# Does Gravity Gravitate?

1. Sep 1, 2014

### Staff: Mentor

[Reposted from my PF Blog; I'll post this in three separate posts, as I did in the original.]

This has come up in enough threads now that I feel the need to give my take on it in a convenient central location, so I can just link to it instead of having to restate it again and again. ;)

Short answer: mu. (The terms of the question are not well-defined, so it doesn't have a well-defined answer.)

Somewhat less short answer: Yes and no. (It depends on how you define the terms "gravity" and "gravitate". There are possible definitions that lead to each answer.)

Ok, enough fun. ;) Let's look into this in more detail. We'll start with the "no" answer.

The fundamental equation of GR is the Einstein Field Equation, which looks like this:

$$G_{ab} = 8 \pi T_{ab}$$

Conceptually, what this equation says is " "spacetime curvature = constant * stress-energy". So the LHS of the equation is "gravity", conceptually, and the RHS is the "source" that produces it. The key point here is that the RHS does not include any stress-energy due to gravity itself. That's because there isn't any; the "source" in the EFE does not include any "energy stored in the gravitational field", because there is no coordinate-free way of defining any such energy, and the EFE is a coordinate-free, tensor equation.

But how did we come up with this equation, you ask? There are a number of ways to derive it (Misner, Thorne, and Wheeler list six of them), but the most direct is the way first derived by David Hilbert in 1915, at the same time that Einstein was completing his derivation (by a different route). We start with a simple action for gravity:

$$S_{G} = \frac{1}{16 \pi} \int d^4 x \sqrt{-g} R$$

where $R$ is the Ricci curvature scalar and $g$ is the determinant of the metric tensor. By the principle of least action, the variation of the action with respect to the metric must be zero, and this gives an "equation of motion" for the physical system described by the action. Varying $S_{G}$ with respect to the metric leads directly to the vacuum Einstein Field Equation (i.e., the EFE with the RHS equal to zero). So if gravity is the only "field" present (we'll come back to what "field" means in a bit), the Einstein tensor (the LHS of the EFE) completely describes the dynamics of the system.

What if there are other "fields" present? We need to know the action for those fields (which are usually called "matter" fields, even though they can include things like electromagnetic radiation that are not normally thought of as "matter"). Call that action (including all other "fields" besides gravity) $S_{M}$. Then the variation of the total action $S = S_{G} + S_{M}$ with respect to the metric must be zero by the principle of least action; and this leads directly to the full EFE, with a tensor on the RHS, obtained by varying $S_{M}$ with respect to the metric, that we call the "stress-energy tensor".

So we have a simple explanation for our "no" answer to "does gravity gravitate": the RHS of the EFE does not include any stress-energy due to gravity, because it comes from varying $S_{M}$, not $S_{G}$. But why do we divide the action up in that particular way? There are actually a couple of answers to that, and here's where it gets interesting, because one of those answers (the one I'll give in a moment) confirms the "no" answer to our question, while the other (which I'll give a bit later on) suggests a "yes" answer. So it's not too surprising that PF threads tend to go on and on when this topic comes up.

The first answer to why we divide up the action the way we do is simple. If we take the covariant divergence of both sides of the EFE, we get zero:

$$\nabla^a G_{ab} = \nabla^a T_{ab} = 0$$

For the LHS, the Einstein tensor $G_{ab}$, this is a geometric identity called the Bianchi identity. It therefore also holds for the stress-energy tensor $T_{ab}$ on the RHS because of the EFE. Physically, $\nabla^a T_{ab} = 0$ means that the "source" of gravity, stress-energy, is conserved: it is neither created nor destroyed in any infinitesimal volume of spacetime. This is the most fundamental GR version of energy-momentum conservation, and it is a highly desirable property for the stress-energy tensor to have. But it only holds if we write the EFE the way we did above, with the SET only containing "fields" other than gravity; and that requires that we split the action up the way we did, into a part $S_{G}$ that only includes gravity, and a part $S_{M}$ that only includes fields other than gravity.

So we write the EFE the way we do in order to ensure automatic conservation of the "source", and that way of writing the EFE requires the "source" to only include fields other than gravity. That, as noted above, seems to reinforce the "no" answer to the question. But there's still something we haven't talked about: where did the action $S_{G}$ come from in the first place?

Hilbert's answer (and Einstein's, when he saw Hilbert's work) was basically that the action $S_{G}$ is the simplest possible action for gravity that is not trivial. However, in the late 1950's and early 1960's, a different approach was developed, based on trying to treat gravity as "just another quantum field", like the electromagnetic field and all the other fields that were then being studied. I won't go into too much detail about this, but the upshot was that the quantum field theory of a massless, spin-two field on a flat spacetime background, when made self-consistent, turns out to have as its classical limit the Einstein-Hilbert action $S_{G}$! (The "massless, spin-two" part comes from the fact that only a massless field can give rise to a long-range interaction, which gravity is, and only a spin-two field can give rise to an interaction which is always attractive *and* couples to all the other known "matter" fields.)

In other words, on the "gravity as just another quantum field" view, classical GR is just a low-energy effective field theory; it is what you get when gravity is too weak for its quantum nature to show up. (Don't be misled by that "too weak", btw; in the sense of the term used here, gravity is "too weak" at, and well inside, the horizon of a stellar-mass black hole.) Which means that the way we wrote the action, and hence the EFE, above is just the natural way to write the classical limit of a theory with gravity present along with other fields.

But here's the punch line: as a massless, spin-two field, gravity not only couples to all the other "matter" fields; it also couples to *itself*. Its field equation, both in the quantum version *and* in its classical limit, the EFE, is nonlinear. At the quantum level, this means gravitons (the quantum particles associated with the massless, spin-two field) interact with other gravitons. At the classical level, it means that, since the EFE is nonlinear, curvature can be present even when the "source" on the RHS of the EFE is zero, i.e., there can be vacuum solutions of the EFE that have curvature present. (Schwarzschild spacetime is an obvious example.) In other words, on this view, the answer to our question is "yes": gravity *does* gravitate!

It's important to note that there is no contradiction between the two answers we have just described. "Gravity" in the two answers means two different things: gravity as a massless, spin-two field (either quantum or classical) does gravitate (the field interacts with itself), but gravity as a tensor satisfying the Bianchi identity doesn't gravitate, because there is nothing "left over", once the Bianchi identity is satisfied, to contribute to the source on the RHS of the EFE.

To sum up what we've said so far: we've talked about two possible ways to answer our title question, and they lead to opposite answers:

(1) In order to ensure conservation of the source, the complete Einstein tensor, including *all* contributions from gravity, must appear on the LHS of the EFE; there is nothing left over to contribute to the "source" on the RHS of the EFE. So in this sense, gravity does *not* gravitate.

(2) Viewed as a quantum field, gravity is a massless, spin-two field, and the classical limit of the quantum theory of such a field is standard GR (based on the Einstein-Hilbert action for gravity). But this field interacts with itself; its field equation, at both the quantum and classical levels, is nonlinear. So in this sense, gravity *does* gravitate.

However, there is still more to come; we haven't even talked about the issues that usually give rise to our title question. But that's for the next post.

Last edited: Sep 24, 2014
2. Sep 1, 2014

### Staff: Mentor

Does Gravity Gravitate: The Sequel

In the first post in this thread, I talked about two ways to answer the title question, one leading to the answer "no" and the other leading to the answer "yes". However, this will leave a lot of people who ask our title question unsatisfied, because the usual motivations for asking the question have little, if anything, to do with the general points I discussed. So let's look at some particular cases to hopefully shed some more light on the subject.

Consider a massive, gravitating body like the Earth. For simplicity, we'll assume that the Earth is perfectly spherically symmetric and is not rotating. Then we can describe gravity in the vacuum region outside the Earth using the Schwarzschild metric:

$$ds^2 = - \left( 1 - \frac{2M}{r} \right) dt^2 + \frac{1}{1 - 2M / r} dr^2 + r^2 d\Omega^2$$

What is this "M" that appears in the line element? Obviously, you say, it's the mass of the Earth. How do we measure it? Well, we put some small test object into orbit about the Earth and measure the orbital parameters. In other words, M is the "externally measured" mass of the Earth.

This heuristic definition of what M means has been formalized. When we measure orbital parameters, we are really measuring components of the metric and comparing them with the line element given above to see what value of M makes the two match. But the results of such measurements, in any real case, can vary depending on the radius from the central body at which we choose to make measurements. What we would really like is a way of capturing the "effect of mass on the metric" that is independent of such considerations, and there is one. For any spacetime which is asymptotically flat (meaning that the metric becomes the Minkowski metric as the radius from the central body goes to infinity), we can define something called the "ADM mass":

$$M_{ADM} = \frac{1}{16 \pi} \lim_{S \rightarrow i^0} \int_{S} g^{ij} \left( \partial_j g_{ik} - \partial_k g_{ij} \right) n^k dS$$

Here $g_{ij}$ and $g^{ij}$ are the 3-metric and inverse 3-metric on a spacelike slice of "constant time" (where "time" means the time coordinate of the Minkowski metric in the asymptotically flat region), $S$ is a 2-sphere surrounding the central body, and $n^k$ is an outward-pointing unit vector that is normal to the 2-sphere. The limit is taken as $S$ goes to spatial infinity, $i^0$.

For the idealized case of Schwarzschild spacetime, it's fairly straightforward to show that $M_{ADM} = M$; that is, the ADM mass as defined above equals the M that appears in the line element. But the ADM mass is applicable to *any* asymptotically flat spacetime, so it applies to objects that aren't spherically symmetric, like the actual Earth, and to systems containing multiple bodies, like the Solar System or a binary pulsar. (We'll come back to the latter case below.)

But the ADM mass only depends on the metric coefficients, and those only in the limit of spatial infinity; in other words, it is purely an "external" measure of mass, as we noted above. It would really be nice if we had a way of measuring the mass of the Earth "internally", by adding up the contributions of all the individual pieces of matter inside the Earth. The "naive" way of doing this is just to integrate the stress-energy tensor over a spacelike slice:

$$M_{naive} = \frac{1}{4 \pi} \int dV T_{ab} u^a u^b$$

where $dV$ is the volume element of a spacelike slice of constant time, and $u^a$ is a unit timelike vector that is normal to the spacelike slice.

However, there are two things wrong with this integral. One is that we have just assumed that $T_{ab}$ itself is what belongs in the integrand, without stopping to think about how $T_{ab}$ is actually *related* to the M that appears in the Schwarzschild line element. What we should do, instead, is to look at the Einstein Field Equation, which reads

$$G_{ab} = R_{ab} - \frac{1}{2} g_{ab} R = 8 \pi T_{ab}$$

We saw in the previous post that putting $G_{ab}$ on the LHS of this equation is needed in order to ensure that the covariant divergence of both sides is zero. However, we don't actually *measure* $G_{ab}$ directly. What we actually measure is $R_{ab}$, the Ricci tensor, which directly measures the inward acceleration of a small ball of test particles at a given event due to the stress-energy present at that event. (Note that here we are talking about spacetime events *inside* the Earth, which is where the Ricci tensor is nonzero.) In other words, the Ricci tensor, *not* the stress-energy tensor, is really the best local measure of the "contribution to gravity" of a small piece of matter.

So what we really want for our purposes here is to re-arrange the field equation to put just $R_{ab}$ on the LHS, with the RHS independent of $R_{ab}$ or $R$ (which is the trace of the Ricci tensor). It turns out that this can be done pretty easily; the result is

$$R_{ab} = 4 \pi \left( 2 T_{ab} - g_{ab} T \right)$$

where now $T$ is the trace of the stress-energy tensor, i.e., $T = T_{00} + T_{11} + T_{22} + T_{33}$.

The RHS of this equation is really what should appear in our integral, instead of just $T_{ab}$. So our next try at the integral looks like this:

$$M_0 = \int dV \left( 2 T_{ab} - g_{ab} T \right) u^a u^b$$

However, there is still something wrong. If we compute $M_{0}$ for an object like the Earth, we will get the wrong answer; our answer will be *larger* than the externally measured mass $M$ that appears in the line element. What gives? Well, we forgot one other thing: spacetime is curved. The volume element $dV$ in the above integral is only correct in flat spacetime; in curved spacetime, we have to correct for the fact that the metric varies from place to place. In this particular case, it's easy to do this because the spacetime is static; I won't go into detail about this, but the upshot is that we just need to include an extra correction factor in the integral, like so:

$$M = \int dV \sqrt{g_{tt}} \left( 2 T_{ab} - g_{ab} T \right) u^a u^b$$

If we compute *this* integral for an object like the Earth, we get the correct answer: the M yielded by the integral is the same M that appears in the line element. In fact, what we've just done is to compute what's called the Komar mass of the system. (This mass is defined for a somewhat different class of spacetimes than the ADM mass; where the latter required the spacetime to be asymptotically flat, the Komar mass requires the spacetime to be stationary. This is an important point, but I'll save further discussion of why it's important until later on.)

The difference between $M_0$ and $M$ is standardly referred to as "gravitational binding energy", and the fact that it is there--that $M_0 - M$ is not zero--is one thing that often prompts people to assert that "gravity gravitates": that gravity "contributes" to the externally observed mass $M$ of a system. (Personally, I confess that I find this interpretation a bit strange: $M$ is *less* than $M_0$, so the "contribution of gravity" is *negative*. But it's a common interpretation.) As long as the system is not changing with time, this doesn't cause any problems. However, when we try to extend this interpretation to systems that *are* changing with time, we run into complications.

First let's consider an object that is emitting radiation with non-zero stress-energy associated with it, such as EM radiation. What will its externally measured mass look like? There is actually an exact solution called the "Vaidya null dust" for this case, but we won't need to go into the details of it here; the key point is that if we are orbiting at some finite radius $r$ outside the body, the mass we observe it to have, by measuring our own orbital parameters, will slowly decrease as radiation passes by on its way out to infinity. ("Null dust" means "incoherent EM radiation that, for the purposes of this problem, can be approximated as a spherically symmetric dust--i.e., a "fluid" with zero pressure--of particles--photons--moving radially outward at the speed of light".)

How will this decrease in mass show up in the integrals we looked at above? It turns out that the decrease will *not* show up in the ADM mass at all. This is because the ADM mass involves taking a limit at spatial infinity, and no matter how far the radiation travels from the original body, it will still be at some finite radius; it will never reach spatial infinity. So the ADM mass integral will always "see" it, even though we, at a finite radius $r$, no longer do.

This issue was recognized quite some time ago, and the solution was fairly straightforward: define an alternative "mass" integral by taking the limit at future null infinity instead of spatial infinity. This is called the Bondi mass, and it allows us to separate out the energy carried by radiation, which escapes to null infinity, from the energy present in what remains behind, the central object. In the case of the Vaidya null dust, the Bondi mass will *not* include the energy carried away by the radiation, so that energy can be quantified as the difference between the ADM mass and the Bondi mass for the spacetime.

What about the case where the radiation being emitted is gravitational waves? That's for the next post. ;)

Last edited: Apr 26, 2015
3. Sep 1, 2014

### Staff: Mentor

Does Gravity Gravitate: The Wave

In the two previous posts, we looked at various ways of interpreting the question "does gravity gravitate?" We left off at the end of the last post with an open question: what do the various "mass integrals" look like in a spacetime where GWs are being emitted? Let's look at that question now.

The key difference between GWs and other types of radiation is that GWs have zero stress-energy. They are oscillations in the curvature of spacetime. This fact makes it tempting to conclude that GWs can't carry energy at all; in fact, many physicists who studied GR in the 1950's and 1960's thought this. The basic argument goes like this: the "source" in the EFE is on the RHS, but GWs can be present when the RHS of the EFE is zero, i.e., when there is no "source". By analogy with EM waves, that should mean that GWs can't carry any "source" along with them, just as EM waves can't carry any charge. But the "charge" associated with gravity is just energy (more precisely, stress-energy), so it seems like GWs should not be able to carry any energy.

However tempting this line of reasoning is, though, it isn't valid. We can see this by a simple thought experiment, inspired by Feynman's response at a gravity conference when he saw so many people struggling with this issue: "This is what comes from looking for conserved tensors, etc., instead of asking, can the waves do work?" Suppose we have gravitational waves passing through a certain region of empty space, which we will idealize as plane waves for this case (we assume we are very far from the source and that the waves are weak enough that no nonlinearities, i.e., no self-interactions, can be detected). We place an object transverse to the direction of wave propagation; for example, we could place a flat plate of some material such that it lies in a plane perpendicular to the wave vector. What will happen? It is straightforward to show that the wave oscillations will induce vibrations in the object which will heat it up; i.e., the waves will do work on the object, adding energy to it. That means the waves themselves must be carrying energy.

This is a local view (or at least, quasi-local--we'll see below that the energy carried by GWs can't really be localized), but there is also a global view. Consider a spacetime containing GWs, such as a binary pulsar spacetime, and assume that there is *no* other radiation present besides the GWs (which is *not* true for a real binary pulsar, but which is a valid idealization for our purposes here). We can still do the same comparison we did for the case of EM radiation in the previous post, and we will find the same result: the ADM mass and Bondi mass of the system will *not* be the same. The ADM mass of the spacetime will include the energy carried by the GWs, but the Bondi mass will not. So we can separate out, globally, the energy carried away by GWs from the gravitating mass that remains behind.

But what about that line of reasoning that we gave above? Where, exactly, does it break down? After all, local conservation of the SET still holds; the covariant divergence of the SET is still zero everywhere. So where does the energy get "transferred" from the matter (which has nonzero SET) to the GWs (which have zero SET)? And how can that happen without violating the local conservation law?

Consider an idealized binary pulsar system again. We have two neutron stars orbiting each other, and as GWs are emitted, the orbital parameters slowly change: the stars get closer to each other and their orbital speeds increase. The net effect is an energy decrease for the system consisting of the two stars only (remember that we assumed for our purposes here that there was no radiation present other than GWs), and this can be understood, qualitatively, in Newtonian terms: the reduction in potential energy due to the stars getting closer is larger than the increase in kinetic energy due to the increase in orbital speeds. (If you actually do the math, you will see that this follows from Kepler's Third Law.) However, the change in orbital parameters also means a change in the RHS of the EFE, because the relative positions of the "sources"--the regions of nonzero SET in the interiors of the two neutron stars--have changed. This shows up as a change in the metric coefficients. This change isn't as simple as a slow reduction of $M$ in the Schwarzschild line element would be, but its effect is similar: if we put a test object in orbit at a large distance from the binary pulsar system, and measure its orbital parameters, the "gravitating mass" of the system that we obtain will slowly decrease as GWs pass by us on their way out to infinity.

If we try to localize exactly "where" the change in metric coefficients is being triggered from, though, there is no answer. Locally, the covariant divergence of the SET is zero at every event; there are time and space variations in the metric, and they are driven by time and space variations in the "source" (because the two pulsars are orbiting each other), but those variations are constrained by the local conservation law. It is only when you look on a larger scale (a scale at least as large as the wavelength of the GWs being emitted, which for a typical binary pulsar is pretty large, on the order of the orbital radius) that you can see a pattern in the time and space variations in the metric that tells you that GWs are being emitted and are carrying away energy, globally, from the binary system.

So we have a third way of interpreting our title question, which leads to the answer "yes": if "gravity" is interpreted as "spacetime", and "gravitate" is interpreted as "can store and propagate energy", then yes, spacetime can store and propagate energy: gravity does gravitate.

Postscript

You may be wondering about one loose end: what about the Komar mass of a system that is emitting radiation? That's where the comment I made in the previous post, about the Komar mass being applicable in a somewhat different class of spacetimes, comes in. Spacetimes where radiation is present can be asymptotically flat, so the ADM and Bondi masses can still be defined and computed. But spacetimes where radiation is present *cannot* be stationary: they cannot have a timelike Killing vector field. That means the Komar mass simply can't be properly defined for such spacetimes. There may be ways to adjust the Komar mass definition, for example by using the total 4-momentum of the spacetime as defined in the asymptotically flat region to find a timelike unit vector to plug in to the Komar mass integral. But that's a subject for (possibly) another post.

4. Sep 2, 2014

### WannabeNewton

I don't quite understand the argument here.

I agree that you cannot localize exactly "where" the change in the metric coefficients is coming from and it is true that $\frac{l}{c}\partial_t h_{\mu\nu} \ll h_{\mu\nu}$ in the near zone but that is not unique to gravitational waves because it is just a problem of characteristic length scales. If I instead have a collection of electric charges that move within some sphere of characteristic size $l$ and the characteristic wavelength of the radiation they emit satisfies $\lambda \gg l$ then the electromagnetic field of the source will vary due to the radiation on a length scale on the order of the characteristic wavelength and we also have for example that $\frac{l}{c}\partial_t \vec{E} \ll \vec{E}$ in the near zone.

Of course it is true that the energy of gravitational waves can't be localized, that is, an energy density for it can't be defined but that is an issue with all gravitational fields in general relativity and this is simply because the equivalence principle prevents a gauge-invariant reformulation of the Einstein equation as an equation of motion for the metric tensor in terms of the usual energy-momentum tensor and the "gravitational energy-momentum" (c.f. Landau Lifshitz formulation of GR which fixes the harmonic gauge in order to achieve this).

EDIT: More precisely in this formulation $\square g_{\mu\nu} = -16\pi \tau_{\mu\nu}$ where $\tau_{\mu\nu}$ includes both non-gravitational and gravitational energy density but this is not a gauge-invariant equation because it is also simultaneously solved with the harmonic gauge condition $\partial^{\mu}g_{\mu\nu} = 0$ which is just a reflection of the fact that the equivalence principle allows us to transform away the gravitational field, and thus its energy density, at any given event in space-time.

Last edited: Sep 2, 2014
5. Sep 2, 2014

### stevendaryl

Staff Emeritus
The part of the argument that I don't understand is why the coupling of gravity to stress-energy must be universal. I don't see, right off the bat, why there couldn't be two kinds of mass/energy, gravitating and non-gravitating, and gravity only couples to the first kind. (In the same way that the electromagnetic field only couples to charged particles.)

Is there some simple argument that non-universal coupling to stress-energy is inconsistent?

6. Sep 2, 2014

### atyy

An argument, not simple, from the point of view of gravity as quantum spin 2 was given by Weinberg.
http://arxiv.org/abs/1007.0435 (section 2.2 and Appendix A)

Last edited: Sep 2, 2014
7. Sep 2, 2014

### atyy

Nice posts! I think I finally have some understanding of what the Komar mass is, especially why it doesn't exist when there are gravity waves. I always dyslexically thought it was some sort of curry! http://en.wikipedia.org/wiki/Korma

8. Sep 2, 2014

### Staff: Mentor

I'm basically trying to expand on "the energy of gravitational waves can't be localized", or more generally, "the energy stored in the gravitational field can't be localized". Put another way, I'm trying to emphasize the differences between GWs and other types of waves, that arise from the fact that gravitational field energy can't be localized in a way that other types of stress-energy can.

9. Sep 2, 2014

### vanhees71

One aegument is that you can interpret GR as a non-abelian gauge theory with the gauge group GL(4). In non-abelian gauge theory the coupling is necessarily universal, because the gaugd bosons cary the charge of the gauge field themselves. In this sense the coupling must be universal.

For details see Ramond's book on QFT or the Feynman lectures on gravitation.

10. Sep 2, 2014

### ohwilleke

* So, if I were to try to sum up a bottom line of this analysis, the take away message seems to be that either does not gravitate, or that if it does gravitate because the question is defined differently, that formulations of GR in which gravity is defined in such a way that it does gravitate have no phenomenologically observable differences from those formulations in which it does not (i.e. that it is just a matter of semantics).

* Is there something wrong in the analysis of A. Deur that suggests the self-gravitation of gravity has significant phenomenological effects in very heavy (galaxy sized or more massive systems) that are not spherically symmetric which is overlooked in spherically symmetric approximations, which he articulates, for example, at http://arxiv.org/abs/0901.4005?

If so, what is wrong with his analysis, which is developed by analogy to QCD equations since gluons like hypothetical gravitons are self-interacting?

Deur reaches the conclusion that there is no net self-gravitation in a spherically symmetric system for which the Schwarzschild metric is an appropriate approximation. This is less obvious in the case of ADM mass, but since it is taken in the limit as the distance at which the mass is measured approaches infinity, it would seem that ADM mass is subtly assuming that the source mass is point-like and hence also spherically symmetric. Komar Mass and Bondi mass also seem to be usually analyzed in spherically symmetric systems, although it isn't clear to me that either definition demands spherically symmetric.

* If the energy of gravity inherently cannot be localized, does that imply that any massless spin-2 graviton based analysis is necessarily inconsistent with the EFE because any carrier boson approach to gravity will necessarily localize gravitational energy?

Similarly, is it correct that if the energy of gravity inherently can't be localized, even classical field theories or theories with a localizable space-time curvature would also necessarily be inconsistent with the EFE?

Phased somewhat differently, is the non-localizablity of gravity energy a corollary of some more familiar GR axiom like background independence or the equivalence principle? Or, is this an independent assumption with practical consequences for theory building that arises from some other source?

If there is an observable difference between a gravitational theory in which gravitational energy cannot be localized and one in which it can be localized, under what sort of circumstances would this manifest?

* Finally, H. Nikolic has argued, at http://arxiv.org/abs/1407.8028 that it is possible to construct an energy-momentum tensor for the gravitational field by resorting to second derivatives (which is basically the Einstein tensor times a constant), rather than merely limiting such a tensor to first derivatives, an assumption that he argues is the reason for the conventional view that only an an energy-momentum pseudo-tensor for the gravitational field is possible. He further argues there is no strong reason in principle to assume that only first derivatives need e present in an energy-momentum tensor of a gravitational field. How does the view that the energy-momentum of a gravitational field must be a pseduo-tensor fit into the analysis in the original three posts?

* To be clear, I'm not pressing a BSM interpretation of any of these matters, I am merely trying to understand if analysis of Deur and/or Nikolic differ from conventional EFE GR and if so, what the key assumption that drives the difference could be. Since neither of these physicists are particularly notable in the field of GR, it appears that neither has garnered much commentary from peers or reactions in the literature to analyze these claims. Both physicists are make very conservative and modest tweaks to the conventional assessments of GR, so if there are important distinctions they must be fairly subtle ones.

Last edited: Sep 2, 2014
11. Sep 2, 2014

### Staff: Mentor

I wouldn't put it this way. I'm only considering one "formulation" of GR, which is just GR. The point is that the words "gravity" and "gravitate" can correspond to different particular aspects of that one formulation; and the answer to the question "does gravity gravitate?" depends on which particular aspects you use those words to refer to. In this sense, I suppose it could be viewed as a matter of "semantics", but that's true of *any* discussion of a physical theory using ordinary language; to really have an unambiguous discussion you need to use math, so that you can refer precisely to particular concepts. But in any case, only one theory and one set of phenomenological predictions is under discussion.

I hadn't seen this before so I'll have to take some time to read through it. On a quick skim, it's not clear to me whether the theory he is using is standard GR, or includes some extra ingredients. If it's the latter, then his analysis and mine are using different theories, so of course it's possible for us to get different answers.

I'll defer more detailed comment on this until I've read the paper through, but I think it's worth noting that, for a spherically symmetric gravitating body such as a star, the Komar mass ends up being equal to what you would get if you just did a "naive" Newtonian integral of the mass density over the Euclidean volume of the body. The two "relativistic" contributions--the fact that pressure also gravitates, so it has to be included in a proper integral (as it is in the Komar mass), and the fact that spacetime is curved, so the integration measure has an extra factor in it (as it does in the Komar mass)--end up cancelling each other out. This could be interpreted as "no net self-gravitation", I suppose; but I'm not sure that's how I would describe it. (Again, this is an example of what happens when you use ordinary language instead of math.)

I don't see this at all; the ADM mass, as you say, involves taking a limit as the 2-sphere surface of integration goes to spatial infinity. That would seem to make *no* assumption about the source mass's dimensions at all, except that they're finite. (Btw, the ADM mass is perfectly well-defined for non-spherically-symmetric systems; the only requirement is that the spacetime is asymptotically flat.)

Neither one does. The Bondi mass only requires asymptotic flatness, just like the ADM mass. The Komar mass only requires a stationary spacetime, i.e., a timelike Killing vector field. (Technically, it also requires that the stress-energy tensor vanishes fast enough as you go to spatial infinity if you want the integral to be finite.)

First of all, carrier bosons don't "localize" the energy except within the limits of the uncertainty principle, which effectively means it's only localized on the scale of one wavelength. That's the same amount of effective localization that the classical GR analysis gives you.

Second, modeling gravity as a massless spin-2 field assumes a background spacetime that is non-dynamical--the usual assumption is a Minkowski background, but one can also do it using, say, Schwarzschild spacetime as a background. If there is an issue with this way of modeling gravity, I think it's this fact, that the background spacetime is non-dynamical, since that's in conflict with the EFE, at least as it's usually interpreted.

What theories are these? Can you give any examples?

I would say it's the former.

I would say it comes from the same underlying features that you referred to above (background independence/equivalence principle).

12. Sep 2, 2014

### Staff: Mentor

On reading through this paper, I see that one consequence of this definition is that gravitational waves carry zero energy-momentum. Since gravitational waves can do work, a definition that assigns them zero energy doesn't seem very useful. (Note the argument along similar lines, due to Feynman, in the third initial post in this thread.)

13. Sep 2, 2014

### ohwilleke

What I am really asking amounts to whether there is a "no go theorem" that states that it is not possible in principle to devise a classical field theory that is exactly equivalent in all respects to EFE of GR, or if the non-localization of gravitational energy makes it inherently impossible to do so.

The reason that I wonder if this is possible is that in a classical field theory, the classical field (in this case, a gravitational field) is generally formulated algebraically and has a well defined value (not necessarily a scalar) at every point in any given reference frame (although it numerical value may appear different in different reference frames pursuant to a set of coordinate transformation that is consistent in any reference frame).

Similarly, is there a "no go theorem" that makes it impossible to formulate a geometrical way of expressing GR that is exactly equivalent in all respects to EFE, in which gravitational energy made manifest in space-time curvature can be localized to any desired degree of precision?

I agree that this is obviously the case in the case of a carrier boson. I am very unclear why classical GR analysis gives you that much effective localization.

My understanding is that Deur is looking at the classical limit, the EFE, of a massless, spin-two field that couple not only to all other "matter" fields, but also to "itself", and is choosing a novel set of simplifying assumptions to produce a set of equations that is mathematically tractable in what naively appears to be a sensible, although unconventional manner.

The only assumptions he seems to make are that gravitons self-couple to each other with the same magnitude that they would to any other matter or energy particle of identical mass-energy, and that gravitational field energy is localized to a "real" graviton to the extend permitted by the uncertainty principle.

I am not convinced, however, that the classical limit of a quantum gravity theory formulated in this way really is the EFE, as commonly claimed, as opposed to looking almost identical but having some subtle difference.

Last edited: Sep 2, 2014
14. Sep 2, 2014

### ohwilleke

Good point.

15. Sep 2, 2014

### WannabeNewton

Right I get that but I don't see exactly how the argument proves this (first paragraph): https://www.physicsforums.com/showpost.php?p=4839192&postcount=4

16. Sep 2, 2014

### Staff: Mentor

GR, with the EFE, *is* a classical field theory. See below.

GR satisfies this requirement: the field is the metric tensor, $g_{\mu \nu}$, which has a well-defined value (a second-rank tensor value) at every point.

I don't see how there could be, since the Nikolic paper you linked to gives an obvious counterexample: just interpret the Einstein tensor as "gravitational energy made manifest in spacetime curvature". The fact that this interpretation has a consequence that seems obviously unrealistic to me (it requires gravitational waves to carry zero energy) doesn't mean it isn't valid mathematically.

The wavelength of a gravitational wave gives the length scale on which it can have measurable physical effects (such as heating up a detector). The measurable effects are what enable us to localize the energy carried by the wave (since they localize the energy manifested in the measurable effects).

I believe this isn't just "commonly claimed"; I think it's proven (by Deser sometime around 1970, based on previous work done throughout the 1960's by a number of people including Feynman--IIRC the Deser paper has been linked to before here on PF, I'll try to find the reference). That's why I think different phenomenological predictions require a different underlying theory.

17. Sep 3, 2014

### ohwilleke

The notion would be that the phenomenological predictions aren't different, and that instead he's just evaluating a question that others have failed to ask and hence have failed to notice has an interesting answer.

The other possibility is that there is a subtle error in how the approximation is done.

18. Sep 10, 2014

### tzimie

Can GW create a BH?

GW carry information: theoretically, we can encode some useful information and send a transmission using GW. Then, if we cross all these transmissions in relatively small region, we can saturate a Bekenstein bound, so an absolutely *empty* area of space-time *should* become a BH. Am I right?

19. Sep 10, 2014

### Staff: Mentor

I'm not sure. Intuitively, as you say, it seems like we ought in principle to be able to focus enough GWs into a small enough space to create a BH. (In practice, of course, this would take technology *far* in excess of what we have now or in the foreseeable future.) However, I don't know if anyone has ever tried to investigate this seriously. I don't think any analytical solutions are known for this (the GW solutions I'm aware of focus on weak GWs, whereas you would need very strong ones to form a BH); I think there have been some numerical simulations of colliding GWs, but not with a view to seeing whether a BH forms. Perhaps others more knowledgeable can chime in here.

20. Sep 10, 2014

### martinbn

This is a bit long to read, but may be relevant.

The Formation of Black Holes in General Relativity

Abstract: The subject of this work is the formation of black holes in pure general relativity, by the focusing of incoming gravitational waves. The theorems established in this monograph constitute the first foray into the long time dynamics of general relativity in the large, that is, when the initial data are no longer confined to a suitably small neighborhood of Minkowskian data. The theorems are general, no symmetry conditions on the initial data being imposed.

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