# Affine connection in general relativity

In The GR lecture my teacher deduced the relation between affine connection and the metric tensor according to the following way: He firstly wrote the relationship of two tensors like this (I understand): $$A^{\mu }{}_{[P\to Q]}=A^{\mu }{}_{[P]}-\Gamma _{\nu \sigma }^{\mu }{}_{[P]}A^{\nu }{}_{[P]}dx^{\sigma } \tag{1}$$

Then he wrote (I understand): $$g_{\mu \nu (Q)}=dx^{\sigma } g_{\mu \nu ,\sigma (P)}+g_{\mu \nu (P)}\tag{2}$$

If the vector is to have the same length after been transported, we have: $$g_{\mu \nu (Q)}A^{\mu }{}_{[P\to Q]}A^{\nu }{}_{[P\to Q]}=g_{\mu \nu (P)}A^{\mu }{}_{[P]}A^{\nu }{}_{[P]}\tag{3}$$

The above three equations give the result which connects the affine connection with metric tensor: $$g_{\mu \nu ,\sigma }-g_{\alpha \nu } \Gamma _{\mu \sigma }^{\alpha }-g_{\mu \alpha } \Gamma _{\nu \sigma }^{\alpha }=0\tag{4}$$

1. $\Gamma$ is symmetric

2. the length should be same.

My questions are:

1. In fact the derivation of $$A^{\mu }{}_{[P\to Q]}=A^{\mu }{}_{[P]}-\Gamma _{\nu \sigma }^{\mu }{}_{[P]}A^{\nu }{}_{[P]}dx^{\sigma } \tag{5}$$ can at the same time give $$\Gamma _{\mu \nu }^{'\lambda }=\frac{\partial x^{'\lambda }}{\partial x^{\rho }}\frac{\partial x^{\tau }}{\partial x^{'\mu }}\frac{\partial x^{\sigma }}{\partial x^{'\nu }}\Gamma _{\tau \sigma }^{\rho }+\frac{\partial x^{'\lambda }}{\partial x^{\rho }}\frac{\partial ^2x^{\rho }}{\partial x^{'\mu }\partial x^{'\nu }}.\tag{6}$$ Since we can always choose a coordinate such that $$\Gamma _{\mu \nu }^{'\lambda }=\frac{\partial x^{'\lambda }}{\partial x^{\rho }}\frac{\partial ^2x^{\rho }}{\partial x^{'\mu }\partial x^{'\nu }},\tag{7}$$ we can see that $\Gamma$ is already symmetric (since we can exchange the indices), and thus we don't need to add this condition. Is my statement here true?
2. I read from Weinberg's Gravitation and Cosmology on P.7 (3.2.4) that he derived
$$\Gamma _{\mu \nu }^{'\lambda }=\frac{\partial x^{'\lambda }}{\partial x^{\rho }}\frac{\partial ^2x^{\rho }}{\partial x^{'\mu }\partial x^{'\nu }}\tag{8}$$ simply from geodesics, while for geodesics the length of the vectors need not be the same. Then Weinberg still derived $$g_{\mu \nu ,\sigma }-g_{\alpha \nu } \Gamma _{\mu \sigma }^{\alpha }-g_{\mu \alpha } \Gamma _{\nu \sigma }^{\alpha }=0 \tag{9}$$ without using the the second condition. But, my teacher did use that one. So what's wrong with the conditions?

3. To explain my last confuse better:

I just want to know the difference between the affine connections defined by my teacher and Weinberg. It seems that both have the same expression. But my teacher's seemed to need another "same length" condition ,while Weinberg's didn't ,to derive the final result of $$g_{\mu \nu ,\sigma }-g_{\alpha \nu } \Gamma _{\mu \sigma }^{\alpha }-g_{\mu \alpha } \Gamma _{\nu \sigma }^{\alpha }=0.\tag{10}$$ What properties can the $\Gamma$given by my teacher have ( parallel? the same length?), and what properties of Weinberg's (parallel? the same length?)？

• Small comment: the notation in Eq.s 1-3 is incomprehensible. Oct 10, 2016 at 14:09
• @Qmechanic I think I get it. If as your Physics SE states, When having this expression $\Gamma _{\mu \nu }^{'\lambda }=\frac{\partial x^{'\lambda }}{\partial x^{\rho }}\frac{\partial ^2x^{\rho }}{\partial x^{'\mu }\partial x^{'\nu }}\tag{8}$, one already chooses $\Gamma$ to be zero. Oct 11, 2016 at 12:18
• Then this is already a flat space. Therefore Weinberg's derivation has already assumed a flat space to deduce everything and the $\Gamma$ is of course a parallel and 'same length' and torsion free connection. While however my teacher's connection is only parallel, therefore he needed 'same length' and torsion free. By the way, is it true that if we have the Riemann curvature to be zero and torsion free then we can have $\Gamma$ equals zero??? Oct 11, 2016 at 12:18
• $\uparrow$ Yes. Oct 11, 2016 at 12:25