Summary: The Dirac delta func- tion: δ(x) Suggested Reading:

Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13
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Summary: The Dirac delta function: δ(x)
1. Introducing the Dirac delta function:
2. The Dirac delta function in curvilinear
coordinates
~ · (r̂/r2)
3. Revisiting ∇
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 8695.
Kreyszig, Chapter 9, Sections 9.3-9.9, pages
515-564.
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13
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The Dirac Delta Function: δ(x)
Some time ago we introduced the Kronecker delta
(
0 if j 6= k,
δjk =
(1)
1 if j = k
The Kronecker delta is used in discrete summations. This should not be confused with the Dirac
delta function which is a similar quantity but which
is used in integrations.
(
0 if x 6= 0,
δ(x) =
(2)
∞ if x = 0
such that
Z
+∞
δ(x) dx = 1
−∞
By analogy, for the Kronecker delta we would have
∞
X
δjk = δjj = 1
k=−∞
How do we define such a “function”? It
actually cannot be considered a function at all. Instead, it is the limit of a sequence of functions such
as
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13
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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
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the product is 0 everywhere except at x = 0. Thus,
Z ∞
Z ∞
f (x) δ(x) dx =
f (0) δ(x) dx
−∞
−∞ Z
∞
= f (0)
δ(x) dx
−∞
= f (0)
since
Z
∞
δ(x) dx = 1
−∞
We can offset the origin (the
delta function is not zero) by
that
(
0
δ(x − a) =
∞
point at which the
an amount ‘a’ such
x 6= a,
x=a
Then,
Z
∞
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,
Z 1
f (x) δ(x) dx = f (0)
−1
Consider the integral
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13
Z
5
∞
f (x) δ(ax) dx
−∞
If we make the substitution y = ax then dy = adx
and the integral becomes
Z
Z ∞
1
1 ∞
f (y/a) δ(y) dy = f (0)
f (y/a) δ(y) dy/a =
a −∞
a
−∞
Thus, we can conclude that
1
δ(ax) = δ(x)
a
In general, we consider two δ-functions D1 and D2
to be equal (equivalent) if
Z
Z
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 δ 3(~r) to be the
product of three one dimensional delta functions
δ 3(~r) = δ(x)δ(y)δ(z)
Then, in volume integrals
Z Z Z ∞
−∞
δ 3(~r) = 1
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13
Z Z Z
6
∞
f (~r) δ 3(~r) dτ = f (~0)
−∞
and
Z Z Z
∞
f (~r) δ 3(~r − ~r0) dτ = f (~r0)
−∞
Singular Density Functions: for example,
point masses or point charges We often encounter 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
Z Z Z
%m(x, y, z) dτ =
Z
ZV Z
dx dy dz m0 δ(x) δ(y) δ(z − a) = m0
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13
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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
X
mi δ(~r − ~ri)
i=1
How would we write such a density distribution 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 definition of the δ-functions
Z
δ(x) dx = 1
so that if dx has dimensions of length, then δ(x)
must have dimensions of (1/length). Similarly,
Z Z Z
δ 3(~r) dτ = 1
V
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13
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requires that δ 3(~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 radians, 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
Z Z Z
Z
Z
Z
%m dτ = dρ ρ dφ dz m0 δ(ρ−ρa)
V
Z
= m0
× δ(φ − φa) δ(z − za)
Z
Z
ρdρ δ(ρ−ρa) dφ δ(φ−φa) dz δ(z−za)
= m 0 ρa
which is incorrect!. Clearly the correct expression
for the mass density in this case is
m
%m =
δ(ρ − ρ0) δ(φ − φ0) δ(z − z0)
ρ
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13
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In curvilinear coordinates (q1, q2, q3) where dτ =
h1 h2 h3 dq1 dq2 dq3 we need to use
δ(q1 − q10) δ(q2 − q20) δ(q3 − q30)
h1
h2
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
%m(q1, q2, q3) =
%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
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~ · (r̂/r2)
Re-examination of ∇
The expression
(
~ · r̂ = 0 r 6= 0,
∇
r2
∞ r=0
Thus, it looks like a δ-function.
Previously we showed that
Z Z Z
Z Z Z
q
~ · r̂ dτ
~ ·E
~ dτ =
∇
∇
r2
T
Z4πZ0
~ · dA
~= q
=
E
0
S
and it appeared that we must be violating the Divergence Theorem, however, if we write
~ · r̂ = 4πδ 3(~r)
∇
r2
then, since
Z Z Z Z Z
r̂
~ · r̂ dτ =
~
· dA
∇
2
2
r
r
T
Z ZS r̂
2
r̂
r
=
·
sin θdθ dφ
2
r
Z 2πS Z π
=
dφ
sin θ dθ
0
= 4π
0
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13
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We obtain,
Z Z Z
Z Z Z
r̂
q
q
q
3
~
∇·
dτ
=
δ
(~
r
)
dτ
=
4π0
r2
0
0
T
T
and the Divergence Theorem holds. Furthermore,
since
Z Z Z
1
q
% dτ =
0
0
T
then, if we consider the charge ‘q’ to be due to a
point charge at the origin,
% = qδ 3(~r)
with
1
0
Z Z Z
T
q δ 3(~r) dτ =
q
0
Examples: Evaluate the integrals
a.)
Z
2
(2x + 3)δ(3x)dx
−2
Make the substitution y = 3x, then dy = 3dx
and
Z
Z 2
1 6 2y
1
( +3)δ(y)dy = ·3 = 1
(2x+3)δ(3x)dx =
3 −6 3
3
−2
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 13
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b.)
Z
2
(x3 + 3x + 2)δ(1 − x)dx
0
Notice that δ(1 − x) = δ(x − 1) since the δfunction is an even function of its argument.
Then it is straightforward to evaluate this integral since it picks out the value f (x = 1)
Z 2
(x3 + 3x + 2)δ(1 − x)dx = (1 + 3 + 2) = 6
0
c.)
Z
1
9x2δ(3x + 1)dx
−1
Here you can again make the substitution y =
3x so that dy = 3dx and
Z 1
Z 3
1
1
9x2δ(3x+1)dx =
y 2δ(y+1)dy/3 = ·1 =
3
3
−1
−3
d.)
Z
a
δ(x − b)dx
−∞
Here the only question is whether a > b or not.
If it is, then
Z a
δ(x − b)dx = 1
−∞
if it isn’t, then the integral is zero.

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