covariant derivatives and christoffel symbols

**oblixps** · March 11th 2011, 11:39 PM

$0=\frac{\partial{g_{mn}}}{\partial{y^p}}+ \Gamma_{pm}^sg_{sn} + \Gamma_{pn}^rg_{mr}$

i am trying to verify this equality so here is what i did so far:

$\frac{\partial{g_{mn}}}{\partial{y^p}}+\frac{1}{2} g^{sd}(\frac{\partial{g_{dm}}}{\partial{y^p}}+\fra c{\partial{g_{dp}}}{\partial{y^m}}-\frac{\partial{g_{pm}}}{\partial{y^d}})g_{sn}+\fra c{1}{2}g^{rz}(\frac{\partial{g_{zn}}}{\partial{y^p }}+\frac{\partial{g_{zp}}}{\partial{y^n}}-\frac{\partial{g_{pn}}}{\partial{y^z}})g_{mr}$

then since $g^{sd}g_{sn}=\delta^d_n$ i now have

$\frac{\partial{g_{mn}}}{\partial{y^p}}+\frac{1}{2} \frac{\partial{g_{nm}}}{\partial{y^p}}+\frac{1}{2} \frac{\partial{g_{np}}}{\partial{y^m}}-\frac{1}{2}\frac{\partial{g_{pm}}}{\partial{y^n}}+ \frac{1}{2}\frac{\partial{g_{mn}}}{\partial{y^p}}+ \frac{1}{2}\frac{\partial{g_{mp}}}{\partial{y^n}}-\frac{1}{2}\frac{\partial{g_{pn}}}{\partial{y^m}}$

and i seem to be stuck here. have what i've been doing so far correct and if so how may i continue. thanks.

**oblixps** · March 12th 2011, 04:05 PM

oops i had a few typos in the christoffel symbols. now that they are fixed i had some nice cancellations.

now i am left with:
$0=2\frac{\partial{g_{mn}}}{\partial{y^p}}$ and since the derivative of g is 0 and since it is a tensor and has a derivative of zero in cartesian coordinates, it must have the same for all coordinates and therefore the identity is 0 = 0. is this the correct reasoning? thanks.

**oblixps** · March 13th 2011, 07:15 PM

i am still confused as i don't know how to get rid of the terms left over which is $2\frac{\partial{g_{mn}}}{\partial{y^p}}$ where g is the metric tensor.

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