"Conservation of pressure" in general relativity?

Our previous blog post brought up a possibility that general relativity actually has a "conservation of pressure" principle. Any volume of positive pressure has to be enclosed into a shell of negative pressure. Then the metric changed by an internal pressure cannot spread out of a spherically symmetric mass M. Birkhoff's theorem is saved.

The stress-energy tensor T

Let us check how T is precisely defined. Could it be that an accelerating mass at the surface of M actually must be counted as negative pressure?

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

The component T¹¹ is defined in Wikipedia as the flux of x₁ momentum across a surface of a constant x₁ coordinate.

  

                     

          pressure P        mass M

              ---------------> ■■■■■■■

                                       ---> 

                                       |   |             |

                                       X   Y            Z    

      ----> x

Let us analyze the flux in the above configuration. What is the flux at X, Y, and Z?

Let us consider the volume element between X and Y. The pressure at X is higher than at Y. The element "eats" pressure.

"Eating" pressure would be defined as a process where a mass dm accelerates inside the volume element.

But this will not work as a definition of negative pressure for M. If P is negative, then M still "eats" pressure.

And the volume containing a positive pressure P may be arbitrarily large. Assigning the right amount to compensate it with a negative pressure within M is hard.

Conservation of pressure in a large block of an elastic material

Suppose that we have a large block of material with constant Young's modulus. If we stretch the material at one position, we have to squeeze it at another position. The sum of pressures is conserved.

We have to analyze this from the perspective of Noether's theorem. Could it be that the Einstein-Hilbert action implies conservation of pressure?

In a rubber sheet model, masses put on the sheet produce negative pressure (stretching). But otherwise, pressure is conserved.

Is the proof of Birkhoff's theorem correct?

https://www.roma1.infn.it/teongrav/VALERIA/TEACHING/RELATIVITA_GENERALE/AA2015inpoi/birkhoff_theorem.pdf

The proof is based on defining new coordinates based on the usual coordinates t and r, and showing that the metric stays as the Schwarschild metric in the new coordinates.

   t

   ^

   |                |        |          • m

   |         ----------------------

   |               /            \

   |              |            | 

   |         ----------------------

   |              |            |

    ------------------------------------>  x

                          ●

                          M

But is it ok to switch the coordinates in an arbitrary way? Let us have test masses which are initially static in the old coordinates. If we define new coordinates which move at an accelerating speed relative the old coordinates, are such coordinates allowed?

If the metric would be Minkowski, reasonable coordinate systems are inertial.

In the diagram we have a case where the spatial metric around M suddenly expands. The x coordinate lines make a turn. If we have test masses m floating around M, their proper distance from M suddenly grows.

But if we define new spatial coordinates, we can make the spatial metric to appear constant in the new coordinates. As if nothing would have happened. In the diagram, the new coordinates would "accelerate" relative to the old coordinates.

Could it be that in the proof of Birkhoff's theorem new coordinates hide the fact that the metric has changed relatively to freely floating test masses?

Conclusions

There seems to be no simple and intuitive way of defining a "negative pressure" of accelerating mass, such that the negative pressure would compensate a positive pressure.

It looks like Birkhoff's theorem is broken by pressure.

Changing pressure in a spherical mass breaks Birkhoff's theorem?

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

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

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.

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

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.

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

The stress-energy tensor of a moving point particle contains a positive pressure:

Could it be that an acceleration somehow creates a negative pressure?

https://en.wikipedia.org/wiki/Penrose%E2%80%93Hawking_singularity_theorems

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

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

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.

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

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.

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

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?

Tolman's paradox: the gravity of pressure

We corrected the calculations of September 9 and October 11, 2023. Now it looks like that gravity satisfies Gauss's law, at least in certain cases.

If we convert a central mass M into an expanding dust shell, then the gravity attraction at a distance R seems to stay constant.

What if the dust particles do not all fly outward, but bump into each other?

That corresponds to a "back and forth" movement. Earlier we have argued that the gravity in that case is equal to γ M, that is, the mass-energy.

A mass M bouncing back and forth radially relative to the test mass m

                                    wall         s          wall

               •                       |         ● --> v      |

              m                     R         M             R'

                        r = distance (m, M)

Let us have a mass M which bounces back and forth between walls at distances R and R'.

After we corrected the September 9, 2023 calculation, the coordinate acceleration of the gravity of M is

       1 / γ⁵  *  G M / r²

       =  (1  -  5/2 v² / c²)  *  G M / r²,

if v << c. That is, the gravity of a moving M is substantially weaker than a for a static M.

Let us calculate the average force from the bouncing of the extra inertia of m. The extra inertia of m "held by" M is

       2 m G M / (c² r).

The impulse from a bounce is

       4 v m G M / (c² r).

Let us assume that s = R' - R is small relative to r.

A cycle back and forth for M takes the time

       t  =  2 s / v.

The net impulse for a cycle is

       p  =  4 v * s m G M / (c² r²)

and it points to the right in the diagram. The force

       p / t  =  2 v² m G M / (c² r²)

                =  2 v² / c² * m G M / r².

The acceleration is

       2 v² / c²  *  G M / r².

That does not seem right. We would like the total gravity to be

       (1  +  1/2 v² / c²)  G M / r².

When we evaluated a sideways bouncing of M relative to m, the key was to remove the acceleration caused by the stretching of the radial spatial metric. That acceleration is "reversible" as M moves back and forth.

Let us calculate the effect of the stretching of the spatial metric. Let m move at a local velocity V a distance Δr toward M. The spatial metric slows down the coordinate velocity V' through the formula:

       V'  ~  V  *  (1 - 1/2 rₛ / r).

The slowdown from the spatial metric is

       Δr * V * G M / (r² c²)

       = Δt V² / c² * G M / r²,

which corresponds to an acceleration of

       V² / c² * G M / r².

Let us sum all these:

       (1 - 5/2 v² / c²) * G M / r²

        + 2 v² / c² * G M / r²

        + v² / c² * G M / r²

       = (1  +  1/2 v² / c²)  G M / r².

The sum is reasonable.

On October 18, 2023 we realized that in certain cases, a linear sum of the metric perturbations is not the correct way to obtain the solution for the combined system. We have to keep this in mind.

Pressure does not cause gravity?

Our calculations above suggest that pressure does not produce a gravity attraction, after all. How is this possible? If we move the test mass m closer to M, the test mass makes the radial metric to stretch, releasing energy from pressure. There should be some attraction.

A possible explanation: manipulating the pressure inside M usually causes an acceleration of mass in M. And we saw on October 17, 2023 that general relativity cannot handle accelerating masses.

                                 _____

                              /   P      \

                 •          |<---------->|   M

                              \______/

              m

Let us consider a shell of mass M, held static by an internal pressure P.  If we suddenly remove the pressure, then the shell starts to contract.

M attracts m less since the pressure was released?

However, we also have to consider the fact that the extra inertia "held" by M starts to accelerate toward the center of M. Since the near side of M holds more inertia than the far side, m accelerates toward M.

Let us calculate the effect. Let M be very small, so that its internal gravity is negligible. We suddenly increase the pressure P from zero to a large value.

                        r                   s

               •                  |        P        |     A = area of

              m             dm / 2         dm / 2         wall

We assume s << r.

If we move m closer a distance dr, we release pressure energy worth

        dW / dr = 1/2 *  1 / (1  -   rₛ / r)³/²  *  rₛ / r²

                           * s A P

                      =  G m / (r² c²)  *  s A P,

where we assumed rₛ / r is small. The acceleration of m is

       a  =  G / (r² c²)  * s A P,

and its is attractive.

In the configuration, the near wall "holds"

       2 s * 1/2 m G dm / (r² c²)

more inertia of m than the far wall. The acceleration of the near wall is 

       2 P A / dm,

which translates to a repulsive acceleration of

       2 G / (r² c²)  *  s A P.

The net effect is

       G / (r² c²)  *  s A P.

If the pressure is uniform, it makes the shell to expand up and down in the diagram, too. And that might cause an acceleration of m. Let us analyze.

                                            ^  a

                                            |

                                            • dm

                                                  ^

              | F                               |   1/2 s

              v                                  v

              •               r

             m            

     ^ y

     |

      -----> x

There dm is an element of M. How does m react to dm being accelerated upward in the diagram? If we imagine that dm would be static, and m would be pushed down by a force F, what would the path of m be?

The test mass m has some extra inertia which is double in the radial direction. The radial component of the acceleration caused by F is approximately

       a  *  1/2 s / r,

and it is reduced by a factor 1 - 1/2 rₛ / r, because the inertia is larger to the radial direction. We get a correction to the x acceleration

       1/4 rₛ / r  *  s / r * 2 P A / dm

       =  1/4 * 2 G dm / (r² c²)  *  s / r  * 2 P A / dm

       =   G / (r² c²)  * s A P.

The acceleration is attractive.

What is the average correction? The pressure component Pₓ produces a repulsion

       1/2 G / (r² c²) * s A P

for an element of the sphere whose projected area is A in the y, z plane.

The repulsion is dominated by the corresponding attractive effects of Py and Pz:

       G  / (r² c²)  * s A P.

But here we forgot the fact that we cannot sum individual perturbations like this. How to correct?

Let us consider the electromagnetic analogue.

                                 ____

                               /          \

            •               |             |

                               \______/

           -q                    +Q

      

If the charged shell Q expands, then some excess energy density between q and Q is erased and some reduced energy density behind Q is erased, too. But is there a reason why this energy shipment should produce a force on q?

We have assumed vaguely that in gravity the field energy is shipped "on the average" from the distance (m, M).

In the case of a static electric field it is more complicated. If Q expands, there is an energy shipment.

Expanding dust shells

In the (corrected) blog post on October 11, 2023 we studied a dust shell which expands at a constant velocity. It looked reasonable that the gravity field of such a shell is equivalent to a static mass M which has the same mass-energy. Birkhoff's theorem would hold in that case.

What about a shell whose expansion accelerates?

That boils down to the question, "where" the extra inertia of the test mass m is "stored". Birkhoff's theorem suggests that it is, on the average, stored at the center of the dust shell. If not, then dust shells of the same mass-energy but different radii would have differing gravity on a test mass m approaching them.

Assumption. The extra inertia of m, "on the average" is stored at the center of the dust shell.

If we accept the assumption above, then the pressure P does generate an extra gravity attraction. Birkhoff's theorem fails, after all.

Conclusion

Our best guess is that pressure does gravitate. That, in turn, means that Birkhoff's theorem fails for gravity.

It may be that general relativity does not understand a pressurized configuration and the metric calculated from it satisfies Birkhoff's theorem. Then we have a failure of the geodesic equation.

The only accelerating solution of general relativity which we know well is the Oppenheimer-Snyder dust collapse (1939). The authors were able to match the solution to a static Schwarzschild vacuum solution outside the collapsing dust ball. In the comoving coordinates the dust does not move, and the pressure component in the stress-energy tensor is zero. However, in semi-static coordinates the dust moves and the component is non-zero.

What a failure of Birkhoff's theorem means?

Let us assume that we really can affect the gravity attraction of a spherical mass M by manipulating the pressure within it. Let us analyze what implications would that have for general relativity.

Let us assume that the metric around M really describes the force on m.

The change of the metric required to reduce the attraction of M

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

                  m   •                          ●   M

      

        ^ y

        |

         ------> x

The test mass m is initially static. What kind of a change in the metric could decrease the acceleration

       d²x / dτ²,

where x is the x coordinate of m, and τ is the proper time of m?

In the geodesic formula, μ is the x coordinate and α and β must be the t coordinate. All the other terms are zero.

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

We have

       Γₓₜₜ  =  1/2  (2 dgₓₜ / dt  -  dgₜₜ / dx).

We can decrease the force on m by making gₜₜ less steep in the x direction, or making gₓₜ to increase with time.

What would that require to happen in the stress-energy tensor T of the vacuum area? The tensor T cannot remain zero because then Schwarzschild metric around M could not change at all.

The element gₜₜ becomes less steep if M ejects mass-energy out and the ejected mass-energy passes m.

We can manipulate gₓₜ by letting a compact mass M' pass by m at a close distance. But how could a spherically symmetric process do that?

Let us calculate the Einstein tensor for the metric g. From that we get the required stress-energy tensor T. Or do we? The component gₜₜ can be made steeper by moving M closer to m. Many different changing stress-energy tensors may be able to bring the desired changes.

This looks complicated.

Ejection of negative mass-energy?

Can we make the metric to change with time? Yes, if we assume that some strange matter whose mass-energy is negative is ejected from M when we increase the pressure.

A shell of negative mass-energy would recede from M, making the effective mass of M larger. But negative mass-energy may facilitate superluminal communication and bring all the time paradoxes. This is not a nice solution.

Also, if we want to reduce the attraction of M, we should eject a shell of positive mass-energy. What kind of matter would that be?

Can small departures from perfect symmetry help?

The system is never perfectly symmetric. Can we utilize small departures from the symmetry to manipulate the metric in such a way that it exerts a varying force on m, depending on the time?

How "stable" is Birkhoff's theorem? Could it be that the metric could otherwise change with time, but the perfect symmetry prevents changes in the metric from "escaping" from M to empty space?

This idea is against our own Minkowski-newtonian gravity model. The gravity field in it is determined in a simple fashion from the sources of gravity. Small departures from symmetry cannot make much of a difference.

Since the Einstein field equations are nonlinear, it is very difficult to prove mathematically that small perturbations cannot work miracles. But we may demand that a physical theory must not be based on almost-miracles.

The Schwarzschild interior solution when the pressure is suddenly released

Karl Schwarzschild in 1916 was able to glue together an interior solution for incompressible fluid ball and the external solution in the vacuum.

What would happen if we would suddenly remove the pressure which is stopping the incompressible fluid from collapsing? The stress-energy tensor T inside the ball suddenly changes. How does that affect the vacuum solution?

Birkhoff's theorem claims that the vacuum solution cannot change. The metric should solely start changing inside the ball. Is that possible?

In the Einstein approximation formula (1916), pressure does contribute to the metric in the vacuum area. The pressure P is "mapped" to the point we are interested in through the formula

       ~  P / r,

where r is the distance from the point to the pressurized element. After trace reversing, the pressure affects gₜₜ, gₓₓ, and so on, at the point.

A pressure which changes with time

The stress-energy tensor of a particle moving in the x direction looks like this:

     Dirac δ function *

         γ m          γ v m          0                  0

         γ v m       γ v² m        0                  0

         0        ... 

         ...

We can "simulate" pressure with moving matter.

         • -->                            <-- •

         <-- •             ...            • -->

         

Pressure T₁₁, T₂₂, or T₃₃ in the stress-energy tensor can come either from matter moving to opposite directions, or from mechanical pressure from a force field.

However, we cannot manipulate T₀₀, but only the pressure and the mass-energy flow components. It could still be that the pressure does not cause any effect on a test mass m which is outside a ball of matter M.

The Ehlers et al. paper (2005) proves that one can preserve the metric outside the ball, if there is a negative pressure in a "crust" of the ball. What happens if we remove the crust?

Let us think. The Schwarzschild interior solution proves that it is possible to find a metric for a pressure which is quite a complicated function of the radius r. The external metric is the Schwarzschild vacuum solution.

Is there any reason why a time-dependent pressure should affect the metric in the vacuum area?

Here we must note the following three things:

1.   Pressure probably does affect the orbit of a pointlike test mass m outside the spherical mass M. We have in this blog repeatedly argued that "tidal" effects in M and a "backreaction" change the orbit of m.

2.   But we have also repeatedly argued that the metric derived from the Einstein field equations does not describe the orbit of the test mass m. That is, the geodesic equation fails when there are tidal effects.

3.   It is possible that a pressure inside M does not affect the vacuum metric calculated from the Einstein field equations, but the pressure does affect the orbit of m.

The Einstein equations may be flexible enough so that they allow the metric inside M to change when the pressure is changed, but do not require the metric outside M to change.

The pressure is always zero at the surface of M. It is not far-fetched to assume that there is no need to change the metric outside M when we change the metric inside M, to suit a pressure change.

In an earlier blog post we suggested that in a spherically symmetric configuration, the test mass m actually has to be a spherical shell. The test mass must not break the spherical symmetry. A spherical shell, apparently, does not distort the spatial metric inside it  –  only the metric of time. Pressure inside a spherical mass M would not attract a spherical shell outside M.

A pressure "focuses" beams of light passing through M: it must bend light also outside M?

In general relativity, mass-energy, and presumably also pressure, "focuses" a narrow beam of light if there is mass or pressure inside the volume of the beam. Empty space does not focus a beam of light.

           --------------------------------

           ----------_______________  beam of light

                  /        \

                   \____/   M

Suppose that M only bends rays of light which touch M or pass through M. Then there is "defocusing" of a beam which touches M. This is not allowed.

In the Schwarzschild solution, the focusing capability is only inside M, but the metric also bends rays of lights which do not pass through M.

Now, if we are able to increase the focusing capability of M temporarily with an increased pressure, we expect that M should also bend rays of light outside M more strongly. But Birkhoff's theorem prohibits that!

In the Einstein approximation formula (1916), pressure does affect the metric also outside M.

If we are able to prove that an increased pressure inside M break's Birkhoff's theorem, that might imply that the Einstein field equations do not have a solution for a realistic physical system.

Conclusions

Our analysis converged on a crucial question: does a changed pressure inside a spherically symmetric mass M require that the metric outside M changes. That would break Birkhoff's theorem.

We have to check what happens if we remove the pressure from the Schwarzschild internal solution. Also, we have to study how focusing/defocusing are precisely defined for a beam of light.

If Birkhoff's theorem is broken, that may imply that the Einstein field equations do not have a solution for any realistic physical system.

The spatial metric around a lightweight shell is very different when summed from perturbations

The shell theorem of Newton states that a spherical shell of a mass M has the same gravity as the equivalent mass M placed at the center of the shell.

The summed spatial metric perturbation is quite different from the one solved from the Einstein field equations

What about general relativity? Let us first calculate the metric perturbation from an initially static lightweight shell.

                M

               ___

             /       •  dm

             \____/  

                             R = distance (center of M, m)

                             r = distance (dm, m)

                 •  m

Each part dm causes a slowdown of time at the test mass m by a factor

       1  -  1/2 rₛ / r

where

       rₛ  =  2 G dm / c²,

and r is the distance of dm and m.

The Newton shell theorem states that the sum of the metric perturbations for gₜₜ is

       1  -  2 G M / (c² R).

Also, dm stretches the radial metric, parallel to r, at m by a factor

       1  +  1/2 rₛ / r.

The stretching of the spatial metric for each dm is to the direction of their individual vector (dm, m). The stretching is not parallel for all dm. If we sum the stretchings in the direction (center of M, m), the sum is not

       2 G M / (c² R),

but less. There is also stretching in the tangential metric.

The stretching of the spatial volume is similar to how much time "shrinks". But the division of the stretching into the polar coordinates r, φ, θ is not at all the same as in the Schwarzschild metric around an equivalent small mass M!

What does this mean? Are the Einstein field equations "magically nonlinear", after all, so that even tiny faraway masses can substantially influence the metric perturbation created by a mass dm?

Or is there a switch of coordinates which brings the perturbation caused by the linear sum close to the metric around M?

Bending of light close to the Sun

We know that the bending of light close to the Sun conforms to the Schwarzschild metric. Do we get the same bending by summing the bending effect for all mass elements dm of the Sun?

Yes. For each element dm, exactly a half of the bending comes from the perturbation of the metric of time, and the other half from the perturbation of the spatial metric.

If we sum the perturbations of time by each dm, we obtain almost exactly the metric of time of the Schwarzschild solution around the Sun. Let α be the deflection angle due to time. Then

       2 α

is the total deflection angle for the sum of all dm, and the total deflection angle is the same for the Schwarzschild metric around the Sun.

Since the deflection angle is the same, it may be difficult to measure which is the correct spatial metric: the one obtained by summing or the one obtained by solving the Einstein field equations.

If we have an initially static mass m close to the shell, its acceleration is solely from the metric of time. Tests with a static mass m will not reveal which is the correct metric.

Can we find coordinates where the summed metric is close to the Schwarzschild metric?

             ___                                 ___    shell

                   -- • --_________------

              dm  

                               β = angle (m, dm) vs. y axis

                            \    |

                              \  |

                                \|

                                 • m  

     ^ y

     |

      ------> x

Close to the shell, the summed spatial metric perturbation differs substantially from the Schwarzschild metric of the equivalent M. The angle β above is quite large for most elements dm.

Suppose that the Schwarzschild metric close to m has r stretched by a factor

       1  +  b.

The summed metric has r stretched by

       1  +  b',

and the tangential metric by

       1  +  c'.

The stretching of the volume tells us that

       b  =  b'  +  2 c.

If m is close to the shell, the value of b' may be substantially smaller than b. The difference might be 10%.

Can we rescale the r coordinate to make the metric look like the Schwarzschild metric?

The radial metric, of course, can be made identical through rescaling. A possible rescale is

       r'  =  r * (1 + b - b').

We may be able to make the metric to look rather similar to the Schwarzschild metric. But if we define the radial coordinate

       r  =  proper circumference / (2 π),

then the metric does differ significantly from the Schwarzschild metric. No coordinate transformation will change that.

Does there exist a solution in general relativity for the shell?

The Einstein field equations have exact solutions for a ball of incompressible fluid (Schwarzschild 1916) and for the collapse of a ball of uniform dust (Oppenheimer and Snyder 1939).

We have to check the literature if there is a solution for a shell of dust. Since the acceleration of the inner surface of the shell is zero, the collapse becomes quite complicated. The outer surface will press against the inner surface.

If there exists a solution, then Birkhoff's theorem dictates that the outer metric has to be Schwarzschild  –  and we showed that it would differ considerably from the summed metric.

Is it plausible that gravity is nonlinear in such a drastic way?

It is a natural assumption that weak interactions are almost linear. But here we seem to have a case where an extremely lightweight shell would fundamentally modify the field of its parts dm. Does this make sense?

Are there other natural phenomena which would be so extremely nonlinear? So that even tiny effects would coordinate between themselves, and modify each other substantially?

The solution to the problem: the metric perturbation comes from the collective perturbation of the time? Or a "rotation" is easy?

The perturbation of time is the same in Schwarzschild and in the sum of perturbations.

We have claimed that the stretching of the radial metric in Schwarzschild comes from shipping of energy around in the field. When the test mass m approaches M, energy comes from, on the average, the distance between M and m and is absorbed by m. The shipping of energy increases the inertia of m, which simulates stretched radial metric.

We have also claimed that the Schwarzschild field allows field energy to rotate around M "easily". If the tangential spatial metric were stretched, that would indicate that field energy would not rotate easily.

          ● M              • m              ● M'

     ^ y

     |

      -------> x

What about the configuration above? We believe that we must sum the radial metric perturbations by M and M' at m. In this configuration, field energy does not "rotate" around, when we move m to the x direction. There is extra inertia, which causes stretching of the x metric at m.

Conclusions

The correct metric around the spherical shell probably is the Schwarzschild metric. The field energy for each pair (m, dm) is not "private" enough to cause the tangential metric to stretch.

We have to investigate if and how this affects frame dragging of a rotating disk.

Also, we have to find a precise definition: when does energy shipping affect inertia of m, and how much?

Electromagnetism and general relativity do "lose" crucial information

We have been writing about "private interactions" of charged particles, meaning that we must calculate them between each individual pair. Using the summed field of many particles would lose crucial information.

We need a longitudinal magnetic field

On October 9, 2023 we derived the Lorentz transformation of the "electric field Eₓ" if we define the "electric field" through the acceleration of a test charge q. We noted that the ordinary electric field Eₓ does not contain enough information to derive the x' acceleration of q in another frame. This is because the acceleration of q depends on the velocity of the charge Q, and that velocity is not coded into the ordinary electric field Eₓ.

We suggested that one has to define a "longitudinal magnetic field" which would reveal the velocity v of Q, and allow the calculation of the effects of the inertia of q inside the electric field of Q.

The ordinary magnetic field B is

       ~ v × E,

but we need a new field C which is something like 

       ~ v • E.

Turning mass flows in general relativity

            1                                 3

                \                           /          ^

                  \                       /          /   v

                    \__________/        mass flow

                             2

                             ^   V

                             |

                             •  m test mass

On September 23, 2023 we showed that the metric of general relativity cannot handle a case where a mass flow turns between the parts 1 and 2, and 2 and 3.

We concluded that if we sum the metric perturbations for all the (very many) particles dm in the part 1, the summed metric is not aware of the way how the metric for each individual particle behaves dynamically with respect to time. The summed metric is time-independent, while the metric for each particle is very much time-dependent in every aspect.

Each particle dm in the part 1 moves toward the test mass m. There are dynamic effects with the increasing inertia as dm approaches m. But these effects are lost in the sum of metric perturbations.

This very much looks like the same problem as we had with Eₓ: the summed field has lost crucial information of the velocity of each charge element dq.

As each dm turns in the bends between 1, 2, and 3, there are further effects since the inertia of m held by dm accelerates in the turn. It is not clear if we can describe this effect with a field at all.

Birkhoff's theorem

Suppose that the gravity acceleration of a test mass m really does change when we manipulate the pressure inside the mass M. The field of M would then undergo a "longitudinal" change  –  something which is impossible in traditional electromagnetism.

This may be associated with the previous two sections. We have to analyze this in detail.

https://www.roma1.infn.it/teongrav/VALERIA/TEACHING/RELATIVITA_GENERALE/AA2015inpoi/birkhoff_theorem.pdf

The Einstein field equations quite simply imply that the metric outside a spherically symmetric object cannot change. But it must change if the pressure trick affects the acceleration of m.

The Schwarzschild metric seems to include all the inertia effects of the field, in contrast to the Coulomb force. But somehow a summed metric then loses this ability.

Conclusions

This is a short note which explains how the Lorentz transformation problem of Eₓ, the gravitomagnetic problem of mass flows, and maybe even Birkhoff's theorem are related.

The traditional magnetic field B contains the required information to know a transverse "movement" of an electric field E. But it does not contain the crucial information which we need in a longitudinal movement of the electric field E.

The summed metric in general relativity, similarly, lacks the crucial information of a longitudinal movement of a gravity field. The information is present if we just have one mass element dm moving. But when we integrate over many such elements, the information may be lost.

Could this information loss explain why Birkhoff's theorem seems to fail?

Gauss's law for a moving mass; Birkhoff's theorem

UPDATE October 22, 2023: The calculation below about m approaching M contains errors.

We also corrected the corresponding calculation from September 9, 2023. The result there for a radial movement is

       1 / γ⁵  * G M / R².

If we believe the tangential calculation below, there

       γ⁴ G M / R².

The "average" over x, y, z directions is

       (-5 + 4 + 4) / 3 = 1.

This would mean that Gauss's law works: the average gravity of a moving mass M is γ M at the distance R.

----

The most common form of Gauss's law states that if we have static charges, then the flux of the electric field through a closed surface is the sum of the charges inside that surface. The electric lines of force should never break in electromagnetism in a volume where there are no charges. This is a form of Gauss's law for moving charges.

In our previous blog post we suggested that the acceleration of a test charge q also depends on how the electric field of Q affects the inertia of q.

Acceleration of a test mass m close to a moving mass M

We want to find out what is the flux of the gravity field through a closed surface in various situations.

                    M

                     ● ---> v

 

                                       R = distance (m, M)

                     ^  ay  

                     |

                     • m

We assume that

       rₛ / R   <<   v² / c²,

and that v is slow. We are interested in how v affects the gravity acceleration. Then, if m and M are both static, we can assume that the acceleration of m is

       G M / R².

In our first post on September 20, 2023 we calculated that the acceleration ay in the above configuration is

       ay  =  γ⁴  G M / R²,

where

       γ  =  1 / sqrt(1  -  v² / c²).

                                         M

           m • --> aₓ              ● ---> v

       ================  ruler

               r

              R = laboratory distance (m, M)

Let us then check the acceleration "on the side" of M. We assume that M is carrying along with it a long ruler and m is at a position r on the ruler at the moment we are interested in. Because of length contraction,

        r  =  γ R.

We have to determine the acceleration in the comoving frame of M.

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

The formula above says that the (pseudo) "kinetic energy" on the left side and the negative "potential energy" G M m / r on the right side have a constant sum. It is like newtonian gravity, with the exception that the time which is used, τ, is the proper time of m.

If v is slow and M / r is small, we can write

       dτ²  =  dt²  * (1  -  v'² / c²)  *  (1 - rₛ / r),

where v' is the speed of m as measured a local static observer, and

       rₛ  =  2 G M / c².

The energy-momentum relation says

       E²  =  p² c²  +  m² c⁴.

The formula in the square brackets [ ] is

       p² / (2 m)  + 1/2 m c² - 1/2 m c²

       =  p² / (2 m),

which for a slow v and small M / r is the kinetic energy of the test mass m if it would fly out of the field of M. In newtonian gravity, the above equation says that the kinetic energy of m is the kinetic energy that m had outside the field of M, plus the energy it gained by descending down in the potential of M.

We obtain

        V  =  dr / dt  =  sqrt(p² / (2 m)  +  2 G M / r)

                                  *  (1  +  1/2 v'² / c²)

                                  *  (1  +  1/2 rₛ / r).

We have to calculate

       dV/ dr

to get the coordinate acceleration aₓ' in the comoving frame of M. Note that

       v  ≅  sqrt(p² / m²  +  2 G M / r),

since v is slow and M / r small.

The second and the third terms contribute the corrections from the change of the rate of the proper time τ of m. When m moves farther away from M, the proper time of m runs faster. The velocity of m increases because of this. The second and the third term contribute equally.

The change dV₃ of V from the third term is

       dV₃ / dr  =  v / (1  -  2 G M / (c² r))³/²

                          *  -1/2  *  -2 G M / (c² r²)

                       ≅  v  *  G M / r²  *  1 / c²

                       =  v  *  m G M / r²  * 1 / (m c²),

that is, the relative change in the total energy of m. It is as if m would lose dW / c² of inertia where dW is the work done by gravity on when r changes by dr.

Then

       dV₃ / dt  =  dV₃ / dr  *  dr / dt

                      ≅  dV₃ / dr  *  v

                      =  v² / c²  *  G M / r².

The first term in the definition of V is exactly like in newtonian gravity.

We get

       aₓ'  =  dV / dt 

              =  G M / r²  *  (1  +  1/2 v² / c²)

                  - 2 v² / c²  *  G M / r²

              =  1 / γ³  *  G M / r²

              =  1 / γ⁵  *  G M / R².

The Lorentz transformation of the acceleration gives

       aₓ  =  γ² aₓ'

             =  1 / γ²  *  G M / R².

Birkhoff's theorem fails?

                    ●  M

                ---------  "gravimetric" sphere

Suppose that we have a static spherical mass shell whose mass, measured from far away, is M. The sphere is very lightweight.

We put a much larger "gravimetric" sphere symmetrically around M and measure the gravity acceleration on that shell.

Then we use some of the mass-energy of M to make its parts to move at a velocity v toward the center of M. Let dm be a moving part. If the gravity of dm would be

       ~  γ dm,

where

       γ  =  1 / sqrt(1  -  v² / c²),

then the gravitational pull on the gravimetric sphere would stay constant. Let v be to the x direction. Our calculations in the previous section suggest that the gravity acceleration is

       ay  =  az  ~  γ⁴ dm,

and

      aₓ  ~  dm / γ².

The "average of these is 

       γ² dm,

not γ dm! This suggests that the gravity M becomes larger by a factor

       γ M,

when the parts of M are made to move toward the center. If that is the case, then Birkhoff's theorem fails.

Birkhoff's theorem is a direct result of general relativity. We calculated the gravity using the Schwarzschild metric for each part which moves at a constant velocity. Thus, our result should be a fairly accurate solution of the Einstein field equations. How can then Birkhoff's theorem fail?

The reason might be that general relativity cannot handle the situation when the parts dm of M are accelerated from a zero velocity to the velocity v. The field outside M maybe turns from one vacuum solution to another?

The ADM formalism can be used to prove some kind of Gauss's law for gravity. We have to check how they handle accelerating masses.

It looks like that Birkhoff's theorem holds, after all. See the next section.

The radial stretching of the spatial metric does not affect the average acceleration?

Our "average" of γ⁴ dm and dm / γ² in the previous section is probably a wrong way to calculate. Let us analyze the acceleration which is due to the distorted spatial metric.

         

                   \|/   v

                --- • ---    dm

                   /|\

                    • m

Let us have a dm which moves to a random direction from the origin, at a velocity v.

We have a test mass m very far away. What is the average effect of the the stretched radial metric of dm on the acceleration of m?

It is probably zero. We may imagine that dm moves along a circle.

                  ___        ^  v

                /        \     |

                 \___ •    dm

                    • m

After dm has made a full round, its effect on the spatial metric around m is restored. When dm traveled around the circle, the stretched spatial metric made m to accelerate to various directions. It is like dm would be carrying a magnifying glass. When we view m through the glass, its position appears to change as the glass travels a full circle  –  but the apparent position of m after the full circle is back to the original.

The various locations of dm on the circle have v pointing to any direction: the process does describe the average for different directions of the velocity v.

This suggests that in the average acceleration, we can ignore the distorted spatial metric. The acceleration solely comes from the metric of time.

This also (partially) solves the "back and forth" problem of our first blog post on September 20, 2023. If we have a large number of particles bouncing around inside a mass M, the velocities of the particles have a random direction. Thus, we can ignore the effect of the stretched radial metric around each particle, and only concentrate on the metric of time.

However, there is more to bouncing than a simple circular movement. In bouncing, particles are accelerated, and general relativity seems to have problems in handling the gravity of accelerating masses M.

What is the "pressure" in the stress-energy tensor?

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

Suppose that we have gas inside a container. Its molecules collide frequently. There is a pressure. But what if we have a single molecule or a small rock bouncing around? Is there "pressure" in the energy-momentum tensor in that case?

What about a cloud of dust moving to one direction? If we add another, overlapping cloud which moves to the opposite direction, is there pressure? We might assume that there are no collisions between dust particles.

Wikipedia does contain the answer to this question. If a particle has an x velocity of vₓ, then the xx "pressure" component is γ m vₓ².

Thus, also one-way movement of matter generates "pressure".

Tolman's paradox of pressure as a source of gravity

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

J. Ehlers, I. Ozsvath, E. L. Schucking, and Y. Shang (2005) write about Tolman's paradox: can we increase the gravitational attraction of a spherically symmetric mass M by converting it to a spherically symmetric configuration where particles move swiftly, i.e., there is a lot of kinetic energy and a lot of pressure?

If this were possible, it would break Birkhoff's theorem.

We may consider a special case where an initially static M is converted into a spherical shell of particles which move at a large velocity v outward from the center of M.

                             ^

                             |

                             •

              <-- •                  • -->

          

                             •  dm

                             |

                             v

A reasonable version of Gauss's theorem for gravity would state that the average gravitational attraction of a particle dm over a spherical shell enclosing the entire system is

       γ dm,

where γ = 1 / sqrt(1  -  v² / c²). That theorem would make Birkhoff's theorem true.

Has anyone calculated the metric? Did it match the static Schwarzschild metric like Birkhoff's theorem claims?

Pressure does not require much energy. In principle, we could change the pressure p inside M vastly, for a short time, with only a minimal change in the composition of M.

It is believed that a positive pressure attracts test mass m. This makes sense: moving m closer to M stretches the radial distances relative to m, and expands the volume of M.

https://journals.aps.org/pr/abstract/10.1103/PhysRev.116.1045

Charles Misner calculated in 1959 that if we have a static configuration where the positive pressure is balanced with a negative pressure in the walls of the container, then the pressure does not have an effect on the attraction. The paper of Jürgen Ehlers et al. (2005) does the same, but also considers the internal effects of gravity inside M.

This is expected. But what happens in a dynamic case?

If gravity is Lorentz covariant, we should be able to derive the gravity of pressure by Lorentz transforming the Schwarzschild solution around each dm, and summing them, assuming that M is very lightweight and not very compact, and assuming that gravity is reasonably linear.

If Birkhoff's theorem is broken by such a setup, then either general relativity is not Lorentz covariant, or there is magical nonlinearity which miraculously restores the behavior to the one described by Birkhoff's theorem.

We already showed that general relativity does not understand accelerating masses M.

Could it be that the Schwarzschild solution around a spherically symmetric M changes if we accelerate the particles dm?

Then Birkhoff's theorem would remain true, but it would not accept any acceleration inside the system M. But gravity itself causes acceleration in the parts m. The theorem would never be applicable?

A dynamic system almost certainly breaks Birkhoff's theorem

                •

            /\/\/\/\  springs

        • /\/\/\/\/\/\ •

            \/\/\/\/

                • dm

                •  m

Consider a spherically symmetric system which is kept from collapsing with a positive pressure p in springs. Besides the mass-energy density ρ, it is commonly believed that the pressure p "generates" gravity and attracts a test mass m.

Let us then release the springs. The pressure is suddenly removed. We expect the attractive force on m to decrease.

Birkhoff's theorem seems to claim that the force must stay constant.

Note that we cannot use our true-and-tested trick of summing Lorentz transformations of Schwarzschild solutions, because the pressure field does not consist of small particles dm.

What are the assumptions in the proof of Birkhoff's theorem?

https://www.roma1.infn.it/teongrav/VALERIA/TEACHING/RELATIVITA_GENERALE/AA2015inpoi/birkhoff_theorem.pdf

The derivation of the theorem is quite simple. If we assume that the metric is spherically symmetric and only depends on t and r, then by coordinate transformations, and demanding two Ricci tensor components to be zero in the vacuum area, we see that the metric has to be time-independent.

The only possible time-independent metric is the Schwarzschild metric.

In the previous section we have an example where the metric seems to change with time. This would be a contradiction. How to resolve this?

The proof of Birkhoff's theorem seems to rest on at least the following assumptions:

1.   The Einstein field equations have a solution for at least some spherically symmetric configurations. If they would have none, the theorem would be empty of content. Then one could claim that "all solutions" satisfy whatever condition we want to define.

2.   The metric of a spherically symmetric configuration is spherically symmetric at all times. This condition is broken in the real world. If we start from a configuration which is not symmetric, then it would take an infinite time to even out all the asymmetry in the metric. Let us assume that the asymmetry in the far field is not significant.

3.   The stress-energy tensor outside the system is zero at all times. There is no mystical "longitudinal radiation" which could carry away mass-energy from the system. Neither is there some subtle asymmetry which could do the same trick.

Birkhoff's theorem in our own Minkowski & newtonian gravity model

We do not claim that there exists a "metric" which can capture the gravity interaction. We neither claim that there exists a "field" (like in electromagnetism) which can capture the collective interaction of many particles.

We calculate the interaction "privately" between any pair of particles.

Gravity must preserve the usual conservation laws. We are not sure how that is implemented technically, though. It is as if nature would perform "transactions" which always keep the grand total of energy and some other parameters constant.

Under these assumptions, Birkhoff's theorem may be true for some configurations. For example, if we use a spring to increase the radius of the central mass M very slowly, the force on a static test mass m outside M will probably stay almost exactly constant.

However, if we change the pressure inside M, as described in the previous sections, the gravity force on m will change. It is no problem for us if Birkhoff's theorem is broken in most cases.

Our model does allow kind of "longitudinal waves" in gravity, if we call a spherically symmetric change in the gravity attraction a wave.

We may imagine a general interaction between m and M as a kind of clockwork around M which grabs m, adds or reduces the inertia of m, and exerts a force on m. There is nothing which requires the interaction to satisfy Birkhoff's theorem. Rather, it would be a surprise if the complex clockwork would function in a way that Birkhoff wants!

Conclusions

We should calculate the gravity flux for a moving M through a spherical surface. The calculation is somewhat complex and we will defer it for now. We do not know if gravity satisfies Gauss's law. Birkhoff's theorem would imply such a law.

Tolman's paradox clashes with Birkhoff's theorem. We claimed that by manipulating the pressure inside M we can break Birkhoff's theorem. We will write a new blog post about what it means to general relativity if Birkhoff's law is broken.