__ "drooping"
/ | E
| v |
| | pipe
| |
Q charge
^
|
●
M
We bring a large mass M close to an end of the pipe. The electric lines of force E will start to "droop". There will be a vertical electric field inside the pipe.
From where does the energy to this vertical field flow?
If we think about an elementary charge q, its electric field E' has potential energy in the gravity field of M. The process of drooping converts some of the gravity potential energy of E' to electric field energy, since the drooping field has slightly more electric field energy than a symmetric field.
However, inside the pipe, the electric field E was essentially zero, and its gravity is essentially zero. Drooping cannot release much gravity potential energy of the field.
A rubber plate analogue: this is very different from the electric field, if we have several charges
q drooping
___----•----___ rubber plate
● M
Let us model the electric field of an elementary charge q with a horizontal rubber plate attached to q.
The drooping in the gravity field of M is caused by gravity pulling each part of the rubber plate. The plate is distorted and will gain some elastic energy, and lose some gravity potential energy.
Let us then imagine that each elementary charge q in the pipe has its own rubber plate.
The "linear sum" of those plates indicates a drooping electric field.
However, if we think about the total electric field E of the pipe, there is essentially no gravitating mass inside the pipe. Will this affect the drooping, and break the linearity of electromagnetism?
Gravity, or acceleration, makes electromagnetism nonlinear?
Maxwell's equations are linear. However, gravity couples to the field energy density
1/2 ε₀ E² + 1/2 * 1 / μ₀ * B².
Is there any reason why electromagnetism should be "linear", in some sense, under gravity?
There is a good reason to assume that the electric field mass-energy density really is in the formula above, also with respect to gravity. If we have an electric field somewhere, we can reset the field to zero with capacitor plates, and harvest the energy just at that location.
Let us take two electric charges q and Q. If each charge exists separately in space, its electric field is spherically symmetric. But if we put both close to each other, then the gravity of each electric field distorts the other field. It is not linear, in this sense.
Electric fields under a constant acceleration
Let us then study static electric fields of static charges under a uniform gravity field. This is equivalent to studying a constellation of uniformly accelerated charges. We ignore the gravity of the electric fields on other electric fields.
If we take charges q and Q, can we determine their electric fields separately, and sum the fields to obtain the field of the combined system q & Q?
If we take both q and Q, then the mass-energy of their combined electric field changes in a complex way. Is there any reason why gravity would bend the combined field in the exact same way as it bent the fields separately?
We encounter the August 24, 2024 problem once again: what is the inertia and the kinetic energy of the combined electric field of several charges?
A small cylinder charge inside a uniformly charged pipe, in uniform gravity: the lowest energy state is not linear
Q
| | ---> E
| | |
| | --> | ----> E + E'
| q E' | W "extra" energy
| |
R = radius of the cylinder
● M
Let us have a vertical cylinder uniformly charged. The charge is Q. How much will the lines of force of its cylindrical electric field "droop" under gravity?
Assumption 1. The drooping angle α of a line of force probably approaches zero close to a very thin cylinder.
Assumption 2. The drooping angle is
α ~ r,
where r is the distance from the cylinder. The line of force is a parabola. This is what people generally assume. This matches the "geometry of spacetime" interpretation of general relativity.
The vertical component of the field of the cylinder is roughly constant: the strength of the field is ~ 1 / r, and the drooping angle is ~ r.
Let E be the field of the pipe. If the pipe is very long, and we are studying its middle part of a length L, then the horizontal component of E inside the pipe is essentially zero. This is because an end part of the pipe of the length L has a negligible field near the midpoint.
Thus, the field E of the pipe, inside the pipe is vertical, and roughly constant. The field does not depend on the radius R of the pipe.
Let q be a thin uniformly charged vertical cylinder at the center of the pipe. Let E' be its field. The field E' is roughly constant inside the cylinder. The vertical component of E' is roughly constant, too.
The energy from the interaction of the fields E and E' inside the cylinder is
~ E E' * π R²
~ R².
Let E' be the field of q when the pipe is not present, and E the field of the pipe when q is not present.
We superpose these fields E and E'. Let us claim that the total energy of the combined system pipe & q is minimized by the sum field E + E'.
We vary the drooping angle α of E' inside the pipe.
We denote by W the energy from the interaction of E and E' outside the cylinder.
The energy saved in the gravity potential of the energy W outside the pipe is approximately
~ α R.
The price paid is the energy added to the combined field E + E' inside the pipe, and is also approximately
~ α R².
Let us assume that when the radius of the pipe is R, an infinitesimal variation of α does change the total energy of the system.
Let us the halve the value of R. We claim that still, E + E' minimizes the total energy of the system.
But that is not true. The price paid with an infinitesimal increase of α is now only 1/4 of the original price, but the energy saved is more than 1/2 of the original saving. More than 1/2, because W is now larger.
We end up in a contradiction. What is wrong in our assumptions?
Because of an equivalence principle, our analysis in a uniform gravity is equivalent to newtonian mechanics in a uniformly accelerating system. Maxwell's equations seem to clash with newtonian mechanics, if we assume that the energy density of an electric field E is locally
1/2 ε₀ E².
Since we firmly believe that newtonian mechanics is correct, the error has to be in Maxwell's equations.
Hypothesis. Maxwell's equations do not have a solution for any system which contains accelerating charges.
The hypothesis is similar to our result that the Einstein field equations do not have a solution for any dynamic system.
It is not known if Maxwell's equations do have any dynamic solution. In the literature we have not seen a proof that they would have solutions. The equations do have solutions for static charges, just like the Einstein equations have the Schwarzschild solution.
The electric field does not bend because of gravity?
Maybe the electric field does not seek the lowest energy state inside a gravity field? It does not care of the weight of the energy W?
That would be strange. We firmly believe that gravity (or acceleration) does bend the field lines of a single charge. Why it would not care about the "weight" of the field of several charges?
A physical system which does not care to seek the minimum energy state is susceptible to a perpetuum mobile.
An option is to use the "private field" concept, which we have discussed many times in our blog. Maybe each elementary charge has a private field which finds its form under gravity independently, regardless of other charges? That would be similar to the rubber plate model which we described above.
In classical electromagnetism, a well-behaved charge is a continuous charge distribution. There is no elementary charge.
Linear Maxwell's equations break energy conservation in a homogeneous gravity field
In our pipe example, the field E of the pipe inside the pipe is roughly constant, regardless of the radius R of the pipe.
Thus the force F on the charge q does not depend much on R. But the weight W which q lifts up does depend on R. Does this break conservation of energy?
Let us try to calculate numeric values. Let g be the acceleration of gravity. A photon will fall a vertical distance
s = 1/2 g t²
in a time interval t. There,
t = R / c,
s = 1/2 g R² / c²,
ds / dR = g R / c².
Let us have a narrow cylinder whose charge density per length is ρ. The radial field at a distance R is
Er = 1 / (4 π ε₀) * ρ / R,
and the "drooping" vertical field component
E = 1 / (4 π ε₀) * ρ / R * g R / c²
= 1 / (4 π ε₀) * g / c² * ρ.
It does not depend on R, as we wrote above. We can as well denote by ρ the charge density of the pipe per length.
The downward force on the cylindrical charge q is
E q = 1 / (4 π ε₀) * g / c² * ρ q.
Let us then calculate the mass-energy of W. It is the potential energy V of q in the field of the pipe. Let L be the length of the pipe. At distances < L, the field is roughly the formula of Er above. The potential is very roughly
V ≈ 1 / (4 π ε₀) * ρ q * ln(L / R),
and the weight of W is
1 / (4 π ε₀) * g / c² * ρ q * ln(L / R).
We assumed that L is much larger than R. Thus, the weight of W is much larger than the electric force pushing q down.
This does not make sense. If we assume that electromagnetism is linear, we can lift a heavy weight W up with a small force on q. It breaks conservation of energy. Or does it? Can we harvest the increased gravity potential energy of W somehow?
Q
| | ---> E
| | |
| | --> | ----> E + E'
| q E' | W "extra" energy
| |
● M
We can do the harvest, if gravity is not uniform everywhere.
____
\ W
\________ gravity potential
Let the schematic gravity potential be like in the diagram. We lower the whole system into the pit in the potential. But during the process, we are able to lift W up with a negligible force. We can harvest the gravity potential from W twice. Energy conservation is broken.
How to repair the model? Let us assume that the field of q somehow exerts a "self-force" on q, and presses q down. Then energy is conserved.
However, then the inertia of q might depend on the potential of q? Or does it?
The spectrum of the hydrogen atom does not change if the atom is in a high potential, which indicates that the electron has the same inertia inside the atom. We have discussed this several times in our blog. Maybe the atom somehow shields the electron from changes in the inertia?
How to interpret energy non-conservation in an accelerating system?
Above we show energy non-conservation under a homogeneous gravity field, whose acceleration is g. An equivalence principle says that it is analogous to a system which is accelerated with the acceleration g.
In the gravity field, we have supported the pipe and the charge q with some structures. In the accelerated analogue, the system is in empty space, and we have a spring, which is accelerating these structures.
W
o / lift up
| ^
/\ | g acceleration
----------------------
| \ structure
v /
F \
/
\ spring
Energy non-conservation in the accelerated analogue means that the sum of kinetic and potential energies does not remain constant.
For the energy to remain constant, we would need a man doing the weightlifting of W. The man would spend some potential energy possessed in his muscles, to lift W. Where would his energy go? His work saves some work done by the spring accelerating the system. When the man starts lifting W up, the spring feels an extra resisting force F. When the man ends the lifting, the spring is relieved by the force -F. Since system is moving faster in the -F phase, the spring saves some work.
But if the inertia of W would be zero, then energy would be conserved?
If momentum is conserved, then moving W up will exert a force F on the spring. The man might be using a motor to lift W up, and will not personally feel the inertia of W, but the spring will feel the inertia of W, regardless.
What could constitute such a "motor" in the system? In the accelerating frame, the field of the pipe and q is static. We do not see how any "motor" could help in lifting q up. If such a motor exists, it probably is not described by Maxwell's equations.
On August 24, 2024 we suspected that Maxwell's equations do not understand the kinetic energy of field energy. Our analysis above suggests that that really is the case.
Do Maxwell's equations understand that energy is not conserved in the accelerating pipe & q system above?
Our analysis suggests that Maxwell's equations break newtonian mechanics in an accelerating system. Could it be that Maxwell's equations actually have no solution in such a case? By Noether's theorem, an extremal point of the electromagnetic action should conserve energy. If energy is not conserved, then it is not an extremal point, and does not satisfy Maxwell's equations.
Q
| | ---> E
| | |
| | --> | ----> E + E'
| q E' | W "extra" energy
| | ^
| | | g
-----------------------
| /
v F \
/
\ spring
Let a man push q up in the diagram. The man needs to do very little work, but there is a significant momentum in W as it moves up.
Momentum is conserved. Therefore the pipe must press the spring down with a significant force F. When the man stops pushing q up, the spring feels a force -F. In the process, the spring saves some work, compared to the process where q is not pushed up.
Let A be a process where q is not pushed up, and B be a process where q is pushed up.
The electromagnetic action (Maxwell's equations) says that at the end, the electromagnetic energy in the system is the same in both cases A and B.
However, in case B, the spring did less work, i.e., the spring retained more potential energy. Energy cannot be conserved in both cases. Maxwell's equations do understand these things, as shown by Poynting's theorem.
An extremal point of the electromagnetic action must conserve energy. We conclude that the process which we described is not an extremal point, and is not a solution of Maxwell's equations. This probably means that Maxwell's equations do not have any solution for the described process B.
If the very simple process above does not have a solution, then Maxwell's equations probably do not have a solution for any dynamic system at all. This complements our May 26, 2024 result that the Einstein field equations do not have a solution for any dynamic system.
We have a tentative proof for the Hypothesis which we stated in an earlier section.
Maxwell's equations might still have solutions for a cylindrically symmetric collapse or an expansion of a charge shell. For gravity, we were able to show that the Oppenheimer-Snyder collapse is a suspicious, probably erroneous, solution.
However, if we have to repair electromagnetism by moving to a rubber plate model, then a spherically symmetric collapse will probably happen in a way which does not agree with Maxwell. There could be oscillation, "dark energy". Rubber plates tend to oscillate.
Various lagrangians for electromagnetism plus gravity
We have to check how various proposed lagrangians (actions) handle the pipe example which we constructed above. Does the weight of W distort the electric field of q?
We do not really need gravity. How does the "weight" of W in a uniformly accelerated system interact with the electric field of q, using a special relativity lagrangian of electromagnetism.
Conclusions
This analysis is very preliminary. We have to check this carefully in future blog posts. If the analysis is correct, then Maxwell's equations seem to have no solution for any dynamic problem at all, if there are at least two charges. The reason is that there is no concept of a "self-force" of the field of a charge q on itself. In our pipe example, the field of q should exert the force of the weight of W on q itself, but Maxwell's equations have no such mechanism.
In the idealized case of a uniformly charged infinitely thin shell expanding or contracting, there may exist a solution to Maxwell's equations.
We want to study if "dark energy" could come from some anomaly in the behavior of expansion or contraction. If we correct Maxwell's equations, can that introduce an anomaly?
We will next look at the Navier-Stokes equations. They are nonlinear. Can we prove that they have no solution, expect in trivial cases?
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