Kusse B.R., Westwig E.A. Mathematical Physics: Applied Mathematics for Scientists and Engineers

Подождите немного. Документ загружается.

116

THE DIRAC

6-FUNCTION

Cartesian coordinates,

this

is simply

S3(F

To)

6(x

x,)S(y

y,)S(z

z,).

(5.55)

Using this notation, the mass density of point mass located at

is, regardless of the

coordinate system chosen, given by

pm(T)

m,63(?

To).

(5.56)

If there are several points with mass

position

Ti,

the density function becomes

a sum

6-functions:

pm(T)

Crni63(T

Ti).

Integrating over a volume

gives

(5.57)

(5.58)

which, when evaluated,

simply the sum

all the masses enclosed in

5.5.2

Sheet

Distributions

Imagine a two-dimensional planar sheet

uniform mass per unit area

uo,

located

the

plane of a Cartesian system,

shown in Figure

5.17.

The intuitive guess

for the mass density

in Cartesian coordinates is

P~(~>Y,Z)

aoa(z

~0)-

(5.59)

The &function makes sure

all

the mass is in the

plane. The dimensions are

correct, because

has the dimensions of mass per area, and the &function adds

another

/length, to give

the dimensions of mass per unit volume. The real check,

Figure

5.17

Simple

Infitute

Planar

Sheet

SINGULAR DENSITY FUNCTIONS

117

however, is the volume integral. In order for Equation

5.59

be correct, an integral

of the surface mass density over some part of the sheet

must give the same total

mass as a volume integral

pm,

over a volume

which encloses

For the case described above, Equation 5.60 expands on both sides to

(5.60)

(5.61)

which is indeed true, because the range of the

integration includes the zero

the &function. Thus the assumption of Equation

5.59

was correct. Notice how the

&function effectively converts the volume integral into a surface integral.

Unfortunately, things

are

not always this easy. The previous example was particu-

larly simple, because the sheet was lying in the plane given by

zo.

Now consider

the same sheet positioned in the plane

as shown in Figure

5.18.

In this case

the mass is only located where

and the intuitive guess for the mass density is

x).

(5.62)

But

our

intuition

incorrect in this case. The surface integral on the

LHS

Equa-

tion 5.60 expands to

crm(T)

ds dz

ao,

(5.63)

where

the differential length on the surface in the xy-plane, as shown in Fig-

ure

5.19.

Notice that

J(dx)*

(dy)2

hdx,

(5.64)

Figure

5.18

Tilted

Planar

Mass

Sheet

118

THE

DIRAC

&FUNCTION

Figure

5.19

The

Differential

Length

Along

the

Surface

the

xy-Plane

where the last step follows because

for

this

problem. Thus Equation

5.63

becomes

(5.65)

If we use the expression in

5.62

for

pm.

the volume integral on the

RHS

of Equa-

tion

5.60

Jdx/dzuo,

(5.66)

where the last step results from performing the integration over

Comparing Equa-

tions

5.66

and

5.65,

we see that they

are

off

by a factor of

fi!

Where does

this

discrepancy come from? The problem was our assumption in

Equation

5.62

to use a 8-function in the form

8(y

x).

We could have very well

chosen

[?]8(y

x),

where

[?I

is some function that we need to determine. This

still

makes all the mass lie in the

plane. The correct choice of

[?I

is the one that

makes both sides

Equation

5.60

equal. For example, when we make our “guess”

for

the distribution function for this example, we write

Pm(x,

y,z>

“?luo6(x

y)-

(5.67)

Then, when we evaluate the

RHS

of Equation

5.60,

we get

(5.68)

In this case, we already showed that the value

[?I

is simply the constant

more complicated problems,

[?]

can be a function of the coordinates. This can happen

SINGULAR

DENSITY

FUNCTIONS

119

if either the mass per unit area of the sheet is not constant, or

the sheet is not flat.

You

will see some examples of this in the next section and in the problems at the end

of this chapter.

5.5.3

Line Distributions

a final example of singular density functions, consider the mass per unit volume

one-dimensional wire of uniform mass per unit length

A,.

The wire is bent to

follow the parabola

Cx2

in the

plane,

shown in Figure

5.20.

The factor

C is

a constant, which has units of l/length. We will follow the same procedure in

constructing the mass density for this wire as we did for the previous example. In this

case, the volume integral of the mass density must collapse to a line integral along

the wire

Ld.rp,(r)

A,(s).

(5.69)

In this equation,

is a variable which indicates parametrically where we are on the

wire, and

A,(s)

is the mass per unit length of the wire at the position

Because all the mass must lie on the wire, we write the mass density function as

Here we have used two &functions. The

6(z)

term ensures all the mass lies in the

plane, while

S(y

Cx2)

makes the mass lie along the parabola.

before, we include

an unknown factor of

[?I,

which we will have to determine using Equation

5.69.

In terms of Cartesian coordinates, the general expression for the differential arc

length

Figure

5.20

Parabolic Line Distribution

120

THE DIRAC 8-FUNCTION

Along the wire,

and

Cx2,

that

4C2x2 dx.

(5.72)

Using Equation 5.70 and the fact that

dx dy dz,

Equation 5.69 becomes

dy /dz

[?]A,S(y

Cx2)S(z)

A,.

(5.73)

Let’s concentrate on the

LHS

this

equation. The integral over

is easy, because

/dZS(Z)

1. (5.74)

Also, when we do the integral over

y, x

is held constant, and

only

one value

makes the argument

the &function vanish,

we have

/dy

s(y

cx2)

Therefore, Equation 5.73 becomes

(5.75)

Jdx

[?]A,

/dx

A,,

(5.76)

and the value of

[?I

clearly must be

[?]

4c2x2.

(5.77)

Notice in this case that

[?]

is a function of position. The mass density for the wire is

given by

pm(x,y,z)

4c2x2

A,S(y

Cx2)S(z).

(5.78)

When we converted the volume integral in Equation 5.73 to a line integral, it was

easier to perform the

integration before the

integration.

This

was because when

we integrated over

holding

fixed, the only value

which made the argument

of the &function zero was

Cx2.

Another way to look at

this

is that there is a

one-to-one relationship between

and

ds,

as shown in Figure 5.21.

instead, we

performed

the

integration first, holding the value

fixed, there

are

two

values

which zero the

&function

argument.

this

case, the integral

becomes

&/dx

[a(.

s(x

m)].

(5.79)

THE INFINITESIMAL ELECTRIC

DIPOLE

121

Figure 5.21

The

Relation Between

and

Along

the

Parabola

Figure

5.22

The Relation Between

and

Along the Parabola

Looking at Figure

5.22

shows what is going on here. There is not a one-to-one

relationship between

and

ds.

Of course, evaluating the integral in this way produces

exactly the same result for

p,(Q,

as you will prove in one of the exercises of this

chapter.

5.6

THE INFINITESIMAL ELECTRIC DIPOLE

The example

the infinitesimal electric dipole is one of the more interesting appli-

cations of the Dirac S-function and makes use of many of its properties.

5.6.1

Moments

a Charge

Distribution

In electromagnetism, the distribution

charge density in space

p,(F)

can be ex-

panded, in what is generally called a multipole expansion, into a sum

its

moments.

These moments are useful for approximating the potential fields associated with com-

plicated charge distributions in the far field limit (that is, far away from the charges).

Each moment is generated by calculating a different volume integral

the charge

distribution over

all

space. Because our goal is not

derive the mathematics of

multipole expansions, but rather to demonstrate the use of the Dirac 6-functions, the

multipole expansion results are stated here without proof. Derivations can be found

122

THE

DIRAC &FUNCTION

in most any intermediate or advanced book on electromagnetism, such as Jackson’s

Classical Electrodynamics.

The lowest term in the expansion is a scalar called the monopole moment. It is

just the total charge of the distribution and is determined by calculating the volume

integral of

pc:

(5.80)

The next highest moment is a vector quantity called the dipole moment, which is

generated from the volume integral of the charge density times the position vector:

(5.81)

The next moment, referred to

the quadrapole moment, is a tensor quantity generated

by the integral

J’

(3TF

lT12T)pc(T).

(5.82)

All

space

In this equation, the quantity

is a dyad, and

is the identity tensor. There are an

infinite number of higher-order moments beyond these three, but they are used less

frequently, usually only in cases where

the

first

three

moments are zero.

Far away from the charges, the electric potential can be approximated by summing

the contributions from each

the moments. The potential field

due to the first few

moments is

(5.83)

It is quite useful to know what charge distributions generate just a single term in

this expansion, and what potentials and electric fields are associated with them. For

example, what charge distribgtion has just a dipole term (that is,

while all

other terms are zero

etc.). The Dirac &function turns out to be quite

useful in describing these particular distributions.

5.6.2

The

Electric

Monopole

The distribution that generates just the

Q/r

term in Equation 5.83 is called the electric

monopole.

you may have suspected, it is simply the distribution of a point charge

at the origin:

THE

INFINITESIMAL ELECTRIC DIPOLE

Its monopole moment,

123

d7q063(~;)

qo,

(5.85)

is simply equal to the total charge. The dipole, quadrapole, and higher moments of

this distribution are all zero:

All

space

&q0F

63(F)

space

(3F

r2T

)

ti3@)

All

space

The electric field

a monopole obeys Coulomb’s law:

(5.86)

(5.87)

(5.88)

5.6.3

The

Electric

Dipole

An electric dipole consists

two equal point charges,

opposite sign, separated by

some finite distance

do,

as shown in Figure

5.23.

The charge density of this system,

expressed using Dirac &functions, is

Pdpl

[s3

+GI)

+GX)]

(5.89)

where in this case, the dipole is oriented along the x-axis.

Indeed both

and

are zero, while the dipole moment is given by

You might be tempted to believe that this distribution has only a dipole moment.

qodogx.

The higher-order moments for the dipole distribution, however, do not vanish.

(5.90)

Figure

5.23

The

Electric

Dipole

124

THE

DIRAC &FUNCTION

The "ideal" dipole refers to a charge dstribution that

has

only a dipole moment,

and no other. It

the limiting case of the "physical" dipole, described above.

this limit,

the

distance between the charges becomes vanishingly small, while the net

dipole moment

is held constant. In other words,

and

---f

such that

doqo

is held constant. The charge distribution for

this

situation is

(5.91)

Expansion

the

6-functions in

terms of

Cartesian coordinates gives

ptacal

po6(y)6(z)

Notice that

this

limit is just the definition of a derivative:

This

means the charge distribution of an ideal dipole, with a dipole moment

magnitude

oriented along the

axis,

can be written

(5.94)

The electric field from the ideal dipole (and also from a physical dipole in the

far

field limit) is the solution of

But we already

know

the solution for the electric monopole is

(5.95)

(5.96)

Operating on both sides of Equation 5.96 with

[

-dod/dx],

shows the electric field in

Equation 5.95 obeys

Taking the derivative gives the result:

(5.97)

(5.98)

RIEMANN INTEGRATION

AND

THE

DIRAC &FUNCTION

125

This is the solution for a dipole oriented along the x-axis.

general, the dipole is a

vector

which can be oriented in any direction, and Equation

5.98

generalizes to

(5.99)

Notice that the magnitude of

this

field drops

off

faster with

than the field of

the monopole. The higher the order of the moment, the faster its field decays with

distance.

5.6.4

Fields from Higher-Order Moments

The technique described above can be used to find the electric field and charge

distributions for the ideal electric quadrapole, as well as for all the higher moments.

The charge distributions can be constructed using &functions, and the fields can be

obtained by taking various derivatives of the monopole field. You can practice this

technique with the quadrapole moment in an exercise at the end of this chapter.

5.7

RIEMANN INTEGRATION

AND

THE DIRAC &FUNCTION

The &function provides a useful, conceptual technique for viewing integration which

will become important when we discuss Green's functions in a later chapter. From

Equation 5.14, the sifting integral definition of the &function, any continuous func-

tion

f(y)

can be written

f(Y)

S(x

y)f(x).

(5.100)

The Riemann definition of integration says that an integral can be viewed as the limit

of a discrete sum of rectangles,

"+Z

(5.101)

where

is the width of the rectangular blocks which subdivide the area being

integrated, and

an integer which indexes each rectangle. The limiting process

increases the number

rectangles,

for well-behaved functions, the approximation

becomes more and more accurate as

---f

The Riemann definition is pictured

graphically in Figures 5.24(a) and

(b).

Using the Riemann definition, Equation

5.100

becomes

f(Y)

/+-m

f(x)@

--m

(5.102)