Tuesday, October 15, 2024

Density variations in a young universe break Lambda-CDM?

The cosmic microwave background contain luminosity differences on the order of 1 / 100,000. In the ΛCDM model, these are believed to reflect density variations of the same order in a very early universe. "Very early" here means the end of the inflation phase at ~ 10⁻³² seconds after the birth of the universe.



The book Galaxy Formation and Evolution by Mo, van den Bosch, and White (2010) is available as a pdf file in the link.

The chapters about Cosmological Perturbations handle the fate of density variations in the early universe.


The mass excess at certain locations in the early universe: the Schwarzschild radius


The current density of baryonic matter in the observable universe is

       ~  3 * 10⁻²⁸ kg/m³.

A large galaxy cluster is 10²⁴ meters in diameter, which makes a baryonic mass of

       ~  3 * 10⁴⁴ kg.

Let us assume that the "excess" mass is 1 / 100,000 of that, or

       M  =  3 * 10³⁹ kg.

The Schwarzschild radius of such a mass is

       R  =  2 G M / c²
         
            =  1.5 * 10⁻²⁷  *  M  m/kg

            =  5 * 10¹² m.

We see that if the scale factor of the universe is

        a  =  10⁻¹¹,

then the excess mass is inside its Schwarzschild radius. (The factor a = 1 is the present universe.)

The excess mass should form a black hole in the early universe. Why would a black hole expand like ΛCDM claims?


Playing the Big Bang in reverse: density variations form black holes


The ΛCDM model claims that the metric of the early universe is very smooth, spatially uniformly expanding metric. If we have two observers at the edge of excess density volume in the early universe, the observers can communicate much like two observers in Minkowski space.

But is that really possible if the dense volume is inside its own Schwarzschild radius, and is a black hole?

If we assume that gravity is time-symmetric, then we can play the Big Bang expansion in reverse. Any (radio) communication between the two observers can happen in reverse, too.

One of the observers is inside the horizon of a black hole, the other is outside it. We see that the observer inside cannot send a signal to the outside.

This is in contradiction with the claim that the observers can communicate without problems.

A small density perturbation in a middle-aged universe amounts to a huge perturbation in the early universe. This is the well-known problem of the instability of an expanding universe.

Note that the Milne model suffers from the instability just like ΛCDM.

A possible solution: the universe does not begin from a singularity, but is relatively "large" at birth. The diameter of the current observable universe might have been much larger than 5 * 10¹² meters at birth. We do not like singularities in this blog. This could be a step forward, actually.


Does the literature address the instability problem in the early universe?






The increasing energy of the inflaton field breaks the Einstein-Hilbert action?









The basic idea in cosmic inflation is that a scalar field somehow obtains random fluctuations in energy density, and these fluctuations are "blown" spatially very large in the expansion of the universe.









Let us have a history H of the universe. Let us make a variation H' where we keep the metric as is, and thus, the action integral of the Ricci scalar R does not change.


              wave
        -------______--------_____-------   tense string
                     /         /          /
                     \         \          \    springs


The lagrangian describes an oscillating tense string which also is attached to springs, which create the potential V(φ).

In the inflation hypothesis, φ varies from place to place, and the potential energy (or kinetic energy) is greatly increased as the universe expands.

Let us have a "disturbance" W of the field.

Suppose that the disturbance W is not moving substantially. It is the springs in the diagram which make the string to oscillate up and down.

The expansion of the universe adds more length to the oscillating area, and more energy.

Let us have a history H where the oscillating area expands. Energy is produced from "nothing".

Let us make a history H' where we keep the metric the same. The time varies within an interval Δt.

We make a Noether time variation which "measures" the energy of the system at the start and the end of the time interval Δt. The process is like we would have a harmonic oscillator whose mass is zero, but the energy non-zero, created from empty space at some time t during the time interval Δt.

The Noether time variation changes the value of the action close to the end of Δt, because the oscillator there has energy, but does not change the action close to the start of Δt, because the oscillator does not exist there. We conclude that H' has a changed value of the action. H was not a stationary point of the action. H is not an allowed history.

Suppose then that the disturbance W is moving. A Noether time variation can measure the energy of W at the start of Δt, and at the end. We end up in the conclusion that H is not an allowed history.


We showed that a scalar inflaton field is not compatible with the Einstein-Hilbert action.

Note that if there is no disturbance W, and the field φ simply has a constant potential V, then our counterexamples do not work. It may be that dark energy is compatible with the Einstein-Hilbert action.




***  WORK IN PROGRESS  ***
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Thursday, October 10, 2024

Lambda-CDM is not a solution of the Einstein-Hilbert action?

The standard ΛCDM model of cosmology explains the black body radiation spectrum of the cosmic microwave background (CMB) by claiming that the radiation pressure does work and CMB loses energy as the FLRW universe expands.

But the Einstein-Hilbert action seems to imply that CMB cannot lose energy because there is no place where that energy could go. Then the history would lose energy, but that is not allowed in a stationary point of the matter lagrangian L in the Einstein-Hilbert action.










Massive particles in the FLRW universe: they do not lose kinetic energy











The lagrangian for a single massive particle outside of potential fields is the familiar kinetic energy formula.

In the ΛCDM model, it is familarly assumed that as the time passes, fast moving particles "lose" kinetic energy relative to the spatial coordinates which are fixed to "static" expanding matter content.


                     newtonian explosion 

  "static" particle                        "static" particle
         <- ●                       ●   • ----> v           ● ->
                                            fast particle


This would really be the case if the expansion of the universe would be equivalent to a newtonian explosion where the "static" matter is expanding its volume in Minkowski space. When the fast particle reaches the rightmost "static" particle, the relative speed of the fast particle compared to the "static" particle is less.

Let us have a history H of the FLRW universe, such that fast particles gradually lose speed as the universe expands.

Let m be a fast body of mass such that its density is extremely small. Then the self-potential of m relative to its own gravity field is negligible.

Let v₁ be the initial velocity of m relative to the standard FLRW coordinates, and v₂ its final velocity in the history H.

Let us make the following variation H' of the history H which runs for a coordinate time interval Δt:

1.   we keep the metric H constant;

2.   we let m move at an almost uniform speed v during Δt, except that the speed at the start is v₁ at the start and v₂ at the end.


How much does the Einstein-Hilbert action change? For the (rest) mass m itself, the volume element







can be taken as essentially constant during Δt, because the stretching of the spatial metric does not affect the rest mass. The same holds for the kinetic energy of m.

The changes in the Einstein-Hilbert action:

1.   The kinetic energy action of m in the matter lagrangian L (which is of the form kinetic energy - potential energy) is reduced because the velocity v is almost constant during Δt.

2.   The position of m is displaced relative to its own gravity field: the volume element sqrt(-g) obtains a slightly larger value because m does not reside in its own gravity potential well. However, we can make the effect on the action value as small as we want by reducing the density of the body m. Thus, we can ignore this item.


We conclude that the variation H' reduces the value of the Einstein-Hilbert action relative to H. The history H was not a stationary point of the action. We proved that the speed of m cannot change as the FLRW universe expands.


Massless scalar field waves in the FLRW model















The massless scalar field (m = 0) has a very simple lagrangian density formula. Let us have a wave packet. We must determine what history of a wave packet is a stationary point of the action in an expanding FLRW universe.

Let us stretch spatial distances by a factor 2. The naive transformation of keeping the function φ(r) values the same for each coordinate vector r, will make the wave packet volume 8-fold and the squares of the derivatives of φ 1/4-fold. The energy of the wave packet would double. Clearly, we must find a better transformation.

Let us have a history H of the universe for a coordinate time interval Δt. We make the following variation H':

1.   keep the metric g the same;

2.   do a Noether time variation of the wave packet.


By making the energy density of the wave packet very small, we can ignore its potential in its own gravity field, just as we did in the previous section.

The Noether time variation at the start time speeds up the time development of the wave packet, and the variation at the end slows down the time variation of the wave packet.

Is it a problem that the wave packet moves faster than c when we speed it up? Let us for the moment assume that it is not a problem.

For the Noether time variation to keep the value of the action the same, the "energy" measured by the time variation must stay constant. The energy of the wave packet cannot change.

But ΛCDM claims that the energy of the packet changes as the universe expands.

What about doing a Noether time variation of the entire radiation content of the universe?


"ACuriousMind" on the Physics Stack Exchange claims that since the universe as a whole is not time translation invariant, a Noether time variation will not show that energy is conserved. Does this mean that the history of the universe, H, is not a stationary point of the action? If not, then how do we decide which is a valid possible history?

Our own argument above only does a local Noether time variation. There is nothing which prevents us from doing that. And the laws of nature which govern a local wave packet are time translation invariant.


A precise variation which changes the action value in the massless scalar field wave example


                                        H history

                   |            |
                   |            |
                    •          •  absorptions
                      \      /
                        \  /  decay into two wave packets
                        /
                      /
                     • --> v
                     m massive particle

   ^  t
   |
    ------> x


Let us have a massive particle m which decays into two wave packets. The particle m initially moves at a speed v and then decays into two light-speed wave packets. Both packets are later absorbed by other particles.

We look at the history H during a time period Δt.

We let the FLRW universe expand so much that the energy of the wave packets is negligible at the end of the period Δt. The absorbing particles can be very lightweight, much lighter than m.

Let us vary the history H in such a way that we let the particle m move to the right at a faster speed than v for a short time after the start of Δt. At the end of the Δt, we let the other particles move for a short time to the right, so that they catch the wave packets. Then we let the other particles move to the left, so that the variation H' has the same end state as H.

Since the other particles have negligible masses and the wave packets have negligible energies at the end of Δt, the variation at the end changes the value of the action negligibly.

But the variation at the start of Δt changes the action substantially since it changes the kinetic energy of m.

We showed that an infinitesimal variation H' of the history H changed the action substantially (linearly with the variation). Then the history H is not a stationary point of the Einstein-Hilbert action.

The proof illustrates that if a "packet" of energy is destroyed almost completely in a history H, then it is very easy to construct a variation H' which changes the value of the action.

We did not need to vary the propagation speed of the wave packets. They propagate at the light speed c. There is no problem with faster-than-light signals in the variation H'.

We used the fact that the value of the action is not changed if we translate the middle part of the history H to the positive x direction. That is, we used the spatial translation symmetry.

Why the same argument does not work in a newtonian balloon model of an expanding universe? It has to be because the interaction of the balloon rubber and particles moving on it is more complicated than in general relativity. The spatial translation symmetry does not hold if the particle made a significant "dent" into the rubber. If we keep the history of the rubber as is, but move the path of the particle, then the action of the middle part changes significantly.


Why does the literature claim that energy in radiation is lost when the universe expands?



Some papers refer to the Richard Tolman book from the year 1934. Let us find out how Tolman derives the result.

On page 385, Tolman writes:











Tolman claims that a particle moving in an expanding universe will slow down. But in the first section of this blog post we proved from the Einstein-Hilbert action that the particle must keep its velocity. Let us analyze how Tolman arrives at a different result.


                     • ----> v


Tolman derives the path of the particle from the geodesic equation. But if the geodesic equation claims that the particle loses kinetic energy, then it clashes with the Einstein-Hilbert action. How can this be?

A possible reason is that the geodesic equation "measures" the velocity v of the particle relative to the location where it started from. Then there is no loss of kinetic energy.

This leads us to a fundamental problem: if we have a particle which has gone a long way, the expanding universe may make it to recede from its original location faster than light. What is the kinetic energy in that case?

In the Einstein-Hilbert action, is the kinetic energy of the particle reduced or not? The matter lagrangian would require the kinetic energy to be the same, but the Einstein field equations probably require it to be reduced?

All these problems stem from the fact that there are no canonical coordinates in general relativity. The kinetic energy of a particle is a fuzzy concept.

Is the geodesic equation itself ambiguous?

The Einstein-Hilbert action is the fundamental equation. The geodesic equation is more like a guess of how a particle would move under a metric. The guess seems to clash with the action.

Is there any way to make a consistent model where the topology of the spatial metric is isomorphic to a 3-sphere and the geodesic equation holds?


In this blog we have been touting the Milne model where space is Minkowski, and we avoid the contradictions produced by the 3-sphere topology.


The balloon analogy
















The balloon analogy of an expanding universe has no problem handling radiation pressure and so on, because it is a model in newtonian mechanics. For example, if there is a pressure from "gas" on the surface of the balloon, the pressure does work as the balloon expands. The gas cools, and the energy is given to the rubber in the balloon, stretching the rubber and giving it kinetic energy outward. The action of this system conserves energy.


Tuomo Suntola's Dynamic Universe has the 3-sphere expanding to a fourth spatial dimension, and the kinetic energy of a particle for the velocity in the fourth dimension is included. It is a step toward the newtonian balloon model. We have to check how Suntola's model handles energy conservation.


A locally Minkowski space?


What about using "canonical" coordinates which are locally Minkowski at each small "patch" of the the universe? We will worry later about gluing these coordinate systems together.

The kinetic energy of a particle would be defined relative to these canonical coordinates. Energy would be conserved.

The problems start when a particle moves from one patch to another. Suddenly, the kinetic energy of the particle changes.


A reverse Oppenheimer-Snyder collapse: make that the new ΛCDM model?



In the Oppenheimer-Snyder collapse, uniform dust collapses to form a black hole. If we run the collapse in reverse, it is much like an expanding FLRW universe. However  there are natural canonical coordinates (the Schwarzschild coordinates), and we avoid the various problem with the topology of the 3-sphere in the FLRW model.

A radiation pressure really does work in that model, accelerating the explosion of the matter outward. Energy is conserved.

Maybe we should abandon the 3-sphere, and claim that the natural expanding universe model in general relativity is the reverse Oppenheimer-Snyder collapse? The spacetime is asymptotically Minkowski. There may be several Big Bangs at various locations of the asymptotic Minkowski space.

In our blog we have been claiming that the Oppenheimer-Snyder collapse "freezes" when a horizon forms. No singularity can form. But we see that our own universe has had a much larger matter density in the past, and we would have been inside the horizon. There was no freezing. The universe kept expanding. Currently, the observable universe would be outside the horizon, if there is not much matter which is farther than the observable volume.

We have supported the Milne model in this blog, to avoid the freezing. If there is an equal amount of negative gravity charge (in dark matter) as there is positive charge, then no horizon forms and the Milne model is the one.

Another option is to design a model where somehow the Oppenheimer-Snyder collapse can go past the forming horizon. We could claim that the Big Bang is the only case where such a behavior is possible, and it cannot happen in an ordinary black hole. As if an infinite time could pass in this special, Big Bang process, and take matter past the horizon. That is very ad hoc.


Conclusions


Let us close this blog post. In subsequent posts, we will further analyze the problem of missing canonical coordinates in the ΛCDM model.

Currently, our best bet is to replace the 3-sphere spatial topology of ΛCDM with a reverse Oppenheimer-Snyder collapse. The Schwarzschild coordinates would act as canonical coordinates against which we measure the kinetic energy of a particle.

If the ADM formalism is correct, then energy is conserved in a reverse Oppenheimer-Snyder collapse. We avoid the problems of energy non-conservation in the 3-sphere spatial topology.

In our blog we have been claiming that the gravity field merely simulates the "metric" of spacetime. The metric is not fundamental, and the true underlying spacetime metric and topology is the flat Minkowski space. We have said that the 3-sphere spatial topology of ΛCDM cannot exist. In this blog post we showed that the 3-sphere spatial topology causes serious, maybe insurmountable, problems for ΛCDM.

On the Internet there is a lot of discussion about energy non-conservation in expanding models of the universe. However, no one seems to have analyzed what the Einstein-Hilbert action says about energy non-conservation: it prohibits that. We are probably the first to recognize the serious problem in the ΛCDM model, and in the 3-sphere spatial topology.

We have noted in this blog that while there is ample empirical evidence that gravity can simulate "curved" spacetime, there is no empirical evidence that gravity could alter the topology of spacetime, like it does in the ΛCDM model. The hypothesis that gravity can alter the topology is a bold one. Maybe the hypothesis is incorrect.

It may be that an action in physics can truly work only if energy is conserved. Energy non-conservation in ΛCDM may be a fatal error in the model.
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Tuesday, September 24, 2024

Maxwell's equations describe charges in conducting objects?

In our September 18, 2024 blog post we had a negatively charged cylinder, and an electron e- moving close to it. We encountered a paradox: the electron did not get any acceleration along the cylinder, even though in the frame comoving with the electron e-, the cylinder creates a magnetic field B which should accelerate the electron in the x direction.


There is no acceleration of the electron in the x direction in the case of a charged electric insulator cylinder


      negatively charged insulator cylinder
      ========================== - 


                        ^ u                     W = F Δy
                        |
             -v <--- • e-      
                        |
                        v F Coulomb's force

              ^        ^
              |        |

              E       E'
    ^ y
    |
     ------> x  (laboratory frame)


Above, E is the electric field of the insulator cylinder and E' is the electric field of the electron. As the electron comes closer to the long cylinder, the electron does not acquire any momentum in the x direction.

The electron loses its kinetic energy as it comes closer to the cylinder. The inertia of the electron grows smaller. But the extra field energy in the field

       E  +  E',

relative to the plain fields E and E', grows as much as the electron loses its kinetic energy. The combination of the electron plus the extra energy in the field retains its inertia. The electron does not acquire any acceleration in the x direction.


The magnetic acceleration in the comoving frame of the electron: the Lorentz magnetic force formula fails for an electric insulator


        negatively charged insulator cylinder
        ========================= ---> v


             ○      ○      ○      ○    B    field lines point
                                                   out of the screen

                           ^  u
                           |
                          • e-
       ^ y
       |
        ------> x'  (comoving x')


In the comoving frame of the electron, the electron sees a magnetic field B which should accelerate it to the negative x' direction.

But this is inconsistent with the fact that the electron has no x acceleration in the laboratory frame.

What is going on? We proved that the Lorentz force formula is wrong for an electric insulator which is charged. The charges in the cylinder cannot move. It is an electric insulator.


A symmetric charged metal wire: the Lorentz magnetic force formula holds


                         midpoint         wire electrons
         ===========|===========  - 
                                  
                                 ^ u
                                 |
                     -v <---- • e- test electron

            |    |    |    |    |    |    |   E''

        ^ y
        |
         -----> x (laboratory frame)


If the charged cylinder is a metal wire, then conducting electrons can freely move inside it. Let the test electron be close to the wire, at the midpoint of the wire. The test electron repels the electrons in the metal. The combined electric field E'' of the system will be roughly constant far away. There is no stronger field at the location of the electron.

Now the electric field does not increase the inertia of the electron in the x direction. The inertia of the electron decreases as it approaches the charged wire. The electron accelerates to the left.

We see that the Lorentz force formula about B holds if the charged cylinder is an electric conductor. The Lorentz force takes into account both Coulomb's force and the backreaction of the electrons in the wire.

If the electron is not approaching the wire at the midpoint, then the situation is more complicated since the extra field energy is stored, on the average, at the midpoint. What happens then?


A single electron inside an electron gas cloud: the inertia is a constant me?


Let us look at the electron gas inside a block of metal. The thermal velocity of electrons is roughly 100 km/s.

If we move one electron e- along a long wire at a speed v, then to keep the field approximately constant, it has to be compensated for by moving a cloud of electron gas of N electrons to the opposite direction at the speed

       v / N.

There N is on the order of 10²³. The energy needed for the movement is

       ~  1/2 N me v² / N²

       =  1/2 me v²  *  1 / N,

where me is the rest mass of the electron.

We conclude that the inertia of the single electron e- does not grow appreciably from the effect on the electron gas. 

The conclusion: when an electron e- is very close to a metal wire, the "evening out" of the electric field done by repulsed electrons in the wire does not increase the inertia of the electron.

Ultimately, this effect should be measured, since it is a quantum mechanical process, and surprises can occur. An Internet search only returns one experiment where the inertia seemed to change. Apparently, in most cases the inertia change is negligible.

Also, the effect of the electrons in the gas moving at 100 km/s should be studied. The hypothetical electron which we "move", actually moves already at 100 km/s. The other electrons, which it repels, move at 100 km/s, too.


The electric influence of an electron close to a wire: a deficit of electrons worth e+


If we add a single new electron e- into an electron cloud, e- "makes room" for itself by repelling the amount e- of charge around it. There is a deficit of charge equivalent to e+ close to the electron.


                  deficit of electrons
                +       +    + + +    +       +
         ============|============  wire
                           midpoint                     length L

                                   • e-


What about a metal wire and a single electron e- close to it? Let us assume that the wire is very long and the electron e- is at the midpoint.

There is an excess of electrons now at the ends of the wire. Does the excess have any effect? The excess probably is ~ 1/2 e-. The charge density of the excess is

       ~  e- / L.

The electric field of the excess close to the center of the wire is negligible if L is very long.

On the other hand, the deficit of electrons close to the center of the wire does have a significant effect on the electric field close to the center.

Conjecture. The electron e- repels the electrons in a long wire in such a way that the deficit of electrons close to e- is equivalent to e+.


If the electron e- moves along the wire, it carries a fuzzy "positron e+" along with itself inside the wire.


An electron moving along a negatively charged metal wire: a more detailed analysis

      
                   deficit of electrons
                               "e+"
           v <---    +  + +++ +  +
             ======================  -  wire
                                                     negative charge

                                 ^ u    R = distance(e-, wire)
                                 | 
                       v <--- • e-


           |      |      |      |      |      |     E


       ^ y
       |
        ------> x


The "electric influence" of the electron e- repels electrons in the wire. A "positron e+" travels along with the electron.

The velocity component u takes the electron closer to the wire, and the velocity component u slows down. The inertia of the electron e- is reduced. But how does the inertia of the field, or the electrons in the wire, behave?

The combined electric field E of the wire and the electron e- at distances r > R from the wire has the lowest energy if E is approximately uniform. The field strength

       ~  1 / r

and the energy density

       ~  1 / r².

The circumference at a distance r from the wire is ~ r. Thus the total field energy is the integral of

       ~ 1 / r²  *  r  =  1 / r,

which diverges.

The field E is the sum of the field of the electron e-, the "positron e+", and a uniform charge density in the wire.

Most of the "interaction energy" of these fields is at distances r > R?


        ====================
                      deficit of
                   field energy        E

                  v <--- • e-


In the diagram we have marked the volume where the field E has less energy than it would have if the electron e- were not present. Does this "bubble"add much to the inertia of the electron e-?

As the electron moves to the left, field energy must flow to the right, so that the bubble moves with the electron. Does this energy flow possess much kinetic energy?

It is the Poynting vector

       1 / μ₀  *  E × B

which describes the energy flow. The flow takes energy from the left to the right along quite a straight path. Is there a lot of kinetic energy in this flow? Can we "grab" the flow?


The kinetic energy of flowing liquid


Suppose that we have a metal object M submerged into a water pool. The mass M of the object adds inertia to the object. The water in the pool starts to move if we move M. How much inertia does water add to M?

This is a complicated question.

In the electron gas cloud section above, we simply guessed that the extra inertia is negligible. The total kinetic energy of the system gives us a hunch of what the inertia might be, because

        1/2 M' v²

is the energy, where M' is the inertia of the object and v is its velocity.

If M causes a very large mass of water to flow very slowly, then the kinetic energy of that very large mass is negligible. If M is streamlined, the flow of water around it may have very little kinetic energy?

Then we can produce a significant momentum into the liquid, even though we only consume very little energy. How is that possible?


                   m     m
                 ■■|■■
                 ■■|■■
                       |
                       |
                       |  --> 
                   lever


The streamlined object may act like a "lever" which gives two large masses m a significant momentum, even though the energy spent is very little. In the lever configuration, there will be momentum both to the left and to the right.


The bubble of a lower field energy density has a constant effect on the inertia of the electron e-?


In the diagram in a preceding section, the electron drags along with it a bubble of a lower field energy density.

Can we harvest energy from the movement of the bubble, and harvest it separately from the movement of the electron?

We can "grab" the bubble at least through gravitation. The bubble is equivalent to a block of negative mass moving superposed to a constant density mass distribution. The flowing liquid example proved that the energy associated with such a movement can be negligible.


       ----> x       ================= - wire
      |                      ^
      v   r                 |  E  bubble

                                         |  E'
                                         v 
                                         • e-   

                                           R = distance(e-, wire)


Let us calculate how much energy is missing from the bubble in the case where there is an electron close to a negatively charged wire.

The field of the wire is

       -E₀ / r.

The missing energy density is

       ~ E E'

in the bubble. Close to the electron,

       E' ~ 1 / (R - r)²,

and E is roughly constant. The integration volume is ~ (R - r)² dr. The integral is

       ~ R - r.

Close to the wire, E ~ 1 / r, E' is roughly constant, and the integration volume is  ~ r dr. The integral is

       ~ r.

Let us calculate a very rough approximate value of the missing energy in the bubble. Let

       E  =  E₀ / r.

The electron is at a distance R. The electron field at the midpoint between the wire and the electron is 

       E'  =  1 / (4 π ε₀)  *  e- / (R - r)².

The missing energy density at the midpoint is

       ε₀ E E'  ≈  ε₀ E₀  *  2 / R 

                        * 1 / (4 π ε₀)  *  e-  *  4 / R²

                    =  2 / π  *  E₀ e- / R³.

The integration volume at the midpoint is something like

      1/4 R² dr.

The value of r varies 0 ... R. We conclude that the missing energy in the bubble is on the order of

       1 / (2 π)  *  E₀

electron volts. This can be compared to the mass of the electron, 511 keV.

In a typical experiment with a charged wire, the field strength at the distance of 1 meter might be at most 100 kV/m. If it affects the inertia of the electron, the effect is somehing like 15 keV, or 3%. The effect does not depend on the distance of the electron. Thus, the effect does not appear as a magnetic effect when the electron approaches the wire.

Conjecture 1. The bubble of missing lower field energy does not have much effect on magnetism.


The electric field at distances larger than R from the wire is roughly uniform, and does not add much inertia to the electron if it moves along the wire.

Conjecture 2. The inertia of the electron in a parallel movement with the wire does not depend appreciably with its distance R from the wire.


If the electron e- approaches the wire at a velocity u, it loses kinetic energy in its velocity component u. Conjecture 2 implies that its inertia is less as it comes closer. But the momentum is constant parallel to the wire. The electron e- accelerates in the direction parallel to the wire. This is seen as a magnetic "force".

Conjecture 3. An electron approaching a charged wire feels a magnetic "force" which is described by Maxwell's equations.


Poincaré stresses


Do Poincaré stresses have any signicance in the case of the charged electric insulator or a wire?


Poincaré stresses counteract the pressure components of the Maxwell stress tensor of the field. In many cases, Poincaré stresses cancel the effects of pressures on the momentum of the system.

Our analyses in the preceding sections have ignored pressures altogether. We have only looked at energy flows. We can justify our choice by claiming that Poincaré stresses cancel all effects of pressures.

But this should be studied in more detail.


Oppositely charged cylinders which are electric insulators: the Lorentz magnetic force formula holds


In the previous sections we studied the magnetic "force" felt by an electron approaching a charged metal wire. The "force" in that case did not change the momentum of the electron, and, therefore, is not a true force. The acceleration happened because the inertia of the electron decreased.

Normally, a magnetic field is produced by an uncharged wire inside which there is a flow of electrons. Let us first treat a simpler case where the electric current is produced by a moving electric insulator. Then there is no electric influence on the charges in the insulator.


                       charged electric
                       insulator cylinders

             v <---  ---------------------------- +  protons
                       ==============  -   electrons
             v <---  ---------------------------- +  protons 

                        ○       ○       ○       ○    B

             attraction
                                ^  F
                                |            |
                                             v  F'
                                                     repulsion

                                     ^ u
                                     |
                           v <--- • e-
    ----> x


The outer cylinder is positively charged and is comoving with the electron e-. The inner cylinder is negatively charged and is static. The charges cancel each other out, so that the electric field is zero outside the cylinders. The magnetic field B is non-zero.

The static positive charge attracts the electron (the force F). The moving negative charge repulses (the force F').

In our September 18, 2024 blog post we were able to derive magnetism by assuming that the energy given to e- by F is "moving at a velocity v" to the left, and the energy taken by F' is "static".

Let us analyze this from the aspect of fields.


                              ^  W' energy packet
                              |
                               
                              ^ u
                              |
                    v <--- • e-
  
                           <- p' surplus momentum

      ----> x
      

In the first section of this blog post, the electron e- loses some of its "u-kinetic" energy

       1/2 me u²

to Coulomb's force. Let us denote this energy packet by W'. The packet does not "move" in the x direction. The electron would retain the x momentum in the u-kinetic energy; let us denote it by

       p'  =  W' / c²  *  v.

But the electron e- has to give p' to the extra field energy in the combined field of the cylinder and e-, and must drag this energy along with itself.

Let us then add the positively charged cylinder to the configuration. One may imagine that the energy packet W' now does not come from the u-kinetic energy of e-, but it comes from the attractive force between e- and the positive cylinder. The packet travels at the velocity v to the left. The momentum p' is now held by e-, and a "magnetic force" accelerates e- to the left.

Is this explanation plausible? Why the surplus momentum p' is not absorbed by the cylinders? Why is it held by the electron e-?


Conclusions


Let us close this long blog post. We will treat later the case where an electron approaches a moving positively charged cylinder.

We would need empirical measurements of the electron behavior close to moving charged insulators. It is difficult to determine the correct model using pure reasoning alone.
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Friday, September 20, 2024

Poynting vector in a coaxial system

Let us look at the problem which we recognized in the previous blog post.


       ----------------------------- -  outer cylinder wall
             |      |      |   E
       ----------------------------- +  cylinder wall

  
       ----------------------------- +  cylinder wall
             |      |      |   E
       ------------------------------ -

                   --->  v

       ----> x


Let us consider a coaxial configuration where the electric field is nicely isolated between the cylinder walls.

The coaxial system moves to the right at the speed v. There is a magnetic field

       B  =  v / c²  * E.

The Poynting vector has the value

       ε₀ c² E  ×  B
  
       =  ε₀ E²  *  v.

A half of the value comes from shipping the energy of the field. The other half comes from shipping the x pressure of the field.

The Poincaré stresses which keep the cylinders from exploding cancel the pressure part of the momentum. The Poincaré stresses correspond to a negative pressure which cancels the positive x pressure in the field E.


                       ^  E'
                       |
                --------------  +      rod
                       |
                       v


Let us then add a static positively charged rod at the center of the cylinder. The field

       E  +  E'

is larger at the location of the rod. The Poynting vector to the right increases in value by

       ε₀ c² E'  ×  B

       =  ε₀ E × E'.

But there is no extra shipping of field energy or pressure in the x direction. How is it possible that the Poynting vector increases?

The answer: as elementary charges in the positively charged cylinder come close to the positively charged rod, they lose kinetic energy to the potential of the field of the rod. The charges the kinetic energy back when they recede from the rod. This is the origin of the energy flow in the field.


Conclusions


There is no paradox in this. We simply forgot to take into account the kinetic energy of the charges which create the magnetic field.

Now that we understand the energy flow in this case, let us return back to the Biot-Savart law.
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Wednesday, September 18, 2024

Biot-Savart derived from Coulomb's force

UPDATE September 23, 2024: Electromagnetism is not "linear" if we take into account the field energy. If electric fields E₁ = -E₂, then the energy density of the sum field E₁ + E₂ is zero, but each individual field has a non-zero energy density. Keeping this in mind resolves many apparent paradoxes.

If an electron e- is close to a conducting object, like a metal wire, that causes a backreaction in the distribution of electrons in the metal. This will significantly affect the inertia which the electron e- feels. In the "cylinder of electrons" in the diagram below we ignored the backreaction. The strong electric field

       Eelectr  +  Ee-

is "diluted" over the entire length of the conducting metal wire. It does not add to the inertia of the electron.

It looks like that Maxwell's equations describe the electromagnetism of charged metal objects. The rules are different for charged electric insulators!

----

Let us try the derivation of the Biot-Savart law once again. This time we try to take into account the momentum of the field, and all other factors.


The Feynman Lectures on Physics (1964) treats the case where a test charge q moves parallel to to a current-carrying wire. A more difficult case is when q directly approaches the wire.


A wire as two overlapping cylinders of charge: a simple explanation of the magnetic force on an electron e- approaching the wire


Let us model a long wire as two uniformly charged cylinders. The protons form a positively charged cylinder which is static in the laboratory frame. The electrons form another cylinder which moves very slowly relative to the laboratory frame: the speed is of the order of micrometers per second.



                  wire with a current

                 ●●●●●●●●●●                protons
                 •••••••••••••••••  ---> v      electrons e-
                 ●●  ^   ●●●●●                protons
                    |   |  W
                    |   |          ^ u
                    |   |          |
                   v              • e-
                  W energy packet
    ^ y
    |
     ------> x


Let us use this simple model:

1.   As the electron e- approaches the wire, it receives an energy "packet" W from its Coulomb attraction to the protons. The packet W has a zero momentum in the x direction.

2.  The electron must give the energy packet W back to the electrons in the wire, because there is a Coulomb repulsion. This time W must comove with the electrons in the wire, to the right at the velocity v. The packet must contain a momentum

        p  =  W / c²  *  v

to the right.

3.  The electron is left with a momentum p to the left. This is the "magnetic" force which accelerates the electron to the left.


Let us next analyze this in detail, trying to make sense of the field energy and momentum.


A uniformly negatively charged static cylinder and an approaching electron e-


       =============== -  cylinder of electrons


                        ^  u                   W = F Δy
                        |
             -v <--- • e-      
                        |
                        v  F   Coulomb's force

              ^       ^
              |       |

            Eelectr  Ee-
    ^ y
    |
     ------> x


The cylinder consists of the electrons inside a wire. As the electron e- approaches the cylinder, it loses a kinetic energy W. There is no change in the x momentum of e-.

Coulomb's force describes the interaction between charges: we ignore the momentum in the electric field. Since e- loses kinetic energy, its inertia is diminished. The electron accelerates to the left. In the frame of the protons in the wire, this acceleration in interpreted as the magnetic force.

If we take into account the field, then the situation looks different. The extra energy W is now in the combined field

       Eelectr  +  Eq,

which is moving to the left. The momentum to the left of the electron e- and of the extra field energy W has grown. The interaction pushed this combined system to the left.

But now we face a dilemma: if the system acquired a momentum to the left, then the cylinder must have acquired a momentum to the right. The Lorentz force says that the cylinder did not acquire any momentum to the right. Is the Lorentz force formula incorrect, after all?


Conclusions


We are able to derive the magnetic force with a very simple model where we only take into account Coulomb's force, and ignore the electric field momentum.

But if we include the electric field momentum, then the Lorentz force formula seems to be broken. We will investigate this more in the next blog post.
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Friday, September 13, 2024

Lorentz covariance of electromagnetism

In the summer of 2023 we suspected that Lorentz covariance does not work correctly in electromagnetism. Let us return to this question.









Moving charge Q and an approaching test charge q


                      Q ● ---> v
        
                                                    W = q E Δy

                          ^ u       ^ q E     Lorentz force
                          |           | 
                          • q        -----> q v × B
                         /
                         \
                         /    spring

                         |   F = -q E
                         v

        ^ y
        |
         ------> x


Let Q be positive and q negative. The charge Q creates an electric field E, and a magnetic field B which steers q to the right, according to the Lorentz force.

The spring and the force F on it cancels out the electric attraction q E between q and Q.

There is no retardation in the field of Q. If Q is to the y direction from q at the time t = 0, then the electric field E of Q points directly to the y direction at the time t = 0.


Switch to a comoving frame of Q


                              ● Q

                                                         W = q E Δy

                              ^ u        ^   q E   Lorentz force
                              |            |   
                   -v <--- • q
                             /
                             \
                             /  spring

                             |   F = -q E
                             v  

            ^ y'
            |
             -----> x'


The electric field E' points to the y' direction, and there is no magnetic field B. The momentum of q does not change with time.

The Lorentz force says that q does not have an acceleration in the x' direction in this case.

This seems to break Lorentz covariance. The problem is that the Lorentz force does not take into account the momentum held by the energy

       W  =  q E Δy,

which q receives from the field of Q. By Δy we denote the distance which q moves to the y direction.

In the first diagram, the energy W holds momentum to the positive x direction (and this also explains the magnetic field B: we do not really need to consider B at all).


Is the Poynting vector aware of the momentum in W?


Maybe the Lorentz force simply is an imprecise description of the electromagnetic action? Let us check if the action understands the momentum held by W.

Does the Poynting vector

        1 / μ₀  *  E  ×  B

understand that in the first diagram, the energy W flowing to q contains momentum to the positive x direction? And that in the second diagram, W does not contain momentum to the x direction?

The energy W flows further to the spring. We assume that in the first diagram, that flow does not contain x momentum. The momentum is retained by q.

Second diagram. The magnetic field of the negative charge q moving to the y direction makes the Poynting vector 1 / μ₀ * E × B to point toward q on both sides of q:


                             ● Q
                             |
                             |  E
                             v


                             ^  u
                             | 
                  -v <--- •  q
                B ×                 ○  B
            
                   --->            <---
                E × B            E × B


Above, × denotes a magnetic field B line produced by q coming up from the screen, and ○ is a field line going into the screen. We ignore the magnetic field associated with the -v velocity vector.

The Poynting vector 1 / μ₀ * E × B brings field energy to q. Is the diagram symmetric versus the energy flows, or does q receive more energy from the left, as q moves to the left at the speed v?

First diagram. 

                        
                         Q ● ---> v
                             |
                             |  E
                             v


                             ^  u
                             | 
                              •  q
                B ×                 ○  B 
            
                   --->            <---
                E × B            E × B


The first diagram looks more symmetric. The Poynting vector seems to bring the same energy flux from the left and from the right. But the charge q should accelerate to the right. There is something wrong.

Does the magnetic field of Q change this somehow?


Replace Q with a part of a wire with an electric current: analysis of the Poynting vector to q


Let us check if the Poynting vector handles a pure magnetic field correctly. We put an opposite static charge -Q close to Q, so that the electric field E of them is essentially zero at q. 


                        -Q ●
                         Q ● ---> v
                           
                    
                     


                             ^  u
                             | 
                              •  q
               Bq ×                 ○  Bq  
            
                   --->            <---
               E × Bq            E × Bq        E = 0

         ^ y
         |
          ------> x


The fields close to q are now:

1.   The electric field produced by Q and -Q is EQ = 0.

2.   The magnetic field BQ produced by Q is significant.

3.   The fields Eq and Bq produced by q are significant.


The Poynting vector Eq × Bq cannot bring any momentum to q. Only

       1 / μ₀  *  Eq × BQ 

might possibly bring momentum. But Eq is almost radial to q, and Eq × BQ is orthogonal to Eq. If q accelerates to the right, is the electric field deformed enough, so that the Poynting vector could give momentum to q?


(Richard Fitzpatrick, 2014)

Conservation of the electromagnetic 4-momentum is this formula:









The term j × B is the force which pushes q to the right, to the positive x direction. We are interested in the coordinate i = 1, which is the x coordinate.

Is -∂ (ε₀ E × B) /  ∂t then non-zero?


Make q very heavy and push it suddenly toward Q and -Q: momentum conservation works correctly


We suddenly push q to the positive y direction, so that it acquires a velocity u. The x coordinate of Q, -Q and q is the same after the push operation.

Can q somehow acquire momentum to the right, from the electromagnetic field? The speed of light is finite. The momentum must come from a close location to q.


                 -Q ●
                  Q ● ---> v
                      |
                      | line L of cylindrical symmetry
                      |
                      |

                      ^ u
                      |
                      •
                      q

      ^ y
      |
      ○ ----> x
     z points out
     of the screen


Let us assume that q is extremely heavy, so that its acceleration to the right is very small, even though it gets a lot of momentum to the right from the magnetic field BQ. Since q moves almost exactly to the positive y direction, then the fields Eq and Bq of q are almost exactly cylindrically symmetric relative to the line L which contains q and runs to parallel to the y axis.

Let us assume that the distance from q to Q and -Q is very large, and that Q and -Q are very close to each other.

Then the electric field EQ-Q of Q and -Q is almost exactly zero close to q. The magnetic field BQ of Q is almost exactly parallel to the z axis, and is almost exactly uniform.


The field momentum is







The field momentum close to q in Eq × Bq points to the y direction, on the average.

Let us study the field Eq of q at some distance r.
















Daniel V. Schroeder (1999) drew the electric field lines if a charge q is initially static, but is pushed and acquires a speed u upward. The diagram shows the field lines after a little while.

What is the vector Eq × BQ like?

Below q in the picture, that vector points to the left, and above q, the vector points to the right. As q moves up, the field will contain more momentum to the left. This may explain how q itself gains momentum to the right?

The Poynting vector does not bring any new energy to q, but the movement of q changes the amount of x momentum in the integrated Poynting vector.

If we stop the y movement of q, then Q and q absorb the x momentum stored in the field.

The electromagnetic momentum conservation probably works correctly in this case.


The Lorentz force does work correctly


Let us look again at the case where the electric attraction of Q is counteracted with a spring, rather than putting a charge -Q close to Q.

We may assume that the spring is attached to the laboratory table.


                              ● Q

                                                         W = q E Δy

                              ^ u        ^   q E   Lorentz force
                              |            |   
                   -v <--- • q
                             /
                             \
                             /  spring
                  -v <--- × attachment

                             |   F = -q E
                             v  

            ^ y'
            |
             -----> x'


In the moving frame, the spring moves to the left along with q. Now we see that the movement of q to the left will slow down as time passes. That is because q receives an energy packet W which is static relative to Q. But q has to pass W to the spring which is moving to the left. Then q has to give up some of its own momentum to the left. We conclude that q accelerates to the right, just as it should do.


Does the energy W which q gains in the field of Q have a zero momentum relative to Q?


                           ● Q

                           |  EQ
                           v   
                                                 W = F Δy

                                      ^ F
                           ^ u     |  Coulomb's force q EQ 
                           |
                -v <--- • q

       ^ y
       |
        ------> x


We still have to check if the energy packet W really is static relative to Q. Or does the velocity -v of q affect this?

If the Lorentz force formula, or Coulomb's force formula is correct, then q does not receive any momentum in the x direction. Then W really has a zero momentum relative to Q.

We can take as an axiom that Coulomb's force points exactly toward a static charge Q. If Q is accelerating, then this gets more complicated.


Conclusions


Classical electromagnetism fared well in this test of ours. Lorentz covariance and momentum conservation were worked ok.

We can take as an axiom that Coulomb's force points directly toward a static charge, and that it gives an energy packet W which is static relative to Q.

Then we can derive the Biot-Savart law for a wire segment. The derivation of Zile and Overduin (2014) can be corrected this way. We will write a new blog post about Biot-Savart. We will also look at the August 28, 2024 overlapping electric fields again, now that we understand 4-momentum conservation better.
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