Graduate Dark energy = cosmological constant, any problems with that?

[Jorrie]

The chosen comoving worldline becomes the "top" of a potential energy "hill".

Thanks, I now understand this for pure de Sitter. For the argument, I have originally used identical spherical masses spread uniformly throughout, making it equivalent to our present $\Lambda$ dominated LCDM universe. The center of any chosen mass becomes a local potential "valley" and the centers of the "voids" surrounding it, local "hills". Choosing any void as origin, it will be the "top of a potential energy hill", because everything is receding from it.

I think this will be the same for any expanding space, $\Lambda = 0$ or not. The question that is still a little puzzling: in the case of a dominant $\Lambda$, can energy for local use in principle be extracted by a suitable converter? I.e. can I warm my food without using any fuel carried along? And if so, does the energy come from "the vacuum" or from the intrinsic global curvature? Ok, I know these are not well defined concepts, but I think the question is suitably general...

 

[kurros]

You are once again leaving out a crucial point: the surface of the Earth, which is what the dropped object hits, is not traveling on a geodesic. Only the Earth's center of mass is. But the Earth's center of mass is thousands of kilometers away from the impact point. The impact point is accelerating upward at 1 g, and upward acceleration is where the work done comes from. In the cases you are proposing, there is no such acceleration, which invalidates the analogy you are trying to draw here.

What? Oh you mean once the mass stops falling? But that wasn't the question here, we were just discussing the falling part.

It is true that, if you take two masses that are momentarily at rest relative to each other, and release them in a converging tidal gravity field (for example, have them oriented tangentially in the field of a spherically symmetric planet), they will converge, which means they will pick up some kinetic energy relative to each other, which you could in principle extract. But this energy is not the same as the gravitational potential energy of either mass in the planet's field; it's much smaller.

Sure, but it can still be pretty big. It depends on the masses involved. Tidal forces can tear apart stars, so they are not universally small.

 

[timmdeeg]

If the object is compressed before it is dropped, there is energy stored in the object, and that energy is not gravitational potential energy.

Two cases regarding free fall towards a central mass M (not compressed or stretched before being dropped and disregarding the impact):

The dropped mass

A) is small or rigid.
B) is flexible and large.

In contrast to A) in case B) work is done to stretch/compress the mass. On whose costs is this work done? What looses the amount of energy equivalent to this work?

 

[PeterDonis]

I think this will be the same for any expanding space

No, it won't. De Sitter spacetime--positive $\Lambda$ but no other stress-energy present--is special because it has timelike Killing vector fields. An FRW spacetime with any stress-energy present other than $\Lambda$ has no timelike Killing vector field, so there's no way to define a potential energy.