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I am currently reading electrodynamics from Feynman. When talking about the energy of the electromagnetic fields, he says that the location of the field energy could be known at least theoretically since all energy is a source of gravitational attraction, and if we could measure the gravitational attraction we could comment on the location of the energy in the fields.

I have come across formulations where gravitational field is associated with an energy density and is said to be always negative. I wanted to know whether this energy can have an associated gravitational field.

Qmechanic
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veke
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  • Pretty close to a duplicate... https://physics.stackexchange.com/questions/170541/non-linearity-and-self-coupling-of-gravity https://physics.stackexchange.com/questions/643765/does-gravity-bend-gravity – hft Oct 23 '23 at 01:10

4 Answers4

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Yes. One interpretation of the fact that Einstein's equations of general relativity are non-linear is that "gravity gravitates." In other words, the gravitational field itself carries energy, and therefore sources a gravitational field. These interactions are encoded in the non-linear curvature terms in Einstein's equations.

See the discussion at the bottom of page 112/top of page 113 of Sean Carroll's lecture notes on General Relativity: arxiv.org/abs/gr-qc/9712019

Andrew
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  • I'm not formally educated in GR so I may be way off, but isn't this like saying in a series expansion: $e^x=1+x+x^2/2...$ that the function "causes" the first term 1, and the 1 plus the function causes the 2nd term $x$, and those two plus the $x$ cause the $x^2/2$.... Isn't it more correct to simply say the entirety of the nonlinear gravitational field is caused by the initial mass-energy? – RC_23 Oct 20 '23 at 02:49
  • @RC_23 this makes sense, but sometimes it's useful to split the linear part and the nonlinear part of the equations to theorize something like a geon, which is much easier to intuit when thought of as a gravitating gravitational wave. – Ruslan Oct 20 '23 at 07:15
  • That's interesting, I will research that idea – RC_23 Oct 21 '23 at 13:38
  • @Andrew - Does one field having another imply a difference in the scales of the fields (maybe in a sort of infinite regress), and, if so, is such a difference in some way verifiable observationally or experimentally? (Sorry if I'm going a little off the beam in this possibly lame "request for clarification.) – Edouard Oct 28 '23 at 04:27
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Yes, gravitational field can gravitate on its own. One example would be to look at trajectory of an accelerated particle which can emit gravitational wave (see Pirani 1959). If the mass/ energy of this particle is significant enough, then the curvature due to these radiation can influence the particle itself. A perturbative analysis is given using the MiSaTaQuWa equation(see here, here). The Electro-magnetic analog is called the Abraham-Lorentz equation.

However, one of the distinguishing factor b/w the gravitational and electro-magnetic counterparts is that gravitational field has no local energy density of its own (while EM does). The negative gravitational energy is usually the case in Newtonian limit. In the above example, the radiation due to accelerated particle represents vacuum solution...only the Weyl curvature encodes local information about the radiation. One useful quantity in this context will be the Bel-Robinson tensor which is symmetric, trace-free and satisfies conservation equation and for vacuum solution, it corresponds exactly to a free Maxwell SET. However, this quantity is not energy/momentum density when considered dimensionally

KP99
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  • I found Mashhoon's (1999 etc.) interpretation of the Bel-Robinson tensor appealing, though I admit I have not studied this topic in depth. He contracts this tensor with two copies of an "observer" 4-velocity $\mathbf u$, to get $T_{\mu\nu\rho\sigma}u^\rho u^\sigma$, which is the gravitational stress-energy tensor for that observer. So it has the correct dimensions, but has the unfamiliar feature of being observer-dependent. – Colin MacLaurin Oct 25 '23 at 01:27
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I agree with Andrew's answer above (+1), that the nonlinearities of Einstein's equations are saying precisely what the OP's intuition says. But it's a little tricky to break things down any further into separate pieces.

There are two related ideas that are actually used in General Relativity. First is the "Post-Minkowskian expansion", which sort of takes an initial guess for what the gravitational field is doing, then plugs that back into Einstein's equations, and finds a slightly better guess, then plugs that back into Einstein's equations, and finds a slightly better guess, then...

Second is the concept of a stress–energy–momentum pseudotensor, which basically encompasses the usual tensor for matter/radiation/etc. and the gravitational field, to act as a source for the gravitational field. In his description, Wald ["General Relativity", Sec. 4.4.] does a good job of explaining why this is hard:

In general relativity, there is no known meaningful notion of local energy density of the gravitational field. The basic reason for this is closely related to the fact that the spacetime metric, $g_{ab}$, describes both the background spacetime structure and the dynamical aspects of the gravitational field, but no natural way is known to decompose it into its "background" and "dynamical" parts. Since one would expect to attribute energy to the dynamical aspect of gravity but not to the background spacetime structure, it seems unlikely that a notion of local energy density could be obtained without a corresponding decomposition of the spacetime metric.

But Wald then goes on to find something like the best known example of a pseudotensor: the Landau-Lifshitz pseudotensor.

Both of these ideas, the post-Minkowski expansion and psuedotensor, only really make sense when you're dealing with some "background" or approximate spacetime that you expect to be somehow "close" to the full spacetime. Ultimately, the only truth — as far as General Relativity is concerned — is contained in Einstein's field equations, which contain all of these effects as limiting cases.

Mike
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  • Side note: the phrase "answer above" should not be used. I can sort by "newest first" and when I'm writing this your answer is above all other answers, so for me, now there is no "Andrew's answer above". The only firm way to refer to another answer is to link to it. – Kamil Maciorowski Oct 20 '23 at 03:13
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TLDR: No, the gravitational field does not have it's own gravitational field. But, it's not too unreasonable to guess that this might happen. I ran into similar ideas in electrodynamics where electric fields create polarizations which creat electric fields which create ... The solution is to find an intermediary concept which relates the quantities you want to study.

In this case we the intermediary concept the relates gravity and mass/energy is the Ricci curvature. In the electric field example it's the electric-displacement field. In both cases there can still be derived quantities which are attributed to the fields, but ultimately these always are the result of the matter which generates the fields away from a zero-field state.

More to the point, Newton's law of gravity can be associated with a gravitational energy density. It would be reasonable to then assume that the energy density of the matter and the field add to create more curvature than just from the mass alone. This is incorrect and in general relativity the associated 'gravitational energy density' must be calculated from the curvature of space-time which is itself determined by the distribution of mass and momentum. So no there is no new gravity generated by gravity.

There is an interesting passage from Griffith's Introduction to Electrodynamics (4e) that this question reminds me of:

the external field will polarize the material, and this polarization will produce its own field, which then contributes to the total field, and this in turn modifies the polarization, which ... (p186)

The idea is different insofar as topic and fields, but the concept of confusion here seems not to be about whether material has an inherent energy density (it does) nor whether gravitational fields also have an associated energy density (they do, just as EM fields do), but whether the energy density of the field creates a new gravitational field that is associated with the additional energy density of the gravitational field.

So I cannot provide a formal answer with quotes, but I will reference something else from Griffith's book. There is a section on page 96 under the subsection (ii) Where is the energy stored?. Here Griffith discusses in the first subsection that for charged point particles there is an associated electric field that is inherent to their creation. Moving around these particles implies a finite work done, but destroying a point particle implies an infinite amount of energy must be released. That is to say the energy stored in the very existence of the field from point particles is infinite, even though Einstein's mass-energy relation is finite. In fact this is true for any inverse-square force such as gravity.

It is on this basis then that I assert that

No, the gravitational field and its associated energy density does not increase the global curvature of space and therefore create additional associated gravitational fields beyond the associated mass-energy and momentum of the constituent mass.

I can simplify my answer a bit. In general relativity, the gravitational field itself and its associated energy density is replaced by the curvature of space and the mass-energy and momentum densities. So a mass in spacetime will tell space how to curve, and the curve will produce a matching gravitational field, but there is no infinite regress of mass -> field -> new-field -> new-new-field.

For further study refer to Carroll's An Introduction to General Relativity or see the Einstein Field equations. These equations directly relate the total energy density of matter to the curvature of space time directly, and introduces no additional intermediate fields to carry energy. Since there is no entry in the index for 'Energy Density - gravity' which points to a gravitational energy density, I take the idea to not be relevant to the equations of motion.

Again I don't think this question is bad. It's frankly a perfectly reasonable question. Energy densities curve space time, curved space time 'is' gravity, gravity has an associated energy density, so energy density + energy density = bigger energy density -> more curvature -> more gravity -> ???

I think that it's also helpful to consider the equivalent question for electrodynamics. Charge densities create fields -> fields have energy density -> (the combined theory of general relativity and EnM has similar ideas to curvature) -> so then does energy density + energy density = more EM fields?

No! Of course not! If the idea held then we would always have this ever-increasing EM field where at each step we calculate the energy density of everything we are considering, add that to the energy density of the previous step and get this infinite feedback loop where the only field that can exist is infinitely strong or something even weirder (maybe the field is non-convergent everywhere?).

Gerald
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    Curvature is represented by the Riemann tensor. This does not need to be zero in the absence of matter. Gravity couples to it's own gravitational energy-momentum tensor, and this differs from electrodynamics. I think this answer as currently worded is incorrect. – Eletie Oct 19 '23 at 08:12
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    @Eletie Well one answer seems to be incorrect because one says yes and the other says no. – Michael Oct 19 '23 at 13:24
  • @Michael See the discussion at the bottom of page 112/top of page 113 of Sean Carroll's lecture notes on General Relativity: https://arxiv.org/abs/gr-qc/9712019 – Andrew Oct 19 '23 at 14:17
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    I think the tldr is misleading. As far as I can understand, the rest of the answer says that yes, there's gravity from the mass, and yes, there's gravity from the gravity, but the equations we use to measure gravity measure both together: the gravitational constants already take this into account. So when you're measuring gravity using the gravitational constants, "there is no new gravity generated by gravity." I think. – Mooing Duck Oct 19 '23 at 20:38
  • @Eletie, yes that is correct. The Reimann Tensor relates to the stress-energy tensor. This is a function of the mass, momentum, and local pressure. For most applications, the local pressure (an energy-density) is zero. Consider the example of a cold interstellar gas where the pressure and momentum are essentially zero. Whatever curvature is present is thus due solely to the mass of the gas. The curvature then presents the matter with a specific spacetime path (calculated through the variety of spacetime calculation methods) consistent with it's boundary conditions. – Gerald Oct 19 '23 at 20:56
  • @Michael, I think that is a complete misreading of my answer. I state that while there is some mathematical relationship between gravity and the associated gravitational energy-density functions, this relationship is replaced in General Relativity. The curvature of spacetime is zero in the absence of mass and momentum energy-densities. This does not discount gravitational waves, which are also formally called gravitational radiation. – Gerald Oct 19 '23 at 21:30
  • @Michael This can be mathematically expressed as spacetime forces on an inertial body, but as Carrol states themselves on page 307: "There is no true local measure of the energy in the gravitational field." If the energy of gravity, the curvature of spacetime, has no consistently definable value then it seems to be that the energy associated with gravity in the Newtonian case is merely a consequence of the weak-field approximation and not a meaningful aspect of General Relativity. So no, there is no feedback effect from an energy-density to gravitational field to new energy density. – Gerald Oct 19 '23 at 21:39
  • "The curvature of spacetime is zero in the absence of mass and momentum energy-densities." No, that's wrong. The $T_{\mu\nu}$ in $G_{\mu\nu}=8\pi T_{\mu\nu}$ is zero in that case, but that doesn't mean $g_{\mu\nu}$ has no curvature — as @Eletie pointed out. Also, the analogies with E&M are misleading precisely because E&M is a linear theory while GR is not, and it is the nonlinearity that makes all the difference to the OP. – Mike Oct 20 '23 at 17:34
  • I specifically mention gravitational waves which are an example of curvature in the absence of matter. But hey, ignore everything else I wrote. – Gerald Oct 20 '23 at 17:39
  • Okay, but the statement I quoted is actually incorrect, and the rest of those comments are too ambiguous for me to really understand. Are you saying that the "This does not discount gravitational waves" statement directly contradicted that wrong statement, but that both should stand? – Mike Oct 20 '23 at 17:49
  • @Mike Could you share a $T_{\mu \nu}$ that is zero with a curved spacetime? Schwarzschild doesn't count, since it corresponds to a delta-function $T_{\mu \nu}$. But it's been a long time since I studied GR, so there's probably another example I'm forgetting. – user196574 Oct 20 '23 at 18:48
  • @Gerald Regarding your flag on this post, you can edit this post or delete it at your discretion. I'm not sure what moderator action you want based on the contents of the flag. – Chris Oct 20 '23 at 19:15
  • @user196574 Well, there are infinitely many of them. Roughly speaking, if you take a slice through spacetime, you can construct a 3-metric on that slice, solve a complicated equation on that slice (which may not work in some cases), and you get a complete 4-d spacetime out of it. This is called the "initial-value formulation of general relativity". If you want exact solutions, there's a small list here, or a whole book here. – Mike Oct 20 '23 at 22:05
  • @Mike Thanks for the links. Maybe I'll need to ask this as a separate question, but do any of those examples in the small list not suffer from curvature singularities? Many of those metrics in the small list I would say actually correspond to delta-function stress-energy tensors, not vanishing stress-energy tensors. I'll need to check them to confirm, but probably the best candidate for me to check would be a vacuum solution without any curvature singularities. Edit: Probably I'll be happy with a gravitational plane wave solution, so I'll let it rest. – user196574 Oct 20 '23 at 23:00