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try to show the two tensorized forms are the same
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Phil's short working calculation dated 1.11.15, part of the May 2015 update to his tensor document. He expands the covariant derivative form of (B)'n using the metric g', the Levi-Civita symbol ε' and the covariant derivatives of B'. He tries to show the coefficients vanish using the tensor character of g'^-1/2 ε'. The attempt stalls, and he notes it casts doubt on a claim in his tensor doc, though both sides vanish in Cartesian space.
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This is the Title PhL 1.11.15
(B)'n = B'n;j;j = δijB'n;j;i = δij g'nmg'jk B'm;k;i = g'nm g'jk B'm;k;j
(B)'n = (B'j;j);n – g'-1/2ε'nab(g'-1/2g'bcε'cdeB'e;d);a (I.5.1) // could say ,a at end
= δijB'j;i;n – g'-1/2ε'nab(g'-1/2g'bcε'cdeB'e;d);a
= δij g'jm g'nsB'm;i;s – g'-1/2ε'nab(g'-1/2g'bcε'cdeB'e;d);a
= g'jm g'nsB'm;j;s – g'-1/2ε'nab(g'-1/2g'bcε'cdeB'e;d);a
So show equal, I have to show that
g'nm g'jk B'm;k;j = g'jm g'nsB'm;j;s – g'-1/2ε'nab(g'-1/2g'bcε'cdeB'e;d);a
where at least all the B' indices are in the same location (down). Rewrite
g'nm g'jk B'm;k;j = g'km g'nsB'm;k;s – g'-1/2ε'nab(g'-1/2g'bcε'cdeB'e;d);a
g'nm g'jk B'm;k;j = g'km g'njB'm;k;j – g'-1/2ε'nab(g'-1/2g'bcε'cdeB'e;d);a
(g'nm g'jk - g'km g'nj )B'm;k;j = – g'-1/2ε'nab(g'-1/2g'bcε'cdeB'e;d);a
Now the mess on the right can be written
(g'-1/2g'bcε'cdeB'e;d);a = (g'-1/2g'bcε'cde);aB'e;d + (g'-1/2g'bcε'cde)B'e;d;a
(g'-1/2g'bcε'cdeB'e;d);a = (g'-1/2g'bcε'cde);aB'e;d + (g'-1/2g'bcε'ckm)B'm;k;a
Then we have
(g'nm g'jk - g'km g'nj )B'm;k;j = – g'-1/2ε'nab *
{(g'-1/2g'bcε'cde);aB'e;d + (g'-1/2g'bcε'ckm)B'm;k;a
or
(g'nm g'jk - g'km g'nj )B'm;k;j = – g'-1/2ε'njb *
{(g'-1/2g'bcε'cde);jB'e;d + (g'-1/2g'bcε'ckm)B'm;k;j }
or
(g'nm g'jk - g'km g'nj )B'm;k;j = – g'-1/2ε'njb(g'-1/2g'bcε'cde);jB'e;d
– g'-1/2ε'njb (g'-1/2g'bcε'ckm)B'm;k;j
Then we put all the B'm;k;j on the left
[g'nm g'jk - g'km g'nj + g'-1/2ε'njb (g'-1/2g'bcε'ckm) ]B'm;k;j
= – g'-1/2ε'njb(g'-1/2g'bcε'cde);j B'e;d
It does not seem very likely! You would think the B'm;k;j would be independent of the B'e;d .
the only rescue would be if both coefficients vanished. The second one vanishing would require that
Qned ≡ ε'njb(g'-1/2g'bcε'cde);j = 0 ?
Write as
(g'-1/2g'bcε'cde);j = [g'-1/2ε'cde];j g'bc + [g'-1/2ε'cde] g'bc;j
= [g'-1/2ε'cde];j g'bc
Then we have to show that
Pned = ε'njb [g'-1/2ε'cde];j g'bc = 0
One way would be to show that ε'njb [g'-1/2ε'cde];j is antisymmetric in b and c, which means we would need to show that
ε'njb [g'-1/2ε'cde];j = - ε'njc [g'-1/2ε'bde];j
Now
(g'-1/2ε'cde) ? g' has weight -2 g'-1/2 has weight +1 ε'cde W = -1
Therefore (g'-1/2ε'cde) has weight 0. Is this then a true tensor? Yes! So define
Q'abc = (g'-1/2ε'abc)
Then
Q'abc;j = Q'abc,j + + ΓanjQnbc + ΓbnjQanc + ΓcnjQabn
Look just at the comma term:
ε'njb [g'-1/2ε'cde],j = Σbc ε'njbε'cde ∂'j(g'-1/2)g'bc does not vanish
So things grind to a halt, I am disturbed by the fact that I cannot show this fact. It casts doubt on my claim in tensor doc. It is true that in Cartesian space both sides vanish since g' and ε' = constants.