I understand that we love to write formula for every possible things in the Universe, but the reality is that, we do not know origination of all force, gravity, energy, mass, motivation of all the galaxies' movement. In other hand GTR, STR and even E=mc^2 is not working with our solar system, and they can not complement quantum mechanics science. we have to think out the box, like Einstein did. best

Stefan Bernhard Rüster,

I find your work interesting, but I don’t think that giving up on general relativity is a good idea. If MOND is a fit of Newtonian theory to the observed rotation speeds of stars in galaxies. Your job is to look for the physical causes that make stars move as they move.

I had similar ideas a few years ago when I found a way to calculate the energy of the gravitational field. Now I consider the gravitational field as an independent substance that fills the universe and is able to interact with the gravitational field of galaxies. Thus, the positive energy of gravity can fulfill the functions of dark matter.

Best regards

Javad Fardaei ,

The world around us is not simple. An exact description of it is impossible without formulas making this description compact visible. GTR successfully copes with its task, accurately and fully describes gravitational interactions. Simple, scalar and vector, theories do not work. The problem of quantum gravity has not yet been solved. No one knows how to do this. I am sure that quantum mechanics will help create a "microscopic" theory of gravity. But this is still a long way off.

Best regards

I adhere to the standard point of view. Motion in a gravitational field (as in an electric field) is determined by the equation of motion, which includes velocities and Christoffel symbols obtained from the field metric.

It is less known to find the forces acting on the body through vacuum tension. If in electrodynamics this problem is solved through boundary conditions. In a gravitational field, boundary conditions cannot be introduced. You can solve the problem only at the level of small bodies or small parts of the body, but the problem is solvable.

 The momentum energy tensor in gravity contains the necessary tension. If we know the tensor of tension, we can find strength. How is this done, for example, in electrostatics.

Preprint Energy-momentum tensor of gravitational field and approximat...

Stefan Bernhard Rüster : "GRT is not complete without a gravitational energy-momentum density tensor E_{mu nu}..."

That's right. However, there is no such tensor in GR. This is embedded in the Einstein equation. Only by abandoning this equation can we correctly introduce the field momentum energy tensor into the theory.

Preprint The General Theory of Relativity, a New Iteration

In order to make the Einstein equation physically accurate, it is necessary to add the field momentum energy tensor to the right side. But then the Einstein equation will become useless in the right side another unknown appeared.

But if it is useless as an equation, then this CORRECT equality allows us to obtain from the metric of space the tensor of momentum energy by anything, including the tensor of energy — the momentum of the field tik if the metric tensor describes empty space.

tik = c4/(8π G) Gik

Getting rid of the Einstein equation, we got a very useful equality.

The story is like that. In the famous article of 1913, written in collaboration with Grossmann, Einstein proposed the equation of the gravitational field in which there were two sources of the gravitational field, the energy of the momentum of matter, and the energy of the momentum of the field. However, it was not possible to obtain the covariant equation. Only two years later, Einstein made the right choice. He removed the second field source from the equation. The basis is the smallness of the energy of the gravitational field in comparison with the energy of matter. The equation became covariant and its solutions corresponded to the known data on the precession of the orbit of Mercury.

The assumption that the energy of the gravitational field is small does not cover almost all known empirical data. In this case, we should expect plausible results from the Einstein equation. Which was brilliantly confirmed over a hundred years.

However, singular solutions and the lack of a local law of conservation of momentum energy caused concern. This can be explained by the inaccuracy of the equation.

Stefan Bernhard Rüster: " I also have a question: Why do you obtain tau proportional to diag(-1, -1, 1, 1) ? Why not prop to diag(1, -1, -1, -1) for example? "

There is an analogy with electrostatics. In electrostatics, the energy-momentum-tension tensor is proportional to (1, 1, -1, -1), i.e., the first term corresponds to positive energy, the second attraction along the lines of force, the third and fourth repulsion in the transverse directions.

CLASSICAL ELECTRICITY and MAGNETISM

PANOFSKY, PHILLIPS

ELEKTRODYNAMIK

von ARNOLD SOMMERFELD

It should be expected that in the case of gravity, the momentum energy tensor will have the opposite sign (-1, -1, 1, 1), however, in fact, I obtained from the equation of the gravitational field a metric with the momentum energy tensor (-1, 3, 1, 1 ) This is strange but does not interfere, since the tension along the lines of force in gravity is not significant.

Best regards

Dear Stefan Bernhard Rüster ,

I found an e-copy of one of the books on the Internet

https://www.fisica.net/ebooks/eletricidade/PANOFSKY%20AND%20PHILIPS%20-%20Classical%20Electricity%20and%20Magnetism%202nd.%20Ed..pdf

This book preceded Jackson's famous book.

R_{mu nu} = 8*pi*G/c^4*(T_{mu nu} - T/2*g_{mu nu}).

This is the Einstein equation in the form that Einstein wrote for the first time. Just now I noticed, there is an analogy with the Maxwell tensor of tension. In electrodynamics, the second term is half the energy density.

The physical meaning of this member was not clear to me. Now a lot has cleared up. Thank.

Dear Stefan Bernhard Rüster ,

The equation R_ {mu nu} = 8 * pi * G / c ^ 4 * (T_ {mu nu} - T / 2 * g_ {mu nu}) is the usual Einstein equation.

Since R = - 8 * pi * G / c ^ 4 * T (Obtained after the convolution of the Einstein equation)

Best regards

In a previous post, I fixed a mistake in the formula.

About terminology

The stress (tension) tensor is the spatial part of the energy-momentum tensor (in Russian literature). Therefore, it is correct to call it the tensor of energy-momentum-tension. Sometimes called him Stress–energy tensor.

https://en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor

Now about energy and tension in electrodynamics. Energy can be estimated from Coulomb's law. Perhaps this was the first time J.J. Thomson. At least he did a lot in that direction. Tension tensor, of course, was invented by Maxwell, but he used the ideas of Faraday. The tensor of tension is not similar to the tensor of tension of ordinary deformable media. This and SRT made it impossible to adequately describe electrodynamics using the ether hypothesis.

The theory of relativity is applicable to mass-dependent movements; it is much more a mathematical construct than a physical one, in empirical terms. Propagation of the theory, in science and in the open public, actually started after A. Einstein received the Nobel prize for the discovery of the law of the photoelectric effect, in 1922.

https://www.nobelprize.org/uploads/2018/06/einstein-lecture.pdf

Einstein made a quantum theory from the Planck hypothesis. He founded the quantum theory of solids, quantum statistics and the formula by which the laser works.

The theory of relativity describes not only the motion of bodies. But it also explains the passage of time, and the gravitational field.

However, in GRT the situation is more complicated.

The gravitational field, possessing energy, should become the source of the gravitational field. But to ensure covariance, Einstein had to exclude the energy of the gravitational field from the field equation. The field equation began to give the right solutions. This determined the success of the theory.

In general relativity there is a reasonable way to calculate the energy-momentum tensor. If we know the metric tensor, then its Einstein tensor gives the energy-momentum tensor. Unfortunately, solutions of the Einstein equation for vacuum give a zero energy-momentum tensor. This makes it impossible to fulfill the usual law of conservation of energy-momentum in GR.

This hole in theory is patched like this. Using the conservation law, a pseudo-tensor is found that helps to write down the energy conservation law for closed systems. There is almost no practical sense in this. Einstein admitted this. The results are highly dependent on the selected coordinate system.

In the article "On Gravitational Waves" (1918), Einstein writes: "There is no reason to force t44 to understand the energy density of the gravitational field, and by (t14, t24, t34) - the components of the gravitational energy flux density."

Dear Stefan Bernhard Rüster ,

"do you see any chance to include the energy-momentum density of the gravitational field into Einstein's field eqs.?"

Not the slightest chance. And this is not only my opinion. This is one of the reasons why it is necessary to abandon the Einstein equation.

Dear Stefan Bernhard Rüster:""This is one of the reasons why it is necessary to abandon the Einstein equation."

What are other reasons?"

There is another way to find the field metric. Sometimes equations are not needed for this at all. I found a general view of the gravitational field metric

ds^2=e^(h(x, y, z)/c^2 ) (c^2 dt^2-dx^2-dy^2-dz^2)

The Einstein tensor from this metric is the momentum energy tensor. For empty space, this is the field momentum energy tensor.

Best regards

Then everything was simple. Knowing the general form of the metric, you can find its simple property. The product of two metrics of a given kind is a metric of the same kind:

 e ^ (h / c ^ 2) (c ^ 2 dt ^ 2-dx ^ 2-dy ^ 2-dz ^ 2) X e ^ (u / c ^ 2) (c ^ 2 dt ^ 2-dx ^ 2 -dy ^ 2-dz ^ 2)

It follows that the parameter h forms an addition group. This means that the equation for the parameter h must be linear. And the simplest linear covariant equation is known

\ Delta h - {\ frac {1} {c ^ {2}}} {\ frac {\ partial ^ {2}} {\ partial t ^ {2}}} h = -4πGT

That’s all GR. The wave equation is solved easily methods for solving it are described in all the textbook. The gravitational field for a concentrated mass is not easily found (Schwarzschild problem). Solutions have the correct energy density and no singularities.

Preprint New version of the general theory of relativity (Initial pri...

Article The General Theory of Relativity, a New Iteration

Valery Borisovich Morozov GRT deal with energy-momentum tensor but material. As I think zero energy-momentum tensor is not the same with zero Ricci tensor. Zero Ricci tensor means vacuum, that is no material in the flux.

Nguyen Le Anh Le Anh : " Zero Ricci tensor means vacuum, that is no material in the flux."

That's right. But this is only true for solutions of the Einstein equation. In my version of general relativity, the Einstein equation is replaced by another. The new equation gives solutions with non-zero tensors of Ricci and Einstein. Then the Einstein tensor is proportional to the tensor of energy of the momentum of the gravitational field in vacuum. The gravitational field has energy and momentum.

Preprint On Einstein equation and energy of gravity field

Article The General Theory of Relativity, a New Iteration

" so that Einstein's field eqs. are correct, but not complete and therefore USELESS."

I fully share this statement Stefan Bernhard Rüster. And I propose a way out of this difficulty.

I have demonstrated that the rejection of the Einstein equation and the construction of the space metric on the principles of general relativity are productive. And it gives results compatible with the known empirical data and the law of conservation of energy.

Dear Stefan Bernhard Rüster ,

Thank you for the good words addressed to my work. Of course, not everything is clear. Solutions should not be limited to isotropic space. However, the main difficulty is the rejection of the results by that part of the physical community that uses GR in astrophysics. Of course, no one is against the momentum energy tensor. But singularities, i.e. black holes are a familiar object of astrophysics, although their mathematical interpretation is not very strict. In the new equation of the gravitational field there are no singular solutions of either black holes or a big bang. A huge number of astrophysicists connected their lives with these objects. Naturally, these astrophysicists do not feel joy about my results. And peer review turns into farce.

The next question is, how much is the energy of the gravitational field attached to the field? Is the field energy always proportional to the square of the field? Solutions for a uniform field and point source follow this rule.

However. An axisymmetric push field solution does not follow this rule. At a large distance from the center of symmetry, the field is still proportional to the square of the acceleration of gravity, but the bodies are pulled out. When approaching the center, the field rapidly decreases in the region of the Schwarzschild radius, but the field energy continues to grow. Thus, it is possible to imagine this formation as a kind of negative energy, which is the source of the buoyant field.

How real is such an education? There are many possible candidates. In space, we observe formations called "bubbles", perhaps this is just the gravitational field that expels the gas. But most likely these push regions correspond to Fermi bubbles. Formations accompanying our galaxy. They are sources of hard radiation, and the hardest radiation (gamma rays) emits a center of formation.

Preprint New version of the general theory of relativity (Initial pri...

Einstein equations describe the structure of gravitational fields in the curved space-time. Tit is evidently that it is impossible to find the general solution of this very complete equations. But we can obtain solutions for different partial case, i.e. for gravitational fields of different objects --- from planets to the Universe. We can consider as objects in the emptiness (Schwarzschild solution for distant object) and objects possessing inner structure (Schwarzscild for inner structure of the Sun as a regular star, space-time of the Universe as for empty models so and for fulled of substance and radiation ...) It is very impossible to remember that gravitation is due to the course of the time: if x^0 = ct, gravitation is absent.

Maxwell tensor for electric field, those (1, 1, -1, -1) if Ex ≠ 0.

The tensor of the gravitational field tension must have the opposite sign (-1, -1, 1, 1). In order to ensure the attraction of bodies with the same "charges". And the gravitational field is negative.

But the tension along the force line does not affect the result. Therefore, this term is not defined (-1, x, 1, 1). Why exactly x = 3, I do not know. But that doesn't bother me.

Respect, Valery

The energy-momentum tensor for various cases.

I - gas with pressure P.

II - Magnetic field.

III - Relativistic particles moving in the x direction.

IV - Relativistic particles moving in the opposite direction .

Dear Stefan Bernhard Rüster

Great, then you're on the right track. I found this by accident in the book of Zeldovich and Novikov. I decided to inform you.

The Maxwell tensor can be written as the tensor of tension (1, -1, -1). so is the stress tensor. (-1, 1, 1) These are just reciprocal quantities. Maxwell-Faraday tension is whiter than old-fashioned. I have a predominance of old literature. http://femto.com.ua/articles/part_1/2140.html I often confuse this. In Russian it sounds like.

Respect, Valery

Dear Stefan Bernhard Rüster

Calculating through the Einstein tensor, I believed that the tensor in the spatial part is worth the tensions. It seems that this is not so. Then, from general considerations, the momentum energy tensor should be (-1, 1, -1, -1). The tensor that I calculated (-1, 3, 1, 1) is the homogeneous field tensor. The physical meaning of this is that two endless plates will be squeezed by an external field. Strange, but it does not contradict anything. Although this must be double-checked.

I looked at the data for other fields. It turned out that everything is complicated. I was not interested in tension, they did not look very clear. First of all, I was interested in energy density. It is important to know what the equations of motion look like. The Christaffel symbols are important here.

Tensions are needed to ensure that the law "action is equal to counteraction." Not everything is clear yet. So far, I know for sure that the law of attraction in the field of weak fields practically coincides with the law of attraction of the Schwarzschild solution. Tension (tension).

While such a thought is ripening - in a vacuum there are not only stresses due to the action of bodies between themselves, but also stresses when the field interacts with itself. This is confirmed by the fact that the stresses at a sufficient distance from the body have a signature (-1, 1, -1, -1), but near the body (there used to be the Schwarzschild radius) the sign of the tension changes. Why not? So far this is too crude a thought. It is necessary to think still.

Respect, Valery

Dear Stefan Bernhard Rüster

It is necessary to supplement the data on the Einstein tensor (and the momentum energy tensor) for the buoyancy solution. Here the signature is (-1, -1, 1, 1), but for r

The energy of the gravitational field. If there is gravitational field energy, it does not necessarily depend on the gravitational field, specifically on the acceleration of gravity. A trivial example of gravitational waves. If there is a tensor of the energy of the momentum of the gravitational field, then this also means that the field has momentum. It follows that the gravitational field can move independently.

Dear Stefan Bernhard Rüster,

I see no problems with Mercury and even with closer systems. The gravitational radius of the Sun rg ≈ 3 km. This is negligible compared to r. Even in a tight system of neutron stars PSR J0737-3039, the orbit is 800 thousand kilometers. So at these distances, we can assume that the Christoffel symbols coincide. Only Γ100 and Γ122 are important here. They are almost the same for Mercury and even PSR J0737-3039.

The only Christoffel who is significantly different from mine is Γ111 it has the opposite sign. This can only matter at significant radial speeds. Maybe this explains the effect of the "Pioneers"?

A rather complicated equation for the deflection of light by the sun. Involved there Γ100 , Γ122 and Γ111. The usual calculation through the coordinate speeds of light is not good here.

Respect, Valery .

So far, I have only talked about the negative energy of the gravitational field. It may seem that gravitational energy is always negative. However, the energy of gravitational waves, judging by some data, is positive.

Approximate solutions for gravitational waves have a nonzero Einstein tensor. Calculations of energy density and flux density give positive values ​​of these quantities.

Another solution for a homogeneous empty space gives a metric that depends only on time. The calculation of the momentum energy tensor gives a positive nonzero energy density that numerically coincides with the critical Friedmann density. Stresses in space in this case correspond to negative pressure. In this case, the metric reproduces the Hubble law, which has an increase in redshift at large z.

Article Cosmology of Uniform Empty Space

Preprint Dark energy as zero energy of gravity field

Stefan Bernhard Rüster "It would be very important to check if your theory gives the correct result for the perihelion shift of mercury (43"/cy)."

From the zoo of metrics that we looked at just metric

ds2 := exp(-rg/r)d(t)2 - exp(-rg/r)*(d(r)2+r2*d(theta)2+r2*sin(theta)2*d(phi)2)

remained untested. Let's check. This will make our article more complete.

Preprint A critical analysis of Schwarzschild-like metrics

Best regards, Valery.

Historical studies of the creation of the Einstein equation often claim that this work ended with the replacement of the Ricci tensor with the Einstein tensor. Formally, this was not the case. In November, Einstein wrote the vacuum equation Rik = 0. Then he tried to solve the problem of the motion of Mercury around the Sun and obtained the correct value for the displacement of the perihelion of its orbit. Einstein realized that he was on the right track.

Then he supplemented the previously obtained equation Rik = κTik. Before the modern equation in the form Rik = κ (Tik - 1 / 2gikT).

Dear Stefan,

Remember how Straumann and I figured out when it was time for Einstein to write his equation? I re-read a large number of Einstein's articles and found a recipe for writing the equation of the gravitational field, taking into account the energy-momentum of the gravitational field. I wrote an equation like this. But there seems to be no analytical solution. You have to count numerically. This requires great care, while I am writing an article. I don’t think this changes general relativity much. The effects are only noticeable in strong fields.

Best regards Valery

Dear Stefan, "General relativity is on the wrong track since 104 years and..."

This is the normal way of development of general relativity, all this time it worked and works perfectly. It just has a limited scope. you just need to improve it or create a new theory. You and I know what needs to be done. It is necessary to introduce the momentum energy tensor into the Einstein equation. Einstein tried to do it. But at the end of 1915, Einstein calculated the displacement of the perihelion of Mercury using the equation Rik = 0, and this inspired him to accept the truncated equation without the field tensor.

Best regards Valery

Dear Stefan, Dear all.

I am sure that it is necessary to figure out what is λ.

Three pages from Møller's book can explain everything.

Best regards Valery

Dear Stefan,

I am sure your result is very important. This either solves the problem of the motion of stars in the galaxy, or at least gives direction for further research.

However, this is not the final solution to the energy problem. The momentum energy tensor must be a quadratic function of the Christoffel symbols and / or the derivative. Everything is easily verified by passing to the Newtonian limit.

This is the task of classical general relativity. But there is also the problem of quantum general relativity. There are absolute zero results here. All the numerous works entitled "Quantum theory of gravity" do not solve anything. Clumsy models.

In the meantime, I am finishing the calculation according to the correct equation of the gravitational field (in which the cosmological term remains in the same form). Everything will be solved after calculating Grr of the Schwarzschild problem. Gtt has already asymptotically coincided with the Schwarzschild solution.

Best regards Valery

I was pretty sure the new equation would bury the singularities forever. But that did not happen. However, the singularities have become "softer" - the components of the metric do not change sign.

The work continues.

Dear Stefan,

The cosmological constant appears on the right-hand side of Einstein's equation (3) as part of operator (4).

An operator property similar to the conservation law (11) is an additional requirement for the operator. Read Moeller carefully, everything will become clear.

Best regards Valery

Dear Stefan,

Another version of the introduction of the cosmological constant is introduced in the course LL-2 §111. Isotropic space.

Best regards Valery

Dear Stefan,

The article is good and thorough. But it leaves room for new research.

The momentum energy tensor (for example, the gravitational field tensor) is always equal to some Einstein tensor. But the opposite has not been proven. This leads to the field of hypotheses.

Best regards Valery

Stefan:"As far as I know Einstein searched his whole life for this conservation law but he never found it. Right?"

Testing the conservation law is an indispensable element of any theory. Unfortunately, today's theory does not have a sufficiently complete conservation law. Equation (11) cannot have such a law as a consequence. This is just a property of the operator M ^ k_i. The equation holds for any Λ.

From equation (11), the value of the coefficient at R is found, while the lambla remains arbitrary.

Best regards Valery

Dear Stefan,

I tried to understand your work. Perhaps I would have done it if I had more time and effort. I got the impression that you yourself do not understand everything clearly enough. My attitude towards the cosmological constant has not changed. I think that even in cosmology one can do without it. For example, I calculated the critical Friedmann density without resorting to either the Einstein equation or the Friedman equation

Article Cosmology of Uniform Empty Space

It is necessary for me and you to think.

Best regards Valery

Dear Stefan,

You will probably be interested in what ends and means led Einstein to the necessity of introducing the cosmological constant. Note that it is entered on the left side of the Einstein equation. Those. as an addition to the operator (Einstein's tensor).

I am attaching the article.

Best regards Valery

Dear Stefan, "Please let me know. Please ask questions."

Perhaps I and many others see no reason for the destruction of modern concepts of the cosmological constant. In particular, I do not see anything in this quantity other than a member of the operator of the gravitational field equation. Which preserves the covariance of the operator and the equality to zero of the divergence of the operator. This is the operator introduced by Einstein, Möller and others.

If possible, explain why this approach is bad.

Best regards Valery

Dear Stefan,

Note that Einstein introduces a constant as a hypothesis in Poisson's equation. But in the equation of the gravitational field, it does not change the covariance and the main property of the operator of the equation, namely, the equality of its divergence to zero.

Best regards Valery

Stefan, I'm sorry, I will be very busy communicating sporadically.

If you have time and effort.

Valery

Dear Stefan,

I am finishing the numerical calculations of the solution of the new Einstein equation, as usual the absurd mistake slowed down the matter. But there is still work to be done. An avalanche of work.

Best regards Valery

Dear Stefan,

Let's not get distracted by hypotheses of dark matter and especially dark energy. This is all far from physics. I prefer testable theories. In this case, it is the theory of gravity. There is no reason to doubt her success. We know that Einstein's equation lacks the source of the gravitational field in the form of the energy of the field itself. Do you think that the term with the cosmological constant is the very energy-momentum tensor of the gravitational field?

But it does not contain the energy of the gravitational field. Since the constant energy of the gravitational field is calculated:

Preprint On energy of static gravity field

Best regards Valery

Dear Stefan,

So you say that the sum

-1 / kappa Gik + Lambda / kappa deltaik

is the tensor of the energy of the momentum of the field?

But Gik + Lambda gik

is the operator of the Einstein equation. How did it happen that it all came down to essentially a renaming of values and introducing a multiplier?

How does this compare with my post, and the article where the energy of the gravitational field is found?

Best regards Valery

Dear Stefan,

Except that your results do not fit into the ideas known to me. I see no chain of related evidence leading to this conclusion. Maybe we will start with the first element of this chain without trying to explain everything at once.

And please react to my comments. A discussion cannot be a monologue.

Best regards Valery

Dear Stefan,

In practice, we solve own equation

Gik + Lambda gik=0

or equivalent

Rik + Lambda gik=0

We must find the fundamental tensor and Lambda. Или это уже не так?

Best regards Valery

Dear Stefan,

"No it isn't. There is no operator in Einstein's field equations."

I am interested in general relativity. I cannot imagine a theory without the gravitational field equation. Have you canceled the equation? How will you solve the problem of gravity? For example, how do you find a point source field?

What is it?

Rik + Lambda gik=0

By the way, if I solve the equation in the standard way, then Lambda=0

Best regards Valery