1. Dec 27, 2012

### 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.

2. Dec 27, 2012

### 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 $\partial _{k}g_{ij} = \partial _{k}(e_{i}\cdot e_{j}) = e_{i}\cdot \partial _{k}e_{j} + e_{j}\cdot \partial _{k}e_{i}$.

3. Dec 27, 2012

### 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 -> $e_{i} = \partial _{i}$.

4. Dec 27, 2012

### 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.

5. Dec 27, 2012

### WannabeNewton

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

6. Dec 27, 2012

### 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.

7. Dec 27, 2012

### WannabeNewton

Yep! Good luck!

8. Dec 27, 2012

Thanks!