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 Exercise [14.27] 
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Joined: 26 Mar 2010, 04:39
Posts: 109
Post Exercise [14.27]
In Exercise 14.8 I showed that:

R_{abc}^{\ \ \ d}\xi^c = {\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}

We have
{\partial(\nabla_b \xi^d)\over{\partial x^a}} = {\partial\over{\partial x^a}} \left({\partial\xi^d\over{\partial x^b}} + \xi^c\Gamma^d_{cb}\right) = {\partial^2\xi^d\over{\partial x^a\partial x^b}} + {\partial\xi^c\over{\partial x^a}}\Gamma^d_{cb} + \xi^c{\partial\over{\partial x^a}} \Gamma^d_{cb}
and hence
{\partial(\nabla_b \xi^d)\over{\partial x^a}} - {\partial(\nabla_a \xi^d)\over{\partial x^b}}={\partial\xi^c\over{\partial x^a}}\Gamma^d_{cb} - <br />{\partial\xi^c\over{\partial x^b}}\Gamma^d_{ca} + <br />\xi^c{\partial\over{\partial x^a}} \Gamma^d_{cb} - <br />\xi^c{\partial\over{\partial x^b}} \Gamma^d_{ca}

We also have
(\nabla_b \xi^c)\Gamma^d_{ca} = <br />{\partial\xi^c\over{\partial x^b}}\Gamma^d_{ca} + \xi^e\Gamma^c_{eb}\Gamma^d_{ca}
and thus
(\nabla_b \xi^c)\Gamma^d_{ca}- (\nabla_a \xi^c)\Gamma^d_{cb} = <br />{\partial\xi^c\over{\partial x^b}}\Gamma^d_{ca} + \xi^e\Gamma^c_{eb}\Gamma^d_{ca} -<br />{\partial\xi^c\over{\partial x^a}}\Gamma^d_{cb} - \xi^e\Gamma^c_{ea}\Gamma^d_{cb}

Putting this all together we get:
R_{abc}^{\ \ \ d}\xi^c = <br />\xi^c{\partial\over{\partial x^a}} \Gamma^d_{cb} - <br />\xi^c{\partial\over{\partial x^b}} \Gamma^d_{ca} +<br /> \xi^e\Gamma^c_{eb}\Gamma^d_{ca} -<br /> \xi^e\Gamma^c_{ea}\Gamma^d_{cb}\\<br />= ({\partial\over{\partial x^a}} \Gamma^d_{cb} - <br />{\partial\over{\partial x^b}} \Gamma^d_{ca} +<br />\Gamma^e_{cb}\Gamma^d_{ea} -<br /> \Gamma^e_{ca}\Gamma^d_{eb})\xi^c
Hence
R_{abc}^{\ \ \ d} = <br /> {\partial\over{\partial x^a}} \Gamma^d_{cb} - <br />{\partial\over{\partial x^b}} \Gamma^d_{ca} +<br />\Gamma^e_{cb}\Gamma^d_{ea} -<br />\Gamma^e_{ca}\Gamma^d_{eb}


08 May 2010, 12:15
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