lecture 0 murderous mathematics

8/14/2019 Lecture 0 Murderous Mathematics

http://slidepdf.com/reader/full/lecture-0-murderous-mathematics 1/47

1

MurderousMathematics

Reference: Chapter 1, “Introduction to Electrodynamics” by David J. Griffiths

Text: Chapter 2, “Electromagnetism” by Gerald L. Pollack and Dainel R. Stump



2

“To those who do not know mathematics it is

difficult to get across a real feeling as to thebeauty, the deepest beauty, of nature. … If youwant to learn about nature, to appreciate nature, itis necessary to understand the language that shespeaks in.”

– R. P. Feynman, The Character of Physical Law



3

Curvilinear coordinates



4

Cartesian Coordinates

In terms of Cartesiancoordinates, the positionvector

of a point P in space, withrespect to a chosen origin:k jix ˆˆˆ z y x ++=

If the coordinate axes are rotated,

but the point P is left fixed inspace. For example, for a rotationby angle θ about :k ˆ

jiii ˆsinˆcosˆˆ θ+θ=′→

ji j j ˆcosˆsinˆˆ θ+θ−=′→

k k k ˆˆˆ

=′

→

( x , y , z )



5

Cartesian Coordinates (cont’)

It follows from k jixk ji ′′+′′+′′==++ ˆˆˆˆˆˆ z y x z y x

( ) ( ) k ji ji ˆˆcosˆsinˆsinˆcos z y x ′+θ+θ−′+θ+θ′=

( ) ( ) k ji ˆˆcossinˆsincos z y x y x ′+θ′+θ′+θ′−θ′=

that or

′

′′

θθθ−θ

=

z

y

x

z

y

x

100

0cossin

0sincos

θθ−θθ

=

′

′′

z

y

x

z

y

x

100

0cossin

0sincos

RT R R =−1

The matrix R depends on the angle and axis of rotation,but not on P.



6

Cylindrical Coordinates

k rx ˆˆ z r +=

In terms of cylindricalcoordinates, the positionvector of a point P in space, withrespect to a chosen origin:

z z r yr x =φ=φ= ,sin,cos

where z z y xr =+= ,22

jir ˆsinˆcosˆ φ+φ= ( ) φφ≡φ+φ−φ= ˆˆcosˆsinˆ d d d jir

It follows that k rs ˆˆˆ dz rd dr d +φφ+=



7

rx ˆr =

222 z y xr ++=

Spherical Coordinates

It follows that

φφθ+θ+= ˆsinˆˆ d r rd dr d θrs

( ) ( ) jik jir ˆcosˆsinsinˆsinˆsincosˆcoscosˆ φ+φ−φθ+θ−φθ+φθθ= d d d

φθ= cossinr x

k jir ˆcosˆsinsinˆcossinˆ θ+φθ+φθ=

In terms of spherical coordinates, theposition vector of a point P in space, withrespect to a chosen origin:

φθ= sinsinr y

θ= cosr z

where

θ≡ φ≡ ˆ



8

Infinitesimal displacement vector

The infinitesimal displacement vector, from ( x , y , z )to ( x + dx , y + dy , z + dz ), is

k jis ˆˆˆ dz dydxd ++=

In cylindricalcoordinates,

k rs ˆˆˆ dz rd dr d +φφ+=

φφθ+θ+=ˆ

sinˆ

ˆ d r rd dr d θrs

In sphericalcoordinates,

In general terms, let u1, u2,

u3 denote three coordinates

that specify the points inthree dimensions:

z u yu xu === 321 ,,

Cartesian coordinates

z uur u =φ==321

,,

cylindrical coordinates

φ=θ== 321 ,, uur u

spherical coordinates



9

Suppose the unit vectors pointingin the directions of independent

positive displacements of u

1,u

2,u3, are respectively:,ˆ,ˆ,ˆ321eee

k e jeie ˆˆ,ˆˆ,ˆˆ321

===

k eere ˆˆ,ˆˆ,ˆˆ321

=φ== φ=== ˆˆ,ˆˆ,ˆˆ321eθere

Cartesian coordinates

cylindrical coordinates spherical coordinates

Infinitesimal displacement vector (cont’)

The infinitesimal displacement vector in space that resultsfrom changing the coordinates by du1, du2, du3 can always

be written in the form:333222111 ˆˆˆ eees duhduhduhd ++=

For Cartesian coordinates, the scale factors h1 = h2 = h3

= 1.1,,1 321 === hr hh

cylindrical coordinates

θ=== sin,,1321

r hr hh

spherical coordinates



10

Vector algebra



11

Definitions

A vector A is a quantity with three components

k jik jiA ˆˆˆˆˆˆ321

A A A A A A z y x ++≡++=

In suffix notation, the vector A is denoted Ai (i = 1, 2,

3).

Note 1: Ai stands for two different things: the i th

component of the vector, and the vector itself. Thecontext must be used to decide which is meant.

that transform underrotation in the sameway as the

coordinates of a point:

=

′′′

z

y

x

z

y

x

A

A

A

R

A

A

A

=

′′′

3

2

1

3

2

1

A

A

A

R

A

A

A

jiji A R A =′Einstein summation convention:

Note 2: A scalar is a quantity that does not change

under rotation of the coordinate axes.



12

Definitions (cont’)

Note 3: Multiplication by a scalar & Addition of twovectors

jiji A R A =′jiji B R B =′

( ) jij jiji A R A R A α=α=′α ( ) jij jiji B R B R B β=β=′β

( ) ( ) ( ) j jij jij jijii B A R B R A R B A β+α=β+α=′β+′α



13

Dot Product

The dot product, or scalar product, of two vectors

k jiA ˆˆˆ z y x A A A ++= k jiB ˆˆˆ

z y x B B B ++=

is a scalar z z y y x x B A B A B A ++≡⋅BA

∑==++=

3

1332211

iii B A B A B A B A

Einstein summation convention: jiijii B A B A δ=≡⋅BA

where the Kronecker delta tensor

≠

=

=δ ji

ji

ij if 0

if 1

It follows that

1ˆˆˆˆˆˆ =⋅=⋅=⋅ k k j jii 0ˆˆˆˆˆˆ =⋅=⋅=⋅ ik k j ji



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Dot Product (cont’)

Geometric meaning:

A dot B is the projection of A on B times the magnitude of B, or the projection of B on A times the magnitude of A.

θ=⋅ cosBABA

where θ is the angle between A and B.

A dot B is a scalar:

k jik ijk ik jijii B A R R B R A R B A ==′′=′⋅′ BA

( ) BA ⋅=δ=== k j jk k j jk

T

k jik

T

ji B A B A R R B A R R



15

Cross Product

The cross product, or vector product, of two vectors

k jiA ˆˆˆ z y x A A A ++= k jiB ˆˆˆ

z y x B B B ++=

is a vector

321

321

ˆˆˆˆˆˆ

B B B

A A A

B B B

A A A

z y x

z y x

k jik ji

BA =≡×

The i th component of A x B,( )

k jijk j k

k jijk i

B A B A

ε=ε≡× ∑∑= =

3

1

3

1

BA

The Levi-Civita alternatingtensor or the completelyantisymmetric tensor for

three dimensions, ε ijk is 0

1312231123

=ε=ε=ε

1321132213 −=ε=ε=ε

It follows that jik ik jk ji ˆˆˆ,ˆˆˆ,ˆˆˆ =×=×=×



16

Cross Product (cont’)

An identity of the Levi-Civita tensor

jl im jmil klmkijklmijk δδ−δδ=εε=εε

( ) t s pqr kt kr js jqipt kt s js pqr kr jqipk jijk i B A R R R R R B R A R R R R B A ε=ε=′′ε′=′×′ BA

( ) ( ) t s pqr rt

T

qs

T

ipt s pqr kt

T

rk js

T

qjip B A R R R R R B A R R R R R ε=ε=

( ) pipr q pqr ipt s pqr rt qsip R B A R B A R BA×=ε=εδδ=

A cross B is a vector:

Geometric meaning:

nBABA ˆsin θ=×

θ is the angle between A and B.

d d i i



17

Vector Product Identities

Scalar triple product:

( ) ( ) ( ) ( ) CBABACACBCBA ⋅×=×⋅=×⋅=×⋅

Proof – ( ) ( ) ( ) k jiijk k jijk iii C B AC B A A ε=ε=×=×⋅ CBCBA

( ) ( ) ( )ACB ×⋅=ε=ε= ik jki jk iijk j AC BC A B

( ) ( ) ( )BAC ×⋅=ε=ε= jikijk jiijk k B AC B AC

Note – ( ) ( ) ( )ABCCABBCA ×⋅=×⋅=×⋅

V P d Id i i ( ’)



18

Vector Product Identities (cont’)

Vector triple product:

( ) ( ) ( )BACCABCBA ⋅−⋅=××

Proof – ( )[ ] ( ) ( )ml klm jijk k jijk i C B A A εε=×ε=×× CBCBA

( ) ml j jl im jmil ml jklmkijml jklmijk C B AC B AC B A δδ−δδ=εε=εε=

( ) ( ) ( ) ( )BACA ⋅−⋅=δδ−δδ= iil j jl mimm j jml il C B B AC C A B



19

Vector differential operators

∇ (“del”), ∇ ⋅ (“del dot”),∇ 2

(“del squared”),∇×

(“delcross”)

G di t f S l F ti



20

Gradient of a Scalar Function

In Cartesian coordinates, “del f ”

k ji ˆˆˆ z

f

y

f

x

f f

∂∂+

∂∂+

∂∂=∇ i

i x

f f e

∂∂=∇

“del f ” is a vector:

( ) ( ) ( ) jij

j

jij

j ji

j

i

i f R f R x

f

x

x

x

f f ∇=∇=∂∂′∂

∂=′∂∂=∇′ ∑∑ ==

3

1

3

1

Here, ( ) ( ) ( ) ij ji

T

ji

i

j

i ji j jiji R R R x

x x R x x R x ===

′∂∂

⇒′=⇒=′ −− 11

Consider the change of f (x) resultingfrom an infinitesimal displacement:

k jix ˆˆˆ dz dydxd ++=

The change of f from x to x + d x is

( ) ( ) xxxx d f dz z

f dy

y

f dx

x

f f d f df ⋅∇=

∂∂+

∂∂+

∂∂=−+=

The “del” operator acts algebraically as a vector.

G di t f S l F ti ( t’)



21

Gradient of a Scalar Function (cont’)

for any arbitrary displacement d x along the surface of constant f .

The direction of “del f ” at a point x is perpendicular tothe surface of constant f that includes the point x:

0==⋅∇ df d f x

“del f ” points in the direction of maximum increase of f:

θ∇=⋅∇= cosxx d f d f df

where θ is the angle between “del f ” and d x.

The change of f is maximum if θ = 0. The magnitude of “del f ” is the rate of change of f in that direction.

G di t f S l F ti ( t’)



22

Gradient of a Scalar Function (cont’)

It follows that

( )

ii

i

u

f

h

f ∂

∂=∇1

i

ii u

f

h

f e

1

∂

∂=∇

Consider the change of f (u1, u2, u3) resulting from an

infinitesimal displacement:333222111ˆˆˆ eees duhduhduhd ++=

The change of f , df duu

f du

u

f du

u

f =∂∂+

∂∂+

∂∂

3

3

2

2

1

1

( ) ( ) ( ) 333222111 duh f duh f duh f d f ∇+∇+∇=⋅∇= s

In cylindrical coordinates, “del f ” k r ˆˆ1ˆ

z

f f

r r

f f

∂∂

+φφ∂

∂+

∂∂

=∇

In spherical coordinates, “del f ” φφ∂

∂

θ+θ∂

∂

+∂

∂

=∇ˆ

sin

1ˆ1

ˆ

f

r

f

r r

f

f θr

E l



23

Example

Show that 3

1

xx

xx

xx ′−

′−=′−

∇−

Solution: ( ) ( ) ( )233

2

22

2

11x x x x x x ′−+′−+′−=′− xx

( )ii

i

x x x

′−′−

=′−∂∂

xxxx

1

It follows that i

i xe

xxxxˆ

11

′−∂

∂−=

′−∇−

i

ii

ii

x x

x exxexxxx ˆˆ

1

32 ′−

′−

=

′−∂

∂

′−=

3

1

xx

xx

xx ′−

′−=

′−−∇∴

E ample



24

Example

Show that ( ) f g g f fg ∇+∇=∇

Solution: ( )[ ] ( )iii

i x f g

x g f fg

x fg

∂∂+

∂∂=

∂∂=∇

Homework: Work through Example 2 @ Page 22

Divergence of a Vector Function (or Field)



25

Divergence of a Vector Function (or Field)

In Cartesian coordinates, “del dot F”

z y x F

z

F

y

F

x ∂

∂+

∂

∂+

∂

∂=⋅∇ F i

i

F

x∂

∂=⋅∇ F

“del dot F” is invariant under rotation of the coordinatesystem:

( ) k

j

ik ijk ik

ji

j

i

i

F x

R R F R x x

x F

x ∂∂=

∂∂

′∂∂

=′′∂

∂=′⋅∇′ F

( ) F⋅∇=∂∂

δ=∂∂

=∂∂

= k

j

jk k

j

jk

T

k

j

ik

T

ji F x

F x

R R F x

R R

“del dot F” is a scalar function.

The Laplacian (or “del squared”) is thedivergence of the gradient:

( ) f f 2∇=∇⋅∇

In Cartesian coordinates,

2

2

2

2

2

22

z

f

y

f

x

f

f ∂

∂

+∂

∂

+∂

∂

=∇

Divergence of a Vector Function (cont’)



26


The divergence is a measure of how the vector functionF(x) diverges, i.e., spreads out from x.

Let dV be an infinitesimal cubic volume centered at x, of size ε x ε x ε , aligned with the Cartesian directions:

∑∫ =ε ε−− ε+=⋅

3

1

2ˆ2

ˆ2i

iiiidS F F d exexAF

( )∑=

ε∂∂=

3

1

3

i

i

i

F x

x

( ) 3ε⋅∇= F

The flux of F outward through theboundary surface dS of dV ,

The divergence is equal to theflux per unit volume through an

infinitesimal closed surface: ∫⋅=⋅∇

→ S V

d

V

AFF1

lim0




27

( )21213213

333

duduhh F hh F uduu

−+ +

( ) ( ) ( ) ( )332211 duhduhduhF⋅∇

( )32321321

111

duduhh F hh F uduu

−= +

( )13132132

222

duduhh F hh F uduu

−+ +

Consider an infinitesimal cubic volumedefined by displacements du1, du2, du3:

( ) ∫ ⋅=⋅∇=dS

d dV AFF


( ) ( ) ( ) 213213

3

132132

2

321321

1

dududuhh F u

dududuhh F u

dududuhh F u ∂

∂+∂∂+

∂∂=

( ) ( ) ( )

∂

∂

+∂

∂

+∂

∂

=⋅∇∴ 2133

1322

3211321

1hh F

uhh F

uhh F

uhhhF




28


In cylindrical coordinates, “del dot F”

( ) z r F z F r rF r r ∂∂

+φ∂∂

+∂∂

=⋅∇ φ

11F

In spherical coordinates,

( ) ( ) φθ φ∂∂

θ+θ

θ∂∂

θ+

∂∂=⋅∇ F

r F

r F r

r r r

sin

1sin

sin

11 2

2F

Recall i

ii u

f

h f e

1

∂∂

=∇

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂=∇⋅∇=∇

33

21

322

13

211

32

1321

2 1u f

hhh

uu f

hhh

uu f

hhh

uhhh f f

It follows that

“del squared f ” 2

2

2

2

2

2 11

z

f f

r r

f r

r r f

∂∂+

φ∂∂+

∂∂

∂∂=∇

2

2

222

2

2

2

sin

1

sinsin

11

φ∂∂

θ+

θ∂∂

θθ∂∂

θ+

∂∂

∂∂

=∇f

r

f

r r

f

r r r f

Example



29

Example

Show that 012 =

′−∇−

xx

Solution: ( ) ( ) ( )2

33

2

22

2

11x x x x x x ′−+′−+′−=′− xx

( )ii

i x x x ′−′−=′−∂∂ xxxx

1

provided |x – x’| is not zero.

32

11

xxxx

xxxx ′−

′−−=

′−∂∂

′−−=

′−∂

∂ ii

ii

x x

x x

( )5

2

32

2

311

xxxxxx ′−′−+

′−−=

′−∂

∂ ii

i

x x

x

Example



30

Example

Show that ( ) GGG ⋅∇+⋅∇=⋅∇ f f f

Solution: ( ) ( ) i

i

i

i

i

i

G x

f G x f G f

x f

∂∂+∂∂=∂∂=⋅∇ G


Curl of a Vector Function (or Field)



31

Curl of a Vector Function (or Field)

In Cartesian coordinates, “del cross F”

z y x F F F z y x ∂∂∂∂∂∂=×∇

k ji

F

ˆˆˆ

( ) k

j

ijk i F x∂∂

ε=×∇ F

“del cross F” is a vector function:

( ) ( )t kt

s j

s

pqr kr jqipk

j

ijk i F R x x

x

R R R F x ∂∂

′∂∂

ε=′′∂∂

ε′=′×∇′ F

t

s

pqr kt

T

rk js

T

qjipt

s

pqr kt kr js jqip F x

R R R R R F x

R R R R R∂∂ε=

∂∂ε=

( ) pipr

q

pqr ipt

s

pqr rt qsip R F x

R F x

R F×∇=∂∂ε=

∂∂εδδ=

The curl is a measure of vorticity, i.e., how the vectorfunction F(x) curls around the point x.

Curl of a Vector Function (cont’)



32

je

ie

Let dS be an infinitesimal squarecentered at x, of size ε x ε ,aligned with the Cartesiandirections:

( )

ε+ε+

ε−ε=⋅∫ i j ji

ijdP F F dl exexF ˆ

2ˆ2

ε−ε−

ε+ε− i j ji F F exex ˆ2

ˆ2

( ) 22 ε×∇=ε

∂∂

−∂∂

= k

j

i

i

j

x

F

x

F F

The circulation of F, i.e.,the line integral of F,counterclockwise aroundthe perimeter dP (ij ) of the

square,

The curl of F is equal to thecirculation of F per unit area

around an infinitesimal loop:

( ) ∫ ⋅=×∇⋅→ C A

dl A

FFn1

limˆ0

Curl of a Vector Function (cont )

Here, C denotes a planar closed curve with area A and

normal vector .n




33


Consider an infinitesimalrectangular area produced bydisplacements du1, du2:

( ) ( ) ( )22113duhduhF×∇

( )11111

222

duh F h F duuu +−= ( )

22222111

duh F h F uduu

−+ +

( )

∫ ⋅=×∇=

C

dl dA FF 3

( ) ( ) 21221

12112

duduh F ududuh F u ∂∂

+∂∂

−=

332211

321

213132321 ˆˆˆ

F h F h F h

uuu

hhhhhh

∂∂∂∂∂∂=×∇∴eee

F




34


In cylindrical coordinates,“del cross F”:

In spherical coordinates,

“del cross F”:

z r F rF F

z r

r r

φ

∂∂φ∂∂∂∂φ

=×∇k r

F

ˆˆˆ

( ) φ∂∂−

φ∂∂=×∇∴ F

z F

r z r

1F

( ) z r F r

F z ∂∂−∂∂=×∇ φF

( ) ( ) φφ φ∂∂−

∂∂=×∇ F

r rF

r r z

11F

φθ θφ∂∂θ∂∂∂∂

φθθ=×∇

F r rF F

r

r r r

r sin

ˆsinˆsinˆ2

θr

F

( ) ( )

φ∂∂−θ

θ∂∂

θ=×∇∴ θφ F F

r r sin

sin

1F

( ) ( )

∂

∂−

φ∂

∂

θ

=×∇ φθ rF

r

F

r

r

sin

11F ( ) ( )

θ∂

∂−

∂

∂=×∇ θφ r F rF

r r

1F

Example



35

Example

Show that ( ) ( ) ( ) ( )[ ]xFxxFx ′×∇=′×∇ f f

Solution: ( ) ( )[ ] ( ) ( )xxxFx ′∂∂ε=′×∇ k

j

ijk i F x

f f

( ) ( )[ ]xx ′∂∂ε= k

j

ijk F f x

( ) ( )[ ] i f xFx ′×∇=

Example



36

Example

Show that

( ) ( ) ( ) ( )[ ] ( ) ( )xFxxFxxFx ′∇−′⋅∇∇=′×∇×∇ f f f 2

Solution:

( ) ( )[ ]( )

( )( )

( )xx

xx

xFx ′∂∂

∂εε=

′∂

∂ε∂∂ε=′×∇×∇ m

l j

klmkijm

l

klm

j

ijk i F x x

f F

x

f

x f

2

( ) ( ) ( )xx ′

∂∂∂δδ−δδ= m

l j

jl im jmil F x x

f 2

( )( ) ( ) ( )xxx

x ′∇−′

∂∂

∂= i j

ji

F f F

x x

f 22



Del Identities – Products of derivatives



37

Del Identities Products of derivatives

( ) FFF2∇−⋅∇∇=×∇×∇

( ) m

l j

klmkijm

l

klm

j

ijk i F x x

F x x ∂∂

∂εε=

∂∂ε

∂∂ε=×∇×∇

2

F

( ) i j ji

ml j

jl im jmil F F

x x F

x x

222

∇−∂∂

∂

=∂∂

∂

δδ−δδ=

The curl of the gradient of a scalar function is identicallyzero:

( ) 002

=∇×∇⇒=

∂∂

∂ε=∇×∇ f

x x

f f

k j

ijk i

The divergence of the curl of a vector function isidentically zero:

( )0

2

=∂∂

∂

ε=

∂

∂

ε∂

∂

=×∇⋅∇ k ji

ijk k j

ijk i F x x F x xF



38

Integral theorems

arl Friedrich Gausspril 30th 1777 – February 23rd 1855) George Gabriel Stoke(August 13th 1819 – February 1st 1903

Gauss’s Theorem (divergence theorem)



39

Gauss s Theorem (divergence theorem)

The flux of a vector quantity outwardthrough a closed surface S is equal tothe integral of the divergence of the

function in the enclosed volume V ,

∫ ∫ ⋅∇=⋅V S

xd d 3FAF

The flux of a vector field F(x) through asurface S with area element dA is thesurface integral of , where is theunit normal vector at the point x on S:

( ) nxF ˆ⋅ n∫ ∫ ⋅=⋅

S S d dA AFnF ˆ

( )

3

ε⋅∇=⋅∫ FAFdS d

Stokes’s Theorem



40

Stokes s Theorem

The circulation of a vector field F(x)around a loop C with length element dl is the line integral of where is theunit tangent vector at the point x on C:

( ) txF ˆ⋅ t ∫ ∫ ⋅=⋅C C

dl dl FtF ˆ

je

ie

( )∫ ∫ ⋅×∇=⋅ S C d dl AFF

( ) ( )

2

ε×∇=⋅∫k ijdP dl FF

Homework: Work throughExamples 5, 6 @ Page 27

The circulation of a vector function around a closed curveC is equal to the flux of vorticity through any surfacebounded by C,



41

The Helmholtz theorem

ermann von Helmholtzugust 31st 1821 – September 8th 1894)

Preliminaries



42

Preliminaries

A vector function (or field) F(x) that has zero curl:

0=×∇ F

is called irrotational.A vector function (or field) G(x) that has zero divergence:

0=⋅∇ G

is called solenoidal.

ψ −∇=F

AG ×∇=

Any vector function (or field) H(x) can be written as thesum of an irrotational function F(x) and a solenoidal

function G(x): GFH += AH ×∇+ψ −∇=

The functions F and G are not necessarily unique.However, in some cases, if suitable boundary conditions

are imposed, then the decomposition is unique.




43


Let H(x) be differentiable at all points in space, with curland divergence:

( )xcH =×∇ ( )xH d =⋅∇

If and c(x) approach 0 faster than r –2 as , andH(x) approaches 0 faster than r –1 , then

( )xd ∞→r

AH ×∇+ψ −∇=where

( )( )

∫ ′′−π

′=ψ xd

d 3

4 xx

xx ( )

( )∫ ′

′−π′

= xd 3

4 xx

xcxA

The integration region is all of space.

The Helmholtz theorem implies that if the divergence andcurl of a vector function (field) is known, then the functioncan be determined uniquely (under the assumptions of

the theorem).



44

Green’s function andthe Dirac delta function

eorge GreenJuly 14th 1793 – May 31st 1841)

Paul Adrien Maurice Dirac

Paul Adrien Maurice Dira(August 8th 1902 – October 20th 1984

The Dirac delta function



45

The Dirac delta function δ ( x ) is a “generalizedfunction” or “Schwartz distribution” with the followingdefining property:

( ) ( ) ( )0 f dx x f x =δ∫ ∞

∞− ( ) 1=δ∫ ∞

∞− dx x

for every continuous function f(x).

The Dirac delta function can

be understood as the limitof a sequence of more andmore sharply peakedfunctions:

δ ( x ) is an extremely

singular function:( )

=≠

≠=δ

0when0

0allfor 0

x

x x

Exercise:

( ) ( ) ( )a f dx x f a x =−δ∫ ∞∞−

The Dirac delta function (cont’)



46

( )

The 3-dimensional Dirac delta function δ 3(x) isdefined analogously:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )0xx f f dxdydz z y x f z y x xd f ==δδδ=δ ∫ ∫ ∫ ∫ ∞

∞−

∞

∞−

∞

∞− 0,0,0,,33

for every continuous function f (x).

( )

=≠

≠=δ

0when0

0allfor 03

x

xx( ) 1

33 =δ

∫ xd x

Exercise:

( ) ( ) ( )axax f xd f =−δ∫ 33

The Green’s Function of 2∇−



47

( )xx

xx′−π

=′−4

1G( ) ( )xxxx ′−δ=′−∇− 32G

along with the boundary condition that G approaches 0 atinfinity.Proof :Consider the integral∫

′−π

∇−V

xd 32

4

1

xx

in a spherical volume V of any radius around x’.

∫ ∫ ⋅

′−

∇π

−=

′−

∇π

−S V

d xd Axxxx

1

4

11

4

1 32 divergencetheorem

14

1

ˆˆ

1

4

1

ˆˆ

1

4

1 2

22 =Ωπ=Ω⋅π=⋅π= ∫ ∫ ∫ S S S d d r r dAr rrrr

Here, we recall rr

xx

xx

xxˆ

11233

r r =≡

′−

′−=

′−∇−

2

lecture 0 murderous mathematics

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