Magnetic field of a rotating sphere calculated by rotating the frame

In August 2023 we tried to calculate the gravity field of a spinning uniform ball, and got varying results. Maybe there were calculation errors? One of the problems was that general relativity, as defined by the Einstein field equations, seems to unable to handle accelerating masses.

On November 6, 2023 we tentatively proved that the Einstein field equations in most cases do not have a solution at all if there are accelerating masses. It might be that general relativity does not determine the field of a rotating body at all.

In literature, it is claimed that the Kerr metric can be derived by studying the Schwarzschild metric in a rotating frame. Let us check what we get if we try to determine the magnetic field of a ball of electric charge by switching to a rotating frame. This is a simple calculation, and we can test experimentally what the magnetic field is like.

(Picture Wikipedia)

The correct magnetic moment of a spinning uniform ball of charge

https://www.toppr.com/ask/question/consider-a-solid-sphere-of-radius-r-and-mass-m-which-has-a-charge-q/

The magnetic dipole moment of a solid, uniformly charged sphere of uniform density is

       m  =  Q L / (2 m)

             =  Q / m * 1/2 I ω

where Q is the charge, L is the angular momentum, and m is the mass of the sphere. The moment of inertia of the sphere is I and ω is its angular velocity. If we define

       Lₑ  =  Σ q v r

as the "charge angular momentum", then

       m  =  1/2 Lₑ

             =  1/2 Q I / m  *  ω

             =  1/2 Q * 2/5 R² ω

             =  1/5 Q R² ω,

where R is the radius of the ball.

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

In the equatorial plane, the magnetic field is

       B  =  μ₀ / (4 π)  *  m / r³

            =  μ₀ / (4 π)  *  1/5 Q R² ω /  r³,

where r is the distance from the center. Very close to the surface of the ball the field is

       B  =  μ₀ / (4 π)  *  1/5 Q ω / R

            =  1 / (4 π ε₀)  *  1 / c²  *  1/5 Q ω / R.

A very naive – and wrong – calculation

Let us compare this to a very naive calculation where we assume that the electric field of the ball "moves" at a velocity

       v  =  ω R

at the surface of the ball. At the surface,

       E  =  1 / (4 π ε₀)  *  Q / R².

If v is slow, the magnetic field is

       B  =  1 / c²  *  v × E

            =  1 / (4 π ε₀)  *  1 / c²  *  Q ω / R.

The very naive method gives a magnetic field which is 5X the correct value.

Another naive calculation

            <-- ω

           ___

          /       \   1/4 Q

          \____/

              •  observer

We may approximate the rotating ball with a current loop. Let us assume that 1/4 of Q flows in a circular loop whose radius is R.

Our observer is at

       r  =  R + r'

from the center. We ignore the far side of the loop and assume that the near side is straight. The magnetic field of a straight wire is

       B  =  μ₀ / (4 π)  *  2 I / r',

where I is the electric current and r' the distance from the wire. The current in our case is

       I  =  1/4 Q / (2 π R)  *  R ω.

We obtain

       B  =  1 / (4 π ε₀)  *  1 / c²

                *  Q / (4 π)  *  ω  /  r'.

A reasonable value for r' might be R / 4. We get

       B  =  1 / (4 π ε₀) * 1 / c²  * 1 / π  *  ω / R.

The estimate is 1.6X the correct value.

Yet another naive calculation

The magnetic field primarily rises from the charge moving very close to the observer. Let us guess that such a charge is Q / 10, and it moves at a velocity

       v  =  ω R

at a distance R / 2 from the observer. The magnetic field is then

       B  =  1 / (4 π ε₀)  *  Q / 10  *  4 / R²

                *  1 / c²  *  ω R

            =  1 / (4 π ε₀) * 1 / c²  *  2/5  *  ω / R.

The value is 2X the correct value.

Estimating the gravitomagnetic moment of a rotating sphere: a very naive calculation

Let us first use a very naive method. We assume that we can switch to the rotating frame of the sphere, and in that frame the metric is Schwarzschild.

                  ___

                /       \   M

                \____/   R = radius

                 ^  V

                   \  

                    •  m

The test mass m approaches M, but goes a little bit sideways because m is not spinning along with M. Let us assume that m gathers some extra inertia as it descends, and that inertia is moving along M.

We assume that m is very close to the surface of M, the radius of M is R, and it is rotating at an angular velocity of ω.

If the gravity of M does the work W on m, then m acquires

       W / c²

of extra inertia. That inertia is moving sideways at a velocity

       v  =  ω R.

Let m descend down for a time t. Then

       W =  t V m G M / R²,

and m gets a sideways velocity

       v  =  t V G / c²  * M / R²  *  ω R,

which corresponds to a sideways acceleration

       a  =  V G / c²  *  5/2  * 1 / R³  *  2/5 M R² ω

           =  V * 5/2 G / c² * 1 / R³ * L,

where L is the angular momentum of the sphere. According to our August 10, 2023 definition, the gravitomagnetic moment is then

       mg = 5/2 L.

The moment is 5X the analogous magnetic moment, just like we obtained with a similar very naive method in electromagnetism.

Gravitomgnetic moment: yet another naive calculation

Let us assume that M / 10 is moving at a distance R / 2 from the observer, at a velocity ω R. We can calculate like in the previous section:

       W =  t V m G M / 10  *  4 / R².

We obtain

       mg  =  L,

that is, 2X the analogous electromagnetic moment.

Conclusions

The very naive method, where the entire electric field of a spherical charge is assumed to rotate "fixed" to the rotating sphere, yields a magnetic moment which is 5X too large.

Similarly, using the very naive method to the Schwarzschild metric gives a gravitomagnetic moment which is 5X the analogous electromagnetic moment.

In this blog we have been claiming that the Kerr metric overestimates the gravitomagnetic moment 4X or more. Maybe people doing the Kerr calculations believe that the gravity field rotates "fixed" to the rotating mass?

Our complicated calculations in August and September 2023 brought varying results. A key question is can we add the effects of rotating mass elements linearly?

Our October 18, 2023 blog post suggested that close to the mass M, one cannot obtain the correct Schwarzschild spatial metric by summing the Schwarzschild metric perturbations for each mass element dm. One has to consider the field of M as one "whole".

But that does not mean that if M rotates, then the "whole" field should rotate along with it. The gravitomagnetic effect calculated above depends only on the metric of time, because we calculate the work W done by gravity, and that work comes exclusively from the metric of time. One does obtain the metric of time by linearly adding the perturbations by each mass element.

However, we believe that the 2X extra inertia of a test mass in a radial motion relative to a mass element dm does affect the gravitomagnetic effect, and we did not include that in the calculations above.

On September 4, 2023 we sketched a "unified field theory", where electromagnetism is equivalent to gravity if we ignore the gravity of kinetic energy, of pressure, changes in the spatial metric, and so on. If that hypothesis is correct, then the gravitomagnetic field of a slowly rotating sphere should be exactly analogous to the corresponding electromagnetic field. But we have to check if the correct way to calculate the magnetic field of a rotating electric charge really is the classical one. Our October 9, 2023 blog post suggests that the classical calculation ignores some effects of inertia inside an electric field.

Does nonlinearity of gravity make sense?

On September 16, 2023 we wrote that nonlinearity of gravity leads to a strange result. Let us analyze this more.

When does the nonlinearity occur? If we have a very large static mass M, then general relativity claims that its gravity is surpringly strong close to M. Let us simulate M with two particles flying to opposite directions at almost the speed of light. The rest mass of the particles is very small, but they have lots of kinetic energy.

                            M

                             

                    v <-- ● 1 

                          2 ● --> v

                             

                              •  m test mass

Let us switch to a comoving frame of the rightmost particle 2. In that frame, the particle 2 is very light and, intuitively, very insignificant. The particle 1 and the test mass m move at almost the speed of light to the left and 1 possesses a lot of energy. We can imagine that m has an infinitesimal rest mass which is infinitesimal even when m moves almost at the speed of light.

But if we would remove the very light particle 2, the acceleration of m would drop to a half. That sounds strange, but is a result of special relativity.

                             ● 2

           v ≅ c    <-- • m

Let us then remove the particle 1. We double the rest mass of 2 and look at the configuration in a comoving frame of 2. In the comoving frame of 2, the particle 2 is very light. Doubling its rest mass certainly cannot have any nonlinear effects.

But in the comoving frame of m, the particle 2 does carry a lot of mass-energy, and there might be nonlinear effects. Do we have a contradiction here?

Conservation of momentum and nonlinearity

How momentum is conserved when fields are retarded? That is an open problem in field theories. Maybe nature performs "transactions" which ensure momentum conservation? Let us check how general relativity is supposed to handle momentum conservation.

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

The ADM formalism implies conservation of momentum. But we in this blog have tentatively shown that the Einstein field equations do not have any solutions for a typical dynamic system. The ADM formalism assumes that a solution exists.

                    F'                     F

                 ● -->                <----- •

                M                              m

Suppose that we have a very large mass M, such that its gravity is significantly larger than the linear (newtonian) gravity at a test mass m. Then the field of the mass M seems to pull m with a force

       F  >  F',

which is larger than the force that the field of m exerts on M!

This is suspicious. Newton's law of action and reaction is broken? The Einstein-Hilbert action is translation independent. By Noether's theorem, we expect it to conserve momentum. Maybe the existence of the strong field of M makes the field of m non-linear?

Let us use canonical Minkowski space coordinates. General relativity recognizes "proper" momentum, which is measured by a local observer. That differs from coordinate momentum.

Gravitational waves will take away some momentum. The Ricci tensor is zero in them. The Einstein-Hilbert action is not aware of their existence. This implies that momentum is not conserved, unless we define the momentum through some pseudotensor.

Relativity of simultaneousness

                   M

           v <-- ●

                    ● --> v

                   M

                    • m

Let v be very close to c. In the frame of m, the two masses M overlap. But in the comoving frame of the right-moving M, there is very little overlap. Is this compatible with nonlinear gravity?

Nonlinearity from polarization

Imagine that a very strong gravity field pulls positive mass-energy from empty space toward itself, and repels negative mass-energy. That would make gravity steeper than the newtonian 1 / r² gravity.

However, the gravity field close to a huge black hole is not very strong. The polarization hypothesis requires that it is a low potential which produces polarization and not the field strength.

In quantum electrodynamics, vacuum polarization makes the interaction stronger at very short distances, or at very high energies. But that requires energies which are much larger than the mass of the electron, 511 keV. In our blog we hold the view that the vacuum itself is not polarized, but a pair which is born from the energy of a particle collision, simulates vacuum polarization in a scattering experiment.

Can we somehow calculate how much positive/negative mass-energy might pop up if we have a very low gravity potential? That assumes that the vacuum itself can become polarized.

Nonlinearity is needed to satisfy an equivalence principle?

On August 23, 2023 we showed that general relativity breaks the weak equivalence principle. Why should we use nonlinearity to satisfy a certain equivalence principle, when general relativity breaks the most fundamental equivalence principle?

Nonlinearity makes everything complicated

        _____        _____    rubber sheet plane

                  \ ● /

                    M

In a rubber sheet model of gravity, nonlinearity means that the rubber becomes weaker against stretching if it is depressed below the sheet plane more than a certain distance. The weakening is not from the stress on the rubber but from the depth of the depression  –  that is, from a low potential.

Why should it become weaker? Is there some kind of a heater which heats the rubber when it is pressed down a lot?

It is obvious that having a rubber sheet whose strength depends in such a way on the depth of the depression makes calculations complicated if the system is dynamic.

Empirical data

We have to check what is known about accretion disks. Is there any empirical proof that the gravity potential must be steeper than the newtonian one?

In September and October 2023 we showed that the Kerr metric probably is wrong for a rotating black hole. We have to calculate a new solution for a rotating mass and compare empirical data against it.

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3703352/

The paper by Cosimo Bambi (2013) leaves an impression that empirical data does not tell much about the structure of black holes, besides the fact that matter does not hit a solid surface as it is devoured by a black hole.

If gravity is linear, then the Schwarzschild radius is replaced by a Newton radius, which is 1/2 of the Schwarzschild radius. Once we get more accurate measurements of gravitational waves, we may be able to determine what the correct radius is.

The M87 central black hole (photo Wikipedia)

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

The Event Horizon Telescope measured the radius of the "photon ring" around the M87 black hole. The margin of uncertainty is given as approximately 10%.

https://physics.stackexchange.com/questions/472005/how-do-you-calculate-the-black-hole-diameter-of-m87

In a Schwarzschild black hole, the lowest circular orbit for a photon is at

       3 G M / c².

In a newtonian black hole, the lowest photon orbit is at

       (1 + sqrt(5))² / 4  * G M / c²

       = 2.62 G M / c².

The data from the Event Horizon Telescope cannot decide which is correct.

Conclusions

There probably is nothing which prevents an interaction from being nonlinear in the sense that a large aggregate charge interacts stronger than the sum of the interactions of its components. We can imagine that the components "help" each other to gain more strength. This certainly is possible in newtonian physics, and we did not find any reason why special relativity would prohibit it.

Nonlinearity creates more "effective charge" for a large mass, when the charge is measured from a short distance. This creation of more charge probably breaks the Einstein equations.

Nonlinearity makes everything complicated and there is no obvious need for nonlinearity. Our own Minkowski-newtonian gravity model works without nonlinearity, though one can add there nonlinearity.

Calculus of variations and a carpet

(Photo Wikipedia)

Let us present a simple example where charge conservation keeps the action such that it can be optimized strictly locally, but a change in the charge makes it impossible to optimize the action locally.

                   

                         ___

            ______/ ■■ \_____________    carpet

                      board  -->

We have a long corridor and a carpet which is almost as long as the corridor. Someone forgot a board under the carpet. The board is normal to the corridor and extends from the wall to wall. The volume of the board is a "charge".

The action is the volume V between the floor and the carpet minus the volume of the board under the carpet. Initially the volume V is 0.

If we slide the board along the corridor (i.e., the charge is conserved), we can find the minimum of the action V strictly locally. Just press the carpet tightly to the floor or to the board. This configuration corresponds a typical field theory with charges, like electromagnetism. We can keep the far parts of the carpet as is: the carpet only needs to be adjusted locally.

But if we magically remove the board, the carpet becomes loose. There is a wrinkle, and the action suddenly is V > 0. The charge was not conserved.

We assume that one cannot move the carpet infinitely fast. The action V must stay > 0 for some time.

Now it is obvious that the carpet cannot be optimized strictly locally. We can make V = 0 again by moving the wrinkle to the end of the carpet, but that takes time.

Note that we could have a spare board B which we insert inside the wrinkle. Then we could again optimize the action strictly locally and slide the spare board B out at the end of the carpet. This shows that we can find a solution to the global optimization problem by restoring charge conservation and sliding the spare charge "far away".

                        _

              ____/   \____    carpet

                 wrinkle

Let us study how we can vary the form of the wrinkle in the carpet if there is no board under the carpet. We might have a very large potential which prevents the carpet from curving into a circle of less than, say, 1 cm of radius. We can minimize the volume V under the wrinkle. The optimum is not unique. The wrinkle can be put at many locations. We have a global optimization problem which has many solutions. The solutions do not determine the elevation of the carpet at a specific location.

General relativity

In general relativity, the analogue of the board is a mass M, and the analogue of the carpet is the metric. Suppose that M is a spherically symmetric. Then the field outside M is the Schwarzschild metric.

Suppose that M is suddenly reduced by magic to a mass M' < M. The optimization of the Einstein-Hilbert action suddenly becomes a global problem.

A new optimum, which would satisfy the Einstein field equations, would be the Schwarzschild metric for M'. But we can never get to that metric since the speed of light is finite.

We could use the spare board trick above to switch to the metric of M'. Make a shell of matter whose mass is M - M'. Let that shell recede from M' at the speed of light. However, we can never complete this operation since Minkowski space is infinite.

Conjecture. It is impossible to satisfy the Einstein equations and switch (locally) from the metric of M to the metric of M', if we require the Ricci tensor to be zero in the vacuum around M'. We have to eject some kind of matter to transition to the metric of M'.

In our previous blog post we sketched a proof for this conjecture. The proof uses the "focusing" of a cube of test masses.

Conclusions

The carpet is a "field" and the board is a "charge". We illustrated what happens if we magically reduce the charge. Nice, local optimization conditions are replaced with a much harder global optimization problem.

A single accelerating particle breaks the Einstein field equations?

UPDATE November 30, 2023: In the analysis of the Oppenheimer-Snyder collapse we assumed that Gauss's law for the gravity of a single particle also holds for many particles. This might be false. The inertia of a test mass m inside the common field of all particles in a spherical shell most probably is not the sum of inertias for each individual particle in the shell. We have to check if this destroys Gauss's law for gravity. We wrote about this on October 18, 2023.

----

UPDATE November 30, 2023: For the argument below to work, we have to prove that the "magnetic gravity induction" does not keep the volume of the cube of the test masses constant, after all. Then there would be no "(de)focusing".

Compare to electromagnetism: we can accelerate a charge, and the lines of force of the electric field never break because of the induced magnetic field: there is no "(de)focusing" caused by the electric field. The field equations are not broken in electromagnetism.

The big difference in gravity is that the pressure term ~ v² creates new gravity. We will investigate this in a new blog post.

It might be that Gauss's law for gravity makes the Einstein equations solvable if the pressure term is strictly dependent on the velocity v of the particle. On the other hand, if the pressure changes for some other reason, then Gauss's law is broken, and the Einstein equations have no solution.

Note that acceleration of a particle typically involves some kind of "pressure", and the velocity of the particle does not respond immediately to that pressure. This suggests that Gauss's law is broken in almost all acceleration mechanisms.

----

UPDATE November 10, 2023: The steepening of gravity (nonlinearity) close to a large mass M seems to break the Einstein field equations, too. If we have two masses 1/2 M at some distance from each other, we can increase their combined "focusing power" by moving them very close to each other. That is, the mass-energy charge of the system seems to increase.

A charge is something which occurs in a linear field theory. Any nonlinear effect in gravity may cause focusing or defocusing in "empty" space, and thus make the Ricci tensor non-zero there.

In our own Minkowski-newtonian gravity model there is no obvious reason why gravity should be nonlinear. Maybe gravity, indeed, is a linear phenomenon?

----

In the previous blog post we argued that the Einstein field equations do not have a solution at all if the pressure of a spherically symmetric mass M changes in a way such that the change in the positive pressure is not accompanied by a compensating change of negative pressure.

Ehlers et al. (2005) showed that one can increase the internal pressure of a spherical vessel M if the wall of the vessel obtains a compensating negative pressure. Then the metric outside M does not change, and Birkhoff's theorem is not broken.

We did not yet analyze the non-spherically symmetric case. We believe that the Einstein equations do not have any solution, if the pressure changes and there is no compensating change of negative pressure. Pressure acts as a "charge" which creates gravity. A change in a charge is not tolerated by typical field equations.

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

The stress-energy tensor T of a moving particle m has a pressure term

       m v²  /  sqrt(1  -  v² / c²)

       * Dirac delta function δ,

where m is the mass of the particle, v is the velocity, and c is the speed of light.

If we speed up the particle, the positive pressure term increases, and there is no compensating change in negative pressure? Is it so that a single accelerating particle breaks the Einstein field equations?

Let us analyze.

Emmy Noether (1882 - 1935)  (photo Wikipedia)

Conservation laws for a lagrangian

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

In electromagnetism, conservation of charge is derived from the gauge symmetry:

where X is an arbitrary differentiable real-valued function on the time and the position.

https://en.wikipedia.org/wiki/Noether%27s_theorem

Can we figure out an infinitesimal variation of the Einstein-Hilbert action which would expose the total sum of pressures in a system?

***  

The diagram is from Wikipedia. A spatial translation at t₀ and t₁ exposes the velocity v (= q-dot) of the particle. The diagram is used to prove conservation of momentum.

A temporary stretching of a gravitating system M: can we prove conservation of the "pressure charge"?

Let us stretch a gravitating system M an infinitesimal amount along the x axis, for a short time, and then return it back to the original dimensions.

Let us list changes in the action. The potential energy associated with a positive pressure is reduced for a short time. A negative pressure contributes more potential energy. Parts of the system move slightly during the stretching and unstretching. They contribute changes in the kinetic energy.

We have a problem: the temporary stretching may change many other things in the behavior of M. How can we know that their contribution to the action is neglible?

The above contributions have nothing to do with gravity. They occur in newtonian mechanics, too. They cannot be used to prove conservation of the "pressure charge".

When the pressures change, it may change the gravity field of M. Can we somehow distill the total pressure charge contained inside M?

The Einstein equations probably imply conservation of pressure, while the action does not

It turns out that we are on a wrong track. We suspect that the Einstein field equations imply "conservation of pressure", because they are derived using erroneous variational calculus. On the other hand, the Einstein-Hilbert action probably does not imply conservation of pressure.

In our own Minkowski-newtonian gravity model there is no conservation of pressure. We do not believe that there is conservation of pressure in nature, either.

A simple system where the "pressure" changes

Let us have two equal masses M which are initially static and close to each other. The initial stress-energy tensor is T.

We use some of the mass-energy of both to accelerate them to a fast motion along the x axis, to opposite directions. The masses decline to M' = M / sqrt(1  -  v² / c²).

        •                                     v <--- ●   ● ---> v

       m                                             M' M'

   ----> x

The mass-energy of the system stays as 2 M. After the acceleration, there is a pressure term in the new stress-energy tensor T':

       M' v²  /  sqrt(1  -  v² / c²)

   

       * sum of 2 Dirac delta functions δ.

The sum of momenta is zero.

The metric close to the system must reflect the mass-energy and pressure terms of the new stress-energy tensor T'. But the metric far away still corresponds to T. Can we show that this configuration is impossible?

Let us have a test mass m far away from the M' system, to the negative x direction from the system..

Let us assume that the masses M' are not large.

Hypothesis 1. We can obtain the acceleration of the test mass m by summing the individual gravity fields of each M'.

Hypothesis 2. The changed gravity field of the system spreads at the speed of light. The field corresponding to T is rapidly replaced by the field corresponding to T'.

Hypothesis 3. A "magnetic gravity induction" is not able to connect the lines of force of the gravity field. This hypothesis may be wrong. We will investigate this in a future blog post.

The corrected calculation of our September 9, 2023 blog post shows that the gravity coordinate acceleration due to each M' is as if the mass would be

      M'  /  γ⁵,

where

       γ  =  1 / sqrt(1  -  v² / c²).

When the two masses M were sitting still, the gravity coordinate acceleration of m was as if a mass

       2 M

would be pulling it. After the masses were launched at a velocity v, the "coordinate" attraction is as if a mass

       2 M / γ⁶

would be pulling m. The attractive gravity force declined.

 pull of gravity drops at this

          point, at time t

              <--- c

                     |

                     v

            •   •   • •

            •   •   • •                      v <--- ●  ● ---> v

            •   •   • •                              M'  M'

        test masses

Let us have an initially static cube of test masses floating freely in space. The pull of gravity on the test masses suddenly declines as the field corresponding to T is replaced by the field of T' after the launch. The border between the T metric and T' metric recedes from the M' at the speed of light c.

Does that cause the volume of the cube to decline? If yes, then there is a "focusing" effect, and the Ricci tensor cannot be zero at the cube.

We must also consider the effect of the launch operation itself. The operation involves a pressure between the two M. What kind of a gravity field does that generate? Maybe we do not need to know?

We have to find out the spatial metric at the cube. Maybe the spatial metric expands and compensates the shrinking of the cube? Since the attractive gravity weakens with the introduction of T', most probably the spatial metric shrinks in the direction of the x axis. This makes the cube to shrink even more.

Assumption. The spatial metric change from T to T' shrinks the cube even more.

             •       •       •       •

             

             •       •       •       •                    v <-- ● ● --> v

                                                                   M' M'

             •       •       •       •

                            

      T metric   |    T' metric

               <----- c

    ---> x

Let us create the cube above at a moment when the border between the T and T' metric is in the middle of the cube. That is, we order initially static test masses m in a configuration where the proper distances between the test masses are approximately some fixed s. It is like a cubic crystal system in crystallography. Since the metric is not flat, the crystal cannot be perfect, though.

The border between T and T' moves at the speed of light farther away from the M' system.

Let us wait for a very short time Δt. The left surface of the cube may accelerate much faster in the gravity than the right surface, if the velocity v of the masses M' is large. Does that guarantee that the proper volume of the cube shrinks?

No. If the spatial metric along the x axis would stretch ever more in the transition area between T and T', then the extra stretching could compensate the shrinking of the cube. However, it would be very strange if the spatial metric perturbation would grow as the transition area moves farther from the M' mass system.

Conjecture. The spatial metric perturbation does not grow when the transition moves farther. Rather, the perturbation decreases.

As the cube falls in the gravity, its proper y and z dimensions shrink.

Since the proper volume of the cube shrinks, it must contain very small cubes whose volume shrinks. The metric "focuses" those small cubes. The Ricci tensor cannot be zero there. But the space is empty there and the local stress-energy tensor zero. The Einstein equations then claim that the Ricci tensor is zero. This is a contradiction.

We have a heuristic proof for:

Conjecture 1. The Einstein equations do not have any solution for a simple system which consists of two accelerated masses.

A further conjecture:

Conjecture 2. The Einstein equations do not have a solution for any real-world dynamic physical system. They only have solutions for static systems where pressures do not change, and for some (unrealistic) symmetric dynamic systems.

Yet another:

Conjecture 3. The Einstein equations do not have a solution for any system where two masses orbit each other.

The Oppenheimer-Snyder collapse (1939) does have a solution because it is symmetric

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

The Oppenheimer-Snyder dust collapse is essentially the only known dynamic solution of the Einstein equations. The solution is spherically symmetric. Oppenheimer and Snyder did not make a calculation error. Let us show that the pressure change (= acceleration of the dust) does not change the Schwarzschild metric outside the collapsing dust ball.

Our updated post on October 11, 2023 contains the conjecture that Gauss's law holds for a moving mass.

Gauss's law conjecture. The average coordinate acceleration of a test mass m at a coordinate distance of r from a moving mass dM is

       G γ dM / r²,

where γ = 1 / sqrt(1  -  v² / c²) and v is the coordinate velocity of dM.

The conjecture says that Gauss's law holds for a moving mass. Its gravitating mass is the mass-energy γ M, as we would expect. The gravity field is not spherically symmetric. The field is weaker in the direction of the movement v and stronger normal to the movement. This is analogous to the field of a moving electric charge.

Let the dust ball start collapsing. At a time t, various dust particles are moving at various velocities v. We can use Gauss's law. The flux of gravity field lines of force through a spherical shell enclosing the collapsing dust cloud is at all times the enclosed mass-energy. The gravity field of the system outside the ball remains the same at all times: we do not encounter the problem of the strength of the field changing.

The field of each dust particle is squeezed in the direction of v, but the configuration is spherically symmetric, and the deformation is not reflected outside the dust ball.

The "simple system" in the previous section is not spherically symmetric. That is why it may break the Einstein equations.

Oppenheimer and Snyder write that they were not able to "integrate" the equations if they add some pressure to the collapsing dust ball. Our conjectures say that the Einstein equations do not have a solution at all in such a case, since they cannot tolerate a change in an "ordinary" pressure where the pressure term does not come from the velocity v of the dust particles. Such pressure would alter gravity outside the dust ball, and break Birkhoff's theorem.

Conclusions

We presented several hypotheses and conjectures about the gravity of various systems. Since the Einstein field equations are nonlinear, it is, in principle, possible that they could work miracles: the nonlinearity could magically restore the integrity, and the Einstein equations would have solutions in many cases. We do not think that they are capable of such magic.

We should present at least heuristic proofs for the conjectures. Since the equations are nonlinear, exact proofs may be hard to construct.

In this blog we have for several years suspected:

1.   general relativity has problems handling changes in pressure;

2.   the Einstein field equations are too strict, and do not have a solution for any realistic physical system.

If our conjectures and arguments are correct, we have shown that we guessed right.

A globally stationary point which is not locally stationary

Suppose that we have a wall-to-wall carpet in a room, but the carpet is slightly too large and there is an annoying wrinkle in it.

         |__________/\___________|

      wall         wrinkle            wall

Let

       S  =  the volume between the carpet

               and the floor.

We can determine the "action" S by integrating the difference of the elevation of the carpet versus the floor over the entire room.

We want to find a minimum of the action to determine how the carpet should settle itself. But the minimum is not unique. We can put the wrinkle at any location of in the room, and S has the same value.

A "locally" stationary point for the carpet is one where the carpet touches the floor. A globally stationary point is such that S is at the minimum. The carpet has no globally stationary point which would be locally stationary everywhere.

Magically adding mass to the Einstein-Hilbert action

Suppose that we have a spherically symmetric mass M and the metric around it is Schwarzschild. We use a magic trick to increase the mass M to M'. What does the Einstein-Hilbert action say about the time development of the system after that?

 

                         

               -------___●___--------    metric

                            M

               ------_             _------     metric

                        --- ● ---

                            M'

After adding the extra mass, the action is not in a locally stationary point. There is a wrinkle in the metric around M'. The wrinkle corresponds to a negative mass M - M'. We cannot get rid of the wrinkle instantaneously because the speed of the light is the limit. Actually, we can never get rid of the wrinkle because we cannot expel it past the infinity.

The system is not locally stationary because the required negative mass does not exist at the wrinkle. It could be globally stationary, though. A globally stationary point is not necessarily locally stationary everywhere.

We do not know if a supposed globally stationary point determines the time development of the system in the Einstein-Hilbert action.

Changing pressure inside M

Our example may describe what happens in the Einstein-Hilbert action if we change the pressure inside M. After that, the system may still be in a globally stationary point, but it is not locally stationary any more. It is like adding a wrinkle to the wall-to-wall carpet by sewing a new patch into the carpet.

This shows that the Einstein-Hilbert action might be able to handle changes in pressure, but then the Einstein field equations would not hold in the entire space.

"Negative mass" propagating from M when the internal pressure of M is increased

              •    •    •

              •    •    •           test masses m

                •  •  •    |

                            v   acceleration

                  

                   ● M

Suppose that we are able to increase the attraction of M from zero to a non-zero value suddenly. Let us have a cube of test masses floating in space. Does the volume of the cube increase?

The lowest surface of the cube starts accelerating toward M. The volume of the cube increases. This shows that a sudden increase of the attraction causes a "defocusing" effect. It is like negative mass-energy would be propagating from M.

The variation which is used to derive the Einstein field equations

David Hilbert (1862 - 1943) (photo Wikipedia)

The Einstein field equations assume that the system is everywhere locally stationary. That is, the local metric matches the local stress-energy tensor. Is this assumption incorrect for dynamically changing systems?

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

Let us check the derivation of the Einstein field equations in Wikipedia.

The derivation does not specify what variations of the metric,

       δg^μν

are allowed. We remarked in an earlier blog post that if we add the Schwarzschild metric of a small mass dm, the volume integral of the change in the metric over the entire 3D space is infinite.

Also, the time development of the system must be such that faster-than-light signals are not allowed. Even if a certain development would be a stationary point of the action, it is not allowed if it involves superluminal signals.

"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 δg^μν, but also on its partial derivatives

...

However when the variation of the metric δg^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"

Wikipedia claims that we can forget about the behavior of the variation δg^μν far away: it does not affect the action integral.

Suppose that we magically add some more mass to a spherically symmetric M. Does the metric afterwards obey the Einstein field equations at every event?

Suppose that we are able to find a stationary point of the action, such that it extends the history after the magic trick.

If there would exist a variation which only adjusts the value of the Ricci tensor R at the event we are looking at, then we could argue that the Ricci tensor R must match the stress-energy tensor T at that event. But there probably exists no such variation. If the variation does not extend over the entire 3D space at the time of the event, then the variation presumably modifies R at the event, as well as in many other locations. There is no guarantee that we can make R to match T at all events. It is like the carpet which we can make to touch the floor at any one location, but cannot make it to touch the floor everywhere.

Another way to explain the problem in the Einstein field equations: optimizing the action requires global information of the system because the metric and its curvature have global dependencies. If we increase the curvature in a certain volume of spacetime, we have to reduce it elsewhere. This because we cannot change the metric simultaneously everywhere in the 3D space. The Einstein field equations are strictly local. They cannot understand the global optimization problem.

Pressure and dynamic systems

Adding pressure obviously requires a change in the metric inside M. There is no guarantee that we can make R and T match everywhere.

The same may be true for almost every dynamic system where there are accelerating masses. There is no proof that a stationary point of the Einstein-Hilbert action is locally stationary, that is, that R and T match at every event.

We know a few solutions where we are able to make R and T to match everywhere. The Schwarzschild exterior and interior solutions (1916) are the best known ones. The Oppenheimer-Snyder collapse (1939) is a known dynamic solution.

Conclusions

The derivation of the Einstein field equations from the Einstein-Hilbert action is incomplete, and probably erroneous. This explains why Birkhoff's theorem seems to clash with Tolman's paradox.

It looks like that most changes in the pressure of a system cause the system to enter a state where a stationary point of the action is not locally stationary everywhere. That is, the Einstein fields equations are not satisfied by a stationary point of the Einstein-Hilbert action. In practice, this means the Einstein field equations do not have a solution for any realistic physical system, because such pressure changes will always happen.

Birkhoff's theorem is probably false for gravity, as gravity occurs in nature. The attractive force of a spherically symmetric mass M does change when we manipulate the pressure inside it.

The Einstein-Hilbert action itself is probably incorrect, too. It is based on a very optimistic hypothesis that a "metric" can capture all the complicated phenomena associated with gravity.

Magically adding more charge to the Coulomb field

Let us study the following problem. We have a magic device which can create an electric charge from nothing. We have a spherically symmetric positive charge Q sitting in Minkowski space and we use the magic device to boost it into a charge

       Q'  >  Q.

How would electromagnetism react to such a magic trick?

The hamiltonian of the Coulomb field

A rudimentary "lagrangian" or, rather, a hamiltonian, calculates the energy of the (arbitrary) electric field in the space, where the energy density is

       1/2 ε₀ E²,

and the electric potential energy of the charge Q is

       -Q V,

where V is the electric potential. Incidentally, -Q has to be negative to simulate the field of a positive charge. The electric field allows Q to fall into a lower potential, releasing energy. The price we have to pay is that the creation of the electric field E consumes energy.

A stationary point of the hamiltonian, presumably, is the standard Coulomb electric field of a charge Q.

Magically adding more charge Q'

Let us then use a magic trick to grow the charge Q to a larger charge Q'. The system no longer is in a stationary point of the hamiltonian.

The Coulomb field of Q' would be a stationary point, but it differs considerably from the Coulomb field of Q, and differs from it in the entire Minkowski space.

How does the system develop after this? Without the magic trick, a dynamic system would pass between "adjacent" stationary points which differ from each other very little. But there is no rule about how the system should find a new stationary point after it was put very far from a stationary point by the magic trick.

We could let the field of Q' expand radially from Q' to all directions. Suppose that it now extends to a radius R. Since fewer field lines exist at r > R than at r < R, the system is as if we would have a shell of negative charge

       Q  -  Q'

expanding.

But there is no such negative charge. The system is locally not at a stationary point.

If we try to put the system locally into a stationary point for all r < R and r > R, then we end up having a discontinuity of the electric field E at R. It is like a singularity, and is not allowed.

Another option is to switch between the Coulomb fields of Q and Q' either instantaneously, or slowly but synchronously, in the entire Minkowski space. That would allow superluminal communication and is forbidden by special relativity.

What is the value of the hamiltonian at different stages? If we simply add the extra charge to Q without changing the electric field E, then the value of the hamiltonian jumps suddenly. If we somehow are able to get Q' to descend down to a low potential, then the value of the hamiltonian decreases. However, this does not make sense in hamiltonian mechanics: the value should stay constant.

Birkhoff's theorem for Coulomb fields

If one assumes that the system only passes between adjacent stationary points, then the field outside Q cannot change. This is similar to Birkhoff's theorem.

If we then use a magic trick to add more charge into Q, there simply is no well defined path that the system should take. We could claim that the system ends up in a singularity, but that is just one option.

The analogy in pressure in gravity

What happens if we have a spherical mass M and we add some pressure into it?

The pressurizing operation, apparently, causes the system to move into a state which is very far from a stationary point.

The "Einstein-Hilbert action mechanics" may break down after that. There is no reasonable way to define how the system develops in time after the pressure changed.

One could try to solve the paradox by claiming that the pressurizing operation somehow requires a lot of energy since it takes the system far away from a stationary point. If the required energy is so large that one cannot change the pressure at all, that breaks newtonian mechanics.

The Ehlers et al. (2005) result

https://arxiv.org/abs/gr-qc/0510041

Ehlers et al. compensate an extra positive pressure inside a spherical vessel M with a negative pressure in the surface of the vessel. In electromagnetism, this corresponds to increasing Q to Q', and compensating the change in total charge by adding a shell of negative charge around Q'.

The change in the metric is isolated into M, and the problem which we described above does not occur. There is no need to update the metric in the entire Minkowski space.

https://archive.org/stream/in.ernet.dli.2015.177229/2015.177229.Relativity-Thermodynamics-And-Cosmology_djvu.txt

The problem with pressure and gravity is known as Tolman's paradox. Ehlers et al. refer to R. C. Tolman's book (1934). In the book, in section 109 on Disordered radiation, Tolman writes:

"we are led to the interesting conclusion that disordered radiation in the interior of a fluid sphere contributes roughly speaking twice as much to the gravitational field of the sphere as the same amount of energy in the form of matter."

Conclusions

A problem in general relativity seems to be that it allows new "charge" to be created: pressure acts as a charge.

It is like the magic trick of creating new electric charge in electromagnetism. After the charge has been created, the hamiltonian or lagrangian does not tell us how the system should develop forward in time.

The conflict between Birkhoff's theorem and a pressure change would be a fundamental one: general relativity cannot describe the behavior of a system where pressure changes. Since there are pressure changes in all realistic physical systems, general relativity fails in every realistic case.

Our own Minkowski-newtonian gravity model does not treat the gravity caused by a pressure as a "field" or a "metric". But we have to check if the problem of a changing force field somehow causes other difficulties. For example, a perpetuum mobile might become possible if we can change the attraction of M. A rubber sheet model probably saves us from a perpetuum mobile.

Einstein-Hilbert action: a flat metric is not a stationary point?

UPDATE November 3, 2023: If we add an infinitesimal mass dM into empty Minkowski space, the change of the metric g is infinitesimal at any one place, but the volume integral of the change is infinite over the whole space!

Thus, adding an infinitesimal mass dM is not equivalent to a "small" change of the metric. We are not allowed to use it in the variational calculus.

What if we add dM, but restrict the metric perturbation within some radius R from dM, and keep the flat Minkowski metric unchanged elsewhere? That is equivalent to adding a shell of negative mass -dM to a distance R from dM. The chance in the Ricci scalar over the entire space is probably zero in that case, because dM - dM = 0.

This may explain why dR / dg⁰⁰ = R⁰⁰, etc. That is, a flat space is a stationary point of the Einstein-Hilbert action, after all.

If we already have a mass M in the space, and add dM at the same location as M, then the redshift from M to R allows us to add a negative mass -dM' whose absolute value is less than dM, at a distance R. The total mass then grows by dM - dM', which probably means that R integrated over the whole space grows.

This reminds us of the problem with Birkhoff's theorem.

If we move a mass dM for a short distance, the change in the metric over the whole 3D space volume is how much?

It is very roughly proportional to the integral of 

       1 / r  -  1 / (r + 1)

       ~  1 / r²

integrated over r² dr. The integral is infinite. We may ask what are the infinitesimal variations which we are allowed to use in the variational calculus. Moving a small mass over a short distance corresponds to a "large" variation? Or do we have a reason to label it as a "small" variation?

----

We are still perplexed about the following:

1.   a positive pressure seems to attract a test mass m in general relativity, but

2.   Birkhoff's theorem seems to prohibit any attraction by pressure outside a spherically symmetric mass M.

In electromagnetism, the source of the field, that is, the electric charge Q, is conserved. In general relativity, the mass-energy T⁰⁰, and the mass flow T⁰¹, T⁰², T⁰³ are conserved.

The pressure T¹¹, T²², T³³ is another source of the gravity field. Is pressure conserved? It would be very strange. But if pressure is not conserved, we end up in a conflict with Birkhoff's theorem.

The Einstein-Hilbert action does not imply a zero Ricci scalar for a vacuum area?

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

The stationary action principle says that the variation of the action S is zero for small variations of the metric g.

Let us calculate a variation in two dimensions. If we are allowed to stretch the x and y distances at most by a factor 1.01 within a square whose side is 1, we can make a "bulge" in the plane, such that its height is roughly 0.07, and the radius of curvature is very roughly r = 14.

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

The scalar curvature is

       2 / r².

We are able to add very roughly 0.01 units of scalar curvature into the square by allowing the metric to be stretched by 0.01.

What about shrinking the metric at most by a factor 0.99?

A positive Ricci scalar corresponds to a metric where the circumference of a circle is surprisingly short, compared to the radius. If we are allowed to shrink distances within a square, we obtain a corresponding circle whose circumference is surprisingly long, compared to the radius. We probably are able to construct a metric which adds roughly -0.01 of scalar curvature to the square.

We showed that a flat metric is not a stationary point of the Einstein-Hilbert action. This means that the Einstein field equations do not describe a lagrangian model of mechanics.

In the Wikipedia derivation of the Einstein field equations, the variation of R is calculated against a variation of single components of the metric, e.g., g⁰⁰. We vary two components of the metric simultaneously. Which is the correct procedure?

In a rubber model of gravity, any deviation from the flat metric increases the elastic energy of rubber. A rubber model is able to be a genuine lagrangian mechanical model.

Varying just a single component of a metric

                      _____

                   /           \

                 |                |

                   \______/

      ^ y

      |

       ------> x

Let us analyze the x, y plane. Suppose that we are allowed to stretch the y metric slightly in a certain area, but not the x metric. Does the ratio

       circumference / radius

of a circle shrink for some circle? And does it shrink linearly relative to the perturbation of the y metric?

Conclusions

A possible way to fix the problem is to put |R| instead of R in the Einstein-Hilbert action. Though the derivative of |R| with respect to g is then not defined at g = η, which may spoil the variational calculus.

We have to check what literature says about stationary points of the Einstein-Hilbert action.

If it turns out that the Einstein field equations do not capture the action, that might explain the mystery around Birkhoff's theorem.