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Imagine I have a setup like this:

the setup

(The objects are not attached to each other. The red objects are much heaver than the black one, and the setup is balanced.)

This setup has a non-zero gravitiational potential energy. The red objects, if the black object was removed, would crash into each other. However, the black object is stopping this.

However, since the expansion of the universe is accelerating, there will come a point when the expansion of the universe starts to move these objects away from each other.

These objects are now moving away from each other. There are two scenarios that I think might happen:

  1. They don't have gravitational potential energy anymore; they can't attract each other because of the expansion of the universe forcing them apart. As such, gravitational potential energy is now 0. However, at the time of the initial setup of these objects, the gravitational potential energy was more than zero. As such, we have destroyed energy.
  2. Their gravitational potential energy increases, because they have further to fall due to the expansion of the universe. However, at the time of the initial setup of these objects, the gravitational potential energy was below what it is at the current time. As such, we have created energy.

Where is the flaw in my logic?

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  • $\begingroup$ The answers say that no, energy is not created because the blocks are gravitationally bound and gravitationally bound things are not affected by dark energy (dubious). I wonder what the case would be if the blocks are not gravitationally bound. Say the blocks are galaxies and the black bar is a string (of negligible mass) connecting the galaxies across cosmological distances. As the galaxies drift apart due to dark energy, the string is pulled taut or pulled along with one of the galaxies, driving an engine that powers something. Is there a question about this already on this SE somewhere? $\endgroup$
    – BMF
    Commented Aug 13, 2019 at 20:31

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Gravitationally bound systems do not expand very much because of cosmological expansion. For example, the predicted general-relativistic effect on the radius of the earth's orbit since the time of the dinosaurs is calculated to be about as big as the diameter of an atomic nucleus; if the earth's orbit had expanded according to the cosmological scaling function a(t), the effect would have been millions of kilometers. For more on this, see this question: Can the Hubble constant be measured locally?

So yes, a local observer can detect strains and stresses due to cosmological expansion, and these are capable of doing work, but in your example, they would never be enough to achieve lift-off of the red blocks.

The fact that these forces can do work does violate conservation of energy. General relativity has local energy conservation, but no global energy conservation. See Is the total energy of the universe zero?

They don't have gravitational potential energy anymore; they can't attract each other because of the expansion of the universe forcing them apart.[...]

As a side issue, this paragraph seems to show a misunderstanding of what potential energy is. We can't define whether potential energy is zero, because it involves an arbitrary additive constant.

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    $\begingroup$ General relativity has local energy conservation, but no global energy conservation Totally false. GR does not have a unique (local or global) conserved energy applicable to every possible situation. But there are a lot of situations (again, local and global) of physical interest where energy of a gravitating system could be a well-defined notion. $\endgroup$
    – A.V.S.
    Commented Aug 12, 2019 at 14:40
  • $\begingroup$ I have a small question. If the Big Rip were to happen, then wouldn't the expansion of the universe rip quarks from each other, causing hadronization? $\endgroup$
    – Aravind Karthigeyan
    Commented Aug 12, 2019 at 16:02
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    $\begingroup$ @A.V.S.: You're jumping the gun. This is expressed in nontechnical language, because the OP is obviously not a specialist. What I'm expressing when I refer to local conservation is the fact that the stress-energy has a zero divergence. The OP would have not have understood that. $\endgroup$
    – user4552
    Commented Aug 12, 2019 at 17:26
  • $\begingroup$ @AravindKarthigeyan: If you click through to the question and answer about "Can the Hubble constant be measured locally?," you'll see that the stresses and strains induced in a bound system by cosmological expansion are proportional to $\ddot{a}/a$, and the secular trend of the system's response goes like $(d/dt)(\ddot{a}/a)$. These quantities are small in standard cosmological models, which is what I'm referring to here, but they diverge in big rip models. $\endgroup$
    – user4552
    Commented Aug 12, 2019 at 17:30
  • $\begingroup$ You're jumping the gun. This is expressed in nontechnical language I am quite certain that it should be possible to stay nontechnical without resorting to false absolute statements (like “There is no <insert Newtonian/SR concept here> in GR”). $\endgroup$
    – A.V.S.
    Commented Aug 12, 2019 at 17:50
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When the Universe expands, the distance between gravitationally bound objects does not increase (see here). That is why the red objects will never start to move away from each other (unless the Big Rip happens).

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