What is the acceleration of expanding space?

[DAirey]

If me and my pet duck, who weighs next to nothing, were placed far away from any gravity source and in such a way that our mutual attraction was negligible, would I see my duck accelerate away from me? If so, how fast?

 

[newjerseyrunner]

If you places the duck 3 million light years away, it would accelerate away from you at about 45 miles per second per second.

 

[PeroK]

If you places the duck 3 million light years away, it would accelerate away from you at about 45 miles per second per second.

It doesn't work like that.

The duck would recede at 45mps, but it would take a long time before that recessional velocity doubled. It would have to recede to 6 million light years for that.

 

[newjerseyrunner]

It doesn't work like that.

The duck would recede at 45mps, but it would take a long time before that recessional velocity doubled. It would have to recede to 6 million light years for that.

Ohh, you're right. Silly me. I was reading it like an acceleration and it's not, nothing is accelerating away, it's simply moving with the universe.

 

[kimbyd]

Ohh, you're right. Silly me. I was reading it like an acceleration and it's not, nothing is accelerating away, it's simply moving with the universe.

You can certainly describe it as an acceleration, provided you use a particular definition for distance and therefore velocity. The second Friedmann equation provides a notion of acceleration that can be used:

$${\ddot{a} \over a} = -{4\pi G \over 3} \left( \rho + {3p \over c^2}\right)$$

Going through this equation is a little complicated, so I'll just roughly go over how you can calculate the answer then provide said answer (hopefully without making an error). Basically, you have to know two things: 1) How to convert each type of matter/energy density to a density fraction, as the density fractions are the values reported by cosmological experiments. 2) How the energy density of each type of matter relates to its pressure.

Once you have these, assuming a flat universe and only matter and a cosmological constant, with some understanding of how the scale factor relates to all of these, you get:

$${\ddot{a} \over a} = H_0^2 \left(-{\Omega_m \over 2a^3} + \Omega_\Lambda\right)$$

For an object at a distance of 3 million light years today (with $a=1$ today by convention), $\Omega_m = 0.32$, $\Omega_\Lambda = 0.68$, and $H_0 = 67$km/s/Mpc, the acceleration of distance today is roughly $7 \times 10^{-14} m/s^2$, or 2 meters per second per million years, if I did my math correctly.

Over time, as the matter dilutes and the object gets further away, this acceleration will increase.

 

[PeroK]

You can certainly describe it as an acceleration, provided you use a particular definition for distance and therefore velocity. The second Friedmann equation provides a notion of acceleration that can be used:

$${\ddot{a} \over a} = -{4\pi G \over 3} \left( \rho + {3p \over c^2}\right)$$

Going through this equation is a little complicated, so I'll just roughly go over how you can calculate the answer then provide said answer (hopefully without making an error). Basically, you have to know two things: 1) How to convert each type of matter/energy density to a density fraction, as the density fractions are the values reported by cosmological experiments. 2) How the energy density of each type of matter relates to its pressure.

Once you have these, assuming a flat universe and only matter and a cosmological constant, with some understanding of how the scale factor relates to all of these, you get:

$${\ddot{a} \over a} = H_0^2 \left(-{\Omega_m \over 2a^3} + \Omega_\Lambda\right)$$

For an object at a distance of 3 million light years today (with $a=1$ today by convention), $\Omega_m = 0.32$, $\Omega_\Lambda = 0.68$, and $H_0 = 67$km/s/Mpc, the acceleration of distance today is roughly $7 \times 10^{-14} m/s^2$, or 2 meters per second per million years, if I did my math correctly.

Over time, as the matter dilutes and the object gets further away, this acceleration will increase.

A simpler and more naive calculation is that an object $1 Mpc$ away is receding at $67km/s$. One second later it is receding at this speed plus the expansion for the extra $67km$.

I get $(67km/s) (67km/1Mpc) = 1.6 \times 10^{-13}m/s$.

That gives an acceleration of $1.6 \times 10^{-13}m/s^2$

 

[PeterDonis]

the acceleration of distance today

You have your units wrong; $\ddot{a} / a$ does not have units of acceleration. It has units of inverse time squared. It has no length dimension because of the $a$ in the denominator. That's what makes it useful--it "factors out" the scale, so it gives us a scale-invariant way of describing how the expansion of the universe is accelerating.

A simpler and more naive calculation is

You are calculating something different from @kimbyd. Roughly speaking, you are multiplying $H$ by $a$ to get a change in velocity over a given time, then dividing by that time to get an acceleration. This does not give you $\ddot{a} / a$.

An easy way to see the difference is to note that your number will be positive if the universe is expanding; but @kimbyd's number is only positive if the expansion is accelerating. If there were no dark energy, we would have $\Omega_{\Lambda} = 0$ in @kimbyd's equation and his number would be negative; but your number would still be the same.

 

[rootone]

View attachment 216691 If me and my pet duck, who weighs next to nothing, were placed far away from any gravity source and in such a way that our mutual attraction was negligible, would I see my duck accelerate away from me? If so, how fast?

If you and your duck have zero mass. then the distance between you and the duck will increase over time.
You don't have zero mass though, and even if you did you would have to live for millions of years to see anything changed.

 

[newjerseyrunner]

If you and your duck have zero mass. then the distance between you and the duck will increase over time.
You don't have zero mass though, and even if you did you would have to live for millions of years to see anything changed.

Even if they had mass, they may still increase distance apart. It would depend on whether the influence of dark energy can overpower the influence of their mutual gravity. At the 3 million light year mark we’ve been using my assumption is that gravity would be even more neglicable than dark energy. The effects of gravity decrease with distance but the effects of dark energy increase with distance.

 

[kimbyd]

You have your units wrong; $\ddot{a} / a$ does not have units of acceleration. It has units of inverse time squared. It has no length dimension because of the $a$ in the denominator. That's what makes it useful--it "factors out" the scale, so it gives us a scale-invariant way of describing how the expansion of the universe is accelerating.

Yes, I multiplied by the 3 million light years suggested by newjerseyrunner to get acceleration, based upon the idea that $d(t) = d_0 a(t)$, so that $\ddot{d} = d_0 \ddot{a}$ for two objects separated by some distance, both at rest with respect to the expansion. I didn't clarify that part.

Also, the scale factor $a$ is usually defined to be dimensionless. This is the convention I was assuming above, where $a=1$ at the current time. A dimensionful scale factor is possible, but seems to be less common. Of course, $\ddot{a}/a$ is dimensionless no matter the convention.

You are calculating something different from @kimbyd. Roughly speaking, you are multiplying $H$ by $a$ to get a change in velocity over a given time, then dividing by that time to get an acceleration. This does not give you $\ddot{a} / a$.

An easy way to see the difference is to note that your number will be positive if the universe is expanding; but @kimbyd's number is only positive if the expansion is accelerating. If there were no dark energy, we would have $\Omega_{\Lambda} = 0$ in @kimbyd's equation and his number would be negative; but your number would still be the same.

Her number.

 

[PeterDonis]

Her number.

Sorry about that, thanks for the correction.

 

[Arman777]

View attachment 216691 If me and my pet duck, who weighs next to nothing, were placed far away from any gravity source and in such a way that our mutual attraction was negligible, would I see my duck accelerate away from me? If so, how fast?

Well we can apply the Hubble's Law. But we can use it only in large scales. Approximately larger then 100Mpc. Since we need to apply cosmological principle so that we can use comoving coordinate system.

If we look the basic equation we will see that

$D=a(t)Δx$ here $D$ is the proper distance and $Δx$ is the comoving distance. If we take second derivative we get accceleration and that's equal to.

$A=\frac {dD^2} {d^2t}=\frac {da^2(t)} {d^2t}Δx$.
Hence,σ $A=\frac {\frac {da^2(t)} {d^2t}} {a(t)}D$

In matter dominant universe a goes like;
$a(t)≅t^{\frac {2} {3}}$

In a distance about $100Mpc$,
$A≅-t^{-2}D$
$A≅\frac {-3.10^{21}km} {2.10^{35}s^2}$. Its as expected, really low and decelerating.

I did this case for matter dominant.In dark energy dominant case which, $a(t)≈e^{√Λt}$
we would get a different answer.

In any case Hubble's Law is not valid on small scales. In my case I used matter dominant model but in other models it would be a bit complex.

 

[Arman777]

What happens If we put our duck near us like a 10pc. Then nothing happens for a long time. Since in a such a small scale it would take billions of years to see an real expansion.

First reason is we can't use friedmann equation or Hubble's Law to calculate the motion of the duck, cause distance or our scale is really low respect to umiverse and we can't talk about cosmological principle.

Second thing is, If such thing was possible then our galaxy would also feel such expansion. But we dont, If you think that our milky way has a radius of 15kpc. And we don't see that stars are getting apart or etc. If we place the duck in example between andromeda and milky way or between two superclusters, I think the result would be the same.

 

[newjerseyrunner]

Second thing is, If such thing was possible then our galaxy would also feel such expansion. But we dont, If you think that our milky way has a radius of 15kpc. And we don't see that stars are getting apart or etc. If we place the duck in example between andromeda and milky way or between two superclusters, I think the result would be the same.

The galaxies are large enough and close enough together to be gravitationally bound. The universe still expands under them, gravity is just so much more dominate. We’re talking about a human and a duck which have essentially no mass one an astronomical scale.

 

[Arman777]

The galaxies are large enough and close enough together to be gravitationally bound. The universe still expands under them, gravity is just so much more dominate. We’re talking about a human and a duck which have essentially no mass one an astronomical scale.

Yes I said the same thing didnt I ? Or you mean duck and human will feel the expansion If we put them close?

I mean of course they will but it will be really slow and undetectable.

Here matter density is important I think or its related to gravity. Even we put them close (less then 50 Mpc or so) the region of the universe is the same, the physics rules are the same. My point is we ll not see a different type (value) of acceleration for duck respect to the galaxies.

For example, let's suppose there's a galaxy 20 Mpc away and our duck.

Does the acceleration due to expansion, will be the same for both objects ? I think they will be the same.

 

[DAirey]

$${\ddot{a} \over a} = -{4\pi G \over 3} \left( \rho + {3p \over c^2}\right)$$

Going through this equation is a little complicated, so

I'd be lying if I said I followed your derivation, but I don't see the lambda constant in the above equation. The discovery that the expansion of the universe is accelerating is due to the negative pressure of the vacuum that produces a repulsive gravity. Intuitively, it seems to me that this anti-gravity would be ubiquitous and constant (for a given epoch), hence, the question about the duck (would it appear like gravity, but in reverse). Have you considered the effects of vacuum energy in your calculation?

 

[PeterDonis]

The universe still expands under them

Careful. The ordinary expansion of the universe is not a "force", and doesn't change anything "under" gravitationally bound systems. The ordinary expansion of the universe is just a statement about the relative motion of comoving objects, i.e., objects that are not gravitationally bound to each other. It doesn't tell you anything about gravitationally bound systems.

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).

 

[PeterDonis]

I don't see the lambda constant in the above equation

$\Lambda$ provides part of the $\rho$ and $p$ that appear in that equation.

Have you considered the effects of vacuum energy in your calculation?

Yes, she did. That's why there is an $\Omega_{\Lambda}$ on the RHS of @kimbyd's equation, and why that term is positive, whereas the matter term involving $\Omega_m$ is negative.

 

[PeterDonis]

Intuitively, it seems to me that this anti-gravity would be ubiquitous and constant (for a given epoch)

See my post #17 just now.

 

[DAirey]

See my post #17 just now.

Ah, yes, I see that now. Thank you.
It would be very helpful if you could edit the above posts down to a single answer. Are you saying that that there's a single, constant acceleration that's too tiny to measure is correct or are you saying that @kimbyd's answer of $7 \times 10^{-13} m$ $s^{-2}$ per every 3 MLy years is correct? Does @kimbyd 's answer imply that you'd have an acceleration of $1.4 \times 10^{-12} m$ $s^{-2}$ every 6 MLy? I would think the negative pressure would be constant and, thus, the acceleration would be constant (for any given epoch).

 

[DAirey]

For an object at a distance of 3 million light years today (with $a=1$ today by convention), $\Omega_m = 0.32$, $\Omega_\Lambda = 0.68$, and $H_0 = 67$km/s/Mpc, the acceleration of distance today is roughly $7 \times 10^{-14} m/s^2$, or 2 meters per second per million years, if I did my math correctly.

Are you sure that's $m$ $s^{-2}$ and not $km$ $s^{-2}$?

 

[PeterDonis]

Are you saying that that there's a single, constant acceleration that's too tiny to measure is correct

As I pointed out earlier in response to @kimbyd , the "acceleration" caused by dark energy does not have units of distance per time squared; it just has units of inverse time squared. In other words, the ordinary acceleration--distance per time squared--caused by dark energy does not have a single constant value; its value depends on how far apart the two objects are whose relative acceleration we are considering.

I would think the negative pressure would be constant and, thus, the acceleration would be constant (for any given epoch).

The negative pressure is constant at a given epoch, but negative pressure corresponds to the scale-invariant "acceleration" $\ddot{a} / a$, i.e., to a thing that has units of inverse time squared. That thing is constant at a given epoch. But, as above, to convert it into an ordinary acceleration, with units of distance per time squared, you need to multiply by the distance between two objects.

If you want to use @kimbyd 's equation to calculate $\ddot{a} / a$ and hence the relative acceleration of you and your duck, you need to obtain a value for $H_0$ that has units of inverse time; that means taking the usually quoted value of 67 km/s/Mpc and converting it to a value that has units of inverse seconds. You do that by dividing 67 by the number of km in a Mpc, which is about 31 trillion. Then you can plug in the result (squared, since the equation for $\ddot{a} / a$ has $H_0^2$) and the other numbers @kimbyd gave, to obtain a value for $\ddot{a} / a$; this is the number that is the same everywhere in the universe at the given epoch (our current epoch has $a = 1$ in the equation, as @kimbyd said). Then you multiply that number by the distance between you and your duck to get your relative acceleration.

 

[kimbyd]

I'd be lying if I said I followed your derivation, but I don't see the lambda constant in the above equation. The discovery that the expansion of the universe is accelerating is due to the negative pressure of the vacuum that produces a repulsive gravity. Intuitively, it seems to me that this anti-gravity would be ubiquitous and constant (for a given epoch), hence, the question about the duck (would it appear like gravity, but in reverse). Have you considered the effects of vacuum energy in your calculation?

I used shorthand to include the cosmological constant in the energy density, to keep things simple. You can always include $\Lambda$ as an energy density which has $p = -\rho$, which also implies it doesn't change over time.

 

[kimbyd]

Ah, yes, I see that now. Thank you.
It would be very helpful if you could edit the above posts down to a single answer. Are you saying that that there's a single, constant acceleration that's too tiny to measure is correct or are you saying that @kimbyd's answer of $7 \times 10^{-13} m$ $s^{-2}$ per every 3 MLy years is correct? Does @kimbyd 's answer imply that you'd have an acceleration of $1.4 \times 10^{-12} m$ $s^{-2}$ every 6 MLy? I would think the negative pressure would be constant and, thus, the acceleration would be constant (for any given epoch).

It's only constant if you are in a universe that only has a cosmological constant, and no matter density. In that special case, the acceleration formula is much simpler:

${\ddot{a} \over a} = H_0^2$

Implying that $\dot{d} = H_0 d$ and $\ddot{d} = H_0^2 d$.

 

[kimbyd]

Are you sure that's $m$ $s^{-2}$ and not $km$ $s^{-2}$?

I may have made a mistake, but I definitely didn't make that mistake.

If you want to check my math, what I did was I plugged the equation I wrote above (multiplied by the distance in post #2) into Google and let it handle the unit conversions. The precise search I used was:

(67 km/s/(1e6 parsecs))^2 * (-0.5*0.32 + 0.68) * 3 million light years

I then multiplied that by seconds per year to get the acceleration per year, which came in at $2 \times 10^{-6}$, giving 2m/s per million years.

 

[PeterDonis]

I may have made a mistake

Here's what I get when I do the calculation. You are calculating $\ddot{a} / a$ and then multiplying that by 3 million light years to get an ordinary acceleration.

From your formula for $\ddot{a} / a$, and using the method I described in post #22 to obtain a value for $H_0$ in appropriate units for your formula, I get $H_0 = 67 / ( 31 \times 10^{18} ) \approx 2.2 \times 10^{-18}$. (Note that the value I posted for km per Mpc previously was wrong; that was for km per parsec, not megaparsec. The formula I gave just now has the correct value for km per megaparsec.) Thus $H_0^2 \approx 4.6 \times 10^{-36}$, and $\ddot{a} / a = H_0^2 \left( 0.68 - 0.32 / 2 \right) \approx 2.4 \times 10^{-36} \text{s}^{-2}$. 3 million light-years is about $2.8 \times 10^{22} \text{m}$, so multiplying that by $\ddot{a} / a$ gives about $6.8 \times 10^{-14} \text{m} \text{s}^{-2}$. This is basically the same as the value you gave.

 

[timmdeeg]

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;

As both, dark energy and matter density contribute to tidal gravity should one distinguish between tidal gravity on large scales where one can assume homogeneous matter density and tidal gravity within gravitationally bound systems where the the matter density is much higher locally than the homogeneous matter density?
Or perhaps this question is more clear: would one expect less tidal gravity within gravitationally bound systems compared to tidal gravity assuming perfect fluid at the same scale (corresponding to gravitational bound system)?

I understand that the Friedmann equations can't be applied in an inhomogeneous universe locally as you have pointed out in another Thread recently (if I remember correctly). From this it seems that if one can't apply the second Friedmann equation locally then one can't discuss tidal gravity locally. Where am I misled?

 

[PeterDonis]

would one expect less tidal gravity within gravitationally bound systems compared to tidal gravity assuming perfect fluid at the same scale

It's not a question of more or less, but of the type of tidal gravity.

Within a homogeneous perfect fluid, the Weyl tensor is zero and all of the tidal gravity is due to the Ricci tensor, which affects the volume of a small ball of test particles without affecting its shape (the ball either shrinks, for tidal gravity due to matter or radiation, or expands, for tidal gravity due to dark energy).

In a gravitationally bound system, the density is not so much higher than in the universe at large, as much more variable--you have regions of very high density (stars and planets) separated by regions of very low density (space between them). In the interiors of stars and planets, tidal gravity is basically Ricci, like in a homogeneous perfect fluid (the interiors of stars and planets aren't precisely homogeneous perfect fluids, but they're close enough for this discussion). In the vacuum regions between them, the Ricci tensor is zero and all of the tidal gravity is due to the Weyl tensor, which affects the shape of a small ball of test particles without affecting its volume (the ball expands radially and shrinks tangentially).

the Friedmann equations can't be applied in an inhomogeneous universe locally

The "locally" is superfluous. The Friedmann equations are based on the assumption of homogeneity; if that assumption is violated the equations aren't valid. That is true regardless of the distance scale involved. In our actual universe, homogeneity is a reasonable approximation on distance scales of roughly hundreds of millions of light years or larger, but not on smaller scales.

From this it seems that if one can't apply the second Friedmann equation locally then one can't discuss tidal gravity locally

I'm not sure why you would think that. You don't have to have a homogeneous system to have tidal gravity. See above.

 

[timmdeeg]

In a gravitationally bound system, the density is not so much higher than in the universe at large, as much more variable--you have regions of very high density (stars and planets) separated by regions of very low density (space between them). In the interiors of stars and planets, tidal gravity is basically Ricci, like in a homogeneous perfect fluid (the interiors of stars and planets aren't precisely homogeneous perfect fluids, but they're close enough for this discussion). In the vacuum regions between them, the Ricci tensor is zero and all of the tidal gravity is due to the Weyl tensor, which affects the shape of a small ball of test particles without affecting its volume (the ball expands radially and shrinks tangentially)..

Lets say the gravitationally bound system is a galaxy. How would tidal gravity due to the Weyl tensor act? If the galaxy (thought as a ball of stars) would be in free fall towards a large mass it would be deformed while keeping the volume constant. But I think this isn't what you are talking about.
When you say "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" I was thinking that you mean tidal gravity due to Ricci curvature here because the evidence of dark energy is based on the perfect fluid model which in our universe is consistent with scales chosen large enough. Then however I don't get how this (tidal gravity due to Ricci curvature) can be applied to a galaxy which consists almost of vacuum.

 

[PeterDonis]

Lets say the gravitationally bound system is a galaxy. How would tidal gravity due to the Weyl tensor act?

I already described it.

If the galaxy (thought as a ball of stars) would be in free fall towards a large mass it would be deformed while keeping the volume constant.

I wasn't talking about the whole galaxy. I was talking about a small ball of test particles somewhere in a vacuum region in the galaxy.

For more background, I suggest reading John Baez' excellent article on the meaning of Einstein's equation; specifically, this part (but the whole thing is worth reading):

http://math.ucr.edu/home/baez/einstein/node3.html

When you say "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" I was thinking that you mean tidal gravity due to Ricci curvature

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).

I don't get how this (tidal gravity due to Ricci curvature) can be applied to a galaxy which consists almost of vacuum

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?

 

[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.