Is there an error in this contracted calculation involving the Riemann and Ricci tensors?
Jay R. Yablon
R^a_buv A_a = [&;_u, &;_v]A_b (2)
which defines the Riemann tensor. The calculation is in the one-page
file
http://jayryablon.files.wordpress.com/2009/10/can-you-find-the-error-in-this-calculation.pdf
The contracted result I end up with is:
R_au A^a = [&;_a, &;u] A^a (2)
where R_au is the Ricci tensor, but the mixed symmetry and antisymmetry
strikes me funny. I would like to know if this is a correct
calculation, and if I am missing anything.
Thanks,
Jay
____________________________
Jay R. Yablon
Email: jya...@nycap.rr.com
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.roadrunner.com/~jry/FermionMass.htm
Oh No
> I have attempted to obtain a contracted version of the relationship:
>
>R^a_buv A_a = [&;_u, &;_v]A_b (2)
>
>which defines the Riemann tensor. The calculation is in the one-page
>file
>
>http://jayryablon.files.wordpress.com/2009/10/can-you-find-the-error-
>in-this-calculation.pdf
>
>The contracted result I end up with is:
>
>R_au A^a = [&;_a, &;u] A^a (2)
>
>where R_au is the Ricci tensor, but the mixed symmetry and antisymmetry
>strikes me funny. I would like to know if this is a correct
>calculation, and if I am missing anything.
>
I think the basic problem is that when you switch indices in R_mu_alpha
you correctly do not change sign, but then you seem to think this
enables you to do the same for the commutator. The bit you think follows
by considering R in isolation does not in fact follow.
Regards
--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)
Jay R. Yablon
"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message
news:x9x$LtAnVZ...@charlesfrancis.wanadoo.co.uk...
R_au A^a = [&;_a, &;u] A^a (2)
a correct result, and if not, at what exact step is there a mistake?
Jay.
Oh No
That looked right to me, depending possibly on sign conventions.