Would you care to explain how did you make this breakthrough. Or put a link to your preprint?

The best version of the presentation in this article:

Preprint New version of General Relativity and the Structure of "Black Holes"

Valery Borisovich Morozov

Thanks, I read the paper, and also the the cited preprint in arxiv (Gravitational Field Equation and the Structure of Black Holes).

What struck me at first glance is the definition $\sqrt(-g) = 1$, i. e. the metric tensor determinant is constant. This is true only in flat space and also in the locality that is modeled by the tangential space to the manifold at a point.

In the above spaces, Einstein's general relativity has no problem with the energy-momentum conservation. Namely, the energy-momentum tensor $T_i^k$ is always conserved, or, in other words, the gradient of the energy-momentum tensor is 0, which means that the matter-energy conservation principle (EMCP) is always fulfilled in such spaces. You can look in Chapter 96 of Landau-Lifshitz vol. 2 Classical Field Theory, or, more detailed, in my preprint Preprint Landau-Lifshitz stress-energy pseudotensor

or the blog post linked to this preprint.

The problem with EMCP is in curved spaces where $g$ is not constant. I will give only the beginning of the blog post which explains why this is so.

===================================================

Energy and momentum conservation principle in general relativity

Two of the most popular definitions of the combined Energy and Momentum Conservation Principle (EMCP) are:

- the energy and momentum of a closed isolated system are conserved

- the stress - energy - momentum tensor of a closed isolated system is a constant quantity

The second definition is preferred in theoretical physics. The stress-energy-momentum (SEM) tensor, $T^k_i$, that combines energy, mass, momentum, their fluxes and stresses (also called "super energy-momentum tensor", etc.), is the content of the Universe. It is both the matter and the electromagnetic fields. SEM is derived by varying the action integral and is defined in eq. LL32,3 as:

$$T_i^k = q_{,i}\frac{\partial \Lambda }{\partial q_{,k}} - \Lambda \delta _i^k$$

where Λ is the Lagrange function and $q$ is a generalised coordinate. EMCP requires that the SEM tensor is constant. Imagine the closed isolated system as a volume (Universe, black hole), enclosed by an impenetrable envelope such that no matter or radiation can pass through. The SEM tensor within the enclosed volume is constant: there is no tensor flow through the envelope. Tensor flow, or tensor divergence, is the sum of partial derivatives of tensor components with respect to coordinates (similar to vector divergence). From the equations of motion it follows that divergence should be zero according to LL32,4:

$$\frac{\partial T_i^k}{\partial x^k} = T_{i,k}^k = 0$$

which is, of course, the EMCP. In the middle of the above equation the ordinary derivatives are given with the simpler comma notation. In a similar mood, covariant derivatives will be marked with semi-colons.

Presence of gravitation spoils this clear picture in several ways. Gravitation curves space. In curved space, tensor (and vector) flows are ill-defined because they change direction in each point. Parallel transfer, which is essential for definition of derivatives, is very complicated on curved paths. Derivatives are only obtained after making corrections for the curvature – one correction for each tensor index. These corrected derivatives are known as covariant derivatives. In curved space, ordinary derivatives are replaced by covariant ones:

$$ T_i^l \Gamma _{kl}^k -\Gamma _{ik}^l T_l^k+\frac{\partial T_i^k}{\partial x^k} \equiv T_{i;k}^k $$

But $T^{ik}$ is a symmetric tensor, and, according to eq. LL86,11 the above becomes (eq. LL96,1):

$$T_{i;k}^k=\frac{1}{\sqrt{-g}} \frac{\partial \sqrt{-g} T_i^k}{\partial x^k} - \frac{1}{2}T^{kl}\frac{\partial g_{kl}}{\partial x^i}$$

Does this equation express the conservation of energy and momentum in curved space-time or, in other words, can it be turned to zero? The answer is emphatically NO and that' s why : Let $T^{ik}$ be the true stress - energy tensor and $P^i$ be the 4-momentum (energy + momentum) of the system that must be a conserved quantity. By an analogy with eq. LL32,6

$$P^i=\frac{1}{c} \int \sqrt{-g} T^{ik} dS_k$$

where $\sqrt{-g}dS_k$ is the element of a hypersphere in a curved plane. In order to make the expression for $P^k$ exactly identical to that in eq. LL32, 6, $T^{ik}$ should be equated to the stress-energy of the system (this is only by analogy, and does not reflect the tensor properties of $T^{ik}$, as the Book footnote warns). Then, in order to conserve the 4-vector $P^i$, the integral (LL32,6) must be conserved over the whole hyperplane, that is, over the whole 3-dimensional space. As proved in Book, Ch. 29 with the help of Gauss theorem, the integrand must fulfill the following condition in order to conserve the integral:

$$\frac{\partial T^{ik}}{\partial x^k} = 0.$$

This condition is not, however, met in eq. LL96,1, where the additional term $-\frac{1}{2} T^{kl} \frac{\partial g_{kl}}{\partial x^i}$ and the presence of $\sqrt{-g}$ around $T^{ik}$ do not allow derivation of a global conservation law in a manner, analogous to the Classical Field Theory. This is so, because the gravitational field is also a component of $T^{ik}$ which should be included inside it, together with pure matter and electromagnetic field. The offending terms cannot be eliminated because the metric tensor $g^{ik}$ is not constant – it can be anything, depending on the spacetime topology. So, in the first term in the right hand side $\sqrt{-g}$ cannot be pulled out from inside the differential operator. However, the real difficulty lies in the term $-\frac{1}{2} T^{kl} \frac{\partial g_{kl}}{\partial x^i}.$

Taking up the above analogy, the volume that contains the SEM tensor itself stretches, shrinks, and moves in every imaginable way. It cannot be bound in an envelope with zero volume. The spacetime itself generates tensor flows. Do what you want, you can’t just ignore or remove the second term. Failure to do so means nothing less than the startling conclusion: in the curved spacetime of general relativity, the EMCP is no longer valid. Whether this is real or apparent remains subject to interpretation. It is easy to see, however, that this result can be expected from what we know from elementary physics. In the heart of the concept of the energy conservation principle is the assumption that space is isotropic while the momentum conservation principle requires symmetric space. Both isotropy and symmetry are no longer assumed in curved spacetime. The energy and momentum conservation principles fail in principle. The very basis on which they are built sinks into a quicksand. Is there a way to salvage something?

Stefan Bernhard Rüster

With all due respect to Levi-Civita who is an excellent geometer, he is wrong in this paper because he does not understand the physical meaning of curved space. To cite from his paper:

"The mechanical [meaning "physical"-L.A.] meaning of the divergence (like the meaning of T, which I neglect to notice, because it is immediate) becomes clear when one goes back to the variables y. With respect to such variables g (ik) = 0 (i != k), g (00) = 1, g (ii) = −1 (i = 1, 2, 3), and the covariant differentiation coincides with the ordinary one."

That is, Levi-Civita thinks that the metric tensor (and, therefore, its determinant) can be made constant by a mere change of the coordinate frame. Then, of course, covariant differentiation coincides with the ordinary one.

However, the curvature of space is a physical quantity responsible for generation of the gravitational field, and not only coordinate-dependent geometrical quantity. Therefore, it exists independently from coordinates and cannot be extinguished by changing them. This was elaborated by Einstein almost simultaneously with the Levi-Civita paper in

Einstein, A., The Foundation of the General Theory of Relativity, Annalen der Physik, 49, (1916)

If you can extinguish the gravitation everywhere by mere change of coordinates, then gravitation could not exist physically. You can extinguish gravitation only locally (in a neighbourhood of a point) by making the metric tensor constant (and, therefore, making the local system energy-momentum conserved).

Dear Stefan Bernhard Rüster

I remember reading your paper about an year ago (not in RG). It impressed me by citing all the authoritative sources, and still extracting from them information that supports an opposite point of view. It intrigued me to such an extent that I decided to make a comparison of all known pseudotensors in regard to their similarities and differences especially as to the degree they fulfill the EMCP. I also found and read Lavenda's works.

Also, one should have in mind the Bel-Robinson tensor, which is a tensor par excellence, and is related to energy conservation.

In an unfinished paper, (link https://lantonov.blogspot.com/2022/08/landau-lifshitz-synge-and-fock.html), I compare LL, Synge, and Fock pseudotensors and find that they are different both mathematically and physically despite the widespread opinion that they are the same. Fock pseudotensor is mixed one and this fact is interesting in the light of your idea giving importance to mixed tensors. Synge pseudotensor has an opposite sign and 4 terms less than LL, and so on. It would be interesting to see how those fit your ideas, and also other papers that require a particular sign, put them in spherical coordinates (black holes), examine them (the LL one) with adding a cosmological constant, etc.

Dear Stefan Bernhard Rüster

I respect your point of view and your choice to not participate in this discussion. But I remain open to all ideas to solve this important question and do not think that I know the ultimate truth.

I am not married to pseudotensors and these bother me, too, because they are not conserved for rotational motions. That's why I think that your paper goes into a right direction looking for a conserved quantity. Still, alluding to a cosmological variable (instead of cosmological constant) introduces more quantities of unclear origin and interactions transiting to the hypothetical dark matter and passing to uncharted and unsure ground. This is not necessarily bad, but I think there are more conventrional ways founded on classical physics (Lagrangian) and not introducing additional variables. In this respect, I think that there is such a conserved tensorial quantity, which is not a pseudotensor but not the energy-momentum tensor either. I must do much work, though, to see if it is appropriate.

Dear Lyudmil Antonov and Stefan Bernhard Rüster ,

One of the principles laid down by Einstein as the basis of general relativity is the principle of correspondence. This principle should also apply to the energy of the gravitational field. The energy density of the field in the Newtonian limit is known (LL §106 Problem 1). It can be calculated without resorting to GR:

Preprint On energy of static gravity field

In my work, the field energy tensor is given, which asymptotically transforms into the field energy density in Newton's theory of gravity. It is clear that the pseudotensor or the Levy-Cevita equation will not replace it.

Preprint New version of General Relativity and the Structure of "Black Holes"

It seems in the discussion, where the posters discuss is Einstein's relativistic theory of gravity complete or not, discussing energies and stresses that in the GR for some mystic reason and by some mystic way some Matter’s “mass” stuffs into the spacetime, curving it, and these energies and stresses again for some mystic reason and by some mystic way really force real bodies to move along the “geodesics” in this curved spacetime.

Really, of course, no mystic exists in Matter; including anything in Matter fundamentally cannot impact on the Matter’s fundamentally absolute, fundamentally flat, and fundamentally “Cartesian”, (here - utmost universal) [5]4D spacetime with metrics (utmost universal) (cτ,X,Y,Z,ct); and fundamentally spacetime cannot impact on anything in Matter. The spacetime is fundamentally nothing else than an fundamentally infinite empty container, where Matter is placed, exist, and constantly changes.

Gravity is fundamentally nothing else than one of the fundamental Nature forces, which is in a few traits similar to the fundamental Electric Force; more see the Shevchenko-Tokarevsky’s 2007 initial model of Gravity and Electric Forces in section “Mediation of the fundamental forces in complex systems” in

https://www.researchgate.net/publication/355361749_The_informational_physical_model_and_fundamental_problems_in_physics

- where it is rigorously scientifically shown that both Forces act by one universal scheme, moreover, it is shown that Nuclear Force acts by the same scheme – more see 2023 SS&VT model of this Force in

https://www.researchgate.net/publication/369357747_The_informational_model_-Nuclear_Force

That are initial models, however in the first link it is pointed how really more reliable Newton Gravity field theory should be modified – as that happened with Coulomb theory; and so if real physical theorists would develop the really scientific whole Gravity theory, that would be much more useful than a wasting time on “corrections of GR”.

Cheers

Lyudmil Antonov

"What struck me at first glance is the definition $\sqrt(-g) = 1$"

This is part of the equation of Einstein's 1915 simplified equation.

Pay attention to the notation Gik is the Ricci tensor, Rik is part of it.

Sergey Shevchenko ,

Я ничего не понял, зачем эта философия 19 века?

Сейчас экспериментально доказано - мы живем в искривленном пространстве.

Preprint Space-time as a physical object. Einstein's testament

Preprint Пространство-время как физический объект. Завещание Эйнштейна

Учитесь. Когда-то люди верили, что земля плоская. Теперь некоторые уверены, что наше пространство плоское. Это пройдет с развитием образования.

Lyudmil Antonov

From Einstein's 1918 paper.

there is undoubtedly a certain benefit from the pseudopotential. For example, you can calculate the flow of energy. But this will never be the momentum energy tensor of the gravitational field.

Valery Borisovich Morozov

The notation in Einstein (1915) is indeed unusual compared with our present textbooks. Gim corresponds to Ricci tensor with Rim its first two terms and Sim its second two terms. (2a) is one of the forms of Einstein equations exactly corresponding to formula (95.8) in Landau and Lifshitz.

t_i^k is known as the Einstein pseudotensor. However, in the 1915 paper, Einstein calls it "energy tensor" of the gravitational field. This is because Einstein makes the definition sqrt(-g) = 1 effectively making the metric constant, i.e. working in flat spacetime. The divergence of a constant metric tensor is 0 and, consequently, the EMCP is fulfilled. This is found in this text from the paper:

"Like always we assume that the divergence of the energy tensor of matter vanishes in the sense of the general differential calculus (Momentum-Energy theorem [i.e. EMCP]). When specializing the coordinate choice in accordance with (3a) [i.e., sqrt(-g) = 1] ..." and goes on to derive the gravitational field as "energy tensor".

In the general case, g is not constant, the divergence of the metric tensor does not vanish and we have a problem with fulfilment of EMCP.

In the gravitational field equation of general relativity, Einstein's formula for calculating the deflection angle of starlight passing through the Sun is

θ = 4GM/Rc2(Radian).

The deflection angle at the perigee (solar surface (1R)) observed during an actual eclipse could not measure starlight because of the bright part of the Sun's surface, and it was possible to measure the deflection angle of starlight at a distance of approximately 2R(1.4×109m) to 10R(7.0×109m). The observed value θ was approximately 0.93" to 2.73".(D.W. Sciama, The Physical Foundation of General Relativity Chap. 7, 1969) According to Einstein's gravitational field equation, when the observation distance r is 2R∼10R, θ is calculated as 0.173"∼0.867".The calculation of the deflection angle of starlight through Einstein's gravitational equation and observations do not match.

We calculate the deflection angle of starlight bending in the direction of starlight under the influence of the gravitational field of the ether(dark matter), which gives mass to a unit volume, using classical mechanics. It can be seen that the deflection angle of 10R(r=7.0×109m) at the perigee distance 2R(r=1.4×109m) is 0.77″ to 3.46″. The above result is the angle of divergence of starlight passing through the Sun, which is consistent with the observation values of 0.93″ to 2.73″. It can be seen that the calculation of the starlight's deflection angle using classical mechanics is an accurate numerical value encompassing the observation results.

Paper: “Beyond the Speed of Light – A Review of the Theory of Relativity and the Speed of Light by the Ether (Dark Matter)”

Lyudmil Antonov

All right. But in my work there is no condition g=-1. But despite this, half of the Ricci tensor is a tensor, and one of these halves is proportional to the momentum energy tensor of the gravitational field.

Einstein realized as early as 1913 that his pseudotensor was not covariant. But in 1915 he writes a conservation law suggesting that t_i^k is a tensor. difficulties with the conservation law are revealed only in 1918.

Valery Borisovich Morozov

I see now. Sorry to being perfunctory on the first reading of your paper. I will read it more deeply.

Valery Borisovich Morozov

For your Theorem 1, it can be shown that the Christoffel symbol $\Gamma_{ki}^i$ is not a vector. The Christoffel symbols are not tensors, but rather connection coefficients that describe how the basis vectors change along a direction on a curved spacetime. The Christoffel symbols depend on the choice of coordinates and do not transform like tensors under general coordinate transformations. However, they do transform like tensors under linear coordinate transformations.

To see why $\Gamma_{ki}^i$ is not a vector, consider its transformation rule under a general coordinate transformation from x to y coordinates:

$$\Gamma_{ki}^i = \frac{\partial y^\alpha}{\partial x^k}\frac{\partial y^\beta}{\partial x^i}\frac{\partial x^i}{\partial y^\gamma}\Gamma_{\alpha\beta}^\gamma + \frac{\partial y^\alpha}{\partial x^k}\frac{\partial y^\beta}{\partial x^i}\frac{\partial^2 x^i}{\partial y^\gamma \partial y^\delta}y_\delta$$

The second term on the right-hand side does not have the form of a vector transformation, since it involves second derivatives of the coordinate functions. Therefore, $\Gamma_{ki}^i$ is not a vector. In fact, it can be shown (on request) that $\Gamma_{ki}^i$ is just the divergence of the coordinate basis vector $\frac{\partial}{\partial x^k}$, which is also not a vector.

You can see this also from Landau-Lifshitz (86.5) $\Gamma_{ki}^i = \frac{1}{2g} \frac{\partial g}{\partial x^k}$ where we have a divergence of the scalar g which is a scalar, not a vector.

The Ricci tensor is defined as the contraction of the Riemann curvature tensor:

$$R_{\mu\nu}=R^\alpha_{\mu\alpha\nu}=g^{\alpha\beta}R_{\mu\beta\nu\alpha}$$

where $g^{\alpha\beta}$ is the inverse metric tensor. The Riemann curvature tensor has four terms, which can be written as:

$$R_{\mu\nu\rho\sigma}=\frac{1}{2}\left(\frac{\partial^2 g_{\mu\sigma}}{\partial x^\nu \partial x^\rho}-\frac{\partial^2 g_{\mu\rho}}{\partial x^\nu \partial x^\sigma}+\frac{\partial^2 g_{\nu\rho}}{\partial x^\mu \partial x^\sigma}-\frac{\partial^2 g_{\nu\sigma}}{\partial x^\mu \partial x^\rho}\right)+g_{\alpha\beta}\left(\Gamma^\alpha_{\mu\rho}\Gamma^\beta_{\nu\sigma}-\Gamma^\alpha_{\mu\sigma}\Gamma^\beta_{\nu\rho}\right)$$

where $\Gamma^\alpha_{\mu\nu}$ are the Christoffel symbols of the second kind. The first two terms are called the second derivatives of the metric, and the last two terms are called the quadratic terms of the Christoffel symbols. When we contract the Riemann curvature tensor to get the Ricci tensor, we only keep the terms that have repeated indices, such as $\mu$ and $\alpha$. Therefore, the third and fourth terms of the Ricci tensor are given by:

$$R_{\mu\nu}=g_{\alpha\beta}\left(\Gamma^\alpha_{\mu\rho}\Gamma^\beta_{\nu\rho}-\Gamma^\alpha_{\mu\nu}\Gamma^\beta_{\rho\rho}\right)$$

These terms are not a tensor by themselves, because they do not transform like a tensor under general coordinate transformations. They depend on the choice of coordinates and the Christoffel symbols, which are not tensors either. However, when we add the second derivatives of the metric, which also do not transform like a tensor, we get a combination that does transform like a tensor. This is because the second derivatives of the metric and the quadratic terms of the Christoffel symbols cancel out some of their non-tensorial behavior when added together. Therefore, only the full Ricci tensor is a tensor, not its individual terms.

The Ricci tensor can be written in terms of Christoffel symbols as:

$$R_{\mu\nu}=\frac{\partial \Gamma^\alpha_{\mu\nu}}{\partial x^\alpha}-\frac{\partial \Gamma^\alpha_{\mu\alpha}}{\partial x^\nu}+\Gamma^\alpha_{\mu\nu}\Gamma^\beta_{\alpha\beta}-\Gamma^\alpha_{\mu\beta}\Gamma^\beta_{\nu\alpha}$$

The third and fourth terms are the same as before, and they are not a tensor by themselves. The first and second terms are the second derivatives of the Christoffel symbols, which also do not transform like a tensor under general coordinate transformations. They depend on the choice of coordinates and the partial derivatives of the metric, which are not tensors either. However, when we add the second derivatives of the Christoffel symbols and the quadratic terms of the Christoffel symbols, we get a combination that does transform like a tensor. This is because the second derivatives of the Christoffel symbols and the quadratic terms of the Christoffel symbols cancel out some of their non-tensorial behavior when added together. Therefore, only the full Ricci tensor is a tensor, not its individual terms.

Dear All,

I don't have your high theoretical knowledge of physics, nor the necessary mathematical background...

Dealing with Earth Sciences led me to focus on understanding the phenomenon of the Earth, instead of mathematics and pure.

Here are my questions:

Why does Earth have a gravitational field?

Why does the moon revolve around the earth?

Why does the Earth spin?

What causes the tides?

Why is it that the asteroids of the asteroid belt are made up of such minerals, which suggests that they were once formed in the crust and mantle of a Earth like rocky planet?

Do you accept that a simple metaphysical explanation related to ancient Chinese philosophy can elegantly explain gravity?

There would still be questions... But it's enough if I get adequate answers to them.

I will also accept answers privately if any of you are afraid of losing your job. his career...

Regards,

Laszlo

Some of the why? questions could go one or two levels lower but fail when digging deeper.

Tides are caused mainly by the gravitational field of the Moon and with a smaller contribution of that of the Sun.

Chinese metaphysics is mostly without proofs and accepting its explanations depends on beliefs and interpretation.

Dear Lyudmil Antonov ,

You did not respond to all my questions... only one:' Tides are caused mainly by the gravitational field of the Moon and with a smaller contribution of that of the Sun.'

In the Natural metaphysical book 'AZ ESZME (THE IDEAL' first there is the idea that it is impossible that the gravitational influence of the Moon Can create tides of Earth! When on the Moon the effect of gravity is so small

All this is clearly described in the following two articles:

Article Tényekkel igazolható a gravitáció valós oka

but there is also a small error in this: which can be improved ]

Article Gravity a paradym shift in reasoning

If I were you, I would not underestimate the ancient Chinese philosophy!

The traditional Chinese medicine and Chinese eating habits are excellent proof of this... And if we consider the metaphysical concept of apriori entity which is a one basic of terms of my new article it is a very good prove for ying yang principle, with it can be extended the gravity theory of before linked articles... exists solution to get explanation to telepathy, and can be demonstrated the existence of graviton:

Research Proposal Physico-metaphysical proof of the existence of graviton (Fiz...

(only abstract)

Using this concept and formulating with other meanings of someone's mathematical gravity formula, perhaps we will have a better mathematical description of gravity... or exist other tools actually...

I have simple answers to the questions asked. They would keep Okcam's razor from breaking...

Regards,

Laszlo

I knew that the questions are loaded :)

Valery Borisovich Morozov

Your attempt to solve the problem of energy-momentum conservation in general relativity is ingenious and I am sorry it doesn't succeed. The curse of GR is the panoply of Christoffel connections which are not tensors and make most combinations pseudotensors for rotational motions. I didn't think of splitting Ricci (or Riemann) tensors but have made other combinations which do not work either. Maybe there is something in the Weyl tensor which is a traceless Riemann tensor and reflects only the gravitational field though it have been treated and interpreted intensively. Another possible aim is the Bel-Robinson tensor, part of which concerns the energy-momentum. Problem is that this part cannot be coerced to tensor.

Maybe there is something more fundamental, going even back to the Hamilton principle and the Lagrangian from which the Einstein equations were derived. Teleparallel gravity and the many modifications around it come to mind ...

Dear Stefan Bernhard Rüster

I wonder if we miss some rotational motion. Everything rotates: planets, stars, galaxies, galactic clusters ... Why can't we think that the Universe as a whole rotates. Then because of the centrifugal forces matter tends to escape to the periphery and the farther it is, the more acceleration it has. Well, we have to postulate then some center of rotation where matter moves more slowly or is somewhat clustered but we'll dispense with the need for dark energy, at least. Such hypothetical rotation of the Universe is not taken into account by the total SEM tensor at present.

Dear Stefan Bernhard Rüster;

Does that mean that the net rotation of the world was not originally zero?

I will be grateful if you answer me.

Dear Stefan Bernhard Rüster

What is your opinion about the paper of Chang, Nester, and Chen Energy-momentum (quasi-)localization for gravitating systems at https://arxiv.org/pdf/gr-qc/9912058.pdf. Is it a viable approach?

I see. You think that the conserved quantity is Tik and not (Tik + tik). And tik (Aik) should also be conserved independently of Tik but it should not be 0.

In order to find the total Lik, you integrate over flat space (no sqrt(gamma) under the integral) and then by non-relativistic time contrary to all textbooks stating that spacetime is curved.

I found Robert Wald's book General Relativity (1984) and especially its chapter 11.2 Energy very informative about the difficulties in finding an energy conservation law for general relativity.

"think what would happen in case there would not be conservation of total energy in the theory of gravitation"

Very interesting and hard question. I think that there should be a total energy conservation law, however difficult to find because of existing global time symmetry (time translation symmetry is necessary for energy conservation, linear space translation symmetry for momentum conservation, rotational translation symmetry for angular momentum conservation, according to the first Noether theorem). I see this symmetry in the fact that in whichever direction we look with the telescope we come to the same past moments with almost the same redshifts. This is something similar to "asymptotic time translation" leading to "asymptotic flatness" of Bondi, van der Burg, and Metzner.

There are circumstances, however, that speak against global energy conservation.

First, I am not sure to what extent the Universe can be regarded as an isolated system.

Then, looking not globally but in sufficiently large volumes, we see these volumes expanding with time so energy in a given volume will not be constant.

Third, the current LambdaCDM cosmological model is not in favour of global energy conservation. At the time of Big Bang, matter and space seemingly sprang out of nothing. After that, inflation, and more space and more matter.

If we look at microscopic scales, elementary particles all the time are created and annihilated (the virtual particles of Feynman) and in this respect energy conservation is valid only on average.

There is, however, an interesting interaction of matter and space curvature (or, equivalently, gravitational field). It seems the less matter-energy structure (bigger entropy) the bigger the curvature, i.e. the more energy in spacetime. Think black holes in which even quarks are not preserved (in fact, nobody knows what the state of matter is inside or is there any matter at all with "infinite" density). This poses hard problems which are not solved at present.

Weinberg divides total action in two (IM + IG) and postulates that the matter action IM is scalar (he gives an example of such scalar action with collisionless plasma). Then varying IM he varies not only the coordinates (as with the classical Lagrange-Euler procedure), but also the metric tensor and vector potentials. Further, he takes the Lie derivatives for the variation of the last two, which add symmetry (Lie derivatives are homeomorphic). Even if all of this can be explained somehow, IG still remains and for it Weinberg does not give a covariant conservation formula.

Weinberg proves in Sec. 12.3 that "Thus the energy-momentum tensor defined by Eq. (12.2.2) is conserved (in the covariant sense) if and only if the matter action is a scalar. " (emphasized). The proof is not so clear (at least to me) because of what I said above about the variation of the action. By including homeomorphic variation of dynamic variables other than coordinates to the matter-energy such as the metric tensor, Tik becomes not strictly matter-energy but include also space elements (metric). I don't know if this is good or bad.

Yes, but instead of varying coordinates, which is how the Lagrangian is obtained in classic mechanics and the Einstein GR, Weinberg varies the metric tensor in eq. (12.2.2) which makes the metric tensor not a dynamical variable of the system itself but an external field. This makes Tμν dependent or function of coordinates, Tμν(x) that is, varying over the coordinates depending on this external field. See the definition of the energy-momentum tensor as Tμν(x) under (12.2.2). This definition is different from that given by Einstein.

An action of external field would not preserve the system stationary. Weinberg says this on page 360: "The action IM will not be stationary with respect to this variation, because for the moment we are regarding gμν(x) not as a dynamical variable like xnμ or Aμ but as an external field." This external field would act as a force that pushes the system in some direction in the general case, that is, action would be a vector in the general case. In such case δIM would not be 0. Only if IM is scalar then δIM = 0 and as a consequence, the covariant derivative of Tμν should be conserved instead of the ordinary derivative.

But to make it clear again, this derivation of the energy-momentum conservation may be perfectly legit which is confirmed by the absence of serious criticism of Weinberg work as far as I know.

Exactly. delta g_{mu nu} is the variation in respect to the metric tensor. sqrt(g) is the Jacobian. But the integration by parts includes also the Jacobian.

But if these are identical, why all the fuss about the ordinary and not covariant derivatives of the stress-energy tensor to be 0.

I understand it very well and, being a theoretical physicist, I am certain that you understand it, too. However, the temptation to refuse it to the last is too great. Before going to Weinberg, and why his derivation is different from that of Einstein, I'll try to first explain why the equation ∇μTμν = 0 is not a conservation law. It's not by chance that I asked first if this is compatible with Noether's theorems.

The equation ∇μTμν = 0 is not a conservation law because it does not imply that the energy-momentum four-tensor Tμν is conserved locally or globally. In flat spacetime, the covariant derivative reduces to the ordinary partial derivative, and the equation implies that ∂μTμν = 0, which means that the energy-momentum four-tensor Tμν is conserved locally. However, in curved spacetime, the covariant derivative involves the Christoffel symbols, which depend on the metric and its derivatives. The equation does not imply that ∂μTμν = 0, and therefore the energy-momentum four-tensor Tμν is not conserved locally. Instead, it implies that there is a non-trivial exchange of energy and momentum between matter and gravity. The equation ∇μTμν = 0 is also not a conservation law globally, because it does not imply that the integral of Tμν over a closed surface is zero. In fact, such an integral is not even well-defined in general relativity, because there is no global notion of simultaneity or spatial infinity. Therefore, the equation ∇μTμν = 0 is not a conservation law in general relativity, but a consistency condition between matter and gravity.

Using Noether's theorem, it states that for every continuous symmetry of the action, there is a corresponding conserved current. The conserved current is also called the Noether current, and it can be derived from the variation of the action under an infinitesimal symmetry transformation. The conservation law implies that the divergence of the Noether current is zero, i.e., ∂μ jμ = 0, where ∂μ is the ordinary partial derivative and jμ is the Noether current. The energy-momentum tensor is a special case of the Noether current, associated with the symmetry of global spacetime translations. The variation of the action under a global spacetime translation is given by δS = ∫d4x δxμ ∂ν Tμν , where δxμ is an infinitesimal constant vector and Tμν is the energy-momentum tensor. By Noether's theorem, this variation must vanish for any δxμ , which implies that ∂ν Tμν = 0. This is the conservation law for the energy-momentum tensor in special relativity. However, in general relativity, the action is not invariant under global spacetime translations, but only under local (or general) coordinate transformations. The variation of the action under a local coordinate transformation is given by δS = ∫ d4x δxμ(x) ∇νTμν , where δxμ(x) is an infinitesimal function of x and ∇ν is the covariant derivative. By Noether's second theorem, this variation must vanish for any δxμ(x), which implies that ∇νTμν = 0. This is not a conservation law, but an identity involving the equations of motion. It does not imply that the energy-momentum four-tensor Tμν is conserved locally or globally, but only that there is a non-trivial exchange of energy and momentum between matter and gravity.

To see why this is not a conservation law, let us compare it with the conservation law for the Noether current associated with a global symmetry. In that case, we have ∂μ jμ = 0, where jμ is the Noether current and ∂μ is the ordinary partial derivative. This equation implies that the integral of jμ over any closed surface is zero, i.e., ∮ jμ dSμ = 0, where dSμ is the surface element. This means that the Noether current is conserved globally, i.e., there is no net flow of the current across any boundary. Moreover, this equation also implies that the integral of jμ over any volume is constant in time, i.e., d dt ∫ jμ dVμ = 0, where dVμ is the volume element. This means that the Noether current is conserved locally, i.e., there is no source or sink of the current inside any region.

However, in the case of ∇νTμν = 0, we do not have these properties. First of all, ∇νTμν is not an ordinary partial derivative, but a covariant derivative, which involves the Christoffel symbols Γλμν , which depend on the metric and its derivatives. The metric encodes the gravitational field, which can vary with time and space. Therefore, ∇νTμν = 0 does not imply that ∮Tμν dSν = 0, because dSν is not a covariant quantity under general coordinate transformations. In fact, such an integral is not even well-defined in general relativity, because there is no global notion of simultaneity or spatial infinity. Therefore, Tμν is not conserved globally in general relativity.

Secondly, ∇νTμν = 0 does not imply that d dt ∫ Tμν dVν = 0 either, because dVν is also not a covariant quantity under general coordinate transformations. Moreover, even if we fix a coordinate system and use ordinary partial derivatives instead of covariant derivatives, we still do not have d dt ∫ Tμν dVν = 0 in general relativity. This is because Tμν does not include the contribution from the gravitational field itself, which can exchange energy and momentum with matter fields. Therefore, Tμν is not conserved locally in general relativity either.

Instead of being a conservation law, ∇νTμν = 0 is an identity involving the equations of motion. It follows from the fact that the action is invariant under local coordinate transformations (or diffeomorphisms), which are symmetries of general relativity. It also implies that the Einstein field equations are consistent with the Bianchi identities, which are differential identities for the curvature tensor. Physically, it means that there is a non-trivial exchange of energy and momentum between matter and gravity in general relativity.

I can convince you with simple examples if you fail to undersand, but because you don't want to be convinced, I desist.

When a substance is accelerated, the energy of the substance is partially converted into the energy of the gravitational field.

The background of the gravitational field constantly transfers its energy to matter. These processes must be in balance.

Dear @Valery Borisovich Morozov

Matter curves spacetime, and curved spacetime affects matter movement, and therefore, the energy of the system. This is exactly the reason why we cannot speak about conservation of energy-momentum of matter alone, independently of spacetime curvature. Relativity differentiates between two types of conservation: conservation (in the proper sense) and covariant conservation or conservation (in the covariant sense).

That's why Weinberg writes in italics "Thus the energy-momentum tensor defined by Eq. (12.2.2) is conserved (in the covariant sense) if and only if the matter action is a scalar. "

By covariant conservation is understood that the covariant derivative of the energy-momentum tensor is conserved. But this derivative includes also the metric tensor and its derivatives (generally, in the form of Christoffel symbols), so spacetime curvature is not separated from matter but both are included together in the conservation law. Mathematically speaking, they co-vary (vary together). That's why it is covariant derivative and covariant conservation.

As an example, you have 1 kg ball of some material put 1 m above the earth surface. The potential energy of this ball is 9.806 Joules. However, if you put this ball 1 m above the Moon, its potential energy is 1.62 Joules. The potential energy of the ball is not conserved. If you drop it on the surface of the earth or moon, you'll see that its kinetic energy is not conserved either (the ball falls slower on the moon). Only if you take into account the difference of gravity on earth or moon (so the difference of spacetime curvature) the energy will be conserved in the frame of the earth-moon system.

In the covariant conservation you always have to specify the boundaries or volume of your system because you have spacetime coming in the covariant conservation law. These cannot be infinite unless you make some assumptions such as asymptotic flatness. Further, you have to specify a proper time valid for the whole system. In the example of earth-moon system, time ticks slower on the earth than on the moon because of time dilation due to gravity. One can speak of conservation of energy-momentum only if time is synchronized in the system.

In theoretical physics, energy-(momentum, angular momentum) conservation is treated by Noether's first law, and local covariant conservation is treated by Noether's second law. For the local covariant conservation, you must have a Killing vector field (e.g. time) and a Cauchy (causal) hypersurface (e.g. the space volume for which you have specified a proper time).

If we talk about conservation law, not only of energy but also of momentum and angular momentum, we can say that with some pseudotensors (such as Landau-Lifshitz pseudotensor), we have conservation (in the proper sense) of momentum for both the gravitatational field and the matter separately because Christoffel symbols are conserved for linear translations. However, energy and angular momentum (rotational conservation) are not conserved separately for matter and gravitational field. They are conserved only if we lump together matter and gravitation.

Some theoreticians take matter and spacetime that is curved around it as an elementary particle that is treated as a basis for their conservation hypotheses. The point is that matter and spacetime cannot be separated when one infers conservation laws, as you implied above.

I recommend the review of Gennady Gorelik

http://ggorelik.narod.ru/Fock/ES_GG_Conservation_Laws.pdf

exactly on the question which expounds the theme without much math, rather from philosophical point of view.

A more mathematical treatment of the question is

International Journal of Modern Physics A VOL. 13, NO. 12

EINSTEIN'S THEORY OF GRAVITATION AS A LAGRANGIAN THEORY FOR TENSOR FIELDS

S. MANOFF Bulgarian Institute of Nuclear Research and Nuclear Energy

https://doi.org/10.1142/S0217751X98000846

Dear frend,

We show

https://www.researchgate.net/publication/372244706_Localization_of_gravitational_energy

that the curved space representing the "gravitational field" has a negative massless energy. The space transferred energy in such quantity to the substance, which curved the space, in the form of mass-energy. But this negative energy of space enters into the gravitational mass of the body. So the gravitational mass is less than the inertial mass, according to Birkhoff's theorem

Встретил статью в разделе «General Relativity and Quantum Cosmology» под выразительным названием Carroll stories. Добрая половина работ по этой тематике о таинственных и ужасных Черных Дырах и параллельных мирах, горизонтах событий и прочей чертовщине.

Вряд ли сообщество астрофизиков с радостью примет известие о том, что черные дыры — банальные потенциальные ямы, глубокие, но ямы.

Preprint New version of General Relativity and the Structure of "Black Holes"

К тому же наш любимый Штерн с ныне покойным В. Рубаковым опубликовали статью (по поводу очередных фриков), что исправить ОТО нельзя, хотя известны ее недостатки и сам Эйнштейн (1955) предлагал исключить сингулярности из теории. Мог ли великий Физик предполагать, что эти самые сингулярностей будут кормить десятки тысяч(!) людей?

Dear Lyudmil Antonov ,

In order to bend space, it is necessary to do work; in Einstein's theory, space is considered as a physical object. Recall the special theory of relativity, there is no matter at all - only the Minkowski metric. The substance in no way affects the Minkowski space. Minkowski space.

In general relativity we consider space as a physical object.

Preprint Space-time as a physical object. Einstein's testament

Einstein planned (1913) to include the energy of the gravitational field in the gravitational field equation. But he didn't succeed.

Preprint New version of General Relativity and the Structure of "Black Holes"

Agree. You cannot describe the dynamical properties of the system ignoring spacetime curvature as a dynamical variable. There is no conservation of the pure matter energy-momentum if you ignore spacetime.

Валерий, раздел «General Relativity and Quantum Cosmology», где Вы встретили статью, - раздел чего?

Rather vivid discussion about some strange incompleteness of the GR, where the authors attempt to “correct” the postulated in the GR interactions in systems “spacetime-mass-spacetime”, continues; while it is quite natural for any normal human to think that such interactions cannot exist and so quite naturally never were really observed;

- and in the Shevchenko-Tokarevsky’s 2007 initial model of Gravity and Electric Forces in

https://www.researchgate.net/publication/365437307_The_informational_model_-_Gravity_and_Electric_Forces

[or a bit more recent version in

https://www.researchgate.net/publication/355361749_The_informational_physical_model_and_fundamental_problems_in_physics section 6. “Mediation of the fundamental forces in complex systems” ] it is rigorously proven that the postulated interactions above are fundamentally impossible; including just so in the GR equation the energy conservation law is violated.

Gravity is absolutely for sure is some fundamental Nature force – as that Electric, Nuclear/Strong, and probably Weak, Forces are, including at least Gravity, Electric, and Nuclear Forces act by the same one scheme [how Nuclear Force act see initial SS&VT 2023 model in https://www.researchgate.net/publication/369357747_The_informational_model_-Nuclear_Force] .

Correspondingly for any normal human it is quite clear that any “corrections”, etc., of the GR are really senseless wasting of time, while instead it is necessary to develop whole theories from the initial models above; which [the theories, not only Gravity, but all other Forces] must be based on that the Forces’ fields fundamentally don’t contain energy, only provided this the energy conservations law acts in Matter.

Though yeah, in this case it is necessary to be able to think scientifically rationally, and to spend time just at the development and not at studying of LaTeX, whereas now the last is seems unique and completely sufficient condition for be published in midstream journals.

Though this

“…In order to bend space, it is necessary to do work; in Einstein's theory, space is considered as a physical object. Recall the special theory of relativity, there is no matter at all - only the Minkowski metric. The substance in no way affects the Minkowski space..…..”

- is too strange claim as well. In the SR just substance just really affects the Minkowski space; what is, of course, too strange for any normal human; but quite adult true SR believers quite frankly claim, say, that muons that are created in Earth atmosphere on tens km heights at their motion reach Earth surface just because just they “contract space” and “dilate time”; etc.

Again, first “substance affecting the space/time/spacetime” was discovered by Minkowski, Einstein in the GR only developed this “discovery” .

Cheers

Radi I. Khrapko ,

Рад Вас видеть"на коне".

«General Relativity and Quantum Cosmology» это раздел arxiv.org.

Валерий, а нет ли кого-нибудь, кто согласится представить меня в arXiv?

There are other possible causes for precession of mercury's perihelion; namely the torque imposed by the galaxy, as I have proposed in my diffusion gravity theory.

Preprint Diffusion Gravity (5): Perihelion Precessions as Indicators ...

Please see Masses of black holes

https://www.researchgate.net/publication/372725342_Masses_of_black_holes

Radi I. Khrapko See

Preprint Einstein's postulate as a correction to Newton’s law of gravity

Radi I. Khrapko

У меня был питерский мужик который меня представил, умер, к сожалению. Не помогло, все равно модераторы зарубили.

Одну статью они согласились, да это публикация, но пошел бы ты на фиг.

Я считаю, что редакторы сами должны предлагать публикацию. Вы будете смеяться, но такое предложение поступило. Что из этого выйдет посмотрим.

Еще работаете? Не обращали внимания, что умирают исключительно люди младше нас? Умер Миша Белоненко недавно, ровесник моего сына (57).

Dear Radi I. Khrapko

Of your arguments, I agree with everything except the last sentence.

===

So the gravitational mass is less than the inertial mass, according to Birkhoff's theorem

===

The last phrase is a bit misleading, as it is written as if the principle of equivalence of inertial mass and gravitational mass is broken.

1. Mass defect due to gravitational binding energy (gravitational potential energy)

● ----- r ----- ●

m -------------m

When two masses m are separated by r, the total energy of the system is

E_T=2mc^2−Gmm/r

In the dimensional analysis of energy, E has kg(m/s)^2, so all energy can be expressed in the form of (mass) X (velocity)^2. So, E=Mc^2 holds true for all kinds of energy. Here, M is the equivalent mass. If we introduce the negative equivalent mass "-m_gp" for the negative gravitational potential energy,

−Gmm/r=(−m_gp)c^2

E_T=2mc^2−Gmm/r=(2m−m_gp)c^2=(m^∗)c^2

The gravitational force acting on a relatively distant third mass m_3 is

F=−G(m^∗)(m_3)/R^2=−G(2m−m_gp)(m_3)/R^2=−G(2m)(m_3)/R^2 + G(m_gp)(m_3)/R^2

That is, when considering the gravitational action of a bound system, not only the mass in its free state but also the binding energy term (-m_gp) should be considered.

When the mass m on the left here acts as a gravitational force, the inertial mass is m, and the gravitational mass is also m. Therefore, m is also included in the gravitational potential term (-Gmm/r).

Now, when a system in which two masses are combined exerts gravitational action on a distant third object,

The inertial mass of a system containing two particles is the total energy of the system divided by c^2, m^*=(2mc^2-Gmm/r)/c^2. Also, since the gravitational mass of a system containing two particles is m^*, the principle of equivalence between inertial mass and gravitational mass still holds.

Dear Radi I. Khrapko

By the way, as you may have realized, we have to consider the negative mass effect, the mass defect effect due to gravitational potential energy in astronomy.

And, considering this negative equivalent mass term, some big problems(The singularity problem of black hole, Dark energy, the problem of the creation of the universe from nothing) in astronomy are solved.

1. Singularity problem of Black hole

In the general case, the value of gravitational potential energy (gravitational self-energy) is small enough to be negligible, compared to mass energy Mc^2. However, as more mass is collected, the ratio of (negative) gravitational potential energy to (positive) mass energy increases.

Gravitational self-energy eq. is

U_gs = -(3/5)(GM^2/R)

Therefore, calculating the gravitational self-energies of several celestial bodies yields surprising results.

In the case of Moon,

U_{gs - Moon} = ( - 1.89 x10^ -11)M_{Moon}(c^2)

In the case of Earth,

U_{gs - Earth} = ( - 4.17 x 10^ -10)M_{Earth}(c^2)

In the case of the Sun,

U_{gs - Sun}} = ( - 1.27 x 10^ -4)(M_{Sun}(c^2)

In case of a Black hole,

U_{gs - Black - hole} = ( - 3.00 x 10^-1){M_{Black - hole}(c^2)

At the event horizon of a black hole,

U_gs=-(3/5)(GM^2/R) = -(3/5)(GM^2/(2GM/c^2)) = -(3/10)Mc^2 = -0.3Mc^ 2

In the case of a black hole, the negative gravitational potential energy is 30% of the positive mass energy. It becomes a size that cannot be ignored.

In the generality of cases, the value of gravitational self-energy is small enough to be negligible, compared to mass energy Mc^2. So generally, there was no need to consider gravitational self-energy. However the smaller R becomes, the higher the absolute value of U_gs. For this reason, we can see that U_gs is likely to offset the mass energy in a certain radius.

Thus, looking for the size in which gravitational self-energy becomes equal to mass energy by comparing both,

U_gs = | - (3/5)(GM^2)/R_gs | = Mc^2

R_gs = (3/5)(GM)/c^2

This equation means that if mass M is uniformly distributed within the radius R_gs, gravitational self-energy for such an object equals mass energy in size. So, in case of such an object, mass energy and gravitational self-energy can be completely offset while total energy is zero. Since total energy of such an object is 0, gravity exercised on another object outside is also 0.

Comparing R_gs with R_S, the radius of Schwarzschild black hole,

R_gs = (3/5)(GM)/c^2 = 0.3R_S

This means that there exists the point where (negative) gravitational self-energy becomes equal to (positive) mass energy within the radius of black hole, and that, supposing a uniform distribution, the value exists at the point 0.3R_S, a 30% level of the black hole radius.

From the equation above, even if some particle comes into the radius of black hole, it is not a fact that it contracts itself infinitely to the point R = 0. From the point R_gs(or R_gs - vir), gravity is 0, and when it enters into the area of R_gs(or R_{gs - vir}), total energy within R_gs(or R_{gs - vir}) region corresponds to negative values enabling anti-gravity to exist. This R_gs(or R_{gs - vir}) region comes to exert repulsive effects of gravity on the particles outside of it, therefore it interrupting the formation of singularity at the near the area R=0.

If you have only the concept of positive energy, please refer to the following explanation.

From the point of view of mass defect, r=R_gs(or R_{gs - vir}) is the point where the total energy of the system is zero. For the system to compress more than this point, there must be an positive energy release from the system. However, since the total energy of the system is zero, there is no positive energy that the system can release. Therefore, the system cannot be more compressed than r=R_gs(or R_{gs - vir}). So black hole doesn't have singularity.

In case of the smallest black hole with three times the solar mass, R_S = 9km. R_gs of this object is as far as 3km. In other words, even in a black hole with smallest size that is made by the contraction of a star, the distribution of internal mass can't be reduced to at least radius 3km. Before reaching the quantum mechanical scale, the singularity problem is solved by gravity itself.

Article Solution of the Singularity Problem of Black Hole

Dear Radi I. Khrapko

2. Dark energy problem

2.1 Logical structure of the standard cosmology

Let's look at the equation expressing (ρ+3P) as the critical density of the universe.

In the second Friedmann equation,

(1/R)(d^2R/dt^2) = -(4πG/3)(ρ+3P)

Matter + Dark Matter (approximately 31.7%) = ρ_m ~ (1/3)ρ_c

Dark energy density (approximately 68.3%) = ρ_Λ ~ (2/3)ρ_c

(Matter + Dark Matter)'s pressure = 3P_m ~ 0

Dark energy’s pressure = 3P_Λ = 3(-(2/3)ρ_c ) = -2ρ_c

ρ+3P≃ ρ_m +ρ_Λ +3(P_m +P_Λ)= (1/3)ρ_c +(2/3)ρ_c +3(−2/3)ρ_c= (+1)ρ_c + (-2)ρ_c = (−1)ρ_c

ρ+3P ≃ (+1)ρ_c + (-2)ρ_c = (−1)ρ_c

Standard cosmology is a universe with a positive mass density of (+1)ρ_c and a negative mass density of (-2)ρ_c. So, finally, the universe has a negative mass density of “(-1)ρ_c”, so accelerated expansion is taking place.

The current universe is similar to a state where the negative mass density is twice the positive mass density. And the total mass of the observable universe is the negative mass state.

2.2. So, what can correspond to this negative mass density?

When mass-energy is present, the negative gravitational potential energy produced by that mass-energy distribution can play a role. The gravitational potential energy model is a model in which +ρ and -ρ_gs exist, and shows that |-ρ_gs| can be larger than +ρ.

For the observable universe 46.5Gly, calculating mass energy and gravitational self-energy

Use a critical density ρ_c=8.50x10^-27[kgm^-3]

Gravitational self-energy = -3.04(Mass-energy)

It is about the same as the dark energy density. Considering additional corrections, the same value as the dark energy density can be obtained.

In addition, this model can explain the inflection point that changed from decelerated expansion to accelerated expansion.

Calculate the inflection point where the negative gravitational potential energy (binding energy) and positive mass energy are equal in magnitude,

Mc^2 = |(-3/5)(GM^2/R)|

R_gs=(5c^2/4πGρ)^(1/2)

If we insert the value of the average density of the universe and roughly calculate it,

We get a value of 26.2 Gly. This is about 5 to 7 billion years before the particle horizon. This is the period when the universe transitions from decelerating expansion to accelerating expansion, which is within the range of observations. Therefore, this model has the potential to solve the dark energy problem as well.

Preprint Dark Energy is Gravitational Potential Energy or Energy of t...

3. Birth and Expansion of the Universe from the Nothing

ΔxΔp≥hbar/2

ΔtΔE≥hbar/2

By the uncertainty principle, quantum fluctuations ΔE can be created, but the problem is that these quantum fluctuations must return to nothing. Therefore, a mechanism is needed to prevent quantum fluctuations from returning to nothing.

Since there is energy ΔE that is the source of gravity and there is a time Δt for gravity to be transmitted, the gravitational self-energy (gravitational potential energy) must be considered. For simple calculations, assuming a spherical uniform distribution, the total energy including the gravitational self-energy is

E_T=Mc^2 - (3/5)GM^2/R

The magnitude at which the negative gravitational self-energy becomes equal to the positive mass energy can be obtained through the following equation.

R_gs=(5c^2/4πGρ)^(1/2)

From the uncertainty principle,

if, Δt≤(3/5)^(1/2)t_p = 0.77t_P, ΔE≥hbar/2Δt =(5/12)^(1/2)m_pc^2

By performing some calculations, we can find the time and energy at which ΔE enters accelerated expansion during Δt, in which quantum fluctuations can exist.

E=4πr^3ρ/3=(5/12)^(1/2)m_pc^2

The meaning of the calculation result is that,

If Δt ≤ ((3/5)^(1/2))t_p, then ΔE ≥ ((5/12)^(1/2))m_pc^2 is possible. By the way, at the moment this Δt exists, the gravitational interaction also proceeds. And, within Δt, the minimum magnitude at which the negative gravitational self-energy exceeds the positive mass energy is ΔE=((5/12)^(1/2))m_pc^2. So when Δt < ((3/5)^(1/2))t_p, the mass distribution expands as the repulsive force dominates the attractive force.

Since the repulsive force due to the negative gravitational self-energy is greater than the attractive force due to the positive mass energy, the corresponding mass distribution expands. Thus, the quantum fluctuations generated by the uncertainty principle cannot return to nothing, but can expand and create the present universe.

===

According to the energy-time uncertainty principle, during Δt, an energy fluctuation of ΔE is possible, but this energy fluctuation should have reverted back to nothing. By the way, there is also a gravitational interaction during the time of Δt, and if the negative gravitational self-energy exceeds the positive mass-energy during this Δt, the total energy of the corresponding mass distribution becomes negative energy, that is, the negative mass state. Because there is a repulsive gravitational effect between negative masses, this mass distribution expands. Thus, it is possible to create an expansion that does not go back to nothing.

===

In this model, the following relationship holds between density and time to enter accelerated expansion:

R_gs=(5c^2/4πGρ)^(1/2) ~ ct

As a powerful constraint, it can be used to verify the correctness of a model.

Please refer to pages 13-16 of the thesis below.

# The Birth Mechanism of the Universe from Nothing and New Inflation Mechanism

Preprint The Birth Mechanism of the Universe from Nothing and New Inf...

It is necessary to study this model because many important problems in astronomy can be solved by considering the negative mass effect due to gravitational potential energy (or gravitational field's energy).

Dear Hyoyoung Choi, I do not understand this:

“By the way, as you may have realized, we have to consider the negative mass effect, the mass defect effect due to gravitational potential energy in astronomy.

And, considering this negative equivalent mass term, some big problems(The singularity problem of black hole, Dark energy, the problem of the creation of the universe from nothing) in astronomy are solved.

1. Singularity problem of Black hole

In the general case, the value of gravitational potential energy (gravitational self-energy) is small enough to be negligible, compared to mass energy Mc^2. However, as more mass is collected, the ratio of (negative) gravitational potential energy to (positive) mass energy increases.

Gravitational self-energy eq. is

U_gs = -(3/5)(GM^2/R)”

What is “U_gs = -(3/5)(GM^2/R)”?

I point out that inertial mass-energy of the apple-Earth-system increase when the apple falls.

Dear Valery Morozov,

"Gravitational field" does not exist within the framework of general relativity. All gravitational phenomena are explained by the curvature of space-time. But no energy or mass is attributed to space itself, and the conservation law of inertial mass-energy is violated because, in reality, the inertial mass includes the energy coming from the curved space. Accordingly, curved space contains negative energy. If we sum the negative energy of space and the inertial mass-energy of matter, we get the conserved gravitational mass.

Dear Radi I. Khrapko

The gravitational self-energy is the total gravitational potential energy possessed by an object, and is also the gravitational binding energy.

Since the gravitational binding energy is defined as the energy that must be supplied to return the bound system to a free state, it is sometimes defined as a + value depending on the document.

https://en.wikipedia.org/wiki/Gravitational_binding_energy

Regarding your claim,

Isn't mass defect due to binding energy completely proven in elementary particle physics?

A body that has a second cosmic speed near the Earth loses speed and, accordingly, kinetic mass-energy as it moves away from the Earth. So the inertial mass of the body-Earth system decreases. And the gravitational mass of the same system does not change. Hence, gravitational mass is not equal to inertial mass.