graduate electrodynamics notes (8 of 9)

7/30/2019 Graduate electrodynamics notes (8 of 9)

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Chapter 8. Radiation from Moving ChargesNotes: Most of the material presented in this chapter is taken from Jackson, Chap. 12 and

14, and Rybicki and Lightman, Chap. 4.

In this chapter, we will be using Gaussian units for the Maxwell equations and otherrelated mathematical expressions.

Latin indices are used for space coordinates only (e.g., i = 1,2,3, etc.), while Greekindices are for space-time coordinates (e.g., ! = 0,1,2,3, etc.).

8.1 Solution of the Wave Equation in Covariant FormIn this chapter, we will solve for the electromagnetic fields, in a covariant manner,

through the use of the four-potential. We will do so in a way similar to what was done inChapter 4 in analyzing the solution of the wave equation for the potentials. The

differences between to the two analyses reside in the facts that in this chapter we will use

a covariant treatment, and solve the wave equation through the (invariant) Green functionmethod (a generalization of what was done in Section 4.3.2).

Our starting point is equation (7.96) for the wave equation of the four-potential A!

in the

Lorenz gauge (i.e., !"A"= 0 ). More precisely, we wish to solve

!A!=

4"

cJ

!, (8.1)

for a general four-current distribution J!!

x( ) . We can accomplish this by first solving for

the Green function D!

x,

!

!x

( )with

!xD

!

x,

!

!x( ) = "!

x #!

!x( ), (8.2)

and then find the components of the four-potential using

A! !x( ) =

4"

cD

!

x,!

#x( )J!!

#x( ) d4 #x$ . (8.3)

Since the problem is spherically symmetric (that is, in four dimensions), then the Green

function can only depend on the difference!

z !!

x "!

#x , and we write therefore

D!

x,!

!x( ) = D!

x "!

!x( ) = D!

z( ) . We now operate on both sides of equation (8.2) with aLaplace transform (see the Appendix on the subject) for the time component, and a three-

dimensional Fourier transform for the three-space components. Doing so, we find

s2+ k

2( )D s,k( ) = 1, (8.4)


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since != !0!0" #

2 . Please note that, in this case, the Laplace transform links the

component z0

(not the time) to the s-domain . Alternatively, we can write

D s,k( ) =1

s2+ k

2

( )

=1

s ! ik( ) s + ik( ), (8.5)

and we are ready to apply the inverse transforms to recover D!

z( ) . By first computing the

inverse Laplace transform (with the residue theorem), we find

D!

z( ) =H z

0( )

2!( )3

d3

k eik"z e

ik0z0

2ik#

e#ik0z0

2ik

$%&

'()*

=

H z0( )

2!( )3

d3

ke

ik"z

ksin k

0z

0( )* ,(8.6)

with H z0( ) the Heaviside step function (see equation (1.14)). We now change to

spherical coordinate (in three-space, this time) where d3k= dkd!d" k2 sin !( ) , then

D!

z( ) =H z

0( )

2!( )2

dk ksin k0z

0( ) d cos "( )#$ %&eikRcos "( )

'1

1

(0)

(

=

H z0( )

2!2R

dk sin k0z

0( )sin kR( )0

)

( ,(8.7)

withR = z = x ! "x

. Expanding the sine functions with complex exponentials we have

D!

z( ) =H z

0( )8!

2R

dk eik R"z0( )

+ e" ik R"z0( ) " eik R+z0( ) " e"ik R+z0( )( )

0

#

$ , (8.8)

and making the change of variable k! "k for the integrals where the imaginaryargument of the corresponding exponential is negative, we then find

D!

x !!

"x( ) =H x

0! "x

0( )8#

2R

dk eik R! x

0! "x

0( )$% &' ! eik R+ x

0! "x

0( )$% &'( )!(

(

)

= H x0

! "x0( )

4#R* R ! x

0! "x

0( )$% &' !* R + x0 ! "x0( )$% &'{ }.(8.9)

But since R > 0 and x0! "x

0> 0 , then the Green function is

D!

x !!

"x( ) =H x

0! "x

0( )4#R

$ R ! x0! "x

0( )%& '( . (8.10)


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Alternatively, equation (8.10) can be transformed with

!!

x "!

#x( )2$

%&' = ! x0 " #x0( )

2

" x " #x2$

%&'

= ! x0 " #x0" R( ) x0 " #x0 + R( )$% &'

=1

2R! x

0" #x

0" R( )+ ! x0 " #x0 + R( )$% &',

(8.11)

where we used equation (1.13)

! f x( )"# $% =! x & xi( )

df

dx

'()

*+,

x=xi

i

- (8.12)

and

d!

x !!

"x( )2#

$%&

d x0! "x

0

x0! "x

0=R

= 2 x0! "x

0x0! "x0 =R

= 2R. (8.13)

Furthermore, since only one of the two delta functions (i.e., the first) in equation (8.11),for reasons discussed earlier, contributes to the Green function, then

D

!

x !!

"x( ) =1

2#H x0 ! "x0( )$

!

x !!

"x( )2

%& '( (8.14)

8.2 The Linard-Wiechert Potentials and Fields for a Point ChargeWe now consider the potentials and fields due to a single moving particle of charge q . If

the position of the particle in a given inertial frame K is r t( ) , then its charge and current

densities in that frame are

! x,t( ) = q" x # r t( )$% &'

J x,t( ) = qu t( )" x # r t( )$% &', (8.15)

where u t( ) = dr t( ) dt is the velocity of the charge in K . Using the definition for the

four-current (see equation (7.91)), equations (8.15) can be combined into a single

covariant relation with


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J! = qc U!

"( )#!

x $!

r "( )%& '( d") , (8.16)

where U!

is the charges four-velocity, and r!

its four-position (c.f., equation (6.13) of

the second problem list). In K , we have

U!=

"c

"u

#$%

&'(

, and r!

)( ) =ct

r t( )

#$%

&'(

. (8.17)

Substituting equations (8.16) and (8.14) into equation (8.3), we have for the four-potential

A!= 2q d" U

!"( ) d4 #x H x0 $ #x0( )%

!

x $!

#x( )2&

'()%

!

#x $!

r "( )&' ()**= 2q d" U

!"( )H x0 $ r0 "( )&' ()%

!

x $!

r "( )&' ()2{ }* ,

(8.18)

where the integral is evaluated at the retarded proper time ! = !0

(since from the

argument of the Heaviside function x0> r

0!( ) ) defined by

!

x !!

r "( )#$ %&2

= 0. (8.19)

We use once more equation (1.13) (or (8.12)) with

d

d!

!

x "!

r !( )#$ %&2{ } = "2 !x " !r !( )#$ %&'U' !( ), (8.20)

and

!!

x "!

r #( )$% &'2{ } =

! #" #0( )+! #+ #0( )

2!

x "!

r #( )$% &'( U( #( )

, (8.21)

where it is understood that these relations are evaluated at !0. Since only the first of the

two delta functions in equation (8.21) contributes to the integral of equation (8.18), the

four-potential becomes

A! !x( ) =

qU! "( )

!

U #!

x $!

r "( )%& '( "="0

(8.22)

These potentials are known as the Linard-Wiechert potentials. In the inertial frame K ,

we can expand the denominator using equations (8.17) to get


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!

U !!

x "!

r #0( )$% &' =U0 x0 " r0 #0( )$% &' " u ! x " r #0( )$% &'

= (cR " (u !nR

= (cR 1" !n( ),

(8.23)

where R ! x0 " r0 #0( ) = x " r #0( ) , and n is the unit vector in the direction of x ! r "0( ) ,and = u "( ) c . We are now in a position to give expressions for the scalar and vector

potentials in K

! x,t( ) =q

1" $ n( )R

%

&'

(

)*ret

A x,t( ) =q

1" $ n( )R

%

&'

(

)*ret

(8.24)

where ret means that the potentials are to be evaluated at the retarded proper time !0,

given by r0

!0( ) = x0 " R . The electromagnetic fields could be evaluated from equations

(8.24) (or equations (8.22)), but we will use instead equation (8.18) as a starting point,

and apply equation (7.96). Thus, we need to evaluate the following

!"A# = 2q d$U# $( ) !"H x0% r

0$( )&' ()*

!

x %!

r $( )&' ()2{ }(+

+H x0% r

0$( )&' ()!

"*!

x %!

r $( )&' ()2{ })

= 2q d$U# $( ) * x0 % r0 $( )&' ()*!

x %!

r $( )&' ()2

{ }(++H x

0% r

0$( )&' ()!

"*!

x %!

r $( )&' ()2{ })

= 2q d$U# $( ) * x0% r

0$( )&' ()* %R

2( )(++H x

0% r

0$( )&' ()!

"*!

x %!

r $( )&' ()2{ }).

(8.25)

We will not, however, consider in our analysis cases where R = 0 (i.e., the observer is

not located at the retarded position of the source), equation (8.25) then simplifies to

!"A# = 2q d$U# $( )H x0% r

0$( )&' ()!

"*!

x %!

r $( )&' ()2{ }+ . (8.26)

We will need to use the following to solve this integral


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!"# f $( )%& '( = !" f $( )%& '(

d

df# f $( )%& '(

= !" f $( )%& '(d$

df

d # f $( )%& '({ }d$

.

(8.27)

With f !( ) =!

x "!

r !( )#$ %&2

we have

!"!

x #!

r $( )%& '(2{ } = !" x) # r)( ) x) # r)( )%& '(= *") x) # r)( )+ *") x) # r)( )= 2 x

" # r"( ),

(8.28)

and using equation (8.20) we find

!"#!

x $!

r %( )&' ()2{ } = $

x" $ r"( )

!

x $!

r %( )&' ()*U* %( )

d

d%#

!

x $!

r %( )&' ()2{ }. (8.29)

Inserting equation (8.29) into equation (8.26) we get

!"A# = $2q d%U# %( )H x0 $ r0 %( )&' ()x" $ r" %( )&' ()

!

x $!

r %( )&' ()* U* %( )

d

d%+

!

x $!

r %( )&' ()2{ },

=

2q d%

d

d%

U# %( ) x" $ r" %( )&' ()H x0 $ r0 %( )&' ()!

x $!

r %( )&' ()* U* %( )

-

.

/

0/

1

2

/

3/+

!

x $

!

r %( )&' ()

2

{ }, ,

(8.30)

after integrating by parts, and

!"A# = 2q d$ %U0

x" % r"( )U#& x0 % r0( )x' % r'( )U'

+d

d$

x" % r"( )U#

x' % r'( )U'(

)**

+

,--

./0

10

230

40&

!

x %!

r( )2(

)+,5

= 2q d$ %U

0U# x" % r"( )& x0 % r0( )

x' % r'( )U'+

d

d$

x" % r"( )U#

x' % r'( )U'(

)**

+

,--

H x0% r

0( )./0

10

230

40

5 &!

x %!

r( )2(

)+,

= 2q d$U

0U#R"& x

0% r

0( )x' % r'( )U'

& %R2( )+d

d$

x" % r"( )U#

x' % r'( )U'(

)**

+

,--

H x0% r

0( )&!

x %!

r( )2(

)+,

./0

10

230

405 ,

(8.31)

but the first part of the integrand does not contribute to integral when R ! 0 , so


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175

!"A# = 2q d$d

d$

x" % r" $( )&' ()U

# $( )

x* % r* $( )&' ()U* $( )

+,-

.-

/0-

1-H x

0% r

0$( )&' ()2

!

x %!

r $( )&' ()2{ }3 . (8.32)

Upon using equation (8.21) to solve this integral, and equation (7.81) for the

electromagnetic tensor, we find that

F!" =q

x# $ r#( )U#

d

d%

x! $ r! %( )&' ()U" %( ) $ x" $ r" %( )&' ()U

! %( )

x* $ r* %( )&' ()U* %( )

+,-

.-

/0-

1- %=%0

(8.33)

With this result, we are now in a position to express the electric and magnetic induction

field. Remembering that R ! x0" r

0#0( ) = x " r #0( ) , then

x!" r

!( ) = R,"Rn( ), (8.34)

and furthermore

U!= "c,#"c( ). (8.35)

Since

d!

d"=

d

d"1# %( )

#12= !

3%

d

d"= !

4%! , (8.36)

with ! = d dt, we have

dU!

d"= c#

4 % !, & c#4 % !( ) c#2 !( )d

d"x

#& r

#( )U#'( )* = &U#U#+ x

#& r

#( )dU

#

d"

= &c2 + x#& r

#( )dU

#

d".

(8.37)

Using equations (8.37) and (8.23), it can be shown from equation (8.33) that

E x, t( ) = qn !

#21! $ n( )

3R

2

%

&''

(

)**ret

+q

c

n + n !( )+ !{ }1! $ n( )

3R

%

&

''

(

)

**ret

B x, t( ) = n + E[ ]ret

(8.38)


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We see that the electromagnetic fields are composed of a velocity and an accelerationcomponent, and only the latter radiates in the far field. Note also that in the radiation field

E, B, and n form a right-handed triad of mutually perpendicular vectors, and that

E = B .

8.3 The Total Power Radiated by an Accelerated ChargeIn order to simplify our derivation for the total power radiated by an accelerated charge,we need to make a short digression to discuss ways to evaluate the time component of

any four-vector as measured by an observer with a given four-velocity.

8.3.1 Evaluating the time components of four-vectorsSince a four-velocity

!

U of a particle is expressed as

U!=

"c

"u

#$%

&'(

(8.39)

when seen by a stationary observer, it takes a particularly simple form in the rest frame of

the particle itself. That is, for an observer moving with the particle we have

!U " =c

0

#$%

&'(

, (8.40)

where 0 is the null vector in three-space. Because of this, the time component of an

arbitrary four-vector!

A as measured in the rest frame of the particle is

!A0=

1

c!U" !A

", (8.41)

but since !U" !A"=

!

U #!

A is an invariant, we can write in general

!A0=

1

c

!

U "!

A (8.42)

It is important to emphasize the fact that if !A 0 is evaluated in the rest frame of the

particle, the quantity

!

U !!

Acan be evaluated in any reference frame. For example, set!

A =!

x in equation (8.42)


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!x0=1

cU

"

x"=1

c

dx"

d#x

"

=1

2c

d

d#x

"

x"( ) =

1

2c

d c2#

2( )d#

= c#,

(8.43)

which is expected, since proper time is the time seen by an observer in the rest frame.

Another example can be worked out if we set!

A =!

k , then

!k 0 =1

cU

"k" = # ck

0 $ u %k( )

= #k0 1$uk

ck0cos &( )

'

()*

+,,

(8.44)

and

!" = #" 1$u

c

cos %( )&

'()

*+. (8.45)

Equation (8.45) is the Doppler shift formula.

8.3.2 Emission from Relativistic ParticlesLets consider a particle moving at relativistic velocities, and we define its instantaneousrest frame !K as the referential where the particle has zero velocity at a given time. We

also define another frame K relative to which the particle is moving at velocity !u (i.e.,u is the instantaneous velocity of !K , or the particle, as seen by K ). We can use

equation (8.42) to relate radiated power as seen in both frames. First, we consider the

four-momentum!

P , and more precisely its time component !P 0 in the instantaneous rest

frame of the particle

!P 0 =1

cU

"P"=

#

cP

0c $ u %p( )

=

#

cP

0c $ #mu2( ) = #P0 1$

u2

c2

&'(

)*+

= P

0

#,

(8.46)

since P!

= mU!

. Similarly, we have


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d !x 0 =1

cU

"dx

"=

#

ccdx

0 $ u %dx( )

= #dx0

1$u

2

c2

&'(

)*+=

dx0

#.

(8.47)

Equations (8.46) and (8.47) can obviously rewritten as

E0= ! "E

0and dt= !d "t . (8.48)

If we consider the amount of energy dW (ord !W ) radiated by the particle in an amount

of time dt (ord !t ), we can evaluate the power radiated P (or !P ) as

P =dW

dtand !P =

d !W

d !t, (8.49)

and from equations (8.48) we find that the total power emitted is a Lorentz invariant.That is,

P = !P . (8.50)

We can use this invariance of the total power to our advantage. If we put ourselves in therest frame !K of the particle, then we can consider the non-relativistic form for the

equations of the electromagnetic fields. From equations (8.38), in the radiation field, wehave

E x, t( ) =q

c

n ! n ! !

( )R#

$%%

&

'((

ret

, (8.51)

which implies that the radiation is polarized in the plane containing the acceleration of

the particle and n . The instantaneous energy flux S is given by

S =c

4!E " B( ) =

c

4!E

2

n. (8.52)

The power radiated per unit solid angle is

dP

d!=

c

4"RE

2

=

q2

4"cn # n # !( )

2

.

(8.53)

If we define ! as the angle between the acceleration !u and n , then


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dP

d!=

q2

4"c3

!u2sin

2#( ), (8.54)

and the total instantaneous power radiated by an accelerated charge, in the non-relativistic limit, is given by the so-called Larmor formula

P =2

3

q2

c3

!u2

(8.55)

It is possible to generalize this result to the fully relativistic case if we remember that the

four-acceleration a ! dU d" is perpendicular to the four-velocity (see equation

(7.105)) with

aU

= 0. (8.56)

Then using equations (8.42) and (8.56) we find that

!a0=

1

ca

U

= 0, (8.57)

and with a = !u

a !a = " #a

#a= "a

a

= "

!

a !!

a. (8.58)

Hence, we can write the generalized relativistic version of the Larmor formula in a

manifestly covariant form as

P = !2

3

q2

c3d!

U

d"#d!

U

d"

$

%&'

()(8.59)

Alternatively, we can write from the first of equations (8.37) that

dU!

d"

dU!

d"= c

2#8 % !( )2

& c2#8 % !( )2

$2 & c2#4 !( )2

& 2c2#6 % !( )2

= c2#6 % !( )2

& c2#4 !( )

2

& 2c2#6 % !( )

2

= & c2#4 !( )2

+ c2#6 % !( )

2'()

*+,

= &c2#6 !( )2

& ( )2 !( )

2

& % !( )2'

()*+,{ }

= &c2#6 !( )2

& - !( )2'

()*+,,

(8.60)


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180

and finally,

P =2

3

q2

c!

6 !( )2

# $ !( )2%

&'()*

(8.61)

We see that the power radiated is composed of parts proportional and perpendicular to theacceleration of the particle.

8.4 Angular Distribution of Radiation Emitted by an AcceleratedCharge

In the non-relativistic approximation, i.e., equation (8.54), the angular distribution of

radiation is sin2 !( ) , where ! is the angle between the acceleration !u and the unit vector

n . In the fully relativistic case, we must use the radiation components of equations

(8.38) for the electromagnetic fields to calculate the energy flux emitted in the directionofn . That is,

S ! n[ ]ret=

c

4"n ! E # B( )$% &'ret

=

c

4"n ! E # n # E[ ]( )$% &'ret

=

c

4"n ! n E

2( E n ! E[ ]( )$% &'ret

=

c

4" E

2$%

&'ret ,

(8.62)

since n !E = 0 for the radiation field (see equations (8.38)). Finally, the flux is

S !n[ ]ret=

q2

4"c

1

R2

n # n $( ) # !{ }1$ !n( )

3

2&

'

((

)

*

++ret

. (8.63)

Although the energy flux expressed in equation (8.63) is evaluated at the retarded time

!t = t" R !t( ) c , it reaches the observer located at x at time t . The energy per unit area

measured by the observer in a period of time going from T1+ R T

1( ) c and T2 + R T2( ) c is given by

S !n[ ]ret

dtt=T1 +R T1( ) c

t=T2 +R T2( ) c

" = S !n( )dt

d #td #t

#t =T1

#t =T2

" , (8.64)


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where !t is the retarded time as measured in the inertial frame of the observer, and notthe time in the frame where the charge is at rest (as it appears in Lorentz transformations,

for example). Since dt= 1! #n( )d $t (see equation (4.63) of the lecture notes, and

equation (6.19) of the second list of problem), we can write the power radiated per unitangle as

dP !t( )

d"= R

2S #n( ) 1$ #n( ). (8.65)

If we further assume that the amplitude of the displacements of the charge are small

compared to R , then both n and R are (approximately) unaffected by its motion and

dP !t( )d"

=

q2

4#c

n $ n %( )$ !{ }2

1% 'n( )5

(8.66)

8.4.1 Acceleration Parallel to the VelocityWhen a charge exhibits a linear motion and its acceleration is parallel to its velocity, then

equation (8.66) becomes

dP! !t( )

d"=

q2 "u2

4#c3sin

2 $( )

1% &cos $( )'( )*5, (8.67)

where ! is, once again, the angle betweenn and

(or!

). Equation (8.67) simplifies toequation (8.54) in the non-relativistic limit, but as !" 1 the angular distribution is

tipped forward and increases in amplitude, as show in Figure 8.1. Setting the!-derivative of equation (8.67) to zero, we find that the angle of maximum intensity !

max

is given by

!max

= cos"1 1

3#1+15#2 "1( )$

%&

'

(). (8.68)

In the extremely relativistic limit this relation becomes

lim!"1

#max

=

1

2$, (8.69)


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Figure 8.1 Radiation pattern for a charge accelerated along its direction of motion. Thetwo patterns are not to scale, the relativistic one (!! 2 ) has been reduced by a factor ofabout 100 for the same acceleration.

since != 1"1 #2( )1

2! 1"1 2#2( ). Because the angular distribution of the radiation is

highly peaked, we can write cos !( ) ! 1"!2 2 , and from equation the previous

approximation for!, we find that

1! "cos #( ) ! 1! 1!1

2$2%&'

()*

1!#2

2

%&'

()*!

1+ $2#2

2$2, (8.70)

and thus

dP!

!t( )

d""

8

#

q2 #u2

c3

$8$2%2

1+ $2%2&' ()5. (8.71)

8.4.2 Acceleration Perpendicular to the VelocityIf the orientation of n, , and ! as a function of the angles ! and " is as defined in

Figure 8.2, then equation (8.66) becomes

dP! "t( )

d#=

q2

4$c

n % n % !( )' n % % !( )2

1' ( n( )5

=q2

4$c

n n ( !( ) ! ' n ( !( )+ ! n (( )2

1' ( n( )5

=

q2 !u2

4$c3sin )( )cos *( ) n ' &ez+, -. ' 1' &cos )( )+, -.ex

2

1' &cos )( )+, -.5

,

(8.72)

and after calculating the required scalar products


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183

Figure 8.2 The orientation of n, , and ! as a function of the angles ! and " for the

case of a charge accelerated in a direction perpendicular to its velocity.

dP! "t( )

d#

=

q2

4$c3

!u2

1% &cos '( )() *+3

1%sin

2 '( )cos2 ,( )

-2

1% &cos '( )() *+2

./0

10

230

40

. (8.73)

Again, if we concentrate on the extreme relativistic limit where !" 1, then inserting

equation (8.70) into equation (8.72) we get

dP! "t( )

d#!

2

$

q2

c3%6

"u2

1+ %2&2'( )*3

1+4%

2&2 cos2 ,( )

1+ %2&2( )2

'

(

--

)

*

.

.. (8.74)

The angular distribution of the radiation is mostly contained within a range of angles

approximately given by !!1 " , as shown in Figure 8.3.

8.4.3 Distribution in Frequency and Angle of Energy Radiated by AcceleratingCharges

Looking back at equation (8.62), we could equation write the following for the

expression of the power radiated per unit solid angle as function of observers time (notthe retarded as was done in the last section)

dP t( )

d!= A t( )

2

, (8.75)

with

A t( ) =c

4!

"#$

%&'

1 2

RE[ ]ret

. (8.76)

Integrating equation (8.75) over time yields the energy radiated per solid angle


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184

Figure 8.3 The angular distribution of radiation emitted by a particle with

perpendicular acceleration and velocity (the acceleration is directed upward).

dW

d!= A t( )

2

dt"#

#

$ . (8.77)

If we now introduce the Fourier transform pair

A !( ) =1

2"A t( )ei!t dt

#$

$

%

A t

( )=

1

2"A !

( )e#i!t

d!#$

$

%,

(8.78)

(please note that if A t( ) is real, then A !( ) = A" #!( ) ), then equation (8.77) can be

transformed to

dW

d!=

1

2"d# d $# A $#( )A% #( ) dt e i $# ( )t

&'

'

(&''

(&''

(

= d# d $# A $#( )A% #( )) $# ( )&'

'

(&''

(

= A #( )2

d#&'

'

(.

(8.79)

The general result that

A !( )2

d!"#

#

$ = A t( )2

dt"#

#

$ (8.80)

is called Parsevals theorem. We now define a new quantity for the radiated energy per

solid unit angle per unit frequency as

d2I !,n( )

d"d!

= A !( )2

+ A #!( )2

, (8.81)

which for a real signal becomes

d2I !,n( )

d"d!= 2 A !( )

2

. (8.82)


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185

Inserting equation (8.81) (or (8.82)) in equation (8.79), we find

dW

d!=

d2I ",n( )

d!d"0

#

$ d". (8.83)

To apply equation (8.83) to the problem of an accelerated charge, we first calculate theFourier transform of the radiation electric field in the far field (from the first of equations(8.38))

A !( ) =q2

8"2c

#$%

&'(

1 2

ei!tn ) n *( ) ) !{ }

1* ,n( )3-

.//

0

122

ret

dt*3

3

4 , (8.84)

where the quantity in between the brackets is to be evaluated at the retarded time

!t = t" R !t( ) c . We now change the variable of integration from t to !t just as we did for

equation (8.64), and we get

A !( ) =q2

8"2c

#$%

&'(

1 2

ei! )t +R )t( ) c*+ ,-

n . n /( ). !*+ ,-1/ 1n( )2

d )t/2

2

3 , (8.85)

and we, again, assume that the amplitude of the displacements of the charge are small

compared to R so that both n and R are (approximately) unaffected by its motion.Furthermore, if we define the distance between the observation point and the origin from

which the charges position r !t( ) is evaluated as x , then we can write

R !t( ) ! x " n # r !t( ), (8.86)

and

A !( ) =q2

8"2c

#$%

&'(

1 2

ei!x c ei! )t *n+r )t( ) c,- ./

n 0 n *( )0 !,- ./1* +n( )2

d )t*2

2

3 . (8.87)

We should note that the exponential term in front of the integral has no physical

significance and can be neglected, since the quantity we seek to evaluate is proportional

toA !

( )

2

. Inserting equation (8.87) into equation (8.82), we get

d2I !,n( )

d!d"=

q2

4#2c

ei! $t %n&r $t( ) c'( )*

n + n %( ) + !'( )*1% &n( )

2d $t

%-

-

.

2

(8.88)


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186

Given the equation for the motion of the charge r !t( ) , this result (i.e., equation (8.88))

can be used to calculate the radiated energy per unit solid angle per unit frequency.Alternatively, we can modify equation (8.87) by integrating by parts, since the ratio

involving n, , and ! is a perfect time differential. More precisely,

d

d !t

n " n "( )

1$ %n

&

'(

)

*+ =

n " n " !( )&' )* 1$ %n( )+ ! %n( ) n " n "( )&' )*1$ %n( )

2

=

n " n " !( ) $ ! %n( )n $ !&' )* %n( )+ %n( )n $&' )*! %n( )

1$ %n( )2

=

n " n " !( ) + ! %n( ) $ ! %n( )1$ %n( )

2

=

n " n " !( ) $ n " " !( )1$ %n( )2

=

n " n $( )" !{ }1$ %n( )

2,

(8.89)

so that we can write

A !( ) =q2

8"2c

#$%

&'(

1 2

ei! )t *n+r )t( ) c,- ./

n 0 n *( )0 !,- ./1* +n( )2

d )t*2

2

3

= *i! q2

8"2c

#$%

&'(

1 2

ei! )t *n+r )t( ) c,- ./ 1* +n( ) n 0 n 0( )

1* +n,-4 .

/5 d )t*2

23

= *i!q2

8"2c

#$%

&'(

1 2

ei! )t *n+r )t( ) c,- ./

n 0 n 0( ),- ./ d )t*22

3 ,

(8.90)

where it was assumed that vanishes at !t = " . Inserting equation (8.90) into equation

(8.82) we finally find that

d2I !,n( )

d!d"=

q2!2

4#2ce

i! $t %n&r $t( ) c'( )*n + n +

( )'(

)*d $t

%-

-

.

2

(8.91)

graduate electrodynamics notes (8 of 9)

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