# Does Gravity Gravitate? Part 2

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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:

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

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

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. 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:

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

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 $M_0$ and $M$ is sometimes 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.) 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. ;)

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30 replies
1. PeterDonis says:

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. :wink:

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:

$$ds^2 = – left[ frac{3}{2} sqrt{1 – frac{8}{3} pi rho R^2} – frac{1}{2} sqrt{1 – frac{8}{3} pi rho r^2} right]^2 dt^2 + frac{1}{1 – frac{8}{3} pi rho r^2} dr^2 + r^2 dOmega^2$$

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

$$ds^2 = – left[ frac{3}{2} sqrt{1 – k R^2} – frac{1}{2} sqrt{1 – k r^2 } right]^2 dt^2 + frac{1}{1 – k r^2} dr^2 + r^2 dOmega^2$$

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!

2. exponent137 says:

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 $xi^axi_a=g_{tt}$ 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.

3. PeterDonis says:

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.

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.

In this example, your $xi^axi_a=g_{tt}$ equals $1-2GM/(rc^2)$ what can give a correct result.

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

4. exponent137 says:

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

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 $M_0$?

5. PeterDonis says:

I need this only for imagination, how binding energy in NP disapears in GR.

I’m sorry, I don’t think the path you are trying to go down is a fruitful one, so I really can’t offer any help. You are trying to use Newtonian concepts in a regime where they are not valid. To me, the correct response is “don’t do that”.

Thus, it is enough, that formula is valid only one moment.

No, it isn’t. The concept of “gravitational binding energy” only makes sense in the first place in a stationary spacetime. It can’t be valid at one moment but not at another.

I obtained $g_{tt}=1-Gm/(c^2r)$

That is $g_{tt}$ for the vacuum region surrounding a spherically symmetric gravitating mass. It is not valid inside the mass. If you are going to do the Komar mass integral, which is what I was doing, you have to do it inside the mass; integrating over the vacuum region just gives zero.

where we have $r^{-1}$, you obtained like $r^2$, if we look dimensionally

I don’t understand what you’re referring to here, but in general, there is no requirement that $g_{tt}$ have a $1/r$ dependence. The solution I wrote down is a well-known exact solution for the interior of a spherically symmetric mass of uniform density.

Your M0−MM_0-M is much smaller than binding energy.

Much smaller than the Newtonian “binding energy” you referred to, which, as I’ve already pointed out several times, is based on Newtonian assumptions and should not be expected to be the same as what you get from a GR calculation. I’ve said this before, and I don’t know that there’s much point in continuing to repeat it. As I said above, I simply don’t think the path you are trying to pursue is a fruitful one; GR is not Newtonian gravity and you can’t just ignore the differences between them.

Is additon of mass because of pressure larger at $M$ than at $M_0$?

I don’t know how to answer this. Perhaps one could try re-doing the $M$ and $M_0$ integrals without the pressure term included, and seeing how they compare. But this would have no physical meaning. Pressure gravitates in GR, and you can’t ignore its contribution.

6. PeterDonis says:

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.)

The $1 + beta^2$ factor in the equation for light bending has nothing to do with the pressure contribution to mass. They’re different things. The analogies you are trying to draw are simply not valid; as far as I can see, the correct response, once again, is “don’t do that”.

7. exponent137 says:

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

Is this correct? According your next formulae for $M$ and $M_0$, I expect still factor $frac{1}{sqrt{1 – k r^2}}$? Because your formule for $M$ and $M_0$ are distinct for the above $sqrt{xi^a xi_a}$ and for $frac{1}{sqrt{1 – k r^2}}$.

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

Substituting for $p$ gives

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

8. exponent137 says:

The $1 + beta^2$ factor in the equation for light bending has nothing to do with the pressure contribution to mass. They’re different things. The analogies you are trying to draw are simply not valid; as far as I can see, the correct response, once again, is “don’t do that”.

I did not try to say so as you understand. Similarly I did not try to say about binding energy, as you understand, but it is no matter, the most important is that you derived above for uniformly distributed sphere, so that I better understand GR.
Can you give any link for your integral for $M$ in general?

9. exponent137 says:

I see that these formulae are included: [URL]https://en.wikipedia.org/wiki/Komar_mass[/URL] and in your 2. blog. So, this is clear.

But, I am interested in formula $S_g =frac{1}{16pi}int d^4xsqrt{-g} R$ in the first blog. It is a simple formula, but how to construct it?

10. vanhees71 says:

This is the Hilbert action, and it’s unique (up to a cosmological constant, which is not included here) in being generally covariant and leading to 2nd-order equations of motion. This is so, because $R$ is the only scalar you can build from the pseudometric tensor that is only linear in the 2nd derivative with coefficients containing no derivatives, which means that via partial integration you can show that the action is a functional of a Lagrangian that only depends on the $g_{mu nu}$ and its first derivatives.