# Insights Does Gravity Gravitate? Part 2 - Comments

1. Sep 23, 2015

### Staff: Mentor

2. Sep 24, 2015

### vanhees71

Thanks for the very clear explanation concerning the mass parameter in the Schwarzschild metric. I've a very small correction, the trace of the energy-momentum tensor is
$$T=g_{\mu \nu} T^{\mu\nu}={T^{\mu}}_{\mu}={T^0}_0+{T^1}_1 + {T^2}_2 + {T^3}_3.$$
This is different from the sum with two lower indices!

3. Sep 24, 2015

### exponent137

Peter,
1,
In the last two equations you wrote M_0=integral..., and M=integral...
$M_0=\int dV..$
$M=\int dV..$
Is this correct? I suppose that both left sides should be the same?
2.
I suppose that M_0 is mass of the earth if gravity is switced off, but, M_{ADM} and M are valid, when gravity is swiched on?
3
More simply said, why gravity does not gravitate: Stress energy tensor is proportional to mass, which has gravity switched on, it is not switced off?
4
Do you writte this blog before my question in the first blog? :)
Best Regards

4. Sep 25, 2015

### Staff: Mentor

Yes, agreed, that's what's in the article.

What sum are you talking about here?

5. Sep 25, 2015

### Staff: Mentor

No, they are two different integrals for two different quantities, and as the article notes, those quantities are not equal.

Not really. "The mass of earth if gravity is switched off" is meaningless, since there is no way to switch gravity off. $M_0$ is a "naive" way of trying to calculate the mass of the Earth by adding up the masses of all the little pieces of matter inside it. If I were to try to describe that calculation in ordinary language, it would be something like this: we take the density of matter in each infinitesimal volume element of the Earth, and multiply it by the volume of that volume element, calculated under the assumption that the geometry of space is Euclidean, to get the mass of matter in the infinitesimal volume, and then add up all the masses to get the total mass of the Earth.

The reason this calculation gives the wrong answer is that the geometry of space is not Euclidean, so we are calculating the physical volume of an infinitesimal volume element incorrectly. Correcting this gives us the integral for $M$ instead of $M_0$.

Yes. Both of my "does gravity gravitate?" posts (plus a third one that will be posted shortly) were written as blog posts back when PF allowed individual users to have blogs. Those old blogs were discontinued a while back, but PF Insights was intended to give a similar venue for people to write articles, so I am posting these there.

6. Sep 26, 2015

### exponent137

1
Yes in reallity, gravity cannot be switched off, but for easier imagination it is useful to do this. I think that your Euclidean space does the same?
2
Let us assume that density of earth is uniform. Does this uniformity remains, when we go into Euclidean space, or oppositely, when we assume this in Euclidean space and the we continue in curved spacetime?
3
Tensor for gravitational energy does not exist and in non-quantum gravity "gravitons" does not interact between themselves. So I think that your explanation will clarify this.

7. Sep 26, 2015

### haushofer

Already at the classical level GR is a selfinteracting field theory,exponent137.

A nice exercise could be to write these mass-definitions in terms of Newton-Cartan geometry and see if you get what you expect.

I must say I'm starting to enjoy these insights more and more. Keep them coming y'all! :D

8. Sep 26, 2015

### exponent137

In Newtonian gravity, self gravitational interaction among different parts of earth gives smaller effective mass of earth because of negative gravitational energy. I suppose that inclusion of GR gives in principle the same result? I suppose that other small corrections are not important according to the principles which are important in this debate? So, what is this principle in GR which says that gravity does not gravitate? Maybe because we can say that curved spacetime replaces the meaning of self-gravitational energy?

I do not know enough Newton-Cartan geometry. https://en.wikipedia.org/wiki/Newton–Cartan_theory Is something better than this link?

9. Sep 26, 2015

### Staff: Mentor

But you can't do it. The laws of physics include gravity; asking what happens if gravity were switched off is asking what happens if you violate the laws of physics. There's no way to answer that, because the only basis we have for giving an answer is the laws of physics.

No. It's just an incorrect assumption that then needs to be corrected. It has no physical meaning.

Again, there is no way to answer this, because it is asking what happens if you violate the laws of physics. The laws of physics for static gravitating bodies say that space is not Euclidean inside and around them.

That's correct. But the stress-energy tensor I talk about in the articles does not include any gravitational energy. It's just the standard SET derived from all non-gravitational fields.

10. Sep 26, 2015

### Staff: Mentor

No, it doesn't. In Newtonian gravity, the mass of the Earth is just put in "by hand"; it's not derived from any of the first principles of the theory. Even if you view the Earth's mass as the sum of the masses of all the little pieces of matter making up the Earth, in Newtonian theory those masses just sum up directly; there's no correction factor in the sum for gravitational binding energy, the way there is in GR--i.e., in Newtonian theory there is no analogue to the integral $M$ as opposed to $M_0$.

The only "negative gravitational energy" in Newtonian theory is the negative gravitational potential energy of a test object moving in the gravitational field of a massive body like the Earth, according to the standard definition where the potential energy is zero at infinity.

Inclusion of GR is the only way to get that result, by adding the correction factor that distinguishes the integral $M$ from the "naive" (incorrect) integral $M_0$. As above, there is no analogue of that correction factor in Newtonian theory.

Have you read the first article in the series? It explains all this.

11. Sep 26, 2015

### vanhees71

It's directly under the formula $R_{ab}=\ldots$. It's of course just a simple typo, nothing really serious.

12. Sep 26, 2015

### Staff: Mentor

Ah, I see now. Fixed. Thanks!

13. Sep 26, 2015

### exponent137

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 ... :)

14. Sep 26, 2015

### Staff: Mentor

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?

15. Sep 26, 2015

### Staff: Mentor

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.

16. Sep 27, 2015

### exponent137

https://en.wikipedia.org/wiki/Mass_...e_Newtonian_limit_for_nearly_flat_space-times
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 do not understand you, when we put little pieces together, they obtain gravitational self energy? Why not?

Last edited: Sep 27, 2015
17. Sep 27, 2015

### Staff: Mentor

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 $M_0$, except that 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 see the binding energy in the calculation. So, relativistically speaking, I wouldn't say we are "replacing" binding energy with curvature; I would say we are equating the 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, minus the 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.

18. Sep 29, 2015

### exponent137

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 $...\xi^a\xi_a...$. 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.

Last edited: Sep 29, 2015
19. Sep 29, 2015

### Staff: Mentor

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 the concepts GR 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 not flat; 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.

20. Sep 30, 2015

### exponent137

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 $R_{\alpha\beta}$? 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 http://arxiv.org/abs/gr-qc/9712019, where this is calculated?

21. Sep 30, 2015

### Staff: Mentor

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.

22. Sep 30, 2015

### Staff: Mentor

Ok, here is the promised follow-up post on the case of a spherically symmetric mass of uniform density. I should say at the outset that digging into this has made me realize that the interpretation of $M_0 - M$ as "gravitational binding energy" is at best heuristic; the reasoning involved has a lot of caveats in it, and I have updated the Insights post accordingly (I have also updated it with some other corrections arising from what follows). However, this also gives me some good material for the next follow-up post in the series, so it all works out in the end.

We are using units in which $G = c = 1$. As I said, this case has a known exact solution for the metric, which is as follows:

$$ds^2 = - \left[ \frac{3}{2} \sqrt{1 - \frac{8}{3} \pi \rho R^2} - \frac{1}{2} \sqrt{1 - \frac{8}{3} \pi \rho r^2} \right]^2 dt^2 + \frac{1}{1 - \frac{8}{3} \pi \rho r^2} dr^2 + r^2 d\Omega^2$$

where $\rho$ is the density (and is constant) and $R$ is the radial coordinate of the object's surface (i.e., for $r > R$ there is vacuum). For convenience, we define a constant $k = \frac{8}{3} \pi \rho$; the metric then becomes

$$ds^2 = - \left[ \frac{3}{2} \sqrt{1 - k R^2} - \frac{1}{2} \sqrt{1 - k r^2 } \right]^2 dt^2 + \frac{1}{1 - k r^2} dr^2 + r^2 d\Omega^2$$

The Killing vector field $\xi^a$ is just $\partial / \partial t$ in these coordinates, and its norm is $\sqrt{\xi^a \xi_a} = \sqrt{g_{ab} \xi^a \xi^b} = \sqrt{g_{tt}}$, so

$$\sqrt{\xi^a \xi_a} = \left[ \frac{3}{2} \sqrt{1 - k R^2} - \frac{1}{2} \sqrt{1 - k r^2 } \right]$$

The 4-velocity $u^a$ is equal to $\xi^a / \sqrt{\xi^a \xi_a}$, so its only nonzero component is

$$u^0 = \frac{1}{\left[ \frac{3}{2} \sqrt{1 - k R^2} - \frac{1}{2} \sqrt{1 - k r^2 } \right]}$$

However, we won't actually need this in this particular case, as we'll see in a moment.

Next, we have the stress-energy tensor $T_{ab}$ and its trace $T$; the quantity that appears in the mass integrals is $\left( 2 T_{ab} - g_{ab}T \right) u^a u^b = \left( \rho + 3 p \right)$, where $p$ is the pressure. (Note that this is really a way of stating the physical definition of $\rho$ and $p$, because we are contracting the stress-energy tensor with the 4-velocity--this is why we don't need to actually know the expression for the 4-velocity components.) However, even though the density $\rho$ is constant, the pressure $p$ is not; it increases as $r$ decreases, from $p = 0$ at the surface to some positive value $p_c$ at the center $r = 0$. So we need the equation for $p(r)$, which turns out to be

$$p = \rho \frac{\sqrt{1 - k r^2} - \sqrt{1 - k R^2}}{3 \sqrt{1 - k R^2} - \sqrt{1 - k r^2}}$$

Finally, we need to look at the volume element $dV$. In flat spacetime, this volume element, after integrating over the angular directions, would simply give $4 \pi r^2 dr$, i.e., the area of a 2-sphere at radius $r$ times the radial differential. However, in curved spacetime, this coordinate volume element is not actually the physical volume element; there is an extra factor of $\sqrt{g_{rr}}$, to account for curvature. (I should have made this clearer in previous posts, since it has implications for the physical meaning of $M_0$; see further comments below.) So we have

$$\int dV = \int 4 \pi r^2 \frac{1}{\sqrt{1 - k r^2}} dr$$

We now have all we need to evaluate the integrals. We will do the integral for $M$ first, for reasons which will shortly become apparent; it is

$$M = \int \sqrt{\xi^a \xi_a} \left( 2 T_{ab} - g_{ab} T \right) u^a u^b dV = \int_0^R \left[ \frac{3}{2} \sqrt{1 - k R^2} - \frac{1}{2} \sqrt{1 - k r^2 } \right] \left( \rho + 3 p \right) 4 \pi r^2 \frac{1}{\sqrt{1 - k r^2}} dr$$

Substituting for $p$ gives

$$M = \int_0^R \left[ \frac{3}{2} \sqrt{1 - k R^2} - \frac{1}{2} \sqrt{1 - k r^2 } \right] \rho \left[ 1 + 3 \frac{\sqrt{1 - k r^2} - \sqrt{1 - k R^2}}{3 \sqrt{1 - k R^2} - \sqrt{1 - k r^2}} \right] 4 \pi r^2 \frac{1}{\sqrt{1 - k r^2}} dr$$

The messy expressions involving the square roots turn out to all cancel, leaving

$$M = \int_0^R 4 \pi \rho r^2 dr = \frac{4}{3} \pi \rho R^3$$

Notice that this is what we would have expected in Newtonian physics! That means I actually told a bit of fib in an earlier post, when I said $M$ was "different" in GR vs. NP; I had forgotten that in GR, pressure gravitates. We'll discuss that further below. However, it was only a bit of a fib, because even though this result formally looks the same as the Newtonian result, the way we arrived at it makes clear that we did not just "naively" integrate $\rho$ over the Euclidean volume of the object. We integrated a much more complicated expression, involving effects of both pressure (in addition to density) and spacetime curvature, in which those effects just happened to cancel in the right way. Again, more on this below.

The fact that $M$ came out this way does illustrate, though, what I said in an earlier post about the difference between $M_0$ and $M$ being negligible if we really adopt the weak field, slow motion limit. In that limit, if we do indeed ignore spacetime curvature, and if we also ignore pressure, because it will be negligible compared to energy density in this limit, then the integral for $M_0$ will in fact look exactly like the above--and in fact it will also be the same as the "naive" integral that we started out with in the Insights post! This is because $T_{ab} u^a u^b = \rho$--this is the physical definition of $\rho$--and so $M_{naive}$ from the post also looks exactly the same as the integral for $M$ above. This is one of the key caveats in trying to interpret what $M_0$ means, physically.

If we don't ignore pressure, but we do leave out the effects of spacetime curvature, we can do the integral for $M_0$ and expect to get something different (and hopefully larger) than what we got for $M$ above; but we have to consider, first, what volume element $dV$ we should use for this case. If we are assuming flat spacetime (or at least assuming that the effects of space curvature are negligible because the field is weak enough), then we should not have the extra factor of $\sqrt{g_{rr}}$ in the volume element. (Again, we'll discuss this further below.) So we have:

$$M_0 = \int \left( 2 T_{ab} - g_{ab} T \right) u^a u^b dV = \int_0^R \left( \rho + 3 p \right) 4 \pi r^2 dr = \int_0^R \rho \left[ 1 + 3 \frac{\sqrt{1 - k r^2} - \sqrt{1 - k R^2}}{3 \sqrt{1 - k R^2} - \sqrt{1 - k r^2}} \right] 4 \pi r^2 dr$$

Here we don't have quite the same nice cancellation, but we can still simplify quite a bit:

$$M_0 = \int_0^R \rho \left[ \frac{2 \sqrt{1 - k r^2}}{3 \sqrt{1 - k R^2}} \frac{1}{1 - \frac{\sqrt{1 - k r^2}}{3 \sqrt{1 - k R^2}}} \right] 4 \pi r^2 dr$$

This is still quite messy, but we can use the fact that, for weak fields and slow motion, $k R^2 < k r^2 << 1$, so we can expand out the denominators and square roots and drop terms of second and higher order in $k$ to obtain, after some algebra:

$$M_0 = M + \frac{4}{27} \frac{M^2}{R}$$

I'm not entirely sure all of the algebra is correct here, but we expect the leading term to be $M$, and the correction term is based on how the various square roots and denominators expand out and then get integrated over $r$. This correction, of course, is quite a bit smaller than the expected Newtonian value of $3 M^2 / 5 R$, which at least shows, as I said in a previous post, that we should not expect the Newtonian values to always be good estimates of what a GR calculation will show, even in the weak field, slow motion limit.

However, let's now discuss what, if anything, the $M_0$ calculation we just did actually means, physically. On its face, the calculation is certainly "wrong", in the sense that we ignored spacetime curvature but did not ignore pressure, even though the effects of both are expected to be of the same order (we know that because those effects cancelled exactly in the integral for $M$ above). But, as we saw, if we ignore both spacetime curvature and pressure, we get the same answer as $M$ above--i.e., we don't even see any "gravitational binding energy" at this order of approximation. And if we include both spacetime curvature and pressure, we get $M$ by the route we got it above, because their effects cancel.

But let's now consider the alternative interpretation of what $M_0$ might mean, as I described in a previous post. Imagine that we have a bound object, with mass $M$ as calculated by the correct integral above, and we now "disassemble" it into its constituent particles, moving each particle out to infinity (meaning, in practice, far enough away from all other particles that gravity between them is negligible). We will obviously have to add energy to the system to do this; and at the end of it, we will have a system composed of a cloud of widely separated particles, each of which has negligible self-gravity. This cloud will have some density $\bar{\rho}$, and will be contained within a spherically symmetric region of some radius $\bar{R}$, and its total mass should be given by

$$\bar{M} = \int_0^\bar{R} 4 \pi \bar{\rho} r^2 dr = \frac{4}{3} \pi \bar{\rho} \bar{R}^3$$

This should be true because the cloud is so dispersed that its spacetime curvature should really be negligible, and its pressure should also really be negligible since the particles are too far apart to interact with each other, so we really can use the "naive" integral in this case to get the right answer.

Then, if $E$ is the total energy we had to add to the original object to "disassemble" it into the widely dispersed cloud, we should have

$$E = \bar{M} - M = \frac{4}{3} \pi \left( \bar{\rho} \bar{R}^3 - \rho R^3 \right)$$

In other words, on this intrepretation, the difference between $M_0$ (which we are calling $\bar{M}$ now) and $M$ arises from the fact that, when we "disassemble" a bound object, the product of its density and its radius cubed increases--the density doesn't decrease as much as the cube of the radius increases. But even considering this interpretation requires relativistic energy-mass equivalence; as I've said before, in Newtonian physics, the energy $E$ being added does not change the total mass, and a Newtonian calculation would predict $\bar{M} = M$, i.e., it would predict that, as an object is "disassembled", its density should decrease in exact proportion to the increase in the cube of its radius. In GR, however, this doesn't happen; adding the energy $E$ really does add mass to the system, and that added mass shows up in the slower decrease of the density as the cube of the radius increases, so $\bar{M} > M$.

This has been a long post, but hopefully it has helped to clarify what is going on in this scenario. @exponent137, your original question has sparked a very good discussion, and that's the mark of a good question!

Last edited: Sep 30, 2015
23. Oct 2, 2015

### exponent137

I think that pressure should be treated differently:

I see in your article a good way to understand why gravitational tensor does not exist. This means that gravitational self energy does not exist, this means that binding gravitational energy in Newtonian physics (NP) should be explained differently in GR. In the same way, as you said - by integration of energy in the curved spacetime.

But, it should be respected that Newtonian gravity is based only on masses, not on energy, momentum. and pressure, as GR is based. This means, when we switch to GR, we should to operate only with energy, not with momentum. Thus we need an example where pressure is not present. Such examples are, for instance, (1) two point masses or (2) spherically distributed mass, distributed only on surface, interior is empty. In this example, your $\xi^a\xi_a=g_{tt}$ equals $1-2GM/(rc^2)$what can give a correct result. $M$ in NG is replaced by rest energy $Mc^2$ in GR. In this examples, your calculation with $M_0$ and $M$ gives a correct result, I suppose. Curvature factor $1-2GM/(rc^2)$ takes over binding energy.

Yes, we should be precise about some terms, but I think that this is a correct direction, otherwise there any other visual explanation does not exist.

24. Oct 2, 2015

### Staff: Mentor

The first example can have negligible pressure, but the second can't. A spherical shell with an empty interior can only support itself against its own gravity through non-negligible pressure.

For the first example, two point masses, the problem is that it's not stationary; two point masses with no other stress-energy present will fall together. But we can only compute the Komar mass in a stationary spacetime.

Fundamentally, the latter point is the reason we probably can't ignore pressure in any relevant example; any static object with non-negligible gravity has to support itself against its own gravity, and that means it has to have non-negligible pressure.

The only possibility I see for a pressure-free example would be some sort of rotating ring of "dust" (i.e., pressure-free matter) that "supports" itself by rotating at a fast enough rate to balance its own gravity. (It would have to be a ring so that it is axially symmetric; otherwise the spacetime won't be stationary). Unfortunately, I don't know of an exact solution for this case.

Which example? If you mean the second (spherically symmetric shell), see above.

25. Oct 3, 2015

### exponent137

I think spherically symmetric shell or two point masses, $2 \times m/2$.
I need this only for imagination, how binding energy in NP disapears in GR. Thus, it is enough, that formula is valid only one moment. Thus $g_{tt}=1-Gm/(c^2r)$, what includes binding energy . In the case of spherical shell, mass is pressureless dust.

I think that inclusion of pressure is similarly general relativistic as inclusion of moving masses. But they include a factor $1+\beta^2$. (This factor is a reason that bending of a ray close to sun in GR has a factor 2 according to NP calculation.)
Thus, your $M_0$ is too relativistic, if a pressure is included.

1. I obtained $g_{tt}=1-Gm/(c^2r)$, where we have $r^{-1}$, you obtained like $r^2$, if we look dimensionally. Why this?
2. Your $M_0-M$ is much smaller than binding energy. How inclusion of pressure causes this, as your derivation shows? Is additon of mass because of pressure larger at $M$ than at $M_0$?