PHZ 3113 Fall 2010
Homework #5, Due Wednesday, October 13
1. Work out in detail the value of ∇ · (∇φ × ∇ψ).
The vector ∇φ × ∇ψ is
∂φ
∂φ
∂φ ∂ψ
∂ψ
∂ψ ∇φ × ∇ψ = x̂
+ ŷ
+ ẑ
+ ŷ
+ ẑ
× x̂
∂x
∂y
∂z
∂x
∂y
∂z
∂φ ∂ψ ∂φ ∂ψ ∂φ ∂ψ ∂φ ∂ψ ∂φ ∂ψ ∂φ ∂ψ = x̂
−
+ ŷ
−
+ ẑ
−
∂y ∂z
∂z ∂y
∂z ∂x ∂x ∂z
∂x ∂y
∂y ∂x
The divergence is then
∇ · (∇φ × ∇ψ) =
=
∂ ∂φ ∂ψ ∂φ ∂ψ ∂ ∂φ ∂ψ ∂φ ∂ψ ∂ ∂φ ∂ψ ∂φ ∂ψ −
−
−
+
+
∂x ∂y ∂z
∂z ∂y
∂y ∂z ∂x
∂x ∂z
∂z ∂x ∂y
∂y ∂x
∂ 2 φ ∂ψ ∂φ ∂ 2 ψ
∂ 2 φ ∂ψ ∂φ ∂ 2 ψ
+
−
−
∂x ∂y ∂z
∂y ∂x ∂z ∂x ∂z ∂y
∂z ∂x ∂y
2
2
2
∂ φ ∂ψ ∂φ ∂ ψ
∂ φ ∂ψ ∂φ ∂ 2 ψ
+
+
−
−
∂y ∂z ∂x
∂z ∂y ∂x ∂y ∂x ∂z
∂x ∂y ∂z
∂ 2 φ ∂ψ ∂φ ∂ 2 ψ
∂ 2 φ ∂ψ ∂φ ∂ 2 ψ
+
+
−
−
= 0.
∂z ∂x ∂y
∂x ∂z ∂y ∂z ∂y ∂x
∂y ∂z ∂x
As long as second partial derivatives taken in either order are the same, terms in the last
expression cancel in pairs: the first and seventh, the second and twelfth, the third and ninth,
the fourth and sixth, the fifth and eleventh, and the eight and tenth. I know that the Table
of Vector Identities Involving ∇ on p. 339 says ∇ · (U × V ) = V · (∇ × U ) − U · (∇ × V )
and other useful things, or even ∇ · (∇φ × ∇φ) = 0, but this calculation (“in detail”) derives
the result.
There is another way to do this, using the ijk ,
∇ · (∇φ × ∇ψ) = ∇i (ijk ∇j φ ∇k ψ) = ijk ∇i (∇j φ ∇k ψ)
= ijk [(∇i ∇j φ) ∇k ψ + ∇i φ (∇j ∇k ψ)] = 0.
This happens again because partial derivatives commute, ∇i ∇j φ = ∇j ∇i φ, and are summed
against the antisymmetric ijk = −jik .
2. Let v(x, y) be the vector field
v=
(x − y)
(x + y)
x̂ +
ŷ,
3
r
r3
p
where r = x2 + y 2 . Let C be the circumference of the circle of radius r = 1 in the x-y
plane centered at the origin.
I
(a) Compute
C
v · n̂ ds two different ways.
To compute the integrals, we need to write everything in terms of some one thing that can
then integrate over. Let a point on the circle be given by angle θ, x = cos θ, y = sin θ. Then,
position and its derivative are
r = x x̂ + y ŷ = cos θ x̂ + sin θŷ,
dr = (− sin θ x̂ + cos θ ŷ) dθ,
ds = |dr| = dθ;
and the unit vector perpendicular to the circle (“unit normal”) is
n̂ = cos θ x̂ + sin θ ŷ.
In terms of θ,
I
I
v · n̂ ds = [(cos θ − sin θ) x̂ + (cos θ + sin θ) ŷ] · (cos θ x̂ + sin θ ŷ) dθ
C
I h
Z 2π
i
2
2
=
cos θ − sin θ cos θ + cos θ sin θ + sin θ dθ =
dθ = 2π.
0
Using the two-dimensional divergence theorem,
I
Z
v · n̂ ds = (∇ · v) d2 a,
C
A
with
∇·v =
∂ (x − y)
∂ (x + y)
1
3x(x − y)
1
3y(x + y)
1
+
= 3−
+ 3−
= − 3,
3
3
5
5
∂x r
∂y r
r
r
r
r
r
would seem to give an answer that is of the opposite sign, not to mention infinite, but that
can’t be: there is something hiding in that divergence. Let the denominator be regulated so
that it cannot vanish: replace r 3 with (x2 + y 2 + a2 )3/2 (or other possibilities of your choice);
then
2a2 − x2 − y 2
∇·v =
,
(x2 + y 2 + a2 )5/2
and
1
2π
=
(∇ · v) d2 a =
2πr dr =
→ 2π.
5/2
3/2
2
2
2
2
(r + a )
(1 + a2 )3/2
A
0 (r + a )
0
Z
Z 1
(2a2 − r 2 )
2πr 2
I
(b) Compute
C
v · dr two different ways.
This time, using the θ-parametrization,
I
Z
v · dr = [(cos θ − sin θ) x̂ + (cos θ + sin θ) ŷ] · (− sin θ x̂ + cos θ ŷ) dθ
C
Z
= [− cos θ sin θ + sin2 θ + cos2 θ + sin θ cos θ] dθ = 2π.
Meanwhile, the curl is
∇×v =
∂v
y
∂x
−
(2a2 − x2 − y 2)
∂vx ẑ =
ẑ
∂y
(x2 + y 2 + a2 )5/2
leading to the same integral:
I
Z
C
v · dr =
A
(∇ × v) · n̂ d2 a =
Z 1
(2a2 − r 2 )
0 (r 2 + a2 )5/2
=
2π
(1 + a2 )3/2
→ 2π.
(c) What happens when the denominator in v is r 2 instead of r 3 ?
The values of the integrals are the same for any power r p . Without regularization, the
divergence or curl for denominator r 2 would appear to vanish entirely, but with the cutoff a
2a2
∇·v = 2
,
(x + y 2 + a2 )2
and
Z
A
(∇ · v) d2 a =
2a2
ẑ,
∇×v = 2
(x + y 2 + a2 )2
1
2a2
2πr 2 2π
2πr
dr
=
=
→ 2π.
2
2
r 2 + a2 0 1 + a2
0 r +a
Z 2
The only contribution to the area integral is from the δ-function hiding in the divergence/curl.
3. (a) Apply the divergence theorem to compute the integral
Z
(φ1 ∇2 φ2 − φ2 ∇2 φ1 ) d3 v
V
as
involving ∂φ1 /∂n = n̂ · ∇φ1 , and ∂φ2 /∂n = n̂ · ∇φ2 . [Hint: start with
R a surface integral
3
V ∇ · (φ1 ∇φ2 ) d v; do it again exchanging φ1 and φ2 , and subtract.]
Following the hint,
∇ · (φ1 ∇φ2 ) = ∇φ1 · ∇φ2 + φ1 ∇2 φ2 ,
and so
Z
Z
I
I
∂φ
3
2
3
2
n̂ · (φ1 ∇φ2 ) d a =
∇ · (φ1 ∇φ2 ) d v =
(∇φ1 · ∇φ2 + φ1 ∇ φ2 ) d v =
φ1 2 d2 a,
∂n
V
V
S
S
known as Green’s first identity. Exchanging φ1 and φ2 gives
Z
Z
I
∂φ
3
2
3
∇ · (φ2 ∇φ1 ) d v =
(∇φ2 · ∇φ1 + φ2 ∇ φ1 ) d v =
φ2 1 d2 a;
∂n
V
V
S
and subtracting leaves Green’s second identity,
Z I ∂φ2
∂φ1 2
2
2
3
φ1 ∇ φ2 − φ2 ∇ φ1 d v =
φ1
− φ2
d a.
∂n
∂n
V
S
(b) Suppose φ1 is the electrostatic potential obtained from charge density ρ1 in the interior
of V with surface charge density σ1 on the surface S of V , such that
ρ
∂φ1 σ1
∇2 φ 1 = − 1 ,
= ;
0
∂n S
0
and φ2 is found similarly from ρ2 and σ2 . Prove the reciprocity theorem,
Z
Z
Z
Z
3
2
3
ρ1 φ2 d v +
σ1 φ2 d a =
ρ2 φ1 d v +
σ2 φ1 d2 a
V
S
V
S
(the energy of ρ1 , σ1 in the potential φ2 is the same as the energy of ρ2 , σ2 in the potential
φ1 ).
Substituting for the Laplacians and normal derivatives in (a),
I Z h
σ2
σ1 2
ρ2
ρ1 i 3
− φ2
φ1 (− ) − φ2 (− ) d v =
φ1
d a,
0
0
0
0
V
S
which rearranges to what is wanted
Z
Z
Z
Z
3
2
3
ρ1 φ2 d v +
σ1 φ2 d a =
ρ2 φ1 d v +
σ2 φ1 d2 a.
V
S
V
S
This is more or less Problem 1.12 at the end of Chapter 1 of Jackson’s Classical Electrodynamics (the standard graduate text). [It isn’t quite that easy; Jackson doesn’t give explicitly
the replacements for ∇2 φ and ∂φ/∂n.]