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PeterDonis submitted a new PF Insights post
Does Gravity Gravitate? Part 2
Continue reading the Original PF Insights Post.
Does Gravity Gravitate? Part 2
Continue reading the Original PF Insights Post.
Peter,vanhees71 said: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
$$T=g_{mu nu} T^{munu}={T^{mu}}_{mu}={T^0}_0+{T^1}_1 + {T^2}_2 + {T^3}_3.$$
This is different from the sum with two lower indices!
vanhees71 said:the trace of the energy-momentum tensor is
$$
T = g_{\mu \nu} T^{\mu \nu} = T^{\mu}{}_{\mu} = T^0{}_0 + T^1{}_1 + T^2{}_2 + T^3{}_3
$$
vanhees71 said:This is different from the sum with two lower indices!
exponent137 said:Is this correct? I suppose that both left sides should be the same?
exponent137 said:I suppose that M_0 is mass of the Earth if gravity is switced off
exponent137 said:Do you writte this blog before my question in the first blog? :)
Already at the classical level GR is a selfinteracting field theory,exponent137.exponent137 said: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.
Maybe, I should ask differently.haushofer said: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
exponent137 said:Yes in reallity, gravity cannot be switched off, but for easier imagination it is useful to do this.
exponent137 said:I think that your Euclidean space does the same?
exponent137 said:Let us assume that density of Earth is uniform. Does this uniformity remains, when we go into Euclidean space,
exponent137 said:Tensor for gravitational energy does not exist
exponent137 said:In Newtonian gravity, self gravitational interaction among different parts of Earth gives smaller effective mass of Earth because of negative gravitational energy.
exponent137 said:I suppose that inclusion of GR gives in principle the same result?
exponent137 said:what is this principle in GR which says that gravity does not gravitate?
It's directly under the formula ##R_{ab}=\ldots##. It's of course just a simple typo, nothing really serious.PeterDonis said:Yes, agreed, that's what's in the article.
What sum are you talking about here?
vanhees71 said:It's directly under the formula Rab=…R_{ab}=\ldots.
Of course, laws of physics are defined by GR much more than with Newtonian physics.PeterDonis said: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.
At all, I needed analogy in Newtonian physics that I will understand math of GR.PeterDonis said:No. It's just an incorrect assumption that then needs to be corrected. It has no physical meaning.
Again, I needed analogy in Newtonian physics that I will understand math of GR.PeterDonis said: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.
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.PeterDonis said: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.
exponent137 said: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?
exponent137 said:GR is also not absolutely physically correct. The next theory is theory of quantum gravity ... :)
I think about this:PeterDonis said: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.)
PeterDonis said:Also, what do you think ##W_G## is? That is, how would you calculate gravitational self-energy in Newtonian physics?
I do not understand you, when we put little pieces together, they obtain gravitational self energy? Why not?PeterDonis said: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 M0.
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.
exponent137 said:I think about this:
https://en.wikipedia.org/wiki/Mass_...e_Newtonian_limit_for_nearly_flat_space-times
This is calculated in spacetime of Minkowski.
exponent137 said:##M_0## means in principle sum of tiny mass pieces on big distances, from which we built earth.
exponent137 said: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.
exponent137 said:I do not understand you, when we put little pieces together, they obtain gravitational self energy? Why not?
PeterDonis said: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.
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 ##.PeterDonis said: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.
exponent137 said:The problem here is only, that I wish to imagine GR with the help of Newtonian physics (NP)
exponent137 said:Gravitational self energy can be defined also in special relativity
exponent137 said:What type of approximation we need that we obtain that ##M_0-M## is something like ##3GM^2/5R## .
exponent137 said:I suppose that ##M_0## formula should remain the same. Is your formula for ##M##, flat space approximation?
PeterDonis said: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.
exponent137 said:Can you, please, calculate ##M-M_0## for one simplest example, maybe spherically distributed mass
exponent137 said: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##.
exponent137 said:Can you show me in http://arxiv.org/abs/gr-qc/9712019, where this is calculated?
exponent137 said: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.
exponent137 said:In this example, your ##\xi^a\xi_a=g_{tt}## equals ##1-2GM/(rc^2)## what can give a correct result.
I think spherically symmetric shell or two point masses, ##2 \times m/2##.PeterDonis said:Which example? If you mean the second (spherically symmetric shell), see above.
exponent137 said:I need this only for imagination, how binding energy in NP disapears in GR.
exponent137 said:Thus, it is enough, that formula is valid only one moment.
exponent137 said:I obtained ##g_{tt}=1-Gm/(c^2r)##
exponent137 said:where we have ##r^{-1}##, you obtained like ##r^2##, if we look dimensionally
exponent137 said:Your M0−MM_0-M is much smaller than binding energy.
exponent137 said:Is additon of mass because of pressure larger at ##M## than at ##M_0##?
exponent137 said: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.)
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}}##.PeterDonis said:$$
\sqrt{\xi^a \xi_a} = \left[ \frac{3}{2} \sqrt{1 - k R^2} - \frac{1}{2} \sqrt{1 - k r^2 } \right]
$$
PeterDonis said:$$
M = \int \sqrt{\xi^a \xi_a} \left( 2 T_{ab} - g_{ab} T \right) u^a u^b dV = \int_0^R \left[ \frac{3}{2} \sqrt{1 - k R^2} - \frac{1}{2} \sqrt{1 - k r^2 } \right] \left( \rho + 3 p \right) 4 \pi r^2 \frac{1}{\sqrt{1 - k r^2}} dr
$$
Substituting for ##p## gives
$$
M = \int_0^R \left[ \frac{3}{2} \sqrt{1 - k R^2} - \frac{1}{2} \sqrt{1 - k r^2 } \right] \rho \left[ 1 + 3 \frac{\sqrt{1 - k r^2} - \sqrt{1 - k R^2}}{3 \sqrt{1 - k R^2} - \sqrt{1 - k r^2}} \right] 4 \pi r^2 \frac{1}{\sqrt{1 - k r^2}} dr
$$
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.PeterDonis said: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".
The concept of "gravity gravitating" refers to the idea that gravity, the force that pulls objects towards each other, may also exert a force on itself. This means that gravity may have an effect on its own strength and behavior.
Currently, there is no direct evidence to support the idea of gravity gravitating. However, some theories, such as general relativity, suggest that gravity may have an effect on itself. Further research and experiments are needed to confirm this concept.
The concept of gravity gravitating challenges our current understanding of the laws of physics, specifically the law of conservation of energy. If gravity can affect itself, it would violate this law as energy would not be conserved. This is why further research and experiments are needed to fully understand this concept.
At this time, there is no evidence to suggest that the concept of gravity gravitating can explain dark matter and dark energy. These phenomena are still poorly understood and require further research and study.
If the concept of gravity gravitating is proven to be true, it would greatly impact our understanding of the universe and the laws of physics. It would require us to revise our current theories and models to account for this new concept and its effects on the behavior of gravity.