Vector

7
1. Gradient, divergence and curl in R
3
in Cartesian coordinates
For the sequel we fix the following notation.
Position vector: r = xi + yj + zk, here i, j, k or e
1
, e
2
, e
3
denote the canonical unit vectors.
|r| = r =
_
x
2
+ y
2
+ z
2
length of r.
Scalar field φ(r) = φ(x, y, z) ∈ R, ψ, . . .
Vector field f(r) = f
1
e
1
+ f
2
e
2
+ f
3
e
3
= f(r)i + g(r)j + h(r)k where f(r) ∈ R
3
. Similarly for
other vector fields g, h etc.
We write “·” for the dot (= scalar) product, and “∧” for the vector product. Recall that
e
i
· e
i
= 1 = |e
i
|
2
, and e
i
· e
j
= 0 for i = j.
e
1
∧ e
2
= e
3
, e
2
∧ e
3
= e
1
, e
3
∧ e
1
= e
2
. (“cyclic permuting is OK”)
a ∧ b = −b ∧ a, a ∧ b ⊥ a, b for general a, b ∈ R
3
, ∧ is not associative!!.
Now we define three differential operators which are natural generalizations from Calculus
in 2 dimensions. It will turn out, that treating the “nabla ∇” as a (formal) vector,
simplifies and unifies some of the expressions. This approach is widely used in the literature.
Althought we will not justify it here rigorously, it is very useful to gain some practice with
this so called “nabla calculus”
Gradient (vector valued)
∇φ =
3

i=1
∂φ
∂x
i
e
i
=
∂φ
∂x
1
e
1
+
∂φ
∂x
2
e
2
+
∂φ
∂x
3
e
3
=
∂φ
∂x
i +
∂φ
∂y
j +
∂φ
∂z
k
Formally,
∇=
3

i=1
∂
∂x
i
e
i
=
∂
∂x
1
e
1
+
∂
∂x
2
e
2
+
∂φ
∂x
3
e
3
=
∂
∂x
i +
∂
∂y
j +
∂
∂z
k
is a vector (technical problem: not in R
3
, so where?). With this agreement we have that
∇φ
∧
= vector ×scalar
∧
= a vector.
8
Divergence (scalar valued)
div f =
3

i=1
e
i
·
∂f
∂x
i
=
3

i=1
e
i
·
∂
∂x
i
[f
1
e
1
+ f
2
e
2
+ f
3
e
3
]
=
3

i=1
e
i
·
_
∂f
1
∂x
i
e
1
+
∂f
2
∂x
i
e
2
+
∂f
3
∂x
i
e
3
_
=
∂f
1
∂x
1
+
∂f
2
∂x
2
+
∂f
3
∂x
3
=
∂f
∂x
+
∂g
∂y
+
∂h
∂z
An alternative notation, often used in the literature is to write formally ∇· f = div f
Indeed,
div f =
3

i=1
∂
∂x
i
f
i
=
_
3

i=1
∂
∂x
i
e
i
_
·
_
3

i=1
f
i
e
i
_
= ∇· f
by formal application of the usual dot product.
Rotation or Curl (vector valued)
curl f =
3

i=1
e
i
∧
∂f
∂x
i
=
3

i=1
e
i
∧
_
∂f
1
∂x
i
e
1
+
∂f
2
∂x
i
e
2
+
∂f
3
∂x
i
e
3
_
=
_
∂f
3
∂x
2
−
∂f
2
∂x
3
_
e
1
+
_
∂f
1
∂x
3
−
∂f
3
∂x
1
_
e
2
+
_
∂f
2
∂x
1
−
∂f
1
∂x
3
_
e
3
=
¸
¸
¸
¸
¸
¸
i j k
∂
∂x
∂
∂y
∂
∂z
f g h
¸
¸
¸
¸
¸
¸
=
¸
¸
¸
¸
¸
¸
i j k
∇
f
¸
¸
¸
¸
¸
¸
considering a formal determinant.
Notice that the same formal determinant is used to define the wedge product of two
vectors in R
3
. So replacing one vector by ∇ we obtain the nabla calculus presentation
curl f = ∇∧ f
In a more direct attempt to justify this we can calculate
_
3

i=1
∂
∂x
i
e
i
_
∧ f =
3

i=1
∂
∂x
i
(e
i
∧ f) =
3

i=1
e
i
∧
∂f
∂x
i
,
using the product rule and that e
i
is constant, hence ∂e
i
/∂x
i
= 0.
9
Now we can derive some basic properties of and between these differential expressions
Theorem 1.1. If φ and f are smooth, then
(i) curl ∇φ = 0
(ii) div ∇φ = ∇
2
φ (≡ ∆φ in some books) called “Laplacian”
(iii) div curl f = 0
Remark:
- (i), (iii) have far reaching generalisation in the theory of differential forms
- Formally, these rules are clear, e.g. for (i) ∇∧ (∇φ) = 0 since the two vectors
are parallel. We prefer, however, valid proofs and give them below.
Proof:
(i)
curl ∇φ =

i
e
i
∧
∂
∂x
i
_

j
∂φ
∂x
j
e
j
_
=
3

i,j=1
∂
2
φ
∂x
i
∂x
j
e
i
∧ e
j
=: c
want to show c = 0. But indeed, just renaming i ↔ j we have
c =
3

j,i=1
∂
2
φ
∂x
j
∂x
i
e
j
∧ e
i
=
3

i,j=1
∂
2
φ
∂x
j
∂x
i
e
j
∧ e
i
, reordering indices of

=
3

i,j=1
∂
2
φ
∂x
i
∂x
j
e
j
∧ e
i
, as
∂
2
∂x
j
∂x
i
=
∂
2
∂x
i
∂x
j
,
=
3

i,j=1
∂
2
φ
∂x
i
∂x
j
(−e
i
∧ e
j
) = −c, as e
i
∧ e
j
= −e
j
∧ e
i
Thus c = −c & c = 0 follows.
(ii)
div ∇φ =

i
∂
∂x
i
(∇φ) · e
i
=

i
∂
∂x
i
∂φ
∂x
i
=

i
∂
2
φ
∂x
2
i
= ∇
2
φ Laplacian
10
(iii)
div curl f =

i
e
i
·
∂
∂x
i
(curl f)
=

i
e
i
·
∂
∂x
i
_

j
e
j
∧
∂f
∂x
j
_
=

i,j
e
i
·
_
e
j
∧
∂
2
f
∂x
i
∂x
j
_
, (∈ R
3
)
=

i,j
∂
2
f
∂x
i
∂x
j
· (e
i
∧ e
j
) = 0
The last equality holds for the same reason as in (i) since again
∂
2
f
∂x
i
∂x
j
=
∂
2
f
∂x
j
∂x
i
.
Example 1.1. Let r(x, y, z) = xi + y j + z k be the position vector.
• div r =
∂x
∂x
+
∂y
∂y
+
∂z
∂z
= 3 (since we are in 3 dimensions! Note that in contrast
div r = 2 in R
2
, this difference has impact e.g. on the shape of the gravitational
potential.)
• curl r =
¸
¸
¸
¸
¸
¸
i j k
∂
∂x
∂
∂y
∂
∂z
x y z
¸
¸
¸
¸
¸
¸
= i (0 −0) −j (0 −0) + k (0 −0) = 0.
_
∂y
∂x
= 0 etc.
_
Now consider
r = |r| =
_
x
2
1
+ x
2
3
+ x
2
3
⇒
∂r
∂x
i
=
x
i
_
x
2
1
+ x
2
3
+ x
2
3
=
x
i
|r|
.
⇒∇r =
∂r
∂x
1
e
1
+
∂r
∂x
2
e
2
+
∂r
∂x
3
e
3
=
x
1
e
1
+ x
2
e
2
+ x
3
e
3
r
=
r
|r|
∇r =
r
|r|
= e
r
unit vector in direction r.
11
Theorem 1.2. “Product rules”. If φ, ψ, f are smooth, then
(i) ∇(φψ) = φ∇ψ + ψ ∇φ vector
(ii) div (φf) = φ div f +∇φ · f scalar,
(iii) curl (φf) = φ curl f +∇φ ∧ f vector
Proof:
(i)
∇(φψ) =

i
∂ (φψ)
∂x
i
e
i
=
3

i=1
_
φ
_
∂ψ
∂x
i
e
i
_
+ ψ
_
∂φ
∂x
i
e
i
__
= φ∇ψ + φ∇ψ
(ii) div (φf) =
3

i=1
∂
∂x
i
(φf · e
i
)
=
3

i=1
∂
∂x
i
(φf
i
) =
3

i=1
__
∂
∂x
i
φ
_
f
i
+ φ
∂f
i
∂x
i
_
= ∇φ · f + φdiv f
(iii) curl (φf) =
3

i=1
e
i
∧
∂
∂x
i
(φf)
=
3

i=1
_
e
i
∧
∂φ
∂x
i
f + e
i
∧ φ
∂f
∂x
i
_
(note
∂φ
∂x
i
∈ R)
=
3

i=1
_
∂φ
∂x
i
e
i
_
∧ f + φ
3

i=1
e
i
∧
∂f
∂x
i
= ∇φ ∧ f + φ curl f
12
Example 1.2. Let φ(x, y, z) = φ(r) = f(r) be a radially symmetric scalar field (r = |r|).
We get “easy” expressions for the gradient and the Laplacian.
•
∂φ
∂x
i
=
df
dr
∂r
∂x
i
=
df
dr
x
i
r
⇒∇φ =
df
dr
r
r
= f

(r) e
r
• ∇
2
φ = div(∇φ) = div
_
1
r
df
dr
r
_
= ∇
_
1
r
df
dr
_
· r +
1
r
df
dr
div r
=
3
r
df
dr
+
_
d
dr
_
1
r
df
dr
_
e
r
_
· r (applying the first part)
=
3
r
df
dr
+
_
−
1
r
2
df
dr
+
1
r
d
2
f
dr
2
_
r
∇
2
φ =
2
r
df
dr
+
d
2
f
dr
2
so
φ(r) =
1
r
⇒∇
2
φ = 0 in R
3
\ {0}
Question: Think about the situation in R
2
, which radially symmetric functions
have zero Laplacian?
Definition 1.1. A vector field f is conservative if there exists a scalar potential field φ
for f, i.e. we have f = ∇φ for some suitable φ.
By Theorem 1.1 (i) we see curl ∇φ = 0, so
curl f = 0 is necessary for f to be conservative.
• Question: Is curl f also sufficient for f to be conservative?
Yes, if we work in all of R
3
. Also in any open ball, and hence yes also locally (as any point
is in a small ball in the domain of f).
NOT sufficient in general, e.g. if domain of f has “tunnels” - see the example on Problem
sheet 1, Q5. We will conider this question once more in Section 2 using line integrals.
To understand for which domains (= regions) the curl-condition is sufficient
was one of the starting points for development of topology as branch of math-
ematics - links to work ofmathematicians like Poincare (and hence Perelman)
start from here.
13
Example 1.3. f(r) = f(r) r is conservative if f is continuous on [0, ∞) (but not only if ).
Indeed,
f(r) = [f(r) r] e
r
= [f(r) r] ∇r = ∇H(r),
if
d
dr
H(r) = f(r) r holds, i.e. if H(r) =
r
_
0
f(x) xdx. Hence, “radial fields” are conservative.
( Indeed, the fact f = −
1
r
2
e
r
= ∇
_
1
r
_
is conservative and has divergence zero plays the
central role in the “inverse square law of gravitation”.)
Example 1.4. c = 0 constant vector. Let f : = c ∧ r, then f is not conservative.
curl f =
3

i=1
e
i
∧
∂
∂x
i
(c ∧ r)
=
3

i=1
e
i
∧
_
c ∧
∂r
∂x
i
_
=
3

i=1
e
i
∧ (c ∧ e
i
)
!
=
3

i=1
[(e
i
· e
i
)c −(e
i
· c)e
i
] = 3c −c = 2c
Compare to the example of “rotation in the plane” as discussed in Lecture 1, there was
c = k.
In the sequel, e.g. when calculating line integrals we will use
Theorem 1.3. If r = r(t), t ∈ (t
0
, t
1
) is the parametric representation of a smooth curve,
φ smooth scalar field, then:
d
dt
φ(r(t)) = ∇φ(r(t)) ·
d
dt
r(t)
Proof: We just need to show the chain rule for the vectorial situation:
14
d
dt
φ(x(t), y(t), z(t)) =
∂φ
∂x
dx
dt
+
∂φ
∂y
dy
dt
+
∂φ
∂z
dz
dt
=
_
∂φ
∂x
i +
∂φ
∂y
j +
∂φ
∂z
k
_
|r(t)
·
_
dx
dt
i +
dy
dt
j +
dz
dt
k
_
|t
= ∇φ(r(t)) ·
d
dt|t
r.