General Relativity/Christoffel symbols

( << Back to General Relativity)

Definition of Christoffel Symbols

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

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

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

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

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

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 ${\displaystyle A^{i}}$ and ${\displaystyle B_{i}}$ respectively. Since ${\displaystyle A^{i}B_{i}}$ is a scalar, ${\displaystyle \delta (A^{i}B_{i})=0}$, one arrives at:

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

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

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

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

Connection Between Covariant And Regular Derivatives

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

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

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

Analogously, for tensors:

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

Calculation of Christoffel Symbols

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

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

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

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

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

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

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

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

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

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

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

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