Dec 3, 2008
One of the basic equations of differential geometry is the 1st Bianchi identity
(1) D Ù Ta = Rab Ù qb .
In their webnote [1] the authors attempt to dualize that identity by claiming the Hodge dual of eq.(1) to be
(1~) D Ù Ta~ = Rab~ Ù qb
as an up to now unknown further basic equation of differential geometry.
To prove that new equation (1~) is transformed to the tensorial equation
(2) Dμ Tκμν = Rκμμν
which is considered to be equivalent to the dualized 1st Bianchi identity (1~). Therefore the tensorial equation (2) should be valid for arbitrary sample manifolds of Riemannian differential geometry. We'll check this assertion (2) by an elementary example below. The reader will find a much deeper treatment of the topic by W.A. Rodrigues Jr. in [4].
Using some basic informations from S.M. Carroll [3] we have the metric
(1.1) ds² = dθ² + sin²θ dφ² = dx1 2 + sin² x1 dx2 2
with the metric tensors
(1.2) (gμν) = diag( 1, sin² x1) , (gμν) = diag( 1, 1/sin² x1) .
using the numbering of indices
(1.3) 1 ~ θ, 2 ~ φ
There are only a few non-vanishing Christoffel coefficients
(1.4) Γ122 = − sin x1 cos x1 , Γ212 = Γ221 = cot x1 ,
while all other Christoffels Γκμν vanish.
The torsion Tκ is given by
(1.5) Tκμν = Γκμν − Γκνμ = 0 ,
i.e. vanishing due to the symmetry of the Christoffels in their lower indices μ,ν.
The Riemann tensor is given by
(2.1) Rκλμν = ∂μΓκλν − ∂νΓκλμ + ΓκμρΓρνλ − ΓκνρΓρμλ
being antisymmetric in μ,ν as is well-known. Therefore we have especially
(2.3) Rκλμν = 0 if μ = ν.
Due to the vanishing of torsion (1.5) the equation (2) to be checked reduces to
(3.1) 0 = Rκμμν = Rκμαβ gμα gνβ .
Therefore, due to the diagonal form (1.2) of (gμρ), we have to check:
(3.2) (Rκ11β g11 + Rκ22β g22) gνβ = (Rκ111 g11 + Rκ221 g22) gν1 + (Rκ112 g11 + Rκ222 g22) gν2 .
This is:
for ν=1: Rκμμ1 = (Rκ111 g11 + Rκ221 g22) g11 + 0 = (Rκ111 g11 + Rκ221 g22) g11 = Rκ221 g22 g11 ,
for ν=2: Rκμμ2 = 0 + (Rκ112 g11 + Rκ222 g22) g22 = (Rκ112 g11 + Rκ222 g22) g22 = Rκ112 g11 g22 ,
i.e. the test reduces to:
(3.3) Rκ221 = 0 ? and Rκ112 = 0 ?
We consider the special case κ = 1 to obtain:
(3.4) R1221 = 0 ? and R1112 = 0 ?
The check 'R1221 = 0 ?' means in detail
(3.5)
R1221 =
∂2Γ121 −
∂1Γ122 +
(Γ121Γ112 +
Γ122Γ212) −
(Γ111Γ122 +
Γ112Γ222) = 0 ?
=o
=o
=o
=o
thus
(3.6) R1221 = − ∂1Γ122 + Γ122Γ212 = − sin² x1 ≠ 0 .
Therefore we have obtained a negative check result: The test equation (2) is not fulfilled for the unit-2-sphere S² which means:
Remark:
Another counter example to eq.(2) is given by the Schwarzschild metric in [2]:
Sect.1.1.4 gives symmetric Christoffel connection, hence the torsion is zero.
However, due to Sect.1.1.12 we have
Roμμo ≠ 0, again contradicting eq.(2).
[1] M.W. Evans, H. Eckardt, Violation of the Dual Bianchi Identity
by Solutions of the Einstein Field Equation
Violation of the Dual Bianchi Identity
[2] M.W. Evans, H. Eckardt, Spherically symmetric metric with perturbation a/r
http://www.atomicprecision.com/blog/wp-filez/a-r.pdf
[3] S.M. Carroll, Lecture Notes on General Relativity, p.60 f.,
http://www2.mathematik.tu-darmstadt.de/~bruhn/Carroll84-85.bmp
[4] W.A. Rodrigues Jr., Differential Forms on Riemannian (Lorentzian) and
Riemann-Cartan Structures and Some Applications to Physics
Ann. Fond. L. de Broglie 32 424-478 (2008)
http://arxiv.org/pdf/0712.3067
[5] G.W. Bruhn, Evans' Duality Experiments
http://www2.mathematik.tu-darmstadt.de/~bruhn/Duality.html