Dark energy = cosmological constant, any problems with that?

[kurros]

All right, so doesn't this mean that if I bring something up to the fourth floor and let it fall is equivalent to bring something in accelerating spacetime and let it stretch?

Aside from Peter's technicality I would say that's the gist of it, yes.

 

[Jorrie]

Aside from Peter's technicality I would say that's the gist of it, yes.

I think 'Peter's technicality' renders timmdeeg's statement completely invalid.

If the "something" is a compressible "mass-less" rod and you could put it in free expanding space, then one of three things could happen. If the expansion is accelerating, the rod will stretch; if the expansion is decelerating, it will compress and if the expansion is doing neither, it will have no stresses. In the first case geodesics of points along the rod are diverging, in the second they are converging and in the third they obviously remain parallel.

This is totally different to what happen to rods that are supported and then dropped in an earth-like gravitational field.

Check out the "Tethered Galaxy Problem", Davis et al.

 

[timmdeeg]

In the dark energy case, nothing is pushing on anything. The effect of dark energy is really like tidal gravity, not the ordinary "gravity" you think of when you think of falling objects on Earth.

Yes agreed, we have to distinguish between gravity and tidal gravity. An inhomogeneous gravitational field or expanding spacetime causes tidal acceleration. If something flexible is exposed to it the created tension is a source of gravity. In the case of the expanding universe I could argue don't worry the energy isn't conserved anyway. But this very vague and perhaps not correct in this specific case. But an inhomogeneous gravitational field caused by a central mass implies static spacetime und thus energy is conserved. Where is the energy to stretch something taken from in either case?

EDIT the case of the forth floor: does the gravitational potential energy of something include the energy for stretching it if released to free fall?

 

[kurros]

I think 'Peter's technicality' renders timmdeeg's statement completely invalid.

If the "something" is a compressible "mass-less" rod and you could put it in free expanding space, then one of three things could happen. If the expansion is accelerating, the rod will stretch; if the expansion is decelerating, it will compress and if the expansion is doing neither, it will have no stresses. In the first case geodesics of points along the rod are diverging, in the second they are converging and in the third they obviously remain parallel.

This is totally different to what happen to rods that are supported and then dropped in an earth-like gravitational field.

Check out the "Tethered Galaxy Problem", Davis et al.

Not if consider the whole system. When you drop something in the Earth's gravitational field, the Earth also falls *up* towards the object. Both are in free fall towards each other, with identical acceleration. So their geodesics are converging.

Put it this way. Attach masses to your massless spring. Depending on the masses and their separation the spring will either be stretched or compressed, depending on whether the gravitational attraction of the masses exceeds the acceleration due to dark energy. The only difference is the sign and r dependence.

 

[Jorrie]

Put it this way. Attach masses to your massless spring. Depending on the masses and their separation the spring will either be stretched or compressed, depending on whether the gravitational attraction of the masses exceeds the acceleration due to dark energy. The only difference is the sign and r dependence.

If you want to 'look at the whole system' that way, you need an infinite number of masses, homogeneously spread. Then there is no gravitational potential gradient drawing masses towards each other - they just go with the expansion dynamics, be it slowing down, remaining constant or increasing faster in terms of proper distance between them.

As soon as you look at what happens in the vicinity of each mass, you will get the local gravitational effects that you refer to. And as Peter has repeatedly said, they are very different than the large scale effects.

 

[kurros]

If you want to 'look at the whole system' that way, you need an infinite number of masses, homogeneously spread. Then there is no gravitational potential gradient drawing masses towards each other - they just go with the expansion dynamics, be it slowing down, remaining constant or increasing faster in terms of proper distance between them.

So? I'm not sure what your point is. The dark energy isn't sourced by the masses, sure, but so what?

As soon as you look at what happens in the vicinity of each mass, you will get the local gravitational effects that you refer to. And as Peter has repeatedly said, they are very different than the large scale effects.

Different how? We are just talking about whether geodesics diverge or converge in both cases. I.e. whether things fall towards or away from each other. No reason to think the situation re. energy conservation is more mysterious in one case or the other.

 

[Jorrie]

So? I'm not sure what your point is. The dark energy isn't sourced by the masses, sure, but so what?

The dark energy is sourced by the cosmological constant. It does not need any masses around to produce curved spacetime and diverging geodesics. So I think discussing two masses misses the point, where the energy available to be released is dependent on the potential energy between the masses. In de Sitter spacetime (cosmological constant only), there is no potential energy differences between points.

Whether we view the energy available in the latter as coming from the uniform negative curvature of spacetime, or from the "energy of empty space" is a matter of preference of which math representation of the same physics you want to use.

I would like to hear Peter's deeper insight into this matter.

 

[PeterDonis]

does the gravitational potential energy of something include the energy for stretching it if released to free fall?

No. If the object is compressed before it is dropped, there is energy stored in the object, and that energy is not gravitational potential energy.

 

[PeterDonis]

In de Sitter spacetime (cosmological constant only), there is no potential energy differences between points.

Actually, there is, as I posted earlier in this thread; de Sitter spacetime has an infinite number of timelike Killing vector fields, and if you pick one (which amounts to picking a particular comoving worldline and centering static de Sitter coordinates on it), you can define a potential energy relative to it. The chosen comoving worldline becomes the "top" of a potential energy "hill".

I think you can actually do something similar, locally, for a case like tidal gravity in a spaceship in a free-fall circular orbit about a spherically symmetric planet (basically by setting up Fermi normal coordinates centered on the ship's center of mass--circular orbit + spherically symmetric planet means the CoM worldline is an integral curve of a timelike KVF), but I haven't had a chance to work out the math.

 

[PeterDonis]

When you drop something in the Earth's gravitational field, the Earth also falls *up* towards the object. Both are in free fall towards each other, with identical acceleration. So their geodesics are converging.

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.

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.

 

[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.