Covariant derivative transformation

[faklif]

Homework Statement

The problem concerns how to transform a covariant differentiation. Using this formula for covariant differentiation and demanding that it is a (1,1) tensor:
$$\nabla_cX^a=\partial_cX^a+\Gamma^a_{bc}X^b$$
it should be proven that
$$\Gamma'^a_{bc}= \frac{{\partial}x'^a}{{\partial}x^d} \frac{{\partial}x^e}{{\partial}x'^b} \frac{{\partial}x^f}{{\partial}x'^c} \Gamma^d_{ef} - \frac{{\partial}x^d}{{\partial}x'^b} \frac{{\partial}x^e}{{\partial}x'^c} \frac{{\partial}^2x'^a}{{{\partial}x^d}{{\partial}x^e}}$$

Homework Equations

$$\partial'_cX'^a= \frac{{\partial}x^f}{{\partial}x'^c}\frac{\partial}{{\partial}x^f} ( \frac{{\partial}x'^a}{{\partial}x^d}X^d )= \frac{{\partial}x'^a}{{\partial}x^d} \frac{{\partial}x^f}{{\partial}x'^c} \frac{\partial}{{\partial}x^f} X^d + \frac{{\partial}^2x'^a}{{\partial}x^f{\partial}x^d} \frac{{\partial}x^f}{{\partial}x'^c} X^d$$

$$X^d = \frac{{\partial}x^d}{{\partial}x'^b} \frac{{\partial}x'^b}{{\partial}x^d} X^d = \frac{{\partial}x^d}{{\partial}x'^b} X'^b$$
$$X^e = \frac{{\partial}x^e}{{\partial}x'^b} X'^b$$

The Attempt at a Solution

I thought I'd simply look at the primed coordinates from two directions and that that should do it... didn't quite happen as planned though.

In primed coordinates
$$\nabla_cX^a' = {\partial}'_cX'^a + \Gamma^a_{bc}'X'^b$$

Transformed using the equations under 2.
$$\nabla_cX^a' = \frac{{\partial}x'^a}{{\partial}x^d} \frac{{\partial}x^f}{{\partial}x'^c} \frac{\partial}{{\partial}x^f} X^d + \frac{{\partial}x'^a}{{\partial}x^d} \frac{{\partial}x^f}{{\partial}x'^c} \Gamma^d_{ef}X^e = {\partial}'_cX'^a - \frac{{\partial}^2x'^a}{{\partial}x^f{\partial}x^d} \frac{{\partial}x^f}{{\partial}x'^c} X^d + \frac{{\partial}x'^a}{{\partial}x^d} \frac{{\partial}x^f}{{\partial}x'^c} \Gamma^d_{ef}X^e = {\partial}'_cX'^a - \frac{{\partial}^2x'^a}{{\partial}x^f{\partial}x^d} \frac{{\partial}x^f}{{\partial}x'^c} \frac{{\partial}x^d}{{\partial}x'^b} X'^b + \frac{{\partial}x'^a}{{\partial}x^d} \frac{{\partial}x^f}{{\partial}x'^c} \Gamma^d_{ef} \frac{{\partial}x^e}{{\partial}x'^b} X'^b$$

So my thought was that anything related to X'^b would be the what I seek which gives

$$\Gamma'^a_{bc}= \frac{{\partial}x'^a}{{\partial}x^d} \frac{{\partial}x^e}{{\partial}x'^b} \frac{{\partial}x^f}{{\partial}x'^c} \Gamma^d_{ef} - \frac{{\partial}x^d}{{\partial}x'^b} \frac{{\partial}x^f}{{\partial}x'^c} \frac{{\partial}^2x'^a}{{\partial}x^f{\partial}x^d}$$

This i kind of close but it's different in the last part where I have f:s instead of e:s. I would really appreciate some help, I've been stuck for a long time.

 

[tiny-tim]

Using this formula for covariant differentiation and demanding that it is a (1,1) tensor:
$$\nabla_cX^a=\partial_cX^a+\Gamma^a_{bc}X^b$$
it should be proven that
$$\Gamma'^a_{bc}= \frac{{\partial}x'^a}{{\partial}x^d} \frac{{\partial}x^e}{{\partial}x'^b} \frac{{\partial}x^f}{{\partial}x'^c} \Gamma^d_{ef} - \frac{{\partial}x^d}{{\partial}x'^b} \frac{{\partial}x^e}{{\partial}x'^c} \frac{{\partial}^2x'^a}{{{\partial}x^d}{{\partial}x^e}}$$
…
$$\Gamma'^a_{bc}= \frac{{\partial}x'^a}{{\partial}x^d} \frac{{\partial}x^e}{{\partial}x'^b} \frac{{\partial}x^f}{{\partial}x'^c} \Gamma^d_{ef} - \frac{{\partial}x^d}{{\partial}x'^b} \frac{{\partial}x^f}{{\partial}x'^c} \frac{{\partial}^2x'^a}{{\partial}x^f{\partial}x^d}$$

This i kind of close but it's different in the last part where I have f:s instead of e:s. I would really appreciate some help, I've been stuck for a long time.

Hi faklif!

Do you mean you're worried about having ∂x^f and /∂x^f instead of ∂x^e and /∂x^e ?

They're "dummy variables" … it doesn't matter what they are!

 

[faklif]

Hi faklif!

Do you mean you're worried about having ∂x^f and /∂x^f instead of ∂x^e and /∂x^e ?

They're "dummy variables" … it doesn't matter what they are!

Hi and thank you!

Now that I think some more about it (should have done that before spending the better part of the weekend , having to much spare time makes me stupid...) it makes sense. For a given a,b,c you'd still get the same implied summations. Thanks again for rescuing my monday!