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.

How can then the Einstein-Hilbert action work at all? The Ricci scalar R is 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

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, 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

***  WORK IN PROGRESS  ***

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.

Shear stress: why do the Einstein equations fail?

On November 5, 2023 we tentatively analyzed why the Einstein field equations "break" in the case of a pressure change. By breaking, we mean that there is no solution for the equations.

On April 22, 24, and 26, 2024 we presented examples where a shear stress inside a cylinder does not allow any solution for the Einstein field equations. Let us analyze why there is no solution.

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

We start from the Einstein-Hilbert action. The state and the development of a physical system should minimize or maximize the action integral. Above, S is the action, R is the Ricci scalar, LM is the lagrangian density of other fields besides gravity, and g is the determinant of the metric g = gμν, det(gμν).

In a cylinder with shear, we were looking for a static solution. There is no kinetic energy in the lagrangian LM. The task is to minimize the potential energy of the system.

On November 5, 2024, the lack of a solution was due to the fact that a "global stationary point" of the action is not a "local stationary point". That is, one cannot find an extremal value of the action through a simple rule which extremizes the action locally at every point. The Einstein field equations assume that we can do the local extremization.

Finding a stationary point is much easier if it can be done locally. That is the advantage of having field equations which are local. However, in this blog we have suggested that the concept of a "field", or a "metric" cannot capture every aspect of the interaction between particles.

Is the strange metric of a thin disk due to the local nature of the Einstein field equations?

In our May 4, 2024 blog post, we wondered if the metric which general relativity gives for a thin disk is correct.

In the Minkowski & newtonian model, the energy shipping happens over a short distance, from the thin disk to the test mass m. But general relativity seems to believe that the field of a thin disk is actually created by a large spherical shell. The metric is like Minkowski & newtonian would predict for a large spherical shell.

The Einstein field equations are local. They cannot recognize what is the source of the field. Minkowski & newtonian does consider the source of the field.

Finding the minimal "potential energy" in the Einstein-Hilbert action S

We are looking for a static solution which minimizes the Einstein-Hilbert action S. There is no kinetic energy. The Ricci scalar R is the "potential energy" density of the gravity field. The matter lagrangian LM is the potential energy density of other fields.

We can save in the potential energy in LM by making

         |g₀₀|

smaller, that is, by slowing down time. That makes

       sqrt( -det( gμν ) )

in the action integral smaller. The price that we pay is that R becomes larger and increases the value of the action integral.

How do we know that the action integral S has a minimum at all?

It could happen that S can have any value larger than some constant C, but no solution has

       S  =  C.

Another possibility is that the values of S have no lower bound C at all.

If we allow singularities, then S might not have a lower bound. Let us assume that pressure prevents the system from falling into a singularity.

If the possible configurations of the system form a closed set in the space of functions, and we assume that the S is continuous, then it is guaranteed that S does attain the smallest value at some point. This is a basic result of topology.

It looks like that with certain restrictions we can make sure that S has the smallest value at certain point.

The Einstein field equations are derived from the Einstein-Hilbert action by varying the metric g. A variation means that a component, say, g₁₁(x) is changed a little by adding another "infinitesimal" function h(x) to g₁₁(x), and

       |h(x)|

is very small at every point x of the spacetime.

The metric g minimizes the action S if the value S cannot be made smaller by adding an infinitesimal function to each component of g.

Note that minimizing the action S here assumes that the coordinate locations of each particle in LM stay constant.

What about minimizing the action S by letting the particles to change their coordinate locations? The global minimum of S has to depend on that, too. It cannot be depend on just the metric g.

In our cylinder examples in April 2024 we tried to make sure that the system already is at a global minimum of LM. The forces on each volume element in the 3D space must add to zero, and the torque must be zero.

The next step is to set the metric g. The metric only differs very little from the flat metric. Can we assume that the position of each particle is constant in the coordinates and that the pressure and the shear are unchanged when g is perturbed a little from the flat metric?

The relative value of the mass density ρ(x), the pressure p(x), and the shear s(x) can only change very little with a small perturbation of the spatial metric. It is a good approximaton to assume that they do not change when g is varied.

The wall-to-wall carpet wrinkle example and the variation of the Einstein-Hilbert action

On November 5, 2023 we analyzed how a changing pressure "breaks" the Einstein field equations.

The metric is initially such that it does satisfy the Einstein field equations. Then the pressure inside the spherical mass changes, which would force the metric to change throughout space. But the metric cannot change faster than light. As the result, there will be a zone around empty space where R is not zero: the Einstein equations are not satisfied.

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

The Wikipedia article derives the Einstein field equations through variation.

https://www.preposterousuniverse.com/spacetimeandgeometry/

The variation procedure in Wikipedia is based on the book by Sean M. Carroll (2004).

Why the Wikipedia variation method does not work in the case of a changing pressure?

The variation method claims that the Ricci tensor R must be zero in empty space. Locally, we can make R zero by varying the metric. But then the zone of a non-zero R just moves to a new location. It does not help. The variation method fails to take into account this displacement of the R ≠ 0 zone?

That is, the variation method does not calculate everything which changes with the infinitesimal variation ∂g^μν of the metric?

Or the method implicitly assumes that the metric which we vary satisfies the Einstein equations everywhere before the time t coordinate at which we vary the metric?

Wikipedia says:

"By Stokes' theorem, this only yields a boundary term when integrated. The boundary term is in general non-zero, because the integrand depends not only on [the variation of the metric] ∂g^μν, but also on its partial derivatives"

"However when the variation of the metric ∂g^μν vanishes in a neighbourhood of the boundary or when there is no boundary, this term does not contribute to the variation of the action. Thus, we can forget about this term and simply obtain  

at events not in the closure of the boundary."

Let us analyze what happens when we try to use the formula above to iron out the "wrinkle" between the different metrics if we (magically) change the mass of a spherically symmetric body.

The metric at r > K + 1 is the Schwarzschild metric for a central mass M, and the metric at r < K is the Schwarzschild metric for

       M'  <  M.

The transition layer of the metric is at K < r < K + 1. We can satisfy the Einstein field equations if we imagine that there is a spherical shell of a uniform density and a mass

       M  -  M'

in the transition layer.

             -------------------------------------> r

                                       K               K + 1

            ●                         |..................|

           M'                              M - M'

  central mass           transition layer (shell)

  

We use the (- + + +) metric signature. The flat metric in spherical coordinates is 

       η  =

              -1             0             0                   0  

               0             1             0                   0

               0             0             r²                  0

               0             0             0     r² sin²(θ).

 

Assuming the imagined uniform density mass, the Ricci tensor R in the transition layer looks like that for a uniform mass density:

       R  ~

               1             0             0                  0

               0             1             0                  0

               0             0             r²                 0

               0             0             0     r² sin²(θ).

The Ricci scalar R is ~ 2 in the layer, and zero outside it.

At r > K + 1, we have

       g⁰⁰  =   1 / g₀₀

              =  -1  -  ε M / r,

where ε is a very small positive constant. Also, we have

       g¹¹  =   1 / g₁₁

              =   1  -  ε M / r.

At r < K, we have

       g⁰⁰  =  -1  -  ε M' / r,

       g¹¹  =   1  -  ε M' / r.

In the transition layer, the metric g transitions smoothly between these two formulae.

Ironing out the wrinkle by making the transition layer thinner. We try to iron out the wrinkle by making g⁰⁰ and g¹¹ to obey the upper formulae (those with M) at an outer layer:

       K + 1 - δ   <   r   <   K + 1,

where δ is a small positive constant. Then the values of g⁰⁰ and g¹¹ decrease a little bit in that layer. The Wikipedia formula

claims that (the integral over the entire spacetime of) the Ricci scalar R then decreases a little bit, and there are no other changes. That claim is false. Ironing out the Ricci scalar R in that outer layer clearly increases the integral of R by the same amount in the remaining transition layer. 

If we reset g in the outer layer, then the transition becomes steeper in the remaining transition layer and the integral of R does not change over the full spacetime.

Ironing out the wrinkle by reducing M. If we would do the ironing by reducing M everywhere, then in the outer layer the values of g⁰⁰ and g¹¹ would grow a little bit, and also for all r > K + 1. The Wikipedia formula claims that the integral of the Ricci scalar increases, while it, in reality, decreases.

There is a "sign error" (?) in Wikipedia, but that is not the main error. The main error is that the integral of R should not change if we use the first ironing method described above.

Analysis of the Wikipedia variation

Variation of the Ricci scalar R. The Ricci scalar R for a Ricci tensor R is defined as

       R  =  ∑  Rij g^ij,

               i, j

where 0 ≤ i ≤ 3 and 0 ≤ j ≤ 3.

The Wikipedia formula

calculates the variation of R, assuming that R stays constant and only the gμν change. This explains the "sign error" in the "ironing out by reducing M" case. The Wikipedia formula ignores most of the changes in R!

In a flat metric, R = 0. A non-zero R is "mostly born" from the curvature which perturbations of the various components of the metric gij cause. The role of multiplying by g^ij is very minor. But the Wikipedia formula claims that the multiplication plays the major role!

Did we misunderstand? Let us check how Wikipedia arrives at the Einstein field equations.

Variation of the determinant sqrt(-det(g)). Wikipedia says that

is the variation. Let us check if the formula is reasonable. Let us assume that the metric g is diagonal and that the flat metric is only slightly perturbed.

Let us assume that g₁₁ is initially 1. We decrease g₁₁ to the value 0.99. The absolute value of the determinant det(g) decreases by 1%, and the value of the square root by 0.5%. The value of g¹¹ increases by 0.01. The formula above says that the square root decreases by 0.5%. That is correct. The variation formula is reasonable.

The Einstein field equations. These come directly from the two variations above:

where 

The final result is the familiar Einstein field equations, but the variation of R was very strange.

https://www.jp-petit.org/nouv_f/videos_liens/illustrations_Francfort/Hilbert-1915-de.pdf

https://www.jp-petit.org/Hilbert-1916-de.pdf

Above we have links to the 1915 and 1916 papers by David Hilbert, Die Grundlagen der Physik. Do they contain the variation procedure?

On page 404 of the first paper, David Hilbert writes that by using the "designation" introduced above to the variation derivation one gets the following form of the equation:

On page 405 David Hilbert writes that the last equation follows simply, without a calculation, since "Rμν is the only second-order tensor besides gμν, which can be built from g^μν and its first and second derivatives." Hilbert uses the letter K instead of R.

Let us look at the variation procedure in other papers. Does anyone appeal to the fact that R is a second-order tensor?

Ok, we solved the mystery. The variation of the metric, g^μν, probably has to be local. We cannot insert a new mass m anywhere, because the integral over the metric perturbation ~ m / r over the whole space would be infinite. We cannot increase the integral of R with local changes.

The variation ∂g^μν produces a very small change to the integral of R. The variation itself is very small, and the Ricci curvatures Rμν are very small. It is small squared – almost negligible.

We can continue our analysis of the "wrinkle", now that we understand the derivation of the Einstein field equations from the Einstein-Hilbert action.

More analysis of the wrinkle in the carpet – or a pressure change in a spherical mass

The Einstein equations imply conservation of mass-energy of a spherically symmetric system. This is Birkhoff's theorem. But the Einstein-Hilbert action is very tolerant to physical phenomena. It cannot dictate any such law. Let us check how the action reacts to a mass-energy change.

Let us imagine that we have a spherical mass M and the Schwarzschild interior and exterior metrics. A magic trick makes M to disappear suddenly. What does the action say about the development of the system?

Suddenly, we can reduce the "potential energy" of the action by letting the metric g become flat at the former location of M. The Ricci scalar R is reset to zero there, maybe instantaneously.

The speed of light prevents resetting the metric to flat in the entire space instantaneously. We may imagine that a "transition layer" escapes from the former location of M, and the metric in the layer is like it would be with the Einstein equations if there were the mass M in the layer.

The Einstein equations then fail because there actually is no mass in the transition layer.

The variation method above proved that the Einstein equations must hold in the system. But they do not hold. What spoils the variation?

The variation procedure above assumes that there was enough time for the system to settle down to the lowest possible value for the action. If the mass M suddenly changes, then that would require faster-than-light communication.

If the conservation law for a charge is broken, then the variation argument does not hold. The Einstein field equations are not satisfied.

This does not explain why the Einstein field equations fail for a cylinder containing shear. Let us try to find out.

If third derivatives of g affect LM, then the Einstein equations may find a wrong solution

In this example we allow also very strong fields. Let us have a large spherically symmetric mass M embedded inside an extremely rigid thick shell of matter N.

               M              N

                ●      ■■■■■■■■■

                 -----------------------> r

Let us assume that we have found a metric g which, when varied only locally with changes in a very small environment U of an arbitrary spacetime point x, keeps the Einstein-Hilbert action S value unchanged.

Let us then consider a variation δg of the metric in the entire volume enclosing both M and N. We may reduce the Ricci scalar R in the entire M, in order to reduce the stretching of N. The elastic energy of N is greatly reduced.

Is there any reason why δg would keep the action S unchanged? If the value of the action S is linear in the sum of small local variations, then we can present δg as a sum of small local variations, and then we know that also δg keeps S unchanged. But, in general, is the action S linear in such small local variations?

Let us assume that the exotic material in N strongly resists adding a varying Ricci scalar R in the volume occupied by N, but falls into a lower energy state if a uniform negative Ricci scalar value is added to the volume of N.

Let us consider local variations δg which add a "positive mass density" at some location, and a "negative" mass density in the volume surrounding that location. The masses cancel each one out. That is, we add a positive Ricci scalar at the location, and a negative Ricci scalar around it.

Now, small local variations increase the energy of N. But a variation enclosing the entire N can reduce the energy of N. Since the Ricci scalar depends on the second derivatives of the metric, the potential energy in the lagrangian LM of our exotic material N depends on the third derivatives of the metric.

The Einstein field equations may claim that the exotic material N is in the lowest energy state, while it is not.

This is probably a known result: if we allow LM to depend on the third derivatives of the metric, then the Einstein field equations do not work. However, such exotic materials are rare in nature.

Question. The stress-energy tensor only measures the variation of dLM / dgij. Could it be that if LM depends on the first or second derivatives of g, that already is enough to confuse the Einstein field equations?

Is it possible to restrict LM in such a way that it does not depend on the third derivatives of g?

Probably not. An exotic material can be built from innocuous parts, each of which does not depend on the third derivatives. But the combination of the parts does depend!

We can build a mechanical device which measures the third derivatives of g, and puts a potential energy V(g''') into a spring according to what the measurement told. Then LM does depend on the third derivatives.

How to define the stress-energy tensor?

The stress-energy tensor T is defined by looking at how the integral of the lagrangian LM reacts to a variation of each gμν.

But what is such a variation like? We increase the value of gμν in a small environment U of a spacetime point x, and measure how much the integral of LM changes?

What if the change in the integral depends on the precise form of the function which we use to increase, say, g₁₁? For example, the precise form of dg₁₁ / dr may affect how much the integral changes.

In that case, the "pressure" component T₁₁ is not well defined.

What if an increase of g₁₁ and g₂₂ has an interaction? The components T₁₁ and T₂₂ do not tell the whole story then.

We have to ban metric variations which introduce a negative Ricci scalar R into empty space

If we do not ban a negative Ricci scalar R, then nothing prevents us from decreasing the Einstein-Hilbert action value without a limit. Let us start from empty space. We simply add negative curvature, as if we would have a negative mass -M sitting in space (though we do not add the mass -M itself to the system – only the metric).

This restricts allowed metric variations δg dramatically.

Let us then consider the following example. We have a uniform density spherically symmetric mass M in otherwise empty space. The metric consists of the Schwarzschild internal and external solutions.

The contribution to the Einstein-Hilbert action consists mostly of the mass density ρ inside, and the corresponding Ricci scalar R = ρ.

The action integral is somewhat reduced because the spacetime volume element sqrt(-det(g)) is somewhat shrunk inside M because of the slowing of time. The mass M in LM contributes less in the integral because it is at a lower potential.

Let us add into the metric g a metric perturbation which would happen if we would move some mass from the outer layer of M to the central part of M. We do not move any mass, we just change the metric.

How does the Einstein-Hilbert action react to this? Does its value grow?

Building a complex dynamic system: does there exist a lowest energy state?

Let us build a complex mechanical machine with springs and rods. The machine is built to form a loop where an "input", say, an impulse hit with a hammer, at a certain location X goes around the loop, and at a later time affects what happens at X.

                          X    -->

           |\_=>/\|||\_=>/\||

           |\\                   \/\

           |\\                   \/\

           |\_<=/\|||\_<=/\||

                        <--  mechanical effects

                               circle to this direction

           complex machine with

           springs and rods

How easy is it to find a "steady state" for this machine?

Probably extremely hard, and a steady state might not exist at all. The system may have an infinite space of states and never returns to the same state again.

What about a static configuration where the machine has the lowest possible energy?

Finding the state may be very hard. Is it guaranteed that such a state exists at all?

If we allow an infinite number of parts in the machine, it might be that it does not possess a lowest energy state. Let the machine consist of all natural numbers and a finite set N which contains n natural numbers. If the energy of the machine is

       1 / n,

then there does not exist the lowest energy state.

In the case of the cylinder with shear, we are looking for the lowest value of the Einstein-Hilbert action, i.e., the lowest "potential energy". How do we know that such a state exists?

Conclusions

Let us close this very long blog post. We were not able to arrive at the shear problem yet. We have to study the variational calculus in general relativity first.

What kind of metric variations δg are allowed?