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Thinking in classical mechanics terms but with the knowledge that $E=mc^2$ let's make the below thought:

Suppose you have a shell of a sphere formed by a mass uniformly distributed over the surface of the sphere. When the radius is large the system has a mass at rest $m_0$. But when we start to lower the radius the system acquires a negative potential energy, thus the mass of the whole system becomes smaller and smaller as the radius diminishes.

If we suppose that the negative mass corresponding to the negative potential energy is inside the sphere than the mass that attracts itself diminishes as the radius gets smaller and smaller, thus allowing for the singularity to be smoothed when $r=0$ and then I may suppose (before doing calculations) that we are never able to arrive to a point where the mass is 0 before the radius becomes 0.

If instead we suppose that the negative mass resides in the space outside the sphere, than that negative mass does not interfere with the gravitational force on the sphere and thus the matter of the surface of the sphere is subjected to the known gravitational force thus there is a point where at a certain radius the energy of the system becomes 0 and thus to my understanding there exists nothing so the matter is annihilated and has been converted to energy during all the contraction trajectory where we have to extract energy to keep the system stationary and stable.

In case we don't keep the system stable, than the system should oscillate around the 0 position and thus should be dampened by the emission of general relativity gravitational waves (I suppose) arriving to be annihilated any way at the end.

So in both cases we have a decay of matter into energy. So the questions are:

a) the potential energy is inside or outside the sphere and what is the outcome of the experiment?

b) what else can we derive?

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  • $\begingroup$ you might want to start with a non infinite radius, otherwise it will need an infinite amount of time to collapse, and the mass will also be infinte. In any case, the mass does not "dissapear "at the singularity, so no way to make that smooth. $\endgroup$
    – Wolphram jonny
    Commented May 8 at 23:43
  • $\begingroup$ Also, what is negative mass, are you talking about some kind of exotic matter? $\endgroup$
    – Wolphram jonny
    Commented May 8 at 23:44
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    $\begingroup$ As the shell collapses, it loses potential energy but gains kinetic. At no point is energy (and hence rest mass) lost, unless you are radiating it somehow (e.g. gravitational waves, but a spherical shell would not do that). $\endgroup$
    – Sten
    Commented May 8 at 23:49
  • $\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$
    – Community Bot
    Commented May 8 at 23:53
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    $\begingroup$ If you want to think about how energy contributes to gravitating mass, in accordance with something like $E=mc^2$, you need to take a fully relativistic view. Otherwise your physics are not self-consistent and will not give a unique answer. $\endgroup$
    – Sten
    Commented May 9 at 21:11

3 Answers 3

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But when we start to lower the radius the system acquires a negative potential energy, thus the mass of the whole system becomes smaller

This is not correct. There is an increase of kinetic energy. This is positive and equal in magnitude to the loss of potential energy. The total energy is unchanged, as is the system mass.

There is no negative mass here, so there is no sense in asking where it resides.

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  • $\begingroup$ Please see the clarification regarding the fact that we will suppose that we will extract mechanical energy during the process, and we will not allow the sphere surface gain any significant velocity. $\endgroup$
    – George Kourtis
    Commented May 9 at 19:29
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    $\begingroup$ I have rolled back the changes. Please do not make changes to a question in a way that invalidates answers already recieved for the original question. Obviously if you extract the energy then your missing mass goes where the extracted energy goes. There is no decay or oscillation or any of this other stuff you mention $\endgroup$
    – Dale
    Commented May 9 at 20:59
  • $\begingroup$ The question did't clearly stated that it was supposed that energy was extracted in the process, neither that the system would had been left free to shrink and convert potential energy to kinetic one. So the question clear only to me but unspecified otherwise. Any way If the system is shrinked slowly extracting energy the mass of it reaches 0 at some specific radius, and thus the system becomes nothing. Instead if the system is left free to shrink it will oscillate around the 0 mass radius if there is no dampening converting back and forth mass at rest to kinetic energy mass. $\endgroup$
    – George Kourtis
    Commented May 9 at 21:20
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    $\begingroup$ “So the question clear only to me but unspecified otherwise”. Yes, that is the issue. You need to write clearly the first time. When you do not do so and people answer the unclear question, then it is too late and the question should remain as written or clarified in a way that validates the received answers. The oscillation around zero mass concept makes no sense either way. If you are extracting energy then when it is all gone there is nothing to oscillate. If you are not extracting energy then there is no mass change anyway $\endgroup$
    – Dale
    Commented May 9 at 21:27
  • $\begingroup$ "If you are not extracting energy then there is no mass change" is obviously correct. Starting from a fixed radius we should make the clarification that if the system allowed to shrink freely, starts to shrink, converts mass at rest to kinetic energy ( the sum is constant as said) arriving to a certain point where there is only kinetic energy, $\endgroup$
    – George Kourtis
    Commented May 9 at 21:42
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As a commenter said, no energy would be lost if a shell started collapsing under its own gravity, unless energy were radiated away. Potential energy is exchanged for kinetic. Then once the matter collides and coalesces into a planet or star, this kinetic energy is converted to thermal or chemical (which is still mass energy). Or, if it is large enough to become a black hole, the all becomes simply mass energy again trapped within a horizon.

Negative mass does not exist, at least in known physics. Negative energy also does not exist. When we say energy is negative, it is a shorthand for saying the difference between two energy states is negative:

$$\Delta E = E_2-E_1$$

Although by convention in Newtonian gravity we often call infinite separation "zero energy" and assign a "negative" potential energy value to any other location in the gravity field, this is really saying those locations are at negative potential relative to the zero point.

To conceptualize what happens when two or more massive objects move closer together under gravity, it is more accurate to say their later potential energy state is lower than the earlier state, rather than saying it acquired negative energy.

One more thought: the name potential energy is a clue that it refers to the energy an object could have, not the energy it actually has. Potential energy does not contribute to mass via $E=mc^2$ because it's not energy the object has.

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    $\begingroup$ Last paragraph is not exactly right: potential energy does contribute (negatively, in gravitational contexts) to the invariant mass $\endgroup$
    – Sten
    Commented May 9 at 4:32
  • $\begingroup$ Can you give an example of gravitational potential energy contributing to mass? I wonder if we are accounting for potential energy and mass in the same way $\endgroup$
    – RC_23
    Commented May 9 at 6:38
  • $\begingroup$ See this answer for a relativist's characterization of the mass associated with potential energy. Note that potential energy must contribute to the gravitating mass purely on conceptual grounds. For the collapsing spherical shell, we know the kinetic energy contributes to gravity, so the potential energy must as well, to avoid making the system "magically" grow in mass over time as potential converts into kinetic energy. (You can also think of nongravitational examples, such as how nuclei are less massive than their constituent nucleons.) $\endgroup$
    – Sten
    Commented May 9 at 19:06
  • $\begingroup$ The term "negative mass" or "negative energy" equivalent for me, does connect with the fact that when the sphere has a big enough radius has as mass the limit "non interaction mass". The gravitational field that is created outside the shrinking sphere, that before it shrunk was null inside of it, is due to the energy that we suck from the system in form of mechanical force during the shrink process. So the loss of mass of the whole system should be attached to the gravitational field created. $\endgroup$
    – George Kourtis
    Commented May 9 at 19:42
  • $\begingroup$ And that gravitational field has a negative mass, that added to the "non interaction mass" of the sphere gives for the whole system the smaller mass that it has when allowed to shrink while not gaining kinetic energy. $\endgroup$
    – George Kourtis
    Commented May 9 at 19:46
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The rest mass of the sphere does not change. The sphere initially starts with its rest mass energy plus potential energy. When the sphere collapses, it gains kinetic energy, but loses potential energy. If energy is extracted from the sphere so that it collapses very slowly then instead of the sphere gaining energy, the extraction device gains energy. Either way, total energy is conserved and there is no annihilation of mass to form energy.

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  • $\begingroup$ We have to differentiate between rest mass (without any contribution thus from kinetic energy) and "not interacting mass" (a term invented for the sake of the discussion) that is the mass at rest when all it's parts are far away from each other. During the slow shrinking process - to my understanding - the rest mass of the whole shell and field lowers ( as energy is sucked away from the system but the "not interacting mass" remains the same. $\endgroup$
    – George Kourtis
    Commented May 9 at 21:30
  • $\begingroup$ The mass of the shell as seen from an observer just outside the shell is it's "non interacting mass", while if seen far away the negative mass of the gravitational field is added to it as it surrounds the shell and then the total mass of the shell and the field, is less given that a negative mass is added to account for the field negative energy. $\endgroup$
    – George Kourtis
    Commented May 9 at 21:30
  • $\begingroup$ In Newtonian physics inertial mass and gravitational mass are the same thing, although Newton didn't see why that should necessarily be the case and I'm not sure they are same thing in general Relativity. Anyway, Birkhoff's theorem clearly demonstrates that the field outside a shell of mass is static and a pulsating star would not emit gravitational waves and would have absolutely no effect on the external field. It follows the same must be true for a collapsing sphere of matter. Externally, it does not matter if the gravitational mass is due to the mass of a static sphere plus ... $\endgroup$
    – KDP
    Commented May 9 at 22:43
  • $\begingroup$ ... potential energy or due to the kinetic energy of the infalling particles minus the lost potential energy. The external field only cares about the total energy inside the sphere whether that energy is mass energy or momentum energy or potential energy and the total energy taking everything into account, does not change in the case of collapsing sphere of particles, as far as I can tell. Noether's theorem also seems to suggest that energy is conserved in the Schwarzschild metric because it is static and symmetrical in time. $\endgroup$
    – KDP
    Commented May 9 at 22:46
  • $\begingroup$ "The external field only cares about the total energy inside the sphere seems correct", albeit, when you speak of potential energy I locate it outside the sphere, in the gravitational field created by the same sphere, thus even when the shell is collapsing, it's mass lowers, but there are two masses a) the mass as seen just outside the shell that is equal to the initial mass, and the mass as seen far away that is smaller due to the negative mass of the gravitational field surrounding the shell. So the mass of the shell as perceived far away diminishes any way also in the collapsing case $\endgroup$
    – George Kourtis
    Commented May 9 at 22:53

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