Where mgh is stored?

1. Jun 13, 2015

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?

Last edited: Jun 13, 2015
2. Jun 13, 2015

HallsofIvy

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

3. Jun 13, 2015

sweet springs

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

4. Jun 13, 2015

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"?

5. Jun 13, 2015

Staff: Mentor

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.

6. Jun 13, 2015

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 enery of me. Best.

7. Jun 13, 2015

sweet springs

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

Best

8. Jun 13, 2015

Staff: Mentor

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.

9. Jun 13, 2015

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.

Last edited: Jun 13, 2015
10. Jun 13, 2015

pervect

Staff Emeritus
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.

11. Jun 13, 2015

Staff: Mentor

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.)

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?

12. Jun 13, 2015

Staff: Mentor

Before the sale, it must exist somewhere else, right?
No. Energy isn't a "thing" that needs to "enter" you to be used. It's just bookkeeping.

13. Jun 13, 2015

pervect

Staff Emeritus
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.

14. Jun 13, 2015

sweet springs

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

15. Jun 13, 2015

Staff: Mentor

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.

16. Jun 14, 2015

sweet springs

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

Best

17. Jun 14, 2015

Staff: Mentor

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.

18. Jun 14, 2015

A.T.

Where do the h and r belong?

19. Jun 14, 2015

sweet springs

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.

20. Jun 14, 2015

Staff: Mentor

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.

21. Jun 15, 2015

sweet springs

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]

Last edited by a moderator: Jun 15, 2015
22. Jun 15, 2015

harrylin

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).

Last edited: Jun 15, 2015
23. Jun 15, 2015

A.T.

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

24. Jun 15, 2015

Staff: Mentor

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.

25. Jun 15, 2015

sweet springs

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.

Last edited: Jun 15, 2015