Confusion about Christoffel Symbols

[tensor33]

In the book Mathematical Methods for Physics and Engineering by Riley, Hobson, and Bence, I came across an equation I just can't seem to understand. In the chapter on tensors, they derive the equation for a Christoffel symbol of the second kind, \Gamma^{m}_{ij}=\frac{1}{2}g^{mk}\left(\frac{ \partial g_{jk}}{\partial u^{i}}+\frac{\partial g_{ki}}{\partial u^{j}}-\frac{\partial g_{ij}}{\partial u^k}\right)
Where the g's are the components of the metric tensor. I understood most of the derivation except for the part where they wrote, \frac{ \partial g_{ij}}{\partial u^{k}}= \frac{\partial e_{i}}{\partial u^{k}} \cdot e_{j}+e_{i} \cdot \frac{\partial e_{j}}{\partial u^{k}}
Where the e's are the basis vectors. I just can't seem to understand how they got this equation.

 

[WannabeNewton]

You can compute the components of the metric tensor with respect to the given basis for the tangent space at hand much like you would the first fundamental form on regular surfaces i.e. g_{ij} = <e_{i}, e_{j}> so we have that \partial _{k}g_{ij} = \partial _{k}(e_{i}\cdot e_{j}) = e_{i}\cdot \partial _{k}e_{j} + e_{j}\cdot \partial _{k}e_{i}.

 

[WannabeNewton]

By the way I assume the basis vectors you have written down are the usual coordinate basis vectors, in terms of a given local chart, for the tangent space at p -> e_{i} = \partial _{i}.

 

[tensor33]

I get the part where g_{ij}=(e_{i}\cdot e_{j}) the part I don't get is \partial_{k}(e_{i}\cdot e_{j})=\partial_{k}e_{i}\cdot e_{j}+e_{i}\cdot \partial_{k}e_{j} I just don't understand how those two are equal.

 

[WannabeNewton]

I get the part where g_{ij}=(e_{i}\cdot e_{j}) the part I don't get is \partial_{k}(e_{i}\cdot e_{j})=\partial_{k}e_{i}\cdot e_{j}+e_{i}\cdot \partial_{k}e_{j} I just don't understand how those two are equal.

It's just the Leibniz rule for dot products, kind of like the usual product rule.

 

[tensor33]

I knew there had to be some rule but I couldn't find it. I looked it up and now it makes sense. Thanks for the reply.

 

[WannabeNewton]

I knew there had to be some rule but I couldn't find it. I looked it up and now it makes sense. Thanks for the reply.

Yep! Good luck!

 

[tensor33]

Thanks!