August 20, 2008

Evans on his blog on 2008/08/11 praising the merits of the authors:

''. . . In paper 93 of the AIAS unified field series Stephen’s suspicions were proven
to be correct and the Einstein-Hilbert field equation was found to be mathematically
incorrect.
. . . ''

And on his blog on 2008/08/19:

''. . . A more thorough understanding of geometry was given by Cartan and is obtainable from the action of the commutator of covariant derivatives (papers 93 onwards). I proved the ECE dual identity in great detail in these papers, and it is this dual identity that finally disposes of any doubts that Einstein’s version of relativity is incorrect because torsion cannot be arbitrarily neglected in geometry. This amounts to disposing of half the geometry, and with hindsight is bound to go badly wrong. The dual identity sharply exposes the way in which it does go wrong. The great problem is that contemporary physics has lost its basic integrity to such an extent that it will not admit that it is pure mythology. . . .''

As mentioned already several times with regard to other Evans' papers the common paper #93 by Evans e.a. lacks from the following Evans' basic flaws:

(i) Evans asserts in Appendix 1 the ''dual transform'' of his field equation

d Ù F^{a}
= μ_{o} j^{a} =
*A*^{o} (R^{a}_{b} Ù q^{b} −
ω^{a}_{b} Ù T^{b})
(4)

to give

d Ù F^{~ a} =
μ_{o} J^{a} =
*A*^{o} (R^{~ a}_{b} Ù q^{b} −
ω^{a}_{b} Ù T^{~ b})
(7)

This assertion is ''proven'' in some detail in Appendix A of paper #93. We shall have a closer look at that Appendix below.

(ii) The ansatz of eqs.(2) and (3) for the electromagnetic field F^{a} in analogy
to the Bianchi identiy (1) yields *four* fields F^{0}, F^{1},
F^{2}, F^{3} since a=0,1,2,3 while in physics only *one*
electromagnetic field F is observed. Evans never explained how the physical 2-form
F could be connected with the vector valued 2-form F^{a} (a=0,1,2,3).
**This is an inadmissible type mismatch.**

(iii) Erroneous use of a non-existing Î-tensor:

. . . A solution of Eq.(10) is:

R^{a}_{b} =
− ½ κ Î^{a}_{bc} T^{c},
ω^{a}_{b} = − ½ κ Î^{a}_{bc} T^{c}
(13)

The eqs.(13) are 4-dimensional tensor equations. However, the tensor Î^{a}_{bc}
cannot be defined in 4-dimensional tensor calculus. **Hence the eqs.(13) are null and void.**

see also Evans' Duality Experiments:

. . .

The Bianchi identity

d Ù T^{a} + ω^{a}_{b}
Ù T^{b} =
**
−** q^{b} Ù R^{a}_{b}
**(wrong sign, index b missing)**
(A12)

is an identity between two-forms. So it remains true for:

d Ù F^{~ a} =
− *A*^{(o)} (R^{~ a}** _{b}
Ùq^{b} missing +**
ω

because F^{~}, R^{~}, T^{~} are two-forms,
antisymmetric in their last two indices
(the suppressed indices μ,ν)

The wrong eqs. should read:

d Ù T^{a} + ω^{a}_{b}
Ù T^{b} =
q^{b} Ù R^{a}_{b} =
R^{a}_{b} Ù q^{b} ,
(A12')

d Ù F^{~ a} =
*A*^{(o)} (R^{~ a}_{b}
Ù q^{b} −
ω^{a}_{b} Ù T^{~ b}) .
(A13')

The above red marked errors are removable. However, not so Evans' misunderstandings
of elementary differential geometry: Evans apparently believes that he could substitute the forms T^{a} and R^{a}
in eq.(A12) by F^{~ a} = *A*^{(o)} T^{~ a} and by R^{~ a} instead of R^{a}.
This argumentation is wrong:

Evidently Dr Evans has not understood the basic idea of the 1st and 2nd Bianchi identity of differential geometry, which shall be repeated here for the reader's information:

The forms T^{a} and R^{a}_{b} have to be generated
from arbitrarily given 1-forms q^{a} and ω^{a}_{b} by

T^{a} := d Ù q^{a} +
ω^{a}_{b} Ù q^{b}
(Def I)

and

R^{a}_{b} := d Ù ω^{a}_{b} +
ω^{a}_{c} Ù ω^{c}_{b},
(Def II)

Then the defined 2-forms
T^{a} and R^{a}_{b} fulfil the *identities*

D Ù T^{a} =
R^{a}_{b} Ù q^{b}
(Id I)

and

D Ù R^{a}_{b} = 0.
(Id II)

**D Ù T ^{a} =**
d Ù T

= d Ù (d Ù q

= 0 + d Ù (ω

= (d Ù ω

d Ù R^{a}_{b} =
d Ù (d Ù ω^{a}_{b} +
ω^{a}_{c} Ù ω^{c}_{b})
= 0 + d Ù (ω^{a}_{c}
Ù ω^{c}_{b})
= ω^{c}_{b} Ù
d Ù ω^{a}_{c}
−
ω^{a}_{c} Ù d Ù ω^{c}_{b}

_{ωac Ù Rcb =
ωac Ù
(d Ù ωcb +
ωcd Ù ωdb)
\}

Þ
ω^{c}_{b} Ù
d Ù ω^{a}_{c}
−
ω^{a}_{c} Ù d Ù ω^{c}_{b}
=
ω^{c}_{b} Ù R^{a}_{c}
−
ω^{a}_{c} Ù R^{c}_{b} ,

^{ωcb Ù Rac =
ωcb Ù
(d Ù ωac +
ωad Ù ωdc)
/}

hence

d Ù R^{a}_{b} =
ω^{c}_{b} Ù R^{a}_{c}
−
ω^{a}_{c} Ù R^{c}_{b} ,

or

**D Ù R ^{a}_{b} :=
d Ù R^{a}_{b}
+
ω^{a}_{c} Ù R^{c}_{b}
−
ω^{c}_{b} Ù R^{a}_{c}
= 0 . **