Page 28 - Intro to Tensor Calculus

24

~
~
2 ~
EXAMPLE 1.1-23. In Cartesian coordinates prove the vector identity ∇× (∇× A)= ∇(∇· A) −∇ A
~
~
Solution: We have for the ith component of ∇× A is given by b e i · [∇× A]= e ijk A k,j and consequently the
~
pth component of ∇× (∇× A)is
~
b e p · [∇× (∇× A)] = e pqr [e rjk A k,j ] ,q
= e pqr e rjk A k,jq .
The e − δ identity produces
~
b e p · [∇× (∇× A)] = (δ pj δ qk − δ pk δ qj )A k,jq
= A k,pk − A p,qq .

~
~
2 ~
Expressing this result in vector form we have ∇× (∇× A)= ∇(∇· A) −∇ A.
Indicial Form of Integral Theorems

The divergence theorem, in both vector and indicial notation, can be written
ZZZ ZZ Z Z
~
~
div · Fdτ = F · b n dσ F i,i dτ = F i n i dσ i =1, 2, 3 (1.1.16)
V S V S
where n i are the direction cosines of the unit exterior normal to the surface, dτ is a volume element and dσ
is an element of surface area. Note that in using the indicial notation the volume and surface integrals are
to be extended over the range specified by the indices. This suggests that the divergence theorem can be
applied to vectors in n−dimensional spaces.
The vector form and indicial notation for the Stokes theorem are

ZZ Z Z Z
~
~
r
(∇× F) · b n dσ = F · d~ e ijk F k,j n i dσ = F i dx i i, j, k =1, 2, 3 (1.1.17)
S C S C
and the Green’s theorem in the plane, which is a special case of the Stoke’s theorem, can be expressed
ZZ Z Z Z
∂F 2 ∂F 1 i
− dxdy = F 1 dx + F 2 dy e 3jk F k,j dS = F i dx i, j, k =1, 2 (1.1.18)
∂x ∂y C S C
Other forms of the above integral theorems are

ZZZ ZZ
∇φdτ = φ b n dσ
V S
~
~
~
~
obtained from the divergence theorem by letting F = φC where C is a constant vector. By replacing F by
~
~
F × C in the divergence theorem one can derive
ZZZ ZZ

~
~
∇× F dτ = − F × ~ndσ.
V S
~
In the divergence theorem make the substitution F = φ∇ψ to obtain
ZZZ ZZ
2
(φ∇ ψ +(∇φ) · (∇ψ) dτ = (φ∇ψ) · b n dσ.
V S