Calculating Christoffel symbol of the second kind in coordinates where basis are orthogonal to each other
Exercise 86,pg-73 of Pavel Grinfeld's Tensor Calculus Book: Derive the Christoffel symbols in spherical and cylindrical coordinates by the equation:$$\Gamma_{ij}^k= \frac12 Z^{km} \left[ \frac{\partial Z_{mj}}{\partial Z^i}+ \frac{\partial Z_{mi}}{\partial Z^j} - \frac{\partial Z_{ij}}{\partial Z^m}\right] \tag{0}$$
This exercise is truly a pain to do directly, it took me one whole page to calculate the christoffel terms for $\Gamma_{11}^k$. I think maybe this was not how the Pavel Grinfeld intends to do the exercise, I considered maybe that there is a better way to do it. Here are my observations so far:
- Given that we are an orthonormal basis only diagonal terms of the metric tensor exist.
- Given that we are in an orthonormal, a relatively simply form exists for the non zero entries of the inverse metric tensor.$$ B_i^i = \frac{1}{A_i^i} \tag{1}$$
I think I figured out how to apply observation-1 onto equation (0), we can write the equation again as (for only non zero terms):
$$ \Gamma_{ij}^k = \frac12 \sum_m \frac{1}{Z_{km} } \left[ \frac{\partial Z_{mj}}{\partial Z^i}+ \frac{\partial Z_{mi}}{\partial Z^j} - \frac{\partial Z_{ij}}{\partial Z^m}\right] $$
Note that I kind of 'broke' the contraction here, but still the summation over $m$ exists. I've kept the symbol outside, to be clear about it. I am confused on how to apply the observation -1, I considered that putting $m=j=i$ but the expression resulting from that doesn't seem too right.
My proof of (2): Consider the identity between matrix and it's inverse in tensor notation $A_k^i B_j^k=\delta_j^i$, for an orthonormal matrix $A$ the only existing terms are only the diagnol ones, this leads to: $A_i^i B_j^i =\delta_j^i$, now we want the non zero entries. For that $i=j$ is the condition, hence we get the non zero entries in $B$ as required.
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$\begingroup$Perhaps if it takes so much effort to calculate Christoffel symbols in an orthogonal basis you are going about this in the wrong way. Let's take an example of spherical coordinates where
$$ Z_{ij} = \begin{bmatrix} 1 &0 &0\\0&r^2 &0\\0&0&r^2\sin^2(\theta) \end{bmatrix} $$ and $$ Z^{ij} = \begin{bmatrix} 1 &0 &0\\0&r^{-2} &0\\0&0&r^{-2}\sin^{-2}(\theta) \end{bmatrix} \text{ .}$$
Rather than writing the Christoffel down as 9 vectors, I would write it as 3 matrices just for convenience (of course the correct way to think about it is as a triple-indexed object). Using your expression $(0)$ for reference, this is how I would think about the $\Gamma_{ij}^1$ terms.
Since $k=1$ and since $(0)$ is multiplied by $Z^{km}$, a nonzero term can only occur when $m=1$ since $Z$ is diagonal. The terms $$\frac{\partial Z_{mj}}{\partial Z^i} \:\text{ and } \: \frac{\partial Z_{mi}}{\partial Z^j}$$ will always be zero because $Z$ is diagonal and $Z_{11}$ has a zero derivative for every coordinate. The term $$-\frac{\partial Z_{ij}}{\partial Z^m}$$will only be nonzero for $(i,j)=(2,2)$ and $(i,j)=(3,3)$ because they are the only components of the metric that depend on $r$ (remember that $m$ has to be 1). Using this we can immediately say that only two of the nine components of $\Gamma_{ij}^1$ are nonzero and they are
$$ \Gamma_{22}^1 = \frac{1}{2}(1)(-2r) = -r $$and$$ \Gamma_{33}^1 = \frac{1}{2}(1)(-2r\sin^2 \theta) = -r\sin^2\theta \text{ .}$$
Rather than spending a page to write down 3 components, we have figured out 9 components with very minimal headache. In fact, when you don't have to explain your reasoning to someone else it can be done in only a few lines.
This is of course not a super formal way of thinking about these computations in the sense that we could compactly write a reduced formula for it. This line of reasoning can however be applied to any orthogonal coordinate system as is the subject of the question. Different lines of reasoning similar to what is presented above can be used for non-orthogonal coordinate systems (especially ones with lots of constant entries).
$\endgroup$ 2 $\begingroup$Note: Everything below , I only have thought through for orthogonal basis. I am not sure if it applies otherwise.
$$ \Gamma_{ij}^k = \frac12 \sum_m \frac{1}{Z_{km} } \left[ \frac{\partial Z_{mj}}{\partial Z^i}+ \frac{\partial Z_{mi}}{\partial Z^j} - \frac{\partial Z_{ij}}{\partial Z^m}\right] $$
The first point is that for non zero entries, $m=k$ this leads to:
$$ \Gamma_{ij}^k = \frac12 \frac{1}{Z_{kk} }\left[ \frac{\partial Z_{kj}}{\partial Z^i}+ \frac{\partial Z_{ki}}{\partial Z^j} - \frac{\partial Z_{ij}}{\partial Z^k}\right] \tag{1}$$
Now, for the next simplification, we can do some combinatorics discussion before jumping into it.
Game plan:
The Christoffel term is determined by a triplet of numbers where each number can take values in $\{1,2,3 \}$. Having three numbers to determine, the next point is to consider the different strings made of numbers considering the number of repetitions of a numbers. Given a string of three numbers, there three main kinds of combination we can get (digit repeating):
(1). All numbers are distinct
(2). Only two numbers in the triplet are same
(3). All numbers in the triplet are same
In case of the Christoffel symbol , if it case (3) hen there are two subcases. The first subcase, is when the lower indices are same and the second case is when one of the lower and one of the upper are same.
Working through the symbols
For (1), it can be shown that the symbol equals zero. Proof: Note that $Z_{kj}=Z_{ki}=Z_{ij}=0$ if $ i \neq j \neq k \neq i$, hence the result.
For (2), for the first sub case case we have that the lower indices are same $i=j\neq k$, we can show that the symbol is given as:
$$ \Gamma_{ii}^k = \frac12 \frac{1}{Z_{kk}} \left[ - \frac{\partial Z_{ii} }{\partial Z^k} \right]$$
The second sub case is when $ k \neq i \neq j=k$, this is given as: $$ \Gamma_{ik}^k = \frac{\partial Z_{kk} }{\partial Z^i} \frac{1}{2Z_{kk} }$$
For (3), $i=j=k$, this leads to:
$$ \Gamma_{kk}^k = \frac{1}{2Z_{kk} } \left[\frac{ \partial Z_{kk}}{ \partial Z_k} \right]$$
Ultimately for writing directly by hand this is useless as it'll take time to write the simplified expressions. However, the important idea is the 'principles' used to simplify.
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