z
giving
φ
ρ
so
eρ = cos φex + sin φey
eφ = − sin φex + cos φey
ez = ez .
and so
ρ = const.
φ = const.
z = const.
are cylinders
are planes through Oz
are planes ⊥ to Oz
∂r ∂r ∂r hρ = = 1 hφ = = ρ hz = = 1 ,
∂ρ
∂φ
∂φ
{eρ , eφ , ez } are orthonormal (check). Note that r = ρ eρ (φ) + z ez ≡ (ρ, 0, z)
and
which are mutually orthogonal surfaces.
dS ρ = hφ hz dφdz eρ = ρdφ eρ ,
• Geometrically, the situation is simple enough that we can find the h-factors
giving
7.2.3
dV = hρ hφ hz dρdφdz = ρdρdφdz .
Spherical polar co-ords: (r, θ, φ)
ρdφ
z
φ
dz
θ
r
O
dρ
hρ = 1
hφ = ρ
hz = 1
This gives curved surface element and volume elements
dS ρ = ρdφeρ ,
dV = dρ (ρdφ) dz = ρ dρdφ dz
• Alternatively, algorithmically, we have
r = ρ cos φ ex + ρ sin φ ey + z ez .
Thus
∂r
= cos φex + sin φey
∂ρ
∂r
= −ρ sin φex + ρ cos φey
∂φ
∂r
= ez
∂z
69
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
with 0 ≤ r < ∞, 0 ≤ θ < π, 0 ≤ φ < 2π. So
r = const.
θ = const.
φ = const.
are spheres
are cones
are planes through Oz
which are mutually orthogonal surfaces.
• Geometrically, the situation is again simple enough that we can find the hfactors giving
70
z
φ
dr
θ
r
rdθ
O
r sin θdφ into paper
hr = 1
hθ = r
hφ = r sin θ
Chapter 8
Vector differential operators, the
Divergence and Stokes’ Theorems
This gives surface element (on the sphere) and volume element of
dS r = (rdθ)(r sin θdφ) er = r 2 sin θ dθdφ er
dV
= dr(rdθ)(r sin θdφ) = r 2 sin θ drdθdφ .
We first repeat, and slightly extend, the definitions and results of section 1.1.6.
8.1
Gradient - grad
• Alternatively, algorithmically, we have
r = r sin θ cos φ ex + r sin θ cos φ ey + r cos θ ez .
Returning to cartesian co-ordinates, we define the differential operator (acting on a
scalar, φ) as
Thus
∂r
= sin θ cos φex + sin θ sin φey + cos θez
∂r
∂r
= r cos θ cos φex + r cos θ sin φey − r sin θez
∂θ
∂r
= −r sin θ sin φex + r sin θ cos φey
∂φ
∂φ
∂φ
∂φ
e +
e +
e .
∂x x ∂y y ∂z z
or using indices,
(∇φ)i =
∂φ
≡ ∂i φ .
∂xi
• This is a vector (with φ a scalar)
giving
so
def
∇φ =
∂r ∂r hr = = 1 hθ = = r
∂r
∂θ
∂r hφ = = r sin θ
∂φ
er = sin θ cos φex + sin θ sin φey + cos θez
eθ = cos θ cos φex + cos θ sin φey − sin θez
eφ = − sin φex + cos φey .
(∇φ)′i =
as x′i = liα xα or xα = x′i liα .
• Other notation grad(φ); alternative name nabla
• Alternatively as
{er , eθ , eφ } are orthonormal (check). Note that r = r er (θ, φ) ≡ (r, 0, 0) and
dS r = hθ hφ dθdφ er = r 2 sin θ dθdφ er
dV
= hr hθ hφ drdθdφ = r 2 sin θ drdθdφ
71
∂φ′
∂xα ∂φ
=
= liα (∇φ)α
∂x′i
∂x′i ∂xα
dφ =
∂φ
∂φ
∂φ
dx +
dy +
dz ,
∂x
∂y
∂z
then
dφ = ∇φ · dr .
72
~n
or using indices,
d~r
∇·a=
φ = const
∂ai
.
∂xi
• This is a scalar. (Proof – see example sheet)
– First let dr be a displacement in the surface φ = const., so dφ = 0. So
∇φ is ⊥ dr, ie
• Alternative notation: div(a).
• Generalising:
The divergence of a tensor of rank n is a tensor of rank n − 1.
∇φ k n .
– Secondly take dr = ndl (ie ds ≡ dl) so dφ = ∇φ · n dl = |∇φ| dl (as
∇φ k n) so
For example, the divergence of Tij is ∂Tij /∂xj is a vector.
~n
8.3
Rotation - curl
dl
|∇φ| =
dφ
.
dl
or using indices,
def
Thus ∇φ is a vector perpendicular to surface φ = const. and with magnitude
equal to rate of change of φ along the normal.
• Generalising:
If Tij···(r) is an m-th rank tensor, ∂Tij··· /∂xn is an m + 1-th rank tensor.
ex ey ez ∂ ∂ ∂ ∇ × a = ∂x ∂y ∂z a a a x
y
z
(∇ × a)i = ǫijk
∂
ak .
∂xj
• If a is a vector, then ∇ × a is a pseudoscalar.
• Alternative notation/names: curl(a), rot(a).
For example
Thus, for example,
If φ = r, then
∇r = r/r
∇(φψ) = φ∇ψ + ψ∇φ
∇ · (φu) = φ∇ · u + (u · ∇)φ
∇ × (u × v) = u(∇ · v) + (v · ∇)u − v(∇ · u) − (u · ∇)v
– vector ,
If Ti = ri (r · a) then
(∇Ti )j =
8.2
∂Ti
= δij (r · a) + ri aj
∂rj
– 2nd rank tensor .
8.4
Three Fundamental Theorems
1. The line integral of ∇φ along a curve,
Divergence - div
∂ax ∂ay ∂az
+
+
.
∇·a =
∂x
∂y
∂z
def
73
Z
P2
dr · ∇φ = φ(P2 ) − φ(P1 ) ,
P1
74
8.4.2
Proof and simple consequences of the divergence theorem
P2
d~r
Take a frame Oxyz such that Ox, Oy, Oz intersect S each in an even number of
points.
P1
y
2. Divergence or Gauss’ Theorem
S
~n
1
3
2
4
V
x
O
Z
∇ · a dV =
V
Z
a · dS .
S
z
S is the closed surface containing volume V . RHS is called flux of a through
S.
Consider
3. Stokes’ Theorem
Z
V
Z
x2
∂ax
dx +
∂x
x4
∂ax
dx + . . .
∂x
∂ax
dxdydz =
∂x
Z Z
dydz
=
Z Z
dydz [ax (x2 , y, z) − ax (x1 , y, z) + . . .]
=
Z
S
C
S
x1
Z
x3
ax ex · dS ,
as
Z
(∇ × a) · dS =
S
I
C
C is a closed curve spanned by the open surface S.
Many (sensible) conditions needed on φ, a, S, C, V for theorems to hold. (eg
differentiable functions, surfaces not infinitely twisted, orientable surfaces, . . .)
8.4.1
ex · dS =
a · dr .
+dydz x2 , x4 , . . .
−dydz x1 , x3 , . . .
Similarly for the other components and hence
Z Z
Z
ax ex + ay ey + az ez · dS =
a · dS .
∇ · a dV =
V
S
S
Proof of the line integral theorem
Simple consequences of the divergence theorem
Z
P2
dr · ∇φ =
P1
Z
P2
dφ = φ(P2 ) − φ(P1 ) .
P1
So integral only depends on end points and not on the path taken between P1 and
P2 (in the one dimensional case there is, of course, only one path from x1 to x2 ).
75
1. Let
a = (u(x, y), v(x, y), 0) ,
and take V to be the cylinder 0 < z < 1.
76