Retardation weakens gravity in an expanding dust ball - retardation explains why the universe is not "frozen"?

We have been struggling to understand what retardation exactly does in a contracting or expanding uniform dust ball which is very large, compared to the "observable universe" of an observer.

In the following, we use a (fuzzy) Minkowski & newtonian gravity model. The speed of light c is the maximum speed of a signal. The gravity force in a static case is as given by Isaac Newton.

Two particles which pull on each other

Let us look at an expansion which is slowing down because of gravity.

          particle 1                           particle 2

      v <---  • --> a                           a <-- • ---> v

                m                   s                     m

  t₀ time                    distance

Let us look the process in global "laboratory" coordinates. We have two dust particles flying away from each other. The global time coordinate is t₀.

Gravity decelerates the particles. Because of retardation, the particles "see" each other farther away than they are "currently" in the global coordinates. The gravity is weaker than in the naive model where we would just look at the current distance of the particles.

Also, the gravity field of each particle is squeezed horizontally in the diagram, because of length contraction (Lorentz correction).

Let us calculate how much retardation and the squeezing affects the force of gravity between the two particles. Let us first assume that v << c. Then Lorentz corrections are negligible.

The particle 1 sees the particle 2 as it was the time

        t  =  s / c

earlier. At that time, the particle 2 was receding faster from the particle 1. The particle 1 extrapolates the location of 2 from the velocity of 2 at that earlier time, and sees the gravity field as if 2 were at that extrapolated location.

But the particle is actually "currently" at a location

       1/2 a t²  =  1/2 a s² / c²

closer. The relative error in the distance is

       1/2 a s / c²,

and the relative error in the gravity force is

       a s / c².

The force is weaker by that factor than the naively calculated on, using "current" distances.

We conclude that the naive gravity force is replaced with one of the form:

      F  =  G m² / s²  -  G m² a / (s c²).

The new force formula affects the expansion of the dust ball in a complicated way?

Effect on an expanding dust ball which is largish but not huge

       

                          largish dust ball A

                                 radius R

           ______

        /             \

       |       •  m   |          ×    center of A

        \_______/

              B 

     environment B                         • m'

     of dust particle m       dust particle in A

                                            s = distance(m, m')

We have a largish expanding dust ball A and an arbitrary dust particle m inside it. Let B be the largest spherical environment of m, which fits inside A.

We assume that the expansion of A is approximately uniform. Because of symmetry, the net force imposed on m by B is zero. Also, because of symmetry, the retardation of the gravity fields of particles within B has no net effect on m.

The gravity of the set difference

       A  -  B 

does pull on m, and the retardation has an effect.

If m is close to the center of A, then all the particles m' in A - B are roughly at the same distance 

       s  ≈  R

from m. The effect of retardation is roughly the same for all m'!

We conclude that the effect of retardation is fairly uniform for any m close to the center.

                           

                       /  A

                     /

                   |O m 

                     \ B              • m'

                       \               Ω

            O = environment B of m

Let then m be relatively close to the edge of A. Particles in A - B pull on m.

Let us look at a narrow solid angle Ω whose tip is at m. Most of the pull on m comes from angles which very roughly point to the right from m. The length of the cone determined by Ω from the edge of B to the edge of A is typically

       2 s  ≈  1.5 R.

The average distance of the particles m' in the cone, weighted by the gravity of m' on m, is

       s  ≈ 0.75 R.

We conclude that the effect of retardation is fairly uniform throughout A, with differences of at most +-15% !

If the dust ball is so huge that the "observable universe" is only a tiny part of it

If the test particle m last saw m' receding at almost the speed of light, then m' had a lot of kinetic energy in the frame of m. Also the field of m' is strongly flattened in a way that the force on m is smaller. Which effect wins?

Let us use the analogy from the electric field. The distance in the frame of m' is

       ~  1 / sqrt(1  -  v² / c²),

and the electric field E strength would be

       ~  1  -  v² / c².

The "gravity charge" is

       ~  1 / sqrt(1  -  v² / c²).

This suggests that we can ignore masses m' which are receding very fast from m?

Retardation interpreted in a naive way would break time symmetry of physics

         o

        /|\                       <---  /\/\/\/\

         /\                             

          A observer               W wave packet

Imagine that a wave packet W approaches an observer A at the speed of light. The observer A is not aware of the gravity field of the packet W because he has not yet received any information about the approaching packet W.

Then the packet W hits the observer A, and is absorbed by his body.

If we reverse time, then in the process, the observer A is aware of the wave packet W that was emitted by his body. Does A feel the gravity field of W now?

We believe that physics is time symmetric. The observer A cannot feel the gravity of W. The gravity field of the wave packet simply cannot extend directly to the front or the back of the packet.

Retardation solves the mystery of why the universe is not frozen, even though it is inside its Schwarzschild radius?

We work in Minkowski space.

Let us make a photon shell to collapse, to form a black hole. The information about the approaching photon shell has not yet reached an observer A inside the shell. Consequently, physics at the location of A will function exactly like before. The proper time of A runs just like the global Minkowski time. There is no "freezing" at A.

The universe looks a lot like this scheme, the time reversed. The matter outside our observable universe is moving away from us very fast.

The universe is not a frozen black hole because no observer has yet received information that the universe is inside its Schwarzschild radius.

If we believe Gauss's law for gravity, then the gravity field of the universe must be immense at its outskirts. The universe would definitely be enclosed inside a black hole. Gauss's law does not hold in this case?

Let us look at a collapse of a very sparse and very large photon cloud. Initially, the cloud is much larger than its Schwarzschild radius. Can any observer know that enough matter at some later time is inside its Schwarzschild radius?

Yes. If the collapse ends into a very compact state, an observer outside the photon cloud will see a very massive compact object. It is a black hole.

However, during the collapse process, no observer inside the photon cloud maybe is aware of this? Then the collapse can proceed without freezing.

Collapse of a photon shell: no gravity felt by photons at all – the end result contains no singularity

We can imagine that the gravity field of a photon is an infinitely thin plane, normal to the velocity vector of the photon. Length contraction has made the field absolutely flat. 

In a collapsing spherical photon shell, no photon feels the gravity field of the other photons. The photons can move as if there would be no gravity at all!

An observer outside the photon shell does feel gravity. He may even see that the shell has collapsed into a black hole.

What is the end result of the collapse? As the photons arrive at the center, thy will collide and produce electron-positron pairs. That is, the photons are converted into massive particles. Since massive particles move slowly, the information about the gravity field of the other nearby particles will reach them. A very strong gravity field slows down everything to an enormous degree. The soup of electrons and positrons will "freeze". No singularity is formed. It is just a very dense, essentially frozen soup of particles.

We are able to avoid the formation of a singularity in this model. The end result is a black hole, which has a frozen soup of particles at the center. The Schwarzschild radius may be large, while the radius of the soup is very small.

If we drop an infinitesimal test mass into this black hole, the test mass will freeze at the horizon? The test mass is aware of the huge gravity field of the soup at the center, or is it?

Let us assume that the photon shell was produced by large lasers in a shell structure far away. As the photons converged into the center, their "flat plane" gravity fields were felt by observers outside the shell? Did the observers have time to know that a photon flew by?

Paradox of the gravity field of a photon

        laser                                   photon

         ===                                       • ---> c

                                                      o

                                                     /|\

                                                      /\

                                               observer

          

There is a paradox: if the observer is directly below the photon in the diagram, he cannot know that the photon exists. He cannot feel the gravity of the photon. But if he is not directly below, the flattening of the gravity field of the photon prevents the observer from feeling the gravity!

Is it so that a photon does not have a gravity field at all? Could this enable a perpetuum mobile?

The solution to the paradox might be this: before some mass m was converted into a photon in the laser, that mass m did possess a gravity field. Maybe this old gravity field gets updated as the photon flies? In that case, the gravity field of a photon is not a flat plane at all, but is mostly the remnant of the old field of m.

Collapse of a photon shell. Let us look again at the collapsing photon shell. If the photons were created from various masses m, then the old fields of the m's get gradually updated in way that the field "knows" the new location of the mass-energy. The fields will eventually know that the mass-energy is now in the electron-positron soup at the center. Thus, an event horizon forms.

Weakening of gravity comes both from deceleration and flattening of the gravity field from velocity

In the case of an expanding universe or a collapsing dust ball, both retardation from the deceleration, and the flattening of the gravity field, reduce the gravity felt by a test mass inside the system.

Friedmann equations and retardation in Minkowski & newtonian gravity

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

In the absence of pressure, and a zero cosmological constant Λ, the second Friedmann equation is exactly analogous to a classical newtonian collapse or expansion (classical = the year 1687 version of newtonian mechanics).

Furthermore, we know that the Friedmann equations explain well the observed baryonic acoustic oscillation (sound horizon) phenomena.

Let us interpret this in a collapsing dust ball model. In the classical newtonian version,

1.   we can use Newton's shell theorem at the center of the dust ball, and ignore any dust outside the sphere that we are calculating,

or

2.   we can use the trick in the second section of this blog post: we look at the gravity of the set difference A - B. We can ignore the gravity of B.

The first is a "local" way to calculate, the second a "global" way. These should yield the same results. Do they yield the same results for retardation, too? If yes, then the shell theorem would hold also for retardation.

However, the formula that we derived in the first section says that the relative retardation correction (in a Minkowski & newtonian model) to the gravity force is

       ~  a s / c²

       ~  s².

The relative correction is much smaller in the "local" calculation alternative 1.

We conclude that retardation phenomena gives us information about the entire collapsing dust ball. In cosmology, this means that we get information of the universe outside the observable universe. Dark energy might give us a clue about how large is the entire universe, if the expanding dust ball is defined as "the universe".

Eliminating coupling between particles through retardation: making the particles free

Our example of the collapsing photon shell is an example of a system, where we have been able to eliminate a very strong coupling between particles through retardation.

However, energy has to be conserved. There has to be a mechanism which keeps track of the energy of the system and prevents perpetuum mobiles.

Paradox in the flattening of the gravity field

          •                                    <--- ● 

         m test mass           v ≈ c    M neutron star

We calculated in a preceding section that the gravity field of an object moving almost at the speed of light is very weak to the direction of the movement. In the diagram we would expect the gravity force of M on m to be weak.

But the field of m does pull M with a very strong force, if we look at the field of m. This is a paradox?

We once again encounter the problem of momentum conservation in a force field.

Calculating the effect of retardation and flattening for a huge dust ball

Let us try to calculate the two effects for a dust ball whose edges are expanding almost at the speed of light, and whose density is close to the "critical density".

The role of the gravity potential is unclear. How to handle it? Clocks tick slower close to the center of the dust ball?

Also, how to handle the stretching of the radial spatial metric?

We believe that slow clocks and length-contracted rulers come from a strong gravity field. If an observer does not yet have the information of that he is surrounded by huge masses that are quite close, then we, maybe, can ignore the change in the metric? It would be similar to the case of the collapsing photon shell.

    edge of dust cloud          edge of dust cloud

           • --->          •                              <--- •

    m'  v ≈ c         m test mass           v ≈ c   m''

The expansion of the universe looks uniform and classical newtonian. The particles at the edge of the cloud have gained a very large kinetic energy. The total energy of the cloud has not increased, though. Energy of the gravity field was converted into kinetic energy of particles, especially at the edges of the cloud.

Gravity is pulling a test mass m. Where is the mass-energy of the cloud located?

Hypothesis of mass distribution. The energy of the dust cloud is still uniformly distributed among the dust particles, regardless of the large kinetic energy of the edges of the cloud.

Hypothesis of retardation. If the particles in the cloud are moving at a constant velocity, then the gravity felt by the test mass m pulls it toward the current position of each dust particle m'. "Current" here means a global laboratory time t₀. The field of m' "knows" where m' is located at the current moment.

If the particle m' is in an accelerating motion, then gravity pulls m toward the "calculated" position of m', where the position is calculated based on the last information m can have about the velocity of m'. This is the traditional retardation hypothesis.

                            |   v ≈ c

                            v 

                   ___________

                /                      \

              /                          \

            |              ×                |

              \                          /

                \____________/

                 almost all the mass at the edge

Exponentially dense outer shells. Let us try to describe the dust ball in the frame of the center. If a particle at a distance r is approaching at a velocity 0.9 c, then a particle at a distance approaches at 0.99 c, and so on. Length contraction makes the shells at radii

       n r ... (n + 1) r

exponentially thinner. At the center is a dust ball where the velocities are nonrelativistic. Let its radius be R. The entire huge dust ball has a radius which is only a few times R, say, 3 R.

Almost all the huge mass is close to the radius 3 R, and approaching, say, at a velocity (1 - 0.1¹⁰⁰) c.

We should find a reason why this system roughly satisfies Newton's shell theorem, but not entirely, because of retardation.

      A                r = distance(A, B)                    B

       •                •                •                •               •

  <- 0.99 c    <- 0.9 c                   0.9 c ->   0.99 c ->

The diagram above is drawn in comoving coordinates of dust particles •, but the velocities are in the frame of the mid particle. Is there some reason why we should claim that the gravity force which A exerts on B is

       F  =  G m * m / r²,

where m is the (rest) mass of each particle, and r is their distance in the comoving coordinates?

If the velocities were slow, then it would be the newtonian gravity formula. But now we are dealing with velocities ≈ c, and extreme length contraction in the frame of the mid particle.

Two particles once again

                                    <--- F

                 ============ ruler --> v

                • -> a                  • --> v

               m                        m'

                 ============ ruler

                              r

Let us assume that the acceleration of m is

       a  =  G m' / r²  *  γ,                                       (1)

where

       γ  =  1 / sqrt(1  -  v² / c²).

Let us Lorentz transform a to the comoving frame of m'. Do we obtain consistent results?

https://en.wikipedia.org/wiki/Acceleration_(special_relativity)

In this case, ux = 0, and we have denoted γv by plain γ. We have

       a'  =  G m' / r²  *  1 / γ².                               (2)

In the comoving frame of the m', m is at a (moving) ruler position r γ. The result would be consistent if m would not be moving fast, at the velocity -v in the moving frame. But if v is large, the acceleration is much less. Our assumption was wrong.

Let us assume that the gravity force on m in the moving frame is F'. We calculate the acceleration a' of m in the moving frame. It depends on the inertia of m.

If m would be flying in empty space, the momentum would be

        p  =  v m / sqrt(1  -  v² / c²).

(But m is inside the gravity field of m'. The inertia of m may be different.) Let us calculate the "inertial mass":

       dp / dv  =  m / sqrt(1  -  v² / c²)

                          +  v m  

                              * -1/2  (1  -  v² / c²)^-3/2

                              * -2 v / c²

                      =   m / sqrt(1  -  v² / c²)

                           * (1 + (v² / c²) / (1  -  v² / c²)).

For v² / c² << 1, we can ignore the second summand above, and

       dp / dv  =  γ m.

Then the formula (1) above is consistent with (2). The "inertial mass" is the same as the gravitating mass.

But we are interested in cases where v² / c² ≈ 1. For v² / c² ≈ 1, we have:

       dp / dv  ≈  γ³ m.

The "inertial mass" is larger than the gravitating mass, which is only γ m.

Let us assume that in this case, the "active gravitating mass" (the active mass pulls other masses) for a very fast particle is m' / γ:

        a  =  G m' / r²  *  1 / γ,

        a'  =  G m' / r²  *  1 / γ⁴.

In the moving frame, the distance is γ r, and the "inertial mass" of m is γ² its "passive gravitational mass" (passive mass pulls m toward other masses). We get the same value for a'. This agrees with our calculation in an earlier section where we used the analogy between gravity and an electric field E. However, we have so far ignored the fact that m and m' are immersed in a gravity field. That may change the inertial masses.

***  WORK IN PROGRESS  ***

A large uniform universe necessarily has the spatial metric flat in Minkowski-newtonian model

UPDATE January 12, 2025: The flatness problem really is not about the flat spatial metric in the universe, but the question: why the mass density of the universe is quite close to the "critical density"?

The critical density means that the velocity of a galaxy at a distance R from us is close to the escape velocity from the mass M contained within the distance R from us.

For example, the universe might have a mass density which is only 1/100 of the current one. Then the velocity of a distant galaxy would be 10X the escape velocity.

Our Minkowski & newtonian model does not explain why the velocity is close to the escape velocity.

A possible explanation: a bounce-back model in which an initially almost static cloud of dust collapses, and then bounces back through some unknown physical mechanism. Then the dust cloud has roughly the critical density.

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

----

In the Friedmann equations, the mass density of the universe has to be set very carefully to the "critical density", in order to ensure that the spatial metric is flat, and will stay flat for a long time. We know that the spatial metric in the observable universe is roughly flat on the large scale.

In our own, Minkowski-newtonian gravity model, the spatial metric is determined by a different inertia of a test mass m to different directions. For example, around a neutron star, the radial metric is stretched because the inertia of a test mass is larger in the radial direction: there is "energy shipping" to the test mass m if it moves radially, which adds extra inertia. The spatial metric bulges at the neutron star, and is not flat.

If the universe is spatially very large (much larger than 15 billion light-years), and almost uniform, then the inertia of a test mass m cannot vary much in any direction, within the observable universe whose radius is only 15 billion light-years. This implies that the spatial metric is almost flat.

We do not need any fine-tuning of the mass density of the universe to a "critical" value. It is enough to demand that the universe is large and almost uniform.

This observation solves the flatness problem of cosmology. We do not need to assume inflation to fine-tune the mass density to the critical one at the beginning.

It may also explain why the expansion looks similar to all directions. If we are dealing with a huge expanding dust ball, the expansion may locally look rather uniform. The uniformity of the CMB would be explained if the dust ball is really large.

Still, we have to assume uniformity of the mass density in a very large dust ball. Why did the uniformity arise?

Also, we do not understand how Minkowski-newtonian handles a dust ball which is much inside its own Schwarzschild radius. Why is it not frozen, like a black hole is?

Retardation in a newtonian explosion of a dust ball

Let us look at a very simple example, to gain understanding of retardation effects in a collapse, or an expansion.

We assume that a uniform ball consisting of dust particles was initially static, and had a very large radius. Then we let it collapse. It is well known that it will stay uniform in newtonian mechanics.

We study the time-reversed process: the dust ball is expanding.

                                                       ^  expansion

                                                       |

       large cloud of dust

             environment of m

                     ______

                   /             \

                 |    • m         |                ×           • m'

                   \_______/                   center

                        R = distance (m, ×)

                       m = test particle

                         r = distance (m', ×)

                                                       |  expansion

                                                       v

     Current laboratory

     time t = 0.

The diagram is an extremely crude description of the process.

In the diagram, the dust ball is large and fills the entire diagram. It is expanding uniformly. Gravity is decelerating the expansion. We have marked a close environment of the test particle m. In that environment m knows the current location of dust particles quite well (current in the laboratory time coordinate).

For the bulk of the dust ball, the gravity field is retarded at m: the field is as if the dust particles would have moved at a constant velocity since m last "saw" them. The field is not aware of the fact that currently (in the laboratory time), the dust particles are located just like in the close environment of m, symmetrically around the center ×.

Let us have a dust shell S whose radius

       r  <  R

at the current laboratory time. Let m' be a dust particle in the shell S.

Let

       s  =  distance(m , m').

The speed of m' at a laboratory time

       t(s)  =  s / c

earlier was larger than now. The deceleration d(t) of the expansion of the dust ball is

       d(t)  =  C M / a(t)²,

where C is a constant, M is the mass of the ball, and a is the scale factor of the ball. We may set a(0) = 1.

On the line going through m and ×, we can calculate an approximate distortion the location of m' between

1.   the true current location in the laboratory time, and

2.   the retarded location of m', which determines the field of m' at m.

The current position of m' differs by 

       ~  -1/2 d(t) r t(s)²

       =  -1/2 d(t) r s² / c²

from the retarded position. There r is negative if m' is between m and ×.

Let us have r fixed. The difference in the position of m' is

       ~  s²,

and its effect on the gravity force of m' on m is at most

       ~  s²  *  2 / s

       ~  s.

                                                 S

                            "cap"                         "cap"

         •   Ω               |      r        ×       r        |

        m                   A           center            B

              R = distance (m, ×)

Let us look at a narrow solid angle Ω drawn from m, such that the angle intersects S. If we draw a straight line through S, then by symmetry, the intersection angle between the line and S is the same at both intersections. The intersection area thus is

       ~  s².

The gravity force of the intersection area is

       ~  s² / s²  =  1.

That is, the intersection "cap" A close to m has the same gravity force as the intersection "cap" B far from m.

Let us then correct the gravities of A and B for retardation. The correction increases the gravity of A, and reduces the gravity of B.

The correction is ~ s, which means that the correction to B wins: the gravity is smaller in the correct, retarded view than in the naive laboratory view.

We can now easily calculate order-of-magnitude estimates for the retardation correction to gravity.

The dependency of retardation correction on R

Does the correction still keep the expansion of the dust ball uniform?

Let us double R, r, and s. Then the acceleration of m doubles. What about the correction?

The correction to the position of m' becomes 8-fold. It must be divided by s to get the relative correction to gravity: the relative correction is 4-fold. We conclude that the relative effect of the correction on gravity is much larger with large R.

That is, m at a large R will feel a surprisingly weak gravity.

Conclusions

We found a very crude formula for calculating the effect of retardation in a collapse, or expansion process where the force field is analogous to the electric field (newtonian gravity with c the maximum speed of a signal). We did not consider "spacetime geometry" effects, or the mass-energy of kinetic energy, at all.

In a collapsing star, or in the universe, spacetime geometry may affect retardation a lot. We need a more detailed analysis.

Suppose that two electric charges are in a free fall, side by side, in a homogeneous gravity field. The electric field certainly is distorted in some way, because of the acceleration, but in which way? An equivalence principle would suggest that in the accelerating frame, the electric field around each charge looks like it would look like in an inertial frame. That is, the equivalence principle holds for the electric field, too.

The gravity field in a spherically symmetric collapse is not homogeneous. The equivalence principle does not hold there.

Retardation spoils the action principle of a global (force) field: loss of information in the sum global field is the problem

The behavior of a global field is often defined through an action integral. We have to find a stationary point of the integral, and that point is then an allowed physical history of the field.

Retardation presents a grave problem to this approach: the behavior of the field at a location A must only depend on what happened inside the light cone of A. If a location B is outside the light cone of A, then nothing that happens at B is allowed to affect A.

But an stationary point of the action is a global minimum, maximum, or an inflection point of the action integral. The value of the global stationary point at A may depend on things which happened outside the light cone of A. That will produce solutions in which faster-than-light communication is possible. And we must not allow faster-than-light communication.

FLRW models

A spherical collapse or expansion history can often be derived as a global stationary point of an integral. The FLRW solutions are (probably) stationary points of the Einstein-Hilbert action.

Does the FLRW solution at a location A depend solely on things that happened inside the light cone of A?

The FLRW universe is perfectly uniform. But suppose that there is a mechanical device which will break the uniformity at a location B at a time t₀. The stationary point of the action integral at a location and time A, t₁ may depend on the events at B, t₀, even if B, t₀ is outside the light cone of A. The time t₀ might even be in the future of A, t₁.

On May 21, 2024 we showed that the Einstein-Hilbert action does not have a solution for any "dynamic" system. That is another problem in the action, but it is different from the retardation problem.

Expansion of a spherical shell of electric charges

On January 3, 2025 we discusses the expansion of a uniform shell of charged particles. The naive solution, which ignores retardation, probably is an stationary point of an action integral. But it is a wrong solution because it allows faster-than-light communication.

A "global field" has to be replaced with "private" interactions between particles?

The way to enforce retardation is to assume that the system consists of particles in Minkowski space, and that the interaction of the particles respects the light speed limit.

We would abandon the concept of a global field.

The self-force of the field of an electron on the electron itself probably cannot be explained with a global field. The concept of a field must be fragmented into the individual fields of each charge carrier. An individual field interacts with another individual particle. There is not much "global" in this.

Since general relativity depends on the existence of a "global" spacetime geometry, it is doomed.

Loss of information in a sum global field

Our arguments above suggest that the global field actually is the collection of the individual fields of charge carriers.

If we try to reduce the global field into a simple sum of individual field strengths, then we lose information about individual fields – and we would need that information in calculating the behavior of the system.

Claim. A "global field", given as the sum of fields of individual charge carriers, is an approximation which simplifies calculations in many cases, but does not handle retardation correctly.

The information loss allows faster-than-light communication

Let us again look at the expanding shell of charges in the January 3, 2025 blog post.

If we want to construct an action integral which prevents faster-than-light signals, we must penalize such signals harshly, so that a history containing such signals cannot be a stationary point of the action integral. Maybe the action integral is not defined at all, if such rogue signals happen. An example is a faster-than-light particle m in a typical action integral. Its contribution would be imaginary:

       m / sqrt(1  -  v² / c²).

But a global field defined as a sum of individual fields loses information. The sum global field can look benign, with no harsh penalty, even though individual fields change faster-than-light!

Thus, a sum global field often allows faster-than-light signals to happen. This is a major shortcoming in the concept of a global field. We must replace it with individual fields of the particles, to avoid loss of information.

A problem with FLRW solutions of general relativity: adjusting the metric using superluminal information

Let us again look at the development of an (approximately) FLRW model at a location A. The solution is not allowed to "know" that the expansion of the universe will slow down uniformly as the time t passes. We have to look at the solution for various possible decelerations of the expansion far away from A. Let a family of possible expansion rates be S(n).

There is a problem in this approach, though. On May 21, 2024 we proved that the Einstein-Hilbert action does not have a stationary point for a "dynamic" system. Thus, there is no solution, unless the expansion rate is the same everywhere! Let us for a while assume that we have been able to correct the action formula, and can find a stationary point.

Setting the metric close to A to some special (different) value for each S(n) may optimize the action, unless the action somehow recognizes that we are using superluminal information, and harshly penalizes such a break of rules. But how could the action recognize that? We are not sending gravitational waves whose energy would be infinite or imaginary. We are simply adjusting the metric in some seemingly innocent way.

An innocent adjustment may amount to a superluminal signal.

Here we again bump into the problem that general relativity does not have canonical coordinates. In Minkowski space, it would be easier to recognize superluminal signals. Though, we still would have to look at the individual field of each particle.

An individual field for each particle is in the spirit of quantum field theory

In quantum field theory, individual particles interact with each other, without any reference to a "global electromagnetic field". It makes sense to introduce an individual force field for each particle.

Conservation of energy and momentum in quantum field theory is implemented through particles exchanging (virtual) quanta. This is a possible solution to the conservation problem in macroscopic fields, though this does not tell us in detail what a macroscopic field does, and how does a macroscopic field implement conservation laws.

Conclusions

We have discovered strong evidence against the traditional global field concept, where the field is understood as the sum of the fields of the individual charges (sum global field). The sum loses information. It cannot recognize and ban faster-than-light signals in some cases.

The self-force of the field on the electron may be hard to describe through a sum global field.

The simple solution to the problem is to split the sum global field into individual fields of each elementary particle. Quantum mechanics likes this solution.

General relativity has major problems, though: there it is not clear what is the field of an individual particle. The mass of the particle acts as a source of spacetime "curvature" at the location of the particle. It affects the curvature also elsewhere, but what is the individual field of a single particle is a fuzzy concept. Nonlinearity of gravity makes this inevitable: how do we assign nonlinear effects to each component field?

Anyway, the individual field of each particle is a useful concept in gravity, too.

The FLRW model is an unusual application of the field concept because the spatial topology is that of a 3-sphere. Can we define the electric field of a single charge in such a topology? Where would the lines of force end? We are not sure if such a topology makes sense at all as a physical model. Is it so that the universe must be a flat Minkowski space?

We will investigate what retardation means in the case of the FLRW model. Does retardation affect the deceleration of the expansion of the universe? Does retardation explain dark energy?

Gauss's law for the electric field fails because of retardation; special relativity conflicts general relativity

UPDATE January 9, 2025: We maybe have to replace the classical law of Gauss with a spacetime version: the average flux of a the electric field E through a closed surface S is a constant times the charge Q enclosed inside S. The instantaneous flux may vary. Conservation of energy requires that the average flu, in some sense, must be constant. Otherwise, we could construct a perpetuum mobile.

Note that in special relativity, time and space cannot be separated. Gauss's law is suspicious because it thinks that space is a separate entity from time. Our retardation argument uncovers the problem.

----

Our analysis on December 30, 2024 proves a surprising result: Gauss's law for the electric field fails!

  E  <------- total electric field 

  E₀ <--  electric field of cap

                                                    ______

                                                 /             \

              •              <--    |                           |

             q               v                 \_______/

      test charge         "cap"     charged shell

Let us have a charged spherical shell which initially is static. Suddenly, the shell starts expanding at a constant speed v. A test charge at some distance learns about the expansion for the nearest part of the shell first. The remaining part of the sphere q still "sees" as static.

The test charge q "sees" the electric field E₀ of the nearest part as it would emanate from a "cap" which is closer to q than the rest of the sphere.

The test charge q sees the total electric field E of the charged sphere now stronger than it was a while ago.

The integral of the electric field E over a spherical surface at the distance of q increased: we broke Gauss's law!

The analysis above uses the retardation law for the electric field: if a charge Q moves at a speed v relative to the laboratory frame, then a test charge which "knows" of the movement of Q, at a laboratory time t, will see Q at the location where Q is at that same laboratory time t. That is, q will see the cap closer.

The cap is moving at a speed v to the left. Let us Lorentz transform its field, in order to make sure that the field E₀ of the cap is now larger than it was when q still saw the cap static.

      electric 

      field

      meter

          O            s          ruler

          -----------------------

           •                <--    |

           q                v     Q cap

                       

     

            -------> x

At a laboratory time t, q's x coordinate is 0 and the x coordinate of the cap is s.

The cap is carrying a ruler which extends over the laboratory distance s to the test charge q. To the ruler there is an electric field meter attached. In the moving frame of the cap, the distance is longer:

       s'  =  s / sqrt(1  -  v² / c²).

The electric field of Q in the moving cap frame is

       E₀'  =  1 / (4 π ε₀)  *  Q / s'²

              =  1 / (4 π ε₀)  *  Q / s²  *  (1  -  v² / c²).

https://www.feynmanlectures.caltech.edu/II_26.html

The Lorentz transformation of E₀' is the identical mapping:

       E₀  =  E₀'.

If the velocity v = c / 10, then q sees the cap 10% closer than the rest of the sphere, and q sees the field E₀ about 20% stronger than when the cap still was static. The correction coefficient in this case is 1 - v² / c² = 0.99, very close to 1. For small v, we can usually ignore Lorentz corrections, since they are ~ v² / c².

The magnetic field of the expanding sphere is zero, because of the symmetry.

What is the effect of the electromagnetic wave which q may "see" from charges in in the sphere suddenly accelerating to the velocity v?

In the analysis above we ignored a possible electromagnetic wave which q may see.

In the laboratory frame, there certainly is no electromagnetic wave from the sudden acceleration. There are no longitudinal waves. Thus, q is not expected to see anything.

But let us analyze in more detail. The test charge q sees that a ring of charges in the shell suddenly accelerates toward q. Does the symmetry eliminate any magnetic field?

       × Be 

       •

      q             

                             v <--  • e

Let e be an elementary charge in the ring. It suddenly accelerates to the left. A magnetic field Be is born at q. In the diagram, the field Be is normal to the screen (hence the symbol ×). When we add the fields Be for each e in the ring, their sum is zero.

It looks like q will not see any electromagnetic wave.

The effect of the extra inertia of q in the field of the sphere

If q and the sphere have a charge of the same sign, then the extra inertia of q is expected to push q to the left. Inertia will not save Gauss's law.

Could there exist a mysterious force which saves Gauss's law? No

The mysterious force should be able to differentiate between the following configurations, drawn in the laboratory frame:

        E 

      <---

                    ______

                  /             \

     <-- |                       |

       v         \_______/

        only the "cap" moves, shell static

                        ^  v

                        |  

                __________

             /                     \

           /                          \

         |                              |

           \                          /

              \___________/

                        |  

                        v   v

        entire shell expands

The mysterious force would compensate the difference of E to what Gauss's law predicts. The force would understand that it must not do anything in the first diagram above, but must do compensation in the second diagram.

For the force to do immediate compensation, it should get information faster than light. Such a force cannot exist if we believe special relativity.

Energy and momentum conservation: an accelerating shell of charges e

We firmly believe that energy and momentum are conserved. The expanding shell creates an electric field E which is larger than what Gauss's law predicts. The impulse to q must somehow eventually have an opposite impulse in the shell.

This is the general problem of conservation of momentum: if particles A and B interact, how does the field store the impulse and eventually deliver it to the other intracting particle?

If we have an accelerating, expanding shell of elementary charges e, then, apparently, any elementary charge e in the shell sees an electric field E which differs from the one given by Gauss's law.

But energy conservation requires that, eventually, the elementary charges e must possess the kinetic energy which we can calculate by integrating E in the Gauss's law solution.

There has to be a compensation mechanism which eventually sets the velocities of elementary charges e to correct ones.

If the acceleration is "too fast" initially, it must be "too slow" later.

How would this compensation mechanism behave? Does it oscillate somehow? In our December 30, 2024 post we speculated that the oscillation between a "too fast" expansion of the universe recently (= dark energy), and "too slow" earlier is due to the compensation mechanism.

Discussion

Gauss's law is one of Maxwell's laws. Let us check the literature, if anyone has realized that retardation spoils Gauss's law.

In our blog we have suspected that Gauss's law fails. We have claimed that the concept of a "field" is too simple to describe complex interactions between particles. The interaction should be calculated individually for each pair of particles. The interaction is "private" between particles. It cannot be simplified into one common "field".

https://physics.stackexchange.com/questions/162744/do-the-integral-forms-of-maxwells-equations-have-limited-applicability-because

In the link people discuss retardation effects in the case where the current through a wire changes. It is much more complicated than our expanding spherical shell.

Is this a miracle that we found a very simple counterexample to Gauss's law? No. Nobody really has claimed that Maxwell's equations are consistent. The self-force of the field on the electron is an open problem. Besides that, there are several well-known, persistent paradoxes.

Birkhoff's theorem fails for gravity: general relativity contradicts special relativity

Weak gravity fields in general relativity should behave much like electric fields. Our argumemt shows that the gravity of a uniform mass shell can vary. The result refutes Birkhoff's theorem which claims that the gravity field of a spherically symmetric object stays constant outside the object.

The refutation uses principles of special relativity. Birkhoff's theorem is derived from the Einstein field equations. We have a proof that the Einstein field equations are incompatible with special relativity.

See our blog post on January 4, 2025 for an analysis why this happens.

Conclusions

We realized on December 30, 2024 that retardation affects collapses and expansions which happen under a force field. A simple consequence is that Gauss's law fails for the electric field.

The speed of light is very large compared to the speed of charge carriers in everyday laboratory experiments. We do not know if we are able to measure the electric field retardation effect in a laboratory.

By far the largest expansion in nature happens in the universe, and the velocities are close to the speed of light. Retardation effects might be important in this huge scale. We will try to calculate an estimate for them in the ΛCDM model. They might explain the peculiarities in the cosmic expansion.