Page 177 - Intro to Tensor Calculus

172

Divergence

r
The divergence of a contravariant tensor A is obtained by taking the covariant derivative with respect
k
to x and then performing a contraction. This produces
r
div A = A r ,r . (2.1.3)

Still another form for the divergence is obtained by simplifying the expression (2.1.3). The covariant deriva-
tive can be represented
∂A r r m
r
A ,k = + A .
∂x k mk
Upon contracting the indices r and k and using the result from Exercise 1.4, problem 13, we obtain

√
∂A r 1 ∂( g) m
r
A ,r = + √ A
∂x r g ∂x m
r √
1 √ ∂A r ∂ g
r
A g + A (2.1.4)
,r = √ r r
g ∂x ∂x
1 ∂ √ r
r
A ,r = √ ( gA ) .
g ∂x r
r
EXAMPLE 2.1-1. (Divergence) Find the representation of the divergence of a vector A in spherical
coordinates (ρ, θ, φ). Solution: In spherical coordinates we have
2
3
1
x = ρ, x = θ, x = φ with g ij =0 for i 6= j and
2
2
2
2
2
2
g 11 = h =1, g 22 = h = ρ , g 33 = h = ρ sin θ.
2
1
3
2
4
2
The determinant of g ij is g = |g ij | = ρ sin θ and √ g = ρ sin θ. Employing the relation (2.1.4) we find

1 ∂ √ ∂ √ ∂ √
r 1 2 3
div A = √ ( gA )+ ( gA )+ ( gA ) .
g ∂x 1 ∂x 2 ∂x 3
In terms of the physical components this equation becomes
1 ∂ √ A(1) ∂ √ A(2) ∂ √ A(3)
r
div A = √ ( g )+ ( g )+ ( g ) .
g ∂ρ h 1 ∂θ h 2 ∂φ h 3
By using the notation
A(1) = A ρ , A(2) = A θ , A(3) = A φ
for the physical components, the divergence can be expressed in either of the forms:
1 ∂ 2 ∂ 2 A θ ∂ 2 A φ
r
div A = (ρ sin θA ρ )+ (ρ sin θ )+ (ρ sin θ ) or
2
ρ sin θ ∂ρ ∂θ ρ ∂φ ρ sin θ
1 ∂ 2 1 ∂ 1 ∂A φ
r
div A = (ρ A ρ )+ (sin θA θ )+ .
2
ρ ∂ρ ρ sin θ ∂θ ρ sin θ ∂φ