Electrodynamics - Solutions to problem set 1
20160905
Vector analysis
We will make use of the following operators:
Gradient: ∇f =
Divergence: ∇ · v =
Curl: ∇ × v =
∂f ∂f ∂f
,
,
∂x ∂y ∂z
(0.1)
∂vy
∂vz
∂vx
+
+
∂x
∂y
∂z
∂vz
∂vy ∂vx
∂vz ∂vy
∂vx
−
,
−
,
−
∂y
∂z ∂z
∂x ∂x
∂y
Laplacian: ∇2 f = ∇ · (∇f ) =
(0.2)
∂2f
∂2f
∂2f
+ 2 + 2
2
∂x
∂y
∂z
(0.3)
(0.4)
From these denitions, one can derive the following identities,
(0.5)
∇ · (∇ × v) = 0
(0.6)
∇ × (∇f ) = 0
(0.7)
2
∇ × (∇ × v) = ∇ (∇ · v) − ∇ v
which we will also use.
1
Show that
∇
1
|r − r 0 |
= −∇0
1
|r − r 0 |
=−
r − r0
|r − r 0 |3
(1.1)
Let r = (x, y, z) and r0 = (x0 , y 0 , z 0 ). We then have
|r − r 0 | =
p
(1.2)
(x − x0 )2 + (y − y 0 )2 + (z − z 0 )2
From 0.1, it follows that
∇
1
|r − r 0 |
=
∂
1
∂
1
∂
1
,
,
∂x |r − r 0 | ∂y |r − r 0 | ∂z |r − r 0 |
1
(1.3)
We start by computing the rst component
∂
1
∂
1
p
=
∂x |r − r 0 |
∂x (x − x0 )2 + (y − y 0 )2 + (z − z 0 )2
=−
1
1
∂
x − x0
0 2
(x
−
x
)
=
−
2 ((x − x0 )2 + (y − y 0 )2 + (z − z 0 )2 )3/2 ∂x
|r − r 0 |3
(1.4)
Similarly, the second and third component are given by
∂
1
y − y0
,
=
−
∂y |r − r 0 |
|r − r 0 |3
∂
1
z − z0
=
−
∂z |r − r 0 |
|r − r 0 |3
(1.5)
The gradient can therefore be written
∇
1
|r − r 0 |
=
y − y0
z − z0
x − x0
,
−
,
−
−
|r − r 0 |3
|r − r 0 |3
|r − r 0 |3
=−
r − r0
|r − r 0 |3
(1.6)
which is one part of our desired result. We now compute the gradient with respect to the primed
coordinate
1
∂
1
∂
1
∂
1
0
=
, 0
, 0
(1.7)
∇
0
0
0
0
0
|r − r |
∂x |r − r | ∂y |r − r | ∂z |r − r |
Once again, we start by looking at the rst component
∂
1
1
1
∂
=−
(x − x0 )2
0
0
3/2
∂x |r − r |
2 ((x − x0 )2 + (y − y 0 )2 + (z − z 0 )2 ) ∂x0
=
x − x0
|r − r 0 |3
(1.8)
Note that this diers by a sign from 1.4. By a similar calculation
∂
1
y − y0
=
,
0
0
∂y |r − r |
|r − r 0 |3
∂
1
z − z0
=
0
0
∂z |r − r |
|r − r 0 |3
(1.9)
and, as a result
∇0
1
|r − r 0 |
=
This implies that
∇
x − x0
y − y0
z − z0
,
,
0
3
0
3
|r − r | |r − r | |r − r 0 |3
1
|r − r 0 |
= −∇0
1
|r − r 0 |
=
r − r0
|r − r 0 |3
(1.10)
(1.11)
which is the other half of our desired result.
2
For arbitrary functions f and g and arbitrary volumes V with boundary S , show that
Z
f ∇2 g + ∇f · ∇g dV =
V
Z
I
f ∇g · da
(2.1)
S
f ∇2 g − g∇2 f dV =
V
I
(f ∇g − g∇f ) · da
S
2
(2.2)
Equation 2.1 is called Green's rst identity and 2.2 is Green's second identity, or Green's
theorem. We start by proving Green's rst identity. Note that
(2.3)
∇ · (f ∇g) = ∇f · ∇g + f ∇ · (∇g) = ∇f · ∇g + f ∇2 g
Using this, we rewrite the LHS of 2.1
Z
2
Z
f ∇ g + ∇f · ∇g dV =
(2.4)
∇ · (f ∇g) dV
V
V
Now we can use Gauss' theorem, which says that
Z
I
∇ · F dV =
V
(2.5)
F · da
S
for arbitrary F and V . This implies that
Z
I
∇ · (f ∇g) dV =
V
(2.6)
f ∇g · da
S
Combining this with 2.4 proves Green's rst identity. In order to prove Green's second identity,
we write down the rst again, with f and g interchanged.
Z
2
I
g∇ f + ∇g · ∇f dV =
V
(2.7)
g∇f · da
S
Subtracting this from 2.1 gives
Z
V
2
2
I
f ∇ g + ∇f · ∇g − g∇ f − ∇g · ∇f dV =
(f ∇g − g∇f ) · da
Z
IS
f ∇2 g − g∇2 f dV =
(f ∇g − g∇f ) · da
V
(2.8)
S
which proves Green's second identity.
Jackson writes Green's identities using the directional derivative along the normal vector n̂,
which can be written as
∂f
= ∇f · n̂
(2.9)
∂n
This is the rate of change of f along n̂. Since da = n̂ da, Green's identities can be written
Z
f ∇ g + ∇f · ∇g dV =
Z
V
I
∂g
da
∂n
V
S
I ∂g
∂f
f ∇2 g − g∇2 f dV =
f
−g
da
∂n
∂n
S
2
f
(2.10)
(2.11)
which is the form given in Jackson.
3
Compute the ux of the vector eld v through the unit sphere, where
v = 3xyex + xz 2 ey + y 3 ez
3
(3.1)
The ux is given by
can use Gauss' theorem
I
v · da, where S is the unit sphere. Since this is a closed surface, we
S
Z
I
∇ · v dV =
(3.2)
v · da
V
S
where V is the volume enclosed by the unit sphere. The divergence of v is given by
∇·v =
∂
∂ 3
∂
(3xy) +
(xz 2 ) +
(y ) = 3y
∂x
∂y
∂z
(3.3)
so the ux is given by
I
Z
v · da = 3
(3.4)
y dV
S
V
In order to compute this integral, we rewrite it in terms of spherical coordinates
(3.5)
(3.6)
dV = r2 sin θdrdθdφ
y = r sin θ sin φ
The ux therefore becomes
I
2π
Z
π
Z
1
Z
r3 sin2 θ sin φ dr dθ dφ
v · da = 3
S
0
0
0
Z π
1
Z
3
=3
r dr
0
since
Z
2π
2
Z
2π
sin θ dθ
0
sin φ dφ = 0
(3.7)
0
sin φ dφ = 0. The ux is therefore zero.
0
4
Formulate conditions under which the divergence and the curl of a vector eld on R3 , taken
together, dene the vector eld uniquely.
We are asked to formulate the Helmholtz theorem, which we will do at the end of this
solution.1 Suppose that we are given the divergence and curl of a vector eld F
∇·F =D
∇×F =C
(4.1)
(4.2)
As a consequence of identity 0.5, ∇ · C = 0. We now make the following construction
F = −∇U + ∇ × W
where
1
4π
Z
D(r 0 )
dV 0
|r − r 0 |
(4.4)
1
W (r) ≡
4π
Z
C(r 0 )
dV 0
|r − r 0 |
(4.5)
U (r) ≡
1
(4.3)
This solution is inspired by (i.e. stolen from) D. Griths, Introduction to Electrodynamics, Appendix B.
4
These integrals are taken over all of space. At the moment, this may seem rather ad hoc, but
it will make sense in a while. What we have to do now is prove that for this F, ∇ · F = D and
∇ × F = C . We start by computing the divergence,
∇ · F = −∇2 U = −
1
4π
Z
D(r 0 )∇2
1
|r − r 0 |
dV 0 =
Z
D(r 0 ) δ 3 (r − r 0 ) dV 0 = D(r)
(4.6)
In the rst step, we have used the identity 0.5, and in the second we have taken the Laplacian
inside the integral, which is allowed since the integration is taken w.r.t the primed coordinate.
In the second-to-last step, we have used
∇2
1
= −4πδ 3 (r)
r
(4.7)
where δ 3 (r) is the three dimensional Dirac delta function.
So we get the correct result for the divergence of F . What about the curl? Using 0.6 and
0.7, we can write
∇ × F = ∇ × (∇ × W ) = −∇2 W + ∇ (∇ · W )
(4.8)
We start with the rst term
−∇2 W = −
1
4π
Z
C(r 0 )∇2
1
|r − r 0 |
dV 0 =
Z
C(r 0 ) δ 3 (r − r 0 ) dV 0 = C(r)
(4.9)
This is our desired result, so we must now prove that the second term vanishes. Consider the
divergence of W
Z
1
dV 0 = − C(r 0 ) · ∇0
dV 0
|r − r 0 |
Z
Z
C(r 0 )
1
0
0
0
0
∇
·
C(r
)
dV
−
∇
·
dV 0
=
|r − r 0 |
|r − r 0 |
Z
4π∇ · W =
C(r 0 ) · ∇
1
|r − r 0 |
(4.10)
In the second step, we have used the result from problem 1, and in the third, we have used
integration by parts. Note that the rst integral on the second line is zero, since ∇ · C = 0. The
second integral can be rewritten using Gauss' theorem, 2.5, yielding the result
I
4π∇ · W = −
C(r 0 )
· da
|r − r 0 |
(4.11)
where the surface is at innity. This means that ∇ · W will vanish if C goes to zero suciently
fast at innity. We'll see in a moment that this is the case.
Note that everything we have done so far assumes that the integrals in 4.4 and 4.5 converge,
otherwise U and W are undened. For large r0 , we have |r − r0 | ≈ r0 , so the integrals look like
Z
∞
X(r0 ) 02 0
r dr =
r0
Z
∞
r0 X(r0 ) dr0
(4.12)
where X is either D or C and we have suppressed the angular integration. In order for this to
1
converge, X must to zero faster than 02 as r0 → ∞. Note that this is more than enough to
r
ensure that the surface integral in 4.11 vanishes.
We have showed that we can express F in terms of its divergence and curl, provided that they
1
go to zero faster than 2 . But the question was under which conditions they uniquely dene F .
r
So, is the F dened by 4.3 unique? The answer is no, as can seen in the following way. Dene
F0 = F + Y
5
(4.13)
where ∇ · Y = 0 and ∇ × Y = 0. Then
(4.14)
(4.15)
∇ · F0 = ∇ · F = D
0
∇×F =∇×F =C
So F is not unique, since on can always add a vector eld that with zero divergence and curl to
it. However, there are no vector elds with zero divergence and curl that goe to zero as r → ∞.
This can be seen by noting that equation 0.7, together with ∇ · Y = 0 and ∇ × Y = 0, imply
that ∇2 Y = 0. This is Laplace's equation. A property of this equation is that its solutions can't
have local maxima or minima, all extrema must be at the boundary. In this case, the boundary
is at innity, so if Y → 0 as r → ∞, then Y = 0 everywhere. In other words, the F dened by
equation 4.3 is unique if we demand that F → 0 as r → ∞.
We can now write down the Helmholtz theorem:
1
If ∇ · F and ∇ × F are specied everywhere, if they both go to zero faster than 2 as r → ∞
r
and if F → 0 as r → ∞, then F is given uniquely by equations 4.3, 4.4 and 4.5.
This answers the original question. As a corollary, it follows that any dierentiable vector
1
eld F that goes to zero faster than as r → ∞ can be expressed as:
r
F (r) = −∇
1
4π
Z
∇0 · F (r 0 ) 0
dV
|r − r 0 |
+∇×
1
4π
Z
∇0 × F (r 0 ) 0
dV
|r − r 0 |
(4.16)
As an example, consider Maxwell's equations for the E -eld in the electrostatic situation,
∇·E =
ρ
,
0
∇×E =0
(4.17)
Combined with equation 4.16, this gives
E(r) = −∇
1
4π0
Z
ρ(r 0 )
dV 0
|r − r 0 |
= −∇Φ(r)
(4.18)
where Φ is the scalar potential. Similarly, in the magnetostatic situation, Maxwell's equations
for the B -eld are
∇·B =0 ,
∇ × B = µ0 j
(4.19)
which, combined with equation 4.16, implies that
B(r) = ∇ ×
µ0
4π
Z
j(r 0 )
dV 0
|r − r 0 |
= ∇ × A(r)
(4.20)
where A is the vector potential. So these familiar expressions for E and B follow from Maxwell's
equations and the Helmholtz theorem.
5
a)
Show that the electric eld inside a conductor vanishes in equilibrium.
Inside a conductor there a a large number of charges that can move around freely,
even if the conductor as a whole is electrically neutral (e.g. free valence electrons in metals). In
equilibrium, there are no moving charges. This means that there cannot any electric eld inside
the conductor, since if there was, charges would start moving.
Answer:
6
b)
Show that a closed hollow conductor shields its interior from electric elds due to charges outside.
If there are no charges in the cavity, then ∇ · E = 0 there (Gauss' law). Since
E = −∇Φ, this means that ∇2 Φ = 0, i.e. the potential satises Laplace's equation inside the
Answer:
cavity. Also, the potential is constant inside the conductor itself (the conductor is an equipotential
volume). This can be seen from the fact that the potential dierence between two points is given
by
Z r2
E · dl
(5.1)
Φ(r1 ) − Φ(r2 ) =
r1
where the integral is path independent. For any two points inside the conductor, we can choose
a path that lies entirely inside the conductor, where E = 0, which means that the potential
dierence if zero. In particular, this means that the potential is constant at the boundary of the
cavity. The potential can't have any local maxima or minima inside the cavity, since it satises
Laplace's equation, and must therefore also be constant in the cavity. This implies that E = 0,
since E = −∇Φ.
c)
Show that a closed hollow conductor does not shield its exterior from electric elds due to charges
inside.
Answer:
Consider Gauss' law on the integral form,
I
E · da =
S
Qencl
0
(5.2)
where Qencl is the charge enclosed by the surface S . Let S be a surface that encloses the
conductor. Since Qencl is not zero, the electric ux through S is not zero. This means that the
electric eld outside the conductor cannot be zero.
d)
Show that the electric eld at the surface of a conductor has a vanishing tangential component.
Answer:
We now use Faraday's law on the integral form,
I
E · dl = 0
(5.3)
l
Let the loop be a small rectangle, with two sides perpendicular to the surface and two sides
tangential to it. One of the tangential sides is inside the conductor, the other outside. Since
E = 0 inside the conductor, that part of the loop gives no contribution to the integral. If we
let the perpendicular components go to zero, their contribution vanishes as well. Hence the only
remaining contribution comes from the tangential side outside the conductor, which must then
be zero. The tangential component of E at the surface is therefore zero.
7