In the context of relativity the conservation of energy comes from the independence of the metric tensor on the time coordinate, i.e. as $g_{\alpha\beta,t} = 0 \quad \forall \quad \alpha, \beta$. The geodesic equation tells us that $t$ component of the energy-moment one-form (covariant vector) is a constant $p_t = E$, and this constant is usually called energy.

For particle of mass $m$ that is momentarily stationary ($\mathbf{p}=0$) at the top of a mountain its energy is calculated from the invariance of the rest mass, i.e.,

\begin{eqnarray}

g^{\alpha\beta} p_\alpha p_\beta & = & -m^2 \\

g^{tt}E^2 & = & -m^2 \\

-\left( 1-\frac{2M_{\earth}}{R_{\earth} + h}\right)^{-1}E^2 & = & -m^2 \\

E &= &m\sqrt{ 1-\frac{2M_{\earth}}{R_{\earth} + h}}

\end{eqnarray}

Note these equations are written in natural units, $G=c=1$.

From the above expression we can see that there is no decoupling of the rest mass energy $m$ and the classical potential energy $V = \frac{mM_{\earth}}{R_{\earth} + h}$. However your magic machine works it can not convert just the rest mass energy of the particle because that can not be decoupled from the ``potential energy''. Hence you will get more energy from your machine at the top of the mountain than the bottom.

I would say there won't be any difference, because you have to use the rest energy of your particles which indicates that they are not in motion. Absence of motion requires the netforce on the particles to be zero: F = 0 (regardless where the are located).

as the gravitational force is compensated by the electromagnetic forces which the particle "feels" when it lies somewhere on the ground, there will be no contribution of that to your resting mass and hence the mass defect will be the same.

Interesting question as ever Derek,

Although ... a whit misleading, because there is no proof that you can actually carry out its premise (convert 100% of a given mass object into E) and as such the question may be a nonstarter.

Second - mass is a catchall word and as such a bit of a misnomer: break down what gives mass into separate components - parts of mass are consequences of the Higgs field, parts are not (such as the rest mass of certain particles.)

Third - where do these separate constituents of overall mass reside? in the quarks, in the surrounding field ....Most of non-Higgs originated mass is circumscribable, but Higgs-generated mass is fuzzy

And finally ... what (parts of the) sundry mass constituents variously convert into energy and how?

Once you have answered all of this you'll have your answer. As you are aware, not everybody agrees on the tentative answers (Matt Strassler runs an interesting blog on kindred subjects ), and in fact, field experimentation of your premise would be interesting.

Regarding the components of mass, as I understand it is only the tiny invariant rest mass of fundamental particles that is generally attributed to the Higgs field - the bound mass of nucleons (for example) is attributed to the confinement of quark kinetic propagation energy by the strong force, mediated by the exchange of gluons among proximal quarks.

I have to guess that the mass-energy bound within a particle would not vary at different altitudes, but if it does, I think the mass-energy of the particle in the valley would be greater that that of the particle on the mountain peak...

Potential energy is always relative to a zero energy reference plane. So, there is no potential energy stored in any of the particles. The potential energy lies within the gravitational field.

Elementary particles are of two sorts: energy particles and massive particles. Massive particles have real mass, but it has different physical nature from inertial or gravitational mass of bodies. In fact in TR mass of these particles is called "electrodynamic mass". Now let us raise two questions:

1. Does electrodynamic mass coincide with inertial or gravitational mass? In TR the answer is no.

2. Do elementary particles, besides electrodynamic mass, have also an inertial or gravitational mass? In TR the answer would be no, but anyway also if elementary particles had an inertial or gravitational mass, it could have altogether negligible physical effects with respect to physical effects produced by electrodynamic mass. The two identical particles placed on the mountain top and down have anyway the same mass even if the weight of the two particles would be different. In that event and in order to answer the raised question, potential energy which resides above all in the distance from the pole besides in mass, has no revealing effect (on this I agree with Dimitrios) because electrodynamic mass and not inertial or gravitational mass participates to the event of transformation of mass to energy.

Dimitrios,

I think your statement may be valid within the context of some specific theory, but not in the context of E=mc^2, or nuclear energy.

In the context of relativity the conservation of energy comes from the independence of the metric tensor on the time coordinate, i.e. as $g_{\alpha\beta,t} = 0 \quad \forall \quad \alpha, \beta$. The geodesic equation tells us that $t$ component of the energy-moment one-form (covariant vector) is a constant $p_t = E$, and this constant is usually called energy.

For particle of mass $m$ that is momentarily stationary ($\mathbf{p}=0$) at the top of a mountain its energy is calculated from the invariance of the rest mass, i.e.,

\begin{eqnarray}

g^{\alpha\beta} p_\alpha p_\beta & = & -m^2 \\

g^{tt}E^2 & = & -m^2 \\

-\left( 1-\frac{2M_{\earth}}{R_{\earth} + h}\right)^{-1}E^2 & = & -m^2 \\

E &= &m\sqrt{ 1-\frac{2M_{\earth}}{R_{\earth} + h}}

\end{eqnarray}

Note these equations are written in natural units, $G=c=1$.

From the above expression we can see that there is no decoupling of the rest mass energy $m$ and the classical potential energy $V = \frac{mM_{\earth}}{R_{\earth} + h}$. However your magic machine works it can not convert just the rest mass energy of the particle because that can not be decoupled from the ``potential energy''. Hence you will get more energy from your machine at the top of the mountain than the bottom.

Nice answer, Sam. So are you saying the particle on top of the mountain actually will have more mass?

Now if you look at the last sentence of my question, it asks what will be the case when seen by a stationary observer versus a moving one. It seems to me the basic mgh notion changes with these observers.

So if it looks different to different observers, where does the actual energy reside? The "reside" part is the heart of the original question.

From one perspective, at least, it is the cars precariously balanced at the top of a roller coaster that contain the potential to accelerate. In the absence of cars atop the roller coaster, there is no potential for gravitational acceleration (except for collapse of the structure, of course). I think this is true for the roller coaster riders as well as observers on the ground...

A common sense answer- The mass at h above has no memory - no means to know it has been raised until it falls back. Therefore the mass is identical to produce identical energies as compared to one at the bottom. The energy mgh could be recovered if there is mass available. Otherwise they continue having equal energies at different levels. A simplistic answer - I know more sophisticated answers are available.

If information was necessary, surely the roller coaster riders are aware of the potential for acceleration! I think the cars and riders at the top together have the potential for being accelerated by gravity - as opposed to those at the bottom, even if they forgot about being raised.

To add to an earlier remark, the problem or paradox results from non invariance of mass. Rest mass has only one measure as dictated by inertia, its energy content base on mc2 is going to be the same irrespective of position. Where is that potential energy like mgh accounted for remains hidden in the energy equivalence at different heights or positions. In a way if E=mc2 is valid, is energy dependent upon position? Whatever we know, it is not. I think the question is very deep. It will be too hasty to say something is invalid.

"Where does mgh potential energy reside exactly?" The answer is nowhere. I think the confusion is caused by a misunderstanding of the idea of energy. Energy is the capacity to perform work and in doing so it is the ability to effect change to a physical system. For example, A moving ball can hit another ball, causing it to move.

To illustrate this better let us perform the following experiment:

Hold a ball in mid-air and drop it, it will lose height, but gain speed. If I drop it over and over, I notice that the higher I drop it from, the faster it gets. Is there some way I can predict the relationship between height (h) before the drop and velocity (v) after the drop? If I repeatedly run this experiment, I'll find that these two quantities always obey the following equation for every ball, and every height:

mass x gravity x height = (1/2) x mass x velocity^2.

This is the classical equality between gravitational potential energy (LHS) of the ball and the kinetics energy (RHS). This equality is just an abstraction that tells me how much velocity I can get out of a certain height above the Earth, or vice versa. This is not something tangible that exist in the physical world.

I hope you can see now why the gravitational potential energy of an object is not located anywhere. It is just a number we assign to it to tell us how fast can an object move when dropped.

Going back to the experiment suggested by the main question of the thread I conclude that the particle at the top of the mountain has the same energy as the one at the bottom of the mountain with only one exception. The particle at the top has the 'potential' to move at velocity v=(2gh)^1/2 if dropped from the top of the mountain (assuming no air resistance and 100% energy conversion). Hence you will get the same energy from your machine at the top of the mountain as the one at the bottom.

Issam Sinjab,

Good point, in strict accordance with the question, but wouldn't "an instrument that can magically convert a particle of mass, m, without moving it, totally into energy via E=mc^2" actually only convert the particle's invariant rest mass, not any "mgh potential energy"?

I think that the proposed experiment simply does not test the problem statement, thus our confusion. Moreover, I can't see how the perspective of any non-relativistic observer can have any effect on gravitational potential energy.

As I understand, as it relates to mgh potential energy, it's considered that some energy at some point in the past had to be expended for the particle on the mountain top to be elevated in relation to to the particle at the bottom - even if it was a portion of the energy that caused the mountain to rise. IMO, that is the source of the elevated particle's increased gravitational potential energy.

This is, of course, clearer in a closed system, such as a roller coaster...

James

The question of this thread is profound, deep and a very interesting one. I have attempted to solve it using simple classical Newtonian physics. I may be wrong, but as V.K. Jindal said above " It will be too hasty to say something is invalid."

As to your question I totally agree such a machine can only convert the particle's invariant rest mass because simply that is the only mass there is. A book on the ground has the same mass as a the same book on the table.

I am trying to work out an answer that includes relativistic considerations.

Dear Arno

Thank you for the advice. You may be right but still I do wish to make an attempt from what we currently know using the simplest of considerations and as little mathematics as possible. I do genuinely want to know if there is a different answer to the main question "where does the gravitational potential energy reside exactly"

Issam,

Based on the definition and use of mgh to account for and evaluate the gravitational potential energy of the water stored behind a dam, for example, IMO the fundamental answer is that the proposed experiment does not evaluate gravitational potential energy.

If the question is what prevents the the kinetic realization of gravitational potential, IMO the answer is a straightforward matter of mechanical opposition.

A more fundamental question is: what is the physical source of gravitational energy? I think there are critical issues that are unresolved. Especially, does gravitational energy physically reside in potential mass-energy or some kinetic energy contained within and represented by the geometric 'curvature' of dimensional spacetime?

Personally, I suspect that gravitation is not strictly a force of matter or a particle interaction, but rather an interaction between accreted potential mass-energy and the spacetime from which it was (energetically) extracted...

But I'm neither a physicist nor a mathematician. Best wishes!

Derek,

Is this a legitimate question to ask? I.e. does this energy have to "reside" somewhere?

Good question, Lev.

For the gravitational (m gh) energy I cannot tell. But I can cive an example, where the same energy can be viewed as being stored either in a finite volume or in all of space. And the two expressions can be converted to each other mathematically. The example is from the field of magnetism. Consider a (finite) volume wich is populated by magnetic dipoles, which have been "brought together" from a state where they all were at infinite mutual distance. The total energy change of bringing the dipoles together can be expressed as the integral over the stray field energy over the entire space. Likewise, it can also be expressed as an integral of Magnetisation times local field (stray field of all other dipoles at the location of a specific one).

The stray field can be, in general, nonzero everywhere. The energy is therefor viewed as being stored in the entire space in the first computation.

Magnetization is zero, though, where there are no dipoles. In the second computation, the energy is therefore residing in the volume containing the dipoles.

I guess that the ambiguity resides in our (various) conceptions of "field" and the very fact that it is a conception of ours, not necessarily a given.

The magnetic example could probably be translated to the gravitational one by pondering about constituents of mass at infinite separation initially and then assembled such that they form a planet with a table and a stone on that table, at altitude h above ground. Do the whole thing a second time, now resulting in the stone underneath the table. Where does the energy reside?

If we look at practical scales, the energy mgh (h can not exceed a few thousand meters or - use exact gravitational energy expression) is insignificantly small in comparison to mc^2. There needs to be an ingenious experiment to probe the difference. But the main question still remains "rest mass does not know the height" therefore mc^2 energy has to be identical. Now the question is energy also a function of position? If yes, may be extra energy mgh is there on top.

V K Jindal,

Sorry to be so uninformed, but I do not follow the reasoning re. "rest mass does not know the height". In comparison, does the relativistic mass of ball thrown from the mountaintop 'know' its height or velocity compared to a ball at rest in the valley that does not? Would the ball's mass 'knowing' anything make any difference to an 'instrument that converts the ball's mass to energy? Can you help by better explaining what seems to be a common euphemism? Thank you in advance...

Kai Fauth,

The magnetic field may be a very useful analog for evaluation, but I think that somehow the concept of potential energy would have to be applied. In the gravitational case, the field's effect on an object of mass is mechanically obstructed: removing that obstruction would result in the gravitational field's energy being imparted to the object - thus in the obstructed state the gravitational energy is constrained, potential energy that would otherwise be applied to the object.

In the magnetic field example, effects are nominally imparted to charged particles. Would placing the test particles in a shielded enclosure similarly produce an analogous energy potential?

James, the rest mass (inertial res mass) does not depend on height or position. Therefore energy from this rest mass - which only depends upon rest mass is independent of its position..

Arno,

I'm not sure what you mean.

My point was related to the possibility that our conception of energy as a *numeric* invariant may not be as complete as we think.

The two energies are seperate, and one should add kinetic energy to the total energy.

Mgh is normally only aplied for small h, otherwise your calculation moves away from the truth too far, because g is dependent on h, i.e. the further you move away from a massive object, the smaller g becomes.

For the calculation of the total energy of the object, you have to add all the energies together i.e. potential (mgh), kinetic (difficult to quantify, because everything in the universe is in motion) and nuclear.

I'm not sure what happens to the existing kinetic and potential energy when mass is converted to energy. I agree with VK Jindal that the magnitude of difference between kinetic- and potential energy compared to nuclear is enormous.

For most calculations it's probably safe to consider the kinetic and potential energies and nuclear energies seperately. If one do have to address the issue, the norm is to state that your assumtion excludes certain aspects because of the enormous difference in magnitudes.

Arno,

For discrete objects, can't rest mass, along with inertial, gravitational and relativistic mass generally be observationally and experimentally identified as being coincident with the material object?

I think gravitational potential energy is a more complex question. However, if one could estimate mass of water stored behind a dam, for example, one could estimate the mgh energy that can be made available to drive turbines attached to electrical generators, correct? In such a case, isn't the potential energy quite real, and isn't it coincident with the stored water? Again, if there is no water stored behind the dam there is no potential for generating electricity...

Also, as I understand, nuclear mass is mostly the product of quarks' kinetic energy being bound by the gluons mediating the strong interaction. Unlike the invariant rest mass of fundamental particles, this bound potential mass-energy (the invariant rest mass of proton and neutron compound particles) is to some extent available through nuclear processes such as fission, fusion and nuclear decay.

Of course, it's not generally available for home use...

Arno,

Perhaps I should better ask, since I'm not understanding, how is

"... the question of "where energy resides"... different from experiment/observation-based calculations?"

Any particle at rest will have a mass equivalent to its total energy including any potential energy. In other words, the particle at the top will have more mass due to its potential energy. If you fully transform mass into energy the top particle will have mgh more. For a moving observer the mass will be greater according to the Lorentz transformation, in other words the kinetic energy is adding to its total mass.

Arno,

Thanks. As a layperson, I may not interpret the same as physicists, but I understood the term "where energy resides" to indicate some entity with which it is physically associated with, or coincident with.

I think the clearest example is that fundamental particles inherently include or contain an invariant rest mass (however, this view disregards the role of any Higgs field that must continuously dynamically impart rest mass to fundamental particles, although IMO a local manifestation of the Higgs mechanism may initially impart rest mass to newly emitted particles). In general usage, some specific invariant rest mass (including zero) is directly associated with each fundamental particle (with the possible exception of oscillating neutrinos).

While a black hole may be the origin of magnetic energy that accelerates a charged particle to relativistic velocities as it is expelled from its polar jets, as I understand each accelerated particle's relativistic mass includes the energy of its imparted momentum, which will be expressed in its interactions with other matter.

As I understand, the gravitational mass (weight) of an elevated object is fractionally less than the that of an identical object located at a lower elevation. However, no such mass effect is detectable as a result of the elevated object's greater gravitational potential energy - that would be expressed if/when the elevated object falls to the lower elevation...

Sorry if I make things overcomplicated - I try to be precise...