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?

For a time-dependent flow, the energy of a parcel is the kinetic energy plus its (very small) potential energy from compression. Maybe we can write a lagrangian density based on that?

***  WORK IN PROGRESS  ***

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.

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

The energy of a photon is

       E  =  h f.

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

         M 

                  f                               f'                     

               o                               o 

              /\                              /\

        observer                   observer

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

       f

for the photon, using his clock.

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

       f'  <  f.

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

       

               4 waves in 1 coordinate second

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

      M

                   o                                                  o

                  /\                                                 /\

           observer                                    observer

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

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

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

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

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

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

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

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

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

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

Why is the radial metric stretched in the Schwarzschild solution?

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

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

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

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

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

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

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

          • --> v                         ●

         m               r               M

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

The kinetic energy of m is

       W  =  G m M / r,

and

       v²  =  2 G M / r.

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

       sqrt(1  -  v² / c²),

and the metric of time for m is

       g₀  =  -c²  *  (1  -  v² / c²)

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

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

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

       g₁  =  1  +  rs / r.

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

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

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

The rulers held the dust particles are contracted by the factor

       sqrt(1  -  v² / c²)  <  1.

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

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

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

       1 /  sqrt(1  -  v² / c²)  >  1

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

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

The Schwarzschild metric of time follows from the gravity potential.

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

Conclusions

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

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

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

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

The form of a line of force for an accelerated charge

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

 

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

                          • q

                          |

                          v   a

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

       x₀ + v (t - t₀).

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

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

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

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

A sparse shell of masses, accelerating radially

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

                                  • m

                                  |

                                  v  a

 

              • --> a          ×          a <-- •

             m                                     m

                                  ^  a

                                  |

                                  • m       R radius

We accelerate the masses inward for a time

       Δt

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

       v  <<  c.

The masses move the distance

       s  =  1/2 a Δt²

closer to the center.

Let the spacing between the masses be L.

   

                 m'                    m

                  •          L          •

                 |                      |   

                 v    v                v    v

Let

       t  =  L / c.

The mass m sees the velocity of m' to be

       v'  =  v  -  a t

and m sees the location m' to be a distance

      1/2 a t²

higher in the diagram than the location of m. 

####

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

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

####

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

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

           ≈  2 m / L³  *  a t²,

and the associated not gained energy in the push is

       Wm  =  F s

                ~  2 m / L³  *  a t²  *  s

                ~  m / L³  *  t²

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

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

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

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

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

Conclusions

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

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

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

       -> E' field of accelerated charges

  E  <------- total electric field 

  E₀ <--  electric field of cap

                                                    ______

                                                 /             \

              •              <--    |                 ×        |

             q               v                 \_______/

      test charge         "cap"     charged shell

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

Gauss's law predicts.

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

The field of the accelerating ring of charges

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

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

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

                    × ×   B transient

                    ---> E' transient

 

                    ○ ○   B transient

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

Bent lines of force

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

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

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

         dEr / dt  =  0.

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

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

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

    

             B                  a  <-- •

            ○○

             • q                                    × center

            ××

             B                  a  <-- •

                        accelerating

                              ring

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

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

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

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

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

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

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

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

Case A

                                      ______

     <-- E                       /             \

     • q              <-- |           ×         |

                         v         \_______/

       only the "cap" and ring move, shell static

Case B

                                          ^  v

                                          |  

                                   __________

                                 /                     \

    <-- E                  /                          \

      • q                |              ×                |

                              \                          /

                                 \___________/

                                          |  

                                          v   v

       entire shell expands

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

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

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

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

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

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

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

Conclusions

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

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