B Where does the energy of gravity come from?

[lordoftheselands]

Since gravity accelerate things (even with some people saying that it's not a force) there must be a consense that it adds energy to things. There are even hydroeletric power plants generating energy everywhere. So where does the energy of gravity come from? Is it spontaneous generation of energy?

 

[Ibix]

If I lift a thing up and drop it, then harvest the kinetic energy to do something, then the energy came from my muscles and was briefly stored as gravitational potential energy before being released and used. Hydro power stations use the Sun to lift water to the clouds which then falls as rain into the reservoirs, so they are an indirect form of solar power.

So it is always with any energy generation scheme that "uses" gravitational potential energy from something moving down. Look to whatever lifted the thing up in the first place.

Since this is in relativity (at time of writing) it's worth noting that conservation of energy doesn't necessarily apply in all spacetimes. It works in Eartylhbound situations, though.

 

[PeterDonis]

gravity accelerate things

No, it doesn't. Objects moving solely under the influence of gravity are in free fall, with zero proper acceleration.

there must be a consense that it adds energy to things

Not at all. For example, water flowing downhill is not gaining any energy in total. It is converting gravitational potential energy to kinetic energy.

 

[Feynstein100]

It is converting gravitational potential energy to kinetic energy.

I think that's what the OP was referring to. Had the earth not been there, the water wouldn't have any gravitational potential energy. Imagine an object outside the earth's gravitational field. It doesn't have any energy. But place it just inside the field and it now gains kinetic energy which it didn't have before.

Ah but wait. You have to move it to get it inside the field and in doing so, you give it kinetic energy. Still, it does speed up once it enters the field (assuming it gets closer to the earth), meaning there is a net gain of energy. And assuming conservation of energy holds true, this extra energy has to come from somewhere, in this case the earth's gravitational field.

One possible resolution to this problem is to say that the earth's gravitational field extends to infinity and thus the object was never outside it, and always had potential energy. However, this feels like cheating, and I can't help but wonder if there's a better resolution.

 

[russ_watters]

The initial gravitational potential energy of the universe comes from its creation, but a hydroelectric dam is a solar powered steam engine.

 

[phinds]

Imagine an object outside the earth's gravitational field.

Since the gravitational field extends to infinity, it's hard to imagine such a place.

 

[member 728827]

No, it doesn't. Objects moving solely under the influence of gravity are in free fall, with zero proper acceleration.

Being "Objects" by definition "any substance that has mass and takes up space by having volume", would that mean that you found an easy analytical solution to the two body problem in GR?

 

[Feynstein100]

Since the gravitational field extends to infinity, it's hard to imagine such a place.

I literally said that in the last part of my comment. Did you even read it?

 

[Dale]

Imagine an object outside the earth's gravitational field. It doesn't have any energy. But place it just inside the field and it now gains kinetic energy which it didn't have before.

The gravitational field goes to infinity. There is no "outside" and no "just inside". All you can do is move from a region of one potential to a region of another potential.

Still, it does speed up once it enters the field (assuming it gets closer to the earth), meaning there is a net gain of energy.

No net gain of energy. There is a loss of PE and a conversion of that PE into the same amount of KE.

One possible resolution to this problem is to say that the earth's gravitational field extends to infinity and thus the object was never outside it, and always had potential energy. However, this feels like cheating

Do you think that you and the universe are childhood friends playing a game where you get some say in the rules? We have no say in the rules. And there is no possible way for the universe to “cheat” since the rules are just descriptions of whatever the universe does, so whatever it does is by definition a rule and not a cheat.

 

[Dale]

Being "Objects" by definition "any substance that has mass and takes up space by having volume", would that mean that you found an easy analytical solution to the two body problem in GR?

Why should it be an easy analytical solution? Do you think that the laws of nature preclude true statements about difficult solutions or numerical solutions?

 

[PeterDonis]

Had the earth not been there, the water wouldn't have any gravitational potential energy.

Not due to the earth, but the earth is not the only gravitating mass in the universe.

Imagine an object outside the earth's gravitational field.

There is no such place, strictly speaking. The earth's field gets weaker as you get farther away from it, but it never vanishes completely. In practical terms, we don't see this because there are other masses in the universe whose fields get much larger than the earth's at distances far enough away from the earth.

 

[Vanadium 50]

Do you think that you and the universe are childhood friends playing a game where you get some say in the rules?

There's always the Whistler quote "Two plus two continue to make four, despite the whine of the amateur for three or the cry of the critic for five."

 

[Feynstein100]

Do you think that you and the universe are childhood friends playing a game where you get some say in the rules? We have no say in the rules. And there is no possible way for the universe to “cheat” since the rules are just descriptions of whatever the universe does, so whatever it does is by definition a rule and not a cheat.

Not quite what I hand in mind I simply meant that at this point, it feels like you're forcing the math to confirm your preconceived notions rather than seeing where it actually leads.

In particular, I find the notion of the gravitational field extending to infinity problematic because a) there are no infinities in real life, and b) it's too newtonian of a view. Perhaps the math is the same in GR, but I find the idea that a mass here can affect things on the other side of the universe to be completely antithetical to the concept of locality, on which SR and GR are based.

 

[Ibix]

I simply meant that at this point, it feels like you're forcing the math to confirm your preconceived notions rather than seeing where it actually leads.

It's the maths that says the field extends to infinity. We didn't pull the notion out of nowhere.

there are no infinities in real life,

Prove it.

it's too newtonian of a view. Perhaps the math is the same in GR, but I find the idea that a mass here can affect things on the other side of the universe to be completely antithetical to the concept of locality, on which SR and GR are based.

I don't think you really understand what "locality" means in this context. And again, you're saying that the maths is inconsistent with the maths, which is obviously nonsense.

 

[PeterDonis]

I find the idea that a mass here can affect things on the other side of the universe to be completely antithetical to the concept of locality, on which SR and GR are based.

That concept just means that interactions propagate at the speed of light. Changes in spacetime geometry due to changes in the configuration of masses meet that requirement in GR. There is no limit on the distance over which such changes can propagate.

 

[Dale]

it feels like you're forcing the math to confirm your preconceived notions rather than seeing where it actually leads

Well, yes. We force the math to conform to our experimental data. It isn’t the math “leading” anything. It is the data. We see what the universe actually does and then write the rules to reflect the facts.

it's too newtonian of a view.

GR contains Newtonian gravity. And far away from any mass is precisely where GR matches Newtonian gravity.

Although, in GR’s FLRW cosmology, energy is not conserved, but not for the reasons mentioned here. There is no “outside of the gravitational field” even there.

 

[PeroK]

a) there are no infinities in real life

Amongst everything else, you are confusing two different ideas of infinite. You cannot have two points an infinite distance apart. But, there is no limit to the possible distance between two points.

 

[Nugatory]

In particular, I find the notion of the gravitational field extending to infinity problematic because a) there are no infinities in real life, and b) it's too newtonian of a view.

”Extending to infinity” is indeed problematic, but the problem is in the very common but imprecise natural language wording. There’s no infinity involved when we state it differently: if the gravitational potential is non-zero at a finite distance $R$ from the gravitating body and $\Delta R$ is some small number, then the gravitational potential will be smaller but still non-zero at the finite distance $R+\Delta R$.

Informally, there is no hard edge detectable even with arbitrarily precise instruments.

 

[member 728827]

Why should it be an easy analytical solution? Do you think that the laws of nature preclude true statements about difficult solutions or numerical solutions?

You're absolutely right, as always.

I think I just "condensed" the remark too much.

What I tried to mean is that if "an object" (which let me say that in physical "terminology" would mean any object, even a piece of a neutron start so to speak), is in free fall and "proper acceleration" is zero (which for an "object" with a measurable size and volume I don't know very much what it means), then it follows a Geodesic trajectory.

Obviously I don't know, but I thought that to calculate a Geodesic from the GR equations is something "mechanical" (I would guess tedious), but easy in the sense that if you know what you're doing, you follow the recipe and you get that Geodesic trajectory.

 

[jbriggs444]

Had the earth not been there, the water wouldn't have any gravitational potential energy.

On the contrary. When the earth is present, the water has less gravitational potential energy than with the earth absent.

Make sure that you are comparing apples with apples. Do not change the zero point of your potential energy calculation when you coallesce the Earth.

 

[DaveC426913]

There's always the Whistler quote "Two plus two continue to make four, despite the whine of the amateur for three or the cry of the critic for five."

Or the Python quote:

"...shalt thou count to [four], no more, no less. [Four] shall be the number thou shalt count, and the number of the counting shall be [four]. [Five] shalt thou not count, neither count thou [three], excepting that thou then proceed to [four]. [Six] is right out.

 

[member 728827]

Since gravity accelerate things (even with some people saying that it's not a force) there must be a consense that it adds energy to things. There are even hydroeletric power plants generating energy everywhere. So where does the energy of gravity come from? Is it spontaneous generation of energy?

I guess that it comes from the mass, as E=mc² and such.

By the way, the other day I saw I video which was something like this: the same way that the binding energy from the proton-electron is part of the mass of one atom, the binding gravitational energy for two massive objects has in itself a mass, and that mass has been demonstrated to obey the strong equivalence principle up to a number of decimal places (in a nutshell, accelerates the same way as the other mass).

 

[PeterDonis]

the other day I saw I video

Videos are usually not valid references here, but the only way we can know for sure is for you to give a link to it.

he binding gravitational energy for two massive objects has in itself a mass

A negative mass--it makes a negative contribution to the total mass of the system.

that mass has been demonstrated to obey the strong equivalence principle up to a number of decimal places

Yes.

 

[PeterDonis]

I guess that it comes from the mass

Don't guess. When you are trying to help someone understand something here, you need to first understand it well enough yourself that you don't have to guess.

In fact this answer is not correct, as other posts in this thread will show.

 

[PeterDonis]

Being "Objects" by definition "any substance that has mass and takes up space by having volume"

No, that's not what "objects" meant in the statement of mine that you quoted.

would that mean that you found an easy analytical solution to the two body problem in GR?

No, because "objects" in the statement of mine that you quoted has a much more limited meaning than you think.

 

[PeterDonis]

if "an object" (which let me say that in physical "terminology" would mean any object, even a piece of a neutron start so to speak), is in free fall and "proper acceleration" is zero (which for an "object" with a measurable size and volume I don't know very much what it means), then it follows a Geodesic trajectory.

This is only an exact statement for test objects, but test objects by definition have zero effect on the spacetime geometry. So they are not "bodies" in the sense of finding a solution to the "two-body problem".

There has been considerable theoretical work on the topic of how good an approximation it is to assume that objects which are not test objects (i.e., which have non-negligible effects on the spacetime geometry), but which have masses that are much smaller than some central mass they are orbiting--for example, the Earth orbiting the Sun--follow geodesic trajectories in the spacetime geometry due to the central mass--for example, just considering the spacetime geometry due to the Sun without taking into account any effects on the spacetime geometry due to the Earth itself. We have had some previous PF threads on this, but it's been a while. Empirically, this does indeed seem to be a pretty good approximation in many cases. But it still has nothing to do with solving the "two-body problem", since the whole point of the approximation is to eliminate all but one of the "bodies" when calculating what the spacetime geometry is.

Obviously I don't know, but I thought that to calculate a Geodesic from the GR equations is something "mechanical" (I would guess tedious), but easy in the sense that if you know what you're doing, you follow the recipe and you get that Geodesic trajectory.

It is if you know the spacetime geometry. But obtaining the spacetime geometry is precisely the part that, for more than one body in GR, has not been solved analytically.

 

[member 728827]

Videos are usually not valid references here, but the only way we can know for sure is for you to give a link to it.

The referenced video is in Spanish. Talks about the retroreflectors left in the Moon, and how they where used to check for the strong equivalence in GR.

 

[member 728827]

In fact this answer is not correct, as other posts in this thread will show.

The concept of "the energy of gravity" is a little bit ambiguous, but don't tell me is that is not mass related!

 

[Nugatory]

but don't tell me is that is not mass related!

No one is trying to tell you that . What they are saying is that "It comes from the mass" is incorrect, which shouldn't surprise you as you were uncertain enough to preface that statement with "I guess"

 

[PeterDonis]

The referenced video is in Spanish.

Even in English it would probably not be the best reference to give. See below for better ones. It only took me a few minutes of web searching to find the references below; that is something you could (and should) have done.

Talks about the retroreflectors left in the Moon, and how they where used to check for the strong equivalence in GR.

This is the Lunar Laser Ranging experiment. The Wikipedia article gives a reasonable overview and a good number of references:

https://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment

This appears to be a good paper on the use of lunar laser ranging to test the equivalence principle:

https://arxiv.org/pdf/gr-qc/0507083.pdf

 

[Vanadium 50]

A few meta-comments:

The OP seems to have departed, and he seems to have a history of doing so. (Looking at past threads) Looking at those self-same threads, I suggest that the thread be relabeled as B-level.

The thread has been taken a little off-track by someone trying to help the OP, but not entirely sure. In this case, it really is best to check things out before posting. If questions remain, you can always start a new thread.

Now, a question for the OP (if he's still around). Why do you think this is a relativity question? Why and how do you think the answer is different in Newtonian gravity? And if the answer is "they are the same", we shuld probably answer this classically and not drag the additional compications of relativity into it.

 

[PeterDonis]

I suggest that the thread be relabeled as B-level.

Fair point. Done.

 

[russ_watters]

In particular, I find the notion of the gravitational field extending to infinity problematic because a) there are no infinities in real life...
[prior post]
Imagine an object outside the earth's gravitational field. It doesn't have any energy.

Note that the concept of gravitational potential energy doesn't really require gravitational force extending to infinity. It's just the integral of gravitational force with distance, and if that force ever drops to zero a million light years away or whatever, it has basically no impact on the gravitational potential energy (it would be too small of a deviation to be noticeable). So even if the first sentence of that prior post were true, the second still wouldn't be.

 

[PeterDonis]

Note that the concept of gravitational potential energy doesn't really require gravitational force extending to infinity. It's just the integral of gravitational force with distance

Since this thread was posted in the relativity forum, it should be noted that this is not quite how things work in relativity. The statement of Feynstein100 that you quoted can certainly be criticized (and I and others have done so elsewhere in this thread), but I'm not sure trying to do so along these particular lines will work.

 

[Dale]

So even if the first sentence of that prior post were true, the second still wouldn't be

That is a really good point. Even if we hypothesize a conservative force with a finite range, the region outside the force would be an equipotential region. The potential would be defined, even outside the finite range of the force.

 

[PeterDonis]

Even if we hypothesize a conservative force with a finite range, the region outside the force would be an equipotential region. The potential would be defined, even outside the finite range of the force.

This is true as a general statement about forces and potentials. But in relativity, gravity is not a force, and formulating a valid concept of "potential energy" for gravity must be done without making use of the concept of force. (It is also worth noting that this can only be done in GR for a particular class of spacetimes, the stationary spacetimes.) Doing this correctly in GR does show that, in spacetimes where there is a valid concept of "gravitational potential energy" at all, it is well defined everywhere in the spacetime (or more precisely, everywhere in the stationary region--which might not be the entire spacetime in a case like a black hole, but that is a whole separate issue that is off topic for this thread). But the logic is different from that expressed in the above quote.

 

[Dale]

formulating a valid concept of "potential energy" for gravity must be done without making use of the concept of force. (It is also worth noting that this can only be done in GR for a particular class of spacetimes, the stationary spacetimes.)

And in that class of spacetimes the concept of a gravitational potential is perfectly legitimate and is reasonable fodder for discussion here. And in those spacetimes the gradient of a potential is still a force and fictitious forces also bear the name of force. I understand your opinion, but my logic is also sound. Particularly as I actually stated the matter.

You are just using a more restrictive sense of the word “force” than I am here. I have no qualms about speaking of gravitational forces, using a broader meaning of the term. Nor do I have qualms about stating that gravity is not a force in GR, using a more narrow meaning of the term. Words have more than one meaning.

 

[PeterDonis]

You are just using a more restrictive sense of the word “force” than I am here.

Fair enough. However:

the gradient of a potential is still a force

This still isn't quite true. Take the simplest example, Schwarzschild spacetime. The potential energy (per unit mass--I'm going to do everything per unit mass here to avoid having to carry around a factor of $m$ in all the formulas that does nothing) is

$$
U = \sqrt{1 - 2M / r} - 1
$$

Note that of course this is negative for any finite value of $r$ (and goes to $0$ at $\infty$).

The force (per unit mass) on an object at $r$ (given by, for example, the proper acceleration necessary to "hover" at rest) is (without worrying about the sign, since that just reflects whether we want the actual force on an object at rest, or the fictitious force in that object's frame on an object freely falling radially)

$$
F = \frac{M}{r^2 \sqrt{1 - 2M / r}}
$$

This is $dU / dr$, but that is not the same as the gradient, because $r$ is not the same as radial distance. The gradient will be the derivative with respect to radial distance, which gives

$$
\frac{dU}{d \ell} = \frac{M}{r^2}
$$

i.e., the "redshifted" acceleration "due to gravity". (A similar discrepancy arises when we compute the work required to lift an object "hovering" at finite $r$ out to infinity.)

 

[member 728827]

Even in English it would probably not be the best reference to give. See below for better ones. It only took me a few minutes of web searching to find the references below; that is something you could (and should) have done.

I didn't look into any particular video to talk about the Equivalence Principles, but just thought (guessed, supposed....) that the concept could fit into the naive question "where does the energy of gravity come from...." question in a Relativity Forum, mentioning a video that the IA of Google put in my playlist, mixed with some cat videos, bird training and other stuff.

Is not the best reference in your opinion, but is an enough clear video for me from the former director of the Theoretical Physics Division of the CERN in his endeavor of outreaching some physic concepts to a broad range of people.

And IMHO, the part that reads "...you could (and should) have done..." is not appropriate, and you "should not" use those type of wording in your answers.

 

[phinds]

Is not the best reference in your opinion, but is an enough clear video for me from the former director of the Theoretical Physics Division of the CERN in his endeavor of outreaching some physic concepts to a broad range of people.

Even with only 119 posts, you should be aware by now that pop-science presentations (designed, as you say, for a broad range of people) are never acceptable references here on PF, regardless of the speaker/writer.

 

[Dale]

pop-science presentations (designed, as you say, for a broad range of people) are never acceptable references here on PF

Well, they can be acceptable references for an OP. But we should use sources that are consistent with the professional scientific literature in our responses.

 

[PeroK]

Is not the best reference in your opinion, but is an enough clear video for me from the former director of the Theoretical Physics Division of the CERN in his endeavor of outreaching some physic concepts to a broad range of people.

This misses the point that PF is not here to present popular-science. Nor to discuss the content as presented in popular science.

Our goal is to provide a community for people (whether students, professional scientists, or hobbyists) to learn and discuss science as it is currently generally understood and practiced by the professional scientific community.

Reference: https://www.physicsforums.com/insights/about-physics-forums/

 

[PeterDonis]

IMHO, the part that reads "...you could (and should) have done..." is not appropriate, and you "should not" use those type of wording in your answers.

I think you need to consider that you should be respectful of other poster's time. You should not depend on other posters to go and find valid references for you. You should be doing it yourself.

 

[PeterDonis]

I didn't look into any particular video to talk about the Equivalence Principles, but just thought (guessed, supposed....) that the concept could fit into the naive question "where does the energy of gravity come from...." question in a Relativity Forum

What does the Equivalence Principle have to do with energy?

 

[vanhees71]

This is true as a general statement about forces and potentials. But in relativity, gravity is not a force, and formulating a valid concept of "potential energy" for gravity must be done without making use of the concept of force. (It is also worth noting that this can only be done in GR for a particular class of spacetimes, the stationary spacetimes.) Doing this correctly in GR does show that, in spacetimes where there is a valid concept of "gravitational potential energy" at all, it is well defined everywhere in the spacetime (or more precisely, everywhere in the stationary region--which might not be the entire spacetime in a case like a black hole, but that is a whole separate issue that is off topic for this thread). But the logic is different from that expressed in the above quote.

Do you have a reference (a review article or textbook?) about the status of this complicated issue. The question of the energy of the gravitational field. It lead to the famous work by Emmy Noether on symmetries and conservation laws, and this work was more or less complete in the sense that it dealt with both, "true symmetries" (global symmetries), which imply conservation laws, i.e., each one-parameter Lie group is generated by a quantity that is conserved (Noether's 1st theorem) as well as local gauge symmetries (in the case of GR it's GL(4), i.e., the general covariance), which leads to the existence of a set of arbitrary functions, which are irrelevant for the physics.

In electrodynamics the four-potential is only determined up to a gradient of a scalar field, but this doesn't matter, because the physics is indeed completely described by the four-potential modulo gauge transformations, and observables must be gauge-invariant. In GR everything is defined by $g_{\mu \nu}$ modulo arbitrary coordinate transformations (general covariance).

Concerning "gravitational field energy" the best I know is the pseudotensor a la Landau and Lifshitz, but I gueess there might be some progress in the matter since then.

Besices this, I think the gravitational interaction still is an interaction described by a local field in GR as all the other interactions. I

In GR gravitation is a true interaction and not only "inertial forces". Forces a la Newton, which is an action-at-a-distance concept, don't exist already in SR in the literal sense. My argument that gravitation in GR is a "true interaction" simply is that you can only transform away the gravitational interaction locally and approximately over spacetime regions, where the gravitational field can be considered as homogeneous. That's always approximate, because for a true gravitational field the curvature tensor is non-vanishing, and that's valid in any reference frame/choice of coordinates, i.e., you cannot get rid of the "tidal forces" also in local free-falling reference frames.

 

[PeterDonis]

I think the gravitational interaction still is an interaction described by a local field in GR

In classical GR, the only "local field" you can use is the metric itself or some tensor derived from it (such as the Riemann tensor or the Einstein tensor). But the metric doesn't describe an "interaction"--it describes the spacetime geometry. That is why it is commonly stated that gravity is not a force in classical GR.

If you view classical GR as the classical limit of the quantum field theory of a massless spin-2 field, then yes, you can view that field as a local field describing a "gravitational interaction". But you will still not be able to construct any local invariant that corresponds to the "energy" of this field. The various pseudotensors that have been used for this purpose in the literature are not local invariants; they are coordinate dependent. That is why there has never been general acceptance of any of them as a proper description of "the energy of the gravitational field" or as describing a classical "gravitational interaction".

There are "energy" concepts that are well defined in particular classes of spacetimes: the ADM energy and Bondi energy in asymptotically flat spacetimes, the Komar energy and the energy at infinity of a test object in stationary spacetimes. (Note that all but the last of these are integral quantities so they are not "local".) In the stationary case, the energy at infinity can be used to define "gravitational potential energy" as I discussed in what you quoted from me. I discussed some further issues involved in post #38 of this thread.

 

[PeterDonis]

My argument that gravitation in GR is a "true interaction" simply is that you can only transform away the gravitational interaction locally and approximately over spacetime regions, where the gravitational field can be considered as homogeneous. That's always approximate, because for a true gravitational field the curvature tensor is non-vanishing

This is not an argument for gravitation being a "true interaction". (Misner, Thorne, & Wheeler, in particular, would vigorously object to any such claim.) It is just an often necessary clarification to what aspect of "gravity" is truly invariant, i.e., independent of any choice of coordinates. That aspect is spacetime curvature, i.e., tidal gravity.

 

[Dale]

In classical GR, the only "local field" you can use is the metric itself or some tensor derived from it (such as the Riemann tensor or the Einstein tensor)

I think you mean, "the only local tensor field" or something similar. You certainly can use many non-covariant fields if you wish.

 

[vanhees71]

This is not an argument for gravitation being a "true interaction". (Misner, Thorne, & Wheeler, in particular, would vigorously object to any such claim.) It is just an often necessary clarification to what aspect of "gravity" is truly invariant, i.e., independent of any choice of coordinates. That aspect is spacetime curvature, i.e., tidal gravity.

My point is that you cannot transform away a true "gravitational field", i.e., a curved spacetime doesn't become flat by any coordinate transformation. In this sense gravity is not merely an "inertial/fictitious force". In MTW you find the expression "tidal gravitational forces" a zillion of times!

 

[PeterDonis]

I think you mean, "the only local tensor field" or something similar.\

Yes.

You certainly can use many non-covariant fields if you wish.

But doing that is open to serious objections, which are discussed in detail in, for example, Misner, Thorne & Wheeler.