A Dark energy = cosmological constant, any problems with that?

[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.

 

[kurros]

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?

Aside from Peter's technicality I would say that's the gist of it, yes.

 

[Jorrie]

Aside from Peter's technicality I would say that's the gist of it, yes.

I think 'Peter's technicality' renders timmdeeg's statement completely invalid.

If the "something" is a compressible "mass-less" rod and you could put it in free expanding space, then one of three things could happen. If the expansion is accelerating, the rod will stretch; if the expansion is decelerating, it will compress and if the expansion is doing neither, it will have no stresses. In the first case geodesics of points along the rod are diverging, in the second they are converging and in the third they obviously remain parallel.

This is totally different to what happen to rods that are supported and then dropped in an earth-like gravitational field.

Check out the "Tethered Galaxy Problem", Davis et al.

 

[timmdeeg]

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.

Yes agreed, we have to distinguish between gravity and tidal gravity. An inhomogeneous gravitational field or expanding spacetime causes tidal acceleration. If something flexible is exposed to it the created tension is a source of gravity. In the case of the expanding universe I could argue don't worry the energy isn't conserved anyway. But this very vague and perhaps not correct in this specific case. But an inhomogeneous gravitational field caused by a central mass implies static spacetime und thus energy is conserved. Where is the energy to stretch something taken from in either case?

EDIT the case of the forth floor: does the gravitational potential energy of something include the energy for stretching it if released to free fall?

 

[kurros]

I think 'Peter's technicality' renders timmdeeg's statement completely invalid.

If the "something" is a compressible "mass-less" rod and you could put it in free expanding space, then one of three things could happen. If the expansion is accelerating, the rod will stretch; if the expansion is decelerating, it will compress and if the expansion is doing neither, it will have no stresses. In the first case geodesics of points along the rod are diverging, in the second they are converging and in the third they obviously remain parallel.

This is totally different to what happen to rods that are supported and then dropped in an earth-like gravitational field.

Check out the "Tethered Galaxy Problem", Davis et al.

Not if consider the whole system. When you drop something in the Earth's gravitational field, the Earth also falls *up* towards the object. Both are in free fall towards each other, with identical acceleration. So their geodesics are converging.

Put it this way. Attach masses to your massless spring. Depending on the masses and their separation the spring will either be stretched or compressed, depending on whether the gravitational attraction of the masses exceeds the acceleration due to dark energy. The only difference is the sign and r dependence.

 

[Jorrie]

Put it this way. Attach masses to your massless spring. Depending on the masses and their separation the spring will either be stretched or compressed, depending on whether the gravitational attraction of the masses exceeds the acceleration due to dark energy. The only difference is the sign and r dependence.

If you want to 'look at the whole system' that way, you need an infinite number of masses, homogeneously spread. Then there is no gravitational potential gradient drawing masses towards each other - they just go with the expansion dynamics, be it slowing down, remaining constant or increasing faster in terms of proper distance between them.

As soon as you look at what happens in the vicinity of each mass, you will get the local gravitational effects that you refer to. And as Peter has repeatedly said, they are very different than the large scale effects.

 

[kurros]

If you want to 'look at the whole system' that way, you need an infinite number of masses, homogeneously spread. Then there is no gravitational potential gradient drawing masses towards each other - they just go with the expansion dynamics, be it slowing down, remaining constant or increasing faster in terms of proper distance between them.

So? I'm not sure what your point is. The dark energy isn't sourced by the masses, sure, but so what?

As soon as you look at what happens in the vicinity of each mass, you will get the local gravitational effects that you refer to. And as Peter has repeatedly said, they are very different than the large scale effects.

Different how? We are just talking about whether geodesics diverge or converge in both cases. I.e. whether things fall towards or away from each other. No reason to think the situation re. energy conservation is more mysterious in one case or the other.

 

[Jorrie]

So? I'm not sure what your point is. The dark energy isn't sourced by the masses, sure, but so what?

The dark energy is sourced by the cosmological constant. It does not need any masses around to produce curved spacetime and diverging geodesics. So I think discussing two masses misses the point, where the energy available to be released is dependent on the potential energy between the masses. In de Sitter spacetime (cosmological constant only), there is no potential energy differences between points.

Whether we view the energy available in the latter as coming from the uniform negative curvature of spacetime, or from the "energy of empty space" is a matter of preference of which math representation of the same physics you want to use.

I would like to hear Peter's deeper insight into this matter.

 

[PeterDonis]

does the gravitational potential energy of something include the energy for stretching it if released to free fall?

No. If the object is compressed before it is dropped, there is energy stored in the object, and that energy is not gravitational potential energy.

 

[PeterDonis]

In de Sitter spacetime (cosmological constant only), there is no potential energy differences between points.

Actually, there is, as I posted earlier in this thread; de Sitter spacetime has an infinite number of timelike Killing vector fields, and if you pick one (which amounts to picking a particular comoving worldline and centering static de Sitter coordinates on it), you can define a potential energy relative to it. The chosen comoving worldline becomes the "top" of a potential energy "hill".

I think you can actually do something similar, locally, for a case like tidal gravity in a spaceship in a free-fall circular orbit about a spherically symmetric planet (basically by setting up Fermi normal coordinates centered on the ship's center of mass--circular orbit + spherically symmetric planet means the CoM worldline is an integral curve of a timelike KVF), but I haven't had a chance to work out the math.

 

[PeterDonis]

When you drop something in the Earth's gravitational field, the Earth also falls *up* towards the object. Both are in free fall towards each other, with identical acceleration. So their geodesics are converging.

You are once again leaving out a crucial point: the surface of the Earth, which is what the dropped object hits, is not traveling on a geodesic. Only the Earth's center of mass is. But the Earth's center of mass is thousands of kilometers away from the impact point. The impact point is accelerating upward at 1 g, and upward acceleration is where the work done comes from. In the cases you are proposing, there is no such acceleration, which invalidates the analogy you are trying to draw here.

It is true that, if you take two masses that are momentarily at rest relative to each other, and release them in a converging tidal gravity field (for example, have them oriented tangentially in the field of a spherically symmetric planet), they will converge, which means they will pick up some kinetic energy relative to each other, which you could in principle extract. But this energy is not the same as the gravitational potential energy of either mass in the planet's field; it's much smaller.