New lagrangian for gravity

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.

The inertial frame field X

Let us try to define an "inertial frame field" X, against which the kinetic energy of a particle has to be determined.

The letter X comes from the fact that we are looking at the distortion of the spatial coordinates x, y, z of a particle.

The kinetic energy of a particle m in the new model is defined as:

       1/2 m v(X)².

In the new model, the velocity v of m has to be taken against the inertial frame field X. If we accelerate the particle, then some of the new kinetic energy goes to accelerating the field X. That reduces 1/2 m v(X)².

In the case of the accelerated mass M, the inertial frame field X is dragged along. How to assign an energy density to X?

In the case of the oscillating quadrupole, the inertial frame field X oscillates throughout space. How to assign an energy density to oscillations of X?

The inertial frame field X reminds us of Mach's principle: the masses of the universe determine which is an inertial frame. If m is very close to a black hole M, then the inertial frame of m is almost totally determined by the movements of M, also if M is accelerated.

But is it possible to formulate the inertial frame system as a "field"?

Let us try to define X first as an oscillating field. We fix some coordinates t, x, y, z, and put initially static infinitesimal test masses m floating freely in space. The field X could be defined by the coordinate acceleration of a test mass m as a gravitational wave passes by. Then the field X would be an "inertia force field" which couples to a mass m:

      F  =  m X.

This is very much like an electromagnetic wave, which couples to a charge q:

       F  =  q E.

The field X may have an energy density:

       ~ 1/2 X².

The field X couples to a mass m through its effect on the kinetic energy of m:

       1/2 m v(X)².

This is the big difference to the electric field E which couples to a charge directly through the electric potential V:

       q V.

However, the potential V is not clearly defined for an electromagnetic wave. A vector potential A has to be added.

Is the inertial frame field X actually the free falling particle field? Close to a mass M, a freely floating particle m is accelerated toward M. In a gravitational wave, m oscillates back and forth.

The inertia force would be the coupling of the ordinary newtonian gravity force inside the kinetic energy formula of the particle. But we have to get a high energy density for a wave in the field X. It cannot be a wave of ordinary newtonian gravity.

***  WORK IN PROGRESS  ***

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

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.

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?

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".

Better rubber sheet model of gravity: prevent horizontal movement of rubber

We have a tense horizontal rubber membrane. We can put weights on it, and springs between weights. The springs model pressure.

    ---|-------|-------|-------|----  tense rubber sheet

 

      steel spikes prevent

      horizontal movement

The new feature is that we have a steel spikes which prevent any horizontal movement of the rubber sheet. Only a vertical movement is allowed.

We may imagine that there are holes in the rubber, reinforced with steel rings. The spikes go through these rings.

The spikes prevent longitudinal waves in the rubber sheet. The model is now closer to general relativity. Only the vertical elevation of the sheet matters.

Birkhoff's theorem now holds? No

 

        ----___          ___----  rubber sheet

                   • • • • 

              particles in 

        circular formation

Recall an example from November 2023. We suddenly increase the pressure inside a circular mass on the rubber sheet.

The pressure starts to stretch the rubber sheet horizontally within the circle, and succeeds in that, too. It pushes the outer parts of the circle upward, trying to make the rubber area within the circle as large as possible.

The rubber sheet is lifted up outside the circle. Birkhoff's theorem fails because the vertical shape of the rubber sheet changes outside the circle.

Can we decouple the metric of space from the metric of time?

The model differs from general relativity in that that the metric of time (vertical elevation) determines the metric of space uniquely. Can we decouple these?

In our previous blog post we have serious problems varying the metric of time and space separately in general relativity. Is it so that they cannot be decoupled?

Pressurized vessel: removing suddenly the negative pressure in the skin of the vessel

Let us have a ring of particles lying on the rubber sheet. We put squeezed springs between them inside the circle, modeling a positive pressure there. Stretched springs between the particles balance the forces and make the system static.

                         particle

        -------•                •--------  rubber sheet

                  \______/

            positive pressure

                   in a "pit"

   

There is a sharp bend in the rubber at the skin of the vessel. In the diagram, the bend is at the particle. The pressure inside the ring pushes the rubber sheet a little down inside the ring. There is a pit.

We suddenly remove the stretched springs, i.e., we remove the negative pressure. The positive pressure starts to accelerate the particles horizontally outward, and also vertically up.

If the particles are very heavy, they accelerate slowly. The pit and the bend may linger there for a long time.

In general relativity, pressure is a "charge". It affects the metric. A sudden loss of a charge confuses the Einstein field equations. They may fail to have a solution.

In our rubber model, a sudden loss of a pressure is no problem. We can calculate the development of the system without any problem.

The negative pressure between the particles simply acted as a force which keeps them static. It was not any "charge" which would affect the shape of the rubber sheet instantaneously.

If we suddenly remove the pressure inside the ring, what happens?

If there are particles also inside the ring, then their movement prevents any abrupt movement of the rubber.

In the rubber sheet model, pressure is not a "charge" which creates a dent in the rubber sheet. The process is more complex.

A "charge" is something which cannot be created or destroyed. The mass of the weights in the rubber sheet model is a charge.

Conclusions

We have worked very hard to find a rubber model which would reasonably approximate general relativity – in vain.

In the previous blog post we were unable to get the Einstein-Hilbert action to work on the Schwarzschild interior metric. This casts a doubt on the correctness of the action.

Our own Minkowski & newtonian model claims that spacetime cannot be "bent" or stretched. The effects of gravity are due to newtonian gravity and the inertia changes of the test mass (or photon) in the gravity field. Then it is unlikely that a rubber model can imitate gravity.

In our own gravity model, the lagrangian density is quite complicated because it has to take into account the changes in inertia, and the kinetic energy stored into the field when the inertia of a test mass m grows.

What metric variations are allowed in general relativity?

UPDATE May 15, 2024: We added the section "An analysis of the Wikipedia formula".

----

In our previous blog post we showed that if a negative Ricci scalar is allowed in empty space, then we can reduce the value of the Einstein-Hilbert action without a limit. Such variations must be banned.

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

We set κ = 8 π G / c⁴ = 1.

Is LM of the form:

       potential energy  -  kinetic energy,

or:

       kinetic energy  -  potential energy?

The value of R inside a mass M is traditionally taken to be positive. It makes sense to take the potential energy to be positive in LM. The curvature R is then "potential energy" of spacetime. In a rubber sheet model of gravity, it is obvious that a positive R represents positive potential energy.

Reducing distances throughout space must be banned

Let us have a positive and a negative electric charge placed in our coordinate system. Let us assume that the metric g is flat. If we shrink all spatial distances equally, the metric stays flat: the Ricci scalar R is zero everywhere.

But the integral of LM decreases because there is less potential energy. The Einstein-Hilbert action has a lower value then.

This does not make sense. We must require that the metric of space is asymptotically 1 far away.

Is this enough? There can be an immense attractive force between the electric charges. An attractive force constitutes a "negative pressure".

We can shrink the distance between the charges by adding a negative Ricci scalar R between them.

Then we can decrease the integral of R and also the integral of LM in the Einstein-Hilbert action S. This does not make sense.

Increasing the volume of a pressurized vessel: an example of the variation procedure

Let us study a practical variation problem. Let us imagine that we have a spherical vessel where the density of the liquid is ρ. The fields are weak. There is a uniform pressure p throughout the vessel.

The Ricci tensor is approximately

       R  =

    1/2 ρ + 3/2 p    0                  0                           0

    0               1/2 ρ - 1/2 p        0                           0

    0                        0       (1/2 ρ - 1/2 p) r²            0

    0                        0                  0      (1/2 ρ - 1/2 p)

                                                           * r² sin²(θ).

The Ricci scalar:

       R  =  ρ  -  3 p.

The variation δg. Let us vary the metric by increasing the radial metric g₁₁ at some r. The volume of the vessel grows.

The Wikipedia article says that

Let us assume that g₁₁ is close to 1. We increase g₁₁ by

       0  <  d  <<  1

at some short segment of r. Let the spatial volume of the spherical shell defined by that segment be U. Let the integration coordinate time interval be T.

Calculating the change in S. 

R / 2.   Then g¹¹ = 1 / g₁₁ decreases by d. Wikipedia says that the Ricci scalar R changes locally by

        

       δR  =  δg^μν  *  Rμν 

              =  δg¹¹  *  R₁₁

              =  -d  (1/2 ρ₀  -  1/2 p).

This changes the integral of R / 2 by

       -d / 2 * (1/2 ρ₀  -  1/2 p)  *  U T.

The volume element sqrt(-det(g)) grows locally by a ratio d / 2. This adds to the integral of R / 2:

       (ρ₀  -  3 p)  *  d / 4  *  U T.

The notation ρ₀ stresses that it is the density before the stretching operation. We are varying the numerical values of the metric g. It does not matter where those numerical values originally came from. We assume that p remains constant.

In total, the integral of R / 2 changes by

       -d / 2  *  p U T.

LM.   Let us assume that the pressure comes from a repulsive force between particles. The potential energy of the repulsion V changes by

       -p d U.

The mass of the particles in U does not change. The change in the integral of LM is

       -p d U T.

The total change in the action S is

       -3/2 p d U T.

This does not make sense.

Practical variation problem: varying the metric of time g₀₀ inside a spherical mass M

Let us calculate the most basic variation in general relativity. We should not be allowed to drop a mass M to an arbitrary low gravity potential. We must be punished by a growing Ricci scalar R. In a rubber sheet model of gravity, a weight M cannot fall arbitrarily low. The elastic energy in the stretched rubber sheet must stop it from falling lower.

The Ricci tensor R is the same as in the previous section, but there is no pressure p.

The variation δg. The initial mass density of the sphere is ρ₀. We assume that g₀₀ is very close to -1. We increase g₀₀ by

       0  <  d  <<  1

in some short segment of r. Let the spatial volume defined by that segment be U. We integrate S over some coordinate time interval T.

Calculating the change in S.

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

        

       δR  =  δg^μν  *  Rμν 

              =  δg⁰⁰  *  R₀₀

              =  -d  *  1/2 ρ₀.

This changes the integral of R / 2 by

       -d / 2 * 1/2 ρ₀ U T.

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

       ρ₀  *  -d / 4  *  U T.

In total, the integral of R / 2 changes by

       -1/2 ρ₀ d U T.

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

       ρ₀  *  -d / 2  *  U T

       =  -1/2 ρ₀ d U T.

The total change in the action S is

       -ρ₀ d U T.

The result would make sense if the lagrangian LM would have the sign flipped. Now it does not make sense.

The Einstein-Hilbert action works correctly only in empty space?

Note that if ρ and p are zero, then the results of the two preceding sections are reasonable: the small variation δg of the metric does not change the value of S.

There has to be a way to make the variational calculus to work because in a rubber sheet model of gravity it does work. In the rubber sheet, it is not possible to vary g₀₀ and g₁₁ separately. If we press a weight M down on the rubber sheet (increase g₀₀ from -1), that inevitably stretches the rubber: g₁₁ must increase from 1.

Could it be that we must restrict variations to "reasonable" ones, which vary both g₀₀ and g₁₁ simultaneously? How do we define which variation is reasonable?

Varying the Schwarzschild interior and exterior solutions

The variation δg. Let us have a sphere of a mass M of a uniform density ρ and of a coordinate radius K. The mass M is measured from far away.

A "reasonable" metric variation might be one where we replace the Schwarzschild metric g produced by ρ, by the metric g' produced by a slightly larger or lower density ρ':

       ρ'  ≠  ρ.

Does that variation keep the value of S unchanged?

Note that ρ' varies the metric. We do not vary the mass M. When the spatial metric of r is stretched, that affects also the real, proper mass density ρ. Should we let K to shrink, so that ρ is kept constant? We will see if that makes a difference.

The variation sounds very sensible because it tries to drop the mass M to a lower gravity potential, and thus save on the integral of the LM. The price we pay is a larger integral of R / 2.

Calculating the change in S?

R / 2. The value of R is a constant ρ' within the sphere. Inside the sphere, we use the Schwarzschild interior solution.

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

where rs = 2 G M / c² and rg is the coordinate radius of the sphere, K. Recall that we set the Einstein constant 8 π G / c⁴ to 1.

Is there a simpler way to calculate?

The contribution of R / 2 to the integral S is

       ρ' / 2  *  the volume element,

and the contribution of LM is

       ρ  *  the volume element.

Is there a simple reason why the optimum is attained at ρ' = ρ?

A new variation δg. Removing the center of the sphere M in the metric g. Let us calculate a simple case. We modify the metric g as if the mass at the center of the sphere M would be missing.

Let the volume of the central area be

        b  *  the volume of M,

where 0 < b << 1.

R / (2 κ).  The Ricci scalar R in SI units is

       R  =  κ  ρ  =  8 π G / c⁴  *  ρ.

The integral of R / (2 κ) over the whole M is

       M / 2.

Removing the center reduces the integral of R / 2 by

       1/2 b M T,

where T is the integration coordinate time interval.

The metric of time g₀₀ drops closer to -1 throughout M, and the radial metric g₁₁ drops closer to 1. The net effect on the volume element sqrt(-det(g)) is probably zero outside the (small) central area.

LM.   The mass M rises to a higher gravity potential. How much higher?

The gravity potential by the mass b M is

       -G b M / r,

where r is the radius from the center. The potential of the sphere M, caused by the central mass b M is

        K

       ∫ -G b M / r  *  ρ  *  4 π r² dr

      a

       ≈  -b * G M ρ  *  2 π K²,

where K >> a > 0 is the radius of the mass b M, K is the radius of the sphere, and ρ is the density of the sphere.

The formula looks very different from the change in R / (2 κ).

What is the problem? Does the radial metric affect this?

Gradually adjusting the Ricci scalar R of the interior Schwarzschild metric

Let us test a very simple variation of the metric. We use spherical coordinates with a metric signature (- + + +). The fields are weak. The metric g is only slightly perturbed from the flat metric θ.

First, let us prove that the Ricci tensor R looks like this inside a mass of a constant mass density ρ:

       R  ≈

            κ  *

               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),

where κ is the Einstein gravity constant.

The Ricci scalar R is then

       R  =  κ  *  (

                 g⁰⁰ * 1/2 ρ  +  g¹¹ * 1/2 ρ

              + g²² * 1/2 ρ r²  +  g³³ * 1/2 ρ r² sin²(θ)

                         )

            =  κ  *  (-1/2 ρ  +  1/2 ρ  +  1/2 ρ  +  1/2 ρ)

            =  κ ρ.

The stress-energy tensor T is

       R  -  1/2 R g  =  T  =

            κ  *

                 ρ     0     0     0

                 0     0     0     0

                 0     0     0     0

                 0     0     0     0.

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

We obtained the correct value. The interior Schwarzschild metric is:

We assumed that the Schwarzschild radius rs is very small. The relevant part of the metric becomes:

          dτ²  =  -( 3/2  -  3/4 rs / rg

                       - 1/2  +  1/2 r² rs / rg)²  * dt²

                       + 1 / c²  *  (1  +  r² rs / rg)  *  dr²

               =  (-1  +  3/4 rs / rg  -  1/2 r² rs / rg) * dt²

                   + 1 / c²  *  ( 1  +  r² rs / rg )  *  dr².

Let us gradually change the metric from flat to the perturbed metric, within the entire sphere. The Ricci curvature for both t and r is within the sphere:

       R₀₀  =  R₁₁  =  κ  *  1/2 ρ.

The Wikipedia formula states:

Let us check that we get reasonable values. Let us calculate the value of R if we gradually make the perturbation in g stronger. The value of both R₀₀ and R₁₁ is then, during the process, on the average,

       1/2 κ ρ / 2.

The rise of g₀₀ up from -1 adds to R the value:

       1/4 κ ρ  *  ( -3/4 rs / rg  +  1/2 r² rs / rg ).

The rise of g₁₁ up from 1 / c² adds to R the value:

        1/4 κ ρ c²  *  -r² rs / rg.

Let us have r = 0. The calculation claims that

       1/4 κ ρ * -3/4 * 2 G M / c² * 1 / rg 

       =  κ ρ

  <=>

       1  =  -3/8 G M / c²  *  1 / rg.

The formula makes no sense.

The notation δR / δg^μν is not too clear. If δg^μν varies the metric g in a large volume, say, inside the entire spherical mass M, then then the equation might mean the integral over M.

Lifting the underlying mass density ρ of the metric a little bit at a time, in a small spatial volume at a time. But why would we be banned from varying the metric in just a small volume at a time? A very small modification of the local metric preserves R₀₀ and R₁₁ almost the same everywhere. We can change the metric in small parts, and we get the entire metric inside M closer to the target.

More precisely, the initial flat metric corresponds to ρ = 0 in the entire volume of M. Then we gradually let ρ grow to its final value. We do that in many rounds. At the end of each round, ρ is constant inside the entire M.

The Ricci tensor R is locally determined by ρ only. If we increase ρ locally a little, the change in the local Ricci scalar R should obey the Wikipedia formula. The local value of R should change as the formula claims.

For example, at the center, r = 0, we should finally have the strange value calculated above.

This still does not make sense.

https://physics.stackexchange.com/questions/547082/einstein-hilbert-action-does-not-yield-the-same-results-as-the-einstein-field-eq

"user195583" complains in the link that he cannot get the same result from varying the Einstein-Hilbert action as he gets from the Einstein field equations.

An analysis of the Wikipedia formula

Let us restrict ourselves to metrics inside a spherical mass whose density is ρ(t, x), and initially uniform, and the metric g is almost flat. The metric signature is (- + + +). We set κ = 1. Then, initially:

       R(t, r)     =  ρ(t, r),

       R₀₀(t, r)  =  1/2 ρ(t, r),

       R₁₁(t, r)  =  1/2 ρ(t, r).

Let us increase the density ρ(t, r) locally in some volume U of the sphere. Then g₀₀(t, r) slightly increases from ≈ -1, and, at least if the volume U is at the center of the sphere, g₁₁(t, r) slightly increases from ≈ 1.

The value of g⁰⁰(t, r) = 1 / g₀₀(t, r) slightly declines and g¹¹(t, r) = 1 / g₁₁(t, r) slightly declines.

The Wikipedia formula above claims that

       dR(t, r)  =  -R₀₀(t, x) dg₀₀(t, r)

                          - R₁₁(t, x) dg₁₁(t, r)

  <=>

       dρ(t, r)  =  -1/2 ρ(t, x) dg₀₀(t, r)

                          -1/2 ρ(t, x) dg₁₁(t, r).

The sign is wrong. The density ρ(t, r) increases, but on the right we have a negative number. Let us fix the sign error, and look at the center of the sphere (r = 0), where g₁₁(t, 0) = 1 always.

The perturbation of g₀₀(t, 0) at the center of the sphere up from -1 is linear in ρ, if the entire sphere has a constant density ρ:

       g₀₀(t, 0)  =  -1  +  C ρ,

for some constant C > 0, because the newtonian gravity potential V is linear in ρ. Let us increase the density by a constant dρ throughout the sphere.

We get:

       dρ(t, 0)  =  1/2 ρ(t, 0)  *  C dρ(t, 0)

  <=>

       1  =  C / 2 ρ(t, 0)

  <=>

        ρ(t, 0)  =  2 / C.

The result is nonsensical. It claims that ρ(t, 0) must be constant.

A toy rubber sheet model – it does not work well

                      pit

        -----___          ___-----  rubber sheet

                    ••••••      h

                       M

              circle of mass

The depth of the pit is h. The surface of the rubber sheet acts roughly like a harmonic spring. The elastic energy of the pit is

       ~  h².

The total energy of the system is

       ~  (h²  +  M)  *  (1  -  h).

This looks like the Einstein-Hilbert action if

       R / 2  ~  h²,

and

       sqrt(-det(g))  ~  (1  -  h).

Are these reasonable?

Then h is the metric perturbation, i.e.,

       g₀₀  =  -1  +  h.

The angle of the rubber sheet α at the edge of the circle must be

       α  ~  M,

for the tension of the sheet to keep M from falling lower.

How much does the rubber sheet stretch horizontally?

The angle α(r), where r is the distance from the center of the mass M, is

       α(r)  ~  M / r

                ~  h'(r),

to keep the rubber sheet circle of a radius r from falling lower.

The curvature radius r of the rubber sheet under the circle is then

       r  ~  1 / α

           ~  1 / M

           ~  1 / h.

The curvature Q of the rubber sheet under the mass circle is

       Q  ~  1 / r²

            ~  h².

We guess that Q is zero outside the mass circle.

The energy of the system can be written roughly as:

       ~  (Q  +  M)  *  (1  -  h).

It looks like the Einstein-Hilbert action. We have a very crude mechanical model for the Einstein-Hilbert action.

The rubber sheet model may help us in understanding why the Einstein-Hilbert action does not seem to work properly if solely the metric of time, g₀₀, or the radial metric g₁₁ is varied.

However, g₀₀ and g₁₁ in this model do not look like those of the external Schwarzschild solution, maybe because our rubber model has fewer dimensions:

        dg₀₀ / dr     ~  1 / r,

        g₁₁(r)  -  1   ~  1 / r².

No one has found a rubber model which would replicate general relativity well.

Our own Minkowski & newtonian model says that gravity is not about the spacetime geometry at all. There is no "spacetime substance" which is being deformed by mass, pressure, and shear. Consequently, there probably does not exist a rubber model which would replicate gravity well.

Conclusions

We are not able to get any calculation right: all indicate that the Einstein-Hilbert action is not at the minimum in the interior Schwarzschild metric. Also, the formula

brings incorrect results.

This is perplexing. When we started to calculate Christoffel symbols in March 2024, we immediately got at least something right, even though most of our calculations were erroneous.

In the literature, we have not seen anyone actually doing variations to the metric g and calculating directly the effect on R / (2 κ) and LM. We have to check if anyone has written about this.

In general relativity, the Einstein field equations are the primary subject of study. The action is something which is added later. David Hilbert in his 1915 paper introduces the action. He claims that the Ricci tensor R is the only tensor which can be formed from second derivatives of g, and that implies that the action is correct. He does not try to calculate variations.

Maybe we have misunderstood something. We will write more blog posts about this mystery.