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Physics 505
Fall 2005
Homework Assignment #6 — Due Thursday, October 27
Textbook problems: Ch. 4: 4.2, 4.6
a
) and
b
), 4.7
a
) and
b
), 4.8
4.2 A point dipole with dipole moment
~
p
is located at the point
~x
0
. From the properties
of the derivative of a Dirac delta function, show that for calculation of the potential
Φ or the energy of a dipole in an external ﬁeld, the dipole can be described by an
eﬀective charge density
ρ
eﬀ
(
~x
) =

~
p
·∇
δ
(
~x

~x
0
)
We ﬁrst consider the potential
Φ(
~x
) =
1
4
π±
0
Z
ρ
(
~x
0
)
1

~x

~x
0

d
3
x
0
=

1
4
π±
0
Z
[
~
p
·
~
∇
x
0
δ
3
(
~x
0

~x
0
)]
1

~x

~x
0

d
3
x
0
=
1
4
π±
0
Z
δ
3
(
~x
0

~x
0
)
~
p
·
~
∇
x
0
±
1

~x

~x
0

²
d
3
x
0
=
1
4
π±
0
~
p
·
~
∇
x
0
±
1

~x

~x
0

²³
³
³
³
~x
0
=
~x
0
=
1
4
π±
0
~
p
·
(
~x

~x
0
)

~x

~x
0

3
This is the expected potential for a dipole.
Similarly, we can work out the energy of the dipole in an external ﬁeld
W
=
Z
ρ
(
~x
)Φ(
~x
)
d
3
x
=

Z
[
~
p
·
~
∇
δ
3
(
~x

~x
0
)]Φ(
~x
)
d
3
x
=
Z
δ
3
(
~x

~x
0
)
~
p
·
~
∇
Φ(
~x
)
d
3
x
=
~
p
·
~
∇
Φ(
~x
0
) =

~
p
·
~
E
(
~x
0
)
Essentially the derivative of a delta function serves to pick out derivatives of the
function that it is multiplied against. Of course, a delta function itself is rather
singular, and its derivatives are even more so. But as far as formal expressions
(or distribution theory expressions) are concerned, these manipulations are in
fact legitimate.
4.6 A nucleus with quadrupole moment
Q
ﬁnds itself in a cylindrically symmetric electric
ﬁeld with a gradient (
∂E
z
/∂z
)
0
along the
z
axis at the position of the nucleus.
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) Show that the energy of quadrupole interaction is
W
=

e
4
Q
±
∂E
z
∂z
²
0
The quadrupole interaction energy is
W
=

1
6
Q
ij
∂E
i
∂x
j
To work out this expression, we ﬁrst note that the cylindrically symmetric charge
distribution for the nucleus gives rise to nonvanishing quadrupole moments
Q
11
=
Q
22
=

1
2
Q
33
(Note that this is obtained by demanding symmetry and tracelessness.) Since the
nuclear quadrupole moment
Q
(without indices) is (1
/e
)
Q
33
we end up with
Q
11
=
Q
22
=

1
2
eQ,
Q
33
=
eQ
The energy is thus
W
=

eQ
6
±

1
2
∂E
x
∂x

1
2
∂E
y
∂y
+
∂E
z
∂z
²
(1)
We now make use of the fact that the (external) electric ﬁeld is divergence free
0 =
~
∇·
~
E
=
∂E
x
∂x
+
∂E
y
∂y
+
∂E
z
∂z
and cylindrically symmetric (so that
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 Fall '08
 Stephens,P
 Magnetism, Work

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