Page 113 - Intro to Tensor Calculus

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§1.4 DERIVATIVE OF A TENSOR

In this section we develop some additional operations associated with tensors. Historically, one of the
basic problems of the tensor calculus was to try and find a tensor quantity which is a function of the metric
2
∂g ij ∂ g ij
tensor g ij and some of its derivatives , , .... A solution of this problem is the fourth order
m
∂x m ∂x ∂x n
Riemann Christoffel tensor R ijkl to be developed shortly. In order to understand how this tensor was arrived
at, we must first develop some preliminary relationships involving Christoffel symbols.

Christoffel Symbols
Let us consider the metric tensor g ij which we know satisfies the transformation law

a
∂x ∂x b
g .
αβ = g ab α β
∂x ∂x
Define the quantity

2 a
c
2 b
a
∂g αβ ∂g ab ∂x ∂x ∂x b ∂ x ∂x b ∂x a ∂ x
(α, β, γ)= γ = γ α + g ab α γ + g ab α
β
c
∂x ∂x ∂x ∂x ∂x β ∂x ∂x ∂x β ∂x ∂x ∂x γ
1
and form the combination of terms [(α, β, γ)+(β, γ, α) − (γ, α, β)] to obtain the result
2
b
2 a
1 ∂g αβ ∂g βγ ∂g γα 1 ∂g ab ∂g bc ∂g ca ∂x ∂x ∂x c ∂x b ∂ x
a

α
α
β
β
2 ∂x γ + ∂x α − ∂x β = 2 ∂x c + ∂x a − ∂x b ∂x ∂x ∂x γ + g ab ∂x ∂x ∂x γ . (1.4.1)
In this equation the combination of derivatives occurring inside the brackets is called a Christoffel symbol
of the first kind and is defined by the notation

1 ∂g ab ∂g bc ∂g ac
[ac, b]= [ca, b]= + − . (1.4.2)
2 ∂x c ∂x a ∂x b
The equation (1.4.1) defines the transformation for a Christoffel symbol of the first kind and can be expressed
as
a
2 a
b
∂x ∂x ∂x c ∂ x ∂x b
[αγ, β]= [ac, b] + g ab . (1.4.3)
α
γ
β
α
∂x ∂x ∂x γ ∂x ∂x ∂x β
Observe that the Christoffel symbol of the first kind [ac, b] does not transform like a tensor. However, it is
symmetric in the indices a and c.
At this time it is convenient to use the equation (1.4.3) to develop an expression for the second derivative
term which occurs in that equation as this second derivative term arises in some of our future considerations.
∂x β
To solve for this second derivative we can multiply equation (1.4.3) by g de and simplify the result to the
∂x d
form
2 e
a
∂ x de ∂x ∂x c ∂x β de
= −g [ac, d] + [αγ,β] g . (1.4.4)
α
α
∂x ∂x γ ∂x ∂x γ ∂x d
d
∂x ∂x e
λµ
de
The transformation g = g allows us to express the equation (1.4.4) in the form
λ
∂x ∂x µ
a
2 e
∂ x de ∂x ∂x c βµ ∂x e
= −g [ac, d] + g [αγ, β] . (1.4.5)
α
α
∂x ∂x γ ∂x ∂x γ ∂x µ