# Does Gravity Gravitate? Part 2

In the first post of this series, 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. In this second post, we’ll 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:

[tex]ds^2 = – \left( 1 – \frac{2M}{r} \right) dt^2 + \frac{1}{1 – 2M / r} dr^2 + r^2 d\Omega^2[/tex]

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”:

[tex]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[/tex]

Here [itex]g_{ij}[/itex] and [itex]g^{ij}[/itex] 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), [itex]S[/itex] is a 2-sphere surrounding the central body, and [itex]n^k[/itex] is an outward-pointing unit vector that is normal to the 2-sphere. The limit is taken as [itex]S[/itex] goes to spatial infinity, [itex]i^0[/itex].

For the idealized case of Schwarzschild spacetime, it’s fairly straightforward to show that [itex]M_{ADM} = M[/itex]; 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:

[tex]M_{naive} = \frac{1}{4 \pi} \int dV T_{ab} u^a u^b[/tex]

where [itex]dV[/itex] is the volume element of a spacelike slice of constant time, and [itex]u^a[/itex] 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 [itex]T_{ab}[/itex] itself is what belongs in the integrand, without stopping to think about how [itex]T_{ab}[/itex] 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

[tex]G_{ab} = R_{ab} – \frac{1}{2} g_{ab} R = 8 \pi T_{ab}[/tex]

We saw in the previous post that putting [itex]G_{ab}[/itex] 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* [itex]G_{ab}[/itex] directly. What we actually measure is [itex]R_{ab}[/itex], 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 [itex]R_{ab}[/itex] on the LHS, with the RHS independent of [itex]R_{ab}[/itex] or [itex]R[/itex] (which is the trace of the Ricci tensor). It turns out that this can be done pretty easily; the result is

[tex]R_{ab} = 4 \pi \left( 2 T_{ab} – g_{ab} T \right)[/tex]

where [itex]T[/itex] is the trace of the stress-energy tensor, i.e., [itex]T = T^0{}_0 + T^1{}_1 + T^2{}_2 + T^3{}_3[/itex].

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

[tex]M_0 = \int dV \left( 2 T_{ab} – g_{ab} T \right) u^a u^b[/tex]

However, there is still something wrong. If we compute [itex]M_{0}[/itex] for an object like the Earth, we will get the wrong answer; our answer will be *larger* than the externally measured mass [itex]M[/itex] that appears in the line element. What gives? Well, we forgot one other thing: spacetime is curved. The volume element [itex]dV[/itex] 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. Also, we have to account for the fact that the “rate of time flow” is not constant everywhere, since we are essentially computing an energy (mass and energy are equivalent in relativity) and the “rate of time flow” is therefore involved.

In this particular case, it’s easy to make the corrections because the spacetime is static; I won’t go into detail about this, but the upshot is that we just need to include two extra correction factors in the integral, like so:

[tex]M = \int \sqrt{\tilde{V}} dV \sqrt{\xi^a \xi_a} \left( 2 T_{ab} – g_{ab} T \right) u^a u^b[/tex]

where ##\tilde{V}## is a “volume correction factor” (in Schwarzschild coordinates in a spherically symmetric spacetime, it will just be ##\sqrt{g_{rr}}##) and ##\xi^a## is the timelike Killing vector field of the spacetime (which must exist because the spacetime is static; in Schwarzschild coordinates the factor ##\sqrt{\xi^a \xi_a}## will just be ##\sqrt{g_{tt}}##).

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 [itex]M_0[/itex] and [itex]M[/itex] is sometimes referred to as “gravitational binding energy”, and the fact that it is there–that [itex]M_0 – M[/itex] is not zero–is one thing that often prompts people to assert that “gravity gravitates”: that gravity “contributes” to the externally observed mass [itex]M[/itex] of a system. (Personally, I confess that I find this interpretation a bit strange: [itex]M[/itex] is *less* than [itex]M_0[/itex], so the “contribution of gravity” is *negative*. But it’s a common interpretation.) There are a number of caveats to interpreting ##M_0 – M## this way (some of which I intend to discuss in a follow-up post in this series), but as long as the system is not changing with time, it doesn’t really 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? We’ll save that for the next post. ;)

Ah, I see now. Fixed. Thanks!

Of course, laws of physics are defined by GR much more than with Newtonian physics.

At all, I needed analogy in Newtonian physics that I will understand math of GR.

Again, I needed analogy in Newtonian physics that I will understand math of GR.

If gravitational field of earth is very weak, ##M## in GR is almost equal to ##M_0-W_G##. At this, ##M_0## is the mass of earth in the Newtonian physics and ##W_G## is gravitational self energy.in Newtonian physics. Is this correct? If it is, then I understand why gravity does not gravitate.

P.:S But, GR is also not absolutely physically correct. The next theory is theory of quantum gravity … :)

I don’t think so; I think the mass of the Earth in Newtonian physics is equal to ##M## in GR, not ##M_0##. That’s because the mass of the Earth in Newtonian physics is defined by what appears in the Newtonian force equation, ##F = G M m / r^2##, and when we go to the GR version of that, in the weak field, slow motion limit, the ##M## that appears is the same as ##M## in GR, not ##M_0##. (Strictly speaking, the ##M## in the GR version of the force equation, in the weak field, slow motion limit, is ##M_{ADM}##, the ADM mass that I describe in the article; but it’s easy to show that, for the case of a body like the Earth, the ADM mass and the Komar mass, which is what ##M## is, are the same.)

Also, what do you think ##W_G## is? That is, how would you calculate gravitational self-energy in Newtonian physics?

True, but that doesn’t affect anything we’re talking about here. We are talking about a domain in which GR is already known, experimentally, to be correct.

I think about this:

[URL]https://en.wikipedia.org/wiki/Mass_in_general_relativity#The_Newtonian_limit_for_nearly_flat_space-times[/URL]

This is calculated in spacetime of Minkowski.

##M_0## means in principle sum of tiny mass pieces on big distances, from which we built earth. This is my definition of ##M_0## and definition in this link.

My question is only, if we calculate in curved spacetime (instead of Minkowski spacetime) by your mentioned procedure, then use of curved spacetime replaces use of binding energy.

I think that this is true, and in that case I understand why GR do not use gravitational energy on RHS – it is compensated by integration in curved space?

Gravitational binding energy is ##3/5 GM^2/R## for a uniform sphere, for instance.

[I]

[/I]

I do not understand you, when we put little pieces together, they obtain gravitational self energy? Why not?

As an approximation, yes.

It means that sum

calculated in the approximation that spacetime is flat. More precisely, it means, as I explained in post #5, that we take each infinitesimal volume of Earth and multiply it by the density of matter in that volume, to get the mass of that little piece of Earth. Then we add up all the little masses. But we calculate the physical volume of each infinitesimal piece in the approximation that spacetime is flat, so the volume is just ##dV##, the coordinate volume element.There is a sense in which this is true, yes. The integral ##M## is defined the same way as ##M_0##,

exceptthat we take spacetime curvature into account; we add up the masses of all the tiny pieces of the Earth (or any other body), the same way we did for ##M_0## above, by multiplying the volume of each little piece by the density of matter in that volume. But taking spacetime curvature into account means the physical volume of each little piece is not ##dV##; it’s ##dV## times an extra factor that accounts for the curvature. That factor is in general less than 1, so we end up with a mass ##M## that is less than ##M_0##.What you are saying is basically that, in the weak field approximation, assuming spacetime is approximately flat, we can get (approximately) the same answer for ##M## by computing the integral ##M_0##, and then subtracting the “Newtonian” gravitational binding energy, which, as you note, would be ##3 GM^2 / 5 R## for a sphere of uniform density). However, that can’t be true as you state it. To see why, let’s unpack and compare the two integrals in question. The integral from the Wikipedia page you linked to is

$$

E = int T_{00} dV = int rho dV

$$

where ##rho## is the energy density (which, since we are talking about the weak field, slow motion limit, is just the ordinary mass density). The integral from the Insights article for ##M_0## is

$$

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

$$

If we adopt appropriate coordinates, we have ##T_{00} = rho## as the only significant component of ##T_{ab}##, and ##u^0 = 1## as the only significant component of ##u^a##, and ##g_{00} T = rho##, so we do indeed get ##E = M_0## in this approximation. So far, so good.

However, now let’s look at the integral for ##M## from the Insights article, which is

$$

M = int sqrt{xi^a xi_a} left( 2 T_{ab} – g_{ab} T right) u^a u^b dV

$$

The problem is, in this approximation, ##xi^a xi_a approx 1##, so we end up with ##M = M_0##! In other words, in this approximation, the Newtonian “gravitational binding energy” is so small compared to the total energy ##E = M_0## that it is negligible–it doesn’t even show up in the calculation! (Try plugging in numbers for the Earth into the equation for gravitational binding energy that you gave; you will see that the answer comes out many orders of magnitude smaller than the rest energy of the Earth.)

In other words, in order to even see the difference between ##M_0## and ##M##, relativistically speaking, you have to go beyond the flat spacetime approximation, because that approximation makes the difference negligible. You have to at least include some curvature in order to even

seethe binding energy in the calculation. So, relativistically speaking, I wouldn’t say we are “replacing” binding energy with curvature; I would say we areequatingthe presence of binding energy with the presence of curvature. No curvature, no binding energy.The point I was trying to make is that, in Newtonian gravity, mass is an inherent property of an object, and energy and mass are not equivalent. So you can’t say that the actual, measured mass of the Earth is the sum of the masses of all the little pieces of Earth,

minusthe gravitational binding energy. You can’t combine masses and energies like that. In Newtonian gravity, the mass of the Earth is just the sum of the masses of all the little pieces of Earth, period. The gravitational binding energy is a separate thing; it is not any kind of “gravitational self-energy” that reduces the mass of the Earth, compared to the sum of the masses of all the little pieces.You can see that in the analysis I gave above; in the Newtonian approximation, the gravitational binding energy disappears. In order to make it appear at all, you have to take spacetime curvature into account. And in order to view the difference between ##M_0## and ##M## as a (negative) contribution of “gravitational self energy” to the total mass of the Earth, you have to view mass and energy as equivalent; but that requires relativity, it isn’t true in Newtonian gravity.

The problem here is only, that I wish to imagine GR with the help of Newtonian physics (NP), to better understand GR.

Gravitational self energy can be defined also in special relativity, it is not necessary to have GR, but anyway, it gravitational self energy can be assumed. Here it is only a question of visualization of NP, not proving of it. NP is easier to be visualized than GR.

If we want to go from NP to GR, we should say to NP good bye, otherwise we will return back. (Good bye = to find comparisons between GR and NP.)

What is a problem with this approximation with ##…xi^axi_a…##. What type of approximation we need that we obtain that ##M_0-M## is something like ##3GM^2/5R ##.

I suppose that ##M_0## formula should remain the same. Is your formula for ##M##, flat space approximation? I do not understand so.

Your blog gives much new information, but this makes me problems.

That’s often not a good strategy. GR is not Newtonian physics, and NP is often not a good guide to intuition for GR. GR has a Newtonian

approximation, i.e., we can use GR in this approximation to explain why the Newtonian equations work as well as they do within their domain of validity; but theconceptsGR uses are very different from the concepts NP uses, so if you are trying to understand the concepts, NP is not a help; it’s a hindrance.No, it can’t. There is no gravity in SR.

I don’t know that there is one. The ##3GM^2 / 5R## formula is derived using Newtonian assumptions, including the assumption that energy does not gravitate. That means that under these assumptions, there is no “gravitational self energy”; the gravitational binding energy ##3GM^2 / 5R## does not contribute to the mass of the gravitating object. So I would not expect there to be any approximation in GR that gives that Newtonian formula as a (negative) contribution to the mass of the object, which is what ##M_0 – M##, in the article, is trying to capture. (See further comments below.)

No, you have it backwards. The formula for ##M## is the correct GR formula. It is not an approximation. The formula for ##M_0## is an approximation assuming that spacetime is flat. However, this approximation is not even valid, really, because assuming that spacetime is flat is equivalent to assuming that there is no gravity and gravitational self-energy is zero. You can’t calculate the “gravitational binding energy” of an object using an approximation that says there is no gravity.

There is another viewpoint one can take about ##M_0##, suggested by the Wikipedia page you linked to. Consider this scenario: we have a very, very large cloud of very, very small particles–each individual particle is so small that its self-gravity is negligible, and the separation between every pair of particles in the cloud is so large (compared to their masses) that gravity between them is negligible. In this scenario, spacetime will be (approximately) flat, and the total mass of the cloud will be given by an integral like ##M_0##–basically, we can just add up the masses of all the particles, with no correction for spacetime curvature because it is negligible.

Now suppose we take this cloud and bring all its particles together to form a single gravitationally bound object. We do this in such a way that no energy is added to the cloud from the outside, and as the particles fall together and gain kinetic energy due to each other’s gravity as they get closer together, we extract that kinetic energy and remove it from the system (for example, by having the particles emit radiation to infinity). We end up with a single final object and a spacetime that is

notflat; it is curved. So if we compute the total mass of the object, we will get an integral like ##M##; we can’t just add up all the masses of the particles any more. We have to include a correction factor for spacetime curvature, since it now is not negligible.When we compare the two integrals, we find that ##M < M_0##, and what's more, we find that the difference is equal to the energy that we extracted and removed from the system and sent out to infinity. Or, equivalently, we could use that same amount of energy to reverse the process of forming the single object, moving each individual particle back out to a very large distance from all the others, and recreating the original condition of a very large cloud with negligible gravity between each particle. This is the process described in the Wikipedia article and used to compute, under Newtonian assumptions, the gravitational binding energy ##3 G M^2 / 5 R##.However, there is a big difference between how the difference ##M_0 -M## is interpreted in NP vs. GR. In NP, the total mass of the system is unchanged, either by the process of forming a single object from the cloud, or of "disassembling" the single object back into a widely dispersed cloud. So NP would predict that, if we had an object in orbit about this system, its orbit would be unchanged during the entire process--we could start the system off as a cloud, collapse it into a single object, then disassemble it into a cloud again, extracting energy and then putting energy back in, and the orbit of an object about the system would be unaffected.But in GR, when we extract energy from the system, we reduce its mass, because mass and energy are equivalent. So if we take a widely dispersed cloud and collapse it into a single bound object, extracting energy from it, we will change the orbit of an object orbiting about the system; the orbital parameters will change to those appropriate for a smaller mass. If we then disassemble the object, adding energy to the system to recreate the original cloud, the orbit of an object about it will change again, the orbital parameters changing back to those appropriate for the original, larger mass.In other words, NP and GR

make different predictions about what will happen in this scenario, and the GR prediction is right and the NP one is wrong. (We haven't run this exact scenario, of course, but we have observed similar processes enough to be sure that GR is correct and NP is wrong in this case.) So I would not expect the NP prediction for gravitational binding energy to be correct, since its prediction for the behavior of the total mass of the system is incorrect to begin with.Can you, please, calculate ##M-M_0## for one simplest example, maybe spherically distributed mass, or two point masses ##m## on a distance ##r##. I am interested in the first approximations, like gravitational field of earth or of sun. Or that you can begin calculation. Their expected binding energies are ##3m^2G/(5r)## or ##m^2G/r##.What are values for ##xi_a## and ##xi^a##? What are their relations with ##R## and ##R_{alphabeta}##? Is ##u^0## the speed of light? If the body is in rest, this is the only component of ##u##.

Can you show me in [URL]http://arxiv.org/abs/gr-qc/9712019[/URL], where this is calculated?

Sure, a good idealized case is a spherically symmetric mass of uniform density. (This case is not very realistic, but it has the advantage of having a known closed form solution for the metric, whereas a spherically symmetric mass of non-uniform density has to be solved numerically.) I’ll do a follow-up post with the details on this case.

The easiest way to answer all these questions is to just work through the solution of the case I gave above. Stand by for a follow-up post.

I don’t think it is. Carroll discusses the Schwarzschild solution, but I don’t think he discusses the mass integrals.