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The title is the question. Here's why it seems like local kinetic energy should increase:

Numerous questions and answers here and elsewhere suggest that the reason the metric expansion of space is not observable locally (even on a galactic scale) is that local forces maintain the metric distance between near objects, with "near" meaning anything from two subatomic particles to two celestial bodies.

My mental picture is of ants on an expanding balloon. The ants get farther apart, but the expansion of the balloon beneath their feet is not going to tear the individual ants apart. They do however have to do a little dance to keep from doing the splits.

It seems to me that this should imply a continuously increasing kinetic energy, presumably generalized as random motion -- i.e. things get hotter.

Here is a simple illustration. A and B could be particles attracted by the electromagnetic force, or whole planets attracted by gravity:

  1. A begin state with an arbitrary distance and energy level:

    T 0

  2. A little bit later, the two objects are farther apart, implying an increase in potential energy:

    T 1

  3. But the force at work brings the objects back to their original positions distance, implying the potential energy was converted to (kinetic?) energy through work W:

    T 2

Obviously A and B don't actually wiggle apart and then back together, rather the force between them acts constantly as the metric expands constantly, implying a smooth increase in kinetic (or some other type) energy while the measurable distance between the objects stays the same.

Scenarios

  1. Gravity at planetary distances: Jerry Schirmer's answer suggests the effect would be too small to measure
  2. What about at subatomic scales? The color force between quarks in a nucleon is a distance-dependent force whose potential energy is orders of magnitude greater than gravity. Should metric expansion not cause an increase in energy of a individual nucleon, detectable as a few extra photons emitted over the course of some time?
Qmechanic
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Joshua Honig
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The notion of kinetic energy is ill-defined in the spacetimes where you have a time-dependent cosmological expansion.

If you somehow attached two galaxies to each other with a spring, however, the expansion of the universe would do "work" against that spring, as there would be a force requied to keep the proper distance of the two galaxies fixed. At solar system distances, however, this would be completely undetectable.

Zo the Relativist
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  • Granted gravity is relatively weak. What about the color force at quantum scales? (I've appended a bit to my question to highlight this) – Joshua Honig Aug 18 '14 at 14:36
  • @JoshuaHonig: that effect would be yet smaller. The force you get will be proportional to the distance between the objects -- you're basically doing work against the velocity given by Hubble's law, after all. – Zo the Relativist Aug 18 '14 at 15:12
  • To the best of our measurments, you can completely ignore gravity when doing any atomic or particle physics. – Zo the Relativist Aug 18 '14 at 15:13
  • The case of an atom is qualitatively different from the case of a solar system. The solar system does theoretically have a secular trend of growth due to cosmological expansion, but the amount is much too small to measure. An atom does not show any such secular trend, even in theory, because it is a quantum-mechanical system with a well-defined ground state. I gave some numerical estimates in this answer: http://physics.stackexchange.com/a/70056/4552 –  Aug 18 '14 at 18:32
  • @BenCrowell: well, it has a well-defined ground state. Technically, the fact that it is in a cosmological background introduces a perturbation Hamiltonian which would change the energy levels, etc, etc. The effect would be ludicrously small, of course. – Zo the Relativist Aug 18 '14 at 20:30
  • @JerrySchirmer: That's different from a secular trend. –  Aug 19 '14 at 03:58
  • @BenCrowell, "The solar system does theoretically have a secular trend of growth due to cosmological expansion" is incorrect; there is no such growth at any time scale. – benrg Aug 19 '14 at 06:55
  • @JerrySchirmer, the (homogeneous, isotropic) FLRW expansion does not affect atomic energy levels even in principle. This follows from the GR version of the shell theorem. There is also no pulling associated with expansion from which you could extract energy. You can extract energy from any relatively moving objects including superclusters by decreasing their relative speed, but they won't return to their old speed unless you spend that energy to push them. – benrg Aug 19 '14 at 06:57
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    @benrg: No. You're wrong. That statement contradicts an analysis of orbits in asymptotically Robertson-walker spacetimes. Orbits ARE static (but definitely altered) in Kerr-de Sitter space. And if you force a particle to move on something other than a geodesic, you definitely can extract work. Requiring macroscopically seperated objects to keep a constant proper distance is definitely forcing them not to move on a geodesic. – Zo the Relativist Aug 19 '14 at 14:57
  • @JerrySchirmer: I may have misinterpreted what you wrote. At any rate, what I was trying to say was that the perturbation of the orbits only depends on $G$ and $\Lambda$, not on $H(t)$ or $a(t)$. The presence or absence of an expanding universe out there doesn't affect the orbit. Likewise, while a force is generally required to keep two objects at a fixed separation, the force doesn't depend on $H(t)$, and you can't extract work from it while keeping the separation constant. (Consider that energy gravitates, and the ΛCDM universe is around forever.) – benrg Aug 19 '14 at 19:04
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    @benrg: but that's not true, and can't be true. $\Lambda$ can easily be hidden inside of $a(t)$, after all. And there is a new force on orbits (actually all orbits are ultimately unstable) in any model where you have a nonzero cosmology. This is manifest if you write the robertson-walker metric in terms of physical coordinates, rather than comoving coordinates. It requires a net force to deviate a particle from constant comoving distance. – Zo the Relativist Aug 19 '14 at 19:45
  • @JerrySchirmer: If you assume FLRW geometry, you're assuming that spacetime is permeated with Hubble-expanding matter. At the atomic or solar-system scale you're assuming that there is Hubble-expanding matter inside the atom or solar system. It probably will give you a net force, but it doesn't actually exist. In the case of the spring-coupled galaxies, any a(t)-dependent force you see is again the effect of this matter. If you assume FLRW is correct over the time you extract the work, you are treating that matter as an unmoved mover, which is why you can get unlimited work from it. – benrg Aug 19 '14 at 20:50
  • @JerrySchirmer: Here's another way of looking at it: the expansion force if it exists must be isotropic. Weyl curvature can't be isotropic; any isotropic gravitational effect must be Ricci. But the Ricci curvature in a region depends only on the matter there, by the GR field equation. There is no long-range Ricci field. This is just a rewording of the shell-theorem argument. – benrg Aug 19 '14 at 21:03
  • @benrg: "The solar system does theoretically have a secular trend of growth due to cosmological expansion" is incorrect; there is no such growth at any time scale. No, you're incorrect. See Cooperstock, Faraoni, and Vollick, "The influence of the cosmological expansion on local systems," http://arxiv.org/abs/astro-ph/9803097v1 , which I've also given a condensed presentation of here: http://physics.stackexchange.com/a/70056/4552 –  Aug 20 '14 at 17:34
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    @benrg: the perturbation of the orbits only depends on G and Λ, not on H(t) or a(t). Not true. See the links in my preceding comment. There is a strain proportional to $\ddot{a}/a$, and a secular trend proportional to $(d/dt)(\ddot{a}/a$. Both of these can be nonvanishing even if $\Lambda=0$; in an FLRW spacetime, the Friedmann equations make the second expression proportional to $\dot{\rho}$. At the atomic or solar-system scale you're assuming that there is Hubble-expanding matter inside the atom or solar system. Again, see the two links. –  Aug 20 '14 at 17:39
  • @BenCrowell: Cooperstock et al say "the analysis of a spherical cavity embedded in an FRW universe is well known: [...] the physics is the same as in flat space". You say the same thing on your web site. You also say that if the cavity is filled with expanding matter then there is a force. This is all correct. What you seem to be missing is that solar systems are not filled with expanding matter, so there is no force. I think Cooperstock et al would back me up on this. – benrg Aug 21 '14 at 00:06
  • @JerrySchirmer and Ben Crowell: I would like to discuss this to some sort of conclusion (probably on chat), because I think it's important, and it affects a number of old questions/answers on this site. – benrg Aug 21 '14 at 00:06
  • @benrg the more I think of this,the more I think this is actually an open research question,mostly because you're appealing to the fact that the cosmology isn't locally important.But at that point,you have to consider the local cosmological field to be some sort of mean field that will be dependent on the particular way that matter from local galaxies is lumpy and inhomogenous, which breaks the local spherical symmetry on which the shell theorem depends. You can either model the cosmology as smooth all the way down, or it becomes lumpy.In the latter case, you can't ignore the lumpiness. – Zo the Relativist Aug 21 '14 at 04:16
  • @JerrySchirmer: I'm happy to forget about the shell theorem and stick to the Ricci argument, which says that the nonlocal influence of matter is limited to tidal forces/spaghettification because there is no long-range Ricci field. This has a Newtonian analogue: $\nabla^2\Phi = 0$ in vacuum, so the potential cannot be concave downward (expansion) or upward (contraction) in vacuum. This means no large-scale matter distribution, isotropic or not, can create an expansion force. (cont'd) – benrg Aug 21 '14 at 05:11
  • (cont'd) But the more fundamental point is that there are no special rules for expanding universes. It's just GR acting locally as always. You can treat spacetime as FLRW plus perturbations, but the universe doesn't care about the leading term, just the sum. At solar system or galactic scales FLRW is so far from correct that it's not a useful perturbative background. You'd be better off perturbing around Minkowski space, and there's no reason you can't. (cont'd) – benrg Aug 21 '14 at 05:13
  • (cont'd) The nonperturbative picture is that the spacetime curvature depends on the matter distribution, which is the case of a galaxy or solar system is mainly the matter in the galaxy or solar system. There is an influence from faraway matter because gravity has infinite range, but the Ricci argument shows that it has to be tidal, not an overall expansion. This isn't an open research question; it's just GR. – benrg Aug 21 '14 at 05:14
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The expansion does lead to a kinetic energy term that can be (at least partially) extracted. For objects that are bound to each other, it would lead to a classic acceleration term, which, of course, is equivalent to a classic pseudo-force. In an expanding universe, any two objects that are bound by a potential, are therefor experiencing an additional (albeit small) repelling potential term.

CuriousOne
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The expansion of the universe is not a force. Forces don't pull things apart at a particular speed; they change the speed by a particular acceleration. The speed itself is just inertia. It's no different in cosmology: there is nothing actively pulling things apart at the speed given by Hubble's law; that speed is just leftover momentum from the big bang, as modified over the intervening time by the attractive gravitational force and the repulsive cosmological constant.

To make this concrete, consider, say, two empty beer cans ($m = 15\,\mathrm g$ each) separated by $d = 1\,\mathrm{km}$ in outer space. Then we have

  • Hubble's law separation speed = $H_0 d \sim 2\cdot10^{-15}\,\mathrm{m/s}$
  • Gravitational attraction = $2Gm/d^2 \sim 2\cdot10^{-18}\,\mathrm{m/s^2}$
  • Cosmological constant repulsion = $(\Lambda/3)d \sim 10^{-32} \,\mathrm{m/s^2}$ (I'm not sure about the factor of ⅓, but this is just an order-of-magnitude estimate. And I'm ignoring factors of $G$ and $c$ because there are no standard units for $\Lambda$.)

Although I quoted the Hubble's law velocity above, it is irrelevant. The only way the beer cans would have that relative velocity is if they coalesced directly from primordial hydrogen and helium, thus inheriting its average velocity, or if you deliberately gave them that relative velocity (which would be kind of like burying fake dinosaur skeletons). If they don't start out with that relative velocity, there is nothing that will make them tend to have it later. Sorry for all the boldface and italics, but this seems to be a very common misconception, possibly because of the misleading inflating-balloon analogy.

Of the two forces actually acting on the cans, the attractive one overwhelmingly dominates the repulsive one. So if these cans are initially at relative rest, they will very gradually start moving toward each other, not away.

These forces do perturb atomic energy levels and such, but the gravitational attraction again overwhelmingly dominates the cosmological repulsion, and even the attractive effect is undetectably small as far as I know. The Hubble recession does not affect atomic energy levels, not even a little bit.

You can extract energy from Hubble recession, but you're limited to the relative kinetic energy, as with any other relative motion. You can extract energy from gravitational attraction, but you're limited to the total potential energy. For cosmological repulsion, I think you can define a similar potential energy in de Sitter static coordinates, implying that the total energy you can extract from that over the lifetime of the universe is limited as well.

benrg
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    The cosmological constant is not a gravitational force between two objects. In fact, pure de Sitter space has no expansion at all -- it has a global timelike killing vector. – Zo the Relativist Aug 19 '14 at 19:47
  • @JerrySchirmer: de Sitter space has no global timelike Killing field. Maybe you're thinking of de Sitter's original metric, which doesn't cover the whole space, but does have a "global" Killing field. The Schwarzschild exterior geometry likewise has a "global" Killing field, yet the central mass does gravitate. The two cases are very similar. So I don't understand how you concluded that "the cosmological constant is not a gravitational force". – benrg Aug 21 '14 at 00:47
  • de Sitter can be written down with the line element $ds^{s} = -\left(1- \lambda r^{2}\right)dt^{2} + \frac{dr^{2}}{1 - \lambda r^{2}} + r^{2}d\Omega^{2}$, which has a manifest timelike killing vector. – Zo the Relativist Aug 21 '14 at 01:07
  • @JerrySchirmer, that's de Sitter's original metric, which doesn't cover the whole space. – benrg Aug 21 '14 at 01:29
  • the space is maximally symmetric, and in that patch, has a manifest timelike killing vector. By the global symmetries, you can rotate that patch wherever you want it. There might be caustics, but every point has a finite neighborhood with a timelike killing vector. – Zo the Relativist Aug 21 '14 at 02:29
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    @JerrySchirmer: True. I'm fine with the statement that "pure de Sitter space has no expansion at all". Likewise Schwarzschild space isn't shrinking. Actually I guess by "The cosmological constant is not a gravitational force between two objects" you meant it's not one can acting on the other can. I didn't mean to suggest otherwise. I only said it's a force, which it is in the same sense gravitational attraction is (geodesic motion in curved spacetime). It's repulsive in that it acts to increase the distance between the cans over time. – benrg Aug 21 '14 at 03:57
  • The factor of 2 was because the force/"force" acts separately on each can, but on second thought the distance should probably have been (d/2) then. I'll change it to 1/3. (The 1/3 is from the dimensionality of space, of course, but I'm still not sure I have this right. Maybe I'm missing a factor of 4π.) – benrg Aug 21 '14 at 03:59