Page 115 - Intro to Tensor Calculus

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CASE I Let a = b = c = i, then the equation (1.4.8) simplifies to

1 ∂g ii
[ab, c]= [ii, i]= (no summation on i). (1.4.9)
2 ∂x i
From this equation we can calculate any of the Christoffel symbols

[11, 1], [22, 2], or [33, 3].

CASE II Let a = b = i 6= c, then the equation (1.4.8) simplifies to the form

1 ∂g ii
[ab, c]= [ii, c]= − (no summation on i and i 6= c). (1.4.10)
2 ∂x c
since, g ic =0 for i 6= c. This equation shows how we may calculate any of the six Christoffel symbols

[11, 2], [11, 3], [22, 1], [22, 3], [33, 1], [33, 2].

CASE III Let a = c = i 6= b, and noting that g ib =0 for i 6= b, it can be verified that the equation (1.4.8)
simplifies to the form
1 ∂g ii
[ab, c]= [ib, i]= [bi, i]= (no summation on i and i 6= b). (1.4.11)
2 ∂x b
From this equation we can calculate any of the twelve Christoffel symbols

[12, 1] = [21, 1] [31, 3] = [13, 3]
[32, 3] = [23, 3] [21, 2] = [12, 2]

[13, 1] = [31, 1] [23, 2] = [32, 2]
CASE IV Let a 6= b 6= c and show that the equation (1.4.8) reduces to

[ab, c]= 0, (a 6= b 6= c.)

This represents the six Christoffel symbols
[12, 3] = [21, 3] = [23, 1] = [32, 1] = [31, 2] = [13, 2] = 0.

From the Cases I,II,III,IV all twenty seven Christoffel symbols of the first kind can be determined. In
practice, only the nonzero Christoffel symbols are listed.

EXAMPLE 1.4-3. (Christoffel symbols of the first kind)Find the nonzero Christoffel symbols of the
first kind in cylindrical coordinates.
3
1
2
Solution: From the results of example 1.4-2 we find that for x = r, x = θ, x = z and
1 2
2
g 11 =1, g 22 =(x ) = r , g 33 =1
the nonzero Christoffel symbols of the first kind in cylindrical coordinates are:
1 ∂g 22 1
[22, 1] = − = −x = −r
2 ∂x 1
1 ∂g 22 1
[21, 2] = [12, 2] = = x = r.
2 ∂x 1