Physics 460 Notes
1
Fall 2006
Susan M. Lea
Dipole radiation
Electromagnetic waves are produced by accelerating charges, as we have seen.
But when lots of charges are involved it is sometimes easier to work with the
charge and current distributions. There is no radiation unless these distributions change in time. Wezll start with a few simple cases. The ¤rst is an ideal
dipole that oscillates in time:
p~ (t) = ~p 0 cos !t
We can create such a source by having a point charge undergo simple harmonic
motion, for example. Following Gri¢ths, we look at the simpler case of two
equal and opposite charges ¤xed in position at z = §d=2, but with a timedependent charge
q (t) = q0 cos !t
Then the dipole moment is
~p = q0 d^
z cos !t = ~p 0 cos !t
Now we calculate the potentials due to this source using equations (9) and (10)
from Notes 4. We will make several simplifying assumptions.
1. d ¿ r: The source dimension is much less than the distance to the source.
2. !d=c ¿ 1 The dipole oscillates slowly, or equivalently, its length is much
less than a wavelength.
3. !r=c À 1: The distance from the source to the observer is much greater
than a wavelength.
Then the potential is
V (~r; t)
=
=
Z
1
½ (~r0 ; t ¡ R=c) 0
d¿
4¼" 0
R
µ
¶
q0
cos ! (t ¡ R+ =c)
cos ! (t ¡ R¡ =c)
¡
4¼" 0
R+
R¡
where, using approximation 1, d ¿ r; we have
R§ =
q0
V (~r; t) =
4¼"0 r
Ã
µ
¶
d
r2 + d2 =4 ¨ rd cos µ ' r 1 ¨
cos µ
2r
p
£
¡
¢¤
£
¡
¢¤ !
d
d
cos !t ¡ ! rc 1 ¡ 2r
cos µ
cos !t ¡ ! rc 1 + 2r
cos µ
¡
¢
¡
¢
¡
d
d
r 1 ¡ 2r
cos µ
r 1 + 2r
cos µ
1
Then we apply the second approximation !d=c ¿ 1 and expand the cosine to
¤rst order in !d=c
µ
¶
µ
¶
³
³
r´
!d
r´
!d
cos ! (t ¡ R+=c) = cos ! t ¡
cos
cos µ ¡ sin ! t ¡
sin
cos µ
c
2c
c
2c
³
³
r ´ !d
r´
' cos ! t ¡
¡
cos µ sin ! t ¡
c
2c
c
Then
V
=
=
=
q0
4¼"0 r
Ã
¡
¢
cos ! t ¡ rc ¡
!d
cos µ sin
2c
d
¡ 2r cos µ
¡
¢
! t ¡ rc
¡
¢
cos ! t ¡ rc +
¡
1
Ã
¡
¢ !d
¡
¢!
d
r
r
cos
µ
cos
!
t
¡
¡
cos
µ
sin
!
t
¡
q0
r
c
c
c
¡d
¢2
4¼"0 r
1 ¡ 2r
cos µ
µ
¶
³
³
p0
1
r´ !
r´
cos µ cos ! t ¡
¡ cos µ sin ! t ¡
4¼"0 r r
c
c
c
1
!d
cos µ sin
2c
d
+ 2r cos µ
In the limit ! ! 0 (static dipole) the potential is
V !
p 0 cos µ
4¼"0 r 2
as expected. However, using approximation 3, we ¤nd that the second term
dominates, and
³
p0 !
r´
V '¡
cos µ sin ! t ¡
(1)
4¼ "0 r c
c
The vector potential is due to the current, which exists everywhere along
the line between the two charges.
I (t) =
dq
= ¡!q0 sin !t
dt
Wezll need the distance
µ
¶
q
z0
R (z 0 ) = r2 + (z 0)2 ¡ 2rz0 cos µ ' r 1 ¡ cos µ
r
2
¡
¢!
! t ¡ rc
Then
~
A
¹0
z^
4¼
=
Z
+ d=2
¡ d=2
Z
+d=2
! sin ! (t ¡ R=c) 0
dz
R
¡d=2
³
´
Z +d=2 sin ! t ¡ r + z 0 cos µ
c
c
¹ q0 !
¡ 0
^z
dz0
z0
4¼ r
1
¡
cos
µ
¡d=2
r
Z +d=2 ·
¸µ
¶
³
³
¹ q0 !
r´
z0
r´
z0
¡ 0
^z
sin ! t ¡
+ ! cos µ cos ! t ¡
1 + cos µ dz 0
4¼ r
c
c
c
r
¡d=2
¾
Z +d=2 ½
³
´
h
³
´
i
0
¹ q0 !
r
z
!r
r
¡ 0
^z
sin ! t ¡
+ cos µ
cos ! t ¡
+ 1 dz 0
4¼ r
c
r
c
c
¡d=2
³
´
³
´
¹ q0 !d
r
¹ p0 !
r
¡ 0
z^ sin ! t ¡
=¡ 0
z^ sin ! t ¡
(2)
4¼ r
c
4¼r
c
= ¡
=
=
=
'
¹0 q0
^z
4¼
I (tre t ) 0
dz
R
The second term integrates to zero (odd integrand over even interval) so the
next non-zero term would be of order (!d=c) (d=r) and we neglect it.
Now that we have the potentials (1) and (2), we calculate the ¤elds.
~
E
~
rV
=
=
'
~
@A
@t
³
³
p0 !
r´
@ ¹0 p 0 !
r´
~
= r
cos µ sin ! t ¡
+
z^ sin ! t ¡
4¼"0 r c
c
@t 4¼r
c
~ ¡
= ¡ rV
@V
1 @V ^
^r +
µ
@r
r @µ
½·
¸
³
³
³
´¾
p0 !
cos µ
r ´ ! cos µ
r´
^µ sin µ sin ! t ¡ r
sin
!
t
¡
+
cos
!
t
¡
r
^
+
4¼" 0 c
r2
c
c r
c
r2
c
³
´
2
p0 ! cos µ
r
cos ! t ¡
r^
4¼" 0 c2 r
c
where the terms we dropped are of order c=!r with respect to the one we kept.
~
@A
@t
=
=
Thus the total ¤eld is
But
³
¹0 p 0 !2
r´
z^ cos ! t ¡
4¼r
c
³
p0 ! 2
r´
¡
z
^
cos
!
t
¡
4¼"0 rc2
c
¡
³
´
2
~ = ¡ p 0 ! cos ! t ¡ r ( r^ cos µ ¡ z^)
E
4¼"0 r c2
c
^z = r^ cos µ ¡ ^µ sin µ
3
so
The magnetic ¤eld is
³
´
2
~ = ¡^µ p 0 ! sin µ cos ! t ¡ r
E
4¼"0 r c2
c
(3)
³
´
~ =r
~ £A
~ = ¡r
~ £ ¹0 p 0 ! z^ sin ! t ¡ r
B
4¼r
c
~ has only r and µ components and each depends only on r and µ; so B
~ has
A
only a Á component:
¸
^ ·
~ = Á @ (rAµ ) ¡ @Ar
B
r @r
@µ
"
¡
¢#
³
³
¹0 p0 ! ^ @
r ´´ 1 @ cos µ sin ! t ¡ rc
=
Á
sin µ sin ! t ¡
+
4¼r
@r
c
r
@µ
·
¸
³
´
³
¹0 p 0 ! ^ !
r
1
r´
= ¡
Á
sin µ cos ! t ¡
+ sin µ sin ! t ¡
4¼r
c
c
r
c
Again the ¤rst term is much larger than the second, so we have
³
´
2
~ = ¡ ¹0 p0 ! ^Á sin µ cos ! t ¡ r
B
4¼rc
c
(4)
Comparing the two expressions (3) and (4) we see that
~ = 1 r^ £ E
~
B
c
as we expect for an EM wave. The Poynting ¢ux is
µ
¶
1 ~
1 ~
1
~
~
~
S =
E£B =
E£
r^ £ E
¹0
¹0
c
µ
¶2 4
³
1
1
p0
!
r´
=
r^E 2 =
^r
sin2 µ cos2 ! t ¡
4
¹0 c
¹0 c
4¼ "0 r
c
c
The 1=r 2 dependence on distance is also what we expect. Now we time average
to get
p 20
!4
~ >= ^r
<S
sin2 µ
2
2
32¼ "0 r c3
Then
¯ ¯
dP
p20 ! 4
¯~ ¯
= r2 < ¯S
sin 2 µ
(5)
¯ >=
d32¼ 2" 0 c3
and, with ¹ = cos µ;
Z
Z
¢
dP
p 20 ! 4 +1 ¡
P =
d- =
1 ¡ ¹2 d¹
d16¼"0 c3 ¡1
=
p20 ! 4 4
p 20 ! 4
=
16¼"0 c3 3
12¼"0 c3
4
(6)
The total power radiated goes like the frequency to the 4th power and the square
of the dipole moment. The angular distribution of power (equation 5) goes as
sin 2 µ: With the polar axis plotted to the right, it looks like this:
1
0.5
0
-0.5
-1
This is a special case of our point charge radiation formula, because we can
model the dipole with an oscillating charge, and so it is no surprise that we get
the same angular dependence.
2
Magnetic dipole radiation
An oscillating current in a circular loop of radius b is a good model for an
oscillating magnetic dipole. With I (t) = I0 cos !t;
m
~ (t) = ¼b2 I (t) ^n = ¼b 2 I0 cos !t n
^=m
~ 0 cos !t
The loop is uncharged, and so we have only a vector potential:
Z
¹0
I0 cos [! (t ¡ R=c)] ^
~
A=
ÁbdÁ0
4¼
R
5
We have to take care with the unit vector, since it is not a constant.
convert to Cartesian unit vectors:
So we
^ = ¡^
Á
x sin Á0 + y^ cos Á0
We also need an expression for R :
R2 = r2 + b2 ¡ 2~r ¢ ~b
We may place the x¡axis so that the observation point is in the x ¡ z¡plane~
as shown in the diagram Then
¡
¢
~r ¢ ~b = r (^
z cos µ + x^ sin µ) ¢ b x^ cos Á0 + y^ sin Á0
= rb sin µ cos Á0
Thus
R=r
r
b2
b
1 + 2 ¡ 2 sin µ cos Á0 ' r
r
r
µ
b
1 ¡ sin µ cos Á0
r
¶
where the analogue of our previous approximation 1 is b ¿ r: Thus
£ ¡
¢¤
Z
¤
¹0 2¼ I0 cos ! t ¡ r=c + bc sin µ cos Á0 £
~
A =
¡
¢
¡ x^ sin Á0 + y^ cos Á0 bdÁ0
b
0
4¼ 0
r 1 ¡ r sin µ cos Á
Z 2¼ ·
¸
¹0 bI0
!b
=
cos ! (t ¡ r=c) ¡ sin ! (t ¡ r=c)
sin µ cos Á0
4¼r 0
c
µ
¶
£
¤
b
£ 1 + sin µ cos Á0 ¡^
x sin Á0 + y^ cos Á0 dÁ0
r
Z 2¼ ·
¸
2
¹0 b I 0
sin µ
!
0
0
=
cos ! (t ¡ r=c)
cos Á ¡ sin ! (t ¡ r=c) sin µ cos Á
4¼r 0
r
c
£
¤
0
0
0
£ ¡ x^ sin Á + y^ cos Á dÁ
where the ¤rst term has
to zero.R The x¡component also integrates
R 2¼integrated
2¼
0
~ has only
to zero, since we have 0 sin Á cos Á0 dÁ0 = 0 sin 2Á0dÁ= = 0: Thus A
a y¡component:
~
A
h
i
¹0 ¼b 2 I0 !
c
sin µ cos ! (t ¡ r=c)
¡ sin ! (t ¡ r=c)
4¼r c
!r
¹0 m 0 !
' ¡ y^
sin µ sin ! (t ¡ r=c)
4¼r c
= y^
(7)
and ¤nally we get the ¤elds:
2
~
~ = ¡ @ A = y^ ¹0 m0 ! sin µ cos ! (t ¡ r=c)
E
@t
4¼ r c
We chose a special location for the x¡axis, but we can see from the diagram,
that for this point
^
y^ = Á
6
So, more generally,
2
~ =Á
^ ¹0 m0 ! sin µ cos ! (t ¡ r=c)
E
4¼ r c
~
B
=
=
'
~ £A
~ = ¡r
~ £ ^Á ¹0 m 0 ! sin µ sin ! (t ¡ r=c)
r
4¼r c
"
#
µ
¶
³
^µ @
¹0 m 0 !
r^
@ sin2 µ
r´
¡
sin ! t ¡
¡
(sin µ sin ! (t ¡ r=c))
4¼ c r sin µ @µ
r
c
r @r
¡
~
¹0 m 0 !2 ^
r^ £ E
µ sin µ cos ! (t ¡ r=c) =
2
4¼r c
c
This time we get
and
dP
¹ ³ m0 ´2 !4
= 0
sin 2 µ
d2 4¼
c2
P =
¹0 m20 ! 4
m 20 !4
=
12¼ c2
12¼" 0 c4
(8)
The expression (8) looks a lot like (6), but actually the magnetic dipole radiation
is much weaker. We can see this by going back to the expressions for p 0 = q0 d
and m0 = ¼b2 I0 ; and remembering that the current in the electric dipole case
had amplitude I0 = !q0 : Thus, if d » b;
p0
q0 d
1
= 2 »
m0
¼b I0
!b
and so
Pe le c tric d ip ole
=
Pm agn e tic d ip o le
µ
p 0c
m0
7
¶2
=
³ c ´2
À1
!b