# 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 $\Gamma _{kl}^{i}$ :

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

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:

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

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

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

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

## Connection Between Covariant And Regular Derivatives

From above, one can obtain the relations between covariant derivatives and regular derivatives:

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

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

Analogously, for tensors:

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

## 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:

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

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

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

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

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

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

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

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

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

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