does negative potential energy reduce the effect of gravity?by Jonnyb302 Tags: astrophysics, general relativity 

#1
Jan2913, 11:13 PM

P: 22

Hello, my question is in the context of modeling static neutron stars via the TOV equation. This is for a 20 week research project for my undergraduate degree. I am creating different equations of state to relate energy density to pressure, I have already used ideal fermi gas models, and now wish to add (very) basic nuclear interactions via an attractive yukawa potential and a repulsive 1/r^n model (modeling Pauli exclusion principle).
I have no formal education in GR, but since I need to use the TOV equation for the modeling I have picked up what I could. I also have no formal education in nuclear physics. Currently as I understand it, all energies cause gravity. But now I am trying to understand what a negative potential energy means. For example, if we assume positrons and electrons have the exact same rest mass in a flat space time at infinity, and we consider a small charge neutral test mass, sufficiently far away, does a positronpositron system exert a greater gravitational "force" on the test mass than an electronpositron system because of the difference in coulomb potential energy? A follow up question would be if there are differences, does a region of negative energy density repel objects? 



#2
Jan2913, 11:50 PM

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PF Gold
P: 5,518

Second, in order to analyze scenarios like the ones you suggest in terms of "potential energy", the system as a whole has to be stationarymeaning, roughly, that its key parameters don't change with time. That's possible for an electronpositron system, since we can imagine them mutually in orbit about each other; but a positronpositron system can't be stationary unless there are other objects involved to hold them in place, and you have to include the energy of those other objects in your analysis. In what follows I'm going to assume that doing that doesn't have any significant impact on the analysis; but that's a questionable assumption. I make it only because you're asking about such a case only as an idealized scenario about potential energy. With those caveats, the answer to your question as you posed it (quoted above) is yes: if we assume that all other relevant parameters are equal (the main one being the distance between the two particles, electronpositron and positronpositron), the positronpositron system will have a larger externally measured mass than the electronpositron system, because of the difference in potential energy. However, that doesn't necessarily mean what you might think it means. Saying that the externally measured mass of the two systems is different is *not* the same as saying that their "energy density" is different. See further comments below. When computing the contribution of potential energy to the externally observed mass of a stationary system, for example, one is actually computing the difference between the mass that the constituents of the system would have if they were all separated "at infinity", and the mass of the actual system itself, with the constituents put together. In other words, the "zero point" of potential energy is taken to be the mass of the separated constituents. In the case of your scenarios, that would be the sum of the rest masses of 1 electron and 1 positron, or 2 positrons (these are the same, obviously). The bound electronpositron system would have *less* mass than a separate electron + a separate positron, because to form the bound system from an electron and a positron "at infinity", one or both of the particles would have to emit radiation equal to the Coulomb binding energy between them in the bound state. Conversely, the positronpositron system would have *more* mass than 2 separate positrons, because one would have to expend energy and do work on the system to push the positrons together and hold them stationary at a finite distance apart. But all of these externally measured masses are positive. Furthermore, if we ask what the energy density is in these systems, it will be the same in all of them. Energy density is a local quantity; we can in principle measure it at a single event in spacetime. Locally, we just have 1 electron and 1 positron, or 2 positrons, and their energy density is the same whether they are in a bound system or not. The difference between these systems and separate electrons or positrons is in the spatial relationships between the particles, not in the energy density of the particles themselves. 



#3
Jan3013, 12:11 AM

P: 22

I think I am lacking a fundamental understanding of what energy density is. Let me describe the exact problem I am running into with my research/modelling.
I am taking my neutron star to be a degenerate fermi gas. This allows me to create an implicit pressure vs. energy density curve. I want to add in a feature to my program which qualitatively (and somewhat quantitatively) captures the influence of nuclear forces. In order to do this I am first approximating the degenerate neutrons as forming an infinite lattice (hexagonal close packed). Then what I do is define a potential energy between 2 neutrons V(r). This potential energy has an attractive term (yukawa potential) and a repulsive term (something like a Leonard Jones) to model the pauli exclusion principle. What I do is I numerically calculate the potential energy between 1 test neutron, and all the neutrons around it in the lattice using V(r). This is an infinite series, but it converges quickly so I truncate it after a few decay constants of the yukawa potential. Then I say that by symmetry every neutron has the same energy, and that every neutron has the same average volume. I divide that energy by volume to get the energy density. However sometimes, because of the way my potential v(r) is defined, I will get a negative energy density. I have defined my v(r) to be 0 as r>infinity. I could add on a constant term to make v(r) > 0 for all r. But do I have the freedom to do so in a GR setting? You said that only the difference between potential energies is meaningful, but here it feels like potential energy no longer has that freedom of an arbitrary shift. 



#4
Jan3013, 07:12 AM

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does negative potential energy reduce the effect of gravity?
In this context negative potential energy should cause less gravity, just as a postive potential energy should cause more.
How are you getting the pressure? You'll need that for the TOV equation as well, it also has an influence on gravity. One other possible issue is that you are doing your hexagonal sphere packing in flat space, I assume. The spatial slice inside your neutron star probably won't be flat, this will affect the number of interacting neutrons that are a distance "r" away. I.e. imagine a 2d hexagonal lattice on a plane, and on a sphere. Then imagine evaluating your series for your potential function in both cases  the concern is it wont be the same number on the flat plane and on the curved sphere. (This is a 2d analogy to your 3d problem). Another issue is that I can imagine using the wrong version of the TOV by not understanding the difference between the stressenergy tensor in a local framefield, and in a coordinate basis. 



#5
Jan3013, 08:58 AM

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PF Gold
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You also seem to be lacking a fundamental understanding of "potential energy". Potential energy is always relative to some arbitrary 0 point. There is no "physical" difference between positive or negative potential energy, only the difference between potential energies. When dealing with motion in the solar system, it is customary to take the 0 point "at infinity" so that all potential energies are negative.




#6
Jan3013, 09:22 AM

P: 22

pervect, I am getting pressure via the method outlined in this paper:
http://arxiv.org/abs/nuclth/0309041 in line 13 of page 9. I am using a numerical derivative. I am also using the TOV equation outlined in line 5 on page 6. I am doing my hexagonal close packing in flat space. Doing it in any other space seems incredibly beyond my ability, I think I have to just punt on that one, and accept an approximation that I hope is not too significant. But I do understand your point. I will point out that the generation of the equation of state isn't exactly in the star. But in an ideal giant reservoir of the gas, where we only account for special relativity. We then hope, that this ideal gas, models the stuff in the star. So in a way, its almost logically consistent. This is somewhat rushed as I have only 3 weeks of the 20 left. Nuclear physics has been incredibly difficult and I am basically trying to come up with my own model that I can explain when I give my final presentation. It may not be perfect, but it needs to make sense to the room full of people I present to at least. Also HallsofIvy I disagree with your statement. To the best of my research in a GR setting, all energy gravitates, including potential energy. That means, potential energy itself shows up in the equations, and the equations are nonlinear, so I do not believe you have the same arbitrary nature of potential energy. 



#7
Jan3013, 09:51 AM

Physics
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PF Gold
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Say we have a total number of neutrons N. Then, if the total externally measured mass of the neutron star, when it is "assembled", is less than N * m_n (the neutron rest mass), the potential energy is negative. But the *total* energy of the neutron star when it is "assembled" is still positive, because it includes the rest energy of all the neutrons. Furthermore, if we look at the whole "assembly" process, we find, as I said before, that to take N free neutrons and put them into a neutron star with net negative potential energy, the neutrons have to emit radiation; so the total energy in the "universe" will be unchanged: it will still be N * m_n. But some portion of that total energy in the "universe" will now be contained in the radiation emitted, instead of in the neutron star. This makes it clear that there is no way to make the total energy of the neutron star (including the potential energy *and* the rest energy of the neutrons) negative; doing that would require emitting more energy in radiation than N * m_n, i.e., converting *all* of the original rest mass of the neutrons to radiationwhich would mean no neutrons left over to form the star*plus* some additional energy that would have to come from nowhere, violating conservation of energy. In other words, the total energy contained in the neutron star is N * m  V, where V is the total (negative) potential energy, summed over all the neutrons, and this total energy must be positive. (And V is equal to the total energy contained in radiation emitted during the star's formation.) [Edit: Regarding changing the "zero point" of potential energy, it should be obvious from the above that that would also require changing the "zero point" for the neutron rest mass; i.e., if we set v(r=infinity) to some nonzero value, we would have to count that nonzero value as part of the total energy of the system of N neutrons when they were all infinitely far away from each other, so that total energy would no longer be N * m_n. But the *difference* between that total energy and the total of the neutron star when it was "assembled" would still be the same: it would still be V, because we would have to add the offset v(r=infinity) to the total energy of the "assembled" star as well.] 



#8
Jan3013, 10:15 AM

P: 22

I had not considered that we also have to set a zero point for the restmass energy (still feels odd to say). That we have to think about the total interaction, and what we measure in our labs, not just break the energy density into a summation of parts, and blindly consider each part separately. There is one final question I have which is probably quite fundamental. Part of my nuclear potential is attractive, while part is repulsive, the repulsive part is me modeling the exclusion principle. However, from what I understand, exclusion is just that, a principle. Not a force, not a potential energy, just a fact of nature. So then, does it make sense to model the exclusion with a potential and let that model produce physical effects in my neutron star? Does it make sense to represent Pauli exclusion as an energy and let that energy generate gravity? 



#9
Jan3013, 10:54 AM

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PF Gold
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#10
Jan3013, 01:19 PM

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PF Gold
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Empirically, this is measured by the PPN [itex]\zeta_3[/itex] parameter, as discussed here (section 3.2): http://relativity.livingreviews.org/.../fulltext.html If negative PE didn't cause a reduction in a body's gravitational field, then [itex]\zeta_3[/itex] would be nonzero, and GR would be falsified. It would also lead to nonconservation of momentum. There is a classic experiment by Kreuzer along these lines, described here: http://www.lightandmatter.com/html_b...tml#Section8.1 There have also been some nice tests involving the earthmoon system and laser lunar ranging. 



#11
Jan3113, 09:29 PM

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I'll try to take a closer look at the paper, but on the first glance it looks fine (I'd trust the paper over my calculations most likely anyway) and you seem to be following it closely. 



#12
Jan3113, 10:15 PM

P: 22

I wouldn't go too deep into the nuclear physics unless you already know a lot about it. The paper is worded very poorly and my adviser believes they incorrectly referenced their work. It appears they worked directly with a current researcher in nuclear physics, and referenced that by simply citing his largest published paper. 


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