How Can We Overcome the Challenges of Locating Gravitational Energy?

[sweet springs]

Hi. To express the difficulties of the localization of gravitational energy, I pose this question.
To high school students or to scholars, how shall we answer?

 

[HallsofIvy]

What makes you think that "mgh" has to be "stored" anywhere? I really do not understand your question.

 

[sweet springs]

Thanks. So gravitational energy mgh or -GMm/r do not belong to anywhere?

 

[HallsofIvy]

What do you mean by "energy" belonging anywhere?

If I understand your question correctly then you can think of the potential energy of an object of mass m at a height h as being "contained in" whatever it is that is holding the object at height h. That's the best I can do.

Suppose you were to buy a house for $100,000 and then sold it for $150,000. Where was that $50,000 profit "stored"?

 

[PeterDonis]

gravitational energy mgh or -GMm/r do not belong to anywhere?

This energy cannot be localized, if that's what you mean. The best you can do, as HallsofIvy implied, is to view it as "stored" in the global configuration of objects that gives rise to the potential energy.

 

[sweet springs]

Thank you. Non localizability of gravitational energy and principle of locality do not contradict? If I get acceleration by gravity of the earth, I woukd think that gravitational energy nearby me is entering into me and is increasing kinetic energy of me. Best.

 

[sweet springs]

Suppose you were to buy a house for $100,000 and then sold it for $150,000. Where was that $50,000 profit "stored"?

Umm... In the bank account of the buyer?

Best

 

[Nugatory]

We had another thread along these lines recently: https://www.physicsforums.com/threads/energy-problem-with-gravity.818750/

What's going on here is that the question "Where is mgh stored?" is ill-formed in GR because it's assuming a conserved and non-local total energy, and that is seriously problematic in GR. (No universally accepted simultaneity makes it hard to compare the total amount of energy in the universe at one moment to the total amount of energy in the universe at the next moment).

Instead, say that energy is locally conserved around the falling body because its kinetic energy is constant in the inertial frame in which the body is at rest. The surface of the Earth is not at rest in this frame, so has substantial kinetic energy that will become available when it and the object collide.

 

[sweet springs]

Thank you. I know the problems now.
Let me ask you one question. The law of energy conservation denies a perpetual motion machine of the first kind. Would GR allow it to exist or keep denying it?
On the Earth we would get energy as much as we want in compensation of graviatational energy decrease around a star far away. Can I wirte such Sci-Fi with the idea on use of non localized gravitational energy? Best.

 

[pervect]

Hi. To express the difficulties of the localization of gravitational energy, I pose this question.
To high school students or to scholars, how shall we answer?

Tell them we don't have a definite answer to that question yet. It's safe and reasonably accurate.

At a higher than high school level, there's more that could be said, for the high school level I'd keep it simple.

 

[PeterDonis]

The law of energy conservation denies a perpetual motion machine of the first kind. Would GR allow it to exist or keep denying it?

GR still does not allow perpetual motion machines. In GR, that is enforced by the fact that the covariant divergence of the stress-energy tensor is zero; that says that stress-energy cannot be created or destroyed at any point in spacetime. (What you are calling "gravitational energy" is not stress-energy and is not locally definable.)

On the Earth we would get energy as much as we want in compensation of graviatational energy decrease around a star far away.

Can you be more specific about what you're imagining here? What sort of mechanism would capture this energy and make it available on Earth?

 

[russ_watters]

Umm... In the bank account of the buyer?

Before the sale, it must exist somewhere else, right?

If I get acceleration by gravity of the earth, I woukd think that gravitational energy nearby me is entering into me and is increasing kinetic energy of me.

No. Energy isn't a "thing" that needs to "enter" you to be used. It's just bookkeeping.

 

[pervect]

BTW, I'd recommend Misner, Thorne, WHeeler's short chapter in "Gravitation" about the impossibility of localizing gravitational energy to someone with some college background. Most high school students just don't have the background to appreciate text - this chapter isn't so bad, so an undergraduate college student has a chance, though the bulk of the text is graduate level. Also, in my experience, high school students often think they know everything already anyways, so they are probablly really listening to the answer anyway. It's just as well try to get them curious about the field so they can learn more about the issues later if they maintain an interest once they get the necessary background.

 

[sweet springs]

GR still does not allow perpetual motion machines. In GR, that is enforced by the fact that the covariant divergence of the stress-energy tensor is zero; that says that stress-energy cannot be created or destroyed at any point in spacetime. (What you are calling "gravitational energy" is not stress-energy and is not locally definable.)

Thanks. The increase of kinetic energy of a falling body seems that stress-energy has been icreasing there. How do you interprete this case?

 

[PeterDonis]

The increase of kinetic energy of a falling body seems that stress-energy has been increasing there.

It may seem that way, but it's not that way. The stress-energy of the falling body is conserved; no stress-energy is entering or leaving the body. The body is in free fall, so no forces are acting on it. (In GR, gravity is not a force.) All that is happening is that the body is following the geometry of spacetime.

 

[sweet springs]

It may seem that way, but it's not that way. The stress-energy of the falling body is conserved; no stress-energy is entering or leaving the body. The body is in free fall, so no forces are acting on it. (In GR, gravity is not a force.) All that is happening is that the body is following the geometry of spacetime.

Thanks. How we should moderate this view with high school teachings that speed or kinetic energy of a ball is increasing?

Best

 

[PeterDonis]

How we should moderate this view with high school teachings that speed or kinetic energy of a ball is increasing?

GR is a bit advanced for high school, but the short answer is that kinetic energy is not the same thing as stress-energy, so kinetic energy can increase while stress-energy is conserved. A longer answer would require a course in GR.

 

[A.T.]

Thanks. So gravitational energy mgh or -GMm/r do not belong to anywhere?

Where do the h and r belong?

 

[sweet springs]

GR is a bit advanced for high school, but the short answer is that kinetic energy is not the same thing as stress-energy, so kinetic energy can increase while stress-energy is conserved.

Thanks. I suppose that kinetic energy of charged particle is included in stress-energy in the theory of electromagnetism. Energy-momentum flows into charged particle from the electromagnetic field there. In counting kinetic energy as part of stress-enegy or not, these cases seem different. Best.

 

[PeterDonis]

I suppose that kinetic energy of charged particle is included in stress-energy in the theory of electromagnetism. Energy-momentum flows into charged particle from the electromagnetic field there.

The electromagnetic field and charged particles can exchange stress-energy, but the particle feels a force in this case, so it is not the same as the case of gravity.

However, even in the EM case, it's still not really correct to equate kinetic energy with stress-energy. You still do not seem to grasp that kinetic energy is frame-dependent, whereas stress-energy is not. More precisely, stress-energy is described by a covariant geometric object, the stress-energy tensor, whereas kinetic energy is not. Only covariant geometric objects have physical meaning in GR.

 

[sweet springs]

However, even in the EM case, it's still not really correct to equate kinetic energy with stress-energy. You still do not seem to grasp that kinetic energy is frame-dependent, whereas stress-energy is not. More precisely, stress-energy is described by a covariant geometric object, the stress-energy tensor, whereas kinetic energy is not. Only covariant geometric objects have physical meaning in GR.

Thanks.
Let me know your point more precisely. For a perfect fluid in thermodynamic equilibrium for an example, the stress–energy tensor takes on a form
T^{\alpha \beta} \, = \left(\rho + {p \over c^2}\right)u^{\alpha}u^{\beta} + p g^{\alpha \beta}
where $\rho$ is the mass–energy density (kilograms per cubic meter), p is the hydrostatic pressure (pascals), $u^{\alpha}$ is the fluid's four velocity, and $g^{\alpha \beta}$ is the metric tensor. I quoted it from Wikipedia , stress-energy tensor, thanks.

First term seems to conatain kinetic energy, doesn't it?

[Mentor's note - A smallish Latex formatting problem in the original has been corrected]

 

[harrylin]

What do you mean by "energy" belonging anywhere?

If I understand your question correctly then you can think of the potential energy of an object of mass m at a height h as being "contained in" whatever it is that is holding the object at height h. That's the best I can do. [..]

If I'm not mistaken, whatever is holding the object is compressed by the work done over the compression distance alone. The potential energy is proportional to h, while at constant g the little compression distance Δh is not a direct function of the total height h; IMHO it merely accounts for the reduced potential energy at the slightly reduced height.

If the object contains a resonator, we can measure the difference in potential energy difference mgh as a difference in resonance frequency. Higher potential energy corresponds to an increase in resonance frequency. That fact as predicted by GR and confirmed by measurements suggests to me that this energy is at least partially located in the masses that were separated from each other (of course this whole discussion relates to situations in which "potential energy mgh" is assumed to be approximately valid).

 

[A.T.]

If I get acceleration by gravity of the earth, I woukd think that gravitational energy nearby me is entering into me and is increasing kinetic energy of me.

How would you explain this to an observer falling along with you, who doesn't observe any increase of your kinetic energy?

 

[Nugatory]

First term seems to contain kinetic energy, doesn't it?

You can read it that way, but you're still missing PeterDonis's point. The values of both that entire expression and the part that you're identifying as the kinetic energy will change according to the coordinate system that you're using at the moment. You can no more assign a frame-independent geometric meaning to these than you could to the value of a single component of a four-momentum vector.

 

[sweet springs]

The values of both that entire expression and the part that you're identifying as the kinetic energy will change according to the coordinate system that you're using at the moment.

Thank you. I think you are talking about a general feature of tensors. Tensors transform according to the coordinate system we use. Things are moving or still according to the coorinates we choose. So what? We can get more than this to know about "genuine", not kinetic, energy in GR? Best.

 

[sweet springs]

How would you explain this to an observer falling along with you, who doesn't observe any increase of your kinetic energy?

Thanks. In the coordinate system that I remain still, he remains still also and we get no energy into us.
In another coordinate that I am falling with increasing velocity, he is falling with increasing velocity and we get energy flowing into us. If we get accerelated by charge we have in the electromagnetic field. it is clear. Why can not we expect similar in gravitational force ? Best.

 

[stedwards]

Energy is not a scalar quantity, so why should it be conserved? P_\mu = E dt + p_i dx^i

 

[harrylin]

Thanks. In the coordinate system that I remain still, he remains still also and we get no energy into us.
In another coordinate that I am falling with increasing velocity, he is falling with increasing velocity and we get energy flowing into us. If we get accerelated by charge we have in the electromagnetic field. it is clear. Why can not we expect similar in gravitational force ? Best.

For the electron, the magnetic field energy seems to correspond rather well with the kinetic energy of motion (approximate analysis in the textbook of Alonso&Finn, Electromagnetism). The classical concept that potential energy is transformed into kinetic energy when an object falls down (from the point of view of a "stationary" reference at infinity) seems to match rather well with that analysis.

 

[A.T.]

Thanks. In the coordinate system that I remain still, he remains still also and we get no energy into us. In another coordinate that I am falling with increasing velocity, he is falling with increasing velocity and we get energy flowing into us.

So energy is a number a frame of reference assigns to you, not a substance that flows into you.

If we get accerelated by charge we have in the electromagnetic field. it is clear.

What is clear?

Why can not we expect similar in gravitational force ?

What is not similar?

 

[sweet springs]

Energy is not a scalar quantity, so why should it be conserved? Pμ=Edt+pidxi

So energy is a number a frame of reference assigns to you, not a substance that flows into you.

I think that momentum-energy is tensor that changes accorging to the applied coordinate systems.

 

[PeterDonis]

I think that momentum-energy is tensor that changes accorging to the applied coordinate systems.

Not momentum-energy, stress-energy. There is something called an energy-momentum 4-vector (or 4-momentum), which is a vector, not a tensor, and energy is one component of that vector. But you can only use a 4-momentum vector to describe an isolated object or system. You can't use it to describe a continuous distribution of matter, or a field. The stress-energy tensor is more general and can be used in those cases.

 

[Nugatory]

I think that momentum-energy is tensor that changes according to the applied coordinate systems.

The momentum four-vector is a rank-one contravariant tensor. As with any tensor, the representation of its components in a particular coordinate system depends on the chosen coordinate system. The invariant properties of the tensor, such as its magnitude, do not.

Similarly, the components of that stress-energy tensor you're looking at (which is a rank two tensor) also depend on the coordinate system that you use to write them down, so the "kinetic energy" term you're looking at changes with them. However, invariant properties such as the divergence do not; the divergence of that tensor is zero no matter what coordinate system you use to calculate it, and that's one way to state conservation of energy in GR - it is inherently a local definition.

 

[stedwards]

All vectors are tensors, but not all tensors are vectors.
Energy-momentum has representations as a vector and as a covector. One can be integrated and one cannot.

 

[sweet springs]

Thanks all.
Let me summarize my understanding so far.

In the case of charged material and electromagnetic field,
\partial_\mu T^{\mu\nu}_{total}=\partial_\mu \{T^{\mu\nu}_{mateial}+T^{\mu\nu}_{e.m.}\}=0
is the law of conservation of momentum and energy. Matial and electromagnetic field exchange momentum and energy. 4-potential of electromagnetic field (\phi, \mathbf{A}) exists physically and it would provide/deduce energy to/from motion of charger materials..

In GR case where gravity requires covariat derivativation to be implemented, the conservation law is
\nabla_\mu T^{\mu\nu}=0
Artificially we can interprete it in the form of ordinary derivataive equation by introducing pseudotensor t ,e.g. Landau-Lifshitz pseudotensor
\partial_\mu \{(-g)T^{\mu\nu}+(-g)t^{\mu\nu} \}=0, Einstein pseudotensor
\partial_\mu \{\sqrt{-g}T^{\mu\nu}+\sqrt{-g}t^{\mu\nu} \}=0

So t is an artificaila tool for this intepretation. t is a fictious thing and does not exist. Gravitaion potential mgh or -GMm/r originated from t, are thus fictious also and do not exist physically.
The high school teaching that conservation of energy, mgh + 1/2mv^2 = const., works practically well but it is not true in this sense. A ball gets kinetic energy during its fall, but NO increase of momentum and energy in the sense of covariant derivation. We require the motion to be interpreted in the same way as in the not curved space-time and thus introduce fictious t for this purpose.

Best

 

[Dale]

Uhh, why are we giving subtle general relativaty answers to high school students? Wouldn't they be better served by a simple answer in Newtonian physics?

In Newtonian gravity, gravitational fields have an associated energy density, and gravitational potential energy (like mgh) is stored in the field.

I think that the general relativity answers would be a disservice to high school students.

 

[stedwards]

Good point Dale. The Newtonian, gravitational Lagrangian has two energy terms. One is the product of the mass density and the gravitational field potential. The other is the field strength squared. Which is potential and, which kinetic?

 

[harrylin]

Thanks all.
Let me summarize my understanding so far.

[..] So t is an artificaila tool for this intepretation. t is a fictious thing and does not exist.[..]

You seem to be saying that resonance frequency is fictitious...

 

[sweet springs]

Ｉ think that the general relativity answers would be a disservice to high school students.

I fully agree with you from the educational point of view that we should not misguide the youth.

Wouldn't they be better served by a simple answer in Newtonian physics?
In Newtonian gravity, gravitational fields have an associated energy density, and gravitational potential energy (like mgh) is stored in the field.

Can we teacn gravity force mM/r^2 in analogy with Coulomb interaction qQ/r^2?
In Newtonian gravity, Like E^2 in e.m., is gravity field energy localised in the space?

Best

 

[sweet springs]

You seem to be saying that resonance frequency is fictitious...

Which resonance frequency are you meaning?

 

[harrylin]

Which resonance frequency are you meaning?

post #22 . [Edit] Gravitational potential energy difference can be detected in that way.

 

[A.T.]

Can we teacn gravity force mM/r^2 in analogy with Coulomb interaction qQ/r^2? In Newtonian gravity, Like E^2 in e.m., is gravity field energy localised in the space?

Is the distance r "localized in the space"?

 

[sweet springs]

Thanks. I do not know very well about the resonance.
As for high and low free falling objects that are moving along the geodestics in other words, could their proper resonances be same ? I suspect so. Best.

 

[sweet springs]

Is the distance r "localized in the space"?

Coulomb static energy is localized as E（r)^2 where E(r)=Q/r with appropriate coefficinets.
Not distance but energy could be localized in the space. I am not sure of it for Newtonian gravity.
Best.

 

[A.T.]

Coulomb static energy is localized as E（r)^2 where E(r)=Q/r with appropriate coefficinets. Not distance but energy could be localized in the space.

So when you have two opposite charges at distance r, where exactly is the potential energy localized in space?

 

[sweet springs]

Thanks. The space volume dxdydz contains Coulomb static energy E(x,y,z)^2 dxdydz　where E is electric field that two charges make there. Your textbook of e.m. can tell you more. Best.　

 

[Dale]

Good point Dale. The Newtonian, gravitational Lagrangian has two energy terms. One is the product of the mass density and the gravitational field potential. The other is the field strength squared. Which is potential and, which kinetic?

They are both related to the potential energy in the field. The kinetic energy would be another term. Remember, this is the Lagrangian density, not the energy density. The energy density is just a term with the field squared.

Here is a discussion on this topic (among others) especially post 8 https://www.physicsforums.com/threads/lagrangian-density-of-Newtonian-gravity.736568/

 

[Dale]

Can we teacn gravity force mM/r^2 in analogy with Coulomb interaction qQ/r^2?
In Newtonian gravity, Like E^2 in e.m., is gravity field energy localised in the space?

I wouldn't make the analogy with Coulombs law. But, yes, the usual energy density is $-1/(8\pi G)g^2$. Of course, there are always tricky details and caveats, but this is a high school class, not a class of PhD candidates.

 

[Nugatory]

Thanks. The space volume dxdydz contains Coulomb static energy E(x,y,z)^2 dxdydz　where E is electric field that two charges make there. Your textbook of e.m. can tell you more. Best.　

The infinitesimal volume element contains an infinitesimal amount of Coulomb static energy, but that doesn't help you "locate" the Coulomb static energy because you have to integrate across a non-local volume of space to get any meaningful number.

The analogy between classical gravity and classical electrostatics is pretty good: They are governed by force equations of identical $1/r^2$ form and potential energy is defined in terms of path-independent line integrals of that force. Because potential energy is defined in terms of line integrals across the entire region in question, it is a property of the entire configuration of mass/charge across that region - there is no sensible way of saying that it is "stored" in one part of the configuration or another.

So the answer for high school students, who are best served by a purely classical treatment, is that $mgh$ isn't "stored" anywhere - it's a property of the entire configuration of objects including both the Earth an dthe object whose height we're talking about.

You started this thread in the Relativity section, so naturally people have been assuming that you are looking for the relativistic answer to the question. Are we mistaken?

 

[stedwards]

They are both related to the potential energy in the field. The kinetic energy would be another term. Remember, this is the Lagrangian density, not the energy density. The energy density is just a term with the field squared.

Here is a discussion on this topic (among others) especially post 8 https://www.physicsforums.com/threads/lagrangian-density-of-Newtonian-gravity.736568/

OK, I see. Energy density is flatlandereeze for integrating the action 4-density over a spatial volume, rather than any other submanifold. V and T come from partitioning into spacelike and timelike parts.

For electromangetism, the partitioning involving charge is \rho \phi | \tilde{J}A.

But I am surprised to find the field might be evaluated for kinetic-like and potential-like energy parts. It's just more obscure.

Take, for instance, {E_x}^2. This is really \left(\frac{\partial \phi}{\partial x}-\frac{\partial A_x}{\partial t}\right)_{tx} {\epsilon_{yz}}^{xt} \left(\frac{\partial \phi}{\partial x}-\frac{\partial A_x}{\partial t} \right)_{tx}, give or take a negative sign and a factor of 2.

After cross multiplying, there doesn't seem to be any logical partitioning. I've been sloppy enough with the signs that I can't tell if the cross terms cancel, but this seems moot.

 

[sweet springs]

So the answer for high school students, who are best served by a purely classical treatment, is that mghmgh isn't "stored" anywhere - it's a property of the entire configuration of objects including both the Earth an dthe object whose height we're talking about.

Thanks. In this sense is Coulomb energy qQ/r not "stored" anywhere either and is also a property of the entire configuration?

E.M. satisfies principle of locality. Energy comes from the nearby field to each charge. I wonder gravity energy or coulomb energy as a property of entire cofiguration would satisfy the principle of locality.