Quantum mechanics requires clocks to tick slower in a low gravity potential

The energy of a photon is

       E  =  h f.

          ●  /\/\/\/\/\/\/\/\/\/\/\/\/\  photon

         M 

                  f                               f'                     

               o                               o 

              /\                              /\

        observer                   observer

Let us assume that a photon is emitted in a low gravity potential close to a star M. An observer close to the birthplace of the photon measures a frequency

       f

for the photon, using his clock.

The photon must lose some energy when it climbs to outer space, against the gravity potential of M. An observer in outer space measures a frequency

       f'  <  f.

Let us use the standard Schwarzschild coordinates around M. The coordinate frequency of the photon cannot change when it climbs up from the potential of M. Imagine that we have a laser which shoots a beam up from M. We assume that the overall shape of the wave in the beam does not change with coordinate time.

       

               4 waves in 1 coordinate second

      ●       /\/\/\/\                                     /\/\/\/\

      M

                   o                                                  o

                  /\                                                 /\

           observer                                    observer

In one coordinate second, the same number N of waves must pass an observer close to M, as in outer space. The coordinate frequency is the same for both observers.

We conclude that the clock of the observer near M must run slower.

If the clocks of both observers would run at the same rate, then both observers would measure the same frequency f for the photon. Then the photon would not have lost energy as it climbed the potential of M. The "implementation" of the gravity potential would be flawed.

If a hydrogen atom decays close to M, then a single photon is created. What about implementing a gravity potential by deleting some percentage of created photons as they climb up from the potential of M? That will not work, since conservation of energy would be violated.

Massive particles do not require us to change the clock rate near M

The basic property of a potential wall is that a particle m climbing up it must lose energy in some way. Massive particles can shed their kinetic energy. There is no need to assume different clock rates.

But a massless particle, like a photon, can only lose energy by lowering its frequency f. That requires different clock rates near M and in outer space.

Can a classical wave packet shed energy without changing its frequency?

Let us have a classical light wave packet which climbs up the potential of M. A way to reduce the energy of the packet is to lower its frequency.

We can also imagine a process which reduces the amplitude of the wave as it propagates. However, such a process may be complicated to implement.

Why is the radial metric stretched in the Schwarzschild solution?

The stretching probably can be derived from an equivalence principle. But can we find a simpler explanation?

Static electric field? A possible explanation: a static electric field "consists of" virtual particles which only carry momentum, no energy.

Let us have spherically symmetric electric E field around a spherical mass M. To keep the energy of E constant, if the metric of time is g₀, the radial metric must be g₁ = -1 / g₀.

Let us have a mass shell. Inside the shell, the metric of time is squeezed, but the spatial metric is 1. Why is there the energy of the electric field E smaller?

Could it be that the matter in the shell can be seen as "polarized" material which reduces the field energy E? Outside M, it is empty space.

The volume element has the same 4-volume outside M, but inside the mass shell, the volume is smaller.

The mechanism to slow down time is the same as in special relativity? In special relativity we can slow down time using a large velocity. Simultaneously, a ruler contracts in the direction of the movement. Maybe the mechanism which nature uses to change the metric in the Schwarzschild solution is copied from special relativity?

          • --> v                         ●

         m               r               M

Let us have a spherical mass M. Let a test mass m start static at the infinity, and fall toward M. Let v << c be the velocity of m.

The kinetic energy of m is

       W  =  G m M / r,

and

       v²  =  2 G M / r.

The metric of time of m is slowed down by a factor

       sqrt(1  -  v² / c²),

and the metric of time for m is

       g₀  =  -c²  *  (1  -  v² / c²)

             =  -c²  *  (1  -  2 G / c²  *  M / r).

             =  -c²  *  (1  -  rs / r),

where rs is the Schwarzschild radius of M. The radial metric is

       g₁  =  1  +  rs / r.

It is the Schwarzschild metric. Is it a coincidence that the metric of a test mass m falling from infinity agrees with the Schwarzschild metric of a static observer?

Comparison to a collapse of a dust ball. Let a uniform dust ball collapse from a very sparse state. Let us assume that the spatial metric in comoving coordinates of dust particles is spatially flat.

Let us have static observers, symmetrically from the center, looking at dust particles flying by. The observers will see the comoving rulers held by dust particles contracted radially. If we convert between the static coordinates and comoving coordinates, do we obtain something like the Schwarzschild metric?

The rulers held the dust particles are contracted by the factor

       sqrt(1  -  v² / c²)  <  1.

These contracted rulers determine a flat spatial metric, by the simultaneousness concept of the dust particles. The static observers have a different view of what is simultaneous.

Let the static observers hold rulers, too. The dust particles see the static rulers length-contracted.

The dust particles figure out that the static observers measure radial distances of static observers by the factor

       1 /  sqrt(1  -  v² / c²)  >  1

longer than the distance measured by dust particles. That is, the radial distances measured by static observers are by that factor longer than the flat metric.

We were able to derive the Schwarzschild spatial metric. Our assumption was that the spatial metric in the comoving coordinates of a collapsing dust ball is flat.

The Schwarzschild metric of time follows from the gravity potential.

From the viewpoint of a dust particle, the clock of a nearby static observer runs at a lower rate.

Conclusions

Quantum mechanics seems to require that clocks run slower in a low gravity potential.

We also found a simple way to derive the stretched radial metric of the Schwarzschild solution, if we assume that the spatial metric is flat for a uniform dust ball which initially is very sparse and has a very large (= infinite) radius, and then collapses. The collapse corresponds to a FLRW solution where the curvature parameter k = 0.

Sparse contracting shell of masses breaks Gauss's law? No!

Our effort to break Gauss's law for the electric field on January 3, 2025 failed. A uniform charged shell satisfies Gauss's law because the accelerating charges that our test charge q "sees", create a correcting electric field E'.

The form of a line of force for an accelerated charge

Suppose that an electric charge q or a mass m with a newtonian-like gravity force is under a constant acceleration a. What form do lines of force take?

 

           -------___    ___-------  line of force

                          • q

                          |

                          v   a

If q is at a laboratory time t₀ at a location x₀, moving at a constant speed v, then the line of force at a later time t at a distance t c from x₀ points to 

       x₀ + v (t - t₀).

This is the regular retardation formula. But q is not moving at a constant speed. The line of force will be bent.

Let us calculate the form of the horizontal line of force in the diagram.

The form is a parabola. The line of force is as if q would have progressed farther that its real position.

We had in this blog been assuming that retardation shows q lagging behind. This means that our efforts to disprove Gauss's law cannot succeed in the way we presented below.

A sparse shell of masses, accelerating radially

Let us try to break the law using a sparse shell of masses m. The masses form a matrix on a spherical shell whose radius is R.

                                  • m

                                  |

                                  v  a

 

              • --> a          ×          a <-- •

             m                                     m

                                  ^  a

                                  |

                                  • m       R radius

We accelerate the masses inward for a time

       Δt

at an acceleration a, so that the radial velocity becomes a nonrelativistic

       v  <<  c.

The masses move the distance

       s  =  1/2 a Δt²

closer to the center.

Let the spacing between the masses be L.

   

                 m'                    m

                  •          L          •

                 |                      |   

                 v    v                v    v

Let

       t  =  L / c.

The mass m sees the velocity of m' to be

       v'  =  v  -  a t

and m sees the location m' to be a distance

      1/2 a t²

higher in the diagram than the location of m. 

####

The simple retardation formula should work in this case? No! We have to derive a more precise formula. The line of force is bent by the acceleration.

In the first section we now derived the formula. It shows that our arguments below do not work.

####

The four m' closest to m will pull m up with a force

       F  ~  4 m / L²  *  sin(1/2 a t² / L)

           ≈  2 m / L³  *  a t²,

and the associated not gained energy in the push is

       Wm  =  F s

                ~  2 m / L³  *  a t²  *  s

                ~  m / L³  *  t²

This means that we gain strangely little energy from the pull of gravity when we contract the shell.

What happens if we halve the value of L? Then m is 1/4 of the old value and t is halved. The value of Wm is halved. But the number of masses m quadruples in the shell. The missing energy W in the entire shell is doubled!

Maybe the missing energy goes to gravitational waves which leave the system?

That is extremely unlikely. If we halve L, then the field of all the masses m on the shell will more closely mimic the field of a uniform massive shell. A uniform massive shell does not generate any gravitational wave at all. That is, the outgoing radiation will have less energy, while the missing energy W is doubled.

Let us calculate the energy which initially might reside in the gravitational waves. The energy of a wave generated by m is directly proportional to m. If we halve L, the sum of the masses m does not change. The initial energy of the gravitational waves does not change appreciably.

Conclusions

We did not succeed in breaking Maxwell's equations yet. We have to look at the Poynting vector and the action of classical electromagnetism. Are there weak points there?

Breach of Gauss's law revisited: the law holds!

In our January 3, 2025 blog post we had a uniform spherical shell of electric charges which very suddenly starts expanding very fast, at a constant speed.

       -> E' field of accelerated charges

  E  <------- total electric field 

  E₀ <--  electric field of cap

                                                    ______

                                                 /             \

              •              <--    |                 ×        |

             q               v                 \_______/

      test charge         "cap"     charged shell

The Coulomb field of the "cap" and the remaining charged shell make the electric field E at q larger than what H

Gauss's law predicts.

The charge q also "sees" some charges in the shell being accelerated radially outward from the center × of the shell. These accelerated charges on the surface of the shell form a ring which is symmetric around the line from q to the center ×.

The field of the accelerating ring of charges

In our January 3, 2025 we compared two configurations. In one, the entire shell is expanding. In the other, only the "cap" moves radially, and the test charge sees the accelerating ring of charges arounf it.

To restore Gauss's law, the field of the ring must produce just the right transient electric field E' which is radial at q.

How can a such a field B arise from a system which is axially symmetric around the line between q and ×?

                    × ×   B transient

                    ---> E' transient

 

                    ○ ○   B transient

The field should satisfy Maxwell's equations. If we calculate B from the transient E', does E' satisfy the corresponding equation for the transient B?

Bent lines of force

https://en.wikipedia.org/wiki/Maxwell%27s_equations

If there is a transient tangential loop of the magnetic field B around the line × to q, then there will be a transient radial electric field E at q. Can the ring of accelerating charges produce such a loop around q?

Let us first use a naive interpretation of the field. Because of symmetry, there obviously cannot be a tangential component of B at q. By the same argument, there cannot be a tangential component of B anywhere. Thus, for the radial component of E:

         dEr / dt  =  0.

But this is a contradiction. The "cap" argument shows that there would be a transient change in Er.

Maybe we must not consider the actual current value of B (current in the laboratory time coordinate), but the value which q "derives" from what q "sees".

The ring of accelerating charges probably produces a tangential loop of the magnetic field B around q:

    

             B                  a  <-- •

            ○○

             • q                                    × center

            ××

             B                  a  <-- •

                        accelerating

                              ring

The tangential component of B is zero at q, but nonzero close to q.

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

The familiar Edward M. Purcell diagram shows the electric field lines of a charge which was suddenly accelerated to the left. The bends in the lines of force, and the "tangential" lines in the diagram, represent the sudden acceleration. The beautiful radial lines represent the old motion (the charge was static) and the new motion with a constant velocity.

The retardation formula and the diagram of Edward M. Purcell are only approximate

Assuming that Maxwell's equations hold, the process described in Purcell's diagram is very complex. There has to be backscattering of the wave which was created by the accelerated charge in the diagram. The lines of force cannot consist of straight line segments.

Approximate retardation formula. If a charge Q moves at a constant velocity v in the laboratory frame, the a test charge q sees the electric field as if Q would be at its current position, where "current" is in the laboratory time coordinate.

We have been using the formula above in our analysis of retardation. Now we realize that it is only approximate. The retardation formula is seen in the straight radial lines of force in the Edward M. Purcell diagram. But the lines cannot be perfectly straight. The formula is only approximate.

Assume that Gauss's law holds: the test charge q must not see any difference between the whole shell or a part expanding

Case A

                                      ______

     <-- E                       /             \

     • q              <-- |           ×         |

                         v         \_______/

       only the "cap" and ring move, shell static

Case B

                                          ^  v

                                          |  

                                   __________

                                 /                     \

    <-- E                  /                          \

      • q                |              ×                |

                              \                          /

                                 \___________/

                                          |  

                                          v   v

       entire shell expands

The speed of light is finite. The test charge q cannot know if the entire uniform shell of charge is expanding, or if just the "cap" and the ring of charges around it are moving.

Let us assume that Gauss's law holds. The electric field measured at q must be the exact same E in cases A and B. Otherwise, q would receive information faster than light.

If we assume Gauss's law, then calculating the electric field E in Case B is trivial.

Case A, actually, is not unique. The acceleration in the ring of charges may happen in many ways. There may be phases of slow and fast acceleration. In all these cases, the electric field E should have the exact same value as in Case B. Why would that be?

It is because q sees in its light cone exactly the same history in A and B. The electric field E has to be the same!

We showed that Gauss's law holds for the electric field E, after all. Our error on January 3, 2025 was a superficial treatment of the charges in the accelerating ring.

Let us suddenly freeze both A and B. The potential of q is then different in A and B. Thus, there is retardation in the potential of q.

Conclusions

We made an elementary error on January 3, 2025. Gauss's law holds for the electric field.

Retardation in gravity is a more complex phenomenon. We will next look at retardation in the rate of clocks.

Gravity is newtonian in comoving coordinates of a dust ball? Retardation as a general law of nature

In the previous blog post we tried to determine what is the gravity field like in a collapsing uniform dust ball. If the ball is lightweight and the velocities are slow, then newtonian gravity is the model for the collapse.

We had problems determining the behavior in a case where the velocities are close to c, and if the ball is heavy, maybe inside its Schwarzschild radius.

Let us study the hypothesis which we raised at the end of the previous blog post: in comoving coordinates, the dust cloud behaves in a "newtonian" way.

If the observable universe is a part of a giant expanding dust ball, then we have empirical evidence that the newtonian hypothesis is correct, to an extent. The expansion of the universe has very crudely followed a newtonian expansion. If we assume a "shell theorem" which says that the masses from a distance > R from us do not affect the behavior inside the radius R, then masses inside R behave in a newtonian way. Though, recall that dark energy breaks this model recently.

We can quite easily calculate the effect of the retardation for a newtonian model. We simply assume that gravity cannot "predict" the acceleration of a collapse, or the deceleration of an expansion. This prediction error makes gravity to oscillate between the naive model (= with a perfect predictive capability) and the retarded model. We believe that the oscillation is able to explain dark energy and other anomalies in the expansion of the universe. Retardation is expected to cause anomalies in the expansion.

Why does gravity have to oscillate? Because conservation of energy forces gravity to have, on the average, the naive strength.

Electric lines of force in a collapsing dust ball: the field E looks just like in newtonian physics

Let us first try to draw electric lines of force for a charge q falling along the dust in a collapsing ball.

  

               ^

          E   |  m        m'

                    •         •  dust particles

                    |       /

                    v    v

                    

                    

                    ● q electric charge

In the comoving frame of the charge q, the lines of force form the usual radial pattern. The electric field E is parallel with the velocity vector of an approaching dust particle m. 

Let us then switch to the comoving frame of m. What does the electric field E look like there?

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

At m, the field E has the same value in the comoving frames of q and m.

If the frame is not accelerated, then Gauss's law for the electric field E holds there.

                            •  m'''

                                           r = distance(q, m''')

                  ^   E

                  |

                      •  m''

                        \    v

                         v

                            • m

                            • q

Each dust particle, in its own comoving frame, sees the electric field E pointing radially toward the charge. Let us take a dust particle m'' very close to m, so that its relative velocity v to m is very slow. Then the Lorentz transformation of E at m'' to the comoving frame of m keeps the direction of E essentially unchanged. The configuration looks locally very simple.

We conclude that we can draw the lines of force of E close to m, radially relative to q, in the comoving frame of m. Let m''' be yet another dust particle on the line from q to m, but a little bit farther from q. The strength of the field E at m''' obeys the usual

       ~  1 / r²

formula, where r is the distance measured in comoving coordinates of dust particles.

In summary, the field E obeys Gauss's law in the comoving coordinates, even though the coordinates are quite exotic compared to standard Minkowski coordinates.

The formula for the electric field is then

       E  =  1 / (4 π ε₀)  *  q / r²,

just as in newtonian physics and coordinates.

The problem with using the proper time of each dust particle as the time coordinate

On May 26, 2024 we observed that using the proper time of each dust particle as the time coordinate leads to very unnatural coordinates: one is able travel to an earlier time coordinate if one approaches the center of a collapsing star! How does that affect our analysis above?

Why clocks do not freeze in a collapsing dust ball? A retardation hypothesis

We can argue that gravity behaves much in the same way as an electric field in a collapsing dust ball.

But how do we explain why the nonlinear effects in gravity are absent in the dust ball? We believe that extremely strong nonlinear effects exist near a neutron star and close to the horizon of a black hole.

Could it be that a dust particle never gets a message that it is inside an extremely strong, forming gravity field? Particles at the edges of the dust ball are falling at almost the speed of light – which reminds us of the collapsing photon shell in the previous blog post.

Hypothesis of the retardation of slowing clocks. A dust particle in the collapsing cloud is aware of the "newtonian" gravity field at all times. It knows that its motion must accelerate. But the dust particle is not immediately aware that it is in a low gravity potential. The information about a low potential must come from the surrounding, asymptotically Minkowski space, and that may take a very long time, possibly an infinite time if the speed of light drops to zero at a horizon.

The slowing of clocks may be somewhat analogous to freezing: the frozen portion begins from the surrounding space and spreads to deeper gravity potentials.

Various authors have remarked that we can only know the location of a horizon afterwards. The redshift of a photon sent from a coordinate location x at a coordinate time t cannot be calculated unless we know also future events. This shows that the rate of a clock at a forming horizon can be quite fast.

On and inside a static neutron star, the gravity potential is known at every location. Clocks know how slowly they should tick. Also, close to the horizon of a static black hole, clocks know that they should tick very slowly.

Retardation as a general natural law: its mechanisms are unknown

                   ~• --> c                      c <-- •~ 

               photon                            photon

Imagine a collision of two photons. Neither one can know that it is inside the gravity field of the other one. A scientist measuring their progress in a laboratory cannot know that either. The photons move as "free particles". There is no interaction.

When the photons collide, massive particles may form and take all the energy of the photons.

These massive particles are inside their common gravity potential. Locally, they have to have more energy than what was in the arriving photons. That is because the massive particles are in a low gravity potential. Energy conservation requires that suddenly, some extra energy appears. How do the new particles know how much extra energy they must possess?

Is it so that the flattened gravity fields of the photons somehow "combine" and release the extra energy? In what time does this happen?

To eliminate outgoing gravitational waves, we may let a spherical photon shell collapse. When and how does the gravity field settle into the Schwarzshild metric?

We realize that retardation in many cases profoundly affects processes in nature. There must exist unknown mechanisms which restore conservation laws. Also, if the end state is static, there must be mechanisms to create the eventual static fields from earlier, dynamic ones.

We do not currently know anything about these mechanisms. Field theory is generally developed for very slow processes, where retardation has a negligible effect.

In general relativity, Birkhoff's theorem (that is, the Einstein field equations) assumes that the field can magically, instantaneously, know that a system is spherically symmetric. Gauss's law for the electric field contains the same assumption, which is very unlikely to be true. See our January 3, 2025 post.

We have uncovered a whole new field of physics: physics of retardation.

Retardation comes from the finite speed of light. Field theory has, so far, only studied few phenomena which arise from the finiteness of the speed.

The universe, as well as a collapse of a star into a black hole, are processes where retardation must have significant effects.

Retardation might have major effects in collisions of elementary particles. We have to think about that.

What is inside a black hole: the observable universe shows us what it is like

The observable universe most probably is inside its Schwarzschild radius. Thus, in the observable universe we probably see what it is like to be inside a collapsing star. Nothing very special. Gravity seems to obey the newtonian law. But what happens when collapsing dust particles collide into each other?

There will be a lot of pressure. That pressure may speed up the collapse. On the other hand, if the particles come to a standstill in collisions, they may have time to receive the information that they are now in a very low gravity potential, and clocks will essentially stop. The end result is a very dense, "frozen" object.

Retardation in the rotation of galaxies: could it explain some of the dark matter?

One rotation of the Milky Way takes 300 million years, and its diameter is 100,000 light-years. Are there retardation effects in the rotation? How does the gravity field know that stars at the other end of the galaxy stay in circular orbits, so that the gravity field does not change with time?

Retardation effects may be significant, but we do not think that they can explain all the dark matter, which makes up 90% of the mass of the Milky Way.

We have to figure out what type of retardation effects does a centrifugal acceleration create. So far, we have only considered an accelerated expansion of a shell of charges.

If we could trust Gauss's law, we could model the lines of force of the gravity field of an individual star with radial "steel wires" or "rubber strings" which start from the star. The centrifugal acceleration would bend the steel wires. The field would probably be "stationary", since the motion of the star is periodic and uniform. 

Let us assume a very simple form of retardation: the gravity field at a test mass m behaves as if the stars of the galaxy would have obeyed straight line orbits at a constant speed, since m last "saw" each star.

Let m be in the plane of the galaxy, at some distance.

Then the gravity field is as if the stars at the far end of the galaxy would be very slightly farther than they are. Gravity at m is reduced slightly, maybe one part in a billion.

Retardation cannot explain dark matter.

Conclusions

We now have some ideas about how gravity might work inside a collapsing dust ball. We have to do a lot more analysis.

Today, we realized that our January 3, 2025 proof of the failure of Gauss's law is erroneous. If we assume as an axiom that the field of each elementary charge at each moment satisfies Gauss'd law in the laboratory frame, then, of course, the entire shell of charges must obey it, too. However, we do not see how a changing magnetic field could create a radial electric field which restores Gauss's law for an expanding shell of charges. We will look at this problem again.

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.

Conclusions

Let us close this very long blog post.

The following might be a good model for the electric field of gravity inside a collapsing, uniform dust ball:

We use the comoving coordinates of the dust particles and assume that the field strength is

       G m / r²

in the comoving coordinates. That is, the field is classical newtonian in these special coordinates. This would explain why the expansion of the universe looks so much like a classical newtonian expansion.

This hypothesis gives us a natural way to estimate the effect of retardation from the decelerating expansion. The effect is similar to one we get from newtonian gravity in Minkowski space, with slow velocities.

In the next blog post we try to argue that the assumption above is the correct one.

Also, we have to analyze how much this model slows down clocks inside the cloud. We do not want a "frozen" model.

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?