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When listing energies for the purposes of keeping track of conservation, or when writing down a Laplacian for a given system, we blithely intermix mass-energy, kinetic energy and potential energy; they are all forms of energy, they all have the same units, and so this looks OK. For example, in the LHC, turning kinetic energy into new particles of mass-energy is routine. We just converted "energy which does not gravitate" (kinetic energy) into "energy which does gravitate". Isn't it a bit peculiar that this same thing called energy can manifest into two different kinds of forms - those forms which gravitate, and those which do not?

How about potential energy? It would be of course ridiculous to calculate your potential in relation to the galactic centre and expect that huge (negative, by convention) quantity of energy to gravitate; and yet if we allow its conversion into kinetic energy, and thence into particle creation, lo and behold we end up with something that does gravitate.

We know that the massless photon gravitates, because it can be "bent" around a star, per GR. A photon also expresses energy in the form E = p c. So clearly finite rest mass is not a requirement for certain forms of energy to gravitate.

So what's the rule here? When does energy gravitate, and why? Isn't it all supposed to be "just energy"?

Then there's the flip side of the equivalence principle - inertia. Do fields have inertia? - they do gravitate, so if they possess no inertia, doesn't that break EEP?

Andrew Palfreyman
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This is a topic many get confused on. Energy(more specifically, the energy-momentum tensor, but in non-relativistic cases, the momentum is negligible compared to the energy aka rest energy + kinetic energy) is what gravitates, NOT mass, a common misconception. If fields carry energy(such as electromagnetic fields), then they gravitate.

ALL types of energy(the non-potential forms, at least) gravitate. According to General Relativity, the current widely accepted theory of gravitation, gravitation is coupled to the energy momentum tensor, which basically includes all forms of energy including contributions from momentum, pressure, rest energy, stress, and kinetic energy. However potential energy is not included in the energy momentum tensor so is non gravitating. So take your example, in LHC, the 'kinetic energy' does gravitate and the mass also gravitates, so there's no change in 'degree of gravitationess'. Note that on such small scales, gravitation is so small that its highly negligible.

Now, I don't think it would be meaningful or is aware of any general definition of inertia for fields, so I suppose it doesn't exist. Answering your question: an equivalent formulation of the EEP is that on local scales, acceleration is indistinguishable from gravitation. As long as the field behaves this way, then it doesn't break the EEP.

resgh
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    Hi, can you refer to some paper or book which proves (or at least mentions) that potential energy doesn't contribute to curvature of stacetime. – user10001 Dec 04 '12 at 13:47
  • Will you accept http://physics.stackexchange.com/questions/45145/potential-energy-in-general-relativity ? – resgh Dec 04 '12 at 14:21
  • I can't see any proof or reference for a proof in your link. I am not saying that your statement is wrong. But if its true then, given that relativity is now a well established subject, there must be some good references where its proved. I actually do not understand how potential energy is a different thing from usual energy. I always thought that energy momentum tensor includes it all. – user10001 Dec 04 '12 at 14:47
  • Well, as I mentioned, the energy momentum tensor includes momentum, pressure, rest energy, stress, and kinetic energy. However, it does not include what we call potential energy. So this fact is inherent in General Relativity. Look up the energy momentum tensor on wiki. You won't find potential energy as one of its components. – resgh Dec 04 '12 at 14:54
  • 00 th component of energy momentum tensor is Hamiltonian which includes both kinetic and potential energy. – user10001 Dec 04 '12 at 15:08
  • @dushya Where did you here that? Quoting wiki The time–time component is the density of relativistic mass, not the Hamiltonian, I'd say. – resgh Dec 04 '12 at 15:14
  • @dushya And even if you were right it could only be a Hamiltonian density in a relativistic variant of Hamiltonian (since Hamiltonians are defined over all of space and the formulation is inherently unrelativistic) – resgh Dec 04 '12 at 15:15
  • I usually prefer to use term Hamiltonian for hamiltonian density and most people do that. Terminology is not the issue. And, yes what i said is right. 00th component of energy momentum tensor is Hamiltonian. – user10001 Dec 04 '12 at 15:21
  • @dushya Where did you see that? I am nearly positively sure that is incorrect. No matter what, the Hamiltonian is non relativistic anyway. – resgh Dec 04 '12 at 15:47
  • "Geometry, Particles and Fields : Bjorn Felsager" is a very well written book, and has a good discussion of energy momentum tensor. – user10001 Dec 04 '12 at 16:19
  • @dushya I don't have the book. Can you quote the relevant parts of the book(like what exactly does it say there)? – resgh Dec 04 '12 at 16:20
  • I am sure you can get it from somewhere online. Chapter 1 and 11 are relevant. Energy momentum tensor has many definitions. A definition at physical level is discussed in chapter 1. Then it can also be defined as a conserved current and as a derivative of action wrt metric. That all these definitions are related (at least in some particular cases) is discussed in chapter 11. – user10001 Dec 04 '12 at 16:33
  • Is there some confusion between the covariant (defined by variational derivative in curved space) and canonical (defined via Noether's theorem for space and time translations in flat space) E.M. tensors? I've heard $T_{00}$ defined as the Hamiltonian in connection with the latter case. – twistor59 Dec 04 '12 at 17:17
  • @dushya@twistor59 Maybe... If so, that explains the thing. Probably the tensor in flat space has to be modified (replacing $\partial$ with $\nabla$?) to accommodate for general manifolds. – resgh Dec 04 '12 at 17:21
  • That causes a problem in defining notion of energy in general relativity but it has nothing to do with treating potential energy as a separate thing, and i don't think it can be separated. – user10001 Dec 04 '12 at 17:29
  • @dushya Well, actually I always thought the concept of potential energy to be especially artificial, jamming in something made up just to make the total energy conserved. Anyways I don't see a problem in separating PE and other forms of energy. Really, I doubt that the energy momentum tensor includes potential energy (except those mentioned by John Rennie). – resgh Dec 04 '12 at 17:33
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    Non-gravitational potential energy does appear in $T_{\mu\nu}$ and does gravitate. The gravitational field also self-gravitates, though it can't be described by a stress-energy tensor and doesn't appear in $T_{\mu\nu}$. – benrg Aug 12 '14 at 18:09
  • potential energy includes MASS terms, so saying potential energy does not gravitate is saying mass does not gravitate. – Kosm Nov 19 '18 at 17:47
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I'm guessing you're asking about what is included in the $T_{00}$ element of the stress-energy tensor. If so, potential energy in the form of pressure or shear stress go into the $T_{11}$ to $T_{33}$ entries. Kinetic energy is a bit more complicated. It does go into $T_{00}$ because that contains the particle energy:

$$ E = \sqrt{p^2c^2 + m_0^2 c^4} $$

but it also goes into the other entries because associated with the kinetic energy is a momentum flux. The Wikipedia article on the stress-energy tensor explains how to calculate it for an isolated (moving) particle.

Re your last question: massless particles can carry momentum (photons do) if this is what you mean by inertia. You obviously can't assign them a mass using Newton's first law because they always travel at $c$ so you can't accelerate them.

John Rennie
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  • So I'm getting a discrepant description of the stress-energy tensor, although you both seem to agree that gravitational potential energy does not in and of itself gravitate, which would be a form of self-action I suppose. I found the commment about the efficacy of inertia in the case of fields/photons moving at c very useful. EEP clearly cannot be applied here. – Andrew Palfreyman Dec 04 '12 at 12:03
  • @AndrewPalfreyman I don't think our descriptions are really that different. I just included the different forms of energies in the form of a list. – resgh Dec 04 '12 at 12:14
  • @JohnRennie I thought kinetic energy went into $T_{00}$!? – resgh Dec 04 '12 at 12:15
  • @namehere - If kinetic energy went into $T_{00}$ we would see a heavy object that was travelling fast turn into a black hole. – John Rennie Dec 04 '12 at 14:56
  • @JohnRennie Really, if it went into momentum the same thing happens. – resgh Dec 04 '12 at 14:59
  • @JohnRennie By the way with even only dimensional analysis we can see that KE should go into $T_{00}$(actually the 00 here already implies a specific coordinate system), with units of energy, not a component with units of momentum. – resgh Dec 04 '12 at 15:01
  • @namehere Let me think about this ... – John Rennie Dec 04 '12 at 15:07
  • @JohnRennie If I recall correctly kinetic energy goes into $T_{00}$ in the form of the relativistic mass($\frac{m}{\sqrt{1-\frac{v^2}{c^2}}}$) increasing. – resgh Dec 04 '12 at 15:10
  • @namehere You're correct of course. I've edited my answer to correct my mistake. – John Rennie Dec 04 '12 at 15:45
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The generator of the gravitational field is the "stress-energy-momentum" tensor $T^{\mu\nu}$, whose components are energy $T^{00}$, co-momentum $T^{0j}$ and co-stress $T^{ij}$ [*]

By symmetry considerations, anything with a non-zero $T^{\mu\nu}$ will feel gravity. In the non-relativistic limit co-momentum and co-stress vanish and energy reduces to $mc^2$, which explains why masses appear in a non-relativistic description of gravity.

All energies contribute to gravitation. I do not know why you believe that kinetic energy "does not gravitate". In fact, the kinetic energy of a photon $E = pc$ is responsible for the light bending effect.

Inertia is an ill-defined term and I cannot answer to your last question without knowing what you mean by inertia.

[*] Many references do not care about dimensions and refer to "momentum and "stress".

juanrga
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We need to realize that radiation has mechanical attributes in the form of momentum and energy(as derived from the momentum), as well as the electrical attributes in the form of an electric field and a magnetic field(as derived from the electric field). When we trap radiation between two mirrors, we have a rest-energy and hence a rest-mass.So rest-mass is trapped momentum/energy.

Elementary particles are trapped and circulating energy/momentum. Momentum is conserved whether moving along as in radiation or circulating as in a particle- like the electron. This then results in the inverse square for gravity and electricity as shown in Bertrand theorem. So gravity and the electric field, or mass and charge, are only a result of energy condensing into elementary particles- like in an electron.

The intrinsic spin, magnetic dipole moment, and the Zitterbewegung clock are all the result of the condensation. These newly created particles can further be trapped inside bigger structure- like the proton. If the constituents move internally at very high speeds, their energy will greatly contribute to the rest-mass of the larger particle- and could form a large part of it- as is the case with the quarks inside a proton.

The proton would also gravitate, and according to its rest mass- as per the same logic given above.The potential energy of the internal constituents and in general also, does not gravitate, mainly because it doesn't carry momentum- shown above to be the source of gravitation and electrification. In any case, the potential energy is a complementary quantity to the kinetic energy, as there is a fixed relation between the two- for a fixed total energy- as given by the virial theorem.

Riad
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