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 Exercise [14.08] 
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Joined: 26 Mar 2010, 04:39
Posts: 109
Post Exercise [14.08]
Recall from the Exercise 14.7 that the torsion tensor is given by:
\tau^{\ \ c}_{ab} = \Gamma^c_{ab}-\Gamma^c_{ba}

Writing
\nabla_b(\nabla_a \xi^d) = {\partial(\nabla_b \xi^d)\over{\partial x^a}} + (\nabla_b \xi^c)\Gamma^d_{ca} - <br />(\nabla_c \xi^d)\Gamma^c_{ba}
we can see immediately that the torsion contribution will only come from the final term and so:
(\nabla_a\nabla_b-\nabla_b\nabla_a) \xi^d =\\<br />\text{\hspace{1cm}} {\partial(\nabla_b \xi^d)\over{\partial x^a}} - {\partial(\nabla_a \xi^d)\over{\partial x^b}} + (\nabla_b \xi^c)\Gamma^d_{ca}- (\nabla_a \xi^c)\Gamma^d_{cb} + <br />\tau^{\ \ c}_{ab}(\nabla_c \xi^d)
Hence the general expression including torsion is:
(\nabla_a\nabla_b-\nabla_b\nabla_a) \xi^d = R_{abc}^{\ \ \ d}\xi^c + <br />\tau^{\ \ c}_{ab}(\nabla_c \xi^d)

Although not required for this question, the curvature tensor can be calculated by simplifying the the first four terms in the previous equation to get:
R_{abc}^{\ \ \ d} =  {\partial\over{\partial x^a}}\Gamma^d_{cb} - {\partial\over{\partial x^b}}\Gamma^d_{ca} + \Gamma^e_{cb}\Gamma^d_{ea}- \Gamma^e_{ca}\Gamma^d_{eb}
(I show the details of how to arrive at this result in my response to Exercise 14.27)


04 May 2010, 08:07

Joined: 23 Aug 2010, 13:12
Posts: 33
Post Re: Exercise [14.08]
I had a lot of trouble with this derivation until I realized that the a and b indices on the left hand side of the second equation (the one following "writing") should be interchanged.


24 Aug 2010, 12:51

Joined: 26 Mar 2010, 04:39
Posts: 109
Post Re: Exercise [14.08]
Yes - well spotted again. It's really easy to mess things up typesetting all those indices.


25 Aug 2010, 05:06
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