypersurfaces are tangent at the point of the observer.

    By actually solving for the equations of motion in Fermi normal coordinates one gets rid of any nagging concerns that the assumptions behind using an inertial frame are "good enough". Instead one demonstrates that in the appropriate limit the inertial frame computations agree with the full computations that don't assume an inertial frame (and don't need to), up to some order.
     
  19. Sep 10, 2012 #18

    bcrowell

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    Thanks, pervect -- that's helpful!
     
  20. Sep 11, 2012 #19

    pervect

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    This may or may not be a bit of a topic drift, but it might be helpful for your book somewhere or other. If you cover the universe with co-moving observers, they will all be drifting apart from each other. If you cover the universe with Fermi observers, they will maintain a constant distance from each other, but only one of them will be at rest relative to the cosmological fluid.

    We can imagine mounting accelerometers on all the Fermi observers, to see what sort of forces they experience required to "hold them in place". We know that the reading will be zero on the observer co-moving with the cosmological fluid, as that's a geodesic. But what about the rest? The detailed calculation is in the paper you reference, you can in fact look at the g_00 term in the metric of the Fermi obsevers and take the gradient of g_00 to find the force, (this is actually an approximation that works when g_00 is nearly unity) just as you do in Newton-Cartan theory.

    I find the Fermi observers, which I semi-secretly imagine as being supported by a lattice of physically rigid rulers, provides a much more intuitive (to me at least) coordinate system for describing the physics of the "effects of expansion", even if the details of the calculation are messy.
     
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