Thursday, March 27, 2025

Retardation of clocks from acceleration in a collapsing dust ball

Let us try to estimate the effect of clock retardation in a collapsing dust ball, or in an expanding universe, if retardation is based on the acceleration of matter. That is, clocks are able to anticipate the effect of a mass shell contracting or expanding at a constant velocity.

We would assume that a clock at the center of the shell "calculates" the radius of the shell based on the latests contraction velocity v that the clock knows of.

Is this a reasonable assumption?

Let us look at a single mass M and a test mass m (or a test clock) in the field of M. The usual retardation rule is that if M moves at a constant speed v, then the test mass or test clock will know the gravity field of M as if m would know the current position of M in the laboratory coordinates.

It makes a lot of sense to assume that a test clock m can "calculate" its own gravity potential based on the assumption that masses M continue their movement at a constant velocity. The clock then adjusts its rate according to that gravity potential.

Another way to look at this is to assume that M is static in the laboratory coordinates, and the test clock m moves around. An atomic clock on the surface of Earth can adjust its ticking based on what is its distance from the center of Earth.

In our March 15, 2025 blog post we calculated the retardation inside a shell assuming that the clock is not aware of the contraction speed v of the shell. That yields a very large retardation effect. If the ignorance of the clock only concerns the acceleration of the shell, the retardation effect is much smaller. But is the effect still large enough to explain dark energy?


A crude calculation based on the expansion of the universe


The "current" radius of the observable universe is estimated to be 46 billion light-years. The age of the universe is 13.7 billion years.

The universe was expanding significantly faster than now, say, 6.9 billion years ago.


The scale factor in the matter-dominated phase is

       a(t)  ~  t^⅔.

The time derivative is

       da / dt  ~  1 / t^⅓.

When the age of the universe was a half of the current age, the expansion speed was

       2^⅓  =  1.26

times the current speed.

If a clock at the center of the observable universe "calculates" its rate by assuming that the universe would still be expanding at that 1.26X speed, then the clock will overestimate its gravity potential and will tick too fast. The speed of light is "too fast" close to the center, which means a repulsive force from the center.

Could it be that the current value of matter and dark matter Ω = 0.3 has something to do with the observed acceleration of the expansion?

Let us try to estimate the repulsion, using the comoving coordinates of the "dust" (= matter and dark matter in the universe). On January 18, 2025 we argued that gravity looks very much newtonian in comoving coordinates. But we did not consider retardation of clocks then.





***  WORK IN PROGRESS  ***

Saturday, March 15, 2025

Retarded slowing down of clocks in a collapse

Let us continue our analysis of the assumption that a clock cannot "know" faster than light how it should tick. It can only receive information of the gravity potential at the local speed of light.


What empirical evidence do we have for the correctness of the Einstein field equations?


We know that the Schwarzschild metric describes gravity phenomena very accurately within the Solar system.

We know that binary pulsars orbit in the way predicted by the Schwarzschild metrics of the components. Also, the power of the produced gravitational waves match calculations where linearized Einstein equations are used, to an accuracy < 1%.

Gravitational waves observed by LIGO match numerical calculations made using numerical programs. However, we have not checked the heuristics used in the numerical programs.

On June 20, 2024 we showed that the Einstein fields equations do not have a solution at all for a two-body system. How do LIGO numerical models handle the nonexistence of solutions? Also, the LIGO measured results have a large margin of error, something like 10%.

The Schwarzschild metric is a static system. Gravitational waves are produced by a quadrupole system. These are configurations which are quite different from a collapse or an expansion of a spherically symmetric dust ball.

The only empirical data which we have about a large collapse or an expansion is what we know about the expansion of the observable universe.

We do know that a star can collapse into a neutron star or a black hole. But we do not have any detailed measured data about the process.

The only data which we have about a large expansion does not match the Einstein field equations. The equations do not predict dark energy, or the Hubble tension. They may also fail to predict the things seen by the James Webb telescope.

Since the Einstein equations fail for our (sparse) empirical data, there is a good chance that the equations describe a collapse or an expansion incorrectly.


The June 20, 2024 result about the nonexistence of "dynamic" solutions to the Einstein equations


The problem in that result seemed to be that general relativity does not possess "canonical coordinates" where we could determine kinetic energy in an unambiguous way. The corresponding problem does not occur in Minkowski space, where any inertial frame can be taken as the canonical coordinates.


Canonical coordinates require that there must be retardation in the rate of clocks?


Suppose that we can determine against a canonical time coordinate how fast a clock ticks. How does the clock know how fast it should tick? Empirically, we know that clocks tick slower in a low gravity potential. But how does the clock know that it is in a low potential?

If we can use canonical coordinates similar to Minkowski space, it is natural to assume that the information about the allowed clock rate can only spread at the speed of light. If the clock does not know that it has fallen into a lower gravity potential, then the clock maintains its rate.

How does general relativity handle this?


How does general relativity decide at which rate a clock should tick?


Empirically, we know that a clock in a static, low gravity potential ticks slower than far away in space. This has been demonstrated with atomic clocks on Earth, as well as in satellites. The phenomenon is also present in the redshift of light when it rises up from the surface of Earth. The redshift is approximately one billionth.

If we have a clock which ticks once per second, then in general relativity, the metric of time determines how many times it will tick in a second of coordinate time. There should be one tick in a second of proper time. The rate of the proper time is

       sqrt(-g₀₀)

times the rate of the coordinate time.


General relativity does not allow the metric to change "faster than light"?


This is a question which we have touched several times in this blog. How fast can a change in the metric propagate in general relativity?

People often seem to assume that it cannot propagate faster than the local speed of light. Some changes in the metric can be detected by an observer. It would open a channel of faster-than-light communication, if changes in the metric can propagate faster than light.

Our example of a collapsing dust shell in the previous blog post seems to contradict this principle. People usually assume that the metric of time slows down instantaneously inside the shell, as the shell contracts.


           # -->                    ×                     <-- #
     dust shell            center            dust shell



In the link, Ajay Mohan cites a book by E. Poisson. The metric inside a thin shell is assumed to be Minkowski.

However, the assumption may not be sound. Suppose that the shell is not exactly symmetric. Then the metric inside the shell should change in a complicated way as the shell contracts. An observer inside the shell can measure the metric close to him. The metric is not flat, but has a complicated form.

If we let the changes in the metric happen instantaneously inside the shell, that can open a faster-than-light communication channel.

If the shell is perfectly spherically symmetric, then slowing down the metric of time inside the shell instantaneously does not enable communication – but that is not a realistic physical configuration.

Maybe we should adopt the rule that the metric cannot change faster than light?

Then we encounter another problem. Inside a spherically symmetric collapsing shell there is no matter inside. Does that require that the metric inside is flat? If yes, then the metric of time will propagate faster than light inside the shell.

Can we find a curved metric inside the shell, such that its Ricci tensor is zero? The Schwarzschild solution is an example of a curved metric for which the Ricci tensor is zero.

Birkhoff's theorem may imply that the metric inside the shell must be flat.

The singularity theorems of Roger Penrose assume that an empty volume of space cannot focus or defocus a beam of light. But if the speed of light is faster at the center of the collapsing shell, then there is defocusing.

Hypothesis. A realistic collapsing shell does not have a solution in general relativity, such that the solution would not allow faster-than-light communication.


Conjecture. Any solution for a spherically symmetric collapsing shell in general relativity requires the metric of time to change faster than light.


Note that we already proved on June 20, 2024 that general relativity probably does not have a solution for any realistic dynamic problem at all. The hypothesis above is probably void. But it might be that any attempt to find a solution will also lead to faster-than-light communication.

In the case of the conjecture above, there may exist a solution where the thickness of the shell is the Dirac delta function. It is not physically realistic.

We may have uncovered yet another fundamental problem in general relativity: it would allow faster-than-light communication if it would have solutions at all!


What implications does faster-than-light communication have?


If we can change the undulating metric inside a shell instantaneously, that probably enables us to transfer energy faster than light. A faster-than-light energy transfer is forbidden in an energy condition.


The dominant energy condition demands that energy can never flow faster than light.

What would happen if we could send signals inside a shell faster than light? Then the physics inside the shell cannot be analogous to Minkowski space, because such signals cannot happen in Minkowski space. This would probably break an equivalence principle.


Can general relativity correctly handle the collapse of a dust ball?


On May 26, 2024 we showed that the Oppenheimer-Snyder 1939 solution is incorrect, since the comoving Tolman coordinates allow one to travel to an "earlier" time coordinate. But maybe there exists a correct solution to the problem?


The Einstein field equations are







where the cosmological constant Λ is zero and the stress-energy tensor is denoted by T. Let us use the standard Schwarzschild coordinates.


Zhang and Yi (2012) write about Birkhoff's theorem.


Willem van Oosterhuit (2019) gives Birkhoff's theorem in the following form:

Birkhoff's theorem. Any C² solution of the vacuum Einstein equations, which is spherically symmetric in an open set U, is locally isometric to the maximally extended Schwarzschild solution in U.


               #                   •  •  •  •                     #
         shell S                ball D                 shell S


We interpret the theorem in this way: let us have a collapsing spherically symmetric dust ball D and a spherically symmetric shell S enclosing D. Then the vacuum solution between S and D stays isometric (= isomorphic) to the Schwarzschild solution for a fixed mass M.

Whatever we do with S, the metric between S and D stays isometric (= isomorphic) to the Schwarzschild metric associated with a fixed M.

This means that D cannot "know" if we let S descend lower or not. The vacuum between S and D prevents any flow of information between S and D.

This implies that if we measure things with proper distances and proper time intervals, the collapse of D happens in the exact same way, regardless of what we do with S.

Let us then compare two histories:

- in history A, the shell S and D form one, almost uniform, dust ball, with an infinitesimal gap between them and we let them collapse freely;

- in history B, we use a tangential pressure within S to slow down its collapse; D collapses freely.


In history B, the metric of time, g₀₀, in the vacuum between S and D, will eventually differ from history A. The metric there will still be isomorphic to the fixed Schwarzschild metric M, but the absolute value of g₀₀ will be different in A and B.

The rate of clocks (i.e., g₀₀) in the vacuum below S depends on how low we let S descend. The gravity potential of a clock depends on how high S is.

The collapse of D happens in the exact same way, measured in proper times and proper lengths, regardless of how high S is. This implies that g₀₀ must change immediately throughout the dust ball D, if we manipulate the shell S.

We proved that the metric of time, g₀₀, changes instantaneously in the vacuum between S and D, and within D. The change in the metric propagates faster than light.

That is, the problem of the infinitely fast metric change remains if we have a dust ball enclosed in a dust shell S.


Discussion


If general relativity has solutions at all for a collapse of a uniform or a slightly nonuniform dust ball, it seems to require infinitely fast changes in the metric of time within the ball. This may even enable faster-than-light communication within the ball.

General relativity seems to break a fundamental principle of special relativity. We conclude that general relativity probably is a wrong model for a dust ball collapse.

The FLRW model of the expanding universe looks very much like a dust ball expansion in general relativity. If general relativity cannot handle a dust ball correctly, why would it handle an expanding universe correctly?

Dark energy is an indication that general relativity fails to treat an expanding universe correctly. If the expansion is accelerated, that seriously contradicts the general relativity model.


What aspects of the FLRW model have been verified empirically?


A.   Nucleosynthesis fits the FLRW model.

B.   The expansion of the universe by a factor 1,100 since the last scattering (cosmic microwave background, CMB) fits FLRW.

C.   Baryon acoustic oscillations (BAO) fit the model where the age of the universe at the last scattering was as in FLRW.


Deviations from FLRW are:

1.   the Hubble constant derived (in a complicated way) from the CMB differs from standard candle observations by 7%;

2.   the James Webb telescope sees "too many" mature galaxies when the age of the universe was just 300 million years;

3.   dark energy seems to be accelerating the expansion of the universe, while the expansion should slow down.


If retardation in the rate of clocks makes the expansion of the universe to oscillate, that might explain items 1, 2, and 3. The average speed of the expansion is correctly predicted by FLRW (or a newtonian gravity model), but an oscillation in the speed of the expansion can produce even large anomalies to the smooth process.

Question. Can retardation explain cosmic inflation?


Retardation when the dust ball approaches its Schwarschild radius


The matter and dark matter density of the observable universe is estimated to be 30% of the "critical density", Ω = 0.30.


The "current" radius of the observable universe is 46 billion light-years and its Schwarzschild radius is 14 billion light-years. Their ratio is approximately 0.30.

Maybe the accelerating expansion is associated with the (dark) matter density falling to 0.3X the critical density?








On the right side is a constant. The density ρ ~ 1 / a³, where a is the scale factor. Thus,

       ρ a²  ~  1 / a.

The value of 1 / Ω - 1 is now roughly 2. When a was 1/2, its value must have been roughly 1, or Ω = 0.5. When a was 1/4, Ω = 0.67. When a was 1/1,000, then Ω = 0.998.

The fine-tuning, or flatness, problem is why Ω was so close to 1 in the early stages of the universe.

Let us try to calculate the effect of retardation when a dust ball collapses close to its Schwarzschild radius. The gravity potential of the edges falls fast. We expect to see a large repulsive force which arises from the retardation of clocks near the center. That is, clocks at the center tick significantly faster than at the edges. A ray of light is bent from the center toward the edge of the ball.

Let us first use newtonian gravity to calculate the retardation potential. The mass of the dust ball is M and the radius R. The gravity potential is 

       -G M / r                                              for r > R,

       -3/2 G M / R  +  1/2 G M r² / R³       for r < R.

The potential at the center is

       V  =  -3/2 G M / R(t).

Let dR(t) / dt = -v. Then

       dV / dt  =  -3/2 G M  *  -1 / R(t)²  * -v

                     =  -3/2 G M v / R(t)².

The delay for the center of the ball to know about the decline in the potential is very crudely a half of the radius R(t) divided by the speed of light c:

       1/2 R(t) / c.

The retardation then would mean that the potential at the center is higher than calculated in newtonian gravity, very roughly by the amount:

       ΔV  =  3/2 G M v / R(t)²  *  1/2 R(t) / c

              =  3/4 G M / R(t)  *  v / c.

We can compare this to the newtonian potential difference between r = 1/2 R(r) and the center:

       1/8 G M / R(t).

We see that if the velocity v = c / 6, then the "retardation force" would approximately cancel the newtonian gravity force when r < R(t) / 2.

When the dust ball is approaching its Schwarzschild radius, the speed of its surface dR(t) / dt is relativistic. We conclude that the retardation force can easily cancel the newtonian gravity force inside the dust ball. The order of magnitude is large enough.

However the retardation potential close to the center is linear in r, while the newtonian gravity potential is ~ r². This would cause the uniform density of the dust ball to be compromised. Would that make the cosmic microwave background (CMB) in the sky nonuniform?

Our dust ball model has hard time explaining the uniformity of the CMB, anyway. Any phenomenon which is associated with the edges of the ball, can easily break the uniformity of the CMB.


The uniformity of the CMB


The cosmic microwave background is uniform in every direction to one part in 100,000. A cosmological model must be able to account for this phenomenon. In ΛCDM, two ad hoc assumptions are introduced in order to explain this:

1.   the spatial topology of the universe is 3D surface of a 4-dimensional sphere, and

2.   inflation.


It is not economical if we have to explain one observed fact with two ad hoc hypotheses. There is no evidence that the spatial topology can differ from a 3D plane, besides the hypothesized FLRW model of the universe. Inflation creates energy from nothing. It runs counter to all the observations we have about nature: energy is conserved.

In our blog we have tried to build a model where the spatial topology is a 3D plane and the observable universe is an explosion of a dust ball. The uniformity should be explained by some mechanism which makes a uniform dust ball to stay uniform when it collapses or expands.

An ad hoc solution would be to claim that in a large dust ball, we can calculate the contraction or expansion speed at a location x simply by looking at some environment of x, and ignoring the rest of the ball. This principle seems to hold for the gravitational attraction: locally, the expansion of the universe seems to obey newtonian gravity (with the exception of dark energy).

But why would retardation obey such a locality principle? And if it obeys that, why should we calculate the retardation based on the radius of the observable universe?


Dark energy is weakening?



Lodha et al. published their results from the Dark Energy Spectroscopic Instrument on March 18, 2025. Dark energy seems to be weakening recently.

If that really is the case, it is consistent with our retardation hypothesis: the expansion rate may even accelerate at times, but on the average, it should obey the formulae of the FLRW model.

Note that if ΛCDM is augmented with an "evolving" dark energy, the model becomes even more ad hoc than it was before. We can explain any deviation from the FLRW expansion rate by adding an evolving dark energy!


Retardation generates "negative mass" inside a collapsing spherical shell


Retardation makes light to bend away from the central volume of a collapsing shell. This is equivalent to putting some negative mass to the central volume

Could it be that this negative mass is relatively uniform throughout the collapsing dust ball? This could explain the uniformity of the CMB.

Let us use comoving coordinates of the dust in a collapsing dust ball. On January 18, 2025 we argued that gravity in those coordinates may look newtonian. We can draw lines of force for the gravity field in a familiar way.

Let us imagine that the collapsing dust ball consists of concentric collapsing dust shells. Could it be that, in the comoving coordinates, these shells create a fairly uniform density of "negative mass" inside the dust ball?

The density of negative mass is zero at the edge of the dust ball, but may be relatively uniform inside the ball. Then in a small subvolume of the ball, it may look almost exactly uniform.

Let us have a contracting uniform shell whose radius is R(t). The first guess for the "retardation potential" for the shell is something like

       V  ~  -r

for r < R(t). That is the, potential is the highest at the center of the shell. However, this does not seem like a good guess, since the negative mass density for this potential is

       ρ  ~  1 / r.

There would be a singularity at the center, which does not look nice.

In a rubber sheet model of gravity, a collapsing shell corresponds to a ring of weights moving toward a center. The rubber sheet in this case can "anticipate" linear processes: the sheet moves downward at a constant speed. But it cannot anticipate an accelerating motion of the weights.

Maybe we should only make a retardation potential based on the acceleration of the masses in the collapsing dust ball?


Conclusions


We have many reasons to believe that there is retardation when a gravity potential adjusts the rate of clocks. It would be strange if a clock at the center of a collapsing spherical shell would immediately know how fast it should tick.

Retardation causes a potential which resists the collapse of the dust ball. A very naive calculation shows that retardation may at times create a repulsive force which is stronger than the attraction of gravity.

The naive retardation model is awkward since the negative mass which would create that potential would have an infinite density at the center.

We will next look at a more sophisticated model in which only the acceleration of the collapse creates retardation. Is retardation large enough to explain dark energy?

The result published on March 18, 2025 makes ΛCMD even more awkward than it was before. Dark energy density can change as time progresses. The predictive power of such a physical model is zero!

Tuesday, March 4, 2025

How is energy conserved in the collapse of a dust ball?

Let us look at the collapse of a uniform "dust" ball. Besides dust particles, the ball may contain photons, or other elementary particles which may move at almost the speed of light.


Collapse of a dust shell on a cloud of photons: conservation of energy?

                       
                              •  dust
                              |
                              v

      • --->            ~  ~  ~            <--- •
                        photons

                              ^
                              |
                              •


In newtonian gravity, dust particles will gain some kinetic energy from the decreasing "gravity potential" of the matter inside the shell.

The declining potential drains energy from a photon by making its frequency lower. General relativity says that "time slows down".

In newtonian gravity, we would have a paradox: the photon would keep its frequency, and could escape to space without losing energy.

In a more advanced model of gravity, we have another problem: if the shell starts to contract very quickly, then the photons inside cannot know that they are in a low gravity potential. Their frequency, measured in global coordinates, cannot slow down. Falling dust particles gain kinetic energy, but that energy is not yet subtracted from the photons.

One could claim that the "negative energy" in the gravity field outside the shell grows, and gives the kinetic energy to the dust. But we do not like the concept of a negative energy density.

Suppose that the knowledge about the gravity potential propagates at the speed of light. The photons will soon learn that they are in a low potential, and their frequency will drop. Conservation of energy is restored, but only afterwards.

We may imagine that all matter consists of waves propagating at the speed of light. Then also massive particles lose energy in a low potential through the same mechanism as photons.

We have here a mechanism which may involve very large retardation effects. If energy cannot be "borrowed" in large quantities, then it has to flow from the photons to the kinetic energy of the dust, and that process can take time.

General relativity works around this problem by claiming that there is no clear definition of energy, or energy conservation.


When are clocks slowed down inside the collapsing dust shell?


One could claim that it does not matter when exactly clocks are slowed down inside the shell. That the end result is the same, as long as the slow period has the same length in various alternative histories.

However, the timing of the slow period does affect the end result. We can compare the ticking rates of clocks at various distances from the center, inside the shell.

For example, if the slowing down of clocks would happen instantaneously as the shell falls down, then clocks inside the shell would stay synchronized. But if the slowing down propagates at the speed of light from the shell, then a clock at the center will show a later time than a clock close to the shell.


Lowering two masses on a third one


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


Suppose that we have three objects whose mass M is the same when measured alone in the space. We let the masses on the left and the right fall freely on the center mass M. We may imagine that the masses can pass through each other, and completely overlap in the end state.

The total mass-energy of the system remains at 3 M at all times, measured by an observer far away.

But measured locally at the center, the two moving masses M possess more mass-energy than the center one. It makes sense to claim that the moving masses have gained energy from the center mass.

Could there be a retardation effect here? If the masses on both sides are rapidly accelerated to approach the center mass, how does the center mass know that it should give up some energy? This would amount to a breach of Gauss's law for gravity, since Gauss's law states that the energy in a spherically symmetric collapse is immediately available, regardless of an acceleration.


Gauss's law does not hold for gravity?


In electromagnetism, an expanding shell can collect energy from the electric field E at the the immediate vicinity of the shell. There is no obvious reason for any retardation.

The expansion of the universe is the only gravitational collapse/expansion for which we have measured data. The data shows that the expansion does not happen in a way which is compatible with Gauss's law: there seems to be dark energy. Thus, the only empirical data, which we possess, suggests that Gauss's law does not hold for gravity.

We can treat the failure of Gauss's law as the primary hypothesis for gravity. 

Do we have empirical data for electromagnetism, such that it would prove that Gauss's law holds? The derivation of the power of electromagnetic waves, by Edward M. Purcell, depends on Gauss's law. The derivation produces the empirically correct formula. We are not aware of any experiments of Gauss's law with static charges, though.

The derivation of the power of gravitational waves is quite different from the derivation in electromagnetism. Also, the power is 16-fold compared to the analogous electromagnetic system. Thus, the derivation might not provide evidence for Gauss's law for gravity. The derivation does provide evidence that the linearized Einstein equations describe the gravity field right for a quadrupole oscillation, in some sense. But a spherically symmetric collapse/expansion is quite a different process from an oscillating quadrupole.


Accelerating a heavy neutron star: Gauss's law probably fails


                  test mass
                        m
                         •
                         ●  ---> a acceleration
                        M
               neutron star


Let us use a rope to accelerate a very heavy neutron star M. A test mass m close to its surface will necessarily follow the movement, since the speed of light is very slow at the surface.

The "force" which moves m is distinct from the newtonian gravity force, and distinct from the possible gravitomagnetic force associated with the newtonian force. Thus, there is no reason why the force moving m would satisfy Gauss's law.


Why would Gauss's law hold in the collapse of a dust ball?


Gauss's law in electromagnetism requires that a varying magnetic field attachs the ends of the lines of force of Coulomb's force.

We do not have any empirical measurements of gravitomagnetism. We do not know if it can attach the lines of force of the newtonian gravity force.

Thus, it may be that Gauss's law does not hold in the collapse of a dust ball, or in the expansion of the observable universe. This opens the possibility that dark energy is a manifestation of Gauss's law failing.

Retardation effects may at some phases of the collapse make gravity surprisingly weak. Conservation of energy requires that gravity, on the average, is as strong as Gauss's law requires. Then the acceleration of the collapse can oscillate in a surprising way.

Can we deduce something about this process? Can we prove that Gauss's law cannot hold?


Pressure breaks Gauss's law for gravity, for a single test mass m


In the fall of 2023 we realized that Tolman's paradox breaks Birkhoff's theorem, and consequently, also Gauss's law.

Pressure arises, for instance, from moving gas molecules. In the collapse of a dust ball, dust particles acquire radial speeds.

If we move a test mass m closer to a collapsing dust ball, the metric stretches in the radial direction from m. The volume of the space increases, and, effecticely, the fall of dust particles slows down in certain directions. Some kinetic energy is released, and that energy pulls m closer to the dust ball?

Looking from far away, the collapsing dust ball is much like a gas cloud with a pressure. The velocities of the dust particles are nicely ordered, though. That is the difference from an ideal gas where particles move randomly. 

It seems likely that the pressure does create a gravitational pull outside the collapsing dust ball.

In the expanding universe, the speeds of galaxies are close to the speed of light. The effects of the pressure may be very large. We have to investigate how this breach of Gauss's law affects the expansion.

The Oppenheimer-Snyder paper (1939) uses the comoving coordinates of Tolman. They can ignore the effects of the pressure which arises from the movement of dust particles. But on May 26, 2024 we showed that the Tolman coordinates will produce flawed results.

If we have a spherical shell falling toward the collapsing dust ball, the shell will not stretch the distances within it. Pressure does not pull a spherical shell?

However, the falling shell does slow down clocks inside it, since the gravity potential falls. Could it be that the slowdown of clocks creates another gravitational pull, besides the newtonian one?


What force slows down the collapse of a dust shell?


When a collapsing dust shell falls into a low gravity potential, its contraction, measured in some static coordinates, slows down because the coordinate speed of light is slow in a very low gravity potential.

If we study the process in the coordinates, the momentum of each side of the shell decreases. What kind of a force is responsible for this slowdown?

The total momentum of the collapsing shell is zero. However, since each side has a substantial momentum, a lot of momentum must flow between the different sides of the shell. How does that happen?

If the slowing down of the mass somehow results from a pull from the environment of the dust ball, then there is a negative pressure inside the dust ball. This might be the negative pressure which is imagined to be the result of dark energy?

Then there would be a positive pressure in a large volume around the dust ball. From the surrounding space, "skyhooks" would slow down the collapse of the dust ball. This is very ad hoc, but let us investigate.

Could the skyhooks cause an oscillation, so that the observed contraction rate inside the dust ball could even slow down?


A rubber sheet model of the collapse of a photon shell


                          ^  y
                          |                    lighter
                          |
          ---------------------------> x
                          | denser
                          |                          rubber sheet


Let us imagine that we have a tense rubber sheet, such that its mass density is larger at the origin. The speed of waves in the sheet is slower near the origin.

We send a circular wave which would collide at the origin. The energy of the wave stays the same as it approaches the origin, but the radial momentum of the wave energy is reduced as the wave moves slower.

This means that the rubber sheet must somehow absorb some of the momentum of the wave.

What effects could this extra momentum have on the system?

The tense rubber sheet can only transfer momentum to the other side of the circular wave by making the tension smaller inside the circle. The momentum would flow in a way where the surrounding sheet pulls the circular wave radially outward from the origin. Could this be the negative pressure of dark energy?


Retarded slowing down of clocks produces a repulsive force in a collapsing dust ball


We had the hypothesis that it takes time for a clock to know that it is in a low gravity potential: the knowledge about the potential can only propagate at the speed of light.

      
           # -->                    ×                     <-- #
     dust shell           center              dust shell


Now we realize that this produces a repulsive force inside a collapsing dust shell. The speed of light is larger at the center of the ball than close to the shell, because the center does not yet know that it is at a low potential.

A wavefront of a photon always steers toward the slower speed of light. That is, a ray of light is bent away from the center. There is a repulsive force. The force can be very large for a dust ball which mimics the universe?

There might be an oscillating process here. There is no repulsive force if the shell is static.


                         •      •
                   •     •     •     •    dust ball
                   •     •     •     •
                         •      •


Let us imagine that the dust ball starts from a static state. The speed of light is slower near the center. That reflects the gravity force which pulls photons toward the center.

In the following, we do the analysis in coordinate time and the radial coordinate. We ignore proper time and proper length.

      
          -----___                ___-----
                      ----___----            gravity potential
                             
                            × center


The potential at the start will fall quickly (in coordinate time) at the edges of a very heavy dust ball, but the information of the fallen potential reaches the center slowly. It is imaginable that the potential graph may "invert" inside the ball. Layers close to the center may then feel a repulsive force from the center.

What happens later? As the edges fall close to the Schwarzschild radius, they will freeze. Maybe the inside of the ball will again feel an attractive force toward the center?

This might explain dark energy, as well as the Hubble tension, and the overmaturity of old galaxies in James Webb photographs.


Conclusions


Let us close this long blog post. We will write a new blog post about our last idea: retarded slowing down of clocks.

Thursday, February 27, 2025

Dark energy comes from bouncing gravitational waves of galaxy groups?

Suppose that we have individual electric charges placed roughly into a configuration of a spherical shell of a radius R. The number of charges is rather small, say 50.

The repulsion between the charges accelerates them. An accelerating charge radiates radio waves, the process which drains energy and slows down the expansion.

But at the distance > 2 R, destructive interference wipes out almost all the energy in the waves. The energy must return to the charges, speeding up the initially slowed down acceleration.


How uneven is the matter content in the universe?


If the matter content in the universe is large-grained enough, the analogous oscillation can substantially alter the speed of the expansion. Can it explain dark energy?


The Boötes Void is one of the largest voids in the universe. It is 330 million light-years across.

The accelerating expansion of the universe is observed in surveys which span several billion light-years.


Gravitational waves


When masses are accelerated in the universe, the uneven distribution produces gravitational waves whose wavelength might be ~ 300 million light years for the largest "pointlike" structures. The waves then travel to various directions, until they are eliminated in destructive interference. The energy in the waves must then be reflected back to the sources which produced the waves.

For the largest structures, the destructive interference may only happen when the waves have traveled ~ 1 billion light-years. Thus, we might have an oscillation whose period is 2 billion years.

If the expanding mass structure is a spherical shell, then the power of its gravitational waves is zero. Only "pointlike" structures produce waves.


The power of gravitational waves from an accelerating gravity monopole









In literature, the power of gravitational waves is always calculated for a quadrupole. The power is 16 times the power of an analogous electric charge.

Let us assume that the power for a pointlike mass M is 16 times the analogous Larmor radiation power:

       P  =  32/3 G / c³  *  M² A²,

where A is the acceleration. 


The scale factor a of the universe in the matter-dominated era is

       a  ~  t^2/3,

and its time derivative is

       da / dt   ~   t^-1/3.

One billion years ago, the age of the universe was 7% less than now. The expansion velocity was 2% larger than now.

Let us have a galaxy cluster M at the distance of two gigaparsecs from us. It recedes from us at the speed of c / 2. The change in the speed is 2% in a billion years, or 3,000 km/s. The acceleration is

       A  =  3 * 10⁶  /  3 * 10¹⁶  m/s²

            =  10⁻¹⁰ m/s².

The radiation power is

       P  =  32/3 * 7  /  3³  *  10⁻⁵⁵  *  M²

           =  3 * 10⁻⁵⁵  M²

watts. Let us have a cluster whose mass is

       M  =  10⁴⁵ kg.

The energy radiated over 1 billion years is

       E  =  3 * 10³⁵  *  3 * 10¹⁶

            =  10⁵²

joules. Let us compare this to the kinetic energy of the galaxy cluster:

       E'  =  1/2 M v²

            =  10⁴⁵  *  10¹⁶

            =  10⁶¹

joules, of which 4% is 4 * 10⁵⁹ joules. We conclude that gravitational waves cannot significantly affect the deceleration.

The result would be completely different if the galaxy cluster would have a mass which is close to the amount of observable matter and dark matter in the universe, 10⁵⁴ kg. Then gravitational waves would have a very large effect.


Conclusions


The uneven distribution of matter and dark matter in the universe cannot explain dark energy. The effect is at least a million times too small.

Saturday, February 15, 2025

Incompressible Navier-Stokes has no dynamic solutions?

On the Internet, people claim that the lagrangian density for the Navier-Stokes equations for an incompressible fluid is of the form:

      T  -  V,

where T is the kinetic energy density of the fluid, and V is the pressure of the fluid. The pressure acts as the potential energy density.

The viscosity of the fluid has to be zero. Otherwise, friction would drain energy, and the traditional lagrangian method would not work.


           ^   ^
           |    |    flow of fluid
           |    |


Suppose that we have a nice smooth solution for the equations. Let us transform it infinitesimally, so that we make the "parcels" of fluid to bump into each other:


         ^    ^
          \   /
          /   \    "bumping" flows of fluid
          \   /
          /   \


The slight variation in the path of the parcels contributes negligibly to T, but the bumping contributes significantly to V.

We proved that the nice smooth solution is not a stationary point of the action?

This might be a negative solution to one of the Millennium Problems:


Our thought experiment also explains turbulence: the action seeks an extremal point, and makes the paths of fluid parcels meandering, so that they bump into each other.

Let us check if we really can modify the total, integrated pressure with this trick. Maybe the pressure of the flow decreases somewhere else, and compensates for the pressure in the bumps?

An incompressible fluid is very unnatural. Can we make a stable physical system if one can create pressure without expending energy? If not, then the Navier-Stokes equations might have no solutions, except in some trivial cases.

In our example, we have a uniform laminar flow, which we replace with a "bumping" flow. Does this mean that even trivial solutions of the equations are unstable if the fluid is incompressible?


Can we make a meandering flow without affecting the kinetic energy T much? Yes


Let us add walls like this to the laminar flow:


            ^      ^      ^
            |       |       |     flow out

          /     \     /     \
          \     /     \     /    "meandering" walls
          /     \     /     \
          \     /     \     /
               
             ^      ^      ^
             |       |       |        flow in


Each wall guides the flow which is close to the wall, making the flow meandering.

We can imagine parcels of the fluid moving along the walls. The parcels bump to each other. Each bump involves an increased pressure.

If the walls differ from a vertical line by an angle ε, then the kinetic energy T may grow by ~ ε², but the pressure grows by ~ ε. We found an infinitesimal variation which changes the value of the action. The original laminar flow was not an extremal point of the action.

There is a problem, however. If the horizontal pressure is increased in the meandering zone, the zone will start to expand outward. Does this spoil the variation? It makes the end state different from the one we started from. The end state should stay the same in the variation.


Preventing the meandering flow from expanding: put it inside a pipe


If the flow is infinitely wide in the horizontal dimension, then the outward pressure would not be a problem. Another way is to enclose the flow into a pipe.

The Millennium Problem does not allow walls, though. It is stated in an infinite pool of water, filling the entire ℝ³.

Let us have an infinitelty rigid pipe in the fluid, preventing the fluid from spreading from the zone where we make the flow meandering.


              meandering
                     flow
                   
              ■    ^     ^    ■
              ■    /      \   ■
              ■    \      /   ■ 
              ■    /      \   ■
     solid                     solid
     wall                      wall


Does the variation above really increase the pressure V, while the kinetic energy T stays essentially the same?

If the original configuration already contains pressure, what happens?


Pressure must be the same in all directions: this spoils our simple idea


The pressure cannot be just horizontal in an ideal fluid. The pressure has to be the same in all directions. This observation spoils our simple idea.


        pipe
                /         /
              /        /  
              \        \   bend
                \         \ 
                       ^
                         \  flow


In the bend, a difference in the horizontal pressure must accelerate a fluid parcel to the right. But the average pressure in the bend must be the same as in the pipe overall. Otherwise, the fluid would slow down in the bend, and its density would grow. That is not allowed as the fluid is incompressible.

This means that our meandering flows are not able to change the average pressure of the flow. The action integral over the pressure V is unchanged.


Turbulence must be a second order phenomenon?


Suppose that the system looks for an extremal point of the action. The analysis above showed that the value of the action cannot change much if we add meandering. However, the action could change a little, and that may be enough to create turbulence.

Hypothesis. Adding turbulence at some length scale L optimizes the action. Adding it at ever shorter scales, L / 2ⁿ, will optimize the action further. The optimization will not stop at any length scale. There is no smooth solution for the Navier-Stokes equations, in most cases.


The potential is created by the motion: this explains turbulence?


Suppose that the potential V in the lagrangian density is an external potential, say, gravity:

       T  -  V.

The orbit of an object in a newtonian gravity field is very regular, for example, an ellipse. It is the very opposite of the chaos of turbulence.

In the Navier-Stokes lagrangian density, the potential V is created by the flow itself, as it is the pressure. This opens the door for a chaotic process.


Adding viscosity creates turbulence


A large viscosity creates turbulence. What happens to our meandering flows if there is a large viscosity?

Viscosity means that there is a shear stress. The fluid is more like a solid substance then. The pressure still is the same in all directions, though.

The lagrangian density cannot be as simple as given above, if viscosity is present.


D'Alembert's paradox
















D'Alembert's paradox states that, in the absence of viscosity, a laminar, time-independent, flow is possible, and there is no turbulence and no drag.

The pressure of the fluid depends on the speed of the fluid. In the diagram, the flow is symmetric on the left and on the right. Consequently, the pressure is the same on the left and on the right. The pressure does not exert any force on the circular cylinder, i.e., there is no drag.

The physicist Ludwig Prandtl suggested in 1904 that there is a "no slip" condition of the fluid on the surface. The viscosity of the fluid is large close to the surface, even if the viscosity would be close to zero in the bulk of the fluid. The hypothesis can explain many empirical phenomena.


From where does the energy for turbulence come?


Let us assume that the fluid has a zero viscosity and is incompressible. In the diagram above, the red cylinder is static. It cannot do any work.

Since the fluid is incompressible, the average horizontal speed component of the parcels of fluid must be constant. If the fluid arrives straight from the left, it cannot lose any of its original kinetic energy.

If there is turbulence on the right, then parcels have a vertical velocity component. They gained kinetic energy. From where did they obtain that energy? The explanation has to be that the pressure is larger on the left than on the right. A parcel slides down this pressure potential and gains kinetic energy.


Bernoulli's principle


For a time-independent (potential) flow, a larger velocity means a lower pressure.


Bernoulli's principle states that for a steady (= time-independent) flow with a zero viscosity, the energy of a parcel of fluid is constant. The pressure serves as the "potential" of the parcel.

Thus, the lagrangian density which we mentioned at the beginning of this blog post, only works for a time-independent flow.

Suppose that the flow is time-independent and we have a flow which is the stationary point of the action

       ∫  T  -  V  dt,

where V is the pressure. For a time-independent flow, Bernoulli's principle holds. Let us assume that we "feed" the flow at a constant pressure p and kinetic energy density E. Then

       T  +  V  =  E  +  p.

Is the stationary point also a stationary point of the action

       ∫  T  dt?

Probably, yes.


Constructing a singularity: can we solve the nonviscous Navier-Stokes starting from a singular flow? We are not able to construct a singularity


If there is no friction, then fluid physics is time-symmetric. Suppose that we are able to construct a solution which contains a singularity at the time

       t  =  0,

but not at later times. If we run time backwards, we can let a smooth flow to develop into a singularity.


                      ^ y
                      |
            ___     |     ____ line of flow
                     \|/
              ---------------------> x
            _____/|\_____  line of flow
           
  t = 0


We probably can construct a flow such that the speed of the flow approaches infinity at x = 0, for y close to zero, when t = 0.

The total kinetic energy is finite because the amount of fluid running very fast is very small.

The solution is smooth and defined for all t > 0, but we cannot define it at t = 0 in the origin.

If Navier-Stokes has solutions for all smooth initial states, then we can solve it starting from any t > 0. But there is no guarantee that we can combine these solutions to obtain one solution which is valid for all t > 0.

Since Navier-Stokes is nonlinear, an interesting question is if nonlinear equations generally develop singularities.

We cannot use a Green's function to construct a singularity because the Navier-Stokes equations are nonlinear. We cannot construct new solutions through summation.


Nonlinear differential equations decomposed into linear equations plus a perturbation


The tools used for linear differential equations do not work for nonlinear equations. If the nonlinear equations are "close" to linear ones, then we might be able to construct approximate – or even exact – solutions for the nonlinear equations by perturbing solutions of the linear equations.

Let

       G(f) = 0

be the approximate linear equation for a function f. Let

       G(f)  =  H(f)

be the nonlinear equation, where H is "small". The function H(f) is the perturbation which acts as the source for the homogeneous linear equation G(f) = 0.

We can use Green's functions to calculate the "response" of f to the perturbation. If the response is very small, we might be able to get f to converge toward a function which solves the nonlinear equation. This might be the method which has previously been used to derive partial existence and smoothness solutions for Navier-Stokes.

Could it be that we can perform Turing machine computations through this iterative process? Then the existence of solutions to certain problems could be undecidable in the axioms of set theory.


Letting two balls of incompressible nonviscous fluid to collide: a singularity forms? No















Let us have the space filled with an incompressible nonviscous fluid. We interpret the diagram above in this way: the red circle is a ball of the fluid moving horizontally relative to the main mass of the fluid.

The fluid speed seems to be the fastest and the pressure the lowest on the equator of the ball, if the axis is horizontal. Thus, the ball starts to flatten in the horizontal direction.

What happens if we let two such balls of fluid collide head-on?

Since the fluid is incompressible, a head-on hard collision would create an infinite pressure.


         R = 1
         solid ball
                  \                r
                   |                | plane
                  /
                  --> v     ρ fluid


Let a rigid solid ball of a radius R = 1 collide at a constant speed v to a solid plane. They are immersed in the fluid of the density

       ρ.

Let r << R be the distance from the contact point. The flow of the fluid to larger radii r at the contact moment is

       v  *  π r²,

and the area through which it flows is

       A  =  1/2 r²  *  2 π r

            =  π r³.

The velocity of the flowing fluid is

       v / r,

and its kinetic energy density is

       1/2 ρ v² / r².

The kinetic energy contained in the fluid in a thin ring dr around the contact point is 

       1/2 ρ v² /  r²

       * π r³  dr

       =  1/2 π ρ v² r  dr

The total kinetic energy E of the fluid at radii less than r is

       E  =  1/4 π ρ v² r².

Let us assume that the density of the ball is ρ. Then the kinetic energy of the part of the ball at radii < r is

       E'  =  1/2 v²  *  ρ  2  *  π r².

            =  ρ π v² r².

The kinetic energy of the ball is 4X the kinetic energy of the fluid. A hard collision might happen.

In the collision between two moving balls of fluid, is there a mechanism which softens the collision?

The speed of the fluid at the contact point would become infinite. The pressure required to accelerate the fluid is there infinite. Does that pressure deform the balls so that they will not touch?

The pressure which makes the fluid between the balls to acquire great speeds, also affects the fluid at the surface of a ball, and accelerates that fluid. The surface of the ball "melts" and flows along with the fluid between the balls.

We conclude that there probably is no singularity in this case.


Why a Green's function does not create a singularity for a nonlinear equation?


Suppose that the fluid is compressible. Then we can form an approximate linear equation which describes sound waves in the fluid.

A Green's function is the response of a linear equation to a singularity impulse – the impulse is a Dirac delta function. The Green's function is obviously singular at the point of the impulse.

But the precise Navier-Stokes equation is not linear. If we imagine a sound wave with an infinite energy density, then the fluid close to the infinite energy density is obviously very much compressed. A simple linear sound wave equation fails if the density of the fluid varies greatly, according to the phase of the sound wave.

Thus, the perturbation of the linear equation becomes very large at a singular point, and we cannot expect an iterative solution method to converge. There is no guarantee that any function f satisfies the nonlinear equation and looks similar to a Green's function.

Is it so that any linear approximation of the Navier-Stokes equation has the property that the approximation is very bad close to a singularity?


Transverse waves in a viscous fluid or elastic solid


If the fluid has a huge viscosity, we might be able to create transverse waves. If the viscosity grows without bounds even at moderate shear speeds, then the fluid might behave much like an elastic solid.

Let us investigate transverse waves in a newtonian solid. Are we able to create a singularity with a Green's function?

A linear wave equation probably only works for small-amplitude waves. A singularity would have a large amplitude. Thus, the wave equation would not be linear. This approach is not going to help us.


The wrinkle in a carpet trick can make solutions nonexistent?


In the fall of 2023 we argued that the Einstein equations cannot have a solution because they behave like we would need to fit an oversized carpet in too small a room. Locally, we can press the carpet optimally to the floor, but a wrinkle will always persist somewhere in the carpet.

Can we utilize this idea if we assume an exotic fluid whose viscosity behaves in a strange way when the shear velocity varies?

Then we may locally have a nice solution to the fluid flow problem, but globally, an annoying "wrinkle" always persists?

Regarding the Millennium Problem, this is cheating. The purpose of the problem is to study turbulence – not oversized carpets.


Exotic viscosity which declines when the shear speed increases: this produces singularities


        F shear force
         <---- 
                   ----------    
                   ----------   pile of papers
                   ----------
                                ---->
                                    F shear force


Suppose that we have a pile of paper sheets on the table. The friction between sheets of paper becomes less if the speed between the sheets is larger. We exert a shear force. Exactly one pair of papers starts to slide relative to each other.

The sliding gap is a singularity. The same trick probably can be used for a viscous fluid. The speed of the fluid will be a non-smooth function at the gap.

Let us check if the Millennium Problem bans exotic viscosity.










The statement of the problem assumes a newtonian fluid, in which the "friction" is linear in the shear velocity. Thus, our trivial singularity is ruled out. The constant ν is the kinematic viscosity.


Viscous fluid and two opposite streams: explosion backward in time


     ^
     |                  --------> v
  0 --                                      2 L transition zone
     |                 <-------  -v
     x


Let us have two streams, between which there is a narrow transition zone where the speed of the fluid varies smoothly between -v and v. The velocity

       v(x)

is a monotonously growing function of x, and symmetric around x = 0.

If we calculate the system backward in time, will the velocity field end up with a discontinuity at x = 0 in a finite time?

The differential equation is 

       dv / dt  =  ν d²v / dx².

Let v = 1 asymptotically when we go to large x, and v = -1 asymptotically when we go to large negative x. Let the viscosity ν be 1.

Let us assume that

       v(t, x)  =  1  -  exp(t  -  x)

at t = 0. Then

       d²v(t, x) / dx²  =  -exp(t  -  x),

and

       dv(t, x)            =  -exp(t  -  x).

We found a solution? No. The value of v(t, 0) should be 0 for all t.

Let

       v(t, x)  =  exp(-a² t)  *  cos(a x),

where a is any real number. Then

       d²v / dx²  =  -a² exp(-a² t)  *  cos(a x),

       dv / dt      =  -a² exp(-a² t)  *  cos(a x).

We found another solution.

Let us choose a sum of the form:

                        ∞ 
       v(t, x)  =  ∑   1 / 2ⁿ  *  exp(-n² t)  * cos(n x)
                     n = 0

The function v(t, x) is defined at t = 0, x =0, but is infinite for t = -1, x = 0.

Are all the derivatives of v(t, x) defined at t = 0? The derivatives contain polynomials of n. They do converge.

We found a function v(t, x) which explodes if we go backward in time. But we are interested in solutions forward in time, starting from a smooth initial condition.

Does the following work: v(t, x) is defined as as an alternating series where negative terms cancel the positive ones. Let us make negative terms such that they shrink when t > 0. The sum becomes infinite. But can we define it for all x at t = 0?

Our example has only two spatial dimensions. Wikipedia says that Navier-Stokes does have smooth solutions in that case. We should construct an example which uses three spatial dimensions, if we wish to prove that a smooth solution does not always exist.


Incompressibility leads to paradoxes?


If the fluid is compressible, we can, in principle, solve the Navier-Stokes equations locally. No change in the state of the fluid can propagate faster than the speed of sound.


                             pipe
             ________________________ ...
    --->   ________________________ ...
    push


Suppose that we have an infinitely long rigid pipe filled with an incompressible fluid. Obviously, one is not able to get the fluid to move at all.

The same holds for fluid between two infinite, rigid planes. If the distance from the person pushing the fluid is r, then the velocity of the fluid would be roughly

       ~  1 / r,

the kinetic energy density would be

       ~  1 / r².

The integral

         ∞
        ∫    1 / r²  *  2 π r  dr
       0

diverges. Pushing the fluid would require infinite energy. In three spatial dimensions, the energy integral does converge.


Conclusions


We were not able to discover anything new about the Navier-Stokes equations. They are nonlinear equations. We do not know a nice action principle for them, so that we could utilize variational calculus to study the existence of a solution.

It might be that the answer to the Millennium Problem is affirmative: there always exists a smooth solution. Our efforts to disprove that failed.

We will next return to studying retardation effects in gravity. Can retardation explain "dark energy"?

Wednesday, February 12, 2025

Gravity and electromagnetism is nonlinear: can Maxwell's equations be linear?

Let us perform the following thought experiment. We have a long uniformly charged pipe, such that the electric field is essentially zero inside the pipe.
       


              __   "drooping"
            /     | E       
           |     v         |  
           |                |     pipe
           |                |
                               Q charge
                    ^
                    |
                    ●
                   M


We bring a large mass M close to an end of the pipe. The electric lines of force E will start to "droop". There will be a vertical electric field inside the pipe.

From where does the energy to this vertical field flow?

If we think about an elementary charge q, its electric field E' has potential energy in the gravity field of M. The process of drooping converts some of the gravity potential energy of E' to electric field energy, since the drooping field has slightly more electric field energy than a symmetric field.

However, inside the pipe, the electric field E was essentially zero, and its gravity is essentially zero. Drooping cannot release much gravity potential energy of the field.


A rubber plate analogue: this is very different from the electric field, if we have several charges


                    q             drooping
          ___----•----___    rubber plate


                    ● M


Let us model the electric field of an elementary charge q with a horizontal rubber plate attached to q.

The drooping in the gravity field of M is caused by gravity pulling each part of the rubber plate. The plate is distorted and will gain some elastic energy, and lose some gravity potential energy.

Let us then imagine that each elementary charge q in the pipe has its own rubber plate.

The "linear sum" of those plates indicates a drooping electric field.

However, if we think about the total electric field E of the pipe, there is essentially no gravitating mass inside the pipe. Will this affect the drooping, and break the linearity of electromagnetism?


Gravity, or acceleration, makes electromagnetism nonlinear?


Maxwell's equations are linear. However, gravity couples to the field energy density

        1/2 ε₀ E²  +  1/2  *  1 / μ₀  *  B².

Is there any reason why electromagnetism should be "linear", in some sense, under gravity?

There is a good reason to assume that the electric field mass-energy density really is in the formula above, also with respect to gravity. If we have an electric field somewhere, we can reset the field to zero with capacitor plates, and harvest the energy just at that location.

Let us take two electric charges q and Q. If each charge exists separately in space, its electric field is spherically symmetric. But if we put both close to each other, then the gravity of each electric field distorts the other field. It is not linear, in this sense.


Electric fields under a constant acceleration


Let us then study static electric fields of static charges under a uniform gravity field. This is equivalent to studying a constellation of uniformly accelerated charges. We ignore the gravity of the electric fields on other electric fields.

If we take charges q and Q, can we determine their electric fields separately, and sum the fields to obtain the field of the combined system q & Q?

If we take both q and Q, then the mass-energy of their combined electric field changes in a complex way. Is there any reason why gravity would bend the combined field in the exact same way as it bent the fields separately?

We encounter the August 24, 2024 problem once again: what is the inertia and the kinetic energy of the combined electric field of several charges?


A small cylinder charge inside a uniformly charged pipe, in uniform gravity: the lowest energy state is not linear


                               Q
          |                 |   --->  E 
          |        |        |
          |        |  --> |  ---->  E + E'  
          |       q  E'   |           W "extra" energy
          |                  |
                                    R = radius of the cylinder

                   ●  M


Let us have a vertical cylinder uniformly charged. The charge is Q. How much will the lines of force of its cylindrical electric field "droop" under gravity?

Assumption 1. The drooping angle α of a line of force probably approaches zero close to a very thin cylinder.


Assumption 2. The drooping angle is

       α  ~  r,

where r is the distance from the cylinder. The line of force is a parabola. This is what people generally assume. This matches the "geometry of spacetime" interpretation of general relativity.


The vertical component of the field of the cylinder is roughly constant: the strength of the field is ~ 1 / r, and the drooping angle is ~ r.

Let E be the field of the pipe. If the pipe is very long, and we are studying its middle part of a length L, then the horizontal component of E inside the pipe is essentially zero. This is because an end part of the pipe of the length L has a negligible field near the midpoint.

Thus, the field E of the pipe, inside the pipe is vertical, and roughly constant. The field does not depend on the radius R of the pipe.

Let q be a thin uniformly charged vertical  cylinder at the center of the pipe. Let E' be its field. The field E' is roughly constant inside the cylinder. The vertical component of E' is roughly constant, too.

The energy from the interaction of the fields E and E' inside the cylinder is

       ~  E E'  *  π R²

       ~  R².

Let E' be the field of q when the pipe is not present, and E the field of the pipe when q is not present.

We superpose these fields E and E'. Let us claim that the total energy of the combined system pipe & q is minimized by the sum field E + E'.

We vary the drooping angle α of E' inside the pipe.

We denote by W the energy from the interaction of E and E' outside the cylinder.

The energy saved in the gravity potential of the energy W outside the pipe is approximately

       ~  α R.

The price paid is the energy added to the combined field E + E' inside the pipe, and is also approximately

       ~  α R².

Let us assume that when the radius of the pipe is R, an infinitesimal variation of α does change the total energy of the system.

Let us the halve the value of R. We claim that still, E + E' minimizes the total energy of the system.

But that is not true. The price paid with an infinitesimal increase of α is now only 1/4 of the original price, but the energy saved is more than 1/2 of the original saving. More than 1/2, because W is now larger.

We end up in a contradiction. What is wrong in our assumptions?

Because of an equivalence principle, our analysis in a uniform gravity is equivalent to newtonian mechanics in a uniformly accelerating system. Maxwell's equations seem to clash with newtonian mechanics, if we assume that the energy density of an electric field E is locally

       1/2 ε₀ E².

Since we firmly believe that newtonian mechanics is correct, the error has to be in Maxwell's equations.

Hypothesis. Maxwell's equations do not have a solution for any system which contains accelerating charges.


The hypothesis is similar to our result that the Einstein field equations do not have a solution for any dynamic system.

It is not known if Maxwell's equations do have any dynamic solution. In the literature we have not seen a proof that they would have solutions. The equations do have solutions for static charges, just like the Einstein equations have the Schwarzschild solution.


The electric field does not bend because of gravity?


Maybe the electric field does not seek the lowest energy state inside a gravity field? It does not care of the weight of the energy W?

That would be strange. We firmly believe that gravity (or acceleration) does bend the field lines of a single charge. Why it would not care about the "weight" of the field of several charges?

A physical system which does not care to seek the minimum energy state is susceptible to a perpetuum mobile.

An option is to use the "private field" concept, which we have discussed many times in our blog. Maybe each elementary charge has a private field which finds its form under gravity independently, regardless of other charges? That would be similar to the rubber plate model which we described above.

In classical electromagnetism, a well-behaved charge is a continuous charge distribution. There is no elementary charge.


Linear Maxwell's equations break energy conservation in a homogeneous gravity field


In our pipe example, the field E of the pipe inside the pipe is roughly constant, regardless of the radius R of the pipe.

Thus the force F on the charge q does not depend much on R. But the weight W which q lifts up does depend on R. Does this break conservation of energy?

Let us try to calculate numeric values. Let g be the acceleration of gravity. A photon will fall a vertical distance

       s  =  1/2 g t²

in a time interval t. There,

       t  =  R / c,

       s  =  1/2 g R² / c²,

       ds / dR  =  g R / c².

Let us have a narrow cylinder whose charge density per length is ρ. The radial field at a distance R is

       Er  =  1 / (4 π ε₀)  *  ρ / R,

and the "drooping" vertical field component

       E  =  1 / (4 π ε₀)  *  ρ / R  *  g R / c²

            =  1 / (4 π ε₀)  *  g / c²  *  ρ.

It does not depend on R, as we wrote above. We can as well denote by ρ the charge density of the pipe per length.

The downward force on the cylindrical charge q is

       E q  =  1 / (4 π ε₀)  *  g / c²  *  ρ q.

Let us then calculate the mass-energy of W. It is the potential energy V of q in the field of the pipe. Let L be the length of the pipe. At distances < L, the field is roughly the formula of Er above. The potential is very roughly

       V  ≈  1 / (4 π ε₀)  *  ρ q  *  ln(L / R),

and the weight of W is

       1 / (4 π ε₀)  *  g / c²  *  ρ q  *  ln(L / R).

We assumed that L is much larger than R. Thus, the weight of W is much larger than the electric force pushing q down.

This does not make sense. If we assume that electromagnetism is linear, we can lift a heavy weight W up with a small force on q. It breaks conservation of energy. Or does it? Can we harvest the increased gravity potential energy of W somehow?


                               Q
          |                 |   --->  E 
          |        |        |
          |        |  --> |  ---->  E + E'  
          |       q  E'   |            W "extra" energy
          |                  |
                                 

                   ●  M


We can do the harvest, if gravity is not uniform everywhere.

      ____
              \  W
                \________   gravity potential


Let the schematic gravity potential be like in the diagram. We lower the whole system into the pit in the potential. But during the process, we are able to lift W up with a negligible force. We can harvest the gravity potential from W twice. Energy conservation is broken.

How to repair the model? Let us assume that the field of q somehow exerts a "self-force" on q, and presses q down. Then energy is conserved.

However, then the inertia of q might depend on the potential of q? Or does it?

The spectrum of the hydrogen atom does not change if the atom is in a high potential, which indicates that the electron has the same inertia inside the atom. We have discussed this several times in our blog. Maybe the atom somehow shields the electron from changes in the inertia?


How to interpret energy non-conservation in an accelerating system?


Above we show energy non-conservation under a homogeneous gravity field, whose acceleration is g. An equivalence principle says that it is analogous to a system which is accelerated with the acceleration g.

In the gravity field, we have supported the pipe and the charge q with some structures. In the accelerated analogue, the system is in empty space, and we have a spring, which is accelerating these structures.

     
                      W 
                   o /    lift up
                   |                     ^
                  /\                    |  g acceleration
               ----------------------   
                 |         \           structure
                 v        /
                 F        \
                           /   
                           \  spring


Energy non-conservation in the accelerated analogue means that the sum of kinetic and potential energies does not remain constant. 

For the energy to remain constant, we would need a man doing the weightlifting of W. The man would spend some potential energy possessed in his muscles, to lift W. Where would his energy go? His work saves some work done by the spring accelerating the system. When the man starts lifting W up, the spring feels an extra resisting force F. When the man ends the lifting, the spring is relieved by the force -F. Since system is moving faster in the -F phase, the spring saves some work.

But if the inertia of W would be zero, then energy would be conserved?

If momentum is conserved, then moving W up will exert a force F on the spring. The man might be using a motor to lift W up, and will not personally feel the inertia of W, but the spring will feel the inertia of W, regardless.

What could constitute such a "motor" in the system? In the accelerating frame, the field of the pipe and q is static. We do not see how any "motor" could help in lifting q up. If such a motor exists, it probably is not described by Maxwell's equations.

On August 24, 2024 we suspected that Maxwell's equations do not understand the kinetic energy of field energy. Our analysis above suggests that that really is the case.


Do Maxwell's equations understand that energy is not conserved in the accelerating pipe & q system above?


Our analysis suggests that Maxwell's equations break newtonian mechanics in an accelerating system. Could it be that Maxwell's equations actually have no solution in such a case? By Noether's theorem, an extremal point of the electromagnetic action should conserve energy. If energy is not conserved, then it is not an extremal point, and does not satisfy Maxwell's equations.


        Q
          |                 |   --->  E 
          |        |        |
          |        |  --> |  ---->  E + E'  
          |       q  E'   |            W "extra" energy
          |                  |    ^ 
          |                  |    |  g
        -----------------------                       
          |        /
          v F     \
                    /
                    \   spring


Let a man push q up in the diagram. The man needs to do very little work, but there is a significant momentum in W as it moves up.

Momentum is conserved. Therefore the pipe must press the spring down with a significant force F. When the man stops pushing q up, the spring feels a force -F. In the process, the spring saves some work, compared to the process where q is not pushed up.

Let A be a process where q is not pushed up, and B be a process where q is pushed up.

The electromagnetic action (Maxwell's equations) says that at the end, the electromagnetic energy in the system is the same in both cases A and B.

However, in case B, the spring did less work, i.e., the spring retained more potential energy. Energy cannot be conserved in both cases. Maxwell's equations do understand these things, as shown by Poynting's theorem.

An extremal point of the electromagnetic action must conserve energy. We conclude that the process which we described is not an extremal point, and is not a solution of Maxwell's equations. This probably means that Maxwell's equations do not have any solution for the described process B.

If the very simple process above does not have a solution, then Maxwell's equations probably do not have a solution for any dynamic system at all. This complements our May 26, 2024 result that the Einstein field equations do not have a solution for any dynamic system.

We have a tentative proof for the Hypothesis which we stated in an earlier section.

Maxwell's equations might still have solutions for a cylindrically symmetric collapse or an expansion of a charge shell. For gravity, we were able to show that the Oppenheimer-Snyder collapse is a suspicious, probably erroneous, solution.

However, if we have to repair electromagnetism by moving to a rubber plate model, then a spherically symmetric collapse will probably happen in a way which does not agree with Maxwell. There could be oscillation, "dark energy". Rubber plates tend to oscillate.


Various lagrangians for electromagnetism plus gravity



We have to check how various proposed lagrangians (actions) handle the pipe example which we constructed above. Does the weight of W distort the electric field of q?

We do not really need gravity. How does the "weight" of W in a uniformly accelerated system interact with the electric field of q, using a special relativity lagrangian of electromagnetism.


Conclusions


This analysis is very preliminary. We have to check this carefully in future blog posts. If the analysis is correct, then Maxwell's equations seem to have no solution for any dynamic problem at all, if there are at least two charges. The reason is that there is no concept of a "self-force" of the field of a charge q on itself. In our pipe example, the field of q should exert the force of the weight of W on q itself, but Maxwell's equations have no such mechanism.

In the idealized case of a uniformly charged infinitely thin shell expanding or contracting, there may exist a solution to Maxwell's equations.

We want to study if "dark energy" could come from some anomaly in the behavior of expansion or contraction. If we correct Maxwell's equations, can that introduce an anomaly?

We will next look at the Navier-Stokes equations. They are nonlinear. Can we prove that they have no solution, expect in trivial cases?