Dark energy = cosmological constant, any problems with that?

[PeterDonis]

Sure, sure, whichever way you want to look at it :).

No; if the rope is being wound onto the spool, the spool has to be doing work on the rope, not the other way around.

 

[kurros]

No; if the rope is being wound onto the spool, the spool has to be doing work on the rope, not the other way around.

Sorry this was just ambiguous grammar on my part. I meant wind the rope onto the spool to set up the experiment, then it can unwind as it is drawn off by the other mass.

 

[PeterDonis]

I meant wind the rope onto the spool to set up the experiment, then it can unwind as it is drawn off by the other mass.

Ok, that clarifies what you were saying.

 

[kurros]

We can't do that, because a rope with zero rest energy density and nonzero tension violates energy conditions; it would have to be made of exotic matter, which is not believed to exist.

However, it might be that the energy stored in the rope is small compared to the potential energy change in the mass. That's one of the things I want to check in the math.

Oh sure, let us use an extremely lightweight but rigid rope, so that it basically doesn't stretch at all and stores basically none of the energy we are talking about in its internal elastic potential energy. Like a chain made out of carbon nanotubes or something :). We just want it to transfer energy from place to place.

 

[Jorrie]

Oh sure, let us use an extremely lightweight but rigid rope, so that it basically doesn't stretch at all and stores basically none of the energy we are talking about in its internal elastic potential energy. Like a chain made out of carbon nanotubes or something :). We just want it to transfer energy from place to place.

I think Peter is right, but I haven't done the math either. Here is another way of looking at it. Let your nanotube chain be much longer than the distance between the two comoving masses (in a LCDM universe), so that both masses slide past it. Now extract some small amount of local energy from this relative motion by whatever means. The two masses will start to recede slower and the relative motion will eventually stop - that is unless the accelerated expansion can counter it. I think this is more or less what needs to be put into math form.

 

[rede96]

You appeared to be describing something that's impossible, and then making a claim about what it would be like if that something happened. That doesn't make sense.

Well we are both going around in circles here as we both agree the scenario is impossible so let's leave it there. I will say however although I'm happy to take it on face value I've never seen any math or what physics prohibits the length of a rope from spanning further than our cosmological horizon. But that's not important right now anyway.

No, there isn't. Once again, you are implicitly assuming that the rope has to move at the same velocity, relative to the spool, as a free-falling, comoving observer at the rope's other end. That's not the case. The rate at which the rope pays out can be as slow as you like.

I am? Ok, so just to check my understanding...

Ignoring cosmological distances for the moment, I have a spool with 2 cables attached, set up in similar way as described by kurros:

Edit: like these: https://www.amazon.com/Go-Travel-...coding=UTF8&psc=1&refRID=XE36FCST24NXZ3XPPHKS

Each end of the cables are attached to 2 separate rocket ships so the spool is in the middle and the rocket ships are both moving away from the spool, at an equal rate wrt to the spool and in opposite directions. As the rocket ships move away it would be like pulling the chords from head phones out at both sides, as in the link above. This would turn the spool as the cables unwind. If the cables were marked every meter, then as the cables unwind I can measure how fast the meters pass me and this would give me the relative velocity of the space ships. (Ignoring any initial stretching of the cable etc.)

So that's simple enough I hope and no issues there.

So replace the rockets with distant galaxies that are moving away from the spool in opposite directions (ignoring what is causing them to move apart for the moment) and I imagine the effect on the spool would be exactly the same. As the galaxies move away, the spool turns as the cables unwind.

So assume those are comoving galaxies, there distance relative to the spool in the middle will increase due to expansion (assuming they are not gravitational bound.) As I understand it, as their distance increases the effect will be the same, the spool will turn as the cables unwind from it. And I can measure how fast the meter intervals pass me and this will give me rate of separation of the galaxies relative to the spool.

So is there any difference between two rocket ships pulling the cable from the spool and two distant galaxies pulling the cable from the spool?

 

[PeterDonis]

I've never seen any math or what physics prohibits the length of a rope from spanning further than our cosmological horizon

It's the physics that tells you that the tension in the rope increases without bound as the distance between its ends approaches the cosmological horizon distance. The precise math is the expansion scalar of the congruence of worldlines that describes the rope.

Each end of the cables are attached to 2 separate rocket ships

Are you assuming flat spacetime? That is, no gravity? No expansion of the universe? If so, the rocket ships will need to fire their engines in order to pull the ends of the cable in opposite directions.

replace the rockets with distant galaxies that are moving away from the spool in opposite directions (ignoring what is causing them to move apart for the moment) and I imagine the effect on the spool would be exactly the same

Not if you're also replacing flat spacetime with an expanding universe containing dark energy. In an expanding universe containing dark energy, the galaxies are in free fall; no rocket engines are attached to them pushing them apart. What makes them move apart is the geometry of spacetime. That is why you can't just assume that the results will be the same.

 

[kurros]

I think Peter is right, but I haven't done the math either. Here is another way of looking at it. Let your nanotube chain be much longer than the distance between the two comoving masses (in a LCDM universe), so that both masses slide past it. Now extract some small amount of local energy from this relative motion by whatever means. The two masses will start to recede slower and the relative motion will eventually stop - that is unless the accelerated expansion can counter it. I think this is more or less what needs to be put into math form.

The whole point is to have this driven by the accelerated expansion. Whether or not this effect is large enough depends on its ratio compared to the mutual gravitational attraction of the masses. But that goes down with the square of the separation, whilst the dark energy acceleration increases linearly with separation, so the dark energy will win at large separations.

 

[JMz]

The whole point is to have this driven by the accelerated expansion. Whether or not this effect is large enough depends on its ratio compared to the mutual gravitational attraction of the masses. But that goes down with the square of the separation, whilst the dark energy acceleration increases linearly with separation, so the dark energy will win at large separations.

The former effect also goes up with the product of the masses, whereas the latter is independent of them. So it "could work" (in some very theoretical sense) even over small distances, as long as the participating masses are small, and the rope's mass is even smaller. Right?

 

[kurros]

The former effect also goes up with the product of the masses, whereas the latter is independent of them. So it "could work" (in some very theoretical sense) even over small distances, as long as the participating masses are small, and the rope's mass is even smaller. Right?

I guess so, the only problem is that it gets swamped by other effects on small scales, like thermal motion. But maybe if you do it in some super-cold environment like a bose Einstein condensate then something could be measurable. I suspect that current condensates are still many orders of magnitude too hot though.

 

[JMz]

I guess so, the only problem is that it gets swamped by other effects on small scales, like thermal motion. But maybe if you do it in some super-cold environment like a bose Einstein condensate then something could be measurable. I suspect that current condensates are still many orders of magnitude too hot though.

Well, I think this thread is well past the point of worrying about a mere few orders of magnitude. ;-)

Anyway, I envisioned some larger "local" region than an Earth-based lab.

 

[Jorrie]

Note: I have corrected a typo in the equation that I gave. It should have been the 2nd derivative of D, not the square of the first derivative.

The whole point is to have this driven by the accelerated expansion. Whether or not this effect is large enough depends on its ratio compared to the mutual gravitational attraction of the masses. But that goes down with the square of the separation, whilst the dark energy acceleration increases linearly with separation, so the dark energy will win at large separations.

To follow on with some simple math:
When talking about both masses comoving (with the Hubble flow) at large scales, there is no gravity (potential gradient) involved. At the present time, with the cosmological constant, there will be a local coordinate acceleration between each mass and the local nano chain of**
$$d^2D/dt^2 = D H_0^2(\Omega_\Lambda-\Omega_m/2)/2$$
in opposite directions. $D$ is the comoving distance between the two masses, $\Omega_\Lambda$ the cosmo-constant and $\Omega_m$ the present matter density parameter. Some form of energy extraction at each mass will enter as another negative term inside the brackets. As long as the factor inside the brackets remains positive, there could in principle be continuous energy extraction, not so? If so, it can even increase as $\Omega_m$ decreases over long periods. Or is it all wishful thinking?

** I have previously used the paper: "On the influence of the global cosmological expansion on the local dynamics in the Solar System" by Matteo Carrera and Domenico Giulini (2006), to calculate cosmic acceleration for a different scenario.

 

[kurros]

Well, I think this thread is well past the point of worrying about a mere few orders of magnitude. ;-)

Anyway, I envisioned some larger "local" region than an Earth-based lab.

Lol fair enough. I'm not quite sure what sort of experiment could be done though, even of the extreme sort. The effect would be so tiny compared to all sorts of other environmental "noise" that it is hard to image what could be done. But ok, let's actually calculate the acceleration and see what we are dealing with.

To follow on with some simple math:
When talking about both masses comoving (with the Hubble flow) at large scales, there is no gravity (potential gradient) involved. At the present time, with the cosmological constant, there will be a local coordinate acceleration between each mass and the local nano chain of**
$$dD^2/dt^2 = D H_0^2(\Omega_\Lambda-\Omega_m/2)/2$$
in opposite directions. $D$ is the comoving distance between the two masses, $\Omega_\Lambda$ the cosmo-constant and $\Omega_m$ the present matter density parameter. Some form of energy extraction at each mass will enter as another negative term inside the brackets. As long as the factor inside the brackets remains positive, there could in principle be continuous energy extraction, not so? If so, it can even increase as $\Omega_m$ decreases over long periods. Or is it all wishful thinking?

Sounds ok, but this form is not so great to use I think since we have to fiddle with H_0 and critical densities and such. I think we can just use the second Friedman equation more directly:
$$\frac{\ddot{a}}{a} = - \frac{4\pi G}{3}\left(\rho + \frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3} $$
Let's then just ignore all matter and focus on the dark-energy dominated case. This is maximum acceleration we are ever going to get with a constant $\Lambda$. So then we can ignore the density and pressure terms, and we are left with just
$$\frac{\ddot{a}}{a} = \frac{\Lambda c^2}{3} $$
$a(t)$ is the scale factor which defines the ratio between proper distance between two comoving objects now $r(t)$, relative to separation at some reference time in the past $r_0$ so
$$r(t) = a(t) r_0$$
Now, I forget the exact meaning of "t" in these coordinates, so I am not totally confident that the acceleration I am about to calculate is the correct one. If anyone sees that it needs fixing then please feel free :). I guess it is the proper time of comoving observers, though our masses are no longer comoving (although the centre of the bound system is comoving, so maybe if we measure the proper acceleration relative to the comoving point of our system then this is correct)

Anyway, suppose that we have a strong cable that just holds the two masses at fixed proper distance. I think the acceleration of the scale factor should then just give us the acceleration we want (up to my possible misuse of the time coordinate), which is the acceleration of the comoving coordinates relative to our proper separation. So, then, taking two deriviatives with respect to $t$ we have
$$\ddot{r}(t) = \ddot{a}(t) r_0$$
We can then substitute these expressions for $a$ and $\ddot{a}$ back into the second Friendman equation:
$$\frac{\ddot{r}(t) / r_0}{r(t) / r_0} = \frac{\Lambda c^2}{3} $$
$$ \ddot{r}(t) = r_0 \frac{\Lambda c^2}{3} $$
Now, $\Lambda \sim 10^{-52} \mathrm{m}^{-2}$ according to Planck data. $c^2/3$ is about $3 \times 10^{16} \mathrm{m}^2 \mathrm{s}^{-2}$. So for 1m separation we are looking at $\ddot{r} \sim 10^{-36} \mathrm{m s}^{-2}$, or $10^{-37} g$. Thus to get to $1\mathrm{g}$ of acceleration we need $r_0 = 10^{37} \mathrm{m}$. The universe is only about $10^{26} \mathrm{m}$ in diameter so this is pretty big. This is why I suggested going to large masses instead; the Earth is about $10^{24}$ kg, so for $1$m separation the Earth would pull on the rope with about $10^{24} \times 10^{-36} = 10^{-12}$ N. 1 Lightyear is $10^{15} \mathrm{m}$, so setting our separation at 1 lightyear gets us about $10^{15} \times 10^{-12} = 10^3$ N of tension, or 1 kN, which is certainly measureable. However, the mutual gravitational attraction of the two Earths is bigger than that I think, even at 1 lightyear separation ($G M^2 / r^2 = 10^{-11} \times (10^{24})^2 / (10^{15})^2 = 10^7 \mathrm{N}$). But the $1/r^2$ kills this pretty quickly so we only need to go a few more tens or hundreds of lightyears apart. Oh and actually I guess we only need one heavy mass, not two. So let's suppose we have the Earth and a 1kg mass. Then the gravitational attraction is just $G M m / r^2 = 10^{-11} \times 10^{24} / (10^{15})^2 = 10^{-17} \mathrm{N}$. So ok we are fine at 1 Ly separation as long as one of the masses is small. Although hmm, that's probably no good, we need both masses to be heavy I think. If one mass is small then the Earth will just follow its comoving path and easily drag the small mass along for the ride. So we only really get tension from accelerating the smaller mass, and lose our factor of $10^{24}$. So ok, back to two large masses, with a few hundred lightyear separation.

On the other hand, if we want to do some smaller experiment with small masses, then we lose this enormous factor of $10^{24}$ which helped us out a lot. The accelerations are super miniscule so I am not sure there is any hope of ever measuring them on small masses.

 

[Jorrie]

Note, I have corrected a crucial typo in the equation.

So ok, back to two large masses, with a few hundred lightyear separation.

In which case I will argue that the equation which I started with is far simpler to use. For a flat de-Sitter spacetime, $\Omega_\Lambda = 1$, I think, so all you need is the Hubble constant and the proper separation (D) between the two ends of the cable at any time:
$d^2D/dt^2 = D H_0^2/2$.

 

[kurros]

In which case I will argue that the equation which I started with is far simpler to use. For a flat de-Sitter spacetime, $\Omega_\Lambda = 1$, I think, so all you need is the Hubble constant and the proper separation (D) between the two ends of the cable at any time:
$dD^2/dt^2 = D H_0^2/2$.

Hmm I don't think that's true, but probably it is defined like $\Omega_\Lambda H_0^2 = \Lambda c^2 / 3$, in which case there is similar complexity. Just depends on whether you rather deal with $\{ \Omega_\Lambda, H_0\}$, or $\{\Lambda, c\}$. I think I prefer the latter :). Actually maybe you are right and it works out that $\Omega_\Lambda$ is 1... well anyway they are pretty equivalent. It seems more physically intuitive to use $\Lambda$ to me, since you would just have to use that to calculate $H_0$ anyway. We also don't have to worry about how $\Omega_\Lambda$ or $H_0$ are defined this way.

 

[Jorrie]

Actually maybe you are right and it works out that $\Omega_\Lambda$ is 1... well anyway they are pretty equivalent. It seems more physically intuitive to use $\Lambda$ to me, since you would just have to use that to calculate $H_0$ anyway.

Actually, we observe $H_0$, and a flat expanding spacetime always has $\Omega_{total} =1$. But yea, the two methods are equivalent and it's a matter of preference.

 

[kurros]

Actually, we observe $H_0$, and a flat expanding spacetime always has $\Omega_{total} =1$. But yea, the two methods are equivalent and is a matter of preference.

Is it actually $H_0$ though? Surely it should be $H(t)$ for some far future $t$, or some such? I don't see why the present-day Hubble constant would be the right number to use if we are looking at the future dark-energy dominated scenario. In which case we have to calculate what $H(\infty)$ will be, from the measured $\Lambda$. Or do you think that just using $H_0$ should give an ok answer for the present-day acceleration?

 

[Jorrie]

Is it actually $H_0$ though? Surely it should be $H(t)$ for some far future $t$, or some such? I don't see why the present-day Hubble constant would be the right number to use if we are looking at the future dark-energy dominated scenario. In which case we have to calculate what $H(\infty)$ will be, from the measured $\Lambda$. Or do you think that just using $H_0$ should give an ok answer for the present-day acceleration?

I do not think we measure $\Lambda$, cosmologically speaking. We measure the present and past expansion rates (the history), with many other parameters as well, and then we can easily calculate H(t). But for the present epoch, using $H_0$ and the (independently) deduced normal + dark matter density, is really all we need to calculate accelerating expansion to a close approximation.

 

[kurros]

I do not think we measure $\Lambda$, cosmologically speaking. We measure the present and past expansion rates (the history), with many other parameters as well, and then we can easily calculate H(t). But for the present epoch, using $H_0$ and the (independently) deduced normal + dark matter density, is really all we need to calculate accelerating expansion to a close approximation.

I guess that's true. I am just taking the value given Planck as it is, but you are right that they have to obtain it via a fit to $\Lambda\mathrm{CDM}$, they don't measure it directly. Though I guess they do not measure expansion rates etc either, they can extract it all from the CMB. Of course this can be combined with the type 1A supernovae and other expansion history data to get better constraints.

 

[JMz]

I think if you measure the stretching of the rope, instead of its tension, you will find that you don't need large masses. Of course, you can't get much work out of infinitesimal tension, but if we've moved on to trying to see how big a rope you'd need, then that doesn't matter.

 

[kurros]

I think if you measure the stretching of the rope, instead of its tension, you will find that you don't need large masses. Of course, you can't get much work out of infinitesimal tension, but if we've moved on to trying to see how big a rope you'd need, then that doesn't matter.

I guess so, the problem then is just the environmental noise. A million other sources of acceleration will dominate lighter masses, since there are a lot of random particles flying about in the intergalactic medium. Still, might be interesting to calculate. I guess the kind of experiment you might imagine is something like LISA, where we put some test masses out in the cosmos some great distance apart from each other, try to shield them from external sources of acceleration like particle winds and so on, and then measure really carefully their positions over time. I wonder if even the photon pressure from measurement lasers would be too much for this to work though.

Edit: Actually there is another problem. For my giant experiment also. With a separation of 1 lightyear, the acceleration is about 10^-21 ms^-2. If our two masses start off at rest relative to each other, then with this acceleration it will take about 30 years before they have even moved 1 millimetre. Our rope has to be so un-stretchable that it will not take up that displacement in slack. And with your proposed lightweight apparatus we still have to measure displacements to sub-millimetre accuracy over lightyears, even if we can make an experiment that is so well shielded that nothing else could move the test masses by this much.

On the plus side, the amount of rope that we need increases incredibly slowly :). What we need instead is an epic gear ratio so that those super tiny displacements can be up-scaled into reasonable generator turning speeds.

 

[JMz]

And they say physicists don't think about practical things. ;-)

 

[rede96]

Not if you're also replacing flat spacetime with an expanding universe containing dark energy. In an expanding universe containing dark energy, the galaxies are in free fall; no rocket engines are attached to them pushing them apart. What makes them move apart is the geometry of spacetime. That is why you can't just assume that the results will be the same.

The thread has moved so I’ll try and keep this brief. But if you imagine I’m stood on a very high cliff top and have a chord wound on a spool, which will turn as the chord is pulled out. I put a weight on the end of the thread and holding the spool I dangle it over he edge. One of two things will happen. If the weight isn’t heavy enough to overcome the friction in the spool it will just hang there as I hold it. Or it will be heavy enough and start to fall unwinding the chord and turning the spool. As the chord unwinds the rotation of the spool will increase. Partly due to the acceleration of the weight falling and partly due to the fact that as the remaining chord gets nearer the centre of the spool it will make the spool have to turn quicker to keep up.

My assumption is that if we are modelling dark energy as something that warps spacetime in the opposite direction to gravity (excuse any poor terminology) then the galaxies will act on the super size spool in exactly the same way. As they ‘fall’ away from the spool in opposite directions they will pull out the cable and turn the spool.

And if that is the case then as the spool unwinds and the cable is released the spool will start to turn faster and faster as the cable that is left gets nearer the centre of the spool. (if I was to assume the galaxies where separating at a constant velocity.)

However we know expansion accelerates the further the distance. So this might (I haven’t worked it out) create a situation where the rotation of the outer edges of the spool start to approach c before the cable snaps.

If that is the case there must be something that gives in the system to stop it before the rotation passes c.

And what I struggle to reconcile is that the laws of physics that would stop the spool rotating > c are different than the laws of physics that prevent something traveling locally to me at speeds >c. If the rope didn’t break then the spool would eventually be rotating > c.

 

[PeterDonis]

if we are modelling dark energy as something that warps spacetime in the opposite direction to gravity (excuse any poor terminology) then the galaxies will act on the super size spool in exactly the same way. As they ‘fall’ away from the spool in opposite directions they will pull out the cable and turn the spool

This is all true if everything is in free fall. But if everything is in free fall, no work can be done.

In the case of the spool on Earth, everything is not in free fall; the spool on the cliff top is not weightless. It has nonzero proper acceleration; it is being pushed upward by the Earth's surface.

In the case of the two galaxies, there is nothing corresponding to the extra "push" provided by the Earth's surface. You might still be able to extract work if you arrange things so that the center of the rope connecting the two galaxies is at the "cliff top" (i.e., the point of maximum potential energy in the de Sitter spacetime potential energy calculation I posted earlier). But at that point the rope is in free fall (zero proper acceleration), so you can't just wave your hands and say it's all the same as the Earth scenario. It isn't. It can still be true that work can be extracted, but you have to actually show that it can; you can't just assume it.

this might (I haven’t worked it out) create a situation where the rotation of the outer edges of the spool start to approach c before the cable snaps

If this is the case (i.e., if the diameter of the spool is large enough--but I haven't calculated how large that would be), then the spool would tear itself apart before the cable snapped. Or else the cable would start slipping on the spool so that the spool rotation rate remained slow enough.

If that is the case there must be something that gives in the system to stop it before the rotation passes c.

Yep. See above.

the laws of physics that would stop the spool rotating > c are different than the laws of physics that prevent something traveling locally to me at speeds >c

No, they aren't. They're all the laws of relativity as applied to materials; basically what all this is telling you is that in relativity there are finite limits to things like the strengths of materials, friction between a cable and a spool it's unwinding from, etc., that show up when you try to push things to the limits you are pushing them. Ultimately it comes down to interactions between things being limited to $c$, which means that in the relativistic limit, no object behaves like a single object any more--its parts start behaving like separate objects instead of parts of one, because the interactions between the parts become too slow compared to whatever else is going on.

 

[rede96]

This is all true if everything is in free fall. But if everything is in free fall, no work can be done.

Right, thanks. I guess this is obviously that part I don’t understand.

So may I ask, what happens if we replace the spool and cable with a very long coil spring and I attach each end of the coil spring to two galaxies that are far enough apart not to be gravitational bound. Does the spring stretch as the galaxies move apart from each other?

From what I understood it must or the distance between galaxies isn’t increasing.

 

[PeterDonis]

I guess this is obviously that part I don’t understand.

I was not making a completely general statement. I was talking about the specific setup you described.

For the cable to turn the spool, there must be friction between the two, sufficient to keep the cable from slipping. But if that is the case, and nothing else is holding the spool in place, the spool won't turn; it will just get pulled along with the cable. (Technically, it will turn as well since the cable's force on the spool is off axis; but the point is that the spool's axis will not remain fixed in space if it isn't held by something.) And if something else is holding the spool in place, you need to add that to your scenario before you start analyzing it; it makes a difference.

Does the spring stretch as the galaxies move apart from each other?

If we assume that the spring is at its unstressed length at the instant it is attached at each end, then the system will reach an equilibrium in which the spring is stretched just enough that its inward pull on each galaxy keeps the two galaxies at a constant distance apart. (Note that it will take a long time for this to happen, since the speed at which the spring adjusts its length is no greater than $c$ and we are postulating a spring that is many, many light-years long.)

 

[kurros]

In the case of the two galaxies, there is nothing corresponding to the extra "push" provided by the Earth's surface. You might still be able to extract work if you arrange things so that the center of the rope connecting the two galaxies is at the "cliff top" (i.e., the point of maximum potential energy in the de Sitter spacetime potential energy calculation I posted earlier). But at that point the rope is in free fall (zero proper acceleration), so you can't just wave your hands and say it's all the same as the Earth scenario. It isn't. It can still be true that work can be extracted, but you have to actually show that it can; you can't just assume it.

You don't have to carefully position anything, de Sitter space-time is translation symmetric. The potential you derived also shows that it doesn't matter, because it gets steeper and steeper with increasing r^2, so even if you "put" both masses on the same side of the "hill" (simply by choice of coordinates, which is already a clue that it isn't important), the mass at larger r^2 sees a steeper potential, so it "falls" outward with larger acceleration than the mass at smaller r^2.

 

[PeterDonis]

You don't have to carefully position anything, de Sitter space-time is translation symmetric.

That's true, but if you have two galaxies--or two test objects, to use the formulation I would prefer, since we're pretending that these "galaxies" have zero gravity--the "maximum potential" point is halfway between them, so that's where the center of the rope has to be. More precisely, it's where the center of the rope has to be at the start of the scenario; see below.

The potential you derived also shows that it doesn't matter, because it gets steeper and steeper with increasing r^2

That assumes that $r = 0$ is a comoving worldline that's halfway between the test objects at the start of the scenario. But the worldline also has to be comoving; otherwise the potential I derived is not valid. That fully determines how the center of the rope must move once the scenario starts.

the mass at larger r^2 sees a steeper potential, so it "falls" outward with larger acceleration than the mass at smaller r^2

But without anything to hold either mass in place, they'll just keep falling. The only place where anything will stay static without being held in place (i.e., without having something like a rocket or a structure holding it in place) is at $r = 0$.

 

[rede96]

If we assume that the spring is at its unstressed length at the instant it is attached at each end, then the system will reach an equilibrium in which the spring is stretched just enough that its inward pull on each galaxy keeps the two galaxies at a constant distance apart.

Right and when it reaches that equilibrium the spring will be slightly stretched from its unstressed length. So there is work done on the spring from expansion.

I was not making a completely general statement. I was talking about the specific setup you described.

Well it wasn’t my set up just my interpretation of the setup described by the OP. But what I had assumed is the spool mechanism had been designed so it only worked if you pulled each cable outward in opposite directions. E.g. two cables wound in opposite directions on to the spool. But you could just as easily imagine we use a fixed length cable to anchor the spool to one galaxy and attached the wound cable to another galaxy. This would allow the cable on the spool to unwind as the galaxies separated. Therefore being able to harness whatever potential energy there is in expansion, as in the coil spring example.

 

[kurros]

That's true, but if you have two galaxies--or two test objects, to use the formulation I would prefer, since we're pretending that these "galaxies" have zero gravity--the "maximum potential" point is halfway between them, so that's where the center of the rope has to be. More precisely, it's where the center of the rope has to be at the start of the scenario; see below.

The potential shape comes from dark energy, it has nothing to do with the positions of the galaxies. All the motion will work out the same if you arbitrary translate the potential. Or perhaps you have to boost it too so that the centre stays comoving. I'm not so sure though, the original comoving frame shouldn't be special if there is no matter. I think we should see this same acceleration in any boosted frame as well. Dark energy doesn't look different depending on velocity, I think.

That assumes that $r = 0$ is a comoving worldline that's halfway between the test objects at the start of the scenario. But the worldline also has to be comoving; otherwise the potential I derived is not valid. That fully determines how the center of the rope must move once the scenario starts.

Well the centre of the rope won't be comoving if the rope is fixed to one of the masses. But sure, there is some comoving point along its length. I'm not really sure what you think is important about this though.

But without anything to hold either mass in place, they'll just keep falling. The only place where anything will stay static without being held in place (i.e., without having something like a rocket or a structure holding it in place) is at $r = 0$.

The masses hold each other in place via the rope. I'm really not sure why you think r=0 is special. The rope can transfer the energy anywhere along its length, that's its whole job in this scenario.

 

[PeterDonis]

The potential shape comes from dark energy, it has nothing to do with the positions of the galaxies.

Yes, it does, because the "shape" depends on which comoving worldline you choose as $r = 0$. There is not one single static coordinate chart on de Sitter spacetime, the way there is on, say, Schwarzschild spacetime. There are an infinite number of them, each one corresponding to a different choice of which comoving worldline is labeled as $r = 0$. Or, to put it in coordinate-free terms, there is not just one timelike Killing vector field on de Sitter spacetime, as there is on Schwarzschild spacetime; there are an infinite number of them, and before you can define "potential energy" at all, you have to pick a single one (which is equivalent to picking which comoving worldline is labeled $r = 0$).

I'm not really sure what you think is important about this though.

I'm really not sure why you think r=0 is special.

Because, as above, in order to define "potential energy" at all, you have to have a single timelike Killing vector field with respect to which you define it. But de Sitter spacetime has an infinite number of them. So you have to pick one. Once you pick one, you have picked out a particular state of motion at a particular point in space--free fall along the comoving worldline that is labeled as $r = 0$--as being the "top" of the potential hill. Once you've made that choice, you can't change it, or the "potential energy" you just defined loses its meaning. And your reasoning about this scenario appears to rely on "potential energy" having a well-defined meaning, which has to be consistent throughout the analysis. That's why I am exploring the implications of that assumption.

An alternative, of course, would be to analyze the scenario without using the concept of "potential energy" at all. But in that case you can't reason on the basis of masses "falling in gravitational fields" or anything like that. All of that reasoning depends on having a well-defined potential energy that is consistent throughout the analysis--or, to go back to the underlying fundamental concept, it depends on having a single timelike Killing vector field that is consistent throughout the analysis. If you don't have that, then you have to use different methods of analysis.

 

[timmdeeg]

I agree also that it isn't clear where the energy comes from, but I think the existence of tension in the tether makes it crystal clear that energy is indeed able to come from somewhere.

Is it supplied free of charge by the vacuum energy which is constant?

 

[kurros]

Is it supplied free of charge by the vacuum energy which is constant?

I don't think so. The vacuum energy accelarates things "for free", but you need mass in the right place to get some energy. Like on Earth, really. Gravity accelerates things for free, but you need to find something high up in the gravitational potential that you can let fall to extract energy. The vacuum energy is pretty similar, it just accelerates things away from each other rather than towards each other.

 

[timmdeeg]

I don't think so.

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

 

[PeterDonis]

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

No. If you bring something up to the fourth floor and let it fall, you and the fourth floor are being accelerated--the surface of the Earth is pushing up on both of you.

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