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Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13 1
Summary: The Dirac delta func-tion: δ(x)
1. Introducing the Dirac delta function:
2. The Dirac delta function in curvilinear
coordinates
3. Revisiting ~∇ · (r̂/r2)
4. Examples: Integrating with the delta
function
Suggested Reading:
Griffiths: Chapter 1, Sections 1.5.1 - 1.5.3, pages
45-52; Sections 1.61-1.62, pages 52-57.
Weber and Arfken, Chapter 2, Section 25, pages
126-135 and Chapter 1, Section 1.14, pages 86-
95.
Kreyszig, Chapter 9, Sections 9.3-9.9, pages
515-564.
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13 2
The Dirac Delta Function: δ(x)
Some time ago we introduced the Kronecker delta
δjk =
{0 if j 6= k,
1 if j = k(1)
The Kronecker delta is used in discrete summa-
tions. This should not be confused with the Dirac
delta function which is a similar quantity but which
is used in integrations.
δ(x) =
{0 if x 6= 0,
∞ if x = 0(2)
such that ∫ +∞
−∞δ(x) dx = 1
By analogy, for the Kronecker delta we would have∞∑
k=−∞
δjk = δjj = 1
How do we define such a “function”? It
actually cannot be considered a function at all. In-
stead, it is the limit of a sequence of functions such
as
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13 3
In the limit, as the width → 0 and the height
→ ∞ (keeping the area constant and equal to 1)
we obtain the delta function δ(x).
If we multiply a continuous function of x, f (x)
by a delta function,
f (x) δ(x)
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13 4
the product is 0 everywhere except at x = 0. Thus,∫ ∞
−∞f (x) δ(x) dx =
∫ ∞
−∞f (0) δ(x) dx
= f (0)
∫ ∞
−∞δ(x) dx
= f (0)
since ∫ ∞
−∞δ(x) dx = 1
We can offset the origin (the point at which the
delta function is not zero) by an amount ‘a’ such
that
δ(x− a) =
{0 x 6= a,
∞ x = a
Then, ∫ ∞
−∞f (x) δ(x− a) dx = f (a)
The limits on the integral need only bracket x = a
since f (x)δ(x − a) is zero everywhere except at
x = a. Thus,∫ 1
−1
f (x) δ(x) dx = f (0)
Consider the integral
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13 5∫ ∞
−∞f (x) δ(ax) dx
If we make the substitution y = ax then dy = adx
and the integral becomes∫ ∞
−∞f (y/a) δ(y) dy/a =
1
a
∫ ∞
−∞f (y/a) δ(y) dy =
1
af (0)
Thus, we can conclude that
δ(ax) =1
aδ(x)
In general, we consider two δ-functions D1 and D2
to be equal (equivalent) if∫f (x) D1(x) dx =
∫f (x) D2(x) dx
for any f (x).
Delta function in three dimensions
In Cartesian coordinates, in three dimensions, we
can consider the delta function δ3(~r) to be the
product of three one dimensional delta functions
δ3(~r) = δ(x)δ(y)δ(z)
Then, in volume integrals∫ ∫ ∫ ∞
−∞δ3(~r) = 1
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13 6∫ ∫ ∫ ∞
−∞f (~r) δ3(~r) dτ = f (~0)
and ∫ ∫ ∫ ∞
−∞f (~r) δ3(~r − ~r0) dτ = f (~r0)
Singular Density Functions: for example,
point masses or point charges We often en-
counter problems where, for simplicity, we assume
that all of the mass of a particle and/or all of the
charge of the particle is concentrated at a single
point in space. These are called point particles or
point charges (the electron actually behaves like a
point particle...it has no measurable radius!)
If we define %m as the mass density (units of
mass/volume) then for a point particle of mass m0
located at the point P : (0, 0, a) we could write
the mass density in terms of delta functions
%m(x, y, z) = m0 δ(x) δ(y) δ(z − a)
If this is the only particle in the system we are
dealing with, then the total mass is just∫ ∫ ∫V
%m(x, y, z) dτ =∫dx
∫dy
∫dz m0 δ(x) δ(y) δ(z − a) = m0
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13 7
if V contains P : (0, 0, a). A point mass at Q :
(a, b, c) would be represented by the density
%m = m0 δ(x− a) δ(y − b) δ(z − c)
and so on. Multiple masses located at coordinates
~ri where i = 1, ...N would have a density
%m(x, y, z) =
N∑i=1
mi δ(~r − ~ri)
How would we write such a density dis-
tribution in cylindrical coordinates?
Say you have a mass ma at position (ρa, φa, za).
The natural reaction would probably be to try
%m?= ma δ(ρ− ρa) δ(φ− φa) δ(z − za)
This seems correct in that it correctly places the
mass at (ρa, φa, za), however, the mass density should
have dimensions of (mass/volume). From our def-
inition of the δ-functions∫δ(x) dx = 1
so that if dx has dimensions of length, then δ(x)
must have dimensions of (1/length). Similarly,∫ ∫ ∫V
δ3(~r) dτ = 1
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13 8
requires that δ3(~r) have dimensions of (1/volume).
In cylindrical coordinates:
• ρ has dimensions of length
• z has dimensions of length
• φ is dimensionless! (it is measured in radi-
ans, but this is not a unit)
Because of this, the expression
%m?= ma δ(ρ− ρa) δ(φ− φa) δ(z − za)
has dimensions of (mass/area), not (mass/volume).
If we were to integrate over a volume containing
(ρa, φa, za) we would get∫ ∫ ∫V
%m dτ =
∫dρ
∫ρ dφ
∫dz m0 δ(ρ−ρa)
× δ(φ− φa) δ(z − za)
= m0
∫ρdρ δ(ρ−ρa)
∫dφ δ(φ−φa)
∫dz δ(z−za)
= m0 ρa
which is incorrect!. Clearly the correct expression
for the mass density in this case is
%m =m
ρδ(ρ− ρ0) δ(φ− φ0) δ(z − z0)
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13 9
In curvilinear coordinates (q1, q2, q3) where dτ =
h1 h2 h3 dq1 dq2 dq3 we need to use
%m(q1, q2, q3) =δ(q1 − q10)
h1
δ(q2 − q20)
h2
δ(q3 − q30)
h3
Obviously, all of the above also holds if we are
talking about charge density%q.
If we have a planar distribution of charge, say
with uniform surface charge density σ0 (charge/area)
distributed over the plane at z = z0, then in three
dimensions we would write
%q = σ0 δ(z − z0)
(notice that it has the correct dimensions of (charge/volume)
and for a uniform line charge density λq (charge/length)
distributed along, for example, the z-axis, we would
write
%q = λq δ(x) δ(y)
or, in cylindrical coordinates,
%q = λq δ(ρ)δ(φ)
ρ
which again have the correct dimensions.
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13 10
Re-examination of ~∇ · (r̂/r2)
The expression
~∇ ·(
r̂
r2
)=
{0 r 6= 0,
∞ r = 0
Thus, it looks like a δ-function.
Previously we showed that∫ ∫ ∫T
~∇ · ~E dτ =q
4πε0
∫ ∫ ∫~∇ · r̂
r2dτ
=
∫ ∫S
~E · d ~A =q
ε0
and it appeared that we must be violating the Di-
vergence Theorem, however, if we write
~∇ ·(
r̂
r2
)= 4πδ3(~r)
then, since∫ ∫ ∫T
~∇ ·(
r̂
r2
)dτ =
∫ ∫S
(r̂
r2
)· d ~A
=
∫ ∫S
(r̂
r2
)· r̂ r2 sin θdθ dφ
=
∫ 2π
0
dφ
∫ π
0
sin θ dθ
= 4π
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13 11
We obtain,
q
4πε0
∫ ∫ ∫T
~∇·(
r̂
r2
)dτ =
q
ε0
∫ ∫ ∫T
δ3(~r) dτ =q
ε0
and the Divergence Theorem holds. Furthermore,
since1
ε0
∫ ∫ ∫T
% dτ =q
ε0
then, if we consider the charge ‘q’ to be due to a
point charge at the origin,
% = qδ3(~r)
with1
ε0
∫ ∫ ∫T
q δ3(~r) dτ =q
ε0
Examples: Evaluate the integrals
a.) ∫ 2
−2
(2x + 3)δ(3x)dx
Make the substitution y = 3x, then dy = 3dx
and∫ 2
−2
(2x+3)δ(3x)dx =1
3
∫ 6
−6
(2y
3+3)δ(y)dy =
1
3·3 = 1
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13 12
b.) ∫ 2
0
(x3 + 3x + 2)δ(1− x)dx
Notice that δ(1 − x) = δ(x − 1) since the δ-
function is an even function of its argument.
Then it is straightforward to evaluate this in-
tegral since it picks out the value f (x = 1)∫ 2
0
(x3 + 3x + 2)δ(1− x)dx = (1 + 3 + 2) = 6
c.) ∫ 1
−1
9x2δ(3x + 1)dx
Here you can again make the substitution y =
3x so that dy = 3dx and∫ 1
−1
9x2δ(3x+1)dx =
∫ 3
−3
y2δ(y+1)dy/3 =1
3·1 =
1
3
d.) ∫ a
−∞δ(x− b)dx
Here the only question is whether a > b or not.
If it is, then ∫ a
−∞δ(x− b)dx = 1
if it isn’t, then the integral is zero.