Hassani S. Mathematical Methods: For Students of Physics and Related Fields
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5.3 Three-Variable Case 163
Let us further simplify the problem by positioning the field point on the x-axis,
i.e., setting z = 0. This reduces the above expressions to
E
x
(x, 0, 0) = k
e
q
x
|x|
3
+2k
e
q
∞
k=1
(−1)
k
x
(x
2
+ k
2
a
2
)
3/2
,
E
y
(x, 0, 0) = 0,
E
z
(x, 0, 0) = k
e
q
∞
k=1
(−1)
k
ka
{x
2
+(ka)
2
}
3/2
+
−ka
{x
2
+(−ka)
2
}
3/2
=0.
At a distance a from the origin on the x-axis, the field strength is
E
x
(x, 0, 0) =
k
e
q
a
2
0
1+2
∞
k=1
(−1)
k
(1 + k
2
)
3/2
=−0.286269
1
=0.42746
k
e
q
a
2
,
E
y
(x, 0, 0) = 0,
E
z
(x, 0, 0) = 0,
where the numerical value for the sum—accurate to six decimal places—is obtained
by adding its first 150 terms.
Another useful quantity is the electrostatic potential which for an arbitrary
charge distribution is given by
Φ(r)=k
e
##
Ω
dq(r
)
|r − r
|
. (5.37)
For the one-dimensional crystal, with the volume charge density of Equation (5.35),
the electrostatic potential at an arbitrary point (x, y, z) in space becomes
Φ(x, y, z)=k
e
##
Ω
ρ
q
(x
,y
,z
) dx
dy
dz
(x − x
)
2
+(y − y
)
2
+(z − z
)
2
= k
e
q
∞
k=−∞
(−1)
k
##
Ω
δ(x
)δ(y
)δ(z
− ka) dx
dy
dz
(x − x
)
2
+(y − y
)
2
+(z − z
)
2
= k
e
q
∞
k=−∞
(−1)
k
x
2
+ y
2
+(z − ka)
2
.
If we are interested in the potential at a specific point such as (x, 0, 0), the
expression simplifies to
Φ(x, 0, 0) = k
e
q
∞
k=−∞
(−1)
k
√
x
2
+ k
2
a
2
= k
e
q
1
√
x
2
+2k
e
q
∞
k=1
(−1)
k
√
x
2
+ k
2
a
2
=
k
e
q
|x|
+2k
e
q
∞
k=1
(−1)
k
√
x
2
+ k
2
a
2
.
164 Dirac Delta Function
For x = a, this further simplifies to
Φ(a, 0, 0) =
k
e
q
a
0
1+2
∞
k=1
(−1)
k
√
1+k
2
=−0.4409
1
=0.1182
k
e
q
a
.
We note that the potential is positive, because the field point is closest to the positive
charge at the origin. To obtain the numerical value of the sum accurate to only four
decimal places, we have to add at least 40,000 terms! This sum is, therefore, much
less convergent than the sum encountered in the evaluation of E
x
above.
An important physical quantity for real crystals is the potential energy U
of the crystal. Physically, this is the amount of energy required to assemble
the charges in their final configuration. A positive potential energy corre-
sponds to positive energy stored in the system, i.e., a tendency for the system
to provide energy to the outside, once disrupted slightly from its equilibrium
position. A negative potential energy is a sign of the stability of the system,
i.e., the tendency for the system to restore its original configuration if dis-
rupted slightly from its equilibrium position.
4
It is shown in electrostatics
that the potential energy of a system located within the region Ω is
U =
1
2
##
Ω
dq(r)Φ(r). (5.38)
Example 5.3.2.
Let us calculate the electrostatic potential energy of the one-electrostatic
potential energy of
aone-dimensional
crystal
dimensional crystal. Let us assume that there are a total of 2N +1 charges stretching
from z = −Na to z =+Na with a positive charge at the origin. Eventually we
shall let N go to infinity, but, in order not to deal explicitly with infinities, we
assume that N is finite but large. Substituting in (5.38) the element of charge in
terms of volume density, and electrostatic potential found in the previous example,
we find
U =
1
2
##
Ω
ρ
q
(x, y, z)Φ(x, y, z) dx dy dz
=
1
2
##
Ω
0
q
N
j=−N
(−1)
j
δ(x)δ(y)δ(z −ja)
1
×
0
k
e
q
N
k=−N
(−1)
k
x
2
+ y
2
+(z − ka)
2
1
dx dy dz
=
k
e
q
2
2
N
j=−N
N
k=−N
k=j
(−1)
j+k
(ja − ka)
2
.
4
A system that has negative potential energy requires some positive energy (such as
kinetic energy of a projectile) to reach a state of zero potential energy corresponding to
dissociation of its parts and their removal to infinity (where potential energy is zero).
5.3 Three-Variable Case 165
The restriction k = j is necessary, because the k = j terms correspond to the
interaction energy of each charge with itself, and should be excluded. Continuing
with the calculation, we write
U =
k
e
q
2
2a
N
j=−N
N
k=−N
k=j
(−1)
j+k
|j − k|
=
k
e
q
2
2a
N
j=−N
⎧
⎨
⎩
j−1
k=−N
(−1)
j+k
j − k
+
N
k=j+1
(−1)
j+k
k −j
⎫
⎬
⎭
.
In the first inner sum, let j − k = m, and in the second let k − j = m.These
substitutions change the limits of the sums, and we get
U =
k
e
q
2
2a
N
j=−N
0
1
m=N+j
(−1)
2j− m
m
+
N−j
m=1
(−1)
2j+m
m
1
=
k
e
q
2
2a
N
j=−N
0
N+j
m=1
(−1)
m
m
+
N−j
m=1
(−1)
m
m
1
.
To evaluate the inner sums, denoted by S, we now assume that N is very large—
compared to j—so that N −j ≈ N ≈ N + j. Then the inner sum yields
5
S =
N+j
m=1
(−1)
m
m
+
N−j
m=1
(−1)
m
m
≈
N
m=1
(−1)
m
m
+
N
m=1
(−1)
m
m
=2
N
m=1
(−1)
m
m
≈−2
∞
m=1
(−1)
m+1
m
= −2ln 2.
Substituting S in the expression for U ,weget
U ≈
k
e
q
2
2a
N
j=−N
(−2ln 2)=−
k
e
q
2
a
ln 2
N
j=−N
1=−(2N +1)
k
e
q
2
a
ln 2.
The negative sign indicates that the one-dimensional salt crystal is stable. A useful
quantity used in solid-state physics is ionization energy per molecule which is defined
to be the potential energy divided by the number of molecules. Noting that the
number of molecules is half the number of particles, we obtain
u ≡ U/N = −
2N +1
N
k
e
q
2
a
ln 2 ≈−
k
e
q
2
a
2ln 2≡−α
k
e
q
2
a
.
A real three-dimensional salt crystal has exactly the same expression. However, Madelung
constantthe constant α, called the Madelung constant has the value of 1.747565 instead
of 2 ln 2 = 1.386294. (See Problem 5.17 for an alternative way of calculating the
potential energy of the one-dimensional ionic crystal.)
5
We are really cheating here! The sum over j indicates that j can assume values close to
N, and therefore, the approximation is not valid for such j’s. However, a careful analysis,
in which one breaks up the sum over j and separates large and small values of j,showsthat
the original approximation is valid as long as N is large enough.
166 Dirac Delta Function
5.4 Problems
5.1. Plot the distribution on the real line of each of the following electric
linear charge densities:
(a) λ(x)=δ(x −2). (b) λ(x)=−δ(x +1).
(c) λ(x)=5δ(x) −3δ(x +3). (d) λ(x)=δ(x +1)+3δ(x − 1).
5.2. Evaluate the following integrals:
(a)
#
∞
0
e
x
sin
πx
2
δ(x
2
− 1) dx. (b)
#
2
−2
e
x
sin
πx
2
δ(x
2
− 1) dx.
(c)
#
∞
0
e
x
sin
πx
2
δ(x
3
+1)dx. (d)
#
∞
−∞
sin
πe
x
2
δ(x
4
+1)dx.
(e)
#
∞
0
sin
−1
(1/x)δ(x
4
− 1) dx. (f)
#
∞
−∞
cos(πx)δ(6x
2
− x − 1) dx.
(g)
#
∞
−0.1
sin
πe
x
2
δ(x
2
+ x) dx. (h)
#
∞
−∞
e
x
sin
πx
2
δ(e
x
sin
πx
2
) dx.
(i)
#
5
0
e
sin x
δ(cos x) dx. (j)
#
∞
0
sin
−1
1
x
δ(x
4
− 4) dx.
(k)
#
∞
−∞
e
x
sin
πx
3
δ(4x
2
− 1) dx. (l)
#
∞
−∞
ln(1 + x)sin
πx
2
δ(x
3
− 1) dx.
(m)
#
∞
−∞
sin
πe
x
2
δ(x
3
+1)dx.
5.3. Show that
#
+∞
−∞
f(x)δ
(x − x
0
) dx = −f
(x
0
)
and
#
+∞
−∞
f(x)δ
(g(x)) dx = −
#
+∞
−∞
f
(x)δ(g(x)) dx.
5.4. Evaluate the following integrals:
(a)
#
∞
0
e
x
sin
πx
2
δ
(x
2
− 1) dx. (b)
#
2
−2
e
x
sin
πx
2
δ
(x
2
− 1) dx.
(c)
#
∞
0
e
x
sin
πx
2
δ
(x
3
+1)dx. (d)
#
∞
−∞
sin
πe
x
2
δ
(x
4
+1)dx.
(e)
#
∞
0
sin
−1
(1/x)δ
(x
4
− 1) dx. (f)
#
∞
−∞
cos(πx)δ
(6x
2
− x − 1) dx.
5.4 Problems 167
(g)
#
∞
−0.1
sin
πe
x
2
δ
(x
2
+ x) dx. (h)
#
∞
−∞
e
x
sin
πx
2
δ
(e
x
sin
πx
2
) dx.
(i)
#
5
0
e
sin x
δ
(cos x) dx. (j)
#
∞
0
sin
−1
1
x
δ
(x
4
− 4) dx.
(k)
#
∞
−∞
e
x
sin
πx
3
δ
(4x
2
− 1) dx. (l)
#
∞
−∞
ln(1 + x)sin
πx
2
δ
(x
3
− 1) dx.
(m)
#
∞
−∞
sin
πe
x
2
δ
(x
3
+1)dx.
5.5. Use integration by parts (or differentiation with respect to x
0
)toshow
that
#
+∞
−∞
f(x)δ
(x − x
0
) dx = f
(x
0
)
and
#
+∞
−∞
f(x)δ
(x − x
0
) dx = −f
(x
0
)
and, in general,
#
+∞
−∞
f(x)δ
(n)
(x − x
0
) dx =(−1)
n
f
(n)
(x
0
)
where δ
(n)
and f
(n)
represent the nth derivatives.
5.6. Derive Equation (5.16). Hint: Use the result of Problem 5.3.
5.7. Six point charges of equal strength q are equally spaced on a circle of
radius a. What is the volume charge density describing such a distribution in
cylindrical coordinates?
5.8. Convince yourself that
σ
q
(x, y)=q
∞
i=−∞
∞
j=−∞
(−1)
i+j
δ(x −ia)δ(y − ja)
indeed describes a two-dimensional ionic crystal. Pay particular attention to
the power of (−1).
5.9. Derive Equations (5.31) and (5.32).
5.10. Plot (or describe) the distribution in space of each of the following
volume charge densities:
ρ
q
(x, y, z)=δ(x)δ(y) {2δ(z) − 3δ(z +3)},
ρ
q
(x, y, z)=5δ(x +1)δ(y − 1) {δ(z − 1) − δ(z +1)},
168 Dirac Delta Function
ρ
q
(ρ, ϕ, z)=−2δ(ρ −3)δ(ϕ −π)δ(z),
ρ
q
(ρ, ϕ, z)=2δ(ϕ − π/4)δ(z)
0
10
k=1
(−1)
k+1
δ(ρ − 0.5k)
1
,
ρ
q
(r, θ, ϕ)=2δ(ϕ − π/4)δ(r − 2)
0
10
k=1
(−1)
k+1
δ
θ −
π
20
k
1
,
ρ
q
(r, θ, ϕ)=2δ(θ − π/4)δ(r − 2)
0
20
k=1
(−1)
k+1
δ
ϕ −
π
10
k
1
.
5.11. Derive Equation (5.36).
5.12. Plot θ(t)θ(1 − t), θ(t) −θ(−t), and θ(t
2
+1)for−∞ <t<+∞.
5.13. Write θ(t
2
− 1) as a product of two step functions.
5.14. For the two-dimensional ionic crystal shown in Figure 5.6:
(a) write the volume charge density describing the distribution (charges are
in the xy-plane);
(b) calculate the electrostatic field at (0, 0,a); and
(c) calculate the electrostatic potential at an arbitrary point in space with
coordinates (x, y, z).
(d) Show that the ionization energy is of the form −αk
e
q
2
/a with α given in
terms of a sum.
(e) Numerically evaluate α.
5.15. For the three-dimensional ionic crystal:
(a) write the volume charge density describing the distribution; and
(b) calculate the electrostatic potential at an arbitrary point in space with
coordinates (x, y, z).
(c) Show that the ionization energy is of the form −αk
e
q
2
/a with α given in
terms of a sum.
(d) Numerically evaluate α.
5.16. Two electric charges +q and −q are located at P
1
and P
2
with position
vectors r
1
and r
2
.
(a) Write the volume charge density describing these charges.
(b) Use (a) to find their dipole moment defined by
r
dq(r
).
5.17. Theelectricchargedensityoftheone-dimensional ionic crystal can be
written as ρ(r)=
N
i=−N
q
i
δ(r −r
i
).
(a) Substitute this in Equation (5.38) and get
U =
1
2
N
i=−N
q
i
Φ(r
i
)
5.4 Problems 169
(b) Assuming that N is very large (infinite), convince yourself that all products
q
i
Φ(r
i
) in the sum are equal (in particular the sign of the charge does not
matter). Therefore, U =
1
2
(2N +1)q
0
Φ(r
0
), where the subscript denotes the
zeroth charge.
(c) Show that Φ(r
0
)=
N
j=−N
k
e
q
j
/|r
j
− r
0
|.
(d) Place the origin at the location of the zeroth charge, and assume that the
this charge is positive. Then, r
0
=0,r
j
= ja
ˆ
e
z
,andq
j
= −(− 1)
j
q.Now
show that
U = −(N +
1
2
)q
2
k
e
N
j=−N
(−1)
j
|aj|
(e) By breaking up the sum into two parts show that
U = −(2N +1)
q
2
k
e
a
N
j=1
(−1)
j
j
5.18. 2N charges of equal sign and magnitude q are arranged equally spaced
on a circle of radius a located in the xy-plane. Assume that the charge num-
bered 2N is at (a, 0, 0).
(a) Write the volume charge density of such a distribution in cylindrical co-
ordinates.
(b) Starting with an integral expression for the electric field, find the cylin-
drical components of the field at an arbitrary point P in space in terms of
a sum. The coordinates of P are (ρ, ϕ, z). Simplify your answer as much as
possible.
(c) Now let P have coordinates (2a, 0, 0). Show that all components of the
field are of the form (k
e
q/a
2
)α.Expresstheα for each component in terms
of a sum. What do you expect the value of α to be? Can you find that value?
(d) For N = 3, i.e., six charges, calculate the numerical value of α in part (c)
for all components.
5.19. 2N + 1 charges of equal sign and magnitude q are arranged on the x-
axis of a Cartesian coordinate system as shown in Figure 5.9, with the zeroth
charge at the origin. The numbers below the axis are labels of the charges.
(a) From the pattern of the figure, determine the location of the kth charge
for −N ≤ k ≤ N.
x
0
12
3
−1 −2 −3
4a
a
9a
a
4a
9a
Figure 5.9: The charges and their distances on the x-axis.
170 Dirac Delta Function
(b) Write a volume charge density in terms of the Dirac delta function de-
scribing such a charge distribution.
(c) Calculate the components of the electric field at a general point P with
coordinates (x, y, z).
(d) Now let P have coordinates (a, a, 0). Show that all components of the
field are of the form (k
e
q/a
2
)α where α is a numerical factor. Find this factor
for each component.
5.20. 2N positive and negative charges of equal magnitude are arranged
equally spaced and alternating in sign on a circle of radius a.
(a) Write the expression of the volume charge density describing this charge
distribution.
(b) Find the ionization energy in the form −αk
e
q
2
/a with α given in terms
of a sum. Simplify this sum as much as possible.
Part II
Algebra of Vectors
Chapter 6
Planar and Spatial Vectors
The preceding chapters made heavy use of vectors in the plane and in space.
The enormous utility of the concept of vectors has prompted mathematicians
and physicists to generalize this concept to include other objects that at first
glance have no resemblance whatsoever with the planar and spatial vectors.
In this chapter, we shall study this generalization in its limited form, i.e.,
only in an algebraic context. Although the analysis of vectors is discussed
in Chapters 12 through 17, it is confined to vectors in space. The analysis
of generalized vectors is the subject of differential geometry and functional
analysis that are beyond the scope of this book.
1
There are many mathematical objects used in physics that allow for the
two operations of addition and multiplication by a number. The collection
of such objects is called a vector space. Thus, a vector space is a bunch
vector spaces
defined
of “things” having the property that when you add two “things” you get a
third one, and if you multiply a “thing” by a number you get another one of
those “things.” Furthermore, the operation of multiplication by a number and
addition of “things” is distributive, and a vector space always has a “thing”
that we call the zero vector.
Using the two operations of multiplication by a number and addition, we
can form a sum,
α
1
a
1
+ α
2
a
2
+ ···+ α
n
a
n
, (6.1)
where α
1
,α
2
,...,α
n
, are real numbers and a
1
, a
2
,...,a
n
are vectors. The
sum in Equation (6.1) is called a linear combination of the n vectors and
linear combination
α
1
,α
2
,...,α
n
are called the coefficients of the linear combinations.
coefficients
1
Hassani, S. Mathematical Physics: A Modern Introduction to Its Foundations,
Springer-Verlag, 1999, discusses differential geometry and functional analysis in some detail.