How Can We Overcome the Challenges of Locating Gravitational Energy?

[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.

 

[PeterDonis]

Energy comes from the nearby field to each charge.

This energy that comes from the field to the charge, to make it accelerate, is not the same as the Coulomb energy qQ/r. The transfer of energy between the field and the charge is described by the covariant divergence of the stress-energy tensor, and is locally conserved. The Coulomb energy qQ/r is not. The same is true of gravity.

 

[sweet springs]

Thanks. Coulomb energy and field energy has the relation
\int E^2 dV=\frac{qQ}{r}
,forgetting about tedious self energy. Does Newtonian gravity has a similar relation? ,e.g.
\int E^2_{gravity} dV=-G\frac{mM}{r}
Best.

 

[PeterDonis]

Coulomb energy and field energy has the relation

$$
\int E^2 dV = \frac{qQ}{r}
$$
​

No, this is wrong. The Coulomb energy (or electrostatic potential energy) is the work required to bring a test charge q from infinity to radius r from the source charge Q. That work is given by the line integral (assuming purely radial motion)

$$
\int^r_{\infty} - q E dr = \int^r_{\infty} - q \frac{Q}{r^2} dr = \frac{qQ}{r}
$$

Note that this is not a local relation; it's an integral. So you can't use this kind of relation to tell anything about local energy conservation.

Does Newtonian gravity has a similar relation?

You can write a similar work equation to the above for Newtonian gravity, yes; instead of E you will have the "acceleration due to gravity" GM / r^2, and instead of q you will have the test mass m. Also the sign will be reversed because gravity is attractive while like charges repel.

 

[sweet springs]

No, this is wrong. The Coulomb energy (or electrostatic potential energy) is the work required to bring a test charge q from infinity to radius r from the source charge Q. That work is given by the line integral (assuming purely radial motion)
$$
\int^r_{\infty} - q E dr = \int^r_{\infty} - q \frac{Q}{r^2} dr = \frac{qQ}{r}
$$
Note that this is not a local relation; it's an integral. So you can't use this kind of relation to tell anything about local energy conservation.

Thanks. I agree with your explanation. But this is not the only way to express energy. I challege your "No, this is wrong." Please taka a look at e.m. textbook for example (8.30) of Feynman Lectures on Physics II
http://www.feynmanlectures.caltech.edu/II_08.html#Ch8-S5
Best.

 

[PeterDonis]

Please taka a look at e.m. textbook for example (8.30) of Feynman Lectures on Physics II

That's an expression for the energy stored in an electric field. It's not the same as Coulomb energy, although they are related.

 

[A.T.]

Not distance but energy could be localized in the space.

The space volume dxdydz contains Coulomb static energy E(x,y,z)^2 dxdydz　where E is electric field that two charges make there.

If that’s what you mean by "localize", then distance can be "localized" as well: The space length dr contains the distance 1 * dr.

 

[sweet springs]

Hi. In further analogy with e.m the gravitation energy of the two partcie U is expressed as volume integral
U=\frac{\epsilon_{gravity}}{2}\int \mathbf{E}^2_{gravity} dV=-G\frac{m_1m_2}{|\mathbf{r}_2-\mathbf{r}_1|}+(self\ energy\ of\ particle\ 1)+ (self\ energy\ of\ particle\ 2)
where at position r
\mathbf{E}_{gravity}=-G (m_1\frac{\mathbf{r}-\mathbf{r}_1}{|\mathbf{r}-\mathbf{r}_1|^3}+m_2\frac{\mathbf{r}-\mathbf{r}_2}{|\mathbf{r}-\mathbf{r}_2|^3}),
\epsilon_{gravity}=-\frac{1}{G} I am not sure whether some numerical coefficient such as pi might be needed.
Thus gravitation energy is distributed in the space with density of
\mathbf{E}_{gravity}^2=G^2 (m^2_1\frac{1}{|\mathbf{r}-\mathbf{r}_1|^4} + m^2_2\frac{1}{|\mathbf{r}-\mathbf{r}_2|^4}
+2m_1m_2\frac{(\mathbf{r}-\mathbf{r}_1)\cdot(\mathbf{r}-\mathbf{r}_2)}{|\mathbf{r}-\mathbf{r}_1|^3|\mathbf{r}-\mathbf{r}_2|^3} )

The first and second terms correspond to self energies. The last cross term corresponds to the interaction energy we are discussing.
Multiplyed by \epsilon <0, the cross term is plus in the sphere whose diameter ends the particles cccupy, minus outside.
It means that the gravitational energy is plus in the sphere between the particles but more minus at out side of the sphere. In sum the field gives kinetic energy to the particles. The particles come closer.

That could be the answer to the question where is mgh ? in Newtnian Mechanics and in analogy with e.m. we give to high school students.
Best

 

[PAllen]

After all the enlightening answers, maybe we need a silly answer. Mgh is unambiguously stored in Massachusetts. I've seen it there many times. Picture shows one of the storage buildings.

 

[DrGreg]

After all the enlightening answers, maybe we need a silly answer. Mgh is unambiguously stored in Massachusetts. I've seen it there many times.

As I live 3,260 miles from Massachusetts I had to use Google to work out what the joke was. But I found it very quickly.

 

[HallsofIvy]

As I live 3,260 miles from Massachusetts I had to use Google to work out what the joke was. But I found it very quickly.

That's better than me. I wanted to say that "mgh is stored in a number Swiss account".