Covariant derivative, timelike vector field (Kerr- Schild Class)

by Lurian
Tags: class, covariant, derivative, field, kerr, schild, timelike, vector
Lurian is offline
Apr12-10, 01:13 PM
P: 5
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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