VECTOR CALCULUS
Prove the expression for∇·ain orthogonal curvilinear coordinates.
Let us consider the sub-expression∇·(a 1 ˆe 1 ). Noweˆ 1 =eˆ 2 ׈e 3 =h 2 ∇u 2 ×h 3 ∇u 3. Therefore
∇·(a 1 ˆe 1 )=∇·(a 1 h 2 h 3 ∇u 2 ×∇u 3 ),
=∇(a 1 h 2 h 3 )·(∇u 2 ×∇u 3 )+a 1 h 2 h 3 ∇·(∇u 2 ×∇u 3 ).
However,∇·(∇u 2 ×∇u 3 ) = 0, from (10.43), so we obtain
∇·(a 1 eˆ 1 )=∇(a 1 h 2 h 3 )·
(
eˆ 2
h 2
×
ˆe 3
h 3
)
=∇(a 1 h 2 h 3 )·
ˆe 1
h 2 h 3
;
letting Φ =a 1 h 2 h 3 in (10.60) and substituting into the above equation, we find
∇·(a 1 ˆe 1 )=
1
h 1 h 2 h 3
∂
∂u 1
(a 1 h 2 h 3 ).
Repeating the analysis for∇·(a 2 ˆe 2 )and∇·(a 3 ˆe 3 ), and adding the results we obtain (10.61),
as required.
Laplacian
In the expression for the divergence (10.61), let
a=∇Φ=
1
h 1
∂Φ
∂u 1
ˆe 1 +
1
h 2
∂Φ
∂u 2
eˆ 2 +
1
h 3
∂Φ
∂u 3
ˆe 3 ,
where we have used (10.60). We then obtain
∇^2 Φ=
1
h 1 h 2 h 3
[
∂
∂u 1
(
h 2 h 3
h 1
∂Φ
∂u 1
)
+
∂
∂u 2
(
h 3 h 1
h 2
∂Φ
∂u 2
)
+
∂
∂u 3
(
h 1 h 2
h 3
∂Φ
∂u 3
)]
,
which is the expression for the Laplacian in orthogonal curvilinear coordinates.
Curl
The curl of a vector fielda=a 1 eˆ 1 +a 2 ˆe 2 +a 3 ˆe 3 in orthogonal curvilinear
coordinates is given by
∇×a=
1
h 1 h 2 h 3
∣
∣
∣
∣
∣
∣
∣
∣
∣
h 1 eˆ 1 h 2 eˆ 2 h 3 ˆe 3
∂
∂u 1
∂
∂u 2
∂
∂u 3
h 1 a 1 h 2 a 2 h 3 a 3
∣
∣
∣
∣
∣
∣
∣
∣
∣
. (10.62)
Prove the expression for∇×ain orthogonal curvilinear coordinates.
Let us consider the sub-expression∇×(a 1 eˆ 1 ). Sinceˆe 1 =h 1 ∇u 1 we have
∇×(a 1 ˆe 1 )=∇×(a 1 h 1 ∇u 1 ),
=∇(a 1 h 1 )×∇u 1 + a 1 h 1 ∇×∇u 1.
But∇×∇u 1 =0,soweobtain
∇×(a 1 eˆ 1 )=∇(a 1 h 1 )×
ˆe 1
h 1