We today completed the blog entry dated to October 19, 2023. The crucial observation was that the "focusing" property of mass-energy or pressure inside a mass M requires M to bend rays of light also when the ray does not touch M at all. If we can alter the focusing power of M by manipulating the pressure inside M, then the metric outside M must change – which breaks Birkhoff's theorem. Birkhoff's theorem states that the metric outside M cannot change.
The Einstein field equations enforce Birkhoff's theorem. How to solve the paradox?
It has to be that general relativity resists the change in the pressure with an infinite force. Or that pressure can only change the metric in certain cases, not all.
In newtonian gravity we are always able to manipulate the pressure inside M. Nothing stops us. It sounds absurd if the Einstein field equations prevent us from doing a very mundane newtonian operation. It also is strange if pressure can alter the metric only in certain cases.
In this blog we have for a long time suspected that the Einstein field equations are too strict: that they do not have a solution for any realistic physical system. If we are able to prove that Birkhoff's theorem is broken for a simple newtonian operation, we are closer to proving that no realistic physical system has a solution at all in general relativity.
A cube of freely falling particles
Let us have test particles in a cubical constellation, initially static. If there is no focusing or defocusing property in the space where test particles fall in unison, then the volume of the cube should stay the same.
s = length of the side
• • • •
• • • • cube s³ of test
particles
• • • •
| free fall
v
r = distance (M, cube)
● M
^ y
|
------> x
What kind of a force keeps the volume of the cube s³ constant in a free fall?
We guess it is the newtonian
~ 1 / r²
force. Let us check it. We ignore perturbations of the spatial metric in this calculation. Let
a(r)
be the acceleration of gravity at a distance r. We assume that the acceleration is lower if r is larger.
The low end of the cube accelerates down faster than the high end. The difference is
-da(r) / dr * s.
That is, the volume of the cube tends to grow in the y direction because of this process.
The sides of the cube are pulled closer together because of a tidal force. The acceleration of the right side relative to the left side is
a(r) * s / r.
The cube shrinks in this process in the x and z directions.
The volume of the cube must stay constant when it is falling in a vacuum: there is no focusing or defocusing effect. We get
-da(r) / dr = 2 a(r) / r.
The separable differential equation is of the form
-da / dr = 2 a / r
<=>
da / a = -2 dr / r.
Integrating the sides yields:
ln(a) = -2 ln(r) + C
<=>
a = 1 / r² * exp(C),
where C is a constant. That is, the acceleration of gravity is
a(r) ~ 1 / r².
If we have a gravity field where the acceleration a(r) falls faster than ~ 1 / r² in r, then the gravity field defocuses.
Proving that an increasing pressure inside M must change the metric outside M
We want to prove that if we have a spherically symmetric mass M, and we increase the pressure inside M, then the metric outside M must change. This would contradict Birkhoff's theorem. To prove that, we need the following steps:
1. An increasing pressure inside M makes the focusing power of M stronger: a test mass m falls faster toward the center of M.
2. If the metric outside M does not change, then M must focus less than before close to the surface of M.
3. Reducing the focusing power is not possible without a negative pressure or negative mass.
4. An acceleration of mass does not create a negative pressure.
Ehlers et al. (2005) proved that if the surface of M holds a compensating negative pressure, then a positive pressure inside M does not affect the metric outside M, compared to a configuration where there is no pressure in M. They used item 3 above to save Birkhoff's theorem.
We are interested in dynamic configurations where there is no negative pressure to compensate an increased positive pressure inside M. Could it be that accelerating matter somehow can defocus in general relativity? That could save Birkhoff's theorem.
The stress-energy tensor of a moving point particle contains a positive pressure:
Could it be that an acceleration somehow creates a negative pressure?
The Penrose singularity theorems rely on an assumption that no matter configuration can defocus a beam of light. In our proof we need a different theorem: the acceleration of matter near the surface of M cannot reduce the focusing power of that matter, relative to the case when the matter is static.
The Komar and ADM masses
The Komar mass is essentially a volume integral which sums mass-energy and a positive pressure into the "gravitating mass" of M. The Komar mass is only defined for a static spacetime. We cannot use it here for our purposes, because we have accelerating mass. But let us check if someone has written something about the dynamic case.
The ADM mass is conserved because it is based on a hamiltonian/lagrangian. Let us check how it handles changes in the pressure. If ADM is consistent with Birkhoff's theorem, and changes in the pressure inside M would change the ADM mass of the system, then conservation of the ADM mass actually is a more general version of Birkhoff's theorem. And we criticized Birkhoff's theorem if the theorem does not allow pressure changes inside M.
If we can change the gravity attraction of a non-spherical mass M with pressure, then we can break conservation of the ADM mass. This would imply that general relativity prohibits almost all changes in the pressure. This, in turn would mean that general relativity does not have a solution for any physically realistic system.
Adding positive and negative mass-energy (or pressure) to a gravity system
Imagine that we have a perpetuum mobile which adds more mass-energy ΔM to a spherically symmetric mass M. Birkhoff's theorem prevents the metric from changing outside M. We end up with strange mass-energy which does not affect the metric of spacetime outside M. The Einstein-Hilbert action in such a case may divide the available Ricci curvature in a "democratic fashion" between the mass M + ΔM.
Another option is that M "emits" a negative mass-energy -ΔM to infinity.
If we add the same amount of positive and negative mass-energy, then we can preserve the metric outside M, and the new mass-energy can change the metric inside M.
The Ehlers et al. (2005) paper shows that we can add an equivalent amount of positive and negative pressure into M, and find a solution of the Einstein field equations.
But what if we only add (temporarily) a positive pressure into M? Is it so that in that case the pressure is unable to change the metric? That would be a strange pressure. We are not familiar with any natural phenomenon which behaves in that way.
Could it be that M in such a case "emits" negative pressure, so that the configuration is balanced?
A general interaction does allow the attraction of M to vary
Let us think about an arbitrary interaction between a spherically symmetric system M and a test particle m outside M. Is there any reason why the attractive force between M and m should not vary if we change the internal state of M?
Conceptually, M can reach at m with a robot hand and pull m toward M. The force of the hand can be adjusted. We just have to make sure that energy is conserved.
Infinite rigidity
Birkhoff's theorem makes the metric outside a spherically symmetric M "infinitely rigid" in the sense that nothing can change the metric. In this blog we noticed a couple of years ago that the spatial metric inside an incompressible ball is infinitely rigid in the sense that pressure cannot force the volume of the inside of the ball to grow.
Do infinitely rigid objects occur in other fields of physics?
A black hole horizon would be an infinitely strong object. Maybe infinitely rigid or infinitely strong objects should not occur in reasonable theories of physics?
Changes in mass-energy and pressure in a rubber sheet model of gravity
• ΔM drops down
|
v
--------____ ____-------- rubber sheet
--●--
M
How does a rubber sheet model of gravity react to abrupt changes in mass-energy? When the extra mass ΔM is lowered down, a "longitudinal" wave starts to spread from M.
"longitudinal" wave
-->
_____------- rubber sheet
depression
The rubber sheet is pressed down, and the extra depression spreads to every direction. The rubber sheet is convex upward in the wave. Thus the wave does carry "negative mass-energy" outward.
How do we model pressure in the rubber sheet model? If there is a high pressure, then energy is released when we stretch the area of the rubber sheet – and we can stretch the area by adding a weight. This suggests that in the rubber sheet model, a pressure is very much like the one in general relativity.
pressure
<------------->
springs M in a form of a ring
● \/\/\/\/\/\/\/\/\/ ● a <--• m test mass
----____________________---- rubber sheet
depression
<----------------->
2 R
<---------------------->
r
----> x
In the diagram, we have a mass M distributed in a form of a ring resting on the rubber sheet. The mass consists of small particles which are not attached to each other. Springs push them and try to make the ring larger.
If we would add a test mass m to the center, its depression would add more length to the springs: we would release elastic energy from the springs. This is very much analogous to general relativity.
What happens if we suddenly increase the pressure? Does that increase the attraction of the test mass m toward the ring M?
If m is suddenly moved toward M a distance s, that makes the measured diameter of the ring larger in the x direction in the diagram, because m stretches the spatial metric. That releases energy from the springs: we get an attractive force on m.
But moving m also pushes the near side of the ring to the left because the near side shares some inertia with m. That squeezes the springs. Let us calculate how much is the squeezing effect versus the stretching effect.
Stretching effect
Let us assume that r is much larger than R. The stretching is
1 + G m / (r c²).
When m is moved s closer, the ring becomes
G m / (r² c²) * s * 2 R
wider in the x direction.
Squeezing effect
Let us have a mass element dm in the near side of the ring. It shares an inertia of
2 G dm m / (r c²)
with m. The difference in the shared inertia between a mass element dm near m and a dm far from m is
2 G dm m / (r² c²) * 2 R.
When we move m the distance s closer, the ring becomes
2 G m / (r² c²) * s * 2 R
narrower in the x direction.
Does the squeezing effect really repel m immediately?
The increased pressure exerts a force which is symmetric around the ring. Our hypothesis has been that a force which makes a sphere to expand does not cause a force on m.
In this case, moving m closer to M exerts a force on each dm. The pressure force F resists the squeezing effect on dm. But the force F does not act directly on m.
F s
• <-----------> • <-- •
dm dm m
Does F have any immediate effect on m?
The force F accelerates both dm's. In the next section we show that the acceleration of dm's repels m.
We conclude that the repulsive mechanism probably is the one presented in the next section.
An expanding shell repels a test mass, after all
We have been claiming that the gravity field of an expanding spherically symmetric shell of mass M does not repel a test mass m, because the gravity field of the shell M stays constant at m.
s F s
<-- • <------------> • --> •
dm dm m
2 R r
But we ignored the fact that the field of m is not spherically symmetric with respect to M. The test mass m increases the inertia of those mass elements dm of M which are close to m. If we have a symmetric configuration of forces F pushing each dm, those dm which are close to m move less. The symmetry of the shell is broken: there is a depression close to m.
The impulse from the forces F must go somewhere: they have to push m outward. Let us calculate the effect.
The presence of m increases the inertia of dm close to m the following amount more than the far dm:
2 G dm m / (r² c²) * 2 R.
Let the force F be in effect for a time t. The test mass m "steals" the impulse
F t * 2 G dm m / (r² c²) * 2 R / dm
= 4 F t G m / (r² c²) * R
from F, which corresponds to an acceleration of m to the right:
4 F G / (r² c²) * R.
Let us compare this to the (hypothetical) attractive force caused by the pressure exerted by F. If we would move m a distance s closer to the dm's, the distance between the two dm's would become
G m / (r² c²) * s * 2 R
wider, releasing a pressure energy
2 F G m / (r² c²) * s * R.
This corresponds to an acceleration of m to the left:
2 F G / (r² c²) * R.
We see that the repelling acceleration wins the attracting acceleration 2 : 1.
A rubber sheet model with pressure
In this blog post we are interested in the metric of general relativity around a pressurized spherical mass. The repulsive effect described in the previous section is a complex dynamic effect. The metric probably does not know about it. However, the metric probably is aware of the attractive effect of pressure. Let us ignore the repulsive effect and consider only the attractive force created by a positive pressure.
In the rubber sheet model it is straightforward to ignore the repulsive effect. Let us continue our rubber sheet analysis.
pressure
<-------------->
● \/\/\/\/\/\/\/\/\/ ● • m test mass
----____________________---- rubber sheet
How can we model the attraction of the pressure to m?
What prevents m from rolling left on the rubber sheet? The sheet is tense. If m moves to the left, it stretches the sheet on the left. It gets a boost by the shrinking of the sheet on the right, though. The net effect is zero. The test mass m does not move.
But if the rubber sheet is less tense on the left than on the right, then m will tend to roll to the left. We could model a positive pressure by reducing the tension of the rubber sheet. A negative pressure means that the tension of the rubber sheet is tighter.
The Ehlers et al. (2005) configuration has the tension reduced inside a sphere (a circle in the rubber sheet model) and the tension tightened along the surface of a sphere. If we would remove the tightening, the loosening of the rubber sheet would spread to a large area. That is, there would be a positive pressure spreading to the environment of M. Does this make sense?
Even if the loosening would be restricted inside M, it would attract the test mass m. There is no reason why the expanding ring M would make the rubber sheet tenser in its environment.
Suppose that we temporarily increase the pressure inside M. The rubber sheet starts to loosen around M. Then we remove the positive pressure from M. The rubber sheet tightens so much that it actually has now negative pressure. A "longitudinal" pressure wave spreads from M. This is quite similar to what happens if we temporarily increase the mass-energy M. In that case the wave is a vertical displacement wave in the rubber sheet, where negative mass-energy is followed by positive mass-energy.
Let us assume that Birkhoff's theorem prohibits any such waves from spreading in the vacuum. What would happen?
Since the effect of the positive pressure is not allowed to escape from M, there has to exist negative pressure around M which compensates it, so that a test mass m at some distance from M does not feel the effect of the pressure.
But this is a paradox: how can a negative pressure arise from nothing? That does not make sense.
Conclusions
Our focusing power argument strongly suggests that a positive pressure should generate an attractive force on a test mass m far away from a spherical mass M, in general relativity.
The same conclusion is true for a rubber sheet model of gravity.
But Birkhoff's theorem says that the metric ouside M cannot change. There cannot be any extra force on m caused by a pressure inside M.
How to solve the paradox? Can we claim that a negative pressure somehow forms around M to compensate the positive pressure?
No. There is no obvious mechanism which would create such a negative pressure.
Let us compare this to the concept of mass-energy in general relativity. Birkhoff's theorem, as well as ADM theorems seem to imply that mass-energy has to be conserved.
Is there a conservation law that the sum of positive and negative pressure must be conserved in general relativity? We have to check what a lagrangian generally can imply. Conservation of energy is associated with time-independence of the lagrangian. What would correspond to conservation of pressure?
Note that we have a conservation law for the mass flow: conservation of momentum.
Maybe we can get a conservation of "pressure" if we add it to the acceleration of mass? The conservation law would be Newton's third law: action and reaction. However, this is vague because how do we classify accerated mass as negative pressure?
Maybe we should classify the mass whose inertia is holding back a positive pressure, such that the acceleration of the mass is negative pressure? Can this vague definition work?
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