How do I proove this? I think it's something related to "index gymnastics"... $$\Gamma^i{}_{ki}=\frac{1}{2} g^{im}\partial_k g_{im}=\frac{1}{2g} \partial_k g =\partial_k \ln \sqrt{|g|} $$ Here's where I get stuck: $$\Gamma^i_{.ki}=g^{im}\Gamma_{mki}=\frac{1}{2}g^{im}(\partial_i g_{mk}+\partial_k g_{mi}-\partial_m g_{ki})$$
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2If I recall correctly the second formula is exactly the definition of the Christoffel symbols right? – Sak Jan 02 '14 at 16:52
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Yes... I don't know how they got the first formula – BinaryBurst Jan 02 '14 at 16:53
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What do you mean you get stuck on the definition of the symbols? – Sak Jan 02 '14 at 16:55
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You're one step away. The first and third terms in the last step of your work cancel (the metric tensor is symmetric, and the sums implied by the Einstein summation notation are hence the same). – Henry T. Horton Jan 02 '14 at 16:58
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I know that the metric tensor is symmetric.. but $\partial_i g_{mk}=\partial_m g_{ik}$ ? – BinaryBurst Jan 02 '14 at 17:01
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I know $g_{ab}=g_{ba}$ :) – BinaryBurst Jan 02 '14 at 17:02
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http://math.stackexchange.com/questions/619853/metric-on-an-open-subset-of-mathbbrd-and-christoffel-symbol-of-the-second/619922#619922 – Stephen Montgomery-Smith Jan 02 '14 at 19:08
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Let's start with your work: $$\Gamma^i_{\phantom{i}ki}=g^{im}\Gamma_{mki}=\frac{1}{2}g^{im}(\partial_i g_{mk}+\partial_k g_{mi}-\partial_m g_{ki}).$$ Notice that \begin{align*} g^{im}\partial_i g_{mk} & = g^{im} \partial_i g_{km} \tag{symmetry of $g$} \\ & = g^{mi} \partial_m g_{ki} \tag{relabeling dummy indices} \\ & = g^{im} \partial_m g_{ki}. \tag{symmetry of $g$} \end{align*} Therefore \begin{align*} \frac{1}{2}g^{im}(\partial_i g_{mk}+\partial_k g_{mi}-\partial_m g_{ki}) & = \frac{1}{2}g^{im}\partial_k g_{mi} \\ & = \frac{1}{2} g^{mi} \partial_k g_{mi} \\ & = \frac{1}{2g} \partial_k g \\ & = \partial_k \ln \sqrt{|g|}. \end{align*}
Henry T. Horton
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Why is $\frac{1}{2} g^{mi} \partial_k g_{mi} = \frac{1}{2g} \partial_k g $ ? – BinaryBurst Jan 02 '14 at 17:50
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