The Road to Reality
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Exercise [14.19]
http://www.roadtoreality.info/viewtopic.php?f=19&t=1505
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Author:  robin [ 06 May 2010, 08:55 ]
Post subject:  Exercise [14.19]

First notice that
\pounds_\xi \eta = [\xi^a\partial_a, \eta^b \partial_b]\\<br /> = \xi^a\partial_a(\eta^b \partial_b) - \eta^a\partial_a(\xi^b \partial_b)\\<br />= \xi^a\partial_a\eta^b\partial_b- \eta^a\partial_a\xi^b\partial_b + \xi^a\eta^b \partial_a\partial_b- \eta^a\xi^b \partial_a\partial_b\\<br />=(\xi^a\partial_a\eta^b- \eta^a\partial_a\xi^b)\partial_b
In other words
(\pounds_\xi \eta)^b =\xi^a\partial_a\eta^b- \eta^a\partial_a\xi^b

Hence
\xi^a \nabla_a \eta^b - \eta^a \nabla_a \xi^b\\<br />= \xi^a \partial_a \eta^b + \xi^a\eta^c\Gamma^b_{ca} - \eta^a \partial_a \xi^b - \eta^a\xi^c\Gamma^b_{ca}\\<br />= (\pounds_\xi \eta)^b  - \eta^a\xi^c\Gamma^b_{ca}+ \xi^a\eta^c\Gamma^b_{ca}\\<br />= (\pounds_\xi \eta)^b  - \eta^c\xi^a\Gamma^b_{ac}+ \xi^a\eta^c\Gamma^b_{ca}\\<br />= (\pounds_\xi \eta)^b  - \xi^a\eta^c \tau_{ac}^{\ \ b}

In particular, if the torsion vanishes we have:
(\pounds_\xi \eta)^b = \xi^a \nabla_a \eta^b - \eta^a \nabla_a \xi^b

It's worth comparing this with the first expression for (\pounds_\xi \eta)^b. Notice that in the torsion-free case we can just swap partial derivatives for covariant derivatives (or vice versa). In other words, the Lie derviative is independent of the torsion-free connection.

Author:  deant [ 24 Oct 2010, 07:23 ]
Post subject:  Re: Exercise [14.19]

Sorry Robin, I don't quite understand how you justify your first line:

\pounds_\xi \eta = [\xi^a\partial_a, \eta^b\partial_b]

This is clearly true if you apply \pounds_\xi\eta to a scalar, but it's not obviously true in general; in fact it's a statement of the result we're trying to show, for the particular choice of the coordinate connection (\nabla = \partial).


I did it this way:

\pounds_\xi\eta = [\xi, \eta] = \xi(\eta(.)) - \eta(\xi(.))

Noting that given any particular connection \nabla, the action of \xi(.) on an arbitrary tensor \phi is \xi(\phi) = \xi^a\nabla_a\phi, we get:

[\pounds_\xi\eta]\phi = \xi(\eta^b\nabla_b\phi) - \eta(\xi^a\nabla_a\phi)
= \xi^a\nabla_a(\eta^b\nabla_b\phi) - \eta^b\nabla_b(\xi^a\nabla_a\phi)
= \xi^a(\nabla_a\eta^b)(\nabla_b\phi) + \xi^a\eta^b(\nabla_a\nabla_b\phi) - \eta^b(\nabla_b\xi^a)(\nabla_a\phi) - \eta^b\xi^a(\nabla_b\nabla_a\phi)
= (\nabla_\xi\eta)\phi - (\nabla_\eta\xi)\phi + \xi^a\eta^b(\nabla_a\nabla_b - \nabla_b\nabla_a)\phi
= [\nabla_\xi\eta - \nabla_\eta\xi + \xi^a\eta^b(\nabla_a\nabla_b - \nabla_b\nabla_a)]\phi

But we have assumed our (given) \nabla is torsion-free, so the third term in the brackets vanishes, and therefore:

\pounds_\xi \eta = \nabla_\xi\eta - \nabla_\eta\xi

for all torsion-free connections \nabla.

Author:  vasco [ 24 Oct 2010, 10:42 ]
Post subject:  Re: Exercise [14.19]

This is deant's solution attached to overcome the Latex problem:
Attachment:
RRpost.pdf [52.27 KiB]
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