Page 114 - Intro to Tensor Calculus

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Define the Christoffel symbol of the second kind as

i i iα 1 iα ∂g kα ∂g jα ∂g jk
= = g [jk, α]= g + − . (1.4.6)
jk kj 2 ∂x j ∂x k ∂x α
This Christoffel symbol of the second kind is symmetric in the indices j and k and from equation (1.4.5) we
see that it satisfies the transformation law

µ ∂x e ∂x ∂x ∂ x
e a c 2 e
= α γ + α γ . (1.4.7)
µ
αγ ∂x ac ∂x ∂x ∂x ∂x
Observe that the Christoffel symbol of the second kind does not transform like a tensor quantity. We can use
the relation defined by equation (1.4.7) to express the second derivative of the transformation equations in
terms of the Christoffel symbols of the second kind. At times it will be convenient to represent the Christoffel
symbols with a subscript to indicate the metric from which they are calculated. Thus, an alternative notation

i i
for is the notation .
jk jk
g
EXAMPLE 1.4-1. (Christoffel symbols) Solve for the Christoffel symbol of the first kind in terms of
the Christoffel symbol of the second kind.
Solution: By the definition from equation (1.4.6) we have

i iα
= g [jk, α].
jk
We multiply this equation by g βi and find

i α
g βi = δ [jk, α]= [jk, β]
β
jk
and so

i 1 N
[jk, α]= g αi = g α1 + ··· + g αN .
jk jk jk

EXAMPLE 1.4-2. (Christoffel symbols of first kind)
Derive formulas to find the Christoffel symbols of the first kind in a generalized orthogonal coordinate
system with metric coefficients

2
g ij =0 for i 6= j and g (i)(i) = h , i =1, 2, 3
(i)
where i is not summed.
Solution: In an orthogonal coordinate system where g ij =0 for i 6= j we observe that

1 ∂g ac ∂g bc ∂g ab
[ab, c]= + − . (1.4.8)
2 ∂x b ∂x a ∂x c
3
Here there are 3 = 27 quantities to calculate. We consider the following cases: