Christoffel symbol

Description

Definition

I consider the Christoffel symbol of the second kind

\[\christoffel,\]

which will appear when I consider the change of basis vectors. Here I assume \(x^l\) is twice-differentiable to find \(\Gamma_{ij}^k = \Gamma_{ji}^k\).

Relation with scale factor

By using the relation for the transformation matrices:

\[\jacobiconv,\]

I have

\[\Gamma_{ij}^k = \frac{1}{H_k H_k} \sum_l \pder{}{X^i} \left( \pder{x^l}{X^j} \right) \pder{x^l}{X^k}.\]

Here I recall the orthogonality:

\[\orthogonal\]

to find

\[\metrictensor,\]

because

\[\begin{split}\vec{E}_i \cdot \vec{E}_j & = \left( \sum_k \pder{x^k}{X^i} \vec{e}_k \right) \cdot \left( \sum_l \pder{x^l}{X^j} \vec{e}_l \right) \\ & = \sum_{kl} \pder{x^k}{X^i} \pder{x^l}{X^j} \vec{e}_k \cdot \vec{e}_l \\ & = \sum_{kl} \pder{x^k}{X^i} \pder{x^l}{X^j} \delta_{kl} \\ & = \sum_k \pder{x^k}{X^i} \pder{x^k}{X^j}.\end{split}\]

Furthermore, I consider its derivative with respect to \(X^l\):

\[\begin{split}\pder{}{X^l} \left( H_i H_j \delta_{ij} \right) & = \pder{}{X^l} \left( \sum_k \pder{x^k}{X^i} \pder{x^k}{X^j} \right) \\ & = \sum_k \pder{}{X^l} \left( \pder{x^k}{X^i} \right) \pder{x^k}{X^j} + \sum_k \pder{}{X^l} \left( \pder{x^k}{X^j} \right) \pder{x^k}{X^i} \\ & = H_j H_j \Gamma_{il}^j + H_i H_i \Gamma_{jl}^i,\end{split}\]

which is equivalent to

\[ \begin{align}\begin{aligned}\newcommand{\csymb}[3]{ H_{#1} H_{#1} \Gamma_{#2#3}^{#1} = \pder{}{X^#3} \left( H_{#1} H_{#2} \delta_{#1#2} \right) - H_{#2} H_{#2} \Gamma_{#1#3}^{#2} } \csymb{i}{j}{l},\\\csymb{j}{l}{i},\\\csymb{l}{i}{j}.\end{aligned}\end{align} \]

Finally I obtain an explicit relation of the Christoffel symbol of the second kind and the scale factors (the indices are interchanged for later convenience):

\[\fromchristoffeltoscalefactor.\]

Useful relations

Assuming \(k = j\) yields

\[\Gamma_{ij}^j = \frac{1}{H_j} \pder{H_j}{X^i}.\]

Now I consider

\[\sum_j \Gamma_{ij}^j \mathcal{Q},\]

where \(\mathcal{Q}\) is an arbitrary order of tensor.

This relation is

\[\sum_j \frac{1}{H_j} \pder{H_j}{X^i} \mathcal{Q} = \sum_j \frac{1}{J} \frac{J}{H_j} \pder{H_j}{X^i} \mathcal{Q}.\]

In summary, I have

\[\sumofchristoffel.\]