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 Exercise [14.07] 
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Joined: 26 Mar 2010, 04:39
Posts: 109
Post Exercise [14.07]
Using the results and notation from my response to exercise 14.4 we have
\nabla_b(\nabla_a \Phi) = {\partial(\nabla_a \Phi)\over{\partial x^b}} - (\nabla_c \Phi)\Gamma^c_{ab}= {\partial^2\Phi\over{\partial x^a \partial x^b}} - {\partial\Phi\over{\partial x^c}}\Gamma^c_{ab}
Hence
(\nabla_a\nabla_b-\nabla_b\nabla_a) \Phi = {\partial\Phi\over{\partial x^a}}(\Gamma^c_{ab}-\Gamma^c_{ba})
Hence the torsion tensor is given by:
\tau_{ab}^{\ \ c} = \Gamma^c_{ab}-\Gamma^c_{ba}


04 May 2010, 07:10
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Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Exercise [14.07]
You don't seem to have explained why the right-hand side must have the general form given - the first part of the exercise.


04 May 2010, 10:19

Joined: 26 Mar 2010, 04:39
Posts: 109
Post Re: Exercise [14.07]
The form can be guessed because it is clear that the derivative term must vanish due to the symmetry of partial derivatives (i.e. \partial_a \partial_b = \partial_b \partial_a).

The derivation that I have shown should also be convincing proof of why the form is correct since it provides the expression for the torsion explicitly.


04 May 2010, 11:09
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