I What is the acceleration of expanding space?

[timmdeeg]

That's correct; dark energy produces Ricci curvature--negative Ricci curvature, which is why it causes a small ball of test particles to expand (positive Ricci curvature, like that due to ordinary matter, causes a small ball of test particles to contract).

Which means that dark energy creates tidal gravity due to Ricci curvature.

The presence of dark energy does create a kind of "force" (more precisely, tidal gravity) that, in principle, is present on all scales and within gravitationally bound systems

The tidal gravity present within gravitationally bound systems you are mentioning here seems to be also due to Ricci curvature because it is created by dark energy.(*) Is this correct?

Where did I say it can? Didn't I say that in a vacuum region in a gravitationally bound system, the relevant curvature is Weyl curvature, not Ricci curvature?

Yes you did. I seem to be mislead here(*) but can't see why.

 

[PeterDonis]

Which means that dark energy creates tidal gravity due to Ricci curvature.

Ricci curvature is (one kind of) tidal gravity.

The tidal gravity present within gravitationally bound systems you are mentioning here seems to be also due to Ricci curvature because it is created by dark energy.(*) Is this correct?

No. Go back and read my previous posts again, carefully.

I seem to be mislead here(*) but can't see why.

Because you aren't paying enough attention to details. There are a total of four cases we are discussing, and you have to take care to keep them distinct:

(1) A homogeneous expanding universe (on large scales) that is dominated by ordinary matter (or radiation, though we haven't really mentioned that here). This creates positive Ricci curvature, which causes the expansion to decelerate. This is our best current model of the early universe.

(2) A homogeneous expanding universe (on large scales) that is dominated by dark energy. This creates negative Ricci curvature, which causes the expansion to accelerate. This is our best current model of the universe on large scales today.

(3) The interior of a gravitating body like a planet or star in a gravitationally bound system like a solar system or galaxy. The stress-energy inside the body creates positive Ricci curvature, because it's ordinary matter, just like case #1 above. But unlike case #1, here the system is static, not expanding, and it's supporting itself against its own gravity by pressure. That makes it very difficult to actually observe any effects of the Ricci curvature; strictly speaking, the effect of Ricci curvature is to require nonzero pressure inside the body, but it takes some analysis to see why (more complicated analysis than the simple one, e.g. as given in the Baez article, that explains cases #1 and #2 above).

(4) The vacuum region outside a gravitating body like a planet or star in a gravitationally bound system like a solar system or galaxy. The curvature here is all Weyl curvature (since Ricci curvature is zero in vacuum). This is the sort of tidal gravity that is usually referred to by that term, the kind that causes tides in the Earth's oceans because of the Weyl curvature due to the Moon and Sun.

 

[timmdeeg]

Thanks for your effort and patience.

(3) The interior of a gravitating body like a planet or star in a gravitationally bound system like a solar system or galaxy. The stress-energy inside the body creates positive Ricci curvature, because it's ordinary matter, just like case #1 above. But unlike case #1, here the system is static, not expanding, and it's supporting itself against its own gravity by pressure. That makes it very difficult to actually observe any effects of the Ricci curvature; strictly speaking, the effect of Ricci curvature is to require nonzero pressure inside the body, but it takes some analysis to see why (more complicated analysis than the simple one, e.g. as given in the Baez article, that explains cases #1 and #2 above).

(4) The vacuum region outside a gravitating body like a planet or star in a gravitationally bound system like a solar system or galaxy. The curvature here is all Weyl curvature (since Ricci curvature is zero in vacuum). This is the sort of tidal gravity that is usually referred to by that term, the kind that causes tides in the Earth's oceans because of the Weyl curvature due to the Moon and Sun.

Yes I understand these points and also the explanations in your previous posts. My confusion goes back to your post #17.

The presence of dark energy does create a kind of "force" (more precisely, tidal gravity) that, in principle, is present on all scales and within gravitationally bound systems; however, its magnitude on the scale of gravitationally bound systems is so tiny that it can't be measured and has no observable effect on the properties of such systems. It is only observable on very large distance scales and very long time scales (billions of light years and billions of years).

My impression here was that the "force" due to dark energy (and hence due to tidal gravity originating from Ricci curvature) is tiny but in principle has an effect on gravitationally bound systems such that they expand. But that impression seems wrong because said "force" instead has an effect on the gravitating bodies belonging to a gravitationally bound system (not on the system as a whole) and then this is consistent with your points (3) and (4).

 

[PeterDonis]

My impression here was that the "force" due to dark energy (and hence due to tidal gravity originating from Ricci curvature) is tiny but in principle has an effect on gravitationally bound systems such that they expand.

Not that they expand; just that in principle they are very, very, very, very slightly larger than they would be in the absence of dark energy. In practice the difference is much too small to measure.

that impression seems wrong because said "force" instead has an effect on the gravitating bodies belonging to a gravitationally bound system (not on the system as a whole) and then this is consistent with your points (3) and (4).

I'm not sure what you mean here. Dark energy is everywhere, and its density is the same everywhere; it's the same inside the Earth or the Sun as in the vacuum between them. But its density is so tiny that its effects are way too small to measure.

 

[timmdeeg]

Not that they expand; just that in principle they are very, very, very, very slightly larger than they would be in the absence of dark energy. In practice the difference is much too small to measure.

I'm not sure what you mean here. Dark energy is everywhere, and its density is the same everywhere; it's the same inside the Earth or the Sun as in the vacuum between them. But its density is so tiny that its effects are way too small to measure.

Yes we can assume that the density of dark energy is the same everywhere and with respect to accelerated expansion that the dark energy density dominates the matter density. The latter however is inhomogeneous.

I wonder how we do handle that. Of course if we apply the average matter density (which can be seen as being homogeneous on very large scales) locally then we would expect a extremely tiny expansion of gravitational bound systems as you said.
Now two questions:

(1) What allows us to apply the average matter density locally? Compared to the average matter density on large scales the average matter density of e.g. a galaxy is much much higher. But nevertheless it seems that the global accelerated expansion is assumed to work the same way on such small scales.

(2) On very large scales we assume homogeneous matter density consistent with the perfect fluid model, Ricci curvature and no Weyl curvature. Locally within gravitationally bound systems vacuum is by far predominant. So we should assume Weyl curvature and no Ricci curvature (with the exception of gravitating bodies). What allows us to consider a very tiny accelerated expansion (which requires Ricci curvature) of the system though?

So in short my concern is that we seem to apply at small scales what is primarily true globally. I didn't get the reasoning behind that respectively how that can be explained heuristically.

 

[PeterDonis]

The latter however is inhomogeneous.

On small enough distance scales, yes. On large enough distance scales, homogeneity is a good approximation. That's why it's used in the models of the universe as a whole in cosmology. Those models aren't intended to apply on distance scales small enough for inhomogeneities to be significant.

Of course if we apply the average matter density (which can be seen as being homogeneous on very large scales) locally

Which we don't. Nobody models a galaxy, let alone a solar system, let alone a star or planet, by assuming it has the average matter density of the entire universe. Why would you think they would?

I'm confused about where you are getting these impressions from.

my concern is that we seem to apply at small scales what is primarily true globally

Which is an unfounded concern, because we don't; why would we, since it would make no sense?

Once again, I'm confused about where you are getting these impressions from. You seem to have some underlying assumptions that are leading you astray, but I don't know what they are.

 

[PeterDonis]

On very large scales we assume homogeneous matter density consistent with the perfect fluid model, Ricci curvature and no Weyl curvature.

Yes.

Locally within gravitationally bound systems vacuum is by far predominant. So we should assume Weyl curvature and no Ricci curvature (with the exception of gravitating bodies).

Yes.

What allows us to consider a very tiny accelerated expansion (which requires Ricci curvature) of the system though?

We don't. The effect of dark energy on these scales is way too small to matter. So nobody considers it in a practical sense.

In principle, if you just look at a theoretical model, I don't see what the problem is with having a tiny amount of Ricci curvature due to dark energy present in a bound system. Its effect wouldn't be to make the system undergo a very tiny accelerated expansion; it would just be to make the system (for example, the average orbital radius of a planet around its sun) larger by a very tiny amount. But there is nothing theoretically wrong with this. So I'm confused about why you apparently see a problem with it.

 

[timmdeeg]

Which we don't. Nobody models a galaxy, let alone a solar system, let alone a star or planet, by assuming it has the average matter density of the entire universe. Why would you think they would?

I didn't like this idea but had the feeling that somehow the global matter density should be "felt" locally. Thanks for clarifying this point. And sorry that I didn't realize that earlier, wasting your time thereby.

In principle, if you just look at a theoretical model, I don't see what the problem is with having a tiny amount of Ricci curvature due to dark energy present in a bound system. Its effect wouldn't be to make the system undergo a very tiny accelerated expansion; it would just be to make the system (for example, the average orbital radius of a planet around its sun) larger by a very tiny amount. But there is nothing theoretically wrong with this. So I'm confused about why you apparently see a problem with it.

If we anticipate zero matter density in a gravitational bound system then this tiny effect is clear. Should one understand it such that the system is stretched isotropically very tiny in accordance with the equilibrium of the "forces"?

Would we expect the same amount of tiny stretching of a gravitationally bound system in the early universe where the amount of average matter density was much higher (and the universe expanding decelerated) but the density of the dark energy (or the cosmological constant respectively) the same as today?

 

[PeterDonis]

If we anticipate zero matter density in a gravitational bound system

I assume you are referring specifically to the vacuum regions between objects like stars and planets? If we include the stars and planets the density is certainly not zero.

Should one understand it such that the system is stretched isotropically very tiny in accordance with the equilibrium of the "forces"?

That's one heuristic way of thinking about the theoretical effect, yes.

Would we expect the same amount of tiny stretching of a gravitationally bound system in the early universe where the amount of average matter density was much higher

In the very early universe there were no gravitationally bound systems; the temperature was too high and there hadn't been time for gravitational "clumping" to occur.

A few billion years ago, before the universe became dark energy dominated, gravitationally bound systems were basically the same as they are now, and the theoretical effect of dark energy was the same as it is now. The average density of the universe as a whole didn't mean individual gravitationally bound systems were denser; it just meant the average distance between them (more precisely, between galaxy clusters, the largest bound systems) was smaller, so there were more of them per unit volume, on average, than there are now.

 

[timmdeeg]

A few billion years ago, before the universe became dark energy dominated, gravitationally bound systems were basically the same as they are now, and the theoretical effect of dark energy was the same as it is now.

Ok, even though the universe was expanding decelerated at that time which confirms that the average matter density has no effect locally on gravitationally bound systems.

To illustrate "the theoretical effect of dark energy" in still another way and taking dark energy as $\Lambda$ we can imagine a super void in our universe with a cloud of comoving particles in the center which therefor are not bound gravitationally. If it is legitim to treat this patch of spacetime de Sitter like because the average matter density has no effect here then this cloud of particles should expand exponentially. Is this reasoning correct?

 

[PeterDonis]

even though the universe was expanding decelerated at that time

Yes.

which confirms that the average matter density has no effect locally on gravitationally bound systems.

The average matter density "has no effect locally" because it has no meaning locally.

we can imagine a super void in our universe with a cloud of comoving particles in the center which therefor are not bound gravitationally

Where does this "therefore" come from? Are you assuming that the comoving particles have negligible stress-energy and therefore don't affect the spacetime geometry? If you are, then that makes the particles comoving and not bound gravitationally, yes.

If it is legitim to treat this patch of spacetime de Sitter like because the average matter density has no effect here

What average matter density are you talking about? And what dynamics are you talking about?

You seem to be confused about how models work. When we talk about the average matter density of the universe as a whole, we do so because we are talking about a model of the universe as a whole. The dynamics of that model, at least to a good approximation, are determined by the average matter density (more precisely the average stress-energy, which might include things other than matter, like radiation or dark energy).

When we talk about a gravitationally bound system, we are talking about a different model with different dynamics, so we have to talk about a different stress-energy: the stress-energy due to the parts of the gravitationally bound system. The average matter density of the entire universe has no meaning in this context. Note, however, that dark energy density still does, because dark energy density really is constant everywhere; it's not an average, it's actually physically constant. That's why we can talk about the (in practice much too small to measure, but present in principle) effect of dark energy on a bound system, whereas it makes no sense to talk about the effect of the average matter density of the universe on a bound system.

In your example here, you appear to be talking about something in between: the dynamics of the super void patch, which is not a bound system but is not the universe as a whole either. If the super void patch really is void, i.e, there is no stress-energy inside it except for dark energy (i.e., the cloud of comoving particles has negligible stress-energy), then it can be treated as a patch of de Sitter spacetime and its dynamics will be the dynamics of such a patch. (This also assumes that the rest of the universe outside the patch is spherically symmetric, as seen from within the patch--that ensures that the spacetime geometry outside the patch has no effect on the geometry inside the patch.)

 

[timmdeeg]

Where does this "therefore" come from? Are you assuming that the comoving particles have negligible stress-energy and therefore don't affect the spacetime geometry?

Yes I was assuming this and appreciate that you are precise. This helps a lot.

You seem to be confused about how models work. When we talk about the average matter density of the universe as a whole, we do so because we are talking about a model of the universe as a whole.

Saying " because the average matter density has no effect here" I meant it can't be applied here, so it makes no sense to mention it at all. I seem to suffer from incorrect wording.

In your example here, you appear to be talking about something in between: the dynamics of the super void patch, which is not a bound system but is not the universe as a whole either. If the super void patch really is void, i.e, there is no stress-energy inside it except for dark energy (i.e., the cloud of comoving particles has negligible stress-energy), then it can be treated as a patch of de Sitter spacetime and its dynamics will be the dynamics of such a patch.

Ok.

Thank you very much, your answers during this discussion led to new insights and may be more important to correct some wrong notions.

 

[PeterDonis]

Saying " because the average matter density has no effect here" I meant it can't be applied here, so it makes no sense to mention it at all.

Ah, ok, yes, that's fine.

 

[PeterDonis]

Thank you very much, your answers during this discussion led to new insights

You're welcome!