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Radiation by Moving Charges

May 19, 20101

1J.D.Jackson, Classical Electrodynamics, 3rd Edition, Chapter 14Radiation by Moving Charges
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Lienard - Wiechert Potentials

The Lienard-Wiechert potential describes the electromagneticeffect of a moving charge.

Built directly from Maxwells equations, this potential describes thecomplete, relativistically correct, time-varying electromagnetic fieldfor a point-charge in arbitrary motion.

These classical equations harmonize with the 20th centurydevelopment of special relativity, but are not corrected forquantum-mechanical effects.

Electromagnetic radiation in the form of waves are a natural result

of the solutions to these equations. These equations were developed in part by Emil Wiechert around

1898 and continued into the early 1900s.


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Lienard - Wiechert Potentials

We will study potentials and fields produced by a point charge, for whicha trajectoryx0(t

)has been defined a priori.

It is obvious that when a charge qis radiating is giving away momentumandenergy, and possiblyangular momentumand this emission affectsthe trajectory. This will be studied later. For the moment, we assumethat the particle is moving with a velocity much smaller than c.The density of the moving charge is given by

(x, t) =q(x x0[t]) (1)and since in general the current density J is v, we also have

J(x, t) =qv(x x0[t]) , where v(t) = dx0dt

(2)

In the Lorentz gauge ( A+ (1/c)/t= 0) the potential satisfy thewave equations (??) and (??) whose solutions are the retarded functions

(x, t) = (x, t |xx|/c)

|x

x

| d3x (3)

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A(x, t) = 1

c

J(x, t |xx|/c)|xx| d

3x (4)

It is not difficult to see that theseretarded potentialstake into accountthe finite propagation speed of the EM disturbances since an effectmeasured atx and twas produced at the position of the source at time

t=t|xx0(t)|c

(5)

Thus, using our expressions for andJ from eqns (1) and (2)and

putting v/c,

(x, t) =q (x x0[t |xx|/c])

|x

x

| d3x (6)

A(x, t) =q

(t |xx|/c)(x x0[t |xx|/c])|xx| d

3x (7)

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Note that for a given space-time point(x, t), there exists only one point

on the whole trajectory, the retarded coordinatex coresponding to theretarded time tdefined in (5) which produces a contribution

x=x0(t) =x0(t |xx0|/c) (8)Let us also define the vector

R(t) =xx0(t) (9)

in the directionn R/R. Then

(x, t) = q

(x x0[t R(t)/c])

R(t) d3x (10)

A(x, t) = q

(t R(t

)/c)(x

x0[t R(t

)/c])R(t)

d3x (11)

Because the integration variablex appears in R(t)we transform it byintroducing a new parameterr, where

x =x

x0[t

R(t)/c] (12)

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The volume elements d3x and d3x are related by the Jacobiantransformation

d3x =Jd3x , where J

1 n(t) (t)

(13)

is the Jacobian (how?).With the new integration variable, the integrals for the potentialtransform to

(x, t) =q

(x) d3x

|xx x0(t)|(1 n )(14)

andA(x, t) =q

(t) (x) d3x|xx x0(t)|(1 n )

(15)

which can be evaluated trivially, since the argument of the Dirac deltafunction restrictsx to a single value

(x, t) =

q

(1 n )|x x|

t

=

q

(1 n )R

t

(16)

A(x, t) = q

(1

n

)|x

x|t

= q

(1

n

)Rt

(17)

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Lienard - Wiechert potentials : radiation fields

The next step after calculating the potentials is to calculate the fields viathe relations

B= A and E= 1c

A

t (22)

and we write the Lienard - Wiechert potentials in the equivalent form

(x, t) = q

(t t R(t)/c)

R(t) dt (23)

A(x, t) = q

(t)(t t+R(t)/c])R(t)

dt (24)

where R(t) |xx0(t)|. This can be verified by using the followingproperty of the Dirac delta function (how?)

g(x)[f(x)]dx=

i

g(x)|df/dx|

f(xi)=0

(25)

which holds for regular functions g(x)and f(x)of the integrationvariable x where xiare the zeros off(x).

The advantage in pursuing this path is that the derivatives in eqn (22)

can be carried out before the integration over thedeltafunction.Radiation by Moving Charges

Thi d i lifi h l i f h fi ld id bl i
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This procedure simplifies the evaluation of the fields considerably since,we do not need to keep track of the retarded time until the last step.We get for the electric field

E(x, t) =

q

(t t+ R(t)/c)R(t

) dt

qc

t

(t)(t t+R(t)/c)R(t)

dt (26)

Thus, differentiating the integrand in the first term, we get (HOW?)

E(x, t) = q

nR2

t t+ R(t

)c ncRt t+ R(t

)c

dt

qc

t

(t)(t t+R(t)/c)R(t)

(27)

But (HOW?)

t t+ R(t)

c

=

t

t t+ R(t

)

c

(28)

E(x, t)= q

n

R2

t t+ R(t

)

c dt+

q

c

t (n )

cR(t)

t t+ R(t

)

c (29)


W l h i l i h Di d l f i d i
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We evaluate the integrals using the Dirac delta function expressed inequation (25). But we need to know the derivatives of the deltafunctions arguments with respect to t. Using the chain rule ofdifferentiation

ddt

t t+ R(t

)c

=

1 n t

(30)

with which we get the result (HOW?):

E(r, t) =q n(1 n )R2t +

q

c

t n

(1 n )Rt (31)Since

R

t =

R

tt

t = nvt

t =c1t

t t

t =

1

(1

n

)(32)

Thus

t

n

(1 n )R

t

= 1

(1 n )

t

n

(1 n )R2

t

(33)

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By using the additional pieces

R|t= cn

t(34)

n|t= c

R

n(n )

t

(35)

d

dt

1 n

t

=

n + n

(36)

and we finally get

E(r, t) =q

(n )(1 2)(1 n )3R2

+n

(n )

c(1 n )3R

t(37)

A similar procedure for Bshows that

B(r, t) = A=n(t) E (38)

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Some observations

When the particle is at rest and unaccelerated with respect to us,the field reduces simply to Coulombs law qn/R2. whatever

corrections are introduced the do not alter the empirical law.

We also see a clear separation into the near field (which falls off as1/R2) and the radiation field (which falls off as1/R)

Unless the particle is accelerated (= 0), the field falls off rapidlyat large distances. But when the radiation field is present, itdominate over the near field far from the source.

As 1with = 0the field displays a bunching effect. Thisbunching is understood as being a retardation effect, resultingfrom the finite velocity of EM waves.

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Power radiated by an accelerated charge

If the velocity of an accelerated charge is small compared to the speed oflight ( 0) then from eqn (37)we get

E= q

c

n (n )

R

ret

(39)

The instantaneous energy flux is given by the Poynting vector

S= c4 E B= c4 |E|2n (40)The power radiated per unit solid angle is

dP

d=

c

4R2|E|2 = q

2

4c|n (n )|2 (41)

and ifis the angle between the acceleration v andn then the powerradiated can be written as

dP

d=

q2

4c3|v|2 sin2 (42)


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Larmor Formula

The total instantaneous power radiatedis obtained by integration overthe solid angle. Thus

P= q2

4c3|v|2

0

2sin3 d =2

3

q2

c3|v|2 (43)

This expression is known as the Larmor formula for a nonrelativisticaccelerated charge.

NOTE : From equation (39) is obvious that the radiation is polarized in

the plane containing v andn.

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Relativistic Extension

Larmors formula (43) has an easy relativistic extension so that can beapplied to charges with arbitrary velocities

P= 23

e2

m2c3|p|2 (44)

where m is the mass of the charged particle and p its momentum.The Lorentz invariant generalization is

P= 2

3

e2

m2c3dp

d

dp

d

(45)

where d=dt/ is the proper time and p is the charged particlesenergy-momentum 4-vector. Obviously for small it reduces to (44)

dp

d

dp

d =dp

d2

1

c2dE

d2

=dp

d2 2

dp

d2

(46)

If (45) is expressed in terms of the velocity & acceleration (E=mc2 &p=mv with = 1/(1 2)1/2), we obtain the Lienard result (HOW?)

P=2

3

e2

c

6 2

2

(47)Radiation by Moving Charges

A li i
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Applications

In the charged-particle accelerators radiation losses are sometimesthe limiting factor in the maximum practical energy attainable.

For a given applied force the radiated power (45) depends inverselyon the square of the mass of the particle involved. Thus these radiativeeffects are largest for electrons. In a linear accelerator the motion is 1-D. From (46) we can findthat the radiated power is

P= 23

e2

m2c3

dpdt

2

(48)

The rate of change of momentum is equal to the rate of change ofthe energy of the particle per unit distance. Thus

P=2

3

e2

m2c3

dE

dx

2

(49)

showing that for linear motion the power radiated depends only on theexternal forces which determine the rate of change of particle energy with

distance, not on the actual energy or momentumofthe particle.Radiation by Moving Charges
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The ratio ofpower radiatedtopower suppliedby external sources is

P

dE/dt =

2

3

e2

m2c31

v

dE

dx 2

3

(e2/mc2)

mc2dE

dx (50)

Which shows that the radiation loss in an electron linear accelerator willbe unimportant unless the gain in energy is of the order ofmc2 = 0.5MeVin a distance ofe2/mc2 = 2.8 1013cm, or of the order of2 1014MeV/meter. Typically radiation losses are completely negligiblein linear accelerators since the gains are less than50MeV/meter.

Can you find out what will happen in circular accelerators likesynchrotron or betatron?In circular accelerators like synchrotron or betatron can change drastically.In this case the momentumpchanges rapidly in direction as the particle

rotates, but the change in energy per revolution is small. This meansthat:

dp

d

=|p| 1c

dE

d (51)

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Then the radiated power, eqn (45), can be written approximately

P=2

3

e2

m2c322|p|2 =2

3

e2c

2 44 (52)

where = (c/), being the orbit radius.The radiative loss per revolution is:

E=2

c

P=4

3

e2

34 (53)

For high-energy electrons ( 1) this gets the numerical value

E(MeV) = 8.85 102 [E(GeV)]4

(meters) (54)

In a 10GeV electron sychrotron (Cornell with 100m) the loss perrevolution is8.85MeV. In LEP (CERN) with beams at 60 GeV

( 4300m) the losses per orbit are about300 MeV.


A l Di t ib ti f R di ti E itt d b
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Angular Distribution of Radiation Emitted by anAccelerated Charge

The energy per unit area per unit time measured at an observation point

at time tof radiation emitter by charge at time t =t R(t)/c is:

Snret

= e2

4c

1

R2

n [(n ) ]

(1 n)3

2

ret(55)

The energy radiated during a finite period of acceleration, say fromt =T1 to t

=T2 is

E=

T2+R(T2)/cT1+R(T1)/c

Sn

ret

dt=

t=T2t=T1

Sn

dtdt

dt (56)

Note that the useful quantity is (Sn)(dt/dt) i.e. the power radiatedper unit area in terms of the charges own time. Thus we define thepower radiated per unit solid angle to be

dP(t)

d =R2

Sn dt

dt =R2

Sn (1 n) (57)

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The angle max for which the intensity is maximum is:

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g y

cos max= 1

3

1 + 152 1

for 1 max 1

2 (60)

For relativistic particles, max is very small, thus the angular distributionis confined to a very narrow cone in the direction of motion. For small angles the angular distribution (59) can be written

dP(t)

d 8

e2v2

c3 8

()2

(1 +22)5 (61)

The peak occures at = 1/2, and the half-power points at = 0.23

and = 0.91.Radiation by Moving Charges

The root mean square angle of emission of radiation in the relativistic
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limit is

21/2 = 1

= mc2

E (62)

The total power can be obtained by integrating (59) over all angles

P(t) =2

3

e2

c3v26 (63)

in agreement with (47) and (48). In other words this is a generalizationof Larmors formula.

It is instructive to express this is terms of the force acting on the

particle.This force is F =dp/dtwherep=mv is the particles relativisticmomentum. For linear motion in the x-direction we have px=mvand

dpx

dt =mv+mv23 =mv3

and Larmors formula can be written as

P=2

3

e2

c3|F|2m2

(64)

This is the total charge radiated by a charge in instantaneous linear

motion.Radiation by Moving Charges

Angular distr of radiation from a charge in circular motion
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Angular distr. of radiation from a charge in circular motion

The angular distribution of radiation for a charge in instantaneous

circular motion with accelerationperpendicular to its velocity is

another example.We choose a coordinate system such

as is in the z-direction and in

the x-direction then the generalformula (58) reduces to (HOW?)

dP(t)

d =

e2

4c3|v|2

(1

cos )3 1

sin2 cos2

2(1

cos )2 (65)

Although, the detailed angular distribution is different from the linearacceleration case the characteristic peaking at forward angles is present.In the relativistic limit ( 1) the angular distribution can be written

dP(t)

d

2e2

c3

6 |v|2

(1 +2

2

)3 1

422 cos2

2

(1+2

2

)2 (66)

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The root mean square angle of emission in this approximation is similarto (62) just as in the 1-dimensional motion. (SHOW IT?)The total power radiated can be found by integrating (65) over allangles or from (47)

P(t) =2

3

e2|v|2c3

4 (67)

Since, for circular motion, the magnitude of the rate of momentum isequal to the force i.e. mvwe can rewrite (67) as

Pcircular(t) =

2

3

e2

m2c32

dp

dt

2(68)

If we compare with the corresponding result (48) for rectilinear

motion, we find that the radiation emitted with a transverseacceleration is a factor 2 larger than with a parallel acceleration.


Radiation from a charge in arbitrary motion

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Radiation from a charge in arbitrary motion

For a charged particle in arbitrary & extremely relativistic motion theradiation emitted is a superposition of contributions coming from

accelerationsparallel to and perpendicular to the velocity. But the radiation from the parallel component is negligible by a factor1/2, compare (48) and (68). Thus we will keeponly the perpendicularcomponentalone. In other words the radiation emitted by a particle in arbitrary motionis the same emitted by a particle in instantaneous circular motion, with aradius of curvature

= v2

v c

2

v(69)

where v is the perpendicular component of the acceleration.

The angular distribution of radiation given by (65) and (66)corresponds to a narrow cone of radiation directed along theinstantaneous velocity vector of the charge.

The radiation will be visible only when the particles velocity isdirected toward the observer.


Since the angular width of the beami 1/ h i l ill l
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is1/the particle will travel adistance of the order of

d=

in a time

t=

v

while illuminating the observerIf we consider that during the illumination the pulse is rectangular , thenin the timetthe front edge of the pulse travels a distance

D=ct=

Since the particle is moving in the same direction with speed vand movesa distance d in timetthe rear edge of the pulse will be a distance

L= D d=

1

1

23 (70)

behind the front edge as the pulse moves off.


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The Fourier decomposition of a finite wave train, we can find that thespectrum of the radiation will contain appreciable frequency components

up to a critical frequency,

c cL

c

3 (71)

For circular motion the termc/ is the angular frequency of rotation0 and even for arbitrary motion plays the role of thefundamentalfrequency. This shows that a relativistic particle emits a broad spectrum offrequencies up to 3 times the fundamental frequency.

EXAMPLE : In a 200MeV sychrotron, max 400, while0 3 10

8

s1

. The frequency spectrum of emitted radiation extendsup to 2 1016s1.


Distribution in Frequency and Angle of Energy Radiated
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Distribution in Frequency and Angle of Energy Radiated ...

The previous qualitative arguments show that for relativistic motionthe radiated energy is spread over a wide range of frequencies.

The estimation can be made precise and quantidative by use ofParsevals theoremof Fourier analysis.The general form of the power radiated per unit solid angle is

dP(t)

d = |A(t)|2 (72)

where A(t) = c

4

2 RE

ret

(73)

and E is the electric field defined in (37). Notice that here we will use the observers time instead of theretarded time since we study the observed spectrum.The total energy radiated per unit solid angle is the time integral of (72):

dW

d =

|A(t)|2dt (74)

This can be expressed via the Fourier transforms as an integral over the

frequency. Radiation by Moving Charges

The Fourier transform is:
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A() = 1

2

A(t)eitdt (75)

and its inverse:

A(t) = 12

A()eitd (76)

Then eqn (74)can be written

dW

d =

1

2

dt

d

dA() A()ei()t (77)

If we interchange the order of integration between tand we see thatthe time integral is the Fourier represendation of the delta function( ). Thus the energy radiated per unit solid angle becomes

dW

d =

|A()

|2d (78)

The equality of equations (74) and(78) is a special case ofParsevalstheorem.

NOTE: It is customary to integrate only over positive frequencies, since

the sign of the frequency has no physical meaning.


The energy radiated per unit solid angle per unit frequency interval is
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dW

d =

0

d2I(, n)

dd d (79)

where d2I

dd= |A()|2 + |A()|2 (80)

IfA(t)is real, form (75) - (76)it is evident that A() =A(). Then

d2

Idd

= 2|A()|2 (81)

which relates the power radiated as a funtion of time to the frequencyspectrum of the energy radiated.NOTE : We rewrite eqn (37) for future use

E(r, t) =e

n 2(1 n )3R2

+n

(n )

c(1 n )3R

ret

(82)


By using (82) we will try to derive a general expression for the energyradiated per unit solid angle per unit frequency interval in terms of an
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radiated per unit solid angle per unit frequency interval in terms of anintegral over the trajectory of the particle.We must calculate the Fourier transform of (73) by using (82)

A() =

e282c

1/2

eit

n [(n ) ]

(1 n)3ret

dt (83)

where ret means evaluated at t=t +R(t)/c. By changing theintegration variable from t to t we get

A() =

e2

82c

1/2

ei(t+[R(t)/c])n [(n ) ]

(1 n)2dt (84)

since the observation point is assumed tobe far away the unit vector n can be

assumed constant in time, while we can usethe approximation

R(t) xn r(t) (85)where x is the distance from the origin Oto the observation point P, and

r(t)is the position of the particle relative to O.


Th (84) b
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Then (84)becomes:

A() = e2

82c1/2

ei(tnr(t)/c)n [(n ) ]

(1 n)

2dt (86)

and the energy radiated per unit solid angle per unit frequency interval(81) is

d2I

dd =

e22

42c

ei(tnr(t)/c)n

[(n

)

]

(1 n)2 dt2

(87)

For a specified motionr(t)is known, (t)and(t)can be computed,

and the integral can be evaluated as a function of and the direction ofn.

If we study more than one accelerated charged particles, a coherent sum

of amplitudes Aj()(one for each particle) must replace must replace the

single amplitude in (87).


If one notices that, the integrand in (86) is a perfect differential(excluding the exponential)
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(excluding the exponential)

n [(n ) ](1

n)3

= d

dt

n (n )

1

n

(88)

then by integration by parts we get to the following relation for theintensity distribution:

d2I

dd=

e22

42c

n (n )ei(tnr(t)/c)dt2

(89)

For a number of charges ej in accelerated motion the integrand in (89)becomes

eei(/c)nr(t) N

j=1ejje

i(/c)nrj(t) (90)

In the limit of a continuous distribution of charge in motion the sum overjbecomes an integral over the current density J(x, t):

eei(/c)nr(t) 1c

d3xJ(x, t)ei(/c)nx (91)

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Then the intensity distribution becomes:

d2I

dd=

2

42c3

dt

d3x ei(tnx/c)n [n J(x, t)]

2

(92)

a result that can be obtained from the direct solution of theinhomogeneous wave equation for the vector potential.


What Is Synchrotron Light?
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When charged particles are accelerated, they radiate. If electrons areconstrained to move in a curved path they will be accelerating toward theinside of the curve and will also radiate what we call synchrotronradiation.

Synchrotron radiation of this type occursnaturally in the distant reaches of outer space. Accelerator-based synchrotron light was seenfor the first time at the GE Research Lab

(USA) in 1947 in a type of accelerator knownas asynchrotron. First considered a nuisance because it causedthe particles to lose energy, it recognized in the1960s as light with exceptional properties.

The light produced at todays light sources isvery bright. In other words, the beam of x-raysor other wavelengths is thin and very intense.Just as laser light is much more intense andconcentrated than the beam of light generatedby a flashlight.


Synchrotrons
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Synchrotronsare particle accelerators - massive (roughly circular)machines built to accelerate sub-atomic particles to almost the speed oflight.

The accelerator componentsinclude an electron gun, one ormore injector accelerators (usuallya linear accelerator and asynchrotron but sometimes just a

large linear accelerator) to increasethe energy of the electrons, and astorage ring where the electronscirculate for many hours. In the storage ring, magnets forcethe electrons into circular paths. As the electron path bends, lightis emitted tangentially to the curvedpath and streams down pipes calledbeamlines to the instruments wherescientists conduct their experiments.

Figure: Components of a synchrotronlight source typically include (1) anelectron gun, (2) a linear accelerator,(3) a booster synchrotron, (4) astorage ring, (5) beamlines, and (6)

experiment stations.Radiation by Moving Charges

Synchrotrons
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The storage ring is specificallydesigned to include special magneticstructures known as insertiondevices (undulators andwigglers). Insertion devices generate speciallyshaped magnetic fields that driveelectrons into an oscillating

trajectory for linearly polarized lightor sometimes a spiral trajectory forcircularly polarized light. Each bend acts like a sourceradiating along the axis of the

insertion device, hence the light isvery intense and in some cases takeson near-laser-like brightness.

They produce synchrotron radiation- an amazing form of light thatresearchers are shining on molecules,atoms, crystals and innovative newmaterials in order to understandtheir structure and behaviour. Itgives researchers unparalleled powerand precision in probing thefundamental nature of matter.


Synchrotrons
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Synchrotron Radiation
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To find the distribution of energy in

frequency and in angle it is necessary tocalculate the integral (89) Because the duration of the pulse isvery short, it is necessary to know thevelocity and the positionr(t)over

only a small arc of the trajectory. The origin of time is chosen so that att= 0the particle is at the origin ofcoordinates. Notice tht only for very small angles there will be appreciable radiationintensity.

Figure: The trajectory lies on the

plane x y with instantaneousradius of curvature . The unitvectorn can be chosen to lie in thex zplane, and is the angle withthe x-axis.


The vector part of the integrand in eqn (89) can be written

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n (n )= sin(vt/) + cos(vt/)sin (93)

=2 is a unit vector in the y-direction, corresponding to the

polarization in the plane of the orbit. =n 2 is the orthogonal polarization vector correspondingapprox. to polarization perpendicular to the orbit plane (for small ). The argument of the exponential is

t n r(t)c

=

t

c sin

vt

cos

(94)

Since we are dealing with small angle and very short time intervalswe can make an expansion to both trigonometric functions to obtain

t n r(t)c

2

12

+2

t+ c232

t3

(95)

where was set to unity wherever possible.

CHECK THE ABOVE RELATIONS


Thus the radiated energy distribution (89) can be written

d2I 2 2

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d2I

dd=

e22

42c

A() + A()2 (96)where the two amplitudes are (How?)

A() c

texp

i

2

1

2+2

t+

c2t3

32

dt (97)

A()

expi

2 1

2+2 t+

c2t3

32 dt (98)by changing the integration variable

x= ct

(1/2 +2)1/2

and introducing the parameter

=

3c

1

2+2

3/2(99)

allows us to transform the integrals into the form


1 2 3 1 3
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A() c

1

2+2

xexp

i3

2

x+

1

3x3

dx (100)

A()

c 1

2+2

1/2

exp i32

x+13

x3 dx(101)These integrals are identifiable as Airy integrals or as modified Besselfunctions (FIND OUT MORE)

0

x sin i32

x+13

x3 dx = 13 K2/3() (102) 0

cos

i3

2

x+

1

3x3

dx = 1

3K1/3() (103)

The energy radiated per unit frequency interval per unit solid angle is:

d2I

dd=

e2

32c

c

2 12

+2

2

K22/3()+ 2

1/2 +2K21/3()

(104)

The1st termcorresponds to radiation polarized in the orbital plane.

The2nd termterm to radiation polarized perpendicularto that plane.


By integration over all frequencies we find the distribution of energy in
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By integration over all frequencies we find the distribution of energy inangle (Can you prove it?)

dI

d= 0

d2I

dd d= 7

16

e2

1

(1/2 +2)5/2

1 +5

7

2

(1/2) +2

(105)This shows the characteristic behavior seen in the circular motion casee.g. in equation (66).This result can be obtained directly, by integrating a slight generalization

of the power formula for circular motion, eqn (65), over all times. Again: The1st termcorresponds polarization parallel to the orbital plane. The2nd termterm to perpendicular polarization.Integration over all angles shows that seven (7) timesas much energy isradiated with parallel polarization as with perpendicular polarization. In

other words:The radiation from a relativistically moving charge is very strongly, but

not completely, polarized in the plane of motion.


The radiation is largely confined to the plane containing the motion,
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being more confined the higher the frequency relative to c/. If gets too large, then will be large atallangles, and then therewill be negligible power emitted at those high frequencies.

Thecritical frequencybeyond which there will be negligible totalenergy emitted at any angle can be defined by = 1/2and = 0(WHY?).Then we find

c=3

2

3c =3

2 E

mc2

3c

(106)

this critical frequency agrees with the qualitative estimate (71). If the motion is circular, thenc/ is the fundamental frequency ofrotation,0. The critical frequency is given by

c=nc0 with harmonic number nc=3

2

E

mc2

3(107)


For 1the radiation is predominantly on the orbital plane and we canevaluate via eqn (104) the angular distribution for = 0.Th f fi d
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Thus for cwe findd2I

dd|=0 e

2

c (2/3)

2

3

41/3

c 2/3

(108)

For cd2I

dd|=0 3

4

e2

c 2

ce/c (109)

These limiting cases show that the spectrum at = 0increases withfrequency roughly as 2/3 well bellow the critical frequency, reaches amaximum in the neighborhood ofc, and then drops exponentially to 0above that frequency. Thespread in angleat a fixed frequency can be estimated bydetermining the angle cat which (c) (0) + 1.

In thelow frequency range (

c), (0)

0so (c)

1which

gives

c

3c

1/3=

1

2c

1/3(110)

We note that the low frequency components are emitted at much wider

angles than the average,

2

1/2

1.


In thehigh frequency limit ( > c), (0) 1and the intensityfalls off in angle as:
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d2I

dd d

2I

dd|=0 e3

22/20 (111)

Thus the critical angle defined by the1/epoint is

c 1

2c3

1/2(112)

This shows that the high-frequency components are confined to anangular range much smaller than average.

Differential frequency spectrum

as a function of angle. For

frequencies comparable to the

critical frequency c, theradiation is confined to angles

of order 1/ For much smaller(larger) frequencies the angular

spread is larger (smaller).


The frequency distribution of the total energy emitted as the particle
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Thefrequency distributionof the total energy emitted as the particlepasses by can be found by integrating (104) over angles

dI

d = 2 /2/2

d2I

ddcos d 2

d2I

dd d (113)

For the low-frequency rangewe can use (95) at= 0and (108) atc, to get

dI

d 2cd2I

dd |=0 e2

c

c1/3

(114)

showing the the spectrum increases as 1/3 for c. This gives avery broad flat spectrum at frequencies below c. For thehigh-frequency limit cwe can integrate (111) overangles to get:

dI

d

3

2

e2

c

c

1/2

e/c (115)

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A proper integration of over angles yields the expression,

dI

d

3

e2

c

c

/c

K5/3(x)dx (116)

In the limit c, this reduces to the form (114) with numericalcoefficient 13/4, while for c it is equal to (115).


Bellow the behavior ofdI/d as function of the frequency. The peakintensity is of the order ofe2/cand the total energy is of the order ofe2 /c = 3e24/ This is in agreement with the value 4e24/3 for
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e c/c= 3e /. This is in agreement with the value 4e /3 forthe radiative loss per revolution (53) in circular accelerators.

Normalized synchrotron radiation spectrum

1

I

dI

dy =

9

3

8 y

K5/3(x)dx

where y=/c and I = 4e24/3.


The radiation represented by (104) and (116) is called synchrotronradiationbecause it was first observed in electron synchrotrons (1948).

F i di i l ti th t i t ll di t b i
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For periodic circular motion the spectrum is actuallydiscrete, beingcomposed of frequencies that are integral multipoles of the fundamentalfrequency 0=c/.

Thus we should talk about the angular distribution of power radiatedin the nth multiple of0 instead of the energy radiated per unit frequencyinterval per passage of the particle. Thus we can write (WHY?)

dPn

d

= 1

2c

2d2I

dd |=n0 (117)

Pn = 1

2

c

2dI

d|=n0 (118)

These results have been compared with experiment at various energysynchrotrons. The angular, polarization and frequency distributions areall in good agreement with theory.

Because of the broad frequency distribution shown in previous Figure,covering the visible, ultraviolet and x-ray regions, synchrotron radiation is

a useful tool for studies in condensed matter and biology.

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Fourier transform o the electric field produced by a charged particle in

circular motion. The plots reveal that the number of relevant harmonicsof the fundamental frequency 0 increases with . And the dominant

harmonic is shifted to higher frequencies. (A. = 1, = 0, c= 1, B.

= 1.2, = 0.55, c= 1.7, C.= 1.4, = 0.7, c= 2.7, D.= 1.6,

= 0.78, c= 4.1,


Thomson Scattering of Radiation
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When a plane wave of monochromatic EM radiation hits a free particle ofcharge eand mass m the particle will be accelerated and so emit

radiation.The radiation will be emittedin directions other than the propagationdirection of the incident wave, but (for non-relativistic motion of theparticle) it will havethe same frequencyas the incident radiation.According to eqn (41) the instantaneous power radiated into polarizationstate by a particle is (How?)

dP

d=

e2

4c3

v2 (119)If the propagation vectork0 and its thepolarization vector0 can be written

E(x, t) =0E0eik0xit


Then from the force eqn (F =qE) the acceleration will be

v(t) =0e

E0eik0xit (120)
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( )m

( )

If we assume that the charge moves a negligible part of a wavelegthduring one cycle of oscillation, the time average of

|v

|2 is 1

2(v

v)Then the averaged power per unit solid angle can be expressed as

dPd

= c8|E0|2

e2

mc2

2| 0|2 (121)

And since the phenomenon is practically scattering then it is convenient

to used the scattering cross section as

d

d=

Energy radiated/unit time/unit solid angle

Incident energy flux in energy/unit area/unit time (122)

The incident energy flux is the time averaging Poynting vector for the

plane wave i.e. c|E0|2/8. Thus from eqn (121) we get the differentialscattering cross section

d

d=

e2

mc2

2| 0|2 (123)


The scattering geometry with a choice ofpolarization vectors for the outgoing wave isshown in the Figure.
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gThe polarization vector1 is in the planecontainingn andk0 ;2 is perpendicular to

it.In terms of unit vectors parallel to thecoordinate axes,1 and2 are:

1 = cos (xcos + ysin ) zsin 2 =

xsin + ycos

For an incident linearly polarized wave with polarization parallel to thex-axis, the angular distribution is(cos2 cos2 + sin2 ).For polarization parallel to the y-axis it is (cos2 sin2 + cos2 ).For unpolarized incident radiation the scattering cross section is

dd

=

e2

mc2

212

1 + cos2

(124)

This is called the Thomson formula for scattering of radiation by a free

charge, and is appropriate for the scattering ofx-rays by electrons or

gamma rays by protons.Radiation by Moving Charges

The total scattering cross section called the Thomson cross section

T =8

3

e2

2

2(125)
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T3

mc2

( )

The Thomson cross section for electrons is0.665 1024 cm2. The unitof length e

2

/mc2

= 2.82 1013

cmis called classical electron radius. This classical Thomson formula is valid only for low frequencies wherethe momentum of the incident photon can be ignored. When the photon momentum /cbecomes comparable to or largerthan mcmodifications occur.

The most important is that the energy or momentum of the scattered

photon is less than the incident energy because the charged particlerecoils during the collision.The outgoing to the incident wave number is given by Compton formula

k/k= 1 + mc

2(1

cos )

1

(126)

In quantum mechanics the scattering of photons by spinlesspointparticles of charge eand mass m yields the cross section:

d

d=

e2

mc22

k

k2

| 0|2 (127)Radiation by Moving Charges
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