Property of Tensor Satisfying Binary Law

Σχετικά έγγραφα
C.S. 430 Assignment 6, Sample Solutions

CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS

6.1. Dirac Equation. Hamiltonian. Dirac Eq.

Congruence Classes of Invertible Matrices of Order 3 over F 2

Every set of first-order formulas is equivalent to an independent set

derivation of the Laplacian from rectangular to spherical coordinates

Phys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)

2 Composition. Invertible Mappings

Second Order Partial Differential Equations

Econ 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1

Srednicki Chapter 55

ΤΕΧΝΟΛΟΓΙΚΟ ΠΑΝΕΠΙΣΤΗΜΙΟ ΚΥΠΡΟΥ ΤΜΗΜΑ ΝΟΣΗΛΕΥΤΙΚΗΣ

Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3

Section 8.3 Trigonometric Equations

Example Sheet 3 Solutions

Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit

Homework 3 Solutions

Generating Set of the Complete Semigroups of Binary Relations

The Simply Typed Lambda Calculus

ΤΕΧΝΟΛΟΓΙΚΟ ΕΚΠΑΙΔΕΥΤΙΚΟ ΙΔΡΥΜΑ ΚΡΗΤΗΣ ΣΧΟΛΗ ΔΙΟΙΚΗΣΗΣ ΚΑΙ ΟΙΚΟΝΟΜΙΑΣ ΤΜΗΜΑ ΛΟΓΙΣΤΙΚΗΣ ΠΤΥΧΙΑΚΗ ΕΡΓΑΣΙΑ

Περίληψη (Executive Summary)

SCITECH Volume 13, Issue 2 RESEARCH ORGANISATION Published online: March 29, 2018

PARTIAL NOTES for 6.1 Trigonometric Identities

Finite Field Problems: Solutions

Section 7.6 Double and Half Angle Formulas

4.6 Autoregressive Moving Average Model ARMA(1,1)

Tridiagonal matrices. Gérard MEURANT. October, 2008

Partial Trace and Partial Transpose

Fourier Series. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics

«ΨΥΧΙΚΗ ΥΓΕΙΑ ΚΑΙ ΣΕΞΟΥΑΛΙΚΗ» ΠΑΝΕΥΡΩΠΑΪΚΗ ΕΡΕΥΝΑ ΤΗΣ GAMIAN- EUROPE

Approximation of distance between locations on earth given by latitude and longitude

EE512: Error Control Coding

3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β

Practice Exam 2. Conceptual Questions. 1. State a Basic identity and then verify it. (a) Identity: Solution: One identity is csc(θ) = 1

Inverse trigonometric functions & General Solution of Trigonometric Equations

Paper Reference. Paper Reference(s) 1776/04 Edexcel GCSE Modern Greek Paper 4 Writing. Thursday 21 May 2009 Afternoon Time: 1 hour 15 minutes

Second Order RLC Filters

Homework 8 Model Solution Section

ΑΓΓΛΙΚΑ Ι. Ενότητα 7α: Impact of the Internet on Economic Education. Ζωή Κανταρίδου Τμήμα Εφαρμοσμένης Πληροφορικής

Math221: HW# 1 solutions

Concrete Mathematics Exercises from 30 September 2016

Statistical Inference I Locally most powerful tests

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 19/5/2007

Other Test Constructions: Likelihood Ratio & Bayes Tests

Main source: "Discrete-time systems and computer control" by Α. ΣΚΟΔΡΑΣ ΨΗΦΙΑΚΟΣ ΕΛΕΓΧΟΣ ΔΙΑΛΕΞΗ 4 ΔΙΑΦΑΝΕΙΑ 1

Instruction Execution Times

Πτυχιακή Εργασία Η ΠΟΙΟΤΗΤΑ ΖΩΗΣ ΤΩΝ ΑΣΘΕΝΩΝ ΜΕ ΣΤΗΘΑΓΧΗ

ΕΠΙΧΕΙΡΗΣΙΑΚΗ ΑΛΛΗΛΟΓΡΑΦΙΑ ΚΑΙ ΕΠΙΚΟΙΝΩΝΙΑ ΣΤΗΝ ΑΓΓΛΙΚΗ ΓΛΩΣΣΑ

Matrices and Determinants

Exercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1.

HOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:

ΕΛΛΗΝΙΚΗ ΔΗΜΟΚΡΑΤΙΑ ΠΑΝΕΠΙΣΤΗΜΙΟ ΚΡΗΤΗΣ. Ψηφιακή Οικονομία. Διάλεξη 7η: Consumer Behavior Mαρίνα Μπιτσάκη Τμήμα Επιστήμης Υπολογιστών

CRASH COURSE IN PRECALCULUS

TMA4115 Matematikk 3

Advanced Subsidiary Unit 1: Understanding and Written Response

Cambridge International Examinations Cambridge International General Certificate of Secondary Education

On a four-dimensional hyperbolic manifold with finite volume

ΕΠΙΧΕΙΡΗΣΙΑΚΗ ΑΛΛΗΛΟΓΡΑΦΙΑ ΚΑΙ ΕΠΙΚΟΙΝΩΝΙΑ ΣΤΗΝ ΑΓΓΛΙΚΗ ΓΛΩΣΣΑ

Problem Set 9 Solutions. θ + 1. θ 2 + cotθ ( ) sinθ e iφ is an eigenfunction of the ˆ L 2 operator. / θ 2. φ 2. sin 2 θ φ 2. ( ) = e iφ. = e iφ cosθ.

Chapter 6: Systems of Linear Differential. be continuous functions on the interval

Partial Differential Equations in Biology The boundary element method. March 26, 2013

Πνευματική ιδιοκτησία και ιατρικά επιστημονικά έργα

Η αλληλεπίδραση ανάμεσα στην καθημερινή γλώσσα και την επιστημονική ορολογία: παράδειγμα από το πεδίο της Κοσμολογίας

Test Data Management in Practice

14 Lesson 2: The Omega Verb - Present Tense

Jesse Maassen and Mark Lundstrom Purdue University November 25, 2013

2. THEORY OF EQUATIONS. PREVIOUS EAMCET Bits.

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 6/5/2006

6.003: Signals and Systems. Modulation

DESIGN OF MACHINERY SOLUTION MANUAL h in h 4 0.

forms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with

ΠΑΡΑΜΕΤΡΟΙ ΕΠΗΡΕΑΣΜΟΥ ΤΗΣ ΑΝΑΓΝΩΣΗΣ- ΑΠΟΚΩΔΙΚΟΠΟΙΗΣΗΣ ΤΗΣ BRAILLE ΑΠΟ ΑΤΟΜΑ ΜΕ ΤΥΦΛΩΣΗ

Απόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.

ΠΑΝΕΠΙΣΤΗΜΙΟ ΚΥΠΡΟΥ ΤΜΗΜΑ ΠΛΗΡΟΦΟΡΙΚΗΣ. ΕΠΛ342: Βάσεις Δεδομένων. Χειμερινό Εξάμηνο Φροντιστήριο 10 ΛΥΣΕΙΣ. Επερωτήσεις SQL

The Nottingham eprints service makes this work by researchers of the University of Nottingham available open access under the following conditions.

Homomorphism in Intuitionistic Fuzzy Automata

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education

Solutions to Exercise Sheet 5

Nowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in

IIT JEE (2013) (Trigonomtery 1) Solutions

Cambridge International Examinations Cambridge International General Certificate of Secondary Education

ECON 381 SC ASSIGNMENT 2

Potential Dividers. 46 minutes. 46 marks. Page 1 of 11

ΠΑΝΕΠΙΣΤΗΜΙΟ ΠΕΙΡΑΙΑ ΤΜΗΜΑ ΝΑΥΤΙΛΙΑΚΩΝ ΣΠΟΥΔΩΝ ΠΡΟΓΡΑΜΜΑ ΜΕΤΑΠΤΥΧΙΑΚΩΝ ΣΠΟΥΔΩΝ ΣΤΗΝ ΝΑΥΤΙΛΙΑ

Solutions to the Schrodinger equation atomic orbitals. Ψ 1 s Ψ 2 s Ψ 2 px Ψ 2 py Ψ 2 pz

Fractional Colorings and Zykov Products of graphs

ANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?

ΤΕΧΝΟΛΟΓΙΚΟ ΠΑΝΕΠΙΣΤΗΜΙΟ ΚΥΠΡΟΥ ΣΧΟΛΗ ΕΠΙΣΤΗΜΩΝ ΥΓΕΙΑΣ ΤΜΗΜΑ ΝΟΣΗΛΕΥΤΙΚΗΣ ΠΤΥΧΙΑΚΗ ΕΡΓΑΣΙΑ ΕΠΗΡΕΑΖΕΙ ΤΗΝ ΠΡΟΛΗΨΗ ΚΑΡΚΙΝΟΥ ΤΟΥ ΜΑΣΤΟΥ

SCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions

Right Rear Door. Let's now finish the door hinge saga with the right rear door

F-TF Sum and Difference angle

Section 1: Listening and responding. Presenter: Niki Farfara MGTAV VCE Seminar 7 August 2016

Physical DB Design. B-Trees Index files can become quite large for large main files Indices on index files are possible.

Orbital angular momentum and the spherical harmonics

Συντακτικές λειτουργίες

Cambridge International Examinations Cambridge International General Certificate of Secondary Education

A summation formula ramified with hypergeometric function and involving recurrence relation

This is an electronic reprint of the original article. This reprint may differ from the original in pagination and typographic detail.

ST5224: Advanced Statistical Theory II

Transcript:

Journal of Modern Physics 017 8 944-963 http://www.scirp.or/journal/jmp ISSN Online: 153-10X ISSN Print: 153-1196 Property of Tensor Satisfyin Binary Law Koji Ichidayama 716-000 Okayama Japan How to cite this paper: Ichidayama K. (017) Property of Tensor Satisfyin Binary Law. Journal of Modern Physics 8 944-963. https://doi.or/10.436/jmp.017.86060 Received: pril 7 017 ccepted: May 8 017 Published: May 31 017 Copyriht 017 by author and Scientific Research Publishin Inc. This work is licensed under the Creative Commons ttribution International License (CC BY 4.0). http://creativecommons.or/licenses/by/4.0/ Open ccess bstract I report the reason why Tensor satisfyin Binary Law has relations toward physics in this article. Q: The n th-order covariant derivative of the Vector : n = 1 satisfyin Binary Law. R: The n th-order covariant deriva- { } ( ) tive of the Vector { } :( n ) satisfyin Binary Law. I have reported in other articles about Q. I report R in this article. I obtained the followin results in this. I ot the conclusion that derived function became 0. The derived function becomin 0 in the n th-order covariant derivative of the covariant vector here in the case of n =. Similarly in the n th-order covariant derivative of the contravariant vector in the case of n = 4. Keywords Tensor Covariant Derivative 1. Introduction Definition 1 1 λ 1 λ λ λ λ λ = + + λ λ x x x 1 λ 1 λ τλ λ τ τλ λ 1 λ τλ λ τ + λ + + λ + τ λ 1 1 1 τλ λ λ τλ λ λ λ λ τλ τ λ τ λ τ λ + x x x x is established [1]. Definition x x x x x = x x = x is established []. I named x x x x x = x x = x Binary Law []. Definition 3 If x x x x x = x x = x is established x = x is established []. Definition 4 If x x x x x = x x = x is established x = x is established []. DOI: 10.436/jmp.017.86060 May 31 017

Definition 5 If x x x x x = x x = x is established x = x is established []. Definition 6 If x x x x x = x x = x is established x = x is established []. Definition 7 If all coordinate systems x x x x λ satisfies x x x x x = x x = x all coordinate systems x x x x λ shifts to only two of x x []. Definition 8 1 τλ λ λ 1 τλ λ λ = τ + λ λ τ x x x x is established [3]. Definition 9 = e e is established [4]. Definition 10 = 1 0: ( ) = is establishment [3]. Definition 11 1 λ ελ λ ε 1 λ ελ λ ε = + x x x + + + λ x x x λ 1 τλ λ λ 1 τλ λ λ 1 λ ελ τλ ετ λ τ λ τ + + x x x x λ τ 1 λ λ τλ τ 1 λ λ τλ τ 1 τλ ελ λ ε + τ λ τ λ λ + x x x x + x x x is established [3]. Definition 1 3 1 κ κ κ λ = + λ λ + κ x x x 1 κ 1 κ κ κ κ κ + + κ + λ λ + κ x x x x x x x x α 1 κ κ κ 1 κ ακ κ α + + κ + λ λ x x x x + α κ α 1 κ ακ κ α + α κ λ 1 1 κ κ ακ κ α ακ κ + λ α κ κ x 1 ακ κ κ 1 κ ακ κ α + λ + κ x x x x + α κ 1 1 κ κ ακ κ α ακ κ + α κ κ λ 1 ακ κ κ 1 ακ κ κ + λ κ x α x + κ α λ 1 1 κ ακ κ κ κ ακ α λ κ α κ x 1 κ κ ακ α 1 ακ κ κ λ + + α κ κ 945

1 1 ακ κ κ κ κ ακ α + κ + α κ λ τ 1 κ τκ λκ τλ + λ τ κ 1 1 κ κ τκ λκ τλ τκ κ + λ τ κ κ 1 1 κ τκ λκ τλ τκ κ κ + + λ τ κ + κ α 1 κ τκ λκ τλ 1 τκ ακ κ α + λ τ κ α κ x x x 1 κ τκ λκ τλ 1 τκ ακ 1 κ α ακ κ κ + + λ τ κ + α κ + κ τ 1 κ τκ λκ τλ 1 ακ κ κ λ τ κ κ α x x x 1 κ τκ λκ τλ 1 ακ κ κ 1 τκ κ ακ α + + λ τ κ κ x x x + α κ 1 τκ κ λκ λ + λ κ τ 1 1 κ τκ κ λκ λ κ τκ τ λ κ τ κ x x x 1 τκ κ λκ λ 1 κ κ τκ τ + λ κ + x x τ κ x α 1 τκ κ λκ λ 1 κ ακ κ α + λ κ α κ τ x x x 1 1 1 τκ κ λκ λ κ ακ κ α ακ κ τκ τ λ κ α κ τ + + x x x κ 1 τκ κ λκ λ 1 ακ τκ κ τ + + λ κ x x x + τ κ α 1 1 1 κ τκ κ λκ λ ακ τκ κ τ κ ακ α + λ κ τ κ α κ x x x 1 τκ κ λκ λ + λ κ τ 1 1 τκ κ λκ λ κ κ κ λ κ τ κ x x x 1 1 κ τκ κ λκ λ κ κ λ κ κ τ α 1 τκ κ λκ λ 1 κ ακ τκ ατ + λ κ x x x + τ α κ 1 1 1 κ κ λκ λ ακ τκ ατ κ λ κ τ α κ κ x x x 1 τκ κ λκ λ 1 ακ κ τκ τ + + λ κ x x + τ κ α 1 1 1 κ κ λκ λ κ τκ τ ακ α + λ κ τ κ α κ x x x τκ κ ακ τκ ακ κ is established [3]. 946

Definition 13 For every coordinate systems there is no immediate reason for preferrin certain systems of co-ordinates to others. Definition 14 The physical law is invariable for all coordinate systems [1]. Definition 15 ll coordinate systems satisfies Definision 13 is established if Definision 14 is established. Definition 16 Definision 14 is established if The physical law is described in Tensor is established [1]. Definition 17 ll coordinate systems satisfies Definision 13 is established if ll coordinate systems satisfies Binary Law is established []..Einstein required establishment of Definision 14 approximately 100 years ao [1]. Furthermore he required establishment of The physical law is described in Tensor based on Definision 16 [1]. However. Einstein does not mention Definision 15 at all [1]. I et the conclusion that ll coordinate systems satisfies Definision 13 must be established if Definision 14 is established accordin to Definision 15. On the other hand I ot that Definision 17 was established []. nd I ot the conclusion that must require establishment of ll coordinate systems satisfies Binary Law if I required establishment of Definision 14 by Definision 17. Scalar and Vector have already satisfied these two demands here []. In other words we can use Scalar and Vector to express a physical law. Therefore I do not mention it for Scalar and Vector. I researched it about the Tensor which had not yet satisfied Binary Law in this article. The first purpose of this article is to rewrite the Tensor which does not satisfy Binary Law in Tensor satisfyin Binary Law. Then the second purpose is to find out the property from Tensor satisfyin Binary Law.. bout Property of Tensor Satisfyin Binary Law: The Second Third Fourth-Order Covariant Derivative of the Vector { } Proposition 1 If x x x x x = x x = x is established = = 0 is established. Proof: I et ( ) = 1 = 0 : (1) from Definision 10 if all coordinate systems x x x x λ satisfies Definision. I et 1 = + x x x 1 1 1 1 + + + 947

1 + 1 1 + + + 1 1 1 = x 1 1 1 1 1 + + = x 1 1 1 λ λ 1 λ 1 1 1 1 λ + + from Definision 1 if all coordinate systems x x x x λ satisfies Definision. By the way we cannot handle () accordin to Definision 7. I simplify () here and et λ 1 1 1 λ =. + (3) However (3) can rewrite 1 1 1 = + (4) if x and x of (3) are chaneable to x or x () each. Because index doesn t exist at all in the third term of the riht side of (3) I can chane dummy index λ of (3) to dummy index. Furthermore (4) can rewrite 1 1 1 =. + (5) Because index doesn t exist at all in the second term of the riht side of (4) I can chane dummy index of (4) to dummy index. nd we can handle (5) accordin to Definision 7. The possible rewrite by or of is x x (6) (7) (8) accordin to Definision 4 Definision 6. Because three covariant Vector of the same index exists in one term I don t handle (6). Two sets are dummy index 948

amon three same index in (7) (8). Therefore we must rewrite () to ( ) = + + 1 1 1 1 1 + 1 1 1 + ( ) = + + 1 1 ( ) 1 1 1 + x x x x 1 1 1 + x x x 1 1 = 1 1 1 + x x x x 1 1 1 + x x x by usin Definision 4 Definision 6 with considerin (7) (8). I et ( ) ( ) = = 1 1 = x in consideration of establishment of = from (9) (10) here. I et in consideration of (1) for (1). nd I et from (13). I et from (9) (10) (11) (1) = 0 (13) = 0 = in consideration of Definision 4 here. I et (14) = (15) = 0 (16) 949

from (14) (15). Therefore I et = (17) x x ( ) = (18) x x ( ) ( ) =. (19) x x from (9) (10) (11) in consideration of (1) (13) (16). nd we can rewrite (17) (18) (19) by usin Definision 4 Definision 6 for = (0) x x. Because the second third term of the riht side of (17) (18) (19) does not exist here we may adopt (17) (18) (19) and (0) description form of which. Furthermore I rewrite (0) by Definision 4 and et ( ) = = 0 in consideration of Proposition. nd I rewrite (1) by Definision 4 and et = = 0. End Proof Because () is established I decide not to handle the third-order covariant derivative of the covariant Vector. (1) () Proposition If x x x x x = x x = x is established S = 0 is established. Proof: I et 1 1 1 1 = x = + (3) from Definision 8 if all coordinate systems x x x x λ satisfies Definision. I et from from (4). I et Definision 3. I et = (4) x x = = (5) 950

from (5) Definision 4. Therefore I et = = (6) = 0 (7) from (3) (6). I et = e e (8) from Definision 9 if all coordinate systems x x x x λ satisfies Definision. I et B = B (9) from B Definision 3 if all coordinate systems x x x x λ satisfies Definision. I et from (9). I et B = B (30) B = B (31) from (8) (30). I et ( B ) B = (3) from covariant derivative of (31). I et S = 0 from (7) (3). End Proof Proposition 3 If x x x x x = x x = x is established = is established. x x Proof: I et 1 = + x x x + 1 1 + x x x x 1 1 + + 1 + + 1 1 + x x x 1 1 1 = + + x 1 1 1 1 1 + + 1 1 1 = + + x λ λ 1 1 1 1 1 λ + + λ x (33) 951

= + + x 1 1 1 λ λ 1 1 1 λ 1 1 λ + + (34) from Definision 11 if all coordinate systems x x x x λ satisfies Definision. By the way we cannot handle (33) (34) accordin to Definision 7. I simplify (33) here and et λ 1 1 1 =. + + + λ x x x (35) However (35) can rewrite 1 1 1 = + + + x x x (36) if x and x of (35) are chaneable to x or x each. Because index doesn t exist at all in the third term of the riht side of (35) I can chane dummy index λ of (35) to dummy index. Furthermore (36) can rewrite 1 1 1 =. + + + x x x (37) Because index doesn t exist at all in the second term of the riht side of (36) I can chane dummy index of (36) to dummy index. nd we can handle (37) accordin to Definision 7. The possible rewrite by or of is x x (38) (39) (40) accordin to Definision 4 Definision 6. Because three contravariant Vector of the same index exists in one term I don t handle (40). Two sets are dummy index amon three same index in (38) (39). Therefore we must rewrite (33) to 1 1 = + + 1 1 1 x x x 1 1 1 + + (41) 95

1 1 1 ( ) = + + x x (4) 1 1 1 1 1 + x x x ( ) = + + 1 1 1 1 1 1 1 1 + x x x x x x by usin Definision 4 Definision 6 with considerin (38) (39). I et ( ) = ( ) = 1 1 = in consideration of establishment of = from (4) (43) here. I et = 0 in consideration of (1) for (44). Therefore I et (43) (44) (45) = (46) x x ( ) = (47) x x ( ) = (48) x x from (41) (4) (43) in consideration of (1) (45). nd we can rewrite (46) (47) (48) by usin Definision 4 Definision 6 for. = (49) x x Because the second third term of the riht side of (46) (47) (48) does not exist here we may adopt (46) (47) (48) and (49) description form of which. Similarly we must rewrite (34) to 1 1 1 = + + x x x x (50) 1 1 1 1 1 + + 953

1 1 1 ( ) = + + x (51) 1 1 1 1 1 + ( ) = + + 1 1 1 1 1 + x x x x x x 1 1 1 by usin Definision 4 Definision 6 with considerin (38) (39). Because (51) includes here I don t handle (51). Therefore I et (46) (48) from (50) (5) in consideration of (1). End Proof Proposition 4 If x x x x x = x x = x is established 3 = = is established. Proof: I et 3 1 = + x x x x x + 1 + x 1 + x 1 + + 1 + x + 1 + 1 1 + x + x x x 1 1 + x + x x x + 1 1 + x x x 1 1 + x x 1 1 x x x x 1 1 x + 1 1 + x x x 1 + (5) 954

1 1 + + x x x x 1 1 + x x x 1 1 + + 1 1 1 + x x x + x x x 1 1 x x x + 1 1 1 x x x + x x x 1 1 x 1 1 1 x x x 1 1 x x x 1 1 1 + x x x + x x x 1 1 + x x x 1 1 1 + + + x x x 1 1 1 + x x x x 1 1 x x x 1 1 x x x 1 1 1 + x x 1 1 + x x x 1 1 1 + + + x x x = + + + x x x x x x 3 1 1 1 955

1 1 1 + + + x x x x x 1 1 1 1 1 1 + + + x x x x x 1 1 1 1 x x x x x x x 1 1 1 1 x 1 + 1 1 1 1 1 1 + + + x x x x 1 1 1 1 1 1 1 1 + x x x x x x 1 1 1 1 1 1 x 1 1 1 1 1 1 + 1 1 1 1 1 + 1 1 1 1 1 1 1 1 1 1 1 1 1 + + = + + + x x x 3 1 1 3 3 1 1 3 + x x x x 5 1 1 1 1 + 1 1 1 1 1 + + = + + + x x x 3 1 1 3 3 λ 1 1 3 + x x x x x λ 5 1 1 1 1 λ + λ x x x x x x λ λ λ λ ε λ λ ε λ 1 1 ε 1 λ 1 ε 1 λ + + λ (53) 956

= + + x x x λ 3 λ 1 + + 1 3 x x λ λ 5 1 λ 1 λ 1 + x x x x x x λ ε λ 1 λ 1 λ 1 ε x x x λ ε λ 1 λ 1 1 ε λ + + 3 1 1 3 (54) from Definision 1 if all coordinate systems x x x x λ satisfies Definision. By the way we cannot handle (53) (54) accordin to Definision 7. I simplify (53) here and et 1 = + + x λ 1 + λ ˆ ε λ 1 1 ε ˆ. + λ 3 ˆ ˆ ˆ However (55) can rewrite 1 = + + x λ 1 + λ ˆ λ 1 1 ˆ + λ 3 ˆ ˆ ˆ (55) (56) if x and x xˆ of (55) are chaneable to x or x each. Because index doesn t exist at all in the fourth term of the riht side of (55) I can chane dummy index ε of (55) to dummy index. Furthermore (56) can rewrite 1 = + + x 1 + ˆ 1 1 ˆ. + 3 ˆ ˆ ˆ Because index doesn t exist at all in the third term of the riht side of (56) I can chane dummy index λ of (56) to dummy index. Furthermore (56) (57) 957

can rewrite = + + 3 1 ˆ ˆ ˆ 1 + ˆ (58) 1 1. ˆ + Because index doesn t exist at all in the second term of the riht side of (57) I can chane dummy index of (57) to dummy index. nd we can handle (58) accordin to Definision 7. The possible rewrite by or of 3 is 3 (59) x x x 3 3 3 x x (60) 3 3 3 3 (61) (6) accordin to Definision 4 Definision 6. Because two covariant Vector of the same index exists in one term I don t handle (59). Because two contravariant Vector of the same index exists in one term I don t handle (61). Because four contravariant Vector of the same index exists in one term I don t handle (6). Therefore we must rewrite (53) to 3 1 1 = + 3 + ( ) + + 3 1 1 3 x 5 1 1 1 + 1 1 1 x x x x x x 1 1 1 + + x x x x x (63) 958

3 = + 3 + x x x x x x x x x ( ) ( ) 1 1 + + 3 1 1 3 x x 5 1 1 1 + 1 1 1 x x x + x 1 1 1 + 1 1 3 = + 3 + + + 3 1 1 3 x 5 1 1 1 + x x x x x 1 1 1 x x x x 1 1 1 + + x x x x by usin Definision 4 Definision 6 with considerin (60). I et ( ) = ( ) 3 3 = 3 3 = x x x x x in consideration of establishment of ˆ = ˆ from (64) (65) here. I et in consideration of (1) for (66). Therefore I et ( ) (64) (65) (66) = 0 (67) 3 = (68) x x x 959

( ) 3 = (69) x x x ( ) 3 = (70) x x x from (63) (64) (65) in consideration of (1) (67). nd we can rewrite (68) (69) (70) by usin Definision 4 Definision 6 for 3 = =. (71) Because the second third term of the riht side of (68) (69) (70) does not exist here we may adopt (68) (69) (70) and (71) description form of which. Similarly we must rewrite (54) to ( ) 1 1 3 = + 3 + ( ) 3 1 + + 1 3 5 1 1 1 + x x x x x x 1 1 1 x x x x x x 1 1 1 + + x x x x x 3 1 1 = + + 3 ( ) 3 1 + + x x x x x 1 3 5 1 1 1 + x x x x x x 1 1 1 x x x 1 1 1 + + 3 1 1 = + + 3 3 1 + + x x x x x (7) (73) 960

1 3 x x x 5 1 1 1 + x x x x x x 1 1 1 x 1 1 1 + + x x x x (74) by usin Definision 4 Definision 6 with considerin (60). Because (7) includes here I don t handle (7). I et ( ) = ( ) 3 3 = 3 3 = (75) in consideration of establishment of ˆ = ˆ from (73) (74) here. I et = 0 (76) in consideration of (1) for (75). Therefore I et (69) (70) from (73) (74) in consideration of (1) (76). End Proof Proposition 5 If x x x x x = x x = x is established = 0 is established. Proof: I et from λε λε = (77) = if all coordinate systems x x x x λ satisfies Definision. nd I et = 0 (78) from (77) Proposition Proposition 4. End Proof Because (78) is established I decide not to handle the fifth-order covariant derivative of the contravariant Vector. Proposition 6 If x x x x x = x x = x is established = sin x is established. Proof: I et 3 1 3 1 = 1 1 1 = 3 3 = 1 1 1 = (79) 961

from (71) if a dimensional number is. I et from (79). nd I et from (80). I et from (81). I et 3 1 3 1 1 1 dx= x d dx = x d 1 1 1 3 3 1 1 dx= x d dx = x d 1 1 1 1 1 1 = x = x 1 1 1 = x = x 1 1 = x = x = from (8) Definision 5. nd I et 1 1 1 1 = = 1 1 = = 1 1 from (83). I et 1 1 1 = sin x = sin x = sin x = sin x 1 (80) (81) (8) (83) (84) (85) from (84). nd I et from (85). = sin x (86) End Proof 3. Discussion bout Proposition 1 In () we can handle as Tensor similarly. Furthermore = 0 is established. I do not handle the derived function of a hiher order because derived function is already 0. bout Proposition 3 In (49) we can handle bout Proposition 4 3 In (71) we can handle Furthermore = S is established. as Tensor similarly. as Tensor similarly. 96

bout Proposition 5 In (78) = 0 is established. I do not handle the derived function of a hiher order because derived function is already 0. bout Proposition 6 3 If = 0 is established in (71) can t have a wave-like property. 3 However has a wave-like property if 0 is established in (71). These remind me of the matter wave in the quantum theory. References [1] Einstein. (1916) nnalen der Physik 354 769-8. https://doi.or/10.100/andp.1916354070 [] Ichidayama K. (017) Journal of Modern Physics 8. [3] Dirac P..M. (1975) General Theory of Relativity. John Wiley and Sons Inc. [4] Fleisch D. (01) Student s Guide to Vectors and Tensors. Cambride University Press. Submit or recommend next manuscript to SCIRP and we will provide best service for you: cceptin pre-submission inquiries throuh Email Facebook LinkedIn Twitter etc. wide selection of journals (inclusive of 9 subjects more than 00 journals) Providin 4-hour hih-quality service User-friendly online submission system Fair and swift peer-review system Efficient typesettin and proofreadin procedure Display of the result of downloads and visits as well as the number of cited articles Maximum dissemination of your research work Submit your manuscript at: http://papersubmission.scirp.or/ Or contact jmp@scirp.or 963