6Div, Grad, Curl and ∇
IA Vector Calculus
6.1 Div, Grad, Curl and ∇
Recalled that
∇f
is given by (
∇f
)
i
=
∂f
∂x
i
. We can regard this as obtained from
the scalar field f by applying
∇ = e
i
∂
∂x
i
for cartesian coordinates
x
i
and orthonormal basis
e
i
, where
e
i
are orthonormal
and right-handed, i.e. e
i
× e
j
= ε
ijk
e
k
(it is left handed if e
i
× e
j
= −ε
ijk
e
k
).
We can alternatively write this as
∇ =
∂
∂x
,
∂
∂y
,
∂
∂z
.
∇ (nabla or del) is both an operator and a vector. We can apply it to a vector
field F(r) = F
i
(r)e
i
using the scalar or vector product.
Definition (Divergence). The divergence or div of F is
∇ · F =
∂F
i
∂x
i
=
∂F
1
∂x
1
+
∂F
2
∂x
2
+
∂F
3
∂x
3
.
Definition (Curl). The curl of F is
∇ × F = ε
ijk
∂F
k
∂x
j
e
i
=
e
1
e
2
e
3
∂
∂x
∂
∂y
∂
∂z
F
x
F
y
F
z
Example. Let F = (xe
z
, y
2
sin x, xyz). Then
∇ · F =
∂
∂x
xe
z
+
∂
∂y
y
2
sin x +
∂
∂z
xyz = e
z
+ 2y sin x + xy.
and
∇ × F =
ˆ
i
∂
∂y
(xyz) −
∂
∂z
(y
2
sin x)
+
ˆ
j
∂
∂z
(xe
z
) +
∂
∂x
(xyz)
+
ˆ
k
∂
∂x
(y
2
sin x) −
∂
∂y
(xe
z
)
= (xz, xe
z
− yz, y
2
cos x).
Note that ∇ is an operator, so ordering is important. For example,
F · ∇ = F
i
∂
∂x
i
is a scalar differential operator, and
F × ∇ = e
k
ε
ijk
F
i
∂
∂x
j
is a vector differential operator.
Proposition.
Let
f, g
be scalar functions,
F, G
be vector functions, and
µ, λ
be constants. Then
∇(λf + µg) = λ∇f + µ∇g
∇ · (λF + µG) = λ∇ · F + µ∇ · G
∇ × (λF + µG) = λ∇ × F + µ∇ × G.
Note that Grad and Div can be analogously defined in any dimension
n
, but
curl is specific to n = 3 because it uses the vector product.
Example.
Consider
r
α
with
r
=
|r|
. We know that
r
=
x
i
e
i
. So
r
2
=
x
i
x
i
.
Therefore
2r
∂r
∂x
j
= 2x
j
,
or
∂r
∂x
i
=
x
i
r
.
So
∇r
α
= e
i
∂
∂x
i
(r
α
) = e
i
αr
α−1
∂r
∂x
i
= αr
α−2
r.
Also,
∇ · r =
∂x
i
∂x
i
= 3.
and
∇ × r = e
k
ε
ijk
∂x
j
∂x
i
= 0.
Proposition. We have the following Leibnitz properties:
∇(fg) = (∇f)g + f(∇g)
∇ · (fF) = (∇f) · F + f(∇ · F)
∇ × (fF) = (∇f) × F + f(∇ × F)
∇(F · G) = F × (∇ × G) + G × (∇ × F) + (F · ∇)G + (G · ∇)F
∇ × (F × G) = F(∇ · G) − G(∇ · F) + (G · ∇)F − (F · ∇)G
∇ · (F × G) = (∇ × F) · G − F · (∇ × G)
which can be proven by brute-forcing with suffix notation and summation
convention.
There is absolutely no point in memorizing these (at least the last three).
They can be derived when needed via suffix notation.
Example.
∇ · (r
α
r) = (∇r
α
)r + r
α
∇ · r
= (αr
α−2
r) · r + r
α
(3)
= (α + 3)r
α
∇ × (r
α
r) = (∇(r
α
)) × r + r
α
(∇ × r)
= αr
α−2
r × r
= 0