Radiation ELETRODYNAMICS

8/18/2019 Radiation ELETRODYNAMICS

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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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Liénard - Wiechert Potentials

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

Built directly from Maxwell’s 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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Liénard - Wiechert Potentials

We will study potentials and fields produced by a point charge, for whicha trajectory x 0(t

) has been defined a priori.

It is obvious that when a charge q is radiating is giving away momentumand energy, and possibly angular momentum and 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

ρ( x , t ) = q δ ( x − x 0[t ]) (1)and since in general the current density J is ρ v , we also have

J ( x , t ) = q v δ ( x − x 0[t ]) , where v (t ) = d x 0dt

(2)

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

Φ( x , t ) = ρ( x , t − | x − x |/c )

| x

− x

| d 3x (3)


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A( x , t ) = 1

c

J ( x , t − | x − x |/c )| x − x | d

3x (4)

It is not difficult to see that these retarded potentials take into accountthe finite propagation speed of the EM disturbances since an effectmeasured at x and t was produced at the position of the source at time

t̃ = t − | x − x 0(t̃ )|c

(5)

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

putting β ≡ v /c ,

Φ( x , t ) = q δ ( x − x 0[t − | x − x |/c ])

| x

− x

| d 3x (6)

A( x , t ) = q

β (t − | x − x |/c )δ ( x − x 0[t − | x − x |/c ])| x − x | 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 coordinate ̃ x coresponding to theretarded time t̃ defined in (5) which produces a contribution

̃ x = x 0(t̃ ) = x 0 (t − | x − x 0|/c ) (8)Let us also define the vector

R (t ) = x − x 0(t ) (9)

in the direction n ≡ R /R . Then

Φ( x , t ) = q

δ ( x − x 0[t − R (t )/c ])

R (t ) d 3x (10)

A( x , t ) = q

β (t − R (t

)/c )δ ( x

− x 0[t − R (t

)/c ])R (t )

d 3x (11)

Because the integration variable x appears in R (t ) we transform it byintroducing a new parameter r ∗, where

x ∗ = x

− x 0 [t

−R (t )/c ] (12)


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The volume elements d 3x ∗ and d 3x are related by the Jacobiantransformation

d 3x ∗ = Jd 3x , where J ≡

1 − n(t ) · β (t )

(13)

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

Φ( x , t ) = q

δ ( x ∗) d 3x ∗

| x − x ∗ − x 0(t̃ )|(1 − n · β )(14)

and A( x , t ) = q

β (t̃ ) δ ( x ∗) d 3x ∗| x − x ∗ − x 0(t̃ )|(1 − n · β )

(15)

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

Φ( x , t ) =

q

(1 − n · β )| x − ̃ x |

t̃

=

q

(1 − n · β )R

t̃

(16)

A( x , t ) = q β

(1−

n·

β )| x −

̃ x |t̃

= q β

(1−

n·

β )R t̃

(17)


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Liénard - Wiechert potentials

Φ( x , t ) =

q

(1 − n · β )| x − ̃ x |

t̃

=

q

(1 − n · β )R

t̃

(18)

A( x , t ) =

q β

(1 − n · β )| x − ̃ x |t̃

=

q β

(1 − n · β )R t̃

(19)

These are the Liénard - Wiechert potentials.

It is worth noticing the presence of the term (1 − n · β ), which clearlyarises from the fact that the velocity of the EM waves is finite, so the

retardation effects must be taken into account in determining the fields.


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Special Note about the shrinkage factor (1 − n · β )Consider a thin cylinder movingalong the x -axis with velocity v .To calculate the field at x when theends of the cylinder are at (x 1, x 2),we need to know the location of theretarded points x̃ 1 and x̃ 2

x 1

−x̃ 1

x − x̃ 1=

v

c and

x 2

−x̃ 2

x − x̃ 2=

v

c (20)

by setting L̃ ≡ x̃ 2 − x̃ 1 and L ≡ x 2 − x 1 and subtracting we get

L̃− L = v c

L̃ → L̃ = L1

−v /c

(21)

That is, the effective length L̃ and the natural length L differ by the

factor (1 − x · β )−1 = (1 − v /c )−1 because the source is moving relativeto the observer and its velocity must be taken into account when

calculating the retardation effects.


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Liénard - 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 − ∇Φ (22)

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

Φ( x , t ) = q

δ (t − t − R (t )/c )

R (t ) dt (23)

A( x , t ) = q

β (t )δ (t − t + R (t )/c ])R (t )

dt (24)

where R (t ) ≡ | x − x 0(t )|. This can be verified by using the followingproperty of the Dirac delta function (how?)

g (x )δ [f (x )]dx =

i

g (x )|df /dx |

f (x i )=0

(25)

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

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

can be carried out before the integration over the delta f unction.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 ∇

δ (t − t + R (t )/c )R (t

) dt

− q c

∂

∂ t

β (t )δ (t − t + R (t )/c )R (t )

dt (26)

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

E ( x , t ) = q

nR 2 δ

t − t + R (t

)c − ncR δ t − t + R (t

)c

dt

− q c

∂

∂ t

β (t )δ (t − t + R (t )/c )R (t )

(27)

But (HOW?)

δ

t − t + R (t )

c

= − ∂

∂ t δ

t − t + R (t

)

c

(28)

E ( x , t ) = q

n

R 2δ

t − t + R (t

)

c dt +

q

c

∂

∂ t ( n − β )

cR (t ) δ

t − t + R (t

)

c (29)


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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 deltafunction’s arguments with respect to t . Using the chain rule of differentiation

d dt

t − t + R (t

)c

=

1 − n · β t̃

(30)

with which we get the result (HOW?):

E ( r , t ) = q n(1 − n · β )R 2t̃ +

q

c

∂

∂ t n −

β

(1 − n · β )R t̃ (31)Since

∂ R

∂ t =

∂ R

∂ t ∂ t

∂ t = − n· v ∂ t

∂ t = c 1 − ∂ t

∂ t ⇒ ∂ ̃t

∂ t =

1

(1−

n·

β )(32)

Thus

∂

∂ t

n − β

(1 − n · β )R

t̃

= 1

(1 − n · β )∂

∂ ̃t

n − β

(1 − n · β )R 2

t̃

(33)


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By using the additional pieces

Ṙ |t̃ = −c n · β

t̃ (34)

̇ n|t̃ = c

R

n( n · β ) − β

t̃

(35)

d

d ̃t

1 − n · β

t̃

= −

n · ̇ β + β · ̇ n

(36)

and we finally get

E ( r , t ) = q

( n − β )(1 − β 2)(1 − n · β )3R 2

+ n ×

( n − β )× ̇ β

c (1 − n · β )3R

t̃ (37)

A similar procedure for B shows that

B ( r , t ) = ∇× A = n(t̃ )× 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 Coulomb’s law q n/R 2. 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/R 2) and the radiation field (which falls off as 1/R )

Unless the particle is accelerated ( β̇ = 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 β → 1 with β̇ = 0 the field displays a “bunching” effect. This“bunching” 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 of light (β → 0) then from eqn (37) we get

E = q

c

n × ( n × ̇ β )

R

ret

(39)

The instantaneous energy flux is given by the Poynting vector

S = c 4π E × B = c 4π | E |2 n (40)The power radiated per unit solid angle is

dP

d Ω =

c

4πR 2| E |2 = q

2

4πc | n × ( n × ̇ β )|2 (41)

and if Θ is the angle between the acceleration ̇ v and n then the powerradiated can be written as

dP

d Ω =

q 2

4πc 3|̇ v |2 sin2 Θ (42)


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Larmor Formula

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

P = q 2

4πc 3|̇ v |2

π0

2π sin3 Θd Θ = 2

3

q 2

c 3|̇ 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 and n.


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Relativistic Extension

Larmor’s formula (43) has an “easy” relativistic extension so that can beapplied to charges with arbitrary velocities

P = 23

e 2

m2c 3|̇ p |2 (44)

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

P = −2

3

e 2

m2c 3dp µ

d τ

dp µ

d τ

(45)

where d τ = dt /γ is the proper time and p µ is the charged particle’senergy-momentum 4-vector. Obviously for small β it reduces to (44)

−dp µ

d τ

dp µ

d τ =d p

d τ 2 −

1

c 2dE

d τ 2

=d p

d τ 2 − β 2

dp

d τ 2

(46)

If (45) is expressed in terms of the velocity & acceleration (E = γ mc 2 & p = γ m v with γ = 1/(1 − β 2)1/2), we obtain the Liénard result (HOW?)

P = 2

3

e 2

c

γ 6 ̇ β 2

− β

× ̇ β

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

e 2

m2c 3

dp dt

2

(48)

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

P = 2

3

e 2

m2c 3

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 momentum of the particle.Radiation by Moving Charges

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The ratio of power radiated to power supplied by external sources is

P

dE /dt =

2

3

e 2

m2c 31

v

dE

dx → 2

3

(e 2/mc 2)

mc 2dE

dx (50)

Which shows that the radiation loss in an electron linear accelerator willbe unimportant unless the gain in energy is of the order of mc 2 = 0.5MeVin a distance of e 2/mc 2 = 2.8 × 10−13cm, or of the order of 2 × 1014MeV/meter. Typically radiation losses are completely negligiblein linear accelerators since the gains are less than 50MeV/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 momentum p changes rapidly in direction as the particle

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

d p

d τ

= γω| p | 1c

dE

d τ (51)




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

P = 2

3

e 2

m2c 3γ 2ω2| p |2 = 2

3

e 2c

ρ2 β 4γ 4 (52)

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

δ E = 2πρ

c β

P = 4π

3

e 2

ρ

β 3γ 4 (53)

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

δ E (MeV) = 8.85 × 10−2 [E (GeV)]4

ρ(meters) (54)

In a 10GeV electron sychrotron (Cornell with ρ ∼ 100m) the loss perrevolution is 8.85MeV. In LEP (CERN) with beams at 60 GeV

(ρ ∼ 4300m) the losses per orbit are about 300 MeV.


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If β and ̇ β are nearly constant (e.g. if the particle is accelerated for short

time) then (57) is proportional to the angular distribution of the energyradiated.For the Poynting vector (55) the angular distribution is

dP (t )

d Ω =

e 2

4πc

| n × {( n − β )× ̇ β }|2(1 − n · β )5

(58)

The simplest example is linear motion in which β and

̇ β are parallel i.e. β × ̇ β = 0 and (HOW?)

dP (t )

d Ω =

e 2v̇ 2

4πc 3sin2 θ

(1 − β cos θ)5 (59)

For β 1, this is the Larmor result(42). But as β → 1, the angulardistribution is tipped forward andincreases in magnitude.


• The angle θmax for which the intensity is maximum is:

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g y

cos θmax = 1

3β

1 + 15β 2 − 1

for β ≈ 1 → θmax ≈ 1

2γ (60)

For relativistic particles, θmax 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 Ω ≈ 8

π

e 2v̇ 2

c 3 γ 8

(γθ)2

(1 + γ 2θ2)5 (61)

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

and γθ = 0.91.Radiation by Moving Charges

• The root mean square angle of emission of radiation in the relativistic



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limit is

θ21/2 = 1γ

= mc 2

E (62)

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

P (t ) = 2

3

e 2

c 3 v̇ 2γ 6 (63)

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

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

particle.This force is F = d p /dt where p = mγ v is the particle’s relativisticmomentum. For linear motion in the x -direction we have p x = mv γ and

dp x

dt = mv̇ γ + mv̇ β 2γ 3 = mv̇ γ 3

and Larmor’s formula can be written as

P = 2

3

e 2

c 3| 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 acceleration ̇ β perpendicular to its velocity β is

another example.We choose a coordinate system such

as β is in the z-direction and ̇ β in

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

dP (t )

d Ω =

e 2

4πc 3|̇ v |2

(1

−β cos θ)3 1 −

sin2 θ cos2 φ

γ 2(1

−β cos θ)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 (γ 1) the angular distribution can be written

dP (t )

d Ω ≈

2e 2

πc 3

γ 6 |̇ v |2

(1 + γ 2

θ2

)3 1 −

4γ 2θ2 cos2 φ

γ 2

(1 + γ 2

θ2

)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

e 2|̇ v |2c 3

γ 4 (67)

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

P circular(t ) =

2

3

e 2

m2c 3γ 2

d p

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 γ 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

accelerations parallel to and perpendicular to the velocity.• But the radiation from the parallel component is negligible by a factor1/γ 2, compare (48) and (68). Thus we will keep only the perpendicularcomponent alone.• 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 ρ

ρ = v 2

v̇ ⊥≈ c

2

v̇ ⊥(69)

where v̇ ⊥ 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 particle’s velocity isdirected toward the observer.


Since the angular width of the beami 1/ h i l ill l



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is 1/γ the particle will travel adistance of the order of

d = ρ

γ

in a time

∆t = ρ

γ v

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

D = c ∆t = ρ

γβ

Since the particle is moving in the same direction with speed v and movesa distance d in time ∆t the rear edge of the pulse will be a distance

L = D − d =

1

β − 1

ρ

γ ≈ ρ

2γ 3 (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,

ωc ∼ c L ∼

c

ρ

γ 3 (71)

• For circular motion the term c /ρ is the angular frequency of rotationω0 and even for arbitrary motion plays the role of the fundamentalfrequency.• This shows that a relativistic particle emits a broad spectrum of frequencies up to γ 3 times the fundamental frequency.

• EXAMPLE : In a 200MeV sychrotron, γ max ≈ 400, whileω0 ≈ 3 × 10

8

s −1

. The frequency spectrum of emitted radiation extendsup to ω ≈ 2 × 1016s −1.


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 of Parseval’s theorem of Fourier analysis.The general form of the power radiated per unit solid angle is

dP (t )

d Ω = | A(t )|2 (72)

where A(t ) = c

4π

2 R E

ret

(73)

and E is the electric field defined in (37).• Notice that here we will use the observer’s 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 Ω =

∞−∞

| A(t )|2dt (74)

This can be expressed via the Fourier transform’s as an integral over the

frequency. Radiation by Moving Charges

The Fourier transform is:

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A(ω) = 1√

2π

∞−∞

A(t )e i ωt dt (75)

and its inverse:

A(t ) = 1√ 2π

∞−∞

A(ω)e −i ωt d ω (76)

Then eqn (74) can be written

dW

d Ω =

1

2π ∞

−∞

dt ∞

−∞

d ω ∞

−∞

d ω A∗(ω) · A(ω)e i (ω−ω)t (77)

If we interchange the order of integration between t and ω we see thatthe time integral is the Fourier represendation of the delta functionδ (ω − ω). Thus the energy radiated per unit solid angle becomes

dW

d Ω = ∞

−∞ | A(ω)

|2d ω (78)

The equality of equations (74) and (78) is a special case of Parseval’stheorem.

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 Ω =

∞0

d 2I (ω, n)

d ωd Ω d ω (79)

where d 2I

d ωd Ω = | A(ω)|2 + | A(−ω)|2 (80)

If A(t ) is real, form (75) - (76) it is evident that A(−ω) = A∗(ω). Then

d 2

I d ωd Ω

= 2| A(ω)|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 − β γ 2(1 − n · β )3R 2

+ n ×

( n − β ) × ̇ β

c (1 − n · β )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(ω) =

e 28π2c

1/2 ∞−∞

e i ωt

n × [( n − β ) × ˙ β ]

(1 − β · 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(ω) =

e 2

8π2c

1/2

∞

−∞

e i ω(t +[R (t )/c ]) n × [( n − β )× ̇ β ]

(1 − β · 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 ) ≈ x − n · r (t ) (85)where x is the distance from the origin O to 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(ω) = e 2

8π2c 1/2

∞

−∞

e i ω(t − n· r (t )/c ) n × [( n − β ) × ̇ β ]

(1 − β · n)

2dt (86)

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

d 2I

d ωd Ω =

e 2ω2

4π2c

∞

−∞ e i ω(t − n· r (t )/c ) n

×[( n

− β )

× ̇ β ]

(1 − β · n)2 dt 2

(87)

For a specified motion r (t ) is known, β (t ) and ̇ β (t ) can be computed,

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

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

of amplitudes A j (ω) (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 − β ) × ̇ β ](1

− β

· n)3

= d

dt

n × ( n × β )

1

− β

· n

(88)

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

d 2I

d ωd Ω =

e 2ω2

4π2c ∞

−∞

n × ( n × β )e i ω(t − n· r (t )/c )dt 2

(89)

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

e β e −i (ω/c ) n· r (t ) →N

j =1e j β j e

−i (ω/c ) n· r j (t ) (90)

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

e β e −i (ω/c ) n· r (t ) → 1c

d 3x J ( x , t )e −i (ω/c ) n· x (91)


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Then the intensity distribution becomes:

d 2I

d ωd Ω =

ω2

4π2c 3

dt

d 3x e i ω(t − n· x /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 a synchrotron.• First considered a nuisance because it causedthe particles to lose energy, it recognized in the1960s as light with exceptional properties.

• The light produced at today’s 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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Synchrotrons are particle accelerators - massive (roughly circular)machines built to accelerate sub-atomic particles to almost the speed of light.

• 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



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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 β and the position r (t ) over

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

Figure: The trajectory lies on the

plane x − y with instantaneousradius of curvature ρ. The unitvector n can be chosen to lie in thex − z plane, and θ is the angle withthe x-axis.


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



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n × ( n × β ) = β − sin(vt /ρ) + ⊥ cos(vt /ρ)sin θ (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 θ).• The argument of the exponential is

ω

t − n · r (t )c

= ω

t − ρ

c sin

vt ρ

cos θ

(94)

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

ω

t − n · r (t )c

≈ ω

2

1γ 2

+ θ2

t + c 23ρ2

t 3

(95)

where β 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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d 2I

d ωd Ω =

e 2ω2

4π2c

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

A(ω) ≈ c ρ

∞−∞

t exp

i ω

2

1

γ 2 + θ2

t +

c 2t 3

3ρ2

dt (97)

A⊥(ω) ≈ θ ∞

−∞

expi ω

2 1

γ 2 + θ2 t +

c 2t 3

3ρ2 dt (98)by changing the integration variable

x = ct

ρ(1/γ 2 + θ2)1/2

and introducing the parameter ξ

ξ = ωρ

3c

1

γ 2 + θ2

3/2(99)

allows us to transform the integrals into the form


ρ

1 2 ∞ 3 1 3

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A(ω) ≈ ρc

1

γ 2 + θ2

−∞

x exp

i 3

2ξ

x +

1

3x 3

dx (100)

A⊥(ω) ≈

ρ

c θ 1

γ 2 + θ2

1/2

∞−∞

exp i 32

ξ x + 13

x 3 dx (101)These integrals are identifiable as Airy integrals or as modified Besselfunctions (FIND OUT MORE)

∞0

x sin i 32

ξ x + 13

x 3 dx = 1√ 3 K 2/3(ξ ) (102) ∞0

cos

i 3

2ξ

x +

1

3x 3

dx = 1√

3K 1/3(ξ ) (103)

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

d 2I

d ωd Ω =

e 2

3π2c

ωρc

2 1γ 2

+ θ2

2

K 22/3(ξ ) + θ2

1/γ 2 + θ2K 21/3(ξ )

(104)

• The 1st term corresponds to radiation polarized in the orbital plane.

• The 2nd term term to radiation polarized perpendicular to 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 Ω = ∞0

d 2I

d ωd Ω d ω = 7

16

e 2

ρ

1

(1/γ 2 + θ2)5/2

1 + 5

7

θ2

(1/γ 2) + θ2

(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:• The 1st term corresponds polarization parallel to the orbital plane.• The 2nd term term to perpendicular polarization.Integration over all angles shows that seven (7) times as 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 /ρ.• If ω gets too large, then ξ will be large at all angles, and then therewill be negligible power emitted at those high frequencies.

• The critical frequency beyond which there will be negligible totalenergy emitted at any angle can be defined by ξ = 1/2 and θ = 0(WHY?).Then we find

ωc = 3

2

γ 3c ρ = 3

2 E

mc 2

3c

ρ

(106)

this critical frequency agrees with the qualitative estimate (71).• If the motion is circular, then c /ρ is the fundamental frequency of rotation, ω0.• The critical frequency is given by

ωc = nc ω0 with harmonic number nc = 3

2

E

mc 2

3(107)


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

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Thus for ω ωc we findd 2I

d ωd Ω|θ=0 ≈ e

2

c Γ(2/3)

π 2

3

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ωρ

c 2/3

(108)

For ω ωc d 2I

d ωd Ω|θ=0 ≈ 3

4π

e 2

c γ 2

ω

ωc e −ω/ωc (109)

These limiting cases show that the spectrum at θ = 0 increases withfrequency roughly as ω2/3 well bellow the critical frequency, reaches amaximum in the neighborhood of ωc , and then drops exponentially to 0above that frequency.• The spread in angle at a fixed frequency can be estimated bydetermining the angle θc at which ξ (θc ) ≈ ξ (0) + 1.

• In the low frequency range (ω

ωc ), ξ (0)

≈0 so ξ (θc )

≈1 which

gives

θc ≈

3c

ωρ

1/3=

1

γ

2ωc

ω

1/3(110)

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

angles than the average,

θ2

1/2

∼γ −1.


• In the high frequency limit (ω > ωc ), ξ (0) 1 and the intensityfalls off in angle as:

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d 2I

d ωd Ω ≈ d

2I

d ωd Ω|θ=0 · e −3ωγ

2θ2/2ω0 (111)

Thus the critical angle defined by the 1/e point is

θc ≈ 1γ

2ωc 3ω

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 ωc , theradiation is confined to angles

of order 1/γ 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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The frequency distribution of the total energy emitted as the particlepasses by can be found by integrating (104) over angles

dI

d ω = 2π π/2−π/2

d 2I

d ωd Ω cos θd θ ≈ 2π ∞−∞

d 2I

d ωd Ω d θ (113)

• For the low-frequency range we can use (95) at θ = 0 and (108) atθc , to get

dI

d ω ∼ 2πθc d 2I

d ωd Ω |θ=0 ∼ e 2

c ωρ

c 1/3

(114)

showing the the spectrum increases as ω1/3 for ω ωc . This gives avery broad flat spectrum at frequencies below ωc .• For the high-frequency limit ω ωc we can integrate (111) overangles to get:

dI

d ω ≈

3π

2

e 2

c γ

ω

ωc

1/2

e −ω/ωc (115)


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A proper integration of over angles yields the expression,

dI

d ω ≈

√ 3

e 2

c γ

ω

ωc

ωω/ωc

K 5/3(x )dx (116)

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


Bellow the behavior of dI /d ω as function of the frequency. The peakintensity is of the order of e 2γ/c and the total energy is of the order of e2γω /c = 3e2γ4/ρ This is in agreement with the value 4πe2γ4/3ρ for

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e γωc /c = 3e γ /ρ. This is in agreement with the value 4πe γ /3ρ forthe radiative loss per revolution (53) in circular accelerators.

Normalized synchrotron radiation spectrum

1

I

dI

dy =

9√

3

8π y

K 5/3(x )dx

where y = ω/ωc and I = 4πe 2γ 4/3ρ.


The radiation represented by (104) and (116) is called synchrotronradiation because 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 actually discrete, beingcomposed of frequencies that are integral multipoles of the fundamentalfrequency ω0 = c /ρ.

• Thus we should talk about the angular distribution of power radiatedin the nth multiple of ω0 instead of the energy radiated per unit frequencyinterval per passage of the particle. Thus we can write (WHY?)

dP n

d Ω

= 1

2πc

ρ

2d 2I

d ωd Ω |ω=nω0 (117)

P n = 1

2π

c

ρ

2dI

d ω |ω=nω0 (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 ω0 increases with γ . And the dominant

harmonic is shifted to higher frequencies. (A. γ = 1, β = 0, ωc = 1, B.

γ = 1.2, β = 0.55, ωc = 1.7, C. γ = 1.4, β = 0.7, ωc = 2.7, D. γ = 1.6,

β = 0.78, ωc = 4.1,


Thomson Scattering of Radiation

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When a plane wave of monochromatic EM radiation hits a free particle of charge e and mass m the particle will be accelerated and so emit

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

dP

d Ω =

e 2

4πc 3

∗ · ̇ v 2 (119)If the propagation vector k 0 and its thepolarization vector 0 can be written

E ( x , t ) = 0E 0e i k 0· x −i ωt


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The scattering geometry with a choice of polarization vectors for the outgoing wave isshown in the Figure.


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gThe polarization vector 1 is in the planecontaining n and k 0 ; 2 is perpendicular to

it.In terms of unit vectors parallel to thecoordinate axes, 1 and 2 are:

1 = cos θ (x cos φ + y sin φ) −z sin θ2 =

−x sin φ + y cos φ

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

d σd Ω

=

e 2

mc 2

212

1 + cos2 θ

(124)

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

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

gamma rays by protons.Radiation by Moving Charges

The total scattering cross section called the Thomson cross section

σT = 8π

3

e 2

2

2(125)

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T 3

mc 2

( )

The Thomson cross section for electrons is 0.665 × 10−24 cm2. The unitof length e

2

/mc 2

= 2.82 × 10−13

cm is 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 ω/c becomes comparable to or largerthan mc modifications 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 + ωmc

2(1

−cos θ)

−1

(126)

In quantum mechanics the scattering of photons by spinless pointparticles of charge e and mass m yields the cross section:

d σ

d Ω =

e 2

mc 22

k

k 2

|∗ · 0|2 (127)Radiation by Moving Charges

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Radiation ELETRODYNAMICS