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

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The discussion centers on the nature of dark energy and its relationship to the cosmological constant in Einstein's equations. A key point raised is whether dark energy should be viewed as a distinct entity or simply a constant that fits within general relativity. The cosmological constant's observed value poses a significant problem, as it is much smaller than expected from quantum field theory, leading to ongoing debates about its physical implications. Some participants argue that labeling it "dark energy" allows for the possibility of variation over time and space, while others question the necessity of seeking alternative explanations. Ultimately, the conversation reflects a deep inquiry into the fundamental nature of the universe and the constants that govern it.
  • #31
George Jones said:
What about the terminology used on pages 860-861 of Misner, Thorne, and Wheeler?

I don't have my copy right now, but I'll take a look when I get a chance. If it's a place where the term "force" is used to mean something other than "causes nonzero proper acceleration", then I agree that the terminology in the literature is not entirely consistent on this point.
 
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  • #32
PeterDonis said:
Sorry, again you are mistaken. Please go back and read my exchange with @Bandersnatch again.

In the case of tides on the Earth, the water in the Earth's tidal bulge moves outward, relative to the Earth, because of the natural geodesic motion due to the spacetime curvature (tidal gravity) created by the Moon (and the Sun, but we'll focus on the Moon here) and by the Earth. That water feels no force due to the tidal gravity. The only force it feels is the net (non-gravitational) force due to hydrostatic pressure exerted by the water (or ocean bottom) underneath it; that force prevents the water from completely following the natural geodesic motion due to the spacetime geometry in its vicinity.

(Note, also, that the water in the Earth's tidal bulge does not have internal tension the way a solid object like a rope would. That makes the analysis of the two cases differ in some respects. See below.)
I agreed that there is tension in the rope; but the reason there is tension in the rope is not that tidal gravity is exerting a force on it, or on the galaxies. It is that the rope is preventing the two galaxies from following the natural geodesic motion due to the spacetime curvature in their vicinity. If the rope were not there, the two galaxies would separate. With the rope there, the two galaxies reach an equilibrium in which their separation is constant and the rope is stretched beyond its unstressed length, just enough so that the tension in the rope imposes the right proper acceleration on the galaxies to keep them at a constant separation.

Once again, gravity in GR is not a force. I don't think you have fully understood the implications of that fact.
I think you have not thought the scenario through. Consider the situation I just described above: the two galaxies are at constant separation, with the rope stretched beyond its unstressed length just enough so that its tension keeps the galaxies at constant separation. In this situation, everything is in equilibrium and there is a constant amount of energy stored in the rope. (If you don't see how this situation is an equilibrium, then stop and convince yourself of this before reading further. It is a crucial point.)

Now, suppose we start in this equilibrium situation, and then let out some more rope. What will happen? First, what does "letting out more rope" mean? It means that we have increased the unstressed length of the rope. That means there will be a new equilibrium configuration now, with the galaxies at a larger separation, and with a different amount of tension in the rope. The new tension in the rope will have to be larger than it was before (because the proper acceleration required to hold the galaxies at constant separation increases with the separation). So yes, you are correct that if we let out rope, the tension will increase.

However, you are claiming something more: that we can somehow extract work from this process. How?

I'll reply in more detail tomorrow, but I am just describing something exactly like a mass hanging from a rope on earth. We can and do extract energy from such a system by allowing the mass to fall and the rope to turn a turbine. I see no reason the exact same thing would not work in the dark energy situation I described. I do not see from your current argument what would prevent such a thing.
 
  • #33
kurros said:
I'll reply in more detail tomorrow, but I am just describing something exactly like a mass hanging from a rope on earth. We can and do extract energy from such a system by allowing the mass to fall and the rope to turn a turbine. I see no reason the exact same thing would not work in the dark energy situation I described. I do not see from your current argument what would prevent such a thing.
That energy would all come from the relative motion of the two galaxies. It would have nothing (directly) to do with dark energy. All that dark energy does is increases the tension on the rope by such a minuscule amount that it likely would have no measurable impact.

The dark energy also does change the relative motion over time between the two galaxies.
 
  • #34
kurros said:
I am just describing something exactly like a mass hanging from a rope on earth.

No, you're not. If a rope is suspended on Earth, its point of suspension has nonzero proper acceleration whether the rope and the mass are present or not. It is that outside source of proper acceleration that provides the force which suspends the mass, and against which work is done if the mass is allowed to descend. But in the scenario we're discussing, in the absence of the thread, both "points of suspension"--the two galaxies--have zero proper acceleration. The only possible source of proper acceleration is the thread itself.

So what you should be imagining as an analogy is a huge spaceship floating in free fall, and two masses connected by a thread floating inside the ship. If the ship is subject to a constant tidal gravity (for example, if it's in a circular orbit about a planet and is large enough for the planet's field to vary detectably inside the ship), what will be the equilibrium state of the two masses and the thread?
 
  • #35
Dr' Whom in post #30 has the right idea - We observe the universe to be expanding, but we don't have an explanation for what the physical cause of that expansion might be? Hence we call it "dark energy".
 
  • #36
PeterDonis said:
No, you're not. If a rope is suspended on Earth, its point of suspension has nonzero proper acceleration whether the rope and the mass are present or not. It is that outside source of proper acceleration that provides the force which suspends the mass, and against which work is done if the mass is allowed to descend. But in the scenario we're discussing, in the absence of the thread, both "points of suspension"--the two galaxies--have zero proper acceleration. The only possible source of proper acceleration is the thread itself.

But there *is* a thread, and it is pulling the masses off their geodesics, both in the Earth and dark energy examples. I don't see any difference except for the direction of "falling".

So what you should be imagining as an analogy is a huge spaceship floating in free fall, and two masses connected by a thread floating inside the ship. If the ship is subject to a constant tidal gravity (for example, if it's in a circular orbit about a planet and is large enough for the planet's field to vary detectably inside the ship), what will be the equilibrium state of the two masses and the thread?

There will be tension, and you should be able to do work with it. It seems the same. It could do so much work that it rips the ship apart, in the extreme case. (edit: and dark energy could rip you apart too, if the cosmological constant was really huge. So it seems like it *can* do work. Of course you could never form in the first place with a huge cosmological constant, but if someone did the work needed to put you together in such a universe then dark energy could certainly then use that stored energy to rip you apart again).
 
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  • #37
kimbyd said:
That energy would all come from the relative motion of the two galaxies. It would have nothing (directly) to do with dark energy. All that dark energy does is increases the tension on the rope by such a minuscule amount that it likely would have no measurable impact.

The dark energy also does change the relative motion over time between the two galaxies.

Haha sure, it is miniscule, but this is an in-principle sort of question :).
 
  • #38
alantheastronomer said:
...we don't have an explanation for what the physical cause of that expansion might be? Hence we call it "dark energy".
How so? We have one perfectly good explanation: It's Einstein's own Λ. It's true that people can come up with other explanations -- and that work is certainly worthwhile.

But you could likewise make the case that the gravitational coupling parameter G is not independent of time, or that the speed of light is not constant for all observers. However, those are the standard explanations, they are in agreement with virtually all current observations, and they don't come with any physical cause either, at least not so far. Yet we don't consider the lack of a cause to be a problem.

@kurros's friend is right: This isn't a problem, it's simply a discovery.

The one good reason to treat it otherwise is --
kimbyd said:
many physicists have been working hard to come up with alternative models that don't contain constants that are so ridiculously tiny.
 
  • #39
alantheastronomer said:
Dr' Whom in post #30 has the right idea - We observe the universe to be expanding, but we don't have an explanation for what the physical cause of that expansion might be? Hence we call it "dark energy".
This is also not correct. The universe could be expanding even if there was no dark energy. Dark energy is needed only for accelerated expansion.
 
  • #40
Ok so here is a better reply to what you said earlier:

PeterDonis said:
Sorry, again you are mistaken. Please go back and read my exchange with @Bandersnatch again.

In the case of tides on the Earth, the water in the Earth's tidal bulge moves outward, relative to the Earth, because of the natural geodesic motion due to the spacetime curvature (tidal gravity) created by the Moon (and the Sun, but we'll focus on the Moon here) and by the Earth. That water feels no force due to the tidal gravity. The only force it feels is the net (non-gravitational) force due to hydrostatic pressure exerted by the water (or ocean bottom) underneath it; that force prevents the water from completely following the natural geodesic motion due to the spacetime geometry in its vicinity.

I think you are getting too caught up in the language here. I'll agree that yes you are strictly correct, but I don't think it is incorrect to call that situation "feeling a tidal force". It is just a frame of reference difference, like feeling a centrifugal force when you ride the gravitron. Yes, you actually feel the walls pushing you off your geodesic, but there is no real reason not to talk about the "fictional" force that exists in the spinning reference frame.

I agreed that there is tension in the rope; but the reason there is tension in the rope is not that tidal gravity is exerting a force on it, or on the galaxies. It is that the rope is preventing the two galaxies from following the natural geodesic motion due to the spacetime curvature in their vicinity. If the rope were not there, the two galaxies would separate. With the rope there, the two galaxies reach an equilibrium in which their separation is constant and the rope is stretched beyond its unstressed length, just enough so that the tension in the rope imposes the right proper acceleration on the galaxies to keep them at a constant separation.

Once again, gravity in GR is not a force. I don't think you have fully understood the implications of that fact.

Exactly, without the rope the galaxies would follow their natural geodesic motion. Exactly like a mass hanging on a rope on Earth would fall downwards along its natural geodesic motion if the rope was not there. Both cases seem completely analogous, and I see no reason why you can turn a turbine in one scenario and not the other.

I think you have not thought the scenario through. Consider the situation I just described above: the two galaxies are at constant separation, with the rope stretched beyond its unstressed length just enough so that its tension keeps the galaxies at constant separation. In this situation, everything is in equilibrium and there is a constant amount of energy stored in the rope. (If you don't see how this situation is an equilibrium, then stop and convince yourself of this before reading further. It is a crucial point.)

It doesn't matter how much energy is stored in the rope. That isn't where the energy comes from. Like on earth, when you create the gravity-battery situation described earlier, it doesn't matter how much energy is stored in the cable. What matters is the mass hanging from the cable and the distance over which you let it fall. That determines the potential energy difference, which is the maximum you can recover in power generation. It should be exactly the same in the dark energy scenario. I'm not 100% sure where the potential energy is stored, but I guess it might just be stored in the configuration of matter you start with. You start with two galaxies "near" each other, and they want to "fall" away from each other. So there is a certain amount of stored potential energy there.

Now, suppose we start in this equilibrium situation, and then let out some more rope. What will happen? First, what does "letting out more rope" mean? It means that we have increased the unstressed length of the rope. That means there will be a new equilibrium configuration now, with the galaxies at a larger separation, and with a different amount of tension in the rope. The new tension in the rope will have to be larger than it was before (because the proper acceleration required to hold the galaxies at constant separation increases with the separation). So yes, you are correct that if we let out rope, the tension will increase.

However, you are claiming something more: that we can somehow extract work from this process. How?

Same as on Earth. Attach our end of the rope to a turbine and let it spin as the rope uncoils. Though I guess in this analogy the turbine is attached to the falling mass. For an even closer analogy you might imagine a space elevator. It takes energy to lift stuff up to our anchor in geostationary orbit, but we can certainly recover energy by attaching a rope to stuff as it falls back down the elevator, and turning a generator as it falls back down. Even more, we could generate power by attaching a tether on the *outside* of our anchor, above geostationary orbit, and allowing the centrifugal force of the orbit to drag a mass outwards while turning a generator. All this seems fine as far as I can see. It can be hard to pinpoint where the energy comes from sometimes, but it can be done.
 
  • #41
kurros said:
there *is* a thread, and it is pulling the masses off their geodesics, both in the Earth and dark energy examples.

No. In the Earth example, the rope is not what is pulling the mass off of a geodesic: it's whatever structure is holding up the rope's attachment point. In the two galaxies example, there is no such attachment point for the rope.

kurros said:
I don't think it is incorrect to call that situation "feeling a tidal force".

It is if it leads you to an incorrect analysis of a scenario, which it is doing.

kurros said:
Both cases seem completely analogous, and I see no reason why you can turn a turbine in one scenario and not the other.

That's because the term "tidal force" is misleading you.

kurros said:
It doesn't matter how much energy is stored in the rope. That isn't where the energy comes from.

This is true in the Earth example. It is not true in the galaxies connected by rope example.

kurros said:
It takes energy to lift stuff up to our anchor in geostationary orbit, but we can certainly recover energy by attaching a rope to stuff as it falls back down the elevator, and turning a generator as it falls back down.

This is not analogous to the galaxy scenario either. The galaxies were not "lifted" to their positions, and you're not recovering any energy by letting them "fall".

Please take a step back and think carefully about the one valid analogy I suggested to you, the one that actually involves tidal gravity, and nothing else. That is the situation with the two galaxies and the rope: there is tidal gravity, and nothing else. No planet, no potential well, nothing. Just tidal gravity and nothing else. To support your claim you need to show that you can extract an arbitrary amount of work by paying out rope in that situation. Nothing else will do.
 
  • #42
alantheastronomer said:
Dr' Whom in post #30 has the right idea - We observe the universe to be expanding, but we don't have an explanation for what the physical cause of that expansion might be? Hence we call it "dark energy".
To expand a little on Orodruin's response, dark energy doesn't cause the expansion at all. It modifies the rate of expansion, making it higher in the late universe than would otherwise be the case.
 
  • #43
PeterDonis said:
No. In the Earth example, the rope is not what is pulling the mass off of a geodesic: it's whatever structure is holding up the rope's attachment point. In the two galaxies example, there is no such attachment point for the rope.

Yes there is, the attachment point is the other galaxy. There is an acceleration off their geodesics occurring. Acceleration times mass times distance is work. Where is the problem?

This is not analogous to the galaxy scenario either. The galaxies were not "lifted" to their positions, and you're not recovering any energy by letting them "fall".

Please take a step back and think carefully about the one valid analogy I suggested to you, the one that actually involves tidal gravity, and nothing else. That is the situation with the two galaxies and the rope: there is tidal gravity, and nothing else. No planet, no potential well, nothing. Just tidal gravity and nothing else. To support your claim you need to show that you can extract an arbitrary amount of work by paying out rope in that situation. Nothing else will do.

I see no problem in the purely tidal situation either. There is a tension in the rope, a force, if you apply it over a distance then that is work. Why wouldn't it be? And we already do extract power from tidal gravity, that's what tidal power on Earth is. The mechanism is a bit different but the power is still generated due to tidal forces. If there is some problem with any of this then you'll have to be much more clear about what it is.

Here's a similar siutation: https://physics.stackexchange.com/q...e-break-due-to-the-tidal-forces-or-not/214363. Are you trying to tell me that you cannot do work by allowing that rope to unspool while turning a turbine? I'm sorry but I see absolutely no problem with that at all.
 
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  • #44
JMz said:
How so? We have one perfectly good explanation: It's Einstein's own Λ. It's true that people can come up with other explanations -- and that work is certainly worthwhile.

But you could likewise make the case that the gravitational coupling parameter G is not independent of time, or that the speed of light is not constant for all observers. However, those are the standard explanations, they are in agreement with virtually all current observations, and they don't come with any physical cause either, at least not so far. Yet we don't consider the lack of a cause to be a problem.

If we accept a premise that there was an early inflationary phase, then it hints that "Λ" is not a constant, it depends on vacuum state you are in - IOW: it probably comes from the vacuum energy density: inflationary vacuum has a high one, our current vacuum has a very tiny, but nonzero one. Heuristically this makes sense, as inflationary vacuum probably has some huge VEVs, and current fields' VEVs are either zero (leptons and photons) or much lower than Planck.

We "only" need to learn how to actually calculate vacuum energy density, correctly in UV limit.
 
  • #45
nikkkom said:
If we accept a premise that there was an early inflationary phase, then it hints that "Λ" is not a constant, it depends on vacuum state you are in - IOW: it probably comes from the vacuum energy density: inflationary vacuum has a high one, our current vacuum has a very tiny, but nonzero one. Heuristically this makes sense, as inflationary vacuum probably has some huge VEVs, and current fields' VEVs are either zero (leptons and photons) or much lower than Planck.

We "only" need to learn how to actually calculate vacuum energy density, correctly in UV limit.

I'm not sure it has anything to do with VEVs of fields, does it? I mean the Higgs field has a VEV of 246 GeV, but this is nothing to do with dark energy. I presume because we can consider the Higgs field to be at the minima of its potential, so it is in fact in a minimum-energy state, despite that state being at non-zero field values. So it is about the potentials of our fields. In contrast, in inflation the inflaton starts off at the top of some big potential energy hill.
 
  • #46
kurros said:
I'm not sure it has anything to do with VEVs of fields, does it? I mean the Higgs field has a VEV of 246 GeV, but this is nothing to do with dark energy.

Directly, no. Indirectly, I assume the math will work out so that large VEVs usually correspond to large vacuum energy density.

In contrast, in inflation the inflaton starts off at the top of some big potential energy hill.

I don't think so. Inflationary state may be a (false) vacuum, i.e. a local _minimum_.
 
  • #47
nikkkom said:
Indirectly, I assume the math will work out so that large VEVs usually correspond to large vacuum energy density.
This assumption is generally not correct.
 
  • #48
nikkkom said:
Directly, no. Indirectly, I assume the math will work out so that large VEVs usually correspond to large vacuum energy density.

I don't see why? 246 GeV is rather high energy, vastly higher than we see for the cosmological constant, so on the face of it it seems like this isn't the case.

I don't think so. Inflationary state may be a (false) vacuum, i.e. a local _minimum_.

Local being the key word. If there is a deeper minimum somewhere else then there is still a potential energy difference between the states, with the false vacuum being higher energy.
 
  • #49
kurros said:
I don't see why? 246 GeV is rather high energy, vastly higher than we see for the cosmological constant, so on the face of it it seems like this isn't the case.

246 GeV is rather low energy compared to the natural scale, the Planck. Basically, our current vacuum state and physics we observe is very "cold", way way down below Planck.

Also, "cosmological constant" nee "dark energy" can not be directly compared with VEVs because they have different units. It's like comparing meters with volts. Dark energy has energy _density_ (energy per unit volume), not energy.

Local being the key word. If there is a deeper minimum somewhere else then there is still a potential energy difference between the states, with the false vacuum being higher energy.

Earlier you said "in inflation the inflaton starts off at the top of some big potential energy hill". That description definitely does not describe a local minimum.
 
  • #50
Orodruin said:
> Indirectly, I assume the math will work out so that large VEVs usually correspond to large vacuum energy density.

This assumption is generally not correct.

Sure. "I have a hunch" is not scientifically valid. Only the mathematically consistent, real theory's predictions are scientifically valid. I'm looking forward to the time we'll have one which can calculate vacuum energy density from first principles.
 
  • #51
nikkkom said:
Earlier you said "in inflation the inflaton starts off at the top of some big potential energy hill". That description definitely does not describe a local minimum.

Well I was describing a different kind of inflation there. The point was only that I assume the potential energy difference is important for the field to contribute to the energy-momentum tensor and thus drive inflation. I don't remember how that calculation actually goes though, but now I am curious so I will probably look it up...
 
  • #52
nikkkom said:
If we accept a premise that there was an early inflationary phase, then it hints that "Λ" is not a constant
I'm not sure that it does give that hint, though I recognize that a full quantum-gravity theory might show how that constant is the limiting value of some dynamical field. (I think this verges on mere terminology, though: "The speed of light c", likewise, is a constant, but light usually does not travel at that speed, c is just a limiting value with a not-quite-appropriate common name.)
 
  • #53
kurros said:
Yes there is, the attachment point is the other galaxy.

You continue to miss the point. The other galaxy has no proper acceleration other than what is imposed by the tension in the rope. The suspension point on the Earth does; the Earth's surface (or whatever structure is built on the Earth's surface to support the suspension point) is imposing a proper acceleration on the suspension point, which gets transmitted through the rope. There is no analogue to that in the two galaxy situation, and it makes a huge difference in the predictions.

At this point you really need to stop waving your hands and do the math. When you do, you will find that the math is different for the two scenarios you think are the same (two galaxies connected by a rope, vs. a mass hanging from a rope on Earth). And you will find that the math is the same for the two scenarios that I have said are the same: two galaxies connected by a rope, and two masses connected by a rope in a spacecraft in a free-fall orbit in the tidal gravity of the Earth.

kurros said:
I see no problem in the purely tidal situation either.

Stop waving your hands and do the math. Show me in the math where the "force" is that corresponds to the force the suspension point on the Earth is exerting. (You won't be able to, because there isn't any, as I said above, but apparently you won't be able to grasp this unless you do the math yourself. So do it.)
 
  • #54
PeterDonis said:
You continue to miss the point. The other galaxy has no proper acceleration other than what is imposed by the tension in the rope.

That's because you haven't made any point, you're just saying it doesn't work, you aren't saying why. Why do we need some other acceleration than that? Any acceleration should do.

The suspension point on the Earth does; the Earth's surface (or whatever structure is built on the Earth's surface to support the suspension point) is imposing a proper acceleration on the suspension point, which gets transmitted through the rope. There is no analogue to that in the two galaxy situation, and it makes a huge difference in the predictions.

At this point you really need to stop waving your hands and do the math. When you do, you will find that the math is different for the two scenarios you think are the same (two galaxies connected by a rope, vs. a mass hanging from a rope on Earth). And you will find that the math is the same for the two scenarios that I have said are the same: two galaxies connected by a rope, and two masses connected by a rope in a spacecraft in a free-fall orbit in the tidal gravity of the Earth.

Stop waving your hands and do the math. Show me in the math where the "force" is that corresponds to the force the suspension point on the Earth is exerting. (You won't be able to, because there isn't any, as I said above, but apparently you won't be able to grasp this unless you do the math yourself. So do it.)

Ok I will do so in a few hours. But I'm not sure where a difference is supposed to turn up. We can compute the tension in the cable in our various scenarios, but the free body diagrams of the "hanging" mass are all the same, and pretty simple. There is just a mass with a tension force from a cable in one direction, and a tidal force pulling it in the other direction (in the rest frame of the cable, say). If we loosen the cable a little by letting it unspool from something then the mass will move in the direction of the fictional tidal force, doing work against friction in the spooling mechanism of the rope, or a generator.
 
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  • #55
kurros said:
you're just saying it doesn't work, you aren't saying why

Yes, I have. I've said it several times.

kurros said:
I will do so in a few hours. But I'm not sure where a difference is supposed to turn up.

That's why I'm telling you to do the math.

kurros said:
There is just a mass with a tension force from a cable in one direction, and a tidal force pulling it in the other direction

The mass is not the only object in the scenario. The tension in the cable has to come from somewhere; a cable attached at only one end has no tension. Where does the tension come from?

kurros said:
If we loosen the cable a little by letting it unspool from something then the mass will move in the direction of the fictional tidal force, doing work against friction in the spooling mechanism of the rope, or a generator.

In order to do work against the spooling mechanism, what state of motion must the spooling mechanism be in, in the frame in which you are working? What keeps it in that state of motion?
 
  • #56
PeterDonis said:
Yes, I have. I've said it several times.

That's why I'm telling you to do the math.

The mass is not the only object in the scenario. The tension in the cable has to come from somewhere; a cable attached at only one end has no tension. Where does the tension come from?

It comes from the same diagram on the other end of the cable but reversed. In the frame of the cable there is a fictional tidal force pulling the masses apart.

In order to do work against the spooling mechanism, what state of motion must the spooling mechanism be in, in the frame in which you are working? What keeps it in that state of motion?

It just needs to move against a force. In this case it moves outwards against the tension in the cable due to the fictional tidal force. As yet another analogy, you could imagine that the tension exists due to a centrifugal force. You could extract energy from two masses attached by a cable and spinning around each other in exactly the same way. The force diagrams are the same in the rotating reference frame.
 
  • #57
kurros said:
It comes from the same diagram on the other end of the cable but reversed.

Not in the Earth case, which is the case in which I agree work will be done. In the Earth case, what does the free body diagram of the cable's suspension point look like?

Or put it this way: suppose that both ends of the cable were free falling towards the Earth (instead of the cable being suspended by a structure), with a mass at each end. Could you extract work by paying out the cable?

kurros said:
In this case it moves outwards against the tension in the cable due to the fictional tidal force

Take a step back to the Earth case, and modify it as I said above, to remove the obvious difference (as compared to the two galaxy case) of there being an extra force on the suspension point due to the Earth (i.e., not due to the tension in the cable). What happens?
 
  • #58
kurros said:
As yet another analogy, you could imagine that the tension exists due to a centrifugal force. You could extract energy from two masses attached by a cable and spinning around each other in exactly the same way.

Analyze this case in an inertial frame in which the center of mass of the system is at rest. Where is the energy you claim is being extracted going to come from?
 
  • #59
PeterDonis said:
Analyze this case in an inertial frame in which the center of mass of the system is at rest. Where is the energy you claim is being extracted going to come from?

From the rotational kinetic energy of the spinning system. This is L^2/(2I). I goes up as the masses separate, and L is conserved, so the kinetic energy goes down.
 
  • #60
kurros said:
From the rotational kinetic energy of the spinning system.

So what happens when all of that rotational energy is extracted?
 

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