 16.03.07: Wrong factors 4 and ¼ removed from Eqs.(14.21') and (14.22') respectively

## É. Cartan mirrored in Evans' Einstein Cartan Evans Theory

### Some simple Maple exercises to Evans and his followers

"Cartan geometry is available in a standard and well known computer package such as Maple, a fact which shows that the attacks on Cartan geometry are pseudo-mathematics. Cartan geometry is actually wired into the circuits, and this computer code can be used to automate calculations. This process is self checking and self consistent. British Civil List Scientist."

"After all the major parts are entered, we can take all rebuttals and criticism from Bruhn, Lakhtakia etc. and show where their faulty logic is. After this is compiled, we can send a letter to the main Wikipedia editors along with Myron’s list of papers and books and his copies of Civil List appointment and medals etc. along with the rebuttals with solutions and examples demonstrating their correctness.

Best regards,

Ross Pawley "

So here are some pseudo-mathematical attacks on "Cartan geometry" by M.W. Evans himself that he should (let) check by Maple. To recognize these errors other people don't need Maple, it belongs to their standard knowledge:

p. 262 of Evans' GCUFT book Vol. 1:

Using ... the Cartan convention 

qaμ qμa = 1 ,                                                                 (14.20)

p.289 of Evans' GCUFT book Vol. 1:

Using the Cartan convention 

qaμ qμa = 1 ,                                                                 (15.80)

### Not only that Evans attributes his wrong equation to Cartan he even falsely refers to S.M. Carroll's Lecture Notes, where - of course - nothing of that nonsense can be found.

Again p.261 f. of Evans' GCUFT book Vol. 1:

Rμν = Raμ qbν ηab,                                                                 (14.13)

. . .

together with the definitions (13) (= eq. (14.13)) . . . we obtain:

R = gμνRμν
= qμa qbν ηabRaμ qbν ηab                                                                 (14.21)
= (ηabηab)(qνb qbν)(qμa Raμ) = 4 qμa Raμ .

a) Show by Maple or anyhow that (14.21) is an invalid equation.

b) The correct equation as a consequence of eq. (14.13) is

R = qaμ Raμ .                                                                 (14.21')

Confirm that by Maple.

Multiply either side of Eq. (21) (= eq. (14.21)) by qaμ to obtain

Raμ = ¼ Rqaμ ,                                                                 (14.22)

This should according to the correct Eq.(14.21') read

Raμ = R qaμ ,                                                                 (14.22')

c) Show by Maple that eq. (14.21') (the correction of eq. (14.21)) cannot be resolved for the 16 quantities Raμ. Thus, eq. (14.22) is invalid.
Hint Confirm by maple that eq. (14.21') is fulfilled by

Raμ = R qaμ + qbμ     where b=b(a):=(a+1) mod 4.

### Thus, eq. (14.22) is no implication of eq. (14.21').

p.317 of Evans' GCUFT book Vol. 1:

Multiplying both sides of Eq (B.3) by qbλ and using:

qbλqλb = 1                                                                 (B.4)

p.319 of Evans' GCUFT book Vol. 1:

In order to show that this is zero use:

qλcqcλ = 1                                                                 (B.25)

quote from p.55:

The symmetric metric tensor is then defined through the line element, a one form of differential geometry:

ω1 = ds² = qij(S)duiduj ,                                                                 (3.12)

and the anti-symmetric metric tensor through the area element, a two form of differential geometry:

ω2 = dA = − ½ qij(A)duiÙduj .                                                                 (3.13)

These results generalize as follows to the four dimensions of any non-Euclidean space-time:

ω1 = ds² = qμν(S)duμduν,                                                                 (3.14)

ω2 = *ω1 = dA = − ½ qμν(A)duμÙduν.                                                                 (3.15)

Try to compute *ω1 with Maple. The Hodge dual operator * is only defined for antisymmetric forms. Is the form ω1 antisymmetric?

In differential geometry the element duσ is dual to the wedge product duμÙduν.

So try to prove the claim

### *duσ = duμÙduν

with Maple. It is completely wrong in 4-D (*duσ is a n−1 form in n-D) and depends on the metric in 3-D.

Again from the same page:

The symmetric metric tensor is:

[ h0² h0h1 h0h2 h0h3 ]
[ h1h0 h1² h1h2 h1h3 ]
qμν(S)   =   [                              ]                                                                (3.16)
[ h2h0 h2h1 h2² h2h3 ]
[ h3h0 h3h1 h3h2 h3² ]

Compute by Maple the determinant of the matrix [qμν(S)] and - if possible - its inverse [qμν(S)]−1. The result will be a vanishing determinant as we have all line vectors being parallel: Each line is parallel to the vector [ h0 h1 h2 h3 ]. Thus, the matrix is singular having no inverse. However, the metric must have a nonsingular matrix, since its inverse is needed in tensor calculus too.

### Thus, eq. (3.16) cannot provide a metric.

Let ds² be defined by eq. (3.14) and the coefficients qμν(S) be given by eq. (3.16). All quantities are assumed reals. Prove with Maple the identiy

ds² = hμhνduμduν = (hμduμ)2 ≥ 0 .

(We have written the indices of the differentials du in the upper position as is use in tensor calculus.)

### The "metric" defined by the eqs. (3.14) cannot attain negative values, it is positive semi-definite. Therefore this "metric" is always spacelike. There do not exist timelike directions duμ.

Compare with the Minkowski-metric (qμν) := diag(−1, +1,+1,+1) which yields timelike directions if du02 > du12+du22+du32 and spacelike directions if du02 < du12+du22+du32 . Therefore equ. (3.16) cannot be used to define the (flat) Minkowski metric.