Cosmic Tidal Force: Measuring CTF With Strain Gauge

[PeterDonis]

In the weak field limit (at least), the answer is yes.

Only if there is no matter in between the two labs.

If the two labs have matter in between them whose average distribution is similar to that of the universe as a whole, then there is no such thing as "gravitational acceleration" and "cosmological acceleration" canceling each other, because what you are calling "cosmological acceleration" already takes the attractive gravity of the matter into account; it is the resultant of the effects of dark energy and matter, not the effect of dark energy alone.

The only way to get a measure of the acceleration that is purely due to dark energy is to find a void that contains no matter (other than the two labs themselves) and is large enough that you can set up the kind of experiment you have in mind.

 

[PAllen]

I think there might be a 'theoretical' way to realize @Jorrie 's latest set up. Consider first, a hypothetical universe actually filled with perfect fluid of matter, energy, dark matter and cosmological constant exactly matching the Einstein tensor of our best current cosmology model. Then, instead of being an emergent, large scale average geometry, the FLRW metric would apply at all scales. You could now even imagine replacing most of this perfect fluid with suitably arranged galaxies except for one intergalactic region of perfect fluid such that the emergent large scale geometry is the same FLRW solution as the small scale geometry of the special region. That is, it seems to me, that if you could somehow arrange a large intergalactic region with just the right contents, you could have the FLRW metric apply on a small scale.

Then, if you posit two bodies small enough to justify quasilinear behavior in GR, you could realize @Jorrie 's proposal - two bodies in equilibrium between their mutual attraction and Ricci curvature making them tend to diverge.

 

[PeterDonis]

two bodies in equilibrium between their mutual attraction and Ricci curvature making them tend to diverge.

But the mutual attraction of the two bodies by themselves would be negligible compared to the effects of the "cosmic fluid" filling the region they occupy. The only way to make the mutual attraction of the two bodies non-negligible would be to make them massive enough that they would affect the overall dynamics and the "cosmic fluid" would no longer be the only significant source of stress-energy.

 

[PAllen]

But the mutual attraction of the two bodies by themselves would be negligible compared to the effects of the "cosmic fluid" filling the region they occupy. The only way to make the mutual attraction of the two bodies non-negligible would be to make them massive enough that they would affect the overall dynamics and the "cosmic fluid" would no longer be the only significant source of stress-energy.

I don't believe that. The Ricci force is microscopic on scales of a light year. I guess, though, only running some numbers would decide this. I'm thinking, start with two marbles a kilometer apart. If they recede a little after a year, retry with them close together. If they approach, retry farther apart. My intuition is that for realistic cosmological parameters you would be talking about something like this because of how infinitesimal Ricci tidal force is over small scales.

 

[PeterDonis]

The Ricci force is microscopic on scales of a light year.

If there is going to be a measurable equilibrium at all of the type you describe, the two forces have to be equal. That means you can't neglect the effects of the two bodies themselves on the overall spacetime geometry.

To put this another way: you appear to me to be trying to treat the bodies as test bodies, with negligible effect on the spacetime geometry (so the "Ricci force" is solely due to the net effect of the "cosmic fluid"), but at the same time to have them have non-negligible mutual gravitational attraction. I don't think that is possible; non-negligible mutual gravitational attraction requires a non-negligible effect on the spacetime geometry.

 

[PAllen]

If there is going to be a measurable equilibrium at all of the type you describe, the two forces have to be equal. That means you can't neglect the effects of the two bodies themselves on the overall spacetime geometry.

To put this another way: you appear to me to be trying to treat the bodies as test bodies, with negligible effect on the spacetime geometry (so the "Ricci force" is solely due to the net effect of the "cosmic fluid"), but at the same time to have them have non-negligible mutual gravitational attraction. I don't think that is possible; non-negligible mutual gravitational attraction requires a non-negligible effect on the spacetime geometry.

I disagree. While GR is nonlinear, there are wide regimes in which it is nearly linear, in practice. My intuition strongly suggests that this is one. Both effects are extremely weak compared to many cases when weak field approximations are used.

 

[PeterDonis]

While GR is nonlinear, there are wide regimes in which it is nearly linear, in practice. My intuition strongly suggests that this is one. Both effects are extremely weak compared to many cases when weak field approximations are used.

The issue isn't nonlinearity or the weak field approximation. The issue is that you can't treat the masses as test bodies and at the same time give them non-negligible mutual gravitational attraction. Even if both effects are extremely weak, they are the same size--otherwise there would be no equilibrium of the type you describe and your proposed experiment wouldn't work. But if they are the same size, then the "Ricci force" is going to be different than what it would be if the masses were test masses and the only non-negligible effect on the spacetime geometry was the "cosmic fluid". So your experiment won't be measuring what you want it to measure: you want it to measure the "Ricci force" in the absence of any other non-negligible masses, but that's not what your experimental setup gives.

 

[PAllen]

The issue isn't nonlinearity or the weak field approximation. The issue is that you can't treat the masses as test bodies and at the same time give them non-negligible mutual gravitational attraction. Even if both effects are extremely weak, they are the same size--otherwise there would be no equilibrium of the type you describe and your proposed experiment wouldn't work. But if they are the same size, then the "Ricci force" is going to be different than what it would be if the masses were test masses and the only non-negligible effect on the spacetime geometry was the "cosmic fluid". So your experiment won't be measuring what you want it to measure: you want it to measure the "Ricci force" in the absence of any other non-negligible masses, but that's not what your experimental setup gives.

I still don't accept your argument. I will try to think about how to quantify this, otherwise we are just disagreeing about intuitions about approximation. Linearity means you can add effects a la Newtonian gravity. So I think everything depends on how close to linear this regime is.

 

[PeterDonis]

I will try to think about how to quantify this

The simplest way to quantify is to estimate the accelerations involved:

For a 1 kg "marble" (way too heavy, but maybe somebody plays with very large marbles made of osmium ) at a distance of 1 light-year, in geometric units (everything in meters), we have $m = 7.42 \times 10^{-28}$ and $r = 9.30 \times 10^{15}$. The (inward) acceleration (in inverse meters) is $m / r^2 = 8.59 \times 10^{-60}$.

For a rough order of magnitude estimate of the effect of the "cosmic fluid", we can use pure dark energy with a density of $0.7$ times our universe's current critical density. In geometric units, this gives a density (in inverse meters squared) of $\rho = 4.92 \times 10^{-54}$. The (outward) acceleration is $\rho r = 4.57 \times 10^{-38}$.

You would need a mass of about [Edit: corrected] $5 \times 10^{21}$ kg, or about one tenth the mass of the Moon, meters, or about $7 \times 10^{48}$ kilograms, or about $3 \times 10^{18}$ solar masses, for the acceleration at 1 light year to be equal to the dark energy acceleration.

 

[PAllen]

The simplest way to quantify is to estimate the accelerations involved:

For a 1 kg "marble" (way too heavy, but maybe somebody plays with very large marbles made of osmium ) at a distance of 1 light-year, in geometric units (everything in meters), we have $m = 7.42 \times 10^{-28}$ and $r = 9.30 \times 10^{15}$. The (inward) acceleration (in inverse meters) is $m / r^2 = 8.59 \times 10^{-60}$.

For a rough order of magnitude estimate of the effect of the "cosmic fluid", we can use pure dark energy with a density of $0.7$ times our universe's current critical density. In geometric units, this gives a density (in inverse meters squared) of $\rho = 4.92 \times 10^{-54}$. The (outward) acceleration is $\rho r = 4.57 \times 10^{-38}$.

You would need a mass of about $5 \times 10^{21}$ meters, or about $7 \times 10^{48}$ kilograms, or about $3 \times 10^{18}$ solar masses, for the acceleration at 1 light year to be equal to the dark energy acceleration.

So bring them much closer. Using your numbers, somewhere around 500,000 km would produce a balance. One would normally consider the attractions of two 1 kilogram balls 500,000 km apart extremely weak. Note, I originally suggested marbles a km apart, and I was thinking more of a few grams, so rather amazingly, I was within a few orders of magnitude (this was pure luck - I did not estimate anything, just 'intuited').

 

[PeterDonis]

Using your numbers

Please note that I just corrected the last paragraph of that post. The corrected mass the "marble" would need to be for equality at one light year is about one tenth the mass of the Moon. Still not negligible, but a lot smaller than my previous incorrect value.

So bring them much closer.

If we stick to a 1 kg mass, the equality point is $r = \left( m / \rho \right)^{1/3}$. That gives $r = 2.53 \times 10^3$ meters. (At least, that's what we get as an order of magnitude; an actual correct calculation looks somewhat different. See below.)

extremely weak

Yes, both forces are extremely weak--they must be, since they're equal! But, as I've said, the point is precisely that they are equal, so you can't say one contributes to the spacetime curvature and the other does not. Both must contribute equally.

To put this another way: what is the correct spacetime metric to use for this hypothetical case? In other words, for a "marble" immersed in a "fluid" of pure dark energy? (I'll add back in the ordinary matter component of the "cosmic fluid" below.) The known solution fo this case is the Schwarzschild-de Sitter metric:

$$
ds^2 = - \left( 1 - \frac{2M}{r} + \frac{\Lambda}{3} r^2 \right) + \frac{1}{1 - \frac{2M}{r} + \frac{\Lambda}{3} r^2} dr^2 + r^2 d\Omega^2
$$

where $M$ is the "marble" mass in meters and $\Lambda$ is what I called $\rho$ above, in inverse meters squared.

The condition for a test object at radius $r$ to "hover" motionless in free fall, i.e., that at this value of $r$ a curve of constant $r$ (and constant angular coordinates) is a geodesic, is $f(r) = 1$, i.e.,

$$
\frac{2M}{r}= \frac{\Lambda}{3} r^2
$$

This gives

$$
r = \left( \frac{6M}{\Lambda} \right)^{\frac{1}{3}}
$$

So there is an extra factor of $6^{1/3}$ from the estimate above when we do the correct calculation. (Edit: I think there's also a factor of $8 \pi$ in the denominator inside the cube root because the correct definition of $\Lambda$ is $8 \pi \rho$, not just $\rho$.)

In your proposed experiment, instead of pure dark energy for the "cosmic fluid", you are proposing a mixture of 0.7 dark energy, 0.3 ordinary matter, with total density equal to the critical density. (In the numbers I used earlier, I was only including the 0.7 dark energy part, so the density I used was 0.7 times critical.) I don't think there is a known exact solution for this case, but heuristically, adding the ordinary matter component to the "cosmic fluid" will make the net acceleration due to the "cosmic fluid" somewhat smaller, so its effect on the net acceleration due to the "cosmic fluid' will be equivalent to decreasing $\Lambda$ somewhat. That in turn will increase somewhat the value of $r$ where a test object will "hover" in free fall. But the contribution of the "marble" to the spacetime geometry will still be significant; the geometry will not be an FRW geometry but a "Schwarzschild-FRW" geometry along the same lines as Schwarzschild-de Sitter.

 

[PAllen]

If we stick to a 1 kg mass, the equality point is $r = \left( m / \rho \right)^{1/3}$. That gives $r = 2.53 \times 10^3$ meters. (At least, that's what we get as an order of magnitude; an actual correct calculation looks somewhat different. See below.)

I agree with your formula but disagree with your result. Using your numbers I get 532,000 km, to 3 significant figures.

Yes, both forces are extremely weak--they must be, since they're equal! But, as I've said, the point is precisely that they are equal, so you can't say one contributes to the spacetime curvature and the other does not. Both must contribute equally.

That is not the point. The point is whether their effects add approximately linearly. There is no claim that one contributes more to spacetime curvature. The analog is how well Newton's law of gravity works in the solar system for many bodies by just adding forces. Jupiter and Saturn both contribute similarly to geometry, yet can be treated as additive to a very good approximation.

To put this another way: what is the correct spacetime metric to use for this hypothetical case? In other words, for a "marble" immersed in a "fluid" of pure dark energy? (I'll add back in the ordinary matter component of the "cosmic fluid" below.) The known solution fo this case is the Schwarzschild-de Sitter metric:

$$
ds^2 = - \left( 1 - \frac{2M}{r} + \frac{\Lambda}{3} r^2 \right) + \frac{1}{1 - \frac{2M}{r} + \frac{\Lambda}{3} r^2} dr^2 + r^2 d\Omega^2
$$

where $M$ is the "marble" mass in meters and $\Lambda$ is what I called $\rho$ above, in inverse meters squared.

The condition for a test object at radius $r$ to "hover" motionless in free fall, i.e., that at this value of $r$ a curve of constant $r$ (and constant angular coordinates) is a geodesic, is $f(r) = 1$, i.e.,

$$
\frac{2M}{r}= \frac{\Lambda}{3} r^2
$$

This gives

$$
r = \left( \frac{6M}{\Lambda} \right)^{\frac{1}{3}}
$$

So there is an extra factor of $6^{1/3}$ from the estimate above when we do the correct calculation.

Cube root of 6 is not a very large correction.

Further, with two equal mass balls, this is not really the correct solution - there actually is none in closed form. You would want a two body static solution with cosmological constant. This would need to be solved numerically for exact results. I retain my belief that for field strengths like those under consideration, a linear approximation could be used. Nothing you have argued undermines this based on my understanding of the regimes where such methods have been validly used.

In your proposed experiment, instead of pure dark energy for the "cosmic fluid", you are proposing a mixture of 0.7 dark energy, 0.3 ordinary matter, with total density equal to the critical density. (In the numbers I used earlier, I was only including the 0.7 dark energy part, so the density I used was 0.7 times critical.) I don't think there is a known exact solution for this case, but heuristically, adding the ordinary matter component to the "cosmic fluid" will make the net acceleration due to the "cosmic fluid" somewhat smaller, so its effect on the net acceleration due to the "cosmic fluid' will be equivalent to decreasing $\Lambda$ somewhat. That in turn will increase somewhat the value of $r$ where a test object will "hover" in free fall. But the contribution of the "marble" to the spacetime geometry will still be significant; the geometry will not be an FRW geometry but a "Schwarzschild-FRW" geometry along the same lines as Schwarzschild-de Sitter.

And I still am wholly unconvinced of the necessity of this. To determine the interactions of Jupiter and Saturn orbiting the sun it is wholly unnecessary to consider the curvature near each body. All that is needed is to determine distant field in a Newtonian approximation and add the effects. I continue to insist that such a method should work here, given the tiny field strengths and large distances.

 

[Jorrie]

The condition for a test object at radius $r$ to "hover" motionless in free fall, i.e., that at this value of $r$ a curve of constant $r$ (and constant angular coordinates) is a geodesic, is $f(r) = 1$, i.e.,

$$
\frac{2M}{r}= \frac{\Lambda}{3} r^2
$$

This gives

$$
r = \left( \frac{6M}{\Lambda} \right)^{\frac{1}{3}}
$$

So there is an extra factor of $6^{1/3}$ from the estimate above when we do the correct calculation. (Edit: I think there's also a factor of $8 \pi$ in the denominator inside the cube root because the correct definition of $\Lambda$ is $8 \pi \rho$, not just $\rho$.)

In your proposed experiment, instead of pure dark energy for the "cosmic fluid", you are proposing a mixture of 0.7 dark energy, 0.3 ordinary matter, with total density equal to the critical density. (In the numbers I used earlier, I was only including the 0.7 dark energy part, so the density I used was 0.7 times critical.) I don't think there is a known exact solution for this case, but heuristically, adding the ordinary matter component to the "cosmic fluid" will make the net acceleration due to the "cosmic fluid" somewhat smaller, so its effect on the net acceleration due to the "cosmic fluid' will be equivalent to decreasing $\Lambda$ somewhat. That in turn will increase somewhat the value of $r$ where a test object will "hover" in free fall. But the contribution of the "marble" to the spacetime geometry will still be significant; the geometry will not be an FRW geometry but a "Schwarzschild-FRW" geometry along the same lines as Schwarzschild-de Sitter.

Hmm, it looks like I will have to find a point at the center of some large void, separate my two labs by the calculated distance D apart and see if they fall towards each other or the other way.
The reason for two labs is that they can continuously observe known emissions from each other and decide whether D changes. And then adjust position until they find D_crit, where they remain at the same distance, with their common center of mass comoving.Since the normal Newtonian attaction is well known, does it give me the cosmological part? Or what analysis need I do on the data to find the cosmological effect?

 

[PeterDonis]

I agree with your formula but disagree with your result.

I'll check my numbers when I get a chance.

That is not the point. The point is whether their effects add approximately linearly.

No, the point is that there are two effects. Yes, in this weak field regime they will add linearly to a good approximation. But there are still two effects to add. That means you're not measuring the "pure" effect of the "Ricci force" due to the "cosmic fluid". To measure that, you would need to set up an experiment where that is the only effect present.

with two equal mass balls, this is not really the correct solution

Yes, agreed, but adding a second ball just adds a third effect to the mix, it doesn't change the main point I'm making. I was just using the one ball scenario to illustrate that point.

To determine the interactions of Jupiter and Saturn orbiting the sun it is wholly unnecessary to consider the curvature near each body.

If you mean determining the interactions between Jupiter and Saturn (not the interactions of either or both with the Sun), you are taking into account the stress-energy due to each body, because that's where the interaction comes from. It's true that you can ignore the tidal gravity due to each body, so the curvature due to each body doesn't directly appear in your analysis. But you can't ignore the effects of each body on the spacetime geometry altogether, because that would require you to treat each body as a test body, and you're not; you're treating them as sources of gravity.

 

[PeterDonis]

Since the normal Newtonian attaction is well known, does it give me the cosmological part?

No. The cosmological "force" is a constant times the distance (i.e., it increases linearly with the distance). But, as I have been saying in response to @PAllen, if you really want to measure the "cosmological force" by itself, with no other effects involved, you need to use test objects, not objects with non-negligible Newtonian gravitational attraction. The simplest way I can see to do that would be to have a comoving lab at the center and two test objects equidistant from it in opposite directions, all in a cosmic void with no other mass present, and measure the rate of change of round-trip light travel time between the lab and each of the objects.

 

[PAllen]

Of course an experiment would settle our disagreement ..

 

[PeterDonis]

Of course an experiment would settle our disagreement ..

Yes, all we need is a large enough void and a long enough time...

 

[PAllen]

Yes, all we need is a large enough void and a long enough time...

Too bad Joe’s Galactic Voids was a cousin of Theranos …

 

[Jorrie]

The simplest way I can see to do that would be to have a comoving lab at the center and two test objects equidistant from it in opposite directions, all in a cosmic void with no other mass present, and measure the rate of change of round-trip light travel time between the lab and each of the objects.

Agreed, but then we have a lab with non-negligible mass in the middle. Why not make the two labs (that I had in my last post) two nano-satellites as the test objects and let them do the measurements. They can have the added advantage of tiny boosters to accurately find the 'critical distance'.

Nothing in the real world of experimental science will be perfect, but I like to think like an engineer: "close enough for all practical purposes".

 

[PeterDonis]

then we have a lab with non-negligible mass in the middle

No, the lab can be a test object as well. It doesn't have to have non-negligible mass in order to be comoving and be able to send and receive light signals. The object is to have the only non-negligible stress-energy in the entire scenario be dark energy.

Why not make the two labs (that I had in my last post) two nano-satellites as the test objects and let them do the measurements.

You could, but it might also be helpful to have a comoving test object in between them to provide a reference (the comoving lab in the middle would be in free fall the whole time, whereas the two nano-satellites at the end points would have to have nonzero proper acceleration in order to adjust their motion to keep the round-trip light travel times constant).

 

[PAllen]

Let's clarify the range of agreement vs. disagreement.

I think we all agree with the following qualitative description:

- If you magically had a universe actually described at all scales by a realistic FLRW solution, then introduced two 1 kg balls a light year apart at mutual rest per constant Fermi-Normal distance (somehow), having equal and opposite peculiar velocity relative to colocated comoving world lines (towards each other - necessary for the constant distance) , and let the balls go, they would behave as test objects and slowly separate (per normal distance).

- Close together (e.g. a meter apart) they simply behave effectively as they would in a vacuum and slowly come together.

- There would exist a balance point where they would naturally (though unstably) remain mutually stationary.

What we disagree on is to what degree you can just add separately computed forces to approximate where the balance point would be versus necessarily needing a more complex treatment.

 

[PeterDonis]

I think we all agree with the following qualitative description:

- If you magically had a universe actually described at all scales by a realistic FLRW solution, then introduced two 1 kg balls a light year apart at mutual rest per constant Fermi-Normal distance (somehow), having equal and opposite peculiar velocity relative to colocated comoving world lines (towards each other - necessary for the constant distance) , and let the balls go, they would behave as test objects and slowly separate (per normal distance).

- Close together (e.g. a meter apart) they simply behave effectively as they would in a vacuum and slowly come together.

- There would exist a balance point where they would naturally (though unstably) remain mutually stationary.

Yes, I agree with this qualitative description. But let me give a somewhat different qualitative description:

1. If you magically had a universe described at all scales by a realistic FRW solution that is the same as our current best-fit Lambda CDM model, in which the expansion is currently accelerating, then two idealized test objects magically placed at mutual rest per constant Fermi-Normal distance and released into free fall at any separation whatever will start moving apart.

2. If we recognize that there ain't no such thing as an exact idealized test object, that every object has some nonzero stress-energy, then we must recognize that for any pair of real objects, such as real 1 kg balls made of real matter, there will be some separations at which the effect of their own stress-energy will outweigh the effect of the "cosmic fluid" in the magical universe described above, and they will start moving towards each other if magically placed at mutual rest per constant Fermi-Normal distance and released into free fall.

3. Therefore, for any pair of real objects, since there are two possible regimes as a function of separation--the regime where they start moving apart, and the regime where they start moving towards each other--there must be a boundary between them, i.e., a separation at which they will (unstably) remain mutually stationary.

What we disagree on is to what degree you can just add separately computed forces to approximate where the balance point would be versus necessarily needing a more complex treatment.

No. What we disagree on (if "disagree" is even the right word, since I think both of the qualitative descriptions given above are correct, and I have already agreed in previous posts that nonlinearity is not significant and that the scenario as you posed it is within the domain of the weak field approximation) is that I have been emphasizing the difference between my 1. and my 2. above, which is not included at all in your qualitative description. Whether one thinks that is something worth emphasizing will depend on exactly why one is formulating the scenario. If one is interested in measuring the properties of the "cosmic fluid", then it seems to me that one would want to do it using test objects, or something as close to that as possible, and that including objects that you explicitly treat as not being test objects and contributing non-negligible stress energy kind of defeats the purpose.

 

[PeterDonis]

What we disagree on is to what degree you can just add separately computed forces to approximate where the balance point would be versus necessarily needing a more complex treatment.

I have already agreed in previous posts that nonlinearity is not significant and that the scenario as you posed it is within the domain of the weak field approximation

Having said all that, I am still working on updating and rechecking my numbers, to address a nagging sensation I have that there is still an additional effect of the 1 kg balls not being treated as test objects that is not negligible. I might be mistaken. I will post an update when I have one.

 

[PeterDonis]

Ok, here is an updated mathematical treatment. I'm going to treat a scenario that somewhat combines scenarios proposed by @Jorrie and @PAllen. The analysis that follows, btw, will show that some of my previous posts were incorrect in some respects.

First, let's verify that the weak field approximation is ok. The simplest way to check is to verify that the maximum possible values of all terms in the metric that would make it different from the flat Minkowski metric are small (absolute value much less than $1$). The two terms of interest are $2M / r$, where $M$ is the mass of the "marbles" we are going to use as our experimental objects, and $K r^2$, where $K$ is an appropriate constant related to dark energy. (In the notations of previous posts, this constant would be $\Lambda / 3$, where $\Lambda$ is the cosmological constant, or $8 \pi \rho / 3$, where $\rho$ is the dark energy density. The small numerical factors here are not important for the order of magnitude analysis we are about to do.)

The maximum value of $2M / r$, if we assume $M$ is 1 kg, will be the one corresponding to the minimum value of $r$, i.e., the minimum distance apart that our experimental objects will be. Let's assume that is 1 meter; i.e., we are only going to test separations of 1 meter or larger. Then we have $2 M / r = 2 \times 7.42 \times 10^{-28} / 1$, which is obviously small. (In fact we could make $r$ as small as the size of an atomic nucleus, about $10^{-15}$ meters, and still have $2M / r$ be small. So we have no issue here.)

The maximum value of $K r^2$ will be the one corresponding to the maximum distance apart that our experimental objects will be. Let's assume that is 1 light year; i.e., we are only going to test separations of 1 light year or smaller. Then we have $K r^2 = (8 \pi / 3) \times 4.92 \times 10^{-54} \times ( 9.30 \times 10^{15} )^2$, which is also obviously small. So we are quite justified in using the weak field approximation.

Before writing down the weak field approximation to the metric, we want to first make a few adjustments to our coordinates. First, we will only consider motion along a single line, so we can eliminate angular coordinates and we can use $x$ instead of $r$ for our coordinate along the line. Second, we will put a "lab" at the point $x = 0$, which is defined to be equidistant between our two experimental objects; the "lab" is always in free fall and serves as a "comoving" reference point. Third, we start our two experimental objects, which are each "marbles" with mass $M$, at coordinates $x = D$ and $x = -D$, and with zero coordinate velocity. We then seek the conditions under which the objects, in free fall, will remain at their starting coordinates.

With these stipulations, the weak field approximation to the metric then is:

$$
ds^2 = - \left( 1 - \frac{2M}{D - x} - \frac{2M}{D + x} - K x^2 \right) dt^2 + \frac{1}{1 - \frac{2M}{D - x} - \frac{2M}{D + x} - K x^2} dx^2
$$

Note that the only use we are actually making of the weak field approximation is to allow us to add both of the $2M / r$ terms together without any nonlinear effects.

The relevant equation for evaluating the motion of our experimental objects is the geodesic equation for the $x$ component of 4-velocity. That equation is:

$$
\frac{d U^x}{d \tau} + \Gamma^x{}_{\mu \nu} U^\mu U^\nu = 0
$$

We are looking for the conditions that will give us $U^x = 0$ for all $\tau$ for the worldlines of our experimental objects, which means we want $d U^x / d\tau = 0$ at the values of $x$ where our experimental objects are. That means the second term above must vanish. Since the only nonzero 4-velocity component under these conditions is $U^t$, for the second term to vanish means we must have $\Gamma^x{}_{tt} = 0$. This gives (including only nonzero terms, of which there is only one since our metric is diagonal and its components are only functions of $x$):

$$
\Gamma^x{}_{tt} = 0 = - \frac{1}{2} g^{xx} g_{tt,x}
$$

which in turn gives, formally,

$$
\left( 1 - \frac{2M}{D - x} - \frac{2M}{D + x} - K x^2 \right) \left( \frac{M}{\left( D + x \right)^2} - \frac{M}{\left( D - x \right)^2} - K x \right) = 0
$$

This will be zero if the factor inside the second parentheses is zero (the factor inside the first parentheses will always be close to $1$ when the weak field approximation is valid, so we don't need to consider that one).

I say "formally" because, as noted above, we want this to hold at the values of $x$ where our objects are, i.e., $x = D$ and $x = - D$, but our expression is obviously singular there. The hand-wavy non-rigorous way we will handle this is by simply ignoring the singular terms (this amounts to assuming that each experimental object does not respond to its own self-field but only to the dark energy and the field of the other experimental object). Doing this, the evaluations at $x = D$ and $x = - D$ both give the same condition:

$$
\frac{M}{4 D^2} = K D
$$

or, using $K = 8 \pi \rho / 3$,

$$
D = \left( \frac{M}{4 K} \right)^{\frac{1}{3}} = \left( \frac{3 M}{32 \pi \rho} \right)^{\frac{1}{3}}
$$

Plugging in numbers gives $D = 2.62 \times 10^8$ meters.

Now, looking at the condition above, it certainly looks like we are just equating the gravitational force (strictly speaking, the acceleration) between the two balls at distance $2D$ with the "dark energy force" (or acceleration). (Although note that the "distance" in the dark energy force term is just $D$, not $2 D$. Why isn't it $2 D$? I'll leave that possible issue for another discussion.)

(Note, however, that there is a subtlety here. What is $D$? Well, it's the $x$ coordinate of the balls. But the $x$ coordinate is not the same as proper distance. The $g_{rr}$ term in the metric is not $1$. So the proper distance between the balls will not be exactly $2 D$, and the Newtonian gravitational acceleration between the balls will not be exactly the same as $M / (2D)^2$. I might have more to say about this in a future post.)

In a follow-up post I'lll compare the above with the case where we use idealized test objects and the only nonzero stress-energy is dark energy.

 

[PeterDonis]

This is a follow-up to post #59 to compare the scenario analyzed there with a scenario where we use idealized test objects.

We have two choices about how to set up this scenario. We retain the "lab" in the center that is comoving, in free fall, and serves as a reference, and where our $x$ coordinate is zero. At coordinates $x = D$ and $x = - D$ we have idealized test objects that have zero stress-energy and have no effect on the spacetime geometry. The choice we have is whether to let these objects fall freely (after being started out at rest relative to the "lab", i.e., not comoving) and measure the effect of dark energy by their coordinate acceleration, or whether to hold them at a constant $x$ and measure the effect of dark energy by the proper acceleration required to do that.

Either way, the key quantity will be the connection coefficient $\Gamma^x{}_{tt}$. Our expression for that from the previous post was:

$$
\Gamma^x{}_{tt} = \left( 1 - \frac{2M}{D - x} - \frac{2M}{D + x} - K x^2 \right) \left( \frac{M}{\left( D + x \right)^2} - \frac{M}{\left( D - x \right)^2} - K x \right)
$$

In the scenario we are now discussing, the only difference is that $M = 0$, so we now have:

$$
\Gamma^x{}_{tt} = \left( 1 - K x^2 \right) \left( - K x \right)
$$

So how different are these?

At the value of $D$ where equilibrium occurs in the previous scenario in post #59, obviously these two aren't the same, since for the $M \neq 0$ case we have $\Gamma^x{}_{tt} = 0$ there. So one obvious measure of difference is the value of $\Gamma^x{}_{tt}$ for the $M = 0$ case at that value of $D$. Plugging in numbers gives $\Gamma^x{}_{tt} = - 1.08 \times 10^{-44}$ inverse meters.

Another possible comparison is to note the extra factor $\left( 1 - K x^2 \right)$ in the above formula for $\Gamma^x{}_{tt}$ that multiplies what we expect to be the "dark energy force" at distance $D$, namely $K D$, as discussed in post #59. This factor differs from $1$ by only about 3 parts in $10^{36}$, so its effect is extremely negligible. Also, one might observe that, if we tried to actually calculate the "Newtonian force" between the two "marbles" in post #59 and compare it with the "dark energy force", we would find a similar factor arising in both, just with the $M$ dependent terms added back in; but since the same factor would appear in both forces, it would cancel out when we equated them in order to find the equilibrium condition.

This does, however, suggest a third comparison: compare the factors with and without the $M$ dependent terms. This gives, if we take the equilibrium value of $D$ and adopt the same policy we used in post #59 of ignoring singular terms:

$$
1 - \frac{M}{D} - K D^2 = 1 - 5.661 \times 10^{-36}
$$

as compared with

$$
1 - K D^2 = 1 - 2.832 \times 10^{-36}
$$

So $M / D$ and $K D^2$ are approximately equal (they differ by about 1 part in 1000), meaning that the correction factor differs from $1$ by about twice as much in the $M \neq 0$ scenario as compared with the $M = 0$ (idealized test objects) scenario. Of course this still leaves the correction factor extremely negligible for these scenarios. (Note that this factor arises from $g^{rr}$, so it is also related to the correction factor for proper distance as compared with coordinate distance, which was mentioned in post #59. So this factor being extremely negligible at least strongly suggests that the correction to proper distance from coordinate distance will also be negligible.)

I think this has gotten rid of the nagging feeling I mentioned in an earlier post. There is an "in principle" effect of the non-negligible masses of the "marbles" in the post #59 scenario that makes the spacetime geometry and the acceleration different; but running the numbers for the cases discussed in this thread shows that the effect in practice is extremely negligible. It might not be negligible for much larger distances, but this would also require much larger masses, going up as the cube of the distance. Exploring that a bit, for a distance of 1 light year, the numbers would look like this:

$D = 9.30 \times 10^{15}$

$M = 3.31 \times 10^{-5}$ (this is about $5 \times 10^{22}$ kg, a little short of the mass of the Moon)

$K D = 3.83 \times 10^{-37}$

$1 - \frac{M}{D} - K D^2 = 1 - 7.123 \times 10^{-21}$

$1 - K D^2 = 1 - 3.560 \times 10^{-21}$

The corrections are still negligible, but you can see that we still have $M / D$ and $K D^2$ approximately equal. Let's try a distance of a million light years:

$D = 9.30 \times 10^{21}$

$M = 3.31 \times 10^{14}$ (this is about $5 \times 10^{40}$ kg, or about 10 billion solar masses, the mass of a small galaxy)

$K D = 3.83 \times 10^{-31}$

$1 - \frac{M}{D} - K D^2 = 1 - 7.123 \times 10^{-9}$

$1 - K D^2 = 1 - 3.560 \times 10^{-9}$

This is starting to get into the range where the correction factors might be measurable.

 

[Jorrie]

Absolutely cool thanks! Just a question, do you assume that the 'lab' at x=0 has negligible mass, or does it not feature in the final analysis?

 

[PeterDonis]

do you assume that the 'lab' at x=0 has negligible mass

Yes.