往届学习供参考远经典物理电动力学

第五章电磁波的辐射电磁波的激发变化的电荷和电流分布激发变化电磁场变化电磁场对源的依赖关系¶tD

=

r¶B·

E

=

-¶tB

=

0·

H

=

J

+

¶DD

=

e

E0B

=

m0

H真空§1

电磁场的矢势和标势1．一般电磁场的势B

=

0B

=

·

A矢势¶t·

E

=

-

¶B¶t

·

E

+

¶A

=

0¶tE

+

¶A

=

-

j标势¶tE

=-

j

-

¶A2．规范变换电磁势的不确定性j¢=

j

-

¶y¶tA¢=

A

+

y规范变换·

y

=

0·

A¢=

·

A-=

-¶t¶

y

=

¶y¶t¶t¶tj¢-

¶A¢j

-

¶A3．电磁势的微分方程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规范

A

=

020c2e¶t

2¶t1

¶2

A

1

¶A

-c2-

j

=

-m0

J2j

=-

rE

=

EL

+

ETL¶AE

=-

j

ET

=-¶tLorentz规范1

¶j=

0¶tA

+c2220c2¶t

22j

-1

¶

j

=

-

r¶t

2

e1

¶2

AA

-c2=

-m0

Jd’Alembert方程有源波动方程例1

平面电磁波的电磁势解法J

=

0

r

=

0222=

0¶t

21

¶2jj

-=

0c

¶t

21

¶2

AA

-c2自由空间波动方程平面波解((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洛伦兹规范下的剩余规范自由度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库仑规范(0i

k

x

-w

tj

(x,

t=

0

A(x,

t

=

A

eA

=

0

k A

=

0B

=

ik

·

AE

=

iw

A§2

推迟势2201

¶2j

r¶t

2¶t

2

=

-

e1

¶2

AA

-c2=

-m0

J

j

-

c200e2j

=-

r2

A

=

-m

J变化电荷和电流分布J

r电磁势A

j静场14π

r4πe

rV

¢

0

V

¢Jr

x¢

dV

¢A(x

)=

m0

x¢

dV

¢

j

(x

)=积分遍及全部电荷和电流的分布区域Poisson方程d’Alembert方程¶tj

-

¶AB

=

·

A E

=

-1．变化点电荷的标势原点处的变化点电荷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球对称性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

2一维波动方程通解u

=

f

t

-

r

+

g

t

+

r

c

c

rf t

-

r

c

g t

+

r

cj

=

+向外发散的球面波r向内汇聚的球面波rt

-

r

cj

(r,

t

)=

f221

¶2jj

-=

0c

¶t

2任意函数形式

f

均满足

x

„

02203Q

(te1

¶2j

j

-dV

=

-c

¶t

2V

=4

πR3

fi

0

x

=

0决定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

点处的变化点电荷r

=

x

-

x¢2．变化电荷和电流分布的电磁势14πe

0

V

¢r

x¢,

t

-

r

c

dVj

(x,

t

)=dQ

x¢,

t

-

r

c

=

r

x¢,

t

-

r

c

dV线性方程的叠加性20c21

¶2j

rj

-=

-¶t

2

e0c2¶t

21

¶2

A2

A

-=

-m

J4π

rV

¢rJ

x¢,

t

-

r

c

dVA(x,

t

)=

m0

推迟势r

=

x

-

x¢电磁场以光速传播，源对场点的物理作用有时间上的推迟恒定电荷电流分布情形，推迟势回复到静电标势和静磁矢势3．推迟势满足洛伦兹规范的证明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

¢=区域边界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¢

=

-电流连续性方程c2¶tA

+

1

¶j

=

0§3

谐振荡电流的电磁场微观变速带电粒子宏观

变化电流分布辐射电磁波1．单频谐振电流的电磁场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单频谐振源激发同频谐振电磁场¶t¶E

=

-iw

E2．推迟势的多极展开(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近区

xm¢ax

R

l(

)00m4π

rV

¢J0x¢

dVA

x

»0r04πe

r0

V

¢x¢

dVj

(x

)»

1

推迟效应可忽略准静（似稳）场源与场关系类似于静场的库仑形式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感应区xm¢ax

R

lR+R

x¢1

=

1

-

x¢r

R

R1

+

=

1

1+

eReikr=

eikR

(1-

ikeRx¢+展开式对应各级同等重要过渡区域远区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辐射场各向异性球面波§4

电偶极辐射(

)(

)(

)10

0

4πR

0

4πRmm

ei(kR

-w

tV

¢V

¢dV

¢=A x,

t

=Jx¢J

x¢,

t¢

dV1．辐射场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

=

-振荡电偶极矩产生的辐射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．辐射能流与功率00Rfw

2

p4πe

c3cos

kR

-wtB

=-

sinqe00Rqw

2

p4πe

c2cos

kR

-wtE

=-

sinqe1p辐射场是横场

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

平均能流辐射功率S

1

R2辐射功率与传播距离无关，电磁能可传播至无限远处。§5

磁偶极和电四极辐射1．辐射场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