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Covariant derivative, timelike vector field (Kerr Schild Class) 
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#1
Apr1210, 01:13 PM

P: 5

Hi!
It seems I have some major problems in understanding covariant derivatives in concrete calculations. I discovered in an article dealing with Kerr Schild classes the following: In calculating the Riemann Tensor from the Christoffel symbols one gets the following expression: [tex]C^{a}_{ac}[/tex]=[tex]1/2[/tex][tex]\left\{[/tex][tex]\partial_{a}\left(k^{a}k_{c}f\right)+\partial_{c}\left(k^{a}k_{a}f\righ t)\partial^{a}\left(k_{c}k_{a}f\right)+fk^{a}\left(k\partial\right)\left( k_{c}k_{a}\right)\right\}[/tex] where since [tex]k^{a}[/tex] is lightlike [tex]k^{a}k_{a}=0[/tex] holds,and since we are dealing with Kerr Schild class the indices are raised and lowered by [tex]\eta^{ab}[/tex] the Lorentz metric and [tex]\left(k^{a}\nabla_a\right)k^{c}=\left(k^{a}\partial_a\right)k^{c}=0[/tex] geodetic and f is a scalar function. This expression should give zero which I cannot reproduce. I used the product rule on every term except the second one which is obviously zero. I get the following result [tex]C^{a}_{ac}=k_{c}k^{a}\partial_{a}f+k_{c}f\partial_{a}k^{a}+fk^{a}\parti al_{a}k_{c}f k_{a}\partial^{a}k_{c}fk_{c}\partial^{a}k_{a}k_{c}k_{a}\partial^{a}f+f^{2}k^{a}k_{c}\left(k\partial\right)k_{a} [/tex] where all terms should cancel to zero. Can someone explain this to me, would be very grateful. I think one need not know much about Kerr Schild classes to explain this. Thank you 


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