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