第五章電磁波的輻射電磁波的激發(fā)變化的電荷和電流分布激發(fā)變化電磁場變化電磁場對源的依賴關(guān)系?tD

=

r?B·

E

=

-?tB

=

0·

H

=

J

+

?DD

=

e

E0B

=

m0

H真空§1

電磁場的矢勢和標勢1．一般電磁場的勢B

=

0B

=

·

A矢勢?t·

E

=

-

?B?t

·

E

+

?A

=

0?tE

+

?A

=

-

j標勢?tE

=-

j

-

?A2．規(guī)范變換電磁勢的不確定性j￠=

j

-

?y?tA￠=

A

+

y規(guī)范變換·

y

=

0·

A￠=

·

A-=

-?t?

y

=

?y?t?t?tj￠-

?A￠j

-

?A3．電磁勢的微分方程D

=

r?t·

H

=

J

+

?D?t2j

+

?

A

=

-

r

e01

?j

-

c2?t?t

21

?2

A·(

·

A)=

m0

J

-

c2(=·

·

AA)-2

A201

?j

-?t

2?t1

?2

AA

-c2A

+c2=

-m

J2201

?2j

?

1

?j

rj

-c

?t

2

+

?t

?t

=

-

eA

+c2Coulomb規(guī)范

A

=

020c2e?t

2?t1

?2

A

1

?A

-c2-

j

=

-m0

J2j

=-

rE

=

EL

+

ETL?AE

=-

j

ET

=-?tLorentz規(guī)范1

?j=

0?tA

+c2220c2?t

22j

-1

?

j

=

-

r?t

2

e1

?2

AA

-c2=

-m0

Jd’Alembert方程有源波動方程例1

平面電磁波的電磁勢解法J

=

0

r

=

0222=

0?t

21

?2jj

-=

0c

?t

21

?2

AA

-c2自由空間波動方程平面波解((00i

k

x

-w

ti

k

x

-w

tA(x,

t

=

A

ej

(x,

t

=

j

ew

=

ck1

?jA

+

=

0c2

?t?fi

ikc2fi

-iw j0

=

k A0

=

cek

A0?t

wB

=

·

A

=

ik

·

A?t?AE

=-

j

-c2w=

-ikj

+iw

A

j0

=

k

A02ic2

ic2

kE

=-

(k

A)k

-

k

A

=

-

k

·(k

·

A)=

-c

·

Bw

w

k洛倫茲規(guī)范下的剩余規(guī)范自由度A0￠=

A0

+a

k

j0￠=

j0

+awk A

=

0B

=

ik

·

Aj

=

0

E

=

iw

A2c2-?t

2?tj

=

02j

=

01

?2

A

1

?A

-c2庫侖規(guī)范(0i

k

x

-w

tj

(x,

t=

0

A(x,

t

=

A

eA

=

0

k A

=

0B

=

ik

·

AE

=

iw

A§2

推遲勢2201

?2j

r?t

2?t

2

=

-

e1

?2

AA

-c2=

-m0

J

j

-

c200e2j

=-

r2

A

=

-m

J變化電荷和電流分布J

r電磁勢A

j靜場14π

r4πe

rV

￠

0

V

￠Jr

x￠

dV

￠A(x

)=

m0

x￠

dV

￠

j

(x

)=積分遍及全部電荷和電流的分布區(qū)域Poisson方程d’Alembert方程?tj

-

?AB

=

·

A E

=

-1．變化點電荷的標勢原點處的變化點電荷221

?2j=

0c

?t

2x

?

0

j

-2203Q

(teQ

t

x

=

01

?2j

j

-dV

=

-c

?t

2

V

=4

πR3

fi

0

x

=

0球?qū)ΨQ性j

=

j

r,

tj

=

j

r,q,f,

t?

+

?q

?

+

?f

?

?

=

?r?xi

?xi2222221sinq1

?2?r

?xi

?q

?xi

?f?21

?

?

=r

++r

?rr

sinq

?q?qr

sin

q

?f1

?21

?2j=

0r

?r

2

(rj

)-

c2

?t

2x

?

0u

r,

t=

rj

r,

tc2?2u

1

?2u-=

0?r

2

?t

2一維波動方程通解u

=

f

t

-

r

+

g

t

+

r

c

c

rf t

-

r

c

g t

+

r

cj

=

+向外發(fā)散的球面波r向內(nèi)匯聚的球面波rt

-

r

cj

(r,

t

)=

f221

?2jj

-=

0c

?t

2任意函數(shù)形式

f

均滿足

x

?

02203Q

(te1

?2j

j

-dV

=

-c

?t

2V

=4

πR3

fi

0

x

=

0決定f202

11Q

(te1

?2

f

f

+r

f

r

+

2dV

=

-f

-c2

r

?t

2

Rfi

0

R21r

Rrrf

(t

-

r

c)

2

1

dV

=

f

(t

)Rfi

0Rfi

0Rfi

02

1

dV

=

f

(t

)

1

dSr2Rrr3RRfi

0Rfi0Rfi0dS

=

-4πr1

dS

=

-dS

=

-lim

14πe0f

(t

)=

Q

t4πe0f

(t

-

r

c)=

Q t

-

r

c4πe0rj

(r,

t

)=

Q t

-

r

cQ

x￠,

t4πe0

rj

(x,

t

)=

Q

x￠,

t

-

r

cx

點處的變化點電荷r

=

x

-

x￠2．變化電荷和電流分布的電磁勢14πe

0

V

￠r

x￠,

t

-

r

c

dVj

(x,

t

)=dQ

x￠,

t

-

r

c

=

r

x￠,

t

-

r

c

dV線性方程的疊加性20c21

?2j

rj

-=

-?t

2

e0c2?t

21

?2

A2

A

-=

-m

J4π

rV

￠rJ

x￠,

t

-

r

c

dVA(x,

t

)=

m0

推遲勢r

=

x

-

x￠電磁場以光速傳播，源對場點的物理作用有時間上的推遲恒定電荷電流分布情形，推遲勢回復(fù)到靜電標勢和靜磁矢勢3．推遲勢滿足洛倫茲規(guī)范的證明c2?tA

+

1

?j

=

01rrdV

￠4πe

0

V

￠m

4π

V

￠r

x￠,

t

-

r

c

dVj

(x,

t

)=J

(x￠,

t

-

r

c)A(x,

t

)=

0

r4π

V

￠J

x￠,

t

-

r

c

dVA(x,

t

)=

m0

1rJ

x￠,

t￠

1=

J

(x￠,

t￠)+

J

(x￠,

t￠)

r

rt￠=

t

-

r

c(

)(t?t￠

x￠

?J

x￠,

t￠J

x￠,

t￠

=f

r

=-

￠f

rr

=

x

-

x￠1r￠￠￠

￠

￠=-

￠r

?t

x￠J

(x￠,

t￠r1

?J

(x￠,

t￠

t

-

J

(x

,

t

)1rr￠￠

￠ ￠

=-

1rJ

x￠,

t￠J

(x

,

t

)￠

J

(x￠,

t￠)￠t

t￠?t￠

x￠

?J

(x￠,

t￠

+￠

J

(x￠,

t￠)=

￠

J

(x￠,

t￠)rt￠J

x￠,

t￠

J=-

￠x￠,

t￠

1r

+

r

￠

J

(x￠,

t￠)r￠V

￠S

￠J

x￠,

t￠J

x￠,

t￠rdS￠=

0dV

￠=區(qū)域邊界J

x￠,

t￠

=

0t￠1

4π

V

￠rA(x,

t

)=

m0

￠

J

(x￠,

t￠)

dV

￠1r?t4πe

?t0

V

￠?

r

x￠,

t

-

r

c

dV?

j

(x,

t

)=?

f

(t￠)=

?

f

(t￠)?t

?t￠t￠=

t

-

r

cdVr

?t?t￠4πe

0

V

￠

x￠1

?r

(x￠,

t￠

?

j

(x,

t

)=

1

(

)t￠

?t￠

x￠

?r

(x￠,

t￠

￠

Jx￠,

t￠

=

-電流連續(xù)性方程c2?tA

+

1

?j

=

0§3

諧振蕩電流的電磁場微觀變速帶電粒子宏觀

變化電流分布輻射電磁波1．單頻諧振電流的電磁場r4π

V

￠J

x￠,

t

-

r

c

dVA(x,

t

)=

m0

(0-iw

tJ

(x￠,

t

=

Jx￠ek

=

w

c(

)

0

0

0

dVrm4π

V

￠J

(x￠

eikrA

x

=(0-iw

tA(x,

t=

A

x

e(0-iw

tr

(x￠,

t

=

rx￠e?r?tJ

=-00iwr

=

JB

=

·

A1r4πe

0

V

￠r

x￠,

t

-

r

c

dVj

(x,

t

)=(0-iw

tj

(x,

t

=

jx

e01

0

dVr4πe

0

V

￠r

(x￠

eikrj

(x

)=?tE

=-

j

-

?A(0-iw

tB

(x,

t

=

Bx

e00B

=

·

A(0-iw

tE

(x,

t

=

Ex

e0E

=-

j0

+iw

A0c2?t·

B

=

1

?EkE

=

i

c

·

B單頻諧振源激發(fā)同頻諧振電磁場?t?E

=

-iw

E2．推遲勢的多極展開(0-iw

tJ

(x￠,

t

=

Jx￠e(0-iw

tA(x,

t=

A

x

e(

)

0

0

0

dVrV

￠m

4π

J

(x￠

eikrA

x

=R

=

xr

=

x

-

x￠Vx

OxrRR+R

x￠1

+

=

1

1+

eR1

=

1

-

x￠r

Rr

=

R

-

x￠R

+

=

R

-

eR

x￠+eikrR=

eikR

(1-

ikex￠+Rxm￠ax

1lxm￠ax

1kl

=

2π近區(qū)

xm￠ax

R

l(

)00m4π

rV

￠J0x￠

dVA

x

?0r04πe

r0

V

￠x￠

dVj

(x

)?

1

推遲效應(yīng)可忽略準靜（似穩(wěn)）場源與場關(guān)系類似于靜場的庫侖形式eikr=

eikR

(1-

ikeRx￠+

?

eikR

?1r

R

RR+R

x￠1

=

1

-

x￠

1

+

=

1

1+

eR(

)04πR

R

RV

￠1+m

e-iwt

ex￠A(x,

t

)=

0

Jx￠+

dVRe-iwt4πe0

R

V

￠ex￠j

(x,

t

)=r0

(x￠)1+

R

+dV感應(yīng)區(qū)xm￠ax

R

lR+R

x￠1

=

1

-

x￠r

R

R1

+

=

1

1+

eReikr=

eikR

(1-

ikeRx￠+展開式對應(yīng)各級同等重要過渡區(qū)域遠區(qū)xm￠ax

l

R1

r

?1

R(

)(0R4πRi(kR

-w

tm0eV

￠A(x,

t

)=Jx￠1-

ikex￠+)dV(

)04πe

Rei(kR-w

tV

￠j

x,

t

=r0

(x￠)(1-

ikeRx￠+)dV(

)ikr4πRV

￠A(x,

t

)?

m0J

x￠,

t

e dV

￠(

)j

1

4πe0

RV

￠x,

t

?r

(x￠,

t

)eikr

dV輻射場各向異性球面波§4

電偶極輻射(

)(

)(

)10

0

4πR

0

4πRmm

ei(kR

-w

tV

￠V

￠dV

￠=A x,

t

=Jx￠J

x￠,

t￠

dV1．輻射場t￠=

t

-

R

cn

n

n

nn

ndtV

￠J

(x￠,

t￠)dV

￠=

Q

v￠(t￠)=

d

Q

x￠(t￠)=

p

(t￠)V

￠p

(t￠

=

x￠r

(x￠,

t￠

dV((0i

kR-w

tt￠p=

p

e00V￠p

=

x￠r

(x￠

dV(

)

0 0

1i4π

Rm

w

p

ei(kR

-w

tA x,

t

=

-振蕩電偶極矩產(chǎn)生的輻射04πeikRRm

w

e-iwtB

=

·

A

=

-i

0

·

p1eikReikReikRikRRR

RRReikR

1

=+e=R

Rik

-

Re

?

ikefi

ikeR04πe

c3

Rw

2e·

p

ei(kR-w

tB=

R

0

ckE

=

i·

B

=

-ceR

·

B00R

RR4πe

c2ei(kR

-w

tw

2e

·(e

·

pE

=-2．輻射能流與功率00Rfw

2

p4πe

c3cos

kR

-wtB

=-

sinqe00Rqw

2

p4πe

c2cos

kR

-wtE

=-

sinqe1p輻射場是橫場

RzkBEq001RR2w

4

p2m16π2e

c3cos2

(kR

-wtS

=

E

·

B

=

0

sin2qe0Rw

4

p2sin2q32π2e

c3

R2S

=

0

eP

=

S

4

πR2S

dσ=π00000w

4

p2w

4

p232π

e

c

R12πe

c32πR2sin3qdq

=P

=2

3 2

平均能流輻射功率S

1

R2輻射功率與傳播距離無關(guān)，電磁能可傳播至無限遠處。§5

磁偶極和電四極輻射1．輻射場t￠=

t

-

R

c(020

RR4πR4πRkm

ei

kR

-w

tV

￠V

￠A

(x,

t

)=

-iJ

(x￠)(e

x￠)dV=

-i

km0

ex￠J

(x￠,

t￠)dV

￠x￠J

=

1

(x￠J

+

Jx￠)+

1

(x￠J

-

Jx￠)2eR

x￠J

-

Jx￠

=

eR2x￠

J

-

eR

Jx￠=

-eR

·

x￠·

J(

)12V

￠t￠=mx￠·

J

(x￠,

t￠)dV((0i

kR

-w

tt￠m=

m

e002V

￠=

1mx￠·

J

(x￠)dVV

￠2

1

x￠J

(x￠,

t￠)+

J

(x￠,

t￠)x￠

dV

￠=

1

(t￠)D6V

￠D

(t￠

=

3x￠x￠r

(x￠,

t￠dV

D

(t￠

=

D0ei(kR-w

t0V

￠D0

=

3x￠x￠r

(x￠

dV2MR4πc

Rei(kR-w

t

)=

i

0

R

0

4πR

V

￠2

A