Lecture 1 the Maxwell Equations

8/14/2019 Lecture 1 the Maxwell Equations

1/44

1mes Clerk Maxwellune 13 th 1831 November 5 th 1879)

The MaxwellEquations


2/44

2

enclosed0

1Qd

S = AE

and God said:

0= S d AB

Gausss law for E

Gausss law for B

Faradays law = S C d dt d

dl ABE

+=

S S C

d dt

d d dl AEAJB 000 Amperes law

and there was light!

displacement current, ID


3/44

3

= 01

E

and God said:

0= B

Gausss law for E

Gausss law for B

Faradays lawBEt

=

EJBt

+= 000 Amperes law



4/44

4

= 01

E

0= B

Gausss law for E

Gausss law for B

Faradays lawBEt

=

EJBt

+= 000 Amperes law

electric field diverges from charges

there are no magnetic monopoles, i.e., lines of B areclosed loops

E curls around the rate of change of B if B varies in time

displacement current density, J D


5/44

5

The Maxwell equationsin vacuum and

the displacement current


6/44

6

The Displacement Current

= 01

E

0= B BEt

=

EJBt

+= 000

The Maxwell equations

Consider ( )

+= EJB t 000

( )EJ += 0000 t

EJBt

+= 000 = 01

E

0=+ Jt continuity equation

This expresses the local conservation of electric charge the charge density and current density J are all

functions of position x and time t: electric charge isconserved throughout space.

+=t 00

J


7/44

7

Example 1 Charging a capacitor, Part 1.Consider a parallel plate capacitor being charged by acurrent I that flows in wires along the z axis:

By Gausss law, the electric field between the plates, i.e.,for r < a , is

( ) ( ) ( )k k E 200 a

t Qt t

== ( ) ( ) ( ) k EJ 2

00 a

t Qt

t t D =

=

By Amperes law, EJB t

+= 000

( )t Q ( )t Q

a a


8/44

8

Example 1 Charging a capacitor, Part 1. (cont)

( ) 01

=

=

= B z

B

z

B

r z r

B

( ) 0=

= z r Br B z B

( ) ( ) ( )

=

= rBr r Br rBr r r z 111

B

( )= r BB

Consider the Amperian loop C1 ,

( )

+=111

000S S S

d t

d d AEAJAB

The curl of B is in the z direction, and B is azimuthal:

+=

111

000 S S C

dAt

dAdl k Ek JB

( )

+= 11 000 S C dAt I rd k EB


9/44

9

Example 1 Charging a capacitor, Part 1. (cont)In the quasistatic approximation,

( ) I r r B 02 = ( )r

I r B

=2

0


( )

+=

222000 S S S

d t

d d AEAJAB

( ) ( ) I aat Q

r r B 02

20

002 ==( )

r I

r B

= 20


( ) += 333 000 S S S d t d d AEAJAB

( ) ( ) 220

002 r at Q

r r B =( ) 2

0

2 a Ir

r B

=


10/44

10


Consider the Amperian loop C,

( ) ( ) I d d d dl aaa S S

DS C 000==+== AJAJJABB

( ) ( ) I d d d dl bbb S DS DS C 000 ==+== AJAJJABB


11/44

11

Scalar and vector potentials


12/44

12

Potential Functions

= 01

E

0= B BEt

=

EJBt

+= 000

Any vector function whose divergence is 0 can be writtenas the curl of another vector:

( )t ,0 xABB ==

( )

=

=

= AAEBE t t t Any vector function whose curl is 0 can be written as thegradient of a scalar function:


vector potential

( )t V t t ,0 xAEAE =

+=

+ scalarpotential


13/44


14/44

14

Gauge Transformations and Gauge InvarianceThe potentials A(x, t ) and V (x, t ) are not uniquelydetermined by the charge and current sources in asystem.Let f (x, t ) be an arbitrary scalar function. Define

f + AAt

f V V

BAAA ==+= f

EAAA ==

+

=

V t t

f V f

t t V

t

Then,

B and E are invariant with respect to the gaugetransformation f + AAA

t f

V V V


15/44

15

Gauge Choices and Equations for V and A

The Coulomb gauge 0= A

This gauge choice makes A unique: f + AA f f 22 =+= AA

( ) ( ) 00A0A === f ,

Imposing the Coulomb gauge, ( )

=

0

2 1V

t

A

= 02 1V reduces to Poissons equation

( )( )

= xxx

xxd t

t V

3

0

,4

1,

It follows thatRecall the Greens Functionof 2 : ( ) xxxx = 4

1G

Demanding would mean requiring or0= A 02 = f constant= f


16/44

16

Gauge Choices and Equations for V and A (cont)The scalar potential is thus also uniquely determined,except for an arbitrary additive constant.

Remark The scalarpotential has an unphysicalaspect:

Imposing the Coulomb gauge,JAAA 0

22

2

0000 =+

+

t V

t

reduces toJAA

0

2

2

2

0000=

+

t V

t The vector potential A(x, t ) will in general be complicatedif the sources vary in time. The Coulomb gauge is thus nota convenient gauge choice for problems involving thegeneration of radiation.

It depends on the charge density at all source points x

at the very same time t .

( ) ( ) = xxxx xd t t V

3

0

,4

1,


17/44

17

Gauge Choices and Equations for V and A (cont)

The Lorentz gauge 000 =+ AV t

Imposing the Lorentz gauge,( ) =

0

2 1V t

A

JAAA 02

2

2

0000 =

+

+

t V t reduce to

=

0

22

2

001

V V

t

JAA 02

2

2

00 =

t


18/44

18

Example 2 V and A for a point charge at rest @ theorigin

( )r

qt V

04,

=x( ) 0, =t xA

It follows that ( ) ( ) rxxAE 4

,, 20 r

qt V t t

==

( ) 0, == t xAB

Consider the gaugetransformation with

( )r

qt t f

04,

=x

( ) ( ) ( ) rxxAxA 4

,,, 20r

qt t f t t

=+=

( ) ( ) ( )0,,, =

= t f t

t V t V xxx


19/44

19

Energy and momentum of electromagnetic fields

hn Henry Poyntingeptember 9 th 1852 March 30 th 1914)


20/44

20

Poyntings Theorem

= 01

E

0= B BEt

=

EJBt

+= 000


( )

=

= i

jijk k k iijk

jk

jijk i E x

B B E x

B x

E BE

( ) i j

kjik k i jik

j

E

x

B B E

x

+

=

( ) ( ) BBBEEBBEt

=+=

Consider

BE

t

=

EEJE t += 000

EJBt

+= 000


21/44

21

Poyntings Theorem (cont)

JEBE =

+

+

0

20

2

0

121

21

E Bt

Therefore, ( ) EEJEBBBEt t

+=

000

JEBEBBEE =

++ 00011

t t

Poyntingstheorem

This expresses the local conservation of energy theenergy density u and Poynting vector (energy flux) S:

20

2

0 2

1

2

1 E Bu +

= BES

=

0

1

are all functions of position x and time t .

h ( )


22/44

22

Poyntings Theorem (cont)If a particle with charge q moves, then the change of itskinetic energy K is equal to the work done on q:

( ) dt qdt qdt dK vEvBvEvF =+==

work-

kineticenergytheorem

vE =qdt dK

For a general continuous distribution of charge, withcharge density (x, t ) and kinetic energy density u K (x, t ),the total kinetic energy in an infinitesimal volume d 3 x at xsatisfies ( ) vE

=

xd xd t

u K 33 ( ) JEvE==

t

u K

E l 3 Ch i i P 2


23/44

23

Example 3 Charging a capacitor, Part 2.

k E 20 aQ=

The energy flux density on the surface is

rk BES 2

21

30

230

20 a

QI a

QI

===

( )t Q ( )t Q

a a

On the cylindrical surface of radius a that encloses thevolume between the plates

=

20

a I B quasistatic

approximation

E l 3 Ch i i P 2 ( )


24/44

24

Example 3 Charging a capacitor, Part 2. (cont)Field energy flows radially into the cylinder as Q increases, i.e., when I > 0. The total rate at which fieldenergy flows into the cylinder

20

30

2 22 aQId ad aQI P ==

E U dt d

d a E dt d

= 22021

Field energy flowing in through the surface builds up in

the field between the plates. The energy stored in acapacitor resides in the electric field.

=

=

d aa

Qdt d

ad Q

dt d 2

2

20

020

2

21

2

E l 3 Ch i i P 2 ( )


25/44

25


Homework: Work through Example 4 @ Page 420

Magnetic field energy between the plates is negligiblecompared to UE: ( )

a

M dr rd BU 02

0

2

2

1

( )

==

= 1642221 20

0

34

20

0

2

20

0

d I dr r

ad I

dr rd a Ir aa

2

22

22

22

002

20

20

81

82

16 Qa I

cQa I

d Qad I

U U

E

M == =

It follows that

So, unless the characteristic time of the charging process

is comparable to the time for light to cross the radius of the capacitor, UM


26/44

26

Field Momentum

= 01

E

0= B BEt

=

EJBt

+= 000


Consider

( ) ( ) ( ) +=+=+= V V xd xd q 33 BJEBvEBvEF

or ( ) BEBEEBJEf

+=+= t 000

1

( ) ( ) BEBEBBEEt t

+= 0000

1

EJBt

+= 000 = 01

E

( ) ( ) EEBEBBEE

= 0000

1

t

BE

t

=


27/44

Fi ld M t ( t)


28/44

28

Field Momentum (cont)

Homework: Read Example 5 @ Page 422

( ) ( )[ ] ( ) ( )[ ]BBBBEEEE ++++

+= 00

2

0

20

12

121 B E

( ) ( )BEf Sf +=

+t t 000

The momentum density of the electromagnetic field is

SP

00em

=dV d

The angular momentum density of the electromagneticfield is

SrP

rL == 00emem

dV

d

dV

d


29/44

29

Electromagnetic wavesin vacuum

einrich Rudolf Hertzebruary 22 nd 1857 January 1 st 1894)

The Maxwell Equations in Vacuum


30/44

30

The Maxwell Equations in Vacuum

0= E

0= B BEt

=

EB

t

= 00

= 01

E

0= B BEt

=

EJBt

+= 000


In vacuum, i.e., without any matter at all: = 0 and J =0; these reduce to

Note 1: There may be static fields in a region where =0 andJ = 0, produced by static charges and currents outside the

region.

Note 2: electromagnetic waves

Derivation of the Wave Equation


31/44

31

Derivation of the Wave Equation

BE

= t

( ) EEE 22

002

t =

0222

00 =+ EEt

0= E

0= B BEt

=

EBt

= 00

The Maxwell equations in vacuum

Consider

BE

t

=

EBt

= 00 0= E

One can similarly show that

02

2

2

00 =+

BBt

wave equation

wave equation 00

1

=c

The General Plane Wave Solution


32/44

32

The General Plane Wave Solution

The most general linearly polarized harmonic plane wave:

( ) ( )[ ]t it = xk BxB exp, 0 ( ) ( )[ ]t it = xk ExE exp, 0

where B 0 and E 0 are constant vectors.

Note: The real parts are understood to be the physicalfields.

== 2k k

= 2

c

k

= =The wave number

k jik z y x k k k ++=

A harmonic plane wave has definite values of wavelength and frequency , related by = c . The wave frontsof a plane wave are infinite planes, perpendicular to thedirection of propagation determined by the wave vector:

satisfies the dispersion

relation:with

phasevelocity

The General Plane Wave Solution (cont)


33/44

33

The General Plane Wave Solution (cont )

E satisfies the wave equation: 0222

00 =+ EEt

( ) ExE 222

, =t

t

( ) ( ) ( ) EExExE 222222

2

2

2

22 ,, k k k k t

z y xt z y x ++=

+

+=

if the k = /c .An Example of a Plane Wave Solution

( ) ( )[ ] jxB exp, 0 t kz i Bt = ( ) ( )[ ]ixE exp, 0 t kz i E t =

wherec

E B 00 =



34/44

34

The General Plane Wave Solution (cont )The Maxwell equations

0= B 0= E

require

EBt

= 00 BE t =

00 =Bk 00 =Ek

0000 EBk = 00 BEk =

E0 and B 0 are perpendicular to each other, as well asperpendicular to k, with E 0 x B 0 parallel to k. The energyflux is in the direction of k: Poynting vector

transverse wave

( )t

=

= xk BEBES 20000

cos11



35/44

35

The General Plane Wave Solution (cont )

0000 EBk = 00 BEk =

0002000001

E c

B E c

kB ==== EBk

Note: How to treat complex wavesAs long as we consider only expressions linear in thefields we may delay taking the real part.But before doing any calculation that is nonlinear in E andB, we must first take the real part, reducing the fields toreal numbers.Homework: Work through Examples 6 & 7 @ Page 429


36/44

and God said:


37/44

37

enclosed0

1Qd

S = AE

and God said:

0= S d AB

Gausss law for E

Gausss law for B

Faradays law = S C d dt d

dl ABE

+= S S C d dt

d d dl AEAJB 000 Amperes law


displacement current, ID

free-enclosed1

Qd S = AE

+= S S C d dt

d d dl AEAJB free

and God said:


38/44

38

( ) free= ED

and God said:

0= B

Gausss law for E

Gausss law for B

Faradays lawBEt

=

( ) DJEJBHt t

+=

+=

freefree

1

Amperes law


Gausss Law


39/44

39

Gauss s LawWhen dielectrics are present the source of E will includebound charge b as well as free charge f :

( ) bf 00

11+== EGausss law

The bound charge density b describes the effect of (atomic) electric dipoles as sources of electric field:

P= b

( ) f 0 =+ DPEGausss law

Here, P(x, t ) is the polarization field or electric dipolemoment density in matter the origin of the electric

dipoles is the separation of positive and negative chargeswithin atoms.

Here, D(x, t ) is the displacement field.

Polarization Current


40/44

40

Polarization CurrentThe polarization current J P is the current associated withchange of electric polarization, which occurs whenpositive charge shifts one way and negative chargeanother:

Since matter is electrically neutral, b and J P satisfy

0P b =+ Jt

It followsthat

( ) 0PP =

+=+

JPJP

t t PJ

t =P


41/44

Linear Constitutive Equations


42/44

42

Linear Constitutive EquationsIn many applications of electromagnetism in matter, thepolarization P is proportional to E

EP 0= e litysusceptibielectric=e

and the magnetization M is proportional to H

HM m= litysusceptibimagnetic=m

For such linear materials, D and E (B and H) satisfy thefollowing constitutive equations:

( ) EEPED+=+= 00

1e

where the constant is the permittivity of the material,and the permeability.

( ) HHMHBMBH +=+= 0001

1m

Boundary Conditions of Fields


43/44

43

Boundary Conditions of Fields

f = D0= B

The Maxwell equations in matter

( ) ===== dAQ xd xd d dA D D V V S nn f free-enclosed3f 312 DAD

( ) 03

12 === V S nn xd d dA B B BAB

Consider the surface S separating two linear media, withpermittivity 1 and permeability 1 on one side of S, and 2 and 2 on the other side.

1

2

G

In the limit 0,

f 12 = nn D D 012 = nn B B

Boundary Conditions of Fields (cont)


44/44

44

Boundary Conditions of Fields (cont )

BE

t

=DJHt

+= f

The Maxwell equations in matter

( ) ( ) ==== dl dl d d dt

d t t C S S 12

0 EEEAEAB

( ) +=+ S S d dt d

d dl ADAJnK f f 0

In the limit 0,

2

1 C012 = t t EE nK HH f 12 = t t

Lecture 1 the Maxwell Equations

Documents