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is the gamma part of B symmetric
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A brief working note by Phil dated 3.25.12 in his tensor documentation support files. It tests symmetry of the gamma part of the covariant derivative of a vector B using three methods: the metric derivative identity, index-raised tensor decomposition, and a direct expansion of the Christoffel symbols. The first two are inconclusive; the third finds a nonzero antisymmetric difference, so the term is not symmetric.
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Question: is the Γ part of Ba;α symmetric? PhL 3.25.12
First, consider that
Ba;α = ∂α Ba + gαβΓaβn Bn (*)
The question is whether or not this object is symmetric in α,a.
gαβΓaβn = gαmΓamn = gαmΓanm
My first two methods deliver no conclusion though suggest maybe not symmetric. The third method I think proves not symmetric.
[ Fact: the Γ part of Ba;α is NOT symmetric, although the Γ part of Ba;α is symmetric. ]
Method 1. Consider the identity
(∂cgab) = – [gam Γ bcm + gbm Γacm]
(∂ngαa) = – [gαm Γ anm + gam Γαnm]
sym ?? ??
This method is inconclusive, we just know that the sum of the two terms shown is symmetric.
Method 2. Consider the fact that Ba;α' is a true rank-2 tensor and that
Ba;α = ∂α Ba – ΓnaαBn
Then,
Ba;α = gaa' gαα' Ba;α' = gaa' gαα' Ba',α' + gaa' gαα'(– Γna'α'Bn)
I think the second term here is symmetric for this reason:
Qaα = gac gαd ΓncdBn = gαd gac ΓncdBn = gαc gad ΓndcBn = gαc gad ΓncdBn = Qαa
The first term here is this
gaa' gαα' Ba',α' = gaa' gαα' ∂α'Ba' = gaa'(∂αBa')
but this is NOT the first term in (*), and therefore Qaα is not the second term in (*) and therefore the conclusion that Qaα is symmetric does not pertain to our question, so no conclusion.
Method 3. Our object of interest here will be
gαβΓaβn → geaΓdab // is this symmetric on e and d ?
Fed = geaΓdab = gea (1/2) gdc [ ∂agbc + ∂bgca – ∂cgab]
Fde = gdaΓeab = gda (1/2) gec [ ∂agbc + ∂bgca – ∂cgab]
= gec (1/2) gda [ ∂agbc + ∂bgca – ∂cgab]
= gec (1/2) gdA [ ∂Agbc + ∂bgcA – ∂cgAb]
= gea (1/2) gdA [ ∂Agba + ∂bgaA – ∂agAb]
= gea (1/2) gdc [ ∂cgba + ∂bgac – ∂agcb]
Now subtract
Fde- Fed = gea (1/2) gdc{[ ∂cgba + ∂bgac – ∂agcb] - [ ∂agbc + ∂bgca – ∂cgab]}
= gea (1/2) gdc { ∂cgba + ∂bgac – ∂agcb - ∂agbc - ∂bgca + ∂cgab}
= gea (1/2) gdc { 2∂cgba – 2∂agcb }
= gea gdc { ∂cgba – ∂agbc } ≠ 0
AS in c,a
Since the leading factor is not SYM on c,a, we don't get 0, so this seems to argue in the negative quite strongly.
Conclusion