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?

***  WORK IN PROGRESS  ***

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?

https://en.wikipedia.org/wiki/Bo%C3%B6tes_Void

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

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

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. 

https://en.wikipedia.org/wiki/Scale_factor_(cosmology)

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.

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:

https://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_existence_and_smoothness

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

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

https://en.wikipedia.org/wiki/Bernoulli%27s_principle

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.

https://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_existence_and_smoothness

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"?

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

https://en.wikipedia.org/wiki/Lagrangian_(field_theory)

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?

Maxwell's equations do not fail with gravity

We have been working very hard, trying to break Maxwell's equations. To no avail, so far. Our best effort was on August 24, 2024 with the "extra energy" W that a small charge carries along with it when put inside a tube whose charge Q is of the same sign and very large.

                         strong field E' + E

                             \       |      /

                                \    |    /              

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

                                      |  weak field E'

                                  q • ---> v

                                      |  weak field E'

                          ---------------------------------- Q

                                /    |    \             

                              /      |      \  --> pW

                        strong field  E' + E

                       "extra" field energy W

                        moves with the charge q

We have recently analyzed the experiment in more detail, looking at the Poynting theorem and the Poynting vector.

As the small charge q moves, it will carry a large amount of field energy W outside the tube. Inside the tube, the strong field E of the pipe is negligible. Only the weak field E' of the charge q is present.

From where does the momentum come to the extra energy W? The inertia of q is not changed

The momentum pW may be much larger than the momentum of q itself.

Let us accelerate q. The field lines of q bend to the left. The Poynting theorem suggests that W gains its momentum from the bent field lines of q which push the charge Q in the pipe to the left.

The inertia of q does not increase from the extra energy W. This solves the question which we have been investigating in this blog for many years! The inertia of a charge q does not depend on its potential, assuming that the charge Q which is causing the potential is held static.

Does W have any kinetic energy as it moves? No

The charge q does not feel the force of Q at all. We do not need to do any extra work as we accelerate q to the right, compared to the case where the pipe Q would not exist.

Thus, even though the extra energy W is accelerated to the right, it does not gain any kinetic energy of its own.

The bent field lines of q do not do any work on Q.

How to prevent a perpetuum mobile in the presence of retardation

Let q and Q have the same sign and q = Q. In this blog we have been perplexed about retardation which makes the repulsion of two accelerated charges q and Q to differ from

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

in laboratory coordinates. If we accelerate q and Q toward each other, they will feel less repulsion. Does this allow us to construct a perpetuum mobile?

     field lines                                 field lines

   bent to left                                 bent to right

                   \ |                                | /

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

                     q                                Q

Since the field lines are sparser on the line from q to Q, the electric field and the repulsion is weaker.

But this does not allow a perpetuum mobile. Squeezing the field lines to the left and to the right requires energy. When q and Q come close, the left side of their joint field gets its energy from the field of q, and the right side from Q. We had to use enough energy to build the joint field of q and Q. We could not save any energy.

The energy which we save in the repulsion is consumed in squeezing and bending the electric lines of force. This is, of course, also a consequence of Poynting's theorem.

If q and Q would have opposite signs, then the attraction between q and Q would be surpfisingly weak, and we could harvest "too little" energy by moving q and Q close to each other. In this case, the surplus energy is radiated away as radio waves, when the joint field of q and Q settles to its low-energy state. No energy is lost.

Use gravity to construct a perpetuum mobile?

We try to prove that the "extra energy" W below does possess kinetic energy if it moves to the right.

In our earlier blog posts we realized that one can "grab" the mass-energy of W directly through gravity. In electromagnetism alone, we cannot "grab" W because it has no electric charge.

           ● M

              \    ≈ c

               v

                                    W

                             \       |      /

                                \    |    /              

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

                                     | 

                                  q • ---> v

                                     | 

                   ---------------------------------- Q

                                /    |    \             

                              /      |      \ 

 

Let us accelerate and move a large mass M at almost the speed of light to the left of the extra energy W. The mass M is stopped then.

When W is aware of the gravity of M, but before q knows anything about M, we move q to the right.

We make q static before the field of M arrives. The electric field of q will force the mass-energy of W to move farther from M. That requires some energy W'. Where does W' come from?

The electric field lines of q will "lift" W up in the gravity field, as the experiment settles down. The Poynting vector is not aware of gravity, and cannot describe the flow of energy to lift W up?

Could the squeezed gravity field of M supply the energy to lift W up? Why would it do that? If W were an ordinary lump of matter, there is no reason why the field of M would lift it up.

The electromagnetic lagrangian density

https://en.wikipedia.org/wiki/Lagrangian_(field_theory)

Can we combine the gravity potential to the lagrangian?

The potential would affect the term

What happens if we simply add the gravity potential to the term? Our analysis in the preceding section suggests that then energy is not conserved in the process which we described. But Noether's theorem says that energy is conserved in a stationary point of the action

       S  =  ∫  L(x) dt.

Simulating gravity with acceleration

We can simulate gravity with acceleration. Can we construct a pathological process without gravity?

                               ^  E

                               |      pipe

            Q  --------------------------   ---> a

                               × B

                       <---  E

                               ○ B'

                            q • ---> a

                               × B'

                         

                      <---  E

                              ○ B

                --------------------------   ---> a

                               |

                               v  E

Initially, the charged pipe Q is static. Its electric field E point outward from the pipe and is essentially zero inside the very long pipe. Also, the small charge q is static.

We suddenly start to accelerate the pipe to the right. The field E bends to the left, and a horizontal field component appears inside the pipe.

Simultaneously, we move q to the right and stop it soon. The electric field of q, E', becomes like the one in the Edward M. Purcell diagram.

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

                             field E' of q

The "tangential" part in the diagram is the electric field induced by a pulse of a magnetic field B' which we created as we moved q.

When the pulse B' meets the forming field E, the Poynting vector

       S  =  1 / μ₀  *   E  ×  B'

indicates that energy is flowing outward from the center of the pipe. The electric field

       E  +  E'

on the circle has a larger energy density because it is stronger. The Poynting vector S describes that extra energy moving outward.

But where did that extra energy come from? We did not need to do much work moving the small charge q. Energy should be conserved locally, according to Poynting's theorem.

The extra energy may come from the magnetic field B associated with the forming field E?

The magnetic fields B and B' point to opposite directions. The extra energy probably comes from the magnetic field

       B  +  B'.

The Poynting vector

       1 / μ₀  *  E'  ×  B

describes the missing energy in the magnetic field moving outward. The total extra energy traveling with the pulse in the field of q would be zero.

Let us then wait until the field E' of q is updated also outside the pipe. In the comoving frame of the pipe, the extra energy W outside the pipe in the field E + E' will be moved to the right, "up", if we think of the acceleration a as gravity. From where does the energy to lift W "up" come?

               <--- "gravity"

                                    W  ---> movement "up"

                             \       |      /   E + E'

                                \    |    /              

    F --->   Q ----------------------------------   ---> a

                                        | E'

                                      q • ---> a

                                        | E'

                   ----------------------------------   ---> a

                                /    |    \             

                              /      |      \   E + E'

 

The strongly bent pulse field E' of q pushes the charge Q in the pipe to the left. The force F accelerating the system has to do extra work when the pulse of E' passes the surface of the pipe, if the pipe is moving in the laboratory frame at that moment.

Some process has to use energy to push W up in the accelerating frame of the pipe. The force propelling the pipe will probably give the rest of the kinetic energy to W. It is like a person standing in an accelerating rocket and pushing a weight W up. The person has to do a little work. The motor propelling the rocket supplies the rest of the large kinetic energy increase of W.

Now we face a problem: the energy to push W up may be considerable. Where does it come from?

When we started accelerating the pipe, a radio wave carried quite a lot of energy away from the system, since the field of Q was deformed. Could it be that some of that energy still lingers and helps to lift W up?

Let us consider the case where q is attached to the pipe. Then the energy to propel W to the right clearly comes from the force pushing Q against the bent electric field of q.

If we suddenly move q up, the field is bent even more, as we accelerate q. When we decelerate q at the end, the field is less bent. Is this enough to give the energy to lift W up?

The kinetic/potential energy of W

Our thought experiments try to find out if behaves like any object which contains the mass-energy worth W.

Does W possess kinetic energy when it moves? Does W have a gravity potential, or an equivalent "gravity" potential in an accelerating frame?

Let us analyze the gravity example further.

                              W

                               |  E + E'

                   -------------------------  +Q

                               |   E'

   ● -->                    • +q   -->

  M                         |   E'

                   ------------------------  +Q

                              |  E + E'

                             W

We move the mass M very quickly close to the left end of the pipe. As M moves to its position, we can harvest from M the energy that M gains as it descends in the gravity potential of W.

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

Wikipedia states:

"the decrease in electromagnetic energy is the work done by the electromagnetic field on the gravity field plus the work on matter".

In this blog we have previously shown that the Einstein equations do not have a solution for any dynamic problem. We will forget about that for now.

Static configuration. Let us assume that M is close to the pipe, and the system has settled down. We move q upward. We have to do the work W' against the bent lines of force of E. How does this energy W' end up being gravity potential energy of W? What does the Poynting vector say?

                 Q

                      ●  

                      | E   ^

                     |       |

                    /     ○  B' pulse

                  v   

                 • --> move    W' = work done

                q

If we move q suddenly to the right in the diagram, a magnetic pulse B' is born. The Poynting vector says that the horizontal component of E, and B' transports energy away from q in the diagram:

       1 / μ₀  *  E  ×  B'.

The electromagnetic wave which is born from the movement, transports the work W' to the far parts of the field of Q. It is just like in the case where we would simply push q directly toward Q.

The tangential part of the field of q points to the same direction as the horizontal component of E. The sum of these two electric fields contains most of the energy W'. The energy of the magnetic pulse B' is small.

Dynamic configuration. Let us then assume that we move q before the field E has had time to bend to the left. Then we have to do much less work. The energy W' is much smaller.

However, when the wave reaches Q, the field E is already settled down. The wave must now be carrying as much energy as in the static case. Where did the energy come from?

The obvious source of the energy is the energy which is deforming E and bending the field lines. The existence of the field of q at the same location means that some of the deformation energy is spent to increase the value of W'.

But how can the wave of q steal energy from the wave associated with E? It is a linear theory. The waveform of E does not change in any way if it encounters the wave of q. Does the wave of q cause a destructive interference to the wave of E?

There probably is no paradox at all in this. If a wave hits any static electric field, then the energy of the electric field in the wave is locally increased, and that is compensated with a decreasing energy at some other location.

Further analysis of the Poynting vector

                       --->  accelerate

                   Q

                      ● -------------- pipe wall

                      | E   

                     |      

                    /        

                  |  ^        ○  B' pulse

                  |    \  E'

                  v    |

                        |

                        |

                     q • --> move   

                 

                 

Let us first forget about q. As the field E inside the pipe settles down to a non-zero value E ≠ 0 , the Poynting vector must indicate energy flowing either from the walls of the pipe, or from the end of the pipe into the pipe.

Let us then assume that E has not yet settled down.

We use an inertial frame which is comoving with the pipe at the moment that the wave of E' meets the wave of E'. Then the only significant magnetic field is B'.

The Poynting vector

       1 / μ₀  *  (E + E')  ×  B'

shows a significant energy flow radially outward from the center of the pipe when the waves of E and E' overlap. Where does that energy come from?

Conclusions

Let us close this long blog post. We need to study in more detail what is the electromagnetic field of a charge q like inside a strong gravity field.

An equivalence principle suggests that the field of a static charge q should look similar to the one when q is being uniformly accelerated. Then there probably is both a magnetic field and an electric field.

Under a uniform acceleration, q must continuously send energy to the far parts of its electric field, to supply them kinetic energy. How is this compatible with the fact that q cannot send any energy when it is static in a gravity field? Is this a contradiction?

Let us imagine an elastic sheet of rubber which is under a constant acceleration because of a rocket pushing it in the middle. There is energy flowing from the middle to the edges of the sheet. But if the rubber sheet is supported by a pole on Earth, there is no energy flow. Does the Poynting vector understand the difference? The "field" looks locally the same in both cases, but in just one there is an energy flow.