9Orthogonal curvilinear coordinates
IA Vector Calculus
9.2 Grad, Div and Curl
Consider f(r(u, v, w)) and compare
df =
∂f
∂u
du +
∂f
∂v
dv +
∂f
∂w
dw,
with df = (∇f) · dr. Since we know that
dr =
∂r
∂u
du +
∂r
∂v
dv +
∂r
∂w
dw = h
u
e
u
du + h
v
e
v
dv + h
w
e
w
dv,
we can compare the terms to know that
Proposition.
∇f =
1
h
u
∂f
∂u
e
u
+
1
h
v
∂f
∂v
e
v
+
1
h
w
∂f
∂w
e
w
.
Example. Take f = r sin θ cos ϕ in spherical polars. Then
∇f = sin θ cos ϕ e
r
+
1
r
(r cos θ cos ϕ) e
θ
+
1
r sin θ
(−r sin θ sin ϕ) e
ϕ
= cos ϕ(sin θ e
r
+ cos θ e
θ
) − sin ϕ e
ϕ
.
Then we know that the differential operator is
Proposition.
∇ =
1
h
u
e
u
∂
∂u
+
1
h
v
e
v
∂
∂v
+
1
h
w
e
w
∂
∂w
.
We can apply this to a vector field
F = F
u
e
u
+ F
v
e
v
+ F
w
e
w
using scalar or vector products to obtain
Proposition.
∇ × F =
1
h
v
h
w
∂
∂v
(h
w
F
w
) −
∂
∂w
(h
v
F
v
)
e
u
+ two similar terms
=
1
h
u
h
v
h
w
h
u
e
u
h
v
e
v
h
w
e
w
∂
∂u
∂
∂v
∂
∂w
h
u
F
u
h
v
F
v
h
w
F
w
and
∇ · F =
1
h
u
h
v
h
w
∂
∂u
(h
v
h
w
F
u
) + two similar terms
.
There are several ways to obtain these formulae. We can
Proof. (non-examinable)
(i) Apply ∇· or ∇× and differentiate the basis vectors explicitly.
(ii)
First, apply
∇·
or
∇×
, but calculate the result by writing
F
in terms of
∇u, ∇v
and
∇w
in a suitable way. Then use
∇×∇f
= 0 and
∇·
(
∇×f
) = 0.
(iii) Use the integral expressions for div and curl.
Recall that
n · ∇ × F = lim
A→0
1
A
Z
∂A
F · dr.
So to calculate the curl, we first find the e
w
component.
Consider an area with
W
fixed and change
u
by
δu
and
v
by
δv
. Then
this has an area of h
u
h
v
δuδv with normal e
w
. Let C be its boundary.
u
δu
v
δv
C
We then integrate around the curve
C
. We split the curve
C
up into 4 parts
(corresponding to the four sides), and take linear approximations by assum-
ing
F
and
h
are constant when moving through each horizontal/vertical
segment.
Z
C
F · dr ≈ F
u
(u, v)h
u
(u, v) δu + F
v
(u + δu, v)h
v
(u + δu, v) δu
− F
u
(u, v + δv)h
u
(u, v + δv) δu − F
v
(u, v)h
v
(u, v) δv
≈
∂
∂u
h
v
F
v
−
∂
∂v
(h
u
F
u
)
δuδv.
Divide by the area and take the limit as area → 0, we obtain
lim
A→0
1
A
Z
C
F · dr =
1
h
u
h
v
∂
∂u
h
v
F
v
−
∂
∂v
(h
u
F
u
)
.
So, by the integral definition of divergence,
e
w
· ∇ × F =
1
h
u
h
v
∂
∂u
(h
v
F
v
) −
∂
∂v
(h
u
F
u
)
,
and similarly for other components.
We can find the divergence similarly.
Example. Let A =
1
r
tan
θ
2
e
ϕ
in spherical polars. Then
∇ × A =
1
r
2
sin θ
e
r
re
θ
r sin θe
ϕ
∂
∂r
∂
∂θ
∂
∂ϕ
0 0 r sin θ ·
1
r
tan
θ
2
=
e
r
r
2
sin θ
∂
∂θ
sin θ tan
θ
2
=
1
r
2
e
r
.