Chapter 8
Radiation by Moving Charges
8.1
Introduction
The problem of radiation of electromagnetic waves by a single charged particle moving at an
arbitrary velocity had correctly been formulated independently by Lienard and Wiechert before
the advent of the special relativity theory. This is because once emitted from a charged particle,
electromagnetic waves propagate at the speed c irrespective of the velocity of the charged particle,
just as sound waves propagate at a speed independent of the source velocity. (The major …nding
made by Einstein was that electromagnetic waves still propagate at the speed c regardless of the
observer’s velocity in contrast to the case of sound waves.)
The scalar and vector potentials due to a moving charge can be found rigorously using the
Green’s function for the wave equation. Then radiation electromagnetic …elds can readily be calculated. In nonrelativistic regime, the radiation power only depends on the acceleration of charged
particles. As the velocity approaches c; however, signi…cant increase in the radiation power occurs.
Furthermore, in highly relativistic limit, radiation occurs primarily along the direction of velocity
within an angular spread of order ' 1= about the velocity irrespective of the direction of acp
2
is the relativity factor with = v=c: Hence radiation frequency
celeration. Here = 1= 1
is subject to strong Doppler shift. For example, in synchrotron radiation due to highly relativistic
electron beam bent or undulated by a magnetic …eld, radiation even in hard x-ray regime can be
created.
In material medium, radiation processes without acceleration on charged particles are possible.
If the velocity of a charged particle exceeds the velocity of electromagnetic waves in the medium
v>p
1
"
;
0
where " is the permittivity, Cherenkov radiation occurs. Furthermore, if a charged particle crosses a
boundary of two dielectric media, the transition radiation occurs even if the condition for Cherenkov
radiation is not met. Transition radiation is due to sudden change in the normalized velocity from
p
p
1 = v "1 0 to 2 = v "2 0 which may be regarded as an e¤ective acceleration even though the
1
particle velicty v remains constant.
8.2
Lienard-Wiechert Potentials
The charge and current densities of a moving point charge are singular and described by
(r; t) = e [r
rp (t)];
J(r; t) = ev(t) [r
(8.1)
rp (t)];
(8.2)
where e is the charge, rp (t) is the instantaneous location of the charge and v(t) = drp (t)=dt is the
instantaneous velocity of the charge which may be changing with time. Exploiting the Green’s
function for the wave equation,
G(r
r0 ; t
t0 ) =
1
4 jr r0 j
t
jr
t0
c
r0 j
;
(8.3)
we can write down solutions for the inhomogeneous wave equations,
1 @2
c2 @t2
r2
in the form
e
(r; t) =
4 "0
A(r; t) =
0e
4
r2
1 @2
c2 @t2
Z
0
Z
dV
dV 0
Z
Z
A=
dt0
dt0
=
1
jr
(8.4)
0 J;
(8.5)
[r0 rp (t0 )] [f (t0 )];
t+
jr
c
r0 j
:
The volume integrations can be carried out immediately with the results
Z
e
1
(r; t) =
[f (t0 )]dt0 ;
4 "0
jr rp (t0 )j
A(r; t) =
0e
4
Z
v(t0 )
[f (t0 )]dt0 ;
jr rp (t0 )j
where f (t0 ) is now
f (t0 ) = t0
(8.6)
v(t0 )
[r0 rp (t0 )] [f (t0 )];
jr r0 j
where
f (t0 ) = t0
r0 j
;
"0
t+
2
jr
rp (t0 )j
:
c
(8.7)
(8.8)
(8.9)
(8.10)
The integral involving the delta function can be simpli…ed as
Z
Z
0
0
0
g(t ) [f (t )]dt =
g(t0 ) [f (t0 )]
g(t0 )
jdf =dt0 j
=
1
df
df (t0 )
dt0
;
(8.11)
f =0
where t0 is now understood as a solution for t0 satisfying f (t0 ) = 0 or
t0
t+
rp (t0 )j
= 0:
c
jr
(8.12)
This is in general an implicit equation for t0 : The time derivative of f (t0 ) is
n(t0 ) vp (t0 )
=1
c
df
=1
dt0
n(t0 )
where n(t0 ) is the unit vector along the relative distance r
time t and time t0 are related through
n(t0 )
[1
or
dt
=1
dt0
(t0 );
rp (t0 ) and
(8.13)
= v=c: The observing
(t0 )]dt0 = dt;
n(t0 )
(t0 ):
After performing time integration, we …nally obtain
(r; t) =
e
4 "0 1
A(r; t) =
0e
4
1
1
n(t0 )
1
n(t0 )
1
e
1
=
;
rp (t0 )j
4 "0 (t0 ) jr rp (t0 )j
(t0 ) jr
v(t0 )
v(t0 )
0e
=
;
(t0 ) jr rp (t0 )j
4
(t0 ) jr rp (t0 )j
(8.14)
(8.15)
where
(t0 ) = 1
n(t0 )
(t0 ):
(8.16)
These retarded potentials, called Lienard-Wiechert potentials, had been formulated in 1898. They
are applicable to arbitrary velocity of the charged particle. Retarded nature of the potentials clearly
appears in the condition that all time varying quantities, rp (t0 ); v(t0 ); n(t0 ); must be evaluated at
t0 ; not at the observing time t because of …nite propagation speed of electromagnetic disturbance.
Having found the retarded potentials, we are now ready to calculate the electromagnetic …elds
due to a moving point charge. The electric …eld is to be found from
E(r; t) =
r
@A
;
@t
(8.17)
where spatial and time derivatives pertain to r (the coordinates of the observing location) and
3
t (observing time). Since the potentials
and A are implicit functions of r and t; it is more
convenient to use the original integral representations, Eqs. (8.8) and (8.9), respectively, for proper
di¤erentiation with respect to r and t: For example, the spatial derivative of the scalar potential
can be performed as follows. Letting R(t0 ) = jr rp (t0 )j ; and introducing a unit vector in the
direction r rp (t0 );
r rp (t0 )
n(t0 ) =
;
(8.18)
jr rp (t0 )j
we …nd
r
=
=
=
=
Z
1
e
r
[f (t0 )] dt0
4 "0
jr rp (t0 )j
Z
e
1
@
n
[f (t0 )] dt0
4 "0
@R R
Z
e
n
n d
[f (t0 )]
[f (t0 )] dt0
2
4 "0
R
cR df
e
n
1 d
n
+
;
2
0
4 "0
R
c dt
R
(8.19)
where use is made of the integration by parts,
Z
d
[f (t0 )]dt0
df
Z
1 d
=
g(t0 )
[f (t0 )]dt0
df dt0
dt0
g (t0 )
d
1
=
df =dt0 dt0 df =dt0
g(t0 )
g (t0 )
(t0 )
1 d
(t0 ) dt0
=
(8.20)
(8.21)
f (t0 )=0
:
(8.22)
f (t0 )=0
Similarly,
e 1 d
@A
=
@t
4 "0 c dt0
;
R
(8.23)
and the electric …eld becomes
E(r; t) =
e
4 "0
n
1 d
+
2
R
c dt0
n
R
:
(8.24)
f (t0 )=0
To proceed further, we need concrete expressions for the derivatives,
dn
dt0
d
dt0
and
The unit vector n along the distance vector R = r
4
1
R
:
rp (t0 ) changes its direction only through the
velocity component perpendicular to R;
v? 0
dt ;
R
dn =
(8.25)
as can be seen in Fig. 8-1. This yields
Figure 8-1: The change in the unit vector n is caused by the perpendicular velocity v? :
dn
=
dt0
v?
:
R
(8.26)
Also,
d
dt0
1
R
=
=
=
1
d
[(1 n )R]
2
( R) dt0
1 h
v?
(n _ )R c(1 n
( R)2
c
1
2
n
R(n _ ) :
2
( R)
c
)n
i
(8.27)
Substitution of Eqs. (8.26) and (8.27) to Eq. (8.24) gives
E(r; t) =
=
e
4 "0
"
n
( R)2
n
( R)2
2
e
1
(n
3 R2
4 "0
)+
2
1
c
3R
1
R(n _ )
c
n
n
[(n
_
_]
)
2R
c
:
#
(8.28)
f (t0 )=0
The …rst term in the RHS,
ECoulomb =
2
e
1
(n
3 R2
4 "0
5
)
;
f (t0 )=0
(8.29)
is the Coulomb …eld corrected for relativistic e¤ects. It is proportional to 1=R2 and thus does not
contribute to radiation of energy. The second term,
Erad =
e
4 "0 c
1
3R
[(n
n
_]
)
;
(8.30)
f (t0 )=0
contains acceleration _ and is proportional to 1=R: This is the desired radiation electric …eld due
to a moving charged particle.
The magnetic …eld can be calculated in a similar manner from
B = r A
1
[n E]f (t0 )=0 :
=
c
(8.31)
Derivation of this result is left for an exercise.
8.3
Radiation from a Charge under Linear Acceleration
If the acceleration is parallel (or anti-parallel) to the velocity, _
reduces to
"
#
e
n (n _ )
Erad =
4 "0
c 3R
0
= 0; the radiation electric …eld
:
(8.32)
f (t )=0
The angular distribution of radiation power at the observing time t is
dP (t)
d
= c"0 jErad j2 R2
"
1 e2 n (n
=
4 "0 4 c
c 3
Denoting the angle between _ and n by ; we have [n
(n
_)
#2
:
(8.33)
f (t0 )=0
_ )]2 = _ 2 sin2 ; and thus
"
#
_ 2 sin2
dP (t)
1 e2
=
d
4 "0 4 c (1
cos )6
:
(8.34)
f (t0 )=0
However, the power P (t) in the above formulation is the rate of energy radiation at t; the observing
time, which is not necessarily equal to the energy loss rate of the charge at the retarded time t0
determined from f (t0 ) = 0: To …nd the radiation power at the retarded time P (t0 ) ; let us consider
the amount of di¤erential energy dE=d radiated during the time interval between t0 and t0 +dt0 : By
de…nition dE = P (t0 )dt0 : The radiation energy dE is sandwitched between two eccentric spherical
surfaces with a volume
dV = R2 cdt0 (1
cos ):
6
Therefore, the di¤erential radiation energy is
dE
1
= "0 jErad j2 R2 cdt0 (1
d
2
cos );
and
dP (t0 )
d
"
_ 2 sin2
1 e2
(1
4 "0 4 c (1
cos )6
"
#
_ 2 sin2
1 e2
4 "0 4 c (1
cos )5
0
=
=
#
cos )
f (t0 )=0
:
(8.35)
f (t )=0
In nonrelativistic limit j j
1; the radiation occurs predominantly in the direction perpendicular
to the acceleration = =2. The total radiation power in this case is
P
'
=
2 Z
1 e2 _
sin2 d
4 "0 4 c
2
1 2e2 _
;
1:
4 "0 3c
(8.36)
This is the well known Larmor’s formula for radiation power due to a nonrelativistic charge. Since
in nonrelativistic limit,
n [(n
) _ ] ' n (n _ );
(8.37)
Larmor’s formula is applicable for acceleration in arbitrary direction relative to the velocity.
For arbitrary magnitude of the velocity ; the radiation power can be found from
2
0
P (t ) =
=
=
1 e2 _
4 "0 4 c
2
1 e2 _
Z
sin2
d
(1
cos )5
Z
sin3
2
d
4 "0 4 c
cos )5
0 (1
2
1 2e2 _ 6
;
4 "0 3c
where
1
=p
1
2
;
(8.38)
(8.39)
is the relativity factor and the following integral is used,
Z
0
sin3
d =
(1
cos )5
Z
1
1
1
(1
7
x2
4
1
dx =
2 3:
5
x)
3 (1
)
(8.40)
The angular dependence of the radiation intensity,
sin2
;
(1
cos )6
peaks at angle
0
where
cos
In highly relativistic limit
0
=
1; the angle
p
0
0
1 + 24
4
2
(8.41)
1
:
(8.42)
becomes of order
'
1
1;
(8.43)
which indicates a very sharp pencil or beam of radiation along the direction of the velocity : (This
is also the case for acceleration perpendicular to the velocity as shown in the following section.)
Angular distribution of radiation intensity I ( )) for = 0; 0:2; 0:9 and 0:999 is shown below for a
p
2
:
common acceleration. Note that the radiation intensity rapidly increases with = 1= 1
y
1
1.0
0.8
0.6
0.5
0.4
0.2
-1 -0.8
-0.4
0
-0.2
0.2 0.4 0.6 0.8 1
-1.0
-0.5
-0.4
0.5
x
-0.5
-0.6
-0.8
-1.0
-1
= 0 ( = 1:0):
= 0:2 ( = 1:02):
y
2000
y
0
2000 4000 6000 8000 10000 12000 14000
x
-2000
= 0:9 ( = 2:3):
9e+7
-9e+7
2e+8
4e+8
6e+8
8e+8
= 0:99 ( = 7:09):
8
1e+9 1.2e+9
x
Polar plot of I( ) for several
factors but with common parallel acceleration.
For linear acceleration, the momentum change of the charged particle is
d
dp
=
dt
dt
mv
p
2
1
!
=m
3
v:
_
(8.44)
Therefore, the radiation power can be rewritten as
dp
dt0
2
P (t ) =
1
2e2
4 "0 3c3 m2
dE
dx0
2
=
1
2e2
4 "0 3c3 m2
0
=
1
2e4
E2
4 "0 3c3 m2 ac
;
(8.45)
where Eac is an external acceleration electric …eld and
dE
;
dx0
is the energy gradient of a linear accelerator which is at most of the order of 100 MeV/m in practice.
The radiation loss in linear accelerators is negligibly small compared with energy gain. This is one
of the advantages of high energy linear accelerators. Note that the radiation power due to linear
acceleration is independent of the particle energy or the relativity factor :
8.4
Radiation from a Charge in Circular Motion
In circular motion, the acceleration is perpendicular to the velocity. Here we consider highly
relativistic motion of a charged particle with
' 1, for nonrelativistic case has already been
discussed in Chapter 4. A geometry convenient for analysis to follow is shown in Fig.8-2. A
particle undergoes circular motion with an orbit radius in the x z plane and it passes the
origin at t0 = 0: At that instant, the acceleration is in the x direction while the velocity is in the z
direction,
_ = _ ex ;
= ez :
The radiation electric …eld can then be written down in terms of cartesian components,
E(r; t) =
=
1
e
n
4 "0 c 3 R
_
e
[(
4 "0 c 3 R
[(n
)
_]
cos ) cos e + (1
9
cos ) sin e ] ;
(8.46)
Figure 8-2: Particle undergoing circular motion in the x
At t = 0; the particle passes the origin.
z plane with radius
and frequency ! 0 :
and the angular distribution of radiation power is given by
2
2 _
dP (t0 )
1 e
=
d
4 "0 4 c (1
1
(1
cos )5
1
cos )2
2
sin2 cos2
:
(8.47)
The total radiation power is
2
P (t0 ) =
2 _
Z
1 e
4 "0 4 c
(1
1
(1
cos )5
cos )2
1
2
sin2 cos2
d
cos )2
1
2
=
=
2 _
Z 2
Z
1 e
sin d
d
4 "0 4 c 0
(1
0
1 2e2 jvj
_ 2
4 "0 3c3
4
1
(1
cos )5
:
Relevant integrals are:
Z
2
sin2 cos2
(8.48)
Z
1
1
1
1
dx
(1
1
(1
x)3
=
2
2 2
)
(1
=2
4
x2
4
1
4
dx =
=
2
5
3
x)
3 (1
3
)
;
6
:
In highly relativistic case, the acceleration may be approximated by
jvj
_ =
v2
10
'
c2
:
(8.49)
Then, the radiation power in terms of the orbit radius
P (t0 ) '
is
1 2e2 c 4
:
4 "0 3 2
(8.50)
To maintain the radiation loss in a circular accelerator at a tolerable level, the orbit radius must
be increased as the particle energy mc2 increases. Note that the radiation power is a sensitive
function of the particle energy in contrast to the case of linear acceleration.
As an example, let us consider the Betatron, the well known inductive electron accelerator
invented by Kerst. In the Betatron, the electron cyclotron orbit is maintained constant,
mc2
= ecB (t) = ecB0 sin !t ' ecB0 !t;
where only the initial phase of the sinusoidal magnetic …eld is useful for acceleration, !t
the rate of electron energy gain is
d
dt
1. Then
mc2 = ec B0 ! = const.
Equating this to the radiation energy loss,
P t0 =
1 2
4 "0 3c3
we …nd
max
where
re =
'
3!eB0 3
2mc2 re
e2
= 2:8
4 "0 mc2
2
c2
4
;
1=4
;
10
15
(8.51)
m;
(8.52)
is the classical radius of electron. If = 50 cm, ! = 2
60 rad/s, and B0 = 0:5 T (5 kG), the
upper limit of is about 400 and the maximum electron energy attainable is approximately 200
MeV.
8.5
Fourier Spectrum of Radiation Fields
The formulae such as Eqs. (8.45) and (8.50) only tell us the total radiation power integrated
over the frequency. Radiation …eld emitted by highly relativistic particle is hardly monochromatic
but cnsists of broad frequency spectrum. Knowing such frequency spectrum of radiation is of
practical importance for identifying radiation source. A typical example is synchrotron radiation
due to highly relativistic electrons bent by, or trapped in, a magnetic …eld. In nonrelativistic limit,
the radiation …elds all have a single frequency component corresponding to the classical electron
cyclotron frequency ! c = eB=m: However, as the relativity factor increases, the radiation …elds
11
consist of harmonics of the fundamental frequency ! c = eB= m. In highly relativistic case
1;
3
2
the frequency spectrum becomes almost continuous peaking at the frequency ! ' ! c = eB=m:
Frequency spectrum of the radiation electric …eld can be formulated by directly applying Fourier
transformation on the …eld in Eq. (8.24),
E(r; !) =
=
Z
1
E(r; t)ei!t dt
1
"
Z
)
1 e 1 n [(n
3
4 "0 c 1
R
Changing the variable from t to t0 and assuming r
E(r; !) =
1 e ei!r=c
4 "0 c r
Z
1
[(n
n
#
ei!t dt:
(8.53)
f (t0 )=0
rp ; we obtain
)
_]
2
1
_]
The unit vector n(t0 ) may be regarded constant since r
Then,
n [(n
) _]
d n
' 0
2
dt
exp i! t0
1
cn
rp (t0 )
dt0 :
(8.54)
rp and thus approximated by n ' r=r:
(n
)
;
(8.55)
and Eq. (8.53) can be integrated by parts,
E(r; !) =
i! e ei!r=c
4 "0 c r
Z
1
n
(n
) exp i! t0
1
1
cn
rp (t0 )
dt0 :
(8.56)
Any time derivatives contained in the physical electric …eld are merely multiplied by i! in the
Fourier space and the disappearance of the acceleration _ is not surprising. Amazing fact about
Eq. (8.56) is that it is applicable to radiation …elds which do not require particle acceleration
such as Cherenkov and transition radiation provided a proper velocity of electromagnetic waves in
dielectrics is substituted for c:
The radiation energy (not power) associated with the electric …eld is
c"0 r
2
Z Z
jE(r; t)j2 dtd ; (J).
(8.57)
However, since the electric …eld E(r; t) and its Fourier transform E(r; !) are related through
Parseval’s theorem:
Z
jE(r; t)j2 dt =
1
2
Z
jE(r; !)j2 d!;
(8.58)
the radiation energy can be written in terms of the Fourier transform E(r; !) as
1
c"0 r2
2
Z Z
jE(r; !)j2 d d!:
12
(8.59)
The quantity
dI(!)
d
1
c"0 r2 jE(r; !)j2 ;
2
(8.60)
can therefore be identi…ed as the radiation energy per unit solid angle per unit frequency. Substituting Eq. (8.56), we …nd
dI(!)
e2 2
1
=
!
d
4 "0 8 2 c
Z
2
(n
n
) exp i! t
0
0
1
cn
rp (t )
dt
0
;
1 < ! < 1;
(8.61)
or
dI(!)
e2 2
1
=
!
d
4 "0 4 2 c
Z
2
n
) exp i! t0
(n
1
cn
rp (t0 )
dt0 ; 0 < ! < 1:
(8.62)
Let us work on a few examples.
8.6
Synchrotron Radiation I
Synchrotron radiation is due to highly relativistic electrons trapped in a magnetic …eld. The
radiation beam rotates together with an electron and is directed along the direction of the velocity
with an angular spread of order
' 1=
1: If the orbiting frequency is ! 0 = eB= me ; the
radiation beam shines a detector for a duration
t0 '
=
!0
1
;
!0
(8.63)
as seen by the electron at the retarded time t0 . Since
t
=1
t0
=1
n
'
1
2
2
;
(8.64)
which is entirely due to Doppler e¤ect, the pulse width detected is of order
t'
t0
2
2
=
2
1
:
3!
0
(8.65)
Therefore, synchrotron radiation is dominated by frequency components in the range
!'
3
!0 =
2 eB
me
:
(8.66)
In ultrarelativistic case, the frequency spectrum of synchrotron radiation can extend to very high
frequencies even for a modest magnetic …eld.
We calculate the amount of energy radiated in one period of cyclotron motion T = 2 =! 0 : Since
the radiation power is constant, the radiated energy is
1
E=
T
Z
0
1
!0
I (!) d! =
2
13
Z
0
1
I (!) d!:
(8.67)
The time integration in the energy spectrum,
Z
e2 2
1
dI (!)
=
!
d
4 "0 4 2 c
2
T =2
(n
n
) exp i! t
0
1
cn
T =2
0
rp (t )
dt
0
; 0 < ! < 1;
(8.68)
can be extended from 1 to 1 since the characteristic frequency of the radiation …eld is much
higher than the fundamental frequency !
!0;
e2 2
dI (!)
1
=
!
d
4 "0 4 2 c
Z
1
2
) exp i! t0
(n
n
1
cn
1
rp (t0 )
dt0 ; 0 < ! < 1:
(8.69)
To perform the integration, we assume the trajectory shown in Fig. (8-2) in which an electron
passes the origin at t = 0: The vector n is assumed to be in the y z plane since the radiation
pro…le is essentially symmetric about the z axis. The trajectory is described by
rp t0 =
and the velocity is
=
cos ! 0 t0 ex + sin ! 0 tez ;
1
!0
(sin ! 0 tex + cos ! 0 tez ) :
c
(8.70)
(8.71)
Then,
1
n rp (t0 ) = sin (! 0 t) cos :
c
c
(8.72)
)=
(8.73)
Since
n
(n
=
?
sin ! 0 tex + cos ! 0 t sin e? ;
where
e? = n
ex ;
(8.74)
is a unit vector perpendicular to both n and x axis, and the radiation lasts for a very short time
and is limited within a small angle ; Eq. (8.73) reduces to
'
?
! 0 tex + e? :
Within the same order of accuracy, the phase function !(t
! t
n rp
c
where v has been approximated by c but 1
1
!
2
1
2
+
by
'
14
n rp =c) can be approximated by
sin ! 0 t cos
c
= ! t
'
(8.75)
1
2
2
:
2
t+
c2 3
t ;
3 2
(8.76)
Then,
Z
h
exp
i! t
?
n rp i
dt
c
1
Z 1
1
c2 3
i!
2
(! 0 tex
e? ) exp
+
t
+
t
=
2
2
3 2
1
Z 1
!
1
c2 3
2
t sin
= 2i! 0
t
+
dt ex
+
t
2
2
3 2
0
Z 1
!
1
c2
2
cos
+ 2 t + 2 t3 dt e? :
2
2
3
0
1
dt
(8.77)
The integrals reduce to the modi…ed Bessel functions of fractional orders (or the Airy’s functions),
Z
1
3
x
2
cos
0
Z
1
3
x
2
sin
0
where
Then
2i! 0
Z
1
=p
t sin
0
2
Z
0
1
cos
!
2
1+
2 2
1
2
1
2
+
+
2
1
3
+
!0t
2
!
2
+
1
3
(8.78)
1
d = p K2=3 (x) ;
3
3
!
1+
3 3!0
; x=
t+
1
d = p K1=3 (x) ;
3
3
1
c2 3
t
3 2
c2
t + 2 t3
3
2 2 3=2
dt = 2i
dt = 2
q
2
+
:
2
!0
1
2
+
(8.79)
2
!0
(8.80)
1
p K2=3 (x) ;
3
1
p K2=3 (x) ;
3
and for I (!) =d ; we obtain
dI (!)
d
where
=
=
1
e2 ! 2
4 "0 3 2 c ! 20
=
1 3e2 2 2
!
^ 1+
4 "0 2 c
1
2
+
2
2
h
n
1
2
1+
2
K2=3
(x) +
2
+
2
2
K2=3
[^
! 1+
2
i
2
K1=3
(x)
2 3=2
]+
2
2
K1=3
[^
! 1+
2 3=2
o
] ;
and !
^ is the normalized frequency,
!
^=
!
:
3 3!0
(8.81)
The modi…ed Bessel functions K2=3 (x) and K1=3 (x) both diverge at x ! 0: However, xK2=3 (x)
2 (x) and x2 K 2 (x)
and xK1=3 (x) are well behaving and vanish at small x: Fig. (8-3) shows 2x2 K2=3
1=3
which represent radiation intensities associated with electric …eld polarization along ex (that is, in
the particle orbit plane) and e? ; respectively.
The energy spectrum I (!) emitted during one revolution (T = 2 =! 0 ) can be found by inte-
15
y 0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0
1
2
3
4
5
x
2 (x) (solid line) and x2 K 2 (x) (dotted line). The factor of 2 in 2x2 K 2 (x)
Figure 8-3: 2x2 K2=3
1=3
2=3
assumes ' 1= :
grating dI (!) =d
=
=
'
=
over the solid angle,
I (!)
Z 2
Z
1 3e2 2 2
0 0
d 0 1+ 2
sin d
!
^
4 "0 2 c
0
0
n
o
3=2
2
2
2 3=2
2
1+
K2=3 [^
! 1+
] + 2 K1=3
[^
! 1+ 2
]
Z =2
n
1 6e2 2 2
2
cos d 1 + 2
1 + 2 K2=3
[^
! 1+
!
^
4 "0
c
=2
Z
n
1 6e2 2 1
3=2
2
!
^
1+ 2
1 + 2 K2=3
[^
! 1+ 2
]+
4 "0 c
1
1 6e2
f (^
!) ;
4 "0 c
2 3=2
2
]+
2
2
K1=3
[^
! 1+
Z
0
1
1+
2
n
1+
2
2
K2=3
[^
! 1+
2 3=2
]+
2
2
K1=3
[^
! 1+
o
] d
o
]
(8.82)
where 0 = 2
is the polar angle from the axis of electron revolution (y axis),
angle about the y axis, and the function f (^
! ) is de…ned by
f (^
! ) = 2^
!2
2 3=2
2
K1=3
[^
! 1+
2 3=2
0
is the azimuthal
2 3=2
o
] d ;
(8.83)
and shown in Fig. (8-4) (linear scale) and Fig. (8-5) (log-log scale). In Fig. (8-5), the straight line
3!
1=3 : f (^
in the low frequency regime !
! ) peaks
0 has a slope of 1=3; and indicates I (!) _ !
3! ;
at !
^ ' 0:14; or ! ' 0:42 3 ! 0 ; and its peak value is about 0.83. In high frequency regime !
0
the spectrum decays exponentially. The energy radiated per revolution can be calculated as
16
0.8
0.6
0.4
0.2
0
1
2
3
Figure 8-4: The function f (x) = f !=3
E=
Z
1
0
I (!) d! '
1 6e2
3
4 "0 c
4
3!
3
0
!0
5
plotted in linear scale.
Z
1
f (^
! ) d^
!;
(8.84)
0
where the integral numerically evaluated is approximately
Z
1
0
Then
E=
Z
1
f (^
! ) d^
! ' 0:713:
I (!) d! =
0
1 4:08e2
4 "0
c
4!
0
; (J)
and the radiation power is
P =E
!0
1 0:65e2 c
=
2
2
4 "0
4
;
(8.85)
which agrees reasonably well with Eq. (8.50),
P =
1 2e2 c
4 "0 3 2
4
'
1 0:667e2 c
2
4 "0
4
; (W).
The discrepancy may be attributed to the various approximations made in the analysis.
8.7
Synchrotron Radiation II
An alternative approach to …nding the frequency spectrum of synchrotron radiation is to apply
discrete Fourier analysis in terms of harmonics of the fundamental frequency ! 0 directly to radiation
17
.1
.1e-1
.1e-2
.1e-3
.1e-2
.1e-1
.1
1.
Figure 8-5: Log-log plot of the function f (x) = f !=3
3!
0
:
…elds. If a charge e is in circular motion with a constant angular frequency ! 0 , the radiation …eld
contains higher harmonics of ! 0 and can be Fourier decomposed as follows. Let us recall the
Lienard-Wiechert vector potential,
A(r; t) =
ev(t0 )
n )R(t0 )
0
4
(1
;
(8.86)
f (t0 )=0
where R = jr rp (t0 )j and the subscript f (t0 ) = 0 indicates that all time dependent quantities
should be evaluated at the retarded time t0 determined from the implicit equation for t0 ,
f (t0 ) = t0
jr
t+
rp (t0 )j
= 0:
c
(8.87)
il! 0 t
(8.88)
The vector potential can be Fourier decomposed as
X
A(r; t) =
Al (r)e
;
l
where
Al (r) =
0e
4
1
T
Z
0
T
ev(t0 )
eil!0 t dt;
(1 n )R
(8.89)
with T = 2 =! 0 being the period of the circular motion. Changing the integration variable from t
to t0 by noting
dt
=1 n ;
(8.90)
dt0
18
leads to
Z
eikl r
Al (r) '
4 r T
0e
T
0
v(t0 )ei(l!0 t
kl n rp )
dt0 ;
(8.91)
0
where kl = l! 0 =c. Note that the period T remains unchanged through the transformation. Let the
particle trajectory be
rp (t0 ) = (cos ! 0 t0 ex + sin ! 0 t0 ey );
(8.92)
v(t0 ) = ! 0 ( sin ! 0 t0 ex + cos ! 0 t0 ey ):
(8.93)
Since all radiation …elds rotate with the charge, the observing point can be chosen at arbitrary
azimuthal angle and we choose = =2; so that n = (1; ; = =2): Then
n rp (t0 ) = sin sin ! 0 t0 :
(8.94)
The velocity in the spherical coordinates is
v(t0 ) = ! 0 (sin cos ! 0 t0 er + cos cos ! 0 t0 e + sin ! 0 t0 e ):
Thus, the
(8.95)
component of Al (r) is given by
Al =
0 e ikl r
4 r
!0
cos
T
e
Letting x = ! 0 t0 ; and noting
Z
T
0
sin sin ! 0 t0 =c)
cos ! 0 t0 eil!0 (t
dt0 :
(8.96)
0
= ! 0 =c; we can rewrite this as
ikl r
Al = e
Z
0e
! 0 cos
4 r 2
2
cos xeil(x
sin sin x)
dx:
(8.97)
0
The integral reduces to
Z
2
cos xeil(x
sin sin x)
dx
0
=
=
[Jl+1 (l sin ) + Jl
2
Jl (l sin );
sin
and thus …nally,
Al =
0 ec ikl r
Similarly,
Al = i
sin )]
cot Jl (l sin ):
(8.98)
! 0 ikl r 0
e Jl (l sin );
4 r
(8.99)
4 r
e
1 (l
0e
where use has been made of the recurrence formula of the Bessel functions,
Jl
1 (x)
Jl+1 (x) = 2Jl0 (x):
19
(8.100)
The far-…eld radiation magnetic …eld can be found from
i
Hl '
Al ;
kl
(8.101)
0
which yields
e ! 0 kl ikl r 0
e Jl (l sin );
4 r
eckl ikl r
e
cot Jl (l sin ):
=i
4 r
Hl =
Hl
(8.102)
(8.103)
The radiation power associated with the l-th harmonic is
Z
2
jHl j2 d
Pl = c 0 r
Z
1 (el! 0 )2
2 02
=
Jl (l sin ) + cot2 Jl2 (l sin ) sin d :
8 0
c
0
Since Pl = P l , the total power is
P =
1
X
Pl ;
(8.104)
(8.105)
l=1
where Pl is now
1 (el! 0 )2
Pl =
4 0
c
Z
2 02
Jl (l
0
In nonrelativistic limit
sin ) + cot2 Jl2 (l sin ) sin d ;
1; the l = 1 term is dominant. For x
l
1:
(8.106)
1;
1
1
J1 (x) ' x; J10 (x) ' :
2
2
(8.107)
Then the lowest order radiation power agrees with the Larmor’s formula,
P1 '
=
1 (e! 0 )2 2
4 0 c
4
2
2
1 2e a
;
4 0 3 c3
Z
(1 + cos2 ) sin d
0
(8.108)
where a = v 2 = is the acceleration.
The integral in Eq. (8.105) cannot be reduced to elementary functions. However, the total
power given in Eq. (8.105) should reduce to Eq. (8.50),
P =
1 2 e2 a2
4 0 3 c3
4
=
1 2 e2 a2
1
2 2:
3
4 0 3 c (1
)
(8.109)
To show this, we modify the integral by noting
Jl2 (x)
=
1
Z
J0 (2x sin ) cos(2l)d ;
0
20
(8.110)
Jl02 (x)
=
1
Z
l2
x2
J0 (2x sin ) cos 2
0
cos(2l)d :
(8.111)
2
(8.112)
Then
2 02
Jl (
2
l sin ) + cot
Jl2 (
Furthermore,
1
l sin ) =
Z
J0 (2 l sin sin x)(
cos 2x
1) cos(2lx)dx:
0
Z
J0 (2 l sin x sin ) sin d =
0
sin(2 l sin x)
;
l sin x
(8.113)
cos 2x
(8.114)
and the power Pl reduces to
1 (e! 0 l)2
Pl =
4 0
c
Z
0
sin(2 l sin x)
(
l sin x
2
1) cos(2lx)dx:
Noting
Z
sin x sin(2 l sin x) cos(2lx)dx
0
=
=
=
1
2
2
Z
[sin(2l + 1)x + sin(1
2l)x] sin(2 l sin x)dx
0
[J2l+1 (2 lx)
J2l
1 (2
l)]
0
(2 l)
J2l
(8.115)
and
Z
sin(2 l sin x) cos(2lx)
dx
sin x
0
Z Z
= 2l
cos(2l sin x) cos(2lx)dxd
0
0
Z
= 2 l
J2l (2l )d ;
(8.116)
0
the power Pl can be rewritten as
1 2(e! 0 l)2
Pl =
4 0 c l
2 0
J2l (2l
and the total power is
P =
)
l(1
2
)
Z
J2l (2l )d
;
(8.117)
0
1
X
Pl :
l=1
Relevant sum formula of the Bessel functions is
1
X
J2l (2lx) =
l=1
21
x2
2(1
x2 )
:
(8.118)
Di¤erentiating by x;
X
0
2lJ2l
(2lx) =
l
Also,
X
l
2
l
Z
x
:
(1 x2 )2
(8.119)
3
J2l (2lx)dx =
0
6(1
2 3:
)
(8.120)
This is one of Kapteyn series formulae. (See for example, Mathematical Formulae (in Japanese),
(Iwanami, Tokyo, 1960), vol. 3, p. 212.) Then, …nally, the total radiation power becomes
P =
1 2e2 a2
4 0 3c3
4
;
a=
v2
(8.121)
which is consistent with the known radiation power from a charge undergoing circular motion.
Analytic expression for the radiation power Pl can be found by exploiting following approximation,
!
l2=3
1
;
(8.122)
J2l (2 l) ' p 1=3 Ai
2
l
where Ai(x) is the Airy function de…ned by
For large x
1
Ai(x) = p
Z
1
f (x) = p
Z
cos
1 3
t + xt dt:
3
cos
1 3
t + xt dt
3
0
20
0
(8.123)
1; the function takes the form
Ai(x) '
Also, Ai0 (0) =
1
1
exp
2x1=4
2 3=2
x
; x
3
1:
(8.124)
0:4587: Using these approximations, we …nd the following approximate formulae,
Pl '
1
4 "0
0:5175
1 e2 ! 20
p
Pl '
4 "0 2 c
s
l
e2 ! 20 1=3
l ; 1
c
exp
2 l
3 3
l
; l
3
;
3
(8.125)
:
(8.126)
Note that the radiation power increases with l in the manner Pl _ l1=3 up to l ' 3 beyond which
Pl decays exponentially. This is consistent with the analysis in the preceding section.
8.8
Free Electron Laser
In a synchrotron radiation source, many beamlines can be installed by bending an electron beam.
In straight sections of race track, no radiation occurs. However, by inserting a device called wiggler,
22
Figure 8-6: In a wiggler, an electron beam is modulated by a periodic magnetic …eld. Electrons
acquire spatially oscillating perpendicular displacement x(z) and velocity vx (z) which together with
the radiation magnetic …eld BRy produces a ponderomotive force vx (z) BRy (z) directed in the z
direction. The force acts to cause electron bunching required for ampli…cation of coherent radiation.
high intensity radiation can be extracted. A wiggler consists of periodically alternating magnets
and gives an electron beam periodic kick perpendicular to both the beam velocity and magnetic
…eld. Electrons receive kicks at an interval
t0 =
where
w
c
;
(8.127)
is the “wavelength”of the periodic wiggler structure. Because of Doppler shift, this time
1
interval is shortened by a factor
' 2 2 for a stationary detector in front of the beam,
1
w
t'
2
w
:
2c
(8.128)
Therefore, the wavelength of resultant radiation is approximately given by
'
2
w
2
w;
(8.129)
:
(8.130)
and the frequency by
!'
4
2c
w
The intensity of free electron laser can be orders of magnitude higher than that of synchrotron
radiation because of coherent ampli…cation through the periodic structure. Electrons tend to be
bunched in the wiggler as the electron beam travels through the periodic structure. In contrast, no
collective interaction between electrons and electromagnetic waves exists in synchrotron radiation.
In electron bunching, the magnetic ponderomotive force plays a major role. Let us assume a
23
periodic wiggler magnetic …eld in y direction,
By = B0 cos
2
z
= B0 cos kw z; kw =
w
2
:
(8.131)
w
The Lorentz force is
F = ev
B;
or
Fx =
evz B0 cos kw z:
Then electron acquires a velocity vx in x direction,
vx =
eB0
sin kw z;
m kw
(8.132)
evx
(8.133)
and a resultant ponderomotive force is
BR ;
where BR is the radiation magnetic …eld propagating in the form ei(kz !t) along the beam. Since
the acceleration due to the wiggler magnetic …eld is in x direction, the radiation electric …eld is
predominantly in x direction and radiation magnetic …eld BR is in y direction. Then the ponderomotive force directed in z direction is proportional to ei[(k+kw )z !t] and propagates at a velocity
!
:
k + kw
When this propagation velocity matches the electron beam velocity c, strong interaction between
the radiation …eld and electron motion takes place and electrons tend to be bunched. This results
in positive feedback for wave ampli…cation. From the condition
!
k
' c; or
= ;
k + kw
k + kw
we readily recover
k=
8.9
1
Radiation Accompanying
kw ' 2
2
kw :
Decay
Equation (8.61) for the angular distribution of radiation energy can be applied to cases in which
particle acceleration is not involved explicitly. In decay, an energetic electron (or positron) is
suddenly released from a nucleus together with neutrino. The situation is equivalent to sudden
acceleration of an electron. The duration of acceleration t is limited by the uncertainty principle
24
t mc2 & ~: Therefore, the upper limit of the frequency spectrum should be of the order of
mc2
:
~
! max '
(8.134)
In
decay, the maximum value of is of order of 30.
Let us assume that an electron suddenly acquires a velocity
= v=c and then travels at a
constant velocity. The integration in Eq. (8.61) is limited from t = 0 to 1;
dI(!)
d
=
=
where
yields
e2 2
1
!
4 "0 4 2 c
1
4 "0 4
1 e2 1
ln
4 "0 c
2
sin
Z
1
2
ei!(1
cos )t
dt
0
2
e2
2 c (1
is the angle between the velocity
I(!) =
2
2
sin
;
cos )2
(8.135)
and the unit vector n: Integration over the solid angle
1+
1
2 = const.;
! . ! max =
mc2
:
~
The frequency spectrum is ‡at up to ! max : Therefore, the total energy radiated through
is approximately given by
E'
1 e2 1
ln
4 "0 c
1+
1
mc2
1
'
ln(4
~
2
2
)
2
mc2 ;
(8.136)
decay
(8.137)
where the dimensionless quantity ,
=
1 e2
1
'
;
4 "0 c~
137
is the …ne structure constant. The energy emitted as radiation through
of the electron energy.
8.10
(8.138)
decay is a small fraction
Cherenkov Radiation
Cherenkov radiation occurs when a charged particle travels faster than electromagnetic waves in
a material medium. It does not require acceleration of charges and the basic mechanism is very
similar to that of sound shock waves in gases. As in the case of
decay, we assume a charge
travelling along a straight line at a velocity = v=c(!); where
1
c(!) = p
"(!)
25
;
0
is the velocity of electromagnetic waves in a dielectric having a permittivity "(!): Eq. (8.61) should
be modi…ed as follows after taking into account the proper de…nition of c(!);
Z
e2
dI(!)
= 0 2
! 2 v 2 sin2
d
4 4 c(!)
Note that the integration limits are from
Z
1
ei!(1
1
2
ei!(1
cos )t
dt :
(8.139)
1
1 to 1: The time integral is singular,
cos )t
dt = 2
[!(1
cos )];
1
and the condition for radiation is
cos =
or
(!) =
1
< 1;
v
> 1:
c(!)
(8.140)
Then,
dI(!)
e2
= 0
! 2 v 2 sin2
d
4 (2 )2 c(!)
2
[!(1
cos )]:
(8.141)
Square of a delta function is not integrable and the radiation energy simply diverges. This is merely
due to the assumption that the charge is radiating forever from t = 1 to 1 which is of course
unphysical. It is more appropriate to consider a radiation power rather than energy. For this
purpose, we consider a thin slab of the dielectric of thickness dz. The transit time over the distance
dz is T = dz=v and we calculate energy radiated during that time,
Z
e2
dI(!)
= 0 2
! 2 v 2 sin2
d
4 4 c(!)
2
T =2
i!(1
e
cos )t
dt :
(8.142)
T =2
The integral can be carried out easily,
Z
Thus
where
T =2
T =2
ei!(1
cos )t
dt =
2
!(1
cos )
dI(!)
e2 ! 2
= 0 2
sin2
d
4 4 c(!)
1
= (1
2
sin (1
sin
cos )! T:
26
cos )
T!
:
2
2
(dz)2 ;
(8.143)
(8.144)
In high frequency regime ! T
1; the function (sin = )2 may be approximated by a delta
function ( ): Integration over the solid angle yields
dI(!)
dz
1
4 "0
e
c0
2
=
1
4 "0
e
c0
2
=
1
! 1
! 1
2
1
"(!) 0 v 2
;
(8.145)
p
where c0 = 1= "0 0 is the speed of light in vacuum. The rate of energy loss due to Cherenkov
emission is given by
dE
1
=
dz
4 "0
e
c0
2Z 1
1
"(!) 0 v 2
! 1
0
d!;
0 < ! < 1:
(8.146)
The result obtained is meaningful only if
> 1;
1
or v > c(!) = p
"(!)
;
(8.147)
0
which is the condition for Cherenkov radiation. The instantaneous radiation power can be estimated
from
Z
d
P =
I(!)d!
dt
Z
1
e 2
1
=
v ! 1
d!:
(8.148)
4 "0 c0
"(!) 0 v 2
In order to …nd the …eld pro…les emitted through Cherenkov radiation, we start from the wave
equations for the potentials in a material medium,
@2
@t2
~" (r; t) =
free ;
(8.149)
@2
@t2
A(r; t) =
0 J;
(8.150)
r2
"
0~
r2
"
0~
where ~" is the dielectric operator containing time derivative,
~" = ~"
@
@t
:
(8.151)
For a charged particle e travelling at a constant velocity v; the charge density and current density
are described by
= e (r vt);
(8.152)
J = ev (r
27
vt):
(8.153)
Then, after Fourier-Laplace transformation, the Fourier potentials can readily be found,
2 e
(k; !) =
"(!)
(!
!2
c2 (!)
k2
2 e
A(k; !) =
0v
(!
!2
c2 (!)
k2
(8.154)
k v);
(8.155)
k v);
where, as before,
1
"(!)
c2 (!) =
;
(8.156)
0
and the transformation
Z Z
(r
vt)ei(!t
k r)
dV dt = 2
(!
k v);
(8.157)
is substituted. Since the physical electric …eld is
E(r; t) =
r (r; t)
@
A(r; t);
@t
(8.158)
the Fourier component of the electric …eld is given by
E(k; !) =
=
ik (k; !) + i!A(k; !)
!
v
k
2
c (!)
2 ie
!2
"(!) k 2
c2 (!)
(!
k v):
(8.159)
Similarly, the Fourier-Laplace component of the magnetic …eld is
B(k; !) = ik
A(k; !)
k v
= 2 ie 0
(!
2
!
k2
c2 (!)
k v):
The physical electromagnetic …elds can then be found through inverse transformations,
Z
Z
1
3
d k d!E(k; !)ei(k r !t) ;
E(r; t) =
(2 )4
B(r; t) =
1
(2 )4
Z
3
d k
Z
d!B(k; !)ei(k r
!t)
:
(8.160)
(8.161)
(8.162)
To proceed further, we assume that the charged particle is travelling along the z axis at a
constant velocity v: The system is symmetric about the axis and we may assume an observing
point in the x z plane without loss of generality. We denote the cylindrical coordinates of the
28
Figure 8-7: Geometry for Fourier inverse transform.
observing point by ( ;
= 0; z) and spherical Fourier coordinates by k =(k; ; ): Then,
k r = k(z cos + sin cos ):
Because of the cylindrical symmetry, radiation of energy is expected in the radial direction ; and
the relevant Poynting vector is
S
= (E
H )
Ez (r; t)H (r; t):
=
(8.163)
The energy radiated per unit length along the particle trajectory (z axis) can be calculated from
dE
dz
=
=
2
Z
Z
Ez (r; t)H (r; t)dt
Ez (r; !)H (r; !)d!;
(8.164)
where Ez (r; !) is the Laplace transform of the electric …eld,
Ez (r; !) =
1
(2 )3
Z
d3 kEz (k; !)eik r ;
(8.165)
and H (r; !) is the Laplace transform of the magnetic …eld,
H (r; !) =
1
(2 )3
Z
29
d3 kH (k; !)eik r :
(8.166)
The Laplace transform of the axial electric …eld Ez (r; !) can be calculated as follows:
Ez (r; !) =
=
=
=
1
(2 )3
Z
ie
(2 )2
i 0 e!
(2 )2
i
0 e!
2
d3 kEz (k; !)eik r
Z
1
k 2 dk
Z
sin d
1
1Z
1
2
d
2
1=
1
Z
i!z
c(!)
exp
!v
c2 (!)
k cos
2
d
k2
"(!)
(
i!
z
exp
+
c(!)
0
0
0
Z
Z
2
0
"
1
1
d
2
1
2
1
1=
where
=
s
! q
c(!)
J0
kc(!)
;
!
!2
c2 (!)
1
2
kv cos )eik(z cos
(!
1
(
)2
1
;
2
cos
)#
d
(8.167)
v
:
c(!)
=
+ sin cos )
(8.168)
Letting
2
1
2
=
2;
(8.169)
we …nally obtain
Ez (r; !) =
=
i
0 e!
2
ie
2
0
exp
! exp
i!z
c(!)
1
i!z
c(!)
1
where
=
!
c (!)
s
1
2
1
2
1
Z
1
0
K0 (
J0
2
!
c(!)
1+
1
d
2
);
1;
2
(8.170)
(8.171)
and use is made of the integral representation of the modi…ed Bessel function K0 (ax);
Z
K0 (ax) =
1
0
In the asymptotic regime j j
1; K0 (
tJ0 (at)
dt:
t2 + x2
(8.172)
) approaches
r
e
2
:
(8.173)
For this to be propagating radially outward in the form eik ; we must choose
s
=
!
c (!)
=
!
i
c (!)
1
2
s
1
30
1
1
2;
> 1:
(8.174)
With this choice for ; the Laplace transform of the axial electric …eld becomes proportional to
"
i!
exp
c(!)
s
1
1
2
+
z
!#
;
(8.175)
and the radial and axial wavenumbers can be identi…ed as
s
!
1
!
1
k =
kz = ;
2;
c(!)
v
(8.176)
respectively. Cherenkov radiation is con…ned in a cone characterized by an angle ;
sin =
c(!)
;
v
(8.177)
as shown in Fig. 8-8.
Figure 8-8: Cherenkov cone. Radiation …elds are con…ned in the cone.
The Laplace transform of the azimuthal magnetic …eld is given by
Z
1
H (r; !) =
d3 kH (k; !)eik r
(2 )3
Z
Z
Z 2
ie
kv sin
2
=
k
dk
sin
d
d
2
!2
(2 )
k2
0
0
2
(!
kv cos )
c (!)
ik(z cos + sin cos )
=
e
e
K1 (
2
) exp
i!
z ;
c(!)
31
(8.178)
where the following integral
Z
0
1
x2 J1 (bx)
dx = aK1 (ab);
x2 + a2
(8.179)
is noted. (Calculation steps are left for an exercise.) Substituting Ez (r; !) and H (r; !) into Eq.
(8.164), we …nd
Z 1
i 0 e2
1
dE
K0 ( )K1 ( )d!:
(8.180)
=
! 1
2
2
dz
(2 )
1
In the asymptotic region j j
1; this reduces to
dE
1 e2
=
dz
4 "0 c20
Z
1
! 1
0
1
2
d!;
(8.181)
in agreement with the earlier result, Eq. (8.146). Eq. (8.180) can be used even when Cherenkov
condition is not satis…ed. In this case, energy loss is through near …eld Coulomb interaction between
a charged particle and ions and electrons in molecules in the dielectric media. We will return to
this problem in Section 8.12.
8.11
Transition Radiation
Transition radiation occurs when a charge crosses a boundary of two dielectric media. No acceleration is required, nor is it necessary for charge to move faster than the speed of light as in Cherenkov
radiation. In this respect, transition radiation is a least demanding radiation mechanism. Radiation emitted from a charge approaching a conductor is an extreme case of transition radiation with
an in…nite permittivity, and may be regarded as the inverse process of radiation accompanying
decay. Disappearance, rather than creation, of charge is responsible for transition radiation.
We …rst consider a simple case: a charge e approaching normally a conducting plate at a velocity
v (> 0). On impact, the charge is assumed to come to rest. A conducting plate is mathematically
equivalent to an in…nitely permissive dielectric plate. An image charge e moving in the opposite
direction in the conducting plate can be introduced so that the current density is
Jz (r; t) =
ev [ (z + vt) + (z
vt)] (x) (y);
1 < t < 0;
(8.182)
where at t = 0 (or z = 0) the particle is brought to rest. Its Laplace transform is
Jz (r; !) =
=
Z
0
Jz (r;t)ei!t dt
1
e exp
i
!
jzj
v
(x) (y):
(8.183)
Let the observing point be at P located at (r; ): (The system is symmetric about the axis and thus
32
Figure 8-9: Radiation from a charge impinging on a metal surface. Sudden deceleration at the
metal surface is the inverse process of radiation accompanying beta decay.
is ignorable.) The vector potential at P can be calculated in the usual manner,
Z
Jz (r0 ; !) ikjr r0 j 0
0
0 ikr
Az (r; !) =
e
dV '
e
Jz (r0 ;!)e ik r dV 0
0
4
jr r j
4 r
Z 1h
i
e
!
!
0
0
0
=
exp
i z 0 e ikz cos + exp
i z 0 eikz cos dz 0
eikr
4 r
v
v
0
e
1
1
1
= i 0 eikr
+
; z > 0;
(8.184)
4 r
k 1 + cos
1
cos
0
Z
and the magnetic …eld from
H (r; !) '
=
ikAz sin = 0
e ikr
1
e
+
4 r
1 + cos
1
33
1
cos
sin :
(8.185)
The angular distribution of radiation energy is thus given by
dI(!)
d
=
=
=
1 2
r c 0 jH (r; !)j2
2
2
e2 2 c 0
1
1
+
sin2
32 3
1 + cos
1
cos
1 e2 v 2
sin2
;
1 < ! < 1:
2
4 "0 2 2 c3 (1
cos2 )2
Integration over the solid angle in the region z > 0 (0 <
I(!) =
1 e2 1
4 "0 2 c
1+
2
ln
1+
1
(8.186)
< =2) yields
2
; ! > 0:
R1
Evidently, the radiation energy 0 I(!)d! diverges. This is due to the assumption of a perfect
conductor. In practice, metals cannot be regarded as perfect conductor. The condition that the
surface impedance of metal
r
i! 0
;
Z=
i!"0 +
p
="0 imposes an upper limit
be su¢ ciently small compared with the free space impedance Z0 =
R !m a x 0
I(!)d!:
of the frequency, !"0
and a cuto¤ emerges in the integral 0
Figure 8-10: Transition radiation emitted by a charge passing through a dielectric boundary.
We now analyze the case of a dielectric slab having a relative permittivity "r = "="0 : In this
34
case, the particle continues to travel after passing the boundary and the current density is now
Jz (r; t) =
ev (x) (y) (z + vt);
1 < t < 1:
Its Laplace transform is
Jz (r; !) =
ev (x) (y)e
!
ivt
(8.187)
:
(8.188)
The contribution to the radiation …elds from the region z > 0 consists of two parts, one directly
from the charge (as in free space) and the other via re‡ection at the dielectric boundary. Denoting
the magnetic re‡ection coe¢ cient by in the Fresnel’s formulae,
p
"
"r cos
p r
=
"r cos + "r
sin 2
;
sin 2
(8.189)
and following the same procedure as in the case of conductor plate, we …nd
H 1 (r; !) '
1
ev ikr
e
4 cr
1+
+
cos
1
sin ;
cos
z > 0:
(8.190)
The contribution from the region z < 0 involves refraction at the boundary, and thus additional
retardation because of the longer path length. In Fig.8-10, for z < 0, we observe
l = r + z 0 tan 00 sin ; z 0 < 0
l0 = z 0 = cos 00 ;
p
sin 0
";
00 =
sin
and thus
kl +
p
kl0 = kr
kz
p
sin 2 :
(8.191)
Then the contribution from the region z < 0 to the integral becomes
H 2 (r; !) =
Note that the factor 1
total magnetic …eld is H
ev
(1
4 cr
)
1+
p
1
"r
sin 2
eikr sin
(8.192)
here indicates the …eld amplitude transmitted into the air region. The
1 + H 2 ; and the angular distribution of the radiation energy is given by
dI(!; )
r2 c
=
d
2
where
A=
1
1+
cos
In the case of ideal conductor "r ! 1;
+
0
jH (r;!)j2 =
1
cos
2 2
0e v
A2 sin 2
32 3 c
1+
1
p
sin 2
"r
= 1; we recover
dI(!; )
1 e2 v 2
=
d
4 0 2 2 c3
35
1
sin
2
cos2
;
2
.
(8.193)
:
(8.194)
When "r = 1; = 0; radiation evidently disappears.
The factor A in Eq. (8.194) does not fully agree with that in the original work by Frank and
Ginzburg,
1+
1
p
+
;
(8.195)
A0 =
1 + cos
1
cos
"r (1 +
"r sin 2 )
although this too vanishes when "r ! 1 and reduces to the case of conducting medium when
"r ! 1:
8.12
Energy Loss of Charged Particles Moving in Dielectrics
The formula derived in Eq. (8.180) yields a physically meaning energy loss rate even when the
Cherenkov condition is not satis…ed,
< 1: A charged particle moving in a dielectric medium
“collides” with atoms and lose its energy through Coulomb interaction with electrons in atoms.
Electrons in an atom are bounded. However, they do respond to electromagnetic disturbance and
absorb energy through the resonance "(!) = 0, where
"(!) = "0
! 2p
1
!2
! 20
!
:
(8.196)
Resonance of the type
1
x
x0
;
(8.197)
can be handled mathematically by introducing an imaginary part,
1
x
x0
=P
1
x
x0
i
(x
x0 );
(8.198)
where P stands for the principal part. This is justi…able because the imaginary part of the function
x
1
x x0
=
x0 + i"
(x x0 )2 + "2
remains …nite even in the limit " ! 0;
lim
"!0
Z
(x
i
(x
"
;
x0 )2 + "2
"
dx = :
x0 )2 + "2
(8.199)
(8.200)
Physically, the resonance leads to absorption of wave energy by charged particles in a material
medium dielectrics, plasmas, etc.
The characteristic scale length of interaction between a charged particle and atoms in a dielectric is evidently of the order of atomic size which indicates that the interaction is of near-…eld,
nonradiating nature dominated by longitudinal (electrostatic) …elds. As we will see, the major
contribution to the energy loss occurs through the pole of the dielectric function, "(!) = 0:
In the near-…eld region
1; the modi…ed Bessel functions K0 ( ); K1 ( ) may be approx-
36
imated by
K0 (
where
E
= 0:5772
)'
ln
K1 (
)'
E;
2
1
(8.201)
;
(8.202)
is the Euler’s constant. The real part of Eq. (8.180) becomes
dE
e2
' 0 2 Re
dz
(2 )
Z
1
1
with
!
=
c(!)
h
1
"(!) 0 v 2
i! 1
s
1
1;
2
ln
+
2
E
i
d!;
< 1:
(8.203)
(8.204)
The dielectric function "(!) is in the form
"(!) = "0
! 2p
1
!2
! 20
!
:
Therefore, the integration can be carried out by evaluating the pole contribution at "(!) = 0; which
occurs at
q
(8.205)
! 2p + ! 20 ;
!=
and exploiting Plemelj’s formula,
lim
!0
x
1
1
=P
a+i
x a
i
(x
a);
where P indicates the principal part of the singular function 1=(x
dE
1
'
dz
4 "0
e! p
v
2
a): The result is
1
1:12v
A:
ln @ q
! 2p + ! 20
0
(8.206)
As the minimum distance ; the intermolecular distance may be substituted because the shell
electrons e¤ectively shield the electric …eld of the charge well inside the atom.
In a plasma, ! 0 is evidently zero (because electrons in a plasma are free). Then,
dE
1
'
dz
4 "0
e! p
v
2
where
min
'
v
ln
3
4 n
min ! p
;
(8.207)
1=3
;
is the average distance between ions. For a Maxwellian electron distribution with a temperature
37
Te ; the average energy loss rate may be estimated from
1 (e! p )2
dE
ln
'
dt
4 "0 3vT e
where
D
=
p
8 2
n
3
3
D
!
;
vT e
;
!p
(8.208)
(8.209)
is the Debye shielding length.
8.13
Bremsstrahlung
Whenever a charged particle collides with another charged particle, electromagnetic radiation occurs
due to acceleration by Coulomb force. Collisions between like particles (e.g., electron-electron)
emit quadrupole radiation while collisions between unlike particles (e.g., electron-ion) emit dipole
radiation. Bremsstrahlung is due to collisions between electrons and ions and provides a basic
mechanism for x-ray production.
Let an electron approach an ion having a charge Ze with a velocity v (
c) and impact
parameter b: The acceleration due to Coulomb force is of the order of
a'
1 Ze2
;
4 "0 mb2
(8.210)
and consequent radiation power can be estimated from the Larmor’s formula,
P '
1 2e2 a2
:
4 "0 3c3
(8.211)
The total energy radiated can in principle be found by integrating the power over time along the
electron trajectory. However, in experiments, one is seldom interested in measuring radiation power
or energy associated with a single electron. What is more relevant is the radiation associated with
a beam of electrons impinging on an ion. In this case, some electrons have impact parameters
vanishingly small. However, the impact parameter has a lower bound imposed by the uncertainty
principle,
~
(8.212)
bmin ' ;
p
where p is the electron momentum. For an impact parameter b; the time duration in which the
acceleration is signi…cant is
b
t' :
(8.213)
v
38
Therefore, energy radiated by a single electron is
E
= P t
1
2
'
3
(4 "0 ) 3c3
2
Ze3
m
1
b3 v
:
(8.214)
For an electron beam having a density n; the radiation power can thus be estimated from
P
= ne v
Z
1
bmin
=
E2 bdb
4
1
(4 "0 )3 3
Ze3
m
2
ne mv
:
c3 ~
(8.215)
If the ion density is ni ; the quantity P ni de…nes the power density,
1
4
3
(4 "0 ) 3
Ze3
m
2
ne ni mv
;
c3 ~
(W/m3 ):
(8.216)
The frequency spectrum of radiation energy may qualitatively be found as follows. Since the
characteristic time of acceleration
b
(8.217)
= ;
v
is short, radiation occurs as an impulse and the spectrum is ‡at in the region 0 < ! < v=b; and
vanishes for ! > v=b: Since the minimum impact parameter is
bmin '
~
;
mv
(8.218)
the upper limit of the frequency spectrum extends to
! max '
mv 2
; or ~! . mv 2 :
~
(8.219)
This is essentially a statement of energy conservation, that is, the maximum photon energy emitted during bremsstrahlung is limited by the incident electron kinetic energy, which is reasonable.
Therefore, the frequency spectrum of bremsstrahlung is
I(!; b) =
8
1
2 (Ze3 )2
v
>
>
>
< (4 "0 )3 3 mc2 mv 2 b2 ; 0 < ! < b ;
>
>
>
: 0;
(8.220)
v
!> :
b
I(!; b) has dimensions of J/frequency. It is convenient to introduce a radiation cross-section (!)
39
de…ned by
Z
(!) =
bmax
I(!; b)2 bdb
bmin
4
1
(4 "0 )3 3
4
1
3
(4 "0 ) 3
=
=
The integral over the frequency,
Z
(Ze3 )2
ln
mc2 mv 2
(Ze3 )2
ln
mc2 mv 2
bmax
bmin
mv 2
:
~!
(8.221)
(!)d!;
(8.222)
evidently diverges at the lower end ! ! 0. To remedy this di¢ culty, Bethe and Heitler recognized
that if the velocity v is understood as the mean value of the initial and …nal velocities, i.e., before
and after emission of a photon,
1
(vinitial + v…nal )
2
p
p
1
p
E + E ~! ;
2m
v =
=
(8.223)
the integral remains …nite,
Z
(!)d! =
=
=
(Ze3 )2
1
4
3
(4 "0 ) 3 mc2 mv 2
1
4
3
(4 "0 ) 3
4
1
3
(4 "0 ) 3
Z
! max
0
Z
(Ze3 )2 1
ln
mc3 ~ 0
(Ze3 )2
:
mc3 ~
0 p
p
E + E ~!
B
ln @
~!
1+
p
1
p
x
x
dx; x =
2
1
C
A d!
~!
;
E
(8.224)
The integral in the intermediate step is unity. Multiplying by the ion density ni ; we thus obtain
the bremsstrahlung rate per unit length,
dE
dz
= ni
=
Z
(!)d!
1
4 (Ze3 )2 ni
;
(4 "0 )3 3 m2 c3 ~
(8.225)
and radiation power density,
P=V =
1
4 (Ze3 )2 ne ni v
; (W m
3
(4 "0 ) 3
mc3 ~
3
):
(8.226)
This agrees with the earlier qualitative estimate in Eq. (8.216).
Relativistic correction to the classical bremsstrahlung formulae can be readily found if we move
40
to the electron frame wherein the electron velocity is nonrelativistic. Since the energy and frequency
are Lorentz transformed in the same manner, and the transverse dimensions are Lorentz invariant,
it follows that the radiation cross-section (!) is Lorentz invariant,
lab (! lab )
=
0
(! 0 );
where the primed quantities are those in the electron frame. The frequencies ! lab and ! 0 are related
through the relativistic Doppler shift,
! 0 = ! lab (1
cos ); ! lab = ! 0 (1 +
cos 0 );
(8.227)
where and 0 are the angles with respect to the electron velocity in each frame. Since in the
electron frame, the radiation is con…ned in a small angle about 0 = =2; we have
!0 '
! lab
:
(8.228)
The collision time is shortened by the factor since the transverse …eld is intensi…ed by the same
factor through Lorentz transformation. Therefore, the maximum impact parameter is modi…ed as
bmax =
2v
v
=
;
0
!
! lab
(8.229)
and the radiation cross-section in the laboratory frame becomes
(!) =
0
(! 0 ) =
1
4 e4 (Ze)2
ln
(4 "0 )3 3 m2 c3 v 2
2 mv 2
~!
:
(8.230)
It is noted that the minimum impact parameter remains unchanged through the transformation
because it is essentially the Compton length based on the uncertainty principle.
8.14
Radiation due to Electron-Electron Collision
In this case the dipole radiation is absent because in the center of mass frame, two electrons stay
at opposite positions, r1 = r2 ; and the dipole moment identically vanishes.. The lowest order
radiation process is that due to electric quadrupole. (The magnetic dipole moment also vanishes.)
The quadrupole moment tensor is
1
Qij = exi xj ;
(8.231)
2
where xi is the i-th component of the relative distance 2r: In Chapter 5, a general formula for the
quadrupole radiation power has been derived. Noting
d3
:::
:: :
: ::
:::
(xi xj ) = x i xj + 3xi xj + 3xi xj + xi x j ;
dt3
41
(8.232)
Figure 8-11: Colliding electrons in the center of mass frame. The impact parameter is b:
::
xi =
:::
xi =
we …nd
1 2e2
xi ;
4 "0 mr3
1 2e2
4 "0 m
d3
1 2e2
(xi xj ) =
3
dt
4 "0 mr3
:::
Q=
1 e3
4 "0 mr3
xi
r3
(8.233)
3xi vr
r4
;
4(vi xj + vj xi )
(8.234)
6vr xi xj
r
6vr xi xj
r
4(vi xj + vj xi )
;
(8.235)
(8.236)
Substituting this into the quadrupole radiation power,
2
1
1 4 X ::: 2
P =
3
Qij
4 "0 60c5
X :::
Qii
ij
i
!2 3
5;
(8.237)
we obtain, after somewhat lengthy calculations,
P =
1
2
4 "0 15c5
e3
m
2
v 2 + 11v 2
r4
;
(8.238)
where v is the component of the velocity related to the initial angular momentum bv0 = r(t)v (t)
with v0 the velocity at r ! 1: Energy conservation reads
1
1
1 e2
mv02 = mv 2 +
:
2
2
4 "0 2r
42
(8.239)
Then
4e2
(bv0 )2
+ 11 2 ;
4 "0 mr
r
v 2 + 11v 2 = v02
(8.240)
and the total radiation energy can be found by integrating the radiation power over time along the
trajectory,
2Z 1
2
e3
1
4e2
(bv0 )2
1
2
v
dt:
(8.241)
+
11
E=
0
4
4 "0 15c5 m
4 "0 mr
r2
1 r
This can be converted to an integral over the distance r by noting
dt =
E=
2
e3
2
1
4 "0 15c5
dr
=s
vr
2
m
dr
v02
Z
;
(bv0 )2
r2
v2
1 0
s
r4
1
rmin
(8.242)
4e2
4 "0 mr
4e2
(bv0 )2
+ 11 2
4 "0 mr
r
(bv0 )2
r2
v02
dr; (J)
(8.243)
4e2
4 "0 mr
where rmin is the distance of the closest approach,
rmin =
0
4e2
1 @
+
m
2v02
s
4e2
m
1
2
+ 4(bv0 )2 A :
(8.244)
However, the radiation energy by a single electron pair is of no practical interest. What is more
relevant is the radiation power emitted by an electron beam impinging on a single electron which
can be evaluated from
Z 1
E(b)bdb
P = 2 nv0
0
=
1
4
4 "0 15c5
2
e3
2 nv0
m
Z
1
bdb
Z
1
rmin
0
v2
1 0
s
r4
4e2
(bv0 )2
+ 11 2
4 "0 mr
r
v02
(bv0 )2
r2
dr; (W). (8.245)
4e2
4 "0 mr
The double integral reduces to
Z
0
=
1
Z
4e2
(bv0 )2
+ 11 2
4 "0 mr
r
v2
1 0
bdb
dr 4 s
r
rmin
25
v0
3
Z
1
rmin
1
v02
1
r5
r
4re2
mv02
2
43
(bv0 )2
r2
3=2
dr =
4e2
4 "0 mr
5 mv03
:
6 e2
(8.246)
Therefore,
P =
1 4 ne4 v04
; (W).
4 "0 9 mc5
(8.247)
This de…nes a radiation cross-section,
=
P
1
nmv03
2
=
8 v 2
r ; (m2 )
9 c e
where 21 nmv03 is the energy ‡ux density of the beam and
re =
1 e2
= 2:85
4 "0 mc2
is the classical radius of electron.
44
10
15
m;
(8.248)
Problems
8.1 A charged particle with charge e and mass m undergoes motion in external electric and
magnetic …elds E and B: Show that the radiation power is given by
1 2e2 6 2
a
(
a)2
4 "0 3c3
2e2 2
1
(E + v B)2
4 "0 3m2 c3
P (t0 ) =
=
where a is the acceleration. Since a2
( E)2 .
(
E)2 ;
(
a)2 is positive de…nite, so is (E + v
Note: In an instantaneous rest frame of the particle, v = = 0 and
power in that frame is thus with respect to the proper time,
B)2
= 1: The radiation
dE 0
1
2e2
= P( ) =
E02 ;
d
4 "0 3m2 c3
where E0 is the electric …eld also in that frame. Since the electric …eld is Lorentz transformed
as
E0k = Ek ; E0? = (E? + v B);
we readily see that
P ( ) = P (t0 );
that is, the proper radiation power is Lorentz invariant. This is not surprising because both
time and energy are Lorentz transformed in a similar manner.
8.2 Carry out the integration in the radiation power due to acceleration perpendicular to the
velocity
2 _
?
1 e
0
P (t ) =
4 "0 4 c
2
Z
(1
1
(1
cos )5
to obtain
2 _
1 2e
0
P (t ) =
4 "0 3c
cos )2
1
2
sin2 cos2
d
2
4
:
Plot the angular distribution of the radiation intensity
f( ) =
for
(1
1
(1
cos )6
cos )2
1
2
sin2 cos2
;
= 0 and =2 in a highly relativistic case.
8.3 A nonrelativistic electron with velocity v collides head-on with a heavy negative ion having
a charge Ze:Find the total energy radiated. What is the dominant frequency component?
45
8.4 A highly relativistic electron passes by a proton at an impact parameter b:Assuming that
the electron trajectory is una¤ected by the proton and radiation, show that the total energy
radiated is
2
4
e6
1
e6 2
1
:
'
E=
2
4 "0 12m2e c3 b3 v 1
4 "0 4m2e c4 b3
What is the dominant frequency component contained in the radiation …eld?
8.5 An electron having an initial energy E(t = 0) = 0 mc2 undergoes cyclotron motion in a
magnetic …eld B: Show that the rate of energy loss to radiation is given by
dE(t)
1 2e4 B 2 2
E
=
dt
4 "0 m4 c5
m2 c4 :
Integrate this and …nd an expression for E(t):
8.6 An electron is placed in a plane wave E0 ei(kx
Show that the radiation power is
!t) :
P = I0
If ~!
mc2 ; electron recoil may be ignored.
8 2
r ;
3 e
where I0 = c"0 E02 (W/m2 ) is the intensity of the incident wave and
re =
e2
= 2:8
4 "0 mc2
is the classical radius of electron.
=
10
15
(m),
8 2
r ;
3 e
is known as Thomson scattering cross section.
Note: For ~!
mc2 ; Klein and Nishina, solving the Dirac equation to take into account
electron recoil, obtained
mc2 1
2~!
' re2
+ ln
:
~! 2
mc2
8.7 The light from Crab nebula is believed to be synchrotron radiation emitted by highly energetic
electrons trapped in a weak intergalactic magnetic …eld. If the electron energy is 2 1011 eV
and B = 10 7 T, what is the dominant wavelength component?
8.8 Perform numerically the integration in the radiation power of l-th harmonic of synchrotron
radiation,
Z
sin(2 l sin x) 2
1 (e! 0 )2 l
Pl =
( cos 2x 1) cos(2lx)dx;
4 0
c
sin x
0
for the case
= 0:99 ( ' 7) and plot the result as a function of l:
8.9 Estimate the bremsstrahlung cooling rate (in W/m3 ) of a hydrogen plasma for an electron
temperature of 10 keV and plasma density of n = 1020 m 3 : If the plasma is con…ned by a
magnetic …eld of 5 T, what is the cooling rate due to cyclotron radiation if reabsorption by
46
the plasma is ignored? (Cyclotron radiation can be reabsorbed by a plasma if its density is
high enough and its contribution to energy loss may be ignored.)
8.10 Assuming a permittivity in the form
" (!) =
! 2p
1
!2
! 20
!
"0 ;
carry out the integration in the rate of Cherenkov radiation,
dE
1 e2
=
dz
4 "0 c20
Z
0
1
! 1
1
2
d!:
8.11 Two identical charges are moving along a circular orbit of radius a at an angular velocity !:
Assume nonrelativistic motion, !a
c: Angular separation of the charges is 0 : If 0 = 0;
the problem reduces to a charge 2e rotating at ! which radiates as a dipole. If 0 = ; the
dipole moment vanishes and higher order quadrupole radiation should be considered. At what
angular separation 0 does the dipole radiation power become equal to quadrupole radiation
power? Is it necessary to consider magnetic dipole radiation?
8.12 Four identical charges are placed along a circle of radius a with equal angular spacing =2:
If the system rotates about the axis at angular velocity !; what is the lowest order radiation
power? What would happen if the number of charges N increases?
8.13 Generalize the quadrupole radiation in electron-electron collision to relativistic velocity.
47