Field Theory
Answer for 3rd Set of Problems
RADIATING SYSTEMS
√
1. With the relation of 0 µ0 = 1/c2 and (a + b)/2 ≥ ab for non-negative values of a and b,
q
~ 2 + c2 |B|
~ 2
|
E|
2
2
~
~
~ ~
~
~
~
0
|B|
0 |E|
~ 2 c2 |B|
~ 2 = 0 c|E||
~ B|
~ = |E||B| ≥ |E × B| = |S| .
u=
+
=
≥ 0 |E|
2µ0
2
2
µ0 c
µ0 c
c
1
2. (a) In the wave zone, the electric dipole fields are
ck 2
~ )),
(~n × p(τ
4πr
~ x, t) × ~n,
= Z0 H(~
~ x, t)
H(~
=
~ x, t)
E(~
p
where k = ω/c, ~n = ~x/r, Z0 = µ0 /0 , and τ = t − r/c is the retarded time. In Cartesian coordinates, we
obtain
~ ~x)
B(t,
=
=
~ ~x)
E(t,
=
=
µ0 ω 2
(~x × p~)
4πcr2
µ0 ω 2 p0
[−z sin(ωτ )~ex + z cos(ωτ )~ey + {x sin(ωτ ) − y cos(ωτ )} ~ez ] ,
4πcr2
c~
B × ~x
r
µ0 ω 2 p0 2
(y + z 2 ) cos(ωτ ) − xy sin(ωτ ) ~ex + (z 2 + x2 ) sin(ωτ ) − xy cos(ωτ ) ~ey
3
4πr
− {yz sin(ωτ ) + zx cos(ωτ )} ~ez ,
In order to change the basis, we use the relation
∂xi j
∂xi j
i
B̃
,
and
E
=
Ẽ ,
∂ x̃j
∂ x̃j
the r component of the electric and magnetic fields are
Bi =
Br
=
Er
=
∂r x
B +
∂x
∂r x
E +
∂x
∂r y
B +
∂y
∂r y
E +
∂y
∂r z
B ,
∂z
∂r z
E .
∂z
With ∂r/∂x = x/r, ∂r/∂y = y/r, and ∂r/∂z = z/r, we can find that B r (t, ~x) = E r (t, ~x) = 0. Similarly we
can describe the below relations;
Bx
=
By
=
Bz
=
Ex
=
Ey
=
Ez
=
1 These
∂x θ ∂x φ
B +
B ,
∂θ
∂φ
∂y θ ∂y φ
B +
B ,
∂θ
∂φ
∂z θ
B ,
∂θ
∂x θ ∂x φ
E +
E ,
∂θ
∂φ
∂y θ ∂y φ
E +
E ,
∂θ
∂φ
∂z θ
E .
∂θ
formulas are written in terms of the physical (real-valued) time-dependent fields and dipole moment, in contrast to the formulas
given in the lecture which apply to the corresponding time-independent complex valued quantities.
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By solving these relations, we get
B θ (t, ~x)
=
=
B φ (t, ~x)
=
=
E θ (t, ~x)
=
=
E φ (t, ~x)
=
=
µ0 ω 2 p0
[cos φ sin(ωτ ) − sin φ cos(ωτ )] ,
4πcr2
2
µ0 ω p0
−
sin(ωτ − φ),
4πcr2
µ0 ω 2 p0
cot(θ) [sin(φ) sin(ωτ ) + cos(φ) cos(ωτ )] ,
4πcr2
µ0 ω 2 p0
cot(θ) cos(ωτ − φ),
4πcr2
µ0 ω 2 p0
cos(θ) [sin(φ) sin(ωτ ) + cos(φ) cos(ωτ )] ,
4πr2
µ0 ω 2 p0
cos(θ) cos(ωτ − φ),
4πr2
2
µ0 ω p0
[cos(φ) sin(ωτ ) − sin(φ) cos(ωτ )] ,
4πr2 sin(θ)
µ0 ω 2 p0
sin(ωτ − φ).
4πr2 sin(θ)
−
With these components the electric and magnetic fields are written as
~ ~x) = B θ (t, ~x)~eθ + B φ (t, ~x)~eφ ,
B(t,
~ ~x) = E θ (t, ~x)~eθ + E φ (t, ~x)~eφ .
E(t,
(b) The power radiated per unit solid angle is given as
dP
(τ, θ, φ)
dΩ
=
c2 Z0 k 4
2
|(~n × p~(τ )) × ~n|
16π 2
Since the time average of the radiated power should be axisymmetric for a dipole rotating around the z-axis,
we only need to compute it for φ = 0, i.e. we can choose choose ~n = cos(θ)~ez + sin(θ)~ex without loosing
generality. Then
|(~n × p~) × ~n| =
=
... |(~n × p~) × ~n|2
=
|~n(~n · p~) − p~|
p0 − cos2 θ cos(ωτ )~ex − sin(ωτ )~ey + sin θ cos θ cos(ωτ )~ez ,
p20 cos2 θ cos2 (ωτ ) + sin2 (ωτ ) .
Thus by averaging over a complete wave cycle,
dP
dΩ
=
=
Z 2π/ω
2
µ0 ω 4 p20 ω
cos θ cos2 (ωτ ) + sin2 (ωτ ) dτ
16π 2 c 2π 0
µ0 ω 4 p20
(1 + cos2 θ).
32π 2 c
(c) In order to obtain the total radiated power < P >, we must integrate over the angles. Using the relation
I
Z
2
1
(1 + cos θ)dΩ = 2π
(1 + cos2 θ)d(cos θ) =
−1
we obtain the radiated power
< P >=
µ0 ω 4 p20 16π
µ0 ω 4 p20
=
.
2
32π c 3
6πc
2
16π
,
3
On the other hand, for the oscillating dipole given by p~(t) = p0 cos(ωt)~ez , the radiated power per unit solid
angle is
dP
µ0 ω 4 p20
=
cos2 (ωτ ) sin2 θ.
dΩ
16π 2 c
Then the radiated power averaged over a complete wave cycle is
dP
µ0 ω 4 p20
=
sin2 θ,
dΩ
32π 2 c
and the total radiated power is
µ0 ω 4 p20
.
12πc
Thus the total radiated power by a rotating dipole is twice of that by an oscillating dipole.
< P >=
3. (a) See Figure ??.
(b) See Figure ??.
Figure 1: Electric field lines of an oscillating dipole aligned with the z-axis at t = 0, 1, 2, and 3 for ω = 1. The vertical
axis corresponds to the z-axis and the horizontal one to the x-axis.
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Figure 2: Electric field lines of a dipole rotating around the z-axis, at t = 0, 1, and 2 for ω = 1. The horizontal and
vertical axes correspond to the x- and y axis. Note that although it seems to change, the electric field is just rotating
around the z-axis. The reason for the apparent change is that the starting points chosen for the integration of the
field lines are not corotating.
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