7.1 、 potentials of electromagnetic field, gauge invariance 7.2 、 d’Alembert equation and retarded potential
7.3 、 electric dipole radiation
7.4 、 EM radiation from arbitrary motion charge
Ch 7 Radiation of Electromagnetic Waves
1. What is EM radiation
EM field is excited by time-dependent charge and currents. It may propagate in form of waves. The problem is usually solved in terms of potentials.
2 . It is a boundary value problem
Source (charge and current) excites EMF, EMF in turn affects source distribution --- boundary value problem!
For convenience, our discussions are limited to a simple case –Distribution of source is known.
特征:与 1/r 正比的电磁场!
§7.1 vector potential and scalar potential
potentials are slightly different from the static cases
since , we can introduce vector
potential as the static field,
0 B
A
AB
1.a ) vector potential
Since , scalar potential can not
be defined as before
0
t
BE
1.b ) scalar potentialAB
t
AA
tE
6
0)(
t
AE
t
AE
Define scalar func
t
AE
22)) .. Gauge invarianceGauge invariance
t
AAA
),( A
Potentials are not uniquely determined,
they differ by a gauge transformation.
Gauge: Given a set of
give identical electric and magnetic fields' '( , )A
Prove: since and , and can not change E
and B, so
A
A
B A A A
AA
t
A
ttt
A
t
AE
)(
A
t
t
规范不变性规范不变性:在规范变换下物理规律满足的动力学方程保持不变的性质(在微观世界是一条物理学基本原理)。
Coulomb gauge condition 0 A
3) . Two typical gauges
transverse ( 横场 ) , longitudinal ( 纵场 ) 。 is determined by instantaneous
distribution of charge density (similar to
static coulomb field)
A
To reduce arbitrariness of potential, we give
some constraint --- Gauge fixing 。
Symmetry or explicit physical interpretation
02 Function satisfies
Lorenz gauge
condition 012
tc
A
satisfy manifest relativistic covariant equations
,A
Function satisfies
01
2
2
22
tc
0)1
()1
(2
2
22
2
tctc
A
0 AA
02 Prove
01
2
2
22
tc
2
2
222
111
tctcA
tcA
prove :
Ludwig Lorenz
22
02 2 2
2
0
1 1( )
( )
AA A J
tc t c
At
Prove : substitute , into
Maxwell eqs
And using
AB
t
AE
0000 ,
EJt
EB
AAA
2)()(
4) . D’Alembert equation
22
02 2 2
2
0
1 1AA J
tc t c
So satisfies Poisson equation as in static case. instantaneous interaction?
4.a) Under coulomb gauge
4.b) Under Lorenz gauge
2 22 2
02 2 2 20
1 1AA J
c t c t
洛仑兹规范下的达朗贝尔方程是两个波动方程,因此由它们求出的 及 均为波动形式,反映了电磁场的波动性。
),( A
),( BE
wave properties
highly symmetric and independent to each other
Get one, get 2nd for free.
Solution of d’alembert eq under Lorenz gauge indicates that EM interaction takes time.
To study radiation, we use Lorenz gauge.
§7.2§7.2 Retarded potentialRetarded potential
AAssumessume is known. We first solve is known. We first solve
point charge problempoint charge problem ,, then use then use
superposition to get general solutionsuperposition to get general solution
),( tx
1. Solve d’Alembert equation1. Solve d’Alembert equation
Assume point charge at Assume point charge at originorigin , ,, , symmetry indicates symmetry indicates is independent of is independent of
, so d’Alembert eq for scalar potential is , so d’Alembert eq for scalar potential is
)()(),( xtQtx
),( tr
,
asas 0r2
22 2
1 1( ) 0r
r r r c t
22
2 2 20
1 1 ( ) ( )( )
Q t rr
r rr c t
**
r
trutr
),(),(
2 2
2 2 2
10
u u
r c t
letlet
rcrtf )(
rcrtg )(
Outward spherical waveOutward spherical wave
Inward waveInward wave
The general solution for 1D wave equation isThe general solution for 1D wave equation is
)()(),(c
rtg
c
rtftru
rcrtg
rcrtf
tr)()(
),(
)0( r
Compared with static potential , we have :rcrtQ
tr04
)(),(
0)( c
rtg
For radiationFor radiation
If point charge is placed at x
rcrtxQ
tx04
),(),(
容易证明上述解的形式满足波动方程*式
For contineous charge distribution
0
( , )( , )
4V
rx t cx t dVr
Vdr
crtxJ
txAV
),(
4),( 0
Since satisfies identical equation as ,
so the solution :A
2.Show the solution 、 satisfies Lorenz conditionA
2
10A
tc
证:令 ( , , )rt t t t x xc
A
Vdr
txJV
),(
40
0 1 1[ ( , ) ( , ) ]
4J x t J x t dV
r r
( , ) 1 ( , ) 1 ( , )( , )
J x t J x t J x tJ x t t r r
t c t c t
r r ( , )
( , ) ( , )
1 ( , )( , )
t c
t c
J x tJ x t J x t t
t
J x tJ x t r
c t
),(),(),( txJtxJtxJct
A
VdtxJr ct
),(1
40
Vd
rtxJtxJ
r]
1),(),(
1[
0 1( , )
4 t c
J x t dVr
( , )J x t
dVr
( , )
0S
J x tdS
r
2 20
1 1 1 ( , )
4 V
x t tdV
t r t tc c
0 1 ( , )
4 V
x tdV
r t
tc
A
2
10 1 ( , )
[ ( , ) ] 04 t c
x tJ x t dV
r t
0 电荷守恒定律
The value of the retarded potential at , depends on charge/current distribution at .
xt
crt
physical excitation at reaches observation point by .
And the speed of signal traveling in vacuum is c.
crt
t
33 .. Physical interpretationPhysical interpretation
Electromagnetic interaction takes time!
§7.3§7.3 Electric dipole Electric dipole radiationradiation
we limit our discussion to charge distribution of periodic motion. Furthermore,size of charge distribution is much smaller than the distance between charge and the observation point.
电磁波是从变化的电荷、电流系统辐射出来的。电磁波是从变化的电荷、电流系统辐射出来的。Antenna with high frequency alternative electric Antenna with high frequency alternative electric current current
Non-uniform moving charged particlesNon-uniform moving charged particles
Substitute into retarded solution ( )ck
0 0( , / ) ( )( , ) [ ]
4 4
ikri tJ x t r c J x e
A x t dV dV er r
let , so0 ( )( )
4
ikrJ x eA x dV
r
( , ) ( ) i tA x t A x e
1. General formula for radiation field1. General formula for radiation field
Compared with static case, there is an additional phase factorikre
( , ) ( )
( , ) ( )
i t
i t
J x t J x e
x t x e
Charge/currentCharge/current :: 随时间正弦随时间正弦或余弦变化或余弦变化
Similarly,
)()(2
xAic
x
Satisfy Lorenz condition
Electromagnetic fields are
tiexBtxAtxB )(),(),(
),(),( txBk
ictxE
( ) 0J
0 0 0E
B Jt
0
( )( , ) [ ]
4
ikri tx e
x t dV er
( , ) ( ) i tx t x e
Assume source to field point distance Assume source to field point distance
(size of charge/current distribution), so(size of charge/current distribution), so
Perform power expansion around
x
lr
rR
xxr
110x
...1
...1
...111
23
R
xn
RR
xR
Rx
RRr
22 .. Multiple expansionMultiple expansion
Where is unit vector along ,n
R
R x
1) . Power expansion for small size of source
(R is distance (R is distance between center of between center of coordinate and field coordinate and field point)point)
xnRR
xnR
RxnR
xnR
Rr
)1()
/1
1(
2
VdxnR
exJxA
xnRik
)(0 )(
4)(
0 ( )( )
4
ikrJ x eA x dV
r
2 n x
Since , so in denominator can be neglected. But it maybe important in phase,
R x n x xn
because is not necessary small, compared with2
Keep first two terms, we have
VdxnikxJR
exA
ikR...)1)((
4)( 0
xl 2
2n x
kn x
if ,
( , ) ( )
( , ) ( )
i t
i t
J x t J x e
x t x e
The radiation field for is
The first term dominates
VdxJR
exA
ikR)(
4)( 0
Electric dipole radiation
Under , , we can further divide into three cases according to and
2) . 与 的关系R Rl l
R
a ) (近区)R2
, 1ikRR ek
R
t Tc
, Time of propagation
EM field is similar to the static case.
c ) (远区,即辐射区)REM waves propagates away from the source. interested
Rb ) (感应区)Very complicate.
0( , )4
ikReA x t p
R
1) . Re-express in terms of dipole momentp
3 . Electric dipole dipole radiationradiation
0( ) ( )4
ikReA x J x dV
R
Vdtxxp ),(
( , )p J x t dV
2) . Electric and magnetic fields
. . .0 0( ) ( )
4 4
ikR ikR ikRe e eB A p p p
R R R
0
3
1 1 1 1( ) ( )
ikRe R nikR ikR ikR ikR ikRe e e ike ik eR R R R R RR
ki
nkk
here
Rk
1
R
11
R
nR
ike
R
e ikRikR
Consider 远区 , ,即 ,
so
0( , )4
ikRikeB x t n p
R
( , )p J x t dV
.. .p i p
it
. ..
/p i p
( )Magnetic induction
Rl
30
( , )4
ikReB x t p n
c R
/ ,k c 0 01/c
30
20
1( , ) sin
4
1( , ) sin
4
ikR
ikR
B x t p e ec R
E x t p e ec R
In spherical coordinates, Let along axisp
z
),(),( txBk
ictxE
20
( , ) ( )4
ikReE x t p n n
c R
using
电场线是经面上的闭合线
Discussion:( 1 ) E oscillates along longitude and B along latitude lines. Direction of propagating, E and B are orthogonal to each other (right-hand).
( 2 ) E,B are proportional to , so they are propagating spherical waves. They are transverse ( TEM 波) and maybe regarded as plane wave as .
( 3 ) without ( ) , it can be shown that E is no longer perpendicular to k, electric lines are not close, but magnetic lines are still close ( TM 波) .
1/ R
R
R1 1
R
4 . Energy flux, angular distribution and power of
radiation
22 40
3012
pP SR d
c
Average power
1) 。与球半径无关,能量可以传播到无穷远。2) 。与电磁波的频率 4 次方成正比。
* 202 3 2
0
1Re( ) sin
2 32
pS E H n
c R
Average energy flux vector ( 平均能流密度矢量 )
角分布
Example: Short antenna
2,)
21(),( 0
lzez
lItzI ti
lIdzzIpl
l
0
2/
2/ 2
1)(
rRIl
Ic
lIP 2
0
220
0
0222
00
2
1
1248
2
197l
Rr 短天线辐射能力不强。通常天线长度与波长同数量级,不能用简单的偶极辐射公式。
例:半波天线(长度为半波长)0 ( )
( )4
ikrJ x eA x dV
r
必须直接用推迟势计算
天线电流要与外场联合作为边值问题求解,一般较复杂。对于细长直天线,电流分布应是驻波,两端是波节。如
kzIzlzlkI
lzzlkIzI l cos
02/),2/(sin
2/0),2/(sin)( 0
2/
0
0
z
ikR
ikzikR
zz
zRrikr
z
ekR
eI
dzkzeR
eIedzkzI
r
eeA
200
cos4/
4/
00cos0
4/
4/
0
sin
)cos2
cos(
2
cos4
cos4
ekR
eIcinBcE
ekR
eIiB
ikR
ikR
sin
)cos2
cos(
2
sin
)cos2
cos(
2
00
00
辐射电磁场
2
2
2
22
200* sin,
sin
)cos2
(cos
8)Re(
2
1tosimilarn
R
IcHES
2.73
444.2
844.2
0
2002
cR
IcdRnSP
r
The energy flux
要得到高度定向的辐射,可利用天线阵的干涉效应。张角
)/(sin 1 Nl
§7.4§7.4 Magnetic dipole radiation Magnetic dipole radiation and electric quadrupole radiation and electric quadrupole radiation
VdxnikxJR
exA
ikR...)1)((
4)( 0
)]()[(2
1)()( xJJxxJJxnJxnxnJ
Dnxxedt
dnxxne
dt
d
xdt
xdn
dt
xdxnexJJxnVd
D
6
1
2
1)(
2
1
])()[(2
1)[(
2
1
mnJxVdn
xJnJxnVdxJJxnVd
2
1
])()[(2
1)(
2
1
]6
1[
40)2( DnmnR
eikA
ikR D
nBcBk
icE
AnikAB radiation
电四极辐射电四极辐射
磁偶极辐射磁偶极辐射
0( , )4
ikReA x t p
R
§7.5§7.5 Radiation from a localized charge in Radiation from a localized charge in arbitrary motion (Bo-p124)arbitrary motion (Bo-p124)
xxrcrttwhere
r
xtjxdxtA
r
xtxdxt
V
V
,/
),(
4),(
),(
4
1),(
30
3
0
method1 : use the retarded potential粒子看作小体积电荷分布,直接积分
, ( )et x t
( )v t
( , )P t xr
0~
,~4~
0
Ar
e
rc
ve
c
vA
r
e
~4~
~4~
20
2
0
method2 : using Lorentz transformation
)(
)]()([
)~~(~
rr
xxttc
ttcr
)(4
)(4
0
0
rrc
eA
rr
e
Lienard-Wiechert potential
At rest frame
where
At Lab frame
1 ( ) 1 ( )1 1
11
1
t r t t r x t t
t c t t cr t tv r t t
cr t t n
,A
E B At
r
vr
t
xrrt
r
txxc
tc
rtt
)(1
))((1 2
Derivative was performed with respective to t , x.However
( ), ( )et r x x t
2 3 3 20 0
2 3 3 3 2 30 0 0 0
|4 4
( ( ))4 4 4 4
t const
A ev r ev rB A t
t c r c r
ev er er eE t r r v
c r r t r c r
c
nt
t
tFor
,1,0
1 1( | )
( / )
t const
rt r r t
c c tr v r r
t tcr cr c r r v c
, ,Let p ex p ev identical to electric dipole
Enc
Bvrrrc
eE
1)),((
4 320
vandrbetweenangle
nrc
veHES
)(sin16
223
02
22
在与加速度垂直方向辐射最强!Q :圆周运动和直线运动时的辐射方向?
Radiation field from a relativistic charged particle
nn
vnn
rc
encES
6230
2
22
0)1(
|))((|
16
当 v 趋于光速,辐射集中于朝前方向,张角为
1
drtd
dtnStP 2)(
Radiation power
Enc
B
n
vnn
rc
eE
1
)1(
))((
4 320
Radiation field for arbitrary moving charge particle
30
226
30
22
2
3
66)(
cm
Fe
c
vetP vmF
Acceleration is parallel to velocity( 轫致辐射)
Bremsstrahlung and Synchrotron radiation
2 2 22
2 3 50
( ) sin
16 (1 cos )
dP t dt e vr S n
d dt c
Angular distribution
32 3/ 22 (1 )1
d mv mvF mv
dt
where
Radiation power
Acceleration is perpendicular to velocity(同步辐射)
2
30
224
30
22
2
66)(
cm
Fe
c
vetP vmF
辐射功率与粒子能量平方正比 !
2 2 2 2 2 2
2 3 50
( ) (1 cos ) (1 )sin cos
16 (1 cos )
dP t e v
d c
Angular distribution
Power
where 21
mvF mv
e.g. BEPC E= 2.8GeV , =5479
高能加速器设计与同步辐射光源