How Can We Overcome the Challenges of Locating Gravitational Energy?

[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".

 

[pervect]

Another non-high school reference that I'm personally fond of that relates to the problem.

E. Noether's Discovery of the Deep Connection Between Symmetries and Conservation Laws

http://www.physics.ucla.edu/~cwp/articles/noether.asg/noether.html

Though the general theory of relativity was completed in 1915, there remained unresolved problems. In particular, the principle of local energy conservation was a vexing issue. In the general theory, energy is not conserved locally as it is in classical field theories - Newtonian gravity, electromagnetism, hydrodynamics, etc.. Energy conservation in the general theory has been perplexing many people for decades. In the early days, Hilbert wrote about this problem as 'the failure of the energy theorem '. In a correspondence with Klein [3], he asserted that this 'failure' is a characteristic feature of the general theory, and that instead of 'proper energy theorems' one had 'improper energy theorems' in such a theory. This conjecture was clarified, quantified and proved correct by Emmy Noether.

The terminology in the above quoe (in particular the usage of 'local energy conservation') is a bit odd, which I attribute to the difference in outlook between physicists and mathematicians. The point I'd like to stress is that the problem of energy conservation in GR have been known for long time, nearly as long as the theory has been around, going back to Hilberet. There is a lot of work that has been done on the problem, and it's rather illuminating, but it' not at high school level. I would state that while not all answers to this quetion rely on Noether's work, that it is one of the better attempts at an answer out there. I will also digress a bit, and state that the way the question is formulated is a bit unfortunate (though very high-schoolish!). It makes a lot of assumptions, ones which impede any sort of really serious discussion necessary to a full answer.

 

[Cecil Tomlinson]

mgh is "stored" in the elevated mass. The known fact that clocks run faster at higher elevations, and the principle of relativity, together imply that mgh is stored in the elevated mass. By the principle of relativity an observer at the elevated mass could not, by any means of local observation, detect that his clock is running faster and that his unit of time is smaller. Neither can he determine that his units of mass and energy are larger. He will see only the "rest mass". Mass and energy are equivalent. Mass is constituted by three kinds of energy: rest mass, kinetic mass, and gravitational mass. As a mass falls freely, gravitational mass is converted to kinetic mass and the total mass remains constant. The quantum is a product of an energy unit and a time unit, and it is universally constant. As the time unit decreases at higher elevations, the energy/mass units increase proportionally, but the increase can only be seen remotely, as the principle of relativity precludes local observation of either kinetic or gravitational mass.

 

[A.T.]

mgh is "stored" in the elevated mass.

Which mass is "elevated" in a system of two equal masses?

 

[Cecil Tomlinson]

Both are elevated fom their commn center of mass.

 

[Jonathan Scott]

Both are elevated fom their commn center of mass.

The idea that the potential energy is caused by time dilation of the mass works fairly well when one is considering a single mass in a static field.

However, if you apply the time dilation to each mass due to the potential caused by the other mass, you find that both source masses have lost the full potential energy, which means that their total energy appears to have decreased by twice the total potential energy. A self-consistent solution to this apparent paradox is that there is positive energy density in the field of g^2 / (8 \pi G) where g is the gravitational field. This is closely analogous to the energy density of an electromagnetic field. When this is integrated over all space, the result is equal and opposite to the (negative) potential energy, so the total energy of the system is correct. It also describes a conserved flow of Newtonian energy and momentum overall through space between the sources.

A specific case in which this scheme is worked out exactly is described in an MNRAS paper "Gravitational field energy density for spheres and black holes" by D Lynden-Bell and J Katz, published in 1985: http://adsabs.harvard.edu/full/1985MNRAS.213P..21L

Note that the Newtonian "energy" in this sense includes gravitational energy and is not exactly the same as the usual General Relativity view of "energy" as a gravitational source term, which has zero density in vacuum.

This model is just one possible approach to the very controversial subject of gravitational energy, but certainly seems to provide a practical and plausible answer to where the energy is stored in Newtonian terms.

 

[Cecil Tomlinson]

Indeed, you are correct. However, nothing you have said negates my simplified statement. For a very small mass elevated from a very large mass, mgh may be accurately discribed as being stored in the smaller mass.

 

[Jonathan Scott]

Indeed, you are correct. However, nothing you have said negates my simplified statement. For a very small mass elevated from a very large mass, mgh may be accurately discribed as being stored in the smaller mass.

I agree that if you can approximate the larger mass as being fixed, the potential energy due to the time dilation on the smaller body matches the mechanical potential energy for that body, and this is a practical simplification.

However, I would not say that being small helps the "accuracy" of the overall picture in that model. If you have two bodies, which could be a flea and the earth, then the potential energy of the Earth due to the time dilation cause by the flea is exactly the same as the potential energy of the flea due to the time dilation caused by the earth, but the difference in the energy of the field due to their interaction (that is, the difference between the total energy that their fields would have separately and the energy of the combined field of the interacting bodies) is equal to the same total but positive.