What are "real" elementary particles ?

Top page (correct Bohr model including helium atoms)
Mott scattering.   S-matrix.
Superstring and Loop quantum gravity ?
Standard model is just wrong.
SUSY is a waste of time and money. (14/5/1)

Table of contents

Most of elementary particles are "virtual" particles made by computers ??

According to the standard model of the particle physics, there are too many elementary particles.
There are three generations (1st, 2nd, 3rd) in quarks.
And each generation contains two different quarks (Total is 3 × 2 = 6 quarks).

Furthermore, each quark needs to have different three colors (red, blue, green) to satisfy Pauli exclusion principle.
( Of course, each quark can NOT be separated, so these are only "speculation". )
Surprisingly, the top quark has the same mass as a gold atom, even though it's an elementary particle !
( "Gold atom" like top quark has only +2/3e fractional charge... )

(Fig.1) "Virtual" particles made by computers ?

As you know, strange quarks are said to have fractional charges such as -1/3e or +2/3e.
An electron is the smallest elementary particle (= 0.5 MeV ), which has -e integer charge. ( NOT fractional !)
( And why the quarks must not have +0.7e, -0,5e, or +0.4e ? This is strange. )

The important point is that quarks can NOT be isolated and Gell-Mann himself did NOT believe quark's existence.
You may have often seen the comments like "quarks have actually been found" in books.

But the experimental results about quarks are completely dependent on the assumption that the standard model and gauge theory are correct.
Unreal quarks and fractional charges are very favorable to only "mathematical" concepts of gauge symmetry such as SU(2), SU(3), which is explained later.

Very short-lifetime particles cannot be confirmed as real.

(Fig.1') Unstable particles in the accelerator are real, or only "Math" ?

Quantum chromodynamics ( QCD ) deals with gluons and quarks, but as you know, these particles cannot be isolated, so QCD itself cannot be confirmed whether it is correct or not.

Most important thing is that various unstable particles such as mesons, quarks, W,Z bosons and Higgs in the accelerator are said to be generated only one in trillions of collisions, as shown on this page.
Furthermore, we cannot observe these unstable particles directly, we only estimate their existences from final products.

Before insisting these unstable particles really exist with the probability of 99.99% or 5-sigma, we have to think about their extremely low probability of generation (= less than 1/1000000000000 ).
( These processes are all computerized, so these may be "virtual particles" created by computers. )
And they should explain this fact more clearly to the ordinary people, I think.

Unstable Muon is just a composite ( NOT elementary ) particle.

(Fig.2) Who ordered the muon ?

Also about strange elementary particle muon, the physicists at that time said "Who ordered the muon ?"
Unstable muon's lifetime is very short (= about 2.2 microsecond ), it breaks into an electron and neutrino.

In spite of this fact, the unstable muon is considered to be an elementary particle. Why ? And we are NOT allowed to ask how these electron and neutrino are packed into the muon.
Because "mathematical" standard model does NOT allow us to ask "concrete" particle's mechanism.

It is natural that we think the high energy from the cosmic rays breaks nuclei, and they generate an electron and longitudinal de Broglie wave (= neutrino), which is treated as a muon.

The electron which is entangled with high energy de Broglie wave may show different mass from an electron for very short time.
And if Relativity is wrong, the longer life time of muon is caused by its higher energy itself ( NOT time dilation ).

"Shut up and calculate ! " about particle physics ?

(Fig.3) Ancient aether theory ??

Higgs particles fill all space, and cause "resistance" (= mass ), according to the standard model.
But they try NOT to admit Higgs is like "aether".
Because Higgs particles are based on relativistic Klein-Gordon equation, which denied ether.

But these infinite partilces filling all space is inconsistent with the "nothing" vacuum of the relativity.
And if Higgs acts as resistance, a moving electron easily stops. This is strange.
On the other hand, heavy proton mass is NOT related to Higgs. How Higgs distinguishes them ??

Very "rare" Higgs is inconsistent with "original" infinite Higgs.

The most important thing is that estimated Higgs particle in the accelerator is thought to appear in only 1 in trillions of collisions.
And we can NOT isolate and observe the Higgs particle itself, we only see a little "excess" such as γ rays.
Of course, these processes are computerized, so all the estimated unstable elementary particles may be "virtual" particles made by computers.

They seem to avoid explaining concrete mechanism of how infinite Higgs particles act on and distinguish various particles.
Unless we stop " Shut up and calculate !" about the particle physics, it means we admit the present standard model is only "mathematical" thing.

Heavy W boson completely violates energy conservation.

(Fig.4) W boson in the accelerators is NOT equal to the beta decay.

When a neutron breaks into a proton and an electron in β- decay, very heavy W boson (= about 80 times proton ) is generated.
Of course, if W boson is 80 times heavier (= 80400 MeV ) than a proton, this clearly violates the basic energy conservation rule.

Because the mass difference between a neutron and a proton is very small ( less than 3 MeV ).
So they try to explain this energy violation using the convenient uncertainty principle of quantum mechanics.
But if we admit this strange phenomenon, we must admit any unreasonable phenomena, which violate basic physical rules.

Of course, if Bohr-Sommerfeld model can replace this quantum mechanics, the uncertainty principle itself is wrong, which means W boson is wrong, too.

All processes in the accelerator violates special relativity.

(Fig. 4') "Virtual" particles contradicts relativity.

As yor know, when a particle (= electron ) and an antiparticle fuse into photon, its process is said to obey special relativity (= mass formula ).
But in fact, all processes in the accelerator involved in producing quarks, Higgs, muons, and W bosons do NOT obey special relativity.
( Though the present particle physics completely depends on relativistic quantum field theory. )

For example, in the center-of-mass system of Fig.4'-B, an electron (= p1 ) and a positron (= p2 ) have the 4-momentums of

"E" denotes the energy and 1-3 components are momentums, which are just opposite to each other in the center-of-mass system.
So according to the energy-momentum conservation, photon's 4-momentum becomes

As you know, photon's mass MUST be zero according to the special relativity. ( q0 = |q| , → q2 = 0 ).
But Eq.B-2 shows "virtual" photon's mass is greater than the electron or positron.
This is clearly inconsistent with special relativity.
( Eq.B-2 is "scalar", so in every reference frame, virtual photon violates relativity. )
And as shown on this page, the quantum electrodynamics takes advantage of this virtual particle to get various values.

In the particle physics, we use the following mass unit ( ex. MeV/c2 ).
(Eq. 1) Mass unit.

So, for example, the mass of the electron is 0.511 MeV, as follows,
(Eq. 2)

And the proton mass is 938 MeV, the weak boson mass is 80400 MeV, and the top quark mass is 174000 MeV (= about gold atom ).

Mass ( energy ) of each elementary particle.

(Fig.5) Higgs are "indispensable" for getting "mass" in electrons, quarks, W bosons ...

As I explain later, Higgs particles (bosons) are indispensable for the electrons, quarks, W, Z bosons having their masses.
( Caution: 98% of proton mass is said to be caused not by Higgs but by gluon energy. )
So if Higgs particles do not exist, all these elementary particles can NOT exist, which means all things in the world vanish !

But in fact, these interpretations originate in only "mathematical" theories of "symmetry".
( These theories have NO "real" images at all, which is explained later. )
Higgs particle's mass is said to be about 125000 MeV (= 125 GeV ), which is about 125 times proton. ( if Higgs exist. )
Strange to say, we can NOT feel these very heavy and infinite Higgs particles at all. Why ?

Higgs hunt is like finding a "needle" from haystick.

About Higgs existence, they "estimate" the state of "no Higgs", excluding a lot of annoying backgrounds, and the probability of finding Higgs ( 99% or something ) is gotten from "excess" with respect to the "estimated" no Higgs state.

But actually, one Higgs particle can be generated with extremely low probability of one in trillions of collisions. ( Is it really filling all space ? )
So depending on the definition of this estimated no Higgs state, this value varies, and furthermore, the standard model cannot predict their masses. See also this page.

Higgs is "spin 0".   How can they confirm it ?

(Fig.5') What is spin 0 ??

As shown on this page, the spin is very strange.
Higgs is said to have spin zero.
But of course, we cannot see rotating ( or no-rotating ) states of Higgs particle, directly.
( First, Higgs particle itself cannot be observed directly, these are only speculation. )

If Higgs has no spin, new particles generated from Higgs decay into various ( uniform ) directions, they insist.
If the particle has some spin, new generated particles decay into some angle, after breaking.

So "spin" is only some decay deviation.
Very abstract spin itself has NO clear concept, but they like to use this word of "spin". Why ?
This misguiding about the word "spin" confuses ordinary people very much, I think.

"Symmetry" is only abstract math, NOT physics.

[ Mathematical "symmetry" has NO evidence, and NOT physics. ]

To understand the elementary particle and standard model, we must understand "symmetry".
Because all four fundamental forces ( except gravity ) are caused by "symmetry".
( You would be disappointed if you expected more clear visualization of the standard model. )

U(1) gauge symmetry causes electromagnetic force.
SU(2) gauge symmetry causes weak force.
SU(3) gauge symmetry causes strong force.
And for the fermions and weak bosons to get masses, "Higgs boson" is indispensable.
( Unless Higgs boson exists, the mass of all these particles becomes zero ! )

By the way, what is the "physical" meaning of this symmetry ?
Unfortunately, there are no concrete images of these symmetries.
This means these are only "mathematical" things.
In this page, we explain about this in detail.

Relativity → Lagrangian invariance → Symmetry.

(Fig.6) Any observers see the same thing = gauge invariance ??

According to the relativity, all physical phenomena must look the same to any observers.
Due to this very strict restriction, Lagrangian in the quantum field theory must be Lorentz invariant scalar, and includes infinite kinds of momentums and energies ( See also this page ).
Of course, if the special relativity is wrong, we do NOT need to use these very abstract math world.

For example, from the viewpoints of observers at the different position and time (= space and time translation ), action S must be invariant.
These invariances under translation are related to the momentum and energy (= Hamiltonian ) in the quantum theory.

So they tried to apply these concepts to other things such as (local) gauge invariance to get the boson-mediated forces, though they are completely different things.

But under the very strict relativistic restriction, all they could do was to find the relation between symmetries (= invariance under various transformations ) and "physical entities".
Eq.3-Eq.8 (= one example of symmetries, time translation ) are moved to this section.

U(1) gauge symmetry causes electromagnetic force ??

( Gauge theory = "Math" or physics ?? )

First we have to understand U(1) gauge transformation of electromagnetic force.
Because all other forces ( weak and strong forces ) are based on this electromagnetic theory.

Unfortunately, QED is only a "mathematical" language and has very "limited" mathematical tools.
So there are no other ways to extend the theory to other forces than this.
About the basic quantum field theory, read this page first.

Lagrangian (of free particle ) which leads to Dirac's equation is

We get Dirac equation from this Largangian.
And Lagrangian must be invariant under Lorentz transformaion.

Most important difference between classical mechanics and quantum field theory (QFT) is QFT has less physical meanings.
( So you need to consider almost all of QFT things as just "mathematical" rules. )

Lagrangian is invariant under global ( phase ) transformation. ← NOT physics.

This Dirac Lagrangian is invariant under the global gauge transformation, as follows,

Because this θ does not include xν (= time and position ) variables.

When we consider the global phase thransformation, the conserved thing is "charge Q" in the stress-energy tensor.
( Of course, these charge and current are based on the definition of the quantum field theory . )

In this way, they came to use these invariance under various transformations as tools for getting some physical entities.
And under the very strict restriction of the relativity, these symmetric "tools" generate the "mathematical" concepts such as SU(2) (= weak force, W bosons ) and SU(3) (= quarks ).

Local gauge transformation is just "math" object.

When this θ includes xν variables, this transformation is called local gauge transformation.
Due to these xν variables of θ, Lagrangian ( differential term ) of Eq.9 changes when it is differentiated, as follows,

So Lagrangian of Eq.9 is not invariant under local gauge transformation.

To make the local gauge transformation invariant (= symmetric), we need to add 4- potential ( Aμ ) to the free particle Lagrangian of Eq.9.

When you deal with minus electron, change this "q" to "-e".
This is called covariant derivative.
And the magnetic ( or electric ) potential Aμ is called "gauge field", which means "photons" according to QFT.
As shown on this page (Eq.2-16), this covariant derivative comes from classical Lorentz force.

Gauge transformation and covariant derivative.

Furthermore, the electric (= E ) and magnetic (= B ) fields do not change, when we take any value of θ as shown in Eq.14.

So this θ can NOT be actually observed. But they considered it as a most important thing in the "symmetry".

Using Eq.12 and Eq.14, the derivative of Lagrangian ( Eq.9, Eq.11 ) changes into

Adjusting the coefficient of θ seems to be a little mathematical trick.
As a result, also under local gauge transformation, it is invariant, as follows,

( Of course, the mass term of Lagrangian Eq.9 is invariant, too. )

Using covariant derivative of Eq.12, Dirac Lagrangian changes into

The third term of Eq.17 is the interaction term between photon (= A ) and fermions (= ψ ), which is often used in QED.
This means if we make Lagrangian invariant under local gauge transformation (= U(1) gauge transfomation ), the electromagnetic force is generated.

But as you notice, there are no concrete images and visualization here. It's only a "mathematical" thing.
And we cannot know what the "local phase θ" actually is.

As shown in this section, all they could do was to bind the invariance under some transformation (= symmetry ) to some physical things.
This is what the relativistic quantum field theory is.

Lagrangian of photon is invariant under gauge transformation ?

As shown on this page (Eq.3-1), antisymmetric tesor ( Fμν ) of magnetic (or electric ) potential is

Substituting the gauge transformation of Eq.14 into Eq.18,

The antisymmetric tensor doesn't change under gauge transformation (= invariant ).

As a result, the Lagrangian of Maxwell equation doesn't change, either, as follows,

( Mass term obstructs "symmetry". )

Photon mass term is NOT invariant. → Photon is massless ?

But if the photon has mass, this Lagrangian needs to include AμAμ term like Lagrangian of Klein-Gordon field.
This term changes under gauge transformation of Eq.14, as follows, (= not invariant )

So if the photon has mass, Lagrangian of Maxwell equation is not invariant under gauge transformation.

They say this means the photon has no mass.
But can you understand this "abstract" reason well ?

Unfortunately, these concepts do NOT link to real objects at all. They are only "mathematical" thing.
The problem is that they extended these "mathematical" concepts to weak and strong forces.
As a result, imaginary particles such as W boson (= violation of energy conservation ) and quarks (= cannot be isolated ) were adopted.

(Fig.7) Summary in photon.

W boson's large mass needs to be given by Higgs ?

In case of very heavy W boson, it must have "mass" term.
( W bosons are charged particles, which mass cannot be zero. )
So under the gauge transformation (= SU(2) ), the Lagrangian of W boson is NOT invariant.

To avoid this situation and keep symmetry, they created Higgs fields and erased W boson's mass-term.
This is the main reason why we need Higgs particles.
( The equation of Higgs includes "mass" term of W boson instead of W boson Lagrangian. )

( W boson mass ? )

Unfortunately, ordinary people do not know about this very abstract reason well.
They should explain to ordinary people about these things, I think.
So "mathematical" symmetries are very important for the present standard model.

Interaction term and Feynman diagram.

When we consider the interation and reaction among the elementary particles, we use the interaction terms such as Eq.17.
( See also this page. )

And we use also the anticommutation relation of particle (= c ) and antiparticle (= d ) of

The combinations other than Eq.24 vanish between the vacuum (= < 0| |0 > ).
For example, a pair of annihilation - annihilation ( c c ), ( d d ), creation - creation ( c c ), ( d d ) operators vanishes.

So the following combination is allowed due to this relation between creation and annihilation operators.

But the following combination is not allowed.

Because the second line is creation - creation operators.

For example, we think about the reaction of beta- (β - ) decay of

where neutron dacayes into proton, electron, and antineutrino.
According to Feynman diagram, this antineutrino cannot be neutrino.
And beta- decay uses W boson instead of photon (A) in the interaction term.

The diagram of Eq.27 is

Due to the reasons similar to Eq.25 and Eq.26, the generation of the antineutrino and electron (= particle ) is allowed.
But the following reaction is not allowed.

Of course, we can not discriminate between neutrino and antineutrino clearly.
This is only due to the "mathematical" reason.

Relativistic QED contradicts special relativity ??

Suppose one electron (= p ) at rest absorbs one photon (= q ) and becomes p' electron.

The energy of an electron at rest (= p ) is only rest mass energy ( E/c = mc ).
( The momentum of this electron at rest is zero. )
And an electron (= p' ) after absorbing photon has the momentums.
According to the relativity, the electrons ( before and after the interaction with photon ) need to satisfy

where (-1,1,1,1) version metric tensor is used.
So Eq.31 must be negative.

From Eq.30 and the conservation of energy and momentum, the photon's energy and momentum is

As you know, photon's mass is zero, but from Eq.32,

Eq.33 means this photon's mass is minus, which violates special relativity !

As a result, the relation between the relativistic energy and momentum includes self-contradiction.
So they introduced the very strange concept of "virtual particle" ( virtual photon in this case ).

Furthermore, as shown on this page, the special relativity breaks down in the electromagnetism.
( Of course, charge and current must be conserved in any inertial frame, so this means Lorentz transformation is wrong. )

Relativistic restriction is "unrealistically" strong !

(Fig.8) Relativistic restriction is too strong ! → "abstract" "mathematical" model.

In fact, the limitation of Lorentz symmetry (= special relativity ) is unrealistically strong.
To keep Lorentz symmetric completely, we must give up usual atomic and molecular models.
All we can use is a very "limited" mathematical model ( See Fig.8, right ).

Its Lagrangian (= L ) must be a scalar under Lorentz transformation, and its function must obey Lorentz symmetry.
In this purely relativistic limited world, one particle needs to be created (or annihilated ) in all space at the same time.

And it always includes all kinds of momentums (from infinity to infinity ) to keep Lorentz symmetric, which causes serious ultraviolet divergence.

Virtual photon contradicts basic principle.

(Fig.9) Non-ether = virtual photon model violates special relativity !

Special relativity forbids "ether", so it needs other things to explain "Coulomb force".
As I said above, even if we obey the very strict relativistic limitation, the virtual particles do NOT satisfy relativity.
And surprisingly, the special relativity adopts this strange virtual photons as "Coulomb force medium".

As shown in Fig.9, the virtual photon with some energy suddenly appears from the vacuum.
This process violates the energy conservation.

And only when this virtual photon arrives at the target charged particle, the charged particle receives "Coulomb force" (= virtual photon carrying energy ).
As a result, even if we try to obey the very strict relativistic limitation, it is impossible.

General relativity cannot be renormalized.

(Fig.10) General relativistic restriction in QFT breaks Einstein's gravity.

As I said, to keep Lorentz symmetry, interaction terms must include infinite kinds of momentums, which leads to divergence ( infinity ).
As shown on this page, in case of QED, if we suppose the infinite bare charge, the divergence can be renormalized into a charge.

Of course, this idea is very unreasonable. ( Pauli and Dirac disliked renormalization. )
As I explain later, the interaction by weak boson with zero mass is similar to photon, so this is also renormalizable.

In general relativity (see this page), "complicated" Christoffel symbols and covariant derivatives cause various divergences in QFT.
( In QFT, the derivative means the momentum (= from infinity to infinity ) with the exponential function. )
If we try to eliminate these troublesome divergences, the new interaction appears (= Fig.10, right ).

This means "macroscopic" gravitational relation is also broken, when we try to renormalize it. This is a serious problem.
So in the general relativity, the renormalization method can not be used.
This leads to strange 10-dimensional superstring theory.

SU(2) gauge symmetry = weak force ?

[ What on earth is weak force ? ]

In the β - decay, the neutron decayes into proton, electron, and antineutrino.
The isolated neutron is metastable (= half-life is about 10 minutes ).
So some weak force is related to this reaction.
Of course, proton and electron attract each other by Coulomb force (= photon ).
In spite of this attractive force, neutron decayes

(Fig.11) β - decay. ( neutron → proton + electron + antineutrino. )

In QED, we have to express all these reactions using the interaction term and Feynman diagram.
W- bosons need to carry negative charge, so we cannot take the interaction term of neutral photon (= Aμ of Eq.23) in beta decay.
This is the reason why "weak" force ( W boson ) is born.
( In short, the "mathematical" rules in QFT need new weak boson. )

According to the standard model, the down quark (= -1/3e ) of neutron changes into up quark (= +2/3e) of proton emitting a minus charge (-e), which is transferred to W- boson.
( Though an elementary particle cannot be divided, the down quark is divided. )
So the total charge is conserved ( -1/3e = +2/3e + (-e) ).
And this heavy W- boson completely disobeys energy conservation rules, as I said in Fig.4.

Neutrino is high-energy de Broglie waves inside nucleus.

(Fig.12) β - decay. ( neutrino = remaining de Broglie waves. )

In Bohr model, the electron cannot be closer to the nucleus than the ground state (= 1 × de Broglie wavelength ).
As the electron's energy becomes greater, de Broglie wavelength becomes shorter.

In the stable nucleus, the trapped electron's very short de Broglie wave is stable due to some reason.
But the isolated neutron, this de Broglie wave of the electron becomes unstable, which unstability becomes more dominant than their Coulomb attractive force.
So finally it decays, I think.

And considering the fact that the neutrino can go through the earth easily, it is natural that we think neutrino ( or antineutrino ) as some remaining de Broglie energy waves.
So we need not introduce a strange new concepts such as weak force and weak boson ( and Higgs particles ), I think.

Left-handed spinor is only "mathematical" product ?

(Fig.13) Electron in β - decay = left-handed ?

Under the space inversion, the vectors such as position and momentum becomes minus.
And the angular momentum (= r × p ) remains unchanged.

The relativistic Lagrangian which looks the same from any observers, should not change even under space inversion.
So irrespective of the direction of the cobalt-60 nuclear spin (= angular momentum ), the probability that electrons of beta- decay flies into both directions with respect to Co-60, should be the same according to the relativistic theory.
( Under the space inversion, Co-60 spin is invariant, the electron's momentum becomes opposite. )

Under the magnetic field (= z direction ), Co-60 spin points to the z direction.
But in this case, the electron tends to fly into minus z direction.
The difference between Co-60 and Ni-60 spins is ħ, so the electron's spin direction is opposite to its momentum (= left-handed ).
( Of course, these spin angular momentums are only quantum mechanical definitions as shown on this page. )
Antineutrino is right-handed, which means neutrino is left-handed in the beta-decay, too.

Left-handed = spinor,   Right-handed = "number" !?

(Eq.35) Division into left-handed and right handed electrons.

So they "artificially" divided the electrons (and neutrinos ) into left-handed and right-handed, as shown in Eq.35.
Strange to say, only left-handed electron is 2 × 1 spinor, it makes these fermions' mass zero to keep gauge invariance.
( And these fermions need Higgs mechanism, too, which is explained later. )

Co60 decay has Nothing to do with left or right-handed.

But I think, from the classical viewpoint, the difference between the flying directions of the electrons ( with respect to the magnetic field ) is quite natural.

If these probabilities are completely the same, it is more unnatural from the classical mechanical viewpoint.
So if we use usual classical mechanics, we don't need to depend on the "mathematical" concept such as 2 × 1 spinor and Higgs mechanism.

Why neutral Z boson is needed.

(Eq.35') Why we need ±W and Z bosons ?

As I explain later, Dirac equation can be divided into left and right-handed spinors within one equation.
So to pick up only left-handed spinor, we need to erase "mass term", and transfer it into Higgs mechanism.

But in this case, symmetric photon Aμ (= electromagnetic force ) also violates the parity like W boson.
To avoid it, they introduced another neutral particle Z boson.
These unstable W and Z bosons cannot be observed directly, we just see final products such as electrons and estimate their existences.

Lagrangian and wavefunction for weak force

To introduce weak force, we have to use matrices.
Because four kinds of forces including photon (= A ), ±W, and Z bosons need to be packed into them.
Of course, this form has to imitate the electromagnetic vector potential A, as I said above.
They tried to extend the beautiful symmetry in the electromagnetic force to other forces.

The spinor for weak force is

Each φ1 and φ2 of it is 4 × 1 matrix, as follows,

So Eq.36 is 8 × 1 matirix !

The conjugate transpose (× γ0 ) of Eq.36 is

This is 1 × 8 matrix.

Lagrangian for weak force is

This is a little complicated "mathematicaly".

(Eq.40) Pauli matrices.

Here Pauli matrices are used.
But there seem to be no physical meanings of Pauli patrices in the weak force.
As you notice, standard model tends to adopt some mathematical expressions without showing ( physical ) reasons.
Pauli matrices are Hermitian and favorable to group theory.

Local gauge transformation for weak force.

The local gauge transformation of electromagnetic force is Eq.11.
The weak force version is

You need to understand there are no reasons why this phase transformation is introduced suddenly.
Different from the elecromagnetic force, Eq.41 is 2 × 2 matrix, because of Pauli matrices.
( First, we suppose only θ doesn't include x variables. )

Due to the Hermitian property (Eq.40) , Eq.41 becomes "unitary", as follows,

This group of Eq.41 is called SU(2).

So the local gauge transformation of weak force is

We aim to get the invariance under the transformation of Eq.43.
The covariant derivative for weak force becomes

where σ is Pauli matrices.

Eq.44 corresponds to Eq.12 of electromagnetic force.

To make Lagrangian of Eq.39 invariant under local gauge transformation, the following relation needs to be satisfied.

This corresponds to Eq.15 of the electromagnetic force.

Using Eq.43 and Eq.44,

To satisfy the relation of Eq.45, W' of Eq.46 needs to satisfy

This is a little complicated, but Eq.47 corresponds to Eq.14 of electromagnetic force gauge transformation.

Using covariant derivetive of Eq.44 in Eq.39, Lagrangian becomes

The last term of Eq.48 means the interaction between fermions and weak bosons.

Lagrangian of weak boson.

Next we consider the Lagrangian of only weak bosons, which corresponds to Eq.20 (= Lagrangian of Maxwell equation ).

This is an inner product.

Most important difference between Maxwell's Lagrangian (= Eq.20 ) and weak force Lagrangian (= Eq.49 ) is only Maxwell's Lagrangian gives real electromagnetic ( Maxwell ) equations solving Euler-Lagrange equation.

Weak force Lagrangian has no meaning as Maxwell's.
In the electromagnetic case, the antisymmetric tensor (= Fμν ) can be derived using the covariant derivative of Eq.12, as follows,

Antisymmetric tensor F of W boson is artificial.

Like Eq.50, we define antisymmetric tensor for weak force, as follows,

where the relation of Eq.40 in Pauli matrices is used.

For example, using Eq.44, in the calcuation of 1 component of Eq.51,

Eq.52 is a little complicated, because it uses Pauli matrices.
( Of course, there is no more meaning than "mathematical" in it. You just calculate it. )
Eq.51 can be simply expressed as

As shown above, when W' satisfies Eq.45, the following relation can be gotten

So the antisymmetric tensor after gauge transformation becomes

Lagrangian of W boson is invariant under gauge transformation ?

Lanrangian after gauge transformation is

where Eq.40 and Eq.54 are used.

As a result, Lagrangian of weak boson is invariant under gauge transformation of Eq.41.
( Of course, to be invariant, Eq.47 and Eq.51 are "artificially" defined. )
But different from the electromagnetic force, the weak boson has to have big mass.
( And charged particle ±W bosons cannot be mass zero. )
Because if it has no mass, the weak force reaches very far away like electromagnetic force.

Mass term of W boson is NOT invariant → W boson is massless ?

So the Lagrangian must include the mass term of

As shown in Eq.56, the mass term is NOT invariant under gauge transformation of Eq.47.
( This is similar to Eq.21 and Eq.22. )
If Eq.47 doesn't include the "excess" part, mass term is invariant.

So the mass of the weak boson needs to be zero according to the symmetry theory.
Instead, Higgs boson is important to give the weak boson mass, which is explained later.
( The mass term of W boson is transferred to Higgs equation. )
As you notice, there are no concrete images in these theories.

Unification of electroweak force.

[ Left-handed and right-handed spinors ]

When the Dirac's particle has no mass, it is separated into left-handed and right-handed. (See also this page or this page )
Here simply, we explain these left and right handed spinors.

" Right-handed " means the directions of spin and momentum are the same, and those of " left-handed " are opposite.
( Of course, these are "mathematical" definitions rather than physical reality. )

The eigenvalue of the left-handed spinor with respect to γ5 matrix is " -1 "
And the eigenvalue of the right-handed spinor is " +1 ".
From Eq.57, when the particle has mass, left and right handed spinors are mixed in one Dirac equation, as follows,

As I said in Fig.13, beta-decay violates parity invariance.
So they divided the fermions into left-handed (= 2 ×1 matrix ) and right-handed (= number ) spinors (= Eq.59, Eq.60 ).

So the forms of gauge transformation of these spinors need to be different.
To be gauge-invariant, their (= fermions ) masses have to be all zero, and they need Higgs mechanism.

Weak force left-handed wavefunction.

Based on these ideas, the wavefunction of weak force is separated like

uL and dL are "left-handed up quark" and "left-handed down quark", respectively.
The important point is that only ψ1 of Eq.59 is 2 × 1 matrix.
(Though this is difficult to imagine.)

And the lepton is

Covariant derivative "D" of weak force.

Instead of Eq.44, we use the next covariant derivative of

where Bμ is related to photon, and Yj is called "weak hypercharge", (which is very "artificial", I think, Sorry.)

Using Eq.61, Lagrangian ( excluding mass term ) becomes

The important point is that Pauli matrices of Eq.61 act only on φ1 of Eq.59 (or Eq.60).
This is only a rule, so we have to obey this rule.

Unification of weak and electromagnetic forces.

Using Pauli matrices of Eq.40,

Here we define W± bosons as follows,

± means opposite charges of W bosons.
This is only a rule, too. There are no clear reason why this ± means opposite charges .

Using Eq.64, Eq.63 becomes

Using Eq.64, the second line of Eq.62 is

Based on this definition, both of W3 and B have no charge.
And like photon, W bosons have no mass.
( Of course, charged particle ±W bosons must have "mass", so Higgs is needed. )

What is the meaning of Weinberg angle θ ?

Due to mass zero, these particles cannot be discriminated.
So we can combine these two similar fields of W3 and B, as follows,

Eq.67 is equal to

The "artificial" definition of Eq.67 is to get the photon interaction term dealing with the remaining gauge W3.
If new neutral Z particle doesn't exist, photon A also violates parity like W boson, though electromagnetic force is symmetric.

But as you see, we did NOT ask what the "local gauge symmetry" really means in the physical sense, so we have to introduce these "mathematical" definitions of Eq.67.

Substituting Eq.68 into the third term of Eq.62,

Here we make new concepts of the third component of the weak isospin (= T3 ).

The weak isospin is different from usual spin.
The third components of the weak isospin of uL and dL are +1/2 and -1/2, respectively.
And weak isospins of right-handed fermions are all zero.

Even though this weak isospin is completely different from usual spin ( for example, right-handed fermion's isospin is zero ), usual Pauli matrices (= σ3 ) is used.
And to incorporate W3 of the 2 × 1 matrix into right-handed fermions, only right-handed isospin is zero.
This is too good to be true.
So you think that the law of nature really obeys these "mathematical" definitions ?

Momentum direction (= left, right-handed ) is so important ?

( Directions of "spin" and "momentum" are so important ?? )

The differences between right-handed and left-handed are just directions of movement.
For example, while the right-handed particle is flying to the north, the left-handed one is flying to the south.
Such a trifling matter really causes the complicated and artificial differences above ??
"Spin" has some mysterious forces influenced by the moving direction ??

Using Eq.70, Eq.69 becomes

Weak isospin zero of the right-handed spinor (= Eq.70 ) is for uniting A and Z particles.

Again we introduce a new definition of

This is only a rule. There are no clear reasons.

From Eq.72,

Weak isospin and hypercharge are "artificial" definitions.

Again we define a new concept of

Qe of Eq.74 means the charge of the fermion. "Y" is weak hypercharge.
For example, T3 and Y of uL are +1/2 and +1/3, respectively, so the charge of up quark is +2/3.

T3 and Y of uR are 0 and +4/3, respectively, so the charge of up quark (R) is +2/3, too.
Do you think these are too good to be true ? And very artificial, I think.

Using Eq.72 and Eq.74, the first term of Eq.71 is

As you notice, Eq.75 is the same as the interaction term of the photon and fermions (see also Eq.17).

And using Eq.72 and Eq.74, the second term of Eq.71 is

Eq.76 means the interaction between Z boson and fermions.
Their binding force is

The angle of θw is defined from various experiments, as follows,

The standard model includes 18 parameters, which can be gotten only from experiments ( NOT from the gauge theory itself ).
This angle θw, the binding constants of g and g' are included in these 18 parameters.

Of course, the masses of Higgs, W and Z bosons, leptons, fermions are all these parameters, which can NOT be gotten from the standard model.
( As I said, these important values cannot be measured directly, we only estimate their existences. )
So the standard model cannot predict any important values.

God particle Higgs boson is only "mathematical" thing ?

[ What is spontaneous symmetry breaking ? ]

Lagrangian for Klein-Gordon equation is

where the first term is the energy and momentum term and the second term is the mass term.
The important point is that the mass term is minus in Lagrangian. (So it's plus in Hamiltonian.)

Lagrangian of Eq.79 is invariant under the trandformation of

Next we consider the complex Klein-Gordon field.
This complex filed (φ) can be expressed by two real fields ( φ1 and φ2 ), as follows,

One example of Lagrangian of complex Klein-Gordon fields is

where μ and λ are real numbers.
The secons term of Eq.82 is plus, so this doesn't mean mass term.

When we do the transformation of

Lagrangian of Eq.82 is invariant. Eq.83 is U(1) transfomation.

Hamiltonian is minus Lagrangian, so the second and third terms of Eq.82 are

These terms exprees potential energy V ( not mass term ).

So the potential energy of Eq.84 becomes the lowest when φ1 and φ2 satisfy

If we define the next variables of

Lagrangian of Eq.82 ( -V term of Eq.84 ) becomes

As shown in Eq.87, mass term appears ( from red line ).
But instead, Lagrangean including Eq.87 is not symmetric.
This is called "spontaneous symmetry breaking", which causes "mass".
But unfortunately, there are no clear images at all here.

Higgs particle is only "math" object.

Next we consider local SU(2) gauge transformation.
Here we define the complex fields, as follows,

Using Eq.88,

And Lagrangian is

Of course, Eq.90 is invariant under global SU(2) transformation of

And then we use the covariant derivative of Eq.92 in Lagrangian of Eq.90.

Here we define the potential energy V is

Like Eq.84 and Eq.85, when the potential energy becomes the lowest, the complex fields satisfy

where Eq.88 and Eq.89 are used.

Here we choose the vacuum of

The electric charge of the vacuum is zero, as follows,

where we use Eq.74.

The third component of weak isospin of dL or eL (= the second line of 2 × 1 matrix of Eq.96 ) is -1/2. ( See also Eq.59 and Eq.60. )
So the hypercharge (Y) of the vacuum is

Using real field h(x), we redefine Φ, as follows,

where this h(x) is a famous Higgs particle.

Substituting Eq.98 into Eq.93 ( using Eq.84 ), the potential energy V is

And using Eq.61, Eq.62 and Eq.65 (and Y=1 )

As a result,

where we use (W-)* = W+.

From Eq.67 and Eq.73,

Using Eq.102, Eq.101 becomes

Substituting Eq.103 into Lagrangian (Eq.90), the following terms,


are important, because these terms mean the mass terms of W± and Z bosons !
( In Lagrangian of Eq.90, these mass terms are minus. )

As a result, the spontaneous symmetry breaking of Eq.98 causes mass in weak bosons !
And h(x) of Eq.98 means Higgs boson !
So other terms of Lagrangian mean the interaction among Higgs and gauge bosons.

This is what Higgs particle really is
Unfortunately, there are no concrete images at all in Higgs mechanism.
( How do you think about it ? )

If heavy Higgs boson ( = more than 100 GeV ) really exists, why we can not feel them in the real world ??
( We need "infinite" Higgs bosons according to the standard model.)
And in the accelerator, as much as trillions of collisions are necessary for generating one Higgs particle.
Can we say Higgs particle has actually been found from this very low probability and vague results ?
Again I want to ask if Higgs particle really exists.

Quark and strong force are real ?

Next we try quark and strong force of gluons.
Quark is said to have three colors of red, blue and green.
And a single quark can not be isolated, so the observed color is always white.
( If quark really exists. )

So the wavefunction is 12 × 1 matrix ( first, 3 × 1 matrix ), as follows,

Local phase transformation is

where λ has eight kinds of

These λ matrices correspond to Pauli matrices of the weak force.

The trace of λ matrices is always zero, and they are Hermitian matrix, as follows,

The property of Eq.109 corresponds to Pauli matrices, too.

Unitary matrix is made from Hermitian matrix, as follows,

So transformation of Eq.107 is unitary matrix. (This is called SU(3) group.)

The covariant derivative of SU(3) version is

where G means "gluon" fields.

The inner product of Eq.111 contains eight terms of

When the Lagrangian is invariant under local phase transformation, the next relation needs to be satisfied

Using Eq.111,

As a result, Eq.113 needs the next condition of

Like Eq.51-Eq.52', the antisymmetric tensor of SU(3) version is

The outer product of Eq.116 is eight dimensions, as follows,

where f is

Like Eq.54 and Eq.55, when we use the antisymmetric tensor of Eq.116, the next Lagrangian of gluons is invariant under SU(3) gauge transformation.

The gluon has no mass, so we do not need to consider the symmetry of the mass term.
In case of gluon, the strange idea of "asymptotic freedom" is important.
But does this strange property really exist ?
The important point is that quark itself can NOT be isolated.
So we can not investigate the realities about quarks and gluons in detail.

What elementary particles real ?

[ Antiparticles are real ? ]

It is said that the anriparticles really exit.
But they are very unstable, and always diappear when they meet their partner "particle".
Does this mean the antiparticle is real ? See also this page,

Here we consider the difference between the beta plus decay and electron capture.

(Fig.14) Which will happen ?

The beta plus ( β + ) decay and the electron capture generate the same nucleus.
Both of them change one proton in the nucleus into a neutron.
Of course, this reaction happens only when the final product is energetically favorable in the nuclei.
As a result, both reactions emit energy.
But there is a difference in them.

(Fig.15) "Where" is the energy emitted ?

In Na (22 11), both the electron capture and beta plus decay can occur.
In the electron capture, when the nucleus changes into the stable one capturing one electron, it emits energy.
But in the beta plus decay, when the nucleus changes into the stable one, it does NOT emit energy but emit a positron.
And then the positron meets a electron somewhere else, annihilating each other, and emit energy.
So there is a self-contradiction in the places where the energy is emitted as the same nucleus is generated.
For example, it is said that in proton-rich nuclei where the energy difference between initial and final states is less than 2mec2, then β+ decay is not energetically possible, and electron capture is the sole decay mode.
This means these reactions are basically the same things ?
( As the positron and electron have some momentums, the annihilation energy is usually greater than 2mec2. )
Of course, these nuclei can NOT predict the electron mass energy flying from somewhere else, when they emit the positron.

[ Why do we need to think many kinds of "unnecessary" antiparticles really exist ?? ]

As a result, the nuclear reaction related to positron is completely equal to that of the electron capture.
For examle, the nuclear reaction in the positron emission tomography (PET) can be replaced by the electron capture, as shown in Fig.14.
( Originally, the concept of "electron capture" was introduced to explain various phenomena in which the beta plus decay broke down. )
Antiparticles such as positrons are very unstable, and we have NOT found the worlds made from antiparticles.
In this real world, Only "particles" exist, and we need only particles. So we don't need antiparticles for the daily living at all.
It is very unnatural that we think many kinds of "unnecessary" antiparticles really exist.

As shown on this page, the positron is generated when high energy electrons crash into heavy metals, though it is said we can generate the antiparticles from the vacuum.
So is it possible that the positrons or other antiparticles are kinds of temporarily excited states by high energy ( related to atomic nuclei ) ?

Also in the accelerators, these heavy metals are used to produce positrons.
In the accelerator, there are many photoelectrons and positive ions as noizes.
Each bunch is made of a lot of produced particles.
So we need to question about the reality of very "unstable" antiparticles again, I think.

"Quark" really exist ??

Almost all elementary particles in the standard model are very unstable.
As you know, the stable ones are only electrons, protons, electromagnetic wave and various nuclei (and neutrino ).
The proton is much heavier than the electron, though they have the same magnitude of charge (+e and -e).

So it is natural that we think the proton is made from some smaller things.
And the nuclei are stable, though they include more than one positive protons.
So it is natural that we think some strong forces bind positive protons in addition to electrons inside the neutrons.

The problem is that what we call quarks really exist ?

(Fig.16) "Quark" really exist ?

For example, the proton is said to be made of two up quarks ( +2/3e ) and one down quark ( -1/3e ).
But each mass of quark is only 3 (or 6 ) MeV.
The mass of proton is 938 MeV !
So it is said that almost all of the proton mass depends on the energy of the strong forces (= gluon energy ).
( Of course, we can NOT separate single quark due to its color, so we can not confirm it directly. )
And do you think the fractional charges such as +2/3e and -1/3e are too good to be true ?

Experimental results of Quark's existence ?

As shown on this page and this page, the cross section depends on the particle's charge ( of target or final ).
Qurak is not observed directly.
It binds to other quarks (or antiquarks) and decays into more stable things, which are observed as "jets".

From the energy of the final products such as electron gamma ray, neutrinos, we can estimate the masses of various unstable particles.
( But the energy of the important neutrino can NOT completely be measured ! )

Approximately, the generation of a pair of quarks is similar to the generation of other hadrons.
Considering the possibility of the production of up (+2/3e), down (-1/3e), strange (-1/3e), and charm (+2/3e), and three colors (red, blue and green), the cross section is proportional to

In some energy range, about 10/3 of R is seen.
It is said that other energy ranges show the generation of other combinations.
But these are only speculations.
For example, we can not know the upper four quarks really occur.

And the relation between "very strong" gluon and the charges ? It is more complicated, I think.
And are their probabilities (of three colors and four quarks) really the same ?

Due to infinite noises, these particles are said to appear, one in trillions of collisions.
Furthermore, the estimation from the final products completely relies on the assumption that the present standard model is right.

If we forget QED (quantum electrodymanics) and QCD (quantum chromodynamics ), more real particles would be found. I think.

"Muon" is an elementary particle ?

The muon is an elementary particle with a negative electric charge (-e) and about 200 × electron mass.
The mean lifetime of this muon is about 2.2 μs ( 2.2 microsecond ), and the muon decays into electron and neutrinos soon.
Though the muon decays into electron and neutrino soon, can we call it an elementary particle ??.

It is natural that we think muon is some combination of one electron + high energy de Broglie wave.
(As I said above, it is natural that we consider the neutrino which can penetrate the earth as "remaining de Broglie energy waves".)
And as the energy of electron' momentum is bigger (= its de Broglie wave becomes shorter and its wave speed is closer to the light speed c ), it is more difficult to accelerate by electric field, which maximum speed is c.

This means we can explain the relativistic mass change using the medium theory naturally.
So what we call "heavy relativistic mass" of the electron causes the mass which is about 200 times heavier than the electron in a cloud chamber.
Or the electron entangled with high energy de Broglie wave may show different property temporarily under the normal electromagnetic fields.

Very interesting and impressive book

Here I recommend one of very interesting and impressive books to you.

Lee Smolin,
Trouble with Physics. The Rise of String Theory and The Fall of a Science, and What comes Next.

This book indicates the possibility that there are some big mistakes in the basic physics (including relativity and quantum theory ) which are accepted now.
He refers to the lack of reality in the quantum mechanics and the quantum field theory, too.
And this book says that the theoretical physics now is trying to manipulate the mathematics "artificially" and make the unreal world such as many dimensions (more than 4) to match the experimental results.

Of course, this author is an expert and has plenty experience in the particle and theoretical physics, and explains the situation of the present physics, objectively, properly and intelligibly.

So I would like you all to read this great book.

Symmerty in time translation = Hamiltonian ?

Here we show the case of the time translation as an example.
About the QED notations of this webpage, see also appendix and Ap.1 of this page.
And if you have not read this page about the basic QFT yet, read it first.

When the infinitesimal coordinate transformation is

The change of the action S is

Of course, this S must be invariant under the relativity.

Different from transformations of field variables (φ), the time and space variables (= x, t ) themselves need to consider the change of d4x, as follows,

where Eq.3 is used, and the second order infinitesimals are neglected in the determimant calculation.

Under the transformation of Eq.3, the change of Lagrangian (L) is

where we use the Euler-Lagrange equation of this page (Eq,1-17').

So the action S' of Eq.4 is

where the integration by parts is used.

(0,0)-component of the stress-energy tensor of Eq.7 under Dirac Lagrangian is

This result is consistent with this page (Eq.5-39).

As a result, according to the relativistic quantum theory, the Hamiltonian is conserved under time translation.
They tried to use this analogy for gauge invariance.


2011/12/25 updated. Feel free to link to this site.