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 $A^i$ at $x^i$, and at a neighbouring point, the vector is $A^i+d{A}^i$ at $x^i+d{x}^i$.

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

$D{A}^i=d{A}^i-\delta A^i$

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

$\delta A^i=-{\mathrm{\Gamma }}_{kl}^i{A}^{k}d{x}^l$

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:

$B_i\delta A^i+A^i\delta B_i=0$

$\Rightarrow A^i\delta B_i={\mathrm{\Gamma }}_{kl}^i{A}^k{B}_{i}d{x}^l$

$\Rightarrow A^i\delta B_i={\mathrm{\Gamma }}_{il}^k{A}^i{B}_{k}d{x}^l$

$\Rightarrow \delta B_i={\mathrm{\Gamma }}_{il}^k{B}_{k}d{x}^l$

Connection Between Covariant And Regular Derivatives

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

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

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

Analogously, for tensors:

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

Calculation of Christoffel Symbols

From $g_{ik}D{A}^k+A^{k}D{g}_{ik}=D\left(g_{ik}{A}^k\right)=D{A}_i=g_{ik}D{A}^k$, 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:

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

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

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

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

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

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

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

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

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

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