Pure magnetic field B: for a wire, the momentum stored is "relative" to the protons

In this blog we have been wondering the relation between the energy and momentum stored in a magnetic field and the Poynting vector. We know that a magnetic field stores a large amount of energy and momentum in a coil. Breaking the circuit in a coil can cause an explosion and the coil can be ejected at a large speed.

Let us have a circular wire loop. Let us use some kind of a mechanical device to accelerate the electrons in the loop, so that a magnetic field is built. At any moment, the electric field outside the loop is negligible, but the magnetic field B stores a lot of energy.

The Poynting vector is defined as

       S  =  1 / μ₀  *  E × B.

If the electric field E is negligible, there is no energy flow, according to the Poynting vector. How does the energy then flow to the magnetic field B, if there is no energy flow?

                      protons

                 +++++++++++

                 -----------------------    --> v  electrons e-

                 +++++++++++

                      protons

            

We can make a wire loop by bending a straight wire and attaching the ends. Maybe the "correct" way to analyze the magnetic field of a wire loop is to study a straight segment of a wire? We assume that the electrons form a negatively charged cylinder which moves relative to a positively charged cylinder of protons.

The total charge of the conducting electrons in the wire is huge. When cylinder of electrons get displaced relative to the protons, even a little, the electric fields at the ends of the wire will have a large energy.

                            wire

                  +  ==========   -

    positive                            negative

    charge                              charge

When the electrons in the diagram slide to the right, a negative net charge appears at the right end of the wire and a positive net charge appears at the left end. The energy in these electric fields is considerable. Let us analyze how the energy flows to those electric fields.

The Poynting vector for two charges

The simplest configuration is the one where we move two single charges away from each other:

                         force F

              ●  <------------------>  ●

            +q                             -q

The force does work on the point charges q. How is this energy transported to the fields of the charges?

Let us assume that +q is static. The charge -q moves slowly to the right. The electric field E is large close to -q, and the movement creates a magnetic field B. The energy seems to flow from the point charge -q to all over the combined field of +q and -q.

Many charges in a wire: acceleration of electrons creates a net electric field outside the wire

Suppose that we have very many charges in a wire, and they are uniformly distributed. Can we accelerate the conducting electrons, and at the same time keep the electric field E negligible outside the wire, so that the Poynting vector S cannot carry the energy to the strengthening magnetic field B?

When we accelerate an electron, its electric field is distorted. The electric field of the protons probably cannot cancel the electric field of the electron in such a case. This allows the Poynting vector to carry the energy to the strengthening magnetic field B.

Is the energy of the magnetic field in the inertia and the movement of conducting electrons?

Since it requires a lot of energy to get the conducting electrons drifting in a wire, then, by definition, those electrons have large inertia. But what is the origin of this inertia, and where is the energy stored?

The electrons close to protons acquire inertia in the field of the protons. Could this extra inertia explain the momentum and the energy of the magnetic field? Once one has got the conducting electrons moving relative to the field of the protons, it is hard to stop the electrons since they have extra inertia inside the field of the protons.

We have a problem: the thermal movement of the electrons is very fast. Why the extra inertia does not slow them down considerably?

Inertia of a sliding charged tube

Let us use the following model: we have a cylinder of positive charge, and a tube of negative charge sliding along with the cylinder:

                   negative charge

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

          ++++++++++++++++++  positive charge

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

                            --> v

How do we explain the inertia of the sliding negative tube?

The tube cancels the electric field of the positive charge in a certain segment. The energy of the magnetic field of the tube is

       m v²,

where m c² is the energy of the electric fiekd canceled by the tube. That is, the magnetic field contains 2X the hypothetical "kinetic energy" of the electric field of the tube.

What is the inertia of the charges in the tube? Let the tube contain 6 * 10²³ electrons. The mass of the electrons is only 6 * 10⁻⁷ kg. Let the radius of the tube be 1 mm and the length 1 meter. The tube has a charge of 10⁵ coulombs. The electric field E close to the tube is something like

        E  =  9 * 10⁹ * 100 / (0.001)² V/m

            =  10¹⁸ V/m,

and the mass-energy density of the electric field is

        1/2 ε₀ E² / c²  =  5 * 10⁷ kg/m³.

The mass-energy density at a distance of 1 meter from the tube is still 50 kg/m³. We conclude that the inertia of the electrons themselves is negligible. Essentially all the inertia comes from the electric field of the tube.

In our December 17, 2023 blog post we present the assumption that the inertia of a test charge q inside the electric field of an opposite charge Q grows by |U| / c² where U is the potential between q and Q. The inertia of the tube is thus twice the mass-energy of the electric field of the tube. We have an explanation for the fact that the energy of the magnetic field of the tube is twice the kinetic energy of the electric field of the tube.

When the charged tube slides, is there energy shipping?

       electric field                electric field

             | | | |                             | | | |

           --------=============--------

             | | | |       tube --> v      | | | |

In our blog we have suggested that the "energy shipping" in the electric field is behind the increased inertia of the charged tube. 

       

              ________  electric field line?

            /

          v       ______

                /             \

               |                 |

                \_______/

                      --> v

                wire loop

But if we let the tube cover the entire cylinder and join the ends of the cylinder and the tube, then we essentially have a wire loop where a current is running (i.e., the charged tube is rotating around) – and there is no substantial electric field outside the tube, even though the magnetic field stored substantial momentum. But is there still significant energy shipping?

If the cylinder is still straight and we have not joined the end, then the magnetic field of the sliding tube acts together with the electric field, and the Poynting vector ships field energy from the right side of the tube to the left side.

Joining the ends of the cylinder and the tube alters the topology of the system. It is no longer a straight wire but a loop. Could it be that the loop topology somehow retains the energy shipping?

The electrons in the loop are in an accelerating motion. Their electric field cannot cancel the electric field of the protons exactly? Could it be that the Poynting vector of this small residual electric field and the magnetic field involves enough energy shipping, so that it explains the momentum held by the electrons?

The electric field lines are "lagging" the movement of the electrons. It is as if there would be a "vortex" of electric field lines running around the loop. However, loops of electric field are forbidden. What is the electric field like?

           ___________ S Poynting vector?

         /            • • •   B magnetic field

        v         ______  points out of the screen

                /             \

               |                 |     ----->  E electric field?

                \_______/

                e- •--> v

                wire loop 

The Poynting vector makes a circle around the loop if the residual electric field points radially from the center of the loop.

Let R be the distance from the center of the loop. The magnetic field and the electric field go like

       B  ~  1 / R³,

       E  ~  1 / R².

The area of a sphere goes as R² and the associated angular momentum goes as R. The integral of B × E for the whole space is over ~ 1 / R². Could this integral be large enough, so that it could store the energy and the angular momentum of the magnetic field?

The energy of the magnetic field is 2X the "kinetic energy" of the electric field of the the conducting electrons, and that electric field has a large "mass" as we calculated above. How can the Poynting vector store both the energy and the momentum of the magnetic field? Maybe it only needs to store the angular momentum? If the energy is located at a large radius R from the loop, then it can store a very large angular momentum.

A paradox: the Poynting vector cannot explain the momentum stored in a magnetic field? Yes it can

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

Above we have the spherical coordinates used in physics. The θ coordinate vector points obliquely down in the diagram.

https://physics-and-stuff.com/media/H7_1.pdf

Marvin Zanke (2019) calculated the magnetic field of a rotating, uniformly charged spherical shell. Note that the B vector in the upper hemisphere points "down" relative to the radial vector. The electric field E points almost precisely to the radial direction. Thus, the Poynting vector S outside the shell points consistently to the same direction with respect to φ. The Poynting vector describes a vortex of energy flow around the rotating shell.

Inside the shell, E is zero and the Poynting vector is zero.

If we superpose a static sphere of the opposite charge on the rotating sphere, we can make E almost exactly zero everywhere. The Poynting vector will have a much smaller value everywhere. The angular momentum L described by the Poynting vector dropped to almost zero.

But the magnetic field B did not change at all. It still contains the same energy as before. Can the angular momentum stored by B drop to almost zero?

Let us assume that the rotating shell is positively charged. We let a concentric, negatively charged shell shrink and descend down on the rotating shell. The magnetic field tries to pull the negative shell to rotate along. To keep B constant, we must add more angular momentum L to the rotating shell.

Thus, once we have neutralized the electric field of the rotating shell, the magnetic field B stores even more angular momentum L than before. At the same time, the Poynting vector S shrinks everywhere to almost zero.

Resolution of the paradox. We confused two things: angular momentum "relative" to the charge in the static sphere and the genuine angular momentum in Minkowski space. Let E be zero. The energy in the field B can apply a torque on the rotating sphere if it simultaneously applies the opposite torque on the static sphere.

Let us have two concentric rings rotating to opposite directions. There is a spring between them, such that it stores the kinetic energy of the rings and stops their motion. The spring stored "relative angular momentum" between the rings. It did not store genuine angular momentum. A pure magnetic field B without an associated field E can be compared to the the spring.

Does the resolution of the paradox work? Suppose that we inside a wire loop have tiny robots pushing the electrons to one direction. The feet of the robots give an angular momentum -L to Earth and their hands give an opposite angular momentum L to the system electrons & protons & magnetic field B. If the angular momentum L goes back to Earth through an interaction with the protons, then B does not need to store any angular momentum.

Let us have a current flowing in a wire loop. We try to reduce the magnetic field by slowing down the electrons. The changing magnetic field B tries to keep the electrons flowing through the induced electric E field from the changing magnetic field. The induced electric field applies an equal but opposite force on the protons. Thus, Earth does not receive any net torque from the magnetic field.

We conclude that there is no paradox. A pure magnetic field B without an electric field E stores energy but not momentum. It stores momentum relative to the protons in the wire, but this is not reflected in the Poynting vector.

Conclusions

There is no paradox with the Poynting vector. However, we want to analyze what actually stores the momentum relative to the protons. It is not the Poynting vector. We will continue the analysis in future blog posts.

Biot-Savart law from the inertia inside the electric field

Our derivation of the Biot-Savart law on November 14, 2023 was erroneous. Let us derive it again, now explicitly stating all the assumptions.

             dl wire segment

             proton(s)

             ------------ 

          v <-- • dq = e-

                   |\

                   |  \

                   |    \

                      α  = angle (r, negative y axis)

                              r = vector (dq, q)

   

                                             Vy     V

                                              ^    ^

                                              |   /

                                              | /

                                               • ---> Vx

                                           q = e-

         

        ^ y

        |

         -----> x

Assumption about the Coulomb force. The Coulomb force on q has to be calculated in the comoving frame of q.

The above assumption claims that the laboratory frame or a comoving frame of dq are "wrong" frames. The Coulomb force would be different in those frames because simultaneousness is different. When q is at a specific location (x, y) at a time t, simultaneousness determines where the electrons in the wire are.

Assumption about the extra inertia inside an electric field. The extra inertia of q relative to dq in a tangential movement relative to dq is

       |U| / c²,

where U is the potential of q in the field of dq, relative to infinity. If there is an opposite charge dq' ≤ dq "close to" dq, then the inertia of q is reduced by

       |U'| /  c²,

where U' is the potential lf q in the field of dq'.

In a radial movement, the extra inertia is double. However, for a "current in a wire" of charge, the doubling does not hold. The extra inertia is like in a tangential movement. See below.

Definition. A current in a wire is a stream of moving charges whose charge is essentially canceled by static charges close to them, when viewed in the laboratory frame.

In an ordinary metal wire where the drift velocity of electrons is only about v = 10⁻⁶ m/s, the protons essentially cancel the charge of the electrons when viewed in the laboratory frame. The length contraction factor γ differs from 1 only by

       ~  1/2 v² / c²  =  5 * 10⁻³⁰.

However, if an observer moves to the direction of v at a reasonable velocity V, he may see a significant net charge in the wire.

Assumption of extra inertia when there is a "current in a wire".  The distance to the charge has to be assumed constant, that is, the distance to the wire segment dl. However, the extra inertia which q "receives" or loses still moves at the drift velocity v of the charge carriers.

The current is viewed as a "flux" and the flux itself does not move, even though its charge carriers move.

Assumption about paradoxical momentum exchange. If q and dq have the same sign, then the momentum that q receives from the inertia inside the field dq is, paradoxically, opposite to what one would expect if the extra "package" of inertia is moving with dq.

The paradoxical momentum exchange is needed to make the negative test charge q to steer to the opposite direction relative to where the electrons in the wire move.

In the diagram, the wire element dl is parallel to the x axis, as well as the average drift velocity v of the electrons. The velocity of q is V.

We assume that

       |v|  <<  |V|  <<  c,

so that we can ignore terms of the form ~ v² and ~ V² v. The interesting terms are ~ V • v.

Question. Conducting electrons in a metal move at ~ 10⁶ m/s, while the drift velocity is only ~ 10⁻⁶ m/s. How does the almost relativistic speed of conducting electrons affect the Coulomb force and the inertia force?

The Coulomb force

The Coulomb force comes from the fact that length contraction in the comoving frame of q makes q to "see" a different amount of negative charge of electrons in the wire segment from the positive charge of protons. In the laboratory frame, the segment is essentially neutral. But in the comoving frame there is considerable length contraction.

The absolute value of the Coulomb force by dq is

       F  =  1 / (4 π ε₀)  *  dq q / r².

and the absolute value of the "residual" force, when the opposite force by the protons in the wire segment is subtracted, is approximately

       Fc  =  F Vₓ v / c².

Let us calculate the x and y components of Fc:

       Fcx  =   F sin(α)  *  Vₓ v / c²,

       Fcy  =  -F cos(α)  *  Vₓ v / c².

The "inertia force"

The test charge q approaches dl at a velocity

       Vr  =  cos(α) * Vy  -  sin(α) * Vₓ.

The inertia of q increases in a time t by

       W / c²  =  Vr t F / c².

The test charge q "picks up" a momentum which corresponds to an "inertia force"

       Fi  =  v Vr F / c².

The x and y components of the inertia force are

       Fix  =  v  *  (cos(α) * Vy  -   sin(α) * Vₓ)

                   * F / c²,

       Fiy  =  0.

The combined force on q is

       Fₓ'  =   Vy v  cos(α)  *  F / c²,

       Fy'  =  -Vₓ v  cos(α)  *  F / c².

The magnetic field calculated from the ordinary Biot-Savart law is

       B  =  1 / q  *  F v cos(α) / c².

The magnetic force calculated from B is:

       Fₓ''  =   Vy v cos(α)  *  F / c²,

       Fy''  =  -Vₓ v cos(α)  *  F / c²,

the same as we calculated as the sum of the Coulomb force and the inertia force.

Does the Coulomb force used above make sense?

If we would calculate the residual Coulomb force in the laboratory frame, or in the comoving frame of the electrons, the Coulomb force would be essentially zero. The equations of classical electromagnetism would have

       B  ~  v

       E  ~  v²,

where E is essentially zero in the setup

       |v|  <<  |V|  <<   c.

However, the magnetic field B would be significant.

Let us have a test charge q approaching a wire from a normal direction. The following method yields a wrong result:

1. Calculate the sideways impulse on the test charge q by the electrons, in the comoving frame of the electrons. This is essentially zero.

2. Add the sideways impulse by the protons in the laboratory frame. This is zero.

3. Conclude that q will approach the wire along a straight path.

In our example, one is not allowed to add the forces by the electrons and the protons linearly. In this sense, classical electromagnetism is not a linear theory.

Conclusions

We are able to derive the Biot-Savart law, but we have to make several ad hoc assumptions. A key question is where and how is the energy and the momentum of the magnetic field B stored if we have a current in a wire. Traditionally, we think that the electric field is zero. The Poynting vector is zero in that case.

Could the field energy and the momentum exist in the microscopic fields between electrons and protons?

Another explanation: if we cut the wire loop and let two cylinders of opposite charge slide past each other, there is a huge energy flux from the kinetic energy of the cylinders to form the electric field at the ends of the system. Could this energy flux somehow exist "hidden" in a current loop?

Edward M. Purcell's derivation of the magnetic field from the Coulomb field

In this blog we have been claiming that one cannot derive the magnetic field from the Coulomb field using special relativity alone. But Edward M. Purcell seemingly did derive the magnetic field in his classic book Electricity and magnetism (1965):

https://en.wikipedia.org/wiki/Electricity_and_Magnetism_(book)

What is going on?

https://physics.weber.edu/schroeder/mrr/mrrtalk.html

Daniel V. Schroeder (1999) presents the argument of Purcell. The key idea is to do the analysis in the comoving frame of the test charge.

                        ---->  v 

     

Purcell argues that the electric field of a moving charge is "squeezed" in the direction of its velocity v.

Above we have a wire where positive charges carry the electric current. The positive test charge approaches the wire directly from a normal direction, in the laboratory frame (upper picture).

In the comoving frame of the test charge, the positive carriers of electricity move obliquely down and slightly to the right. The deformation of their electric field produces a net force to the left, a "magnetic" force (lower picture).

              positive carriers

                 +   +   +   +   +

                           ^   V

                           |

                           |

                   v <-- • + test charge

                           | 

                           v   F

But something is wrong. In the comoving frame of the positive charge carriers, the test charge sees a force F which pushes it directly downward. The velocity of the test charge upward is slowed down. If the inertia of the test charge would decline with its kinetic energy, then the test charge would move faster than v to the left.

But the momentum p of the test charge in the horizontal direction does not change. And since the negative charges in the wire cancel the repulsion of the positive charges, the kinetic energy of the test charge does not actually change. Thus, there should be no acceleration in the horizontal direction.

The model of Edward M. Purcell probably fails to recognize that there is no change in the kinetic energy of the test charge. If there were, then we would get the right horizontal acceleration to match the calculated magnetic field B.

Conclusions

The model of Edward M. Purcell can be used to derive the magnetic field B from the Coulomb force, but then we must ignore the effect of the opposite charges on the test charge  –  and there is no obvious reason why we would be allowed to do so.

Big Bang and Penrose-Hawking singularity theorems

Roger Penrose proved in 1965 that if in a collapse there forms a "trapped surface" then general relativity develops the system toward a "singularity". A weakness of the result is that no one has been able to prove mathematically that a realistic collapse of a star produces a trapped surface.

A singularity at the start of the Big Bang?

What about the hypothetical singularity in the Big Bang? Do we know that there has to be a singularity in out past?

https://royalsocietypublishing.org/doi/pdf/10.1098/rspa.1970.0021

In the link we have the 1970 paper by Hawking and Penrose about singularity theorems.

Let us assume that the universe at a very large scale is close to Minkowski space. The universe has an infinite spatial volume. Let us reverse time and develop the universe backward in time. Does our local observable universe collapse into a singularity?

The density of visible matter and dark matter seems to be so low that we are not yet inside a trapped surface. Thus, the problem is qualitatively equivalent to a collapse of a star. It looks likely that the collapse leads to a trapped surface, but there is no mathematical proof.

Hawking and Penrose also prove that if the universe contains a spacelike compact hypersurface, then a singularity must exist in our past. Such a surface means that the "current" spatial volume of the universe is finite. This assumption is close to assuming the existence of a trapped surface, because for someone living inside a trapped surface, the volume of space that he can reach is finite.

The Milne model

In this blog we have suggested that the Milne model describes the observable universe. That is, the amount of dark matter whose gravity charge is negative, exactly cancels the positive gravity charge of other matter.

Then there is no need for a singularity to exist in the history of observable universe. We may live inside a "Milne explosion" cloud in a universe which is essentially Minkowski. Note that our Milne model breaks the weak energy condition which Hawking and Penrose assume, since we assume the existence of negative gravity charges.

Conclusions

The existence of a singularity in our past is a problem quite similar to the existence of a singularity in a star collapse. There is no mathematical proof in general relativity of the existence of a singularity for a realistic case where matter is not uniformly distributed.

Birkhoff's theorem and Oppenheimer-Snyder

We wrote on November 6, 2023 that Gauss's law for gravity explains why Robert Oppenheimer and Hartland Snyder were able to find a solution for the Einstein field equations in 1939:

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

Our claim was too simple. Gravity has nonlinear effects. Even if a single particle moving at a constant velocity would satisfy Gauss's law, that does not prove that the law would hold for a collapsing dust ball.

The pressure term in the stress-energy tensor of the dust ball grows as it contracts. The stress-energy tensor T looks schematically like this for a single dust particle.

           m           m v             ...

           m v        m v² / c²

       

            ...

If the momentum components m v were zero, then we would have a "pure" pressure besides the mass density. The interior Schwarzschild solution (1916),

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

as well the result by Charles W. Misner and Peter Putnam (1955),

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

and the paper by Jürgen Ehlers et al. (2005) show that pressure "gravitates": a positive pressure tends to pull a test mass to the center of a pressurized spherical vessel. The force comes from the fact that the metric of time is slower at the center. Pressure "focuses" a cube of test masses.

But these three results are for a "pure pressure" with no momentum components in the stress-energy tensor.

The bounce-back mechanism of how pressure gravitates?

                          

                   •             |         • -->        <-- •      |

                  m        wall    dm               dm wall

Consider the above configuration where particles dm bounce between two walls back and forth. We believe that dm "holds" some inertia dI of m, and when dm bounces back from the wall close to the test mass m, then m receives an impulse

       2 v dI

to the right. When dm bounces from the wall far from m, then m receives a smaller impulse

       2 v dI'

to the left.

Conjecture. If the pressure term in the stress-energy tensor T comes from a locally uniform mass flow, where all the matter flows at a certain velocity v and does not "bounce back", then the pressure term does not "gravitate". That is, we cannot increase the gravity of a system by making its parts to move in a uniform fashion.

The conjecture prohibits Tolman's paradox in such cases.

How "uniform" the flow has to be? Is it enough that each particle is moving uniformly with itself? In the early stages of the Oppenheimer-Snyder collapse, the movement is extremely uniform, such that the dust cloud contracts through uniform contraction of all distances. If we have a single particle moving at a constant velocity, that is very uniform, too.

Corollary. The Oppenheimer-Snyder collapse does not clash with Birkhoff's theorem. The gravity, measured from outside the collapsing dust ball, stays constant.

However, just before the collapse, a positive "pure" pressure is removed from the dust ball, allowing the collapse to start. We believe that this pressure change reduces the gravity outside the dust ball, and does clash with Birkhoff's theorem.

Could we remove the concept of pressure altogether from a model of gravity? It is not possible to model negative pressures with an ideal gas of particles bouncing around. We need a concept of pressure in a theory of gravity.

Does the bounce-back always involve an impulse which COMPENSATES the gravity of pressure? No, this model does not work

In the diagram above, when the particle dm bounces from a wall, the wall bounces from the particle. The wall "holds" some inertia of the test mass m. Thus, the wall gives an opposite impulse

       -2 v dI

to m.

Is it so that the impulse always compensates the "gravity" of pressure? Then pressure would not gravitate at all if all pressure really "consists" of particles bouncing back and forth!

That would solve Tolman's paradox and preserve Birkhoff's theorem.

This may hold for a test mass m which is static relative to the pressurized system. But what if m moves closer? We believe that m stretches the spatial metric of the pressurized system and "frees" pressure energy. We would expect m to be attracted toward the pressurized system.

Also, the results of Schwarzschild, Misner, Putnam, Ehlers, et al. show that "pure pressure" gravitates in general relativity. The bounce-back mechanism cannot describe the pressure of general relativity.

The bounce-back is a different thing from pressure. Note that if we have an electric test charge q which shares some inertia with another charge Q, and Q bounces from an electrically neutral wall, then the bounce does give an impulse to q.

The role of "privacy"

If we have a single particle moving at a constant velocity v, we believe that Gauss's law holds for its gravity. If we would be allowed to linearly sum the metric perturbations caused by such particles, then there would be no gravity effect of pressure at all. Thus, pressure must produce gravity through a nonlinear, or "public" effect.

Our derivation of the Biot-Savart law for electromagnetism on November 14, 2023 required "privacy" with respect to a charge which has the opposite sign. Could it be that privacy does not apply when the charges have the same sign?

Gravity by a pressure is a "reaction" in the pressurized matter?

If the test mass m approaches pressurized gas, m stretches the spatial metric and the pressure is lowered. This frees energy which is used to accelerate m toward the gas. This is a "reaction" which m causes on the gas.

But if we calculate the interaction of m and a gas particle privately, we assume that m causes no reaction on the gas particle whatsoever. This is an incorrect assumption? If the reaction is

       ~ m,

then we cannot ignore it even if we set m very small.

We still face the problem why the pressure term in the Oppenheimer-Snyder collapse does not seem to cause gravity.

Varying the metric in the presence of a mass-flow element in the stress-energy tensor

If Gauss's law holds for a moving particle in gravity, then there must be some reason why the Einstein-Hilbert action "resists" the attraction that the pressure element in the stress-energy tensor causes on a test mass m.

         •                      <-- •

        m                     v   dm

     -----> x

If we move m closer to the moving particle dm, then the stretching of the radial metric increases the kinetic energy of dm. If the matter lagrangian is of the form

      kinetic energy  -  potential energy,

the effect of increasing kinetic energy is like the effect of decreasing potential energy: the system tends to move to that direction. If the system would contain ordinary pressure, then moving m closer would reduce the potential energy in the pressure, because of the stretching. An equivalent effect comes from a kinetic energy term: the stretching increases the kinetic energy.

But is there an effect from moving m closer, such that it somehow reduces the kinetic energy of a moving particle or particles?

In the diagram, the stress-energy tensor of the particle looks something like this:

         dm       -dm v     ...

       -dm v      dm v²

         ...

The metric around m does not contain cross terms between dt and dx if m is static, but it does contain them if m moves. If we move m closer to dm, does that counteract the attraction that the pressure element dm v² causes on m?

If yes, that might explain why in the Oppenheimer-Snyder collapse, the pressure does not increase the gravity. 

The metric around a moving particle is complicated

 

                                      • m test mass 2

        •                            ● --> v

      m test mass 1      M particle

Consider the following configuration. Initially the particle M is static. The test masses m are accelerated uniformly toward M, and from their acceleration we can deduce that Gauss's law for gravity holds in this case.

Let us then use some of the mass-energy of M to give it kinetic energy and M starts to move right in the diagram. We believe that Gauss's law still holds: the "flux" of the lines of force of gravity through a surface is still equivalent to the mass-energy enclosed inside the surface. We ignore the steepening of strong gravity fields here.

The acceleration of the test mass 1 becomes smaller and the acceleration of the test mass 2 becomes larger. We can easily calculate this behavior in the comoving frame of M, but we have hard time finding an intuitive explanation for the behavior in the laboratory frame. How does the momentum flow component in the stress-energy tensor cause the gravity field to become flattened like this?

Maybe we simply have to accept this as a fact without a further intuitive explanation: the momentum components in the stress-energy tensor "cancel" the gravity of the pressure component.

Momentum means that the gravitating system is moving, and that we can in many cases cancel the pressure component by switching to a comoving frame of the gravitating system. Then the pressure does not seem to cause extra gravity in any frame.

If there is a "pure" pressure, then we cannot cancel the pressure component in any frame. In this case, the pressure seems to cause extra gravity.

The metric around "pure" pressure

Let us have a volume of "pure" pressure p where the momentum flow is zero. The stress-energy tensor looks schematically like this:

          M       0        ...

          0        p

          ...

The metric is such that both the mass-energy M and the positive pressure p tend to slow down the metric of time at the center of the system. That is, the pressure p "gravitates". The pressure "focuses" a cube of test masses so that the volume of the cube starts to shrink.

On October 28, 2023 we argued that the focusing effect inside a pressurized volume must necessarily cause gravity attraction also outside that volume. To cancel the attraction, we would need an equivalent negative pressure wrapped around the volume. But in the preceding section we noted that also momentum components can cancel the extra gravity of pressure.

The problem of particle granularity. Suppose that a pure pressure is caused by microscopic particles, like in an ideal gas. Can we do the following: switch to the comoving frame for each particle, calculate the gravity, and then linearly sum the gravity effects? Then the pure pressure would not cause extra gravity, since for each individual particle, it is not pure pressure. It is just a moving particle.

If the granularity would significantly affect the gravity of matter, then general relativity would be a very badly behaving theory. We expect granularity at a fine level not to have much effect on the gravity field. Proving this property for general relativity is very hard, though, since it is a nonlinear theory.

Conjecture. Fine granularity does not affect the gravity field of a system much in general relativity. One cannot reduce pure pressure into a sum of moving particles.

Why the Oppenheimer-Snyder collapse maybe can be reduced to moving particles?

The pressure component in the Oppenheimer-Snyder collapse comes from a very uniform movement of mass. It is not a fine granularity thing. We can cancel the pressure component by switching to a comoving frame of the dust particles. This may be the reason why the pressure in this case does not seem to cause extra gravity.

Conclusions

Pure pressure certainly produces extra gravity attraction in general relativity. This is proved by the results Schwarzschild and others. But why the pressure component in the Oppenheimer-Snyder collapse does not cause extra gravity?

The reason may be that we can cancel the pressure term by switching to the comoving frame of the dust particles. However, we were not able to prove that this is the reason. It is just a guess.