Synchrotron Radiation

G. Wang

What is synchrotron radiation

Static field for a charge at rest When a particle moves with a constant velocity, field moves with particle.

When a particle gets accelerated,some part of the field moves awayfrom the particle to infinity:radiation.

The electromagnetic radiation emitted when charged particles are accelerated radially, , is called synchrotron radiation.a v⊥

Some history of Synchrotron radiation• Synchrotron radiation was named after its

discovery in Schenectady, New York from aGeneral Electric synchrotron acceleratorbuilt in 1946 and announced in May 1947by Frank Elder, Anatole Gurewitsch, RobertLangmuir .

• Synchrotron radiation is the main constraintto accelerate electrons to very high energyand hence is bad for high energy physicsapplication, such as colliders.

• However, it was then realized that theradiation can be so helpful for otherbranches of science such as biology,material science and medical applications.As a result, dedicated storage rings havebeen built to generate synchrotronradiation, which are called light sources.

Application of Synchrotron radiation

Plant Architectural view of NSLS-II

Theoretical Model: wave equation

22

02 21A A A Jc t

α α α αμ∂≡ − ∇ =∂

,A Ac

α Φ =

t

x

yq

Observation point

and ?E B Traditionally, the equation is solved by finding the Green

function

( ) ( ) ( ) ( ) ( )2

422 2

1, , , ,D x t D x t D x t x tc t

δ∂≡ − ∇ =∂

through Fourier transformation:

( )

( )( )

( )( )

0 0

220

4 304 4 22

0

1

1 12 2

ik xik z ik x

D kk k

eD x d kD k e d ke dkk k

μμ

π π

∞ −− ⋅

−∞

= − −

= = −−

( ) ( ) ( )44 ' ' 'A x d x D x x J xc

α απ= −

The singularity is treated by treating k0 as a complex number, which make it NOT Fourier transformation…

To better understand how the synchrotron radiation is quantitatively investigated, we will try to derive formulas from ‘first principle’. (refer to ‘Accelerator physics’ by S.Y. Lee and ‘ classical electrodynamics’ by J.D. Jackson)

Laplace Transformation

( ) ( ) ( )0

for Re 0sxF s e f x dx s∞

−= >

The Laplace transform of the function f(x), denoted by F(s), is defined by the integral

The inversion of the Laplace transform is accomplished for analytic function F(s) by means of the inversion integral*

( ) ( ) ( )1 for Re 02

isx

i

f x e F s ds si

γ

γπ

+ ∞

− ∞

= >

where is a real constant that exceeds the real part of all the singularities of F(s).

Analytic Continuation

( ) 0 for 0f x x= <

*Note that the definition of inverse Laplace transform implies causality, i.e. .

Theoretical Model I: wave equation2

202 2

1A A A Jc t

α α α αμ∂≡ − ∇ =∂

,A Ac

α Φ =

t

x

yq

Observation point

and ?E B

( ) ( )0 2 21, ,A k s J k s

sα αμ

κ= ⋅

+

( ) ( ) ( ) ( )30 0 0 0

0

, exp exp ,A k s dx sx d x ik x A x xα α τ∞ ∞

−∞

≡ − − ⋅ +

0τ−

( ) ( ) ( ) ( )30 0 0 0

0

, exp exp ,J k s dx sx d x ik x J x xα α τ∞ ∞

−∞

≡ − − ⋅ +

( ) ( ) ( ) ( )1 ,x xF k G k x f x g dξ ξ ξ∞

−

−∞

= − F

( ) ( ) ( ) ( )0

10

0

x

F s G s f x g dξ ξ ξ− = − L

( ) ( )( ) ( )0

00 0 0 0

0

sin, ,

x xA k x J k dα ακ ξ

τ μ ξ τ ξκ

−+ = +

Solution in Fourier-Laplace domain

Convolution theory of Laplace transformation:

Solution in Fourier-time domain

Convolution theory of Fourier transformation:

2 2 2 21 2 3k k kκ ≡ + +

Theoretical Model II: wave equation

( ) ( ) ( )30 0 0 0, ' ' ' ', 'rA x x dx d x D x x J x xα αμ

∞ ∞

−∞ −∞

= −

( ) ( ) ( )( )20 0

1' ' '2rD x x H x x x xδπ

− ≡ − −

( )( )

( )( ) ( ) ( ) ( )0 30 03

sin1 1exp42

xf x ik x d k x x x x

xκ η

δ η δ ηκ ππ

∞

−∞

− = − ⋅ = − − − − +

( ) ( ) ( )0 0 3

0 0 0 00

, ,4

x x xA x x d J d

xα α

δ η ξτ μ η ξ η τ ξ

π ξ

∞

−∞

− − −+ = +

−

( ) ( )( ) ( )20 0 0 0' ' ' '2 '

' 'x x x x x x x x

x xx x x x

δ δδ

− − − − + −= − −

− −

Theoretical Model III: Solution for point charge (Lienard-Wiechert Potential)

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )( )4 3 3,J x ec d U x r ec x r t ev t x r tα ατ τ δ τ δ δ∞

−∞

= − = − −

( ) ( ) ( )( ) ( )( )( )200 0 0,

2e cA x x U H x r x r dα αμ τ τ δ τ τ

π

∞

−∞

= − −

( ) ( )( ) ( ) ( )

00,

4 1r

r r r

te cA x xR t n t t

βμπ β

= − ⋅

( )( ) ( ) ( )0

0

1,4 1r r r

ex xR t n t tπε β

Φ = − ⋅

( ) ( )0 00

1r r

rt x R t

c cτ

= = −

( )( )( ) ( )

( )( )( )

( ) ( ) ( )0

2 0 0

2 0 0 01x r

cR nd x rd τ τ

δ τ τ δ τ τδ τ

γ τ τ β τ γτ

τ =

− −− = =

− ⋅ −

( ) ( )r rR t x r t= −

Lienard-Wiechert Potential:

Theoretical Model III: E&M field

( ) ( ) ( ), , ,xE x t x t A x tt

∂= −∇ Φ −∂

( ) ( ), ,xB x t A x t= ∇ ×

( ) ( )( ) ( )( )

( ) ( ) ( )

( ) ( )( ) ( )

( ) ( ) ( )( ) ( )

3 32 20 0

,4 41 1

, ,

r r rr r

r r r r r r r

static rad

n n t t tn t te eE x tt cR t n t t R t n t t

E x t E x t

β ββπε γ πεβ β

× − ×− = + − ⋅ − ⋅

= +

( ) ( )1, ,B x t n E x tc

= ×

The electric and magnetic field can be directly obtained from the following relation (notice that depends on .rt ( ),x t

( ) ( )1

1r

r r

dtdt n t tβ

=− ⋅

( ) ( )( ) ( )rx r r r x r

n tt n t t t

cβ∇ = ⋅ ∇ −

( ) ( )( ) ( )1

rr

r r

n t cd R tdt n t t

ββ

− ⋅=

− ⋅

( ) ( )

( ) ( )1r

x rr r

n tR t

n t tβ∇ =

− ⋅

Note: Jackson follows a differentapproach but directly takingderivatives generate the same result.

Taking the radiation part of the field

and the energy flow is determined by the Poynting vector

Radiation Power I

( ) ( )( ) ( )

( ) ( ) ( )3

04 1

r r r

rad

r r r

n n t t teEc R t n t t

β β

πε β

× − × = − ⋅

( ) ( )1, ,rad radB x t n E x tc

= ×

( ) ( ) ( ) ( )2

0 0

1 1, , , ,rad rad radS x t E x t B x t E x t ncμ μ

= × =

( ) ( ) ( )( ) ( ) ( )( ) ( )

( ) ( )

2

22

50

14 4 1

r r r rr

rr r r

n t n t t tdP t dt en S R td dt c n t t

β β

πε π β

× − × = ⋅ =Ω − ⋅

The radiated power per solid angle is then given by

Time interval difference between radiation and observation. See the next slide

Time interval at radiation point and the observation point

0rt =β

/rt L c=β

L

Lβ

( )1rct L L t tβ β= − = −

ct

Time interval at observation point Time interval at the radiation point

Radiation Power II( ) ( ) ( ) ( )( ) ( )

( ) ( )

2

2

50

14 4 1

r r r rr

r r

n t n t t tdP t ed c n t t

β β

πε π β

× − × =Ω − ⋅

( ) ( ) ( ) ( )2 26 2

0

1 2sin4 3

r rr

dP t dP t eP t d d dd d c

θ θ φ γ β β βπε

= Ω = = − × Ω Ω

Note: Jackson uses Lorentz transformation to derive this from non-relativistic result. Here, we take a more tedious but straightforward approach.

Parallel acceleration (Linac)

( )2

6 2

0

1 24 3r

eP tc

γ βπε

=

( )2

2 // 3 /

r

e

P t dE dxdE dt mc rβ

=2

1415

0.55 1.9 102.8 10e

mc MeV MeVr m m−= = ×

⋅

The state of art accelerating rate at the moment is below 100 MeV/m and hence synchrotron radiation is negligible in linear accelerators.

Circular orbit2 2

ˆ ˆv ca βρ β ρρ ρ

= − = −

( ) ( )2 2 4 4

6 2 22

0 0

1 2 1 214 3 4 3r

e e cP tc

β γγ β βπε πε ρ

= − =

For a storage ring, the energy loss per turn: ( ) ( )2 3 4

0 20

1 1 2 14 3r r

C C C

eU P t dt P t ds dsc

β γβ πε ρ

= = =

If all dipoles in the storage ring has the same bending radius (iso-magnetic case):

2 3 4 2 3 4

0 20 0

1 2 24 3 3

e eU β γ πρ β γπε ρ ε ρ

= =

Power radiated by a beam of average current Ib:3 4

003

bbeam b

I eP U Ie

β γε ρ

= =

Compare parallel with perpendicular

( )2 2

6////

0

1 24 3r

eP tcβ γ

πε=

( )2 4 4 2 2

42

0 0

1 2 1 24 3 4 3r

e c eP tc

β γ β γπε ρ πε

⊥⊥ = =

2a cc

ββρ

⊥⊥ = =

It looks as if the longitudinal acceleration cause more radiation for the same values of acceleration… However what really matters is the force.

( ) 3//// // / /

//

1dp d dF c m cm m cdt dt dt

γγβ γ ββ

= = = =

( ) / /dp dF c m m cdt dt

γβ γ β⊥⊥ ⊥= = =

( )22 6 2

2//// / /3 2 3

0 0

1 2 1 24 3 4 3r

e F eP t Fc mc m cγ

πε γ πε = =

( )22 4 2 2

23

0 0

1 2 1 24 3 4 3r

e F eP t Fc m c mcγ γ

πε γ πε⊥

⊥ ⊥ = =

Therefore, for similar accelerating force, the radiation power from perpendicular acceleration is larger than that from parallel acceleration by a factor of . 2γ

For and , it can be shown that the angular spread of the radiation power is ~ . (Homework problem)

Angular distribution

( ) ( ) ( ) ( )( ) ( )

( ) ( )

2

2

50

14 4 1

r r r rr

r r

n t n t t tdP t ed c n t t

β β

πε π β

× − × =Ω − ⋅

( )1ˆ ˆˆ sin cos cos cos sinnβ βε β θ φ θ θ φ φ φ= = + −

( )3ˆˆ cos sinnβ βε β θ θ θ= = −

ˆn r=

( ) ( ) ( )ˆ ˆcos cos sin 1 cosn n β β β θ β φθ β φ β θ φ × − × = − − + −

( )( ) ( )

2 2 2 2

3 220

1 sin cos14 4 1 cos 1 cos

rdP t ed c

β θ φπε π β θ γ β θ

= −

Ω − −

4 1γ θ− << << 1γ >>

1γ −

Angular distributionβ β

β

β

β

β

0.002β = 0.2β =0.8β =

( ) ( )2 2

3 22

1 sin cos11 cos 1 cos

r θ φβ θ γ β θ

= −

− −

* These plots show how the length of a vector, r, depends on its direction . Since the length has the same directional dependanceas the power, we can see the angular distribution of power by looking at the length of the vector along all directions. (Spherical 3D plot in Mathematica)

( ),θ φ

Spectrum

( ) ( ) ( ) ( ) ( )* '1, , ' , , '2

i t i tdP ta x t a x t d d a x a x e e

dω ωω ω ω ω

π

∞−

−∞

= ⋅ = ⋅Ω

( ) ( ) ( )22

0 0

2 ,dP t d IdW dt a x d d

d d d dω

ω ω ωω

∞ ∞ ∞

−∞

= = ≡Ω Ω Ω

( ) ( ) ( )0, ,r rada x t c R t E x tε= ( ) ( )1, ,

2i ta x t a x e dωω ω

π

∞−

−∞

=

( ) ( )2

22 ,

d Ia x

d dω

ωω

=Ω

In order to get the frequency contents of the radiation, or the spectrum, we need to do Fourier transformations.

Now we calculate the total energy per solid angle received at the observation point

Spectrum intensity: energy received per solid angle per frequency interval

Spectrum II

( )( ) ( ) ( )( ) ( )

( ) ( )( )( )

( ) ( ) ( )( ) ( )( )

/2

0

//

0

,4 2 1

4 2

r r

r r

r r r r i t R t cr

r r

i t n r t ci x cr r r r

n t n t t tea x e dtc n t t

i e e n t n t t e dtc

ω

ωω

β βω

π πε β

ω βπ πε

∞+

−∞

∞− ⋅

−∞

× − × = − ⋅

= − × ×

( ) ( )( ) ( )( ) ( ) ( ) ( )2 2 00 0 0 01 2 r

n rR x r x r x R x n r t

xτ

τ τ τ τ ⋅

= − ⋅ − ≈ − ≈ − ⋅

( )( )

( )( )( )21 1r

n nn nddt n n

β ββ

β β

× − × × × ≈ − ⋅ − ⋅

( ) ( ) ( )( ) ( )( )22 2 22 /

20

12 ,4 4

r ri t n r t cr

d I ea x n n e dtd d c

ωω ωω βω πε π

∞− ⋅

−∞

= = × ×Ω

Far field approximation, ( )0r xτ <<

To proceed, we need to calculate the Fourier components of the electric field: ( ) ( )1, ,

2i ta x a x t e dtωω

π

∞

−∞

=

( )( ) ( )( ) ( )

( ) ( ) ( )3

0

,4 1

r r r

rad

r r r

n n t t tea x t Ec R t n t t

β β

πε β

× − × ∝ = − ⋅

Spectrum III

( ) ( ) ( ) 0 / / 0 / /ˆ ˆ ˆ ˆsinr r r r rn t n t t t tβ β θε βω ε βθε βω ε⊥ ⊥ × × = − ≈ − 1 1θ

γ<<

( ) ( ) ( ) ( ) ( )3 20 0

0 1 0 2 2 0 1 20 0 0 0

ˆ ˆ ˆ ˆ ˆsin cos6 2

r rr r r r

t tc c c cr t t t tω ωβ β β βω ε ω ε ε ω ε ε

ω ω ω ω

= − + ≈ − +

( ) ( ) ( )302

0 20

1/2 3

rr r r r

tt n t r t c t

ωωω ω θω γ

− ⋅ ≈ + +

( ) ( ) ( )( ) ( ) ( )

/ / / /

/ / / /

ˆ ˆ, , ,

ˆ ˆ, , ,

a x a x a x

E x E x E x

ω ω ε ω ε

ω ω ε ω ε

⊥ ⊥

⊥ ⊥

= +

= +

Polarization of electric field is decomposed into

0 1rtω <<

/ /ˆ : inside the orbit plane of particleˆ : nearly perpendicular to the orbit planeεε⊥

( ) ( ) ( ) ( ) ( )2 2 2 2 2 222 2 2 2

1 22 2 2 20 3 3

1 3 14 4 1c

d I e K Kd d c

ω γ ω θ γγ θ η ηω πε π ω θ γ

= + + Ω +

For , using the asymptotic approximation of Bessel function,

Spectrum IV( ) ( ) ( )

( ) ( )

232 2 2

2 00 // 02 2

0 0

22 22 2

// //2 2 2 20 0

1 1ˆ ˆ exp4 4 2 3

1 1 1ˆ ˆ4 4

rr r r

d I te t i t dtd d c

e I Ic

ω ωω ωβθε βω ε ω θω πε π ω γ

ω θ βθε η θ βε ηπε π ω γ γ

∞

⊥−∞

⊥ ⊥

= − + + Ω

= + − +

( ) ( ) ( )313

2exp 32 3

I i x x dx Kηη η∞

⊥−∞

≡ + = ( ) ( ) ( )3// 2

3

2exp 32 3

I x i x x dx Ki

ηη η∞

−∞

≡ + = −

( )3

322 2 2 2

20

1 1 13 2 c

ω ωη θ γ θω γ ω

= + = +

Contribution from ( ) ˆ,E x ω ε⊥ ⊥

Contribution from ( )// //ˆ,E x ω ε

3 30

3 32 2c

cω γ γ ωρ

= ≈

Critical frequency

γθc

ωω

0θ =

( ) ( )

1 222 2 3 3

202 2 2 2

222 2

0 3 2 2

0

1 3 4 2 ; <<4 4 3

1 34 4

1 3 ; 4 4

c

cc

c

cc

ec

d I e Kd d c

e ec

ωω

γ ω ω ωπε π ω

ω γ ω ηω πε π ω

γ ω ω ωπε π ω

−

Γ = = Ω

>>

Energy spectrum V• The total energy spectrum is obtained by integrating over the

solid angle:

( ) ( ) ( )

( ) ( ) ( ) ( )

2 22 2

2 2

3 32 2 222 2 2 2 22 21 22 2

0 3 3

22 cos

1 3 1 1 14 2 2 21c c c

d I d IdW d dd d d d d

e yy K y K y dyc y

π πγ

π πγ

ω ωππ θ θ γθω ω γ ω

γ ω ω ωπε π ω ω ω

− −

∞

−∞

= =Ω Ω

≈ + + + + +

( )2

50 3/

1 34

cc

dW e K x dxd c ω ω

γ ωω πε ω

∞

=

A more concise and popularexpression for the energyspectrum:

Homework

• Consider an electron storage ring at an energy of1 GeV, a circulating current of 200 mA and abending radius of . Calculate the energyloss per turn, the critical photon energy, and thetotal synchrotron radiation power.

• Make a short argument about why the trajectoryof a charged particle can not intersect with lightcone more than once (see slide #8).

2.22mρ =

Homework

• As shown in slide #15, the angular distribution of radiation power is

Show that for and , the angular spread of the radiation power is in the order of .

( )( ) ( )

2 2 2 2

3 220

1 sin cos14 4 1 cos 1 cos

rdP t ed c

β θ φπε π β θ γ β θ

= −

Ω − −

4 1γ θ− << << 1γ >>

1γ −