Prof. David R. JacksonDept. of ECE

Spring 2021

Notes 13Divergence

ECE 3318 Applied Electricity and Magnetism

1

2

ˆ

ˆ

4

S

S

rS

r

D n dS

D r dS

D dS

D r

ψ

π

= ⋅

= ⋅

=

=

∫

∫

∫

Divergence – The Physical ConceptFind the flux going outward through a sphere of radius r .

2

x

y

z

Sr

0vρa

Spherical region of uniform volume charge density

30

43 v

aψ π ρ= (r > a)

(r < a)

From Gauss’s law:

30

43 v

rψ π ρ=

20 C/m3v

rrD ρ =

23

02 C/m3

vr

aDr

ρ =

Divergence – The Physical Concept

(r < a)

(r > a)

Hence

3

(increasing with distance inside sphere)

(constant outside sphere)

x

y

z

Sr

0vρa

24rD rψ π=Also, from last slide:

More flux lines are added as the radius increases (as long as we stay inside the charge).

Observation:

Divergence -- Physical Concept (cont.)

ˆ 0S

D n dSψ∆

∆ = ⋅ >∫

The net flux out of a small volume ∆V inside the charge is not zero.

Divergence is a mathematical way of describing this.

4

S∆

V∆

0

1 ˆlimV

S

div D D n dSV∆ → ∆

≡ ⋅∆ ∫

Definition of divergence:

Divergence Definition

Note: The limit exists independent of the shape of the volume (proven later).

A region with a positive divergence acts as a “source” of flux lines. A region with a negative divergence acts as a “sink” of flux lines.

5

n̂ = outward normalV∆

Divergence at a given point

0

1 ˆlimV

S

div D D n dSV∆ → ∆

≡ ⋅∆ ∫

( )ˆ encl vS

D n dS Q r Vρ∆

⋅ = ≈ ∆∫

( ) ( )( )( )

0

1lim vV

v

div D r r VV

r

ρ

ρ∆ →

= ∆∆

=

Apply divergence definition to a small volume ∆V inside a region of charge:

Gauss’s Law -- Differential Form

Gauss's law:

Hence

6

V∆

S∆ r

( )v rρ

Gauss’s Law -- Differential Form (cont.)

( ) ( )vdiv D r rρ=

The electric Gauss law in point (differential) form:

This is one of Maxwell’s equations.

7

Calculation of Divergence

ˆ ,0,02

,0,02

0, ,02

0, ,02

0,0,2

0,0,2

xS

x

y

y

z

z

xD n dS D y z

xD y z

yD x z

yD x z

zD x y

zD x y

∆ ⋅ ≈ ∆ ∆

∆ − − ∆ ∆

∆ + ∆ ∆

∆ − − ∆ ∆

∆ + ∆ ∆

∆ − − ∆ ∆

∫

Assume that the point of interest is at the origin for simplicity (the center of the cube).

The integrals over the 6 faces are approximated by “sampling” at the

centers of the faces.

0

1 ˆlimV

S

div D D n dSx y z∆ →

≡ ⋅∆ ∆ ∆ ∫

8

( )/ 2,0,0x∆

x

y

z

x∆y∆

z∆( )0,0,0

Calculation of Divergence (cont.)

1 ˆ

,0,0 ,0,02 2

0, ,0 0, ,02 2

0,0, 0,0,2 2

S

x x

y y

z z

D n dSx y z

x xD D

xy yD D

yz zD D

z

⋅ ≈∆ ∆ ∆

∆ ∆ − −

∆∆ ∆ − −

+∆

∆ ∆ − − +

∆

∫ˆ ,0,02

,0,02

0, ,02

0, ,02

0,0,2

0,0,2

xS

x

y

y

z

z

xD n dS D y z

xD y z

yD x z

yD x z

zD x y

zD x y

∆ ⋅ ≈ ∆ ∆

∆ − − ∆ ∆

∆ + ∆ ∆

∆ − − ∆ ∆

∆ + ∆ ∆

∆ − − ∆ ∆

∫

0

1 ˆlimV

S

div D D n dSx y z∆ →

≡ ⋅∆ ∆ ∆ ∫

9

Calculation of Divergence (cont.)

0

,0,0 ,0,02 2

0, ,0 0, ,02 2lim

0,0, 0,0,2 2

x x

y y

V

z z

x xD D

xy yD D

div Dy

z zD D

z

∆ →

∆ ∆ − −

∆∆ ∆ − −

≡ +∆

∆ ∆ − − +

∆

Hence

yx zDD Ddiv D

x y z∂∂ ∂

= + +∂ ∂ ∂

10

Calculation of Divergence (cont.)

yx zDD D

div Dx y z

∂∂ ∂= + +

∂ ∂ ∂

Final result in rectangular coordinates:

11

The “del” operator

ˆ ˆ ˆx y zx y z∂ ∂ ∂

∇ ≡ + +∂ ∂ ∂

( ): sin cosd d x xdx dx

=

Examples of derivative operators:

scalar

vector

This is a vector operator.

( )

( ) ( )

( ) ( )

ˆ ˆ ˆ: sin cos

ˆ ˆ ˆ ˆsin sin cos

ˆ ˆ ˆ ˆ ˆsin sin cos

d dx x x x xdx dx

d dx x x x x x xdx dxd dx y x x y x z xdx dx

=

⋅ = ⋅ = × = × =

12

Note: The del operator is only defined in

rectangular coordinates.

ˆˆ ˆ

ˆ ˆˆ

zz

rr

ρ φρ φ

θ φθ φ

∂ ∂ ∂∇ ≠ + +

∂ ∂ ∂∂ ∂ ∂

∇ ≠ + +∂ ∂ ∂

Divergence Expressed with del OperatorNow consider:

( )ˆ ˆ ˆ ˆˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆ

x y z

yx z

D x y z x D y D z Dx y z

DD Dx x y y z zx y z

∂ ∂ ∂∇ ⋅ = + + ⋅ + + ∂ ∂ ∂

∂∂ ∂= ⋅ + ⋅ + ⋅

∂ ∂ ∂

yx zDD DDx y z

∂∂ ∂∇ ⋅ = + +

∂ ∂ ∂Hence

This is the same as the divergence.

13

D div D∇ ⋅ =

VV

Φ =⋅ =

× =

∇∇

∇

Note that the dot the del operator is important; any

symbol following it tells us how it is to be used

and how it is read: "gradient"

"divergence""curl"

after

14

Divergence with del Operator (cont.)

Summary of Divergence FormulasRectangular:

Cylindrical:

Spherical:

yx zDD DDx y z

∂∂ ∂∇ ⋅ = + +

∂ ∂ ∂

( )1 1 zD DD Dz

φρρρ ρ ρ φ

∂∂ ∂∇ ⋅ = + +

∂ ∂ ∂

( ) ( )221 1 1sin

sin sinrD

D r D Dr r r r

φθ θθ θ θ φ

∂∂ ∂∇ ⋅ = + +

∂ ∂ ∂

See Appendix A.2 in the Hayt & Buck book for a

general derivation that holds in any coordinate system.

15

Note on ∇ Operator

16

( )ˆ ˆ ˆ ˆˆ ˆ rdiv D r rD D Dr θ φθ φ θ φθ φ ∂ ∂ ∂

≠ + + ⋅ + + ∂ ∂ ∂

ˆˆ ˆ

ˆ ˆˆ

zz

rr

ρ φρ φ

θ φθ φ

∂ ∂ ∂∇ ≠ + +

∂ ∂ ∂∂ ∂ ∂

∇ ≠ + +∂ ∂ ∂

For example, in spherical coordinates:

ˆ ˆ ˆx y zx y z∂ ∂ ∂

∇ ≡ + +∂ ∂ ∂

Divergence is defined in any coordinated system, but the ∇ operator is only defined in rectangular coordinates:

Note on ∇ Operator (cont.)

17

We can also write divergence like this:

( )ˆ ˆ ˆ ˆˆ ˆx y zD x y z x D y D z Dx y z ∂ ∂ ∂

∇ ⋅ = + + ⋅ + + ∂ ∂ ∂

( )( )1 ˆ ˆˆ ˆˆ ˆ zD z A A zAz ρ φρ φ ρ ρ φρ ρ φ ∂ ∂ ∂

∇ ⋅ = + + ⋅ + + ∂ ∂ ∂

( ) ( ) ( )( )22 1 ˆ ˆ ˆ ˆˆ ˆ sin sinsin rD r r r A r A rAr r θ φθ φ θ θ θ φθ θ φ ∂ ∂ ∂

∇ ⋅ = + + ⋅ + + ∂ ∂ ∂

Rectangular

Cylindrical

Spherical

(But there is really no advantage to using these over the previous formulas.)

Electric Gauss Law(Point or Differential Form)

18

( ) ( ), , , ,vD x y z x y zρ∇⋅ =

We now have, in the notation of the “del” operator:

vD ρ∇⋅ =

Putting back the coordinate variables in the notation, it looks like:

Note: There is only one form of this equation, which has volume charge density. There is no

form that has surface charge density or line charge density.

Electric Gauss law (point form)

Maxwell’s Equations

0v

BEtDH Jt

DB

ρ

∂∇× = −

∂∂

∇× = +∂

∇ ⋅ =

∇ ⋅ =

Faraday’s law

Ampere’s law

Electric Gauss law

Magnetic Gauss law

Divergence appears in two of Maxwell’s equations.

19

(Maxwell’s equations in point or differential form)

Note: There is no magnetic charge density!

(Magnetic lines of flux must therefore form closed loops.)

Example

r < a

0vD ρ∇⋅ =

This agrees with the electric Gauss law.

( )

( )

22

2 02

302

202

1

13

131

r

v

v

v

D r Dr r

rrr r

rr r

rr

ρ

ρ

ρ

∂∇ ⋅ =

∂ ∂ = ∂ ∂

=∂

=

Evaluate the divergence of the electric flux vector inside and outside a sphere of uniform volume charge density, and verify that the answer is

what is expected from the electric Gauss law.

0ˆ3v rD r ρ =

20

x

y

z

r a

0vρ

Example (cont.)

r > a3

02ˆ 3

v aD rr

ρ =

0D∇⋅ =

This agrees with the electric Gauss law.

( )223

2 02 2

30

2

1

13

1 03

r

v

v

D r Dr r

arr r r

ar r

ρ

ρ

∂∇ ⋅ =

∂ ∂

= ∂ ∂

= = ∂

21

x

y

z

ra

0vρ

Divergence Theorem

ˆV S

A dV A n dS∇⋅ = ⋅∫ ∫

The volume integral of "flux per volume" equals the total flux!

In words:

22

A = arbitrary vector function

n̂ = outward normal

n̂V

S

Please see the Appendix for a proof.

Example( )ˆ 3A x x=

Verify the divergence theorem using this region.

( )( )( ) ( ) ( )( )( )ˆ ˆ ˆ ˆ ˆ3 3 2 3 0 2

18S

A n dS x x x x⋅ = ⋅ + ⋅ −

=

∫

( )3 3

yx zAA AAx y z

xx

∂∂ ∂∇ ⋅ = + +

∂ ∂ ∂∂

= =∂

( )3 3 3 1 2 3 18V V

A dV dV V∇⋅ = = = ⋅ ⋅ =∫ ∫

Given:

23

Note:A is constant on the front

and back faces, and the area of these faces is 2 [m2].

so

x

y

z

231

Dimensions in meters

Note on Divergence Definition

0

1 ˆlimV

S

div D D n dSV∆ → ∆

≡ ⋅∆ ∫

( )( )0

0

1lim

1lim

VV

rrV

div D D dVV

D V DV

∆ →∆

∆ →

= ∇ ⋅∆

= ∇⋅ ∆ = ∇⋅∆

∫

Is this limit independent of the shape of the volume?

Hence, the limit is the same regardless of the shape of the limiting volume.

ˆS V

D n dS D dV∆ ∆

⋅ = ∇ ⋅∫ ∫

Use the divergence theorem for RHS:

24

Small arbitrary-shaped volume

n̂

V∆S∆

r

Gauss’s Law (Differential to integral form)

ˆ enclS

D n dS Q⋅ =∫

vD ρ∇⋅ =

Apply the divergence theorem to the LHS:

25

We can convert the differential form into the integral form byusing the divergence theorem.

vV V

D dV dVρ∇ ⋅ =∫ ∫

ˆ vS V

D n dS dVρ⋅ =∫ ∫

Integrate both sides over a volume:

Use the definition of Qencl :

Gauss’s Law (Summary of two forms)

ˆ enclS

D n dS Q⋅ =∫

vD ρ∇⋅ =

Divergence theorem

Integral (volume) form of Gauss’s law

Differential (point) form of Gauss’s law

Definition of divergence

26

Note: All of Maxwell’s equations have both a point (differential) and an integral form.

Appendix:Proof of Divergence Theorem

Proof

( )0 1

limn

N

rV nV

A dV A V∆ →

=

∇ ⋅ = ∇ ⋅ ∆∑∫

Note: The point rn is the center of cube n.

The volume is divided up into

many small cubes.

27

V∆

nr

From the definition of divergence:

( )0

1 ˆlim

1 ˆ

n

n

n

r VS

S

A A n dSV

A n dSV

∆ →∆

∆

∇ ⋅ = ⋅∆

≈ ⋅∆

∫

∫

Hence:

( )0 0 01 1 1

1 ˆ ˆlim lim limn

n n

N N N

rV V Vn n nV S S

AdV A V A n dS V A n dSV∆ → ∆ → ∆ →= = =∆ ∆

∇ ⋅ = ∇ ⋅ ∆ = ⋅ ∆ = ⋅ ∆

∑ ∑ ∑∫ ∫ ∫

28

nS n∆ = surface of cube

rn

Proof of Divergence Theorem (cont.)

Hence, the surface integral cancels on all INTERIOR faces.

0 1

ˆlimn

N

V nV S

AdV A n dS∆ →

= ∆

∇ ⋅ = ⋅∑∫ ∫

2n̂ 1n̂1 2

ˆA n⋅

Consider two adjacent cubes:

is opposite on the two faces.

29

V∆

nr

Proof of Divergence Theorem (cont.)

0

outsideface

1

0s

ˆlim

ˆlimn

n

N

V nV S

VS

A dV A n dS

A n dS

∆ →= ∆

∆ →∆

∇ ⋅ = ⋅

= ⋅

∑∫ ∫

∑ ∫

ˆV S

AdV A n dS∇⋅ = ⋅∫ ∫

But

outside0faces

ˆ ˆlimn

VS S

A n dS A n dS∆ →

∆

⋅ = ⋅∑ ∫ ∫

(proof complete)

30

Hence

V∆

nr

Proof of Divergence Theorem (cont.)

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