General Relativity/Christoffel symbols

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Definition of Christoffel Symbols
Consider an arbitrary contravariant vector field defined all over a Lorentzian manifold, and take $$A^{i}$$ at $$x^{i}$$, and at a neighbouring point, the vector is $$A^{i}+dA^{i}$$ at $$x^{i}+dx^{i}$$.

Next parallel transport $$A^{i}$$ from $$x^{i}$$ to $$x^{i}+dx^{i}$$, and suppose the change in the vector is $$\delta A^{i}$$. Define:

$$DA^{i} = dA^{i} -\delta A^{i}$$

The components of $$\delta A^{i}$$ must have a linear dependence on the components of $$A^{i}$$. Define Christoffel symbols $$\Gamma^{i}_{kl}$$:

$$\delta A^{i} = -\Gamma^{i}_{kl} A^{k} dx^{l}$$

Note that these Christoffel symbols are:  dependent on the coordinate system (hence they are NOT tensors) functions of the coordinates 

Now consider arbitrary contravariant and covariant vectors $$A^{i}$$ and $$B_{i}$$ respectively. Since $$A^{i}B_{i}$$ is a scalar, $$\delta (A^{i}B_{i}) = 0$$, one arrives at:

$$B_{i}\delta A^{i} + A^{i}\delta B_{i} = 0$$

$$\Rightarrow A^{i}\delta B_{i} = \Gamma^{i}_{kl} A^{k} B_{i} dx^{l}$$

$$\Rightarrow A^{i}\delta B_{i} = \Gamma^{k}_{il} A^{i} B_{k} dx^{l}$$

$$\Rightarrow \delta B_{i} = \Gamma^{k}_{il} B_{k} dx^{l}$$

Connection Between Covariant And Regular Derivatives
From above, one can obtain the relations between covariant derivatives and regular derivatives:

$${A^{i}}_{;l} = \frac{\partial A^{i}}{\partial x^{l}} + \Gamma^{i}_{kl} A^{k}$$

$$A_{i;l} = \frac{\partial A_{i}}{\partial x^{l}} - \Gamma^{k}_{il} A_{k}$$

Analogously, for tensors:

$${A^{ik}}_{;l} = \frac{\partial A^{ik}}{\partial x^{l}} + \Gamma^{i}_{ml}A^{mk} + \Gamma^{k}_{ml}A^{im}$$

Calculation of Christoffel Symbols
From $$g_{ik}DA^{k} + A^{k}Dg_{ik} = D\left(g_{ik}A^{k}\right) = DA_{i} = g_{ik} DA^{k}$$, one can conclude that $$g_{ik;l} = 0$$.

However, since $$g_{ik}$$ is a tensor, its covariant derivative can be expressed in terms of regular partial derivatives and Christoffel symbols:

$$g_{ik;l} = \frac{\partial g_{ik}}{\partial x^{l}} - g_{mk}\Gamma^{m}_{il} - g_{im}\Gamma^{m}_{kl} = 0$$

Rewriting the expression above, and then performing permutation on i, k and l:

$$\frac{\partial g_{ik}}{\partial x^{l}} = g_{mk}\Gamma^{m}_{il} + g_{im}\Gamma^{m}_{kl}$$

$$\frac{\partial g_{kl}}{\partial x^{i}} = g_{ml}\Gamma^{m}_{ki} + g_{km}\Gamma^{m}_{li}$$

$$-\frac{\partial g_{li}}{\partial x^{k}} = - g_{mi}\Gamma^{m}_{lk} - g_{lm}\Gamma^{m}_{ik}$$

Adding up the three expressions above, one arrives at (using the notation $$\frac{\partial A^{i}}{\partial x^{j}} = {A^{i}}_{,j}$$):

$$2 g_{mk}\Gamma^{m}_{il} = g_{ik,l} + g_{kl,i} - g_{li,k}$$

Multiplying both sides by $$\frac{1}{2} g^{kn}$$:

$$\Rightarrow \Gamma^{n}_{il} = \delta^{n}_{m}\Gamma^{m}_{il} = \frac{1}{2} g^{kn}\left(g_{ik,l} + g_{kl,i} - g_{li,k}\right)$$

Hence if the metric is known, the Christoffel symbols can be calculated.