Page 124 - Intro to Tensor Calculus

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The equations (1.4.46) and (1.4.47) enable us to write

~ 1 " ~ ~ #
~ ∂E j ~ ∂E j ~ ∂E k
[jk, m]=E m · = E m · + E m ·
∂x k 2 ∂x k ∂x j
~
~
" #
1 ∂ ∂ ∂E m ∂E m
~
~
~
~
~
~
= E m · E j + E m · E k − E j · − E k ·
2 ∂x k ∂x j ∂x k ∂x j
" #
~
~
1 ∂ ∂ ∂E k ∂E j
~
~
~
~
~
~
= E m · E j + E m · E k − E j · − E k ·
2 ∂x k ∂x j ∂x m ∂x m

1 ∂ ∂ ∂
~
~
~
~
~
~
= E m · E j + E m · E k − E j · E k
2 ∂x k ∂x j ∂x m

1 ∂g mj ∂g mk ∂g jk
= + − =[kj, m]
2 ∂x k ∂x j ∂x m
which again agrees with our previous result.
~
j ~
~
For future reference we make the observation that if the vector A is represented in the form A = A E j ,
involving contravariant components, then we may write
!
~
~
∂A k ∂A j j ∂E j k
~
~
dA = dx = E j + A dx
∂x k ∂x k ∂x k
j
∂A j i k
~
~
= E j + A E i dx (1.4.48)
∂x k jk
j
∂A j m k j k ~
~
= + A E j dx = A ,k dx E j .
∂x k mk
~
~
~ j
Similarly, if the vector A is represented in the form A = A j E involving covariant components it is left as
an exercise to show that
~
k ~ j
dA = A j,k dx E (1.4.49)
Ricci’s Theorem
Ricci’s theorem states that the covariant derivative of the metric tensor vanishes and g ik,l =0.
Proof: We have

∂g ik m m
g ik,l = − g im − g mk
∂x l kl il
∂g ik
g ik,l = − [kl, i] − [il, k]
∂x l

∂g ik 1 ∂g ik ∂g il ∂g kl 1 ∂g ik ∂g kl ∂g il
g ik,l = − + − − + − =0.
∂x l 2 ∂x l ∂x k ∂x i 2 ∂x l ∂x i ∂x k
Because of Ricci’s theorem the components of the metric tensor can be regarded as constants during covariant
differentiation.
EXAMPLE 1.4-7. (Covariant differentiation) Show that δ i =0.
j,k
Solution
∂δ i j i σ i i
δ i = + δ σ − δ i = − =0.
j,k k j σ
∂x σk jk jk jk