We need a "magnetic" Higgs field? The Higgs boson spin is 1?

UPDATE July 6, 2022: A scalar field can be Lorentz covariant. See our note in the next blog entry.

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In the update to the previous blog post we noted that if a fast moving observer in his frame displaces a volume S of the Higgs field to a value z, then a static observer will see the volume smaller, and see the energy in  the displacement smaller than the fast moving observer. This contradicts special relativity.

Can a scalar field be Lorentz covariant?

Let us try to form a scalar field theory from the electric Coulomb potential. That is, we assume that the magnetic vector potential is zero.

One runs into difficulties in making the scalar field theory Lorentz covariant. This suggests that a scalar field theory cannot be Lorentz covariant. One must introduce a magnetic form of the field to obtain Lorentz covariance.

If a field has an electric and a magnetic component, then it has an associated 4-vector and the spin of the quantum should be 1.

https://atlas.web.cern.ch/Atlas/GROUPS/PHYSICS/PAPERS/HIGG-2013-01/

The LHC measured that the spin of the Higgs particle is 0. How do we explain this discrepancy? Could the very short lifetime of the particle explain this?

If the spin of the Higgs particle is 1, then its angular momentum may have values +1, 0, -1. Could it be that the LHC experiment for some reason only produces the 0 variety?

In quantum mechanics, scalar field theories are used as toy models in teaching. They are probably ok if we study nonrelativistic phenomena.

Symmetry breaking would break the Galilei symmetry and Lorentz covariance

We think that symmetry breaking is something which is not Lorentz covariant. If we have a crystal, then its lattice defines preferred directions in space, which contradicts the Galilei symmetry.

In a previous blog post we explained that, in a sense, symmetry is not broken in the Higgs model. Thus, the Higgs model does not have this problem.

Conclusions

The Higgs field in the Standard Model is strange: it is the only scalar field and its quantum has the spin 0. Ockham's razor suggests that we should not introduce a field which differs from all the other fields that we know.

If a scalar field cannot be Lorentz covariant, then we must introduce a magnetic Higgs field. The spin of the quantum should then be 1.

Does a magnetic Higgs field spoil its function as the generator of mass?

Can we add a magnetic Higgs field? The "velocity" of the Higgs field is not defined. How do we define the magnetic Higgs field?

Removing Higgs field defects: set the potential infinite at the origin of the complex plane

UPDATE June 30, 2022: the Mexican hat potential may contradict quantum mechanics. Let us create a very high energy Higgs particle. If the Higgs potential is not the harmonic potential

      k (r - v)²,

where v is the minimum energy value and k is a constant, then the wave of the particle may be disturbed by the strange form of the potential function V(x). The wave would be scattered from itself. Is it possible in quantum mechanics that a particle scatters itself?

A possible solution: the wave which makes the field slip over the center of the Mexican hat necessarily contains a lot of energy, and very many Higgs particles. We can treat the wave as a classical wave. The next question is if the classical wave conserves energy and momentum?

If the action is conserved in translations of position and time, then momentum and energy are conserved. This is true for the Higgs lagrangian. Thus, the classical wave does conserve energy and momentum in newtonian mechanics.

However, is the Higgs system Lorenz covariant? Suppose that we have on observer X moving at a very high speed. He has displaced the Higgs field in a volume S to a value z. He used the energy

       E = (S, z).

A static observer thinks that the volume is only 1/2 S, because of length contraction. He thinks that the energy is less. Does this break Lorentz covariance?

The concept of a scalar field is suspicious from the point of view of Lorentz covariance. The electromagnetic field has to be Lorentz transformed. Can a scalar field be Lorentz covariant?

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In our previous blog post we suggested that one can eliminate sources in the Higgs field by using a Mexican hat potential which is infinite at the origin of the complex plane.

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

If the Higgs field originally is without sources, then it (probably) cannot obtain sources even at a high temperature because the value of the field cannot "slip" over the center of the hat. Then no defects or crystal boundaries can appear in the field.

The value of the field is then given by

       φ(x) = r exp(i α),

where the radius r > 0 is real and the angle α is real.

One may identify the space (r, α) with the Archimedes' screw where the origin r = 0 has been removed.

But is it sensible to remove the origin from the space of the values of the Higgs field?

https://www.theorie.physik.uni-muenchen.de/lsfrey/teaching/archiv/sose_09/rng/higgs_mechanism.pdf

Let us look at the article by Kien Nguyen (2009):

The coupling between the Higgs field φ and the gauge boson 4-potential field A looks like the minimal coupling of electrodynamics.

The gauge transformation

The value of the lagrangian without the Higgs field is conserved by adding the gradient of an arbitrary function η(x) to the 4-potential A. This is just like in electrodynamics. The addition changes the gauge.

When changing the gauge of A, we have to transform the Higgs field, too, to conserve the value of the lagrangian. The full gauge transformation is the lowest pair of equations above.

The gauge transformation of A and φ conserves the value of the lagrangian: the physics of the system is not affected by the gauge transformation.

Since φ(x) is transformed by multiplying it by

       exp(i e η(x))

we do not need to include the origin 0 among the possible values of the Higgs field.

Removing the origin from the possible values of the Higgs field

Is it an ugly modification to remove the origin 0 from the possible values of the Higgs field?

The motivation for the introduction of the Higgs field is to create masses for the W and Z bosons.

The Higgs field has to have an almost fixed, non-zero vacuum expectation value to serve this function.

But it is not required that the value of the Higgs field must be able to "slip" over the center of the Mexican hat. We can claim that the "slip" feature is superfluous and makes the model more ugly than a model where slipping is prohibited.

In the gauge transformation, we have to transform φ(x) by rotating its value around the origin of the complex plane. Slipping is something very different. Why should slipping be allowed?

Higgs field defects as cosmological objects

https://www.researchgate.net/publication/262953976_Tom_Kibble_and_the_early_universe_as_the_ultimate_high_energy_experiment

Neil Turok (2014) writes about defects in cosmology.

No defects have been observed. Do they exist at all?

If defects would exist, they would be a form of matter which is radically different from the particle matter which we are familiar with.

Ockham's razor suggests that defects do not exist. If defects would exist, why we have seen no such matter so far?

Wormholes in general relativity can be regarded as "defects". They differ in the topology from flat Minkowski space. In this blog we have argued that Minkowski is the true structure of spacetime and wormholes cannot exist. In the case of the Higgs field, defects would exist in the combination of Minkowski space plus the complex number space of the Higgs field values.

We do not like defects in general relativity. It is logical that we do not like defects in the Higgs field either.

Defects in crystals and superfluids

Defects do exist is crystal structures of solid matter, as well as in the vortex structure of superfluids. Solid matter is a complex structure which has many parts, atoms. We do not know the structure of a superfluid. The structure might be complex.

Defects in the case of crystals are emergent objects which "live" above a complex substructure.

If there were defects in the Higgs field, they would live in a very simple and fundamental field of nature. Is it a plausible assumption that fundamental fields can contain defects?

The LHC particle accelerator and defects in the Higgs field

We need to check if the conventional Mexican hat potential would allow defects to be created at the 7 TeV energy of the LHC.

In LHC, individual particles can obtain the 246 GeV energy of the center of the Mexican hat. Is that enough to create a defect.

The LHC has not observed any defects.

The Higgs field is "almost" real-valued

The value of the radius r in the value of the Higgs field

       φ(x) = r exp(i α)

must be almost constant. Otherwise, the mass of W and Z bosons would vary.

Why r cannot be exactly constant? We do not know. The Higgs boson would not exist if r would have a strictly constant value. The Higgs boson consists of the oscillation of r around the minimum potential.

Most of the variability of the value of the Higgs field is in the angle α. In this sense, the value of the Higgs field is almost a real number.

Conclusions

We have argued that defects in the Higgs field would be an ugly feature.

We have to check if anyone has calculated the minimum energy of a defect in the Higgs field, and if the LHC would be able to create defects.

There is no "symmetry breaking" in the Higgs field, after all; galaxy clusters

UPDATE June 27, 2022: Suppose that we define the Higgs field as a "helix":

       φ(x) = (r, α),

where r >= 0 is real and the angle α is any real number.

The "value" of φ in an equation is the complex number defined by r and α, but after one round around the origin of the complex plane we do not return to the same point. It is like an Archimedes' screw.

We require the Higgs field to be continuous in spacetime. Could it be that then there will be no "crystal boundaries" nor defects?

But in this case there is a singularity at φ(0, 0).

Another attempted solution: require that the vector field defined by φ never contains a source. We can ensure that it stays that way by putting the Higgs potential infinite for φ(x) = 0. Is there any reason why the Higgs potential should be finite there?

----

The classical complex Higgs field is assumed to fill the entire spacetime, and its vacuum expectation value in the minimum energy state is a constant complex number

       v exp(i α),

where v is the radius of the groove, or the circular valley, of the Mexican hat potential, and α is an angle from the real axis.

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

The field rolls to some angle α in the groove of the Mexican hat potential.

We in this blog are somewhat worried of the following possibilities:

A. Even in the minimum energy state, spacetime seems to contain information: the angle α.

B. If the field rolls down from an arbitratry state in a large spatial volume, then the angle α will differ from place to place. There will probably be something like boundaries of "crystals" or defects in the state of the field. Could there even be singularities?

https://www.theorie.physik.uni-muenchen.de/lsfrey/teaching/archiv/sose_09/rng/higgs_mechanism.pdf

Kien Nguyen (2009) has written a simple introduction to symmetry breaking.

Problem A: the information in "empty space"

We may define empty space, or the vacuum, as the lowest energy state of the classical fields.

For most fields, that means that they are identically zero everywhere.

However, the Higgs field is strange: it is not zero, and also seems to contain information, the angle α.

This would be ugly, but it looks like that we cannot in any way find out the value of α. We can only see that α is the same throughout space.

If α is not the same constant everywhere, then the electroweak 4-potential A is not zero everywhere, and we should see some kind of "matter" in space.

Only differences in α are observable.

In the paper of Kien Nguyen the angle α is eliminated altogether, and the 4-potential A will do its job.

Thus, do we have a broken symmetry or not?

Not in empty space. But if the space is not empty? Then the existence of matter breaks the symmetry.

https://arxiv.org/abs/1102.0468

Ward Struyve (2011) argues that there is no symmetry breaking.

He writes that Peter Higgs himself in his 1966 paper presented an interpretation where the symmetry is not broken.

Problem B: are singularities possible in the Higgs field? No

For low energies, the Higgs field is approximately determined by the angle α at each location.

Since α can vary from place to place, it may happen that we will have something which is analogous to crystal boundaries or defects.

We are saved from singularities, because the potential of the Higgs field at zero, or at the center of the Mexican hat, is finite.

There could still exist crystal boundaries. We need to check the work of Tom Kibble. He has written about defects.

Conclusions

Empty space is still "empty", even with the Higgs field. There is no genuine symmetry breaking where the Higgs field in empty space would determine a preferred direction in the complex plane.

The Higgs field is "scalar": it does not determine any direction in spacetime.

We do not like singularities. Fortunately, singularities cannot form.

We do not like that the Higgs field in empty space has a non-zero value nor that it has a charge which fills all the space. The Higgs field is different from all the other fields in that respect. But we have not found a way to build a model where the value of the Higgs field is zero: we cannot find a way to give a mass to the W and Z bosons if the field is zero.

Have astronomers found anything which might be crystal boundaries in the Higgs field? The large-scale structure of galaxy clusters looks like "filaments". Could they come from defects in the Higgs field?

Our work removes the motivation for supersymmetry: there are no divergences

UPDATE May 13, 2022: We made an elementary error by assuming that the Higgs particle is free. It is not free: its lifetime is very short. The particle exists as an internal line of a Feynman diagram. In an internal line we do allow the loop diagrams below.

Do we have a problem with divergences? We have to check if the machinery which we developed in the last fall is able to remove the divergences.

A further observation: for very short wavelength particles, a Higgs particle in an internal line is "essentially free" because short wavelength phenomena are not "aware" of the interaction where the Higgs particle is taking part. We could argue that destructive interference removes all very short wavelength particles. This is a new way to understand the natural cutoff of high frequencies.

We need to check what kind of divergences occur in the diagrams below.

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The main motivation for supersymmetry is the "hierarchy problem": why are self-energy corrections to the Higgs boson mass so small compared to the Planck scale.

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

In this blog we have argued that the Feynman diagrams above are forbidden for a free particle. A particle cannot do a Baron Munchausen trick and affect its own state alone, without any external interaction.

Our "sharp hammer" model explains this. There is perfect destructive interference for the reactions above. The energy, momentum, and the speed of the center of mass of the particle have to be conserved. There is nothing to witness that the reaction happened, and perfect destructive interference is possible.

Thus, there is no need to cancel anything.

Conjecture. We will never find any superpartners of particles.

Elastic time crystal: a model of superconductivity

In our blog post on April 22, 2022 we tried to explain a superconducting current in a linear wire which forms a loop with an ordinary wire.

An energy gap model explains permanent currents in an isolated time crystal

Let us have a block of a metal at 0 K, and no external current flowing through the block. The block is at the lowest energy state and cannot radiate. The motion of electrons in the block forms a time crystal.

In this case we can appeal to an energy gap to explain the loss of resistivity. If an internal current inside the time crystal would slow down, that would require more energy.

This effect is true for all isolated systems at 0 K. There cannot be any resistance in the movement of electrons.

In a permanent magnet there is a considerable ordered current inside. We argued in our previous blog post that the magnetic field of the magnet must be static. If the magnetic field would change periodically, there would be electromagnetic radiation.

For a superconductor we have an additional requirement that we can add an arbitrary current which flows through the superconductor, and there is no resistance.

All energy gap or potential wall models of superconductivity are broken: a perpetuum mobile would exist?

In our previous blog post we assumed that a single electron joins a flow of many electrons which carries the new electron to the other end of the wire.

We ran into problems. The model would make a perpetuum mobile possible.

The perpetuum mobile problem may exist in all models which try to explain superconductivity through an energy gap or a potential wall. That is, there would be a potential wall which prevents the system from decaying from a nonzero current state to a zero current state.

Empty space has zero resistance

We can implement zero resistance by sending electrons through empty space. Then there is no potential wall between different magnitudes of the current - and no perpetuum mobile is possible.

To prevent the electrons from dispersing, we can load them on a "ship" which takes many electrons at a time through empty space.

A charged ship in empty space is a nice model for superconductivity. It certainly satisfies energy and momentum conservation.

The mystery in superconductivity is how this ship can sail through the lattice, impurities, crystal borders, and thermal phonons in a superconductor - and still feel (almost) no friction.

The smoothness of the electron cloud of a time crystal may explain superconductivity

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmmic.html

The standard free path model of electrical resistivity suggests that it is the collisions of individual electrons to phonons, impurity atoms, and grain (crystal) boundaries which are responsible for resistance.

If we can eliminate these collisions, then the resistance can be extremely small.

For a molecule in the lowest energy state, a static observer cannot see any individual electron motion because if he did, he then the molecule would radiate.

What about a slowly moving observer? The molecule may be electrically polarized in a complicated pattern. A slowly moving observer can see the "bumps" in the electric field of the molecule. He probably cannot see any individual electrons.

Seeing an individual electron in a molecule requires considerable energy, because of uncertainty principles. Maybe they are invisible for a slowly moving observer?

Thus, the molecule appears like a smooth, elastic object to a slowly moving observer.

The problem of grain boundaries: a "flux" model cannot explain zero resistance

Conducting electrons in a superconductor have to cross grain boundaries to enable the electron drift, and the existence of an electric current.

Let us analyze this in a model where the time crystal of conducting electrons and lattice vibrations initially stays at the same place. Then we apply a voltage for a short time to make the time crystal to move.

The grain size in a typical metal is of the order 10 micrometers.

           Fermi velocity

                     ^                              #

                    /                                #

                  /                                  #

                /                                    #

              /                                      #

            e- -->                             

                drift                         grain boundary

The Fermi velocity is ~ 10⁶ m/s and the drift velocity is only ~ 10⁻⁶ m/s.

The de Broglie wavelength of a Fermi electron is

       λ = h / p

          ~ 1 nm.

The "drift flux" of conducting electrons is ~ 10⁻¹² on top of the "Fermi flux" of electrons.

If the size of a grain is 10 micrometers, that corresponds to 100,000 atoms. Bloch's theorem suggests that the reflected flux of electrons at a boundary might be even much less than 1 / 100,000. A conducting electron passes

        ~ 10⁶ m / 10 μm = 10¹¹

grain boundaries per second. It might be reflected ~ 1 million times per second.

We calculated on April 25, 2022 that 10¹¹ reflections would drain the momentum in the magnetic field of the current.

That many reflections for each conducting electron might happen in 100,000 seconds, or in a day.

The "flux" model cannot explain a current which flows for 30,000 years without adding more energy.

The speed of sound in the time crystal

The time crystal forms at a temperature ~ 10 K, which suggests that the binding energy is around 1 millielectronvolt.

The binding energy of an atom of a metal in a crystal is 0.1 - 1 eV.

The rest mass of conducting electrons is ~ 1 / 10,000 of the atoms.

The speed of sound may be proportional to

       sqrt(E / m),

where E is the binding energy and m is the mass.

We conclude that the speed of sound in the time crystal is ~ 30 km/s.

However, we know that the speed of a potential change in a wire is close to the speed of light. How can we reconcile this with the speed of sound?

The speed of sound assumes that the zones of extra negative charge move in unison with the zones of extra positive charge in the lattice. This is very different from just moving the electrons.

How does a system with the lowest resonant frequency f react to a smooth disturbance which lasts for a long time t?

The system may, for example, be point masses joined together with springs, or an elastic body.

Let us assume that the system moves slowly with a velocity v and meets a smooth potential pit or a smooth potential hill.

    

        part of the system

                     v

                ● ----->                         time t

                ______               ______

                             \_____/

                smooth potential well

If the part moves very slowly, the system has plenty of time to adjust to the extra force which is pulling or pushing the part.

Let t be the time under which the part passes the potential pit. If

        t >> 1 / f,

then the potential pit cannot produce significant vibration to the system. The force field of the potential pit is in that case conservative, and kinetic energy is not lost to vibration.

An elastic time crystal meets a grain boundary

     elastic time crystal of electrons

                 |/\/\/\/\/\/\/\/\/\/\/\|     -->   drift

                ●          ●    ●             ●  

   

                  grain boundaries

Let us consider the drift as a movement of the time crystal of electrons. We ignore the movement of individual electrons because we assume that the time crystal is at its lowest energy state in its comoving frame.

The elastic time crystal will be deformed as it slides past grain boundaries.

Deformations create vibrations of the time crystal. How much energy do these vibrations drain from the movement?

The de Broglie wavelength of Fermi electrons is ~ 10 nm. We may assume that the smallest "feature size" in the time crystal is of that order.

How long does it take for a 10 nm feature to slide over a grain boundary? The width of the boundary is one atom, or 0.1 nm.

The velocity of the time crystal is 1,000 nm/s. We conclude that the time of the transit is

       t ~ 0.01 s.

Let the time crystal be of a size 1 meter. We calculated above that the speed of sound in the crystal is ~ 50 km/s, or might even be close to the speed of light. Then the lowest resonant frequency of the time crystal is

      f ~ 50 kHz,

or higher.

We have

       t >> 1 / f.

The time crystal cannot lose much of its kinetic energy to vibration because it moves so slowly.

Can we calculate an upper bound for the energy loss?

Almost all of the momentum of the time crystal is stored in its collective magnetic field. The magnetic field wants to keep the current constant. If some part of the time crystal slows down, other parts must make up for the lost current.

Let the part lose the momentum p in the collision. The magnetic field must put the momentum p back to the system. That should be easy.

We guess that the momentum p lost in the transit over the potential pit is much less than the momentum of the electrons in that part.

A part collides with a grain boundary every 10 seconds. The momentum p which it loses is much less than

       10⁻¹¹

of the corresponding momentum in the magnetic field. The current will last much longer than

       10¹² seconds

       = 30,000 years

if we just look at the dissipation at grain boundaries.

There is a grain boundary for each 10 micrometers. But there might be an impurity atom for each 10 nanometers, or 100 atoms.

Our calculation may explain the insignicant dissipation at impurity atoms, too.

Phonons and the elastic time crystal

Let us then look at the interaction of phonons and the elastic time crystal.

On April 25, 2022 we calculated that the black body radiation power at 4 K is

       1.4 * 10⁻⁵ J/m².

A typical "large energy" quantum at that temperature is

      1 meV = 1.6 * 10⁻²² J.

Since the speed of sound is ~ 5 km/s, we have one quantum in each cube whose side is 40 micrometers.

The quantum is a phonon in the lattice.

We need to do a sanity check. The number of phonons grows as ~ T³ with the temperature. At 293 K we would have one phonon in a cube whose side is

        40 micrometers / 73

        = 550 nm.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmmic.html

In the link it is calculated that the free path in copper at 293 K is 39 nm.

Why is the free path so short? Does a conducting electron scatter many times from a single phonon? Or does it scatter from low-frequency phonons?

Phonons will certainly make the time crystal to vibrate. The time crystal will be in an equilibrium with its environment. But can phonons rob significant momentum and energy from the slow movement of the time crystal?

Let us write a new blog post about this difficult problem.

Electron movement in a wire: some crude numbers

Let us have a 1 cm thick and 1 meter long wire made of a metal. Its weight is roughly 1 kg, and it contains some 10²⁵ atoms.

There are

       ~ 10²⁵ conducting electrons

in the wire. The mass of these electrons is

       ~ 10⁻⁵ kilograms.

The charge of these electrons is

       ~ 10⁶ coulombs.

We want to create a 1 ampere current in the wire.

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

The inductance of the wire is

       L ~ 1 microhenry.

The voltage to increase the current I in the wire is

       V = L dI / dt.

We put a voltage V of 1 microvolt over the wire for 1 second. After that time, the current is 1 ampere.

The Coulomb force of 1 microvolt / meter on the charge 10⁶ coulombs is

        F ~ 1 newton.

The energy of the magnetic field after the operation is

       1/2 L I² ~ 1 microjoule.

The "impulse" which the voltage exerted on the conducting electrons was

       p = 1 newton second.

The collective velocity of the electrons is only

       v = 1 micrometer / second.

It is as if the "inertial mass" of the electrons were a whopping

       10⁶ kilograms.

The kinetic energy and the momentum calculated from the rest mass of the electrons is very small. The momentum is only

       10⁻¹¹ newton seconds,

and the kinetic energy is

       10⁻¹⁷ joules.

Radiation pressure at 4 K is negligible on an object moving 1 micrometer per second

Suppose that we have an object carrying those one million coulombs of charge and moving at v = 1 micrometer per second. That corresponds to a current of one ampere.

We want to calculate the frictional force that the reflection of black body radiation at 4 kelvins imposes on the object.

Let us assume that the area A of the object is one square meter. The power of black body radiation is

       P = A σ T⁴,

where σ = 5.67 * 10⁻⁸ W/(m² K⁴) is the Stefan-Boltzmann constant and T is the temperature. We have

       P = 1.4 * 10⁻⁵ W

at 4 K. The radiation pressure force by the power P is

       P / c = 5 * 10⁻¹⁴ N.

The pressure is almost the same on the each side of the object, except for the Doppler shift caused by the tiny velocity v = 1 μm/s.

The effect of the Doppler shift is

       4 v / c,

because the reflection adds a factor of two, and the effect is on both sides of the object.

We conclude that the frictional force on the object by black body radiation is

       F ~ 4 v / c * P / c

           = 7 * 10⁻²⁸ N.

We calculated above that, to create the magnetic field for a 1 ampere current in a one meter wire, we have to expend 1 newton second of impulse. The force F is negligible relative to that.

The friction from black body radiation is negligible at 4 K. The resistivity at a low temperature has to come from collisions with the lattice. Those collisions move a large amount of impulse from the lattice to the electrons.

Problems with time crystals and superconductivity: a perpetuum mobile is created?

In an ordinary crystal, the lowest energy state is static, except of zero-point vibrations of the atoms. In a time crystal, the lowest energy state involves movement of particles. A moving lowest energy state is hard to grasp from the classical point of view. In classical mechanics we are used to configurations where one can reduce the energy of the system by slowing down the motion.

The simplest time crystal: a particle in a box in the lowest energy state

Let us consider the simplest textbook example of quantum mechanics at work: a particle in a box.

                                                   |

                                                   |   spoon

                                                   0

                               particle

               |               <-- • -->               |

                           box, length L

Classically, the "crystal" would be a state where the particle has been stopped at a certain location. In quantum mechanics, the particle bounces back and forth, so that its de Broglie wavelength is double the length of the box.

Suppose that we insert a spoon in the box and try to slow down the particle. Why this cannot succeed?

                               ________

                             /                 \ 

                      jumping rope rotates

The wave function of the particle is like the jumping rope of children rotating around its ends fixed at the walls of the box.

Any disturbance to the wave moves the system to a higher energy state. When we try to insert the spoon into the box, we feel "pressure" resisting the insertion procedure. We must do work against the pressure.

If the particle is at the lowest energy level, then the time crystal of its movement is the slowest possible movement. The time crystal is stable simply because the particle cannot move any slower.

A time crystal involves "frame-dragging": any change in the motion requires additional energy

From general relativity we know frame-dragging around a rotating neutron star or a black hole. The lowest energy state of an approaching test mass is one where the test mass moves along with the rotating body. If we want to keep the approaching mass static relative to the global frame, we have to supply "kinetic energy" to it.

The lowest energy state of a time crystal drags the frame of electrons.

Let us compare this to an ordinary crystal. There the lowest energy state is static (except for zero-point oscillation). A time crystal probably behaves just like an ordinary crystal, if we take into account that the frame is dragged here and there.

If we push a single atom in an ordinary crystal, we have to do work. Furthermore, the momentum which we gave to the atom quickly disperses to the entire crystal.

An ordinary crystal may act as an easy-to-grasp analogue of a time crystal.

The time crystal of conducting electrons plus lattice vibrations in a superconductor

At a high temperature, conducting electrons behave much like a gas. A gas is not a time crystal since it is not ordered.

             lattice of ions

                    +        +

           o     o o     o o     o    <----

           o     o o     o o     o    <---- vibration

           o     o o     o o     o    <----

           o     o o     o o     o    <----

                    ^        ^

                    |        |

                    e-       e-

         conducting electrons

We conjecture that in a superconductor many, or all, conducting electrons form a time crystal together with lattice vibrations. Electrons like to move in zones where a vibration has concentrated positive charge.

The electrons move some 1,000 km/s. It cannot be an ordinary crystal where electrons would be confined each to a small space.

The orbitals of a molecule are an analogue for the time crystal in the lattice. Electrons must keep on moving. Some electrons may have orbits which ship them around for the whole length of the molecule.

We conjecture that a superconductor is like a giant molecule where denser zones of the lattice ions are like nuclei in a molecule, except that these denser zones move at the speed of 5 km/s. Electrons move 200 times faster and will order and synchronize their own movement according to these zones.

How does the superconductor time crystal deal with impurities, or borders of crystals of the metal?

An ordinary crystal grows in a way where it tries to avoid impurity atoms or molecules. A time crystal probably behaves in the same way. It avoids impurities in the lattice.

How about borders of metal crystals? In a non-superconducting metal, borders of crystals cause scattering of conducting electrons. For the time crystal there are two options at a border:

1. The structure of the time crystal is so strong that it prevents scattering. Any vibration caused in the structure by crossing the border is a part of the time crystal itself. There cannot be any dissipation because it is the lowest energy state. The vibration cannot escape from the time crystal to the outside world: it remains as a part of the time crystal itself.

2. Some of the electrons are scattered at the border, but they remain as a part of the time crystal. There cannot be any dissipation of phonons or electromagnetic waves because it is the lowest energy state.

How does a thermal phonon behave in the time crystal of a superconductor?

Let us have a time crystal, attached to the center of the laboratory floor, in a thermodynamic equilibrium with its environment. The environment is a cavity filled with black body radiation.

Photons in surrounding space are turned into phonons as they enter the time crystal.

Since it is an equilibrium, the movements in the time crystal cannot systematically change, e.g., the angular momentum relative to the center of the laboratory. The time crystal should behave just like an ordinary crystal in a thermodynamic equilibrium.

An ordinary crystal can refract or reflect photons. It can absorb energy from photons, as long as the total flux of energy in and out of the crystal is zero.

Let us try to model how a time crystal reacts to a phonon which enters the crystal at a random location.

       "diffraction grating" by the phonon

                    e-  electron flow

                    |

                    v

            +         +         +         +    zones of positive charge 

                                        ^

                                        |

                                        e-   electron flow

The phonon concentrates positive charge in the lattice into zones marked with symbols +. It is like a diffraction grating for the approaching electrons.

The phonon contains a small amount of energy. The energy, or a part of it, is absorbed as vibrations in the time crystal.

The vibrations may be longitudinal in the electron flows, or they may be transverse, making the flows move sideways.

If the time crystal would not be in the lowest energy state, then there would be free kinetic energy in the electron flows, and the vibrations would steal kinetic energy: there would be resistance.

But what exactly is the reason why we cannot disturb much the movement in the electron flows? A hydrogen atom may appear completely neutral to the outside world, even though the electron is moving rapidly around the proton. Could it be that the electron flow plus associated lattice vibrations are not coupled to the thermal phonon at all?

If the negative charge in the electron flows is completely canceled by the positive charge concentration in the lattice vibrations, then the system might appear neutral.

Theorem. Any system of charges in its lowest energy state must have a static electromagnetic field if observed from a distance. Otherwise, the system would radiate.

The theorem suggests that the diffraction grating meets a continuous flow of electrons. It must not see individual electrons, because if it would, there would be an electromagnetic wave radiating from the system.

   

                              + charge

                   e- --->

               o o o o o o o o o o o o 

                     long molecule          

               |                                  |

            ======================  frame

An analogous setup: we have a long molecule where an electron orbits from end to end. The molecule is attached to a frame. We put a positive charge close to the molecule. There probably will be some dipole force between the molecule and the electron. The force has to be constant. If it were periodic, then the molecule would radiate electromagnetic waves.

We conclude that the electron flows appear to the diffraction grating as essentially continuous flows of charge. The grating disturbs the flows only a little.

Electron flows in a time crystal are like "continuous" flows of charge

Hypothesis. Electron flows in a time crystal look like continuous flows of charge to a disturbance like a thermal phonon. There is no scattering of individual electrons.

                                                 \   scattered wave

                                            \

                                       \

               ------------------      ^

               ------------------      |

               +    +    +    +              periodic potential

               ------------------      ^

               ------------------      | e- free electron

The scattering of a free electron looks like the diagram above. The periodic potential perturbs the electron wave and creates a weak scattered wave.

If there can be no scattering, the diagram above cannot be the right depiction of the process. What kind of a wave might describe a continuous flow of charge?

The perpetuum mobile problem if we assume a potential wall between different states of the current in a superconductor

Suppose that we have a superconducting wire in the lowest energy state. It is a time crystal.

          e- ---->                                               e- ---->

       ---------------- ================== -------------

      ordinary       superconductor

      wire

Then we add an electron to the left end of the superconductor, or alternatively, remove an electron from the right end.

Let us assume that the added electron "joins" a flow of electrons in the time crystal.

There are problems, though: how does the electron know to join a flow which goes exactly from end to end? Also, the claim that an extra electron can "join" a flow without spoiling the properties of the time crystal, is very ad hoc. Why should it be true?

                            resistor

                      ------- ### -------

                    |                           |  

                     =============

                             e-  ----> 

                      superconductor

                              <-------

                                I = ε

Let us consider the following setup. There is a superconductor in a loop with an ordinary conductor whose resistance is extremely small. We put a small initial current 

       I = ε

into the loop.

Let us assume that the electron flow in the superconductor now is in some kind of a local lowest energy state with the current ε flowing. There is a potential wall which prevents the system from decaying to an I = 0 state.

Then we would have a perpetuum mobile which generates heat in the resistor for ever.

How to prevent the existence of a perpetuum mobile? Since there is now a small voltage over the superconductor, maybe that voltage creates the opposite current -ε within the superconductor, so that the total current is zero?

But that does not work. If there is a potential wall which prevents the decay ε -> 0, then there probably is a potential wall which prevents the transition 0 -> -ε.

Question. If we try to explain superconductivity by a (local) energy minimum argument, does that always lead to the existence of a perpetuum mobile?

Conclusions

Energy minimum arguments may explain the behavior of a time crystal in the lowest energy state. The time crystal in this case is a closed system.

But a superconductor in a loop with an ordinary conductor is a not a closed system if there is a current ε flowing in the loop. A local energy minimum argument for the current ε seems to lead to the existence of a perpetuum mobile. That is, conservation of energy is broken.

We have to investigate this more. In earlier blog posts we tried to explain superconductivity with Bloch's theorem for amorphous matter. Maybe that is the way. Above we imagined that a single electron moves along an electron flow to the other end of the semiconductor. Maybe the current is a collective movement of the time crystal? The hypothetical "condensate" in BCS theory moves the charge with a collective motion.

How would we explain the magnetic flux quantum? If the fraction part of the flux is canceled by a superconducting current, is that compatible with our earlier claim that the time crystal of electrons circulating around the loop prefer to have a flux an integer times Φ₀ through the loop?

Since the fraction part of the flux is canceled by a superconducting current, that means that the magnetic field of a superconducting current is not quantized!