A Cosmic Tidal Force: Measuring CTF With Strain Gauge

[Jorrie]

TL;DR Summary

I want to investigate/discuss/quantify the strength of the cosmological tidal force caused by accelerating cosmic expansion

In order to make it as simple and non-controversial as possible, I propose a setup where inhomogeneity and change of reference frames play as little role as possible, i.e. in free space and with its center of mass comoving. I'll call that point the origin and set up my lab there. Rockets deliver two identical masses a distance D/2 in opposite directions, attach a cable from each to the lab and allow the situation to stabilize as something like this, with the cables taking some cosmological stress due to accelerating expansion.

We can measure the cosmological 'tidal' force (CTF) on the cable in the lab by inserting a strain gauge. But before that we should calculate the OOM, so that we can insert a properly scaled strain gauge. Using standard cosmological equations, I get the CTF:

$$CTF = D H_0^2 \left(\Omega_{\Lambda} - \Omega_M/(2a^3)\right)$$
(From Peebles 1993, eq. 13.3, p 312)
At present, with a=1 and the Omegas 0.7 and 0.3, with Ho roughly 0.072/Gyr², it works out to:^[1]

CTF ~ 2.9x10^-11 D Newton/kg, with D in Glyr, or in easier terms:

29 nano-Newton per kg mass per Glyr separation.

Unless I have made some serious blunders in conversion, this is a small force, really small, unless we make the two masses very large...

Note [1] Since I often play around with Gyr, meaning acceleration is geometrically in the units Gyr^-1, I find it very convenient that
1 Gyr^-1 is very close to 1 nano-g (Precisely 0.970735785686955 ng).

​ ​

 

[anuttarasammyak]

I am afraid that the wire is loosened by the approach of the two massive bodies to each other due to universal gravitation force of
-G\frac{M^2}{D^2}.

 

[PeroK]

Summary: I want to investigate/discuss/quantify the strength of the cosmological tidal force caused by accelerating cosmic expansion

In order to make it as simple and non-controversial as possible, I propose a setup where inhomogeneity and change of reference frames play as little role as possible, i.e. in free space and with its center of mass comoving. I'll call that point the origin and set up my lab there. Rockets deliver two identical masses a distance D/2 in opposite directions, attach a cable from each to the lab and allow the situation to stabilize as something like this, with the cables taking some cosmological stress due to accelerating expansion.

View attachment 313372

We can measure the cosmological 'tidal' force (CTF) on the cable in the lab by inserting a strain gauge.

​ ​

Why would space be expanding in a lab?

 

[PAllen]

Why would space be expanding in a lab?

This is an excellent point, with a key caveat. In general, FLRW metric is an emergent, large scale average of local geometry that would not be expected to governed by FLRW metric at all. The exception is the cosmological constant. IF the cosmological constant correctly models physical reality, then it does apply at arbitrarily small scales (though with undetectable effects for realistic parameter choices).

 

[Bandersnatch]

I am afraid that the wire is loosened by the approach of the two massive bodies to each other due to universal gravitation force

Not here. The gravitational interactions between (all the) masses (in the universe) are already taken into account in the evolution of the scale factor, together with the influence of the cosmological constant. As you can see from the equation, with sufficiently diluted mass density the contribution from the cosmological constant takes over and keeps the string taut.

Why would space be expanding in a lab?

It's not the point of the setup that the space should or shouldn't expand in the lab. The setup has the two masses at cosmological distances attached to strings, and the ends of those strings are attached to a strain gauge in the lab so that the force can be measured. The tension in the string is then due to the dynamics of the expansion, and the lab is there just to measure it unambiguously.

 

[PeroK]

If there is a cable of cosmological scale under strain, then does that represent a non negligible stress energy within the experimental setup?

 

[Jorrie]

I apologize to all who have misunderstood the scenario that I have presented. I should have made it more clear that I was talking about cosmological distances and I should have done a better job of the schematic.

That said, I should also have stressed that the equation that I've used is a weak field approximation and not a full relativistic one. This implies that the two masses should not be excessively large and the distances large, but not something near a Hubble radius apart.

The equation should however hold for galaxy (or even cluster) sizes and masses and distances into the hundreds of millions of light years, where we can assume weak gravitational fields and Hubble rates well below the speed of light.

This is where the effect of the accelerating expansion is simple enough to be very closely approximated by the (Peebles) equation given in the OP. If anyone knows a reference for a more general relativistic equation, I would be grateful.

 

[Jorrie]

If there is a cable of cosmological scale under strain, then does that represent a non negligible stress energy within the experimental setup?

The idea was that the cables are only attached to a strain gauge in the relatively small lab; so the lab itself is under negligible strain.

 

[PeroK]

The idea was that the cables are only attached to a strain gauge in the relatively small lab; so the lab itself is under negligible strain.

My point was that you no longer have a vacuum. Even if you neglect the mass of the cable, there is still the strain, which cannot necessarily be neglected. But ought to be calculated.

 

[Jorrie]

My point was that you no longer have a vacuum. Even if you neglect the mass of the cable, there is still the strain, which cannot necessarily be neglected. But ought to be calculated.

Correct, but I was thinking that the very symmetrical situation would enable a fairly reliable OOM determination. As always, the "devil" might well be in the details...

 

[PeterDonis]

If there is a cable of cosmological scale under strain, then does that represent a non negligible stress energy within the experimental setup?

This is an interesting question. The key thing is energy density, and what we need to compare is the energy density of dark energy, since that is what is causing the strain, and the energy density of the cable itself. (Since the OP has specified the weak field case, the strain in the cable should be small compared to the cable's energy density, so the latter is what we need to estimate.)

Even if you neglect the mass of the cable, there is still the strain, which cannot necessarily be neglected.

If we are in the weak field regime, then, as above, the strain in the cable must be small compared to its energy density, so you can't neglect the mass of the cable.

 

[anuttarasammyak]

Not here. The gravitational interactions between (all the) masses (in the universe) are already taken into account in the evolution of the scale factor, together with the influence of the cosmological constant. As you can see from the equation, with sufficiently diluted mass density the contribution from the cosmological constant takes over and keeps the string taut.

Thanks. I think of a similar thought experiment which we do not have to introduce mass.

Let us prepare a ideally light, long retractable measure. At initial time the end of the tape is fixed at a point A and the holder is fixed at point B. Here "fixed" means fixed to points where CMG is observed homogeneous. When distance AB is our daily scale nothing happens. As distance AB becomes longer, we start noticing the tape is pulled out from the holder. Longer the distance AB, faster the speed of pull out and it would become beyond c, when physical mechanism of the measure cannot follow.

 

[PAllen]

Thanks. I think of a similar thought experiment which we do not have to introduce mass.

Let us prepare a ideally light, long retractable measure. At initial time the end of the tape is fixed at a point A and the holder is fixed at point B. Here "fixed" means fixed to points where CMG is observed homogeneous. When distance AB is our daily scale nothing happens. As distance AB becomes longer, we start noticing the tape is pulled out from the holder. Longer the distance AB, faster the speed of pull out and it would become beyond c, when physical mechanism of the measure cannot follow.

This is different because if A and B are comoving, we don't have a tethered problem at all. When AB distance is large and they are comoving, they are already separating by initial conditions and this has nothing to do with tidal forces - the subject of this thread. Instead, B can be replaced with a free fall body whose initial state is such that the initial motion of the measure end is zero on both sides and there is no tension. This is possible for very large distances even in a universe with accelerated expansion. Then (in a universe with accelerated expansion), the tape measure will start being pulled (put under tension) despite the initial set up. But you can start with conditions where the initial tension is zero.

 

[PAllen]

We can dodge any discussion of material of the string or measure by using mathematical constructs and proper acceleration rather than force. Just consider some comoving body A. Extend from it a spacelike geodesic 4-orthogonal to it. Now, consider a timelike geodesic, B, that is initially 4-orthogonal to the other end of this "mathematical string". This, by construction, guarantees that initial rate of change of distance between A and B per this distance definition, is zero. If there is accelerating expansion, a spacelike geodesic from a later moment along A, that is 4-orthogonal to it and intersects the B geodesic will:

1) be longer than it was at the starting moment.
2) will no longer be 4-orthogonal to the B geodesic.

If, instead, we replace the B geodesic with world line that preserves distance measured as above, it will have a proper acceleration in the direction from B to A. This corresponds to tension in the string without any debates about string properties.

 

[PeterDonis]

When AB distance is large and they are comoving, they are already separating by initial conditions and this has nothing to do with tidal forces - the subject of this thread.

The "tidal forces" the OP is referring to are the changes in the rate of separation of comoving objects, in the particular case of a dark energy dominated universe: i.e., the relative acceleration away from each other of comoving geodesics. In a matter dominated universe in which the expansion was decelerating, the relative acceleration of comoving geodesics would be towards each other, but it would still be relative acceleration. Both of these are manifestations of spacetime curvature, and in GR "tidal gravity" is synonymous with spacetime curvature, so both of these could be termed "tidal forces" (or "tidal gravity", which is a better term since comoving geodesic objects feel no force).

The actual forces in the OP's scenario, the tension in the cables, is due to the fact that the cables prevent the masses from following geodesic paths, so the masses feel nonzero force (proper acceleration). Such forces are commonly referred to as "tidal forces", but this is really a misnomer.

 

[PeterDonis]

We can dodge any discussion of material of the string or measure by using mathematical constructs and proper acceleration rather than force.

The expression the OP gives from Peebles actually is for proper acceleration, since it is in units of Newtons per kilogram, i.e., force per unit mass, i.e., acceleration.

I agree that it is generally simpler and easier to look at proper acceleration rather than force. However, one of the questions raised in this thread has been whether the cables themselves will have non-negligible stress-energy and will therefore affect the observations. This question cannot be answered just by looking at proper accelerations.

 

[PAllen]

The "tidal forces" the OP is referring to are the changes in the rate of separation of comoving objects, in the particular case of a dark energy dominated universe: i.e., the relative acceleration away from each other of comoving geodesics. In a matter dominated universe in which the expansion was decelerating, the relative acceleration of comoving geodesics would be towards each other, but it would still be relative acceleration. Both of these are manifestations of spacetime curvature, and in GR "tidal gravity" is synonymous with spacetime curvature, so both of these could be termed "tidal forces" (or "tidal gravity", which is a better term since comoving geodesic objects feel no force).

The actual forces in the OP's scenario, the tension in the cables, is due to the fact that the cables prevent the masses from following geodesic paths, so the masses feel nonzero force (proper acceleration). Such forces are commonly referred to as "tidal forces", but this is really a misnomer.

But why have both bodies be comoving? That would cause stretching of a connector between them that has nothing to do with tidal forces. It would be present in the Milne cosmology. My understanding of tethered galaxy problems is to have at least one of them not be comoving, so the there is initially no stretching of a tether.

 

[PAllen]

The expression the OP gives from Peebles actually is for proper acceleration, since it is in units of Newtons per kilogram, i.e., force per unit mass, i.e., acceleration.

I agree that it is generally simpler and easier to look at proper acceleration rather than force. However, one of the questions raised in this thread has been whether the cables themselves will have non-negligible stress-energy and will therefore affect the observations. This question cannot be answered just by looking at proper accelerations.

I don't see that in the OP at all. That was introduced later, without fully answering the quantitative questions in the idealized scenario of the OP.

 

[PAllen]

The actual forces in the OP's scenario, the tension in the cables, is due to the fact that the cables prevent the masses from following geodesic paths, so the masses feel nonzero force (proper acceleration). Such forces are commonly referred to as "tidal forces", but this is really a misnomer.

I agree, and that is what I described, but you definitely do not want to start with both bodies being comoving. In fact, if you fix the length of the connecting geodesic 4- orthogonal to one that is comoving, the other end will never be comoving, as I'm sure you know. (And I don't have any issue with calling such things tidal forces - basically you measure deviation between two geodesics that are initially parallel by some construction, or proper acceleration of the non-geodesic world line that prevents the deviation).

 

[anuttarasammyak]

This is different because if A and B are comoving, we don't have a tethered problem at all. When AB distance is large and they are comoving, they are already separating by initial conditions and this has nothing to do with tidal forces - the subject of this thread. Instead, B can be replaced with a free fall body whose initial state is such that the initial motion of the measure end is zero on both sides and there is no tension. This is possible for very large distances even in a universe with accelerated expansion. Then (in a universe with accelerated expansion), the tape measure will start being pulled (put under tension) despite the initial set up. But you can start with conditions where the initial tension is zero.

Thanks. Now I understand the essence of the setting of OP that only Lab, which is located at the middle of the wire, is comoving and the two ends are free so that the ends are observed moving and under acceleration at their local comoving frames of reference. The tension of wire is measured at the lab whose division by the end mass would show tidal force ( per unit mass ).

I find an analogy in rotation. The center lab is rotation center and the ends undertake centrifugal force per ends mass of $D/2\ \omega^2$. The center observe tension / mass of $D \omega^2$. $D \omega < 2c$ is required to keep the mechanism of the experiment.

I feel mechanism of transmitting tension should be investigated further in OP setting as well as my previous setting of retractable tape measure. The speed of tension cannot go beyond c. I am not sure for such a long D the distance between the ends are kept constant by wire tension.

 

[Jorrie]

If we are in the weak field regime, then, as above, the strain in the cable must be small compared to its energy density, so you can't neglect the mass of the cable.

It is easy to calculate the mass of the cable and also how it is distributed. What is not so easy is to calculate is the force contributed by the distributed mass. The cable mass will contribute zero at the lab and more and more for the more distant parts being forced off their geodesics.

H. Nikolic's Relativistic contraction of an accelerated rod analyzes a different problem, but his method described in section 5 pages 7 to 9, applicably adapted, might be useful here, e.g.

 

[PAllen]

Thanks. Now I understand the essence of the setting of OP that only Lab, which is located at the middle of the wire, is comoving and the two ends are free so that the ends are observed moving and under acceleration at their local comoving frames of reference. The tension of wire is measured at the lab whose division by the end mass would show tidal force ( per unit mass ).

I find an analogy in rotation. The center lab is rotation center and the ends undertake centrifugal force per ends mass of $D/2\ \omega^2$. The center observe tension / mass of $D \omega^2$. $D \omega < 2c$ is required to keep the mechanism of the experiment.

I feel mechanism of transmitting tension should be investigated further in OP setting as well as my previous setting of retractable tape measure. The speed of tension cannot go beyond c. I am not sure for such a long D the distance between the ends are kept constant by wire tension.

I think the limiting factor in size of tethered experiment for accelerated expansion is better viewed in terms of infinite proper acceleration rather the superluminal speed or tension. An SR analogy is that if the right end of a rod has proper acceleration to the right of 1 g, then the maximum size of constant length rod is 1 ly. As this limit is reached, the left end of the rod would need unbounded proper acceleration. I believe that with accelerated expansion there is some similar limit, where the proper acceleration of of a body maintaining constant 4-orthogonal distance from a comoving body would approach infinity as the length approaches some maximum.

 

[Jorrie]

I believe that with accelerated expansion there is some similar limit, where the proper acceleration of of a body maintaining constant 4-orthogonal distance from a comoving body would approach infinity as the length approaches some maximum.

Yes, I think that limit would be the Hubble radius, relative to the position of the lab. But I also think that the low speed/weak field equations will break down long before that, as the peculiar radial velocity grows to a significant fraction of the speed of light.

This is all rather academic, but is a tasty challenge to amateur-armchair relativists-cosmologists like myself.

 

[PeterDonis]

why have both bodies be comoving?

The OP's proposal did not. But depending on what the OP actually wants to measure, having both bodies be comoving could be useful.

That would cause stretching of a connector between them that has nothing to do with tidal forces.

I disagree; the stretching is due to the divergence of comoving geodesics in a dark energy dominated universe, which is tidal gravity (spacetime curvature). (It is not "tidal forces" because comoving objects feel no force, and the "connector" would have to have zero structural strength and would therefore register zero strain, which of course is not possible with any real material.)

It would be present in the Milne cosmology.

Yes, and in that particular edge case, spacetime curvature is zero so there is no tidal gravity. But that does not generalize to a dark energy dominated universe (accelerating expansion), which is what the OP was considering.

 

[PeterDonis]

I think that limit would be the Hubble radius

More precisely, the cosmological horizon distance. The cosmological horizon is the point at which the proper acceleration diverges.

 

[PeterDonis]

I also think that the low speed/weak field equations will break down long before that

I agree.

 

[PeterDonis]

I am not sure for such a long D the distance between the ends are kept constant by wire tension.

They will be eventually, but it will take at least as much time as the light travel time from one end to the other for the situation to reach equilibrium. Until equilibrium is reached, the overall motion of the cables will indeed not be Born rigid--the various parts of the cable will be in motion relative to each other. The strain measurement the OP wants to make cannot be made until equilibrium is reached.

 

[Jorrie]

They will be eventually, but it will take at least as much time as the light travel time from one end to the other for the situation to reach equilibrium.

Sure, but the two spaceships that hook up the cables with the lab will take much longer than that. The cables could presumably be rolled out from the two spaceships with just the right tension to keep the two masses at the required constant proper distance from the lab.

It will be an interesting challenge to program the spaceship computers to achieve that accurately. Both the thrust and the brakes on the cable drums will need to have an increasing profile of just the right shape. Cool homework problem?

 

[PAllen]

I disagree; the stretching is due to the divergence of comoving geodesics in a dark energy dominated universe, which is tidal gravity (spacetime curvature). (It is not "tidal forces" because comoving objects feel no force, and the "connector" would have to have zero structural strength and would therefore register zero strain, which of course is not possible with any real material.)

And I disagree. To measure geodesic deviation or focusing, you start with initially parallel geodesics and see whether they diverge or converge after this. The tidal force arises because, with accelerated expansion, a geodesic initially 4-orthogonal to the spacelike geodesic extended 4-orthogonally to a reference comoving observer will diverge - increase in distance defined the same way. Having comoving observers on both ends does not measure curvature efffects at all - it is totally dominated by recession from initial conditions.

Yes, and in that particular edge case, spacetime curvature is zero so there is no tidal gravity. But that does not generalize to a dark energy dominated universe (accelerating expansion), which is what the OP was considering.

No, there is a perfect analogy. Two distant comoving observers in Milne model have initial high relative velocity, and this has nothing to do with presence or absence of curvature. Instead, in the Milne model, corresponding to some reference comoving observer, there is a distant time like geodesic which remains at constant distance to the reference, measured along 4-othogonal spacelike geodesics. This is the feature that requires flatness. The corresponding case in realistic cosmology is exactly as described in my post #14, and it is properties (1) and (2) there described that indicate tidal forces = curvature.

 

[PAllen]

The "tidal forces" the OP is referring to are the changes in the rate of separation of comoving objects, in the particular case of a dark energy dominated universe: i.e., the relative acceleration away from each other of comoving geodesics. In a matter dominated universe in which the expansion was decelerating, the relative acceleration of comoving geodesics would be towards each other, but it would still be relative acceleration. Both of these are manifestations of spacetime curvature, and in GR "tidal gravity" is synonymous with spacetime curvature, so both of these could be termed "tidal forces" (or "tidal gravity", which is a better term since comoving geodesic objects feel no force).

As stated, what you say is correct, but, IMO, not a useful way to get at the essence of tidal gravity. The key to understanding force on a string that maintains, e.g. Born rigidity, is to compare its path to a geodesic initially tangent to the string end world line (which would be one 4-orthogonal to the spacelike geodesic representing the string at some chosen point). These diverge when there is curvature, while they don't diverge when there is no curvature. Comparing the string end to a colocated comoving observer is irrelevant because the dominant effect here is that this geodesic has enormous relative velocity to the colocated geodesic I construct above.

The actual forces in the OP's scenario, the tension in the cables, is due to the fact that the cables prevent the masses from following geodesic paths, so the masses feel nonzero force (proper acceleration). Such forces are commonly referred to as "tidal forces", but this is really a misnomer.

I agree with this, and I don't see any problem calling this tidal acceleration.

 

[PeterDonis]

To measure geodesic deviation or focusing, you start with initially parallel geodesics and see whether they diverge or converge after this.

That's a common way of doing it, and the way it is usually explained in textbooks, but it is not the only way of detecting that spacetime curvature is present. Anything that shows a nonzero Riemann tensor is sufficient.

The tidal force arises because, with accelerated expansion, a geodesic initially 4-orthogonal to the spacelike geodesic extended 4-orthogonally to a reference comoving observer will diverge

And with decelerated expansion, a geodesic initially 4-orthogonal to the spacelike geodesic extended 4-orthogonally to a reference comoving observer will converge (another version of a tethered galaxy scenario that was discussed in a recent thread). So either way we have nonzero tidal force--it's just divergence in one case and convergence in the other.

Some physicists would object to calling this "tidal force" on the grounds that that term should only be used for Weyl curvature, which is zero in both cases--all of the spacetime curvature is Ricci curvature. I don't personally agree with that objection.

Having comoving observers on both ends does not measure curvature efffects at all - it is totally dominated by recession from initial conditions.

I disagree: "recession from initial conditions" cannot explain either acceleration or deceleration. The only case where "recession from initial conditions" explains the entire motion is the Milne universe, since that is just flat Minkowski spacetime in disguise and there is zero spacetime curvature.

No, there is a perfect analogy.

I don't think "perfect analogy" is a good term to use, because I disagree with your claim that the relative motion of comoving observers does not measure curvature effects; see above.

I do understand the general scenario you want to set up: you have a reference object which is comoving, you take a spacelike geodesic orthogonal to its worldline and extend it for some fixed proper distance, and you put a test object at the other end. You are comparing two possible choices for what state of motion to put the test object in: "comoving" vs. "orthogonal". So consider the following:

The spacelike geodesic in question does not lie in a single spacelike hypersurface of constant comoving time (i.e., a surface of constant FRW coordinate time). In fact, the FRW coordinate time at the test object end of the spacelike geodesic is earlier than the FRW coordinate time at the reference object end.

This means that, if we consider the relative velocity of the two possible test objects at the other end of the spacelike geodesic, the "comoving" and "orthogonal" ones, it will not necessarily be the same as the peculiar velocity of a test object at rest relative to the reference object at the same proper distance in a spacelike hypersurface of constant FRW coordinate time. (Btw, I think this means that the formula given in the OP from Peebles is not exactly correct for the cases being considered, since I think that formula assumes that everything is evaluated at the same FRW coordinate time.) In fact, if we call the two velocities $v_r$ (relative velocity of test object at the end of the orthogonal spacelike geodesic with proper length $D$) and $v_p$ (peculiar velocity of test object at relative rest at distance $D$ in constant FRW coordinate time surface), we can distinguish [Edit: crossed out part that is not correct; see post #34 below] three cases:

(1) $v_r < v_p$: this indicates accelerating expansion.

(2) $v_r = v_p$: this indicates the Milne universe.

(3) $v_r > v_p$: this indicates decelerating expansion.

In other words, (2) is what you would expect if "recession from initial conditions" were the only effect present. So if you have (1) or (3) instead, you know that spacetime curvature is present.

These three cases also correspond to the three possible effects on a cable (or rod) between the reference object and the test object whose proper length is stipulated to be constant:

(1) Cable is under tension (this is the case described in the OP of this thread);

(2) Cable/rod has zero stress/strain;

(3) Rod (a better term than "cable" for this case) is under compression.

As I understand it, you are using these last three effects to distinguish the cases.

 

[PAllen]

I disagree: "recession from initial conditions" cannot explain either acceleration or deceleration. The only case where "recession from initial conditions" explains the entire motion is the Milne universe, since that is just flat Minkowski spacetime in disguise and there is zero spacetime curvature.

What I mean here is that in the proposal of using an extensible tape measure, the initial motion of the tape measure end using comoving observers has nothing to do with curvature. I agree that evolution over time does give curvature information. However, it seems much cleaner to me to eliminate the initial conditions issue, and start with zero motion for the tape measure connected to an appropriately selected inertial test object. Then, accelerated expansion implies the tape measure will start extending, decelerating expansion implies it will 'retract' (even if the expansion itself never ceases - this is a possible case), while the Milne edge case will have the tape measure neither extend nor retract.

I do understand the general scenario you want to set up: you have a reference object which is comoving, you take a spacelike geodesic orthogonal to its worldline and extend it for some fixed proper distance, and you put a test object at the other end. You are comparing two possible choices for what state of motion to put the test object in: "comoving" vs. "orthogonal". So consider the following:

The spacelike geodesic in question does not lie in a single spacelike hypersurface of constant comoving time (i.e., a surface of constant FRW coordinate time). In fact, the FRW coordinate time at the test object end of the spacelike geodesic is earlier than the FRW coordinate time at the reference object end.

I'm well aware of this. In fact, I was considering giving another view of why the tether scenario becomes impossible for sufficiently large distances in an accelerating expansion case. The issue is that there exist events such that all 4-orthogonal geodesics from the distant comoving world line intersect the initial 'big bang surface' without reaching such an event, even out to infinite proper time along the comoving world line. In other terms, there are regions in the causal future of the entire Fermi-Normal coordinate patch constructed from a comoving world line - when you have accelerated expansion (and only then). [For decelerated expansion there is no problem with global coverage by FN coordinates, and also obviously not for the Milne case].

This means that, if we consider the relative velocity of the two possible test objects at the other end of the spacelike geodesic, the "comoving" and "orthogonal" ones, it will not necessarily be the same as the peculiar velocity of a test object at rest relative to the reference object at the same proper distance in a spacelike hypersurface of constant FRW coordinate time. (Btw, I think this means that the formula given in the OP from Peebles is not exactly correct for the cases being considered, since I think that formula assumes that everything is evaluated at the same FRW coordinate time.)

I agree and fully understood this. However, given my focus on the tether scenario, I was interested in pointing out the huge relative velocity between a free fall test object that initially does not 'extend or retract a tape measure' and a colocated comoving observer. I would certainly call this peculiar velocity even though it is at an earlier cosmological time than other end of the tether.

In fact, if we call the two velocities $v_r$ (relative velocity of test object at the end of the orthogonal spacelike geodesic with proper length $D$) and $v_p$ (peculiar velocity of test object at relative rest at distance $D$ in constant FRW coordinate time surface), we can distinguish three cases:

(1) $v_r < v_p$: this indicates accelerating expansion.

(2) $v_r = v_p$: this indicates the Milne universe.

(3) $v_r > v_p$: this indicates decelerating expansion.

I think it may be more complicated than this. Consider the Milne case. Consider the reference comoving observer to be the t axis of a standard Minkowski coordinate system (by symmetry, there is no loss of generality in this assumption). Pick some event on the t axis. If I understand your $v_r$, you would take a horizontal line in these coodinates some distance D from this event, and compute the relative velocity between a vertical world line at the end, and a world line from the origin to this same end event.

Meanwhile, for $v_p$, you would extent a family of hyperbolas from the t axis, and consider the relative velocity of the the locus of constant D measured along these versus a world line from the end directly to the origin - with this comparison being made at the event D along a hyperbola from the same t axis event as the prior paragraph.

A priori, I see no reason for your claim of equality between these. I haven't calculated it, but I certainly don't see any reason for your (2) above to be true for this case.

These three cases also correspond to the three possible effects on a cable (or rod) between the reference object and the test object whose proper length is stipulated to be constant:

(1) Cable is under tension (this is the case described in the OP of this thread);

(2) Cable/rod has zero stress/strain;

(3) Rod (a better term than "cable" for this case) is under compression.

As I understand it, you are using these last three effects to distinguish the cases.

Yes, this last is the way I think is most useful for understanding the tethered galaxy problem

 

[PAllen]

Since @PeterDonis has brought the following issue up, it is worth discussion:

Why use a spacelike geodesic 4-orthogonal to a world line as a model for rigid rod rather than a geodesic of the submanifold of a surface of constant cosmological time (with metric induced from the overall manifold)? Note these are not the same, even in the Milne special case. I don't know of any FLRW case where they are the same.

The reason is that it is a known theorem that any Born-rigid congruence is also a Ferm-Normal congruence (the converse is not generally true). Meanwhile, the second option above violates local rigidity.

 

[PeterDonis]

I mean here is that in the proposal of using an extensible tape measure, the initial motion of the tape measure end using comoving observers has nothing to do with curvature.

Actually, it does, because if you define the initial motion as "the motion of a comoving observer at the far end of the spacelike geodesic", then exactly what that motion is, if we hold the proper distance along the spacelike geodesic fixed, will depend on which universe you are in. More precisely, the spacetime angle between the spacelike geodesic and the initial 4-velocity of the test object, if the test object is comoving, will be different for the three cases I described.

it seems much cleaner to me to eliminate the initial conditions issue, and start with zero motion for the tape measure connected to an appropriately selected inertial test object.

I don't necessarily disagree with this, and it does correspond better with the original scenario in the OP.

there exist events such that all 4-orthogonal geodesics from the distant comoving world line intersect the initial 'big bang surface' without reaching such an event, even out to infinite proper time along the comoving world line.

For the accelerated case, yes, agreed.

If I understand your $v_r$, you would take a horizontal line in these coodinates some distance D from this event, and compute the relative velocity between a vertical world line at the end, and a world line from the origin to this same end event.

Yes.

for $v_p$, you would extent a family of hyperbolas from the t axis, and consider the relative velocity of the the locus of constant D measured along these versus a world line from the end directly to the origin

I'm not sure I understand, so let me describe $v_p$ from scratch in my own words. Take the hyperbola that meets the reference object's worldline at the same reference event where the spacelike geodesic (horizontal line) meets it. Find the event on that hyperbola at arc length $D$ from the reference event. $v_p$ is then the relative velocity of a worldline from the origin to that $D$ event, as compared to a vertical worldline at the same event.

I see now, though, that your suspicion that we might not have $v_r = v_p$ for the Milne case is correct (if $T$ is the comoving proper time of the reference event, then we will have $v_r = D / T$ but $v_p = \tanh D / T$), so that particular trichotomy that I gave won't work. So we would need to use the tension/zero stress/compression trichotomy instead.

 

[Jorrie]

This is an interesting question. The key thing is energy density, and what we need to compare is the energy density of dark energy, since that is what is causing the strain, and the energy density of the cable itself. (Since the OP has specified the weak field case, the strain in the cable should be small compared to the cable's energy density, so the latter is what we need to estimate.)If we are in the weak field regime, then, as above, the strain in the cable must be small compared to its energy density, so you can't neglect the mass of the cable.

After this reply, the thread went into a "highly A" level, so I thought about a way to simplify it somewhat. Is it possible to do away with the cable and all its complexities? Perhaps.

If we (one day) have the technology to put two labs, each of non-negligible mass M, somewhere in the intergalactic void, separated by a proper distance D, is there a 'critical distance' D_critical, call it D_c for brevity, where the gravitational acceleration and the cosmological acceleration cancel each other?

In the weak field limit (at least), the answer is yes. They will each create an attractive acceleration of (geometric) magnitude $\frac{M}{(0.5D)^2}$ in the frame of their common center of gravity, for a total of $\frac{0.5M}{D^2}$ when added together. This is to be compatible with the measurement of the OP's strain gauge at the midpoint.

So we can find D_c by equating this acceleration to the OP's CTF, i.e.
$$\frac{0.5M}{D^2} = D H_0^2 \left(\Omega_{\Lambda} - \Omega_M/2\right)$$
and I found
$$D_c = \left(\frac{M}{H_0^2} (\Omega_{\Lambda}-\Omega_M/2)\right)^{1/3}$$

If this is correct for the present time, does it give us a potential way to measure the Cosmic Tidal Force (CTF)? That is, by precisely measuring that they stay at the constant distance D_c for long enough?

 

[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.