![Introduction
In the beginning was the Word.
Johnn, 1.1
It is known that any physical theory consists of basic notions and statements received
from these basic elements by the classic logic. We have the sequence of theories each of
which explained by the preceding one according to the logic. That means that basic notions
and statements of every subsequent theory are more logical than basic notions and axioms
of the preceding one. When these basic elements of the theory become absolutely logic, i.e.
when they become notions and rules of classical logic, the theoretical physics will come to
an end, it will rather be logic than physics. It seems that such situation will take place in
the nearest future. I’ll prove it in this book.
The first known ”theory of everything” was created by the ancient Greek philosopher
Aristotle1 in IV BC.
Aristotle related each of the four elements proposed earlier by Empedocles, Earth, Wa-
ter, Air, and Fire, to two of the four sensible qualities, hot, cold, wet, and dry. Aristotle’s
scheme added the heavenly Aether, the divine substance of the heavenly spheres, stars and
planets [1]. Aristotle maked experiments in optics using a camera obscura [2]. In astron-
omy, Aristotle refuted Democritus’s claim that the Milky Way was made up of ”those stars
which are shaded by the earth from the sun’s rays,” pointing out correctly that if ”the size
of the sun is greater than that of the earth and the distance of the stars from the earth many
times greater than that of the sun, then... the sun shines on all the stars and the earth screens
none of them.” [3]. The influence of the geocentric cosmology of Aristotle persisted until
Copernicus. More detai read: [4], [5].
Sir Isaac Newton 2 is the author of the fundamental work Mathematical Principles of
Natural Philosophy (1687) [6], in which he outlined the law of world wideness and the
three laws of mechanics, which became the basis of classical mechanics. He developed
differential and integral calculus, the theory of color, laid the foundations of modern phys-
ical optics, created many other mathematical and physical theories It was the First Great
Absorption – Newton’s work combined the laws of mechanics and the gravity.
1 Aristotle, Greek Aristoteles, (born 384 BCE, Stagira, Chalcidice, Greece died 322, Chalcis, Euboea),
ancient Greek philosopher and scientist, one of the greatest intellectual figures of Western history.
2Sir Isaac Newton PRS FRS (25 December 1642 20 March 1726) Newton is an English physicist, mathe-
matician, mechanic and astronomer, one of the founders of classical physics
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![Second Great Absorption was maded by the James Clerk Maxwell equations in XIX
centery. 3 These equations combine electricity, magnetism and optics [7].
Next, it was necessary to coordinate the laws of Newton and Maxwell’s equations. It
was made in end of XIX centery by Hendrik Antoon Lorentz 4, Jules Henri Poincar 5 and
Albert Einstein 6 They builded Special Theory Relativity (STR) [8].
Subsequently (at the beginning of XX), Albert Einstein constructed a theory of curved
space-time (General Theory of Relativity) on the basis of STR [9]. This theory absorbed
the Newtonian theory of gravity. This was Third Great Absorptipn.
Now everything fell into place. Everything got its explanation. Everything worked ... It
was ”The Second Theory of Everything”.
But in 1900 this beautiful continuous world crash down – Max Karl Ernst Ludwig
Planck discovered that our world is discrete - not continuous [10] 7
In the 30s of the 20th century, Erwin Rudolf Josef Alexander Schrdinger 8, Werner Karl
3James Clerk Maxwell; (13 June 1831 5 November 1879) was a Scottish scientist in the field of mathemat-
ical physics.
4 (Hendrik Antoon Lorentz; 18 July 1853 4 February 1928) was a Dutch physicist
5Jules Henri Poincar 29 April 1854 17 July 1912) was a French mathematician, theoretical physicist,
engineer, and philosopher of science.
6Albert Einstein; 14 March 1879 18 April 1955) was a German-born theoretical physicis
7Max Karl Ernst Ludwig Planck, 23 April 1858 4 October 1947) was a German theoretical physicist whose
discovery of energy quanta
8Erwin Rudolf Josef Alexander Schrdinger (12 August 1887 4 January 1961), was a Austrian physicist who
developed a number of fundamental results in the field of quantum theory
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![Heisenberg 9 and Paul Adrien Maurice Dirac 10 built a theory, for this discreteness. It was
the Quantum Mechanics - a new paradigm of physics. It put physics on a probabilistic basis
[11]. This is Fourth Great Absorption.
Fifth Great Absorption (mid 20th century) is the Feynman Quantum electrodynamics
[12]. This theory (Quantum Field Theory) has swallowed Quantum Mechanics and the
Maxwell equations.
And finally, in the 60s of the 20th century, Sheldon Lee Glashow united the electromag-
netic and weak interactions into one gauge theory. 11 This theory became the basis of the
Standard Model - the theory of all elementary particles.
Before the Theory of Everything, it remained to combine the Standard Model with the
theory of gravity
9Werner Karl Heisenberg (; 5 December 1901 1 February 1976) was a German theoretical physicist and
one of the key pioneers of quantum mechanics.
10Paul Adrien Maurice Dirac (8 August 1902 20 October 1984) was an English theoretical physicist who is
regarded as one of the most significant physicists of the 20th century.
11Sheldon Lee Glashow (born December 5, 1932) is an American theoretical physicist
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![For example, sentence It is raining is true if and only if it is raining1 [16].
Def. 1.1.2: Sentence Θ is false if and only if there is not that Θ.
It is clear that many neither true nor false sentences exist. For example, There is rainy
21 august 3005 year in Chelyabinsk .
Still an example: Obviously, the following sentence isn’t true and isn’t false [17]:
The sentence which has been written on this line, is false.
Those sentences which can be either true, or false, are called as meaningful sentences.
The previous example sentence is meaningless sentence.
Further we consider only meaningful sentences which are either true, or false.
Def. 1.1.3: Sentences A and B are equal (design.: A = B) if A is true, if and only if B is
true.
Further I’m using ordinary notions of the classical propositional logic [18].
Def. 1.1.4: A sentence C is called conjunction of sentences A and B (design.: C =
(A&B)) if C is true, if and only if A and B are true.
Def. 1.1.5: A sentence C is called negation of sentences A (design.: C = (¬A)) if C is
true, if and only if A is not true.
Def. 1.1.6: A sentence C is called disjunction of sentences A and B (design.: C =
(A∨B)) if C is true, if and only if A is true or B is true or both A and B are true.
Def. 1.1.7: A sentence C is called implication of sentences A and B (design.: C =
(A ⇒ B)) if C is true, if and only if B is true and/or B is false.
A sentence is called a simple sentence if it isn’t neither conjunction, nor a disjunction,
neither implication, nor negation.
Th. 1.1.1:
1) (A&A) = A; (A∨A) = A;
2) (A&B) = (B&A); (A∨B) = (B∨A);
3) (A&(B&C)) = ((A&B)&C); (A∨(B∨C)) = ((A∨B)∨C);
4) if T is a true sentence then for every sentence A: (A&T) = A and (A∨T) = T.
5) if F is a false sentence then (A&F) = F and (A∨F) = A.
Proof of Th. 1.1.1: This theorem directly follows from Def. 1.1.1, 1.2, 1.3, 1.4, 1.6.
Further I set out the version of the Gentzen Natural Propositional calculus2 (NPC) [15]:
Expression ”Sentence C is a logical consequence of a list of sentences Γ.” will be wrote
as the following: ”Γ C”. Such expressions are called sequences. Elements of list Γ are
called hypothesizes.
Def. 1.1.8
1. A sequence of form C C is called NPC axiom.
2. A sequence of form Γ A and Γ B is obtained from sequences of form Γ (A&B)
by a conjunction removing rule (design.: R&).
3. A sequence of form Γ1,Γ2 (A&B) is obtained from sequence of form Γ1 A and a
sequence of form Γ2 B by a conjunction inputting rule (design: I&).
4. A sequence of form Γ (A∨B) is obtained from a sequence of form Γ A or from
a sequence of form Γ B by a disjunction inputting rule (design.: I∨).
1Alfred Tarski (January 14, 1901 October 26, 1983) was a Polish logician and mathematician
2Gerhard Karl Erich Gentzen ( November 24, 1909, Greifswald, Germany August 4, 1945, Prague,
Czechoslovakia) was a German mathematician and logician.
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![5. A sequence of form Γ1 [A],Γ2 [B],Γ3 C is obtained from sequences of form Γ1 C,
Γ2 C, snd Γ3 (A∨B) by a disjunction removing rule (design.: R∨) (Here and further:
Γ1 [A] is obtained from Γ1 by removing of sentence A, and Γ2 [B] is obtained from Γ2 by
removing of sentence B).
6. A sequence of form Γ1,Γ2 B is obtained from a sequence of form Γ1 A and from
a sequence of form Γ2 (A ⇒ B) by a implication removing rule (design.: R⇒).
7. A sequence of form Γ[A] (A ⇒ B) is obtained from a sequence of form Γ B by
an implication inputting rule (design.: I⇒).
8. A sequence of form Γ C is obtained from a sequence of form Γ (¬(¬C)) by a
negation removing rule (design.: R¬).
9. A sequence of form Γ1 [C],Γ2 [C] (¬C) is obtained from a sequence of form Γ1 A
and from a sequence of form Γ2 (¬A) by negation inputting rule (design.: I¬).
10. A finite string of sequences is called a propositional natural deduction if every
element of this string either is a NPC axioms or is received from preceding sequences by
one of the deduction rules (R&, I&, I∨, R∨, R⇒, I⇒, R¬, I¬).
Actually, these logical rules look naturally in light of the previous definitions.
Example 1: Let us consider the following string of sequences:
1.((R&S)&(R ⇒ G)) ((R&S)&(R ⇒ G)) - NPC axiom.
2.((R&S)&(R ⇒ G)) (R&S) - R& from 1.
3.((R&S)&(R ⇒ G)) (R ⇒ G) - R& from 1.
4.((R&S)&(R ⇒ G)) R - R& from 2. (1.1)
5.((R&S)&(R ⇒ G)) G - R ⇒ from 3. and 4.
6.((R&S)&(R ⇒ G)) S - R& from 2.
7.((R&S)&(R ⇒ G)) (G&S) - I& from 5. and 6.
This string is a propositional natural deduction of sequence
((R&S)&(R ⇒ G)) (G&S).
since it fulfills to all conditions of Def. 1.1.8.
Hence sentence (G&S) is logical consequence from sentence
((R&S)&(R ⇒ G)).
Th. 1.1.2:
(A∨B) = (¬((¬A)&(¬B))), (1.2)
(A ⇒ B) = (¬(A&(¬B))). (1.3)
Proof of Th. 1.1.2:
The following string is a deduction of sequence
(A∨B) (¬((¬A)&(¬B))):
1. ((¬A)&(¬B)) ((¬A)&(¬B)), NPC axiom.
2. ((¬A)&(¬B)) (¬A), R& from 1.
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![4. (¬B) (¬B) - NPC axiom.
5. (¬B),(A ⇒ B) (¬A) - I¬ from 3. and 4.
6. (A ⇒ B) ((¬B) ⇒ (¬A)) - I⇒ from 5.
7. ((A ⇒ B) ⇒ ((¬B) ⇒ (¬A))) - I⇒ from 6.
This string is a deduction of sentence of form
((A ⇒ B) ⇒ ((¬B) ⇒ (¬A)))
from the empty list of sentences. I.e. sentences of such form are logicaly provable.
Th. 1.1.3: If sequence Γ → C is deduced and C is false then some false sentence is
contained in Γ.
Proof of Th. 1.1.3: is received by induction of number of sequences in the deduction
of sequence Γ → C.
The recursion Basis: Let the deduction of sequence Γ → C contains single sentence.
In accordance the definition of propositional natural deduction this deduction must be of
the following type: C → C. Obviously, in this case the lemma holds true.
The recursion Step:The recursion assumption: Let’s admit that the lemma is carried
out for any deduction which contains no more than n sequences.
Let deduction of Γ → C contains n+1 sequence. In accordance with the propositional
natural deduction definition sequence Γ → C can be axiom NPC or can be received by the
deduction rules from previous sequence.
a) If Γ → C is the NPC axiom then see the recursion basis.
b) Let Γ →C be received by R&. In this case sequence of type Γ → (C&B) or sequence
of type Γ → (B&C) is contained among the previous sequences of this deduction. Hence,
deductions of sequences Γ → (C&B) and Γ → (B&C) contains no more than n sequences.
In accordance with the recursion assumption, these deductions submit to the lemma. Be-
cause C is false then (C&B) is false and (B&C) is false in accordance with the conjunction
definition. Therefore, Γ contains some false sentence by the lemma. And in this case the
lemma holds true.
c) Let Γ → C be received by I&. In this case sequence of type Γ1 → A and sequence of
type Γ2 → B is contained among the previous sequences of this deduction, and C = (A&B)
and Γ = Γ1,Γ2. Deductions of sequences Γ1 → A and Γ2 → B contains no more than n
sequences. In accordance with the recursion assumption, these deductions submit to the
lemma. Because C is false then A is false or B is false in accordance with the conjunction
definition. Therefore, Γ contains some false sentence by the lemma. And in this case the
lemma holds true.
d) Let Γ → C be received by R∨. In this case sequences of type Γ1 → (A∨B), Γ2 [A] →
C, and Γ3 [B] → C are contained among the previous sequences of this deduction, and Γ =
Γ1,Γ2,Γ3. Because these previous deductions contain no more than n sequences then in
accordance with the recursion assumption, these deductions submit to the lemma. Because
C is false then Γ2 [A] contains some false sentence, and Γ3 [B] contains some false sentence.
If A is true then the false sentence is contained in Γ2. If B is true then the false sentence is
contained in Γ3. I.e. in these case some false sentence is contained in Γ. If A is false and
B is false then (A∨B) is false in accordance with the disjunction definition. In this case Γ1
contains some false sentence. And in all these cases the lemma holds true.
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![e) Let Γ → C be received by I∨. In this case sequence of type Γ → A or sequence of
type Γ → B is contained among the previous sequences of this deduction, and C = (A∨B).
Deductions of sequences Γ → A and Γ → B contains no more than n sequences. In accor-
dance with the recursion assumption, these deductions submit to the lemma. Because C is
false then A is false and B is false in accordance with the disjunction definition. Therefore,
Γ contains some false sentence by the lemma. And in this case the lemma holds true.
f) Let Γ →C be received by R⇒. In this case sequences of type Γ1 → (A ⇒ C), Γ2 → A
are contained among the previous sequences of this deduction, and Γ = Γ1,Γ2. Because
these previous deductions contain no more than n sequences then in accordance with the
recursion assumption, these deductions submit to the lemma. If A is false then Γ2 contains
some false sentence. If A is true then (A ⇒ C) is false in accordance with the implication
defination since C is false. And in all these cases the lemma holds true.
g) Let Γ → C be received by I⇒. In this case sequences of type Γ[A] → B is contained
among the previous sequences of this deduction, and C = (A ⇒ B). Because deduction of
Γ[A] → B contains no more than n sequences then in accordance with the recursion assump-
tion, this deduction submit to the lemma. Because C is false then A is true in accordance
with the implication definition. Hence, some false sentence is contained in Γ. Therefore, in
this case the lemma holds true.
i) Let Γ →C be received by R¬. In this case sequence of type Γ → (¬(¬C)) is contained
among the previous sequences of this deduction. This previous deduction contains no more
than n sequences then in accordance with the recursion assumption, this deduction submit
to the lemma Because C is false then (¬(¬C)) is false in accordance with the negation
definition. Therefore, Γ contains some false sentence by the lemma. And in this case the
lemma holds true.
j) Let Γ → C be received by I¬. In this case sequences of type Γ1 [A] → B, and Γ2 [A] →
(¬B) are contained among the previous sequences of this deduction, and Γ = Γ1,Γ2 and
C = (¬A). Because these previous deductions contain no more than n sequences then in
accordance with the recursion assumption, these deductions submit to the lemma. Because
C is false then A is true. Hence, some false sentence is contained in Γ because B is false or
(¬B) is false in accordance with the negation definition. Therefore, in all these cases the
lemma holds true.
The recursion step conclusion: If the lemma holds true for deductions containing n
sequences then the lemma holds true for deduction containing n+1 sequences.
The recursion conclusion: Lemma holds true for all deductions .
Def. 1.1.9 A sentence is naturally propositionally provable if there exists a prpositional
natural deduction of this sentence from the empty list.
In accordance with Th. 1.1.3 all naturally propositionally provable sentences are true
because otherwise the list would appear not empty.
But some true sentences are not naturally propositionally provable.
Alphabet of Propositional Calculations:
1. symbols pk with natural k are called PC-letters;
2. symbols ∩, ∪, ⊃, ˆ are called PC-symbols;
3. (, ) are called brackets.
Formula of Propositional Calculations:
1. any PC-letter is PC-formula.
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![sequence in accordance with the recursion assumption because this deduction contains no
more than n elements.
If g ϕ−1(A) = 0 then g ϕ−1(¬(¬A)) = 0 in accordance with the Boolean function
definition. Hence there exists sentence C such that C ∈ Γ and g ϕ−1(C) = 0.
Hence the lemma holds true for the deduction of sequence Γ A.
d) Let Γ A be obtained from previous sequences by I¬.
In that case forms of these previous sequences are Γ1 B and Γ2 (¬B) with Γ =
Γ1 [G],Γ2 [G] and A = (¬G) in accordance with the definition of deduction.
The lemma holds true for the deductions of sequences Γ1 B and Γ2 (¬B) in ac-
cordance with the recursion assumption because these deductions contain no more than n
elements.
If g ϕ−1(A) = 0 then g ϕ−1(G) = 1 in accordance with the Boolean function defi-
nition.
Either g ϕ−1(B) = 0 or g ϕ−1(¬B) = 0 by the same definition. Hence there exists
sentence C such that either C ∈ Γ1 [G] or C ∈ Γ2 [G] andg ϕ−1(C) = 0 in accordance with
the recursion assumption.
Hence in that case the lemma holds true for the deduction of sequence Γ A.
e) Let Γ A be obtained from a previous sequence by I∨.
In that case a form of A is (B∨G) and a form of this previous sequence is either Γ B or
Γ G in accordance with the definition of deduction. The lemma holds true for this previous
sequence deduction in accordance with the recursion assumption because this deduction
contains no more than n elements.
If g ϕ−1(A) = 0 then g ϕ−1(B) = 0 and g ϕ−1(G = 0 in accordance with
the Boolean function definition. Hence there exists sentence C such that C ∈ Γ and
g ϕ−1(C) = 0.
Hence in that case the lemma holds true for the deduction of sequence Γ A.
f) Let Γ A be obtained from previous sequences by R∨.
Forms of these previous sequences are Γ1 A, Γ2 A, and Γ3 (B∨G) with Γ =
Γ1 [B],Γ2 [G],Γ3 in accordance with the definition of deduction. The lemma holds true for
the deductions of these sequences in accordance with the recursion assumption because
these deductions contain no more than n elements.
If g ϕ−1(A) = 0 then there exists sentence C1 such that C1 ∈ Γ1 and
g ϕ−1(C1) = 0, and there exists sentence C2 such that C2 ∈ Γ2 and g ϕ−1(C2) = 0 in
accordance with the lemma.
If g ϕ−1(B∨G) = 0 then there exists sentence C such that C ∈ Γ3 and g ϕ−1(C) = 0
in accordance with the lemma. Hence in that case the lemma holds true for the deduction
of sequence Γ A.
If g ϕ−1(B∨G) = 1 then either g ϕ−1(B) = 1 or g ϕ−1(G) = 1 in accordance with
the Boolean function definition.
If g ϕ−1(B) = 1 then C1 ∈ Γ1 [B]. Hence in that case the lemma holds true for the
deduction of sequence Γ A.
If g ϕ−1(G) = 1 then a result is the same.
Hence the lemma holds true for the deduction of sequence Γ A in all these cases.
g) Let Γ A be obtained from previous sequences by R⇒.
10](https://image.slidesharecdn.com/fn-190805113833/85/TOE-24-320.jpg)
![Forms of these previous sequences are Γ1 (B ⇒ A) and Γ2 (B) with Γ = Γ1,Γ2
in accordance with the definitions of deduction. Hence the lemma holds true for these
deduction in accordance with the recursion assumption because these deductions contain
no more than n elements.
If g ϕ−1(B ⇒ A) = 0 then there exists sentence C such that C ∈ Γ1 and g ϕ−1(C) = 0
in accordance with the lemma. Hence in that case the lemma holds true for the deduction
of sequence Γ A.
If g ϕ−1(B ⇒ A) = 1 then g ϕ−1(B) = 0 in accordance with the Boolean function
definition. Hence there exists sentence C such that C ∈ Γ2 and g ϕ−1(C) = 0.
Hence the lemma holds true for sequence Γ A in all these cases.
h) Let Γ A be obtained from a previous sequence by I⇒.
In that case a form of sentence A is (B ⇒ G) and a form of this previous sequence is
Γ1 G with Γ = Γ1 [B] in accordance with the definition of deduction. The lemma holds
true for the deduction of this sequence in accordance the recursion assumption because this
deduction contain no more than n elements.
If g ϕ−1(A) = 0 then g ϕ−1(G) = 0 and g ϕ−1(B) = 1 in accordance with the
Boolean function definition. Hence there exists sentence C such that C ∈ Γ1 [B] and
g ϕ−1(C) = 0.
The recursion step conclusion: Therefore, in each possible case, if the lemma holds
true for a deduction, containing no more than n elements, then the lemma holds true for a
deduction contained n+1 elements.
The recursion conclusion: Therefore the lemma holds true for a deduction of any
length
Th. 1.1.4: Each naturally propositionally proven sentence is a tautology.
Proof of Th. 1.1.4: If a sentence A is naturally propositionally proven then there exists
a natural propositional deduction of form A in accordance with Def. 1.1.9. Hence for
every Boolean function g: g ϕ−1(A) = 1 in accordance with Lm. 1.1.1. Hence sentence
A is a tautology in accordance with the tautology definition (Def. 1.1.13)
Designation 1: Let g be a Boolean function. In that case for every sentence A:
Ag
:=
A if g ϕ−1(A) = 1,
(¬A) if g ϕ−1(A) = 0.
Lm. 1.1.2: Let B1,B2,..,Bk be the simple sentences making sentence A by PC-symbols
(¬, &, ∨, ⇒).
Let g be any Boolean function.
In that case there exist a propositional natural deduction of sequence
Bg
1,Bg
2,..,Bg
k Ag
.
Proof of Lm. 1.1.2: is received by a recursion on a number of PC-symbols in sentence
A.
Basis of recursion Let A does not contain PC-symbols . In this case the string of one
sequence:
1. Ag Ag, NPC-axiom.
is a fit deduction.
11](https://image.slidesharecdn.com/fn-190805113833/85/TOE-25-320.jpg)
![1.1. Recorders
Any information, received from physical devices, can be expressed by a text, made of sen-
tences.
Let a be some object which is able to receive, save, and/or transmit an information [22].
A set a of sentences, expressing an information of an object a, is called a recorder of this
object. Thus, statement: ”Sentence A is an element of the set a” denotes : ”a has
information that the event, expressed by sentence A , took place.” In short: ”a knows
that A.” Or by designation: ”a• A ”.
Obviously, the following conditions are satisfied:
I. For any a and for every A: false is that a• (A&(¬A)), thus, any recorder doesn’t
contain a logical contradiction.
II. For every a, every B, and all A: if B is a logical consequence from A, and a•A, then
a•B.
*III. For all a, b and for every A: if a• b•A then a•A.
For example, if device a has information that device b has information that mass of
particle ←−χ equals to 7 then device a has information that mass of particle ←−χ equal to 7.
1.2. Time
There is only a moment between past and
future.
L. Derbenev, ”Sannikov Land”, (1973)
”Do not expect answers before you have found clear meanings”
Hans Reichenbach, The Direction of Time, (1953)
There are many concepts of the theory of ”time” - in particular, quantum mechanical,
relativistic, thermodynamic, causal, etc. All of them are based on unclearly defined con-
cepts and in most cases contain a vicious circle.
The thermodynamic concept has the greatest favor. But if a sergeant has a platoon in a
line, does this sergeant’s wristwatch change direction?
Any subjects, connected with an information is called informational objects. For ex-
ample, it can be a physics device, or computer disks and gramophone records, or people,
carrying memory on events of their lifes, or trees, on cuts which annual rings tell on past
climatic and ecological changes, or stones with imprints of long ago extincted plants and
bestials, or minerals, telling on geological cataclysms, or celestial bodies, carrying an in-
formation on a remote distant past Universe, etc., etc.
It is clearly that an information, received from such information object, can be expressed
by a text which made of sentences.
Let’s consider finite (probably empty) path of symbols of form q•.
Def. 1.3.1 A path α is a subpath of a path β (design.: α β) if α can be got from β by
deletion of some (probably all) elements.
17](https://image.slidesharecdn.com/fn-190805113833/85/TOE-31-320.jpg)
![Designation: (α)1
is α, and (α)k+1
is α(α)k
.
Therefore, if k ≤ l then (α)k
(α)l
.
Def. 1.3.2 A path α is equivalent to a path β (design.: α ∼ β) if α can be got from β by
substitution of a subpath of form (a•)k
by a path of the same form ((a•)s
).
In this case:
III. If β α or β ∼ α then for any K:
if a•K then a• (K&(αA ⇒ βA)).
Obviously, III is a refinement of condition *III.
Def. 1.3.3 A natural number q is instant, at which a registrates B according to κ-clock
{g0,A,b0} (design.: q is [a•B ↑ a,{g0,A,b0}]) if:
1. for any K: if a•K then
a•
(K&(a•
B ⇒ a•
(g•
0b•
0)q
g•
0A))
and
a•
K& a•
(g•
0b•
0)q+1
g•
0A ⇒ a•
B .
2. a• a•B& ¬a• (g•
0b•
0)q+1
g•
0A .
Lm. 1.3.1 If
q is [a•
αB ↑ a,{g0,A,b0}], (1.15)
p is [a•
βB ↑ a,{g0,A,b0}], (1.16)
α β, (1.17)
then
q ≤ p.
Proof of Lm. 1.3.1: From (1.16):
a•
(a•
βB)& ¬a•
(g•
0b•
0)(p+1)
g•
0A . (1.18)
From (1.17) according to III:
a•
a•
βB& ¬a•
(g•
0b•
0)(p+1)
g•
0A &(a•
βB ⇒ a•
αB) . (1.19)
Let us designate:
R := a•βB,
S := ¬a• (g•
0b•
0)(p+1)
g•
0A ,
G := a•αB.
In that case a shape of formula (1.18) is
a•
(R&S),
18](https://image.slidesharecdn.com/fn-190805113833/85/TOE-32-320.jpg)
![and a shape of formula (1.19) is
a•
((R&S)&(R ⇒ G)).
Sentence (G&S) is a logical consequence from sentence
((R&S)&(R ⇒ G)) (1.1). Hence
a•
(G&S),
in accordance with II.
Hence
a•
a•
αB& ¬a•
(g•
0b•
0)(p+1)
g•
0A
in accordance with the designation.
Hence from (1.15):
a•
a•
αB& ¬a•
(g•
0b•
0)(p+1)
g•
0A &(a•
αB ⇒ a•
(g•
0b•
0)q
g•
0A) .
According to II:
a•
¬a•
(g•
0b•
0)(p+1)
g•
0A &a•
(g•
0b•
0)q
g•
0A (1.20)
If q > p, i.e. q ≥ p+1, then from (1.20) according to III
a•
¬a• (g•
0b•
0)(p+1)
g•
0A &a• (g•
0b•
0)q
g•
0A &
a• (g•
0b•
0)q
g•
0A ⇒ a• (g•
0b•
0)(p+1)
g•
0A
.
According to II:
a•
¬a•
(g•
0b•
0)(p+1)
g•
0A &a•
(g•
0b•
0)(p+1)
g•
0A .
It contradicts to condition I. Therefore, q ≤ p .
Lemma 1.3.1 proves that if
q is [a•
B ↑ a,{g0,A,b0}],
and
p is [a•
B ↑ a,{g0,A,b0}]
then
q = p.
That’s why, expression ”q is [a•B ↑ a,{g0,A,b0}]” is equivalent to expression ”q =
[a•B ↑ a,{g0,A,b0}].”
Def. 1.3.4 κ-clocks {g1,B,b1} and {g2,B,b2} have the same direction for a if the
following condition is satisfied:
if
19](https://image.slidesharecdn.com/fn-190805113833/85/TOE-33-320.jpg)
![r = [a• (g•
1b•
1)q
g•
1B ↑ a,{g2,B,b2}],
s = [a• (g•
1b•
1)p
g•
1B ↑ a,{g2,B,b2}],
q < p,
then
r ≤ s.
Th. 1.3.1 All κ-clocks have the same direction.
Proof of Th. 1.3.1:
Let
r := [a•
(g•
1b•
1)q
g•
1B ↑ a,{g2,B,b2}],
s := [a•
(g•
1b•
1)p
g•
1B ↑ a,{g2,B,b2}],
q < p.
In this case
(g•
1b•
1)q
(g•
1b•
1)p
.
Consequently, according to Lm. 1.3.1
r ≤ s
Consequently, a recorder orders its sentences with respect to instants. Moreover, this
order is linear and it doesn’t matter according to which κ-clock it is set.
Def. 1.3.5 κ-clock {g2,B,b2} is k times more precise than κ-clock
{g1,B,b1} for recorder a if for every C the following condition is satisfied: if
q1 = [a•C ↑ a,{g1,B,b1}],
q2 = [a•C ↑ a,{g2,B,b2}],
then
q1 < q2
k < q1 +1.
Lm. 1.3.2 If for every n:
qn−1 <
qn
kn
< qn−1 +1, (1.21)
then the series
q0 +
∞
∑
n=1
qn −qn−1kn
k1 ...kn
(1.22)
20](https://image.slidesharecdn.com/fn-190805113833/85/TOE-34-320.jpg)
![(q− p)+(u−s) ≥ r − p
Thus, all four axioms of the metrical space [19] are accomplished for a,H in an
internally stationary systeminternally stationary system of recorders.
Consequently, a,H is a distance length similitude in this space.
Def. 1.4.5 A set ℜ of recorders is degenerated into a beam ab1 and point a1 if there
exists C such that the following conditions are satisfied:
1) For any sequence α, made of elements of set ℜ, and for any K: if a•K then
a• (K&(αa•
1C ⇒ αb•
1a•
1C)).
2) There is sequence β, made of elements of the setℜ, and there exist sentence S such
that a• (βb•
1C&S) and it’s false that a• (βa•
1b•
1C&S)
Further we’ll consider only not degenerated sets of recorders.
Def. 1.4.6: B took place in the same place as a1 for a (design.: (a)(a1,B)) if for every
sequence α and for any sentence K the following condition is satisfied:
if a•K then a• (K&(αB ⇒ αa•
1B)).
Th. 1.4.4:
(a)(a1,a•
1B).
Proof: Since αa•
1 ∼ αa•
1a•
1 then according to III: if a•
1K then
a•
1 (K&(αa•
1B ⇒ αa•
1a•
1B))
Th. 1.4.5: If
(a)(a1,B), (1.25)
(a)(a2,B), (1.26)
then
(a)(a2,a•
1B).
Proof: Let a•K.
In this case from (1.26):
a•
(K&(αa•
1B ⇒ αa•
1a•
2B)).
From (1.25):
a•
((K&(αa•
1B ⇒ αa•
1a•
2B))&(αa•
1a•
2B ⇒ αa•
1a•
2a•
1B)).
According to II:
26](https://image.slidesharecdn.com/fn-190805113833/85/TOE-40-320.jpg)
![ℜaH (a2,b1) = ℜaH (a2,a1). (1.72)
Again in accordance with Theorem 1.4.3:
ℜaH (a2,a1) = ℜaH (a1a2).
From (1.72), (1.71), (1.70):
ℜaH (b1,b2) = ℜaH (a1a2)
According to Urysohn‘s theorem5 [20]: any homogeneous space is homeomorphic to
some set of points of real Hilbert space. If this homeomorphism is not Identical transfor-
mation, then ℜ will represent a non- Euclidean space. In this case in this ”space-time”
corresponding variant of General Relativity Theory can be constructed. Otherwise, ℜ is
Euclidean space. In this case there exists coordinates system Rµ such that the following
condition is satisfied: for all elements a1 and a2 of set ℜ there exist points x1 and x2 of
system Rµ such that
a,H (ak,as) = ∑
µ
j=1 xs,j −xk,j
2 0.5
.
In this case Rµ is called a coordinates system of frame of reference ℜaH and numbers
xk,1,xk,2,...,xk,µ are called coordinates of recorder ak in Rµ.
A coordinates system of a frame of reference is specified accurate to transformations of
shear, turn, and inversion.
Def. 1.5.4: Numbers x1,x2,...,xµ are called coordinates of B in a coordinate system
Rµ of a frame of reference ℜaH if there exists a recorder b such that b ∈ ℜ, (a)(b,B)
and these numbers are the coordinates in Rµ of this recorder.
Th. 1.5.4: In a coordinate system Rµ of a frame of reference ℜaH : if z is a dis-
tance length between B and C, coordinates of B are (b1,b2,..,bn), coordinates of C are
(c1,c2,..,c3), then
z =
µ
∑
j=1
(cj −bj)2
0.5
.
Proof came out straight from Definition 1.5.4
Def. 1.5.5: Numbers x1,x2,...,xµ are called coordinates of the recor-der b in the
coordinate system Rµ at the instant t of the frame of reference ℜaH if for every B the
condition is satisfied: if
t = b•
B | ℜaH
5Pavel Samuilovich Urysohn, Pavel Uryson (February 3, 1898, Odessa - August 17, 1924, Batz-sur-Mer)
was a Jewish mathematician who is best known for his contributions in the theory of dimension, and for de-
veloping Urysohn’s Metrization Theorem and Urysohn’s Lemma, both of which are fundamental results in
topology.
35](https://image.slidesharecdn.com/fn-190805113833/85/TOE-49-320.jpg)
![then coordinates of b•B in coordinate system Rµ of frame of reference ℜaH
are the following:
x1,x2,...,xµ .
Lm. 1.5.1 If
τ := [b•
C ↑ b,{g0,B,b0}], (1.73)
p := a•
b•
(g•
0b•
0)τ
g•
0B ↑ a,{g1,A,b1} , (1.74)
q := a•
b•
(g•
0b•
0)τ+1
g•
0B ↑ a,{g1,A,b1} , (1.75)
t := [a•
b•
C ↑ a,{g1,A,b1}] (1.76)
then
p ≤ t ≤ q.
Proof
1) From (1.75):
a•
a•
b•
(g•
0b•
0)τ+1
g•
0B& ¬a•
(g•
1b•
1)q+1
g•
1A . (1.77)
Hence from (1.73):
b• (g•
0b•
0)τ+1
g•
0B ⇒ b•C
then from (1.77) according to II:
a• a•b•C& ¬a• (g•
1b•
1)q+1
g•
1A .
According to II, since from (1.76):
a•b•C ⇒ a• (g•
1b•
1)t
g•
1A
then
a•
a•
(g•
1b•
1)t
g•
1A& ¬a•
(g•
1b•
1)q+1
g•
1A . (1.78)
If t > q then t ≥ q+1. Hence according to III from (1.78):
a• a• (g•
1b•
1)q+1
g•
1A& ¬a• (g•
1b•
1)q+1
g•
1A ,
it contradicts to I. So t ≤ q.
2) From (1.76):
a•
a•
b•
C& ¬a•
(g•
1b•
1)t+1
g•
1A . (1.79)
Since from (1.73):
36](https://image.slidesharecdn.com/fn-190805113833/85/TOE-50-320.jpg)
![b•C ⇒ b• (g•
0b•
0)τ
g•
0B
then from (1.79) according to II:
a•
a•
b•
(g•
0b•
0)τ
g•
0B& ¬a•
(g•
1b•
1)t+1
g•
1A . (1.80)
Since from (1.74):
a•b• (g•
0b•
0)τ
g•
0B ⇒ a• (g•
1b•
1)p
g•
1A
then according to II from (1.80):
a•
a•
(g•
1b•
1)p
g•
1A& ¬a•
(g•
1b•
1)t+1
g•
1A . (1.81)
If p > t then p ≥ t +1. In that case from (1.81) according to III:
a• a• (g•
1b•
1)t+1
g•
1A& ¬a• (g•
1b•
1)t+1
g•
1A ,
it contradicts to I. So p ≤ t
Th. 1.5.5 In a coordinates system Rµ of a frame of reference ℜaH : if in every instant
t: coordinates of6:
b: xb,1 +v·t,xb,2,xb,3,...,xb,µ ;
g0: x0,1 +v·t,x0,2,x0,3,...,x0,µ ;
b0: x0,1 +v·t,x0,2 +l,x0,3,...,x0,µ ; and
tC = b•C | ℜaH ;
tD = b•D | ℜaH ;
qC = [b•C ↑ b,{g0,A,b0}];
qD = [b•D ↑ b,{g0,A,b0}],
then
lim
l→0
2·
l
(1−v2)
·
qD −qC
tD −tC
= 1.
Proof: Let us designate:
t1 := b•
(g0
•
b0
•
)qC
g0
•
B | ℜaH , (1.82)
t2 := b•
(g0
•
b0
•
)qC+1
g0
•
B | ℜaH , (1.83)
t3 := (g0
•
b0
•
)qC
g0
•
B | ℜaH , (1.84)
t4 := (g0
•
b0
•
)qC+1
g0
•
B | ℜaH . (1.85)
In that case coordinates of:
6below v is a real positive number such that |v| < 1)
37](https://image.slidesharecdn.com/fn-190805113833/85/TOE-51-320.jpg)
![1.5. Probability
”The two greatest tyrants on earth: case
and time”
Johann G. Herder
The first significant result in probability theory was obtained by the Swiss mathemati-
cian Jacob Bernoulli8 in 1713 [37] (the Bernoulli Large Number law). Further, the de-
velopment of the theory of probability went in two ways: using the axiomatic method,
Soviet mathematician, Andrei Kolmogorov9 embed this theory into mathematical analysis
[25], and the American physicist Edwin Thompson Jaynes10 began to develop the theory of
probability from logic [26]. And we continue this way.
There is the evident nigh affinity between the classical probability function and the
Boolean function of the classical propositional logic [23]. These functions are differed by
the range of value, only. That is if the range of values of the Boolean function shall be
8Jacob Bernoulli (also known as James or Jacques; 6 January 1655 [O.S. 27 December 1654] 16 August
1705) was one of the many prominent mathematicians in the Bernoulli family.
9Andrey Nikolaevich Kolmogorov, 25 April 1903 20 October 1987) was a 20th-century Soviet mathemati-
cian who made significant contributions to the mathematics of probability theory
10Edwin Thompson Jaynes (July 5, 1922 April 30, 1998) was the Wayman Crow Distinguished Profes-
sor of Physics at Washington University in St. Louis. He wrote extensively on statistical mechanics and on
foundations of probability and statistical inference
51](https://image.slidesharecdn.com/fn-190805113833/85/TOE-65-320.jpg)
![expanded from the two-elements set {0;1} to the segment [0;1] of the real numeric axis
then the logical analog of the Bernoulli Large Number Law [37] can be deduced from the
logical axioms. These topics is considered in this article.
Further we consider set of all meaningfull sentences.
1.5.1. Events
Def. 1.6.1.1: A set B of sentences is called event, expressed by sentence C, if the following
conditions are fulfilled:
1. C ∈ B;
2. if A ∈ B and D ∈ B then A = D;
3. if D ∈ B and A = D then A ∈ B.
In this case denote: B := ◦C.
Def. 1.6.1.2: An event B occurs if here exists a true sentence A such that A ∈ B.
Def. 1.6.1.3: Events A and B equal (denote: A = B) if A occurs if and only if B
occurs.
Def. 1.6.1.4: Event C is called product of event A and event B (denote: C = (A ·B)) if
C occurs if and only if A occurs and B occurs.
Def. 1.6.1.5: Events C is called complement of event A (denote: C = (#A)) if C occurs
if and only if A does not occur.
Def. 1.6.1.6: (A +B) := (#((#A)·(#B))). Event (A +B) is called sum of event A and
event B.
Therefore, the sum of event occurs if and only if there is at least one of the addends.
Def. 1.6.1.7: The authentic event (denote: T ) is the event which contains a tautology.
Hence, T occurs in accordance Def. 1.6.1.2:
The impossible event (denote: F ) is event which contains negation of a tautology.
Hence, F does not occur.
1.5.2. B-functions
Def. 1.6.2.1: Let b(X) be a function defined on the set of events.
And let this function has values on he real numbers segment [0;1].
Let there exists an event C0 such that b(C0) = 1.
Let for all events A and B: b(A ·B)+b(A ·(#B)) = b(A).
In that case function b(X) is called B-function.
By this definition:
b(A ·B) ≤ b(A). (1.118)
Hence, b(T · C0) ≤ b(T ). Because T · C0 = C0 (by Def.1.6.1.4 and Def.1.6.1.7) then
b(C0) ≤ b(T ). Because b(C0) = 1then
b(T ) = 1. (1.119)
From Def.1.6.2.1: b(T · B) + b(T · (#B)) = b(T ). Because T D = D for any D then
b(B)+b(#B) = b(T ). Hence, by (1.119): for any B:
53](https://image.slidesharecdn.com/fn-190805113833/85/TOE-67-320.jpg)
![1.5.3. Independent tests
Definition 1.6.3.1: Let st(n) be a function such that st(n) has domain on the set of natural
numbers and has values in the events set.
In this case event A is a [st]-series of range r with V- number k if A, r and k fulfill to
some one amongst the following conditions:
1) r = 1 and k = 1, A = st (1) or k = 0, A = (#st (1));
2) B is [st]-series of range r −1 with V-number k −1 and
A = (B ·st (r)),
or B is [st]-series of range r −1 with V-number k and
A = (B ·(#st (r))).
Let us denote a set of [st]-series of range r with V-number k as [st](r,k).
For example, if st (n) is a event Bn then the sentences:
(B1 ·B2 ·(#B3)), (B1 ·(#B2)·B3), ((#B1)·B2 ·B3)
are the elements of [st](3,2), and
B1 ·B2 ·(#B3)·B4 ·B5 ∈ [st](5,3).
Definition 1.6.3.2: Function st(n) is independent for B-function b if for A: if
A ∈ [st](r,r) then:
b(A) =
r
∏
n=1
b(st (n)).
Definition 1.6.3.3: Let st(n) be a function such that st(n) has domain on the set of
natural numbers and has values in the set of events.
In this case sentence A is [st]-disjunction of range r with V-number k (denote: t[st](r,k))
if A is the disjunction of all elements of [st](r,k).
For example, if st (n) is event Cn then:
((#C1)·(#C2)·(#C3)) = t[st](3,0),
t[st](3,1) = ((C1 ·(#C2)·(#C3))+((#C1)·C2 ·(#C3))+((#C1)·(#C2)·C3)),
t[st](3,2) = ((C1 ·C2 ·(#C3))+((#C1)·C2 ·C3)+(C1 ·(#C2)·C3)),
(C1 ·C2 ·C3) = t[st](3,3).
Definition 1.6.3.4: A rational number ω is called frequency of sentence A in the [st]-
series of r independent for B-function b tests (designate: ω = νr [st](A)) if
1) st(n) is independent for B-function b,
2) for all n: b(st (n)) = b(A),
3) t[st](r,k) is true and ω = k/r.
Theorem: 1.6.3.1: (the J.Bernoulli11 formula [37]) If st(n) is independent for B-
function b and there exists a real number p such that for all n: b(st (n)) = p then
b(t [st](r,k)) =
r!
k!·(r −k)!
· pk
·(1− p)r−k
.
11Jacob Bernoulli (also known as James or Jacques) (27 December 1654 16 August 1705) was one of the
many prominent mathematicians in the Bernoulli family.
55](https://image.slidesharecdn.com/fn-190805113833/85/TOE-69-320.jpg)
 then:
b(B) = pk
·(1− p)r−k
.
Since [st](r,k) contains r!/(k!·(r −k)!) elements then by the Theorems (1.127),
(1.128) and (1.126) this Theorem is fulfilled.
Definition 1.6.3.5: Let function st(n) has domain on the set of the natural numbers and
has values in the set of the events.
Let function f(r,k,l) has got the domain in the set of threes of the natural numbers and
has got the range of values in the set of the events.
In this case f(r,k,l) = T[st](r,k,l) if
1) f(r,k,k) = t[st](r,k),
2) f(r,k,l +1) = (f(r,k,l)+t[st](r,l +1)).
Definition 1.6.3.6: If a and b are real numbers and k−1 < a ≤ k and l ≤ b < l +1 then
T[st](r,a,b) = T[st](r,k,l).
Theorem: 1.6.3.2:
T[st](r,a,b) =◦ a
r
≤ νr [st](A) ≤
b
r
.
Proof of the Theorem 1.6.3.2: By the Definition 1.6.3.6: there exist natural numbers r
and k such that k −1 < a ≤ k and l ≤ b < l +1.
The recursion on l:
1. Let l = k.
In this case by the Definition 1.6.3.4:
T[st](r,k,k) = t[st](r,k) =◦
νr [st](A) =
k
r
.
2. Let n be any natural number.
The recursive assumption: Let
T[st](r,k,k +n) =◦ k
r
≤ νr [st](A) ≤
k +n
r
.
By the Definition 1.6.3.5:
T[st](r,k,k +n+1) = (T[st](r,k,k +n)+t[st](r,k +n+1)).
By the recursive assumption and by the Definition 1.6.3.4:
T[st](r,k,k +n+1) =
= (◦ k
r
≤ νr [st](A) ≤
k +n
r
+◦
νr [st](A) =
k +n+1
r
).
Hence, by the Definition 2.10:
T[st](r,k,k +n+1) =◦ k
r
≤ νr [st](A) ≤
k +n+1
r
.
56](https://image.slidesharecdn.com/fn-190805113833/85/TOE-70-320.jpg)
) = ∑
a≤k≤b
r!
k!·(r −k)!
· pk
·(1− p)r−k
.
Proof of the Theorem 1.6.3.3: This is the consequence from the Theorem 1.6.3.1 by
the Theorem 3.6.
Theorem: 1.6.3.4 If st(n) is independent for the B-function b and there exists a real
number p such that b(st (n)) = p for all n then
b(T[st](r,r ·(p−ε),r ·(p+ε))) ≥ 1−
p·(1− p)
r ·ε2
for every positive real number ε.
Proof of the Theorem 1.6.3.4: Because
r
∑
k=0
(k −r · p)2
·
r!
k!·(r −k)!
· pk
·(1− p)r−k
= r · p·(1− p)
then if
J = {k ∈ N|0 ≤ k ≤ r ·(p−ε)}∩{k ∈ N|r ·(p+ε) ≤ k ≤ r}
then
∑
k∈J
r!
k!·(r −k)!
· pk
·(1− p)r−k
≤
p·(1− p)
r ·ε2
.
Hence, by (1.120) this Theorem is fulfilled.
Hence
lim
r→∞
b(T[st](r,r ·(p−ε),r ·(p+ε))) = 1 (1.129)
for all tiny positive numbers ε.
1.5.4. The logic probability function
Definition 1.6.4.1: B-function P is P-function if for every event ◦ Θ :
If P(◦ Θ ) = 1 then Θ is true sentence.
Hence from Theorem 1.6.3.2 and (1.129): if b is a P-function then the sentence
(p−ε) ≤ νr [st](A) ≤ (p+ε)
is almost true sentence for large r and for all tiny ε. Therefore, it is almost truely that
νr [st](A) = p
for large r.
Therefore, it is almost true that
57](https://image.slidesharecdn.com/fn-190805113833/85/TOE-71-320.jpg)
for large r.
Therefore, the function, defined by the Definition 1.6.4.1 has got the statistical meaning.
That is why I’m call such function as the logic probability function.
1.5.5. Conditional probability
Definition 1.6.5.1: Conditional probability B for C is the following function:
b(B/C) :=
b(C ·B)
b(C)
. (1.130)
Theorem 1.6.5.1 The conditional probability function is a B-function.
Proof of Theorem 1.6.5.1 From Definition 1.6.5.1:
b(C/C) =
b(C ·C)
b(C)
.
Hence by Theorem 1.1.1:
b(C/C) =
b(C)
b(C)
= 1.
Form Definition 1.6.5.1:
b((A ·B)/C)+b((A ·(#B))/C) =
b(C ·(A ·B))
b(C)
+
b(C ·(A ·(#B)))
b(C)
.
Hence:
b((A ·B)/C)+b((A ·(#B))/C) =
b(C ·(A ·B))+b(C ·(A ·(#B)))
b(C)
.
By Theorem 1.1.1:
b((A ·B)/C)+b((A ·(#B))/C) =
b((C ·A)·B)+b((C ·A)·(#B))
b(C)
.
Hence by Definition 1.6.2.1:
b((A ·B)/C)+b((A ·(#B))/C) =
b(C ·A)
b(C)
.
Hence by Definition 1.6.5.1:
b((A ·B)/C)+b((A ·(#B))/C) = b(A/C)
58](https://image.slidesharecdn.com/fn-190805113833/85/TOE-72-320.jpg)
![Moreover in accordance with (1.120): P(#A) = 1 − P(A) since the function P is a B-
function.
If event A occurs then (A·B) = B and (A·(#B)) = (#B) Hence, P(A·B)+P(A·(#B)) =
P(A) = P(B)+P(#B) = 1.
Consequently, if an element A of B occurs then P(A) = 1. If does not occurs then (#A)
occurs. Hence, P(#A) = 1 and because P(A) + P(#A) = 1 then P(A) = 0 . Therefore, on
B the range of values of is the two-element set {0;1} similar the Boolean function range
of values. Hence, on set B the probability function obeys definition of a Boolean function
(Def.1.1.10).
The logic probability function is the extension of the logic B-function. Therefore, the
probability is some generalization of the classic propositional logic. That is the proba-
bility is the logic of events such that these events do not happen, yet.
1.5.8. THE NONSTANDARD NUMBERS
Here some modification of the Robinson12 NONSTANDARD NUMBERS [30] is con-
sidered.
Let us consider the set N of natural numbers.
Definition A.1: The n-part-set S of N is defined recursively as follows:
1) S1 = {1};
2) S(n+1) = Sn ∪{n+1}.
Definition A.2: If Sn is the n-part-set of N and A ⊆ N then A∩Sn is the quantity
elements of the set A∩Sn, and if
ϖn (A) =
A∩Sn
n
,
then ϖn (A) is the frequency of the set A on the n-part-set Sn.
Theorem A.1:
12Abraham Robinson (born October 6, 1918 April 11, 1974) was a mathematician who is most widely
known for development of non-standard analysis
60](https://image.slidesharecdn.com/fn-190805113833/85/TOE-74-320.jpg)
![1) ϖn(N) = 1;
2) ϖn(/0) = 0;
3) ϖn(A)+ϖn(N−A) = 1;
4) ϖn(A∩B)+ϖn(A∩(N−B)) = ϖn(A).
Proof of the Theorem A.1: From Definitions A.1 and A.2.
Definition A.3: If ”lim” is the Cauchy-Weierstrass ”limit” then let us denote:
ix = A ⊆ N| lim
n→∞
ϖn(A) = 1 .
Theorem A.2: ix is the filter [29], i.e.:
1) N ∈ ix,
2) /0 /∈ ix,
3) if A ∈ ix and B ∈ ix then (A∩B) ∈ ix ;
4) if A ∈ ix and A ⊆ B then B ∈ ix.
Proof of the Theorem A.2: From the point 3 of Theorem A.1:
lim
n→∞
ϖn(N−B) = 0.
From the point 4 of Theorem A.1:
ϖn(A∩(N−B)) ≤ ϖn(N−B).
Hence,
lim
n→∞
ϖn (A∩(N−B)) = 0.
Hence,
lim
n→∞
ϖn (A∩B) = lim
n→∞
ϖn(A).
In the following text we shall adopt to our topics the definitions and the proofs of the
Robinson Nonstandard Analysis [31]:
Definition A.4: The sequences of the real numbers rn and sn are Q-equivalent (de-
note: rn ∼ sn ) if
{n ∈ N|rn = sn} ∈ ix.
Theorem A.3: If r,s,u are the sequences of the real numbers then
1) r ∼ r,
2) if r ∼ s then s ∼ r;
3) if r ∼ s and s ∼ u then r ∼ u.
Proof of the Theorem A.3: By Definition A.4 from the Theorem A.2 is obvious.
Definition A.5: The Q-number is the set of the Q-equivalent sequences of the real
numbers, i.e. if a is the Q-number and r ∈ a and s ∈ a, then r ∼ s; and if r ∈ a and r ∼ s
then s ∈ a.
Definition A.6: The Q-number a is the standard Q-number a if a is some real number
and the sequence rn exists, for which: rn ∈ a and
61](https://image.slidesharecdn.com/fn-190805113833/85/TOE-75-320.jpg)
![Therefore, for all sequences tn of real numbers: if tn ∈ u then by Definition A.5:
tn ∼ ψ(xn,yn,zn) .
Hence, tn ∈ v; and if tn ∈ v then tn ∼ ϕ(xn,yn,zn) ; hence, tn ∈ u.
Therefore, u = v.
Theorem A.7: If for all real numbers a, b, c:
f(a,ϕ(b,c)) = ψ(a,b,c)
then for all Q-numbers x, y, z:
f(x,ϕ(y,z)) = ψ(x,y,z).
Consequences from Theorems A.6 and A.7: [32]: For all Q-numbers x, y, z:
1: (x+y) = (y+x),
2: (x+(y+z)) = ((x+y)+z),
3: (x+0) = x,
5: (x·y) = (y·x),
6: (x·(y·z)) = ((x·y)·z),
7: (x·1) = x,
10: (x·(y+z)) = ((x·y)+(x·z)).
Proof of the Theorem A.7: Let wn ∈ w, f(x,w) = u, xn ∈ x, yn ∈ y, zn ∈ z,
ϕ(y,z) = w, ψ(x,y,z) = v.
By the condition of this Theorem: f(xn,ϕ(yn,zn)) = ψ(xn,yn,zn).
By Definition A.9: ψ(xn,yn,zn) ∈ v, ϕ(xn,yn) ∈ w, f(xn,wn) ∈ u.
For all sequences tn of real numbers:
1) If tn ∈ v then by Definition A.5: tn ∼ ψ(xn,yn,zn) .
Hence tn ∼ f(xn,ϕ(yn,zn)) .
Therefore, by Definition A.4:
{n ∈ N|tn = f(xn,ϕ(yn,zn))} ∈ ix
and
{n ∈ N|wn = ϕ(yn,zn)} ∈ ix.
Hence, by Theorem A.2:
{n ∈ N|tn = f(xn,wn)} ∈ ix.
Hence, by Definition A.4:
tn ∼ f(xn,wn) .
Therefore, by Definition A.5: tn ∈ u.
2) If tn ∈ u then by Definition A.5: tn ∼ f(xn,wn) .
Because wn ∼ ϕ(yn,zn) then by Definition A.4:
{n ∈ N|tn = f(xn,wn)} ∈ ix,
64](https://image.slidesharecdn.com/fn-190805113833/85/TOE-78-320.jpg)
![{n ∈ N|wn = ϕ(yn,zn)} ∈ ix.
Therefore, by Theorem A.2:
{n ∈ N|tn = f(xn,ϕ(yn,zn))} ∈ ix.
Hence, by Definition A.4:
tn ∼ f(xn,ϕ(yn,zn)) .
Therefore,
tn ∼ ψ(xn,yn,zn) .
Hence, by Definition A.5: tn ∈ v.
From above and from 1) by Definition A.7: u = v.
Theorem A.8: 4: For every Q-number x the Q-number y exists, for which:
(x+y) = 0.
Proof of the Theorem A.8: If xn ∈ x then y is the Q-number, which contains −xn .
Theorem A.9: 9: There is not that 0 = 1.
Proof of the Theorem A.9: is obvious from Definition A.6 and Definition A.7.
Definition A.10: The Q-number x is Q-less than the Q-number y (denote: x < y) if the
sequences xn and yn of real numbers exist, for which: xn ∈ x, yn ∈ y and
{n ∈ N|xn < yn} ∈ ix.
Theorem A.10: For all Q-numbers x, y, z: [33]
1: there is not that x < x;
2: if x < y and y < z then x < z;
4: if x < y then (x+z) < (y+z);
5: if 0 < z and x < y, then (x·z) < (y·z);
3 : if x < y then there is not, that y < x or x = y and vice versa;
3 : for all standard Q-numbers x, y, z: x < y or y < x or x = y.
Proof of the Theorem A.10: is obvious from Definition A.10 by the Theorem A.2.
Theorem A.11: 8: If 0 < |x| then the Q-number y exists, for which (x·y) = 1.
Proof of the Theorem A.11: If xn ∈ x then by Definition A.10: if
A = {n ∈ N|0 < |xn|}
then A ∈ ix.
In this case: if for the sequence yn : if n ∈ A then yn = 1/xn
- then
{n ∈ N|xn ·yn = 1} ∈ ix.
Thus, Q-numbers are fulfilled to all properties of real numbers, except Ω3 [34]. The
property Ω3 is accomplished by some weak meaning (Ω3’ and Ω3”).
65](https://image.slidesharecdn.com/fn-190805113833/85/TOE-79-320.jpg)
![Definition A.11: The Q-number x is the infinitesimal Q-number if the sequence of real
numbers xn exists, for which: xn ∈ x and for all positive real numbers ε:
{n ∈ N||xn| < ε} ∈ ix.
Let the set of all infinitesimal Q-numbers be denoted as I.
Definition A.12: The Q-numbers x and y are the infinite closed Q-numbers (denote:
x ≈ y) if |x−y| = 0 or |x−y| is infinitesimal.
Definition A.13: The Q-number x is the infinite Q-number if the sequence rn of real
numbers exists, for which rn ∈ x and for every natural number m:
{n ∈ N|m < rn} ∈ ix.
1.5.9. Model
Let us define the propositional calculus like to ([28]), but the propositional forms shall be
marked by the script greek letters.
Definition C1: A set ℜ of the propositional forms is a U-world if:
1) if α1,α2,...,αn ∈ ℜ and α1,α2,...,αn β then β ∈ ℜ,
2) for all propositional forms α: it is not that (α&(¬α)) ∈ ℜ,
3) for every propositional form α: α ∈ ℜ or (¬α) ∈ ℜ.
Definition C2: The sequences of the propositional forms αn and βn are Q-
equivalent (denote: αn ∼ βn ) if
{n ∈ N|αn ≡ βn} ∈ ix.
Let us define the notions of the Q-extension of the functions for like as in the Definitions
A.5, A.2, A.9, A.5, A.6.
Definition C3: The Q-form α is Q-real in the U-world ℜ if the sequence αn of the
propositional forms exists, for which: αn ∈ α and
{n ∈ N|αn ∈ ℜ} ∈ ix.
Definition C4: The set ℜ of the Q-forms is the Q-extension of the U-world ℜ if ℜ is
the set of Q-forms α, which are Q-real in ℜ.
Definition C5: The sequence ℜk of the Q-extensions is the S-world.
Definition C6: The Q-form α is S-real in the S-world ℜk if
k ∈ N|α ∈ ℜk ∈ ix.
Definition C7: The set A (A ⊆ N) is the regular set if for every real positive number ε
the natural number n0 exists, for which: for all natural numbers n and m, which are more or
equal to n0:
|wn(A)−wm(A)| < ε.
Theorem C1: If A is the regular set and for all real positive ε:
66](https://image.slidesharecdn.com/fn-190805113833/85/TOE-80-320.jpg)
![Chapter 2
Quants
If quantum physics did not frighten you, it
means that you did not understand
anything in it.
Niels Bohr
Quantum theory evolved as a new branch of theoretical physics during the first few
decades of the 20th century in an endeavour to understand the fundamental properties of
matter. It began with the study of the interactions of matter and radiation. Certain radiation
effects could neither be explained by classical mechanics, nor by the theory of electromag-
netism.
Quantum theory was not the work of one individual, but the collaborative effort of
some of the most brilliant physicists of the 20th century, among them Niels Bohr1, Erwin
Schrodinger2, Wolfgang Pauli3, and Max Born4, Max Planck5 and Werner Heisenberg6.
Quantum Field Theory (QFT) is the mathematical and conceptual framework for
contemporary elementary particle physics (Eugene Wigner7, Hans Bethe8, Tomonaga9,
Schwinger10, Feynman11, Dyson12, Yang13 and Mills14).
1Niels Henrik David Bohr (7 October 1885 - 18 November 1962) was a Danish physicist
2Erwin Rudolf Josef Alexander Schrodinger (12 August 1887 - 4 January 1961) was an Austrian physicist
and theoretical biologist who was one of the fathers of quantum mechanics
3Wolfgang Ernst Pauli (25 April 1900 15 December 1958) was an Austrian theoretical physicist
4Max Born (11 December 1882 5 January 1970) was a German-born physicist and mathematician
5Max Karl Ernst Ludwig Planck (April 23, 1858 October 4, 1947) was a German physicist
6Werner Karl Heisenberg (5 December 1901 1 February 1976) was a German theoretical physicist
7Eugene Paul Wigner (Hungarian Wigner Jeno Pal; November 17, 1902 - January 1, 1995) was a Hungarian
American physicist and mathematician.
8Hans Albrecht Bethe (July 2, 1906 - March 6, 2005) [1] was a German-American nuclear physicist,
9Sin-Itiro Tomonaga (March 31, 1906 July 8, 1979) was a Japanese physicist
10Julian Seymour Schwinger (February 12, 1918 - July 16, 1994) was an American theoretical physicist.
11Richard Phillips Feynman (May 11, 1918 - February 15, 1988)[2] was an American physicist
12Freeman John Dyson FRS (born December 15, 1923) is a British-born American theoretical physicist and
mathematician
13Chen-Ning Franklin Yang (born October 1, 1922) is a Chinese-American physicist
14Robert L. Mills (April 15, 1927 – October 27, 1999) was an English physicist](https://image.slidesharecdn.com/fn-190805113833/85/TOE-83-320.jpg)
![Probability, for which u2
1 +u2
2 +u2
3 ≤ c are called traceable probability.
Denote:
12 :=
1 0
0 1
, 02 :=
0 0
0 0
, β[0]
:= −
12 02
02 12
= −14,
the Pauli matrices
σ1 =
0 1
1 0
, σ2 =
0 −i
i 0
, σ3 =
1 0
0 −1
.
A setC of complex n×n matrices is called a Clifford set 15 of rank n [39] if the following
conditions are fulfilled:
if αk ∈ C and αr ∈ C then αkαr +αrαk = 2δk,r;
if αkαr +αrαk = 2δk,r for all elements αr of set C then αk ∈ C.
If n = 4 then a Clifford set either contains 3 matrices (a Clifford triplet) or contains 5
matrices (a Clifford pentad).
Here exist only six Clifford pentads [39]: one light pentad β:
β[1] :=
σ1 02
02 −σ1
, β[2] :=
σ2 02
02 −σ2
,
β[3] :=
σ3 02
02 −σ3
,
(2.1)
γ[0]
:=
02 12
12 02
, (2.2)
β[4]
:= i·
02 12
−12 02
; (2.3)
three chromatic pentads:
the red pentad ζ:
ζ[1]
=
−σ1 02
02 σ1
,ζ[2]
=
σ2 02
02 σ2
,ζ[3]
=
−σ3 02
02 −σ3
, (2.4)
γ
[0]
ζ
=
02 −σ1
−σ1 02
, ζ[4]
= i
02 σ1
−σ1 02
; (2.5)
the green pentad η:
η[1]
=
−σ1 02
02 −σ1
,η[2]
=
−σ2 02
02 σ2
,η[3]
=
σ3 02
02 σ3
, (2.6)
γ
[0]
η =
02 −σ2
−σ2 02
, η[4]
= i
02 σ2
−σ2 02
; (2.7)
15William Kingdon Clifford (4 May 1845 3 March 1879) was an English mathematician and philosopher.
71](https://image.slidesharecdn.com/fn-190805113833/85/TOE-85-320.jpg)
![the blue pentad θ:
θ[1]
=
σ1 02
02 σ1
,θ[2]
=
−σ2 02
02 −σ2
,θ[3]
=
−σ3 02
02 σ3
, (2.8)
γ
[0]
θ =
02 −σ3
−σ3 02
,θ[4]
= i
02 σ3
−σ3 02
; (2.9)
two gustatory pentads:
the sweet pentad ∆:
∆[1]
=
02 −σ1
−σ1 02
,∆[2]
=
02 −σ2
−σ2 02
,∆[3]
=
02 −σ3
−σ3 02
,
∆[0]
=
−12 02
02 12
,∆[4]
= i
02 12
−12 02
;
the bitter pentad Γ:
Γ[1]
= i
02 −σ1
σ1 02
,Γ[2]
= i
02 −σ2
σ2 02
,Γ[3]
= i
02 −σ3
σ3 02
,
Γ[0]
=
−12 02
02 12
,Γ[4]
=
02 12
12 02
.
Further we do not consider gustatory pentads since these pentads are not used yet in the
contemporary physics.
Let κ :=
3
∑
s=0
β[s]xs.
If
U0,1 (σ) :=
coshσ −sinhσ 0 0
−sinhσ coshσ 0 0
0 0 coshσ sinhσ
0 0 sinhσ coshσ
(2.10)
then
U†
0,1 (σ)κU0,1 (σ) =
β[0]
(x0 cosh2σ+x1 sinh2σ)
+β[1]
(x1 cosh2σ+x0 sinh2σ)
+β[2]
x2 +β[3]
x3.
Hence, U0,1 makes the Lorentz transformaton x0,x1 :
x0 → x0 := x0 cosh2σ+x1 sinh2σ,
x1 → x1 := x1 cosh2σ+x0 sinh2σ,
x2 → x2 := x2,
x3 → x3 := x3.
72](https://image.slidesharecdn.com/fn-190805113833/85/TOE-86-320.jpg)
![Similarly,
U0,2 (φ) :=
coshφ isinhφ 0 0
−isinhφ coshφ 0 0
0 0 coshφ −isinhφ
0 0 isinhφ coshφ
(2.11)
makes the Lorentz transformaton x0,x2 and
U0,3 (ι) :=
eι 0 0 0
0 e−ι 0 0
0 0 e−ι 0
0 0 0 eι
(2.12)
makes the Lorentz transformaton x0,x3 .
If
U1,3 (ϑ) :=
cosϑ sinϑ 0 0
−sinϑ cosϑ 0 0
0 0 cosϑ sinϑ
0 0 −sinϑ cosϑ
(2.13)
then U1,3 (ϑ) makes the cartesian turn x1,x3 :
U†
1,3 (ϑ)κU1,3 (ϑ)
= β[0]
x0
+β[1]
(x1 cos2ϑ+x3 sin2ϑ)
+β[2]
x2
+β[3]
(x3 cos2ϑ−x1 sin2ϑ)
Similarly,
U1,2 (ς) :=
e−iς 0 0 0
0 eiς 0 0
0 0 e−iς 0
0 0 0 eiς
(2.14)
makes the cartesian turn x1,x2 and
U2,3 (α) :=
cosα isinα 0 0
isinα cosα 0 0
0 0 cosα isinα
0 0 isinα cosα
(2.15)
makes the cartesian turn x2,x3 .
Let us consider the following set of four real equations with eight real unknowns: b2
with b > 0, α, β, χ, θ, γ, υ, λ:
73](https://image.slidesharecdn.com/fn-190805113833/85/TOE-87-320.jpg)
![6. Let us raise to the second power the second equation and add this equation to the
previous one:
sin2
(2β)cos2 (2α) = −u1
c
2
+ −u2
c
2
,
cos2 (2α)cos2 (2β) = −u3
c
2
sin2
(2β)+cos2
(2β) cos2
(2α) = −
u1
c
2
+ −
u2
c
2
+ −
u3
c
2
,
cos2
(2α) = −
u1
c
2
+ −
u2
c
2
+ −
u3
c
2
, (2.17)
We receive cos2 (2α) (for a trackeable probabilities).
7. From
cos2
(2α)cos2
(2β) = −
u3
c
2
we receive cos2 (2β).
8. From
cos2
(2α)cos2
(θ−γ)sin2
(2β) = −
u1
c
2
we receive cos2 (θ−γ).
———————————————-
If
ϕ1 := bexp(iγ)cos(β)cos(α),
ϕ2 := bexp(iθ)sin(β)cos(α),
ϕ3 := bexp(iλ)cos(χ)sin(α), (2.18)
ϕ4 := bexp(iυ)sin(χ)sin(α)
then you can calculate that
ρ =
4
∑
s=1
ϕ∗
s ϕs, (2.19)
jα
c
= −
4
∑
k=1
4
∑
s=1
ϕ∗
s β
[α]
s,kϕk
2.2. Entanglement
The presence in our universe of Planck’s constant gives reason to presume that our world is
in a confined space: |x| ≤ πc/h. Indeed, the eigenvalues of physical operators are multiples
of Planck constant integers. Consequently, the basic eigenfunctions of these operators con-
stitute the discrete spectrum. Consequently. physical functions are representable by Fourier
series. Hence, functions
75](https://image.slidesharecdn.com/fn-190805113833/85/TOE-89-320.jpg)
![Because p and q are natural then shift
πc
h
−πc
h
dx·a∗
p
πc
h
eih
c p(x+r)
aq
πc
h
e−ih
c q(x+r)
= c∗
pcqeih
c (p−q)r
δp,q.
Hence,
πc
h
−πc
h
dx·a∗
p
πc
h
eih
c p(x+r)
aq
πc
h
e−i h
c q(x+r)
= c∗
pcp.
Therefore, here a scalar product is invariant for an inner shift.
2.3. Equations of moving
If ϕ := U0,2 (φ)ϕ then
ρ = ϕ †
ϕ = ϕ†
U†
0,2 (φ)U0,2 (φ)ϕ = ρcosh2φ+
j2
c
sinh2φ
and
j2
c
= −ϕ †
s β[2]
ϕk = −ϕ†
U†
0,2 (φ)β[2]
U0,2 (φ)ϕ =
j2
c
cosh2φ+ρsinh2φ.
Similarly U0,1 and U0,3 transform the 3+1 vector cρ,j by the Lorentz formulas and
U1,2, U1,3, U2,3 transform this vector by the cartesian formulas.
Because
∂j0
∂x0
=
∂4F
∂x0∂x1∂x2∂x3
= −
∂j1
∂x1
= −
∂j2
∂x2
=
∂ j3
∂x3
then (Continuity equation ):
∂ρ
∂x0
+
∂ j1
∂x1
+
∂ j2
∂x2
+
∂j3
∂x3
= 0 (2.20)
In that case:
∂ ϕ†ϕ
∂x0
−
∂ ϕ†β[1]ϕ
∂x1
−
∂ ϕ†β[2]ϕ
∂x2
−
∂ ϕ†β[1]ϕ
∂x3
= 0
∂ ϕ†
∂x0
ϕ+ϕ† ∂(ϕ)
∂x0
−
∂ ϕ†β[1]
∂x1
ϕ−ϕ† ∂ β[1]ϕ
∂x1
−
∂ ϕ†β[2]
∂x2
ϕ−ϕ† ∂ β[2]ϕ
∂x2
−
∂ ϕ†β[3]
∂x3
ϕ−ϕ† ∂ β[3]ϕ
∂x3
= 0
78](https://image.slidesharecdn.com/fn-190805113833/85/TOE-92-320.jpg)
![ϕ† ∂
∂x0
−β[1] ∂
∂x1
−β[2] ∂
∂x2
−β[3] ∂
∂x3
†
ϕ
+ϕ† ∂
∂x0
−β[1] ∂
∂x1
−β[2] ∂
∂x2
−β[3] ∂
∂x3
ϕ
= 0
Let
Q :=
∂
x0
−
3
∑
s=1
β[s] ∂
xs
(2.21)
Hence,
ϕ†
Q†
+Q ϕ = 0
Q†
= −Q (2.22)
Therefore, for every function ϕj here exists an operator Qj,k such that a dependence of
ϕj on t is described by the following differential equations 16:
∂tϕj = c
4
∑
k=1
β
[1]
j,k∂1 +β
[2]
j,k∂2 +β
[3]
j,k∂3 +Qj,k ϕk. (2.23)
and Q∗
j,k = −Qk,j.
In that case if
Hj,k := ic β
[1]
j,k∂1 +β
[2]
j,k∂2 +β
[3]
j,k∂3 +Qj,k
then H is called a Hamiltonian17 of a moving with equation (2.23).
A matrix form of formula (2.23) is the following:
∂tϕ = c β[1]
∂1 +β[2]
∂2 +β[3]
∂3 +Q ϕ (2.24)
with
ϕ =
ϕ1
ϕ2
ϕ3
ϕ4
and
Q =
iϑ1,1 iϑ1,2 −ϖ1,2 iϑ1,3 −ϖ1,3 iϑ1,4 −ϖ1,4
iϑ1,2 +ϖ1,2 iϑ2,2 iϑ2,3 −ϖ2,3 iϑ2,4 −ϖ2,4
iϑ1,3 +ϖ1,3 iϑ2,3 +ϖ2,3 iϑ3,3 iϑ3,4 −ϖ3,4
iϑ1,4 +ϖ1,4 iϑ2,4 +ϖ2,4 iϑ3,4 +ϖ3,4 iϑ4,4
(2.25)
16This set of equations is similar to the Dirac equation with the mass matrix [40], [41], [42]. I choose a form
of this set of equations in order to describe the behavior of ρ℘(t,x) by spinors and by Clifford’s set elements.
17Sir William Rowan Hamilton (4 August 1805 2 September 1865) was an Irish physicist, astronomer, and
mathematician, who made important contributions to classical mechanics, optics, and algebra.
79](https://image.slidesharecdn.com/fn-190805113833/85/TOE-93-320.jpg)
![with ϖs,k = Re(Qs,k) and ϑs,k = Im(Qs,k). Matrix ϕ is called a state vector of the event
A probability.
An operator U (t,t0) with a domain and with a range of values on the set of state vectors
is called an evolution operator if each state vector ϕ fulfils the following condition:
ϕ(t) = U (t,t0)ϕ(t0). (2.26)
Let us denote:
Hd := c
3
∑
s=1
iβ[s]
∂s.
In that case
H = Hd +icQ
according the Hamiltonian definition:
H = ic β[1]
∂1 +β[2]
∂2 +β[3]
∂3 +Q .
From (2.24):
i∂tϕ = Hϕ.
.
Hence:
i∂tϕ = Hd +icQ ϕ.
This differential equation has the following solution:
∂ϕ
ϕ
= −i Hd +icQ ∂t,
t
t=t0
∂ϕ
ϕ
= −i
t
t=t0
Hd +icQ ∂t,
lnϕ(t)−lnϕ(t0) = −i
t
t=t0
Hd∂t −iic
t
t=t0
Q∂t .
Since Hd does not depend on time then
t
t=t0
Hd∂t = Hd (t −t0).
Hence, according logarithm properties:
ln
ϕ(t)
ϕ(t0)
= −iHd (t −t0)+c
t
t=t0
Q∂t .
80](https://image.slidesharecdn.com/fn-190805113833/85/TOE-94-320.jpg)
![An operator
K (t −t0,x−x0,t,t0) :=
h
2πc
3
∑p exp −iHd (t −t0)+c t
t=t0
Q∂t ·
·exp −ih
c p(x−x0)
is called propagator of the event A probability.
Hence:
ϕ(t,x) =
(Ω)
dx0 ·K (t −t0,x−x0,t,t0)ϕ(t0,x0). (2.27)
A propagator has the following property:
K (t −t0,x−x0,t,t0) = dx1 ·K (t −t1,x−x1,t,t1)K (t1 −t0,x1−x0,t1,t0).
2.4. Double-Slit Experiment
In a vacuum (Figure 1, Figure 2, Figure 3): Here transmitter s of electrons, wall w, and the
electrons detecting black screen d are placed[45].
Electrons are emitted one by one from the source s. When an electron hits against
screen d then a bright spot arises in the hit place of d..
1. Let slit a be opened in wall w (Figure 1). An electron flies out from s, passes by a,
and is detected by d.
If such operation will be reiterated N of times then N bright spots shall arise on d against
slit a in the vicinity of point ya.
2. Let slit b be opened in wall w (Figure 2). An electron flies out from s, passes by b,
and is detected by d.
If such operation will be reiterated N of times then N bright spots shall arise on d against
slit b in the vicinity of point yb.
3. Let both slits be opened. In that case do you expect a result as on fig. 3? But no. We
get result as on Figure 419[46].
For instance, such experiment was realized at Hitachi by A. Tonomura, J. Endo, T. Mat-
suda, T. Kawasaki and H. Ezawa in 1989. Here was presumed that interference fringes are
produced only when two electrons pass through both slits simultaneously. If there were
two electrons from the source s at the same time, such interference might happen. But
this cannot occur, because here is no more than one electron from this source at one time.
Please keep watching the experiment a little longer. When a large number of electrons is
accumulated, something like regular fringes begin to appear in the perpendicular direction
as Figure 5(c) shows. Clear interference fringes can be seen in the last scene of the ex-
periment after 20 minutes (Figure 5(d)). It should also be noted that the fringes are made
19Single-electron events build up over a 20 minute exposure to form an interference pattern in this double-
slit experiment by Akira Tonomura and co-workers. Figure 5(a) 8 electrons; Figure 5(b) 270 electrons; Fig-
ure 5(c) 2000 electrons; Figure 5(d) 60,000. A video of this experiment will soon be available on the web
(www.hqrd.hitachi.co.jp/em/doubleslit.html).
82](https://image.slidesharecdn.com/fn-190805113833/85/TOE-96-320.jpg)
![the third and the fourth columns:
iM0 +iMθ,0 −M4 −Mθ,4 iMζ,0 −iMη,4 −Mζ,4 +Mη,0
iMζ,0 +iMη,4 −Mζ,4 −Mη,0 iM0 −iMθ,0 −M4 +Mθ,4
−iΘ0 −iΘ3 +iϒ0 +iϒ3 −iΘ1 +iϒ1 −Θ2 +ϒ2
−iΘ1 +iϒ1 +Θ2 −ϒ2 −iΘ0 +iΘ3 +iϒ0 −iϒ3
.
Hence,
Q =
= iΘ0β[0] +iϒ0β[0]γ[5]+
+iΘ1β[1] +iϒ1β[1]γ[5]+
+iΘ2β[2] +iϒ2β[2]γ[5]+
+iΘ3β[3] +iϒ3β[3]γ[5]+
+iM0γ[0] +iM4β[4]−
−iMζ,0γ
[0]
ζ
+iMζ,4ζ[4]−
−iMη,0γ
[0]
η −iMη,4η[4]+
+iMθ,0γ
[0]
θ +iMθ,4θ[4].
Therefore, from (2.24):
1
c
∂tϕ− iΘ0β[0]
+iϒ0β[0]
γ[5]
ϕ =
3
∑
ν=1
β[ν] ∂ν +iΘν +iϒνγ[5] +
+iM0γ[0] +iM4β[4]−
−iMζ,0γ
[0]
ζ
+iMζ,4ζ[4]−
−iMη,0γ
[0]
η −iMη,4η[4]+
+iMθ,0γ
[0]
θ +iMθ,4θ[4]
ϕ. (2.29)
with
γ[5]
:=
12 02
02 −12
. (2.30)
Because
ζ[k]
+η[k]
+θ[k]
= −β[k]
93](https://image.slidesharecdn.com/fn-190805113833/85/TOE-107-320.jpg)
![with k ∈ {1,2,3} then from (2.29):
− ∂0 +iΘ0 +iϒ0γ[5] +
3
∑
k=1
β[k] ∂k +iΘk +iϒkγ[5]
+2 iM0γ[0] +iM4β[4]
ϕ+
+
− ∂0 +iΘ0 +iϒ0γ[5] −
3
∑
k=1
ζ[k] ∂k +iΘk +iϒkγ[5]
+2 −iMζ,0γ
[0]
ζ
+iMζ,4ζ[4]
ϕ+
+
∂0 +iΘ0 +iϒ0γ[5] −
3
∑
k=1
η[k] ∂k +iΘk +iϒkγ[5]
+2 −iMη,0γ
[0]
η −iMη,4η[4]
ϕ+
+
− ∂0 +iΘ0 +iϒ0γ[5] −
3
∑
k=1
θ[k] ∂k +iΘk +iϒkγ[5]
+2 iMθ,0γ
[0]
θ +iMθ,4θ[4]
ϕ = 0.
In (2.29) summands
−iMζ,0γ
[0]
ζ
+iMζ,4ζ[4]−
−iMη,0γ
[0]
η −iMη,4η[4]+
+iMθ,0γ
[0]
θ +iMθ,4θ[4]
contain elements of chromatic pentads and
3
∑
k=1
β[k]
∂k +iΘk +iϒkγ[5]
+iM0γ[0]
+iM4β[4]
contains only elements of the light pentads. The following sum
Hl := c
3
∑
k=1
β[k]
i∂k −Θk −ϒkγ[5]
−cM0γ[0]
−cM4β[4]
(2.31)
is called lepton Hamiltonian.
And the following equation:
3
∑
k=0
β[k]
i∂k −Θk −ϒkγ[5]
−M0γ[0]
−M4β[4]
ϕ = 0 (2.32)
is called lepton moving equation
If like to (2.19):
ϕ†γ[0]ϕ := − j5
c and ϕ†β[4]ϕ := − j4
c
and:
ρu4 := j4 and ρu5 := j5 (2.33)
94](https://image.slidesharecdn.com/fn-190805113833/85/TOE-108-320.jpg)
![then from (2.18):
−
u5
c
= sin2α
sinβsinχcos(−θ+υ)
+cosβcosχcos(γ−λ)
,
−
u4
c
= sin2α
−sinβsinχsin(−θ+υ)
+cosβcosχsin(γ−λ)
.
Hence, from (2.16):
u2
1 +u2
2 +u2
3 +u2
A,4 +u2
5 = c2
.
Thus, of only all five elements of a Clifford pentad lends an entire kit of velocity com-
ponents and, for completeness, yet two ”space” coordinates x5 and x4 should be added to
our three x1,x2,x3. These additional coordinates can be selected such that
−
πc
h
≤ x5 ≤
πc
h
,−
πc
h
≤ x4 ≤
πc
h
.
Coordinates x4 and x5 are not of any events coordinates. Hence, our devices do not
detect of its as space coordinates.
Let us denote:
ϕ(t,x1,x2,x3,x5,x4) := ϕ(t,x1,x2,x3)·
·(exp(i(x5M0 (t,x1,x2,x3)+x4M4 (t,x1,x2,x3)))).
In this case equation of moving with lepton Hamiltonian (2.31) shape is the following:
3
∑
k=0
β[0]
i∂k −Θk −ϒkγ[5]
−γ[0]
i∂5 −β[4]
i∂4 ϕ = 0 (2.34)
Let g1 be the positive real number and for µ ∈ {0,1,2,3}: Fµ and Bµ be the solutions of
the following system of the equations:
−0.5g1Bµ +Fµ= −Θµ −ϒµ;
−g1Bµ +Fµ= −Θµ +ϒµ.
Let charge matrix be denoted as the following:
Y := −
12 02
02 2·12
. (2.35)
Thus20:
20If products ABj,s exist for all j, s then
A
B0,0 B0,1 ··· B0,n
B1,0 B1,1 ··· B1,n
··· ··· ··· ···
Bm,0 Bm,1 ··· Bm,n
:=
AB0,0 AB0,1 ··· AB0,n
AB1,0 AB1,1 ··· AB1,n
··· ··· ··· ···
ABm,0 ABm,1 ··· ABm,n
and if products Bj,sA exist for all j, s then
95](https://image.slidesharecdn.com/fn-190805113833/85/TOE-109-320.jpg)
![−Θµ −ϒµγ[5]
=
= −Θµ14 −ϒµγ[5]
=
= −Θµ
12 02
02 12
−ϒµ
12 02
02 −12
=
= −
Θµ12 02
02 Θµ12
+
ϒµ12 02
02 −ϒµ12
=
=
(−Θµ −ϒµ)12 02
02 (−Θµ +ϒµ)12
=
=
(−0.5g1Bµ +Fµ)12 02
02 (−g1Bµ +Fµ)12
.
And
B0,0 B0,1 ··· B0,n
B1,0 B1,1 ··· B1,n
··· ··· ··· ···
Bm,0 Bm,1 ··· Bm,n
A :=
B0,0A B0,1A ··· B0,nA
B1,0A B1,1A ··· B1,nA
··· ··· ··· ···
Bm,0A Bm,1A ··· Bm,nA
. (2.36)
If A and all Bj,s are k ×k matrices then
A+
B0,0 B0,1 B0,2 ··· B0,n
B1,0 B1,1 B1,2 ··· B1,n
B2,0 B2,1 B2,2 ··· B2,n
··· ... ··· ··· ···
Bn,0 Bn,1 Bn,2 ··· Bn,n
:=
:= A1nk +
B0,0 B0,1 B0,2 ··· B0,n
B1,0 B1,1 B1,2 ··· B1,n
B2,0 B2,1 B2,2 ··· B2,n
··· ... ··· ··· ···
Bn,0 Bn,1 Bn,2 ··· Bn,n
=
=
B0,0 +A B0,1 B0,2 ··· B0,n
B1,0 B1,1 +A B1,2 ··· B1,n
B2,0 B2,1 B2,2 +A ··· B2,n
··· ... ··· ··· ···
Bn,0 Bn,1 Bn,2 ··· Bn,n +A
. (2.37)
96](https://image.slidesharecdn.com/fn-190805113833/85/TOE-110-320.jpg)
![Fµ +0.5g1YBµ =
= Fµ14 +0.5g1YBµ
= Fµ
12 02
02 12
+0.5g1 −
12 02
02 2·12
Bµ =
=
Fµ12 02
02 Fµ12
−
0.5g1Bµ12 02
02 0.5g1Bµ2·12
=
=
Fµ12 −0.5g1Bµ12 02
02 Fµ12 −g1Bµ ·12
.
Hence,
−Θµ −ϒµγ[5]
= Fµ +0.5g1YBµ
and from (2.34):
3
∑
k=0
β[k]
(i∂k +Fk +0.5g1YBk)−γ[0]
i∂5 −β[4]
i∂4 ϕ = 0 (2.38)
Let χ(t,x1,x2,x3) be the real function and:
U (χ) :=
exp iχ
2 12 02
02 exp(iχ)12
. (2.39)
In that case for µ ∈ {0,1,2,3}:
∂µU = ∂µ
exp iχ
2 12 02
02 exp(iχ)12
=
∂µ exp iχ
2 12 ∂µ02
∂µ02 ∂µ exp(iχ)12
=
i
∂µχ
2 exp iχ
2 12 02
02 i∂µχexp(iχ)12
= i
∂µχ
2
exp iχ
2 12 02
02 2exp(iχ)12
,
and
YU = −
12 02
02 2·12
exp iχ
2 12 02
02 exp(iχ)12
= −
exp iχ
2 12 02
02 2exp(iχ)12
.
Hence:
97](https://image.slidesharecdn.com/fn-190805113833/85/TOE-111-320.jpg)
![∂µU = −i
∂µχ
2
YU. (2.40)
Moreover you can calculate that:
U†
γ[0]
U = γ[0]
cos
χ
2
+β[4]
sin
χ
2
,
U†
β[4]
U = β[4]
cos
χ
2
−γ[0]
sin
χ
2
,
U†
U = 14,
U†
YU = Y,
β[k]
U = Uβ[k]
for k ∈ {0,1,2,3}
Let
x4 = x4 cos
χ
2
−x5 sin
χ
2
,
x5 = x5 cos
χ
2
+x4 sin
χ
2
.
In that case by the partial derivate definition for any function u:
∂4u = ∂4u·∂4x4 +∂5u·∂4x5 = ∂4u·cos
χ
2
+∂5u·sin
χ
2
, (2.41)
∂5u = ∂4u·∂5x4 +∂5u·∂5x5 = ∂4u· −sin
χ
2
+∂5u·cos
χ
2
.
Let ∂4χ = 0 and ∂5χ = 0; hence, ∂4U = U∂4 and ∂5U = U∂5.
From (2.38):
3
∑
s=0
β[s]
(i∂s +Fs +0.5g1YBs)−γ[0]
i∂5 −β[4]
i∂4 ϕ = 0. (2.42)
Let
Bµ = Bµ −
1
g1
∂µχ.
According to (2.41) and since U†U = 14 and U†YU = Y then
∑3
s=0 β[s] i∂s +Fs +0.5g1U†YU Bs + 1
g1
∂sχ −
−γ[0]i −sin χ
2 ∂4 +cos χ
2 ∂5 −β[4]i cos χ
2 ∂4 +sin χ
2 ∂5
U†
Uϕ = 0.
Hence:
∑3
s=0 β[s] i∂s +Fs +0.5g1U†YU Bs + 1
g1
∂sχ −
− −γ[0] sin χ
2 +β[4] cos χ
2 i∂4 − γ[0] cos χ
2 +β[4] sin χ
2 i∂5
U†
Uϕ = 0.
98](https://image.slidesharecdn.com/fn-190805113833/85/TOE-112-320.jpg)
![Since U is a linear operator then
∑3
s=0 β[s] i∂s +Fs +0.5g1U†YU Bs + 1
g1
∂sχ U†−
− −γ[0] sin χ
2 +β[4] cos χ
2 i∂4U† − γ[0] cos χ
2 +β[4] sin χ
2 i∂5U†
Uϕ = 0
and since ∂4U = U∂4 and ∂5U = U∂5 then
∑3
s=0 β[s] i∂sU† +FsU† +0.5g1U†YUU† Bs + 1
g1
∂sχ −
− −γ[0]U† sin χ
2 +β[4]U† cos χ
2 i∂4
− γ[0]U† cos χ
2 +β[4]U† sin χ
2 i∂5
Uϕ = 0. (2.43)
Since
U†
γ[0]
U = γ[0]
cos
χ
2
+β[4]
sin
χ
2
,
U†
β[4]
U = β[4]
cos
χ
2
−γ[0]
sin
χ
2
then
U†
γ[0]
UU†
= γ[0]
U†
cos
χ
2
+β[4]
U†
sin
χ
2
,
U†
β[4]
UU†
= β[4]
U†
cos
χ
2
−γ[0]
U†
sin
χ
2
,
Hence,
U†
γ[0]
= γ[0]
U†
cos
χ
2
+β[4]
U†
sin
χ
2
,
U†
β[4]
= β[4]
U†
cos
χ
2
−γ[0]
U†
sin
χ
2
.
Therefore,
γ[0]U† = U†γ[0] cos χ
2 −U†β[4] sin χ
2 ,
β[4]U† = U†γ[0] sin χ
2 +U†β[4] cos χ
2 .
Thus, from (2.43):
∑3
s=0 β[s] i∂sU† +FsU† +0.5g1U†YUU† Bs + 1
g1
∂sχ −
−
− U†γ[0] cos χ
2 −U†β[4] sin χ
2 sin χ
2
+ U†γ[0] sin χ
2 +U†β[4] cos χ
2 cos χ
2
i∂4
−
U†γ[0] cos χ
2 −U†β[4] sin χ
2 cos χ
2
+ U†γ[0] sin χ
2 +U†β[4] cos χ
2 sin χ
2
i∂5
Uϕ = 0.
99](https://image.slidesharecdn.com/fn-190805113833/85/TOE-113-320.jpg)
![Hence:
∑3
s=0 β[s] i∂sU† +FsU† +0.5g1U†Y Bs + 1
g1
∂sχ −
−U†β[4]i∂4 − U†γ[0]i∂5
Uϕ = 0. (2.44)
Since (2.40):
∂µU = −i
∂µχ
2
YU
then for s ∈ {0,1,2,3}:
∂sU†
= i
∂sχ
2
U†
Y†
= i
∂sχ
2
YU†
.
Therefore,
∂s U†Uϕ = ∂s U† Uϕ =
= ∂sU† Uϕ +U†∂s Uϕ = i∂sχ
2 YU† Uϕ +U†∂s Uϕ =
= i∂sχ
2 YU† +U†∂s Uϕ .
Since YU† = U†Y then
i
∂sχ
2
YU†
+U†
∂s = U†
i
∂sχ
2
Y +U†
∂s.
Hence,
i∂sU†
= −U† ∂sχ
2
Y +U†
i∂s.
Therefore, from (2.44):
∑3
s=0 β[s] −U† ∂sχ
2 Y +U†i∂s +FsU† +0.5g1U†Y Bs + 1
g1
∂sχ −
−U†β[4]i∂4 − U†γ[0]i∂5
Uϕ = 0.
Hence:
∑3
s=0 β[s] U†i∂s +U†Fs +0.5g1U†YBs −
−U†β[4]i∂4 − U†γ[0]i∂5
Uϕ = 0
with Fs := UFsU†.
Since β[s]U = Uβ[s] for s ∈ {0,1,2,3} then
∑3
s=0U†β[s] i∂s +U†Fs +0.5g1U†YBs −
−U†β[4]i∂4 − U†γ[0]i∂5
Uϕ = 0.
Hence, if denote: ϕ := Uϕ then since U is a linear operator then:
100](https://image.slidesharecdn.com/fn-190805113833/85/TOE-114-320.jpg)
![U†
3
∑
s=0
β[s]
i∂s +Fs +0.5g1YBs −β[4]
i∂4 − γ[0]
i∂5 ϕ = 0.
That is
3
∑
s=0
β[s]
i∂s +Fs +0.5g1YBs −β[4]
i∂4 − γ[0]
i∂5 ϕ = 0.
Compare with (2.42).
Thus, this Equation of moving is invariant under the following transformations:
x4 → x4 = x4 cos
χ
2
−x5 sin
χ
2
;
x5 → x5 = x5 cos
χ
2
+x4 sin
χ
2
;
xµ → xµ = xµ for µ ∈ {0,1,2,3}; (2.45)
ϕ → ϕ = Uϕ,
Bµ → Bµ = Bµ −
1
g1
∂µχ,
Fµ → Fµ = UFsU†
.
Therefore, Bµ is like to the B-boson field of Standard Model21 [44]. field.
2.6. Masses
Let
ε1 =
1
0
0
0
, ε2 =
0
1
0
0
, ε3 =
0
0
1
0
, ε4 =
0
0
0
1
. (2.46)
Functions of type :
h
2πc
exp −i
h
c
(sx4 +nx5) εk (2.47)
with an integer n and s form orthonormal basis of some unitary space ℑ with scalar product
of the following shape:
(ϕ,χ) :=
πc
h
−πc
h
dx5
πc
h
−πc
h
dx4 ·ϕ†
χ (2.48)
In that case from (2.19):
21Sheldon Lee Glashow (born December 5, 1932) is a American theoretical physicist.
101](https://image.slidesharecdn.com/fn-190805113833/85/TOE-115-320.jpg)
![(ϕ,ϕ) = ρA, (2.49)
ϕ,β[s]
ϕ = −
jk
c
.
for s ∈ {1,2,3}
Let22
Nϑ (t,x1,x2,x3) := trunc
cM0
h
, Nϖ (t,x1,x2,x3) := trunc
cM4
h
.
Hence, functions Nϑ (t,x1,x2,x3) and Nϖ (t,x1,x2,x3) have got integer values.
In that case to high precision:
ϕ = ϕ(t,x1,x2,x3)·exp −i x5
h
c
Nϑ (t,x1,x2,x3)+x4
h
c
Nϖ (t,x1,x2,x3)
and Fourier series for ϕ is of the following form:
ϕ(t,x1,x2,x3,x5,x4) = ϕ(t,x1,x2,x3)·∑
n,s
δ−n,Nϑ(t,x)δ−s,Nϖ(t,x) exp −i
h
c
(nx5 +sx4)
with
δ−n,Nϑ =
h
2πc
πc
h
−πc
h
exp i
h
c
(nx5) exp iNϑ
h
c
x5 dx5 =
sinπ(n+Nϑ)
π(n+Nϑ)
,
δ−s,Nϖ =
h
2πc
πc
h
−πc
h
exp i
h
c
(sx4) exp iNϖ
h
c
x4 dx4 =
sinπ(s+Nϖ)
π(s+Nϖ)
with integer n and s.
If denote:
f (t,x,−n,−s) := ϕ(t,x)δn,Nϑ(t,x)δs,Nϖ(t,x)
then
ϕ(t,x,x5,x4) =
= ∑n,s f (t,x,n,s)exp −ih
c (nx5 +sx4) .
(2.50)
The integer numbers n and s are denoted mass numbers.
From properties of δ: in every point t,x : either
ϕ(t,x,x5,x4) = 0
22Function trunc(x) returns the integer part of a real number x by removing the fractional part. For example:
trunc(−2.0857) = −2.
102](https://image.slidesharecdn.com/fn-190805113833/85/TOE-116-320.jpg)
![or integer numbers n0 and s0 exist for which:
ϕ(t,x,x5,x4) =
= f (t,x,n0,s0)exp −ih
c (n0x5 +s0x4) .
(2.51)
Here if
m0 := n2
0 +s2
0
and
m :=
h2
c2
m0 (2.52)
then m is denoted mass of ϕ.
That is for every space-time point: either this point is empty or single mass is placed in
this point.
Figure 7:
Equation of moving (2.38) under the transformation (2.45) has the following form:
∑n ,s
β[0] i1
c ∂t +Fµ +0.5g1YBµγ[5]
+∑3
µ=1 β[µ] i∂µ +Fµ +0.5g1YBµγ[5] +γ[0]i∂5 +β[4]i∂4
·
·exp −ih
c (n x5 +s x4) U f = 0
with:
n = ncos χ
2 −ssin χ
2 ,
s = nsin χ
2 +scos χ
2 .
But s and n are integer numbers and s and n must be integer numbers, too. .
103](https://image.slidesharecdn.com/fn-190805113833/85/TOE-117-320.jpg)
![Figure 8:
A triplet λ;n,s of integer numbers23 is called a Pythagorean triple [43] if
λ2
= n2
+s2
.
Let ε be the tiny positive real number. I call an integer number λ as a father number
with precise ε if for each real number χ and for every Pythagorean triple λ;n,s here exists
a Pythagorean triple λ;n ,s such that:
−ssin
χ
2
+ncos
χ
2
−n < λε,
scos
χ
2
+nsin
χ
2
−s < λε.
For example: number 325 is a father number for the following Pythagorean triples
(Figure 7):
23For each natural number n, there exist at least n different Pythagorean triples with the same hypotenuse.
Sierpinski, 2003, . 31. Dover, 2003.ISBN 978-0-486-43278-6. John F. Goehi Jr. TRIPLES, QUARTETS,
PENTADS :Mathematics teacher,ISSN0025-5769,Vol. 98, N? 9, 2005,pag.580
104](https://image.slidesharecdn.com/fn-190805113833/85/TOE-118-320.jpg)
![325;323,36 ,
325;315,80 ,
325;312,91 ,
325;300,125 ,
325;280,165 ,
325;260,195 ,
325;253,204 ,
Here ε is maximal ratio value of difference between adjacent s values to father number.
That is here ε = 253−204
325 = 0.15. But for any value of precise ε here exists a fitting father
number in long distant domain of the natural numerical line. But I can not calculate it since
more high-end machine than my computer is needed for such calculation.
The nearest-neighbors to 325 father numbers are numbers 333 and 337. But these father
numbers have got one at a time triple. Hence, fathers, having many ”children”, are isolated
numbers on the natural numerical line. I suspect that these numbers are fathers of particles
families.
Here are three families (generations) according to the Standard Model of particle
physics [44]:
νe
e−
νµ
µ−
ντ
τ−
u
d
c
s
t
b
.
Each generation is divided into two leptons:
νe
e− ,
νµ
µ− ,
ντ
τ− ,
and two quarks:
u
d
,
c
s
,
t
b
.
The two leptons may be divided into one electron-like (e− - electron, µ− - µ–lepton, τ−
- τ-lepton ) and neutrino (νe, νµ, ντ); the two quarks may be divided into one down-type (d,
s, b) and one up-type (u, c, t). The first generation consists of the electron, electron neutrino
and the down and up quarks. The second generation consists of the muon, muon neutrino
and the strange and charm quarks. The third generation consists of the tau lepton, tau
neutrino and the bottom and top quarks. Each member of a higher generation has greater
mass than the corresponding particle of the previous generation. For example: the first-
generation electron has a mass of only 0.511 MeV, the second-generation muon has a mass
of 106 MeV, and the third-generation tau lepton has a mass of 1777 MeV (almost twice
as heavy as a proton). All ordinary atoms are made of particles from the first generation.
Electrons surround a nucleus made of protons and neutrons, which contain up and down
105](https://image.slidesharecdn.com/fn-190805113833/85/TOE-119-320.jpg)
![quarks. The second and third generations of charged particles do not occur in normal matter
and are only seen in extremely high-energy environments. Neutrinos of all generations
stream throughout the universe but rarely interact with normal matter.
2.7. One-Mass State
Let form of (2.50) be the following:
ϕ(t,x,x5,x4) = exp −i
h
c
nx5
4
∑
k=1
fk (t,x,n,0)εk.
In that case the Hamiltonian has the following form (from (2.38)):
H = c
3
∑
k=1
β[k]
i∂k +
h
c
nγ[0]
+G
with
G :=
3
∑
µ=0
β[µ]
(Fµ +0.5g1YBµ).
Let
ω(k) := k2 +n2 = k2
1 +k2
2 +k2
3 +n2
and
e1 (k) :=
1
2 ω(k)(ω(k)+n)
ω(k)+n+k3
k1 +ik2
ω(k)+n−k3
−k1 −ik2
. (2.53)
Let
H0 := c
3
∑
s=1
β[s]
i∂s +hnγ[0]
. (2.54)
Since (2.52):
hn = m
c2
h
then equation of moving with Hamiltonian H0 has the following form:
1
c i∂tϕ = ∑3
s=1 β[s]i∂s +mc
h γ[0] ϕ. (2.55)
This is the Dirac equation (Paul Dirac24 formulated it in 1928).
Let us denote
24Paul Adrien Maurice Dirac (1902 – 1984) was an English theoretical physicist who made fundamental
contributions to the early development of both quantum mechanics and quantum electrodynamics.
106](https://image.slidesharecdn.com/fn-190805113833/85/TOE-120-320.jpg)
![γ[s]
:= γ[0]
β[s]
for s = 0.
Let us calculate:
γ[s]
γ[j]
+γ[j]
γ[s]
= γ[0]
β[s]
γ[0]
β[j]
+γ[0]
β[j]
γ[0]
β[s]
=
= −γ[0]
γ[0]
β[s]
β[j]
−γ[0]
γ[0]
β[j]
β[s]
=
= − β[s]
β[j]
+β[j]
β[s]
= −2δj,s
for s = 0 and j = 0.
and
γ[s]
γ[0]
+γ[0]
γ[s]
= γ[0]
β[s]
γ[0]
+γ[0]
γ[0]
β[s]
= −β[s]
+β[s]
= 0
for s = 0.
From (2.55):
1
c
iγ[0]
∂t −
3
∑
s=1
γ[s]
i∂s −m
c
h
ϕ = 0.
Let us multiply both parts of this equation on
1
c
iγ[0]
∂t −
3
∑
s =1
γ[s ]
i∂s +m
c
h
:
1
c
iγ[0]
∂t −
3
∑
s =1
γ[s ]
i∂s +m
c
h
1
c
iγ[0]
∂t −
3
∑
s=1
γ[s]
i∂s −m
c
h
ϕ = 0.
Hence,
− 1
c2 ∂2
t
−∑3
s=1
1
c iγ[0]∂tγ[s]i∂s −∑3
s =1 γ[s ]i∂s
1
c iγ[0]∂t
−1
c iγ[0]∂tmc
h +mc
h
1
c iγ[0]∂t
+∑3
s =1 γ[s ]i∂s ∑3
s=1 γ[s]i∂s
+∑3
s =1 γ[s ]i∂s mc
h −∑3
s=1 mc
h γ[s]i∂s
−m2c2
h2
ϕ = 0.
Hence,
− 1
c2 ∂2
t
+∑3
s =1 γ[s ]i∂s ∑3
s=1 γ[s]i∂s
−m2c2
h2
ϕ = 0
Since
107](https://image.slidesharecdn.com/fn-190805113833/85/TOE-121-320.jpg)
![3
∑
s =1
γ[s ]
i∂s
3
∑
s=1
γ[s]
i∂s
= −
3
∑
s=1
3
∑
s =1
γ[s ]
γ[s]
∂s ∂s =
= −
γ[1]γ[1]∂1∂1 +γ[2]γ[1]∂2∂1 +γ[3]γ[1]∂3∂1
+γ[1]γ[2]∂1∂2 +γ[2]γ[2]∂2∂2 +γ[3]γ[2]∂3∂2
+γ[1]γ[3]∂1∂3 +γ[2]γ[3]∂2∂3 +γ[3]γ[3]∂3∂3
=
= −
−∂1∂1
+γ[2]γ[1]∂2∂1 +γ[1]γ[2]∂1∂2
+γ[3]γ[1]∂3∂1 +γ[1]γ[3]∂1∂3
−∂2∂2
+γ[3]γ[2]∂3∂2 +γ[2]γ[3]∂2∂3
−∂3∂3
.
Hence,
3
∑
s =1
γ[s ]
i∂s
3
∑
s=1
γ[s]
i∂s = ∂1∂1 +∂2∂2 +∂3∂3 =
3
∑
s=1
∂2
s .
Thus,
− 1
c2 ∂2
t +∑3
s=1 ∂2
s − m2c2
h2 ϕ = 0.
(2.56)
This is the Klein-Gordon2526 equation for a free particle with mass m.
Let us calculate:
H0e1 (k) h
2πc
3
2
exp −ih
c kx =
= c∑3
s=1 β[s]i∂s +hnγ[0] h
2πc
3
2
e1 (k)exp −ih
c kx =
== c∑3
s=1 β[s]i∂se1 (k) h
2πc
3
2
exp −ih
c kx +
+hnγ[0]e1 (k) h
2πc
3
2
exp −ih
c kx =
= c∑3
s=1 β[s]ie1 (k)∂s
h
2πc
3
2
exp −ih
c kx +
+hn h
2πc
3
2
exp −ih
c kx γ[0]e1 (k) =
= c∑3
s=1 β[s]ie1 (k) −ih
c ks
h
2πc
3
2
exp −ih
c ks +
+hn h
2πc
3
2
exp −ih
c kx γ[0]e1 (k) =
= ∑3
s=1 (−ihks)β[s]ie1 (k) h
2πc
3
2
exp −ih
c kx +
+hn h
2πc
3
2
exp −ih
c kx γ[0]e1 (k) =
= h h
2πc
3
2
exp −ih
c kx ∑3
s=1 ksβ[s] +nγ[0] e1 (k) =
25Oskar Klein, 1894-1977
26Walter Gordon, 1893-1939
108](https://image.slidesharecdn.com/fn-190805113833/85/TOE-122-320.jpg)
![Na (x0)
Ψ
= ϕ∗
a (t,x0)ϕa (t,x0).
(2.65)
That is operator Na (x0) brings the a-component of the event probability density.
Let Ψa (t,x) := ψa (x0)Ψ(t,x).
In that case
Na (x0)
Ψa
=
Ω
dx
Ω
dx ·Ψ†
(t,x)ψ†
a (x0)ψ†
a (x0)
ψa (x0)ψa (x0)ψa (x0)Ψ(t,x).
Since
ψa (x0)ψa (x0) = 0
then
Na (x0)
Ψa
= 0.
Therefore ψa (x0) ”annihilates” the a of the event-probability density.
2.9. Particles and Antiparticles
Operator H obeys the following condition:
H =
2πc
h
3
∑
k
hω(k)
2
∑
r=1
b†
r,kbr,k −
4
∑
r=3
b†
r,kbr,k .
This operator is not positive defined and in this case
EΨ F0 =
2πc
h
3
∑
p
hω(p)
2
∑
r=1
|cr (t,p)|2
−
4
∑
r=3
|cr (t,p)|2
.
This problem is usually solved in the following way [49, p.54]:
Let:
v1 (k) : = γ[0]
e3 (k),
v2 (k) : = γ[0]
e4 (k),
d1,k : = −b†
3,−k,
d2,k : = −b†
4,−k.
In that case:
e3 (k) = −v1 (−k),
e4 (k) = −v2 (−k),
b3,k = −d†
1,−k,
b4,k = −d†
2,−k.
114](https://image.slidesharecdn.com/fn-190805113833/85/TOE-128-320.jpg)
![Therefore,
ψs (x) : = ∑
k
2
∑
r=1
br,ker,s (k)exp −i
h
c
kx +
+d†
r,kvr,s (k)exp i
h
c
kx
H =
2πc
h
3
∑
k
hω(k)
2
∑
r=1
b†
r,kbr,k +d†
r,kdr,k
−2∑
k
hω(k)1.
The first term on the right side of this equality is positive defined. This term is taken
as the desired Hamiltonian. The second term of this equality is infinity constant. And this
infinity is deleted (?!) [49, p.58]
But in this case dr,kF0 = 0. In order to satisfy such condition, the vacuum element F0
must be replaced by the following:
F0 → Φ0 := ∏
k
4
∏
r=3
2πc
h
3
b†
r,kF0.
But in this case:
ψs (x)Φ0 = 0.
And condition (2.64) isn’t carried out.
In order to satisfy such condition, operators ψs (x) must be replaced by the following:
ψs (x) → φs (x):=
:=∑
k
2
∑
r=1
br,ker,s (k)exp −i
h
c
kx +dr,kvr (k)exp i
h
c
kx .
Hence,
H = dx·H (x) = dx·φ†
(x)H0φ(x) =
=
2πc
h
3
∑
k
hω(k)
2
∑
r=1
b†
r,kbr,k −d†
r,kdr,k .
And again we get negative energy.
Let’s consider the meaning of such energy: An event with positive energy transfers this
energy photons which carries it on recorders observers. Observers know that this event
occurs, not before it happens. But event with negative energy should absorb this energy
from observers. Consequently, observers know that this event happens before it happens.
This contradicts Theorem 1.5.2. Therefore, events with negative energy do not occur.
Hence, over vacuum Φ0 single fermions can exist, but there is no single antifermions.
115](https://image.slidesharecdn.com/fn-190805113833/85/TOE-129-320.jpg)
![A two-particle state is defined the following field operator [52]:
ψs1,s2 (x,y) :=
φs1 (x) φs2 (x)
φs1 (y) φs2 (y)
.
In that case:
H = 2h
2πc
h
6
Ha +Hb
where
Ha : = ∑
k
∑
p
(ω(k)−ω(p))
2
∑
r=1
2
∑
j=1
×
× v†
j (−k)vj (−p)e†
r (p)er (k)×
× +b†
r,pd†
j,−kdj,−pbr,k +
+ +d†
r,−pb†
j,kbj,kdr,−p +
+v†
j (−p)vj (−k)e†
r (k)er (p)×
× −b†
r,kd†
j,−pdj,−kbr,p +
+ −b†
r,pd†
j,−kdj,−kbr,p
and
Hb : = ∑
k
∑
p
(ω(k)+ω(p))
2
∑
r=1
2
∑
j=1
×
× v†
j (−p)vj (−k)v†
r (−k)vr (−p)×
× −d†
r,−kd†
j,−pdj,−kdr,−p +
+ −d†
r,−pd†
j,−kdj,−kdr,−p
+e†
r (k)er (p)e†
j (p)ej (k)×
× +b†
r,kb†
j,pbj,kbr,p +
+ +b†
r,pb†
j,kbj,kbr,p .
If velosities are small then the following formula is fair.
H = 4h
2πc
h
6
Ha +Hb
where
Ha : = ∑
k
∑
p
(ω(k)−ω(p))×
×
2
∑
r=1
2
∑
j=1
d†
j,−pb†
r,kbr,kdj,−p −b†
j,pd†
r,−kdr,−kbj,p
116](https://image.slidesharecdn.com/fn-190805113833/85/TOE-130-320.jpg)
![Chapter 3
Fields
No group of people can claim power over
the thinking and views of others.
- Friedrich von Hayek
3.1. Electroweak Fields
In 1963 American physicist Sheldon Glashow1 [68] proposed that the weak nuclear force
and electricity and magnetism could arise from a partially unified electroweak theory. But
”... there is major problem: all the fermions and gauge bosons are massless, while exper-
iment shows otherwise. Why not just add in mass terms explicitly? That will not work,
since the associated terms break SU(2) or gauge invariances. For fermions, the mass term
should be mψψ?
mψψ = mψ(PL +PR)ψ =
= m(ψ(PLPL)ψ+ψ(PRPR)ψ)
= m(ψRψL +ψLψR).
However, the left-handed fermion are put into SU(2) doublets and the right-handed ones
into SU(2) singlets, so ψRψL and ψLψR are not SU(2) singlets and would not give an SU(2)
invariant Lagrangian. Similarly, the expected mass terms for the gauge bosons,
1
2
m2
BBµ
Bµ
plus similar terms for other, are clearly not invariant under gauge transformations Bµ →
Bµ = Bµ −∂µχ/g, The only direct way to preserve the gauge invariance and SU(2) invariance
of Lagrangian is to set m = 0 for all quarks, leptons and gauge bosons:. There is a way to
solve this problem, called the Higgs mechanism” [59].
1Sheldon Lee Glashow (born December 5, 1932) is an American theoretical physicist.](https://image.slidesharecdn.com/fn-190805113833/85/TOE-133-320.jpg)
![No. The Dirac Lagrangian for a free fermion can have of the following form:
Lf := ψ†
β[0]
∂0 +β[1]
∂1 +β[2]
∂2 +β[3]
∂3 +imγ[0]
ψ.
Indeed, this Lagrangian is not invariant under the SU(2) transformation. But it is beautiful
and truncating its mass term is not good idea.
Further you will see, how it is possible to keep this beauty.
3.1.1. The Bi-mass State
Let us consider [47], [48] the subspace ℑJ of the space ℑ spanned of the following subbasis
(2.47):
J :=
h
2πc exp −ih
c (s0x4) ε1, h
2πc exp −ih
c (s0x4) ε2,
h
2πc exp −ih
c (s0x4) ε3, h
2πc exp −ih
c (s0x4) ε4,
h
2πc exp −ih
c (n0x5) ε1, h
2πc exp −ih
c (n0x5) ε2,
h
2πc exp −ih
c (n0x5) ε3, h
2πc exp −ih
c (n0x5) ε4
(3.1)
with some integer numbers s0 and n0.
Let U be any linear transformation of space ℑJ such that for every ϕ: if ϕ ∈ ℑJ then
(2.48, 2.49):
(Uϕ,Uϕ) = ρA, (3.2)
Uϕ,β[s]
Uϕ = −
jA,s
c
for s ∈ {1,2,3}.
In that case:
U†β[µ]U = β[µ]
for µ ∈ {0,1,2,3}.
Such transformation has a matrix of the following shape:
U :=
(a”+b”i)12 02 (c”+ig”)12 02
02 (a‘+b‘i)12 02 (c‘+ig‘)12
(u”+iv”)12 02 (k”+is”)12 02
02 (u‘+iv‘)12 02 (k‘+is‘)12
.
with real functions
a”(t,x), b”(t,x), c”(t,x), g”(t,x), u”(t,x), v”(t,x), k”(t,x), s”(t,x),
a‘(t,x), b‘(t,x), c‘(t,x), g‘(t,x), u‘(t,x), v‘(t,x), k‘(t,x), s‘(t,x).
These functions fulfil the following conditions:
v”2 +b”2 +u”2 +a”2 = 1,
c”2 +g”2 +k”2 +s”2 = 1,
s” = −
a”g”u”−u”b”c”+a”c”v”+b”g”v”
u”2 +v”2
,
120](https://image.slidesharecdn.com/fn-190805113833/85/TOE-134-320.jpg)
![U1U2U3U4 = U
and
U2 =
exp(iθ2)12 02 02 02
02 exp(−iθ2)12 02 02
02 02 exp(iθ2)12 02
02 02 02 exp(−iθ2)12
.
Besides
U1U2 =
ei(θ1+θ2) 0 0 0
0 ei(θ1−θ2) 0 0
0 0 ei(θ1+θ2) 0
0 0 0 ei(θ1−θ2)
.
Let χ and ς be thesolution of the following set of equations:
0.5χ+ς = θ1 +θ2,
χ+ς = θ1 −θ2,
i.e.:
χ = −4θ2,
ς = θ1 +3θ2.
Let
U[e]
:= exp(iς)
and (2.39)
U =
exp iχ
2 12 02
02 exp(iχ)12
.
In that case:
UU[e]
18 = U1U2.
Here real functions
a(t,x), b(t,x), c(t,x), g(t,x), u(t,x), v(t,x), k(t,x),s(t,x)
exist such that:
U3U4 =
(a+ib)12 02 (c+ig)12 02
02 (u+iv)12 02 (k +is)12
(−c+ig)12 02 (a−ib)12 02
02 (−k +is)12 02 (u−iv)12
and
122](https://image.slidesharecdn.com/fn-190805113833/85/TOE-136-320.jpg)
![a2 +b2 +c2 +g2 = 1,
u2 +v2 +r2 +s2 = 1.
If
U(+)
:=
12 02 02 02
02 (u+iv)12 02 (k +is)12
02 02 12 02
02 (−k +is)12 02 (u−iv)12
(3.3)
and
U(−)
:=
(a+ib)12 02 (c+ig)12 02
02 12 02 02
(−c+ig)12 02 (a−ib)12 02
02 02 02 12
(3.4)
then
U3U4 = U(−)
U(+)
= U(+)
U(−)
.
Let us consider U(−).
Let:
◦ :=
1
2 (1−a2)
b+ (1−a2) 14 (q−ic)14
(q+ic)14 (1−a2)−b 14
(3.5)
and
∗ :=
1
2 (1−a2)
(1−a2)−b 14 (−q+ic)14
(−q−ic)14 b+ (1−a2) 14
. (3.6)
These operators are fulfilled to the following conditions:
◦ ◦ = ◦, ∗ ∗ = ∗;
◦ ∗ = 0 = ∗ ◦,
( ◦ − ∗)( ◦ − ∗) = 18,
◦ + ∗ = 18,
◦γ[0] = γ[0]
◦, ∗γ[0] = γ[0]
∗,
◦β[4] = β[4]
◦, ∗β[4] = β[4]
∗
and
U(−)†γ[0]U(−) = aγ[0] −( ◦ − ∗)
√
1−a2β[4],
U(−)†β[4]U(−) = aβ[4] +( ◦ − ∗)
√
1−a2γ[0].
(3.7)
From (2.38) the lepton equation of motion is the following:
3
∑
µ=0
β[µ]
(i∂µ +Fµ +0.5g1YBµ)+γ[0]
i∂5 +β[4]
i∂4 U(−)†
U(−)
ϕ = 0.
123](https://image.slidesharecdn.com/fn-190805113833/85/TOE-137-320.jpg)
![If
∂kU(−)†
= U(−)†
∂k (3.8)
for k ∈ {0,1,2,3,4,5} then
U(−)†i∑3
µ=0 β[µ] (i∂µ +Fµ +0.5g1YBµ)
+γ[0]U(−)†i∂5 +β[4]U(−)†i∂4
U(−)
ϕ = 0.
Hence, from (3.7):
U(−)†
∑3
µ=0 β[µ] (i∂µ +Fµ +0.5g1YBµ)
+γ[0]i a∂5 −( ◦ − ∗)
√
1−a2∂4
+β[4]i
√
1−a2 ( ◦ − ∗)∂5 +a∂4
U(−)
ϕ = 0.
Thus, if denote:
x4 = ( ◦ + ∗)ax4 +( ◦ − ∗) 1−a2x5,
x5 = ( ◦ + ∗)ax5 −( ◦ − ∗) 1−a2x4
then
3
∑
µ=0
β[µ]
(i∂µ +Fµ +0.5g1YBµ)+ γ[0]
i∂5 +β[4]
i∂4 ϕ = 0 (3.9)
with
ϕ = U(−)
ϕ.
That is the lepton Hamiltonian is invariant for the following global transformation:
ϕ → ϕ = U(−)
ϕ,
x4 → x4 = ( ◦ + ∗)ax4 +( ◦ − ∗) 1−a2x5, (3.10)
x5 → x5 = ( ◦ + ∗)ax5 −( ◦ − ∗) 1−a2x4,
xµ → xµ = xµ.
3.1.2. Neutrino
Wolfgang Pauli postulated the neutrino in 1930 to explain the energy spectrum of beta de-
cays, the decay of a neutron into a proton and an electron. Clyde Cowan, Frederick Reines
found the neutrino experimentally in 1955. Enrico Fermi2 developed the first theory de-
scribing neutrino interactions and denoted this particles as neutrino in 1933. In 1962 Leon
M. Lederman, Melvin Schwartz and Jack Steinberger showed that more than one type of
neutrino exists. Bruno Pontecorvo3 suggested a practical method for investigating neutrino
2Enrico Fermi (29 September 1901 28 November 1954) was an Italian-born, naturalized American physicist
particularly known for his work on the development of the first nuclear reactor, Chicago Pile-1, and for his
contributions to the development of quantum theory, nuclear and particle physics, and statistical mechanics.
3Bruno Pontecorvo (Marina di Pisa, Italy, August 22, 1913 – Dubna, Russia, September 24, 1993) was an
Italian-born atomic physicist, an early assistant of Enrico Fermi and then the author of numerous studies in
high energy physics, especially on neutrinos.
124](https://image.slidesharecdn.com/fn-190805113833/85/TOE-138-320.jpg)
![masses in 1957, over the subsequent 10 years he developed the mathematical formalism
and the modern formulation of vacuum oscillations...
Let ℑeν be the unitary space, spanned by the following basis:
Jeν :=
h
2πc
2πn0
sinh(2n0π) cosh h
c n0x4 +sinh h
c n0x4 ε1,
h
2πc
2πn0
sinh(2n0π) cosh h
c n0x4 +sinh h
c n0x4 ε2,
h
2πc
2πn0
sinh(2n0π) cosh h
c n0x4 −sinh h
c n0x4 ε3,
h
2πc
2πn0
sinh(2n0π) cosh h
c n0x4 −sinh h
c n0x4 ε4,
h
2πc exp −ih
c (n0x5) ε1, h
2πc exp −ih
c (n0x5) ε2,
h
2πc exp −ih
c (n0x5) ε3, h
2πc exp −ih
c (n0x5) ε4
. (3.11)
Let ℑe be the subspace of the space ℑeν such that if ϕ ∈ ℑe then ϕ has the following
shape:
ϕ(t,x,x5,x4) = exp −i
h
c
n0x5
4
∑
k=1
fk (t,x,n0,0)εk
That is ϕ has the following matrix in the basis Jeν:
ϕ =
0
0
0
0
f1
f2
f3
f4
. (3.12)
Let us consider the following Hamiltonian on ℑe:
H0,4 := c
3
∑
r=1
β[r]
i∂r +γ[0]
i∂5 +β[4]
i∂4 : (3.13)
H0,4ϕ = c ∑3
r=1 β[r]i∂r +γ[0]i∂5 +β[4]i∂4 ϕ =
= ∑3
r=1 β[r]ci∂rϕ+
+γ[0]ci∂5 exp −ih
c n0x5 ∑4
k=1 fk (t,x,n0,0)εk+
+β[4]ci∂4 exp −ih
c n0x5 ∑4
k=1 fk (t,x,n0,0)εk =
= ∑3
r=1 β[r]ci∂rϕ+
+γ[0]ci −ih
c n0 exp −ih
c n0x5 ∑4
k=1 fk (t,x,n0,0)εk+
+0 =
= ∑3
r=1 β[r]ci∂rϕ+hn0γ[0] exp −ih
c n0x5 ∑4
k=1 fk (t,x,n0,0)ε. =
= ∑3
r=1 β[r]ci∂rϕ+hn0γ[0]ϕ.
Hence, on this space:
125](https://image.slidesharecdn.com/fn-190805113833/85/TOE-139-320.jpg)
![H0,4 = H0 := c
3
∑
r=1
β[r]
i∂r +hn0γ[0]
. (3.14)
Let ℑ◦ be the subspace of the space ℑeν such that if ϕ◦ ∈ ℑ◦ then
ϕ◦ = ◦ ϕ and ϕ ∈ ℑe, and if ϕ ∈ ℑe then ( ◦ϕ) ∈ ℑ◦. If ϕ◦ = ◦ϕ then in the basis Jeν:
ϕ◦ =
1
2 (1−a2)
−(−q+ic) f1
−(−q+ic) f2
−(−q+ic) f3
−(−q+ic) f4
− − (1−a2)+b f1
− − (1−a2)+b f2
− − (1−a2)+b f3
− − (1−a2)+b f4
.
Let us consider the Hamiltonian H0,4 mode of behavior on the space ℑ◦:
Hence,
H0,4ϕ◦ = c∑3
r=1 β[r]i∂rϕ◦+
+γ[0]ic (q−ic)
2
√
(1−a2)
h
2πc
2πn0
sinh(2n0π)×
×
∂5
f1 cosh h
c n0x4 +sinh h
c n0x4 ε1+
+f2 cosh h
c n0x4 +sinh h
c n0x4 ε2+
+f3 cosh h
c n0x4 −sinh h
c n0x4 ε3+
+f4 cosh h
c n0x4 −sinh h
c n0x4 ε4
+
+ (1−a2)−b h
2πc∂5 exp −ih
c (n0x5) ·
·(f1ε1 + f2ε2 + f3ε3 + f4ε4)
+
+β[4]ic (q−ic)
2
√
(1−a2)
h
2πc
2πn0
sinh(2n0π)×
×
∂4
f1 cosh h
c n0x4 +sinh h
c n0x4 ε1+
+f2 cosh h
c n0x4 +sinh h
c n0x4 ε2+
+f3 cosh h
c n0x4 −sinh h
c n0x4 ε3+
+f4 cosh h
c n0x4 −sinh h
c n0x4 ε4
+
+ (1−a2)−b ∂4
h
2πc exp −ih
c (n0x5) ·
·(f1ε1 + f2ε2 + f3ε3 + f4ε4)
.
126](https://image.slidesharecdn.com/fn-190805113833/85/TOE-140-320.jpg)
![H0,4ϕ◦ = c∑3
r=1 β[r]i∂rϕ◦+
+γ[0]ic (q−ic)
2
√
(1−a2)
h
2πc
2πn0
sinh(2n0π)×
×
∂5
f1 cosh h
c n0x4 +sinh h
c n0x4 ε1+
+f2 cosh h
c n0x4 +sinh h
c n0x4 ε2+
+f3 cosh h
c n0x4 −sinh h
c n0x4 ε3+
+f4 cosh h
c n0x4 −sinh h
c n0x4 ε4
+
+ (1−a2)−b h
2πc∂5 exp −ih
c (n0x5) ·
·(f1ε1 + f2ε2 + f3ε3 + f4ε4)
+
+β[4]ic (q−ic)
2
√
(1−a2)
h
2πc
2πn0
sinh(2n0π)×
×
∂4
f1 cosh h
c n0x4 +sinh h
c n0x4 ε1+
+f2 cosh h
c n0x4 +sinh h
c n0x4 ε2+
+f3 cosh h
c n0x4 −sinh h
c n0x4 ε3+
+f4 cosh h
c n0x4 −sinh h
c n0x4 ε4
+
+ (1−a2)−b ∂4
h
2πc exp −ih
c (n0x5) ·
·(f1ε1 + f2ε2 + f3ε3 + f4ε4)
.
Therefore,
H0,4ϕ◦ = c∑3
r=1 β[r]i∂rϕ◦+
+γ[0]ic
√
1−a2−b
2
√
1−a2
×
× 0+ h
2πc∂5 exp −ih
c (n0x5) (f1ε1 + f2ε2 + f3ε3 + f4ε4) +
+β[4]ic q−ic
2
√
1−a2
h
2πc
2πn0
sinh(2n0π)×
×
f1 ∂4 cosh h
c n0x4 +∂4 sinh h
c n0x4 ε1+
+f2 ∂4 cosh h
c n0x4 +∂4 sinh h
c n0x4 ε2+
+f3 ∂4 cosh h
c n0x4 −∂4 sinh h
c n0x4 ε3+
+f4 ∂4 cosh h
c n0x4 −∂4 sinh h
c n0x4 ε4
+
+0
.
Hence,
H0,4ϕ◦ = c∑3
r=1 β[r]i∂rϕ◦+
+γ[0]ic −ih
c n0
√
1−a2−b
2
√
1−a2
h
2πc exp −ih
c n0x5 ×
×(f1ε1 + f2ε2 + f3ε3 + f4ε4)+
+β[4]ich
c n0
q−ic
2
√
1−a2
h
2πc
2πn0
sinh(2n0π)×
×
f1 sinh h
c n0x4 +cosh h
c n0x4 ε1+
+f2 sinh h
c n0x4 +cosh h
c n0x4 ε2+
+f3 sinh h
c n0x4 −cosh h
c n0x4 ε3+
+f4 sinh h
c n0x4 −cosh h
c n0x4 ε4
.
Therefore,
127](https://image.slidesharecdn.com/fn-190805113833/85/TOE-141-320.jpg)
![H0,4ϕ◦ = c∑3
r=1 β[r]i∂rϕ◦+
+hn0γ[0]
√
1−a2−b
2
√
1−a2
h
2πc exp −ih
c n0x5 ×
×(f1ε1 + f2ε2 + f3ε3 + f4ε4)+
+hn0β[4]i q−ic
2
√
1−a2
h
2πc
2πn0
sinh(2n0π)×
×
f1 cosh h
c n0x4 +sinh h
c n0x4 ε1+
+f2 cosh h
c n0x4 +sinh h
c n0x4 ε2−
−f3 cosh h
c n0x4 −sinh h
c n0x4 ε3−
−f4 cosh h
c n0x4 −sinh h
c n0x4 ε4
.
Hence, in basis Jeν:
H0,4ϕ◦ = c∑3
r=1 β[r]i∂rϕ◦ +hn0×
×
γ[0]
√
1−a2−b
2
√
1−a2
0
0
0
0
f1
f2
f3
f4
+β[4]i q−ic
2
√
1−a2
f1
f2
−f3
−f4
0
0
0
0
=
= c∑3
r=1 β[r]i∂rϕ◦ +hn0×
×
γ[0]
√
1−a2−b
2
√
1−a2
0
0
0
0
f1
f2
f3
f4
+β[4]i q−ic
2
√
1−a2
γ[5]
f1
f2
f3
f4
0
0
0
0
.
with
γ[5]
:=
1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1
.
Since
β[4]
iγ[5]
= γ[0]
then
128](https://image.slidesharecdn.com/fn-190805113833/85/TOE-142-320.jpg)
![H0,4ϕ◦ = c∑3
r=1 β[r]i∂rϕ◦ +hn0×
×
γ[0] 1
2
√
1−a2
0
0
0
0
√
1−a2 −b f1
√
1−a2 −b f2
√
1−a2 −b f3
√
1−a2 −b f4
+γ[0] 1
2
√
1−a2
(q−ic) f1
(q−ic) f2
(q−ic) f3
(q−ic) f4
0
0
0
0
.
Therefore,
H0,4ϕ◦ = c
3
∑
r=1
β[r]
i∂rϕ◦ +hn0γ[0] 1
2
√
1−a2
−(−q+ic) f1,
−(−q+ic) f2,
−(−q+ic) f3,
−(−q+ic) f4,
− −
√
1−a2 +b f1,
− −
√
1−a2 +b f2,
− −
√
1−a2 +b f3,
− −
√
1−a2 +b f4
.
Hence,
H0,4ϕ◦ = c
3
∑
r=1
β[r]
i∂rϕ◦ +hn0γ[0]
ϕ◦.
Thus, in space ℑe:
H0,4 = H0 = c
3
∑
r=1
β[r]
i∂r +hn0γ[0]
,
too.
Let ℑ∗ be the subspace of the space ℑeν such that if ϕ∗ ∈ ℑ∗ then
ϕ∗ = ∗ ϕ and ϕ ∈ ℑe, and if ϕ ∈ ℑe then ( ∗ϕ) ∈ ℑ∗. If ϕ∗ = ∗ϕ (3.12) then in the basis
Jeν:
129](https://image.slidesharecdn.com/fn-190805113833/85/TOE-143-320.jpg)
![ϕ∗ =
1
2 (1−a2)
(−q+ic) f1
(−q+ic) f2
(−q+ic) f3
(−q+ic) f4
b+
√
1−a2 f1
b+
√
1−a2 f2
b+
√
1−a2 f3
b+
√
1−a2 f4
.
Similarly to ϕ◦ you can calculate that
H0,4ϕ∗ = H0ϕ∗ = c
3
∑
r=1
β[r]
i∂rϕ∗ +hn0γ[0]
ϕ∗.,
too.
Let
e1L (k) :=
ω(k)+n0 +k3
k1 +ik2
, e1R (k) :=
ω(k)+n0 −k3
−k1 −ik2
,
e2L (k) :=
k1 −ik2
ω(k)+n0 −k3
, e2R (k) :=
−k1 +ik2
ω(k)+n0 +k3
,
e3L (k) := −e1R (k), e3R (k) := e1L (k),
e4L (k) := −e2R (k), e4R (k) := e2L (k).
with
ω(k) := n2
0 +k2
1 +k2
2 +k2
3
( n0, k1, k2,k3 are real numbers).
In this case:
es (k) =
1
2 ω(k)(ω(k)+n0)
esL (k)
esR (k)
.
Let
es (k) :=
−→
0 4
es (k)
here s ∈ {1,2,3,4}.
And let:
130](https://image.slidesharecdn.com/fn-190805113833/85/TOE-144-320.jpg)
![e◦s (k) := ◦es (k)
=
1
2
√
1−a2 −b
√
1−a2
(q−ic)es (k)
√
1−a2 −b es (k)
,
e∗s (k) := ∗es (k)
=
1
2
√
1−a2 +b
√
1−a2
(−q+ic)es (k)
b+
√
1−a2 es (k)
.
Denote
H0 (k) :=
3
∑
r=1
β[r]
kr =
k3 k1 −ik2 n0 0
k1 +ik2 −k3 0 n0
n0 0 −k3 −k1 +ik2
0 n0 −k1 −ik2 k3
.
In that case
H0e◦1 (k)
h
2πc
3
2
exp i
h
c
=
= hH0 (k)e◦1 (k)
h
2πc
3
2
exp i
h
c
= hω(k)e◦1 (k)
h
2πc
3
2
exp i
h
c
.
Therefore, e◦1 (k) h
2πc
3
2
exp ih
c is an eigenvector of H0 with the eigenvalue hω(k).
Similarly you can calculate that
e◦2 (k) h
2πc
3
2
exp ih
c , e∗1 (k) h
2πc
3
2
exp ih
c , e∗2 (k) h
2πc
3
2
exp ih
c ,
are eigenvectors of H0 with the same eigenvalue, and
e◦3 (k) h
2πc
3
2
exp ih
c , e◦4 (k) h
2πc
3
2
exp ih
c ,
e∗3 (k) h
2πc
3
2
exp ih
c , e∗4 (k) h
2πc
3
2
exp ih
c
are an eigenvectors of H0 with the eigenvalue (−hω(k)).
Vectors e◦s (k) h
2πc
3
2
exp ih
c , e∗s (k) h
2πc
3
2
exp ih
c with s ∈ {1,2,3,4} form an or-
thonormalized basis in the space ℑeν (3.11) and
4
∑
s=1
e∗
∗s,r (k)e∗s,r (k)+e∗
◦s,r (k)e◦s,r (k) = δr,r (3.15)
for r,r ∈ {1,2,3,4,5,6,7,8}.
131](https://image.slidesharecdn.com/fn-190805113833/85/TOE-145-320.jpg)
![ln0 (x) := ∑
k
e−ih
c kx
4
∑
s=1
ln0,(s) (k)bs (k).
Hence, in basis Jeν:
χ(x) =
νn0 (x)
ln0 (x)
. (3.23)
Let:
Hl,0 := c
3
∑
r=1
β[r]
i∂r +hn0 aγ[0]
−bβ[4]
,
Hν,0 := c
3
∑
r=1
β[r]
i∂r +hn0 aγ[0]
+bβ[4]
,
Hν,l := (c+iq)
0 0 n0 0
0 0 0 n0
−n0 0 0 0
0 −n0 0 0
,
Hl,ν := (c−iq)
0 0 −n0 0
0 0 0 −n0
n0 0 0 0
0 n0 0 0
.
In that case in basis Jeν:
H0 =
Hν,0 Hν,l
Hl,ν Hl,0
.
Let
Hl,0 (k) :=
3
∑
r=1
β[r]
kr +n0 aγ[0]
−bβ[4]
,
Hν,0 (k) :=
3
∑
r=1
β[r]
kr +n0 aγ[0]
+bβ[4]
.
In that case
H0 (k) =
Hν,0 (k) Hν,l
Hl,ν Hl,0 (k)
An neutrino and it’s lepton are tied by the follows equations:
Hν,0 (k)νn0,(s) (k)+Hν,lln0,(s) (k) = ω(k)νn0,(s) (k)
for s ∈ {1,2} and
141](https://image.slidesharecdn.com/fn-190805113833/85/TOE-155-320.jpg)
![Hν,0 (k)νn0,(s) (k)+Hν,lln0,(s) (k) = −ω(k)νn0,(s) (k)
for s ∈ {3,4}.
I suppose that such neutrino can fly 1.5 cm. [60] and give birth to it’s leptons.
3.1.3. Electroweak Transformations
During the 1960s Sheldon Lee Glashow discovered that they could construct a gauge-
invariant theory of the weak force, provided that they also included the electromagnetic
force.
The existence of the force carriers, the neutral Z particles and the charged W parti-
cles, was verified experimentally in 1983 in high-energy proton-antiproton collisions at the
European Organization for Nuclear Research (CERN).
Let (3.8) does not hold true, that is U(−) depends on x. And let denote:
K :=
3
∑
µ=0
β[µ]
(Fµ +0.5g1YBµ). (3.24)
In that case from (2.38) the equation of moving is of following form:
K +
3
∑
µ=0
β[µ]
i∂µ +γ[0]
i∂5 +β[4]
i∂4 ϕ = 0. (3.25)
Let us consider for this Hamiltonian the following transformations:
ϕ → ϕ := U(−)
ϕ,
x4 → x4 := ( ◦ + ∗)ax4 +( ◦ − ∗) 1−a2x5,
x5 → x5 := ( ◦ + ∗)ax5 −( ◦ − ∗) 1−a2x4, (3.26)
xµ → xµ := xµ, for µ ∈ {0,1,2,3},
K → K = U(−)
KU(−)†
−i
3
∑
µ=0
β[µ]
∂µU(−)
U(−)†
with
∂4U(−) = U(−)∂4 and ∂5U(−) = U(−)∂5:
Since
( ◦ − ∗)( ◦ − ∗) = 18
then
x4 = ax4 −( ◦ − ∗) 1−a2x5 and
x5 = ( ◦ − ∗) 1−a2x4 +ax5.
Since for any f:
142](https://image.slidesharecdn.com/fn-190805113833/85/TOE-156-320.jpg)
![∂4 f = ∂4 f ·∂4x4 +∂5 f ·∂4x5,
∂5 f = ∂4 f ·∂5x4 +∂5 f ·∂5x5
then
∂4 f = ∂4 f ·a+∂5 f ·( ◦ − ∗) 1−a2,
∂5 f = ∂4 f · −( ◦ − ∗) 1−a2 +∂5 f ·a.
Therefore, if
K +
3
∑
µ=0
β[µ]
i∂µ +γ[0]
i∂5 +β[4]
i∂4 U(−)
ϕ = 0
then
U(−)KU(−)† −i∑3
µ=0 β[µ] ∂µU(−) U(−)†
+∑3
µ=0 β[µ]i∂µ +γ[0]i −( ◦ − ∗)
√
1−a2 ∂4 +a∂5
+β[4]i a∂4 +( ◦ − ∗)
√
1−a2∂5
U(−)
ϕ = 0.
Hence,
U(−)KU(−)†U(−) −i∑3
µ=0 β[µ] ∂µU(−) U(−)†U(−)
+∑3
µ=0 β[µ]i∂µU(−) +γ[0]U(−)i −( ◦ − ∗)
√
1−a2 ∂4 +a∂5
+β[4]U(−)i a∂4 +( ◦ − ∗)
√
1−a2∂5
ϕ = 0
since U(−) is a linear operator such that ∂4U(−) = U(−)∂4 and ∂5U(−) = U(−)∂5.
Since
U(−)†
U(−)
= 18,
for any f:
∂µ U(−)
f = ∂µU(−)
f + U(−)
∂µ f = ∂µU(−)
+U(−)
∂µ f,
and
γ[0]
U(−)
= U(−)
aγ[0]
−( ◦ − ∗) 1−a2β[4]
,
β[4]
U(−)
= U(−)
aβ[4]
+( ◦ − ∗) 1−a2γ[0]
then
143](https://image.slidesharecdn.com/fn-190805113833/85/TOE-157-320.jpg)
![
U(−)K −i∑3
µ=0 β[µ] ∂µU(−)
+∑3
µ=0 β[µ]i ∂µU(−) +U(−)∂µ
+U(−) aγ[0] −( ◦ − ∗)
√
1−a2β[4] ×
×i −( ◦ − ∗)
√
1−a2 ∂4 +a∂5
+U(−) aβ[4] +( ◦ − ∗)
√
1−a2γ[0] ×
×i a∂4 +( ◦ − ∗)
√
1−a2∂5
ϕ = 0.
Therefore,
U(−)K −i∑3
µ=0 β[µ] ∂µU(−)
+∑3
µ=0 β[µ]i ∂µU(−) +U(−)∂µ
+iU(−)
aγ[0] −( ◦ − ∗)
√
1−a2β[4] ×
× −( ◦ − ∗)
√
1−a2∂4 +a∂5
+ aβ[4] +( ◦ − ∗)
√
1−a2γ[0] ×
× a∂4 +( ◦ − ∗)
√
1−a2∂5
ϕ = 0,
U(−)
K +
3
∑
µ=0
β[µ]
iU(−)
∂µ +iU(−)
+γ[0]
∂5 +β[4]
∂4 ϕ = 0,
Hence,
U(−)
K +
3
∑
µ=0
β[µ]
i∂µ +i +γ[0]
∂5 +β[4]
∂4 ϕ = 0
since β[µ]U(−) = U(−)β[µ] for µ ∈ {0,1,2,3}. Compare with (3.25).
Therefore, this equation of moving is invariant under the transformation (3.26).
Let g2 be some positive real number.
If design (here: a,b,c,q form U(−) in (3.4)):
W0,µ := −2 1
g2q
q(∂µa)b−q(∂µb)a+(∂µc)q2+
+a(∂µa)c+b(∂µb)c+c2 (∂µc)
W1,µ := −2 1
g2q
(∂µa)a2 −bq(∂µc)+a(∂µb)b+
+a(∂µc)c+q2 (∂µa)+c(∂µb)q
W2,µ := −2 1
g2q
q(∂µa)c−a(∂µa)b−b2 (∂µb)−
−c(∂µc)b−(∂µb)q2 −(∂µc)qa
and
Wµ :=
W0,µ12 02 (W1,µ −iW2,µ)12 02
02 02 02 02
(W1,µ +iW2,µ)12 02 −W0,µ12 02
02 02 02 02
(3.27)
then
−i ∂µU(−)
U(−)†
=
1
2
g2Wµ, (3.28)
144](https://image.slidesharecdn.com/fn-190805113833/85/TOE-158-320.jpg)
![and from (3.24), (3.25):
∑3
µ=0 β[µ]i ∂µ −i0.5g1BµY −i1
2 g2Wµ −iFµ
+γ[0]i∂5 +β[4]i∂4
ϕ = 0. (3.29)
Let (3.4) a (t,x), b (t,x), c (t,x), q (t,x) are real functions and:
U :=
(a +ib )12 02 (c +ig )12 02
02 12 02 02
(−c +ig )12 02 (a −ib )12 02
02 02 02 12
.
In this case if
U := U U(−)
then there exist real functions a (t,x), b (t,x), c (t,x), q (t,x) such thatU has the similar
shape:
U =
(a +ib )12 02 (c +ig )12 02
02 12 02 02
(−c +ig )12 02 (a −ib )12 02
02 02 02 12
.
Let:
Wµ := −
2i
g2
∂µ U U(−)
U U(−)
†
,
Hence,
Wµ = −
2i
g2
∂µU U(−)
U U(−)
†
−
2i
g2
U ∂µU(−)
U U(−)
†
= −
2i
g2
∂µU U(−)
U(−)†
U †
−
2i
g2
U ∂µU(−)
U(−)†
U †
= −
2i
g2
∂µU U †
−
2i
g2
U ∂µU(−)
U(−)†
U †
.
Since from (3.28):
Wµ = −i
2
g2
∂µU(−)
U(−)†
then
Wµ = −
2i
g2
∂µU U †
−
2i
g2
U ∂µU(−)
U(−)†
U †
= −
2i
g2
∂µU U †
+U WµU †
.
Therefore, if
145](https://image.slidesharecdn.com/fn-190805113833/85/TOE-159-320.jpg)
![◦ :=
1
2 (1−a 2)
b + (1−a 2) 14 (q −ic )14
(q +ic )14 (1−a 2)−b 14
,
∗ :=
1
2 (1−a 2)
(1−a 2)−b 14 (−q +ic )14
(−q −ic )14 b + (1−a 2) 14
.
then under the following transformation
ϕ → ϕ := U ϕ,
x4 → x4 := ◦ + ∗ a x4 + ◦ − ∗ 1−a 2x5,
x5 → x5 := ◦ + ∗ a x5 − ◦ − ∗ 1−a 2x4, (3.30)
xµ → xµ := xµ, for µ ∈ {0,1,2,3},
K → K :=
3
∑
µ=0
β[µ]
Fµ +0.5g1YBµ +
1
2
g2Wµ
fields Wµ and Wµ are tied by the following equation
Wµ = U WµU † − 2i
g2
(∂µU )U †
(3.31)
like in Standard Model.
From (3.28):
Wµ = −i
2
g2
∂µU(−)
U(−)†
.
Let us calculate:
∂µWν −∂νWµ =
= ∂µ −i
2
g2
∂νU(−)
U(−)†
−∂ν −i
2
g2
∂µU(−)
U(−)†
=
= −i
2
g2
∂µ∂νU(−) U(−)† + ∂νU(−) ∂µU(−)†
− ∂ν∂µU(−) U(−)† − ∂µU(−) ∂νU(−)† .
Since
∂µ∂νU(−)
= ∂ν∂µU(−)
then
∂µWν −∂νWµ = (3.32)
= −i
2
g2
∂νU(−)
∂µU(−)†
− ∂µU(−)
∂νU(−)†
.
146](https://image.slidesharecdn.com/fn-190805113833/85/TOE-160-320.jpg)
![and with additional terms of the W0,µ interactions with others components of W. You can
receive similar equations for W1,µ and for W2,µ.
If
W0 :=
W0 − v
cWk
1− v
c
2
, Wk :=
Wk − v
cW0
1− v
c
2
, Wk := Wk, if s = k
then
W 2
0 −
3
∑
s=1
W 2
s
=
W0 − v
cWk
2
1− v
c
2
−
Wk − v
cW0
2
1− v
c
2
− ∑
s=k
W 2
s
=
W2
0 + v
c
2
W2
k −W2
k − v
c
2
W2
0
1− v
c
2
− ∑
s=k
W 2
s
=
1− v
c
2
W2
0 − 1− v
c
2
W2
k
1− v
c
2
− ∑
s=k
W 2
s .
Hence,
W 2
0 −
3
∑
s=1
W 2
s = W2
0 −
3
∑
s=1
W2
s .
Therefore, such ”mass” (3.39) is invariant for the Lorentz transformations:
You can calculate that it is invariant for the transformations of turns, too:
Wr = Wr cosλ−Ws sinλ.
Ws = Wr sinλ+Ws cosλ;
with a real number λ, and r ∈ {1,2,3}, s ∈ {1,2,3}.
That is in (3.38) the form
m =
h
c
g2 W2
0 −
3
∑
s=1
W2
s
varies in space, but locally acts like a mass - i.e. it does not allow to particles of this field to
behave as a massless ones.
A mass of the W-boson was measured, between 1996 and 2000 at LEP5 [50].
Let6
5The Large Electron-Positron Collider (LEP) is largest particles accelerator (ring with a circumference of
27 kilometers built in a tunnel under the border of Switzerland and France.)
6here α is the Weinberg Angle. The experimental value of sin2 α = 0.23124±0.00024 [51].
152](https://image.slidesharecdn.com/fn-190805113833/85/TOE-166-320.jpg)
![α := arctan g1
g2
,
Zµ := (W0,µ cosα−Bµ sinα),
Aµ := (Bµ cosα+W0,µ sinα).
(3.40)
In that case:
∑ν gν,ν∂ν∂νW0,µ = cosα·∑ν gν,ν∂ν∂νZµ +sinα·∑ν gν,ν∂ν∂νAµ.
If
∑
ν
gν,ν∂ν∂νAµ = 0
then
mZ =
mW
cosα
with mW from (3.39). It is like Standard Model.
The equation of moving (3.29) under Fµ = 0 has the following form:
∑3
µ=0 β[µ]i ∂µ −i0.5g1BµY −i1
2 g2Wµ
+γ[0]i∂5 +β[4]i∂4
ϕ = 0. (3.41)
Hence, in accordance with (3.27) and (2.35):
∑3
µ=0 β[µ]i×
×
∂µ −i0.5g1Bµ −
12 02
02 2·12
−
−i1
2 g2
W0,µ12 02 (W1,µ −iW2,µ)12 02
02 02 02 02
(W1,µ +iW2,µ)12 02 −W0,µ12 02
02 02 02 02
+γ[0]i∂5 +β[4]i∂4
·ϕ = 0.
In accordance with (3.40) [53]:
Bµ =
−Zµ
g1
g2
1 +g2
2
+Aµ
g2
g2
1 +g2
2
,
W0,µ =
Zµ
g2
g2
1 +g2
2
+Aµ
g1
g2
1 +g2
2
.
Let (e is the elementary charge7: e = 1.60217733×10−19 C).
7Sir Joseph John ”J. J.” Thomson, (18 December 1856 - 30 August 1940) was a British physicist. He is
credited for the discovery of the electron and of isotopes, and the invention of the mass spectrometer.
153](https://image.slidesharecdn.com/fn-190805113833/85/TOE-167-320.jpg)
![e :=
g1g2
g2
1 +g2
2
,
and let
Zµ := Zµ
1
g2
2 +g2
1
g2
2 +g2
1 12 02 02 02
02 2g2
112 02 02
02 02 g2
2 −g2
1 12 02
02 02 02 2g2
112
,
Wµ := g2
02 02 (W1,µ −iW2,µ)12 02
02 02 02 02
(W1,µ +iW2,µ)12 02 02 02
02 02 02 02 ·12
,
Aµ := Aµ
02 02 02 02
02 12 02 02
02 02 12 02
02 02 02 12
.
In that case from (3.41):
∑3
µ=0 β[µ]i ∂µ +ieAµ −i0.5 Zµ +Wµ +γ[0]i∂5 +β[4]i∂4 ϕ = 0.
(3.42)
Let in basis (3.11) (3.23) :
ϕ =
ϕν
−→
0 2
ϕe,L
ϕe,R
.
In that case
∑3
µ=0 β[µ]i ∂µϕ+iAµe
ϕe,L
ϕe,R
−i0.5 Zµ +Wµ ϕ
+ γ[0]i∂5 +β[4]i∂4 ϕ
= 0.
(3.43)
Here the vector field Aµ is the electromagnetic potential 8. And Zµ +Wµ is the weak
interaction potential Evidently neutrinos do not involve in the electromagnetic interactions.
8James Clerk Maxwell of Glenlair (13 June 1831 - 5 November 1879) was a Scottish physicist and mathe-
matician. His most prominent achievement was formulating classical electromagnetic theory.
154](https://image.slidesharecdn.com/fn-190805113833/85/TOE-168-320.jpg)
![3.1.4. Dimension of physical space
Further I use Cayley-Dickson algebras [72, 73]:
Let 1,i, j,k,E,I,J,K be basis elements of a 8-dimensional algebra Cayley (the octavians
algebra) [72, 73]. A product of this algebra is defined the following way [72]:
1. for every basic element e:
ee = −1;
2. If u1, u2, v1, v2 are real number then
(u1 +u2i)(v1 +v2i) = (u1v1 −v2u2)+(v2u1 +u2v1)i.
3. If u1, u2, v1, v2 are numbers of shape w = w1 + w2i (ws, and s ∈ {1,2} are real
numbers, and w = w1 −w2i) then
(u1 +u2j)(v1 +v2j) = (u1v1 −v2u2)+(v2u1 +u2v1)j (3.44)
and ij = k.
4. If u1, u2, v1, v2 are number of shape w = w1 +w2i+w3j+w4k (ws, and s ∈ {1,2,3,4}
are real numbers, and w = w1 −w2i−w3j−w4k) then
(u1 +u2E)(v1 +v2E) = (u1v1 −v2u2)+(v2u1 +u2v1)E (3.45)
and
iE = I,
jE = J,
kE = K.
Therefore, in according with point 2.: the real numbers field (R) is extended to the
complex numbers field (R), and in according with point 3.: the complex numbers field
is expanded to the quaternions field (K), and point 4. expands the quaternions fields to
the octavians field (O). This method of expanding of fields is called a Dickson doubling
procedure [72].
If
u = a+bi+cj+dk+AE+BI+CJ+K
with real a,b,c,d,A,B,C,D then a real number
u
def
=
√
uu = a2
+b2
+c2
+d2
+A2
+B2
+C2
+D2 0.5
is called a norm of octavian u [72].
For each octavians u and v:
uv = u v . (3.46)
Algebras with this conditions are called normalized algebras [72, 73].
155](https://image.slidesharecdn.com/fn-190805113833/85/TOE-169-320.jpg)
![Any 3+1-vector of a probability density can be represented by the following equations
in matrix form (2.19)
ρ = ϕ†ϕ,
jk = ϕ†β[k]ϕ
with k ∈ {1,2,3}.
There β[k] are complex 2-diagonal 4 × 4-matrices of Clifford’s set of rank 4, and ϕ is
matrix columns with four complex components. The light and colored pentads of Clifford’s
set of such rank contain in threes 2-diagonal matrices, corresponding to 3 space coordinates
in according with Dirac’s equation. Hence, a space of these events is 3-dimensional.
Let ρ(t,x) be a probability density of event A(t,x), and
ρc(t,x|t0,x0)
be a probability density of event A(t,x) on condition that event
B(t0,x0).
In that case if function q(t,x|t0,x0) is fulfilled to condition:
ρc(t,x|t0,x0) = q(t,x|t0,x0)ρ(t,x), (3.47)
then one is called a disturbance function B to A.
If q = 1 then B does not disturbance to A.
A conditional probability density of event A(t,x) on condition that event B(t0,x0) is
presented as:
ρc = ϕ†
cϕc
like to a probability density of event A(t,x).
Let
ϕ =
ϕ1,1 +iϕ1,2
ϕ2,1 +iϕ2,2
ϕ3,1 +iϕ3,2
ϕ4,1 +iϕ4,2
and
ϕc =
ϕc,1,1 +iϕc,1,2
ϕc,2,1 +iϕc,2,2
ϕc,3,1 +iϕc,3,2
ϕc,4,1 +iϕc,4,2
(all ϕr,s and ϕc,r,s are real numbers).
In that case octavian
u = ϕ1,1 +ϕ1,2i+ϕ2,1j+ϕ2,2k+ϕ3,1E+ϕ3,2I+ϕ4,1J+ϕ4,2K
is called a Caylean of ϕ. Therefore, octavian
uc = ϕc,1,1 +ϕc,1,2i+ϕc,2,1j+ϕc,2,2k+ϕc,3,1E+ϕc,3,2I+ϕc,4,1J+ϕc,4,2K
156](https://image.slidesharecdn.com/fn-190805113833/85/TOE-170-320.jpg)
![is Caylean of ϕc.
In accordance with the octavian norm definition:
uc
2
= ρc
u 2
= ρ
(3.48)
Because the octavian algebra is a division algebra [72, 73] then for each octavians u and
uc there exists an octavian w such that
uc = wu,
Because the octavians algebra is normalized then
uc
2
= w 2
u 2
.
Hence, from (3.47) and (3.48):
q = w 2
.
Therefore, in a 3+1-dimensional space-time there exists an octavian-Caylean for a dis-
turbance function of any event to any event.
In order to increase a space dimensionality the octavian algebra can be expanded by a
Dickson doubling procedure:
Another 8 elements should be added to basic octavians:
z1,z2,z3,z4,z5,z6,z7,z8,
such that:
z2 = iz1,
z3 = jz1,
z4 = kz1,
z5 = Ez1,
z6 = Iz1,
z7 = Jz1,
z8 = Kz1,
and for every octavians u1, u2, v1, v2:
(u1 +u2z1)(v1 +v2z1) = (u1v1 −v2u2)+(v2u1 +u2v1)z1
(here: if w = w1 + w2i + w3j + w4k + w5E + w6I + w7J + w8K with real ws then w =
w1 −w2i−w3j−w4k−w5E−w6I−w7J−w8K).
It is a 16-dimensional Cayley-Dickson algebra.
In according with [58]: for any natural number z there exists a Clifford set of rank 2z.
In considering case for z = 3 there is Clifford’s seven:
157](https://image.slidesharecdn.com/fn-190805113833/85/TOE-171-320.jpg)
![β[1]
=
β[1] 04
04 −β[1] , β[2]
=
β[2] 04
04 −β[2] ,
β[3]
=
β[3] 04
04 −β[3] , β[4]
=
β[4] 04
04 −β[4] , (3.49)
β[5]
=
γ[0] 04
04 −γ[0] ,
β[6]
=
04 14
14 04
, β[7]
= i
04 −14
14 04
, (3.50)
Therefore, in this seven five 4-diagonal matrices (3.49) define a 5-dimensio-nal space
of events, and two 4-antidiagonal matrices (3.50) defined a 2-dimensi-onal space for the
electroweak transformations.
It is evident that such procedure of dimensions building up can be continued endlessly.
But in accordance with the Hurwitz theorem9 and with the generalized Frobenius theorem10
a more than 8-dimensional Cayley-Dickson algebra does not a division algebra. Hence,
there in a more than 3-dimensional space exist events such that a disturbance function be-
tween these events does not hold a Caylean. I call such disturbance supernatural.
Therefore, supernatural disturbance do not exist in a 3-dimensional space, but in a more
than 3-dimensional space such supernatural disturbance act.
3.2. Quarks and Gluons
The quark model was independently proposed by physicists Murray Gell-Mann11 and
George Zweig12 in 1964.
The first direct experimental evidence of gluons was found in 1979 when three-jet
events were observed at t he electron-positron collider PETRA. However, just before PE-
TRA13 appeared on the scene, the PLUTO experiment at DORIS14 showed event topologies
suggestive of a three-gluon decay.
9Every normalized algebra with unit is isomorphous to one of the following: the real numbers algebra R,
the complex numbers algebra C, the quaternions algebra K, the octavians algebra O [72]
10A division algebra can be only either 1 or 2 or 4 or 8-dimensional [73]
11Murray Gell-Mann (born September 15, 1929) is an American physicist and linguist
12George Zweig (born on May 30, 1937 in Moscow, Russia into a Jewish family) was originally trained as a
particle physicist under Richard Feynman and later turned his attention to neurobiology. He spent a number of
years as a Research Scientist at Los Alamos National Laboratory and MIT, but as of 2004, has gone on to work
in the financial services industry.
13PETRA (or the Positron-Electron Tandem Ring Accelerator) is one of the particle accelerators at DESY in
Hamburg, Germany.
14DORIS (Doppel-Ring-Speicher, ”double-ring storage”), built between 1969 and 1974, was DESY’s second
circular accelerator and its first storage ring with a circumference of nearly 300 m.
158](https://image.slidesharecdn.com/fn-190805113833/85/TOE-172-320.jpg)
![The following part of (2.29):
3
∑
k=0
β[k] −i∂k +Θk +ϒkγ[5] −
−Mζ,0γ
[0]
ζ
+Mζ,4ζ[4] +
−Mη,0γ
[0]
η −Mη,4η[4] +
+Mθ,0γ
[0]
θ +Mθ,4θ[4]
ϕ = 0. (3.51)
is called the chromatic equation of moving.
Here (2.5), (2.7), (2.9):
γ
[0]
ζ
= −
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
, ζ[4]
=
0 0 0 i
0 0 i 0
0 −i 0 0
−i 0 0 0
are mass elements of red pentad;
γ
[0]
η =
0 0 0 i
0 0 −i 0
0 i 0 0
−i 0 0 0
, η[4]
=
0 0 0 1
0 0 −1 0
0 −1 0 0
1 0 0 0
are mass elements of green pentad;
γ
[0]
θ =
0 0 −1 0
0 0 0 1
−1 0 0 0
0 1 0 0
, θ[4]
=
0 0 −i 0
0 0 0 i
−i 0 0 0
0 i 0 0
are mass elements of blue pentad.
I call:
• Mζ,0, Mζ,4 red lower and upper mass members;
• Mη,0, Mη,4 green lower and upper mass members;
• Mθ,0, Mθ,4 blue lower and upper mass members.
The mass members of this equation form the following matrix sum:
M :=
−Mζ,0γ
[0]
ζ
+Mζ,4ζ[4] −
−Mη,0γ
[0]
η −Mη,4η[4] +
+Mθ,0γ
[0]
θ +Mθ,4θ[4]
=
=
0 0 −Mθ,0 Mζ,η,0
0 0 M∗
ζ,η,0 Mθ,0
−Mθ,0 Mζ,η,0 0 0
M∗
ζ,η,0 Mθ,0 0 0
+i
0 0 Mθ,4 M∗
ζ,η,4
0 0 Mζ,η,4 −Mθ,4
−Mθ,4 −M∗
ζ,η,4 0 0
−Mζ,η,4 Mθ,4 0 0
159](https://image.slidesharecdn.com/fn-190805113833/85/TOE-173-320.jpg)
![with Mζ,η,0 := Mζ,0 −iMη,0 and Mζ,η,4 := Mζ,4 −iMη,4.
Elements of these matrices can be turned by formula of shape [52]:
cos θ
2 isin θ
2
isin θ
2 cos θ
2
Z X −iY
X +iY −Z
cos θ
2 −isin θ
2
−isin θ
2 cos θ
2
=
=
Z cosθ−Y sinθ X −i
Y cosθ
+Z sinθ
X +i
Y cosθ
+Z sinθ
−Z cosθ+Y sinθ
.
Hence, if:
U2,3 (α) :=
cosα isinα 0 0
isinα cosα 0 0
0 0 cosα isinα
0 0 isinα cosα
and
M :=
−Mζ,0γ
[0]
ζ
+Mζ,4ζ[4]−
−Mη,0γ
[0]
η −Mη,4η[4]+
+Mθ,0γ
[0]
θ +Mθ,4θ[4]
:= U†
2,3 (α)MU2,3 (α)
then
Mζ,0 = Mζ,0 ,
Mη,0 = Mη,0 cos2α+Mθ,0 sin2α,
Mθ,0 = Mθ,0 cos2α−Mη,0 sin2α,
Mζ,4 = Mζ,4 ,
Mη,4 = Mη,4 cos2α+Mθ,4 sin2α,
Mθ,4 = Mθ,4 cos2α−Mη,4 sin2α.
Therefore, matrix U2,3 (α) makes an oscillation between green and blue chromatics.
Let us consider equation (2.29) under transformation U2,3 (α) where α is an arbitrary
real function of time-space variables (α = α(t,x1,x2,x3)):
U†
2,3 (α)
1
c
∂t +iΘ0 +iϒ0γ[5]
U2,3 (α)ϕ =
= U†
2,3 (α)
3
∑
ν=1
β[ν] ∂ν +iΘν +iϒνγ[5] +
+iM0γ[0] +iM4β[4] +M
U2,3 (α)ϕ.
160](https://image.slidesharecdn.com/fn-190805113833/85/TOE-174-320.jpg)
![Because
U†
2,3 (α)U2,3 (α) = 14 ,
U†
2,3 (α)γ[5]U2,3 (α) = γ[5] ,
U†
2,3 (α)γ[0]U2,3 (α) = γ[0] ,
U†
2,3 (α)β[4]U2,3 (α) = β[4] ,
U†
2,3 (α)β[1] = β[1]U†
2,3 (α) ,
U†
2,3 (α)β[2] = β[2] cos2α+β[3] sin2α U†
2,3 (α) ,
U†
2,3 (α)β[3] = β[3] cos2α−β[2] sin2α U†
2,3 (α) ,
then
1
c
∂t +U†
2,3 (α)
1
c
∂tU2,3 (α)+iΘ0 +iϒ0γ[5]
ϕ =
=
β[1] ∂1 +U†
2,3 (α)∂1U2,3 (α)+ iΘ1 +iϒ1γ[5] +
+β[2]
(cos2α·∂2 −sin2α·∂3)
+U†
2,3 (α)(cos2α·∂2 − sin2α·∂3)U2,3 (α)
+i(Θ2 cos2α−Θ3 sin2α)
+i ϒ2γ[5] cos2α−ϒ3γ[5] sin2α
+β[3]
(cos2α·∂3 +sin2α·∂2)
+U†
2,3 (α)(cos2α·∂3 + sin2α·∂2)U2,3 (α)
+i(Θ2 sin2α+Θ3 cos2α)
+i ϒ3γ[5] cos2α+ϒ2γ[5] sin2α
+iM0γ[0] +iM4β[4] +M
ϕ. (3.52)
Let x2 and x3 be elements of other coordinate system such that 2.15:
∂2 : = (cos2α·∂2 −sin2α·∂3), (3.53)
∂3 : = (cos2α·∂3 +sin2α·∂2).
Therefore, from (3.52):
1
c
∂t +U†
2,3 (α)
1
c
∂tU2,3 (α)+iΘ0 +iϒ0γ[5]
ϕ =
=
β[1] ∂1 +U†
2,3 (α)∂1U2,3 (α)+iΘ1 +iϒ1γ[5]
+β[2] ∂2 +U†
2,3 (α)∂2U2,3 (α)+iΘ2 +iϒ2γ[5]
+β[3] ∂3 +U†
2,3 (α)∂3U2,3 (α)+iΘ3 +iϒ3γ[5]
+iM0γ[0] +iM4β[4] +M
ϕ.
161](https://image.slidesharecdn.com/fn-190805113833/85/TOE-175-320.jpg)
![with
Θ2 := Θ2 cos2α−Θ3 sin2α,
Θ3 := Θ2 sin2α+Θ3 cos2α,
ϒ2 := ϒ2 cos2α−ϒ3 sin2α,
ϒ3 := ϒ3 cos2α+ϒ2 sin2α.
Therefore, the oscillation between blue and green chromatics curves the space in the x2,
x3 directions.
Similarly, matrix
U1,3 (ϑ) :=
cosϑ sinϑ 0 0
−sinϑ cosϑ 0 0
0 0 cosϑ sinϑ
0 0 −sinϑ cosϑ
with an arbitrary real function ϑ(t,x1,x2,x3) describes the oscillation between blue and red
chromatics which curves the space in the x1, x3 directions. And matrix
U1,2 (ς) :=
e−iς 0 0 0
0 eiς 0 0
0 0 e−iς 0
0 0 0 eiς
with an arbitrary real function ς(t,x1,x2,x3) describes the oscillation between green and red
chromatics which curves the space in the x1, x2 directions.
Now, let
U0,1 (σ) :=
coshσ −sinhσ 0 0
−sinhσ coshσ 0 0
0 0 coshσ sinhσ
0 0 sinhσ coshσ
.
and
M :=
−Mζ,0γ
[0]
ζ
+Mζ,4ζ[4]−
−Mη,0γ
[0]
η −Mη,4η[4]+
+Mθ,0γ
[0]
θ +Mθ,4θ[4]
:= U†
0,1 (σ)MU0,1 (σ)
then:
Mζ,0 = Mζ,0 ,
Mη,0 = (Mη,0 cosh2σ−Mθ,4 sinh2σ) ,
Mθ,0 = Mθ,0 cosh2σ+Mη,4 sinh2σ,
Mζ,4 = Mζ,4 ,
Mη,4 = Mη,4 cosh2σ+Mθ,0 sinh2σ,
Mθ,4 = Mθ,4 cosh2σ−Mη,0 sinh2σ.
Therefore, matrix U0,1 (σ) makes an oscillation between green and blue chromatics with
an oscillation between upper and lower mass members.
162](https://image.slidesharecdn.com/fn-190805113833/85/TOE-176-320.jpg)
![Let us consider equation (2.29) under transformation U0,1 (σ) where σ is an arbitrary
real function of time-space variables (σ = σ(t,x1,x2,x3)):
U†
0,1 (σ)
1
c
∂t +iΘ0 +iϒ0γ[5]
U0,1 (σ)ϕ =
= U†
0,1 (σ)
3
∑
ν=1
β[ν] ∂ν +iΘν +iϒνγ[5] +
+iM0γ[0] +iM4β[4] +M
U0,1 (σ)ϕ.
Since:
U†
0,1 (σ)U0,1 (σ) = cosh2σ−β[1]
sinh2σ ,
U†
0,1 (σ) = cosh2σ+β[1]
sinh2σ U−1
0,1 (σ),
U†
0,1 (σ)β[1]
= β[1]
cosh2σ−sinh2σ U−1
0,1 (σ),
U†
0,1 (σ)β[2]
= β[2]
U−1
0,1 (σ),
U†
0,1 (σ)β[3]
= β[3]
U−1
0,1 (σ),
U†
0,1 (σ)γ[0]
U0,1 (σ) = γ[0]
,
U†
0,1 (σ)β[4]
U0,1 (σ) = β[4]
,
U−1
0,1 (σ)U0,1 (σ) = 14 ,
U−1
0,1 (σ)γ[5]
U0,1 (σ) = γ[5]
,
U†
0,1 (σ)γ[5]
U0,1 (σ) = γ[5]
cosh2σ−β[1]
sinh2σ ,
then
U−1
0,1 (σ) cosh2σ· 1
c ∂t +sinh2σ·∂1 U0,1 (σ)
+ cosh2σ· 1
c ∂t +sinh2σ·∂1
+i(Θ0 cosh2σ+Θ1 sinh2σ)
+i(ϒ0 cosh2σ+sinh2σ·ϒ1)γ[5]−
−β[1]
U−1
0,1 (σ) cosh2σ·∂1 +sinh2σ· 1
c ∂t U0,1 (σ)
+ cosh2σ·∂1 +sinh2σ· 1
c ∂t
+i(Θ1 cosh2σ+Θ0 sinh2σ)
+i(ϒ1 cosh2σ+ϒ0 sinh2σ)γ[5]
−β[2] ∂2 +U−1
0,1 (σ)(∂2U0,1 (σ))+ iΘ2 +iϒ2γ[5]
−β[3] ∂3 +U−1
0,1 (σ)(∂3U0,1 (σ))+ iΘ3 +iϒ3γ[5]
−iM0γ[0] −iM4β[4] −M
ϕ = 0. (3.54)
163](https://image.slidesharecdn.com/fn-190805113833/85/TOE-177-320.jpg)
![Let t and x1 be elements of other coordinate system such that:
∂x1
∂x1
= cosh2σ
∂t
∂x1
=
1
c
sinh2σ
∂x1
∂t
= csinh2σ
∂t
∂t
= cosh2σ
∂x2
∂t
=
∂x3
∂t
=
∂x2
∂x1
=
∂x3
∂x1
= 0
. (3.55)
Hence:
∂t :=
∂
∂t
=
∂
∂t
∂t
∂t
+
∂
∂x1
∂x1
∂t
+
∂
∂x2
∂x2
∂t
+
∂
∂x3
∂x3
∂t
=
= cosh2σ·
∂
∂t
+csinh2σ·
∂
∂x1
=
= cosh2σ·∂t +csinh2σ·∂1 ,
that is
1
c
∂t =
1
c
cosh2σ·∂t +sinh2σ·∂1
and
∂1 :=
∂
∂x1
=
=
∂
∂t
∂t
∂x1
+
∂
∂x1
∂x1
∂x1
+
∂
∂x2
∂x2
∂x1
+
∂
∂x3
∂x3
∂x1
=
= cosh2σ·
∂
∂x1
+sinh2σ·
1
c
∂
∂t
=
= cosh2σ·∂1 +sinh2σ·
1
c
∂t .
Therefore, from (3.54):
β[0] 1
c ∂t +U−1
0,1 (σ) 1
c ∂tU0,1 (σ)+ iΘ0 +iϒ0γ[5]
+β[1] ∂1 +U−1
0,1 (σ)∂1U0,1 (σ)+ iΘ1 +iϒ1γ[5]
+β[2] ∂2 +U−1
0,1 (σ)∂2U0,1 (σ)+ iΘ2 +iϒ2γ[5]
+β[3] ∂3 +U−1
0,1 (σ)∂3U0,1 (σ)+ iΘ3 +iϒ3γ[5]
+iM0γ[0] +iM4β[4] +M
ϕ = 0
164](https://image.slidesharecdn.com/fn-190805113833/85/TOE-178-320.jpg)
![with
Θ0 := Θ0 cosh2σ+Θ1 sinh2σ,
Θ1 := Θ1 cosh2σ+Θ0 sinh2σ,
ϒ0 := ϒ0 cosh2σ+sinh2σ·ϒ1 ,
ϒ1 := ϒ1 cosh2σ+ϒ0 sinh2σ.
Therefore, the oscillation between blue and green chromatics with the oscillation be-
tween upper and lower mass members curves the space in the t, x1 directions.
Similarly, matrix
U0,2 (φ) :=
coshφ isinhφ 0 0
−isinhφ coshφ 0 0
0 0 coshφ −isinhφ
0 0 isinhφ coshφ
with an arbitrary real function φ(t,x1,x2,x3) describes the oscillation between blue and red
chromatics with the oscillation between upper and lower mass members curves the space in
the t, x2 directions. And matrix
U0,3 (ι) :=
eι 0 0 0
0 e−ι 0 0
0 0 e−ι 0
0 0 0 eι
with an arbitrary real function ι(t,x1,x2,x3) describes the oscillation between green and red
chromatics with the oscillation between upper and lower mass members curves the space in
the t, x3 directions.
Now let
U (χ) :=
eiχ 0 0 0
0 eiχ 0 0
0 0 e2iχ 0
0 0 0 e2iχ
and
M :=
−Mζ,0γ
[0]
ζ
+Mζ,4ζ[4] −
−Mη,0γ
[0]
η −Mη,4η[4] +
+Mθ,0γ
[0]
θ +Mθ,4θ[4]
:= U†
(χ)MU (χ)
then:
Mζ,0 = Mζ,0 cosχ−Mζ,4 sinχ ,
Mζ,4 = Mζ,4 cosχ+Mζ,0 sinχ ,
Mη,4 = (Mη,4 cosχ−Mη,0 sinχ),
Mη,0 = (Mη,0 cosχ+Mη,4 sinχ),
Mθ,0 = (Mθ,0 cosχ+Mθ,4 sinχ),
Mθ,4 = (Mθ,4 cosχ−Mθ,0 sinχ).
165](https://image.slidesharecdn.com/fn-190805113833/85/TOE-179-320.jpg)
![Therefore, matrix U (χ) makes an oscillation between upper and lower mass members.
Let us consider equation (3.51) under transformation U (χ) where χ is an arbitrary real
function of time-space variables (χ = χ(t,x1,x2,x3)):
U†
(χ)
1
c
∂t +iΘ0 +iϒ0γ[5]
U (χ)ϕ =
= U†
(χ)
3
∑
ν=1
β[ν]
∂ν +iΘν +iϒνγ[5]
+M U (χ)ϕ.
Because
γ[5]U (χ) = U (χ)γ[5] ,
β[1]U (χ) = U (χ)β[1] ,
β[2]U (χ) = U (χ)β[2] ,
β[3]U (χ) = U (χ)β[3] ,
U† (χ)U (χ) = 14 ,
then
1
c
∂t +
1
c
U†
(χ) ∂tU (χ) +iΘ0 +iϒ0γ[5]
ϕ =
=
3
∑
ν=1
β[ν] ∂ν +U† (χ) ∂νU (χ) + iΘν +iϒνγ[5]
+U† (χ)MU (χ)
ϕ.
Now let:
U (κ) :=
eκ 0 0 0
0 eκ 0 0
0 0 e2κ 0
0 0 0 e2κ
and
M :=
−Mζ,0γ
[0]
ζ
+Mζ,4ζ[4]−
−Mη,0γ
[0]
η −Mη,4η[4]+
+Mθ,0γ
[0]
θ +Mθ,4θ[4]
:= U−1
(κ)MU (κ)
then:
Mθ,0 = (Mθ,0 coshκ−iMθ,4 sinhκ),
Mθ,4 = (Mθ,4 coshκ+iMθ,0 sinhκ),
Mη,0 = (Mη,0 coshκ−iMη,4 sinhκ),
Mη,4 = (Mη,4 coshκ+iMη,0 sinhκ),
Mζ,0 = Mζ,0 coshκ+iMζ,4 sinhκ ,
Mζ,4 = Mζ,4 coshκ−iMζ,0 sinhκ .
166](https://image.slidesharecdn.com/fn-190805113833/85/TOE-180-320.jpg)
![Therefore, matrix U (κ) makes an oscillation between upper and lower mass members,
too.
Let us consider equation (3.51) under transformation U (κ) where κ is an arbitrary real
function of time-space variables (κ = κ(t,x1,x2,x3)):
U−1
(κ)
1
c
∂t +iΘ0 +iϒ0γ[5]
U (κ)ϕ =
= U−1
(κ)
3
∑
ν=1
β[ν]
∂ν +iΘν +iϒνγ[5]
+ M U (κ)ϕ.
Because
γ[5]U (κ) = U (κ)γ[5] ,
U−1 (κ)β[1] = β[1]U−1 (κ) ,
U−1 (κ)β[2] = β[2]U−1 (κ),
U−1 (κ)β[3] = β[3]U−1 (κ),
U−1 (κ)U (κ) = 14 ,
then
1
c
∂t +U−1
(κ)
1
c
∂tU (κ) +iΘ0 +iϒ0γ[5]
ϕ =
=
3
∑
ν=1
β[ν] ∂ν +U−1 (κ) ∂νU (κ) + iΘν +iϒνγ[5] +
+U−1 (κ)MU (κ)
ϕ.
If denote:
Λ1 :=
0 −1 0 0
−1 0 0 0
0 0 0 1
0 0 1 0
,
Λ2 :=
0 i 0 0
i 0 0 0
0 0 0 i
0 0 i 0
,
Λ3 :=
0 1 0 0
−1 0 0 0
0 0 0 1
0 0 −1 0
,
Λ4 :=
0 i 0 0
−i 0 0 0
0 0 0 −i
0 0 i 0
,
167](https://image.slidesharecdn.com/fn-190805113833/85/TOE-181-320.jpg)
![U−1 (κ)Λ6U (κ) = Λ6
U−1
1,2 (ς)Λ6U1,2 (ς) = Λ6
U−1
0,2 (φ)Λ6U0,2 (φ) = Λ6 cosh2φ+Λ2 sinh2φ
U−1
1,3 (ϑ)Λ6U1,3 (ϑ) = Λ6 cos2ϑ−Λ1 sin2ϑ
U−1
2,3 (α)Λ6U2,3 (α) = Λ6 cos2α+Λ4 sin2α
U−1
0,1 (σ)Λ6U0,1 (σ) = Λ6 cosh2σ−Λ3 sinh2σ
========
U−1 (χ) Λ7U (χ) = Λ7
U−1
0,3 (ι) Λ7U0,3 (ι) = Λ7
U−1
1,2 (ς) Λ7U1,2 (ς) = Λ7
U−1
0,2 (φ) Λ7U0,2 (φ) = Λ7
U−1
1,3 (ϑ) Λ7U1,3 (ϑ) = Λ7
U−1
2,3 (α) Λ7U2,3 (σ) = Λ7
U−1
0,1 (σ) Λ7U0,1 (σ) = Λ7
=========
U−1
0,1 (σ) Λ8U0,1 (σ) = Λ8
U−1
2,3 (α) Λ8U2,3 (α) = Λ8
U−1
1,3 (ϑ) Λ8U1,3 (ϑ) = Λ8
U−1
0,2 (φ) Λ8U0,2 (φ) = Λ8
U−1
1,2 (ς) Λ8U1,2 (ς) = Λ8
U−1
0,3 (ι) Λ8U0,3 (ι) = Λ8
U−1 (κ) Λ8U (κ) = Λ8
then for every product U of `U’s elements real functions Gr
s (t,x1,x2,x3) exist such that
U−1
(∂sU) =
g3
2
8
∑
r=1
ΛrGr
s
with some real constant g3 (similar to 8 gluons).
The chrome states equations of moving are the following [71, p.86], (3.43):
∑3
µ=0 −ζ[k] ∂µ + −γ
[0]
ζ
∂
ζ
y +ζ[4]∂
ζ
z
+∑3
µ=0 β[µ] −sAµ +0.5 Zµ +Wµ
ξ = 0, (3.56)
∑3
µ=0 −η[k] ∂µ + −γ
[0]
η ∂η
y −η[4]∂η
z
+∑3
µ=0 β[µ] −sAµ +0.5 Zµ +Wµ
ξ = 0,
∑3
µ=0 −θ[k] ∂µ + γ
[0]
θ ∂θ
y +θ[4]∂θ
z
+∑3
µ=0 β[µ] −sAµ +0.5 Zµ +Wµ
ξ = 0;
170](https://image.slidesharecdn.com/fn-190805113833/85/TOE-184-320.jpg)
![here:
ξ :=
dL
dR
uL
uR
(
dL
dR
is a lower chrome state, and
uL
uR
is a upper chrome state);
s is an electric charge, and [71, p.145]
Aµ := Aµ
02 02 02 02
02 12 02 02
02 02 12 02
02 02 02 12
.
Hence,
sAµξ = sAµ
02 02 02 02
02 12 02 02
02 02 12 02
02 02 02 12
dL
dR
uL
uR
= sAµ
−→
0 2
dR
uL
uR
,
−→
0 2 : =
0
0
.
Sum of the equations (3.56) is the following:
∑3
µ=0 − ζ[k] +η[k] +θ[k] ∂µ+
+
−γ
[0]
ζ
∂
ζ
y +ζ[4]∂
ζ
z
−γ
[0]
η ∂η
y −η[4]∂η
z
+γ
[0]
θ ∂θ
y +θ[4]∂θ
z
+
∑3
µ=0 β[µ] (−3s)Aµ
+0.5·3∑3
µ=0 β[µ] Zµ +Wµ
ξ = 0.
Because β[k] = − ζ[k] +η[k] +θ[k] rhen
∑3
µ=0 β[k]∂µ+
+
−γ
[0]
ζ
∂
ζ
y +ζ[4]∂
ζ
z
−γ
[0]
η ∂η
y −η[4]∂η
z
+γ
[0]
θ ∂θ
y +θ[4]∂θ
z
+
∑3
µ=0 β[µ] (−3s)Aµ
+0.5·3∑3
µ=0 β[µ] Zµ +Wµ
ξ = 0.
Hence, from (3.42):
(−3s) =
g1g2
g2
1 +g2
2
171](https://image.slidesharecdn.com/fn-190805113833/85/TOE-185-320.jpg)
![that is:
s = −
1
3
g1g2
g2
1 +g2
2
= −
1
3
e
(e is a lepton electric charge).
Because
sAµξ = −
1
3
eAµ
−→
0 2
dR
uL
uR
then a charge of
uL
uR
is −2
3 e and a module of charge of
−→
0 2
dR
is 1
3 e.
3.3. Asymptotic Freedom, Confinement, Gravitation
The Quarks Asymptotic Freedom phenomenon and the Quarks Confinement phenomenon
has been was discovered by J. Friedman15, H. Kendall16, R. Taylor17 at SLAC in the late
1960s and early 1970s.
Researches of the phenomenon of gravitation were spent by Galileo Galilei18 in the late
16th and early 17th centuries, by Isaac Newton19 in 17th centuries, by A. Einstein20 in 20th
centuries.
From (3.55):
∂t
∂t
= cosh2σ, (3.57)
∂x
∂t
= csinh2σ.
Hence, if v is the velocity of a coordinate system {t ,x } in the coordinate system {t,x}
then
sinh2σ =
v
c
1− v
c
2
, cosh2σ =
1
1− v
c
2
.
Therefore,
15Jerome Isaac Friedman (born March 28, 1930) is an American physicist.
16Henry Way Kendall (December 9, 1926 – February 15, 1999) was an American particle physicist
17Richard Edward Taylor (born November 2, 1929 in Medicine Hat, Alberta) is a Canadian-American pro-
fessor (Emeritus) at Stanford University.
18Galileo Galilei ( 15 February 1564[4] – 8 January 1642), was an Italian physicist, mathematician, as-
tronomer, and philosopher
19Sir Isaac Newton PRS (25 December 1642 – 20 March 1727 was an English physicist, mathematician,
astronomer, natural philosopher, alchemist, and theologian
20Albert Einstein ( 14 March 1879 – 18 April 1955) was a German-born theoretical physicist
172](https://image.slidesharecdn.com/fn-190805113833/85/TOE-186-320.jpg)
![Figure 29:
• interval from S to ∞ the Newton Gravity Zone,
• interval from B to C the the Zone,
• and interval from C to D the Confinement Force Zone.
3.3.1. Dark Energy
In 1998 observations of Type Ia supernovae suggested that the expansion of the universe
is accelerating [61]. In the past few years, these observations have been corroborated by
several independent sources [62]. This expansion is defined by the Hubble21 rule [63]:
V (r) = Hr, (3.60)
here V (r) is the velocity of expansion on the distance r, H is the Hubble’s constant
(H ≈ 2.3×10−18c−1 [64]).
Let a black hole be placed in a point O. Then a tremendous number of quarks oscillate
in this point. These oscillations bend time-space and if t has some fixed volume, x > 0, and
Λ := λt then
v(x) = ctanh
Λ
x2
. (3.61)
A dependency of v(x) (light years/c) from x (light years) with Λ = 741.907 is shown in
Figure 30.
Let a placed in a point A observer be stationary in the coordinate system {t,x}. Hence,
in the coordinate system {t ,x } this observer is flying to the left to the point O with velocity
−v(xA). And point X is flying to the left to the point O with velocity −v(x).
21Edwin Powell Hubble (November 20, 1889 September 28, 1953)[1] was an American astronomer
174](https://image.slidesharecdn.com/fn-190805113833/85/TOE-188-320.jpg)
![with a number that was 400 times his original estimate. This discrepancy in the observed
and computed masses is now known as ”the missing mass problem.” Nobody did much with
Zwicky’s finding until the 1970’s, when scientists began to realize that only large amounts
of hidden mass could explain many of their observations. Scientists also realize that the
existence of some unseen mass would also support theories regarding the structure of the
universe. Today, scientists are searching for the mysterious dark matter not only to explain
the gravitational motions of galaxies, but also to validate current theories about the origin
and the fate of the universe” [65] (Figure 33 [66], Figure 34 [67]).
Figure 33: A rotation curve for a typical spiral galaxy. The solid line shows actual mea-
surements (Hawley and Holcomb., 1998, p. 390) [66]
Some oscillations of chromatic states bend space-time as follows (2.15):
∂
∂x
= cos2α·
∂
∂x
−sin2α·
∂
∂y
, (3.65)
∂
∂y
= cos2α·
∂
∂y
+sin2α·
∂
∂x
.
Let
z : = x+iy,z = reiθ
.
z : = x +iy .
Because linear velocity of the curved coordinate system x ,y into the initial system
x,y is the following23:
v =
•
x
2
+
•
y
2
23
•
x := ∂x
∂t
,
•
y := ∂y
∂t
.
177](https://image.slidesharecdn.com/fn-190805113833/85/TOE-191-320.jpg)
![Figure 36 shows the dependence of velocity v on the radius r at large t ∼ 104 and at
θ = 0.98π
Hence, Dark Matter and Dark Energy can be mirages in the space-time, which is curved
by oscillations of chromatic states.
3.3.3. Baryon Chrome
According to the quark model, [70] the properties of hadrons are primarily determined
by their so-called valence quarks. For example, a proton is composed of two up quarks
and one down quark. Although quarks also carry color charge, hadrons must have zero
total color charge because of a phenomenon called color Confinement. That is, hadrons
must be “colorless” or “white”. These are the simplest of the two ways: three quarks
of different colors, or a quark of one color and an antiquark carrying the corresponding
anticolor. Hadrons with the first arrangement are called baryons, and those with the second
arrangement are mesons.
Let α be any real number and
x0 := x0,
x1 := x1 cos(α)−x2 sin(α);
x2 := x1 sin(α)+x2 cos(α); (3.66)
x3 := x3;
Since jA is a 3+1-vector then from [71, p.59]:
jA,0 = −ϕ†
β[0]
ϕ,
jA,1 = −ϕ†
β[1]
cos(α)−β[2]
sin(α) ϕ; (3.67)
jA,2 = −ϕ†
β[1]
sin(α)+β[2]
cos(α) ϕ;
jA,3 = −ϕ†
β[3]
ϕ.
Hence if for ϕ :
jA,0 = −ϕ †β[0]ϕ ,
jA,1 = −ϕ †β[1]ϕ ;
jA,2 = −ϕ †β[2]ϕ ;
jA,3 = −ϕ †β[3]ϕ ,
and
ϕ := U1,2 (α)ϕ
then
U†
1,2 (α)β[0]
U1,2 (α) = β[0]
,
U†
1,2 (α)β[1]
U1,2 (α) = β[1]
cosα−β[2]
sinα;
U†
1,2 (α)β[2]
U1,2 (α) = β[2]
cosα+β[1]
sinα; (3.68)
U†
1,2 (α)β[3]
U1,2 (α) = β[3]
;
181](https://image.slidesharecdn.com/fn-190805113833/85/TOE-195-320.jpg)
![from [71, p.62]: because
ρA = ϕ†
ϕ = ϕ †
ϕ ,
then
U†
1,2 (α)U1,2 (α) = 14. (3.69)
If
U1,2 (α) := cos
α
2
·14 −sin
α
2
·β[1]
β[2]
i.e. 2.14:
U1,2 (α) =
e−i1
2 α 0 0 0
0 ei1
2 α 0 0
0 0 e−i1
2 α 0
0 0 0 ei1
2 α
(3.70)
then U1,2 (α) fulfils to all these conditions (3.68), (3.69).
Then let
x0 := x0,
x1 := x1 cos(α)−x3 sin(α),
x2 := x2, (3.71)
x3 := x1 sin(α)+x3 cos(α).
Let
U1,3 (α) := cos
α
2
·14 −sin
α
2
·β[1]
β[3]
.
In rhis case 2.13:
U1,3 (α) =
cos 1
2α sin 1
2α 0 0
−sin 1
2 α cos 1
2 α 0 0
0 0 cos 1
2 α sin 1
2 α
0 0 −sin 1
2α cos 1
2α
(3.72)
and
U†
1,3 (α)β[0]
U1,3 (α) = β[0]
,
U†
1,3 (α)β[1]
U1,3 (α) = β[1]
cosα−β[3]
sinα, (3.73)
U†
1,3 (α)β[2]
U1,3 (α) = β[2]
,
U†
1,3 (α)β[3]
U1,3 (α) = β[3]
cosα+β[1]
sinα.
If
ϕ := U1,3 (α)ϕ
and
182](https://image.slidesharecdn.com/fn-190805113833/85/TOE-196-320.jpg)
![jA,k := ϕ †
β[k]
ϕ
where (k ∈ {0,1,2,3}) then
jA,0 = jA,0, (3.74)
jA,1 = jA,1 cosα− jA,3 sinα, (3.75)
jA,2 = jA,2,
jA,3 = jA,3 cosα+ jA,1 sinα,
Then let
x0 : = x0,
x1 : = x1,
x2 = cosα·x2 +sinα·x3, (3.76)
x3 = cosα·x3 −sinα·x2.
Let
U3,2 (α) = cos
α
2
·14 −sin
α
2
·β[3]
β[2]
In this case:
U3,2 (α) =
cos 1
2α isin 1
2 α 0 0
isin 1
2α cos 1
2 α 0 0
0 0 cos 1
2 α isin 1
2 α
0 0 isin 1
2 α cos 1
2 α
, (3.77)
and
U†
3,2 (α)β[0]
U3,2 (α) = β[0]
,
U†
3,2 (α)β[1]
U3,2 (α) = β[1]
,
U†
3,2 (α)β[0]
U3,2 (α) = β[0]
cosα+β[3]
sinα, (3.78)
U†
3,2 (α)β[3]
U3,2 (α) = β[3]
cosα−β[2]
sinα
If
ϕ := U3,2 (α)ϕ
and
jA,k := ϕ †
β[k]
ϕ
where (k ∈ {0,1,2,3}) then
jA,0 = jA,0,
jA,1 = jA,1, (3.79)
jA,2 = jA,2 cosα+ jA,3 sinα,
jA,3 = jA,3 cosα− jA,1 sinα,
183](https://image.slidesharecdn.com/fn-190805113833/85/TOE-197-320.jpg)
![Let v be any real number such that −1 < v < 1.
And let:
α :=
1
2
ln
1−v
1+v
.
In this case:
coshα =
1
√
1−v2
,
sinhα = −
v
√
1−v2
. (3.80)
Let
x0 : = x0 coshα−x1 sinhα, (3.81)
x1 : = x1 coshα−x0 sinhα,
x2 : = x2,
x3 : = x3.
Let
U1,0 (α) = cosh
α
2
·14 −sinh
α
2
·β[1]
β[0]
.
That is 2.10:
U1,0 (α) :=
cosh 1
2α sinh 1
2α 0 0
sinh 1
2 α cosh 1
2 α 0 0
0 0 cosh 1
2α −sinh 1
2 α
0 0 −sinh 1
2 α cosh 1
2 α
. (3.82)
In rhis case:
U†
1,0 (α)β[0]
U1,0 (α) = β[0]
coshα−β[1]
sinhα, (3.83)
U†
1,0 (α)β[1]
U1,0 (α) = β[1]
coshα−β[0]
sinhα,
U†
1,0 (α)β[2]
U1,0 (α) = β[2]
,
U†
1,0 (α)β[3]
U1,0 (α) = β[3]
.
If
ϕ := U1,0 (α)ϕ
and
jA,k := ϕ †
β[k]
ϕ
184](https://image.slidesharecdn.com/fn-190805113833/85/TOE-198-320.jpg)
![where (k ∈ {0,1,2,3}) then
jA,0 = jA,0 coshα− jA,1 sinhα, (3.84)
jA,1 = jA,1 coshα− jA,0 sinhα,
jA,2 = jA,2,
jA,3 = jA,3.
Then let
x0 : = x0 coshα−x2 sinhα, (3.85)
x1 : = x1,
x2 : = x2 coshα−x0 sinhα,
x3 : = x3.
Let
U2,0 (α) := cosh
α
2
·14 −sinh
α
2
·β[2]
β[0]
. (3.86)
That is:
U2,0 (α) =
cosh 1
2α −isinh 1
2 α 0 0
isinh 1
2 α cosh 1
2 α 0 0
0 0 cosh 1
2α isinh 1
2α
0 0 −isinh 1
2α cosh 1
2 α
.
In rhis case:
U†
2,0 (α)β[0]
U2,0 (α) = β[0]
coshα−β[2]
sinhα, (3.87)
U†
2,0 (α)β[1]
U1,0 (α) = β[1]
,
U†
2,0 (α)β[2]
U1,0 (α) = β[2]
coshα−β[0]
sinhα,
U†
2,0 (α)β[3]
U2,0 (α) = β[3]
.
If
ϕ := U2,0 (α)ϕ
and
jA,k := ϕ †
β[k]
ϕ
where (k ∈ {0,1,2,3}) then
185](https://image.slidesharecdn.com/fn-190805113833/85/TOE-199-320.jpg)
![jA,0 = jA,0 coshα− jA,1 sinhα, (3.88)
jA,1 = jA,1,
jA,2 = jA,2 coshα− jA,0 sinhα,
jA,3 = jA,3.
Then let
x0 : = x0 coshα−x3 sinhα, (3.89)
x1 : = x1,
x2 : = x2,
x3 : = x3 coshα−x0 sinhα.
Let
U3,0 (α) := cosh
α
2
·14 −sinh
α
2
·β[3]
β[0]
.
That is:
U3,0 (α) =
e
1
2 α 0 0 0
0 e−1
2 α 0 0
0 0 e−1
2 α 0
0 0 0 e
1
2 α
. (3.90)
In rhis case:
U†
3,0 (α)β[0]
U3,0 (α) = β[0]
coshα−β[3]
sinhα, (3.91)
U3,0 (α)β[1]
U3,0 (α) = β[1]
,
U3,0 (α)β[2]
U3,0 (α) = β[2]
,
U3,0 (α)β[3]
U3,0 (α) = β[3]
coshα−β[0]
sinhα.
If
ϕ := U3,0 (α)ϕ
and
jA,k := ϕ †
β[k]
ϕ
where (k ∈ {0,1,2,3}) then
jA,0 = jA,0 coshα− jA,3 sinhα, (3.92)
jA,1 = jA,1,
jA,2 = jA,2.
jA,3 = jA,3 coshα− jA,0 sinhα.
186](https://image.slidesharecdn.com/fn-190805113833/85/TOE-200-320.jpg)
![Function ϕ submits to the following equation:[71, p.82]
1
c ∂tϕ− iΘ0β[0] +iϒ0β[0]γ[5] ϕ =
=
(
3
∑
ν=1
β[ν] ∂ν +iΘν +iϒνγ[5] +
+iM0γ[0] +iM4β[4]−
−iMζ,0γ
[0]
ζ
+iMζ,4ζ[4]−
−iMη,0γ
[0]
η −iMη,4η[4]+
+iMθ,0γ
[0]
θ +iMθ,4θ[4])ϕ
.
That is:
(
3
∑
ν=0
β[ν] ∂ν +iΘν +iϒνγ[5] +
+iM0γ[0] +iM4β[4]−
−iMζ,0γ
[0]
ζ
+iMζ,4ζ[4]−
−iMη,0γ
[0]
η −iMη,4η[4]+
+iMθ,0γ
[0]
θ +iMθ,4θ[4])ϕ = 0.
(3.93)
Like coordinates x5 and x4 [71, p.83] here are entered new coordinates yβ, zβ, yζ, zζ, yη,
zη, yθ, zθ such that
−
πc
h
≤ yβ
≤
πc
h
,−
πc
h
≤ zβ
≤
πc
h
,
−
πc
h
≤ yζ
≤
πc
h
,−
πc
h
≤ zζ
≤
πc
h
,
−
πc
h
≤ yη
≤
πc
h
,−
πc
h
≤ zη
≤
πc
h
,
−
πc
h
≤ yθ
≤
πc
h
,−
πc
h
≤ zθ
≤
πc
h
.
and like ϕ, [71, p.83] let:
[ϕ] t,x,yβ
,zβ
,yζ
,zζ
,yη
,zη
,yθ
,zθ
:= (3.94)
: = ϕ(t,x)×exp(i(yβ
M0 +zβ
M4 +yζ
Mζ,0 +zζ
Mζ,4 +
+yη
Mη,0 +zη
Mη,4 +yθ
Mθ,0 +zθ
Mθ,4)),
In fhis case if
([ϕ],[χ]) :=
:=
πc
h
−πc
h
dyβ
πc
h
−πc
h
dzβ
πc
h
−πc
h
dyζ
πc
h
−πc
h
dzζ
×
×
πc
h
−πc
h
dyη
πc
h
−πc
h
dzη
πc
h
−πc
h
dyθ
πc
h
−πc
h
dzθ
×
× [ϕ]†
[χ]
(3.95)
187](https://image.slidesharecdn.com/fn-190805113833/85/TOE-201-320.jpg)
![then
([ϕ],[ϕ]) = ρA, (3.96)
[ϕ],β[s]
[ϕ] = −
jA,k
c
.
and in this case from (3.93 ):
(
3
∑
ν=0
β[ν] ∂ν +iΘν +iϒνγ[5] +
+γ[0]∂
β
y +β[4]∂
β
z −
−γ
[0]
ζ
∂
ζ
y +ζ[4]∂
ζ
z −
−γ
[0]
η ∂η
y −η[4]∂η
z +
+γ
[0]
θ ∂θ
y +θ[4]∂θ
z )[ϕ] = 0
. (3.97)
Because
γ
[0]
η =
0 0 0 i
0 0 −i 0
0 i 0 0
−i 0 0 0
, η[4]
= i
0 0 0 −i
0 0 i 0
0 i 0 0
−i 0 0 0
; (3.98)
γ
[0]
θ =
0 0 −1 0
0 0 0 1
−1 0 0 0
0 1 0 0
, θ[4]
= i
0 0 1 0
0 0 0 −1
−1 0 0 0
0 1 0 0
(3.99)
γ
[0]
ζ
=
0 0 0 −1
0 0 −1 0
0 −1 0 0
−1 0 0 0
, ζ[4]
= i
0 0 0 1
0 0 1 0
0 −1 0 0
−1 0 0 0
; (3.100)
then from (3.97):
3
∑
ν=0
β[ν] ∂ν +iΘν +iϒνγ[5] [ϕ]
+γ[0]∂
β
y [ϕ]+β[4]∂
β
z [ϕ]+
(
0 0 −∂θ
y ∂
ζ
y −i∂η
y
0 0 ∂
ζ
y +i∂η
y ∂θ
y
−∂θ
y ∂
ζ
y −i∂η
y 0 0
∂
ζ
y +i∂η
y ∂θ
y 0 0
+
i
0 0 ∂θ
z ∂
ζ
z +i∂η
z
0 0 ∂
ζ
z −i∂η
z −∂θ
z
−∂θ
z −∂
ζ
z −i∂η
z 0 0
−∂
ζ
z +i∂η
z ∂θ
z 0 0
)
×[ϕ] = 0.
(3.101)
188](https://image.slidesharecdn.com/fn-190805113833/85/TOE-202-320.jpg)
![Let a Fourier transformation of
[ϕ] t,x,yβ
,zβ
,yζ
,zζ
,yη
,zη
,yθ
,zθ
be the following;
[ϕ] t,x,yβ
,zβ
,yζ
,zζ
,yη
,zη
,yθ
,zθ
=
= ∑
w,p1,p2,p3,nβ,sβ,nζ,sζ,nη,sη,nθ,sθ
c(w, p1, p2, p3,nβ
,sβ
,
nζ
,sζ
,nη
,sη
,nθ
,sθ
)×
×exp(−i
h
c
(wx0 + p1x1 + p2x2 + p3x3 + (3.102)
+nβ
yβ
+sβ
zβ
+nζ
yζ
+sζ
zζ
+
+nη
yη
+sη
zη
+nθ
yθ
+sθ
zθ
)).
Let in (3.101) Θν = 0 and ϒν = 0.
Let us designe:
G0 := (
3
∑
ν=0
β[ν]∂ν +γ[0]∂
β
y +β[4]∂
β
z −
−γ
[0]
ζ
∂
ζ
y +ζ[4]∂
ζ
z −
−γ
[0]
η ∂η
y −η[4]∂η
z +
+γ
[0]
θ ∂θ
y +θ[4]∂θ
z ).
(3.103)
that is:
G0 =
−∂0 +∂3 ∂1 −i∂2 ∂
β
y −∂θ
y ∂
ζ
y −i∂η
y
∂1 +i∂2 −∂0 −∂3 ∂
ζ
y +i∂η
y ∂
β
y +∂θ
y
∂
β
y −∂θ
y ∂
ζ
y −i∂η
y −∂0 −∂3 −∂1 +i∂2
∂
ζ
y +i∂η
y ∂
β
y +∂θ
y −∂1 −i∂2 −∂0 +∂3
+i
0 0 ∂
β
z +∂θ
z ∂
ζ
z +i∂η
z
0 0 ∂
ζ
z −i∂η
z ∂
β
z −∂θ
z
−∂
β
z −∂θ
z −∂
ζ
z −i∂η
z 0 0
−∂
ζ
z +i∂η
z −∂
β
z +∂θ
z 0 0
(3.104)
189](https://image.slidesharecdn.com/fn-190805113833/85/TOE-203-320.jpg)
![G0 [ϕ] = −i
h
c ∑
w,p1,p2,p3,nβ,sβ,nζ,sζ,nη,sη,nθ,sθ
ˇg(w,
p1, p2, p3,nβ
,sβ
,nζ
,sζ
,nη
,sη
,nθ
,sθ
)
3
∑
k=0
ck(w, p1, p2, p3,nβ
,sβ
,nζ
,sζ
,nη
,sη
,nθ
,sθ
)×
×exp(−i
h
c
(wx0 + p1x1 + p2x2 + p3x3 + (3.105)
+nβ
yβ
+sβ
zβ
+nζ
yζ
+sζ
zζ
+
+nη
yη
+sη
zη
+nθ
yθ
+sθ
zθ
)).
here
ck(w, p1, p2, p3,nβ
,sβ
,nζ
,sζ
,nη
,sη
,nθ
,sθ
)
is an eigenvector of
ˇg(w, p1, p2, p3,nβ
,sβ
,nζ
,sζ
,nη
,sη
,nθ
,sθ
)
and
ˇg(w, p1, p2, p3,nβ
,sβ
,nζ
,sζ
,nη
,sη
,nθ
,sθ
) := (3.106)
:= β[0]
w+β[1]
p1 +β[2]
p2 +β[3]
p3 +
+γ[0]
nβ
+β[4]
sβ
−γ
[0]
ζ
nζ
+ζ[4]
sζ
−
−γ
[0]
η nη
−η[4]
sη
+γ
[0]
θ nθ
+θ[4]
sθ
.
Here
{c0,c1,c2,c3}
is an orthonormalized basis of the complex4-vectors space.
Functions
ck(w, p1, p2, p3,nβ
,sβ
,nζ
,sζ
,nη
,sη
,nθ
,sθ
)× (3.107)
×exp(−i
h
c
(wx0 + p1x1 + p2x2 + p3x3 +
+nβ
yβ
+sβ
zβ
+
+nζ
yζ
+sζ
zζ
++nη
yη
+sη
zη
+nθ
yθ
+sθ
zθ
))
are eigenvectors of operator G0.
ϕζ
y := c(w,p, f)exp(−i
h
c
(wx0 +px+γ
[0]
ζ
fyζ
))
is a red lower chrome function,
ϕζ
z := c(w,p, f)exp(−i
h
c
(wx0 +px−iζ[4]
fzζ
))
190](https://image.slidesharecdn.com/fn-190805113833/85/TOE-204-320.jpg)
![is a red upper chrome function,
ϕη
y := c(w,p, f)exp(−i
h
c
(wx0 +px+γ
[0]
η fyη
))
is a green lower chrome function,
ϕη
z := c(w,p,, f)exp(−i
h
c
(wx0 +px−iη[4]
fzη
))
is a green upper chrome function,
ϕθ
y := c(w,p, f)exp(−i
h
c
(wx0 +px+γ
[0]
θ fyθ
))
is a blue lower function,
ϕθ
z := c(w,p,sθ
)exp(−i
h
c
(wx0 +px−iθ[4]
fzθ
))
is a blue upper chrome function.
Operator −∂
ζ
y∂
ζ
y is called a red lower chrome operator, −∂
ζ
z ∂
ζ
z is a red upper chrome
operator,
−∂η
y ∂η
y is called a green lower chrome operator, −∂η
z ∂η
z is a green upper chrome oper-
ator,
−∂θ
y∂θ
y is called a blue lower chrome operator, −∂θ
z ∂θ
z is a blue upper chrome operator
For example, if ϕ
ζ
z is a red upper chrome function then
−∂ζ
y∂ζ
yϕζ
z = −∂η
y ∂η
y ϕζ
z = −∂η
z ∂η
z ϕζ
z =
= −∂θ
y∂θ
yϕζ
z = −∂θ
z ∂θ
z ϕζ
z = 0
but
−∂ζ
z ∂ζ
z ϕζ
z = −
h
c
f
2
ϕζ
z .
Because
G0 [ϕ] = 0
then
UG0U−1
U [ϕ] = 0
If U = U1,2 (α) then G0 → U1,2 (α)G0U−1
1,2 (α) and [ϕ] → U1,2 (α)[ϕ].
In this case:
∂1 → ∂1 := (cosα·∂1 −sinα·∂2),
∂2 → ∂2 := (cosα·∂2 +sinα·∂1),
∂0 → ∂0 := ∂0,
∂3 → ∂3 := ∂3,
∂
β
y → ∂
β
y := ∂
β
y ,
∂
β
z → ∂
β
z := ∂
β
z ,
∂
ζ
y → ∂
ζ
y := cosα·∂
ζ
y −sinα·∂η
y ,
191](https://image.slidesharecdn.com/fn-190805113833/85/TOE-205-320.jpg)
![∂η
y → ∂η
y := cosα·∂η
y +sinα·∂
ζ
y ,
∂
ζ
z → ∂
ζ
z := cosα·∂
ζ
z +sinα·∂η
z ,
∂η
z → ∂η
z := cosα·∂η
z −sinα·∂
ζ
z ,
∂θ
y → ∂θ
y := ∂θ
y,
∂θ
z → ∂θ
z := ∂θ
z .
Therefore,
−∂ζ
z ∂ζ
z ϕζ
z = f
h
c
cosα
2
·ϕζ
z ,
−∂η
z ∂η
z ϕζ
z = −sinα· f
h
c
2
ϕζ
z .
If α = −π
2 then
−∂ζ
z ∂ζ
z ϕζ
z = 0,
−∂η
z ∂η
z ϕζ
z = f
h
c
2
ϕζ
z .
That is under such rotation the red state becomes the green state.
If U = U3,2 (α) then G0 → U3,2 (α)G0U−1
3,2 (α) and [ϕ] → U3,2 (α)[ϕ].
In this case:
∂0 → ∂0 := ∂0,
∂1 → ∂1 := ∂1,
∂2 → ∂2 := (cosα·∂2 +sinα·∂3),
∂3 → ∂3 := (cosα·∂3 −sinα·∂2),
∂
β
y → ∂
β
y := ∂
β
y ,
∂
ζ
y → ∂
ζ
y := ∂
ζ
y,
∂η
y → ∂η
y := cosα·∂η
y −sinα·∂θ
y ,
∂θ
y → ∂θ
y := cosα·∂θ
y +sinα·∂η
y ,
∂
β
z → ∂
β
z := ∂
β
z ,
∂
ζ
z → ∂
ζ
z := ∂
ζ
z ,
∂η
z → ∂η
z := cosα·∂η
z −sinα·∂θ
z ,
∂θ
z → ∂θ
z := cosα·∂θ
z +sinα·∂η
z ,
Therefore, if ϕη
y is a green lower chrome function then
−∂η
z ∂η
z ϕη
y =
h
c
cosα· f
2
·ϕη
y ,
−∂θ
y ∂θ
y ϕη
y =
h
c
sinα· f
2
·ϕη
y .
If α = π/2 then
192](https://image.slidesharecdn.com/fn-190805113833/85/TOE-206-320.jpg)
![−∂η
z ∂η
z ϕη
y = 0,
−∂θ
y ∂θ
y ϕη
y =
h
c
f
2
·ϕη
y .
That is under such rotation the green state becomes blue state.
If U = U3,1 (α) then G0 → U3,1 (α)G0U−1
3,1 (α) and [ϕ] → U3,1 (α)[ϕ].
In this case:
∂0 → ∂0 := ∂0,
∂1 → ∂1 := (cosα·∂1 −sinα·∂3),
∂2 → ∂2 := ∂2,
∂3 → ∂3 := (cosα·∂3 +sinα·∂1),
∂
β
y → ∂3 := ∂
β
y ,
∂
ζ
y → ∂
ζ
y := cosα·∂
ζ
y +sinα·∂θ
y ,
∂η
y → ∂η
y := ∂η
y ,
∂θ
y → ∂θ
y := cosα·∂θ
y −sinα·∂
ζ
y ,
∂
β
z → ∂
β
z := ∂
β
z ,
∂
ζ
z → ∂
ζ
z := cosα·∂
ζ
z −sinα·∂θ
z ,
∂η
z → ∂η
z := ∂η
z ,
∂θ
z → ∂θ
z := cosα·∂θ
z +sinα·∂
ζ
z .
Therefore,
−∂ζ
z ∂ζ
z ϕζ
z = − f
h
c
cosα
2
·ϕζ
z ,
−∂θ
z ∂θ
z ϕζ
z = − sinα· f
h
c
2
ϕζ
z .
If α = π/2 then
−∂ζ
z ∂ζ
z ϕζ
z = 0,
−∂θ
z ∂θ
z ϕζ
z = − f
h
c
2
ϕζ
z .
That is under such rotation the red state becomes the blue state. Thus at the Cartesian
turns chrome of a state is changed.
One of ways of elimination of this noninvariancy consists in the following. Calculations
in [71, p.156] give the grounds to assume that some oscillations of quarks states bend time-
space in such a way that acceleration of the bent system in relation to initial system submits
to the following law (Fig. 1):
g(t,x) = cλ/ x2
cosh2
λt/x2
.
Here the acceleration plot is line (1) and the line (2) is plot of λ/x2.
193](https://image.slidesharecdn.com/fn-190805113833/85/TOE-207-320.jpg)
![Figure 37:
Hence, to the right from point C and to the left from poin C the Newtonian gravitation
law is carried out.
AA is the Asymptotic Freedom Zone.
CB and B C is the Confinement Zone.
Let in the potential hole AA there are three quarks ϕ
ζ
y, ϕη
y , ϕθ
y. Their general state
function is determinant with elements of the following type: ϕ
ζηθ
y := ϕ
ζ
yϕη
y ϕθ
y. In this case:
−∂ζ
y∂ζ
yϕζηθ
y =
h
c
f
2
ϕζηθ
y
and under rotation U1,2 (α):
−∂ζ
y ∂ζ
y ϕζηθ
y =
=
h
c
f
2
γ
[0]
ζ
cosα−γ
[0]
η sinα
2
ϕζ
yϕη
y ϕθ
y =
=
h
c
f
2
ϕζηθ
y .
That is at such turns the quantity of red chrome remains.
As and for all other Cartesian turns and for all other chromes.
Baryons ∆− = ddd, ∆++ = uuu, Ω− = sss belong to such structures.
If U = U1,0 (α) then G0 → U−1‡
1,0 (α)G0U−1
1,0 (α) and [ϕ] → U1,0 (α)[ϕ].
In this case:
∂0 → ∂0 := (coshα·∂0 +sinhα·∂1),
∂1 → ∂1 := (coshα·∂1 +sinhα·∂0),
∂2 → ∂2 := ∂2,
∂3 → ∂3 := ∂3,
∂
β
y → ∂
β
y := ∂
β
y ,
∂
ζ
y → ∂
ζ
y := ∂
ζ
y,
∂η
y → ∂η
y := coshα·∂η
y −sinhα·∂θ
z ,
194](https://image.slidesharecdn.com/fn-190805113833/85/TOE-208-320.jpg)
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TOE
- 1. In this book, a brief history of physics from Aristotle to the present day is presented. The Gentzen variant of the propositional logic is used to substantiate the space-time rela- tions, including the Lorentz transformations. The logical foundations of probability theory, including Jacob Bernulli’s Big Numbers Law and the statistical definition of probability, are also derived from this logic. All concepts and statements of the Standard Model (except for the Higgs) are obtained as concepts and theorems of probability theory. The masses, spins, moments, energies of fermions are the parameters of the distribution of such a probability. The masses of the W and Z bosons are the results of the interaction of the probability flows into space-time. Quark-gluon relations, including the phenomena of confinement and asymptotic free- dom, are also a consequence of the properties of this probability. The phenomenon of gravity with dark matter and dark energy is a continuation of these quark-gyonic relations. For understanding of the maintenance of this book elementary knowledge in the field of linear algebra and the mathematical analysis is sufficient.
- 2. THEOTY OF EVERETHING GUNN QUZNETSOV 2018 PRESS
- 3. Contents Introduction v 1 Time, space, and probability 1 1.1. Recorders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 1.2. Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 1.3. Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 1.4. Relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 1.5. Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 1.5.1. Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 1.5.2. B-functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 1.5.3. Independent tests . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 1.5.4. The logic probability function . . . . . . . . . . . . . . . . . . . . 57 1.5.5. Conditional probability . . . . . . . . . . . . . . . . . . . . . . . . 58 1.5.6. Classical probability . . . . . . . . . . . . . . . . . . . . . . . . . 59 1.5.7. Probability and logic . . . . . . . . . . . . . . . . . . . . . . . . . 59 1.5.8. THE NONSTANDARD NUMBERS . . . . . . . . . . . . . . . . 60 1.5.9. Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 2 Quants 69 2.1. Physical events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 2.2. Entanglement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 2.3. Equations of moving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 2.4. Double-Slit Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 2.5. Lepton Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 2.6. Masses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 2.7. One-Mass State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 2.8. Creating and Annihilation Operators . . . . . . . . . . . . . . . . . . . . . 110 2.9. Particles and Antiparticles . . . . . . . . . . . . . . . . . . . . . . . . . . 114 3 Fields 119 3.1. Electroweak Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 3.1.1. The Bi-mass State . . . . . . . . . . . . . . . . . . . . . . . . . . 120 3.1.2. Neutrino . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 3.1.3. Electroweak Transformations . . . . . . . . . . . . . . . . . . . . 142 3.1.4. Dimension of physical space . . . . . . . . . . . . . . . . . . . . . 155 iii
- 4. 3.2. Quarks and Gluons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 3.3. Asymptotic Freedom, Confinement, Gravitation . . . . . . . . . . . . . . . 172 3.3.1. Dark Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 3.3.2. Dark Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 3.3.3. Baryon Chrome . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 Conclusion 197 Epilogue 199 References 201 Index 207 iv
- 5. Introduction In the beginning was the Word. Johnn, 1.1 It is known that any physical theory consists of basic notions and statements received from these basic elements by the classic logic. We have the sequence of theories each of which explained by the preceding one according to the logic. That means that basic notions and statements of every subsequent theory are more logical than basic notions and axioms of the preceding one. When these basic elements of the theory become absolutely logic, i.e. when they become notions and rules of classical logic, the theoretical physics will come to an end, it will rather be logic than physics. It seems that such situation will take place in the nearest future. I’ll prove it in this book. The first known ”theory of everything” was created by the ancient Greek philosopher Aristotle1 in IV BC. Aristotle related each of the four elements proposed earlier by Empedocles, Earth, Wa- ter, Air, and Fire, to two of the four sensible qualities, hot, cold, wet, and dry. Aristotle’s scheme added the heavenly Aether, the divine substance of the heavenly spheres, stars and planets [1]. Aristotle maked experiments in optics using a camera obscura [2]. In astron- omy, Aristotle refuted Democritus’s claim that the Milky Way was made up of ”those stars which are shaded by the earth from the sun’s rays,” pointing out correctly that if ”the size of the sun is greater than that of the earth and the distance of the stars from the earth many times greater than that of the sun, then... the sun shines on all the stars and the earth screens none of them.” [3]. The influence of the geocentric cosmology of Aristotle persisted until Copernicus. More detai read: [4], [5]. Sir Isaac Newton 2 is the author of the fundamental work Mathematical Principles of Natural Philosophy (1687) [6], in which he outlined the law of world wideness and the three laws of mechanics, which became the basis of classical mechanics. He developed differential and integral calculus, the theory of color, laid the foundations of modern phys- ical optics, created many other mathematical and physical theories It was the First Great Absorption – Newton’s work combined the laws of mechanics and the gravity. 1 Aristotle, Greek Aristoteles, (born 384 BCE, Stagira, Chalcidice, Greece died 322, Chalcis, Euboea), ancient Greek philosopher and scientist, one of the greatest intellectual figures of Western history. 2Sir Isaac Newton PRS FRS (25 December 1642 20 March 1726) Newton is an English physicist, mathe- matician, mechanic and astronomer, one of the founders of classical physics v
- 6. vi
- 7. Second Great Absorption was maded by the James Clerk Maxwell equations in XIX centery. 3 These equations combine electricity, magnetism and optics [7]. Next, it was necessary to coordinate the laws of Newton and Maxwell’s equations. It was made in end of XIX centery by Hendrik Antoon Lorentz 4, Jules Henri Poincar 5 and Albert Einstein 6 They builded Special Theory Relativity (STR) [8]. Subsequently (at the beginning of XX), Albert Einstein constructed a theory of curved space-time (General Theory of Relativity) on the basis of STR [9]. This theory absorbed the Newtonian theory of gravity. This was Third Great Absorptipn. Now everything fell into place. Everything got its explanation. Everything worked ... It was ”The Second Theory of Everything”. But in 1900 this beautiful continuous world crash down – Max Karl Ernst Ludwig Planck discovered that our world is discrete - not continuous [10] 7 In the 30s of the 20th century, Erwin Rudolf Josef Alexander Schrdinger 8, Werner Karl 3James Clerk Maxwell; (13 June 1831 5 November 1879) was a Scottish scientist in the field of mathemat- ical physics. 4 (Hendrik Antoon Lorentz; 18 July 1853 4 February 1928) was a Dutch physicist 5Jules Henri Poincar 29 April 1854 17 July 1912) was a French mathematician, theoretical physicist, engineer, and philosopher of science. 6Albert Einstein; 14 March 1879 18 April 1955) was a German-born theoretical physicis 7Max Karl Ernst Ludwig Planck, 23 April 1858 4 October 1947) was a German theoretical physicist whose discovery of energy quanta 8Erwin Rudolf Josef Alexander Schrdinger (12 August 1887 4 January 1961), was a Austrian physicist who developed a number of fundamental results in the field of quantum theory vii
- 8. Heisenberg 9 and Paul Adrien Maurice Dirac 10 built a theory, for this discreteness. It was the Quantum Mechanics - a new paradigm of physics. It put physics on a probabilistic basis [11]. This is Fourth Great Absorption. Fifth Great Absorption (mid 20th century) is the Feynman Quantum electrodynamics [12]. This theory (Quantum Field Theory) has swallowed Quantum Mechanics and the Maxwell equations. And finally, in the 60s of the 20th century, Sheldon Lee Glashow united the electromag- netic and weak interactions into one gauge theory. 11 This theory became the basis of the Standard Model - the theory of all elementary particles. Before the Theory of Everything, it remained to combine the Standard Model with the theory of gravity 9Werner Karl Heisenberg (; 5 December 1901 1 February 1976) was a German theoretical physicist and one of the key pioneers of quantum mechanics. 10Paul Adrien Maurice Dirac (8 August 1902 20 October 1984) was an English theoretical physicist who is regarded as one of the most significant physicists of the 20th century. 11Sheldon Lee Glashow (born December 5, 1932) is an American theoretical physicist viii
- 9. ix
- 10. ”The road to success and the road to failure are almost exactly the same.” – Colin R. Davis The Manhattan Project began on September 17, 1943. It was attracted many outstand- ing physicists, many of whom were refugees from Europe. By the summer of 1945, the Americans had managed to build 3 atomic bombs, 2 of which were dropped on Hiroshima and Nagasaki, and a third had been tested shortly before. And the atomic race began. In the following years, the governments of many states allocated enormous sums of money to scientific organizations. Following these money, huge masses of easy luck seekers moved to physics. They invented SUSY12, WIMP13, BIG BANG14, HIGGS15 and other theories of the same kind. Giant laboratory facilities were built and enormous human resources were attracted to experimentally confirm these theories. All of these enterprises have worked without success for several decades. For examples, Large Hadron Collider (LHC) worked since 10 September 2008 till 14 February 2013 – RUNI. RUNII works from June 2015 for today. Huge resources have been spent, but did not receive any fundamentally new results - no superpartners, no extra dimensions, or gravitons, or black holes. no dark matter or dark energy, etc. etc .. As for the Higgs, then the prehistory of its ”discovery” is the follows: By December 2011, it became clear that there is nothing substantial - neither of a su- persymmetry, or a black holes, or a dark energy, etc., nothing. ”Higgs mechanism” - that is moisture condensation scalar field and the simulta- neous formation of the mass of the gauge bosons - Was coined by Ginzburg and Landau in 1950 in the same article about phenomenological theory of super- conductivity, for which the overhead line Ginsburg, many years later won the Nobel Prize see. http://en.wikipedia.org/wiki/GinzburgAnother great physi- cist, American Phil Anderson, also marked with the stamp of genius, in 1962 proposed use the mechanism of Ginzburg Landau for the relativistic particle physics (from the word theory of relativity). But he did not bother develop- ing small parts. The modern version, which adds the relativistic invariance and ”non-Abelian” field, ie, not necessarily a photon - a paltry improvement in essence - was proposed simultaneously and independently in the works of the three groups of authors in 1964: 1) Robert Brout and Francois En- glert, 2) Peter Higgs, 3) Gerald Guralnik, CR Hagen, and Tom Kibble, see. 12the superstring model: by this theory elementary particles are pieces of strings. From this theory, the existence of a superpartner particle for each elementary particle, the existence of additional spatial dimensions, and so on, follows. 13 hypothetical dark matter particles 14the hypothesis that the expansion of the universe is the result of an initial explosion 15hypothetical boson, whose field provides the mass of all elementary particles x
- 11. http://en.wikipedia.org/wiki/Higgs mechanism. Of these, all but actually Pe- ter Higgs, known by many other useful contribution to science; Higgs did not know anything more. However, he personally always behaves very modestly, knowing that it is not selected anything other than short names. ”Higgs mech- anism” primitive like a broom, and I would be greatly surprised if God went to implement it, because she would have had to cover their tracks elsewhere. So I rather think that the ”Higgs boson” is actually a kind of strongly inter- acting composite object with a mass greater than where looking for him, and also strongly unstable, so that the width of the order of half-life mass. We with young (then) kid Greg Carter showed how it is, though, just in a different con- text: http://arxiv.org/abs/hep-ph/0001318. In short, I would bet 100 that the Higgs boson is not found, if Hawking has not done it. Maybe next Tuesday or when CERN fanfare announced the capture Higgs at 2.5 - 3 sigma (who knows), and it will not, unfortunately, mean nothing. ”My” pentaquark eight laboratories opened at 3-4 sigma, and then the other ”closed”. However, it was again opened at the level of 5.9 standard deviations, and that few believe. I was recently at CERN and observed there is not very healthy atmosphere. Imagine thousands of physicists and engineers, who are living under constant pressure ”to find the Higgs.” LHC - absolutely amazing, unique machine, where ev- ery two weeks can make wonderful discoveries, but they are ”looking for the Higgs,” throwing out 99 percent or more of the data that current notions that do not contribute. This is the first time in history - the physics is usually written to disk or tape completely all the events - you never know what will come in handy later, but there is so much information that no drives are not able to save her. Therefore, 99how Schliemann, in the heat, thrown into dump Troy until dug up the Neolithic. In this atmosphere, they desperately need a great discov- ery, and they will do it! At CERN as an institution is not a very good reputation: recently they vociferously, with television and the press, announced the open- ing of so-called. Quark–gluon plasma, which turned out to be its opposite - the most strongly-interacting system, which is known in nature. In general, I certainly wish my colleagues, many of whom are friends, good luck in hunting, but we must be prepared for the fact that on Tuesday, nothing special except the PR action will not happen.”16 And they did this great discovery: On the one hand, they have spent a lot of money and a lot of work to search for the Higgs. On the other hand, they do not have anything but only single new particle (124.5 - 126 GeV). And they announced that particle as the Higgs boson. Naturally, the scientific community has embraced the euphoria that has not subsided yet. But stop: The firstly, there is no argument in favor of the fact that the particle 124.5 - 126 GeV 16Dmitri Igorevich Diakonov, Scientific.ru, Bytije rossitskoy nauki, 12.12.2011 04:26. Dmitrii Igorevich Dyakonov (March 30, 1949 - December 26, 2012) - Soviet and Russian theoretical physicist, doctor of physical and mathematical sciences. Specialist in particle physics and quantum field theory. xi
- 12. xii
- 13. has some relation to the Higgs mechanism. Secondly, the Higgs field permeates the vacuum of space, which means that the mass of the Higgs vacuum and stability are closely linked. For a particle of mass near 126 GeV - enough to destroy the cosmos. The Standard Model of particle physics has not given an answer to the question of why the universe did not collapse after the Big Bang. Third, Nothing in Standart Model gives a precise value for the Higgss own mass, and calculations from first principles, based on quantum theory, suggest it should be enormous- roughly a hundred million billion times higher than its measured value. Physicists have therefore introduced an ugly fudge factor into their equations (a process called fine-tuning) to sidestep the problem. The fourth, all the known elementary bosons are gauge - it is photons, W- and Z-bosons and gluons. It is likely that the 125-126 particle is of some hadron multiplet. In the study of the logical foundations of probability theory, I found that the terms and equations of the fundamental theoretical physics represent terms and theorems of the classical probability theory, more precisely, of that part of this theory, which considers the probability of dot events in the 3 + 1 space-time. In particular, all Standard Model’s formulas (higgs ones except) turn out theorems of such probability theory. And the masses, moments, energies, spins, etc. turn out of parameters of probability distributions such events. The terms and the equations of the electroweak and of the quark-gluon theories turn out the theoretical-probabilistic terms and theorems. Here the relation of a neutrino to his lepton becomes clear, the W and Z bosons masses turn out dynamic ones, the cause of the asymmetry between particles and antiparticles is the impossibility of the birth of single antiparticles. In addition, phenomena such as confinement and asymptotic freedom receive their probabilistic explanation. And here we have the logical foundations of the gravity theory with phenomena dark energy and dark matter. Thus, physics is a game of probabilities in space-time. But what is time? What is space? What is probability? xiii
- 15. Chapter 1 Time, space, and probability Logic as a scientific method is used for evidence of obvious and clear things that do not need any proof in view of their obviousness. The more obvious the thing, the more logic and less all the other in it, do not belong to the logic. It is clear that logic as such is most obvious of all things, since there is nothing at all except logic in it. It is for this reason that pure logic does not need any proofs or explanations and does not require any additional logic to understand it. A. S. Shlenski@, Short treatise on logic Let’s consider affirmative sentences of any languages. Def. 1.1.1: Sentence Θ is true if and only if Θ. 1
- 16. For example, sentence It is raining is true if and only if it is raining1 [16]. Def. 1.1.2: Sentence Θ is false if and only if there is not that Θ. It is clear that many neither true nor false sentences exist. For example, There is rainy 21 august 3005 year in Chelyabinsk . Still an example: Obviously, the following sentence isn’t true and isn’t false [17]: The sentence which has been written on this line, is false. Those sentences which can be either true, or false, are called as meaningful sentences. The previous example sentence is meaningless sentence. Further we consider only meaningful sentences which are either true, or false. Def. 1.1.3: Sentences A and B are equal (design.: A = B) if A is true, if and only if B is true. Further I’m using ordinary notions of the classical propositional logic [18]. Def. 1.1.4: A sentence C is called conjunction of sentences A and B (design.: C = (A&B)) if C is true, if and only if A and B are true. Def. 1.1.5: A sentence C is called negation of sentences A (design.: C = (¬A)) if C is true, if and only if A is not true. Def. 1.1.6: A sentence C is called disjunction of sentences A and B (design.: C = (A∨B)) if C is true, if and only if A is true or B is true or both A and B are true. Def. 1.1.7: A sentence C is called implication of sentences A and B (design.: C = (A ⇒ B)) if C is true, if and only if B is true and/or B is false. A sentence is called a simple sentence if it isn’t neither conjunction, nor a disjunction, neither implication, nor negation. Th. 1.1.1: 1) (A&A) = A; (A∨A) = A; 2) (A&B) = (B&A); (A∨B) = (B∨A); 3) (A&(B&C)) = ((A&B)&C); (A∨(B∨C)) = ((A∨B)∨C); 4) if T is a true sentence then for every sentence A: (A&T) = A and (A∨T) = T. 5) if F is a false sentence then (A&F) = F and (A∨F) = A. Proof of Th. 1.1.1: This theorem directly follows from Def. 1.1.1, 1.2, 1.3, 1.4, 1.6. Further I set out the version of the Gentzen Natural Propositional calculus2 (NPC) [15]: Expression ”Sentence C is a logical consequence of a list of sentences Γ.” will be wrote as the following: ”Γ C”. Such expressions are called sequences. Elements of list Γ are called hypothesizes. Def. 1.1.8 1. A sequence of form C C is called NPC axiom. 2. A sequence of form Γ A and Γ B is obtained from sequences of form Γ (A&B) by a conjunction removing rule (design.: R&). 3. A sequence of form Γ1,Γ2 (A&B) is obtained from sequence of form Γ1 A and a sequence of form Γ2 B by a conjunction inputting rule (design: I&). 4. A sequence of form Γ (A∨B) is obtained from a sequence of form Γ A or from a sequence of form Γ B by a disjunction inputting rule (design.: I∨). 1Alfred Tarski (January 14, 1901 October 26, 1983) was a Polish logician and mathematician 2Gerhard Karl Erich Gentzen ( November 24, 1909, Greifswald, Germany August 4, 1945, Prague, Czechoslovakia) was a German mathematician and logician. 2
- 17. 3
- 18. 5. A sequence of form Γ1 [A],Γ2 [B],Γ3 C is obtained from sequences of form Γ1 C, Γ2 C, snd Γ3 (A∨B) by a disjunction removing rule (design.: R∨) (Here and further: Γ1 [A] is obtained from Γ1 by removing of sentence A, and Γ2 [B] is obtained from Γ2 by removing of sentence B). 6. A sequence of form Γ1,Γ2 B is obtained from a sequence of form Γ1 A and from a sequence of form Γ2 (A ⇒ B) by a implication removing rule (design.: R⇒). 7. A sequence of form Γ[A] (A ⇒ B) is obtained from a sequence of form Γ B by an implication inputting rule (design.: I⇒). 8. A sequence of form Γ C is obtained from a sequence of form Γ (¬(¬C)) by a negation removing rule (design.: R¬). 9. A sequence of form Γ1 [C],Γ2 [C] (¬C) is obtained from a sequence of form Γ1 A and from a sequence of form Γ2 (¬A) by negation inputting rule (design.: I¬). 10. A finite string of sequences is called a propositional natural deduction if every element of this string either is a NPC axioms or is received from preceding sequences by one of the deduction rules (R&, I&, I∨, R∨, R⇒, I⇒, R¬, I¬). Actually, these logical rules look naturally in light of the previous definitions. Example 1: Let us consider the following string of sequences: 1.((R&S)&(R ⇒ G)) ((R&S)&(R ⇒ G)) - NPC axiom. 2.((R&S)&(R ⇒ G)) (R&S) - R& from 1. 3.((R&S)&(R ⇒ G)) (R ⇒ G) - R& from 1. 4.((R&S)&(R ⇒ G)) R - R& from 2. (1.1) 5.((R&S)&(R ⇒ G)) G - R ⇒ from 3. and 4. 6.((R&S)&(R ⇒ G)) S - R& from 2. 7.((R&S)&(R ⇒ G)) (G&S) - I& from 5. and 6. This string is a propositional natural deduction of sequence ((R&S)&(R ⇒ G)) (G&S). since it fulfills to all conditions of Def. 1.1.8. Hence sentence (G&S) is logical consequence from sentence ((R&S)&(R ⇒ G)). Th. 1.1.2: (A∨B) = (¬((¬A)&(¬B))), (1.2) (A ⇒ B) = (¬(A&(¬B))). (1.3) Proof of Th. 1.1.2: The following string is a deduction of sequence (A∨B) (¬((¬A)&(¬B))): 1. ((¬A)&(¬B)) ((¬A)&(¬B)), NPC axiom. 2. ((¬A)&(¬B)) (¬A), R& from 1. 4
- 19. 3. A A, NPC axiom. 4. A (¬((¬A)&(¬B))), I¬ from 2. and 3. 5. ((¬A)&(¬B)) (¬B), R& from 1. 6. B B, NPC axiom. 7. B (¬((¬A)&(¬B))), I¬ from 5. and 6. 8. (A∨B) (A∨B), NPC axiom. 9. (A∨B) (¬((¬A)&(¬B))), R∨ from 4., 7. and 8. A deduction of sequence (¬((¬A)&(¬B))) (A∨B) is the following: 1. (¬A) (¬A), NPC axiom. 2. (¬B) (¬B), NPC axiom. 3. (¬A),(¬B) ((¬A)&(¬B)), I& from 1. and 2. 4. (¬((¬A)&(¬B))) (¬((¬A)&(¬B))), NPC axiom. 5. (¬((¬A)&(¬B))),(¬B) (¬(¬A)), I¬ from 3. and 4. 6. (¬((¬A)&(¬B))),(¬B) A, R¬ from 5. 7. (¬((¬A)&(¬B))),(¬B) (A∨B), i∨ from 6. 8. (¬(A∨B)) (¬(A∨B)), NPC axiom. 9. (¬((¬A)&(¬B))),(¬(A∨B)) (¬(¬B)), I¬ from 7. and 8. 10. (¬((¬A)&(¬B))),(¬(A∨B)) B, R¬ from 9. 11. (¬((¬A)&(¬B))),(¬(A∨B)) (A∨B), I∨ from 10. 12. (¬((¬A)&(¬B))) (¬(¬(A∨B))), I¬ from 8. and 11. 13. (¬((¬A)&(¬B))) (A∨B), R¬ from 12. Therefore, (¬((¬A)&(¬B))) = (A∨B). A deduction of sequence (A ⇒ B) (¬(A&(¬B))) is the following: 1. (A&(¬B)) (A&(¬B)), NPC axiom. 2. (A&(¬B)) A, R& from 1. 3. (A&(¬B)) (¬B), R& from 1. 4. (A ⇒ B) (A ⇒ B), NPC axiom. 5. (A&(¬B)),(A ⇒ B) B, R⇒ from 2. and 4. 6. (A ⇒ B) (¬(A&(¬B))), I¬ from 3. and 5. A deduction of sequence (¬(A&(¬B))) (A ⇒ B) is the following: 1. A A, NPC axiom. 2. (¬B) (¬B), NPC axiom. 3. A,(¬B) (A&(¬B)), I& from 1. and 2. 4. (¬(A&(¬B))) (¬(A&(¬B))), NPC axiom. 5. A,(¬(A&(¬B))) (¬(¬B)), I¬ from 3. and 4. 6. A,(¬(A&(¬B))) B, R¬ from 5. 7. (¬(A&(¬B))) (A ⇒ B), I⇒ from 6. Therefore, (¬(A&(¬B))) = (A ⇒ B) Example 2: 1. A A - NPC axiom. 2. (A ⇒ B) (A ⇒ B) - NPC axiom. 3. A,(A ⇒ B) B - R⇒ from 1. and 2. 5
- 20. 4. (¬B) (¬B) - NPC axiom. 5. (¬B),(A ⇒ B) (¬A) - I¬ from 3. and 4. 6. (A ⇒ B) ((¬B) ⇒ (¬A)) - I⇒ from 5. 7. ((A ⇒ B) ⇒ ((¬B) ⇒ (¬A))) - I⇒ from 6. This string is a deduction of sentence of form ((A ⇒ B) ⇒ ((¬B) ⇒ (¬A))) from the empty list of sentences. I.e. sentences of such form are logicaly provable. Th. 1.1.3: If sequence Γ → C is deduced and C is false then some false sentence is contained in Γ. Proof of Th. 1.1.3: is received by induction of number of sequences in the deduction of sequence Γ → C. The recursion Basis: Let the deduction of sequence Γ → C contains single sentence. In accordance the definition of propositional natural deduction this deduction must be of the following type: C → C. Obviously, in this case the lemma holds true. The recursion Step:The recursion assumption: Let’s admit that the lemma is carried out for any deduction which contains no more than n sequences. Let deduction of Γ → C contains n+1 sequence. In accordance with the propositional natural deduction definition sequence Γ → C can be axiom NPC or can be received by the deduction rules from previous sequence. a) If Γ → C is the NPC axiom then see the recursion basis. b) Let Γ →C be received by R&. In this case sequence of type Γ → (C&B) or sequence of type Γ → (B&C) is contained among the previous sequences of this deduction. Hence, deductions of sequences Γ → (C&B) and Γ → (B&C) contains no more than n sequences. In accordance with the recursion assumption, these deductions submit to the lemma. Be- cause C is false then (C&B) is false and (B&C) is false in accordance with the conjunction definition. Therefore, Γ contains some false sentence by the lemma. And in this case the lemma holds true. c) Let Γ → C be received by I&. In this case sequence of type Γ1 → A and sequence of type Γ2 → B is contained among the previous sequences of this deduction, and C = (A&B) and Γ = Γ1,Γ2. Deductions of sequences Γ1 → A and Γ2 → B contains no more than n sequences. In accordance with the recursion assumption, these deductions submit to the lemma. Because C is false then A is false or B is false in accordance with the conjunction definition. Therefore, Γ contains some false sentence by the lemma. And in this case the lemma holds true. d) Let Γ → C be received by R∨. In this case sequences of type Γ1 → (A∨B), Γ2 [A] → C, and Γ3 [B] → C are contained among the previous sequences of this deduction, and Γ = Γ1,Γ2,Γ3. Because these previous deductions contain no more than n sequences then in accordance with the recursion assumption, these deductions submit to the lemma. Because C is false then Γ2 [A] contains some false sentence, and Γ3 [B] contains some false sentence. If A is true then the false sentence is contained in Γ2. If B is true then the false sentence is contained in Γ3. I.e. in these case some false sentence is contained in Γ. If A is false and B is false then (A∨B) is false in accordance with the disjunction definition. In this case Γ1 contains some false sentence. And in all these cases the lemma holds true. 6
- 21. e) Let Γ → C be received by I∨. In this case sequence of type Γ → A or sequence of type Γ → B is contained among the previous sequences of this deduction, and C = (A∨B). Deductions of sequences Γ → A and Γ → B contains no more than n sequences. In accor- dance with the recursion assumption, these deductions submit to the lemma. Because C is false then A is false and B is false in accordance with the disjunction definition. Therefore, Γ contains some false sentence by the lemma. And in this case the lemma holds true. f) Let Γ →C be received by R⇒. In this case sequences of type Γ1 → (A ⇒ C), Γ2 → A are contained among the previous sequences of this deduction, and Γ = Γ1,Γ2. Because these previous deductions contain no more than n sequences then in accordance with the recursion assumption, these deductions submit to the lemma. If A is false then Γ2 contains some false sentence. If A is true then (A ⇒ C) is false in accordance with the implication defination since C is false. And in all these cases the lemma holds true. g) Let Γ → C be received by I⇒. In this case sequences of type Γ[A] → B is contained among the previous sequences of this deduction, and C = (A ⇒ B). Because deduction of Γ[A] → B contains no more than n sequences then in accordance with the recursion assump- tion, this deduction submit to the lemma. Because C is false then A is true in accordance with the implication definition. Hence, some false sentence is contained in Γ. Therefore, in this case the lemma holds true. i) Let Γ →C be received by R¬. In this case sequence of type Γ → (¬(¬C)) is contained among the previous sequences of this deduction. This previous deduction contains no more than n sequences then in accordance with the recursion assumption, this deduction submit to the lemma Because C is false then (¬(¬C)) is false in accordance with the negation definition. Therefore, Γ contains some false sentence by the lemma. And in this case the lemma holds true. j) Let Γ → C be received by I¬. In this case sequences of type Γ1 [A] → B, and Γ2 [A] → (¬B) are contained among the previous sequences of this deduction, and Γ = Γ1,Γ2 and C = (¬A). Because these previous deductions contain no more than n sequences then in accordance with the recursion assumption, these deductions submit to the lemma. Because C is false then A is true. Hence, some false sentence is contained in Γ because B is false or (¬B) is false in accordance with the negation definition. Therefore, in all these cases the lemma holds true. The recursion step conclusion: If the lemma holds true for deductions containing n sequences then the lemma holds true for deduction containing n+1 sequences. The recursion conclusion: Lemma holds true for all deductions . Def. 1.1.9 A sentence is naturally propositionally provable if there exists a prpositional natural deduction of this sentence from the empty list. In accordance with Th. 1.1.3 all naturally propositionally provable sentences are true because otherwise the list would appear not empty. But some true sentences are not naturally propositionally provable. Alphabet of Propositional Calculations: 1. symbols pk with natural k are called PC-letters; 2. symbols ∩, ∪, ⊃, ˆ are called PC-symbols; 3. (, ) are called brackets. Formula of Propositional Calculations: 1. any PC-letter is PC-formula. 7
- 22. 2. if q and r are PC-formulas then (q∩r), (q∪r),(q ⊃ r),(ˆq) are PC-formulas; 3. except listed by the two first points of this definition no PC-formulas are exist. Def. 1.1.10 Let function g has values on the double-elements set {0;1} and has the set of PC-formulas as a domain. And let 1) g(ˆq) = 1−g(q) for every sentence q; 2) g(q∩r) = g(q)·g(r) for all sentences q and r; 3) g(q∪r) = g(q)+g(r)−g(q)·g(r) for all sentences q and r; 4) g(q ⊃ r) = 1−g(q)+g(q)·g(r) for all sentences q and r. In this case a function g is called a Boolean function 3. Hence if g is a Boolean function then for every sentence q: (g(q))2 = g(q). A Boolean function can be defined by a table: g(q) g(r) g(q∩r) g(q∪r) g(q ⊃ r) g(ˆq) 0 0 0 0 1 1 0 1 0 1 1 1 1 0 0 1 0 0 1 1 1 1 1 0 . Such tables can be constructed for any sentence. For example: g(q) g(r) g(s) g(ˆ((r ∩(ˆs))∩(ˆq))) 0 0 0 1 0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 1 , or: g(r) g(s) g(q) g(((r ∩s)∩(r ⊃ q)) ⊃ (q∩s)) 0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 1 . (1.4) Def. 1.1.11 A PC-formula q is called a t-formula if for any Boolean function g: g(q) = 1. 3George Boole (2 November 1815 8 December 1864) was an English mathematician and philosopher. 8
- 23. For example, formula (((r ∩s)∩(r ⊃ q)) ⊃ (q∩s)) is a t-formula by the table (1.4). Def. 1.1.12 Function ϕ(x) which is defined on the PC-formulas set and which has the sentences set as a range of values, is called an interpretation function if the following conditions are carried out: 1. if pk is a PC-letter then ϕ(pk) = A and here A is a simple sentence and if ϕ(ps) = B then if s = k then A = B; 2. ϕ(r ∩s) = (ϕ(r)&ϕ(s)), ϕ(r ∪s) = (ϕ(r)∨ϕ(s)), ϕ(r ⊃ s) = (ϕ(r) ⇒ ϕ(s)), ϕ(ˆr) = (¬ϕ(r)). Def. 1.1.13 A sencence C is called tautology if the following condition is carried out: if ϕ(q) = C then q is a t-formula. Lm. 1.1.1: If g is a Boolean function then every natural propositional deduction of se- quence Γ A satisfy the following condition: if g ϕ−1(A) = 0 then there exists a sentence C such that C ∈ Γ and g ϕ−1(C) = 0. Proof of Lm. 1.1.1: is maked by a recursion on a number of sequences in the deduction of Γ A: 1. Basis of recursion: Let the deduction of Γ A contains 1 sequence. In that case a form of this sequence is A A in accordance with the propositional natural deduction definition (Def. 1.1.8). Hence in this case the lemma holds true. 2. Step of recursion: The recursion assumption: Let the lemma holds true for every deduction, containing no more than n sequences. Let the deduction of Γ A contains n+1 sequences. In that case either this sequence is a NPC-axiom or Γ A is obtained from previous sequences by one of deduction rules. If Γ A is a NPC-axiom then the proof is the same as for the recursion basis. a) Let Γ A be obtained from a previous sequence by R&. In that case a form of this previous sequence is either the following Γ (A&B) or is the following Γ (B&A) in accordance with the definition of deduction. The deduction of this sequence contains no more than n elements. Hence the lemma holds true for this deduction in accordance with the recursion assumption. If g ϕ−1(A) = 0 then g ϕ−1(A&B) = 0 and g ϕ−1(B&A) = 0 in accordance with the Boolean function definition (Def. 1.1.10). Hence there exists sentence C such that C ∈ Γ and g ϕ−1(C) = 0 in accordance with the lemma. Hence in that case the lemma holds true for the deduction of sequence Γ A. b) Let Γ A be obtained from previous sequences by I&. In that case forms of these previous sequences are Γ1 B and Γ2 G with Γ = Γ1,Γ2 and A = (B&G) in accordance with the definition of deduction. The lemma holds true for deductions of sequences Γ1 B and Γ2 G in accordance with the recursion assumption because these deductions contain no more than n elements. In that case if g ϕ−1(A) = 0 then g ϕ−1(B) = 0 or g ϕ−1(G) = 0 in accordance with the Boolean function definition. Hence there exists sentence C such that g ϕ−1(C) = 0 and C ∈ Γ1 or C ∈ Γ2. Hence in that case the lemma holds true for the deduction of sequence Γ A. c) Let Γ A be obtained from a previous sequence by R¬. In that case a form of this previous sequence is the following: Γ (¬(¬A)) in accor- dance with the definition of deduction. The lemma holds true for the deduction of this 9
- 24. sequence in accordance with the recursion assumption because this deduction contains no more than n elements. If g ϕ−1(A) = 0 then g ϕ−1(¬(¬A)) = 0 in accordance with the Boolean function definition. Hence there exists sentence C such that C ∈ Γ and g ϕ−1(C) = 0. Hence the lemma holds true for the deduction of sequence Γ A. d) Let Γ A be obtained from previous sequences by I¬. In that case forms of these previous sequences are Γ1 B and Γ2 (¬B) with Γ = Γ1 [G],Γ2 [G] and A = (¬G) in accordance with the definition of deduction. The lemma holds true for the deductions of sequences Γ1 B and Γ2 (¬B) in ac- cordance with the recursion assumption because these deductions contain no more than n elements. If g ϕ−1(A) = 0 then g ϕ−1(G) = 1 in accordance with the Boolean function defi- nition. Either g ϕ−1(B) = 0 or g ϕ−1(¬B) = 0 by the same definition. Hence there exists sentence C such that either C ∈ Γ1 [G] or C ∈ Γ2 [G] andg ϕ−1(C) = 0 in accordance with the recursion assumption. Hence in that case the lemma holds true for the deduction of sequence Γ A. e) Let Γ A be obtained from a previous sequence by I∨. In that case a form of A is (B∨G) and a form of this previous sequence is either Γ B or Γ G in accordance with the definition of deduction. The lemma holds true for this previous sequence deduction in accordance with the recursion assumption because this deduction contains no more than n elements. If g ϕ−1(A) = 0 then g ϕ−1(B) = 0 and g ϕ−1(G = 0 in accordance with the Boolean function definition. Hence there exists sentence C such that C ∈ Γ and g ϕ−1(C) = 0. Hence in that case the lemma holds true for the deduction of sequence Γ A. f) Let Γ A be obtained from previous sequences by R∨. Forms of these previous sequences are Γ1 A, Γ2 A, and Γ3 (B∨G) with Γ = Γ1 [B],Γ2 [G],Γ3 in accordance with the definition of deduction. The lemma holds true for the deductions of these sequences in accordance with the recursion assumption because these deductions contain no more than n elements. If g ϕ−1(A) = 0 then there exists sentence C1 such that C1 ∈ Γ1 and g ϕ−1(C1) = 0, and there exists sentence C2 such that C2 ∈ Γ2 and g ϕ−1(C2) = 0 in accordance with the lemma. If g ϕ−1(B∨G) = 0 then there exists sentence C such that C ∈ Γ3 and g ϕ−1(C) = 0 in accordance with the lemma. Hence in that case the lemma holds true for the deduction of sequence Γ A. If g ϕ−1(B∨G) = 1 then either g ϕ−1(B) = 1 or g ϕ−1(G) = 1 in accordance with the Boolean function definition. If g ϕ−1(B) = 1 then C1 ∈ Γ1 [B]. Hence in that case the lemma holds true for the deduction of sequence Γ A. If g ϕ−1(G) = 1 then a result is the same. Hence the lemma holds true for the deduction of sequence Γ A in all these cases. g) Let Γ A be obtained from previous sequences by R⇒. 10
- 25. Forms of these previous sequences are Γ1 (B ⇒ A) and Γ2 (B) with Γ = Γ1,Γ2 in accordance with the definitions of deduction. Hence the lemma holds true for these deduction in accordance with the recursion assumption because these deductions contain no more than n elements. If g ϕ−1(B ⇒ A) = 0 then there exists sentence C such that C ∈ Γ1 and g ϕ−1(C) = 0 in accordance with the lemma. Hence in that case the lemma holds true for the deduction of sequence Γ A. If g ϕ−1(B ⇒ A) = 1 then g ϕ−1(B) = 0 in accordance with the Boolean function definition. Hence there exists sentence C such that C ∈ Γ2 and g ϕ−1(C) = 0. Hence the lemma holds true for sequence Γ A in all these cases. h) Let Γ A be obtained from a previous sequence by I⇒. In that case a form of sentence A is (B ⇒ G) and a form of this previous sequence is Γ1 G with Γ = Γ1 [B] in accordance with the definition of deduction. The lemma holds true for the deduction of this sequence in accordance the recursion assumption because this deduction contain no more than n elements. If g ϕ−1(A) = 0 then g ϕ−1(G) = 0 and g ϕ−1(B) = 1 in accordance with the Boolean function definition. Hence there exists sentence C such that C ∈ Γ1 [B] and g ϕ−1(C) = 0. The recursion step conclusion: Therefore, in each possible case, if the lemma holds true for a deduction, containing no more than n elements, then the lemma holds true for a deduction contained n+1 elements. The recursion conclusion: Therefore the lemma holds true for a deduction of any length Th. 1.1.4: Each naturally propositionally proven sentence is a tautology. Proof of Th. 1.1.4: If a sentence A is naturally propositionally proven then there exists a natural propositional deduction of form A in accordance with Def. 1.1.9. Hence for every Boolean function g: g ϕ−1(A) = 1 in accordance with Lm. 1.1.1. Hence sentence A is a tautology in accordance with the tautology definition (Def. 1.1.13) Designation 1: Let g be a Boolean function. In that case for every sentence A: Ag := A if g ϕ−1(A) = 1, (¬A) if g ϕ−1(A) = 0. Lm. 1.1.2: Let B1,B2,..,Bk be the simple sentences making sentence A by PC-symbols (¬, &, ∨, ⇒). Let g be any Boolean function. In that case there exist a propositional natural deduction of sequence Bg 1,Bg 2,..,Bg k Ag . Proof of Lm. 1.1.2: is received by a recursion on a number of PC-symbols in sentence A. Basis of recursion Let A does not contain PC-symbols . In this case the string of one sequence: 1. Ag Ag, NPC-axiom. is a fit deduction. 11
- 26. Step of recursion: The recursion assumption: Let the lemma holds true for every sentence, containing no more than n PC-symbols. Let sentence A contains n+1 PC-symbol. Let us consider all possible cases. a) Let A = (¬G). In that case the lemma holds true for G in accordance with the recursion assumption because G contains no more than n PC-symbols. Hence there exists a deduction of sequence Bg 1,Bg 2,..,Bg k Gg , (1.5) here B1,B2,..,Bk are the simple sentences, making up sentence G. Hence B1,B2,..,Bk make up sentence A. If g ϕ−1(A) = 1 then Ag = A = (¬G) in accordance with Designation 1. In that case g ϕ−1(G) = 0 in accordance with the Boolean function definition. Hence Gg = (¬G) = A in accordance with Designation 1. Hence in that case a form of sequence (1.5) is the following: Bg 1,Bg 2,..,Bg k Ag . Hence in that case the lemma holds true. If g ϕ−1(A) = 0 then Ag = (¬A) = (¬(¬G)). in accordance with Designation 1. In that case g ϕ−1(G) = 1 in accordance with the Boolean function definition. Hence Gg = G in accordance with Designation 1. Hence in that case a form of sequence (1.5) is Bg 1,Bg 2,..,Bg k G. Let us continue the deduction of this sequence in the following way: 1. Bg 1,Bg 2,..,Bg k G. 2. (¬G) (¬G), NPC-axiom. 3. Bg 1,Bg 2,..,Bg k (¬(¬G)), I¬ from 1. and 2. It is a deduction of sequence Bg 1,Bg 2,..,Bg k Ag . 12
- 27. Hence in that case the lemma holds true. b) Let A = (G&R). In that case the lemma holds true both for G and for R in accordance with the recur- sion assumption because G and R contain no more than n PC-symbols. Hence there exist deductions of sequences Bg 1,Bg 2,..,Bg k Gg (1.6) and Bg 1,Bg 2,..,Bg k Rg , (1.7) here B1,B2,..,Bk are the simple sentences, making up sentences G and R. Hence B1,B2,..,Bk make up sentence A. If g ϕ−1(A) = 1 then Ag = A = (G&R) in accordance with Designation 1. In that case g ϕ−1(G) = 1 and g ϕ−1(R) = 1 in accordance with the Boolean func- tion definition. Hence Gg = G and Rg = R in accordance with Designation 1. Let us continue deductions of sequences (1.6) and (1.7) in the following way: 1. Bg 1,Bg 2,..,Bg k G, (1.6). 2. Bg 1,Bg 2,..,Bg k R, (1.7). 3. Bg 1,Bg 2,..,Bg k (G&R), I& from 1. and 2. It is deduction of sequence Bg 1,Bg 2,..,Bg k Ag. Hence in that case the lemma holds true. If g ϕ−1(A) = 0 then Ag = (¬A) = (¬(G&R)) in accordance with Designation 1. In that case g(G) = 0 or g(R) = 0 in accordance with the Boolean function definition. Hence Gg = (¬G) or Rg = (¬R) in accordance with Designation 1. Let Gg = (¬G). In that case let us continue a deduction of sequence (1.6) in the following way: 1. Bg 1,Bg 2,..,Bg k (¬G), (1.6). 2. (G&R) (G&R), NPC-axiom. 3. (G&R) G, R& from 2. 4. Bg 1,Bg 2,..,Bg k (¬(G&R)), I¬ from 1. and 3. It is a deduction of sequence Bg 1,Bg 2,..,Bg k Ag. Hence in that case the lemma holds true. The same result is received if Rg = (¬R). c) Let A = (G∨R). 13
- 28. In that case the lemma holds true both for G and for R in accordance with the recursion assumption because G and R contain no more than n PC-symbols. Hence there exist a deductions of sequences Bg 1,Bg 2,..,Bg k Gg (1.8) and Bg 1,Bg 2,..,Bg k Rg , (1.9) here B1,B2,..,Bk are the simple sentences, making up sentences G and R. Hence B1,B2,..,Bk make up sentence A. If g ϕ−1(A) = 0 then Ag = (¬A) = (¬(G∨R)) in accordance with Designation 1. In that case g ϕ−1(G) = 0 and g ϕ−1(R) = 0 in accordance with the Boolean func- tion definition. Hence Gg = (¬G) and Rg = (¬R) in accordance with Designation 1. Let us continue deductions of sequences (1.8) and (1.9) in the following way: 1. Bg 1,Bg 2,..,Bg k (¬G), (1.8). 2. Bg 1,Bg 2,..,Bg k (¬R), (1.9). 3. G G, NPC-axiom. 4. R R, NPC-axiom. 5. (G∨R) (G∨R), NPC-axiom. 6. G,Bg 1,Bg 2,..,Bg k (¬(G∨R)), I¬ from 1. and 3. 7. R,Bg 1,Bg 2,..,Bg k (¬(G∨R)), I¬ from 2. and 4. 8. (G∨R),Bg 1,Bg 2,..,Bg k (¬(G∨R)), R∨ from 5., 6., and 7. 9. Bg 1,Bg 2,..,Bg k (¬(G∨R)), I¬ from 7. and 8. It is a deduction of sequence Bg 1,Bg 2,..,Bg k Ag. Hence in that case the lemma holds true. If g ϕ−1(A) = 1 then Ag = A = (G∨R) in accordance with Designation 1. In that case g ϕ−1(G) = 1 or g ϕ−1(R) = 1 in accordance with the Boolean function definition. Hence Gg = G or Rg = R in accordance with Designation 1. If Gg = G then let us continue deduction of sequence (1.8) in the following way: 1. Bg 1,Bg 2,..,Bg k G, (1.8). 2. Bg 1,Bg 2,..,Bg k (G∨R), I∨ from 1. It is deduction of sequence Bg 1,Bg 2,..,Bg k Ag. Hence in that case the lemma holds true. The same result is received if Rg = R. d) Let A = (G ⇒ R). 14
- 29. In that case the lemma holds true both for G and for R in accordance with the recur- sion assumption because G and R contain no more than n PC-symbols. Hence there exist deductions of sequences Bg 1,Bg 2,..,Bg k Gg (1.10) and Bg 1,Bg 2,..,Bg k Rg , (1.11) here B1,B2,..,Bk are the simple sentence, making up sentences G and R. Hence B1,B2,..,Bk make up sentence A. If g ϕ−1(A) = 0 then Ag = (¬A) = (¬(G ⇒ R)) in accordance with Designation 1. In that case g ϕ−1(G) = 1 and g ϕ−1(R) = 0 in accordance with the Boolean func- tion deduction. Hence Gg = G and Rg = (¬R) in accordance with Designation 1. Let us continue deduction of sequences (1.10) and (1.11) in the following way: 1. Bg 1,Bg 2,..,Bg k G, (1.10). 2. Bg 1,Bg 2,..,Bg k (¬R), (1.11). 3. (G ⇒ R) (G ⇒ R), NPC-axiom. 4. (G ⇒ R),Bg 1,Bg 2,..,Bg k R, R⇒ from 1. and 3. 5. Bg 1,Bg 2,..,Bg k (¬(G ⇒ R)), I¬ from 2. and 4. It is deduction of sequence Bg 1,Bg 2,..,Bg k Ag. Hence in that case the lemma holds true. If g ϕ−1(A) = 1 then Ag = A = (G ⇒ R) in accordance with Designation 1. In that case g ϕ−1(G) = 0 or g ϕ−1(R) = 1 in accordance with the Boolean function definition. Hence Gg = (¬G) or Rg = R in accordance with Designation 1. If Gg = (¬G) then let us continue a deduction of sequence (1.10) in the following way: 1. Bg 1,Bg 2,..,Bg k (¬G), (1.10). 2. G G, NPC-axiom. 3. G,Bg 1,Bg 2,..,Bg k (¬(¬R)), I¬ from 1. and 2. 4. G,Bg 1,Bg 2,..,Bg k R, R¬ from 3. 5. Bg 1,Bg 2,..,Bg k (G ⇒ R), I⇒ from 4. It is deduction of sequence Bg 1,Bg 2,..,Bg k Ag. Hence in that case the lemma holds true. If Rg = R then let us continue a deduction of sequence (1.11) in the following way: 1. Bg 1,Bg 2,..,Bg k R, (1.11). 2. Bg 1,Bg 2,..,Bg k (G ⇒ R), I⇒ from 1. 15
- 30. It is deduction of sequence Bg 1,Bg 2,..,Bg k Ag. Hence in that case the lemma holds true. The recursion step conclusion: If the lemma holds true for sentences, containing no more than n PC-symbols, then the lemma holds true for sentences, containing n + 1 PC- symbols. The recursion conclusion: The lemma holds true for sentences, containing any number PC-symbols Th. 1.1.5 (Laszlo Kalmar)4: Each tautology is a naturally propositionally proven sentence. Proof of Th. 1.1.5: Let sentence A be a tautology. That is for every Boolean function g: g ϕ−1(A) = 1 in accordance with Def. 1.1.13. Hence there exists a deduction for sequence Bg 1,Bg 2,..,Bg k A (1.12) for every Boolean function g in accordance with Lm. 1.1.2. There exist Boolean functions g1 and g2 such that g1 ϕ−1(B1) = 0, g2 ϕ−1(B1) = 1, g1 ϕ−1(Bs) = g2 ϕ−1(Bs) for s ∈ {2,..,k}. Forms of sequences (1.12) for these Boolean functions are the following: (¬B1),Bg1 2 ,..,Bg1 k A, (1.13) B1,Bg2 2 ,..,Bg2 k A. (1.14) Let us continue deductions these sequence in the following way: 1. (¬B1),Bg1 2 ,..,Bg1 k A, (1.13). 2. B1,Bg1 2 ,..,Bg1 k A, (1.14). 3. (¬A) (¬A), NPC-axiom. 4. (¬A),Bg1 2 ,..,Bg1 k (¬(¬B1)), I¬ from 1. and 3. 5. (¬A),Bg1 2 ,..,Bg1 k (¬B1), I¬ from 2. and 3. 6. Bg1 2 ,..,Bg1 k (¬(¬A)), I¬ from 4. and 5. 7. Bg1 2 ,..,Bg1 k A, R¬ from 6. It is deduction of sequence Bg1 2 ,..,Bg1 k A. This sequence is obtained from sequence (1.12) by deletion of first sentence from the hypothesizes list. All rest hypothesizes are deleted from this list in the similar way. Final sentence is the following: A. Therefore, in accordance with Th. 1.1.3, all tautologies are true sentences. Therefore the natural propositional logic presents by Boolean functions. 4Laszlo Kalmar (March 27, 1905 August 2, 1976) was a Hungarian mathematician and Professor at the Uni- versity of Szeged. Kalmar is considered the founder of mathematical logic and theoretical Computer Science in Hungary. 16
- 31. 1.1. Recorders Any information, received from physical devices, can be expressed by a text, made of sen- tences. Let a be some object which is able to receive, save, and/or transmit an information [22]. A set a of sentences, expressing an information of an object a, is called a recorder of this object. Thus, statement: ”Sentence A is an element of the set a” denotes : ”a has information that the event, expressed by sentence A , took place.” In short: ”a knows that A.” Or by designation: ”a• A ”. Obviously, the following conditions are satisfied: I. For any a and for every A: false is that a• (A&(¬A)), thus, any recorder doesn’t contain a logical contradiction. II. For every a, every B, and all A: if B is a logical consequence from A, and a•A, then a•B. *III. For all a, b and for every A: if a• b•A then a•A. For example, if device a has information that device b has information that mass of particle ←−χ equals to 7 then device a has information that mass of particle ←−χ equal to 7. 1.2. Time There is only a moment between past and future. L. Derbenev, ”Sannikov Land”, (1973) ”Do not expect answers before you have found clear meanings” Hans Reichenbach, The Direction of Time, (1953) There are many concepts of the theory of ”time” - in particular, quantum mechanical, relativistic, thermodynamic, causal, etc. All of them are based on unclearly defined con- cepts and in most cases contain a vicious circle. The thermodynamic concept has the greatest favor. But if a sergeant has a platoon in a line, does this sergeant’s wristwatch change direction? Any subjects, connected with an information is called informational objects. For ex- ample, it can be a physics device, or computer disks and gramophone records, or people, carrying memory on events of their lifes, or trees, on cuts which annual rings tell on past climatic and ecological changes, or stones with imprints of long ago extincted plants and bestials, or minerals, telling on geological cataclysms, or celestial bodies, carrying an in- formation on a remote distant past Universe, etc., etc. It is clearly that an information, received from such information object, can be expressed by a text which made of sentences. Let’s consider finite (probably empty) path of symbols of form q•. Def. 1.3.1 A path α is a subpath of a path β (design.: α β) if α can be got from β by deletion of some (probably all) elements. 17
- 32. Designation: (α)1 is α, and (α)k+1 is α(α)k . Therefore, if k ≤ l then (α)k (α)l . Def. 1.3.2 A path α is equivalent to a path β (design.: α ∼ β) if α can be got from β by substitution of a subpath of form (a•)k by a path of the same form ((a•)s ). In this case: III. If β α or β ∼ α then for any K: if a•K then a• (K&(αA ⇒ βA)). Obviously, III is a refinement of condition *III. Def. 1.3.3 A natural number q is instant, at which a registrates B according to κ-clock {g0,A,b0} (design.: q is [a•B ↑ a,{g0,A,b0}]) if: 1. for any K: if a•K then a• (K&(a• B ⇒ a• (g• 0b• 0)q g• 0A)) and a• K& a• (g• 0b• 0)q+1 g• 0A ⇒ a• B . 2. a• a•B& ¬a• (g• 0b• 0)q+1 g• 0A . Lm. 1.3.1 If q is [a• αB ↑ a,{g0,A,b0}], (1.15) p is [a• βB ↑ a,{g0,A,b0}], (1.16) α β, (1.17) then q ≤ p. Proof of Lm. 1.3.1: From (1.16): a• (a• βB)& ¬a• (g• 0b• 0)(p+1) g• 0A . (1.18) From (1.17) according to III: a• a• βB& ¬a• (g• 0b• 0)(p+1) g• 0A &(a• βB ⇒ a• αB) . (1.19) Let us designate: R := a•βB, S := ¬a• (g• 0b• 0)(p+1) g• 0A , G := a•αB. In that case a shape of formula (1.18) is a• (R&S), 18
- 33. and a shape of formula (1.19) is a• ((R&S)&(R ⇒ G)). Sentence (G&S) is a logical consequence from sentence ((R&S)&(R ⇒ G)) (1.1). Hence a• (G&S), in accordance with II. Hence a• a• αB& ¬a• (g• 0b• 0)(p+1) g• 0A in accordance with the designation. Hence from (1.15): a• a• αB& ¬a• (g• 0b• 0)(p+1) g• 0A &(a• αB ⇒ a• (g• 0b• 0)q g• 0A) . According to II: a• ¬a• (g• 0b• 0)(p+1) g• 0A &a• (g• 0b• 0)q g• 0A (1.20) If q > p, i.e. q ≥ p+1, then from (1.20) according to III a• ¬a• (g• 0b• 0)(p+1) g• 0A &a• (g• 0b• 0)q g• 0A & a• (g• 0b• 0)q g• 0A ⇒ a• (g• 0b• 0)(p+1) g• 0A . According to II: a• ¬a• (g• 0b• 0)(p+1) g• 0A &a• (g• 0b• 0)(p+1) g• 0A . It contradicts to condition I. Therefore, q ≤ p . Lemma 1.3.1 proves that if q is [a• B ↑ a,{g0,A,b0}], and p is [a• B ↑ a,{g0,A,b0}] then q = p. That’s why, expression ”q is [a•B ↑ a,{g0,A,b0}]” is equivalent to expression ”q = [a•B ↑ a,{g0,A,b0}].” Def. 1.3.4 κ-clocks {g1,B,b1} and {g2,B,b2} have the same direction for a if the following condition is satisfied: if 19
- 34. r = [a• (g• 1b• 1)q g• 1B ↑ a,{g2,B,b2}], s = [a• (g• 1b• 1)p g• 1B ↑ a,{g2,B,b2}], q < p, then r ≤ s. Th. 1.3.1 All κ-clocks have the same direction. Proof of Th. 1.3.1: Let r := [a• (g• 1b• 1)q g• 1B ↑ a,{g2,B,b2}], s := [a• (g• 1b• 1)p g• 1B ↑ a,{g2,B,b2}], q < p. In this case (g• 1b• 1)q (g• 1b• 1)p . Consequently, according to Lm. 1.3.1 r ≤ s Consequently, a recorder orders its sentences with respect to instants. Moreover, this order is linear and it doesn’t matter according to which κ-clock it is set. Def. 1.3.5 κ-clock {g2,B,b2} is k times more precise than κ-clock {g1,B,b1} for recorder a if for every C the following condition is satisfied: if q1 = [a•C ↑ a,{g1,B,b1}], q2 = [a•C ↑ a,{g2,B,b2}], then q1 < q2 k < q1 +1. Lm. 1.3.2 If for every n: qn−1 < qn kn < qn−1 +1, (1.21) then the series q0 + ∞ ∑ n=1 qn −qn−1kn k1 ...kn (1.22) 20
- 35. converges. Proof of Lm. 1.3.2: According to (1.21): 0 ≤ qn −qn−1kn < kn. Consequently, series (1.22) is positive and majorizes next to q0 +1+ ∞ ∑ n=1 1 k1 ...kn , convergence of which is checked by d’Alambert’s criterion Def. 1.3.6 A sequence H of κ-clocks: {g0,A,b0}, {g1,A,b2},.., gj,A,bj , .. is called an absolutely precise κ-clock of a recorder a if for every j exists a natural number kj so that κ-clock gj,A,bj is kj times more precise than κ-clock gj−1,A,bj−1 . In this case if qj = a• C ↑ a, gj,A,bj and t = q0 + ∞ ∑ j=1 qj −qj−1 ·kj k1 ·k2 ·..·kj , then t is a•C ↑ a,H . Lm. 1.3.3: If q := q0 + ∞ ∑ j=1 qj −qj−1 ·kj k1 ·k2 ·..·kj (1.23) with qn−1 ≤ qn kn < qn−1 +1, and d := d0 + ∞ ∑ j=1 dj −dj−1 ·kj k1 ·k2 ·..·kj (1.24) with dn−1 ≤ dn kn < dn−1 +1 then if qn ≤ dn then q ≤ d. Proof of Lm. 1.3.3: A partial sum of series (1.23) is the following: 21
- 36. Qu := q0 + q1 −q0k1 k1 + q2 −q1k2 k1k2 +···+ qu −qu−1ku k1k2 ···ku , Qu = q0 + q1 k1 −q0 + q2 k1k2 − q1 k1 +···+ qu k1k2 ···ku − qu−1 k1k2 ···ku−1 , Qu = qu k1k2 ···ku . A partial sum of series (1.24) is the following: Du = du k1k2 ···ku . Consequently, according to the condition of Lemma: Qn ≤ Dn Lm. 1.3.4 If q is a• αC ↑ a,H , d is a• βC ↑ a,H , and α β then q ≤ d. Proof of Lm. 1.3.4 comes out of Lemmas 1.3.1 and 1.3.3 immediately Therefore, if α ∼ β then q = d. 1.3. Space Def. 1.4.1 A number t is called a time, measured by a recorder a according to a κ-clock H, during which a signal C did a path a•αa• (design.: t := m aH (a•αa•C)), if t = a• αa• C ↑ a,H − a• C ↑ a,H . Th. 1.4.1 m aH (a• αa• C) ≥ 0. Proof comes out straight of Lemma 1.3.4 22
- 37. Thus, any ”signal”, ”sent” by the recorder, ”will come back” to it not earlier than it was ”sent”. Def. 1.4.2 1) for every recorder a: (a•)† = (a•); 2) for all paths α and β: (αβ)† = (β)† (α)† . Def. 1.4.3 A set ℜ of recorders is an internally stationary system for a recorder a with a κ-clock H (design.: ℜ is ISS a,H ) if for all sentences B and C, for all elements a1 and a2 of set ℜ, and for all paths α, made of elements of set ℜ, the following conditions are satisfied: 1) a•a• 2a• 1C ↑ a,H − a•a• 1C ↑ a,H = = a•a• 2a• 1B ↑ a,H − a•a• 1B ↑ a,H ; 2) m aH (a•αa•C) = m aH a•α†a•C . Th. 1.4.2 {a}−ISS a,H . Proof: 1)As a• ∼ a•a• then, according to Lemma 1.3.4 : if we symbolize p := a• a• B ↑ a,H , q := a• a• a• B ↑ a,H , r := a• a• C ↑ a,H , s := a• a• a• C ↑ a,H , then q = p and s = r. That’s why q− p = s−r. 2) Since any series α, made of elements of set {a} coincides with α† then m aH (a•αa•C) = m aH a•α†a•C . Thus every singleton is an internally stationary systeminternally stationary system. Lm. 1.4.1: If {a,a1,a2} isISS a,H then a•a• 2a• 1a• 2C ↑ a,H − a•a• 2C ↑ a,H = = a•a• 1a• 2a• 1B ↑ a,H − a•a• 1B ↑ a,H Proof: Let’s symbolize 23
- 38. p := a• a• 1B ↑ a,H , q := a• a• 1a• 2a• 1B ↑ a,H , r := a• a• 2C ↑ a,H , s := a• a• 2a• 1a• 2C ↑ a,H , u := a• a• 2a• 1B ↑ a,H , w := a• a• 1a• 2C ↑ a,H . Thus, according to statement 1.4.3 u− p = s−w,w−r = q−u. Thus, s−r = q− p Def. 1.4.4 A number l is called an aH(B)-measure of recorders a1 and a2 (design.: l = a,H,B (a1,a2) if l= 0.5· a•a• 1a• 2a• 1B ↑ a,H − a•a• 1B ↑ a,H . Lm. 1.4.2 If {a,a1,a2} is ISS a,H then for all B and C: a,H,B (a1,a2) = a,H,C (a1,a2). Proof: Let us designate: Let us design: p := a• a• 1B ↑ a,H , q := a• a• 1a• 2a• 1B ↑ a,H , r := a• a• 1C ↑ a,H , s := a• a• 1a• 2a• 1C ↑ a,H , u := a• a• 2a• 1B ↑ a,H , w := a• a• 2a• 1C ↑ a,H . 24
- 39. Thus, according to Def. 1.4.3: u− p = w−r,q−u = s−w. Thus, q− p = s−r Therefore, one can write expression of form ” a,H,B (a1,a2)” as the following: ” a,H (a1,a2)”. Th. 1.4.3: If {a,a1,a2,a3} is ISS a,H then 1) a,H (a1,a2) ≥ 0; 2) a,H (a1,a1) = 0; 3) a,H (a1,a2) = a,H (a2,a1); 4) a,H (a1,a2)+ a,H (a2,a3) ≥ a,H (a1,a3). Proof: 1) and 2) come out straight from Lemma 1.3.4 and 3) from Lemma 1.4.2. Let’s symbolize p := a• a• 1C ↑ a,H , q := a• a• 1a• 2a• 1C ↑ a,H , r := a• a• 1a• 3a• 1C ↑ a,H , s := a• a• 2a• 1C ↑ a,H , u := a• a• 2a• 3a• 2a• 1B ↑ a,H , w = a• a• 1a• 2a• 3a• 2a• 1C ↑ a,H . Thus, according to statement 1.4.3 w−u = q−s. Therefore, w− p = (q− p)+(u−s). According to Lemma 1.3.4 w ≥ r. Consequently, 25
- 40. (q− p)+(u−s) ≥ r − p Thus, all four axioms of the metrical space [19] are accomplished for a,H in an internally stationary systeminternally stationary system of recorders. Consequently, a,H is a distance length similitude in this space. Def. 1.4.5 A set ℜ of recorders is degenerated into a beam ab1 and point a1 if there exists C such that the following conditions are satisfied: 1) For any sequence α, made of elements of set ℜ, and for any K: if a•K then a• (K&(αa• 1C ⇒ αb• 1a• 1C)). 2) There is sequence β, made of elements of the setℜ, and there exist sentence S such that a• (βb• 1C&S) and it’s false that a• (βa• 1b• 1C&S) Further we’ll consider only not degenerated sets of recorders. Def. 1.4.6: B took place in the same place as a1 for a (design.: (a)(a1,B)) if for every sequence α and for any sentence K the following condition is satisfied: if a•K then a• (K&(αB ⇒ αa• 1B)). Th. 1.4.4: (a)(a1,a• 1B). Proof: Since αa• 1 ∼ αa• 1a• 1 then according to III: if a• 1K then a• 1 (K&(αa• 1B ⇒ αa• 1a• 1B)) Th. 1.4.5: If (a)(a1,B), (1.25) (a)(a2,B), (1.26) then (a)(a2,a• 1B). Proof: Let a•K. In this case from (1.26): a• (K&(αa• 1B ⇒ αa• 1a• 2B)). From (1.25): a• ((K&(αa• 1B ⇒ αa• 1a• 2B))&(αa• 1a• 2B ⇒ αa• 1a• 2a• 1B)). According to II: 26
- 41. a• (K&(αa• 1B ⇒ αa• 1a• 2a• 1B)). According to III: a• ((K&(αa• 1B ⇒ αa• 1a• 2a• 1B))&(αa• 1a• 2a• 1B ⇒ αa• 2a• 1B)). According to II: a• (K&(αa• 1B ⇒ αa• 2a• 1B)) Lm. 1.4.3: If (a)(a1,B), (1.27) t = a• αB ↑ a,H , (1.28) then t = a• αa• 1B ↑ a,H . Proof: Let’s symbolize: tj := a• αB ↑ a, gj,A,bj . Therefore, a• a• αB& ¬a• g• jb• j tj+1 g• jA , from (1.27): a• a• αB& ¬a• g• jb• j tj+1 g• jA &(a• αB ⇒ a• αa• 1B) . According to II: a• a• αa• 1B& ¬a• g• jb• j tj+1 g• jA , (1.29) Let a•K. In this case from (1.28): a• K& a• αB ⇒ a• g• jb• j tj g• jA . Therefore, according to III: a• K& a• αB ⇒ a• g• jb• j tj g• jA &(a• αa• 1B ⇒ a• αB) . According to II: a• K& a• αa• 1B ⇒ a• g• jb• j tj g• jA . (1.30) 27
- 42. From (1.27): a• K& a• g• jb• j tj+1 g• jA ⇒ a• αB &(a• αB ⇒ a• αa• 1B) . according to II: a• K& a• g• jb• j tj+1 g• jA ⇒ a• αa• 1B . From (1.29), (1.30) for all j: tj = a• αa• 1B ↑ a, gj,A,bj . Consequently, t = a• αa• 1B ↑ a,H Th. 1.4.6: If {a,a1,a2} is ISS a,H , (a)(a1,B), (1.31) (a)(a2,B), (1.32) then a,H (a1,a2) = 0. Proof: Let’s symbolize: t := a• B ↑ a,H . According to Lemma 1.4.3: from (1.31): t = a• a• 1B ↑ a,H , from (1.32): t = a• a• 1a• 2B ↑ a,H , again from (1.31): t = a• a• 1a• 2a• 1B ↑ a,H . Consequently, a,H (a1,a2) = 0.5·(t −t) = 0 28
- 43. Th. 1.4.7: If {a1,a2,a3} is ISS a,H and there exists sentence B such that (a)(a1,B), (1.33) (a)(a2,B), (1.34) then a,H (a3,a2) = a,H (a3,a1). Proof: According to Theorem 1.4.6 from (1.33) and (1.34): a,H (a1,a2) = 0; (1.35) according to Theorem 1.4.3: a,H (a1,a2)+ a,H (a2,a3) ≥ a,H (a1,a3), therefore, from (1.35): a,H (a2,a3) ≥ a,H (a1,a3), i.e. according to Theorem 1.4.3: a,H (a3,a2) ≥ a,H (a1,a3). (1.36) From a,H (a3,a1)+ a,H (a1,a2) ≥ a,H (a3,a2): and from (1.35): a,H (a3,a1) ≥ a,H (a3,a2). From (1.36): a,H (a3,a1) = a,H (a3,a2) Def. 1.4.7 A real number t is an instant of a sentence B in frame of reference ℜaH (design.: t = B | ℜaH ) if 1) ℜ is ISS a,H , 2) there exists a recorder b so that b ∈ ℜ and (a)(b,B), 3) t = a•B ↑ a,H − a,H (a,b). Lm. 1.4.4: a• B ↑ a,H = a• B | ℜaH . 29
- 44. Proof: Let ℜ is ISS a,H , a1 ∈ ℜ and (a)(a1,a• B). (1.37) According to Theorem 1.4.4: (a)(a,a• B). From (1.37) according to Theorem 1.4.6: a,H (a,a1) = 0, therefore a• B | ℜaH = a• B ↑ a,H − a,H (a,a1) = a• B ↑ a,H Def. 1.4.8 A real number z is a distance length between B and C in a frame of reference ℜaH (design.: z = ℜaH (B,C)) if 1) ℜ is ISS a,H , 2) there exist recorders a1 and a2 so that a1 ∈ ℜ, a2 ∈ ℜ, (a)(a1,B)) and (a)(a2,C)), 3) z = a,H (a2,a1). According to Theorem 1.4.3 such distance length satisfies conditions of all axioms of a metric space. 1.4. Relativity Def. 1.5.1: Recorders a1 and a2 equally receive a signal about B for a recorder a if (a)(a2,a• 1B) = (a)(a1,a• 2B) . Def. 1.5.2: Set of recorders are called a homogeneous space of recorders, if all its elements equally receive all signals. Def. 1.5.3: A real number c is an information velocity about B to the recorder a1 in a frame of reference ℜaH if c = ℜaH (B,a• 1B) a• 1B | ℜaH − B | ℜaH . Th. 1.5.1: In all homogeneous spaces: c = 1. 30
- 45. Proof: Let c represents information velocity about B to a recorder a1 in a frame of reference ℜaH . Thus, if ℜ is ISS a,H , z := ℜaH (B,a• 1B), (1.38) t1 := B | ℜaH , (1.39) t2 := a• 1B | ℜaH , (1.40) then c = z t2 −t1 . (1.41) According to (1.38) there exist elements b1 and b2 of set ℜ such that: (a)(b1,B), (1.42) (a)(b2,a• 2B), (1.43) z = a,H (b1,b2). (1.44) According to (1.39) and (1.40) there exist elements b1 and b2 of set ℜ such that: (a) b1,B , (1.45) (a) b2,a• 2B , (1.46) t1 = a• B ↑ a,H − a,H a,b1 , (1.47) t2 = a• a• 2B ↑ a,H − a,H a,b2 . (1.48) From (1.38), (1.42), (1.45) according to Theorem 1.4.7: a,H (a,b1) = a,H a,b1 . (1.49) Analogously from (1.38), (1.43), (1.46): a,H (a,b2) = a,H a,b2 . (1.50) Analogously from (1.47), (1.42), (1.49)according to Lemma 1.4.3: 31
- 46. t1 = a• b• 1B ↑ a,H − a,H (a,b1). (1.51) From (1.43) according to Lemma 1.4.3: a• a• 2B ↑ a,H = a• b• 2a• 2B ↑ a,H . (1.52) According to Lemma 1.3.4: a• b• 2a• 2B ↑ a,H ≥ a• b• 2B ↑ a,H . (1.53) From (1.43): (a)(a2,b• 2B). According to Lemma 1.4.3 a• a• 2b• 2B ↑ a,H = a• b• 2B ↑ a,H . (1.54) Again according to Lemma 1.3.4: a• a• 2b• 2B ↑ a,H ≥ a• a• 2B ↑ a,H . From (1.54), (1.52), (1.53): a• a• 2B ↑ a,H ≥ a• b• 2B ↑ a,H ≥ a• a• 2B ↑ a,H , therefore, a• a• 2B ↑ a,H = a• b• 2B ↑ a,H . From (1.48), (1.50): t2 = a• b• 2B ↑ a,H − a,H (a,b2). From (1.42) according to Lemma 1.4.3 t2 = a• b• 2b• 1B ↑ a,H − a,H (a,b2). (1.55) Let’s symbolize 32
- 47. u := a• C ↑ a,H , (1.56) d := a• b• 1a• C ↑ a,H , (1.57) w := a• b• 2a• C ↑ a,H , (1.58) j := a• b• 2b• 1a• C ↑ a,H , (1.59) q := a• b• 1b• 2a• C ↑ a,H , p := a• b• 1b• 2b• 1a• C ↑ a,H , (1.60) r := a• b• 2b• 1b• 2a• C ↑ a,H . Since ℜ is ISS a,H then q−w = p− j, (1.61) j = q. (1.62) From (1.55), (1.51), (1.57), (1.59): t2 + a,H (a,b2) − t1 + a,H (a,b1) = j −d, therefore t2 −t1 = j −d − a,H (a,b2)+ a,H (a,b1). (1.63) From (1.56), (1.57), (1.58) according to Lemma 1.3.4: a,H (a,b2) = 0.5·(w−u), a,H (a,b1) = 0.5·(d −u). From (1.61), (1.62), (1.63): t2 −t1 = 0.5·((j −d)+(j −w)) = 0.5·(j −d + p− j) = 0.5·(p−d). From (1.60), (1.57), (1.44): z = 0.5·(p−d). Consequently z = t2 −t1 That is in every homogenous space a propagation velocity of every information to every recorder for every frame reference equals to 1. Th. 1.5.2: If ℜ is a homogeneous space, then 33
- 48. a• 1B | ℜaH ≥ B | ℜaH . Proof comes out straight from Theorem 1.5.1. Consequently, in any homogeneous space any recorder finds out that B ”took place” not earlier than B ”actually take place”. ”Time” is irreversible. Th. 1.5.3 If a1 and a2 are elements of ℜ, ℜisISS a,H , (1.64) p := a• 1B | ℜaH , (1.65) q := a• 2a• 1B | ℜaH , (1.66) z := ℜaH (a1,a2), then z = q− p. Proof: In accordance with Theorem 1.5.1 from (1.64), (1.65), (1.66): q− p = ℜaH (a• 1B,a• 2a• 1B), thus in accordance with Definition 1.4.8 there exist elements b1 and b2 of ℜ such that (a)(b1,a• 1B), (1.67) (a)(b2,a• 2a• 1B), (1.68) q− p = ℜaH (b1,b2). Moreover, in accordance with Theorem 1.4.4 (a)(a• 1,a• 1B), (1.69) (a)(a• 2,a• 2a• 1B). From (1.68) in accordance with Theorem 1.4.7: ℜaH (b1,b2) = ℜaH (b1,a2). (1.70) In accordance with Theorem 1.4.3: ℜaH (b1,a2) = ℜaH (a2,b1). (1.71) Again in accordance with Theorem 1.4.7 from (1.69), (1.67): 34
- 49. ℜaH (a2,b1) = ℜaH (a2,a1). (1.72) Again in accordance with Theorem 1.4.3: ℜaH (a2,a1) = ℜaH (a1a2). From (1.72), (1.71), (1.70): ℜaH (b1,b2) = ℜaH (a1a2) According to Urysohn‘s theorem5 [20]: any homogeneous space is homeomorphic to some set of points of real Hilbert space. If this homeomorphism is not Identical transfor- mation, then ℜ will represent a non- Euclidean space. In this case in this ”space-time” corresponding variant of General Relativity Theory can be constructed. Otherwise, ℜ is Euclidean space. In this case there exists coordinates system Rµ such that the following condition is satisfied: for all elements a1 and a2 of set ℜ there exist points x1 and x2 of system Rµ such that a,H (ak,as) = ∑ µ j=1 xs,j −xk,j 2 0.5 . In this case Rµ is called a coordinates system of frame of reference ℜaH and numbers xk,1,xk,2,...,xk,µ are called coordinates of recorder ak in Rµ. A coordinates system of a frame of reference is specified accurate to transformations of shear, turn, and inversion. Def. 1.5.4: Numbers x1,x2,...,xµ are called coordinates of B in a coordinate system Rµ of a frame of reference ℜaH if there exists a recorder b such that b ∈ ℜ, (a)(b,B) and these numbers are the coordinates in Rµ of this recorder. Th. 1.5.4: In a coordinate system Rµ of a frame of reference ℜaH : if z is a dis- tance length between B and C, coordinates of B are (b1,b2,..,bn), coordinates of C are (c1,c2,..,c3), then z = µ ∑ j=1 (cj −bj)2 0.5 . Proof came out straight from Definition 1.5.4 Def. 1.5.5: Numbers x1,x2,...,xµ are called coordinates of the recor-der b in the coordinate system Rµ at the instant t of the frame of reference ℜaH if for every B the condition is satisfied: if t = b• B | ℜaH 5Pavel Samuilovich Urysohn, Pavel Uryson (February 3, 1898, Odessa - August 17, 1924, Batz-sur-Mer) was a Jewish mathematician who is best known for his contributions in the theory of dimension, and for de- veloping Urysohn’s Metrization Theorem and Urysohn’s Lemma, both of which are fundamental results in topology. 35
- 50. then coordinates of b•B in coordinate system Rµ of frame of reference ℜaH are the following: x1,x2,...,xµ . Lm. 1.5.1 If τ := [b• C ↑ b,{g0,B,b0}], (1.73) p := a• b• (g• 0b• 0)τ g• 0B ↑ a,{g1,A,b1} , (1.74) q := a• b• (g• 0b• 0)τ+1 g• 0B ↑ a,{g1,A,b1} , (1.75) t := [a• b• C ↑ a,{g1,A,b1}] (1.76) then p ≤ t ≤ q. Proof 1) From (1.75): a• a• b• (g• 0b• 0)τ+1 g• 0B& ¬a• (g• 1b• 1)q+1 g• 1A . (1.77) Hence from (1.73): b• (g• 0b• 0)τ+1 g• 0B ⇒ b•C then from (1.77) according to II: a• a•b•C& ¬a• (g• 1b• 1)q+1 g• 1A . According to II, since from (1.76): a•b•C ⇒ a• (g• 1b• 1)t g• 1A then a• a• (g• 1b• 1)t g• 1A& ¬a• (g• 1b• 1)q+1 g• 1A . (1.78) If t > q then t ≥ q+1. Hence according to III from (1.78): a• a• (g• 1b• 1)q+1 g• 1A& ¬a• (g• 1b• 1)q+1 g• 1A , it contradicts to I. So t ≤ q. 2) From (1.76): a• a• b• C& ¬a• (g• 1b• 1)t+1 g• 1A . (1.79) Since from (1.73): 36
- 51. b•C ⇒ b• (g• 0b• 0)τ g• 0B then from (1.79) according to II: a• a• b• (g• 0b• 0)τ g• 0B& ¬a• (g• 1b• 1)t+1 g• 1A . (1.80) Since from (1.74): a•b• (g• 0b• 0)τ g• 0B ⇒ a• (g• 1b• 1)p g• 1A then according to II from (1.80): a• a• (g• 1b• 1)p g• 1A& ¬a• (g• 1b• 1)t+1 g• 1A . (1.81) If p > t then p ≥ t +1. In that case from (1.81) according to III: a• a• (g• 1b• 1)t+1 g• 1A& ¬a• (g• 1b• 1)t+1 g• 1A , it contradicts to I. So p ≤ t Th. 1.5.5 In a coordinates system Rµ of a frame of reference ℜaH : if in every instant t: coordinates of6: b: xb,1 +v·t,xb,2,xb,3,...,xb,µ ; g0: x0,1 +v·t,x0,2,x0,3,...,x0,µ ; b0: x0,1 +v·t,x0,2 +l,x0,3,...,x0,µ ; and tC = b•C | ℜaH ; tD = b•D | ℜaH ; qC = [b•C ↑ b,{g0,A,b0}]; qD = [b•D ↑ b,{g0,A,b0}], then lim l→0 2· l (1−v2) · qD −qC tD −tC = 1. Proof: Let us designate: t1 := b• (g0 • b0 • )qC g0 • B | ℜaH , (1.82) t2 := b• (g0 • b0 • )qC+1 g0 • B | ℜaH , (1.83) t3 := (g0 • b0 • )qC g0 • B | ℜaH , (1.84) t4 := (g0 • b0 • )qC+1 g0 • B | ℜaH . (1.85) In that case coordinates of: 6below v is a real positive number such that |v| < 1) 37
- 52. b• (g0 • b0 • )qC g0 • B : xb,1 +v·t1,xb,2,xb,3,...,xb,µ , (1.86) b• (g0 • b0 • )qC+1 g0 • B : xb,1 +v·t2,xb,2,xb,3,...,xb,µ , (1.87) (g0 • b0 • )qC g0 • B : x0,1 +v·t3,x0,2,x0,3,...,x0,µ , (1.88) (g0 • b0 • )qC+1 g0 • B : x0,1 +v·t4,x0,2,x0,3,...,x0,µ , (1.89) b• C : xb,1 +v·tC,xb,2,xb,3,...,xb,µ . (1.90) According to Theorem 1.5.1 and Lemma 1.4.4 from (1.82), (1.86), (1.83), (1.87), (1.90): a•b• (g0 •b0 • )qC g0 •B | ℜaH = a•b• (g0 •b0 • )qC g0 •B ↑ a,H = t1 + (xb,1 +vt1)2 +∑ µ j+2 x2 b,j 0.5 , a•b• (g0 •b0 • )qC+1 B | ℜaH = a•b• (g0 •b0 • )qC+1 B ↑ a,H = t2 + (xb,1 +vt2)2 +∑ µ j=2 x2 b,j 0.5 . According to Lemma 1.5.1: t1 + (xb,1 +vt1)2 + µ ∑ j=2 x2 b,j 0.5 ≤ tC + (xb,1 +vtC)2 + µ ∑ j=2 x2 b,j 0.5 (1.91) ≤ t2 + (xb,1 +vt2)2 + µ ∑ j=2 x2 b,j 0.5 . According to Theorem 1.5.1 from (1.82), (1.84), (1.86), (1.88): t1 = t3 + (x0,1 +vt3 −xb,1 −vt1)2 +∑ µ j=2 x0,j −xb,j 2 0.5 . From (1.83), (1.85), (1.87), (1.89): t2 = t4 + (x0,1 +vt4 −xb,1 −vt2)2 +∑ µ j=2 x0,j −xb,j 2 0.5 . Hence: 38
- 53. (t1 −t3)2 = v2 (t1 −t3)2 −2v(t1 −t3)(x0,1 −xb,1)+∑ µ j=2 x0,j −xb,j 2 , (t2 −t4)2 = v2 (t2 −t4)2 −2v(t2 −t4)(x0,1 −xb,1)+∑ µ j=2 x0,j −xb,j 2 . Therefore, t2 −t4 = t1 −t3. (1.92) Let us designate: t5 := b0 • (g0 • b0 • )qC g0 • B | ℜaH . (1.93) In that case coordinates of: b0 • (g0 •b0 • )qC g0 •B : x0,1 +v·t5,x0,2 +l,x0,3,...,x0,µ . hence from (1.84), (1.88) according to Theorem 1.5.1: t5 −t3 = (x0,1 +vt5 −x0,1 −vt3)2 +(x0,2 +l −x0,2)2 +∑ µ j=3 (x0,j −x0,j)2 0.5 , hence: t5 −t3 = l √ 1−v2 . (1.94) Analogously from (1.93), (1.85), (1.89): t4 −t5 = l √ 1−v2 . From (1.94): t4 −t3 = 2l √ 1−v2 . From (1.92): t2 −t1 = 2l √ 1−v2 . Hence from (1.91): t1 + (xb,1 +vt1)2 + µ ∑ j=2 x2 b,j 0.5 ≤ tC + (xb,1 +vtC)2 + µ ∑ j=2 x2 b,j 0.5 ≤ t1 + 2l √ 1−v2 + xb,1 +v t1 + 2l √ 1−v2 2 + µ ∑ j=2 x2 b,j 0.5 . 39
- 54. Or if l → 0 then t2 → t1, and lim l→0 t1 + (xb,1 +vt1)2 + µ ∑ j=2 x2 b,j 0.5 = tC + (xb,1 +vtC)2 + µ ∑ j=2 x2 b,j 0.5 . Since, if v2 < 1 then function f (t) = t + (xb,1 +vt)2 + µ ∑ j=2 x2 b,j 0.5 is a monotonic one, then lim l→0 t1 = tC, hence lim l→0 b• (g0 • b0 • )qC g0 • B | ℜaH = tC. (1.95) Analogously, lim l→0 b• (g0 • b0 • )qD g0 • B | ℜaH = tD. (1.96) According to Theorem 1.5.1 from (1.82) and (1.83): b• (g0 • b0 • )qD g0 • B | ℜaH − b• (g0 • b0 • )qC g0 • B | ℜaH = t1 + 2l √ 1−v2 (qD −qC) −t1 = 2l (qD −qC) √ 1−v2 . From (1.95) and (1.96): lim l→0 2l (qD −qC) tD −tC = 1−v2 Corollary of Theorem 1.5.5: If designate: qst D := qD and qst C := qC for v = 0, then lim l→0 2l qst D −qst C tD −tC = 1, hence: 40
- 55. lim l→0 qD −qC qst D −qst C = 1−v2. For an absolutely precise κ-clock: qst D −qst C = qD −qC √ 1−v2 Consequently, moving at speed v κ-clock are times slower than the one at rest. Th. 1.5.6 Let: v (|v| < 1) and l be real numbers and ki be natural ones. Let in a coordinates system Rµ of a frame of reference ℜaH : in each instant t coor- dinates of: b: xb,1 +v·t,xb,2,xb,3,...,xb,µ , gj: yj,1 +v·t,yj,2,yj,3,...,yj,µ , uj: yj,1 +v·t,yj,2 +l/(k1 ·...·kj),yj,3,...,yj,µ , for all bi: if bi ∈ ℑ, then coordinates of bi: xi,1 +v·t,xi,2,xi,3,...,xi,µ , T is {g1,A,u1}, {g2,A,u2},.., gj,A,uj , .. . In that case: ℑ is ISS b,T . Proof 1) Let us designate: p := b• b• 1B ↑ b,T , q := b• b• 2b• 1B ↑ b,T , r := b• b• 1C ↑ b,T , s := b• b• 2b• 1C ↑ b,T , tp := b• b• 1B | ℜaH , (1.97) tq := b• b• 2b• 1B | ℜaH , (1.98) tr := b• b• 1C | ℜaH , (1.99) ts := b• b• 2b• 1B | ℜaH . (1.100) According to Corollary of Theorem 1.5.5: tq −tp = q− p √ 1−v2 , (1.101) ts −tr = s−r √ 1−v2 . (1.102) 41
- 56. From (1.97-1.100) coordinates of: b• b• 1B : xb,1 +vtp,xb,2,xb,3,...,xb,µ , (1.103) b• b• 2b• 1B : xb,1 +vtq,xb,2,xb,3,...,xb,µ , b• b• 1C : xb,1 +vtr,xb,2,xb,3,...,xb,µ , (1.104) b• b• 2b• 1C : xb,1 +vts,xb,2,xb,3,...,xb,µ . Let us designate: t1 := b• 1B | ℜaH , (1.105) t2 := b• 1C | ℜaH . (1.106) Consequently, coordinates of: b• 1B : x1,1 +vt1,x1,2,x1,3,...,x1,µ , b• 1C : x1,1 +vt2,x1,2,x1,3,...,x1,µ . According to Theorem 1.5.1 from (1.104), (1.106), (1.99): tr −t2 = (xb,1 +vtr −x1,1 −vt2)2 + µ ∑ j=2 xb,j −x1,j 2 0.5 . Analogously from (1.103), (1.105), (1.97): tp −t1 = (xb,1 +vtp −x1,1 −vt1)2 + µ ∑ j=2 xb,j −x1,j 2 0.5 . Hence, tr −t2 = tp −t1. (1.107) Let us denote: t3 := b2 • b• 1B | ℜaH , t4 := b2 • b• 1C | ℜaH . Hence, coordinates of: 42
- 57. b2 • b• 1B : x2,1 +vt3,x2,2,x2,3,...,x2,µ , b2 • b• 1C : x2,1 +vt4,x2,2,x2,3,...,x2,µ . According to Theorem 1.5.1: t3 −t1 = (x2,1 +vt3 −x1,1 −vt1)2 + µ ∑ j=2 (x2,j −x1,j)2 0.5 . t4 −t2 = (x2,1 +vt4 −x1,1 −vt2)2 + µ ∑ j=2 (x2,j −x1,j)2 0.5 . Hence: t3 −t4 = t1 −t2. (1.108) And analogously: tq −t3 = ts −t4. (1.109) From (1.108), (1.109), (1.107): tq −tp = ts −tr. From (1.102), (1.101): q− p = s−r. (1.110) 2) Let us designate: p := b• C ↑ b,T , q := b• αb• C ↑ b,T , r := b• α† b• C ↑ b,T ; here α is b• 1b• 2 ...b• kb• k+1 ...b• N. Hence according Definition 1.4.1: m bT (b• αb• C) = q − p , (1.111) m bT b• α† b• C = r − p . (1.112) Let us designate: 43
- 58. t0 := b• C | ℜaH , t1 = b• 1b• C | ℜaH , t2 := b• 2b• 1b• C | ℜaH , ···, tk := b• k ...b• 2b• 1b• C | ℜaH , (1.113) tk+1 := b• k+1b• k ...b• 2b• 1b• C | ℜaH , ···, tN := b• N ...b• k+1b• k ...b• 2b• 1b• C | ℜaH , tN+1 := b• α† b• C | ℜaH . Hence in accordance with this theorem condition coordinates of: b• C : xb,1 +vt0,xb,2,xb,3,...,xb,µ , b• 1b• C : x1,1 +vt1,x1,2,x1,3,...,x1,µ , b• 2b• 1b• C : x2,1 +vt2,x2,2,x2,3,...,x2,µ , ···, b• k ···b• 2b• 1b• C : xk,1 +vtk,xk,2,xk,3,...,xk,µ , b• k+1b• k ···b• 2b• 1b• C : xk+1,1 +vtk+1,xk+1,2,xk+1,3,...,xk+1,µ , ···, b• N ···b• k+1b• k ···b• 2b• 1b• C : xN,1 +vtN,xN,2,xN,3,...,xN,µ , b• α† b• C : xN+1,1 +vtN+1,xN+1,2,xN+1,3,...,xN+1,µ . Hence from (1.113) according Theorem 1.5.1: 44
- 59. t1 −t0 = (x1,1 +vt1 −xb,1 −vt0)2 + µ ∑ j=2 x1,j −xb,j 2 0.5 , t2 −t1 = (x2,1 +vt2 −x1,1 −vt1)2 + µ ∑ j=2 (x2,j −x1,j)2 0.5 , ..., tk+1 −tk = (xk+1,1 +vtk+1 −xk,1 −vtk)2 + µ ∑ j=2 xk+1,j −xk,j 2 0.5 , ..., tN+1 −tN = (xb,1 +vtN+1 −xN,1 −vtN)2 + µ ∑ j=2 xb,j −xN,j 2 0.5 . If designate: ρ2 a,b := µ ∑ j=1 (xb,1 −xa,1)2 , then for every k: tk+1 −tk = v 1−v2 (xk+1,1 −xk,1) + 1 1−v2 ρ2 k,k+1 −v2 µ ∑ j=2 xk+1,j −xk,j 2 0.5 . Hence: tN+1 −t0 = = 1 1−v2 ρ2 b,1 −v2 ∑ µ j=2 x1,j −xb,j 2 0.5 + ρ2 N,b −v2 ∑ µ j=2 xb,j −xN,j 2 0.5 +∑N−1 k=1 ρ2 k,k+1 −v2 ∑ µ j=2 xk+1,j −xk,j 2 0.5 . Analogously, if designate: τN+1 := b•αb•C | ℜaH 45
- 60. then τN+1 −t0 = = 1 1−v2 ρ2 1,b −v2 ∑ µ j=2 xb,j −x1,j 2 0.5 + ρ2 b,N −v2 ∑ µ j=2 xN,j −xb,j 2 0.5 +∑N−1 k=1 ρ2 k+1,k −v2 ∑ µ j=2 xk,j −xk+1,j 2 0.5 , hence tN+1 −t0 = τN+1 −t0. (1.114) According to Theorem 1.5.5: τN+1 −t0 = q −p√ 1−v2 and tN+1 −t0 = r −p√ 1−v2 . From (1.114), (1.111), (1.112): m bT (b•αb•C) = m bT b•α†b•C . From (1.110) according to Definition 1.4.3: ℑ is ISS b,T Therefore, a inner stability survives on a uniform straight line motion. Th. 1.5.7 Let: 1) in a coordinates system Rµ of a frame of reference ℜaH in every instant t: b : xb,1 +v·t,xb,2,xb,3,...,xb,µ , gj: yj,1 +v·t,yj,2,yj,3,...,yj,µ , uj: yj,1 +v·t,yj,2 +l/(k1 ·...·kj),yj,3,...,yj,µ , for every recorder qi: if qi ∈ ℑ then coordinates of qi : xi,1 +v·t,xi,2,xi,3,...,xi,µ , T is {g1,A,u1}, {g2,A,u2},.., gj,A,uj , .. . C : C1,C2,C3,...,Cµ , D : D1,D2,D3,...,Dµ , tC = C | ℜaH , tD = D | ℜaH ; 2) in a coordinates system Rµ of a frame of reference ℑbT : C : C1,C2,C3,...,Cµ , D : D1,D2,D3,...,Dµ , tC = C | ℑbT , tD = D | ℑbT . In that case: 46
- 61. tD −tC = (tD −tC)−v(D1 −C1) √ 1−v2 , D1 −C1 = (D1 −C1)−v(tD −tC) √ 1−v2 . Proof: Let us designate: ρa,b := µ ∑ j=1 (bj −aj)2 0.5 . According to Definition 1.4.8 there exist elements qC and qD of set ℑ such that (b)(qC,C)), (b)(qD,D) and ℑbT (C,D) = b,T (qC,qD). In that case: tC = C | ℑbT = q• CC | ℑbT , tD = D | ℑbT = q• DD | ℑbT . According to Corollary of Theorem 1.5.5: q• CC | ℜaH = C | ℜaH = tC, q• DD | ℜaH = D | ℜaH = tD. Let us designate: 47
- 62. τ1 := b• C ↑ b,T , τ2 := b• D ↑ b,T , t1 := b• C | ℜaH , t2 := b• D | ℜaH , t3 := b• B | ℜaH , t4 := q• Cb• B | ℜaH , t5 := b• q• Cb• B | ℜaH , t6 := q• Dq• Cb• B | ℜaH , t7 := q• Cq• Dq• Cb• B | ℜaH , t8 := b• q• Cq• Dq• Cb• B | ℜaH . Under such designations: t8 −t7 = t5 −t4 hence: t8 −t5 = t7 −t4 and ℑbT (C,D) = 0.5 b• q• Cq• Dq• Cb• B ↑ b,T − b• q• Cb• B ↑ b,T , hence: ℑbT (C,D) = 0.5(t8 −t5) √ 1−v2 = 0.5(t7 −t4) √ 1−v2, (t7 −t6)2 = (xC,1 +vt7 −xD,1 −vt6)2 +∑ µ j=2 (xC,j −xD,j)2 , (t6 −t4)2 = (xD,1 +vt6 −xC,1 −vt4)2 +∑ µ j=2 (xC,j −xD,j)2 , hence: (t7 −t6)2 = v2 (t7 −t6)2 +2v(xC,1 −xD,1)(t7 −t6)+ρ2 qC,qD , (t6 −t4)2 = v2 (t6 −t4)2 +2v(xD,1 −xC,1)(t6 −t4)+ρ2 qD,qC . Sequencely: t7 −t4 = 2√ 1−v2 v2 (xD,1 −xC,1)2 + 1−v2 ρ2 qC,qD 0.5 . 48
- 63. Let us designate: Ra,b := ρ2 a,b −v2 µ ∑ j=2 (aj −bj)2 0.5 . Under such designation: ℑbT (C,D) = RqC,qD √ 1−v2 . Since C1 = xC,1 +vtC, D1 = xD,1 +vtD, Cj+1 = xC,j+1, Dj+1 = xD,j+1 then RqC,qD = v2 (D1 −vtD −C1 +vtC)2 + 1−v2 (D1 −vtD −C1 +vtC)2 +∑ µ j=2 (Dj −Cj)2 0.5 , hence: RqC,qD = v2 (tD −tC)2 −2v(tD −tC)(D1 −C1) +ρ2 C,D −v2 ∑ µ j=2 (Dj −Cj)2 0.5 . (1.115) Moreover, according to Definition 1.4.7: tD −tC = (τ2 −τ1)− − b,T (b,qD)− b,T (b,qC) (1.116) According to Theorem 1.5.5: τ2 −τ1 = (t2 −t1) 1−v2. (1.117) According to Theorem 1.5.3: (t1 −tC)2 = (xb,1 +vt1 −C1)2 +∑ µ j=2 (xb,j −Cj)2 , (t2 −tD)2 = (xb,1 +vt2 −D1)2 +∑ µ j=2 (xb,j −Dj)2 . Therefore, (t1 −tC)2 = v2 (t1 −tC)2 +2v(xb,1 −xC,1)(t1 −tC)+ρ2 b,qC , (t2 −tD)2 = v2 (t2 −tD)2 +2v(xb,1 −xD,1)(t2 −tD)+ρ2 b,qD . 49
- 64. Hence, t2 −t1 = = (tD −tC)+ v 1−v2 (xC,1 −xD,1) + 1 1−v2 (Rb,qD −Rb,qC ). Because b,T (b,qD) = Rb,qD √ 1−v2 , b,T (b,qC) = Rb,qC √ 1−v2 , then from (1.116), (1.117), (1.118): tD −tC = (tD −tC) 1−v2 − v √ 1−v2 (xD,1 −xC,1), hence: tD −tC = (tD −tC) 1−v2 − v √ 1−v2 ((D1 −C1)−v(tD −tC)), hence: tD −tC = (tD −tC)−v(D1 −C1) √ 1−v2 , D1 −C1 = (D1 −C1)−v(tD −tC) √ 1−v2 . It is the Lorentz spatial-temporal transformations7 . Thus, if you have some set of objects, dealing with information, then time and space are inevitable. And it doesnt matter whether this set is part our world or some other worlds, which dont have a space-time structure initially. I call such Time the Informational Time. Since, we get our time together with our information system all other notions of time (ther- modynamical time, cosmological time, psychological time, quantum time etc.) should be defined by that Informational Time. 7Hendrik Antoon Lorentz (18 July 1853 - 4 February 1928) was a Dutch physicist who shared the 1902 Nobel Prize in Physics with Pieter Zeeman for the discovery and theoretical explanation of the Zeeman effect. He also derived the transformation equations subsequently used by Albert Einstein to describe space and time. 50
- 65. 1.5. Probability ”The two greatest tyrants on earth: case and time” Johann G. Herder The first significant result in probability theory was obtained by the Swiss mathemati- cian Jacob Bernoulli8 in 1713 [37] (the Bernoulli Large Number law). Further, the de- velopment of the theory of probability went in two ways: using the axiomatic method, Soviet mathematician, Andrei Kolmogorov9 embed this theory into mathematical analysis [25], and the American physicist Edwin Thompson Jaynes10 began to develop the theory of probability from logic [26]. And we continue this way. There is the evident nigh affinity between the classical probability function and the Boolean function of the classical propositional logic [23]. These functions are differed by the range of value, only. That is if the range of values of the Boolean function shall be 8Jacob Bernoulli (also known as James or Jacques; 6 January 1655 [O.S. 27 December 1654] 16 August 1705) was one of the many prominent mathematicians in the Bernoulli family. 9Andrey Nikolaevich Kolmogorov, 25 April 1903 20 October 1987) was a 20th-century Soviet mathemati- cian who made significant contributions to the mathematics of probability theory 10Edwin Thompson Jaynes (July 5, 1922 April 30, 1998) was the Wayman Crow Distinguished Profes- sor of Physics at Washington University in St. Louis. He wrote extensively on statistical mechanics and on foundations of probability and statistical inference 51
- 66. 52
- 67. expanded from the two-elements set {0;1} to the segment [0;1] of the real numeric axis then the logical analog of the Bernoulli Large Number Law [37] can be deduced from the logical axioms. These topics is considered in this article. Further we consider set of all meaningfull sentences. 1.5.1. Events Def. 1.6.1.1: A set B of sentences is called event, expressed by sentence C, if the following conditions are fulfilled: 1. C ∈ B; 2. if A ∈ B and D ∈ B then A = D; 3. if D ∈ B and A = D then A ∈ B. In this case denote: B := ◦C. Def. 1.6.1.2: An event B occurs if here exists a true sentence A such that A ∈ B. Def. 1.6.1.3: Events A and B equal (denote: A = B) if A occurs if and only if B occurs. Def. 1.6.1.4: Event C is called product of event A and event B (denote: C = (A ·B)) if C occurs if and only if A occurs and B occurs. Def. 1.6.1.5: Events C is called complement of event A (denote: C = (#A)) if C occurs if and only if A does not occur. Def. 1.6.1.6: (A +B) := (#((#A)·(#B))). Event (A +B) is called sum of event A and event B. Therefore, the sum of event occurs if and only if there is at least one of the addends. Def. 1.6.1.7: The authentic event (denote: T ) is the event which contains a tautology. Hence, T occurs in accordance Def. 1.6.1.2: The impossible event (denote: F ) is event which contains negation of a tautology. Hence, F does not occur. 1.5.2. B-functions Def. 1.6.2.1: Let b(X) be a function defined on the set of events. And let this function has values on he real numbers segment [0;1]. Let there exists an event C0 such that b(C0) = 1. Let for all events A and B: b(A ·B)+b(A ·(#B)) = b(A). In that case function b(X) is called B-function. By this definition: b(A ·B) ≤ b(A). (1.118) Hence, b(T · C0) ≤ b(T ). Because T · C0 = C0 (by Def.1.6.1.4 and Def.1.6.1.7) then b(C0) ≤ b(T ). Because b(C0) = 1then b(T ) = 1. (1.119) From Def.1.6.2.1: b(T · B) + b(T · (#B)) = b(T ). Because T D = D for any D then b(B)+b(#B) = b(T ). Hence, by (1.119): for any B: 53
- 68. b(B)+b(#B) = 1. (1.120) Therefore, b(T )+b(#T ) = 1. Hence, in accordance (1.119) : 1+b(F ) = 1. Therefore, b(F ) = 0. (1.121) In accordance with Def.1.6.2.1, Def.1.6.1.6, and (1.120): b(A ·(B +C)) = b(A ·(#((#B)·(#C)))) = = b(A)−b((A ·(#B))·(#C)) = b(A)−b(A ·(#B))+b((A ·(#B))·C) = = b(A)−b(A)+b(A ·B))+b((#B)·(A ·C)) = = b(A ·B))+b(A ·C)−b(B ·A ·C). And b((A ·B)+(A ·C)) = b(#((#(A ·B))·(#(A ·C)))) = = 1−b((#(A ·B))·(#(A ·C))) = = 1−b(#(A ·B))+b((#(A ·B))·(A ·C)) = = 1−1+b(A ·B)+b((#(A ·B))·(A ·C)) = = b(A ·B)+b((A ·C))−b((A ·B)·(A ·C)) = = b(A ·B)+b((A ·C))−b(A ·B ·C) because A ·A = A . Therefore: b(A ·(B +C)) = b(A ·B)+(A ·C)−b(A ·B ·C)) (1.122) and b((A ·B)+(A ·C)) = b(A ·B))+b(A ·C)−b(A ·B ·C). (1.123) Hence (distributivity): b(A ·(B +C)) = b((A ·B)+(A ·C)). (1.124) If A = T then from (1.122) and (1.123) (the addition formula of probabilities): b(B +C) = b(B)+b(C)−b(B ·C). (1.125) Def. 1.6.2.2– 19: Events B and C are antithetical events if (B · C) = F . From (1.125) and (1.121) for antithetical events B and C: b(B +C) = b(B)+b(C). (1.126) Def. 1.6.2.3-20: Events B and C are independent for B-function b events if b(B ·C) = b(B)·b(B). If events B and C are independent for B-function b events then: b(B ·(#C)) = b(B)−b(B ·C) = b(B)−b(B)·b(C) = b(B)·(1−b(C)) = b(B)·b(#C). Hence, if events B and C are independent for B-function b events then: b(B ·(#C)) = b(B)·b(#C). (1.127) Let calculate: b(A ·(#A)·C) = b(A ·C)−b(A ·A ·C) = b(A ·C)−b(A ·C) = 0. (1.128) 54
- 69. 1.5.3. Independent tests Definition 1.6.3.1: Let st(n) be a function such that st(n) has domain on the set of natural numbers and has values in the events set. In this case event A is a [st]-series of range r with V- number k if A, r and k fulfill to some one amongst the following conditions: 1) r = 1 and k = 1, A = st (1) or k = 0, A = (#st (1)); 2) B is [st]-series of range r −1 with V-number k −1 and A = (B ·st (r)), or B is [st]-series of range r −1 with V-number k and A = (B ·(#st (r))). Let us denote a set of [st]-series of range r with V-number k as [st](r,k). For example, if st (n) is a event Bn then the sentences: (B1 ·B2 ·(#B3)), (B1 ·(#B2)·B3), ((#B1)·B2 ·B3) are the elements of [st](3,2), and B1 ·B2 ·(#B3)·B4 ·B5 ∈ [st](5,3). Definition 1.6.3.2: Function st(n) is independent for B-function b if for A: if A ∈ [st](r,r) then: b(A) = r ∏ n=1 b(st (n)). Definition 1.6.3.3: Let st(n) be a function such that st(n) has domain on the set of natural numbers and has values in the set of events. In this case sentence A is [st]-disjunction of range r with V-number k (denote: t[st](r,k)) if A is the disjunction of all elements of [st](r,k). For example, if st (n) is event Cn then: ((#C1)·(#C2)·(#C3)) = t[st](3,0), t[st](3,1) = ((C1 ·(#C2)·(#C3))+((#C1)·C2 ·(#C3))+((#C1)·(#C2)·C3)), t[st](3,2) = ((C1 ·C2 ·(#C3))+((#C1)·C2 ·C3)+(C1 ·(#C2)·C3)), (C1 ·C2 ·C3) = t[st](3,3). Definition 1.6.3.4: A rational number ω is called frequency of sentence A in the [st]- series of r independent for B-function b tests (designate: ω = νr [st](A)) if 1) st(n) is independent for B-function b, 2) for all n: b(st (n)) = b(A), 3) t[st](r,k) is true and ω = k/r. Theorem: 1.6.3.1: (the J.Bernoulli11 formula [37]) If st(n) is independent for B- function b and there exists a real number p such that for all n: b(st (n)) = p then b(t [st](r,k)) = r! k!·(r −k)! · pk ·(1− p)r−k . 11Jacob Bernoulli (also known as James or Jacques) (27 December 1654 16 August 1705) was one of the many prominent mathematicians in the Bernoulli family. 55
- 70. Proof of the Theorem 1.6.3.1: By the Definition 1.6.3.2 and formula (1.127): if B ∈ [st](r,k) then: b(B) = pk ·(1− p)r−k . Since [st](r,k) contains r!/(k!·(r −k)!) elements then by the Theorems (1.127), (1.128) and (1.126) this Theorem is fulfilled. Definition 1.6.3.5: Let function st(n) has domain on the set of the natural numbers and has values in the set of the events. Let function f(r,k,l) has got the domain in the set of threes of the natural numbers and has got the range of values in the set of the events. In this case f(r,k,l) = T[st](r,k,l) if 1) f(r,k,k) = t[st](r,k), 2) f(r,k,l +1) = (f(r,k,l)+t[st](r,l +1)). Definition 1.6.3.6: If a and b are real numbers and k−1 < a ≤ k and l ≤ b < l +1 then T[st](r,a,b) = T[st](r,k,l). Theorem: 1.6.3.2: T[st](r,a,b) =◦ a r ≤ νr [st](A) ≤ b r . Proof of the Theorem 1.6.3.2: By the Definition 1.6.3.6: there exist natural numbers r and k such that k −1 < a ≤ k and l ≤ b < l +1. The recursion on l: 1. Let l = k. In this case by the Definition 1.6.3.4: T[st](r,k,k) = t[st](r,k) =◦ νr [st](A) = k r . 2. Let n be any natural number. The recursive assumption: Let T[st](r,k,k +n) =◦ k r ≤ νr [st](A) ≤ k +n r . By the Definition 1.6.3.5: T[st](r,k,k +n+1) = (T[st](r,k,k +n)+t[st](r,k +n+1)). By the recursive assumption and by the Definition 1.6.3.4: T[st](r,k,k +n+1) = = (◦ k r ≤ νr [st](A) ≤ k +n r +◦ νr [st](A) = k +n+1 r ). Hence, by the Definition 2.10: T[st](r,k,k +n+1) =◦ k r ≤ νr [st](A) ≤ k +n+1 r . 56
- 71. Theorem: 1.6.3.3 If st(n) is independent for B-function b and there exists a real number p such that b(st (n)) = p for all n then b(T[st](r,a,b)) = ∑ a≤k≤b r! k!·(r −k)! · pk ·(1− p)r−k . Proof of the Theorem 1.6.3.3: This is the consequence from the Theorem 1.6.3.1 by the Theorem 3.6. Theorem: 1.6.3.4 If st(n) is independent for the B-function b and there exists a real number p such that b(st (n)) = p for all n then b(T[st](r,r ·(p−ε),r ·(p+ε))) ≥ 1− p·(1− p) r ·ε2 for every positive real number ε. Proof of the Theorem 1.6.3.4: Because r ∑ k=0 (k −r · p)2 · r! k!·(r −k)! · pk ·(1− p)r−k = r · p·(1− p) then if J = {k ∈ N|0 ≤ k ≤ r ·(p−ε)}∩{k ∈ N|r ·(p+ε) ≤ k ≤ r} then ∑ k∈J r! k!·(r −k)! · pk ·(1− p)r−k ≤ p·(1− p) r ·ε2 . Hence, by (1.120) this Theorem is fulfilled. Hence lim r→∞ b(T[st](r,r ·(p−ε),r ·(p+ε))) = 1 (1.129) for all tiny positive numbers ε. 1.5.4. The logic probability function Definition 1.6.4.1: B-function P is P-function if for every event ◦ Θ : If P(◦ Θ ) = 1 then Θ is true sentence. Hence from Theorem 1.6.3.2 and (1.129): if b is a P-function then the sentence (p−ε) ≤ νr [st](A) ≤ (p+ε) is almost true sentence for large r and for all tiny ε. Therefore, it is almost truely that νr [st](A) = p for large r. Therefore, it is almost true that 57
- 72. b(A) = νr [st](A) for large r. Therefore, the function, defined by the Definition 1.6.4.1 has got the statistical meaning. That is why I’m call such function as the logic probability function. 1.5.5. Conditional probability Definition 1.6.5.1: Conditional probability B for C is the following function: b(B/C) := b(C ·B) b(C) . (1.130) Theorem 1.6.5.1 The conditional probability function is a B-function. Proof of Theorem 1.6.5.1 From Definition 1.6.5.1: b(C/C) = b(C ·C) b(C) . Hence by Theorem 1.1.1: b(C/C) = b(C) b(C) = 1. Form Definition 1.6.5.1: b((A ·B)/C)+b((A ·(#B))/C) = b(C ·(A ·B)) b(C) + b(C ·(A ·(#B))) b(C) . Hence: b((A ·B)/C)+b((A ·(#B))/C) = b(C ·(A ·B))+b(C ·(A ·(#B))) b(C) . By Theorem 1.1.1: b((A ·B)/C)+b((A ·(#B))/C) = b((C ·A)·B)+b((C ·A)·(#B)) b(C) . Hence by Definition 1.6.2.1: b((A ·B)/C)+b((A ·(#B))/C) = b(C ·A) b(C) . Hence by Definition 1.6.5.1: b((A ·B)/C)+b((A ·(#B))/C) = b(A/C) 58
- 73. 1.5.6. Classical probability Let P be P-function. Definition 1.6.6.1: {B1,B2,...,Bn} is called as complete set if the following conditions are fulfilled: 1. if k = s then (Bk ·Bs) is a false sentence; 2. (B1 +B2 +...+Bn) is a true sentence. Definition 1.6.6.2: B is favorable for A if (B ·(#A)) is a false sentence, and B is unfavorable for A if (B ∧A) is a false sentence. Let 1. {B1,B2,...,Bn} be complete set; 2. for k ∈ {1,2,...,n} and s ∈ {1,2,...,n}: P(Bk) = P(Bs); 3. if 1 ≤ k ≤ m then Bk is favorable for A, and if m+1 ≤ s ≤ n then Bs is unfavorable for A. In that case from Theorem 1.1.1 and from (1.119) and (1.120): P((#A)·Bk) = 0 for k ∈ {1,2,...,m} and P(A ·Bs) = 0 for s ∈ {m+1,m+2,...,n}. Hence from Definition 1.6.2.1: P(A ·Bk) = P(Bk) for k ∈ {1,2,...,n}. By point 4 of Theorem 1.1.1: A = (A ·(B1 +B2 +...+Bm +Bm+1 ...+Bn)). Hence by formula (1.124): P(A) = P(A ·B1)+P(A ·B2)+...+ +P(A ·Bm)+P(A ·Bm+1)+...+P(A ·Bn) = = P(B1)+P(B2)+...+P(Bm). Therefore P(A) = m n . 1.5.7. Probability and logic Let P be the probability function and let B be the set of events A such that either A occurs or (#A) occurs. In this case if P(A) = 1 then A occurs, and (A·B) = B in accordance with Def. 1.6.1.4. Consequently, if P(B) = 1 then P(A·B) = 1. Hence, in this case P(A·B) = P(A)·P(B). If P(A) = 0 then P(A·B) = P(A) · P(B) because P(A·B) ≤ P(A) in accordance with (1.118). 59
- 74. Moreover in accordance with (1.120): P(#A) = 1 − P(A) since the function P is a B- function. If event A occurs then (A·B) = B and (A·(#B)) = (#B) Hence, P(A·B)+P(A·(#B)) = P(A) = P(B)+P(#B) = 1. Consequently, if an element A of B occurs then P(A) = 1. If does not occurs then (#A) occurs. Hence, P(#A) = 1 and because P(A) + P(#A) = 1 then P(A) = 0 . Therefore, on B the range of values of is the two-element set {0;1} similar the Boolean function range of values. Hence, on set B the probability function obeys definition of a Boolean function (Def.1.1.10). The logic probability function is the extension of the logic B-function. Therefore, the probability is some generalization of the classic propositional logic. That is the proba- bility is the logic of events such that these events do not happen, yet. 1.5.8. THE NONSTANDARD NUMBERS Here some modification of the Robinson12 NONSTANDARD NUMBERS [30] is con- sidered. Let us consider the set N of natural numbers. Definition A.1: The n-part-set S of N is defined recursively as follows: 1) S1 = {1}; 2) S(n+1) = Sn ∪{n+1}. Definition A.2: If Sn is the n-part-set of N and A ⊆ N then A∩Sn is the quantity elements of the set A∩Sn, and if ϖn (A) = A∩Sn n , then ϖn (A) is the frequency of the set A on the n-part-set Sn. Theorem A.1: 12Abraham Robinson (born October 6, 1918 April 11, 1974) was a mathematician who is most widely known for development of non-standard analysis 60
- 75. 1) ϖn(N) = 1; 2) ϖn(/0) = 0; 3) ϖn(A)+ϖn(N−A) = 1; 4) ϖn(A∩B)+ϖn(A∩(N−B)) = ϖn(A). Proof of the Theorem A.1: From Definitions A.1 and A.2. Definition A.3: If ”lim” is the Cauchy-Weierstrass ”limit” then let us denote: ix = A ⊆ N| lim n→∞ ϖn(A) = 1 . Theorem A.2: ix is the filter [29], i.e.: 1) N ∈ ix, 2) /0 /∈ ix, 3) if A ∈ ix and B ∈ ix then (A∩B) ∈ ix ; 4) if A ∈ ix and A ⊆ B then B ∈ ix. Proof of the Theorem A.2: From the point 3 of Theorem A.1: lim n→∞ ϖn(N−B) = 0. From the point 4 of Theorem A.1: ϖn(A∩(N−B)) ≤ ϖn(N−B). Hence, lim n→∞ ϖn (A∩(N−B)) = 0. Hence, lim n→∞ ϖn (A∩B) = lim n→∞ ϖn(A). In the following text we shall adopt to our topics the definitions and the proofs of the Robinson Nonstandard Analysis [31]: Definition A.4: The sequences of the real numbers rn and sn are Q-equivalent (de- note: rn ∼ sn ) if {n ∈ N|rn = sn} ∈ ix. Theorem A.3: If r,s,u are the sequences of the real numbers then 1) r ∼ r, 2) if r ∼ s then s ∼ r; 3) if r ∼ s and s ∼ u then r ∼ u. Proof of the Theorem A.3: By Definition A.4 from the Theorem A.2 is obvious. Definition A.5: The Q-number is the set of the Q-equivalent sequences of the real numbers, i.e. if a is the Q-number and r ∈ a and s ∈ a, then r ∼ s; and if r ∈ a and r ∼ s then s ∈ a. Definition A.6: The Q-number a is the standard Q-number a if a is some real number and the sequence rn exists, for which: rn ∈ a and 61
- 76. {n ∈ N|rn = a} ∈ ix. Definition A.7: The Q-numbers a and b are the equal Q-numbers (denote: a = b) if a a ⊆ b and b ⊆ a. Theorem A.4: Let f(x,y,z) be a function, which has got the domain in R×R×R, has got the range of values in R (R is the real numbers set). Let y1,n , y2,n , y3,n , z1,n , z2,n , z3,n be any sequences of real numbers. In this case if zi,n ∼ yi,n then f(y1,n,y2,n,y3,n) ∼ f(z1,n,z2,n,z3,n) . Proof of the Theorem A.4: Let us denote: if k = 1 or k = 2 or k = 3 then Ak = {n ∈ N|yk,n = zk,n}. In this case by Definition A.4 for all k: Ak ∈ ix. Because (A1 ∩A2 ∩A3) ⊆ {n ∈ N|f(y1,n,y2,n,y3,n) = f(z1,n,z2,n,z3,n)}, then by Theorem A.2: {n ∈ N|f(y1,n,y2,n,y3,n) = f(z1,n,z2,n,z3,n)} ∈ ix. Definition A.8: Let us denote: QR is the set of the Q-numbers. Definition A.9: The function f, which has got the domain in QR × QR × QR, has got the range of values in QR, is the Q-extension of the function f, which has got the domain in R×R×R, has got the range of values in R, if the following condition is accomplished: Let xn , yn , zn be any sequences of real numbers. In this case: if xn ∈ x, yn ∈ y, zn ∈ z, u = f(x,y,z), then f(xn,yn,zn) ∈ u. Theorem A.5: For all functions f, which have the domain in R×R×R, have the range of values in R, and for all real numbers a, b, c, d: if f is the Q-extension of f; a, b, c, d are standard Q-numbers a, b, c, d, then: if d = f(a,b,c) then d = f(a,b,c) and vice versa. Proof of the Theorem A.5: If rn ∈ a, sn ∈ b, un ∈ c, tn ∈ d then by Definition A.6: {n ∈ N|rn = a} ∈ ix, {n ∈ N|sn = b} ∈ ix, {n ∈ N|un = c} ∈ ix, {n ∈ N|tn = d} ∈ ix. 1) Let d = f(a,b,c). In this case by Theorem A.2: 62
- 77. {n ∈ N|tn = f(rn,sn,un)} ∈ ix. Hence, by Definition A.4: tn ∼ f(rn,sn,un) . Therefore by Definition A.5: f(rn,sn,un) ∈ d. Hence, by Definition A.9: d = f(a,b,c). 2) Let d = f(a,b,c). In this case by Definition A.9: f(rn,sn,un) ∈ d. Hence, by Definition A.5: tn ∼ f(rn,sn,un) . Therefore, by Definition A.4: {n ∈ N|tn = f(rn,sn,un)} ∈ ix. Hence, by the Theorem A.2: {n ∈ N|tn = f(rn,sn,un),rn = a,sn = b,un = c,tn = d} ∈ ix. Hence, since this set does not empty, then d = f(a,b,c). By this Theorem: if f is the Q-extension of the function f then the expression ”f(x,y,z)” will be denoted as ”f(x,y,z)” and if u is the standard Q-number then the expression ”u” will be denoted as ”u”. Theorem A.6: If for all real numbers a, b, c: ϕ(a,b,c) = ψ(a,b,c) then for all Q-numbers x, y, z: ϕ(x,y,z) = ψ(x,y,z). Proof of the Theorem A.6: If xn ∈ x, yn ∈ y, zn ∈ z, u = ϕ(x,y,z), then by Defi- nition A.9: ϕ(xn,yn,zn) ∈ u. Because ϕ(xn,yn,zn) = ψ(xn,yn,zn) then ψ(xn,yn,zn) ∈ u. If v = ψ(x,y,z) then by Definition A.9: ψ(xn,yn,zn) ∈ v, too. 63
- 78. Therefore, for all sequences tn of real numbers: if tn ∈ u then by Definition A.5: tn ∼ ψ(xn,yn,zn) . Hence, tn ∈ v; and if tn ∈ v then tn ∼ ϕ(xn,yn,zn) ; hence, tn ∈ u. Therefore, u = v. Theorem A.7: If for all real numbers a, b, c: f(a,ϕ(b,c)) = ψ(a,b,c) then for all Q-numbers x, y, z: f(x,ϕ(y,z)) = ψ(x,y,z). Consequences from Theorems A.6 and A.7: [32]: For all Q-numbers x, y, z: 1: (x+y) = (y+x), 2: (x+(y+z)) = ((x+y)+z), 3: (x+0) = x, 5: (x·y) = (y·x), 6: (x·(y·z)) = ((x·y)·z), 7: (x·1) = x, 10: (x·(y+z)) = ((x·y)+(x·z)). Proof of the Theorem A.7: Let wn ∈ w, f(x,w) = u, xn ∈ x, yn ∈ y, zn ∈ z, ϕ(y,z) = w, ψ(x,y,z) = v. By the condition of this Theorem: f(xn,ϕ(yn,zn)) = ψ(xn,yn,zn). By Definition A.9: ψ(xn,yn,zn) ∈ v, ϕ(xn,yn) ∈ w, f(xn,wn) ∈ u. For all sequences tn of real numbers: 1) If tn ∈ v then by Definition A.5: tn ∼ ψ(xn,yn,zn) . Hence tn ∼ f(xn,ϕ(yn,zn)) . Therefore, by Definition A.4: {n ∈ N|tn = f(xn,ϕ(yn,zn))} ∈ ix and {n ∈ N|wn = ϕ(yn,zn)} ∈ ix. Hence, by Theorem A.2: {n ∈ N|tn = f(xn,wn)} ∈ ix. Hence, by Definition A.4: tn ∼ f(xn,wn) . Therefore, by Definition A.5: tn ∈ u. 2) If tn ∈ u then by Definition A.5: tn ∼ f(xn,wn) . Because wn ∼ ϕ(yn,zn) then by Definition A.4: {n ∈ N|tn = f(xn,wn)} ∈ ix, 64
- 79. {n ∈ N|wn = ϕ(yn,zn)} ∈ ix. Therefore, by Theorem A.2: {n ∈ N|tn = f(xn,ϕ(yn,zn))} ∈ ix. Hence, by Definition A.4: tn ∼ f(xn,ϕ(yn,zn)) . Therefore, tn ∼ ψ(xn,yn,zn) . Hence, by Definition A.5: tn ∈ v. From above and from 1) by Definition A.7: u = v. Theorem A.8: 4: For every Q-number x the Q-number y exists, for which: (x+y) = 0. Proof of the Theorem A.8: If xn ∈ x then y is the Q-number, which contains −xn . Theorem A.9: 9: There is not that 0 = 1. Proof of the Theorem A.9: is obvious from Definition A.6 and Definition A.7. Definition A.10: The Q-number x is Q-less than the Q-number y (denote: x < y) if the sequences xn and yn of real numbers exist, for which: xn ∈ x, yn ∈ y and {n ∈ N|xn < yn} ∈ ix. Theorem A.10: For all Q-numbers x, y, z: [33] 1: there is not that x < x; 2: if x < y and y < z then x < z; 4: if x < y then (x+z) < (y+z); 5: if 0 < z and x < y, then (x·z) < (y·z); 3 : if x < y then there is not, that y < x or x = y and vice versa; 3 : for all standard Q-numbers x, y, z: x < y or y < x or x = y. Proof of the Theorem A.10: is obvious from Definition A.10 by the Theorem A.2. Theorem A.11: 8: If 0 < |x| then the Q-number y exists, for which (x·y) = 1. Proof of the Theorem A.11: If xn ∈ x then by Definition A.10: if A = {n ∈ N|0 < |xn|} then A ∈ ix. In this case: if for the sequence yn : if n ∈ A then yn = 1/xn - then {n ∈ N|xn ·yn = 1} ∈ ix. Thus, Q-numbers are fulfilled to all properties of real numbers, except Ω3 [34]. The property Ω3 is accomplished by some weak meaning (Ω3’ and Ω3”). 65
- 80. Definition A.11: The Q-number x is the infinitesimal Q-number if the sequence of real numbers xn exists, for which: xn ∈ x and for all positive real numbers ε: {n ∈ N||xn| < ε} ∈ ix. Let the set of all infinitesimal Q-numbers be denoted as I. Definition A.12: The Q-numbers x and y are the infinite closed Q-numbers (denote: x ≈ y) if |x−y| = 0 or |x−y| is infinitesimal. Definition A.13: The Q-number x is the infinite Q-number if the sequence rn of real numbers exists, for which rn ∈ x and for every natural number m: {n ∈ N|m < rn} ∈ ix. 1.5.9. Model Let us define the propositional calculus like to ([28]), but the propositional forms shall be marked by the script greek letters. Definition C1: A set ℜ of the propositional forms is a U-world if: 1) if α1,α2,...,αn ∈ ℜ and α1,α2,...,αn β then β ∈ ℜ, 2) for all propositional forms α: it is not that (α&(¬α)) ∈ ℜ, 3) for every propositional form α: α ∈ ℜ or (¬α) ∈ ℜ. Definition C2: The sequences of the propositional forms αn and βn are Q- equivalent (denote: αn ∼ βn ) if {n ∈ N|αn ≡ βn} ∈ ix. Let us define the notions of the Q-extension of the functions for like as in the Definitions A.5, A.2, A.9, A.5, A.6. Definition C3: The Q-form α is Q-real in the U-world ℜ if the sequence αn of the propositional forms exists, for which: αn ∈ α and {n ∈ N|αn ∈ ℜ} ∈ ix. Definition C4: The set ℜ of the Q-forms is the Q-extension of the U-world ℜ if ℜ is the set of Q-forms α, which are Q-real in ℜ. Definition C5: The sequence ℜk of the Q-extensions is the S-world. Definition C6: The Q-form α is S-real in the S-world ℜk if k ∈ N|α ∈ ℜk ∈ ix. Definition C7: The set A (A ⊆ N) is the regular set if for every real positive number ε the natural number n0 exists, for which: for all natural numbers n and m, which are more or equal to n0: |wn(A)−wm(A)| < ε. Theorem C1: If A is the regular set and for all real positive ε: 66
- 81. {k ∈ N|wk(A) < ε} ∈ ix. then lim k→∞ wk(A) = 0. Proof of theTheorem C1: Let be lim k→∞ wk(A) = 0. That is the real number ε0 exists, for which: for every natural number n the natural number n exists, for which: n > n and wn(A) > ε0. Let δ0 be some positive real number, for which: ε0 −δ0 > 0. Because A is the regular set then for δ0 the natural number n0 exists, for which: for all natural numbers n and m, which are more or equal to n0: |wm(A)−wn(A)| < δ0. That is wm(A) > wn(A)−δ0. Since wn(A) ≥ ε0 then wm(A) ≥ ε0 −δ0. Hence, the natural number n0 exists, for which: for all natural numbers m: if m ≥ n0 then wm(A) ≥ ε0 −δ0. Therefore, {m ∈ N|wm(A) ≥ ε0 −δ0} ∈ ix. and by this Theorem condition: {k ∈ N|wk(A) < ε0 −δ0} ∈ ix. Hence, {k ∈ N|ε0 −δ0 < ε0 −δ0} ∈ ix. That is /0 /∈ ix. It is the contradiction for the Theorem 2.2. Definition C8: Let ℜk be a S-world. In this case the function W(β), which has got the domain in the set of the Q-forms, has got the range of values in QR, is defined as the following: If W(β) = p then the sequence pn of the real numbers exists, for which: pn ∈ p and pn = wn k ∈ N|β ∈ ℜk . 67
- 82. Theorem C2: If k ∈ N|β ∈ ℜk is the regular set and W(β) ≈ 1 then β is S-resl in ℜk . Proof of the Theorem C2: Since W(β) ≈ 1 then by Definitions.2.12 and 2.11: for all positive real ε: n ∈ N|wn k ∈ N|β ∈ ℜk > 1−ε ∈ ix. Hence, by the point 3 of the Theorem 2.1: for all positive real ε: n ∈ N| N−wn k ∈ N|β ∈ ℜk < ε ∈ ix. Therefore, by the Theorem C1: lim n→∞ N−wn k ∈ N|β ∈ ℜk = 0. That is: lim n→∞ wn k ∈ N|β ∈ ℜk = 1. Hence, by Definition.2.3: k ∈ N|β ∈ ℜk ∈ ix. And by Definition C6: β is S-real in ℜk . Theorem C3: The P-function exists. Proof of the Theorem C3: By the Theorems C2 and 2.1: W(β) is the P-function in ℜk . 68
- 83. Chapter 2 Quants If quantum physics did not frighten you, it means that you did not understand anything in it. Niels Bohr Quantum theory evolved as a new branch of theoretical physics during the first few decades of the 20th century in an endeavour to understand the fundamental properties of matter. It began with the study of the interactions of matter and radiation. Certain radiation effects could neither be explained by classical mechanics, nor by the theory of electromag- netism. Quantum theory was not the work of one individual, but the collaborative effort of some of the most brilliant physicists of the 20th century, among them Niels Bohr1, Erwin Schrodinger2, Wolfgang Pauli3, and Max Born4, Max Planck5 and Werner Heisenberg6. Quantum Field Theory (QFT) is the mathematical and conceptual framework for contemporary elementary particle physics (Eugene Wigner7, Hans Bethe8, Tomonaga9, Schwinger10, Feynman11, Dyson12, Yang13 and Mills14). 1Niels Henrik David Bohr (7 October 1885 - 18 November 1962) was a Danish physicist 2Erwin Rudolf Josef Alexander Schrodinger (12 August 1887 - 4 January 1961) was an Austrian physicist and theoretical biologist who was one of the fathers of quantum mechanics 3Wolfgang Ernst Pauli (25 April 1900 15 December 1958) was an Austrian theoretical physicist 4Max Born (11 December 1882 5 January 1970) was a German-born physicist and mathematician 5Max Karl Ernst Ludwig Planck (April 23, 1858 October 4, 1947) was a German physicist 6Werner Karl Heisenberg (5 December 1901 1 February 1976) was a German theoretical physicist 7Eugene Paul Wigner (Hungarian Wigner Jeno Pal; November 17, 1902 - January 1, 1995) was a Hungarian American physicist and mathematician. 8Hans Albrecht Bethe (July 2, 1906 - March 6, 2005) [1] was a German-American nuclear physicist, 9Sin-Itiro Tomonaga (March 31, 1906 July 8, 1979) was a Japanese physicist 10Julian Seymour Schwinger (February 12, 1918 - July 16, 1994) was an American theoretical physicist. 11Richard Phillips Feynman (May 11, 1918 - February 15, 1988)[2] was an American physicist 12Freeman John Dyson FRS (born December 15, 1923) is a British-born American theoretical physicist and mathematician 13Chen-Ning Franklin Yang (born October 1, 1922) is a Chinese-American physicist 14Robert L. Mills (April 15, 1927 – October 27, 1999) was an English physicist
- 84. 2.1. Physical events Denote: x : = (x1,x2,x3), x : = (x0,x), d3+1 x : = dx0 dx1 dx2 dx3, d3 y : = dy1 dy2 dy3, t : = x0 c . Sentence of type: Event A occurs in point x will be written the followig way: A (x) ”. Events of type ◦ A (x) are called dot events. All dot events and all events received from dot events by operations of addition, multiplication and addition, are physical events. A(D) means: (A (x)&◦ (x) ∈ D ). Let P be the probability function. Let XA,0,XA,1,XA,2,XA,3 be random coordinates of event A. Let FA be a Cumulative Distribution Function i.e.: FA (x0,x1,x2,x3) = P((XA,0 < x0)·(XA,1 < x1)·(XA,2 < x2)·(XA,3 < x3)). If j0 : = ∂3F ∂x1∂x2∂x3 , j1 : = − ∂3F ∂x0∂x2∂x3 , j2 : = − ∂3F ∂x0∂x1∂x3 , j3 : = ∂3F ∂x0∂x1∂x2 then j0, j1, j2, j3 is a probability current vector of event. If ρ := j0/c then ρ is a a probability density function. If uA := jA/ρA then vector uA is a velocity of the probability of A propagation. (for example for u2: u2 = j2 ρ = − ∂3F ∂x0∂x1∂x3 c ∂3F ∂x1∂x2∂x3 = − ∆013F ∆123F ∆x2 ∆x0 c ) 70
- 85. Probability, for which u2 1 +u2 2 +u2 3 ≤ c are called traceable probability. Denote: 12 := 1 0 0 1 , 02 := 0 0 0 0 , β[0] := − 12 02 02 12 = −14, the Pauli matrices σ1 = 0 1 1 0 , σ2 = 0 −i i 0 , σ3 = 1 0 0 −1 . A setC of complex n×n matrices is called a Clifford set 15 of rank n [39] if the following conditions are fulfilled: if αk ∈ C and αr ∈ C then αkαr +αrαk = 2δk,r; if αkαr +αrαk = 2δk,r for all elements αr of set C then αk ∈ C. If n = 4 then a Clifford set either contains 3 matrices (a Clifford triplet) or contains 5 matrices (a Clifford pentad). Here exist only six Clifford pentads [39]: one light pentad β: β[1] := σ1 02 02 −σ1 , β[2] := σ2 02 02 −σ2 , β[3] := σ3 02 02 −σ3 , (2.1) γ[0] := 02 12 12 02 , (2.2) β[4] := i· 02 12 −12 02 ; (2.3) three chromatic pentads: the red pentad ζ: ζ[1] = −σ1 02 02 σ1 ,ζ[2] = σ2 02 02 σ2 ,ζ[3] = −σ3 02 02 −σ3 , (2.4) γ [0] ζ = 02 −σ1 −σ1 02 , ζ[4] = i 02 σ1 −σ1 02 ; (2.5) the green pentad η: η[1] = −σ1 02 02 −σ1 ,η[2] = −σ2 02 02 σ2 ,η[3] = σ3 02 02 σ3 , (2.6) γ [0] η = 02 −σ2 −σ2 02 , η[4] = i 02 σ2 −σ2 02 ; (2.7) 15William Kingdon Clifford (4 May 1845 3 March 1879) was an English mathematician and philosopher. 71
- 86. the blue pentad θ: θ[1] = σ1 02 02 σ1 ,θ[2] = −σ2 02 02 −σ2 ,θ[3] = −σ3 02 02 σ3 , (2.8) γ [0] θ = 02 −σ3 −σ3 02 ,θ[4] = i 02 σ3 −σ3 02 ; (2.9) two gustatory pentads: the sweet pentad ∆: ∆[1] = 02 −σ1 −σ1 02 ,∆[2] = 02 −σ2 −σ2 02 ,∆[3] = 02 −σ3 −σ3 02 , ∆[0] = −12 02 02 12 ,∆[4] = i 02 12 −12 02 ; the bitter pentad Γ: Γ[1] = i 02 −σ1 σ1 02 ,Γ[2] = i 02 −σ2 σ2 02 ,Γ[3] = i 02 −σ3 σ3 02 , Γ[0] = −12 02 02 12 ,Γ[4] = 02 12 12 02 . Further we do not consider gustatory pentads since these pentads are not used yet in the contemporary physics. Let κ := 3 ∑ s=0 β[s]xs. If U0,1 (σ) := coshσ −sinhσ 0 0 −sinhσ coshσ 0 0 0 0 coshσ sinhσ 0 0 sinhσ coshσ (2.10) then U† 0,1 (σ)κU0,1 (σ) = β[0] (x0 cosh2σ+x1 sinh2σ) +β[1] (x1 cosh2σ+x0 sinh2σ) +β[2] x2 +β[3] x3. Hence, U0,1 makes the Lorentz transformaton x0,x1 : x0 → x0 := x0 cosh2σ+x1 sinh2σ, x1 → x1 := x1 cosh2σ+x0 sinh2σ, x2 → x2 := x2, x3 → x3 := x3. 72
- 87. Similarly, U0,2 (φ) := coshφ isinhφ 0 0 −isinhφ coshφ 0 0 0 0 coshφ −isinhφ 0 0 isinhφ coshφ (2.11) makes the Lorentz transformaton x0,x2 and U0,3 (ι) := eι 0 0 0 0 e−ι 0 0 0 0 e−ι 0 0 0 0 eι (2.12) makes the Lorentz transformaton x0,x3 . If U1,3 (ϑ) := cosϑ sinϑ 0 0 −sinϑ cosϑ 0 0 0 0 cosϑ sinϑ 0 0 −sinϑ cosϑ (2.13) then U1,3 (ϑ) makes the cartesian turn x1,x3 : U† 1,3 (ϑ)κU1,3 (ϑ) = β[0] x0 +β[1] (x1 cos2ϑ+x3 sin2ϑ) +β[2] x2 +β[3] (x3 cos2ϑ−x1 sin2ϑ) Similarly, U1,2 (ς) := e−iς 0 0 0 0 eiς 0 0 0 0 e−iς 0 0 0 0 eiς (2.14) makes the cartesian turn x1,x2 and U2,3 (α) := cosα isinα 0 0 isinα cosα 0 0 0 0 cosα isinα 0 0 isinα cosα (2.15) makes the cartesian turn x2,x3 . Let us consider the following set of four real equations with eight real unknowns: b2 with b > 0, α, β, χ, θ, γ, υ, λ: 73
- 88. b2 = ρ, b2 cos2 (α)sin(2β)cos(θ−γ)−sin2 (α)sin(2χ)cos(υ−λ) = − j1 c , b2 cos2 (α)sin(2β)sin(θ−γ)−sin2 (α)sin(2χ)sin(υ−λ) = − j2 c , b2 cos2 (α)cos(2β)−sin2 (α)cos(2χ) = − j3 c . (2.16) This set has solutions for any traceable ρ and jA,k. For example one of these solutions is the following: 1. A value of b2 obtain from first equation. 2. Since uk = jk ρ then cos2 (α)sin(2β)cos(θ−γ)−sin2 (α)sin(2χ)cos(υ−λ) = −u1 c , cos2 (α)sin(2β)sin(θ−γ)−sin2 (α)sin(2χ)sin(υ−λ) = −u2 c , cos2 (α)cos(2β)−sin2 (α)cos(2χ) = −u3 c . 3. Let β = χ. In that case: cos2 (α)cos(θ−γ)−sin2 (α)cos(υ−λ) sin(2β) = −u1 c , cos2 (α)sin(θ−γ)−sin2 (α)sin(υ−λ) sin(2β) = −u2 c , cos2 (α)−sin2 (α) cos(2β) = −u3 c . 4. Let (θ−γ) = (υ−λ). In that case: cos(2α)cos(θ−γ)sin(2β) = −u1 c , cos(2α)sin(θ−γ)sin(2β) = −u2 c , cos(2α)cos(2β) = −u3 c . 5. Let us raise to the second power the first and the second equations: cos2 (2α)cos2 (θ−γ)sin2 (2β) = −u1 c 2 , cos2 (2α)sin2 (θ−γ)sin2 (2β) = −u2 c 2 , cos(2α)cos(2β) = −u3 c . and let us summat these two equations: sin2 (2β)cos2 (2α) cos2 (θ−γ)+sin2 (θ−γ) = −u1 c 2 + −u2 c 2 , cos(2α)cos(2β) = −u3 c . Hence: sin2 (2β)cos2 (2α) = −u1 c 2 + −u2 c 2 , cos(2α)cos(2β) = −u3 c . 74
- 89. 6. Let us raise to the second power the second equation and add this equation to the previous one: sin2 (2β)cos2 (2α) = −u1 c 2 + −u2 c 2 , cos2 (2α)cos2 (2β) = −u3 c 2 sin2 (2β)+cos2 (2β) cos2 (2α) = − u1 c 2 + − u2 c 2 + − u3 c 2 , cos2 (2α) = − u1 c 2 + − u2 c 2 + − u3 c 2 , (2.17) We receive cos2 (2α) (for a trackeable probabilities). 7. From cos2 (2α)cos2 (2β) = − u3 c 2 we receive cos2 (2β). 8. From cos2 (2α)cos2 (θ−γ)sin2 (2β) = − u1 c 2 we receive cos2 (θ−γ). ———————————————- If ϕ1 := bexp(iγ)cos(β)cos(α), ϕ2 := bexp(iθ)sin(β)cos(α), ϕ3 := bexp(iλ)cos(χ)sin(α), (2.18) ϕ4 := bexp(iυ)sin(χ)sin(α) then you can calculate that ρ = 4 ∑ s=1 ϕ∗ s ϕs, (2.19) jα c = − 4 ∑ k=1 4 ∑ s=1 ϕ∗ s β [α] s,kϕk 2.2. Entanglement The presence in our universe of Planck’s constant gives reason to presume that our world is in a confined space: |x| ≤ πc/h. Indeed, the eigenvalues of physical operators are multiples of Planck constant integers. Consequently, the basic eigenfunctions of these operators con- stitute the discrete spectrum. Consequently. physical functions are representable by Fourier series. Hence, functions 75
- 90. φn (x) := h 2πc 3 2 exp −i h c (nx) . wiith natural n form an orthonormal basis of this space with scalar product of the following shape: (ϕ(t),χ(t)) := πc h −πc h dx1 πc h −πc h dx2 πc h −πc h dx3 ·ϕ(t,x)† χ(t,x). For the state vector: (ϕA (t),ϕA (t)) = P(A(t)). Let ϕA (t,x) = ∑ n aA,n (t)φn (x) be a Fourier series of ϕA (t,x) . That is: P(A(t)) = ∑ n a† A,n (t)aA,n (t) = ∑ n P(An (t)). Hence, A(t) = ∑ n An (t). there An (t) are incompatible, independent events:Therefore, if P(A(t)) = 1 then A(t) oc- cures. Hence, one among An (t) occures. Operator ˇr, defined on the state function ϕ set and has values on this set, is an realization operator if (ˇrϕ, ˇrϕ) = 1. This operator can act in the measurement process or as a result of some other external disturbance. Let ϕA,B ((t,x)(τ,y)) be a state vector of event A(t,x)and event B(τ,y). In this case the basis of space present the following functions: φn,s (x,y) := h 2πc 3 exp −i h c (nx+sy) . The scalar product has the following shape: (ϕ(t,τ),χ(t,τ)) := πc h −πc h dx1 πc h −πc h dx2 πc h −πc h dx3 πc h −πc h dy1 πc h −πc h dy2 πc h −πc h dy3 · ϕ(t,τ,x,y)† χ(t,τ,x,y). The Fourier series: 76
- 91. ϕA,B ((t,x)(τ,y)) = ∑ n ∑ s aA,B,n,s (t,τ)φn,s (x,y). Hence, (ϕA,B (t,τ),ϕA,B (t,τ)) = ∑ n ∑ s a† A,B,n,s (t,τ)aA,B.n,s (t,τ), (ϕA,B (t,τ),ϕA,B (t,τ)) = P(A(t)·B(τ)), P(A(t)·B(τ)) = ∑ n ∑ s P An (t)·Bs (τ), . For example: Let ϕA,B ((t,x)(τ,y)) = aA,B,1,4 (t,τ)φ1,4 (x,y)+aA,B,2,3 (t,τ)φ2,3 (x,y)+ aA,B,3,2 (t,τ)φ3,2 (x,y)+aA,B,4,1 (t,τ)φ4,1 (x,y). Such event called entangled events if aA,B,n,s are not factorized. In that case: (ϕA,B (t,τ),ϕA,B (t,τ)) = a† A,B,1,4 (t,τ)aA,B.1,4 (t,τ)+a† A,B,2,3 (t,τ)aA,B.2,3 (t,τ)+ a† A,B,3,2 (t,τ)aA,B.3,2 (t,τ)+a† A,B,4,1 (t,τ)aA,B.4,1 (t,τ). P(A(t)·B(τ)) = P A1 (t)·B4 (τ), +P A2 (t)·B3 (τ), + P A3 (t)·B2 (τ), +P A4 (t)·B1 (τ), . Let to ϕA,B ((t,x)(τ,y)) acts realization operator: (ˇrϕ, ˇrϕ) = 1. Then A(t)·B(τ) occures. Therefore, in accordance with the sum definition, one among events An (t)·Bs (τ) occures. Let us consider a shift x → x +r inside the region (−πc/h, πc/h). In this case a scalar product is sum of addends of type: πc h −πc h dx·a∗ p πc h ei h c p(x+r) aq πc h e−ih c q(x+r) with natural p and q. That is: πc h −πc h dx·a∗ p πc h ei h c p(x+r) aq πc h e−ih c q(x+r) = c∗ pcqeih c (p−q)r sin(π(p−q)) π(p−q) . 77
- 92. Because p and q are natural then shift πc h −πc h dx·a∗ p πc h eih c p(x+r) aq πc h e−ih c q(x+r) = c∗ pcqeih c (p−q)r δp,q. Hence, πc h −πc h dx·a∗ p πc h eih c p(x+r) aq πc h e−i h c q(x+r) = c∗ pcp. Therefore, here a scalar product is invariant for an inner shift. 2.3. Equations of moving If ϕ := U0,2 (φ)ϕ then ρ = ϕ † ϕ = ϕ† U† 0,2 (φ)U0,2 (φ)ϕ = ρcosh2φ+ j2 c sinh2φ and j2 c = −ϕ † s β[2] ϕk = −ϕ† U† 0,2 (φ)β[2] U0,2 (φ)ϕ = j2 c cosh2φ+ρsinh2φ. Similarly U0,1 and U0,3 transform the 3+1 vector cρ,j by the Lorentz formulas and U1,2, U1,3, U2,3 transform this vector by the cartesian formulas. Because ∂j0 ∂x0 = ∂4F ∂x0∂x1∂x2∂x3 = − ∂j1 ∂x1 = − ∂j2 ∂x2 = ∂ j3 ∂x3 then (Continuity equation ): ∂ρ ∂x0 + ∂ j1 ∂x1 + ∂ j2 ∂x2 + ∂j3 ∂x3 = 0 (2.20) In that case: ∂ ϕ†ϕ ∂x0 − ∂ ϕ†β[1]ϕ ∂x1 − ∂ ϕ†β[2]ϕ ∂x2 − ∂ ϕ†β[1]ϕ ∂x3 = 0 ∂ ϕ† ∂x0 ϕ+ϕ† ∂(ϕ) ∂x0 − ∂ ϕ†β[1] ∂x1 ϕ−ϕ† ∂ β[1]ϕ ∂x1 − ∂ ϕ†β[2] ∂x2 ϕ−ϕ† ∂ β[2]ϕ ∂x2 − ∂ ϕ†β[3] ∂x3 ϕ−ϕ† ∂ β[3]ϕ ∂x3 = 0 78
- 93. ϕ† ∂ ∂x0 −β[1] ∂ ∂x1 −β[2] ∂ ∂x2 −β[3] ∂ ∂x3 † ϕ +ϕ† ∂ ∂x0 −β[1] ∂ ∂x1 −β[2] ∂ ∂x2 −β[3] ∂ ∂x3 ϕ = 0 Let Q := ∂ x0 − 3 ∑ s=1 β[s] ∂ xs (2.21) Hence, ϕ† Q† +Q ϕ = 0 Q† = −Q (2.22) Therefore, for every function ϕj here exists an operator Qj,k such that a dependence of ϕj on t is described by the following differential equations 16: ∂tϕj = c 4 ∑ k=1 β [1] j,k∂1 +β [2] j,k∂2 +β [3] j,k∂3 +Qj,k ϕk. (2.23) and Q∗ j,k = −Qk,j. In that case if Hj,k := ic β [1] j,k∂1 +β [2] j,k∂2 +β [3] j,k∂3 +Qj,k then H is called a Hamiltonian17 of a moving with equation (2.23). A matrix form of formula (2.23) is the following: ∂tϕ = c β[1] ∂1 +β[2] ∂2 +β[3] ∂3 +Q ϕ (2.24) with ϕ = ϕ1 ϕ2 ϕ3 ϕ4 and Q = iϑ1,1 iϑ1,2 −ϖ1,2 iϑ1,3 −ϖ1,3 iϑ1,4 −ϖ1,4 iϑ1,2 +ϖ1,2 iϑ2,2 iϑ2,3 −ϖ2,3 iϑ2,4 −ϖ2,4 iϑ1,3 +ϖ1,3 iϑ2,3 +ϖ2,3 iϑ3,3 iϑ3,4 −ϖ3,4 iϑ1,4 +ϖ1,4 iϑ2,4 +ϖ2,4 iϑ3,4 +ϖ3,4 iϑ4,4 (2.25) 16This set of equations is similar to the Dirac equation with the mass matrix [40], [41], [42]. I choose a form of this set of equations in order to describe the behavior of ρ℘(t,x) by spinors and by Clifford’s set elements. 17Sir William Rowan Hamilton (4 August 1805 2 September 1865) was an Irish physicist, astronomer, and mathematician, who made important contributions to classical mechanics, optics, and algebra. 79
- 94. with ϖs,k = Re(Qs,k) and ϑs,k = Im(Qs,k). Matrix ϕ is called a state vector of the event A probability. An operator U (t,t0) with a domain and with a range of values on the set of state vectors is called an evolution operator if each state vector ϕ fulfils the following condition: ϕ(t) = U (t,t0)ϕ(t0). (2.26) Let us denote: Hd := c 3 ∑ s=1 iβ[s] ∂s. In that case H = Hd +icQ according the Hamiltonian definition: H = ic β[1] ∂1 +β[2] ∂2 +β[3] ∂3 +Q . From (2.24): i∂tϕ = Hϕ. . Hence: i∂tϕ = Hd +icQ ϕ. This differential equation has the following solution: ∂ϕ ϕ = −i Hd +icQ ∂t, t t=t0 ∂ϕ ϕ = −i t t=t0 Hd +icQ ∂t, lnϕ(t)−lnϕ(t0) = −i t t=t0 Hd∂t −iic t t=t0 Q∂t . Since Hd does not depend on time then t t=t0 Hd∂t = Hd (t −t0). Hence, according logarithm properties: ln ϕ(t) ϕ(t0) = −iHd (t −t0)+c t t=t0 Q∂t . 80
- 95. Therefore,18: ϕ(t) = ϕ(t0)exp −iHd (t −t0)+c t t=t0 Q∂t . Hence, from (2.26): U (t,t0) = exp −iHd (t −t0)+c t t=t0 Q∂t A Fourier series for ϕj (t,x) in ℜΩ has the following shape: ϕj (t0,x) = ∑ p cj,p (t0)ςp (t0,x) with ςp (x) := h 2πc 3 2 exp −ih c px if x ∈ Ω; 0, otherwise and with cj,p (t0) = ςp (x)∗ϕj (t0,x). That is in a matrix form: cp (t0) = (Ω) dx0 · h 2πc 3 2 exp i h c px0 ϕ(t0,x0) Hence, ϕ(t0,x) = ∑ p ( ) dx0 · h 2πc 3 2 exp i h c px0 ϕ(t0,x0) h 2πc 3 2 exp −i h c px . That is: ϕ(t0,x) = (Ω) dx0 · ∑ p h 2πc 3 exp −i h c p(x−x0) ϕ(t0,x0). Therefore, ϕ(t,x) = (Ω) dx0 · h 2πc 3 ∑p exp −iHd (t −t0)+c t t=t0 Q∂t · ·exp −ih c p(x−x0) ϕ(t0,x0). 18For an operator S: exp S := 1+S+ 1 2 S2 + 1 3! S3 +···+ 1 n! Sn +··· with S2 := SS and Sr+1 := Sr S. Here 1 is the unit operator such that for every u: 1u = u. 81
- 96. An operator K (t −t0,x−x0,t,t0) := h 2πc 3 ∑p exp −iHd (t −t0)+c t t=t0 Q∂t · ·exp −ih c p(x−x0) is called propagator of the event A probability. Hence: ϕ(t,x) = (Ω) dx0 ·K (t −t0,x−x0,t,t0)ϕ(t0,x0). (2.27) A propagator has the following property: K (t −t0,x−x0,t,t0) = dx1 ·K (t −t1,x−x1,t,t1)K (t1 −t0,x1−x0,t1,t0). 2.4. Double-Slit Experiment In a vacuum (Figure 1, Figure 2, Figure 3): Here transmitter s of electrons, wall w, and the electrons detecting black screen d are placed[45]. Electrons are emitted one by one from the source s. When an electron hits against screen d then a bright spot arises in the hit place of d.. 1. Let slit a be opened in wall w (Figure 1). An electron flies out from s, passes by a, and is detected by d. If such operation will be reiterated N of times then N bright spots shall arise on d against slit a in the vicinity of point ya. 2. Let slit b be opened in wall w (Figure 2). An electron flies out from s, passes by b, and is detected by d. If such operation will be reiterated N of times then N bright spots shall arise on d against slit b in the vicinity of point yb. 3. Let both slits be opened. In that case do you expect a result as on fig. 3? But no. We get result as on Figure 419[46]. For instance, such experiment was realized at Hitachi by A. Tonomura, J. Endo, T. Mat- suda, T. Kawasaki and H. Ezawa in 1989. Here was presumed that interference fringes are produced only when two electrons pass through both slits simultaneously. If there were two electrons from the source s at the same time, such interference might happen. But this cannot occur, because here is no more than one electron from this source at one time. Please keep watching the experiment a little longer. When a large number of electrons is accumulated, something like regular fringes begin to appear in the perpendicular direction as Figure 5(c) shows. Clear interference fringes can be seen in the last scene of the ex- periment after 20 minutes (Figure 5(d)). It should also be noted that the fringes are made 19Single-electron events build up over a 20 minute exposure to form an interference pattern in this double- slit experiment by Akira Tonomura and co-workers. Figure 5(a) 8 electrons; Figure 5(b) 270 electrons; Fig- ure 5(c) 2000 electrons; Figure 5(d) 60,000. A video of this experiment will soon be available on the web (www.hqrd.hitachi.co.jp/em/doubleslit.html). 82
- 97. up of bright spots, each of which records the detection of an electron. We have reached a mysterious conclusion. Although electrons were sent one by one, interference fringes could be observed. These interference fringes are formed only when electron waves pass through on both slits at the same time but nothing other than this. Whenever electrons are observed, they are always detected as individual particles. When accumulated, however, interference fringes are formed. Please recall that at any one instant here was at most one electron from s. We have reached a conclusion which is far from what our common sense tells us. 4. But nevertheless, across which slit the electron had slipped? Let (Figure 6) two detectors da and db and a photon source sf be added to devices of Figure 4. An electron, slipped across slit a, is lighten by source sf, and detector da snaps into action. And an electron, slipped across slit b, is lighten by source sf, and detector db snaps into action. If photon source sf lights all N electrons, slipped across slits, then we received the picture of Figure 3. If source sf is faint then only a little part of N electrons, slipped across slits, are noticed by detectors da and db. In that case electrons, noticed by detectors da and db, make picture of Figure 3, and all unnoticed electrons make picture of Figure 4. In result here the Figure 6 picture is received. Figure 1: ——————————– Let us try to interpret these experiments by events and probabilities. Denote the source s coordinates as x0,y0 , the slit a coordinates as xa,ya , the slit b coordinates as xb,yb . Here xa = xb, and the wall w equation is x = xa. Denote the screen d equation as x = xd. Denote 83
- 100. Figure 6: an event, expressed by sentence: electron is detected in point t,x,y , as C (t,x,y), an event, expressed by sentence slit a is open , as A, and an event, expressed by sentence slit b is open , as B. Let t0 be an time instant of an electron emission from source s. Since s is a dotlike source then a state vector ϕC in instant t0 has the following form: ϕC (t,x,y)|t=t0 = ϕC (t0,x,y)δ(x−x0)δ(y−y0). (2.28) Let tw be an time instant such that if event C (t,x,y) occurs in that instant then C (t,x,y) occurs on wall w. Let td be an time instant of a electron detecting by screen d. 1. Let slit a be opened in wall w (Figure 1). In that case the C (t,x,y) probabilities propagator KCA (t −t0,x−xs,y−ys) in instant tw should be of the following shape: KCA (t −t0,x−xs,y−ys)|t=tw = KCA (tw −t0,x−xs,y−ys)δ(x−xa)δ(y−ya). According the propagator property: K (t −t0,x−xs,y−ys) = = R dx1 R dy1 ·K (t −t1,x−x1,y−y1)K (t1 −t0,x1 −xs,y1 −ys). Hence: 86
- 101. KCA (td −t0,xd −xs,yd −ys) = = R dx R dy·KCA (td −tw,xd −x,yd −y) KCA (tw −t0,x−xs,y−ys)δ(x−xa)δ(y−ya). Therefore, according properties of δ-function: KCA (td −t0,xd −xs,yd −ys) = = KCA (td −tw,xd −xa,yd −ya)KCA (tw −t0,xa −xs,ya −ys). The state vector for the event C (t,x,y) in condition A probability has the following form (2.27): ϕCA (td,xd,yd) = dxs dys ·KCA (td −t0,xd −xs,yd −ys)ϕC (t0,xs,ys). Hence, from (2.28): ϕCA (td,xd,yd) = dxs dys ·KCA (td −t0,xd −xs,yd −ys) ϕC (t0,xs,ys)δ(xs −x0)δ(ys −y0). That is: ϕCA (td,xd,yd) = dxs dys ·KCA (td −tw,xd −xa,yd −ya)KCA (tw −t0,xa −xs,ya −ys) ϕC (t0,xs,ys)δ(xs −x0)δ(ys −y0). Hence, according properties of δ-function: ϕCA (td,xd,yd) = KCA (td −tw,xd −xa,yd −ya)KCA (tw −t0,xa −x0,ya −y0)ϕC (t0,x0,y0). In accordance with (2.19): ρCA (td,xd,yd) = ϕ† CA (td,xd,yd)ϕCA (td,xd,yd). Therefore, a probability to detect the electron in vicinity ∆x∆y of point xd,yd in instant t in condition A equals to the following: Pa (td,xd,yd) := P(C (td,∆x∆y)/A) = ρCA (td,xd,yd)∆x∆y. 87
- 102. 2. Let slit b be opened in wall w (Figure 2). In that case the C (t,x,y) probabilities propagator KCB (t −t0,x−xs,y−ys) in instant tw should be of the following shape: KCB (t −t0,x−xs,y−ys)|t=tw = KCB (tw −t0,x−xs,y−ys)δ(x−xb)δ(y−yb). Hence, according the propagator property:: KCB (td −t0,xd −xs,yd −ys) = = R dx R dy·KCB (td −tw,xd −x,yd −y) KCB (tw −t0,x−xs,y−ys)δ(x−xb)δ(y−yb). Therefore, according properties of δ-function: KCB (td −t0,xd −xs,yd −ys) = = KCB (td −tw,xd −xb,yd −yb)KCB (tw −t0,xb −xs,yb −ys). The state vector for the event C (t,x,y) in condition B probability has the following form (2.27): ϕCB (td,xd,yd) = dxs dys ·KCB (td −t0,xd −xs,yd −ys)ϕC (t0,xs,ys). Hence, from (2.28): ϕCB (td,xd,yd) = dxs dys ·KCB (td −t0,xd −xs,yd −ys) ϕC (t0,xs,ys)δ(xs −x0)δ(ys −y0). That is: ϕCB (td,xd,yd) = dxs dys ·KCB (td −tw,xd −xb,yd −yb)KCB (tw −t0,xb −xs,yb −ys) ϕC (t0,xs,ys)δ(xs −x0)δ(ys −y0). Hence, according properties of δ-function: ϕCB (td,xd,yd) = KCB (td −tw,xd −xb,yd −yb)KCB (tw −t0,xb −x0,yb −y0)ϕC (t0,x0,y0). 88
- 103. In accordance with (2.19): ρCB (td,xd,yd) = ϕ† CB (td,xd,yd)ϕCB (td,xd,yd). Therefore, a probability to detect the electron in vicinity ∆x∆y of point xd,yd in instant t in condition B equals to the following: Pb (td,xd,yd) := P(C (td,∆x∆y)/B) = ρCB (td,xd,yd)∆x∆y. 3. Let both slits and a and b are opened (Figure 4). In that case the C (t,x,y) probabilities propagator KCAB (t −t0,x−xs,y−ys) in instant tw should be of the following shape: KCAB (t −t0,x−xs,y−ys)|t=tw = = KCAB (tw −t0,x−xs,y−ys)(δ(x−xa)δ(y−ya)+δ(x−xb)δ(y−yb)). Hence, according the propagator property:: KCAB (td −t0,xd −xs,yd −ys) = = R dx R dy·KCAB (td −tw,xd −x,yd −y) KCAB (tw −t0,x−xs,y−ys)· ·(δ(x−xa)δ(y−ya)+δ(x−xb)δ(y−yb)). Hence, KCAB (td −t0,xd −xs,yd −ys) = R dx R dy·KCAB (td −tw,xd −x,yd −y) KCAB (tw −t0,x−xs,y−ys)· ·δ(x−xa)δ(y−ya) + R dx R dy·KCAB (td −tw,xd −x,yd −y) KCAB (tw −t0,x−xs,y−ys)· ·δ(x−xb)δ(y−yb). Hence, according properties of δ-function: KCAB (td −t0,xd −xs,yd −ys) = KCAB (td −tw,xd −xa,yd −ya) KCAB (tw −t0,xa −xs,ya −ys) +KCAB (td −tw,xd −xb,yd −yb) KCAB (tw −t0,xb −xs,yb −ys) . The state vector for the event C (t,x,y) in condition A and B probability has the follow- ing form (2.27): ϕCAB (td,xd,yd) = dxs dys ·KCAB (td −t0,xd −xs,yd −ys)ϕC (t0,xs,ys). Hence, from (2.28): ϕCAB (td,xd,yd) = dxs dys ·KCAB (td −t0,xd −xs,yd −ys) ϕC (t0,xs,ys)δ(xs −x0)δ(ys −y0). 89
- 104. That is: ϕCAB (td,xd,yd) = dxs dys· · KCAB (td −tw,xd −xa,yd −ya) KCAB (tw −t0,xa −xs,ya −ys) +KCAB (td −tw,xd −xb,yd −yb) KCAB (tw −t0,xb −xs,yb −ys) ϕC (t0,xs,ys)δ(xs −x0)δ(ys −y0). Hence, according properties of δ-function: ϕCAB (td,xd,yd) = = KCAB (td −tw,xd −xa,yd −ya) KCAB (tw −t0,xa −x0,ya −y0) +KCAB (td −tw,xd −xb,yd −yb) KCAB (tw −t0,xb −x0,yb −y0) ϕC (t0,x0,y0). That is: ϕCAB (td,xd,yd) = = KCAB (td −tw,xd −xa,yd −ya) KCAB (tw −t0,xa −x0,ya −y0)ϕC (t0,x0,y0) +KCAB (td −tw,xd −xb,yd −yb) KCAB (tw −t0,xb −x0,yb −y0)ϕC (t0,x0,y0). Therefore, ϕCAB (td,xd,yd) = ϕCA (td,xd,yd)+ϕCB (td,xd,yd). And in accordance with (2.19): ρCAB (td,xd,yd) = ϕ† CAB (td,xd,yd)ϕCAB (td,xd,yd), i.e. ρCAB = (ϕCA +ϕCB)† (ϕCA +ϕCB) Since state vectors ϕCA and ϕCB are not numbers with the same number signs then in the general case: (ϕCA +ϕCB)† (ϕCA +ϕCB) = ϕ† CAϕCA +ϕ† CBϕCB. Therefore, since a probability to detect the electron in vicinity ∆x∆y of point xd,yd in instant t in condition AB equals: Pab (td,xd,yd) := P(C (td,∆x∆y)/AB) = ρCAB (td,xd,yd)∆x∆y then Pab (td,xd,yd) = Pa (td,xd,yd)+Pb (td,xd,yd). Hence, we have the fig.23 picture instead of the Figure 3 picture. 4. Let us consider devices of Figure 6. 90
- 105. Denote event, expressed by sentence ”detector da snaps into action”, as Da and event, expressed by sentence ”detector db snaps into action”, as Db. Since event C (t,x,y) is a dotlike event then events Da and Db are exclusive events. According the property 10 of operations on events: (Da +Db)+(Da +Db) = T , according the property 6 of operations on events: (Da +Db) = DaDb, Hence: Da +Db +DaDb = T . According the property 5 of operations on events: C = CT = C Da +Db +DaDb . According the property 3 of operations on events: C = CDa +CDb +CDaDb. Therefore, according the probabilities addition formula for exclusive events: P(C (td)) = P(C (td)Da)+P(C (td)Db)+P C (td)DaDb . But P(C (td)Da) = Pa (td), P(C (td)Db) = Pb (td), P C (td)DaDb = Pab (td), and we receive the Figure 6 picture. Thus, here are no paradoxes for the event-probability interpretation of these experi- ments. We should depart from notion of a continuously existing electron and consider an elementary particle an ensemble of events connected by probability. Its like the fact that physical particle exists only at the instant when it is involved in some event. A particle doesnt exist in any other time, but theres a probability that something will happen to it. Thus, if nothing happens with the particle between the event of creating it and the event of detecting it the behavior of the particle is the behavior of probability between the point of creating and the point of detecting it with the presence of interference. But what is with Wilson cloud chamber where the particle has a clear trajectory and no interference? In that case these trajectories are not totally continuous lines. Every point of ionization has neighboring point of ionization, and there are no events between these points. 91
- 106. Consequently, physical particle is moving because corresponding probability propa- gates in the space between points of ionization. Consequently, particle is an ensemble of events, connected by probability. And charges, masses, moments, etc. represent statistical parameters of these probability waves, propagated in the space-time. It explains all para- doxes of quantum physics. Schrodingers cat lives easy without any superposition of states until the micro event awaited by all occures. And the wave function disappears without any collapse in the moment when an event probability disappears after the event occurs. Hence, entanglement concerns not particles but probabilities. That is when event of the measuring of spin of Alices electron occurs then probability for these entangled electrons is changed instantly on whole space. Therefore, nonlocality acts for probabilities, not for particles. But probabilities can not transmit any information 2.5. Lepton Hamiltonian Let ϑs,k and ϖs,k be terms of Q (2.25) and let Θ0, Θ3, ϒ0 and ϒ3 be a solution of the following equations set: −Θ0 +Θ3 −ϒ0 +ϒ3= ϑ1,1; −Θ0 −Θ3 −ϒ0 −ϒ3= ϑ2,2; −Θ0 −Θ3 +ϒ0 +ϒ3= ϑ3,3; −Θ0 +Θ3 +ϒ0 −ϒ3= ϑ4,4 , and Θ1, ϒ1, Θ2, ϒ2, M0, M4, Mζ,0, Mζ,4, Mη,0, Mη,4, Mθ,0, Mθ,4 be solutions of the following sets of equations: Θ1 +ϒ1= ϑ1,2; −Θ1 +ϒ1= ϑ3,4; −Θ2 −ϒ2= ϖ1,2; Θ2 −ϒ2= ϖ3,4; M0 +Mθ,0= ϑ1,3; M0 −Mθ,0= ϑ2,4; M4 +Mθ,4= ϖ1,3; M4 −Mθ,4= ϖ2,4; Mζ,0 −Mη,4= ϑ1,4; Mζ,0 +Mη,4= ϑ2,3; Mζ,4 −Mη,0= ϖ1,4; Mζ,4 +Mη,0= ϖ2,3 . Thus the columns of Q are the following: the first and the second columns: −iΘ0 +iΘ3 −iϒ0 +iϒ3 iΘ1 +iϒ1 +Θ2 +ϒ2 iΘ1 +iϒ1 −Θ2 −ϒ2 −iΘ0 −iΘ3 −iϒ0 −iϒ3 iM0 +iMθ,0 +M4 +Mθ,4 iMζ,0 +iMη,4 +Mζ,4 +Mη,0 iMζ,0 −iMη,4 +Mζ,4 −Mη,0 iM0 −iMθ,0 +M4 −Mθ,4 , 92
- 107. the third and the fourth columns: iM0 +iMθ,0 −M4 −Mθ,4 iMζ,0 −iMη,4 −Mζ,4 +Mη,0 iMζ,0 +iMη,4 −Mζ,4 −Mη,0 iM0 −iMθ,0 −M4 +Mθ,4 −iΘ0 −iΘ3 +iϒ0 +iϒ3 −iΘ1 +iϒ1 −Θ2 +ϒ2 −iΘ1 +iϒ1 +Θ2 −ϒ2 −iΘ0 +iΘ3 +iϒ0 −iϒ3 . Hence, Q = = iΘ0β[0] +iϒ0β[0]γ[5]+ +iΘ1β[1] +iϒ1β[1]γ[5]+ +iΘ2β[2] +iϒ2β[2]γ[5]+ +iΘ3β[3] +iϒ3β[3]γ[5]+ +iM0γ[0] +iM4β[4]− −iMζ,0γ [0] ζ +iMζ,4ζ[4]− −iMη,0γ [0] η −iMη,4η[4]+ +iMθ,0γ [0] θ +iMθ,4θ[4]. Therefore, from (2.24): 1 c ∂tϕ− iΘ0β[0] +iϒ0β[0] γ[5] ϕ = 3 ∑ ν=1 β[ν] ∂ν +iΘν +iϒνγ[5] + +iM0γ[0] +iM4β[4]− −iMζ,0γ [0] ζ +iMζ,4ζ[4]− −iMη,0γ [0] η −iMη,4η[4]+ +iMθ,0γ [0] θ +iMθ,4θ[4] ϕ. (2.29) with γ[5] := 12 02 02 −12 . (2.30) Because ζ[k] +η[k] +θ[k] = −β[k] 93
- 108. with k ∈ {1,2,3} then from (2.29): − ∂0 +iΘ0 +iϒ0γ[5] + 3 ∑ k=1 β[k] ∂k +iΘk +iϒkγ[5] +2 iM0γ[0] +iM4β[4] ϕ+ + − ∂0 +iΘ0 +iϒ0γ[5] − 3 ∑ k=1 ζ[k] ∂k +iΘk +iϒkγ[5] +2 −iMζ,0γ [0] ζ +iMζ,4ζ[4] ϕ+ + ∂0 +iΘ0 +iϒ0γ[5] − 3 ∑ k=1 η[k] ∂k +iΘk +iϒkγ[5] +2 −iMη,0γ [0] η −iMη,4η[4] ϕ+ + − ∂0 +iΘ0 +iϒ0γ[5] − 3 ∑ k=1 θ[k] ∂k +iΘk +iϒkγ[5] +2 iMθ,0γ [0] θ +iMθ,4θ[4] ϕ = 0. In (2.29) summands −iMζ,0γ [0] ζ +iMζ,4ζ[4]− −iMη,0γ [0] η −iMη,4η[4]+ +iMθ,0γ [0] θ +iMθ,4θ[4] contain elements of chromatic pentads and 3 ∑ k=1 β[k] ∂k +iΘk +iϒkγ[5] +iM0γ[0] +iM4β[4] contains only elements of the light pentads. The following sum Hl := c 3 ∑ k=1 β[k] i∂k −Θk −ϒkγ[5] −cM0γ[0] −cM4β[4] (2.31) is called lepton Hamiltonian. And the following equation: 3 ∑ k=0 β[k] i∂k −Θk −ϒkγ[5] −M0γ[0] −M4β[4] ϕ = 0 (2.32) is called lepton moving equation If like to (2.19): ϕ†γ[0]ϕ := − j5 c and ϕ†β[4]ϕ := − j4 c and: ρu4 := j4 and ρu5 := j5 (2.33) 94
- 109. then from (2.18): − u5 c = sin2α sinβsinχcos(−θ+υ) +cosβcosχcos(γ−λ) , − u4 c = sin2α −sinβsinχsin(−θ+υ) +cosβcosχsin(γ−λ) . Hence, from (2.16): u2 1 +u2 2 +u2 3 +u2 A,4 +u2 5 = c2 . Thus, of only all five elements of a Clifford pentad lends an entire kit of velocity com- ponents and, for completeness, yet two ”space” coordinates x5 and x4 should be added to our three x1,x2,x3. These additional coordinates can be selected such that − πc h ≤ x5 ≤ πc h ,− πc h ≤ x4 ≤ πc h . Coordinates x4 and x5 are not of any events coordinates. Hence, our devices do not detect of its as space coordinates. Let us denote: ϕ(t,x1,x2,x3,x5,x4) := ϕ(t,x1,x2,x3)· ·(exp(i(x5M0 (t,x1,x2,x3)+x4M4 (t,x1,x2,x3)))). In this case equation of moving with lepton Hamiltonian (2.31) shape is the following: 3 ∑ k=0 β[0] i∂k −Θk −ϒkγ[5] −γ[0] i∂5 −β[4] i∂4 ϕ = 0 (2.34) Let g1 be the positive real number and for µ ∈ {0,1,2,3}: Fµ and Bµ be the solutions of the following system of the equations: −0.5g1Bµ +Fµ= −Θµ −ϒµ; −g1Bµ +Fµ= −Θµ +ϒµ. Let charge matrix be denoted as the following: Y := − 12 02 02 2·12 . (2.35) Thus20: 20If products ABj,s exist for all j, s then A B0,0 B0,1 ··· B0,n B1,0 B1,1 ··· B1,n ··· ··· ··· ··· Bm,0 Bm,1 ··· Bm,n := AB0,0 AB0,1 ··· AB0,n AB1,0 AB1,1 ··· AB1,n ··· ··· ··· ··· ABm,0 ABm,1 ··· ABm,n and if products Bj,sA exist for all j, s then 95
- 110. −Θµ −ϒµγ[5] = = −Θµ14 −ϒµγ[5] = = −Θµ 12 02 02 12 −ϒµ 12 02 02 −12 = = − Θµ12 02 02 Θµ12 + ϒµ12 02 02 −ϒµ12 = = (−Θµ −ϒµ)12 02 02 (−Θµ +ϒµ)12 = = (−0.5g1Bµ +Fµ)12 02 02 (−g1Bµ +Fµ)12 . And B0,0 B0,1 ··· B0,n B1,0 B1,1 ··· B1,n ··· ··· ··· ··· Bm,0 Bm,1 ··· Bm,n A := B0,0A B0,1A ··· B0,nA B1,0A B1,1A ··· B1,nA ··· ··· ··· ··· Bm,0A Bm,1A ··· Bm,nA . (2.36) If A and all Bj,s are k ×k matrices then A+ B0,0 B0,1 B0,2 ··· B0,n B1,0 B1,1 B1,2 ··· B1,n B2,0 B2,1 B2,2 ··· B2,n ··· ... ··· ··· ··· Bn,0 Bn,1 Bn,2 ··· Bn,n := := A1nk + B0,0 B0,1 B0,2 ··· B0,n B1,0 B1,1 B1,2 ··· B1,n B2,0 B2,1 B2,2 ··· B2,n ··· ... ··· ··· ··· Bn,0 Bn,1 Bn,2 ··· Bn,n = = B0,0 +A B0,1 B0,2 ··· B0,n B1,0 B1,1 +A B1,2 ··· B1,n B2,0 B2,1 B2,2 +A ··· B2,n ··· ... ··· ··· ··· Bn,0 Bn,1 Bn,2 ··· Bn,n +A . (2.37) 96
- 111. Fµ +0.5g1YBµ = = Fµ14 +0.5g1YBµ = Fµ 12 02 02 12 +0.5g1 − 12 02 02 2·12 Bµ = = Fµ12 02 02 Fµ12 − 0.5g1Bµ12 02 02 0.5g1Bµ2·12 = = Fµ12 −0.5g1Bµ12 02 02 Fµ12 −g1Bµ ·12 . Hence, −Θµ −ϒµγ[5] = Fµ +0.5g1YBµ and from (2.34): 3 ∑ k=0 β[k] (i∂k +Fk +0.5g1YBk)−γ[0] i∂5 −β[4] i∂4 ϕ = 0 (2.38) Let χ(t,x1,x2,x3) be the real function and: U (χ) := exp iχ 2 12 02 02 exp(iχ)12 . (2.39) In that case for µ ∈ {0,1,2,3}: ∂µU = ∂µ exp iχ 2 12 02 02 exp(iχ)12 = ∂µ exp iχ 2 12 ∂µ02 ∂µ02 ∂µ exp(iχ)12 = i ∂µχ 2 exp iχ 2 12 02 02 i∂µχexp(iχ)12 = i ∂µχ 2 exp iχ 2 12 02 02 2exp(iχ)12 , and YU = − 12 02 02 2·12 exp iχ 2 12 02 02 exp(iχ)12 = − exp iχ 2 12 02 02 2exp(iχ)12 . Hence: 97
- 112. ∂µU = −i ∂µχ 2 YU. (2.40) Moreover you can calculate that: U† γ[0] U = γ[0] cos χ 2 +β[4] sin χ 2 , U† β[4] U = β[4] cos χ 2 −γ[0] sin χ 2 , U† U = 14, U† YU = Y, β[k] U = Uβ[k] for k ∈ {0,1,2,3} Let x4 = x4 cos χ 2 −x5 sin χ 2 , x5 = x5 cos χ 2 +x4 sin χ 2 . In that case by the partial derivate definition for any function u: ∂4u = ∂4u·∂4x4 +∂5u·∂4x5 = ∂4u·cos χ 2 +∂5u·sin χ 2 , (2.41) ∂5u = ∂4u·∂5x4 +∂5u·∂5x5 = ∂4u· −sin χ 2 +∂5u·cos χ 2 . Let ∂4χ = 0 and ∂5χ = 0; hence, ∂4U = U∂4 and ∂5U = U∂5. From (2.38): 3 ∑ s=0 β[s] (i∂s +Fs +0.5g1YBs)−γ[0] i∂5 −β[4] i∂4 ϕ = 0. (2.42) Let Bµ = Bµ − 1 g1 ∂µχ. According to (2.41) and since U†U = 14 and U†YU = Y then ∑3 s=0 β[s] i∂s +Fs +0.5g1U†YU Bs + 1 g1 ∂sχ − −γ[0]i −sin χ 2 ∂4 +cos χ 2 ∂5 −β[4]i cos χ 2 ∂4 +sin χ 2 ∂5 U† Uϕ = 0. Hence: ∑3 s=0 β[s] i∂s +Fs +0.5g1U†YU Bs + 1 g1 ∂sχ − − −γ[0] sin χ 2 +β[4] cos χ 2 i∂4 − γ[0] cos χ 2 +β[4] sin χ 2 i∂5 U† Uϕ = 0. 98
- 113. Since U is a linear operator then ∑3 s=0 β[s] i∂s +Fs +0.5g1U†YU Bs + 1 g1 ∂sχ U†− − −γ[0] sin χ 2 +β[4] cos χ 2 i∂4U† − γ[0] cos χ 2 +β[4] sin χ 2 i∂5U† Uϕ = 0 and since ∂4U = U∂4 and ∂5U = U∂5 then ∑3 s=0 β[s] i∂sU† +FsU† +0.5g1U†YUU† Bs + 1 g1 ∂sχ − − −γ[0]U† sin χ 2 +β[4]U† cos χ 2 i∂4 − γ[0]U† cos χ 2 +β[4]U† sin χ 2 i∂5 Uϕ = 0. (2.43) Since U† γ[0] U = γ[0] cos χ 2 +β[4] sin χ 2 , U† β[4] U = β[4] cos χ 2 −γ[0] sin χ 2 then U† γ[0] UU† = γ[0] U† cos χ 2 +β[4] U† sin χ 2 , U† β[4] UU† = β[4] U† cos χ 2 −γ[0] U† sin χ 2 , Hence, U† γ[0] = γ[0] U† cos χ 2 +β[4] U† sin χ 2 , U† β[4] = β[4] U† cos χ 2 −γ[0] U† sin χ 2 . Therefore, γ[0]U† = U†γ[0] cos χ 2 −U†β[4] sin χ 2 , β[4]U† = U†γ[0] sin χ 2 +U†β[4] cos χ 2 . Thus, from (2.43): ∑3 s=0 β[s] i∂sU† +FsU† +0.5g1U†YUU† Bs + 1 g1 ∂sχ − − − U†γ[0] cos χ 2 −U†β[4] sin χ 2 sin χ 2 + U†γ[0] sin χ 2 +U†β[4] cos χ 2 cos χ 2 i∂4 − U†γ[0] cos χ 2 −U†β[4] sin χ 2 cos χ 2 + U†γ[0] sin χ 2 +U†β[4] cos χ 2 sin χ 2 i∂5 Uϕ = 0. 99
- 114. Hence: ∑3 s=0 β[s] i∂sU† +FsU† +0.5g1U†Y Bs + 1 g1 ∂sχ − −U†β[4]i∂4 − U†γ[0]i∂5 Uϕ = 0. (2.44) Since (2.40): ∂µU = −i ∂µχ 2 YU then for s ∈ {0,1,2,3}: ∂sU† = i ∂sχ 2 U† Y† = i ∂sχ 2 YU† . Therefore, ∂s U†Uϕ = ∂s U† Uϕ = = ∂sU† Uϕ +U†∂s Uϕ = i∂sχ 2 YU† Uϕ +U†∂s Uϕ = = i∂sχ 2 YU† +U†∂s Uϕ . Since YU† = U†Y then i ∂sχ 2 YU† +U† ∂s = U† i ∂sχ 2 Y +U† ∂s. Hence, i∂sU† = −U† ∂sχ 2 Y +U† i∂s. Therefore, from (2.44): ∑3 s=0 β[s] −U† ∂sχ 2 Y +U†i∂s +FsU† +0.5g1U†Y Bs + 1 g1 ∂sχ − −U†β[4]i∂4 − U†γ[0]i∂5 Uϕ = 0. Hence: ∑3 s=0 β[s] U†i∂s +U†Fs +0.5g1U†YBs − −U†β[4]i∂4 − U†γ[0]i∂5 Uϕ = 0 with Fs := UFsU†. Since β[s]U = Uβ[s] for s ∈ {0,1,2,3} then ∑3 s=0U†β[s] i∂s +U†Fs +0.5g1U†YBs − −U†β[4]i∂4 − U†γ[0]i∂5 Uϕ = 0. Hence, if denote: ϕ := Uϕ then since U is a linear operator then: 100
- 115. U† 3 ∑ s=0 β[s] i∂s +Fs +0.5g1YBs −β[4] i∂4 − γ[0] i∂5 ϕ = 0. That is 3 ∑ s=0 β[s] i∂s +Fs +0.5g1YBs −β[4] i∂4 − γ[0] i∂5 ϕ = 0. Compare with (2.42). Thus, this Equation of moving is invariant under the following transformations: x4 → x4 = x4 cos χ 2 −x5 sin χ 2 ; x5 → x5 = x5 cos χ 2 +x4 sin χ 2 ; xµ → xµ = xµ for µ ∈ {0,1,2,3}; (2.45) ϕ → ϕ = Uϕ, Bµ → Bµ = Bµ − 1 g1 ∂µχ, Fµ → Fµ = UFsU† . Therefore, Bµ is like to the B-boson field of Standard Model21 [44]. field. 2.6. Masses Let ε1 = 1 0 0 0 , ε2 = 0 1 0 0 , ε3 = 0 0 1 0 , ε4 = 0 0 0 1 . (2.46) Functions of type : h 2πc exp −i h c (sx4 +nx5) εk (2.47) with an integer n and s form orthonormal basis of some unitary space ℑ with scalar product of the following shape: (ϕ,χ) := πc h −πc h dx5 πc h −πc h dx4 ·ϕ† χ (2.48) In that case from (2.19): 21Sheldon Lee Glashow (born December 5, 1932) is a American theoretical physicist. 101
- 116. (ϕ,ϕ) = ρA, (2.49) ϕ,β[s] ϕ = − jk c . for s ∈ {1,2,3} Let22 Nϑ (t,x1,x2,x3) := trunc cM0 h , Nϖ (t,x1,x2,x3) := trunc cM4 h . Hence, functions Nϑ (t,x1,x2,x3) and Nϖ (t,x1,x2,x3) have got integer values. In that case to high precision: ϕ = ϕ(t,x1,x2,x3)·exp −i x5 h c Nϑ (t,x1,x2,x3)+x4 h c Nϖ (t,x1,x2,x3) and Fourier series for ϕ is of the following form: ϕ(t,x1,x2,x3,x5,x4) = ϕ(t,x1,x2,x3)·∑ n,s δ−n,Nϑ(t,x)δ−s,Nϖ(t,x) exp −i h c (nx5 +sx4) with δ−n,Nϑ = h 2πc πc h −πc h exp i h c (nx5) exp iNϑ h c x5 dx5 = sinπ(n+Nϑ) π(n+Nϑ) , δ−s,Nϖ = h 2πc πc h −πc h exp i h c (sx4) exp iNϖ h c x4 dx4 = sinπ(s+Nϖ) π(s+Nϖ) with integer n and s. If denote: f (t,x,−n,−s) := ϕ(t,x)δn,Nϑ(t,x)δs,Nϖ(t,x) then ϕ(t,x,x5,x4) = = ∑n,s f (t,x,n,s)exp −ih c (nx5 +sx4) . (2.50) The integer numbers n and s are denoted mass numbers. From properties of δ: in every point t,x : either ϕ(t,x,x5,x4) = 0 22Function trunc(x) returns the integer part of a real number x by removing the fractional part. For example: trunc(−2.0857) = −2. 102
- 117. or integer numbers n0 and s0 exist for which: ϕ(t,x,x5,x4) = = f (t,x,n0,s0)exp −ih c (n0x5 +s0x4) . (2.51) Here if m0 := n2 0 +s2 0 and m := h2 c2 m0 (2.52) then m is denoted mass of ϕ. That is for every space-time point: either this point is empty or single mass is placed in this point. Figure 7: Equation of moving (2.38) under the transformation (2.45) has the following form: ∑n ,s β[0] i1 c ∂t +Fµ +0.5g1YBµγ[5] +∑3 µ=1 β[µ] i∂µ +Fµ +0.5g1YBµγ[5] +γ[0]i∂5 +β[4]i∂4 · ·exp −ih c (n x5 +s x4) U f = 0 with: n = ncos χ 2 −ssin χ 2 , s = nsin χ 2 +scos χ 2 . But s and n are integer numbers and s and n must be integer numbers, too. . 103
- 118. Figure 8: A triplet λ;n,s of integer numbers23 is called a Pythagorean triple [43] if λ2 = n2 +s2 . Let ε be the tiny positive real number. I call an integer number λ as a father number with precise ε if for each real number χ and for every Pythagorean triple λ;n,s here exists a Pythagorean triple λ;n ,s such that: −ssin χ 2 +ncos χ 2 −n < λε, scos χ 2 +nsin χ 2 −s < λε. For example: number 325 is a father number for the following Pythagorean triples (Figure 7): 23For each natural number n, there exist at least n different Pythagorean triples with the same hypotenuse. Sierpinski, 2003, . 31. Dover, 2003.ISBN 978-0-486-43278-6. John F. Goehi Jr. TRIPLES, QUARTETS, PENTADS :Mathematics teacher,ISSN0025-5769,Vol. 98, N? 9, 2005,pag.580 104
- 119. 325;323,36 , 325;315,80 , 325;312,91 , 325;300,125 , 325;280,165 , 325;260,195 , 325;253,204 , Here ε is maximal ratio value of difference between adjacent s values to father number. That is here ε = 253−204 325 = 0.15. But for any value of precise ε here exists a fitting father number in long distant domain of the natural numerical line. But I can not calculate it since more high-end machine than my computer is needed for such calculation. The nearest-neighbors to 325 father numbers are numbers 333 and 337. But these father numbers have got one at a time triple. Hence, fathers, having many ”children”, are isolated numbers on the natural numerical line. I suspect that these numbers are fathers of particles families. Here are three families (generations) according to the Standard Model of particle physics [44]: νe e− νµ µ− ντ τ− u d c s t b . Each generation is divided into two leptons: νe e− , νµ µ− , ντ τ− , and two quarks: u d , c s , t b . The two leptons may be divided into one electron-like (e− - electron, µ− - µ–lepton, τ− - τ-lepton ) and neutrino (νe, νµ, ντ); the two quarks may be divided into one down-type (d, s, b) and one up-type (u, c, t). The first generation consists of the electron, electron neutrino and the down and up quarks. The second generation consists of the muon, muon neutrino and the strange and charm quarks. The third generation consists of the tau lepton, tau neutrino and the bottom and top quarks. Each member of a higher generation has greater mass than the corresponding particle of the previous generation. For example: the first- generation electron has a mass of only 0.511 MeV, the second-generation muon has a mass of 106 MeV, and the third-generation tau lepton has a mass of 1777 MeV (almost twice as heavy as a proton). All ordinary atoms are made of particles from the first generation. Electrons surround a nucleus made of protons and neutrons, which contain up and down 105
- 120. quarks. The second and third generations of charged particles do not occur in normal matter and are only seen in extremely high-energy environments. Neutrinos of all generations stream throughout the universe but rarely interact with normal matter. 2.7. One-Mass State Let form of (2.50) be the following: ϕ(t,x,x5,x4) = exp −i h c nx5 4 ∑ k=1 fk (t,x,n,0)εk. In that case the Hamiltonian has the following form (from (2.38)): H = c 3 ∑ k=1 β[k] i∂k + h c nγ[0] +G with G := 3 ∑ µ=0 β[µ] (Fµ +0.5g1YBµ). Let ω(k) := k2 +n2 = k2 1 +k2 2 +k2 3 +n2 and e1 (k) := 1 2 ω(k)(ω(k)+n) ω(k)+n+k3 k1 +ik2 ω(k)+n−k3 −k1 −ik2 . (2.53) Let H0 := c 3 ∑ s=1 β[s] i∂s +hnγ[0] . (2.54) Since (2.52): hn = m c2 h then equation of moving with Hamiltonian H0 has the following form: 1 c i∂tϕ = ∑3 s=1 β[s]i∂s +mc h γ[0] ϕ. (2.55) This is the Dirac equation (Paul Dirac24 formulated it in 1928). Let us denote 24Paul Adrien Maurice Dirac (1902 – 1984) was an English theoretical physicist who made fundamental contributions to the early development of both quantum mechanics and quantum electrodynamics. 106
- 121. γ[s] := γ[0] β[s] for s = 0. Let us calculate: γ[s] γ[j] +γ[j] γ[s] = γ[0] β[s] γ[0] β[j] +γ[0] β[j] γ[0] β[s] = = −γ[0] γ[0] β[s] β[j] −γ[0] γ[0] β[j] β[s] = = − β[s] β[j] +β[j] β[s] = −2δj,s for s = 0 and j = 0. and γ[s] γ[0] +γ[0] γ[s] = γ[0] β[s] γ[0] +γ[0] γ[0] β[s] = −β[s] +β[s] = 0 for s = 0. From (2.55): 1 c iγ[0] ∂t − 3 ∑ s=1 γ[s] i∂s −m c h ϕ = 0. Let us multiply both parts of this equation on 1 c iγ[0] ∂t − 3 ∑ s =1 γ[s ] i∂s +m c h : 1 c iγ[0] ∂t − 3 ∑ s =1 γ[s ] i∂s +m c h 1 c iγ[0] ∂t − 3 ∑ s=1 γ[s] i∂s −m c h ϕ = 0. Hence, − 1 c2 ∂2 t −∑3 s=1 1 c iγ[0]∂tγ[s]i∂s −∑3 s =1 γ[s ]i∂s 1 c iγ[0]∂t −1 c iγ[0]∂tmc h +mc h 1 c iγ[0]∂t +∑3 s =1 γ[s ]i∂s ∑3 s=1 γ[s]i∂s +∑3 s =1 γ[s ]i∂s mc h −∑3 s=1 mc h γ[s]i∂s −m2c2 h2 ϕ = 0. Hence, − 1 c2 ∂2 t +∑3 s =1 γ[s ]i∂s ∑3 s=1 γ[s]i∂s −m2c2 h2 ϕ = 0 Since 107
- 122. 3 ∑ s =1 γ[s ] i∂s 3 ∑ s=1 γ[s] i∂s = − 3 ∑ s=1 3 ∑ s =1 γ[s ] γ[s] ∂s ∂s = = − γ[1]γ[1]∂1∂1 +γ[2]γ[1]∂2∂1 +γ[3]γ[1]∂3∂1 +γ[1]γ[2]∂1∂2 +γ[2]γ[2]∂2∂2 +γ[3]γ[2]∂3∂2 +γ[1]γ[3]∂1∂3 +γ[2]γ[3]∂2∂3 +γ[3]γ[3]∂3∂3 = = − −∂1∂1 +γ[2]γ[1]∂2∂1 +γ[1]γ[2]∂1∂2 +γ[3]γ[1]∂3∂1 +γ[1]γ[3]∂1∂3 −∂2∂2 +γ[3]γ[2]∂3∂2 +γ[2]γ[3]∂2∂3 −∂3∂3 . Hence, 3 ∑ s =1 γ[s ] i∂s 3 ∑ s=1 γ[s] i∂s = ∂1∂1 +∂2∂2 +∂3∂3 = 3 ∑ s=1 ∂2 s . Thus, − 1 c2 ∂2 t +∑3 s=1 ∂2 s − m2c2 h2 ϕ = 0. (2.56) This is the Klein-Gordon2526 equation for a free particle with mass m. Let us calculate: H0e1 (k) h 2πc 3 2 exp −ih c kx = = c∑3 s=1 β[s]i∂s +hnγ[0] h 2πc 3 2 e1 (k)exp −ih c kx = == c∑3 s=1 β[s]i∂se1 (k) h 2πc 3 2 exp −ih c kx + +hnγ[0]e1 (k) h 2πc 3 2 exp −ih c kx = = c∑3 s=1 β[s]ie1 (k)∂s h 2πc 3 2 exp −ih c kx + +hn h 2πc 3 2 exp −ih c kx γ[0]e1 (k) = = c∑3 s=1 β[s]ie1 (k) −ih c ks h 2πc 3 2 exp −ih c ks + +hn h 2πc 3 2 exp −ih c kx γ[0]e1 (k) = = ∑3 s=1 (−ihks)β[s]ie1 (k) h 2πc 3 2 exp −ih c kx + +hn h 2πc 3 2 exp −ih c kx γ[0]e1 (k) = = h h 2πc 3 2 exp −ih c kx ∑3 s=1 ksβ[s] +nγ[0] e1 (k) = 25Oskar Klein, 1894-1977 26Walter Gordon, 1893-1939 108
- 123. = h h 2πc 3 2 exp −i h c kx k3 k1 −ik2 n 0 k1 +ik2 −k3 0 n n 0 −k3 −k1 +ik2 0 n −k1 −ik2 k3 · · 1 2 ω(k)(ω(k)+n) ω(k)+n+k3 k1 +ik2 ω(k)+n−k3 −k1 −ik2 = h h 2πc 3 2 exp −i h c kx k3ω(k)+k2 3 +k2 1 +k2 2 +nω(k)+n2 k1ω(k)+ik2ω(k) nω(k)+n2 −k3ω(k)+k2 3 +k2 1 +k2 2 −k1ω(k)−ik2ω(k) = ω(k) 1 2 ω(k)(ω(k)+n) k3 +n+ω(k) k1 +ik2 n+ω(k)−k3 −k1 −ik2 h h 2πc 3 2 exp −i h c kx . Therefore, H0e1 (k) h 2πc 3 2 exp −i h c kx = hω(k)e1 (k) h 2πc 3 2 exp −i h c kx . (2.57) Hence, function e1 (k) h 2πc 3 2 exp −ih c kx is an eigenvector of H0 with eigenvalue hω(k) = h k2 +n2. Similarly, function e2 (k) h 2πc 3 2 exp −ih c kx with e2 (k) := 1 2 ω(k)(ω(k)+n) k1 −ik2 ω(k)+n−k3 −k1 +ik2 ω(k)+n+k3 (2.58) is eigenvector of H0 with eigenvalue hω(k) = h √ k2 +n2, too, and functions e3 (k) h 2πc 3 2 exp −i h c kx and e4 (k) h 2πc 3 2 exp −i h c kx with e3 (k) := 1 2 ω(k)(ω(k)+n) −ω(k)−n+k3 k1 +ik2 ω(k)+n+k3 k1 +ik2 (2.59) 109
- 124. and e4 (k) := 1 2 ω(k)(ω(k)+n) k1 −ik2 −ω(k)−n−k3 k1 −ik2 ω(k)+n−k3 (2.60) are eigenvectors of H0 with eigenvalue −hω(k). Here eµ (k) with µ ∈ {1,2,3,4} form an orthonormal basis in the space spanned on vectors εµ (2.46). 2.8. Creating and Annihilation Operators Let H be some unitary space. Let 0 be the zero element of H. That is any element F of H obeys to the following conditions: 0F = 0, 0+F = F, 0† = 0. Let 0 be the zero operator on H. That is any element F of H obeys to the following condition: 0F = 0F, and if b is any operator on H then 0+b = b+0 = b, 0b = b0 = 0. Let 1 be the identy operator on H. That is any element F of H obeys to the following condition: 1F = 1F = F , and if b is any operator on H then 1b = b1 = b. Let linear operators bs,k (s ∈ {1,2,3,4}) act on all elements of this space. And let these operators fulfill the following conditions: b† s,k,bs ,k := b† s,kbs ,k +bs ,k b† s,k = h 2π 3 δk,k δs,s 1, {bs,k,bs ,k } = bs,kbs ,k +bs ,k bs,k = b† s,k,b† s ,k = 0. Hence, bs,kbs,k = b† s,kb† s,k = 0. There exists element F0 of H such that F† 0 F0 = 1 and for any bs,k: bs,kF0 = 0. Hence, F† 0 b† s,k = 0. Let ψs (x) := ∑ k 4 ∑ r=1 br,ker,s (k)exp −i h c kx . 110
- 125. Because 4 ∑ r=1 er,s (k)er,s (k) = δs,s and ∑ k exp −i h c k x−x = 2πc h 3 δ x−x then ψ† s (x),ψs x := ψ† s (x)ψs x +ψs x ψ† s (x) = δ x−x δs,s 1. And these operators obey the following conditions: ψs (x)F0 = 0, {ψs (x),ψs (x )} = ψ† s (x),ψ† s (x ) = 0. Hence, ψs (x)ψs (x ) = ψ† s (x)ψ† s (x ) = 0. Let Ψ(t,x) := 4 ∑ s=1 ϕs (t,x)ψ† s (x)F0. (2.61) These function obey the following condition: Ψ† t,x Ψ(t,x) = ϕ† t,x ϕ(t,x)δ x−x . Hence, dx ·Ψ† t,x Ψ(t,x) = ρ(t,x). (2.62) Let a Fourier series of ϕs (t,x) has the following form: ϕs (t,x) = ∑ p 4 ∑ r=1 cr (t,p)er,s (p)exp −i h c px . In that case: Ψ(t,p) := 2πc h 3 4 ∑ r=1 cr (t,p)b† r,pF0. If H0 (x) := ψ† (x)H0ψ(x) (2.63) then H0 (x) is called a Hamiltonian H0 density. Because H0ϕ(t,x) = i ∂ ∂t ϕ(t,x) 111
- 126. then dx ·H0 x Ψ(t,x) = i ∂ ∂t Ψ(t,x). (2.64) Therefore, if H := dx ·H0 x then H acts similar to the Hamiltonian on space H. And if EΨ F0 := ∑ p Ψ† (t,p)HΨ(t,p) then EΨ F0 is an energy of Ψ on vacuum F0. Let us consider operator Na (x0) := ψ† a (x0)ψa (x0). Let us calculate an average value of this operator: Na (x0) Ψ := Ω dx·Na (x0)ρ(t,x). In accordance with (2.62): Na (x0) Ψ = Ω dx Ω dx ·Ψ† t,x ψ† a (x0)ψa (x0)Ψ(t,x). Since in accordance with (2.61): Ψ(t,x) = 4 ∑ j=1 ϕj (t,x)ψ† j (x)F0. then Na (x0) Ψ = = Ω dx Ω dx · 4 ∑ s=1 ϕ∗ s t,x F† 0 ψs x ψ† a (x0)ψa (x0) 4 ∑ j=1 ϕj (t,x)ψ† j (x)F0 = = Ω dx Ω dx · 4 ∑ s=1 4 ∑ j=1 ϕ∗ s t,x ϕj (t,x)F† 0 ψs x ψ† a (x0)ψa (x0)ψ† j (x)F0. Since ψ† a (x0)ψs x +ψs x ψ† a (x0) = δ x0−x δs,a1 then 112
- 127. Na (x0) Ψ = Ω dx Ω dx · 4 ∑ s=1 4 ∑ j=1 ϕ∗ s t,x ϕj (t,x)· ·F† 0 δ x0−x δs,a1−ψ† a (x0)ψs x ψa (x0)ψ† j (x)F0 = Ω dx Ω dx · 4 ∑ s=1 4 ∑ j=1 ϕ∗ s t,x ϕj (t,x)· · δ x0−x δs,aF† 0 1−F† 0 ψ† a (x0)ψs x ψa (x0)ψ† j (x)F0. Since F† 0 1 = F† 0 and F† 0 ψ† a (x0) = 0 then Na (x0) Ψ = = Ω dx Ω dx · 4 ∑ s=1 4 ∑ j=1 ϕ∗ s t,x ϕj (t,x)δ x0−x δs,aF† 0 ψa (x0)ψ† j (x)F0. According with properties of δ-function and δ: Na (x0) Ψ = Ω dx· 4 ∑ j=1 ϕ∗ a (t,x0)ϕj (t,x)F† 0 ψa (x0)ψ† j (x)F0. Since ψ† j (x)ψa (x0)+ψa (x0)ψ† j (x) = δ(x0−x)δj,a1 then Na (x0) Ψ = = Ω dx· 4 ∑ j=1 ϕ∗ a (t,x0)ϕj (t,x)F† 0 δ(x0−x)δj,a1−ψ† j (x)ψa (x0) F0 = Ω dx· 4 ∑ j=1 ϕ∗ a (t,x0)ϕj (t,x) δ(x0−x)δj,aF† 0 1F0 −F† 0 ψ† j (x)ψa (x0)F0 = Ω dx· 4 ∑ j=1 ϕ∗ a (t,x0)ϕj (t,x) δ(x0−x)δj,aF† 0 F0 −0† 0 . = Ω dx· 4 ∑ j=1 ϕ∗ a (t,x0)ϕj (t,x)(δ(x0−x)δj,a1−0) = Ω dx· 4 ∑ j=1 ϕ∗ a (t,x0)ϕj (t,x)δ(x0−x)δj,a. Thus: 113
- 128. Na (x0) Ψ = ϕ∗ a (t,x0)ϕa (t,x0). (2.65) That is operator Na (x0) brings the a-component of the event probability density. Let Ψa (t,x) := ψa (x0)Ψ(t,x). In that case Na (x0) Ψa = Ω dx Ω dx ·Ψ† (t,x)ψ† a (x0)ψ† a (x0) ψa (x0)ψa (x0)ψa (x0)Ψ(t,x). Since ψa (x0)ψa (x0) = 0 then Na (x0) Ψa = 0. Therefore ψa (x0) ”annihilates” the a of the event-probability density. 2.9. Particles and Antiparticles Operator H obeys the following condition: H = 2πc h 3 ∑ k hω(k) 2 ∑ r=1 b† r,kbr,k − 4 ∑ r=3 b† r,kbr,k . This operator is not positive defined and in this case EΨ F0 = 2πc h 3 ∑ p hω(p) 2 ∑ r=1 |cr (t,p)|2 − 4 ∑ r=3 |cr (t,p)|2 . This problem is usually solved in the following way [49, p.54]: Let: v1 (k) : = γ[0] e3 (k), v2 (k) : = γ[0] e4 (k), d1,k : = −b† 3,−k, d2,k : = −b† 4,−k. In that case: e3 (k) = −v1 (−k), e4 (k) = −v2 (−k), b3,k = −d† 1,−k, b4,k = −d† 2,−k. 114
- 129. Therefore, ψs (x) : = ∑ k 2 ∑ r=1 br,ker,s (k)exp −i h c kx + +d† r,kvr,s (k)exp i h c kx H = 2πc h 3 ∑ k hω(k) 2 ∑ r=1 b† r,kbr,k +d† r,kdr,k −2∑ k hω(k)1. The first term on the right side of this equality is positive defined. This term is taken as the desired Hamiltonian. The second term of this equality is infinity constant. And this infinity is deleted (?!) [49, p.58] But in this case dr,kF0 = 0. In order to satisfy such condition, the vacuum element F0 must be replaced by the following: F0 → Φ0 := ∏ k 4 ∏ r=3 2πc h 3 b† r,kF0. But in this case: ψs (x)Φ0 = 0. And condition (2.64) isn’t carried out. In order to satisfy such condition, operators ψs (x) must be replaced by the following: ψs (x) → φs (x):= :=∑ k 2 ∑ r=1 br,ker,s (k)exp −i h c kx +dr,kvr (k)exp i h c kx . Hence, H = dx·H (x) = dx·φ† (x)H0φ(x) = = 2πc h 3 ∑ k hω(k) 2 ∑ r=1 b† r,kbr,k −d† r,kdr,k . And again we get negative energy. Let’s consider the meaning of such energy: An event with positive energy transfers this energy photons which carries it on recorders observers. Observers know that this event occurs, not before it happens. But event with negative energy should absorb this energy from observers. Consequently, observers know that this event happens before it happens. This contradicts Theorem 1.5.2. Therefore, events with negative energy do not occur. Hence, over vacuum Φ0 single fermions can exist, but there is no single antifermions. 115
- 130. A two-particle state is defined the following field operator [52]: ψs1,s2 (x,y) := φs1 (x) φs2 (x) φs1 (y) φs2 (y) . In that case: H = 2h 2πc h 6 Ha +Hb where Ha : = ∑ k ∑ p (ω(k)−ω(p)) 2 ∑ r=1 2 ∑ j=1 × × v† j (−k)vj (−p)e† r (p)er (k)× × +b† r,pd† j,−kdj,−pbr,k + + +d† r,−pb† j,kbj,kdr,−p + +v† j (−p)vj (−k)e† r (k)er (p)× × −b† r,kd† j,−pdj,−kbr,p + + −b† r,pd† j,−kdj,−kbr,p and Hb : = ∑ k ∑ p (ω(k)+ω(p)) 2 ∑ r=1 2 ∑ j=1 × × v† j (−p)vj (−k)v† r (−k)vr (−p)× × −d† r,−kd† j,−pdj,−kdr,−p + + −d† r,−pd† j,−kdj,−kdr,−p +e† r (k)er (p)e† j (p)ej (k)× × +b† r,kb† j,pbj,kbr,p + + +b† r,pb† j,kbj,kbr,p . If velosities are small then the following formula is fair. H = 4h 2πc h 6 Ha +Hb where Ha : = ∑ k ∑ p (ω(k)−ω(p))× × 2 ∑ r=1 2 ∑ j=1 d† j,−pb† r,kbr,kdj,−p −b† j,pd† r,−kdr,−kbj,p 116
- 131. and Hb : = ∑ k ∑ p (ω(k)+ω(p))× × 2 ∑ j=1 2 ∑ r=1 b† j,pb† r,kbr,kbj,p −d† j,−pd† r,−kdr,−kdj,−p . Therefore, in any case events with pairs of fermions and events with fermion- antifermion pairs can occur, but events with pairs of antiftrmions can not happen. Therefore, an antifermion can exists only with a fermion. 117
- 133. Chapter 3 Fields No group of people can claim power over the thinking and views of others. - Friedrich von Hayek 3.1. Electroweak Fields In 1963 American physicist Sheldon Glashow1 [68] proposed that the weak nuclear force and electricity and magnetism could arise from a partially unified electroweak theory. But ”... there is major problem: all the fermions and gauge bosons are massless, while exper- iment shows otherwise. Why not just add in mass terms explicitly? That will not work, since the associated terms break SU(2) or gauge invariances. For fermions, the mass term should be mψψ? mψψ = mψ(PL +PR)ψ = = m(ψ(PLPL)ψ+ψ(PRPR)ψ) = m(ψRψL +ψLψR). However, the left-handed fermion are put into SU(2) doublets and the right-handed ones into SU(2) singlets, so ψRψL and ψLψR are not SU(2) singlets and would not give an SU(2) invariant Lagrangian. Similarly, the expected mass terms for the gauge bosons, 1 2 m2 BBµ Bµ plus similar terms for other, are clearly not invariant under gauge transformations Bµ → Bµ = Bµ −∂µχ/g, The only direct way to preserve the gauge invariance and SU(2) invariance of Lagrangian is to set m = 0 for all quarks, leptons and gauge bosons:. There is a way to solve this problem, called the Higgs mechanism” [59]. 1Sheldon Lee Glashow (born December 5, 1932) is an American theoretical physicist.
- 134. No. The Dirac Lagrangian for a free fermion can have of the following form: Lf := ψ† β[0] ∂0 +β[1] ∂1 +β[2] ∂2 +β[3] ∂3 +imγ[0] ψ. Indeed, this Lagrangian is not invariant under the SU(2) transformation. But it is beautiful and truncating its mass term is not good idea. Further you will see, how it is possible to keep this beauty. 3.1.1. The Bi-mass State Let us consider [47], [48] the subspace ℑJ of the space ℑ spanned of the following subbasis (2.47): J := h 2πc exp −ih c (s0x4) ε1, h 2πc exp −ih c (s0x4) ε2, h 2πc exp −ih c (s0x4) ε3, h 2πc exp −ih c (s0x4) ε4, h 2πc exp −ih c (n0x5) ε1, h 2πc exp −ih c (n0x5) ε2, h 2πc exp −ih c (n0x5) ε3, h 2πc exp −ih c (n0x5) ε4 (3.1) with some integer numbers s0 and n0. Let U be any linear transformation of space ℑJ such that for every ϕ: if ϕ ∈ ℑJ then (2.48, 2.49): (Uϕ,Uϕ) = ρA, (3.2) Uϕ,β[s] Uϕ = − jA,s c for s ∈ {1,2,3}. In that case: U†β[µ]U = β[µ] for µ ∈ {0,1,2,3}. Such transformation has a matrix of the following shape: U := (a”+b”i)12 02 (c”+ig”)12 02 02 (a‘+b‘i)12 02 (c‘+ig‘)12 (u”+iv”)12 02 (k”+is”)12 02 02 (u‘+iv‘)12 02 (k‘+is‘)12 . with real functions a”(t,x), b”(t,x), c”(t,x), g”(t,x), u”(t,x), v”(t,x), k”(t,x), s”(t,x), a‘(t,x), b‘(t,x), c‘(t,x), g‘(t,x), u‘(t,x), v‘(t,x), k‘(t,x), s‘(t,x). These functions fulfil the following conditions: v”2 +b”2 +u”2 +a”2 = 1, c”2 +g”2 +k”2 +s”2 = 1, s” = − a”g”u”−u”b”c”+a”c”v”+b”g”v” u”2 +v”2 , 120
- 135. k” = −u”a”c”−u”b”g”+v”a”g”−b”c”v” u”2 +v”2 . v‘2 +b‘2 +u‘2 +a‘2 = 1, c‘2 +g‘2 +k‘2 +s‘2 = 1, s‘ = − a‘g‘u‘−u‘b‘c‘+a‘c‘v‘+b‘g‘v‘ u‘2 +v‘2 , k‘ = −u‘a‘c‘−u‘b‘g‘+v‘a‘g‘−b‘c‘v‘ u‘2 +v‘2 . U has 4 eigenvalues: exp(iα1), exp(iα2), exp(iα3), exp(iα4) for 8 orthonormalized eigenvectors: ε1,1,ε1,2,ε2,1,ε2,2,ε3,1,ε3,2,ε4,1,ε4,2. Let K := ε1,1 ε1,2 ε2,1 ε2,2 ε3,1 ε3,2 ε4,1 ε4,2 . Let θ1, θ2, θ3, θ4 be solution of the following system of equations: θ1 +θ2 +θ3 +θ4 = α1, θ1 +θ2 −θ3 −θ4 = α1, θ1 −θ2 +θ3 −θ4 = α1, θ1 −θ2 −θ3 +θ4 = α1. and U1 := exp(iθ1), U2 := K exp(iθ2)14 04 04 exp(−iθ2)14 K† , U3 := K exp(iθ3)12 02 02 02 02 exp(−iθ3)12 02 02 02 02 exp(iθ3)12 02 02 02 02 exp(−iθ3)12 K† , U4 := K exp(iθ4)12 02 02 02 02 exp(−iθ4)12 02 02 02 02 exp(−iθ4)12 02 02 02 02 exp(iθ4)12 K† . In this case: 121
- 136. U1U2U3U4 = U and U2 = exp(iθ2)12 02 02 02 02 exp(−iθ2)12 02 02 02 02 exp(iθ2)12 02 02 02 02 exp(−iθ2)12 . Besides U1U2 = ei(θ1+θ2) 0 0 0 0 ei(θ1−θ2) 0 0 0 0 ei(θ1+θ2) 0 0 0 0 ei(θ1−θ2) . Let χ and ς be thesolution of the following set of equations: 0.5χ+ς = θ1 +θ2, χ+ς = θ1 −θ2, i.e.: χ = −4θ2, ς = θ1 +3θ2. Let U[e] := exp(iς) and (2.39) U = exp iχ 2 12 02 02 exp(iχ)12 . In that case: UU[e] 18 = U1U2. Here real functions a(t,x), b(t,x), c(t,x), g(t,x), u(t,x), v(t,x), k(t,x),s(t,x) exist such that: U3U4 = (a+ib)12 02 (c+ig)12 02 02 (u+iv)12 02 (k +is)12 (−c+ig)12 02 (a−ib)12 02 02 (−k +is)12 02 (u−iv)12 and 122
- 137. a2 +b2 +c2 +g2 = 1, u2 +v2 +r2 +s2 = 1. If U(+) := 12 02 02 02 02 (u+iv)12 02 (k +is)12 02 02 12 02 02 (−k +is)12 02 (u−iv)12 (3.3) and U(−) := (a+ib)12 02 (c+ig)12 02 02 12 02 02 (−c+ig)12 02 (a−ib)12 02 02 02 02 12 (3.4) then U3U4 = U(−) U(+) = U(+) U(−) . Let us consider U(−). Let: ◦ := 1 2 (1−a2) b+ (1−a2) 14 (q−ic)14 (q+ic)14 (1−a2)−b 14 (3.5) and ∗ := 1 2 (1−a2) (1−a2)−b 14 (−q+ic)14 (−q−ic)14 b+ (1−a2) 14 . (3.6) These operators are fulfilled to the following conditions: ◦ ◦ = ◦, ∗ ∗ = ∗; ◦ ∗ = 0 = ∗ ◦, ( ◦ − ∗)( ◦ − ∗) = 18, ◦ + ∗ = 18, ◦γ[0] = γ[0] ◦, ∗γ[0] = γ[0] ∗, ◦β[4] = β[4] ◦, ∗β[4] = β[4] ∗ and U(−)†γ[0]U(−) = aγ[0] −( ◦ − ∗) √ 1−a2β[4], U(−)†β[4]U(−) = aβ[4] +( ◦ − ∗) √ 1−a2γ[0]. (3.7) From (2.38) the lepton equation of motion is the following: 3 ∑ µ=0 β[µ] (i∂µ +Fµ +0.5g1YBµ)+γ[0] i∂5 +β[4] i∂4 U(−)† U(−) ϕ = 0. 123
- 138. If ∂kU(−)† = U(−)† ∂k (3.8) for k ∈ {0,1,2,3,4,5} then U(−)†i∑3 µ=0 β[µ] (i∂µ +Fµ +0.5g1YBµ) +γ[0]U(−)†i∂5 +β[4]U(−)†i∂4 U(−) ϕ = 0. Hence, from (3.7): U(−)† ∑3 µ=0 β[µ] (i∂µ +Fµ +0.5g1YBµ) +γ[0]i a∂5 −( ◦ − ∗) √ 1−a2∂4 +β[4]i √ 1−a2 ( ◦ − ∗)∂5 +a∂4 U(−) ϕ = 0. Thus, if denote: x4 = ( ◦ + ∗)ax4 +( ◦ − ∗) 1−a2x5, x5 = ( ◦ + ∗)ax5 −( ◦ − ∗) 1−a2x4 then 3 ∑ µ=0 β[µ] (i∂µ +Fµ +0.5g1YBµ)+ γ[0] i∂5 +β[4] i∂4 ϕ = 0 (3.9) with ϕ = U(−) ϕ. That is the lepton Hamiltonian is invariant for the following global transformation: ϕ → ϕ = U(−) ϕ, x4 → x4 = ( ◦ + ∗)ax4 +( ◦ − ∗) 1−a2x5, (3.10) x5 → x5 = ( ◦ + ∗)ax5 −( ◦ − ∗) 1−a2x4, xµ → xµ = xµ. 3.1.2. Neutrino Wolfgang Pauli postulated the neutrino in 1930 to explain the energy spectrum of beta de- cays, the decay of a neutron into a proton and an electron. Clyde Cowan, Frederick Reines found the neutrino experimentally in 1955. Enrico Fermi2 developed the first theory de- scribing neutrino interactions and denoted this particles as neutrino in 1933. In 1962 Leon M. Lederman, Melvin Schwartz and Jack Steinberger showed that more than one type of neutrino exists. Bruno Pontecorvo3 suggested a practical method for investigating neutrino 2Enrico Fermi (29 September 1901 28 November 1954) was an Italian-born, naturalized American physicist particularly known for his work on the development of the first nuclear reactor, Chicago Pile-1, and for his contributions to the development of quantum theory, nuclear and particle physics, and statistical mechanics. 3Bruno Pontecorvo (Marina di Pisa, Italy, August 22, 1913 – Dubna, Russia, September 24, 1993) was an Italian-born atomic physicist, an early assistant of Enrico Fermi and then the author of numerous studies in high energy physics, especially on neutrinos. 124
- 139. masses in 1957, over the subsequent 10 years he developed the mathematical formalism and the modern formulation of vacuum oscillations... Let ℑeν be the unitary space, spanned by the following basis: Jeν := h 2πc 2πn0 sinh(2n0π) cosh h c n0x4 +sinh h c n0x4 ε1, h 2πc 2πn0 sinh(2n0π) cosh h c n0x4 +sinh h c n0x4 ε2, h 2πc 2πn0 sinh(2n0π) cosh h c n0x4 −sinh h c n0x4 ε3, h 2πc 2πn0 sinh(2n0π) cosh h c n0x4 −sinh h c n0x4 ε4, h 2πc exp −ih c (n0x5) ε1, h 2πc exp −ih c (n0x5) ε2, h 2πc exp −ih c (n0x5) ε3, h 2πc exp −ih c (n0x5) ε4 . (3.11) Let ℑe be the subspace of the space ℑeν such that if ϕ ∈ ℑe then ϕ has the following shape: ϕ(t,x,x5,x4) = exp −i h c n0x5 4 ∑ k=1 fk (t,x,n0,0)εk That is ϕ has the following matrix in the basis Jeν: ϕ = 0 0 0 0 f1 f2 f3 f4 . (3.12) Let us consider the following Hamiltonian on ℑe: H0,4 := c 3 ∑ r=1 β[r] i∂r +γ[0] i∂5 +β[4] i∂4 : (3.13) H0,4ϕ = c ∑3 r=1 β[r]i∂r +γ[0]i∂5 +β[4]i∂4 ϕ = = ∑3 r=1 β[r]ci∂rϕ+ +γ[0]ci∂5 exp −ih c n0x5 ∑4 k=1 fk (t,x,n0,0)εk+ +β[4]ci∂4 exp −ih c n0x5 ∑4 k=1 fk (t,x,n0,0)εk = = ∑3 r=1 β[r]ci∂rϕ+ +γ[0]ci −ih c n0 exp −ih c n0x5 ∑4 k=1 fk (t,x,n0,0)εk+ +0 = = ∑3 r=1 β[r]ci∂rϕ+hn0γ[0] exp −ih c n0x5 ∑4 k=1 fk (t,x,n0,0)ε. = = ∑3 r=1 β[r]ci∂rϕ+hn0γ[0]ϕ. Hence, on this space: 125
- 140. H0,4 = H0 := c 3 ∑ r=1 β[r] i∂r +hn0γ[0] . (3.14) Let ℑ◦ be the subspace of the space ℑeν such that if ϕ◦ ∈ ℑ◦ then ϕ◦ = ◦ ϕ and ϕ ∈ ℑe, and if ϕ ∈ ℑe then ( ◦ϕ) ∈ ℑ◦. If ϕ◦ = ◦ϕ then in the basis Jeν: ϕ◦ = 1 2 (1−a2) −(−q+ic) f1 −(−q+ic) f2 −(−q+ic) f3 −(−q+ic) f4 − − (1−a2)+b f1 − − (1−a2)+b f2 − − (1−a2)+b f3 − − (1−a2)+b f4 . Let us consider the Hamiltonian H0,4 mode of behavior on the space ℑ◦: Hence, H0,4ϕ◦ = c∑3 r=1 β[r]i∂rϕ◦+ +γ[0]ic (q−ic) 2 √ (1−a2) h 2πc 2πn0 sinh(2n0π)× × ∂5 f1 cosh h c n0x4 +sinh h c n0x4 ε1+ +f2 cosh h c n0x4 +sinh h c n0x4 ε2+ +f3 cosh h c n0x4 −sinh h c n0x4 ε3+ +f4 cosh h c n0x4 −sinh h c n0x4 ε4 + + (1−a2)−b h 2πc∂5 exp −ih c (n0x5) · ·(f1ε1 + f2ε2 + f3ε3 + f4ε4) + +β[4]ic (q−ic) 2 √ (1−a2) h 2πc 2πn0 sinh(2n0π)× × ∂4 f1 cosh h c n0x4 +sinh h c n0x4 ε1+ +f2 cosh h c n0x4 +sinh h c n0x4 ε2+ +f3 cosh h c n0x4 −sinh h c n0x4 ε3+ +f4 cosh h c n0x4 −sinh h c n0x4 ε4 + + (1−a2)−b ∂4 h 2πc exp −ih c (n0x5) · ·(f1ε1 + f2ε2 + f3ε3 + f4ε4) . 126
- 141. H0,4ϕ◦ = c∑3 r=1 β[r]i∂rϕ◦+ +γ[0]ic (q−ic) 2 √ (1−a2) h 2πc 2πn0 sinh(2n0π)× × ∂5 f1 cosh h c n0x4 +sinh h c n0x4 ε1+ +f2 cosh h c n0x4 +sinh h c n0x4 ε2+ +f3 cosh h c n0x4 −sinh h c n0x4 ε3+ +f4 cosh h c n0x4 −sinh h c n0x4 ε4 + + (1−a2)−b h 2πc∂5 exp −ih c (n0x5) · ·(f1ε1 + f2ε2 + f3ε3 + f4ε4) + +β[4]ic (q−ic) 2 √ (1−a2) h 2πc 2πn0 sinh(2n0π)× × ∂4 f1 cosh h c n0x4 +sinh h c n0x4 ε1+ +f2 cosh h c n0x4 +sinh h c n0x4 ε2+ +f3 cosh h c n0x4 −sinh h c n0x4 ε3+ +f4 cosh h c n0x4 −sinh h c n0x4 ε4 + + (1−a2)−b ∂4 h 2πc exp −ih c (n0x5) · ·(f1ε1 + f2ε2 + f3ε3 + f4ε4) . Therefore, H0,4ϕ◦ = c∑3 r=1 β[r]i∂rϕ◦+ +γ[0]ic √ 1−a2−b 2 √ 1−a2 × × 0+ h 2πc∂5 exp −ih c (n0x5) (f1ε1 + f2ε2 + f3ε3 + f4ε4) + +β[4]ic q−ic 2 √ 1−a2 h 2πc 2πn0 sinh(2n0π)× × f1 ∂4 cosh h c n0x4 +∂4 sinh h c n0x4 ε1+ +f2 ∂4 cosh h c n0x4 +∂4 sinh h c n0x4 ε2+ +f3 ∂4 cosh h c n0x4 −∂4 sinh h c n0x4 ε3+ +f4 ∂4 cosh h c n0x4 −∂4 sinh h c n0x4 ε4 + +0 . Hence, H0,4ϕ◦ = c∑3 r=1 β[r]i∂rϕ◦+ +γ[0]ic −ih c n0 √ 1−a2−b 2 √ 1−a2 h 2πc exp −ih c n0x5 × ×(f1ε1 + f2ε2 + f3ε3 + f4ε4)+ +β[4]ich c n0 q−ic 2 √ 1−a2 h 2πc 2πn0 sinh(2n0π)× × f1 sinh h c n0x4 +cosh h c n0x4 ε1+ +f2 sinh h c n0x4 +cosh h c n0x4 ε2+ +f3 sinh h c n0x4 −cosh h c n0x4 ε3+ +f4 sinh h c n0x4 −cosh h c n0x4 ε4 . Therefore, 127
- 142. H0,4ϕ◦ = c∑3 r=1 β[r]i∂rϕ◦+ +hn0γ[0] √ 1−a2−b 2 √ 1−a2 h 2πc exp −ih c n0x5 × ×(f1ε1 + f2ε2 + f3ε3 + f4ε4)+ +hn0β[4]i q−ic 2 √ 1−a2 h 2πc 2πn0 sinh(2n0π)× × f1 cosh h c n0x4 +sinh h c n0x4 ε1+ +f2 cosh h c n0x4 +sinh h c n0x4 ε2− −f3 cosh h c n0x4 −sinh h c n0x4 ε3− −f4 cosh h c n0x4 −sinh h c n0x4 ε4 . Hence, in basis Jeν: H0,4ϕ◦ = c∑3 r=1 β[r]i∂rϕ◦ +hn0× × γ[0] √ 1−a2−b 2 √ 1−a2 0 0 0 0 f1 f2 f3 f4 +β[4]i q−ic 2 √ 1−a2 f1 f2 −f3 −f4 0 0 0 0 = = c∑3 r=1 β[r]i∂rϕ◦ +hn0× × γ[0] √ 1−a2−b 2 √ 1−a2 0 0 0 0 f1 f2 f3 f4 +β[4]i q−ic 2 √ 1−a2 γ[5] f1 f2 f3 f4 0 0 0 0 . with γ[5] := 1 0 0 0 0 1 0 0 0 0 −1 0 0 0 0 −1 . Since β[4] iγ[5] = γ[0] then 128
- 143. H0,4ϕ◦ = c∑3 r=1 β[r]i∂rϕ◦ +hn0× × γ[0] 1 2 √ 1−a2 0 0 0 0 √ 1−a2 −b f1 √ 1−a2 −b f2 √ 1−a2 −b f3 √ 1−a2 −b f4 +γ[0] 1 2 √ 1−a2 (q−ic) f1 (q−ic) f2 (q−ic) f3 (q−ic) f4 0 0 0 0 . Therefore, H0,4ϕ◦ = c 3 ∑ r=1 β[r] i∂rϕ◦ +hn0γ[0] 1 2 √ 1−a2 −(−q+ic) f1, −(−q+ic) f2, −(−q+ic) f3, −(−q+ic) f4, − − √ 1−a2 +b f1, − − √ 1−a2 +b f2, − − √ 1−a2 +b f3, − − √ 1−a2 +b f4 . Hence, H0,4ϕ◦ = c 3 ∑ r=1 β[r] i∂rϕ◦ +hn0γ[0] ϕ◦. Thus, in space ℑe: H0,4 = H0 = c 3 ∑ r=1 β[r] i∂r +hn0γ[0] , too. Let ℑ∗ be the subspace of the space ℑeν such that if ϕ∗ ∈ ℑ∗ then ϕ∗ = ∗ ϕ and ϕ ∈ ℑe, and if ϕ ∈ ℑe then ( ∗ϕ) ∈ ℑ∗. If ϕ∗ = ∗ϕ (3.12) then in the basis Jeν: 129
- 144. ϕ∗ = 1 2 (1−a2) (−q+ic) f1 (−q+ic) f2 (−q+ic) f3 (−q+ic) f4 b+ √ 1−a2 f1 b+ √ 1−a2 f2 b+ √ 1−a2 f3 b+ √ 1−a2 f4 . Similarly to ϕ◦ you can calculate that H0,4ϕ∗ = H0ϕ∗ = c 3 ∑ r=1 β[r] i∂rϕ∗ +hn0γ[0] ϕ∗., too. Let e1L (k) := ω(k)+n0 +k3 k1 +ik2 , e1R (k) := ω(k)+n0 −k3 −k1 −ik2 , e2L (k) := k1 −ik2 ω(k)+n0 −k3 , e2R (k) := −k1 +ik2 ω(k)+n0 +k3 , e3L (k) := −e1R (k), e3R (k) := e1L (k), e4L (k) := −e2R (k), e4R (k) := e2L (k). with ω(k) := n2 0 +k2 1 +k2 2 +k2 3 ( n0, k1, k2,k3 are real numbers). In this case: es (k) = 1 2 ω(k)(ω(k)+n0) esL (k) esR (k) . Let es (k) := −→ 0 4 es (k) here s ∈ {1,2,3,4}. And let: 130
- 145. e◦s (k) := ◦es (k) = 1 2 √ 1−a2 −b √ 1−a2 (q−ic)es (k) √ 1−a2 −b es (k) , e∗s (k) := ∗es (k) = 1 2 √ 1−a2 +b √ 1−a2 (−q+ic)es (k) b+ √ 1−a2 es (k) . Denote H0 (k) := 3 ∑ r=1 β[r] kr = k3 k1 −ik2 n0 0 k1 +ik2 −k3 0 n0 n0 0 −k3 −k1 +ik2 0 n0 −k1 −ik2 k3 . In that case H0e◦1 (k) h 2πc 3 2 exp i h c = = hH0 (k)e◦1 (k) h 2πc 3 2 exp i h c = hω(k)e◦1 (k) h 2πc 3 2 exp i h c . Therefore, e◦1 (k) h 2πc 3 2 exp ih c is an eigenvector of H0 with the eigenvalue hω(k). Similarly you can calculate that e◦2 (k) h 2πc 3 2 exp ih c , e∗1 (k) h 2πc 3 2 exp ih c , e∗2 (k) h 2πc 3 2 exp ih c , are eigenvectors of H0 with the same eigenvalue, and e◦3 (k) h 2πc 3 2 exp ih c , e◦4 (k) h 2πc 3 2 exp ih c , e∗3 (k) h 2πc 3 2 exp ih c , e∗4 (k) h 2πc 3 2 exp ih c are an eigenvectors of H0 with the eigenvalue (−hω(k)). Vectors e◦s (k) h 2πc 3 2 exp ih c , e∗s (k) h 2πc 3 2 exp ih c with s ∈ {1,2,3,4} form an or- thonormalized basis in the space ℑeν (3.11) and 4 ∑ s=1 e∗ ∗s,r (k)e∗s,r (k)+e∗ ◦s,r (k)e◦s,r (k) = δr,r (3.15) for r,r ∈ {1,2,3,4,5,6,7,8}. 131
- 146. Let e∗s (k) := U(−) e∗s (k) = 1 2 √ 1−a2 +b √ 1−a2 1 2 ω(k)(ω(k)+n0) · a−i √ 1−a2 (−q+ic)esL (k) (−q+ic)esR (k) a−i √ 1−a2 √ 1−a2 +b esL (k) (1−a2)+b esR (k) and e◦s (k) := U(−) e◦s (k) = 1 2 √ 1−a2 −b √ 1−a2 1 2 ω(k)(ω(k)+n0) · a+i √ 1−a2 (q−ic)esL (k) (q−ic)esR (k) a+i √ 1−a2 √ 1−a2 −b esL (k) (1−a2)−b esR (k) . For these vectors: 4 ∑ r=1 e ∗ ∗r,j (k)e∗r,j (k)+e ∗ ◦r,j (k)e◦r,j (k) = δj,j and since U(−)†U(−) = 18 then e◦s (k) and e∗s (k) form an orthonormalized basis in the space ℑeν, too. Let er (k) := U(−) er (k) = 1 2 ω(k)(ω(k)+n0) (c+iq)erL (k) −→ 0 2 (a−ib)erL (k) erR (k) . (3.16) In that case: er (k) = 1 √ 2 1− b √ 1−a2 e◦r (k)+ 1+ b √ 1−a2 e∗r (k) . Let for j, j ∈ {1,2,3,4,5,6,7,8}: 132
- 147. ψ† j (y),ψj (x) = δ(y−x)δj ,j1, ψ† j (y),ψ† j (x) = 0 = ψj (y),ψj (x) and let b◦r,k := h 2πc 3 (Ω) dx·ei h c kx 8 ∑ j =1 e∗ ◦r,j (k)ψj (x), b∗r,k := h 2πc 3 (Ω) dx·ei h c kx 8 ∑ j =1 e∗ ∗r,j (k)ψj (x). In that case: ∑ k e−ih c kx 4 ∑ r=1 e◦r,j (k)b◦r,k + 4 ∑ r=1 e∗r,j (k)b∗r,k = ∑ k e−ih c kx ∑4 r=1 e◦r,j (k) h 2πc 3 · · (Ω) dx ·eih c kx ∑8 j =1 e∗ ◦r,j (k)ψj (x ) +∑4 r=1 e∗r,j (k) h 2πc 3 · · (Ω) dx ·eih c kx ∑8 j =1 e∗ ∗r,j (k)ψj (x ) = h 2πc 3 ∑ k (Ω) dx ·ei h c kx e−ih c kx · · 4 ∑ r=1 8 ∑ j =1 e◦r,j (k)e∗ ◦r,j (k)+e∗r,j (k)e∗ ∗r,j (k) ψj x . In accordance with (3.15): ∑ k e−ih c kx 4 ∑ r=1 e◦r,j (k)b◦r,k + 4 ∑ r=1 e∗r,j (k)b∗r,k = h 2πc 3 (Ω) dx ·∑ k e−ih c k(x−x ) 8 ∑ j =1 δj,j ψj x . Hence, since ∑ k ei h c k(x −x) = h 2πc 3 δ x −x and according properties of δ: 133
- 148. ∑ k e−ih c kx 4 ∑ r=1 e◦r,j (k)b◦r,k + 4 ∑ r=1 e∗r,j (k)b∗r,k = h 2πc 3 (Ω) dx · h 2πc 3 δ x −x ψj x = (Ω) dx ·δ x −x ψj x = ψj (x). Thus: ∑k e−ih c kx ∑4 r=1 e◦r,j (k)b◦r,k +∑4 r=1 e∗r,j (k)b∗r,k = ψj (x). (3.17) Let ψ(x) := h 2πc× × 2πn0 sinh(2n0π) cosh h c n0x4 + +sinh h c n0x4 ∑2 r=1 ψr (x)εr+ + cosh h c n0x4 − −sinh h c n0x4 ∑4 r=3 ψr (x)εr + +exp −ih c (n0x4) ∑4 r=1 ψr+4 (x)εr . (3.18) That is in basis Jeν (3.11): ψ(x) = ψ1 (x) ψ2 (x) ψ3 (x) ψ4 (x) ψ5 (x) ψ6 (x) ψ7 (x) ψ8 (x) . That is in this basis: b◦r,k := h 2πc 3 (Ω) dx·eih c kx e† ◦r,j (k)ψ(x), b∗r,k := h 2πc 3 (Ω) dx·eih c kx e† ∗r,j (k)ψ(x). Let ψ (x) := U(−) ψ(x). In that case: 134
- 149. b◦r,k := h 2πc 3 (Ω) dx·ei h c kx e † ◦r,j (k)ψ (x), b∗r,k := h 2πc 3 (Ω) dx·ei h c kx e † ∗r,j (k)ψ (x). Hence: b◦r,k = h 2πc 3 (Ω) dx·eih c kx U(−) e◦r,j (k) † U(−) ψ(x) , b∗r,k = h 2πc 3 (Ω) dx·eih c kx U(−) e∗r,j (k) † U(−) ψ(x) . Since U(−)†U(−) = 18 then b◦r,k = h 2πc 3 (Ω) dx·eih c kx e† ◦r,j (k)ψ(x), b∗r,k = h 2πc 3 (Ω) dx·eih c kx e† ∗r,j (k)ψ(x). That is: b◦r,k = b◦r,k and b∗r,k = b∗r,k. And from (3.17): ψj (x) = ∑ k e−ih c kx 4 ∑ r=1 e◦r,j (k)b◦r,k +e∗r,j (k)b∗r,k . (3.19) For operators b◦r,k and b∗r,k: b† ◦r ,k ,b◦r,k = h 2πc 3 δr,r δk,k 1, b† ∗r ,k ,b∗r,k = h 2πc 3 δr,r δk,k 1, b† ◦r ,k ,b∗r,k = 0, b† ∗r ,k ,b◦r,k = 0, b† ◦r ,k ,b† ◦r,k = 0, b† ∗r ,k ,b† ∗r,k = 0, {b◦r ,k ,b◦r,k} = 0, {b∗r ,k ,b∗r,k} = 0, b† ∗r ,k ,b† ◦r,k = 0. (3.20) 135
- 150. Let br,k := √ 2 1−a2 1 4 1 √ 1−a2 −b b◦r,k + 1 √ 1−a2 +b b∗r,k . In that case: e◦r (k)b◦r,k +e∗r (k)b∗r,k = = 1 √ 2 1− b √ 1−a2 e◦r (k)+ 1+ b √ 1−a2 e∗r (k) br,k − b− √ 1−a2 b+ √ 1−a2 e◦r (k)b∗r,k − b+ √ 1−a2 b− √ 1−a2 e∗r (k)b◦r,k And from (3.16): e◦r (k)b◦r,k +e∗r (k)b∗r,k = er (k)br,k − b− √ 1−a2 b+ √ 1−a2 e◦r (k)b∗r,k − b+ √ 1−a2 b− √ 1−a2 e∗r (k)b◦r,k. For br,k: b† r ,k ,br,k = 4b2+c2+q2 c2+q2 h 2πc 3 δr,r δk,k 1, b† r ,k ,b† r,k = 0, {br ,k ,br,k} = 0. (3.21) From (3.19): ψj (x) = ∑ k e−ih c kx 4 ∑ r=1 er,j (k)br,k − b− √ 1−a2 b+ √ 1−a2 ∑ k e−ih c kx 4 ∑ r=1 e◦r,j (k)b∗r,k − b+ √ 1−a2 b− √ 1−a2 ∑ k e−ih c kx 4 ∑ r=1 e∗r,j (k)b◦r,k. Let: 136
- 151. χ(x) := ∑ k e−ih c kx 4 ∑ r=1 er (k)br,k, (3.22) χ∗ j (x) := b− √ 1−a2 b+ √ 1−a2 ∑ k e−ih c kx 4 ∑ r=1 e◦r,j (k)b∗r,k χ◦j (x) := b+ √ 1−a2 b− √ 1−a2 ∑ k e−ih c kx 4 ∑ r=1 e∗r,j (k)b◦r,k. In that case: ψj (x) = χj (x)−χ∗ j (x)−χ◦j (x). Let H0 := U(−) H0U(−)† . For this Hamiltonian: (Ω) dx·χ† ∗ (x)H0ψ (x) = (Ω) dx· b− √ 1−a2 b+ √ 1−a2 ∑ k eih c k x 4 ∑ r =1 b† ∗r ,k e † ◦r k · ·H0 ∑ k e−ih c kx 4 ∑ r=1 e◦r (k)b◦r,k +e∗r (k)b∗r,k = (Ω) dx· b− √ 1−a2 b+ √ 1−a2 ∑ k ∑ k eih c k x 4 ∑ r =1 b† ∗r ,k e † ◦r k · ·H0e−ih c kx ∑2 r=1 (e◦r (k)b◦r,k +e∗r (k)b∗r,k)+ +∑4 r=3 (e◦r (k)b◦r,k +e∗r (k)b∗r,k) = (Ω) dx· b− √ 1−a2 b+ √ 1−a2 ∑ k ∑ k eih c k x e−ih c kx 4 ∑ r =1 b† ∗r ,k e † ◦r k · ·hH0 (k) ∑2 r=1 (e◦r (k)b◦r,k +e∗r (k)b∗r,k)+ +∑4 r=3 (e◦r (k)b◦r,k +e∗r (k)b∗r,k) . Hence: 137
- 152. (Ω) dx·χ† ∗ (x)H0ψ (x) = (Ω) dx· b− √ 1−a2 b+ √ 1−a2 ∑ k ∑ k eih c k x e−ih c kx 4 ∑ r =1 b† ∗r ,k e † ◦r k · ·h ω(k)∑2 r=1 (e◦r (k)b◦r,k +e∗r (k)b∗r,k)− −ω(k)∑4 r=3 (e◦r (k)b◦r,k +e∗r (k)b∗r,k) = (Ω) dx· b− √ 1−a2 b+ √ 1−a2 ∑ k hω(k)∑ k eih c k x e−i h c kx · · 4 ∑ r =1 b† ∗r ,k e † ◦r k ∑2 r=1 (e◦r (k)b◦r,k +e∗r (k)b∗r,k)− −∑4 r=3 (e◦r (k)b◦r,k +e∗r (k)b∗r,k) = b− √ 1−a2 b+ √ 1−a2 ∑ k hω(k)∑ k dx·e−ih c (k−k )x · · 4 ∑ r =1 b† ∗r ,k ∑2 r=1 e † ◦r (k )e◦r (k)b◦r,k +e † ◦r (k )e∗r (k)b∗r,k − −∑4 r=3 e † ◦r (k )e◦r (k)b◦r,k +e † ◦r (k )e∗r (k)b∗r,k . Since (Ω) dx·e−ih c (k−k )x = 2πc h 3 δk,k then (Ω) dx·χ† ∗ (x)H0ψ (x) = b− √ 1−a2 b+ √ 1−a2 ∑ k hω(k)∑ k 2πc h 3 δk,k · · 4 ∑ r =1 b† ∗r ,k ∑2 r=1 e † ◦r (k )e◦r (k)b◦r,k +e † ◦r (k )e∗r (k)b∗r,k − −∑4 r=3 e † ◦r (k )e◦r (k)b◦r,k +e † ◦r (k )e∗r (k)b∗r,k . In accordance with properties of δ: (Ω) dx·χ† ∗ (x)H0ψ (x) = b− √ 1−a2 b+ √ 1−a2 ∑ k hω(k) 2πc h 3 · · 4 ∑ r =1 b† ∗r ,k ∑2 r=1 e † ◦r (k)e◦r (k)b◦r,k +e † ◦r (k)e∗r (k)b∗r,k − −∑4 r=3 e † ◦r (k)e◦r (k)b◦r,k +e † ◦r (k)e∗r (k)b∗r,k . 138
- 153. Since e † ◦r k e◦r (k) = δr,r , e † ◦r k e∗r (k) = 0 then (Ω) dx·χ† ∗ (x)H0ψ (x) = b− √ 1−a2 b+ √ 1−a2 ∑ k hω(k) 2πc h 3 · · 4 ∑ r =1 b† ∗r ,k 2 ∑ r=1 (δr,r b◦r,k +0b∗r,k)− 4 ∑ r=3 (δr,r b◦r,k +0b∗r,k) = 2πc h 3 b− √ 1−a2 b+ √ 1−a2 ∑ k hω(k)· · 4 ∑ r =1 b† ∗r ,k 2 ∑ r=1 δr,r b◦r,k − 4 ∑ r=3 δr,r b◦r,k = 2πc h 3 b− √ 1−a2 b+ √ 1−a2 · ·∑ k hω(k) 2 ∑ r=1 b† ∗r,kb◦r,k − 4 ∑ r=3 b† ∗r,kb◦r,k . Therefore, (Ω) dx·χ† ∗ (x)H0ψ (x) = 2πc h 3 b− √ 1−a2 b+ √ 1−a2 ∑ k hω(k) 2 ∑ r=1 b† ∗r,kb◦r,k − 4 ∑ r=3 b† ∗r,kb◦r,k . Similarly you can calculate that dx·χ† ◦ (x)H0ψ (x) = 2πc h 3 b− √ 1−a2 b+ √ 1−a2 ∑ k hω(k) 2 ∑ r=1 b† ◦r,kb∗r,k − 4 ∑ r=3 b† ◦r,kb∗r,k . Since 139
- 154. Ψ(t,p) = 2πc h 3 4 ∑ r=1 cr (t,p)b† r,pF0 and (3.20) b† ∗r ,k ,b◦r,k = 0, b† ∗r ,k ,b† ∗r,k = 0, b† ∗r ,k ,b† ◦r,k = 0. then b† ∗r,kb◦r,kΨ = −b◦r,k 2πc h 3 4 ∑ r=1 cr (t,p)b† ∗r,kb† r,pF0 = 0. Similarly b† ◦r,kb∗r,kΨ = 0. Hence, (Ω) dx·ψ † (x)H0ψ (x)Ψ(t,x0) = (Ω) dx·χ† (x)H0χ(x)Ψ(t,x0). Thus, the function ψ (x) can be substituted for the function χ(x) in calculations of a probabilities evolution. Let νn0,(s) (k) := (c+iq)esL (k) −→ 0 2 , ln0,(s) (k) := (a−ib)esL (k) esR (k) . Hence, from (3.16): es (k) = h 2πc × × 2πn0 sinh(2n0π) cosh h c n0x4 + +sinh h c n0x4 (c+iq)esL (k)+ + cosh h c n0x4 − −sinh h c n0x4 −→ 0 2 + +exp −ih c (n0x4) ln0,(s) (k) . Therefore, in basis Jeν: es (k) = νn0,(s) (k) ln0,(s) (k) . Therefore, from (3.22): χ(x) = ∑ k e−ih c kx 4 ∑ s=1 νn0,(s) (k) ln0,(s) (k) bs,k, Let νn0 (x) := ∑ k e−ih c kx 2 ∑ s=1 νn0,(s) (k)bs (k), 140
- 155. ln0 (x) := ∑ k e−ih c kx 4 ∑ s=1 ln0,(s) (k)bs (k). Hence, in basis Jeν: χ(x) = νn0 (x) ln0 (x) . (3.23) Let: Hl,0 := c 3 ∑ r=1 β[r] i∂r +hn0 aγ[0] −bβ[4] , Hν,0 := c 3 ∑ r=1 β[r] i∂r +hn0 aγ[0] +bβ[4] , Hν,l := (c+iq) 0 0 n0 0 0 0 0 n0 −n0 0 0 0 0 −n0 0 0 , Hl,ν := (c−iq) 0 0 −n0 0 0 0 0 −n0 n0 0 0 0 0 n0 0 0 . In that case in basis Jeν: H0 = Hν,0 Hν,l Hl,ν Hl,0 . Let Hl,0 (k) := 3 ∑ r=1 β[r] kr +n0 aγ[0] −bβ[4] , Hν,0 (k) := 3 ∑ r=1 β[r] kr +n0 aγ[0] +bβ[4] . In that case H0 (k) = Hν,0 (k) Hν,l Hl,ν Hl,0 (k) An neutrino and it’s lepton are tied by the follows equations: Hν,0 (k)νn0,(s) (k)+Hν,lln0,(s) (k) = ω(k)νn0,(s) (k) for s ∈ {1,2} and 141
- 156. Hν,0 (k)νn0,(s) (k)+Hν,lln0,(s) (k) = −ω(k)νn0,(s) (k) for s ∈ {3,4}. I suppose that such neutrino can fly 1.5 cm. [60] and give birth to it’s leptons. 3.1.3. Electroweak Transformations During the 1960s Sheldon Lee Glashow discovered that they could construct a gauge- invariant theory of the weak force, provided that they also included the electromagnetic force. The existence of the force carriers, the neutral Z particles and the charged W parti- cles, was verified experimentally in 1983 in high-energy proton-antiproton collisions at the European Organization for Nuclear Research (CERN). Let (3.8) does not hold true, that is U(−) depends on x. And let denote: K := 3 ∑ µ=0 β[µ] (Fµ +0.5g1YBµ). (3.24) In that case from (2.38) the equation of moving is of following form: K + 3 ∑ µ=0 β[µ] i∂µ +γ[0] i∂5 +β[4] i∂4 ϕ = 0. (3.25) Let us consider for this Hamiltonian the following transformations: ϕ → ϕ := U(−) ϕ, x4 → x4 := ( ◦ + ∗)ax4 +( ◦ − ∗) 1−a2x5, x5 → x5 := ( ◦ + ∗)ax5 −( ◦ − ∗) 1−a2x4, (3.26) xµ → xµ := xµ, for µ ∈ {0,1,2,3}, K → K = U(−) KU(−)† −i 3 ∑ µ=0 β[µ] ∂µU(−) U(−)† with ∂4U(−) = U(−)∂4 and ∂5U(−) = U(−)∂5: Since ( ◦ − ∗)( ◦ − ∗) = 18 then x4 = ax4 −( ◦ − ∗) 1−a2x5 and x5 = ( ◦ − ∗) 1−a2x4 +ax5. Since for any f: 142
- 157. ∂4 f = ∂4 f ·∂4x4 +∂5 f ·∂4x5, ∂5 f = ∂4 f ·∂5x4 +∂5 f ·∂5x5 then ∂4 f = ∂4 f ·a+∂5 f ·( ◦ − ∗) 1−a2, ∂5 f = ∂4 f · −( ◦ − ∗) 1−a2 +∂5 f ·a. Therefore, if K + 3 ∑ µ=0 β[µ] i∂µ +γ[0] i∂5 +β[4] i∂4 U(−) ϕ = 0 then U(−)KU(−)† −i∑3 µ=0 β[µ] ∂µU(−) U(−)† +∑3 µ=0 β[µ]i∂µ +γ[0]i −( ◦ − ∗) √ 1−a2 ∂4 +a∂5 +β[4]i a∂4 +( ◦ − ∗) √ 1−a2∂5 U(−) ϕ = 0. Hence, U(−)KU(−)†U(−) −i∑3 µ=0 β[µ] ∂µU(−) U(−)†U(−) +∑3 µ=0 β[µ]i∂µU(−) +γ[0]U(−)i −( ◦ − ∗) √ 1−a2 ∂4 +a∂5 +β[4]U(−)i a∂4 +( ◦ − ∗) √ 1−a2∂5 ϕ = 0 since U(−) is a linear operator such that ∂4U(−) = U(−)∂4 and ∂5U(−) = U(−)∂5. Since U(−)† U(−) = 18, for any f: ∂µ U(−) f = ∂µU(−) f + U(−) ∂µ f = ∂µU(−) +U(−) ∂µ f, and γ[0] U(−) = U(−) aγ[0] −( ◦ − ∗) 1−a2β[4] , β[4] U(−) = U(−) aβ[4] +( ◦ − ∗) 1−a2γ[0] then 143
- 158. U(−)K −i∑3 µ=0 β[µ] ∂µU(−) +∑3 µ=0 β[µ]i ∂µU(−) +U(−)∂µ +U(−) aγ[0] −( ◦ − ∗) √ 1−a2β[4] × ×i −( ◦ − ∗) √ 1−a2 ∂4 +a∂5 +U(−) aβ[4] +( ◦ − ∗) √ 1−a2γ[0] × ×i a∂4 +( ◦ − ∗) √ 1−a2∂5 ϕ = 0. Therefore, U(−)K −i∑3 µ=0 β[µ] ∂µU(−) +∑3 µ=0 β[µ]i ∂µU(−) +U(−)∂µ +iU(−) aγ[0] −( ◦ − ∗) √ 1−a2β[4] × × −( ◦ − ∗) √ 1−a2∂4 +a∂5 + aβ[4] +( ◦ − ∗) √ 1−a2γ[0] × × a∂4 +( ◦ − ∗) √ 1−a2∂5 ϕ = 0, U(−) K + 3 ∑ µ=0 β[µ] iU(−) ∂µ +iU(−) +γ[0] ∂5 +β[4] ∂4 ϕ = 0, Hence, U(−) K + 3 ∑ µ=0 β[µ] i∂µ +i +γ[0] ∂5 +β[4] ∂4 ϕ = 0 since β[µ]U(−) = U(−)β[µ] for µ ∈ {0,1,2,3}. Compare with (3.25). Therefore, this equation of moving is invariant under the transformation (3.26). Let g2 be some positive real number. If design (here: a,b,c,q form U(−) in (3.4)): W0,µ := −2 1 g2q q(∂µa)b−q(∂µb)a+(∂µc)q2+ +a(∂µa)c+b(∂µb)c+c2 (∂µc) W1,µ := −2 1 g2q (∂µa)a2 −bq(∂µc)+a(∂µb)b+ +a(∂µc)c+q2 (∂µa)+c(∂µb)q W2,µ := −2 1 g2q q(∂µa)c−a(∂µa)b−b2 (∂µb)− −c(∂µc)b−(∂µb)q2 −(∂µc)qa and Wµ := W0,µ12 02 (W1,µ −iW2,µ)12 02 02 02 02 02 (W1,µ +iW2,µ)12 02 −W0,µ12 02 02 02 02 02 (3.27) then −i ∂µU(−) U(−)† = 1 2 g2Wµ, (3.28) 144
- 159. and from (3.24), (3.25): ∑3 µ=0 β[µ]i ∂µ −i0.5g1BµY −i1 2 g2Wµ −iFµ +γ[0]i∂5 +β[4]i∂4 ϕ = 0. (3.29) Let (3.4) a (t,x), b (t,x), c (t,x), q (t,x) are real functions and: U := (a +ib )12 02 (c +ig )12 02 02 12 02 02 (−c +ig )12 02 (a −ib )12 02 02 02 02 12 . In this case if U := U U(−) then there exist real functions a (t,x), b (t,x), c (t,x), q (t,x) such thatU has the similar shape: U = (a +ib )12 02 (c +ig )12 02 02 12 02 02 (−c +ig )12 02 (a −ib )12 02 02 02 02 12 . Let: Wµ := − 2i g2 ∂µ U U(−) U U(−) † , Hence, Wµ = − 2i g2 ∂µU U(−) U U(−) † − 2i g2 U ∂µU(−) U U(−) † = − 2i g2 ∂µU U(−) U(−)† U † − 2i g2 U ∂µU(−) U(−)† U † = − 2i g2 ∂µU U † − 2i g2 U ∂µU(−) U(−)† U † . Since from (3.28): Wµ = −i 2 g2 ∂µU(−) U(−)† then Wµ = − 2i g2 ∂µU U † − 2i g2 U ∂µU(−) U(−)† U † = − 2i g2 ∂µU U † +U WµU † . Therefore, if 145
- 160. ◦ := 1 2 (1−a 2) b + (1−a 2) 14 (q −ic )14 (q +ic )14 (1−a 2)−b 14 , ∗ := 1 2 (1−a 2) (1−a 2)−b 14 (−q +ic )14 (−q −ic )14 b + (1−a 2) 14 . then under the following transformation ϕ → ϕ := U ϕ, x4 → x4 := ◦ + ∗ a x4 + ◦ − ∗ 1−a 2x5, x5 → x5 := ◦ + ∗ a x5 − ◦ − ∗ 1−a 2x4, (3.30) xµ → xµ := xµ, for µ ∈ {0,1,2,3}, K → K := 3 ∑ µ=0 β[µ] Fµ +0.5g1YBµ + 1 2 g2Wµ fields Wµ and Wµ are tied by the following equation Wµ = U WµU † − 2i g2 (∂µU )U † (3.31) like in Standard Model. From (3.28): Wµ = −i 2 g2 ∂µU(−) U(−)† . Let us calculate: ∂µWν −∂νWµ = = ∂µ −i 2 g2 ∂νU(−) U(−)† −∂ν −i 2 g2 ∂µU(−) U(−)† = = −i 2 g2 ∂µ∂νU(−) U(−)† + ∂νU(−) ∂µU(−)† − ∂ν∂µU(−) U(−)† − ∂µU(−) ∂νU(−)† . Since ∂µ∂νU(−) = ∂ν∂µU(−) then ∂µWν −∂νWµ = (3.32) = −i 2 g2 ∂νU(−) ∂µU(−)† − ∂µU(−) ∂νU(−)† . 146
- 161. And let us calculate: WµWν −WνWµ = = −i 2 g2 ∂µU(−) U(−)† −i 2 g2 ∂νU(−) U(−)† − − −i 2 g2 ∂νU(−) U(−)† −i 2 g2 ∂µU(−) U(−)† = − 4 g2 2 ∂µU(−) U(−)† ∂νU(−) U(−)† − ∂νU(−) U(−)† ∂µU(−) U(−)† . Since U(−) U(−)† = 18 then ∂µ U(−) U(−)† = 0, and ∂ν U(−) U(−)† = 0, Hence, ∂µU(−) U(−)† +U(−) ∂µU(−)† = 0, and ∂νU(−) U(−)† +U(−) ∂νU(−)† = 0 Hence, ∂µU(−) U(−)† = −U(−) ∂µU(−)† and ∂νU(−) U(−)† = −U(−) ∂νU(−)† . Therefore, WµWν −WνWµ = = − 4 g2 2 − ∂µU(−) U(−)†U(−)∂νU(−)†+ + ∂νU(−) U(−)†U(−)∂µU(−)† = − 4 g2 2 − ∂µU(−) ∂νU(−)† + ∂νU(−) ∂µU(−)† since U(−)† U(−) = 18. Therefore, in accordance with (3.32): ∂µWν −∂νWµ = ig2 2 (WµWν −WνWµ). (3.33) In accordance with (3.27) matrix WµWν −WνWµ has the following columns: the first and the second columns are the following: 147
- 162. 2iW1,µW2,ν −2iW2,µW1,ν 02 02 02 2W0,νW1,µ +2iW0,νW2,µ −2W0,µW1,ν −2iW0,µW2,ν 02 02 02 , the third and the fourth columns are the following: 2W0,µW1,ν −2iW0,µW2,ν −2W0,νW1,µ +2iW0,νW2,µ 02 02 02 −2iW1,µW2,ν +2iW2,µW1,ν 02 02 02 . And matrix ∂µWν −∂νWµ has the following columns: the first and the second ones are the following: ∂µW0,ν −∂νW0,µ 02 02 02 ∂µW1,ν +i∂µW2,ν −∂νW1,µ −i∂νW2,µ 02 02 02 , the third and the fourth columns are the following: ∂µW1,ν −i∂µW2,ν −∂νW1,µ +i∂νW2,µ 02 02 02 −∂µW0,ν +∂νW0,µ 02 02 02 . Therefore, in accordance with (3.33): i g2 2 (2iW1,µW2,ν −2iW2,µW1,ν) = ∂µW0,ν −∂νW0,µ, i g2 2 (2W0,νW1,µ +2iW0,νW2,µ −2W0,µW1,ν −2iW0,µW2,ν) = ∂µW1,ν +i∂µW2,ν −∂νW1,µ −i∂νW2,µ, i g2 2 (2W0,µW1,ν −2iW0,µW2,ν −2W0,νW1,µ +2iW0,νW2,µ) = ∂µW1,ν −i∂µW2,ν −∂νW1,µ +i∂νW2,µ, i g2 2 (−2iW1,µW2,ν +2iW2,µW1,ν) = −∂µW0,ν +∂νW0,µ. Hence, ∂νW0,µ = ∂µW0,ν −g2 (W1,µW2,ν −W1,νW2,µ), (3.34) ∂νW1,µ = ∂µW1,ν −g2 (W2,µW0,ν −W2,νW0,µ), (3.35) ∂νW2,µ = ∂µW2,ν −g2 (W0,µW1,ν −W0,νW1,µ). (3.36) 148
- 163. The derivative of (3.34) with respect to xν is of the following form: ∂ν∂νW0,µ = ∂µ∂νW0,ν− −g2 (∂νW1,µ)W2,ν +W1,µ (∂νW2,ν) −(∂νW1,ν)W2,µ −W1,ν (∂νW2,µ) . Let us substitute ∂νW1,µ and ∂νW2,µ for its expressions from (3.35) and (3.36): ∂ν∂νW0,µ = ∂µ∂νW0,ν− −g2 (∂µW1,ν −g2 (W2,µW0,ν −W2,νW0,µ))W2,ν +W1,µ (∂νW2,ν)−(∂νW1,ν)W2,µ −W1,ν (∂µW2,ν −g2 (W0,µW1,ν −W0,νW1,µ)) = = ∂µ∂νW0,ν −g2 (∂µW1,ν)W2,ν −g2 (W2,µW0,νW2,ν −W2,νW0,µW2,ν) +W1,µ (∂νW2,ν)−(∂νW1,ν)W2,µ −W1,ν∂µW2,ν +g2 (W1,νW0,µW1,ν −W1,νW0,νW1,µ) = = −g2 2 (W1,νW1,ν +W2,νW2,ν)W0,µ+ +g2 2 (W1,νW1,µ +W2,µW2,ν)W0,ν −g2 (∂µW1,ν)W2,ν −W1,ν∂µW2,ν +W1,µ (∂νW2,ν)−(∂νW1,ν)W2,µ + +∂µ∂νW0,ν Hence, ∂ν∂νW0,µ = = −g2 2 (W1,νW1,ν +W2,νW2,ν)W0,µ+ +g2 2 (W1,νW1,µ +W2,µW2,ν)W0,ν −g2 (∂µW1,ν)W2,ν −W1,ν∂µW2,ν +W1,µ (∂νW2,ν)−(∂νW1,ν)W2,µ + +∂µ∂νW0,ν. Therefore: ∂ν∂νW0,µ = = −g2 2 (W0,νW0,ν +W1,νW1,ν +W2,νW2,ν)W0,µ+ +g2 2W0,νW0,νW0,µ +g2 2 (W1,νW1,µ +W2,µW2,ν)W0,ν −g2 (∂µW1,ν)W2,ν −W1,ν∂µW2,ν +W1,µ (∂νW2,ν)−(∂νW1,ν)W2,µ + +∂µ∂νW0,ν. Thus, ∂ν∂νW0,µ = = −g2 2 (W0,νW0,ν +W1,νW1,ν +W2,νW2,ν)W0,µ+ +g2 2 (W0,νW0,µ +W1,νW1,µ +W2,µW2,ν)W0,ν −g2 (∂µW1,ν)W2,ν −W1,ν∂µW2,ν +W1,µ (∂νW2,ν)−(∂νW1,ν)W2,µ + +∂µ∂νW0,ν. (3.37) 149
- 164. Since W2 ν := W0,νW0,ν +W1,νW1,ν +W2,νW2,ν and Wν|Wµ := W0,νW0,µ +W1,νW1,µ +W2,µW2,ν = Wν|Wµ for Wµ = W0,µ W1,µ W2,µ and Wν = W0,ν W1,ν W2,ν then ∂ν∂νW0,µ = − g2Wν 2 W0,µ+ +g2 2 Wν|Wµ W0,ν −g2 (∂µW1,ν)W2,ν −W1,ν∂µW2,ν +W1,µ (∂νW2,ν)−(∂νW1,ν)W2,µ + +∂µ∂νW0,ν. Hence, ∂0∂0W0,µ = − g2W0 2 W0,µ+ +g2 2 W0|Wµ W0,0 −g2 (∂µW1,0)W2,0 −W1,0∂µW2,0 +W1,µ (∂0W2,0)−(∂0W1,0)W2,µ + +∂µ∂0W0,0. Since ∂0 = 1 c ∂t then 1 c2 ∂2 t W0,µ = − g2W0 2 W0,µ+ +g2 2 W0|Wµ W0,0 −g2 (∂µW1,0)W2,0 −W1,0∂µW2,0 +W1,µ (∂0W2,0)−(∂0W1,0)W2,µ + +∂µ∂0W0,0. And for s ∈ {1,2,3}: ∂s∂sW0,µ = − g2Ws 2 W0,µ +g2 2 Ws|Wµ W0,s −g2 (∂µW1,s)W2,s −W1,s∂µW2,s +W1,µ (∂sW2,s)−(∂sW1,s)W2,µ +∂µ∂sW0,s. 150
- 165. Therefore, − 1 c2 ∂2 t W0,µ +∑3 s=1 ∂2 sW0,µ = − − g2W0 2 W0,µ +g2 2 W0|Wµ W0,0 −g2 (∂µW1,0)W2,0 −W1,0∂µW2,0 +W1,µ (∂0W2,0)−(∂0W1,0)W2,µ +∂µ∂0W0,0 + + ∑3 s=1 − g2Ws 2 W0,µ +g2 2 Ws|Wµ W0,s −g2 (∂µW1,s)W2,s −W1,s∂µW2,s +W1,µ (∂sW2,s)−(∂sW1,s)W2,µ +∂µ∂sW0,s .. Hence, − 1 c2 ∂2 t W0,µ +∑3 s=1 ∂2 sW0,µ = g2W0 2 W0,µ −∑3 s=1 g2Ws 2 W0,µ +g2 2 ∑3 s=1 Ws|Wµ W0,s −g2 2 W0|Wµ W0,0 +g2 (∂µW1,0)W2,0 −W1,0∂µW2,0 +W1,µ (∂0W2,0)−(∂0W1,0)W2,µ −∑3 s=1 (∂µW1,s)W2,s −W1,s∂µW2,s +W1,µ (∂sW2,s)−(∂sW1,s)W2,µ +∂µ ∑3 s=1 ∂sW0,s −∂µ∂0W0,0 −. Hence, − 1 c2 ∂2 t +∑3 s=1 ∂2 s W0,µ = g2 2 W2 0 −∑3 s=1W2 s W0,µ+ +g2 2 ∑3 s=1 Ws|Wµ W0,s − W0|Wµ W0,0 +g2 (∂µW1,0)W2,0 −W1,0∂µW2,0 +W1,µ (∂0W2,0)−(∂0W1,0)W2,µ −∑3 s=1 (∂µW1,s)W2,s −W1,s∂µW2,s +W1,µ (∂sW2,s)−(∂sW1,s)W2,µ +∂µ ∑3 s=1 ∂sW0,s −∂µ∂0W0,0. (3.38) This equation looks like to the Klein-Gordon equation4 of field W0,µ with mass m = h c g2 W2 0 − 3 ∑ s=1 W2 s (3.39) 4(2.56) − 1 c2 ∂2 t + 3 ∑ s=1 ∂2 s ϕ = m2c2 h2 ϕ 151
- 166. and with additional terms of the W0,µ interactions with others components of W. You can receive similar equations for W1,µ and for W2,µ. If W0 := W0 − v cWk 1− v c 2 , Wk := Wk − v cW0 1− v c 2 , Wk := Wk, if s = k then W 2 0 − 3 ∑ s=1 W 2 s = W0 − v cWk 2 1− v c 2 − Wk − v cW0 2 1− v c 2 − ∑ s=k W 2 s = W2 0 + v c 2 W2 k −W2 k − v c 2 W2 0 1− v c 2 − ∑ s=k W 2 s = 1− v c 2 W2 0 − 1− v c 2 W2 k 1− v c 2 − ∑ s=k W 2 s . Hence, W 2 0 − 3 ∑ s=1 W 2 s = W2 0 − 3 ∑ s=1 W2 s . Therefore, such ”mass” (3.39) is invariant for the Lorentz transformations: You can calculate that it is invariant for the transformations of turns, too: Wr = Wr cosλ−Ws sinλ. Ws = Wr sinλ+Ws cosλ; with a real number λ, and r ∈ {1,2,3}, s ∈ {1,2,3}. That is in (3.38) the form m = h c g2 W2 0 − 3 ∑ s=1 W2 s varies in space, but locally acts like a mass - i.e. it does not allow to particles of this field to behave as a massless ones. A mass of the W-boson was measured, between 1996 and 2000 at LEP5 [50]. Let6 5The Large Electron-Positron Collider (LEP) is largest particles accelerator (ring with a circumference of 27 kilometers built in a tunnel under the border of Switzerland and France.) 6here α is the Weinberg Angle. The experimental value of sin2 α = 0.23124±0.00024 [51]. 152
- 167. α := arctan g1 g2 , Zµ := (W0,µ cosα−Bµ sinα), Aµ := (Bµ cosα+W0,µ sinα). (3.40) In that case: ∑ν gν,ν∂ν∂νW0,µ = cosα·∑ν gν,ν∂ν∂νZµ +sinα·∑ν gν,ν∂ν∂νAµ. If ∑ ν gν,ν∂ν∂νAµ = 0 then mZ = mW cosα with mW from (3.39). It is like Standard Model. The equation of moving (3.29) under Fµ = 0 has the following form: ∑3 µ=0 β[µ]i ∂µ −i0.5g1BµY −i1 2 g2Wµ +γ[0]i∂5 +β[4]i∂4 ϕ = 0. (3.41) Hence, in accordance with (3.27) and (2.35): ∑3 µ=0 β[µ]i× × ∂µ −i0.5g1Bµ − 12 02 02 2·12 − −i1 2 g2 W0,µ12 02 (W1,µ −iW2,µ)12 02 02 02 02 02 (W1,µ +iW2,µ)12 02 −W0,µ12 02 02 02 02 02 +γ[0]i∂5 +β[4]i∂4 ·ϕ = 0. In accordance with (3.40) [53]: Bµ = −Zµ g1 g2 1 +g2 2 +Aµ g2 g2 1 +g2 2 , W0,µ = Zµ g2 g2 1 +g2 2 +Aµ g1 g2 1 +g2 2 . Let (e is the elementary charge7: e = 1.60217733×10−19 C). 7Sir Joseph John ”J. J.” Thomson, (18 December 1856 - 30 August 1940) was a British physicist. He is credited for the discovery of the electron and of isotopes, and the invention of the mass spectrometer. 153
- 168. e := g1g2 g2 1 +g2 2 , and let Zµ := Zµ 1 g2 2 +g2 1 g2 2 +g2 1 12 02 02 02 02 2g2 112 02 02 02 02 g2 2 −g2 1 12 02 02 02 02 2g2 112 , Wµ := g2 02 02 (W1,µ −iW2,µ)12 02 02 02 02 02 (W1,µ +iW2,µ)12 02 02 02 02 02 02 02 ·12 , Aµ := Aµ 02 02 02 02 02 12 02 02 02 02 12 02 02 02 02 12 . In that case from (3.41): ∑3 µ=0 β[µ]i ∂µ +ieAµ −i0.5 Zµ +Wµ +γ[0]i∂5 +β[4]i∂4 ϕ = 0. (3.42) Let in basis (3.11) (3.23) : ϕ = ϕν −→ 0 2 ϕe,L ϕe,R . In that case ∑3 µ=0 β[µ]i ∂µϕ+iAµe ϕe,L ϕe,R −i0.5 Zµ +Wµ ϕ + γ[0]i∂5 +β[4]i∂4 ϕ = 0. (3.43) Here the vector field Aµ is the electromagnetic potential 8. And Zµ +Wµ is the weak interaction potential Evidently neutrinos do not involve in the electromagnetic interactions. 8James Clerk Maxwell of Glenlair (13 June 1831 - 5 November 1879) was a Scottish physicist and mathe- matician. His most prominent achievement was formulating classical electromagnetic theory. 154
- 169. 3.1.4. Dimension of physical space Further I use Cayley-Dickson algebras [72, 73]: Let 1,i, j,k,E,I,J,K be basis elements of a 8-dimensional algebra Cayley (the octavians algebra) [72, 73]. A product of this algebra is defined the following way [72]: 1. for every basic element e: ee = −1; 2. If u1, u2, v1, v2 are real number then (u1 +u2i)(v1 +v2i) = (u1v1 −v2u2)+(v2u1 +u2v1)i. 3. If u1, u2, v1, v2 are numbers of shape w = w1 + w2i (ws, and s ∈ {1,2} are real numbers, and w = w1 −w2i) then (u1 +u2j)(v1 +v2j) = (u1v1 −v2u2)+(v2u1 +u2v1)j (3.44) and ij = k. 4. If u1, u2, v1, v2 are number of shape w = w1 +w2i+w3j+w4k (ws, and s ∈ {1,2,3,4} are real numbers, and w = w1 −w2i−w3j−w4k) then (u1 +u2E)(v1 +v2E) = (u1v1 −v2u2)+(v2u1 +u2v1)E (3.45) and iE = I, jE = J, kE = K. Therefore, in according with point 2.: the real numbers field (R) is extended to the complex numbers field (R), and in according with point 3.: the complex numbers field is expanded to the quaternions field (K), and point 4. expands the quaternions fields to the octavians field (O). This method of expanding of fields is called a Dickson doubling procedure [72]. If u = a+bi+cj+dk+AE+BI+CJ+K with real a,b,c,d,A,B,C,D then a real number u def = √ uu = a2 +b2 +c2 +d2 +A2 +B2 +C2 +D2 0.5 is called a norm of octavian u [72]. For each octavians u and v: uv = u v . (3.46) Algebras with this conditions are called normalized algebras [72, 73]. 155
- 170. Any 3+1-vector of a probability density can be represented by the following equations in matrix form (2.19) ρ = ϕ†ϕ, jk = ϕ†β[k]ϕ with k ∈ {1,2,3}. There β[k] are complex 2-diagonal 4 × 4-matrices of Clifford’s set of rank 4, and ϕ is matrix columns with four complex components. The light and colored pentads of Clifford’s set of such rank contain in threes 2-diagonal matrices, corresponding to 3 space coordinates in according with Dirac’s equation. Hence, a space of these events is 3-dimensional. Let ρ(t,x) be a probability density of event A(t,x), and ρc(t,x|t0,x0) be a probability density of event A(t,x) on condition that event B(t0,x0). In that case if function q(t,x|t0,x0) is fulfilled to condition: ρc(t,x|t0,x0) = q(t,x|t0,x0)ρ(t,x), (3.47) then one is called a disturbance function B to A. If q = 1 then B does not disturbance to A. A conditional probability density of event A(t,x) on condition that event B(t0,x0) is presented as: ρc = ϕ† cϕc like to a probability density of event A(t,x). Let ϕ = ϕ1,1 +iϕ1,2 ϕ2,1 +iϕ2,2 ϕ3,1 +iϕ3,2 ϕ4,1 +iϕ4,2 and ϕc = ϕc,1,1 +iϕc,1,2 ϕc,2,1 +iϕc,2,2 ϕc,3,1 +iϕc,3,2 ϕc,4,1 +iϕc,4,2 (all ϕr,s and ϕc,r,s are real numbers). In that case octavian u = ϕ1,1 +ϕ1,2i+ϕ2,1j+ϕ2,2k+ϕ3,1E+ϕ3,2I+ϕ4,1J+ϕ4,2K is called a Caylean of ϕ. Therefore, octavian uc = ϕc,1,1 +ϕc,1,2i+ϕc,2,1j+ϕc,2,2k+ϕc,3,1E+ϕc,3,2I+ϕc,4,1J+ϕc,4,2K 156
- 171. is Caylean of ϕc. In accordance with the octavian norm definition: uc 2 = ρc u 2 = ρ (3.48) Because the octavian algebra is a division algebra [72, 73] then for each octavians u and uc there exists an octavian w such that uc = wu, Because the octavians algebra is normalized then uc 2 = w 2 u 2 . Hence, from (3.47) and (3.48): q = w 2 . Therefore, in a 3+1-dimensional space-time there exists an octavian-Caylean for a dis- turbance function of any event to any event. In order to increase a space dimensionality the octavian algebra can be expanded by a Dickson doubling procedure: Another 8 elements should be added to basic octavians: z1,z2,z3,z4,z5,z6,z7,z8, such that: z2 = iz1, z3 = jz1, z4 = kz1, z5 = Ez1, z6 = Iz1, z7 = Jz1, z8 = Kz1, and for every octavians u1, u2, v1, v2: (u1 +u2z1)(v1 +v2z1) = (u1v1 −v2u2)+(v2u1 +u2v1)z1 (here: if w = w1 + w2i + w3j + w4k + w5E + w6I + w7J + w8K with real ws then w = w1 −w2i−w3j−w4k−w5E−w6I−w7J−w8K). It is a 16-dimensional Cayley-Dickson algebra. In according with [58]: for any natural number z there exists a Clifford set of rank 2z. In considering case for z = 3 there is Clifford’s seven: 157
- 172. β[1] = β[1] 04 04 −β[1] , β[2] = β[2] 04 04 −β[2] , β[3] = β[3] 04 04 −β[3] , β[4] = β[4] 04 04 −β[4] , (3.49) β[5] = γ[0] 04 04 −γ[0] , β[6] = 04 14 14 04 , β[7] = i 04 −14 14 04 , (3.50) Therefore, in this seven five 4-diagonal matrices (3.49) define a 5-dimensio-nal space of events, and two 4-antidiagonal matrices (3.50) defined a 2-dimensi-onal space for the electroweak transformations. It is evident that such procedure of dimensions building up can be continued endlessly. But in accordance with the Hurwitz theorem9 and with the generalized Frobenius theorem10 a more than 8-dimensional Cayley-Dickson algebra does not a division algebra. Hence, there in a more than 3-dimensional space exist events such that a disturbance function be- tween these events does not hold a Caylean. I call such disturbance supernatural. Therefore, supernatural disturbance do not exist in a 3-dimensional space, but in a more than 3-dimensional space such supernatural disturbance act. 3.2. Quarks and Gluons The quark model was independently proposed by physicists Murray Gell-Mann11 and George Zweig12 in 1964. The first direct experimental evidence of gluons was found in 1979 when three-jet events were observed at t he electron-positron collider PETRA. However, just before PE- TRA13 appeared on the scene, the PLUTO experiment at DORIS14 showed event topologies suggestive of a three-gluon decay. 9Every normalized algebra with unit is isomorphous to one of the following: the real numbers algebra R, the complex numbers algebra C, the quaternions algebra K, the octavians algebra O [72] 10A division algebra can be only either 1 or 2 or 4 or 8-dimensional [73] 11Murray Gell-Mann (born September 15, 1929) is an American physicist and linguist 12George Zweig (born on May 30, 1937 in Moscow, Russia into a Jewish family) was originally trained as a particle physicist under Richard Feynman and later turned his attention to neurobiology. He spent a number of years as a Research Scientist at Los Alamos National Laboratory and MIT, but as of 2004, has gone on to work in the financial services industry. 13PETRA (or the Positron-Electron Tandem Ring Accelerator) is one of the particle accelerators at DESY in Hamburg, Germany. 14DORIS (Doppel-Ring-Speicher, ”double-ring storage”), built between 1969 and 1974, was DESY’s second circular accelerator and its first storage ring with a circumference of nearly 300 m. 158
- 173. The following part of (2.29): 3 ∑ k=0 β[k] −i∂k +Θk +ϒkγ[5] − −Mζ,0γ [0] ζ +Mζ,4ζ[4] + −Mη,0γ [0] η −Mη,4η[4] + +Mθ,0γ [0] θ +Mθ,4θ[4] ϕ = 0. (3.51) is called the chromatic equation of moving. Here (2.5), (2.7), (2.9): γ [0] ζ = − 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 , ζ[4] = 0 0 0 i 0 0 i 0 0 −i 0 0 −i 0 0 0 are mass elements of red pentad; γ [0] η = 0 0 0 i 0 0 −i 0 0 i 0 0 −i 0 0 0 , η[4] = 0 0 0 1 0 0 −1 0 0 −1 0 0 1 0 0 0 are mass elements of green pentad; γ [0] θ = 0 0 −1 0 0 0 0 1 −1 0 0 0 0 1 0 0 , θ[4] = 0 0 −i 0 0 0 0 i −i 0 0 0 0 i 0 0 are mass elements of blue pentad. I call: • Mζ,0, Mζ,4 red lower and upper mass members; • Mη,0, Mη,4 green lower and upper mass members; • Mθ,0, Mθ,4 blue lower and upper mass members. The mass members of this equation form the following matrix sum: M := −Mζ,0γ [0] ζ +Mζ,4ζ[4] − −Mη,0γ [0] η −Mη,4η[4] + +Mθ,0γ [0] θ +Mθ,4θ[4] = = 0 0 −Mθ,0 Mζ,η,0 0 0 M∗ ζ,η,0 Mθ,0 −Mθ,0 Mζ,η,0 0 0 M∗ ζ,η,0 Mθ,0 0 0 +i 0 0 Mθ,4 M∗ ζ,η,4 0 0 Mζ,η,4 −Mθ,4 −Mθ,4 −M∗ ζ,η,4 0 0 −Mζ,η,4 Mθ,4 0 0 159
- 174. with Mζ,η,0 := Mζ,0 −iMη,0 and Mζ,η,4 := Mζ,4 −iMη,4. Elements of these matrices can be turned by formula of shape [52]: cos θ 2 isin θ 2 isin θ 2 cos θ 2 Z X −iY X +iY −Z cos θ 2 −isin θ 2 −isin θ 2 cos θ 2 = = Z cosθ−Y sinθ X −i Y cosθ +Z sinθ X +i Y cosθ +Z sinθ −Z cosθ+Y sinθ . Hence, if: U2,3 (α) := cosα isinα 0 0 isinα cosα 0 0 0 0 cosα isinα 0 0 isinα cosα and M := −Mζ,0γ [0] ζ +Mζ,4ζ[4]− −Mη,0γ [0] η −Mη,4η[4]+ +Mθ,0γ [0] θ +Mθ,4θ[4] := U† 2,3 (α)MU2,3 (α) then Mζ,0 = Mζ,0 , Mη,0 = Mη,0 cos2α+Mθ,0 sin2α, Mθ,0 = Mθ,0 cos2α−Mη,0 sin2α, Mζ,4 = Mζ,4 , Mη,4 = Mη,4 cos2α+Mθ,4 sin2α, Mθ,4 = Mθ,4 cos2α−Mη,4 sin2α. Therefore, matrix U2,3 (α) makes an oscillation between green and blue chromatics. Let us consider equation (2.29) under transformation U2,3 (α) where α is an arbitrary real function of time-space variables (α = α(t,x1,x2,x3)): U† 2,3 (α) 1 c ∂t +iΘ0 +iϒ0γ[5] U2,3 (α)ϕ = = U† 2,3 (α) 3 ∑ ν=1 β[ν] ∂ν +iΘν +iϒνγ[5] + +iM0γ[0] +iM4β[4] +M U2,3 (α)ϕ. 160
- 175. Because U† 2,3 (α)U2,3 (α) = 14 , U† 2,3 (α)γ[5]U2,3 (α) = γ[5] , U† 2,3 (α)γ[0]U2,3 (α) = γ[0] , U† 2,3 (α)β[4]U2,3 (α) = β[4] , U† 2,3 (α)β[1] = β[1]U† 2,3 (α) , U† 2,3 (α)β[2] = β[2] cos2α+β[3] sin2α U† 2,3 (α) , U† 2,3 (α)β[3] = β[3] cos2α−β[2] sin2α U† 2,3 (α) , then 1 c ∂t +U† 2,3 (α) 1 c ∂tU2,3 (α)+iΘ0 +iϒ0γ[5] ϕ = = β[1] ∂1 +U† 2,3 (α)∂1U2,3 (α)+ iΘ1 +iϒ1γ[5] + +β[2] (cos2α·∂2 −sin2α·∂3) +U† 2,3 (α)(cos2α·∂2 − sin2α·∂3)U2,3 (α) +i(Θ2 cos2α−Θ3 sin2α) +i ϒ2γ[5] cos2α−ϒ3γ[5] sin2α +β[3] (cos2α·∂3 +sin2α·∂2) +U† 2,3 (α)(cos2α·∂3 + sin2α·∂2)U2,3 (α) +i(Θ2 sin2α+Θ3 cos2α) +i ϒ3γ[5] cos2α+ϒ2γ[5] sin2α +iM0γ[0] +iM4β[4] +M ϕ. (3.52) Let x2 and x3 be elements of other coordinate system such that 2.15: ∂2 : = (cos2α·∂2 −sin2α·∂3), (3.53) ∂3 : = (cos2α·∂3 +sin2α·∂2). Therefore, from (3.52): 1 c ∂t +U† 2,3 (α) 1 c ∂tU2,3 (α)+iΘ0 +iϒ0γ[5] ϕ = = β[1] ∂1 +U† 2,3 (α)∂1U2,3 (α)+iΘ1 +iϒ1γ[5] +β[2] ∂2 +U† 2,3 (α)∂2U2,3 (α)+iΘ2 +iϒ2γ[5] +β[3] ∂3 +U† 2,3 (α)∂3U2,3 (α)+iΘ3 +iϒ3γ[5] +iM0γ[0] +iM4β[4] +M ϕ. 161
- 176. with Θ2 := Θ2 cos2α−Θ3 sin2α, Θ3 := Θ2 sin2α+Θ3 cos2α, ϒ2 := ϒ2 cos2α−ϒ3 sin2α, ϒ3 := ϒ3 cos2α+ϒ2 sin2α. Therefore, the oscillation between blue and green chromatics curves the space in the x2, x3 directions. Similarly, matrix U1,3 (ϑ) := cosϑ sinϑ 0 0 −sinϑ cosϑ 0 0 0 0 cosϑ sinϑ 0 0 −sinϑ cosϑ with an arbitrary real function ϑ(t,x1,x2,x3) describes the oscillation between blue and red chromatics which curves the space in the x1, x3 directions. And matrix U1,2 (ς) := e−iς 0 0 0 0 eiς 0 0 0 0 e−iς 0 0 0 0 eiς with an arbitrary real function ς(t,x1,x2,x3) describes the oscillation between green and red chromatics which curves the space in the x1, x2 directions. Now, let U0,1 (σ) := coshσ −sinhσ 0 0 −sinhσ coshσ 0 0 0 0 coshσ sinhσ 0 0 sinhσ coshσ . and M := −Mζ,0γ [0] ζ +Mζ,4ζ[4]− −Mη,0γ [0] η −Mη,4η[4]+ +Mθ,0γ [0] θ +Mθ,4θ[4] := U† 0,1 (σ)MU0,1 (σ) then: Mζ,0 = Mζ,0 , Mη,0 = (Mη,0 cosh2σ−Mθ,4 sinh2σ) , Mθ,0 = Mθ,0 cosh2σ+Mη,4 sinh2σ, Mζ,4 = Mζ,4 , Mη,4 = Mη,4 cosh2σ+Mθ,0 sinh2σ, Mθ,4 = Mθ,4 cosh2σ−Mη,0 sinh2σ. Therefore, matrix U0,1 (σ) makes an oscillation between green and blue chromatics with an oscillation between upper and lower mass members. 162
- 177. Let us consider equation (2.29) under transformation U0,1 (σ) where σ is an arbitrary real function of time-space variables (σ = σ(t,x1,x2,x3)): U† 0,1 (σ) 1 c ∂t +iΘ0 +iϒ0γ[5] U0,1 (σ)ϕ = = U† 0,1 (σ) 3 ∑ ν=1 β[ν] ∂ν +iΘν +iϒνγ[5] + +iM0γ[0] +iM4β[4] +M U0,1 (σ)ϕ. Since: U† 0,1 (σ)U0,1 (σ) = cosh2σ−β[1] sinh2σ , U† 0,1 (σ) = cosh2σ+β[1] sinh2σ U−1 0,1 (σ), U† 0,1 (σ)β[1] = β[1] cosh2σ−sinh2σ U−1 0,1 (σ), U† 0,1 (σ)β[2] = β[2] U−1 0,1 (σ), U† 0,1 (σ)β[3] = β[3] U−1 0,1 (σ), U† 0,1 (σ)γ[0] U0,1 (σ) = γ[0] , U† 0,1 (σ)β[4] U0,1 (σ) = β[4] , U−1 0,1 (σ)U0,1 (σ) = 14 , U−1 0,1 (σ)γ[5] U0,1 (σ) = γ[5] , U† 0,1 (σ)γ[5] U0,1 (σ) = γ[5] cosh2σ−β[1] sinh2σ , then U−1 0,1 (σ) cosh2σ· 1 c ∂t +sinh2σ·∂1 U0,1 (σ) + cosh2σ· 1 c ∂t +sinh2σ·∂1 +i(Θ0 cosh2σ+Θ1 sinh2σ) +i(ϒ0 cosh2σ+sinh2σ·ϒ1)γ[5]− −β[1] U−1 0,1 (σ) cosh2σ·∂1 +sinh2σ· 1 c ∂t U0,1 (σ) + cosh2σ·∂1 +sinh2σ· 1 c ∂t +i(Θ1 cosh2σ+Θ0 sinh2σ) +i(ϒ1 cosh2σ+ϒ0 sinh2σ)γ[5] −β[2] ∂2 +U−1 0,1 (σ)(∂2U0,1 (σ))+ iΘ2 +iϒ2γ[5] −β[3] ∂3 +U−1 0,1 (σ)(∂3U0,1 (σ))+ iΘ3 +iϒ3γ[5] −iM0γ[0] −iM4β[4] −M ϕ = 0. (3.54) 163
- 178. Let t and x1 be elements of other coordinate system such that: ∂x1 ∂x1 = cosh2σ ∂t ∂x1 = 1 c sinh2σ ∂x1 ∂t = csinh2σ ∂t ∂t = cosh2σ ∂x2 ∂t = ∂x3 ∂t = ∂x2 ∂x1 = ∂x3 ∂x1 = 0 . (3.55) Hence: ∂t := ∂ ∂t = ∂ ∂t ∂t ∂t + ∂ ∂x1 ∂x1 ∂t + ∂ ∂x2 ∂x2 ∂t + ∂ ∂x3 ∂x3 ∂t = = cosh2σ· ∂ ∂t +csinh2σ· ∂ ∂x1 = = cosh2σ·∂t +csinh2σ·∂1 , that is 1 c ∂t = 1 c cosh2σ·∂t +sinh2σ·∂1 and ∂1 := ∂ ∂x1 = = ∂ ∂t ∂t ∂x1 + ∂ ∂x1 ∂x1 ∂x1 + ∂ ∂x2 ∂x2 ∂x1 + ∂ ∂x3 ∂x3 ∂x1 = = cosh2σ· ∂ ∂x1 +sinh2σ· 1 c ∂ ∂t = = cosh2σ·∂1 +sinh2σ· 1 c ∂t . Therefore, from (3.54): β[0] 1 c ∂t +U−1 0,1 (σ) 1 c ∂tU0,1 (σ)+ iΘ0 +iϒ0γ[5] +β[1] ∂1 +U−1 0,1 (σ)∂1U0,1 (σ)+ iΘ1 +iϒ1γ[5] +β[2] ∂2 +U−1 0,1 (σ)∂2U0,1 (σ)+ iΘ2 +iϒ2γ[5] +β[3] ∂3 +U−1 0,1 (σ)∂3U0,1 (σ)+ iΘ3 +iϒ3γ[5] +iM0γ[0] +iM4β[4] +M ϕ = 0 164
- 179. with Θ0 := Θ0 cosh2σ+Θ1 sinh2σ, Θ1 := Θ1 cosh2σ+Θ0 sinh2σ, ϒ0 := ϒ0 cosh2σ+sinh2σ·ϒ1 , ϒ1 := ϒ1 cosh2σ+ϒ0 sinh2σ. Therefore, the oscillation between blue and green chromatics with the oscillation be- tween upper and lower mass members curves the space in the t, x1 directions. Similarly, matrix U0,2 (φ) := coshφ isinhφ 0 0 −isinhφ coshφ 0 0 0 0 coshφ −isinhφ 0 0 isinhφ coshφ with an arbitrary real function φ(t,x1,x2,x3) describes the oscillation between blue and red chromatics with the oscillation between upper and lower mass members curves the space in the t, x2 directions. And matrix U0,3 (ι) := eι 0 0 0 0 e−ι 0 0 0 0 e−ι 0 0 0 0 eι with an arbitrary real function ι(t,x1,x2,x3) describes the oscillation between green and red chromatics with the oscillation between upper and lower mass members curves the space in the t, x3 directions. Now let U (χ) := eiχ 0 0 0 0 eiχ 0 0 0 0 e2iχ 0 0 0 0 e2iχ and M := −Mζ,0γ [0] ζ +Mζ,4ζ[4] − −Mη,0γ [0] η −Mη,4η[4] + +Mθ,0γ [0] θ +Mθ,4θ[4] := U† (χ)MU (χ) then: Mζ,0 = Mζ,0 cosχ−Mζ,4 sinχ , Mζ,4 = Mζ,4 cosχ+Mζ,0 sinχ , Mη,4 = (Mη,4 cosχ−Mη,0 sinχ), Mη,0 = (Mη,0 cosχ+Mη,4 sinχ), Mθ,0 = (Mθ,0 cosχ+Mθ,4 sinχ), Mθ,4 = (Mθ,4 cosχ−Mθ,0 sinχ). 165
- 180. Therefore, matrix U (χ) makes an oscillation between upper and lower mass members. Let us consider equation (3.51) under transformation U (χ) where χ is an arbitrary real function of time-space variables (χ = χ(t,x1,x2,x3)): U† (χ) 1 c ∂t +iΘ0 +iϒ0γ[5] U (χ)ϕ = = U† (χ) 3 ∑ ν=1 β[ν] ∂ν +iΘν +iϒνγ[5] +M U (χ)ϕ. Because γ[5]U (χ) = U (χ)γ[5] , β[1]U (χ) = U (χ)β[1] , β[2]U (χ) = U (χ)β[2] , β[3]U (χ) = U (χ)β[3] , U† (χ)U (χ) = 14 , then 1 c ∂t + 1 c U† (χ) ∂tU (χ) +iΘ0 +iϒ0γ[5] ϕ = = 3 ∑ ν=1 β[ν] ∂ν +U† (χ) ∂νU (χ) + iΘν +iϒνγ[5] +U† (χ)MU (χ) ϕ. Now let: U (κ) := eκ 0 0 0 0 eκ 0 0 0 0 e2κ 0 0 0 0 e2κ and M := −Mζ,0γ [0] ζ +Mζ,4ζ[4]− −Mη,0γ [0] η −Mη,4η[4]+ +Mθ,0γ [0] θ +Mθ,4θ[4] := U−1 (κ)MU (κ) then: Mθ,0 = (Mθ,0 coshκ−iMθ,4 sinhκ), Mθ,4 = (Mθ,4 coshκ+iMθ,0 sinhκ), Mη,0 = (Mη,0 coshκ−iMη,4 sinhκ), Mη,4 = (Mη,4 coshκ+iMη,0 sinhκ), Mζ,0 = Mζ,0 coshκ+iMζ,4 sinhκ , Mζ,4 = Mζ,4 coshκ−iMζ,0 sinhκ . 166
- 181. Therefore, matrix U (κ) makes an oscillation between upper and lower mass members, too. Let us consider equation (3.51) under transformation U (κ) where κ is an arbitrary real function of time-space variables (κ = κ(t,x1,x2,x3)): U−1 (κ) 1 c ∂t +iΘ0 +iϒ0γ[5] U (κ)ϕ = = U−1 (κ) 3 ∑ ν=1 β[ν] ∂ν +iΘν +iϒνγ[5] + M U (κ)ϕ. Because γ[5]U (κ) = U (κ)γ[5] , U−1 (κ)β[1] = β[1]U−1 (κ) , U−1 (κ)β[2] = β[2]U−1 (κ), U−1 (κ)β[3] = β[3]U−1 (κ), U−1 (κ)U (κ) = 14 , then 1 c ∂t +U−1 (κ) 1 c ∂tU (κ) +iΘ0 +iϒ0γ[5] ϕ = = 3 ∑ ν=1 β[ν] ∂ν +U−1 (κ) ∂νU (κ) + iΘν +iϒνγ[5] + +U−1 (κ)MU (κ) ϕ. If denote: Λ1 := 0 −1 0 0 −1 0 0 0 0 0 0 1 0 0 1 0 , Λ2 := 0 i 0 0 i 0 0 0 0 0 0 i 0 0 i 0 , Λ3 := 0 1 0 0 −1 0 0 0 0 0 0 1 0 0 −1 0 , Λ4 := 0 i 0 0 −i 0 0 0 0 0 0 −i 0 0 i 0 , 167
- 182. Λ5 := −i 0 0 0 0 i 0 0 0 0 −i 0 0 0 0 i , Λ6 := 1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 1 , Λ7 := 1 0 0 0 0 1 0 0 0 0 2 0 0 0 0 2 , Λ8 := i 0 0 0 0 i 0 0 0 0 2i 0 0 0 0 2i , then U−1 0,1 (σ) ∂sU0,1 (σ) = Λ1∂sσ, U−1 2,3 (α) ∂sU2,3 (α) = Λ2∂sα, U−1 1,3 (ϑ) ∂sU1,3 (ϑ) = Λ3∂sϑ, U−1 0,2 (φ) ∂sU0,2 (φ) = Λ4∂sφ, U−1 1,2 (ς) ∂sU1,2 (ς) = Λ5∂sς, U−1 0,3 (ι) ∂sU0,3 (ι) = Λ6∂sι, U−1 (κ) ∂sU (κ) = Λ7∂sκ, U−1 (χ) ∂sU (χ) = Λ8∂sχ. Let `U be the following set: `U := U0,1,U2,3,U1,3,U0,2,U1,2,U0,3,U,U . Because U−1 2,3 (α)Λ1U2,3 (α) = Λ1 U−1 1,3 (ϑ)Λ1U1,3 (ϑ) = (Λ1 cos2ϑ+Λ6 sin2ϑ) U−1 0,2 (φ)Λ1U0,2 (φ) = (Λ1 cosh2φ−Λ5 sinh2φ) U−1 1,2 (ς)Λ1U1,2 (ς) = Λ1 cos2ς−Λ4 sin2ς U−1 0,3 (ι)Λ1U0,3 (ι) = Λ1 cosh2ι+Λ3 sinh2ι U−1 (κ)Λ1U (κ) = Λ1 168
- 183. U−1 (χ)Λ1U (χ) = Λ1 ======== U−1 (χ)Λ2U (χ) = Λ2 U−1 (κ)Λ2U (κ) = Λ2 U−1 0,3 (ι)Λ2U0,3 (ι) = Λ2 cosh2ι−Λ4 sinh2ι U−1 1,2 (ς)Λ2U1,2 (ς) = Λ2 cos2ς−Λ3 sin2ς U−1 0,2 (φ)Λ2U0,2 (φ) = Λ2 cosh2φ+Λ6 sinh2φ U−1 1,3 (ϑ)Λ2U1,3 (ϑ) = Λ2 cos2ϑ+Λ5 sin2ϑ U−1 0,1 (σ)Λ2U0,1 (σ) = Λ2 ========= U−1 0,1 (σ) Λ3U0,1 (σ) = Λ3 cosh2σ−Λ6 sinh2σ U−1 2,3 (α) Λ3U2,3 (α) = Λ3 cos2α−Λ5 sin2α U−1 0,2 (φ) Λ3U0,2 (φ) = Λ3 U−1 1,2 (ς) Λ3U1,2 (ς) = Λ3 cos2ς+Λ2 sin2ς U−1 0,3 (ι) Λ3U0,3 (ι) = Λ3 cosh2ι+Λ1 sinh2ι U−1 (κ) Λ3U (κ) = Λ3 U−1 (χ) Λ3U (χ) = Λ3 ========== U−1 (χ) Λ4U (χ) = Λ4 U−1 (κ) Λ4U (κ) = Λ4 U−1 0,3 (ι) Λ4U0,3 (ι) = Λ4 cosh2ι−Λ2 sinh2ι U−1 1,2 (ς) Λ4U1,2 (ς) = Λ4 cos2ς+Λ1 sin2ς U−1 1,3 (ϑ) Λ4U1,3 (ϑ) = Λ4 U−1 2,3 (α) Λ4U2,3 (α) = Λ4 cos2α−Λ6 sin2α U−1 0,1 (σ) Λ4U0,1 (σ) = Λ4 cosh2σ+Λ5 sinh2σ ========== U−1 0,1 (σ) Λ5U0,1 (σ) = Λ5 cosh2σ+Λ4 sinh2σ U−1 2,3 (α) Λ5U2,3 (α) = Λ5 cos2α+Λ3 sin2α U−1 1,3 (ϑ) Λ5U1,3 (ϑ) = (Λ5 cos2ϑ−Λ2 sin2ϑ) U−1 0,2 (φ) Λ5U0,2 (φ) = Λ5 cosh2φ−Λ1 sinh2φ U−1 0,3 (ι) Λ5U0,3 (ι) = Λ5 U−1 (κ) Λ5U (κ) = Λ5 U−1 (χ)Λ5U (χ) = Λ5 =========== U−1 (χ)Λ6U (χ) = Λ6 169
- 184. U−1 (κ)Λ6U (κ) = Λ6 U−1 1,2 (ς)Λ6U1,2 (ς) = Λ6 U−1 0,2 (φ)Λ6U0,2 (φ) = Λ6 cosh2φ+Λ2 sinh2φ U−1 1,3 (ϑ)Λ6U1,3 (ϑ) = Λ6 cos2ϑ−Λ1 sin2ϑ U−1 2,3 (α)Λ6U2,3 (α) = Λ6 cos2α+Λ4 sin2α U−1 0,1 (σ)Λ6U0,1 (σ) = Λ6 cosh2σ−Λ3 sinh2σ ======== U−1 (χ) Λ7U (χ) = Λ7 U−1 0,3 (ι) Λ7U0,3 (ι) = Λ7 U−1 1,2 (ς) Λ7U1,2 (ς) = Λ7 U−1 0,2 (φ) Λ7U0,2 (φ) = Λ7 U−1 1,3 (ϑ) Λ7U1,3 (ϑ) = Λ7 U−1 2,3 (α) Λ7U2,3 (σ) = Λ7 U−1 0,1 (σ) Λ7U0,1 (σ) = Λ7 ========= U−1 0,1 (σ) Λ8U0,1 (σ) = Λ8 U−1 2,3 (α) Λ8U2,3 (α) = Λ8 U−1 1,3 (ϑ) Λ8U1,3 (ϑ) = Λ8 U−1 0,2 (φ) Λ8U0,2 (φ) = Λ8 U−1 1,2 (ς) Λ8U1,2 (ς) = Λ8 U−1 0,3 (ι) Λ8U0,3 (ι) = Λ8 U−1 (κ) Λ8U (κ) = Λ8 then for every product U of `U’s elements real functions Gr s (t,x1,x2,x3) exist such that U−1 (∂sU) = g3 2 8 ∑ r=1 ΛrGr s with some real constant g3 (similar to 8 gluons). The chrome states equations of moving are the following [71, p.86], (3.43): ∑3 µ=0 −ζ[k] ∂µ + −γ [0] ζ ∂ ζ y +ζ[4]∂ ζ z +∑3 µ=0 β[µ] −sAµ +0.5 Zµ +Wµ ξ = 0, (3.56) ∑3 µ=0 −η[k] ∂µ + −γ [0] η ∂η y −η[4]∂η z +∑3 µ=0 β[µ] −sAµ +0.5 Zµ +Wµ ξ = 0, ∑3 µ=0 −θ[k] ∂µ + γ [0] θ ∂θ y +θ[4]∂θ z +∑3 µ=0 β[µ] −sAµ +0.5 Zµ +Wµ ξ = 0; 170
- 185. here: ξ := dL dR uL uR ( dL dR is a lower chrome state, and uL uR is a upper chrome state); s is an electric charge, and [71, p.145] Aµ := Aµ 02 02 02 02 02 12 02 02 02 02 12 02 02 02 02 12 . Hence, sAµξ = sAµ 02 02 02 02 02 12 02 02 02 02 12 02 02 02 02 12 dL dR uL uR = sAµ −→ 0 2 dR uL uR , −→ 0 2 : = 0 0 . Sum of the equations (3.56) is the following: ∑3 µ=0 − ζ[k] +η[k] +θ[k] ∂µ+ + −γ [0] ζ ∂ ζ y +ζ[4]∂ ζ z −γ [0] η ∂η y −η[4]∂η z +γ [0] θ ∂θ y +θ[4]∂θ z + ∑3 µ=0 β[µ] (−3s)Aµ +0.5·3∑3 µ=0 β[µ] Zµ +Wµ ξ = 0. Because β[k] = − ζ[k] +η[k] +θ[k] rhen ∑3 µ=0 β[k]∂µ+ + −γ [0] ζ ∂ ζ y +ζ[4]∂ ζ z −γ [0] η ∂η y −η[4]∂η z +γ [0] θ ∂θ y +θ[4]∂θ z + ∑3 µ=0 β[µ] (−3s)Aµ +0.5·3∑3 µ=0 β[µ] Zµ +Wµ ξ = 0. Hence, from (3.42): (−3s) = g1g2 g2 1 +g2 2 171
- 186. that is: s = − 1 3 g1g2 g2 1 +g2 2 = − 1 3 e (e is a lepton electric charge). Because sAµξ = − 1 3 eAµ −→ 0 2 dR uL uR then a charge of uL uR is −2 3 e and a module of charge of −→ 0 2 dR is 1 3 e. 3.3. Asymptotic Freedom, Confinement, Gravitation The Quarks Asymptotic Freedom phenomenon and the Quarks Confinement phenomenon has been was discovered by J. Friedman15, H. Kendall16, R. Taylor17 at SLAC in the late 1960s and early 1970s. Researches of the phenomenon of gravitation were spent by Galileo Galilei18 in the late 16th and early 17th centuries, by Isaac Newton19 in 17th centuries, by A. Einstein20 in 20th centuries. From (3.55): ∂t ∂t = cosh2σ, (3.57) ∂x ∂t = csinh2σ. Hence, if v is the velocity of a coordinate system {t ,x } in the coordinate system {t,x} then sinh2σ = v c 1− v c 2 , cosh2σ = 1 1− v c 2 . Therefore, 15Jerome Isaac Friedman (born March 28, 1930) is an American physicist. 16Henry Way Kendall (December 9, 1926 – February 15, 1999) was an American particle physicist 17Richard Edward Taylor (born November 2, 1929 in Medicine Hat, Alberta) is a Canadian-American pro- fessor (Emeritus) at Stanford University. 18Galileo Galilei ( 15 February 1564[4] – 8 January 1642), was an Italian physicist, mathematician, as- tronomer, and philosopher 19Sir Isaac Newton PRS (25 December 1642 – 20 March 1727 was an English physicist, mathematician, astronomer, natural philosopher, alchemist, and theologian 20Albert Einstein ( 14 March 1879 – 18 April 1955) was a German-born theoretical physicist 172
- 187. v = ctanh2σ. (3.58) Let 2σ := ω(x) t x with ω(x) = λ |x| , where λ is a real constant with positive numerical value. In that case v(t,x) = ctanh λ |x| t x . (3.59) and if g is an acceleration of system {t ,x1} as respects to system {t,x1} then g(t,x1) = ∂v ∂t = cω(x1) cosh2 ω(x1) t x1 x1 . Figure 28: Figure 28 shows the dependency of a system {t ,x1} velocity v(t,x1) on x1 in system {t,x1}. This velocity in point A is not equal to one in point B. Hence, an oscillator, placed in B, has a nonzero velocity in respect to an observer, placed in point A. Therefore, from the Lorentz transformations, this oscillator frequency for observer, placed in point A, is less than own frequency of this oscillator (red shift). Figure 29 shows a dependency of a system {t ,x1} acceleration g(t,x1) on x1 in system {t,x1}. If an object immovable in system {t,x1} is placed in point K then in system {t ,x1} this object must move to the left with acceleration g and g λ x2 1 . I call: 173
- 188. Figure 29: • interval from S to ∞ the Newton Gravity Zone, • interval from B to C the the Zone, • and interval from C to D the Confinement Force Zone. 3.3.1. Dark Energy In 1998 observations of Type Ia supernovae suggested that the expansion of the universe is accelerating [61]. In the past few years, these observations have been corroborated by several independent sources [62]. This expansion is defined by the Hubble21 rule [63]: V (r) = Hr, (3.60) here V (r) is the velocity of expansion on the distance r, H is the Hubble’s constant (H ≈ 2.3×10−18c−1 [64]). Let a black hole be placed in a point O. Then a tremendous number of quarks oscillate in this point. These oscillations bend time-space and if t has some fixed volume, x > 0, and Λ := λt then v(x) = ctanh Λ x2 . (3.61) A dependency of v(x) (light years/c) from x (light years) with Λ = 741.907 is shown in Figure 30. Let a placed in a point A observer be stationary in the coordinate system {t,x}. Hence, in the coordinate system {t ,x } this observer is flying to the left to the point O with velocity −v(xA). And point X is flying to the left to the point O with velocity −v(x). 21Edwin Powell Hubble (November 20, 1889 September 28, 1953)[1] was an American astronomer 174
- 189. Figure 30: Dependence of v (light year/c) on x (light year) with Λ = 741.907 Figure 31: Dependence of VA (r) on r with xA = 25×103 l.y. 175
- 190. Figure 32: Dependence of H on r Consequently, the observer A sees that the point X flies away from him to the right with velocity VA (x) = ctanh Λ x2 A − Λ x2 (3.62) in accordance with the relativistic rule of addition of velocities. Let r := x−xA (i.e. r is distance from A to X), and VA (r) := ctanh Λ x2 A − Λ (xA +r)2 . (3.63) In that case Figure 31 demonstrates the dependence of VA (r) on r with xA = 25 × 103 l.y. Hence, X runs from A with almost constant acceleration: VA (r) r = H. (3.64) Figure 32 demonstrates the dependence of H on r. (the Hubble constant.). Therefore, the phenomenon of the accelerated expansion of Universe is explained by oscillations of chromatic states. 3.3.2. Dark Matter ”In 1933, the astronomer Fritz Zwicky22 was studying the motions of distant galaxies. Zwicky estimated the total mass of a group of galaxies by measuring their brightness. When he used a different method to compute the mass of the same cluster of galaxies, he came up 22Fritz Zwicky (February 14, 1898 – February 8, 1974) was a Swiss astronomer. 176
- 191. with a number that was 400 times his original estimate. This discrepancy in the observed and computed masses is now known as ”the missing mass problem.” Nobody did much with Zwicky’s finding until the 1970’s, when scientists began to realize that only large amounts of hidden mass could explain many of their observations. Scientists also realize that the existence of some unseen mass would also support theories regarding the structure of the universe. Today, scientists are searching for the mysterious dark matter not only to explain the gravitational motions of galaxies, but also to validate current theories about the origin and the fate of the universe” [65] (Figure 33 [66], Figure 34 [67]). Figure 33: A rotation curve for a typical spiral galaxy. The solid line shows actual mea- surements (Hawley and Holcomb., 1998, p. 390) [66] Some oscillations of chromatic states bend space-time as follows (2.15): ∂ ∂x = cos2α· ∂ ∂x −sin2α· ∂ ∂y , (3.65) ∂ ∂y = cos2α· ∂ ∂y +sin2α· ∂ ∂x . Let z : = x+iy,z = reiθ . z : = x +iy . Because linear velocity of the curved coordinate system x ,y into the initial system x,y is the following23: v = • x 2 + • y 2 23 • x := ∂x ∂t , • y := ∂y ∂t . 177
- 192. Figure 34: Rotation curve of NGC 6503. The dotted, dashed and dash-dotted lines are the contributions of gas, disk and dark matter, respectively. Figure 35: For t = 10000, θ = 13π/14: 178
- 193. Figure 36: For t = 10000, θ = 0.98π: then in thic case: v = • z . Let function z be a holomorphic function. Hence, in accordance with the Cauchy- Riemann conditions the following equations are fulfilled: ∂x ∂x = ∂y ∂y , ∂x ∂y = − ∂y ∂x . Therefore, in accordance with (3.65): dz = e−i(2α) dz where 2α is an holomorphic function, too. For example, let 2α := 1 i ((x+y)+i(y−x))2 . In this case: z = exp ((x+y)+i(y−x))2 t dx+i exp ((x+y)+i(y−x))2 t dy. 179
- 194. Let k := y/x. Hence, z = exp ((x+kx)+i(kx−x))2 t dx+i exp y k +y +i y− y k 2 t dy. Calculate: exp ((x+kx)+i(kx−x))2 t dx = 1 2 √ π erf x −1 t (2ik2 +4k −2i) −1 t (2ik2 +4k −2i) , i exp y k +y +i y− y k 2 t dy = 1 2 i √ π erf y − 1 k2t (2ik2 +4k −2i) − 1 k2t (2ik2 +4k −2i) . Calculate: ∂z ∂t = 1 −8 √ ti(k −i)3 √ −2i −4y(k −i)2 −1 t 2i(k −i)2 exp 1 k2t y22i(k −i)2 +4ikx(k −i)2 − 1 k2t 2i(k −i)2 exp 1 t x22i(k −i)2 +i √ πk2t2i(k −i)2 1 t 1 k2 erf y − 1 k2t 2i(k −i)2 + √ πkt2i(k −i)2 1 t2 1 k2 erf x −1 t 2i(k −i)2 . For large t: ∂z ∂t ≈ 1 −8 √ ti(k −i)3 √ −2i i √ πk2 t2i(k −i)2 1 t 1 k2 erf y − 1 k2t 2i(k −i)2 Hence, v ≈ 1 8 (1−i)k √ π 1 k −i erf x − 1 t 2i(k −i)2 . Because k = tanθ, x = rcosθ then v ≈ 1 8 (1−i)(tanθ) √ π 1 tanθ−i erf r(cosθ) − 1 t 2i((tanθ)−i)2 . Figure 35 shows the dependence of velocity v on the radius r at large t ∼ 104 and at θ = 13π/14 180
- 195. Figure 36 shows the dependence of velocity v on the radius r at large t ∼ 104 and at θ = 0.98π Hence, Dark Matter and Dark Energy can be mirages in the space-time, which is curved by oscillations of chromatic states. 3.3.3. Baryon Chrome According to the quark model, [70] the properties of hadrons are primarily determined by their so-called valence quarks. For example, a proton is composed of two up quarks and one down quark. Although quarks also carry color charge, hadrons must have zero total color charge because of a phenomenon called color Confinement. That is, hadrons must be “colorless” or “white”. These are the simplest of the two ways: three quarks of different colors, or a quark of one color and an antiquark carrying the corresponding anticolor. Hadrons with the first arrangement are called baryons, and those with the second arrangement are mesons. Let α be any real number and x0 := x0, x1 := x1 cos(α)−x2 sin(α); x2 := x1 sin(α)+x2 cos(α); (3.66) x3 := x3; Since jA is a 3+1-vector then from [71, p.59]: jA,0 = −ϕ† β[0] ϕ, jA,1 = −ϕ† β[1] cos(α)−β[2] sin(α) ϕ; (3.67) jA,2 = −ϕ† β[1] sin(α)+β[2] cos(α) ϕ; jA,3 = −ϕ† β[3] ϕ. Hence if for ϕ : jA,0 = −ϕ †β[0]ϕ , jA,1 = −ϕ †β[1]ϕ ; jA,2 = −ϕ †β[2]ϕ ; jA,3 = −ϕ †β[3]ϕ , and ϕ := U1,2 (α)ϕ then U† 1,2 (α)β[0] U1,2 (α) = β[0] , U† 1,2 (α)β[1] U1,2 (α) = β[1] cosα−β[2] sinα; U† 1,2 (α)β[2] U1,2 (α) = β[2] cosα+β[1] sinα; (3.68) U† 1,2 (α)β[3] U1,2 (α) = β[3] ; 181
- 196. from [71, p.62]: because ρA = ϕ† ϕ = ϕ † ϕ , then U† 1,2 (α)U1,2 (α) = 14. (3.69) If U1,2 (α) := cos α 2 ·14 −sin α 2 ·β[1] β[2] i.e. 2.14: U1,2 (α) = e−i1 2 α 0 0 0 0 ei1 2 α 0 0 0 0 e−i1 2 α 0 0 0 0 ei1 2 α (3.70) then U1,2 (α) fulfils to all these conditions (3.68), (3.69). Then let x0 := x0, x1 := x1 cos(α)−x3 sin(α), x2 := x2, (3.71) x3 := x1 sin(α)+x3 cos(α). Let U1,3 (α) := cos α 2 ·14 −sin α 2 ·β[1] β[3] . In rhis case 2.13: U1,3 (α) = cos 1 2α sin 1 2α 0 0 −sin 1 2 α cos 1 2 α 0 0 0 0 cos 1 2 α sin 1 2 α 0 0 −sin 1 2α cos 1 2α (3.72) and U† 1,3 (α)β[0] U1,3 (α) = β[0] , U† 1,3 (α)β[1] U1,3 (α) = β[1] cosα−β[3] sinα, (3.73) U† 1,3 (α)β[2] U1,3 (α) = β[2] , U† 1,3 (α)β[3] U1,3 (α) = β[3] cosα+β[1] sinα. If ϕ := U1,3 (α)ϕ and 182
- 197. jA,k := ϕ † β[k] ϕ where (k ∈ {0,1,2,3}) then jA,0 = jA,0, (3.74) jA,1 = jA,1 cosα− jA,3 sinα, (3.75) jA,2 = jA,2, jA,3 = jA,3 cosα+ jA,1 sinα, Then let x0 : = x0, x1 : = x1, x2 = cosα·x2 +sinα·x3, (3.76) x3 = cosα·x3 −sinα·x2. Let U3,2 (α) = cos α 2 ·14 −sin α 2 ·β[3] β[2] In this case: U3,2 (α) = cos 1 2α isin 1 2 α 0 0 isin 1 2α cos 1 2 α 0 0 0 0 cos 1 2 α isin 1 2 α 0 0 isin 1 2 α cos 1 2 α , (3.77) and U† 3,2 (α)β[0] U3,2 (α) = β[0] , U† 3,2 (α)β[1] U3,2 (α) = β[1] , U† 3,2 (α)β[0] U3,2 (α) = β[0] cosα+β[3] sinα, (3.78) U† 3,2 (α)β[3] U3,2 (α) = β[3] cosα−β[2] sinα If ϕ := U3,2 (α)ϕ and jA,k := ϕ † β[k] ϕ where (k ∈ {0,1,2,3}) then jA,0 = jA,0, jA,1 = jA,1, (3.79) jA,2 = jA,2 cosα+ jA,3 sinα, jA,3 = jA,3 cosα− jA,1 sinα, 183
- 198. Let v be any real number such that −1 < v < 1. And let: α := 1 2 ln 1−v 1+v . In this case: coshα = 1 √ 1−v2 , sinhα = − v √ 1−v2 . (3.80) Let x0 : = x0 coshα−x1 sinhα, (3.81) x1 : = x1 coshα−x0 sinhα, x2 : = x2, x3 : = x3. Let U1,0 (α) = cosh α 2 ·14 −sinh α 2 ·β[1] β[0] . That is 2.10: U1,0 (α) := cosh 1 2α sinh 1 2α 0 0 sinh 1 2 α cosh 1 2 α 0 0 0 0 cosh 1 2α −sinh 1 2 α 0 0 −sinh 1 2 α cosh 1 2 α . (3.82) In rhis case: U† 1,0 (α)β[0] U1,0 (α) = β[0] coshα−β[1] sinhα, (3.83) U† 1,0 (α)β[1] U1,0 (α) = β[1] coshα−β[0] sinhα, U† 1,0 (α)β[2] U1,0 (α) = β[2] , U† 1,0 (α)β[3] U1,0 (α) = β[3] . If ϕ := U1,0 (α)ϕ and jA,k := ϕ † β[k] ϕ 184
- 199. where (k ∈ {0,1,2,3}) then jA,0 = jA,0 coshα− jA,1 sinhα, (3.84) jA,1 = jA,1 coshα− jA,0 sinhα, jA,2 = jA,2, jA,3 = jA,3. Then let x0 : = x0 coshα−x2 sinhα, (3.85) x1 : = x1, x2 : = x2 coshα−x0 sinhα, x3 : = x3. Let U2,0 (α) := cosh α 2 ·14 −sinh α 2 ·β[2] β[0] . (3.86) That is: U2,0 (α) = cosh 1 2α −isinh 1 2 α 0 0 isinh 1 2 α cosh 1 2 α 0 0 0 0 cosh 1 2α isinh 1 2α 0 0 −isinh 1 2α cosh 1 2 α . In rhis case: U† 2,0 (α)β[0] U2,0 (α) = β[0] coshα−β[2] sinhα, (3.87) U† 2,0 (α)β[1] U1,0 (α) = β[1] , U† 2,0 (α)β[2] U1,0 (α) = β[2] coshα−β[0] sinhα, U† 2,0 (α)β[3] U2,0 (α) = β[3] . If ϕ := U2,0 (α)ϕ and jA,k := ϕ † β[k] ϕ where (k ∈ {0,1,2,3}) then 185
- 200. jA,0 = jA,0 coshα− jA,1 sinhα, (3.88) jA,1 = jA,1, jA,2 = jA,2 coshα− jA,0 sinhα, jA,3 = jA,3. Then let x0 : = x0 coshα−x3 sinhα, (3.89) x1 : = x1, x2 : = x2, x3 : = x3 coshα−x0 sinhα. Let U3,0 (α) := cosh α 2 ·14 −sinh α 2 ·β[3] β[0] . That is: U3,0 (α) = e 1 2 α 0 0 0 0 e−1 2 α 0 0 0 0 e−1 2 α 0 0 0 0 e 1 2 α . (3.90) In rhis case: U† 3,0 (α)β[0] U3,0 (α) = β[0] coshα−β[3] sinhα, (3.91) U3,0 (α)β[1] U3,0 (α) = β[1] , U3,0 (α)β[2] U3,0 (α) = β[2] , U3,0 (α)β[3] U3,0 (α) = β[3] coshα−β[0] sinhα. If ϕ := U3,0 (α)ϕ and jA,k := ϕ † β[k] ϕ where (k ∈ {0,1,2,3}) then jA,0 = jA,0 coshα− jA,3 sinhα, (3.92) jA,1 = jA,1, jA,2 = jA,2. jA,3 = jA,3 coshα− jA,0 sinhα. 186
- 201. Function ϕ submits to the following equation:[71, p.82] 1 c ∂tϕ− iΘ0β[0] +iϒ0β[0]γ[5] ϕ = = ( 3 ∑ ν=1 β[ν] ∂ν +iΘν +iϒνγ[5] + +iM0γ[0] +iM4β[4]− −iMζ,0γ [0] ζ +iMζ,4ζ[4]− −iMη,0γ [0] η −iMη,4η[4]+ +iMθ,0γ [0] θ +iMθ,4θ[4])ϕ . That is: ( 3 ∑ ν=0 β[ν] ∂ν +iΘν +iϒνγ[5] + +iM0γ[0] +iM4β[4]− −iMζ,0γ [0] ζ +iMζ,4ζ[4]− −iMη,0γ [0] η −iMη,4η[4]+ +iMθ,0γ [0] θ +iMθ,4θ[4])ϕ = 0. (3.93) Like coordinates x5 and x4 [71, p.83] here are entered new coordinates yβ, zβ, yζ, zζ, yη, zη, yθ, zθ such that − πc h ≤ yβ ≤ πc h ,− πc h ≤ zβ ≤ πc h , − πc h ≤ yζ ≤ πc h ,− πc h ≤ zζ ≤ πc h , − πc h ≤ yη ≤ πc h ,− πc h ≤ zη ≤ πc h , − πc h ≤ yθ ≤ πc h ,− πc h ≤ zθ ≤ πc h . and like ϕ, [71, p.83] let: [ϕ] t,x,yβ ,zβ ,yζ ,zζ ,yη ,zη ,yθ ,zθ := (3.94) : = ϕ(t,x)×exp(i(yβ M0 +zβ M4 +yζ Mζ,0 +zζ Mζ,4 + +yη Mη,0 +zη Mη,4 +yθ Mθ,0 +zθ Mθ,4)), In fhis case if ([ϕ],[χ]) := := πc h −πc h dyβ πc h −πc h dzβ πc h −πc h dyζ πc h −πc h dzζ × × πc h −πc h dyη πc h −πc h dzη πc h −πc h dyθ πc h −πc h dzθ × × [ϕ]† [χ] (3.95) 187
- 202. then ([ϕ],[ϕ]) = ρA, (3.96) [ϕ],β[s] [ϕ] = − jA,k c . and in this case from (3.93 ): ( 3 ∑ ν=0 β[ν] ∂ν +iΘν +iϒνγ[5] + +γ[0]∂ β y +β[4]∂ β z − −γ [0] ζ ∂ ζ y +ζ[4]∂ ζ z − −γ [0] η ∂η y −η[4]∂η z + +γ [0] θ ∂θ y +θ[4]∂θ z )[ϕ] = 0 . (3.97) Because γ [0] η = 0 0 0 i 0 0 −i 0 0 i 0 0 −i 0 0 0 , η[4] = i 0 0 0 −i 0 0 i 0 0 i 0 0 −i 0 0 0 ; (3.98) γ [0] θ = 0 0 −1 0 0 0 0 1 −1 0 0 0 0 1 0 0 , θ[4] = i 0 0 1 0 0 0 0 −1 −1 0 0 0 0 1 0 0 (3.99) γ [0] ζ = 0 0 0 −1 0 0 −1 0 0 −1 0 0 −1 0 0 0 , ζ[4] = i 0 0 0 1 0 0 1 0 0 −1 0 0 −1 0 0 0 ; (3.100) then from (3.97): 3 ∑ ν=0 β[ν] ∂ν +iΘν +iϒνγ[5] [ϕ] +γ[0]∂ β y [ϕ]+β[4]∂ β z [ϕ]+ ( 0 0 −∂θ y ∂ ζ y −i∂η y 0 0 ∂ ζ y +i∂η y ∂θ y −∂θ y ∂ ζ y −i∂η y 0 0 ∂ ζ y +i∂η y ∂θ y 0 0 + i 0 0 ∂θ z ∂ ζ z +i∂η z 0 0 ∂ ζ z −i∂η z −∂θ z −∂θ z −∂ ζ z −i∂η z 0 0 −∂ ζ z +i∂η z ∂θ z 0 0 ) ×[ϕ] = 0. (3.101) 188
- 203. Let a Fourier transformation of [ϕ] t,x,yβ ,zβ ,yζ ,zζ ,yη ,zη ,yθ ,zθ be the following; [ϕ] t,x,yβ ,zβ ,yζ ,zζ ,yη ,zη ,yθ ,zθ = = ∑ w,p1,p2,p3,nβ,sβ,nζ,sζ,nη,sη,nθ,sθ c(w, p1, p2, p3,nβ ,sβ , nζ ,sζ ,nη ,sη ,nθ ,sθ )× ×exp(−i h c (wx0 + p1x1 + p2x2 + p3x3 + (3.102) +nβ yβ +sβ zβ +nζ yζ +sζ zζ + +nη yη +sη zη +nθ yθ +sθ zθ )). Let in (3.101) Θν = 0 and ϒν = 0. Let us designe: G0 := ( 3 ∑ ν=0 β[ν]∂ν +γ[0]∂ β y +β[4]∂ β z − −γ [0] ζ ∂ ζ y +ζ[4]∂ ζ z − −γ [0] η ∂η y −η[4]∂η z + +γ [0] θ ∂θ y +θ[4]∂θ z ). (3.103) that is: G0 = −∂0 +∂3 ∂1 −i∂2 ∂ β y −∂θ y ∂ ζ y −i∂η y ∂1 +i∂2 −∂0 −∂3 ∂ ζ y +i∂η y ∂ β y +∂θ y ∂ β y −∂θ y ∂ ζ y −i∂η y −∂0 −∂3 −∂1 +i∂2 ∂ ζ y +i∂η y ∂ β y +∂θ y −∂1 −i∂2 −∂0 +∂3 +i 0 0 ∂ β z +∂θ z ∂ ζ z +i∂η z 0 0 ∂ ζ z −i∂η z ∂ β z −∂θ z −∂ β z −∂θ z −∂ ζ z −i∂η z 0 0 −∂ ζ z +i∂η z −∂ β z +∂θ z 0 0 (3.104) 189
- 204. G0 [ϕ] = −i h c ∑ w,p1,p2,p3,nβ,sβ,nζ,sζ,nη,sη,nθ,sθ ˇg(w, p1, p2, p3,nβ ,sβ ,nζ ,sζ ,nη ,sη ,nθ ,sθ ) 3 ∑ k=0 ck(w, p1, p2, p3,nβ ,sβ ,nζ ,sζ ,nη ,sη ,nθ ,sθ )× ×exp(−i h c (wx0 + p1x1 + p2x2 + p3x3 + (3.105) +nβ yβ +sβ zβ +nζ yζ +sζ zζ + +nη yη +sη zη +nθ yθ +sθ zθ )). here ck(w, p1, p2, p3,nβ ,sβ ,nζ ,sζ ,nη ,sη ,nθ ,sθ ) is an eigenvector of ˇg(w, p1, p2, p3,nβ ,sβ ,nζ ,sζ ,nη ,sη ,nθ ,sθ ) and ˇg(w, p1, p2, p3,nβ ,sβ ,nζ ,sζ ,nη ,sη ,nθ ,sθ ) := (3.106) := β[0] w+β[1] p1 +β[2] p2 +β[3] p3 + +γ[0] nβ +β[4] sβ −γ [0] ζ nζ +ζ[4] sζ − −γ [0] η nη −η[4] sη +γ [0] θ nθ +θ[4] sθ . Here {c0,c1,c2,c3} is an orthonormalized basis of the complex4-vectors space. Functions ck(w, p1, p2, p3,nβ ,sβ ,nζ ,sζ ,nη ,sη ,nθ ,sθ )× (3.107) ×exp(−i h c (wx0 + p1x1 + p2x2 + p3x3 + +nβ yβ +sβ zβ + +nζ yζ +sζ zζ ++nη yη +sη zη +nθ yθ +sθ zθ )) are eigenvectors of operator G0. ϕζ y := c(w,p, f)exp(−i h c (wx0 +px+γ [0] ζ fyζ )) is a red lower chrome function, ϕζ z := c(w,p, f)exp(−i h c (wx0 +px−iζ[4] fzζ )) 190
- 205. is a red upper chrome function, ϕη y := c(w,p, f)exp(−i h c (wx0 +px+γ [0] η fyη )) is a green lower chrome function, ϕη z := c(w,p,, f)exp(−i h c (wx0 +px−iη[4] fzη )) is a green upper chrome function, ϕθ y := c(w,p, f)exp(−i h c (wx0 +px+γ [0] θ fyθ )) is a blue lower function, ϕθ z := c(w,p,sθ )exp(−i h c (wx0 +px−iθ[4] fzθ )) is a blue upper chrome function. Operator −∂ ζ y∂ ζ y is called a red lower chrome operator, −∂ ζ z ∂ ζ z is a red upper chrome operator, −∂η y ∂η y is called a green lower chrome operator, −∂η z ∂η z is a green upper chrome oper- ator, −∂θ y∂θ y is called a blue lower chrome operator, −∂θ z ∂θ z is a blue upper chrome operator For example, if ϕ ζ z is a red upper chrome function then −∂ζ y∂ζ yϕζ z = −∂η y ∂η y ϕζ z = −∂η z ∂η z ϕζ z = = −∂θ y∂θ yϕζ z = −∂θ z ∂θ z ϕζ z = 0 but −∂ζ z ∂ζ z ϕζ z = − h c f 2 ϕζ z . Because G0 [ϕ] = 0 then UG0U−1 U [ϕ] = 0 If U = U1,2 (α) then G0 → U1,2 (α)G0U−1 1,2 (α) and [ϕ] → U1,2 (α)[ϕ]. In this case: ∂1 → ∂1 := (cosα·∂1 −sinα·∂2), ∂2 → ∂2 := (cosα·∂2 +sinα·∂1), ∂0 → ∂0 := ∂0, ∂3 → ∂3 := ∂3, ∂ β y → ∂ β y := ∂ β y , ∂ β z → ∂ β z := ∂ β z , ∂ ζ y → ∂ ζ y := cosα·∂ ζ y −sinα·∂η y , 191
- 206. ∂η y → ∂η y := cosα·∂η y +sinα·∂ ζ y , ∂ ζ z → ∂ ζ z := cosα·∂ ζ z +sinα·∂η z , ∂η z → ∂η z := cosα·∂η z −sinα·∂ ζ z , ∂θ y → ∂θ y := ∂θ y, ∂θ z → ∂θ z := ∂θ z . Therefore, −∂ζ z ∂ζ z ϕζ z = f h c cosα 2 ·ϕζ z , −∂η z ∂η z ϕζ z = −sinα· f h c 2 ϕζ z . If α = −π 2 then −∂ζ z ∂ζ z ϕζ z = 0, −∂η z ∂η z ϕζ z = f h c 2 ϕζ z . That is under such rotation the red state becomes the green state. If U = U3,2 (α) then G0 → U3,2 (α)G0U−1 3,2 (α) and [ϕ] → U3,2 (α)[ϕ]. In this case: ∂0 → ∂0 := ∂0, ∂1 → ∂1 := ∂1, ∂2 → ∂2 := (cosα·∂2 +sinα·∂3), ∂3 → ∂3 := (cosα·∂3 −sinα·∂2), ∂ β y → ∂ β y := ∂ β y , ∂ ζ y → ∂ ζ y := ∂ ζ y, ∂η y → ∂η y := cosα·∂η y −sinα·∂θ y , ∂θ y → ∂θ y := cosα·∂θ y +sinα·∂η y , ∂ β z → ∂ β z := ∂ β z , ∂ ζ z → ∂ ζ z := ∂ ζ z , ∂η z → ∂η z := cosα·∂η z −sinα·∂θ z , ∂θ z → ∂θ z := cosα·∂θ z +sinα·∂η z , Therefore, if ϕη y is a green lower chrome function then −∂η z ∂η z ϕη y = h c cosα· f 2 ·ϕη y , −∂θ y ∂θ y ϕη y = h c sinα· f 2 ·ϕη y . If α = π/2 then 192
- 207. −∂η z ∂η z ϕη y = 0, −∂θ y ∂θ y ϕη y = h c f 2 ·ϕη y . That is under such rotation the green state becomes blue state. If U = U3,1 (α) then G0 → U3,1 (α)G0U−1 3,1 (α) and [ϕ] → U3,1 (α)[ϕ]. In this case: ∂0 → ∂0 := ∂0, ∂1 → ∂1 := (cosα·∂1 −sinα·∂3), ∂2 → ∂2 := ∂2, ∂3 → ∂3 := (cosα·∂3 +sinα·∂1), ∂ β y → ∂3 := ∂ β y , ∂ ζ y → ∂ ζ y := cosα·∂ ζ y +sinα·∂θ y , ∂η y → ∂η y := ∂η y , ∂θ y → ∂θ y := cosα·∂θ y −sinα·∂ ζ y , ∂ β z → ∂ β z := ∂ β z , ∂ ζ z → ∂ ζ z := cosα·∂ ζ z −sinα·∂θ z , ∂η z → ∂η z := ∂η z , ∂θ z → ∂θ z := cosα·∂θ z +sinα·∂ ζ z . Therefore, −∂ζ z ∂ζ z ϕζ z = − f h c cosα 2 ·ϕζ z , −∂θ z ∂θ z ϕζ z = − sinα· f h c 2 ϕζ z . If α = π/2 then −∂ζ z ∂ζ z ϕζ z = 0, −∂θ z ∂θ z ϕζ z = − f h c 2 ϕζ z . That is under such rotation the red state becomes the blue state. Thus at the Cartesian turns chrome of a state is changed. One of ways of elimination of this noninvariancy consists in the following. Calculations in [71, p.156] give the grounds to assume that some oscillations of quarks states bend time- space in such a way that acceleration of the bent system in relation to initial system submits to the following law (Fig. 1): g(t,x) = cλ/ x2 cosh2 λt/x2 . Here the acceleration plot is line (1) and the line (2) is plot of λ/x2. 193
- 208. Figure 37: Hence, to the right from point C and to the left from poin C the Newtonian gravitation law is carried out. AA is the Asymptotic Freedom Zone. CB and B C is the Confinement Zone. Let in the potential hole AA there are three quarks ϕ ζ y, ϕη y , ϕθ y. Their general state function is determinant with elements of the following type: ϕ ζηθ y := ϕ ζ yϕη y ϕθ y. In this case: −∂ζ y∂ζ yϕζηθ y = h c f 2 ϕζηθ y and under rotation U1,2 (α): −∂ζ y ∂ζ y ϕζηθ y = = h c f 2 γ [0] ζ cosα−γ [0] η sinα 2 ϕζ yϕη y ϕθ y = = h c f 2 ϕζηθ y . That is at such turns the quantity of red chrome remains. As and for all other Cartesian turns and for all other chromes. Baryons ∆− = ddd, ∆++ = uuu, Ω− = sss belong to such structures. If U = U1,0 (α) then G0 → U−1‡ 1,0 (α)G0U−1 1,0 (α) and [ϕ] → U1,0 (α)[ϕ]. In this case: ∂0 → ∂0 := (coshα·∂0 +sinhα·∂1), ∂1 → ∂1 := (coshα·∂1 +sinhα·∂0), ∂2 → ∂2 := ∂2, ∂3 → ∂3 := ∂3, ∂ β y → ∂ β y := ∂ β y , ∂ ζ y → ∂ ζ y := ∂ ζ y, ∂η y → ∂η y := coshα·∂η y −sinhα·∂θ z , 194
- 209. ∂θ y → ∂θ y := coshα·∂θ y +sinhα·∂η z , ∂ β z → ∂ β z := ∂ β z , ∂ ζ z → ∂ ζ z := ∂ ζ z , ∂η z → ∂η z := coshα·∂η z +sinhα·∂θ y , ∂θ z → ∂θ z := coshα·∂θ z −sinhα·∂η y . Therefore, −∂η y ∂η y ϕη y = 1+sinh2 α · h c f 2 ϕη y , −∂θ z ∂θ z ϕη y = sinh2 α· h c f 2 ϕη y . Similarly chromes and grades change for other states and under other Lorentz transfor- mation. One of ways of elimination of this noninvariancy is the following: Let ϕζηθ yz := ϕζ yϕη y ϕθ yϕζ z ϕη z ϕθ z . Under transformation U1,0 (α): −∂θ z ∂θ z ϕζηθ yz = − i h c f 2 ϕζηθ yz . That is a magnitude of red chrome of this state doesn’t depend on angle α. This condition is satisfied for all chromes and under all Lorentz’s transformations. Pairs of baryons {p = uud,n = ddu}, Σ+ = uus,Ξ0 = uss , ∆+ = uud,∆0 = udd belong to such structures. Baryons represent one of ways of elimination of the chrome noninvariancy under Carte- sian’s and under Lorentz’s transformations. , 195
- 211. Conclusion Models are not needed, because fundamental theoretical physics is part of probability the- ory. Physics is a game of probabilities in space-time. Irreversible unidirectional time and metric space is an essential attribute of any information system, and probability is the logic of events that have not yet occurred.
- 213. Epilogue ”... They sawed dumb-bells ... ”What’s the matter?” Balaganov said suddenly, stopping work. ”I’ve been sawing away for three hours, and still it isn’t gold!” Panikovsky did not reply. He had made the discovery a half hour before, and had continued to move the saw only for the sake of appearance. ”Well, let’s saw some more,” redhaired Shura said gallantly. ”Of course we must saw,” remarked Panikovsky, trying to defer the moment of reckoning as long as possible. ... ”I can’t make it out,” said Shura, when he had sawed the dumbbell into two halves. ”This is not gold!” ”Go on sawing! Go on!” gabbled Panikovsky...” Ilya Ilf, Yevgeny Petrov. ”The Little Golden Calf”. M., 1987.
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- 221. Index B-boson field, 101, 153, 154 W-boson, 151–154 Z-boson, 153, 154 3+1 vector, 78 3+1-vector, 156, 181 algebra Cayley, 155 antithetical events, 54 Asymptotic Freedom, 174, 194 Asymptotic Freedomin, 172 Baryon, 181, 194, 195 basis, 101, 110, 120, 125, 126, 128, 129, 131, 132, 134, 140, 141, 154 Bernoulli, Jacob, 55 Bethe, Hans, 69 Bohr, Niels, 69 Boole, George, 8 Boolean function, 8–16, 51, 60 Born, Max, 69 boson, 152 cartesian, 73, 78 Cartesian turns, 193, 194 Cauchy-Weierstrass ”limit” , 61 Caylean, 156–158 Cayley-Dickson algebras, 155, 157, 158 charge, 95, 106, 153 chromatic, 71, 94, 159, 176, 177, 181 chromatic pentad, 71, 94 chrome, 170, 171, 190–193, 195 Clifford pentad, 71, 95 Clifford set, 71 Clifford, W. , 71 clock, 18–23, 41 Confinement, 172, 174, 194 confinement, 181 conjunction, 2, 6 continuity, 78 coordinates, 83, 95 Cowan, Clyde, 124 Cumulative Distribution Function, 70 density, 114 direction, 19, 20 disjunction, 2, 6, 7, 55 distance length, 26, 30, 35 disturbance, 156–158 division algebra, 157, 158 DORIS, 158 dot event, 70 Dyson, Freeman, 69 eigenfunction, 75 eigenvalue, 75, 109, 110, 121, 131 eigenvector, 109, 110, 121, 131 Einstein, Albert, 172 electric charge, 171, 172 electromagnetic, 154 elementary charge, 153 energy, 106, 124 entangle, 77, 92 equation of motion, 123 Equation of moving, 101, 103 equation of moving, 142, 144, 151, 153, 159 event, 17, 53–57, 59, 60, 70, 80, 82, 83, 86–89, 91, 95, 114, 115, 117, 158 father number, 104 Fermi, E., 124 Fermi, Enrico, 124 Feynman, Richard, 69 207
- 222. Fourier, 75, 76, 81, 102, 111 Fourier series, 81 frame of reference, 29, 30, 33, 35, 37, 41, 46 frequency, 60 Friedman, Jerome, 172 Galilei, Galileo, 172 Gell-Mann, Murray, 158 Glashow, S. L., 101 Glashow, Sheldon, 119, 142 gluon, 158, 170 Gordon, W, 108 gravitation, 172, 177, 194 gustatory pentad, 72 Hamilton, W., 79 Hamiltonian, 79, 80, 92, 94, 95, 106, 124–126, 137, 142 Heisenberg, Werner, 69 Hubble, Edwin, 174 implication, 2, 4, 7 independent, 54, 55, 57 infinite, 66 infinitesimal, 66 instant, 18, 20, 29, 35, 37, 41, 46, 83, 86, 87, 89–91 internally stationary system, 23 Kalmar, Laszlo, 16 Kendall, Henry, 172 Klein, O, 108 Klein-Gordon equation, 108, 151 Large Number Law, 53 lepton, 92, 94, 95, 105, 123, 124, 141, 172 light pentad, 71, 94 logical rules, 2, 4, 6, 9 Lorentz, 72, 73 Lorentz transformation, 50, 152, 173 Lorentz, Hrndrik, 50 lower, 171 mass, 79, 101, 103, 105, 106, 108, 125, 151, 152 mass number, 102 matrix, 79–81, 95, 96, 120, 125, 147, 148 Maxwell, J. C., 154 meaningful, 2, 53 metrical space, 26 Mills, Robert, 69 negation, 2, 4, 7, 53 neutrino, 105, 106, 124, 141, 154 Newton Gravity, 174 Newton, Isaak, 172 NONSTANDARD NUMBERS, 60 normalized, 121, 131, 132 octavian, 155–158 operator, 79–82, 99, 100, 112, 114, 123, 135, 143 orthonormal, 76, 101, 110 part-set, 60 Pauli matrices, 71 Pauli, W., 124 Pauli, Wolfgang, 69, 71 PETRA, 158 Planck, Max, 69, 75 PLUTO, 158 Pontecorvo, Bruno, 124 probabilities addition formula, 91 probability density, 70 probability function, 59, 60, 70 Pythagorean triple, 104 Q-equivalent, 61, 66 Q-extension, 63, 66 Q-number, 61–66 quark, 105, 119, 158, 172, 174, 181, 193, 194 rank, 71 realization operator, 76, 77 red chrome, 194, 195 Reines, Frederick, 124 Robinson, 61 scalar product, 76, 101 Schrodinger, Erwin, 69 Schwinger, Julian, 69 208
- 223. sentence, 1, 2, 4, 6–17, 19, 20, 23, 26, 29, 53, 55, 57, 59, 70, 86, 91 sequence, 2, 4–7, 9–16, 26, 61, 62, 64–67 Sierpinski, Waclaw Franciszek, 103 standard, 61, 63, 65 standard , 62 Standard Model, 101, 105, 146, 153 state function, 76 state vector, 76, 80, 86–90 supernatural, 158 tautology, 9, 11, 16, 53 Taylor, Richard, 172 Thomson, J. J., 153 time, 80, 82, 83, 86, 103 Tomonaga, Sin-Itiro, 69 trackeable, 74, 75 truth, 1, 2, 7, 16, 53, 55, 57, 59 unit operator, 81 unitary, 101 unitary space, 101 upper, 171 Urysohn, Pavel, 35 weak interaction, 154 Weinberg, S., 152 Wigner, Eugene, 69 Yang, Chen-Ning, 69 Zweig, George, 158 Zwicky, Fritz, 176 209