General Relativity/Christoffel symbols

From Wikibooks, open books for an open world
< General Relativity
Jump to: navigation, search

( << Back to General Relativity)

Definition of Christoffel Symbols[edit]

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[edit]

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[edit]

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.