Taking the available definitions in Wikipedia as source we find:

The Lagrangian of matter is the differential of all potential energy of the matter and the kinetic energy of the matter. L = T-V. Where L is the Lagrangian, T is the kinetic energy and V is the potential energy.

Potential energy is a value that is dependent on the involved amount of mass. In the case of a universe as we have in the question the density of matter is constant and can be given with the value Rho. To see what is the potential energy of a test mass at a point in this mentioned universe we can’t use the normal function E=-G*m*M/R. This function is only valid as long as we stay with our test mass m outside of the large mass M. In our case we are inside the mass. There we only count the mass in a sphere below us towards the centre. The function becomes E=-G*m* 4/3* PI * Rho * R2.

Every point of space is in the centre of its own light cone. That means that for every point the potential energy is zero. The sum of all potential energy of all points of matter in space under our condition of distribution is zero.

Our definition of homogeneous an isotropic distribution also has the consequence that from the viewpoint of space there is no movement. Any particle that would move would need to be counteracted with another particle moving in the opposite direction in order to maintain uniformity. Because we talk about energy density we also can say all photons moving in one direction are balanced by photons movies in opposite direction so that for space there is no movement. This results in the conclusion that the sum of all kinetic energy also is zero.

The Lagrangian of matter in our universe is according to this deduction zero.

With that the energy-momentum tensor Tuv is zero, so space is as flat as if it is when it is empty.

> Is this correct or not?

Not.

Why don't you start by studying some introductory material on standard cosmology and the Friedman equations? Including f.i. this page: https://en.wikipedia.org/wiki/Friedmann_equations.

Hello Kåre Olaussen. I have studied these pages and some more. These equations are derived from by putting the FWRL metric in Einstein's GR equations. But when you study the derivation as is given by Leonard Suskind then you see how he tries to derive that with Newtonian gravity theory. His derivation makes some assumptions that are not true. From there I studied how large cosmological simulation programs calculate the gravitational effects. When you follow these methods than you come to the conclusion that space remains flat for any energy density in it.

In this question here I try to understand better how you can define the Lagrangian of matter under isotropic and homogeneous conditions. I have had more discussion about what I stated above about movent. I think now that the condition that space is isotropic excludes movement. This means that on the scale where we can describe space as isotropic we have movements that can be ignored because they are too small compared to the scale. As a result the Lagrangian has the value 0.

I know that this is in contrast with the Friedmann equations. But either the gravity calculations with the large space simulators are wrong or space is flat and only the inhomogeneities give rise to local curvature.

You can read the document that I have uploaded here on my project or wait to see it soon published in the Journal of High Energy Physics, Gravitation and Cosmology. Also there they don't agree that I am right. But just a 'no' is not enough. Show me, after reading and understanding my deductions in chapter 3, where these deductions are wrong and then I will thank you that you helped me lose a wrong view of reality. Until now all scientists that have responded on my document have not pointed me on any error. In case you prove my deductions to be wrong we can make a paper that all simulations of the cosmos are invalid and otherwise we have to conclude that there might be a problem with current cosmology models of expansion.

I don't know which derivation by Susskind you are referring to, but I suspect it is kept consistently within Newtonian mechanics and gravity. It must (in some limit) agree with a consistent derivation within General Relativity. But you cannot threat the two approaches as a smørgås-bord where you pick a little bit from here and a little bit from there.

In a direct Lagrangian GR approach, based on fields, you can have L = T - V, with T and V varying in time only (slowly, hence with T

Kåre Olaussen,

I am very happy with your response. You seem to know the GR very well. In my article I don't use the GR to derive the conclusion. I have studied the derivation of the Friedmann equations and from the mathematical point there is little you can object to. Only in the Minkowski space the time has no restriction in its direction. When you go in that space from one event U1 using a relative step U2 to arrive in U3, then from the mathematics using -U2 to go back to U1 is no problem. Here the minus sign means the opposite direction. But in real space once I leave U1 there is no path back. So when you make the = sign then you mean left is equal to right and right is equal to left. So U1 + U2 = U3. Then also U3 - U2 = U1.

But you should have the operator like in a programming language := .

The you would say U3 := U1+ U2. And then U4 := U3 - U2. You see I don't come back to U1 but end up in U4.

Since it is not clear to me how that is taken care of in GR, I used the algorithmic approach of the space simulations.

Now I want to see if from the approach with Lagrangian I come to the same result.

Thank you,

Paul Gradenwitz

No. You have to take into account invariant volume element sqrt(-g). It depends on g^{uv}.

So I take the Lagrangian of space filled with 10112 erg/cm3 of vacuum energy. That would be no potential energy and I also don't see kinetic energy. Then I multiply that with the square root of minus the determinant of the metric. To my understanding that result value is zero. When I then differentiate I still get zero. After that I multiply it with a number.

In case you say that the Lagrangian is not zero you have the -g that is a function of the guv so I would be interested what is that in a homogeneous and isotropic universe. Is there any change from one place to another place or time? Is there a change that could give a value after differentiating?

I only can find that the diagonal of the guv is (-1 1 1 1) or (1 -1 -1 -1) and all other values are 0. There are other metrics like the Schwarzschild metric that are different but they don't fit with homogeneity.

I am eager to learn if there is a different view.

@Paul

The Lagrangian of a space filled with a non-zero amount of vacuum energy is not zero, regardless of how one decides to classify its individual contributions as kinetic or potential energy, or as a cosmological constant term. Hence, your reasoning on this point is wrong.

For the Friedman model (which describes a spatially homogeneous and isotropic universe) the diagonal terms of the metric are, in Robertson-Walker coordinates (corresponding to spherical coordinates), with your choice of metric signature,

(-1, a2(t)/(1-kr2), a2(t)*r2, a2(t)*r2*sin2(θ) ).

All off-diagonal terms are zero, and k one of 1, 0, -1.

For k = 0, which is the observationally preferred case, the diagonal terms of the metric can equally simple be expressed in cartesian coordinates,

(-1, a2(t), a2(t), a2(t)),

with all off-diagonal terms being zero.

There is a causal structure in GR, so that two space-time points (events) is separated by either a time-like, light-like, or space-like interval. This has physical consequences, which can be rather intriguing in a universe with varying expansion rate. But I don't see how the mathematical rules you provide can helpful for the exploration of such consequences.

I made a fast read through your paper. As I understand it, the main point is that a particle at rest at a given point, say described by spherical (Robertson-Walker) coordinates r, θ, φ, will continue to stay at rest at the same point, due to rotational symmetry around that point. This is correct, but the conclusions you draw from this are not correct. The Friedman model provides an explicit counter-example.

Kåre Olaussen,

Thank you very much for the patience you have to explain me this. May I ask you to explain a little bit further. You say "There is a causal structure in GR,.. "

In my calculations I make sure that I keep track what points can have a relation to what other points. I understand that you state that this is taken care of in the GR. Could you explain me that. When I see how the GR equations are developed in the lecture notes of Sean Carroll, then I note that he starts with the theorem of Gauss. Wherever I look I see that we need to step out of the centre and take the mass in a sphere below towards the centre. I have shown that inside that sphere there is no matter that has a relation with a point outside that sphere. So what matter is interacting with what space points and is it correct taken care of that matter only can influence space points that are in the future of that matter?

I can't find this speciality. I can put restrictions on the distribution of matter such that without this time direction restriction the equations will give perfect results. But then I have to break the homogeneity.

You see I am not satisfied with your answer because I stand on the point that I can't see that the special condition for time is taken care of. Your answer is only hinting that there is something and as such is no answer to my problem.

If guv = (-1, a2(t), a2(t), a2(t)) then g would be -a6(t) That is only dependent on time. So all the space directions have have a derivation to zero. In Tuv then only the term T00 would have a value if the Lagrangian is not zero. Is that correct?

Thank you for your explanations.

Do not mix a Lagrangian and a solution to equations of motion. A Lagrangian is a function, not a quantity, and it depends of the matter fields, the metric and all their space-time derivative. But when considering a homogeneous and isotropic Universe, we consider a space coordinates independent solution. Thus, while calculating the energy-momentum tensor you should use the full Lagrangian with full space-time dependence. But after that you may set space derivative equal to zero, substitute the energy-momentum tersor to the Einstein equations and solve them for space-independent metric. That's all.

Michael Lashkevich,

In general the Lagrangian is a function, but special in this case where we have a distribution that is independent from the space coordinates. When I find that the energy-momentum tensor is defined as the derivation to space and time of the Lagrangian and we have no dependency to space then I conclude that derivation to space will give a zero. In order to get curvature I have to find that a geodesic has non-zero coefficients in its space dimensions for its second derivative. If I have a space independent situation then I think there can't be a solution that becomes space dependent.

When I then take a detour and start to compute with the method that space simulators use for cosmological evolution studies, then I get as a solution that space is flat in this situation. This strengthens my opinion that space doesn't depend on the constant part of the density for curvature. It is only dependent on the part that is inhomogeneous.

Till now nobody has shown me a complete calculation where is shown that all conditions are covered. Also your answer points to the equations in big steps but it doesn't, in detail, state the solution in my case.

I ended my last writing with a clear question about T00.

When I get an answer with again a general computation method without filled in values then again I only can go back to the documentation that I can find online and get these general values.

If you agree with me that T00 is the only component in this case that could have a value different from zero because I have not restricted that space is expanding or not, then I am interested how that could be enough to curve space.

Thank you for your response.

Paul Gradenwitz

Paul> When I find that the energy-momentum tensor is defined as the derivation to space and time of the Lagrangian...

But that is not how the energy-momentum tensor is defined. It is not defined according to how the metric changes with respect to space and time, but according to how the action -- the space-time integral of the Lagrangian density -- changes when you make a general first order change in the metric (away from a given metric, f.i. the metric of the Friedman model).

I.e., it seems to me that you confuse two very different concepts. Unfortunately, the definition in the Wikipedia article you link to is not quite correct, but this makes a difference only where the Lagrangian density depends on space-time derivatives of the metric.

Paul> In order to get curvature I have to find that a geodesic has non-zero coefficients in its space dimensions for its second derivative.

Yes, that is one possible starting point. But then you must consider a general class of geodesics, not only those corresponding to particle at rest in an expanding Friedman universe. Your symmetry argument is only valid for such geodesics.

Kåre Olaussen,

You are right that I should do that. But in my case I have a sphere as reference with the centre in my point and the surface at the start of space. Outside of space there is no movement. So when the geodesic of my point is congruent with the geodesic of the sphere then in this case there is only one geodesic. Any moving object inside space would break homogeneity. So I would need to differentiate between the part that is homogeneous and the part that is not. In the second part I have movement. In the first part I have no movement. Can you agree with that?

I have asked my question about the Lagrangian as a real question. You point me on more details that clarifies things for me.

In the earlier post you state how "the action changes.." This is for a better understanding of the equations. Could you then also tell me how this change is in the homogeneous case? What can change in a situation that is defined constant?

I thank you for your patience.

Paul Gradenwitz

Paul> the geodesic of my point is congruent with the geodesic of the sphere then in this case there is only one geodesic.

It is very likely that what you mean to say is very different from how I interpret your words, but I cannot agree with your argumentation. Rotation symmetry alone is not enough to deduce that all geodesics through a point are equivalent. In most cases they differ, dependent on their velocity at that point. There are some exceptions in space-times with additional symmetries (Minkowski space, de Sitter space, anti de Sitter space), but they do not quite fit with the currently best model for our universe. And even (anti) de Sitter space has curvature.

Paul> Could you then also tell me how this change is in the homogeneous case?

Assume you start with a metic which describes a space-time whose spatial part is homogeneous and isotropic (in a preferred class of coordinate systems). To find the energy momentum tensor, you change the metric by an arbitrary amount εδgμν(x), where δgμν(x) are the 16 components of a symmetric 4x4 matrix (hence there are only 10 independent components) which depend on the space-time coordinates x is a twice differentiable way. Next consider the function S(ε), defined as the space-time integral of the Lagrangian density, and its derivative S' ≡ dS(ε)/dε evaluated at ε=0. The latter can be manipulated, sometimes by performing partial integrations, to a form like

S' = -½ ∫ d4x √-g Tμν(x) δgμν(x)

From this the components of energy momentum tensor Tμν(x) can be read out. In a post above I indicated the result of such calculations: Tμν(x) turns out to be a diagonal matrix, whose entries (interpreted as energy density and pressure) can only depend on time. A little more about this can f.i. be found in the Article Lecture Notes on General Relativity

Kåre Olaussen,

I am impressed with the detailed answer you gave me. (I didn't find yet a method to get all the characters here that you use, I have to learn).

I did read the lecture notes of Sean Carroll. I find in there how he explains how the left hand side and the right hand side of the equation come together. I have no discussion with your mathematics. These mathematics are proven correct. I have problems with the application of them to the reality. In his deductions he uses the Poison theorem that is a derivation from the Theorem of Gauss.

When I look to that, and try to collect what event of space-time communicates with what other event, to see if there can be a relation that can result in a functional value, then I notice that the volume, that is enclosed by a surface, always holds events that are in the future of events outside of it. So then I can't find a result. So I change to a different method and calculate direct without the GTR equations. That is how cosmological simulators work. Their result is accepted so my approach should lead to the same result. However it disagrees with GRT.

In the first point is the situation that I am not talking about any mass inside space at my rotational symmetric point but about space itself. For different masses with different speeds there can be different geodesics. For the centre of my light cone I have to focus what space will do there. My space coordinate will not move with respect to other coordinates. There might be an expansion or contraction but nothing else. So all your other geodesics are irrelevant for space. They are only relevant for a mass inside space.

You show me how to start with the homogeneous case in the classical way. But then you give the answer in parameters and I then lose the overview what parameters do change so that they give a value in differential operations because I started to set every change to be zero.

Thank you so much that you took time to answer.

Paul Gradenwitz

Paul> I didn't find yet a method to get all the characters here that you use

There is a webpage, http://math.typeit.org, where you can type in greek letters, mathematical symbols, etc. And copy/paste the results from there to here.

Paul> In his deductions he uses the Poison theorem that is a derivation from the Theorem of Gauss.

I cannot see anything like that in his discussions of cosmology (ch. 8 in the notes), which is the topic of relevance for for your question.

The Einstein equations are a set of partial differential equations with a (locally) causal structure for the cases of interest. It is not so easy to see when considering only spatially constants solutions. Which is found by inserting a symmetric ansatz into the equations, and solving the resulting simplified equations (time-dependent only). One may try to rewrite the equations in integral form, to see how physical properties at one point only depends on what happened on or inside its backward light cone. But I am unsure how straightforward that would be, and how much extra insight that could provide. The Maxwell equations, with prescribed currents, can be rewritten in such a way.

One way to see the effects of causal behaviour is to verify that small disturbances evolve with finite propagation speed (the speed of light, or slower). That in fact have important consequences when the universe expands very rapidly. The observed fluctuations in the cosmic microwave spectrum are interpreted to be a fingerprint of initial quantum fluctuations, with a form that is preserved during expansion, because of causal evolution.

Your verbal arguments are not precise enough to pin down potential fallacies. They must be formulated mathematically. The basic ingredients are symmetries and causality, from which various realisations of the Friedman models can be deduced. No more, no less. From what you write I would expect to find a k=0 (because you implicitly sneak in a flat space assumption), radiation dominated (with pressure equal to one third of the energy density) universe.

Paul> So I change to a different method and calculate direct without the GTR equations.

If your mathematical model don't agree with GTR, you should not be surprised that its results also differ. To some extent. I doubt that the cosmological simulators you mention disagree with GRT within their expected region of validity.

Paul> I have to focus what space will do there

In GR that means to calculate how the metric tensor will change.

Kåre Olaussen,

See Sean M. Carroll (1997, page 109), the derivation of the right-hand side of the GR equation starts with Poissons's equation (4.35)

∇2Φ= 4πGρ

That is what I refer to. Thank you for the link. Only it works bad from China behind the wall...

When I go to Wiki then I find there this is derived from Gauss's theorem.

When I look how Leonard Suskind derived the Friedman equations I see that he uses a Newtonian approach. But in there he has to observe how a test mass out of the centre interacts with the mass towards the centre.

When I want to see gravity effect then I need to look not only in the centre of a globe but also any from the centre. This makes always sense when I have a limited extension of the mass. Then I can talk about a centre of gravity about outside etc.

Yes I assume that k=0. The most recent information I have is that space is more flat than we can measure.

Kåre

The measurement method is like PKDGRAV3 Potter et al. (2017).

That is a simulation of matter in space with more than 4 terra points.

The GTR equations are in the slow low gravity field limit like Newton. The GTR equations give me that two bodies attract each other. It gives me that a spherical shell has no gravity inside from that shell. So when I calculate point for point the gravity effect of the whole light cone on the centre point and I come to a conclusion that this point has no gravity, then I move to the next point and come to the same conclusion, then I see that for any deviation over space there is no change. Now I have to see over time. I see that each light cone has a surface equal to the start of space. Where do you want to move the start of space? I then can move that start of space back in time as far as I want and still have the same result. So the curvature in time only can come from an initial curvature and does not depend on the density.

I think you understand the method I do to come to this conclusion. Now show me what difference makes that the GTR equations have a different outcome. When you just quote the equations then I can go to Carroll and see them there. What I see is that they use Minkowski space where space and time are treated equal. When I use my derivation then I treat time not equal to space dimensions. Because the GTR equations are defined to be equal to the Newtonian under the conditions that I defined in this case I need to get the right result when I use the Newtonian calculation. But I go even further. I only use symmetry. On that situation there is no difference with GTR.

So I think that you have to go so far back that you have to see how this above mentioned Poisson equation works on the light cone.

Thank you, and all the best wishes,

Paul Gradenwitz

It seems to me that the heart-of-the-matter is more related to the so-called Jeans swindle than any GR definition of the energy-momentum tensor. Consider the equation

∇2Φ= 4πGρ (1)

in a homogeneous and isotropic space. How can we have Φ anything but constant in such a universe, which in turn implies that ρ must be zero? Apparently, Jeans proceeded with a finite ρ without being stopped by this inconsistency.

The Friedman equations provide a different solution to this problem, but one which says that the universe must be expanding at a rate proportional to ρ. This seems to imply a change in equation (1), as f.i. discussed in https://arxiv.org/pdf/1210.3363.pdf (and other places, which I can't find back to just now). One must have in mind that the (Robertson-Walker co-moving) coordinate system, used to describe the Friedman model, is different from the one in which equation (1) is (approximately) valid.

Kåre Olaussen,

Thank you so much for this link. I didn't know that link and would certainly have used that in my article. What I see there is that there is something called as the Jeans Swindle where is assumed that a homogeneous and isotropic density distribution does not add to the gravity potential. Then they show how you can do without that swindle.

In all these calculations I see that they integrate to infinity. But you can't integrate to infinity. The universe is assumed to be of limited age. Assume a universe that is 1 hour old. Then each light cone is not larger than our current solar system, but the universe is infinite. For each light cone the centre is space, the rest is past. Integrating to infinity is wrong.

For each point of space you aren't allowed to integrate beyond the end of the light cone. If you do that you cheat and as a result you can the curvature.

Why? Because when you assume you can integrate to infinity that you can assume that the gravity from the density distribution outside a sphere doesn't add to the gravity potential and only the inside density distribution adds to the potential. This is the shell theorem. But it's invalid if you only can integrate till the edge of the light cone. Then, as soon as you move to another point you leave on light cone and move to the other light cone. So your integral boundaries shift.

All your mathematical operations mean that you look how the field changes when you mover over space and or time. But your, on any event point (space time), valid space is the light cone. It moves with your mathematical movement. Your centre and your outer limits move with you. It is like looking in the mirror. You go to the right the mirror moves with you. you turn the mirror image turns. You can't look your self in the mirror to the back of your head. In many cases you can ignore that. But for whole space, where 13.8GLY is a small part of infinity you simply can't ignore that.

You see I don't argue that the calculations give that result. I argue that these calculations don't calculate reality.

So to me your extremely welcome argumentation make me still come to the following conclusion. In the start you showed that you understand my argumentation but you insist that, because the GTR is right and because the GTR has a different solution, I have to be wrong.

Until now you haven't shown how in the GTR equations we take care about the time direction and here you come with an article where they integrate over the wrong boundaries. Where you helped me a lot is that you made me understand the Lagrangian better and see that even there we sit with the same time directionality.

With thanks and regards,

Paul Gradenwitz

Maybe a simple example will be helpful. We can derive the equations of motion of a particle in a metric from the Lagrangian L= gμνdxμ/dτ dxν/dτ, τ being the proper time.

This Lagrangian is constant, because the metric is given by

ds2 = gμνdxμ dxν = c2 dτ2, so L = c2.

Nevertheless, you get the correct equations of motion from it. Let us say, we are in Minkowski space, then ds2= c2dt2-dx2-dy2-dz2 and L= c2(dt/dτ)2-(dx/dτ)2-(dy/dτ)2-(dz/dτ)2. And this is, of course, equal to c2. But to obtain the equations of motion, you have to write the Euler-Lagrange equations obtained by taking the appropriate derivatives of the functional form of the Lagrangian. You can of course not set, say, the partial derivative with respect to dt/dτ equal to zero. Partial derivatives of constants need not be zero!

So we have ∂L/∂ (dt/dτ) = 2 c (dt/dτ), and the corresponding equation of motion becomes d(∂L/∂ (dt/dτ) /dτ) = 2c dt2/dτ2 = 0. Since L is cyclic in all four variables t,x,y,z, you finally get as equations of motion d(xμ)2/dτ2=0 for μ=0,1,2,3, corresponding to time and the three cartesian coordinates and stating that the four-velocity of your particle is constant in Minkowski space in the absence of forces, which is the result to be expected.

To summarize: The value of L is immaterial. (I guess somebody said that before.) What is important is the functional dependence, which can be nontrivial for a constant L, too.

Dear K. Kassner,

Thank you for looking into my questions. You state that the Lagrangian is a constant. Does that also mean that the rest of my question and my conclusion that :

With that Tuv for above mentioned space is zero for any value of Rho.

is a correct conclusion?

Maybe you have seen the derivation in my article. From there I derive that for any density value space will not change its expansion rate. The famous function a(t) in the Friedmann equations for that reason should be independent of a constant Rho. I am not fluent enough in all the GRT mathematic to simply make a sure statement there. I can imagine that inside the mathematical framework that describes the connection between the Lagrangian and Tuv there is no reason to agree with my statement. That is the reason why I put it as a question here.

The way you write your conclusion is not clear enough for me.

Kassner> To summarize: The value of L is immaterial.

Based on the fact that I have a different scientific background these words might have a complete different meaning for me than you intended. When a doctor gives you a diagnosis he also uses the technical words of his trade like Pneumatic Obstipation, then the patient might ask is that infectious? So forgive me that I ask you some more detail explanation.

Regards,

Paul Gradenwitz