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.