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Lecture 15: Vector Operator Identities (RHB 8.8)
There are a large number of identities for div, grad, and curl. It’s not necessary to know all
of these, but you are advised to be able to produce from memory expressions for ∇r, ∇ · r,
∇ × r, ∇φ(r), ∇(a · r), ∇ × (a × r), ∇(f g), and 1 2 3 4 below. You should be familiar with
the rest and to be able to derive and use them when necessary!
Most importantly you should be at ease with div, grad and curl. This only comes through
practice and deriving the various identities gives you just that. In these derivations the
advantages of suffix notation, the summation convention and ijk
will become apparent.
In what follows, φ(r) is a scalar field; A(r) and B(r) are vector fields.
- Distributive Laws
1. ∇ · (A + B) = ∇ · A + ∇ · B
2. ∇ × (A + B) = ∇ × A + ∇ × B
The proofs of these are straightforward using suffix or ‘x y z’ notation and follow from the
fact that div and curl are linear operations.
- Product Laws
The results of taking the div or curl of products of vector and scalar fields are predictable
but need a little care:-
- ∇ · (φ A) = φ ∇ · A + A · ∇φ
- ∇ × (φ A) = φ (∇ × A) + (∇φ) × A = φ (∇ × A) − A × ∇φ
Proof of (4): first using ijk
∇ × (φ A) = e i
ijk
∂x j
(φ A k
= e i
ijk
φ
∂x j
A
k
+ A
k
∂x j
φ
= φ (∇ × A) + (∇φ) × A
or avoiding ijk
and using ‘x y z’ notation: ∇ × (φ A) =
ex ey ez
∂x
∂y
∂z
φAx φAy φAz
The x component is given by
∂(φAz )
∂y
∂(φAy)
∂z
= φ(
∂Az
∂y
∂Ay
∂z
∂φ
∂y
A
z
∂φ
∂z
A
y
= φ(∇ × A) x
[
(∇φ) × A
]
x
A similar proof holds for the y and z components.
Although we have used Cartesian coordinates in our proofs, the identities hold in all coor-
dinate systems.
- Products of Two Vector Fields
Things start getting complicated!
5. ∇ (A · B) = (A · ∇) B + (B · ∇)A + A × (∇ × B) + B × (∇ × A)
6. ∇ · (A × B) = B · (∇ × A) − A · (∇ × B)
7. ∇ × (A × B) = A (∇ · B) − B (∇ · A) + (B · ∇) A − (A · ∇) B
Proof of (6):
∇ · (A × B) =
∂x i
ijk
A
j
B
k
ijk
∂A
j
∂x i
B
k
ijk
A
j
∂B
k
∂x i
= B
k
kij
∂A
j
∂x i
− A
j
jik
∂B
k
∂x i
The proofs of (5) and (7) involve the product of two epsilon ijks. For example, this is why
there are four terms on the rhs of (7).
All other results involving one ∇ can be derived from the above identities.
Example: If a is a constant vector, and r is the position vector, show that
∇ (a · r) = (a · ∇) r = a
In lecture 13 we showed that ∇ (a · r) = a for constant a. Hence, we need only evaluate
(a · ∇) r = a i
∂x i
e j
x j
= a i
e j
δ ij
= a i
e i
= a (1)
and the identity holds.
Example: Show that ∇ · (ω × r) = 0 where ω is a constant vector.
Using 6. ∇ · (ω × r) = ω · (∇ × r) − r · (∇ × ω) = 0 − 0
Example: Show that ∇ · (r
− 3 r) = 0, (where r = |r| as usual).
Using identity (3), we have
∇ · (r
− 3 r) = r
− 3 (∇ · r) + r · ∇(r
− 3 )
Before commencing with integral vector calculus we review here polar co-ordinate systems.
Polar Co-ordinate Systems Here dV indicates a volume element and dA an area
element. Note that different conventions, e.g. for the angles φ and θ, are sometimes used,
in particular in the Mathematics ‘Several Variable Calculus’ Module.
Plane polar co-ordinates
Cylindrical polar co-ordinates
Spherical polar co-ordinates
