Negative potential energy and negative mass

[Vicara]

So, for example, $m$ is the mass of a book and $mgh$ is the potential energy of a system consisting of the book and planet Earth, with $h$ being the height above some arbitrarily chosen height of zero. If you lift the book from the floor to a shelf that's a height $h$ above the floor you've increased the potential energy of the system by $mgh$. This is true whether you choose your zero of height to be the floor or the shelf or the ceiling.

So potential energy is negative only because you are asuming that h = ∞ - r so the distance is negative?

 

[Vicara]

As Nugatory already explained, Ep it is not the potential energy of m but of the system consisting of m and M. The total energy of this system is

E = \left( {m + M} \right) \cdot c^2 - G \cdot \frac{{m \cdot M}}{r}

As it decreases when the objects get closer, your idea sounds reasonable so far. However, to get a negative total energy the objects would need to get so close that the Newtonian gravitational potential is no longer an acceptable approximation. It needs to be replaced by a relativistic potential energy V (which doesn’t need to be limited to gravity). If I understand correctly then you are asking whether

E = \left( {m + M} \right) \cdot c^2 + V < 0

is possible in general relativity or not. My intuitive answer is No. I guess that the system will collapse to a black hole (which cannot release energy anymore, exept by Hawking radiatiation) before its total energy drops to zero or even below. But I have no idea how to prove it.

My original question was referred to a Newtonian system, except the energy-mass equivalence (simply because I haven't studied GR hahaha) but that's a very interesting point of view

 

[Herman Trivilino]

So potential energy is negative only because you are asuming that h = ∞ - r so the distance is negative?

Your equation makes no sense. Potential energy, like anything else, is negative only because it's less than zero. The thing that's confusing you is the situation where the choice of zero potential energy was made in such a way that the potential energy is always less than zero. Make a different choice for where it's zero and then it doesn't have to be always negative.

 

[PAllen]

As Nugatory already explained, Ep it is not the potential energy of m but of the system consisting of m and M. The total energy of this system is

E = \left( {m + M} \right) \cdot c^2 - G \cdot \frac{{m \cdot M}}{r}

As it decreases when the objects get closer, your idea sounds reasonable so far. However, to get a negative total energy the objects would need to get so close that the Newtonian gravitational potential is no longer an acceptable approximation. It needs to be replaced by a relativistic potential energy V (which doesn’t need to be limited to gravity). If I understand correctly then you are asking whether

E = \left( {m + M} \right) \cdot c^2 + V < 0

is possible in general relativity or not. My intuitive answer is No. I guess that the system will collapse to a black hole (which cannot release energy anymore, exept by Hawking radiatiation) before its total energy drops to zero or even below. But I have no idea how to prove it.

I think a way to make this (nearly) rigorous in GR and get away from any issues of the conventionality of potential energy (as well as the fact that potential energy is not generally definable in GR, certainly not in this case for an initial condition that is dynamically unstable thus cannot be consistently made part of a stationary spacetime) is conceptually as follows. Each step is non-trivial, but within the state of the art:

1) Define a series of asymptotically flat spacetimes evolved from series of initial cauchy surfaces as follows: each cauchy 'melds' 2 SC geometries of chosen mass parameters, such that they are some initially stationery distance apart per some plausible harmonic coordinate system (to use ADM evolution). This is highly non-trivial, but is what is done e.g. when the final inspiral stages of BH are simulated from an arbitrary starting point. The series uses smaller and smaller distances. Note that there is no really consistent past continuation leading to such a state, but the BH inspiral experts work around this (it is easy to reason that any prior state includes gravitational radiation that we are not including in our initial cauchy surface).

2) Compute the ADM mass of each such (incomplete - unspecified before the initial cauchy surface) manifold. For given chosen mass parameters, this should decrease as the initial distance decreases. This ADM mass is physically the total gravitational mass of the system measured at spatial infinity. It will include the effects, per full GR, of the mass discount from binding energy.

I am not aware of anyone actually doing such computations, but agree the likely outcome is that you would be dealing with infinitesimally separated apparent horizons without total ADM mass reaching zero.

Note, that extant BH inspiral calculations that I know of, are not relevant because they simply follow one history from some initial state, and the GW emitted includes only some of the initial potential energy. Part of it becomes kinetic energy and and angualar momentum 'captured' by the final BH.

This would be pretty close to a proof that you can never get negative total energy (when it is definable) by virtue of objects being very close together, in the sense that a mixture of Newtonian potential and SR might seem to imply.

 

[PeterDonis]

If I understand correctly then you are asking whether

$$
E = \left( {m + M} \right) \cdot c^2 + V < 0
$$

is possible in general relativity or not. My intuitive answer is No

A better answer would be that the quantity $E$ is not always well-defined; that is, we can't always define a "total energy" for a system in GR. In particular, whenever there is more than one gravitating mass present, strictly speaking, the spacetime is not stationary, and in a non-stationary spacetime there is no way to define a potential $V$.

In practice, for cases where gravity is weak everywhere and everything is moving slowly compared to the speed of light, the expression you give for $E$ is a reasonable approximation if we define $V = - GmM / r$. But note that for that case, we are no longer free to redefine the "zero point" of $V$, because $E$ is now the total energy of the system as measured from the outside, and that measurement will have a definite result; we can't just arbitrarily adjust it by changing the zero point of $V$. We have to use the zero point of $V$ that correctly predicts what the outside measurement will be, and that zero point is $V \rightarrow 0$ as $r \rightarrow \infty$.

Also, since this approximation is restricted to weak fields and slow motion, it won't work for, e.g., an object falling into a black hole, or two black holes coming together. It is still possible to define a total energy $E$ as measured from the outside for such systems, but it requires more care.

 

[DrStupid]

whenever there is more than one gravitating mass present, strictly speaking, the spacetime is not stationary

I forgot to mention that I was speakting about the stationary case where the objects are fixed in their positions. Otherwise the equation for the total energy wouldn't be complete and the problem would be much ore complex.

In practice, for cases where gravity is weak everywhere and everything is moving slowly compared to the speed of light, the expression you give for $E$ is a reasonable approximation if we define $V = - GmM / r$.

The situation we are talking about (E<0) would require a distance of

r_0 &lt; \frac{{r_s }}{8}

between the objects, where r_s is the Schwarzschild radius of the system. That means that $V = - GmM / r$ is definitely not a reasonable approximation.

 

[PAllen]

I forgot to mention that I was speakting about the stationary case where the objects are fixed in their positions. Otherwise the equation for the total energy wouldn't be complete and the problem would be much ore complex.
The situation we are talking about (E<0) would require a distance of

r_0 &lt; \frac{{r_s }}{8}

between the objects, where r_s is the Schwarzschild radius of the system. That means that $V = - GmM / r$ is definitely not a reasonable approximation.

You can have the objects or BH initially stationary but you could not keep them so. Thus the spacetime could not be stationary while satisfying the EFE - which imply the equations of motion. Only for a stationary spacetime is gravitational potential function possible in GR. Also only with BH could you have prayer of getting close enough to be interesting. That is why I felt what I suggested is necessary to approach this in GR

 

[SiennaTheGr8]

In practice, for cases where gravity is weak everywhere and everything is moving slowly compared to the speed of light, the expression you give for $E$ is a reasonable approximation if we define $V = - GmM / r$. But note that for that case, we are no longer free to redefine the "zero point" of $V$, because $E$ is now the total energy of the system as measured from the outside, and that measurement will have a definite result; we can't just arbitrarily adjust it by changing the zero point of $V$. We have to use the zero point of $V$ that correctly predicts what the outside measurement will be, and that zero point is $V \rightarrow 0$ as $r \rightarrow \infty$.

If I'm correctly reading between the lines of OP's post, then I think this is probably the most helpful response so far. I asked a similar question here a few months ago precisely because of the inconsistency between a "fixed" amount of rest energy on the one hand (a quantity that includes potential energies internal to the system!) and the "arbitrary" zero-point for potential energy on the other. Your replies in that thread cleared things up for me.

Maybe this will help OP, too:

$E = E_0 + E_k$

where $E$ is a system's total energy, $E_0$ is its rest energy (an "invariant" quantity that all observers agree on), and $E_k$ is its kinetic energy (energy related to the system's aggregate motion).

Rest energy is the total energy a system has as measured in its rest frame (the frame in which it has no momentum [or, equivalently, no kinetic energy]). It is simply the sum of all rest energy, kinetic energy, and potential energy that's "internal" to the system.

Einstein showed that rest energy and mass are the same thing, just measured in different units: $E_0 = mc^2$.

 

[PeterDonis]

I forgot to mention that I was speaking about the stationary case where the objects are fixed in their positions.

As @PAllen pointed out, the problem with this is that it's not a solution of the equations of motion. If you have gravitating bodies, they will fall towards each other unless something holds them in place. If something holds them in place, it must be exerting a force and must have energy, and that energy needs to be included.

 

[Vicara]

Einstein showed that rest energy and mass are the same thing, just measured in different units: $E_0 = mc^2$.

I didn't know dat. So, if two objects get closer, their mass is modified?

 

[SiennaTheGr8]

I didn't know dat. So, if two objects get closer, their mass is modified?

Yes, if by "their mass" you mean "the mass of the two-object system." The mass of each individual object remains fixed.

Actually, there's an additional subtlety: if by "get closer" you mean that the objects accelerate toward each other because of an attractive force (e.g., electrostatic [opposite charges] or gravitational*), then the mass of the whole system (including the field associated with the force) doesn't change, since the kinetic energy of the objects increases so as to counterbalance the decrease in the system's potential energy.

But if you're asking about comparing two different configurations in which the objects start at rest relative to one another (say), then the configuration with the objects closer together will have less mass, assuming that the force in question is indeed attractive. If the force is repulsive (e.g., the objects have the same charge), then the configuration with the objects closer together will have more mass. As @PeterDonis said, for these purposes we must define the zero-point for potential energy $V \rightarrow 0$ as $r \rightarrow \infty$. For an attractive force, $V < 0$. For a repulsive force, $V > 0$.

*technically I shouldn't include gravity here, but I think it's okay as long as the gravitational field isn't so strong that we need general relativity to model it. Someone correct me if I'm wrong.

 

[Vicara]

Actually, there's an additional subtlety: if by "get closer" you mean that the objects accelerate toward each other because of an attractive force (e.g., electrostatic [opposite charges] or gravitational*), then the mass of the whole system (including the field associated with the force) doesn't change, since the kinetic energy of the objects increases so as to counterbalance the decrease in the system's potential energy.

I was talking about a system without kinetic energy
And what do you mean when you say that the entire mass of the system decreases but the individual masses don't? If somebody inside the system measures the mass of une object individually in diferent distances, there's no difference?

 

[SiennaTheGr8]

I was talking about a system without kinetic energy
And what do you mean when you say that the entire mass of the system decreases but the individual masses don't? If somebody inside the system measures the mass of une object individually in diferent distances, there's no difference?

Mass isn't additive. The mass of a system is not the sum of the masses of the system's constituents.

Mass is nothing but rest energy (expressed in different units). And rest energy is the total energy of a system as measured in the system's rest frame (the frame in which the system has no aggregate momentum). Rest energy includes all energy "internal" to the system: the rest energies (masses) of the system's constituents, the kinetic energy associated with their motion, and the potential energy associated with their relative positions. In everyday situations, the rest energies (masses) of the system's constituents are by far the greatest contributors to the system's rest energy (mass), so much so that the kinetic and potential contributions are often completely negligible or even undetectable. So mass is (approximately) additive in the classical limit.

But there are certainly situations in which the kinetic and potential contributions to a system's rest energy are not negligible. This mainly happens at atomic and nuclear scales.

 

[Herman Trivilino]

I was talking about a system without kinetic energy
And what do you mean when you say that the entire mass of the system decreases but the individual masses don't? If somebody inside the system measures the mass of une object individually in diferent distances, there's no difference?

Look at Post #30 again. The mass of the system you introduced is given by $$m + M - G \frac{{mM}}{rc^2}.$$

The energy that binds the two constituents together lowers the mass of the system. Thus mass is not a measure of the quantity of matter in a system.

 

[DrStupid]

You can have the objects or BH initially stationary but you could not keep them so.

Why not?

 

[DrStupid]

If something holds them in place, it must be exerting a force and must have energy, and that energy needs to be included.

That's why I added the comment, that V doesn't need to be limited to gravity. It also includes the potential energy resulting from any additional interaction involved.

 

[PAllen]

Why not?

If you evolve the initial state using the EFE, the objects would move closer together. If you add extra elements to keep them apart, you have a different problem scenario than the intended one. In any case, I don't see any down side to treating as initially stationary rather than stationary. ADM mass is conserved from initial conditions. You keep the whole problem as close as possible to Newtonian case

 

[DrStupid]

If you evolve the initial state using the EFE, the objects would move closer together.

Even in a co-rotating frame of reference?

If you add extra elements to keep them apart, you have a different problem scenario than the intended one.

Are additional interactions excluded in the intended scenario? I didn't read the original question that way.

 

[PAllen]

Even in a co-rotating frame of reference?
Are additional interactions excluded in the intended scenario? I didn't read the original question that way.

I don't see any relevance to a corotating frame - there wouldn't be orbital motion in the idealized set up. The BH would move directly toward each other.

If you add e.g. a repulsive force field to keep the bodies static, that would add a positive potential function, cancelling the very effect the OP was trying to get at.

 

[DrStupid]

If you add e.g. a repulsive force field to keep the bodies static, that would add a positive potential function, cancelling the very effect the OP was trying to get at.

Can you prove that?

 

[PAllen]

Can you prove that?

See post #41

 

[DrStupid]

See post #41

I don't see the prove in post #41. Could you be more specific please?

 

[PAllen]

I don't see the prove in post #41. Could you be more specific please?

What is it you want proved? That potential has to be positive for a repulsive force? I thought 41 did that perfectly well. Note, I did not state the potentials would exactly cancel, only that they would have cancellation I.e. work in opposite direction so as to make the problem not just different but much harder to possibly realize the effect the OP was after.

Can you answer why on Earth you want to introduce this at all, since it only obscures the issue?

 

[jbriggs444]

What is it you want proved? That potential has to be positive for a repulsive force?

The potential does not have to be positive for a force field that happens to be repulsive near radius r.

 

[DrStupid]

What is it you want proved?

This:

If you add e.g. a repulsive force field to keep the bodies static, that would add a positive potential function, cancelling the very effect the OP was trying to get at.

The fact that the potential of the repulsive force is always positive is not sufficient to prove this claim. As a counter example let's assume the total potential

V = \frac{k}{6} \cdot \left( {\frac{{18}}{{r^2 }} - \frac{{22}}{{r^3 }} + \frac{9}{{r^4 }}} \right) - \frac{k}{r}

where the first term is the potential of the repulsive interaction (which is always positive) and the second the classical gravitational potential (which is always negative). The total potential has two local minima. When the system switches from the initial stable state r=3 to the final stable state r=1 the potential and therefore the total energy will be reduced. To my understanding the OP asked if something like this can result in a negative total energy.

Can you answer why on Earth you want to introduce this at all, since it only obscures the issue?

PeterDonis already answered that question in #35:

[...] whenever there is more than one gravitating mass present, strictly speaking, the spacetime is not stationary, and in a non-stationary spacetime there is no way to define a potential $V$.

The repulsive interaction is requiered to keep the system static.

 

[PAllen]

This:
The fact that the potential of the repulsive force is always positive is not sufficient to prove this claim. As a counter example let's assume the total potential

V = \frac{k}{6} \cdot \left( {\frac{{18}}{{r^2 }} - \frac{{22}}{{r^3 }} + \frac{9}{{r^4 }}} \right) - \frac{k}{r}

where the first term is the potential of the repulsive interaction (which is always positive) and the second the classical gravitational potential (which is always negative). The total potential has two local minima. When the system switches from the initial stable state r=3 to the final stable state r=1 the potential and therefore the total energy will be reduced. To my understanding the OP asked if something like this can result in a negative total energy.
PeterDonis already answered that question in #35:
The repulsive interaction is requiered to keep the system static.

I already answered the points before the last previously. Perhaps I should have said obscure or reduce rather than cancel, because I never meant or said exact cancellation.

[ deleted misunderstanding]

Peter's point is if you want to introduce a GR valid potential function, you need this. But the closer you posit the bodies, the stronger the opposing potential you need, making it impossible to say you've proved anything about impossibility of total energy becoming negative. I argued that all of this is pointless because ADM mass includes the effects of system gravitational potential energy, globally. Then you only need initially stationary. What you lose is a localizable potential function. But what you gain is much more important - not having to introduce major extraneous elements that work against the effect you are trying to maximize.

 

[Vicara]

When I asked the question I was thinking about a system of two static bodies without kinetic energy (and I know that can't be possible or estable without any oter force or structure to keep them apart, but it was the simplest way to explain how "negative mass could exist")

 

[PAllen]

When I asked the question I was thinking about a system of two static bodies without kinetic energy (and I know that can't be possible or estable without any oter force or structure to keep them apart, but it was the simplest way to explain how "negative mass could exist")

Right, and what I answered about was two bodies initially stationary, considering this situation from ever closer starting points. This gets at your goal in pure form in a way that is possible, without having to introduce extra elements that add to total energy.

 

[DrStupid]

But the closer you posit the bodies, the stronger the opposing potential you need, making it impossible to say you've proved anything about impossibility of total energy becoming negative.

As I generalized the question to any potential (not only resulting from gravity), it doesn't matter how strong the opposing potential is. The original scenario with the gravitational potential energy is included in this question as a special case.

Then you only need initially stationary.

Then you need to explain how to reach such a state. Fixing the system with an external field and then releasing it doesn't work, because this field cannot be switched off instantaneous.

 

[Herman Trivilino]

Then you need to explain how to reach such a state. Fixing the system with an external field and then releasing it doesn't work, because this field cannot be switched off instantaneous.

Then switch it off gradually while they move apart. Gravity will then slow them down until they reach a maximum separation distance before they start to approach each other.