sfgsfdgsdfg

7/23/2019 sfgsfdgsdfg

http://slidepdf.com/reader/full/sfgsfdgsdfg 1/21

Maxwell Equations, Macroscopic Electromagnetism,

Conservation Laws

§ Maxwell’s Displacement Current; Maxwell Equations

Ampere’s Law: JB 0 µ =×∇ ⇒ 0=⋅∇ J for steady state

In general case 00 =

∂

∂+⋅∇=

∂

∂+⋅∇

t t

EJJ ε

ρ

Then Maxwell replaced J in Ampere’s law by its generalization

J → t ∂

∂+

EJ 0ε

for time-dependent fields Th!s Ampere’s law became

∂∂+=×∇ t

EJB 00 ε µ

The Maxwell e"!ations can be written as follows:

∂∂+=×∇=⋅∇

∂∂−=×∇=⋅∇

t

t

EJBB

BEE

00

0

#i$% &0 #iii%

#ii% & #i%

ε µ

ε ρ

§ Vector an !calar "otential

0=⋅∇ B ⇒ #B ×∇=

t ∂

∂−=×∇ BE ⇒ 0=

∂∂+×∇t

#E ⇒

t ∂∂−Φ−∇= #

E

Then the inhomogeneo!s e"!ations can be written in terms of the potentials

'



as

0( )%# ε ρ −=⋅∇

∂∂+Φ∇ #t

#'%

J### 0((

(

(

( '' µ −=

∂Φ∂+⋅∇∇−

∂∂−∇

t ct c #(%

*e can find that the electromagnetic fields are in$ariant !nder the gauge

transformations: Λ∇+=′→ ###

t ∂

Λ∂−Φ=Φ′→Φ

§ $auge %rans&ormations, Lorent' $auge, Coulom( $auge

)i* Lorent' $auge+ 0'(

=∂Φ∂

+⋅∇t c

#

The e"!ations #'% and #(% becomes

0(

(

(

( )' ε ρ −=∂Φ∂−Φ∇t c

#+%

J#

# 0(

(

(

( ' µ −=

∂∂−∇t c

#,%

)ii* Coulom( $auge+ 0=⋅∇ #

rom #'% we see that the scalar potential satisfies the .oisson e"!ation&

0( )ε ρ −=Φ∇

with sol!tion& xd t

t ′′−

′=Φ ∫ +

0

%&#

,

'

xx

x*)x,

ρ

πε

The $ector potential satisfies the inhomogeneo!s wa$e e"!ation&

∂Φ∂∇+−=

∂∂−∇

t ct c (0(

(

(

( ''J

## µ

(



§ %e -elmolt' %eorem

Theorem+ .artial Integral xd f da f xd f V S V

+

+

#/%# #*n## ⋅∇−⋅=∇⋅ ∫ ∫ ∫ where ##x%

is a $ector field and f #x% a scalar field

Proof +

se the 1a!ss’s theorem da f xd f S V

/#

+

n##* ⋅=⋅∇ ∫ ∫

and the $ector form!la xd f xd f xd f

V V V

+

+

+

#%## #*##* ⋅∇+∇⋅=⋅∇

∫ ∫ ∫

inally we obtain the partial integral

xd f da f xd f V S V

+

+

#/%# #*n## ⋅∇−⋅=∇⋅ ∫ ∫ ∫

The Helmholtz Theorem

Let -#x% be differentiable at all points in space& with di$ergence ∇ 2 - 3

d #x% and c!rl ∇ × - 3 c#x% If d #x% and c#x% approach to 0 faster than r −2 as r

→ ∞& and -#x% → 0 as r → ∞& then - 3 −∇φ 4 ∇ × # where

xd d ′

′−′

= ∫ + %#

,

'%#

xx

xx

π φ

xd ′′−

′= ∫ +

,

'%#

xx

*xc)x#

π

Proof +

+



( )

-)x*nxx

-

xx

-

xx*x-)xx

.*x-)

xx

.*x-)

xx

.*x-)

xx

*x-)

xx

*x-)

xx

*x-)

π

δ π

,/

,

+

++

+(+

+(++

+

′⋅

′−+′

′−⋅∇′

−−∇=

′′−⋅′+′

′−∇′⋅′−∇=

′

′−

∇⋅′−′

′−

∇⋅′∇=

′′−′

∇−

′

′−′

⋅∇∇=′′−′

×∇×∇

∫ ∫ ∫ ∫

∫ ∫

∫ ∫ ∫

ad xd

xd xd

xd xd

xd xd xd

ad xd

xd xd

xd xd

′×′−′

×∇+′′−′×∇′

×∇=

′

′−′

×∇′−′−′×∇′

×∇=′

′−

∇′×′×∇=

′

′−

∇×′×−∇=′′−′

×∇×∇

∫ ∫ ∫ ∫

∫ ∫

/ +

++

++

nxx

*x-)

xx

*x-)

xx

*x-)

xx

*x-)

xx

.*x-)

xx

.*x-)

xx

*x-)

Let ∞→=′− r xx &

then we ha$e

xd xd ′′−′×∇′

×∇+′′−

⋅∇′−∇= ∫ ∫ ++

,

,

xx

*x-)

xx

--)x*

π π

5o that -#x% 3 −∇φ 4 ∇ × #

where xd d ′

′−′

= ∫ + %#

,

'%#

xx

xx

π φ and xd ′

′−′

= ∫ + ,

'%#

xx

*xc)x#

π

The $ector field -#x% can be written as the s!m of two terms&

-#x% 3 /#x% 4 $#x%

where /#x% is call the longit!dinal or irrotational field and has ∇ × / 3 0&

while $#x% is call the trans$erse or solenoidal field and has ∇ 2 $ 3 0

§ Maxwell’s Equations in Matter

*e consider the static case that an electric polarization " prod!ces a bo!nd

,



charge density

"⋅−∇=b ρ

Li6ewise& a magnetic polarization #or 7magnetization8% M res!lts in a bo!nd

c!rrent

MJ ×∇=b

There’s 9!st one new feat!re to consider in the non-static case: Any change in

" in$ol$es a flow of bo!nd charge #call it J p%& which m!st be incl!ded in the

total c!rrent If " increases a bit& the charge on each end increases

accordingly& gi$ing a net c!rrent

⊥⊥∂

∂=

∂

∂= dA

t

P dAt

dI bσ

The c!rrent density& therefore& is

t p ∂

∂

=

"

J

This polarization c!rrent J p has nothing whate$er to do with the bo!nd

c!rrent Jb *e co!ld chec6 that the abo$e e"!ation for J p is consistent with

the contin!ity e"!ation:

( )t t t

b p

∂

∂−=⋅∇

∂

∂=

∂

∂⋅∇=⋅∇

ρ "

"J

In $iew of all this& the total charge density can be separated into two parts:

"⋅∇−=+= f b f ρ ρ ρ ρ & and the c!rrent density into

three parts:

t

f pb f ∂

∂+×∇+=++=

"MJJJJJ

1a!ss’s law can now be written as

"

−σ b

+σ bdA

⊥



( ) ( )"E ⋅∇−=+=⋅∇ f b f ρ ε

ρ ρ ε 00

''

or f ρ =⋅∇ D

where "ED +≡ 0ε Meanwhile& Amp;re’s law becomes

( ) ( ) ( )

t t f pb f

∂

+∂+×∇+=

∂

∂+++=×∇

"EMJ

EJJJB

000000

ε µ µ ε µ µ

ort

f ∂∂+=×∇ D

J-

where MB- −≡ 0) µ Therefore the Maxwell’s e"!ations in matter can be

written as follows:

t

t

f

f

∂∂+=×∇=⋅∇

∂∂−=×∇=⋅∇

DJ-B

BED

#i$% &0 #iii%

#ii% & #i% ρ

5ome people regard these as the 7tr!e8 Maxwell’s e"!ations& b!t please

!nderstand that they are 9!st only the 7approximate form!la8& they are in no

way more 7general8 than the original Maxwell’s e"!ations

§ Bounar0 Conitions

#i% f ρ =⋅∇ D ⇒ f S

Qd =⋅∫ aD

⇒ f D D σ =− ⊥⊥ ('

#ii% 0=⋅∇ B ⇒ 0=⋅∫ aB d S

<



⇒ 0('

=− ⊥⊥ B B

#iii%t ∂

∂−=×∇ BE ⇒ ∫ ∫ ⋅−=⋅

S cd

dt

d d aBE l

⇒ 0))())' =− E E

#i$%t

f ∂∂+=×∇ D

J-

⇒ ∫ ∫ ⋅+=⋅S

f c

d dt

d I d aD- l

⇒ f I =⋅−⋅ l l (' --

⇒ n1 -- /))())' ×=− f

§ "o0nting’s %eorem an Conservation o& Energ0

According to the Lorentz force law& the wor6 done on a charge q with an

infinitesimal displacement dl is

dt qdt qd dW vEvB*v)E/ ⋅=⋅×+=⋅= l

If there exists a contin!o!s distrib!tion of charge and c!rrent& the total rate of

doing wor6 by the =M fields in a finite $ol!me V is

xd t

xd dt

dW

V V

+

+

∂∂⋅−×∇⋅=⋅= ∫ ∫

DE-*)EEJ

>



If we now employ the $ector identity&

-*)EE*)--E ×∇⋅−×∇⋅=×⋅∇ %#

we ha$e

xd t t

xd V V

+

+

∂∂⋅+

∂∂⋅+×⋅∇−=⋅ ∫ ∫ B

-D

E-*)EEJ #%

The energy density stored in the =M fields is

%#(

'em -BDE ⋅+⋅=u

If we denote the total energy density of the charged particles within the

$ol!me V as umech #the mechanical energy density%& so that

+mech

+mech

xd

t

u xd u

dt

d

dt

dW

V V ∂

∂== ∫ ∫

Then ="#% can be written

xd t

u

xd t

u

V V

+em

+mech

×⋅∇+∂∂

−=∂∂

∫ ∫ -*)E

5ince the $ol!me V is arbitrary& this leads to the contin!ity e"!ation of

energy&

( ) !⋅−∇=+∂∂

emmech uut

where ! 2 E × - is called the Poynting vector

The .oynting’s theorem expresses the conser$ation of energy for the system

of charged particles mo$ing in =M fields as

a! d E E dt

d

dt

dE ⋅−=+= ∫ %# fieldmech

where

?



xd dt

dE

V

+

mech EJ ⋅= ∫ and xd u E V

+em

field ∫ =

Example

Ass!ming that the c!rrent density is

!niform& the electric field parallel to the wire is

L

V E =

The magnetic field at the s!rface has the $al!e

a

I B

π

µ

(

0=

Accordingly& the magnit!de of the .oynting $ector is

aL

VI S

π (=

Therefore the energy per !nit time passing in thro!gh the s!rface of the wire

is

VI aLS d ==⋅∫ %(# π a!

@



§ Maxwell !tress %ensor an Conservation o& Momentum

The conser$ation of linear moment!m can be similarly considered *e can

write& from the Lorentz force and ewton’s second law&

xd dt

d

V

+

mech %# BJE"

×+= ∫ ρ #<%

*e !se the Maxwell e"!ations to eliminate ρ and J from ="#<%:

[ ] B*)EB*BE*EB*BE*E)

B*)EB*BB

EE*E)

BEB*E*E)BJE

×∂∂−×∇×−×∇×−⋅∇+⋅∇=

×∂∂−

×∇×−

∂∂×+⋅∇=

× ∂∂−×∇+⋅∇=×+

t cc

t c

t

t

0((

0

0(

0

00

0

###

#

#'

ε ε

ε ε

ε µ

ε ρ

The rate of change mechanical moment!m ="#<% can now be written

[ ] xd cc

xd dt

d

dt

d

V

V

+((

0

+

mech

### B*BB*BE*EE*E)

B*)E

"3

×∇×−⋅∇+×∇×−⋅∇=×+

∫ ∫

ε

ε #>%

*e may identify the total =M moment!m "field in the $ol!me V :

xd xd V V

+

0

+

field !BE" 33 ∫ ∫ =×= µ ε ε

5o that the density of moment!m in the =M fields is

(0em c

!!3 == µ ε p

If we employ the $ector identity&

[ ] %## (

('

ij ji

j ji E E E

xδ −

∂

∂=×∇×−⋅∇ ∑E*EE*E)

the right hand side of ="#>% has the form of a di$ergence of a second ran6

'0



tensor *ith the definition of the Maxwell stress tensor T ij as

ij ji jiij Bc E B Bc E E T δ ε %#(((

('(

0 +−+=

Therefore we can write ="#>% as

%#

+

fieldmech da xd

dt

d

S V n%%"" ⋅=⋅∇=+ ∫ ∫

#?%

where ( ) ( ) jij

j

iij j j

i nT T x ∑∑ =⋅∂∂=⋅∇ n%%

and

.hysically& % is the force per !nit area #or stress% acting on the s!rface More

precisely& T ij is the force per !nit area in the i-th direction acting on an

element of s!rface oriented in the j-th direction 7diagonal8 elements #T xx&

T yy& T zz % represent press!res& and 7off-diagonal8 elements #T xy& T yz & T zx&% are

shears ="#?% can be written in the differential form as follow:

%# emmech %

⋅∇=+∂∂

p pt

Example

The fields are

ρ ρ

λ

πε eE /

(

'

0

= &

φ ρ π

µ eB /(

0 I =

''



The .oynting $ector is therefore

z

I e! /

, (0

( ρ ε π

λ =

The power transported is

IV ab I

d I

d P b

a===⋅= ∫ ∫ %)ln#

( (

'

, 0(

0( πε

λ ρ πρ

ρ ε π

λ a!

The moment!m in the fields is

z

b

a z ab

Il d

Il d ee!" /%)ln#

( (

'/

, 0

(

(

000field

π

λ µ ρ πρ

ρ π

λ µ τ ε µ === ∫ ∫

5!ppose now that we t!rn !p the resistance& so the c!rrent decrease The

changing magnetic field will ind!ce an electric field:

z cdt

dI eE /ln

(

0

+= ρ

π

µ

The field exerts a force on ±λ :

z z z abdt

dI l cb

dt

dI l ca

dt

dI l eee/ /%)ln#

(/ln

(/ln

(

000

π

λ µ

π

µ λ

π

µ λ −=

+−

+=

The total moment!m imparted to the cable& as the c!rrent drops from I to 0&

is therefore

z ab Il

dt e/" /%)ln#(0mechπ

λ µ

== ∫

Example+ %e /e0nman is4 paraox5

In ig an ins!lator dis6 is free to rotate

on its axis Attached to the dis6

coaxially there are: #i% a solenoid coil #(%

'(



a ring of positi$e charge fixed on the dis6 Initially the battery is not

connected to the coil& no c!rrent flows& and the system is at rest *hen the

switch is closed there is an imp!lsi$e tor"!e on the dis6 Is the ang!lar

moment!m conser$edB

§ #ngular Momentum

The =M fields carry the energy density

%#(

'em -BDE ⋅+⋅=u &

and the moment!m density

%#000em BE! ×== ε µ ε p &

and& for that matter& the ang!lar moment!m density:

[ ]%#0emem BErr ××=×= ε pl

Example 657

Cefore the c!rrent was switched off& there was an

electric field&

%# /'

( 0bal

Q

<<= ρ ρ πε ρ eE &

'+



and a magnetic field&

%# /0 RnI z <= ρ µ eB

The moment!m density was therefore

%# /(

0em Ra

l

nIQ<<−= ρ

ρ π

µ φ e p

The ang!lar moment!m density was

%# /(

0emem Ra

l

nIQ z <<−=×= ρ

π

µ er pl

The total ang!lar moment!m in the fields was

( ) z a RnIQ eL /(

' ((0em −−= µ

*hen the c!rrent is t!rned off& the changing magnetic field ind!ces a electric

field& gi$en by araday’s law:

<−

>−=

%# &/(

'

%&# &/('

0

(

0

Rdt

dI n

R

R

dt

dI

n

ρ ρ µ

ρ ρ

µ

φ

φ

e

e

E

Th!s the tor"!e on the o!ter cylinder is

z b dt

dI nQRQ eEr8 /

(

'%#

(

0 µ =−×=

and it pic6s !p an ang!lar moment!m

z I z

t

z b nIQRdI nQRdt dt

dI nQR eeeL /

(

'/

(

'/

(

' (0

0

(0

0

(0

0

µ µ µ −=== ∫ ∫

5imilarly& the tor"!e on the inner cylinder is

z adt dI nQa e8 /

(' (

0 µ −=

',



and it pic6s !p an ang!lar moment!m

z a nIQa eL /(

' (0 µ =

5o it all wor6s o!t: Lem 3 La 4 Lb

#ppenices+

)9* Multi:pole Expansion

+

′⋅+=

′− +

''

x

xx

xxx

%&#%&#'(

',

' D

'0

φ θ φ θ π lll

l l

l l

! ! r

r

l ′′

+=

′− +>

<

−=

∞

=∑∑

xx #A'%

The electrostatic potential for the charge density ρ #x’% is

xd ′′−′

=Φ ∫ +

0

%#

,

'%#

xx

xx

ρ

πε #A(%

5!bstit!tion #A'% into #A(% leads to

'00

%&#

'(

'%#

+−=

∞

= +=Φ ∑∑ l

lll

l l r

!

l

q φ θ

ε x

where xd r ! q l ll ′′′′′= ∫

+D %#%&# x ρ φ θ

#i% The electric monopole moment # the total charge % is q

#ii% The electric dipole moment is

xd ′′′= ∫ +

%#xx ρ

'



#iii% The electric "!adr!pole moment tensor is

xd r x xQ ij jiij ′′′−′′= ∫ +( %#%+# x ρ δ

Then we ha$e

++⋅+=Φ ∑

:+0 (

'

,

'%#

r

x xQ

r r

q jiij

ij

xpx

πε

)99* Elementar0 %reatment o& Electrostatics wit "onera(le

Meia

*hen an a$eraging is made of the microscopic e"!ation& 0micro =×∇ E & the

macroscopic e"!ation& namely&

0=×∇ E

holds for the a$eraged Eere E is the macroscopic electric field

The electric polarization in a medi!m is gi$en by

∑=i

ii " px" %#

The charge density at the macroscopic le$el will be

excess%# ρ ρ += ∑i

ii # " x

Th!s the charge of ∆V is ρ #x′%∆V and the dipole moment of ∆V is "#x′%∆V If

there are no higher macroscopic m!ltipole moment densities& the potential

ca!sed by the config!ration of moments in ∆V is gi$en by

V ∆

′−

′−⋅′+

′−′

=′∆Φ %#%#

,

'%#

+0 xx

*x)xx"

xx

xxx,

ρ

πε

Then

′−

∇′⋅′+′−′

′=Φ ∫ xxx"

xx

xx

'%#

%#

,

'%# +

0

ρ

πε xd

V

An integration by part for the second term yields

'<



[ ]

ad xd S V

′′−

′⋅′+′

′−

′⋅∇′−′=Φ ∫ ∫

%#

,

'

%#%#

,

'%#

0

+

0 xx

nx"

xx

x"xx

πε

ρ

πε

*ith E 3 −∇Φ & the first Maxwell e"!ation therefore reads

[ ]":E ⋅∇=⋅∇ ρ ε 0

'

*e can define the $ol!me bo!nd charge density

"⋅−∇=b ρ

And the s!rface bo!nd charge density

n" ⋅=bσ

Fefine the electric displacement D: "ED += 0ε

*e ha$e

ρ =⋅∇ D

*e ass!me that the medi!m is isotropic Then

E" # χ ε 0=

The constant χ # is called the electric s!sceptibility of the medi!m The

displacement D is therefore proportional to E

ED ε = and ε ρ )=⋅∇ E

where the electric permitti$ity ε = ε 0(1+ χ #)

'>



)999* Magnetic /iels o& a Locali'e Current Distri(ution,

Magnetic Moment

The Ciot-5a$art law:

∫

∫ ∫ ′

′−

′=⇒

×∇=′′−′

×∇=′′−

′−×′=

xd

xd xd

+0

+0+

+0

%#

,

%#

, %#

,

xx

xJ#)x*

#xx

xJ

xx

xxxJ)x*B

π

µ

π

µ

π

µ

#A+%

The Taylor expansion

+′⋅

+=′− +

''

x

xx

xxx

5!bstit!tion of this into #A+% yield

+′′′⋅+′′= ∫ ∫ xd $ xd $ A iii

+

+

+0 %#%#

'

, xxx

xxx)x* π

µ

#A,%

If J#x′% is localized b!t not necessarily di$ergenceless ha$e the identity

( ) ( ) 0+ =′′⋅=′⋅∇′ ∫ ∫ ad f% xd f% S

nJJ

Then

( ) 0+ =′⋅∇′+⋅∇′⋅+∇′⋅∫ xd f% f % % f JJJ #A%

#i% *ith f 3 '& % 3 x′ i and ∇′2J 3 0& #A% yields

0%#+ =′′∫ xd $ i x

'?



#ii% *ith f 3 x′ i& % 3 x′ j and ∇′2J 3 0& #A% yields

0%#+ =′′+′∫ xd $ x $ x i j ji

The integral in the second term of #A,% can be therefore written

( )

( ) ( )i

& j

& jij&

j

i j ji j

j

i j ji

xd xd x

xd $ x $ x x xd $ x x xd $

′×′×−=′×′−=

′′−′⋅−=′′⋅=′′⋅

∫ ∑ ∫

∑ ∫ ∑ ∫ ∫ +

&

+

+++

(

'

(

'

(

'

JxxJx

xx

ε

Fefine the magnetic moment m :

∫ ′′×′= xd + #(

'*xJxm #A<%

Then the $ector potential from the second term in #A,% is the magnetic

dipole $ector potential&

+

0

, x

xm#)x*

×=π

µ

The magnetic B ind!ction can be calc!lated directly

+−⋅= %#

+

?+

, +

0 xmx

mm*n)nB)x* δ

π

π

µ

If the c!rrent I flows in a closed circ!it whose line element is d l & #A<%

becomes

l d I ×= ∫ xm(

Therefore the magnetic moment has magnit!de&

%Area#×= I m

'@



)9V* Macroscopic Equations o& B an -

The a$eraging of the microscopic e"!ation& 0micro =⋅∇ B & leads to the same

macroscopic e"!ation 0=⋅∇ B

The large n!mber of molec!les per !nit $ol!me& each with its molec!lar

magnetic moment mi& gi$e rise to an a$erage macroscopic magnetization or

magnetic moment density&

∑=i

ii " mxM %#

Then the $ector potential from a small $ol!me ∆V at the point x′ will be

V ∆

′−

′−×′+

′−′

=∆ %#%#

,%#

+

0

xx

*x)xxM

xx

xJx#

π

µ

Then

xd V

′

′−

′−×′+

′−′

= ∫ +

+

0 %#%#

,%#

xx

*x)xxM

xx

xJx#

π

µ

An integration by part for the second term yields

xd xd ad V V S

′

′−

′−×′−

′−′×∇′

=′

′−′

×∇′=′

′−′

×′ ∫ ∫ ∫ +

+

+ %#%#%#%#

/xx

*x)xxM

xx

xM

xx

xM

xx

xMn

Then

ad xd S V

′′−′×′

+′

′−

′×∇′+′= ∫ ∫ xx

nxM

xx

xMxJx#

/%#

,

%#%#

,%# 0+0

π

µ

π

µ

The magnetization is seen to contrib!te an effective current density

MJ ×∇= '

5o that

[ ]MJB ×∇+=×∇ 0 µ

*e can define a new macroscopic field -&

MB- −=0

' µ

(0



Then the macroscopic e"!ations are

J- =×∇

0=⋅∇ B

*e ass!me that the medi!m is isotropic Then

-B µ = #?,%

The parameter µ is called the magnetic permeability # µ > µ 0 for

paramagnetic s!bstances and µ < µ 0 for diamagnetic s!bstances%

or the ferromagnetic s!bstances& #?,% m!st

be replaced by a nonlinear relationship&

-*/B #=

('

sfgsfdgsdfg

Documents