An important e-mail from MWE to the AIAS members and supporters

February 11, 2005 12:22 CET
PROOF OF THE TETRAD POSTULATE FROM FUNDAMENTAL MATRIX INVERTIBILITY OF THE TETRAD.

Note that this proof is valid for the vielbein of any dimension, in four dimensions it becomes the vierbein or tetrad. The Leibnitz Theorem holds for covariant differentiation as discussed by Carroll. It is nice to see that the Leibnitz Institute of the Bavarian Academy of Sciences in Munich has visited www.aias.us many times. The tetrad postulate is more properly the Cartan tetrad postulate. It is the geometrical basis of the Evans wave equation and Evans Lemma of geneally covariant unified field theory. Fuller details of the different proof of the tetrad postulate is given in the appendices of my new book.

MWE


Myron W. Evans' handwritten paper

http://www.atomicprecision.com/new/atetradpostulatefromfirstprinciples.pdf


PROOF OF THE TETRAD POSTULATE FROM FIRST PRINCIPLES

by Myron W. Evans

with comments (in blue) by Gerhard W. Bruhn, Darmstadt University of Technology

Consider the following basic properties of the tetrad:

qνb qbν = 1                                                 (1)

WRONG! The result of that double summation is dimension dependent. In spacetime of dimension 4 we obtain:
qνb qbν = q0b qb0 + q1b qb1 + q2b qb2 + q3b qb3 = 1 + 1 + 1 + 1 = 4. Without consequences.

qμa qaμ = 1                                                 (2)

WRONG! see (1). Again without consequences.

qaμ qνa = δνμ                                                 (3)

qμa qbμ = δba                                                 (4)

where δνμ and δba are the Kronecker deltas. We now differentiate eqs. (1) to (4) covariantly using the Leibnitz Theorem:

qbν Dρqνb + qνb Dρqbν = 0                                                 (5)

qμa Dρqaμ + qaμ Dρqμa = 0                                                 (6)

qaμ Dρqνa + qνa Dρqaμ = 0                                                 (7)

qμa Dρqbμ + qbμ Dρqμa = 0                                                 (8)

Rearranging dummy indices in eqn. (5) (a → b, μ → ν)
Really executed: (b → a, ν → μ) in the left summand of (5). OK.

qaμ Dρqμa + qνb Dρqbν = 0                                                 (9)

Rearranging dummy indices in eqn. (8) (μ → ν)
Really executed: (μ → ν) in the right summand of (8), then both summands swapped. OK.

qbμ Dρqμa + qνa Dρqbν = 0                                                 (10)

Multiply equ. (9) by qμa

Dρqμa + qμa qνb Dρqbν = 0                                                 (11)

Multiply equ. (10) by qμb

Dρqμa + qμb qνa Dρqbν = 0                                                 (12)

It is seen that equ. (11) is of the form:

x + a y = 0                                                                 (13)

and equ. (12) is of the form

x + b y = 0                                                                 (14)

where

a ≠ b .                                                                 (14)

The only possible solution is:

x = 0 .                                                                 (15)

y = 0 .                                                                 (16)

This means

Dρqμa = 0                                                                 (17)

Dρqνb = 0 .                                                                 (18)

Equ. (18) is the tetrad postulate of Cartan.
It is true for all connections because no restriction on the connection is used in deriving it.

It's true: That «proof» contains no restrictions on the connections. However, regrettably, it contains a (nice!) flaw of thinking. Why? So think about the question which is the value of y. You cannot answer that question, because the terms ay and by are no single products but sums of 16 (in the case of 4 dimensions) summands each of which contains a factor Dρqbν (ν, b = 0, 1, 2, 3). Therefore the eqns. (13),(14) are not two equations for two unknowns, but 16 unknowns are involved. And 2 equations for 16 unknowns do not determine the unknowns uniquely. Hence solutions ≠0 exist. Sorry, Myron.