Cosmological models: is there any solution for the Einstein equations?

FLRW models typically assume a perfectly uniform mass density ρ and pressure p throughout an expanding, spatially finite universe.

https://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric

A realistic universe must contain variations in the mass density, though. Let us investigate if the Einstein equations have a solution at all for a realistic universe.

If the universe contains differences in the mass density ρ, and those differences are not made "static" by a pressure, then a rogue metric variation spoils any possible solution of the Einstein equations.

A static uniform sphere M in an FLRW universe

https://articles.adsabs.harvard.edu//full/1956MNRAS.116..678G/0000678.000.html

C. Gilbert (1956) presents a solution. Gilbert has to assume that the universe had a uniform mass density ρ > 0, and that the sphere M condensed from that mass, making a hole in the mass distribution. Then the gravity pull at the edge of the hole is equal from all directions. An infinitesimal rogue metric variation cannot move matter to a lower gravity potential? That is correct. This explains why Gilbert was only able to find the solution in the case where M is exactly the mass of the hole, condensed.

If we assume that a condensation process did take place, then the sphere M was born in a dynamic process, and we can use a rogue variation of the metric to show that there is no solution of the Einstein equations for that process.

Very rogue metric variations break also the FLRW models: we have to ban such variations

Recall from May 26, 2024 that a very rogue variation δg makes a particle to go backward in the time coordinate (of the old coordinates). In the Oppenheimer-Snyder collapse, we used a very rogue metric variation to show that their solution is erroneous. We wrote that particles can "legally" travel to a smaller time coordinate (smaller in the old coordinate), because clocks tick at different rates there, and the clock time is used as the time coordinate.

In an FLRW model, a very rogue variation genuinely takes a particle back in time. It is not an artefact of coordinates.

A very rogue variation can make a particle to have three copies of itself at a single time coordinate. The variation keeps the spacetime geometry constant: therefore, the action integral S of the Ricci scalar R is unchanged. But the integral of LM is changed since a single particle has several copies of itself at the same time of the old coordinates.

What is the problem here? The FLRW metric is a solution of the Einstein field equations. How can we have a very rogue metric variation δg which changes S?

The explanation is that the variation of LM becomes very complicated if we allow travel back in time. The stress-energy tensor is defined as the variation of the integral of LM, when we vary the metric g. The stress-energy tensor is no longer the simple

        T  =

              ρ     0     0     0

              0     0     0     0

              0     0     0     0

              0     0     0     0,

if we allow very rogue variations.

Very rogue variations must in this case be banned in general relativity.

Conclusions

The FLRW model with a uniform density ρ can be called a "dynamic" system, and it does have a solution for the Einstein equations. We can even embed into the FLRW universe static stars which reside is a spherical hole in the otherwise uniform mass distribution. The mass of the star must be equal to the mass missing from the hole. Otherwise, a rogue variation would spoil this solution of the Einstein equations.

Changing the pressure inside a sphere: a new analysis why Einstein equations fail

On November 5, 2023 we gave a tentative proof that the Einstein equations do not have any solution if we suddenly change the pressure inside a uniform sphere. Now that we are familiar with variations of the metric, let us analyze the setup in more detail.

We assume that we start from the static Schwarzschild interior and exterior metrics for a uniform sphere of incompressible fluid. The sphere does not collapse under gravity because it has an internal pressure.

A change in the pressure makes the system dynamic: hence, no solution for the Einstein equations

Suppose that there would exist a solution g for the Einstein field equations. Like in the previous blog post, we can apply an infinitesimal rogue metric variation δg, which does not change the action integral S of the Ricci scalar R, but raises a mass m to a higher gravity potential. Such a variation decreases the action integral S of LM.

Then g is not a stationary point of the Einstein-Hilbert action, and cannot be a solution of the Einstein field equations. We have a contradiction.

One could try to remedy the problem by introducing canonical coordinates and a canonical kinetic energy to LM.

But that might not help in the pathological problem of a non-conserved "pressure charge" in general relativity.

Raising the pressure suddenly inside a spherical uniform mass M

Let the radius of a uniform spherical mass M be r₀, its proper density ρ₀, and its proper pressure p₀. Initially, the system is static. The pressure exactly cancels the gravity force.

The system has the Schwarzschild interior and exterior metric g. We assume weak fields. The metric g is only slightly perturbed from the flat metric η. We use the signature (- + + +).

The Ricci tensor inside the mass M must be approximately

       R  ≈

          κ  *  1/2  *

                  ρ + 3 p    0               0                0 

                  0              ρ - p         0                0

                  0              0              ρ - p           0

                  0              0              0           ρ - p.

The Ricci curvature R₀₀ can be determined with a cube of test masses m which are initially static and then fall freely. Let U₀ be the initial proper volume of the cube.

The proper volume U(t) of the cube decreases according to R₀₀:

       U(t)  =  U₀ * (1  -  C R₀₀ t²),

where C is a constant and t is the time that the test masses have fallen freely.

Let us put test masses m everywhere inside M, and also outside it, up to a radial coordinate r₁ > r₀. We then have a spherical constellation of test masses. The volume of the constellation decreases with time because infinitesimal cubes in the constellation shrink inside M (where R₀₀ > 0). The cubes outside M do not shrink (there R₀₀ = 0).

Since

       R₀₀(ρ, p)  =  1/2 κ (ρ  +  3 p),

we can modify the value of R₀₀ as we wish by changing the pressure p inside M.

                     •    •

                •    •    •    •

                •    •    •    • m test masses 

                     •    •

               <---- 2 r --->

If the sphere is static and we release the test masses, then the outer radius r of the test masses must decrease at a speed which is consistent with the shrinking of the cubes U inside the mass M.

Birkhoff's theorem says that the metric outside M cannot change. The outermost test masses fall the same regardless of p.

Let us suddenly raise the pressure p by Δp.

The shrink rate of the cubes U inside M increases. As the test masses outside M still fall at the same rate, does this lead to a contradiction? The proper volume of the test mass constellation would shrink faster, even though its surface moves as before.

Because of the increased pressure, the average density of the sphere M starts to decline:

       ρ(t)  =  (1  -  C₂ t²) ρ₀,

where C₂ is a constant.

Let us update the formula:

       U(t)  =  U₀  *  ( 1  -  C  *  R₀₀((1 - C₂ t²) ρ₀, p)  *  t² ).

If t is small, we can ignore the change in the density ρ, because the effect on U(t) is only ~ t⁴.

Is it possible that the pressure increase causes the radial metric g₁₁ to shrink, so that it could explain away the shrinking of the cubes U?

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

In the interior Schwarzschild solution, the spatial metric of a slice of the sphere M in the x, y plane is the surface of a sphere. The radius ℛ of this surface only depends on the density ρ, not on the pressure p at all.

Thus, we believe that the change in the pressure does not affect the spatial metric, and we have a contradiction. But we need an exact proof.

Let us raise the pressure by a fixed amount Δp at radii

       r  <  r₂  <  r₀.

At radii

       r₂  <  r   <  r₀,

we let the extra pressure Δp to fall off, so that the pressure is zero at the surface of M.

It is a good guess that the new metric g will imitate the metric inside a static larger sphere whose density is ρ. The added outer layers in a larger sphere cause a uniform extra pressure.

But the pressure falls off rapidly near the surface of M. A rogue variation prevents the Einstein field equations from having a solution there, because we can reduce the potential energy of the extra pressure by moving particles to a larger radius r.

Let us assume that we are able to fix the lagrangian LM by introducing a canonical kinetic energy into it. We might pretend that the Δp is 0 at

       r₂  <  r   < r₀,

and ignore the sharp jump of the pressure at r₂.

Then the Schwarzschild interior solution says that the spatial metric is unchanged at r₂ < r  < r₀. The spatial metric is unchanged in the entire sphere M. This leads to a contradiction because after raising the pressure, the proper volume of the test mass constellation shrinks faster. The radius of the outermost test masses should shrink faster, but it does not. The outermost test masses fall as fast as before.

Raising the pressure in M and Gauss's law for gravity

The Ricci curvature component R₀₀ describes the "focusing power" of gravity.

Outside the spherical mass M, we have R₀₀ = 0, and the lines of force of gravity must be continuous: Gauss's law holds there.

Let us have a cube of test masses inside M. The component R₀₀ tells us how much "source" of gravity is inside the cube. The Komar mass formula gives the total source of gravity inside a static M.

But we can increase the pressure p inside M. Then R₀₀ grows, and suddenly we have more source of gravity inside M. More lines of force start there.

This necessarily breaks Gauss's law, because Birkhoff's theorem says that outside M, we cannot increase the number of lines of force. General relativity has a fundamental problem here.

Pressure generates gravity also in our Minkowski & newtonian gravity model. We do not have Birkhoff's theorem. Therefore, this is not a problem for us.

The Einstein equations do not have a solution for a rotating mass M – the Kerr metric

If M is held static by a pressure, then a rogue metric variation does not change the value of the Einstein-Hilbert action S.

But if the mass is kept "stationary" through rotation, then we obviously can change the value of S by raising some mass m to a higher gravity potential, or lowering m to a lower gravity potential.

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

It has been an open problem if any rotating mass can generate the Kerr metric. We here solved the problem, in the negative.

It may be possible to remedy general relativity by switching to canonical coordinates and adding the kinetic energy to LM.

In August and September 2023 we noted that general relativity "has problems handling accelerating mass flows". Here we proved a major problem: the Einstein equations have no solution for them.

Is the Kerr metric itself reasonable? We were not able to solve the question in the Fall of 2023, using our own Minkowski & newtonian gravity model.

Conclusions

Pressure does generate gravity – also in our own Minkowski & newtonian gravity model. Since one can change pressure, there is no conservation of the "pressure charge". We have to abandon Gauss's law for the gravity generated by pressure.

General relativity implies Birkhoff's theorem, which, in turn, implies that Gauss's law must hold for weak gravity fields.

If we want to repair general relativity, radical changes are needed in it. On the other hand, our own gravity model has no problems handling pressure changes.

We will next look at cosmological FLRW models. Our hypothesis is that only a universe with a totally uniform mass density ρ satisfies the Einstein equations. This means that general relativity does not have a solution for any realistic cosmological model.

The error in the Oppenheimer-Snyder collapse 1939

In this blog we have been suspicious of the Oppenheimer-Snyder collapse paper (1939):

https://journals.aps.org/pr/abstract/10.1103/PhysRev.56.455

Our analysis on May 21, 2024 about "rogue" variations in the Einstein-Hilbert action finally uncovered what is the problem in the paper.

The authors define the metric of time as identically -1 during the collapse. In this text we think in terms of the metric signature (- + + +), while the authors use (+ - - -).

But time, as measured by an observer far away, progresses slower at the center of the star. Setting the metric of time a constant -1 makes the time coordinate unnatural: a particle can travel backward in the time coordinate if it approaches the center of the star. Clocks at the center show a significantly earlier time than clocks at the surface.

https://en.wikipedia.org/wiki/Einstein%E2%80%93Hilbert_action

When we work with the matter lagrangian LM in the Einstein-Hilbert action, we subconsciously assume that the coordinate time always progresses for a particle. If a particle can zigzag in the time coordinate, its mass will contribute in a surprising way to the integral of LM in the Einstein-Hilbert action. The particle can appear as many particles at certain coordinate times.

Thus, variations of the particle path in LM must include paths which travel backward in the coordinate time.

A "very rogue" metric variation

In our May 21, 2024 blog post we defined a rogue variation of a metric: a variation which keeps the spacetime geometry constant, but moves coordinate lines spatially relative to the spacetime geometry (or relative to the old coordinate lines if we interpret the old coordinate lines fixed to the spacetime geometry).

Such variations have surprising results.

Let us define a very rogue metric variation: it moves coordinate lines in such a way that a particle moves back in the coordinate time of the old coordinate lines. The particle will zigzag in the time coordinate of the old coordinates.

The collapse in ordinary static coordinates: a rogue variation

Let us have a spherical uniform cloud of dust (pressure = zero) at the start of the collapse.

                      •    •

                 •    •    •    •

                 •    •    •    • --->  move coordinates

                      •    •    m (t, x, y, z)

                       M

Let us assume that we have a solution g for the Einstein field equations for the collapse, for a coordinate time interval T.

We assume that a particle m in the cloud at a coordinate time t is at certain spatial coordinates x, y, z.

Let us keep the overall spacetime geometry intact, but vary the metric g in such a way that the spatial location x, y, z moves to a larger radial distance r at the time t. The variation takes the particle m along with it.

Another variation will bring m back to its original path at a later coordinate time.

Since we never changed anything in the original spacetime geometry, the Einstein-Hilbert action integral S of the Ricci scalar R did not change at all.

But the mass m was at a higher gravity potential for some coordinate time: the value of the volume element sqrt(-det(g)) was larger for it, because the metric of time was closer to -1 at a larger r.

Therefore the action integral S of LM is smaller than it originally was.

If we temporarily move m to a smaller radius r, we can make the action integral S larger.

We conclude that the collapsing dust cloud has no stationary point in the Einstein-Hilbert action.

We still have to analyze if this proves that the Einstein equations cannot have a solution for a collapsing dust cloud. Probably it does.

The collapse in the Tolman (1934) comoving coordinates: a very rogue variation

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1076370/

The comoving coordinates used by Oppenheimer and Snyder in 1939 are from a 1934 paper by Richard C. Tolman.

The coordinates co-move which each particle of dust. The particles stay at a fixed space coordinates, but their spatial metric distance shrinks as the the coordinate time progresses.

The metric of time is a fixed -1.

The rogue variation of the preceding section refuted any attempt to find a stationary point for the Einstein-Hilbert action. But our analysis required that the metric of time is closer to -1 at a larger r. In the Tolman coordinates, the metric is a constant -1. Is it so that the analysis is not correct in the new coordinates?

In the Tolman coordinates, we can make the particle m to go backward in the coordinate time.

                ^

                |

                |     __

                |  /   /

                |/   /

                    /

                |

                |

                • m particle

    ^ t (coordinate)

    |

     -----> r

    

The path can zigzag like above. Then the contribution of the mass m makes LM smaller than originally, since the particle m is counted as three particles for some time. The action integral S is smaller.

The problem in the paper of Oppenheimer and Snyder seems to be the following: if a particle m can zigzag in the time coordinate, then the variation of LM against changes in the metric of time, g₀₀, becomes complicated.

A small change in the metric of g₀₀ can make three copies of the particle to appear, if we use the old coordinates in the calculation of the S integral of LM.

But Oppenheimer and Snyder assume that the variation of the S integral of LM can be calculated from the simple form of the stress-energy tensor T:

       ρ    0    0    0

       0    0    0    0

       0    0    0    0

       0    0    0    0.

That simple form cannot handle the appearance of several copies of m.

Our result probably means that there is no solution for the Einstein field equations of the dust cloud, even if we use the Tolman coordinates.

The problems stem from the missing kinetic energy when the metric is stretched

In a lagrangian and action of standard newtonian mechanics, rogue paths like above are not a problem because the kinetic energy of the particle increases if it tries to deviate from its calculated path.

In general relativity, one can make a particle to move by manipulating the metric.

We could maybe fix the problem in general relativity if we had canonical coordinates against which kinetic energy is measured. But the spirit of general relativity is that such canonical coordinates do not exist.

In our own Minkowski & newtonian gravity model, standard Minkowski coordinates are the canonical coordinates.

The big picture: an action integral S must be able to determine the kinetic energy, in order to work

The lagrangian density, and the action, of newtonian mechanics:

        kinetic energy  -  potential energy

works correctly. But general relativity has made it difficult, or impossible, to determine the kinetic energy. That is one of the reasons why the Einstein-Hilbert action cannot work for dynamic systems. The other reason is non-conservation of pressure.

The Oppenheimer-Snyder collapse might be the only known dynamic solution of the Einstein field equations in an asymptotic Minkowski space. Our analysis suggests that it is incorrect. In our blog we have speculated that the Einstein equations have no dynamic solutions at all. This result provides more evidence that the hypothesis is true.

The Vaidya metric

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

What about the Vaidya metric where a shell of light expands around a star?

A rogue variation of the metric can move the shell of light farther from the star without changing the overall spacetime geometry. A photon rises to a higher gravity potential, which makes the integral of LM smaller. The action integral S grows smaller. This suggests that the Vaidya metric is incorrect. We have to check it.

The photon moves "faster than light" in the rogue variation. Superluminal movement is not prohibited in general relativity. A change in the spatial metric can cause distances to grow faster than light.

Cosmological models

If a cosmological model contains variations in mass density, and these variations are not supported by a pressure, then a rogue metric variation will change the value of the Einstein-Hilbert action S.

Let us have an expanding universe whose mass density is uniform. The metric of time is the same at every location. A rogue metric variation then cannot change the value of the Einstein-Hilbert action S.

                     "annihilation"

                  / \

                /     \

             m₁     m₂

    ^ t

    |

     ------> x

What about very rogue variations? We can ban metric variations where an observer could see two particles m₁ and m₂ to "annihilate" each other, without emitting any energy. Then a particle cannot turn back in time. Neither can it turn back in the time coordinate of the expanding universe.

Above we wrote that one could maybe correct the Einstein-Hilbert action by introducing canonical coordinates, against which the kinetic energy of a particle is measured. In the case of a cosmological model, defining these canonical coordinates is challenging. There is no asymptotic Minkowski metric far away, into which we could fix the coordinates. Also, an expanding universe can expand "faster than light", in which case it is difficult to define canonical coordinates.

Conclusions

In this blog we have speculated that the Einstein field equations do not have a solution for any dynamic system.

A counterexample to that claim was the Oppenheimer-Snyder collapse (1939). Our analysis above suggests that their solution is erroneous. The reason is that in the Tolman coordinates, particles can move backward in the time coordinate. Then varying LM is complicated, but Oppenheimer and Snyder assumed that it is straightforward.

Our analysis about rogue variations suggest that the Einstein equations cannot have a solution for any dynamic system, with the exception of a totally uniform expanding FLRW universe. Such a universe is unrealistic.

We have to check very carefully how the Einstein equations handle a rogue variation. Above we assumed that they handle it correctly.

It has been an open problem if the Einstein equations have a solution for a planet orbiting a star. Our analysis answers this question: there is no solution.

The fundamental reason for the failure of the Einstein equations is that general relativity does not contain "canonical coordinates", or an "absolute spacetime", against which we could measure the kinetic energy. The problem does not appear in newtonian mechanics or in Minkowski space, because they are built on canonical coordinates.

Our own Minkowski & newtonian gravity model probably handles this without problems because we use Minkowski space.

Our observation provides further evidence for the claim that a spatially finite, expanding or contracting, universe is not a reasonable physical model. That is because defining canonical coordinates is problematic in it.

Lability of the Einstein-Hilbert action

We finally found the correct interpretatiom for the Einstein-Hilbert action, and realized that a variation of the metric δg always has to be in a finite volume of space. Then variations seem to work in the Einstein-Hilbert action.

https://en.wikipedia.org/wiki/Einstein%E2%80%93Hilbert_action

The "lagrangian" inside the action is strange: the "energy" of the Ricci scalar R is either negative potential energy, or kinetic energy, since the sign of R is the same as kinetic energy in LM. The signs go like this:

       ∫  (R  -  V)  *  sqrt(-det(g))  *  dx⁴,

where the Ricci scalar R is positive and the positive potential energy of the lagrangian LM is V. In most cases, V is predominantly the energy of the mass M in the system.

The lagrangian for a static system is not about the minimization of potential energy. Rather, it is like a lagrangian of a dynamical, moving system with both kinetic and potential energy. The task is to find a path which gets the "dynamics" right.

We have been interested in how the Einstein-Hilbert action reacts in case a "charge", like a mass M, or a pressure p, suddenly changes. We suspected that the action is not able to handle such a situation at all. There is no solution for the Einstein field equations then.

Cylinders with a shear

Now we can analyze our cylinder with a shear examples. If the Einstein-Hilbert action would treat a positive R as positive potential energy, then we could try to find the minimum energy state for the cylinder. There probably exists one.

But since a positive R is "kinetic" energy or "positive potential energy", such a procedure is not possible. The action may simply fail to have a stationary point at all.

Looking for a stationary point of the Einstein-Hilbert action S, starting from a wrong metric g

Let us have a spherically symmetric uniform mass M of a density ρ. The interior and exterior Schwarzschild solution is a "stationary" point of S. A variation δg of the metric does not change the value of S. Let the metric be g.

What happens if we suddenly and magically, increase the mass M and the mass density ρ:

       M'  >  M,   ρ'  >  ρ ?

We try to find a stationary point of S by applying finite variations δg to the metric g.

A finite variation can never transform g into the new solution of the Einstein equations, g'.

How does the optimization of g proceed? Toward a singularity?

                                    negative

                                    R

                                    |

                                    v r₀

            ------__              __-----

                       \_     _/

                           •••   positive R

Let us consider the following very simple model where we adjust the metric g --> g' at short radial distances to match the Schwarzschild metric of M'. But we cannot adjust the infinite space to M'. The metric far away is stuck to M < M'.

The transition between the two metrics is at r₀, where the Ricci scalar R is negative.

The variation δg. At 

       r  <  r₀,

we increase g₀₀ --> g₀₀' upward (≈ -1), and g₁₁ --> g₁₁' upward (≈ 1). The Ricci curvatures there are

       R₀₀  ≈  1/2 κ ρ,

       R₁₁  ≈  1/2 κ ρ.

The Wikipedia formula

says that the variation δg decreases the integral of R. The negative R around the transition radius r₀ wins over the positive R that we add inside the mass M'.

... (to be continued).

Manipulating the distance between two masses

              -->  

             ●                 ●

            m                m

Let us assume that g is the metric for two masses m, given by the Einstein field equations. Let us now change the coordinates in such a way that the geometry of spacetime stays as is, but the coordinate lines move a little bit and transport the masses closer to each other.

What happens to the action S? The R part stays as is, but the LM part changes. We found a small variation of the metric, such that it does not keep S constant?

How can the variation of the action S work at all if we always can find a variation which changes the value of S?

Is it so that the variation is always assumed to be a simple variation which only stretches locally some components of g, and does not contract them in the same variation?

Can we somehow ban "bad" variations? Or define the kinetic energy of a particle in such a way that a bad variation is not able to change the value of LM?

Defining the velocity of a particle in general relativity

Literature is silent about how we define the velocity of a particle P in general relativity. An obvious choice is to take the coordinate velocity of P and convert it with the metric g to a proper velocity.

Definition of the velocity of a particle. We have an observer sitting static at certain spatial coordinates x, y, z. The observer measures the proper velocity of the particle.

A "rogue" variation δg which changes the value of the Einstein-Hilbert action S

                        ^  move

                        | P (x, y, z)

                  •    •  

             •    •    •    • 

             •    •    •    •   particles

                  •    •                                  metric g

Let us have particles initially static in a uniform spherical constellation. Let the metric be g from the Einstein field equations.

Let P be a particle at the edge of the sphere, at coordinates (t, x, y, z).

Let us apply an infinitesimal metric variation δg which keeps the spacetime geometry as constant, but "moves" the coordinate location (x, y, z) away from the sphere. Then the mass of P goes to a higher gravity potential, that is, in the action integral S, the volume element sqrt(-det(g)) has a larger value when we integrate over the mass m of the particle P.

The action integral of the Ricci scalar R does not change at all, because the spacetime geometry does not change in any way. But the integral of LM becomes smaller, because the "potential energy" of m is now larger. The lagrangian LM is of the form:

       kinetic energy  -  potential energy.

Thus, the infinitesimal variation δg changes the value of the action integral S, even though we had a solution g of the Einstein equations, and S should not change.

A lagrangian of newtonian mechanics should prevent such a movement of a particle P by having the kinetic energy of P change in the movement.

Could we define "canonical coordinates", which are fixed to the spacetime metric. Then the rogue particle would have a positive kinetic energy relative to standard coordinates, which would restore the value of LM to its original value?

However, e.g., in an expanding universe, it is a hopeless task to define such canonical coordinates.

If the spherical constellation of particles above is held static with pressure, then the potential energy of the particle P does not change if it is moved infinitesimally. This is because the gravity force and the pressure force are in an equilibrium. This may explain why the Schwarzschild interior solution is reasonable.

Analyzing the pressure change in a spherical mass

Our November 5, 2023 counterexample to the Einstein equations involves a changing pressure inside a sphere which is not in a static equilibrium.

                                ^

                            <--|-->              metric g

                                v

             positive pressure p inside

        \_____________________________/

        ---->  negative pressure -P  <----

                 in surface

Let us start from a spherical vessel where a negative pressure in the surface compensates a positive pressure inside. We assume that the metric g is a solution of the Einstein field equations.

What happens if we suddenly (or slowly) remove the negative pressure -P from the surface of the vessel?

Is it possible to evolve the metric g in such a way that the change of R still balances a radically changed LM?

Our November 5, 2023 analysis suggested that the metric outside the vessel should be changed, in order to adapt the metric to the sudden lack of a negative pressure. But we cannot change the external metric with a variation which is in a finite spatial volume. At the edge of such a variation there will be an area which has the Ricci scalar R ≠ 0. That is not allowed because the edge is in empty space.

On the other hand, if we allow variations of the metric throughout space, then the Einstein-Hilbert action does not work at all, as we showed on May 10, 2024. Such variations would also break the speed of light, and lead to all the problems with superluminal communication.

Conclusions

Let us close this blog post.

We have to figure out if there is a way to ban "rogue" variations of the metric. The Einstein-Hilbert action S does change in a bad variation, even if the metric g would be a solution of the Einstein field equations.

If a system is static and in an equilibrium, then LM does not change if particles move an infinitesimal distance. In that case, also "rogue" variations are benign.

We have to analyze further, why the Einstein field equations do not have solutions for cylinders with a shear (if we calculated right in March 2024). We have to check the correctness of the 1939 paper by Oppenheimer and Snyder.

New lagrangian for gravity

UPDATE May 21, 2024: The Einstein-Hilbert action seems to work correctly if we define the matter lagrangian as "kinetic energy - potential energy" and require that a variation of the metric only alters a finite volume of space. We are investigating what happens in the case of a changing pressure in a spherical mass M.

----

Let us try to write a new lagrangian for gravity. We sketched the idea on May 14, 2024. The metric of time is directly from the newtonian potential.

We have to get a lot of energy into gravitational waves. To that end, we introduce a private field for each particle, such that the field is "attached" to the particle which "owns" it. The field acts in a different way on other particles than the one which owns the field.

                         |

                         |  particle

                 ------ ● ---------------  "steel wire"

                         |

                         |

The private field can be interpreted as a physical body which is attached to the particle – like steel wires attached to the particle. The wires interact with other particles, but they have a special particle which they are attached to.

In a rubber sheet model of gravity, the depression in the rubber sheet is not specifically assigned to any particle. It is a "public" field.

For a public field, the energy of a wave in it is controlled by the energy that a static field has around a mass M. We cannot arbitrarily increase the energy of a wave. But a gravitational wave contains 16 times the energy density of an analogous electromagnetic wave.

The wire model differs from a traditional field theory. In this blog we have been suggesting that gravity cannot be accurately modeled as a field, or a metric, because the interaction is too complicated.

Self-force of an electron: the rubber plate model

In this blog we have tried to find a model for the self-force of an electron when the electron is moved back and forth. The elecric field of the electron itself exerts a force on the electron.

We had the "rubber plate" model where we imagine that the electric field of the electron is a rubber disk attached to it. Electromagnetic waves are mechanical waves in the rubber disk.

See, for example, our post on December 20, 2020.

The "steel wire" model above is similar to the rubber plate model. We do not know if it is possible to write a traditional lagrangian for the rubber plate model.

Particles embedded into a block of rubber

If we fuse together the "rubber disk" of each particle, then we have a model where particles are embedded into a solid block of rubber.

But there is a problem with this model: particles moving at a constant speed must be allowed to move undisturbed. Only accelerating particles interact with the rubber, sending waves into it.

Is there any way to implement this with a public field?

Teleportation through the Huygens principle

If the particles are otherwise static, but oscillate back and forth, then the rubber block model of the preceding section works better.

             |       | 

        ---- • ---- • ----

             |       |

        ---- • ---- • ----

             |       |

 

       particles with

       interactions

Rather than rubber, we can imagine a matrix of particles coupled to their neighbors through a gravity-like interaction.

The Huygens principle says that each point in space acts as a "source" for a new wave. This is a simple way to explain how a complex interaction can be teleported from a mass M to another mass light-years away.

We want to teleport the stretching of the spatial metric.

Can we make a public field from the "inertial frame"? A strange rubber block

                            <--->

                  ● /\/\/\/\/\/\ ●

                 M                     M

             oscillating quadrupole

Let us look again at the oscillating quadrupole. The acceleration of the masses M interacts with what is considered the inertial frame for a test mass m.

This is in the spirit of general relativity. We must couple an acceleration of a mass to a field.

It is like particles moving in a strange compressible liquid. A particle P can move without friction at a constant velocity. We may imagine that there is a laminar flow of the liquid past the particle P.

But if P is accelerated relative to the liquid, then P interacts with the liquid. The liquid is compressed, and sound waves are sent through it.

Since gravitational waves are transverse, we cannot really use a liquid. The particles have to be immersed into a strange rubber substance which allows the particles to move at a constant velocity without friction. Only when a particle is accelerated, does it disturb the rubber.

The tense water hose model

             <----------  tension  ------------>  

            ------------------------------------------ tense hose

                          ^  acceleration

                          |                                  • P'

                          •  P            water

            ------------------------------------------

            <~~~  waves  ~~~>

Above we have a real-world example of a system where the acceleration of a particle disturbs the inertial frame far away. A particle is accelerated vertically inside a tense water hose. Waves propagate along the tense hose to both directions. A wave changes what is the "inertial frame" of a particle P' far away. The particle P' starts to oscillate up and down.

If P moves at a constant speed, then it does not disturb the hose.

Frame-dragging

                        m

                         •

                        ● ---> a

                        M

We have written in this blog about the problem of how general relativity treats accelerating masses. In the diagram, if we accelerate the large mass M, then it "drags the inertial frame" and the test mass m, too, starts to accelerate.

If we try to describe the process with a metric around M, it is not clear how the metric behaves if M is accelerated.

Quantum field theory approach: treat the fields in a gravitational wave differently from static fields

In quantum field theory, we imagine that a field of sine waves can be "excited" by interactions with particles. An excitation can, for example, be a photon. It is born and absorbed through an interaction with a charge, e.g., an electron.

 

The field of excitations, or sine waves, is separate from static fields, which are carried by virtual particles.

Can we in this model explain why the energy density of a gravitational wave is 16 times the density of the analogous electromagnetic wave?

If we make a mass to M oscillate, it creates 16 gravitons while the analogous charge Q only creates one photon?

The 16-fold energy must, in the end, be due to a coupling which is 16 times as strong as in the electromagnetic analogue. A gravitational wave does couple with a mass in many ways.

Can we write a lagrangian where the metric of time, g₀₀, imitates Coulomb's potential, but the energy of the wave is set 16X the energy of an analogous electromagnetic wave?

Also, the effects of a gravitational wave are diverse: they replicate many of the effects of the static gravity field of an oscillating mass M. The coupling of a test mass m to the wave is complicated.

Then we would have a lagrangian which is not very precisely defined, but would work at least in some cases.

The static gravity field of a mass M has to be written separately from the field of a gravitational wave. The energy density of a static field is like in the analogous Coulomb field.

The coupling of a static gravity field to a test mass m is complicated, too, like the coupling of a gravitational wave to a mass.

Can we decompose a field into a static field and a wave propagating at the speed of light?

Can we really differentiate between waves and static fields? Does the Fourier decomposition of a field uniquely differentiate between static fields and waves propagating at the speed of light?

We can certainly simulate waves propagating at a speed less than light by letting a chain of masses M fly at a large speed past the observer.

A wave whose wavelength is very large locally looks like a static field.

This is a general question about waves. If we have a tense drum skin, can we uniquely divide its disturbances into waves, and into static depressions caused by weights lying on the skin?

Let us keep hitting the drum skin with a sharp hammer at the same point, at a large frequency. We obtain a static depression in the skin. An individual hit does produce also sine waves which would carry energy away. But there is a total destructive interference of such waves if we hit the skin at an infinite frequency, i.e., have a weight lying on the skin.

Let us guess that one can separate static fields from sine waves.

Changes in the spatial metric and a lagrangian

A complete lagrangian should, in principle, describe all the effects of a gravity field. But how do we make the lagrangian to shrink a measuring rod if it is turned radially pointing to a large mass M? The Einstein-Hilbert action is supposed to do that feat by redefining the metric at the rod.

Another option is to use the newtonian gravity field to determine the spatial metric, and mention in the matter lagrangian LM that LM has to be calculated in that metric.

A traditional lagrangian describes the movements of particles. It does not describe how a measuring rod shrinks. The shrinking is due to the force fields behaving in a different way. Thus, a lagrangian which refers to the "metric of space", describes how force fields "move" or change.

ADM formalism

The ADM formalism is a hamiltonian approach to a system sitting in an asymptotically Minkowski space. We have to check if ADM contains the same errors as the Einstein-Hilbert action.

If ADM works, then the Einstein equations should have a solution for a cylinder with shear? They should find the lowest "energy" state. We have to check if the ADM formalism is correct.

The generalized stress-energy tensor

The stress-energy tensor in general relativity measures how a specific field, a perturbation of the flat metric, changes the value of a matter lagrangian LM.

Analogously, we can define a generalized stress-energy tensor, which measures how an arbitrary field K changes the value of a lagrangian LM. In this way we can include complex interactions of the field K with matter.

For example, K could be a wave in a drum skin, where a complex mechanical system of levers and springs is embedded into the skin. Or, K can be a gravitational wave, which meets masses and rigid bodies.

Making the energy of an electromagnetic wave 16-fold

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

The electromagnetic tensor in the (- + + +) metric signature is:

The lagrangian is usually written:

It has a more intuitive vector form:

Let us analyze it. We set the polarization P and the magnetization M to zero. The energy density of the electric field E is the same in a static electric field and in an electromagnetic wave. The interaction with matter is assumed to be extremely simple: the inner product of the four-potential A and the the four-current field J.

Now, let us separate the field belonging to waves from static fields. We obtain a lagrangian density:

       L  =  8 (ε₀ E²wave  -  1 / μ₀ *  B²wave)

               + 1/2 (ε₀ E²static  -  1 / μ₀ *  B²static)

               - φ ρfree 

               + A • Jfree.

The energy of an electromagnetic wave is 16X the normal energy. Could this work?

The lagrangian density for gravity

The lagrangian density would be completely analogous to the electromagnetic lagrangian with a 16-fold energy density of waves.

       L  =  8 (E²gwave  -  B²gwave)

               + 1/2 (E²gstatic  -  B²gstatic)

               + LM.

We have set the appropriate natural constants to 1.

      Eg 

is the gravitoelectric field and Bg is the gravitomagnetic field, completely analogous with the electric counterparts.

The matter lagrangian LM depends in a complex way on both the static gravity field and the field of a gravitational wave. The "metric" comes to play when we calculate the value of LM.

If all the energies in the lagrangian above are counted like an observer far away sees them, then the action integral of the lagrangian L above is the simple integral over the cartesian coordinate space and time.

Can we do like in the Einstein-Hilbert action, and use "local" values for energies, and integrate over the volume element which is "discounted" by the redshift:

       sqrt(-g₀₀) * dx⁴?

A problem in this is that the binding energy of a mass M is not calculated right if we discount with the full redshift.

For purely static fields with no pressure or shear, our lagrangian works like the lagrangian of newtonian gravity. It will produce reasonable results, while the Einstein-Hilbert action calculated everything wrong inside a mass M.

The reaction of the new gravity lagrangian to pressure or an oscillating mass M

How does our lagrangian react to a positive pressure? If we have a pressurized vessel, then our new lagrangian can free potential energy from the pressure by making the newtonian gravity field stronger, which, in turn, stretches the radial metric?

The stretched radial metric is an indirect consequence of a newtonian gravity field. Is it plausible that a positive pressure can create a newtonian gravity field on its own?

In a rubber sheet model of gravity, pressure stretches the rubber sheet, and makes the sheet to sag lower. Pressure does create a certain gravity field.

In general relativity, pressure is unable to stretch the radial metric.

Our new lagrangian optimizes the gravity field around a mass M in such a way that the potential of M is lowered and we pay the price of increasing the energy of the gravity field. Now, if there is a positive pressure around M, it is logical that the gravity field can become stronger. Thus, pressure does create gravity, and is able to stretch the radial metric.

An alternative model of gravity would claim that the gravity field of M is "inherent" to M, and nothing else can create such a field. That model does not allow a lagrangian, because in a lagrangian model, anything can be adjusted. The mass M cannot dictate what happens in the gravity field around it.

Is there a similar problem in electromagnetism? Let us have an oscillating charge Q. If the electric field of Q becomes weaker, then the system can save in the energy of the radiated waves. Could this weaken the apparent charge of Q?

In a rubber sheet model of gravity, a vigorous oscillation of a weight M raises M higher: it weakens the gravity field of M.

Conclusions

We now have a tentative lagrangian for gravity, such that:

1.   it gets right the newtonian gravity field around static or slowly moving masses, and

2.   it correctly predicts that the energy density of a gravitational wave is 16 times the analogous electromagnetic wave.

Our lagrangian is somewhat fuzzy. How exactly do we define the effect of a gravity field on the matter lagrangian LM?

The lagrangian is our first attempt as the lagrangian for the Minkowski & newtonian gravity model.

Gibbons-Hawking-York boundary term: can it save the Einstein-Hilbert action?

UPDATE May 21, 2024: The form of the lagrangian LM has to be

       kinetic energy  -  potential energy

and the variation δg has to be restricted to a finite area of space. Then the variation of the metric of time g₀₀ produces a correct result.

----

Wikipedia says that a special "boundary term" has to be added to the Einstein-Hilbert action if the spacetime is not "compact"; that is, if the spacetime volume of the spacetime is infinite.

https://en.wikipedia.org/wiki/Gibbons%E2%80%93Hawking%E2%80%93York_boundary_term

Wikipedia tells us that James W. York first realized that a boundary term is required. It was probably published in his 1971 paper.

Could it be that the boundary term rescues the Einstein-Hilbert action?

The boundary term

The Wikipedia page contains complicated instructions about how to calculate the boundary term.

The boundary term is the last two terms in the formula above, and its is denoted SGHY,0.

https://en.wikipedia.org/wiki/Einstein%E2%80%93Hilbert_action

Wikipedia seems to claim that if the boundary term is added, then the following formula holds:

Restrictions on the variation δg

Restriction 1 for δg. Finiteness. Let us restrict ourselves to spherically symmetric variations which only update g in a finite volume of space.

Restriction 2 for δg. Weak energy condition. Let us ban any variation which makes the Ricci scalar R negative anywhere.

----

UPDATE May 21, 2024: Restriction 2 is not required, and, actually, cannot be allowed. In a flat metric in empty space, a variation of the metric will produce zones where R < 0.

----

Spacetime integration volume in the Einstein-Hilbert action

We work in an asymptotically Minkowski space. We integrate the Einstein-Hilbert action for some spatial coordinate ball

       r  <  r₀,

and for some coordinate time interval T:

       t₀  <  t  <  t₁.

Our spacetime manifold has boundaries at those coordinates. To avoid the use of a boundary term, we do not vary the metric g at all at the boundaries, but keep g fixed there.

We assume that the boundary term only depends on what is at the boundary – not on what is in the interior.

Varying the metric of time inside a spherical mass

Let us recapitulate this example from May 10, 2024. We have a uniform sphere M of a mass density ρ. The metric g is the Schwarzschild interior and exterior metric. The metric g is almost flat.

https://en.wikipedia.org/wiki/Einstein%E2%80%93Hilbert_action

We set κ = 1. The Ricci tensor in spherical coordinates is approximately

       R  =

             1/2 ρ      0            0                               0

   

             0            1/2 ρ      0                               0

             0            0            1/2 ρ r²                     0

             0            0            0         1/2 ρ r² sin²(θ).

The Ricci scalar:

       R  =  ρ.

These together yield the stress-energy tensor:

       T  =  R  -  1/2 R g

            =

                  ρ    0    0    0

                  0    0    0    0

                  0    0    0    0

                  0    0    0    0.

The variation δg. We assume that g₀₀ is very close to -1. We increase g₀₀ by

       0  ≤  b(r)  <<  1

in some short segment of r. The function b(r) must be smooth enough, so that the Ricci scalar stays > 0 inside M.

Let the spatial volume defined by that segment be U and the average value of b(r) in that volume U be

       B.

Calculating the change in S.

Integral of R / 2. Wikipedia says that the Ricci scalar R changes locally by

        

       dR  =  dg^μν  *  Rμν 

              =  dg⁰⁰    *  R₀₀

              =  -b(r)   *  1/2 ρ.

This changes the integral of R / 2 by

       -B / 2  *  1/2 ρ U T.

The volume element sqrt(-det(g)) shrinks locally by a ratio b(r) / 2. This changes the integral of R / 2 by:

       ρ  *  -B / 4  *  U T.

In total, the integral of R / 2 changes by

       -1/2 ρ B U T.

Integral of LM. The mass in the volume U does not change. But the volume element shrinks because time runs slower. This changes the integral of the mass-energy in LM by

       ρ  *  -B / 2  *  U T

       =  -1/2 ρ B U T.

The total change in the action S is

       -ρ B U T.

This does not make sense. The change should be zero because we started from a solution of the Einstein field equations.

Conclusions

The Gibbons-Hawking-York boundary term does not help. The Einstein-Hilbert action still produces only nonsensical results.

If there is no matter content in space, then the Einstein-Hilbert action may be correct. All the problems stem from matter. Calculating the energy of the (newtonian) gravity field cannot succeed from mass density alone. But if the mass density is zero, then the Einstein-Hilbert lagrangian might work?

Let us consider electromagnetic waves. They are schematically of the form:

       E  ~  sin(ω (t  -  x / c)),

       B  ~  cos(ω (t  -  x / c)).

A "curvature" corresponds to the second derivative. We can as well calculate the energy of a wave from second derivatives as from first derivatives, because the second derivatives are nice sine or cosine functions. This may explain why the Ricci scalar is a suitable lagrangian density for gravitational waves!

There does not exist a lagrangian for the Einstein field equations?

UPDATE May 21, 2024: We found some of the problems. See the update of the May 10, 2024 post.

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In the previous blog posts, we were not able to get any calculation right using the Einstein-Hilbert action. What is the problem?

The nonexistence of solutions of the Einstein equations for shear suggests that a lagrangian for the Einstein equations does not exist at all

Let us reconsider the cylinder with a shear. The lagrangian for a static solution is the "potential energy" of the system. We have to minimize the potential energy, in order to find the minimum of the action integral.

The Einstein field equations in this case must follow from the formula of the potential energy, through variational calculus.

Then the Einstein field equations would be satisfied in the state with the minimal potential energy.

On April 22, 24, and 26, 2024 we calculated various examples where the Einstein field equations do not seem to have an "approximate" solution for a cylinder with a shear. We can explain the nonexistence of a solution, if either:

1.   the potential energy in the supposed lagrangian has no minimum value,

or

2.   there does not exist a lagrangian at all.

Is it possible that we have a lagrangian where the cylinder with a shear does not have a state which has the lowest potential energy in gravity plus other fields? How could that happen?

1.  The cylinder collapses into a singularity? That does not occur in nature. Cylinders on Earth do not collapse!

2.  The cylinder approaches a minimum potential energy, but can never attain it? That would be strange. If we have a sequence of states of the cylinder, and the potential energy decreases in that sequence, then we would expect that there is a "limit" state which has the lowest potential energy.

It could happen that a sequence has no limit state. But that would be surprising. We have observed in nature that systems with potential energy in them, tend to approach an equilibrium state, and that equilibrium state is the "limit" of the successive states of the system.

An example: a complex oscillating mechanical system. It constantly loses energy in friction, and approaches a state where the potential energy is at a minimum. The minimum is the limit of the successive states of the system.

Our reasoning actually proves that the Einstein field equations are an incorrect theory of nature in the case of a cylinder with a shear. We have observed in nature that there does exist a minimum potential energy state for them. That state would satisfy the correct theory of gravity. But it does not satisfy the Einstein field equations. They are an incorrect theory of gravity.

The lagrangian of newtonian gravity, or static electric fields

https://en.wikipedia.org/wiki/Lagrangian_(field_theory)

Above, we have the lagrangian of newtonian gravity. There, Φ is the potential, and ρ is the mass density.

The lagrangian L(x, t) calculates the negative potential energy of the system. That is, the lagrangian is of the familiar type

       T  -  V,

where T is the kinetic energy and V is the potential energy.

We have to maximize the integral of L(x, t), in order to find the equilibrium state. Then the potential energy is at the minimum.

Let us have a mass M sitting in space. We start from a flat zero potential Φ. If we make Φ(x, t) negative at the location of M, we can increase the integral of the last term:

       ρ(x, t) Φ(x, t),

i.e., increase the value of the action integral.

But then the integral of the first term,

       -( ∇Φ(x, t) )²

grows smaller. The integral over the whole lagrangian density attains the minimum when the potential Φ is the familiar newtonian gravity potential.

Let us analyze the logic of optimizing the integral to the maximum value:

1.   we reduce the energy of the mass M by dropping it into a lower gravity potential Φ; the last term in L(x, t) describes this fall into a lower potential;

2.   the price we have to pay is that the "energy of the field", ( ∇Φ(x, t) )², increases. At some point, it no longer pays to drop M any lower.

A rubber sheet model of gravity describes the optimization process qualitatively. The potential Φ is the height of the surface of the rubber sheet. The energy of the field, ( ∇Φ(x, t) )², is the elastic energy of the rubber.

If we think of static electric fields, then Φ is the electric potential, ∇Φ is the electric field strength, and (∇Φ)² is the energy of the electric field. An electric charge distribution Q digs a potential pit for itself, to get itself into a lower electric potential. Actually, if the field outside looks like the field of a positive charge, then Q is really a negative charge!

Can we describe the energy of a field through a "curvature" or charge density? No

In the Einstein-Hilbert action, the energy of the field is given in a different form: the scalar curvature R of the metric g. Can that idea work?

Something similar in newtonian gravity or electromagnetism would be the second derivative of the potential,

       ∇²Φ.

In electromagnetism, ∇²Φ is the charge density. Similarly, the Ricci scalar R gives the mass density ρ, in the absence of pressure or shear.

If we have a uniform spherical charge density Q, then we certainly can find some formula which, in terms of  

∇²Φ, gives the total energy of the electric field around Q.

But is that possible for more complex charge distributions?

If the Einstein-Hilbert action would work for gravity with weak fields, then we could use it to determine the newtonian gravity potential (from g₀₀), and we would then have an "alternative" lagrangian for static electric fields, where the lagrangian density is in terms of charge density and the electric potential. Is it possible to form such a lagrangian?

The precise problem: can we determine the energy of an electric field "directly" from the charge density, without going indirectly through the calculation of the electric field E, and integrating E²?

It is not possible to determine the energy of the electric field from the charge density alone. Consider a thin spherical shell of charge. We keep the total charge Q constant and slowly shrink the radius in the shell. The energy of the electric field outside the shell grows. But we can let the thickness of the shell grow in such a way that the density of the charge per volume stays constant in the shell.

It does not help either, if we are allowed to calculate a "boundary term" from the electric field at some large radial distance. The electric field far away stays constant. It does not tell us how much the energy of the electric field is at small radial distances.

How can then the Einstein-Hilbert action work at all? The Ricci scalar R is essentially linear in the mass density ρ.

If there is no error in this reasoning, it explains why all our calculations with variations last week led to an error. The Einstein-Hilbert action does not contain the lagrangian for gravity.

The Einstein field equations lead to the Schwarzschild metric, and do describe gravity fairly well. But the Einstein-Hilbert action is a completely unrelated formula which does not describe gravity.

Formulating the correct lagrangian for static or dynamic solutions of the Einstein field equations

If the Einstein-Hilbert action is totally wrong for the Einstein field equations in a static setting, we would like to formulate one which works for static mass configurations.

Our November 5, 2023 result suggests that the Einstein field equations are not satisfied if there is a change in pressure, i.e., in a dynamic setting. Our April 22, 24, and 26, 2024 results suggest that there is no static solution if there is shear inside a cylinder.

It is still possible that a lagrangian exists for a pure mass distribution, both in a static and dynamic setting.

The first candidate, of course, is the lagrangian of newtonian gravity. It does calculate the approximate metric of time correctly.

The question is then what is the role of the spatial metric? In the Einstein-Hilbert action, the spatial metric has an essential role in producing some of Ricci curvature, and contributing to the Ricci scalar. Is it possible that we could ignore the spatial metric in the lagrangian of gravity?

In our own Minkowski & newtonian gravity model, both the slowing down of time and stretching of the spatial metric are side effects of the gravity field. It does make sense to ignore the spatial metric altogether in the lagrangian!

The energy density of a gravitational wave: the inertia interaction

We empirically know that the energy density of a gravitational wave is correctly calculated by the Einstein field equations. Their energy density is 16 times the analogous electromagnetic wave. What should the lagrangian be like for gravitational waves?

Is the "kinetic" energy of the field 16X the corresponding electromagnetic lagrangian? Could that simple trick work?

Is there any reason why a "moving" field, which propagates at a speed of light, should have the same energy as a static field? In the case of electromagnetism, that holds. But for gravity, the energy is 16X. We have earlier, on December 29, 2021, estimated the energy of a gravitational wave from the work it can do against a negative pressure. We can harvest a lot of energy from the stretching of the spatial metric.

In the case of a static gravity field, it is not clear what it would mean to harvest energy from the distorted spatial metric. It might be that the energy in the distorted spatial metric is only is harvestable from a moving field.

For static fields, we can change the energy of the field simply by moving masses closer or farther from each other. The energy is manifest in the work we have to do to move the masses.

For a gravitational wave, the energy in the field is more complicated to determine. We can create gravitational waves by moving masses around. The wave can be absorbed by letting it to move masses in a receiving "antenna". The movement in the antenna cancels a part of the original wave. But how much?

                 

                     oscillation

                         <---->

            ● \/\/\/\/\/\/\/\/\/ ●

           M         spring          M

Practical gravitational waves are emitted by a quadrupole. We can have a string between two masses, and the masses oscillate until they have lost all kinetic energy into gravitational waves. The system sends quadrupole waves.

Note that the oscillation above also causes a positive and negative pressure alternating in the spring. The pressure accelerates the masses back and forth. Our December 2021 calculation with a negative pressure might not be very far from the calculation with a quadrupole.

Let us consider a rubber sheet model of gravity. We let a weight M slide back and forth on the sheet. The energy which the weight M sends in waves depends strictly on the energy in the static field of M. A wave is a part of its static field "escaping". Thus, in a rubber model, the energy of a gravitational wave probably could not be 16X the energy of the analogous electromagnetic wave. If this reasoning is correct, then a rubber model cannot explain gravity. This would be additional evidence for our claim that there is no spacetime geometry which can be bent and stretched.

A question then is what is this 16X energy density of the wave, if it is not energy density of the static field?

           

                                    ---->  wave

            ...  /\/\/\ ● /\/\/\ ● /\/\/\ ● /\/\/\ ...

                           M           M           M

                        springs and masses

If we have masses attached with springs to each other, we can send a wave along the chain. We can increase the energy of the wave by making the springs stiffer.

But if we make the springs stiffer, then we increase the strength of the attraction between the masses?

Not necessarily. If a test mass m in the field of a mass M acquires more inertia, then there is an additional interaction between m and M besides their gravity attraction – an inertia interaction. Thus, the extra inertia acquired by m can explain why the wave carries a surprisingly amount of energy.

If we let a mass M oscillate up and down in the diagram, the inertia interaction will also make the test mass m to oscillate up and down. The inertia interaction can transfer energy along surprising routes.

The inertia interaction between electric charges

This new insight may help us to determine what is the inertia of an electric test charge q in the field of an electric charge Q. We have been confused about that. We know that placing a hydrogen atom into a low or a high electric potential does not alter its spectrum, as if the electric potential would not change the inertia of the electron. On the other hand, it is hard to understand why an energy flow in the electric field would not increase the inertia of a charge.

Could it be that the energy flow only happens relatively "slowly" and does not have time to affect the inertia of the orbiting electron, or have time to affect the energy density of an electromagnetic wave?

The inertia interaction to a negative pressure: "thermal" expansion

         =========   very rigid structure

         ||               ||

         =========

                  ^

                  |

                 ● M

Let us have a very rigid structure close to a mass M. We move M closer to the structure. The metric of space changes, and we can harvest energy from the deformed structure.

In what sense this was an "inertia interaction"?

In the Minkowski & newtonian model, we claim that the inertia of a test mass m is larger in the radial direction relative to M. That causes radially placed rods, relative to M, to be squeezed when they come close to M.

What is the route of energy if we push M closer to the structure, and harvest energy from the structure? Energy somehow moved from our hand to the harvesting apparatus.

                   rod = springs and masses

                          • extra inertia

                          | "spring"

                          |

          ...  /\/\/\ ● /\/\/\ ● /\/\/\ ● /\/\/\ ...

                          m           m           m

                                        

                                   ^   

                                   |  

                                  ● M

The rod might be modeled with the diagram which we drew above. The masses m are atoms which oscillate back and forth. The springs are the electromagnetic interaction. When we move the large mass M closer to the rod, the masses m gain inertia (marked with •) and move slower. Consequently, their oscillation amplitude is reduced, and the rod shrinks in length. The thermal expansion of the rod is less when it is "cooled" by bringing it closer to the large mass M.

If the rod is "cooled" by M, then where does the "heat" from the rod go? Increasing the inertia of a moving mass m releases kinetic energy from m. Could it be that the excess heat is stored into a "spring" which connects the extra inertia to m?

"Teleportation" of the cooling system

We have written about a hypothesis that waves in empty space "teleport" the transmitting antenna close to the "receiving" antenna.

Above we described how the shrinking of the length of a rod can be explained by a cooling effect by the field of M.

Gravitational waves transport the shrinking effect over large distances of space. How exactly is this transport process implemented?

We have also written about the hypothesis that a test mass m moving inside a gravitational wave ships field energy around, which would explain the increase of the inertia of m, and the cooling effect on a rod.

But the fact is that a gravitational wave is transporting 16X the energy of the analogous electromagnetic wave. This large amount of energy must be stored in the wave. Energy cannot be a "side effect" of the wave. The teleportation hypothesis can explain the large energy content. It cannot be a side effect.

The electromagnetic lagrangian: it is a rubber sheet model

https://en.wikipedia.org/wiki/Lagrangian_(field_theory)

A simple form is:

In the tensor notation:

Let us analyze what is the relevant part for a static electric field, and what is the part for an electromagnetic wave.

Static electric field. We will analyze the first equation above. Then the current j is zero, the vector potential A is zero, and the magnetic field B is zero. The lagrangian is essentially the lagrangian of newtonian gravity, which we analyzed above.

Electromagnetic wave.  Then we have ρ = 0 and j = 0. The remaining part of the lagrangian is:

We could interpret E² as the potential energy density and B² as the kinetic energy density. The lagrangian does not directly tell us the energy density of the wave, since if an action S is at an extremal value, and C is a constant, then is also

       C S

at an extremal value. The energy density of an electromagnetic wave has to be derived from its interaction with charges. The energy density happens to be:

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

Is it a coincidence? Why is the energy the same as for a static field? In a rubber sheet model, the energy density of the stretching of the rubber is

       ~  1/2 N²,

where we have denoted by N the strain

       N  =  ΔL / L

of the rubber sheet. The elastic energy of stretching is the same for a static field as well as for a wave. In a wave, a half of the total energy is kinetic energy. The electromagnetic formula is analogous if we identify E with the strain and B with the kinetic energy. We see that a rubber sheet model describes electromagnetism well!

Let us make a weight M oscillate back and forth on a rubber sheet. A way to describe the birth of a wave is that a part of the field of M "escapes" when M is accelerated abruptly. Then it makes sense that the energy density formula for N is the same in a static field and in a wave.

But in gravity, the energy density is a whopping 16X the energy of the analogous electromagnetic wave. Is there any way to simulate this large energy density in a rubber model?

                                      

                 ----___       ====  bicycle chain

                           ••••

                            pit

                        mass M

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

Maybe we can attach "bicycle chains" (roller chains) to the mass M. The chains have the axes of the links horizontal, so that they can effortlessly adapt to the vertical shape of the pit in the rubber sheet when the mass M is static. But if we move M rapidly back and forth on the rubber surface, then the chains transport a lot of energy away. A test mass m may be attached to one of the chains, and receives a lot of energy.

The chains implement another channel of energy transport besides the rubber sheet itself. We found an example where the static field of M contains relatively little energy, but a produced wave contains a lot of energy.

Adding a term which is linear in the stretching of the spatial metric?

The shrinking of of the metric of time pulls masses together: they can sink into a lower potential and can reduce their common potential energy.

If we, in addition, add a term which assigns a lot of energy to the stretching of the spatial metric, then we have a field which has a high energy density but only weakly pulls masses together. Then the energy density of a gravitational wave could be large, but the gravity pull weak.

This makes a lot of sense. If we make a mass M to oscillate, the bulk of the energy from the gravitational wave can be harvested from the oscillation of the spatial metric.

But how can we add the energy to the spatial stretching field? The energy cannot come from the mass M falling into a lower gravity potential. There is too little energy available there. The spatial stretching field is like a "physical body" around M. Sitting in space, not born from anything else. It is like a cloud of substance which M always carries along. That substance gets its form from the newtonian gravity field of M. How to write that into a lagrangian?

The spatial field has to be "private" for M? The field acts in a different way on the particle which creates it? How to write that into a lagrangian?

The trick is to introduce a new separate field for each particle in the universe? It is not as bad as it sounds. Every particle is, in a sense its own separate field. Add another separate field on top of that.

Conclusions

Let us close this blog post.

1.   If we calculated right in April 2024, then the Einstein field equations lack a solution for a cylinder with a shear. This suggests that the Einstein equations do not have a lagrangian at all. Otherwise the minimum "energy" state of the cylinder would satisfy the Einstein equations.

2.   The Einstein-Hilbert action tries to calculate the "energy" of the gravity field from the mass density alone. This cannot succeed. The action is badly incorrect. The action might work in empty space. We have to investigate.

3.   The lagrangian of newtonian gravity, or the lagrangian of electromagnetism, calculates the gravity field of a mass distribution correctly, if the field is weak. Thus, it is a better lagrangian than the incorrect Einstein-Hilbert lagrangian.

4.   The Einstein equations calculate the energy density of gravitational waves correctly, which is 16 times the energy density of the analogous electromagnetic wave.

A lagrangian may require that each particle has a field which acts differently on the particle creating the field from how it acts on other particles. The field is "private".