Today’s Outline - January 23, 2018phys.iit.edu/~segre/phys570/18S/lecture_05.pdf · Today’s...

Today’s Outline - January 23, 2018

• Fundamental wavelength from an undulator

• Higher harmonics

• On and off-axis spectrum

• Undulator to wiggler comparison

• Undulator harmonics

• Undulator coherence

• Emittance

• Time structure

• ERLs and FELs

Reading Assignment: Chapter 3.1–3.3

Homework Assignment #01:Chapter 2: 2,3,5,6,8due Thursday, January 25, 2018

C. Segre (IIT) PHYS 570 - Spring 2018 January 23, 2018 1 / 20








• Emittance

• Time structure

• ERLs and FELs











• Emittance

• Time structure

• ERLs and FELs











• Emittance

• Time structure

• ERLs and FELs











• Emittance

• Time structure

• ERLs and FELs











• Emittance

• Time structure

• ERLs and FELs











• Emittance

• Time structure

• ERLs and FELs











• Emittance

• Time structure

• ERLs and FELs











• Emittance

• Time structure

• ERLs and FELs











• Emittance

• Time structure

• ERLs and FELs











• Emittance

• Time structure

• ERLs and FELs











• Emittance

• Time structure

• ERLs and FELs




Undulator review

For an undulator of period λu we have derived the following undulatorparameters and their relationships

the K parameter, a dimension-less quantity which represents the“strength” of the undulator

the electron path length throughthe undulator, Sλu

K =e

2πmcλuB0

= 0.934λu[cm]B0[T]

Sλu ≈ λu(

1 +1

4

K 2

γ2

)Now let’s calculate the wavelength of the radiation generated by theundulator in the laboratory frame


Undulator review




K =e

2πmcλuB0

= 0.934λu[cm]B0[T]

Sλu ≈ λu(

1 +1

4

K 2

γ2



Undulator review




K =e

2πmcλuB0

= 0.934λu[cm]B0[T]

Sλu ≈ λu(

1 +1

4

K 2

γ2



Undulator review




K =e

2πmcλuB0

= 0.934λu[cm]B0[T]

Sλu ≈ λu(

1 +1

4

K 2

γ2



Undulator review




K =e

2πmcλuB0

= 0.934λu[cm]B0[T]

Sλu ≈ λu(

1 +1

4

K 2

γ2



Undulator review




K =e

2πmcλuB0

= 0.934λu[cm]B0[T]

Sλu ≈ λu(

1 +1

4

K 2

γ2

)

Now let’s calculate the wavelength of the radiation generated by theundulator in the laboratory frame


Undulator review




K =e

2πmcλuB0

= 0.934λu[cm]B0[T]

Sλu ≈ λu(

1 +1

4

K 2

γ2



Undulator wavelength

Consider an electron traveling through the undulator and emittingradiation at the first maximum excursion from the center.

λu

The emitted wave travels slightlyfaster than the electron.It moves cT ′ in the time the elec-tron travels a distance λu along theundulator.

The observer sees radiation with acompressed wavelength, along withharmonics which satisfy the samecondition.

nλn = cT ′ − λu




λu







λu







λu







λu

The emitted wave travels slightlyfaster than the electron.

It moves cT ′ in the time the elec-tron travels a distance λu along theundulator.













The observer sees radiation with acompressed wavelength,

along withharmonics which satisfy the samecondition.

λ1

= cT ′ − λu





The observer sees radiation with acompressed wavelength,

along withharmonics which satisfy the samecondition.

λ1 = cT ′ − λu




λu

cT’

2λ2





The fundamental wavelength

The fundamental wavelength must becorrected for the observer angle θ

λ1 = cT ′ − λu cos θ

= λu

(S

c

v− cos θ

)

= λu

([1 +

K 2

4γ2

]1

β− cos θ

)

Over the time T ′ the electronactually travels a distance Sλu,so that

T ′ =Sλu

v

S ≈ 1 +K 2

4γ2

Since γ is large, the maximum observation angle θ is small so

λ1 ≈ λu(

1

β+

K 2

4γ2β− 1 +

θ2

2

)=

λu2γ2

(2γ2

β+

K 2

2β− 2γ2 + γ2θ2

)





= λu

(S

c

v− cos θ

)

= λu

([1 +

K 2

4γ2

]1

β− cos θ

)


T ′ =Sλu

v

S ≈ 1 +K 2

4γ2


λ1 ≈ λu(

1

β+

K 2

4γ2β− 1 +

θ2

2

)=

λu2γ2

(2γ2

β+

K 2

2β− 2γ2 + γ2θ2

)





= λu

(S

c

v− cos θ

)

= λu

([1 +

K 2

4γ2

]1

β− cos θ

)


T ′ =Sλu

v

S ≈ 1 +K 2

4γ2


λ1 ≈ λu(

1

β+

K 2

4γ2β− 1 +

θ2

2

)=

λu2γ2

(2γ2

β+

K 2

2β− 2γ2 + γ2θ2

)





= λu

(S

c

v− cos θ

)

= λu

([1 +

K 2

4γ2

]1

β− cos θ

)


T ′ =Sλu

v

S ≈ 1 +K 2

4γ2


λ1 ≈ λu(

1

β+

K 2

4γ2β− 1 +

θ2

2

)=

λu2γ2

(2γ2

β+

K 2

2β− 2γ2 + γ2θ2

)





= λu

(S

c

v− cos θ

)

= λu

([1 +

K 2

4γ2

]1

β− cos θ

)


T ′ =Sλu

v

S ≈ 1 +K 2

4γ2


λ1 ≈ λu(

1

β+

K 2

4γ2β− 1 +

θ2

2

)=

λu2γ2

(2γ2

β+

K 2

2β− 2γ2 + γ2θ2

)





= λu

(S

c

v− cos θ

)

= λu

([1 +

K 2

4γ2

]1

β− cos θ

)


T ′ =Sλu

v

S ≈ 1 +K 2

4γ2


λ1 ≈ λu(

1

β+

K 2

4γ2β− 1 +

θ2

2

)=

λu2γ2

(2γ2

β+

K 2

2β− 2γ2 + γ2θ2

)





= λu

(S

c

v− cos θ

)

= λu

([1 +

K 2

4γ2

]1

β− cos θ

)


T ′ =Sλu

v

S ≈ 1 +K 2

4γ2


λ1 ≈ λu(

1

β+

K 2

4γ2β− 1 +

θ2

2

)

=λu2γ2

(2γ2

β+

K 2

2β− 2γ2 + γ2θ2

)





= λu

(S

c

v− cos θ

)

= λu

([1 +

K 2

4γ2

]1

β− cos θ

)


T ′ =Sλu

v

S ≈ 1 +K 2

4γ2


λ1 ≈ λu(

1

β+

K 2

4γ2β− 1 +

θ2

2

)=

λu2γ2

(2γ2

β+

K 2

2β− 2γ2 + γ2θ2

)



λ1 ≈λu2γ2

(2γ2

β+

K 2

2β− 2γ2 + γ2θ2

)

≈ λu2γ2

(2γ2

[1

β− 1

]+

K 2

2β− (γθ)2

)

≈ λu2γ2

(2

1

1− β2

[1− ββ

]+

K 2

2β− (γθ)2

)

≈ λu2γ2

(2

β(1 + β)+

K 2

2β− [γθ]2

)

regrouping terms

γ =

√1

1− β2

1− β2 = (1 + β)(1− β)



λ1 ≈λu2γ2

(2γ2

β+

K 2

2β− 2γ2 + γ2θ2

)

≈ λu2γ2

(2γ2

[1

β− 1

]+

K 2

2β− (γθ)2

)

≈ λu2γ2

(2

1

1− β2

[1− ββ

]+

K 2

2β− (γθ)2

)

≈ λu2γ2

(2

β(1 + β)+

K 2

2β− [γθ]2

)

regrouping terms

γ =

√1

1− β2

1− β2 = (1 + β)(1− β)



λ1 ≈λu2γ2

(2γ2

β+

K 2

2β− 2γ2 + γ2θ2

)

≈ λu2γ2

(2γ2

[1

β− 1

]+

K 2

2β− (γθ)2

)

≈ λu2γ2

(2

1

1− β2

[1− ββ

]+

K 2

2β− (γθ)2

)

≈ λu2γ2

(2

β(1 + β)+

K 2

2β− [γθ]2

)

regrouping terms

γ =

√1

1− β2

1− β2 = (1 + β)(1− β)



λ1 ≈λu2γ2

(2γ2

β+

K 2

2β− 2γ2 + γ2θ2

)

≈ λu2γ2

(2γ2

[1

β− 1

]+

K 2

2β− (γθ)2

)

≈ λu2γ2

(2

1

1− β2

[1− ββ

]+

K 2

2β− (γθ)2

)

≈ λu2γ2

(2

β(1 + β)+

K 2

2β− [γθ]2

)

regrouping terms

γ =

√1

1− β2

1− β2 = (1 + β)(1− β)



λ1 ≈λu2γ2

(2γ2

β+

K 2

2β− 2γ2 + γ2θ2

)

≈ λu2γ2

(2γ2

[1

β− 1

]+

K 2

2β− (γθ)2

)

≈ λu2γ2

(2

1

1− β2

[1− ββ

]+

K 2

2β− (γθ)2

)

≈ λu2γ2

(2

β(1 + β)+

K 2

2β− [γθ]2

)

regrouping terms

γ =

√1

1− β2

1− β2 = (1 + β)(1− β)



λ1 ≈λu2γ2

(2γ2

β+

K 2

2β− 2γ2 + γ2θ2

)

≈ λu2γ2

(2γ2

[1

β− 1

]+

K 2

2β− (γθ)2

)

≈ λu2γ2

(2

1

1− β2

[1− ββ

]+

K 2

2β− (γθ)2

)

≈ λu2γ2

(2

β(1 + β)+

K 2

2β− [γθ]2

)

regrouping terms

γ =

√1

1− β2

1− β2 = (1 + β)(1− β)



λ1 ≈λu2γ2

(2γ2

β+

K 2

2β− 2γ2 + γ2θ2

)

≈ λu2γ2

(2γ2

[1

β− 1

]+

K 2

2β− (γθ)2

)

≈ λu2γ2

(2

1

1− β2

[1− ββ

]+

K 2

2β− (γθ)2

)

≈ λu2γ2

(2

β(1 + β)+

K 2

2β− [γθ]2

)

regrouping terms

γ =

√1

1− β2

1− β2 = (1 + β)(1− β)



If we assume that β ∼ 1 for these highly relativistic electrons

λ1 ≈λu2γ2

(2

β(1 + β)+

K 2

2β− (γθ)2

)

≈ λu2γ2

(1 +

K 2

2β− (γθ)2

)and directly on axis

λ1 ≈λu2γ2

(1 +

K 2

2

)for a typical undulator γ ∼ 104, K ∼ 1, and λu ∼ 2cm so we estimate

λ1 ≈2× 10−2

2 (104)2

(1 +

(1)2

2

)= 1.5× 10−10m = 1.5A

This corresponds to an energy E1 ≈ 8.2keV but as the undulator gap iswidened, B0 decreases, K decreases, λ1 decreases, and E1 increases.




λ1 ≈λu2γ2

(2

β(1 + β)+

K 2

2β− (γθ)2

)

≈ λu2γ2

(1 +

K 2

2β− (γθ)2


λ1 ≈λu2γ2

(1 +

K 2

2


λ1 ≈2× 10−2

2 (104)2

(1 +

(1)2

2

)= 1.5× 10−10m = 1.5A





λ1 ≈λu2γ2

(2

β(1 + β)+

K 2

2β− (γθ)2

)≈ λu

2γ2

(1 +

K 2

2β− (γθ)2

)

and directly on axis

λ1 ≈λu2γ2

(1 +

K 2

2


λ1 ≈2× 10−2

2 (104)2

(1 +

(1)2

2

)= 1.5× 10−10m = 1.5A





λ1 ≈λu2γ2

(2

β(1 + β)+

K 2

2β− (γθ)2

)≈ λu

2γ2

(1 +

K 2

2β− (γθ)2


λ1 ≈λu2γ2

(1 +

K 2

2

)

for a typical undulator γ ∼ 104, K ∼ 1, and λu ∼ 2cm so we estimate

λ1 ≈2× 10−2

2 (104)2

(1 +

(1)2

2

)= 1.5× 10−10m = 1.5A





λ1 ≈λu2γ2

(2

β(1 + β)+

K 2

2β− (γθ)2

)≈ λu

2γ2

(1 +

K 2

2β− (γθ)2


λ1 ≈λu2γ2

(1 +

K 2

2


λ1 ≈2× 10−2

2 (104)2

(1 +

(1)2

2

)

= 1.5× 10−10m = 1.5A





λ1 ≈λu2γ2

(2

β(1 + β)+

K 2

2β− (γθ)2

)≈ λu

2γ2

(1 +

K 2

2β− (γθ)2


λ1 ≈λu2γ2

(1 +

K 2

2


λ1 ≈2× 10−2

2 (104)2

(1 +

(1)2

2

)= 1.5× 10−10m = 1.5A





λ1 ≈λu2γ2

(2

β(1 + β)+

K 2

2β− (γθ)2

)≈ λu

2γ2

(1 +

K 2

2β− (γθ)2


λ1 ≈λu2γ2

(1 +

K 2

2


λ1 ≈2× 10−2

2 (104)2

(1 +

(1)2

2

)= 1.5× 10−10m = 1.5A

This corresponds to an energy E1 ≈ 8.2keV but as the undulator gap iswidened

, B0 decreases, K decreases, λ1 decreases, and E1 increases.




λ1 ≈λu2γ2

(2

β(1 + β)+

K 2

2β− (γθ)2

)≈ λu

2γ2

(1 +

K 2

2β− (γθ)2


λ1 ≈λu2γ2

(1 +

K 2

2


λ1 ≈2× 10−2

2 (104)2

(1 +

(1)2

2

)= 1.5× 10−10m = 1.5A

This corresponds to an energy E1 ≈ 8.2keV but as the undulator gap iswidened, B0 decreases

, K decreases, λ1 decreases, and E1 increases.




λ1 ≈λu2γ2

(2

β(1 + β)+

K 2

2β− (γθ)2

)≈ λu

2γ2

(1 +

K 2

2β− (γθ)2


λ1 ≈λu2γ2

(1 +

K 2

2


λ1 ≈2× 10−2

2 (104)2

(1 +

(1)2

2

)= 1.5× 10−10m = 1.5A

This corresponds to an energy E1 ≈ 8.2keV but as the undulator gap iswidened, B0 decreases, K decreases

, λ1 decreases, and E1 increases.




λ1 ≈λu2γ2

(2

β(1 + β)+

K 2

2β− (γθ)2

)≈ λu

2γ2

(1 +

K 2

2β− (γθ)2


λ1 ≈λu2γ2

(1 +

K 2

2


λ1 ≈2× 10−2

2 (104)2

(1 +

(1)2

2

)= 1.5× 10−10m = 1.5A

This corresponds to an energy E1 ≈ 8.2keV but as the undulator gap iswidened, B0 decreases, K decreases, λ1 decreases

, and E1 increases.




λ1 ≈λu2γ2

(2

β(1 + β)+

K 2

2β− (γθ)2

)≈ λu

2γ2

(1 +

K 2

2β− (γθ)2


λ1 ≈λu2γ2

(1 +

K 2

2


λ1 ≈2× 10−2

2 (104)2

(1 +

(1)2

2

)= 1.5× 10−10m = 1.5A



Higher harmonics

yx

z

ψ

φ

θ

n

dt

dt ′= 1− ~n · ~β(t ′)

≈ 1− β[αφ+

(1− θ2

2− α2

2

)]

Recall that we developed an expres-sion for the Doppler time compres-sion of the emission from a movingelectron as a function of the ob-server angle.

This can be rewritten in terms ofthe coordinates in the figure usingthe vector of unit length in the ob-server direction:

~n ≈{φ, ψ, (1− θ2/2)

}~β ≈ β

{α, 0, (1− α2/2)

}dt

dt ′≈ 1−

(1− 1

2γ2

)(1 + αφ− θ2

2− α2

2

)


Higher harmonics

yx

z

ψ

φ

θ

n

dt

dt ′= 1− ~n · ~β(t ′)

≈ 1− β[αφ+

(1− θ2

2− α2

2

)]

Recall that we developed an expres-sion for the Doppler time compres-sion of the emission from a movingelectron as a function of the ob-server angle.

This can be rewritten in terms ofthe coordinates in the figure usingthe vector of unit length in the ob-server direction:

~n ≈{φ, ψ, (1− θ2/2)

}~β ≈ β

{α, 0, (1− α2/2)

}dt

dt ′≈ 1−

(1− 1

2γ2

)(1 + αφ− θ2

2− α2

2

)


Higher harmonics

yx

z

ψ

φ

θ

n

dt

dt ′= 1− ~n · ~β(t ′)

≈ 1− β[αφ+

(1− θ2

2− α2

2

)]

Recall that we developed an expres-sion for the Doppler time compres-sion of the emission from a movingelectron as a function of the ob-server angle.This can be rewritten in terms ofthe coordinates in the figure usingthe vector of unit length in the ob-server direction:

~n =

{φ, ψ,

√1− θ2

}~β = β

{α, 0,

√1− α2

}

dt

dt ′≈ 1−

(1− 1

2γ2

)(1 + αφ− θ2

2− α2

2

)


Higher harmonics

yx

z

ψ

φ

θ

n

dt

dt ′= 1− ~n · ~β(t ′)

≈ 1− β[αφ+

(1− θ2

2− α2

2

)]


~n ≈{φ, ψ, (1− θ2/2)

}~β ≈ β

{α, 0, (1− α2/2)

}

dt

dt ′≈ 1−

(1− 1

2γ2

)(1 + αφ− θ2

2− α2

2

)


Higher harmonics

yx

z

ψ

φ

θ

n

dt

dt ′= 1− ~n · ~β(t ′)

≈ 1− β[αφ+

(1− θ2

2− α2

2

)]


~n ≈{φ, ψ, (1− θ2/2)

}~β ≈ β

{α, 0, (1− α2/2)

}

dt

dt ′≈ 1−

(1− 1

2γ2

)(1 + αφ− θ2

2− α2

2

)


Higher harmonics

yx

z

ψ

φ

θ

n

dt

dt ′= 1− ~n · ~β(t ′)

≈ 1− β[αφ+

(1− θ2

2− α2

2

)]


~n ≈{φ, ψ, (1− θ2/2)

}~β ≈ β

{α, 0, (1− α2/2)

}dt

dt ′≈ 1−

(1− 1

2γ2

)(1 + αφ− θ2

2− α2

2

)C. Segre (IIT) PHYS 570 - Spring 2018 January 23, 2018 7 / 20

Higher harmonics

dt

dt ′≈ 1−

(1− 1

2γ2

)(1 + αφ− θ2

2− α2

2

)

≈ 1− 1− αφ+θ2

2+α2

2+

1

2γ2=

1

2

(θ2 + α2 +

1

γ2

)− αφ

This differential equation can be solved, realizing that φ and θ areconstant while α(t ′) varies as the electron moves through the insertiondevice, and gives:

ω1t = ωut ′− K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)− 2Kγ

1 + (γθ)2 + K 2/2φ sin (ωut ′)

ω1 � ωu as expected because of the Doppler compression , but they arenot proportional because of the second and third terms.

The motion of the electron, sinωut ′, is always sinusoidal, but because ofthe additional terms, the motion as seen by the observer, sinω1t, is not.


Higher harmonics

dt

dt ′≈ 1−

(1− 1

2γ2

)(1 + αφ− θ2

2− α2

2

)≈ 1− 1− αφ+

θ2

2+α2

2+

1

2γ2

=1

2

(θ2 + α2 +

1

γ2

)− αφ


ω1t = ωut ′− K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)− 2Kγ

1 + (γθ)2 + K 2/2φ sin (ωut ′)




Higher harmonics

dt

dt ′≈ 1−

(1− 1

2γ2

)(1 + αφ− θ2

2− α2

2

)≈ 1− 1− αφ+

θ2

2+α2

2+

1

2γ2=

1

2

(θ2 + α2 +

1

γ2

)− αφ


ω1t = ωut ′− K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)− 2Kγ

1 + (γθ)2 + K 2/2φ sin (ωut ′)




Higher harmonics

dt

dt ′≈ 1−

(1− 1

2γ2

)(1 + αφ− θ2

2− α2

2

)≈ 1− 1− αφ+

θ2

2+α2

2+

1

2γ2=

1

2

(θ2 + α2 +

1

γ2

)− αφ


ω1t = ωut ′− K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)− 2Kγ

1 + (γθ)2 + K 2/2φ sin (ωut ′)




Higher harmonics

dt

dt ′≈ 1−

(1− 1

2γ2

)(1 + αφ− θ2

2− α2

2

)≈ 1− 1− αφ+

θ2

2+α2

2+

1

2γ2=

1

2

(θ2 + α2 +

1

γ2

)− αφ


ω1t = ωut ′

− K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)− 2Kγ

1 + (γθ)2 + K 2/2φ sin (ωut ′)

ω1 � ωu as expected because of the Doppler compression

, but they arenot proportional because of the second and third terms.



Higher harmonics

dt

dt ′≈ 1−

(1− 1

2γ2

)(1 + αφ− θ2

2− α2

2

)≈ 1− 1− αφ+

θ2

2+α2

2+

1

2γ2=

1

2

(θ2 + α2 +

1

γ2

)− αφ


ω1t = ωut ′− K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)

− 2Kγ

1 + (γθ)2 + K 2/2φ sin (ωut ′)

ω1 � ωu as expected because of the Doppler compression , but they arenot proportional because of the second

and third terms.



Higher harmonics

dt

dt ′≈ 1−

(1− 1

2γ2

)(1 + αφ− θ2

2− α2

2

)≈ 1− 1− αφ+

θ2

2+α2

2+

1

2γ2=

1

2

(θ2 + α2 +

1

γ2

)− αφ


ω1t = ωut ′− K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)− 2Kγ

1 + (γθ)2 + K 2/2φ sin (ωut ′)




Higher harmonics

dt

dt ′≈ 1−

(1− 1

2γ2

)(1 + αφ− θ2

2− α2

2

)≈ 1− 1− αφ+

θ2

2+α2

2+

1

2γ2=

1

2

(θ2 + α2 +

1

γ2

)− αφ


ω1t = ωut ′− K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)− 2Kγ

1 + (γθ)2 + K 2/2φ sin (ωut ′)




On-axis undulator characteristics

ω1t = ωut ′ − K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)

Suppose we have K = 1 and θ = 0(on axis), then

ω1t = ωut ′ +1

6sin (2ωut ′)

Plotting sinωut ′ and sinω1t showsthe deviation from sinusoidal.

Similarly, for K = 2 and K =5, the deviation becomes more pro-nounced. This shows how higherharmonics must be present in the ra-diation as seen by the observer.

0 π/2 π

Phase Angle (radians)

0

0.5

1

Fra

ctio

na

l D

isp

lace

me

nt

pure sine

K=1

on axis



ω1t = ωut ′ − K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)


ω1t = ωut ′ +1

6sin (2ωut ′)



0 π/2 π


0

0.5

1

Fra

ctio

na

l D

isp

lace

me

nt

pure sine

K=1

on axis



ω1t = ωut ′ − K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)


ω1t = ωut ′ +1

6sin (2ωut ′)



0 π/2 π


0

0.5

1

Fra

ctio

na

l D

isp

lace

me

nt

pure sine

K=1

on axis



ω1t = ωut ′ − K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)


ω1t = ωut ′ +1

6sin (2ωut ′)


Similarly, for K = 2

and K =5, the deviation becomes more pro-nounced. This shows how higherharmonics must be present in the ra-diation as seen by the observer.

0 π/2 π


0

0.5

1

Fra

ctio

na

l D

isp

lace

me

nt

pure sine

K=1

K=2

on axis



ω1t = ωut ′ − K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)


ω1t = ωut ′ +1

6sin (2ωut ′)


Similarly, for K = 2 and K =5, the deviation becomes more pro-nounced.

This shows how higherharmonics must be present in the ra-diation as seen by the observer.

0 π/2 π


0

0.5

1

Fra

ctio

na

l D

isp

lace

me

nt

pure sine

K=1

K=2

on axis

K=5



ω1t = ωut ′ − K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)


ω1t = ωut ′ +1

6sin (2ωut ′)



E 3E 5E

Energy

Inte

nsity (

arb

. units)

K=1

K=2

K=5

pure sine


Off-axis undulator characteristics

ω1t = ωut ′− K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)− 2Kγ

1 + (γθ)2 + K 2/2φ sin (ωut ′)

0 π/2 π


0

0.5

1

Fra

ctional D

ispla

cem

ent

θ=0

K=2

φ=θ=1/γ

When K = 2 and θ = φ = 1/γ, wehave

ω1t = ωut ′+1

4sin (2ωut ′) + sinωut ′

The last term introduces an antisym-metric term which skews the func-tion and leads to the presence offorbidden harmonics (2nd , 4th, etc)in the radiation from the undulatorcompared to the on-axis radiation.



ω1t = ωut ′− K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)− 2Kγ

1 + (γθ)2 + K 2/2φ sin (ωut ′)

0 π/2 π


0

0.5

1

Fra

ctional D

ispla

cem

ent

θ=0

K=2

φ=θ=1/γ


ω1t = ωut ′+1





ω1t = ωut ′− K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)− 2Kγ

1 + (γθ)2 + K 2/2φ sin (ωut ′)

0 π/2 π


0

0.5

1

Fra

ctio

na

l D

isp

lace

me

nt

θ=0

K=2

φ=θ=1/γ


ω1t = ωut ′+1


The last term introduces an antisym-metric term which skews the func-tion

and leads to the presence offorbidden harmonics (2nd , 4th, etc)in the radiation from the undulatorcompared to the on-axis radiation.



ω1t = ωut ′− K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)− 2Kγ

1 + (γθ)2 + K 2/2φ sin (ωut ′)

E 3E 5E

Energy

Inte

nsity (

arb

. units)

K=1 θ=0

K=1 φ=θ=1/γ

K=5 φ=θ=1/γ

even harmonics


ω1t = ωut ′+1




Spectral comparison

• Brilliance is 6 orders largerthan a bending magnet

• Both odd and evenharmonics appear

• Harmonics can be tuned inenergy (dashed lines)


Spectral comparison





Spectral comparison





Spectral comparison





Diffraction grating

An N period undulator is basically like a diffraction grating, only in thetime domain rather than the space domain.

A diffraction grating consists of N coherent sources whose emission isdetected at a single point.

δL

The radiation from each slit has totravel a slightly different distance toget to the detector. For consecu-tive slits this path length difference,Lm+1 − Lm = δL, gives rise to aphase shift, 2πε = 2πδL/λ. So at thedetector, we have a sum of waves:

N−1∑m=0

e i(~k·~r+2πmε) = e i

~k·~rN−1∑m=0

e i2πmε


Diffraction grating

An N period undulator is basically like a diffraction grating, only in thetime domain rather than the space domain.A diffraction grating consists of N coherent sources whose emission isdetected at a single point.

δL


N−1∑m=0


~k·~rN−1∑m=0

e i2πmε


Diffraction grating


δL

The radiation from each slit has totravel a slightly different distance toget to the detector.

For consecu-tive slits this path length difference,Lm+1 − Lm = δL, gives rise to aphase shift, 2πε = 2πδL/λ. So at thedetector, we have a sum of waves:

N−1∑m=0


~k·~rN−1∑m=0

e i2πmε


Diffraction grating


δL

The radiation from each slit has totravel a slightly different distance toget to the detector. For consecu-tive slits this path length difference,Lm+1 − Lm = δL,

gives rise to aphase shift, 2πε = 2πδL/λ. So at thedetector, we have a sum of waves:

N−1∑m=0


~k·~rN−1∑m=0

e i2πmε


Diffraction grating


δL

The radiation from each slit has totravel a slightly different distance toget to the detector. For consecu-tive slits this path length difference,Lm+1 − Lm = δL, gives rise to aphase shift, 2πε = 2πδL/λ.

So at thedetector, we have a sum of waves:

N−1∑m=0


~k·~rN−1∑m=0

e i2πmε


Diffraction grating


δL


N−1∑m=0

e i(~k·~r+2πmε)

= e i~k·~r

N−1∑m=0

e i2πmε


Diffraction grating


δL


N−1∑m=0


~k·~rN−1∑m=0

e i2πmε


Geometric series

The sum is simply a geometric series, SN with k = e i2πε

SN =N−1∑m=0

km = 1 + k + k2 + · · ·+ kN−2 + kN−1

We can develop a recursion relation by writing the expression for SN−1

SN−1 =N−2∑m=0

km = 1 + k + k2 + · · ·+ kN−2

so we can write that SN−1 = SN − kN−1 and

SN = 1 + kSN−1 = 1 + k(SN − kN−1) = 1 + kSN − kN

Solving for SN , we have

SN − kSN = 1− kN −→ SN =1− kN

1− k


Geometric series


SN =N−1∑m=0

km

= 1 + k + k2 + · · ·+ kN−2 + kN−1


SN−1 =N−2∑m=0

km = 1 + k + k2 + · · ·+ kN−2




SN − kSN = 1− kN −→ SN =1− kN

1− k


Geometric series


SN =N−1∑m=0

km = 1 + k + k2 + · · ·+ kN−2 + kN−1


SN−1 =N−2∑m=0

km = 1 + k + k2 + · · ·+ kN−2




SN − kSN = 1− kN −→ SN =1− kN

1− k


Geometric series


SN =N−1∑m=0

km = 1 + k + k2 + · · ·+ kN−2 + kN−1


SN−1 =N−2∑m=0

km = 1 + k + k2 + · · ·+ kN−2




SN − kSN = 1− kN −→ SN =1− kN

1− k


Geometric series


SN =N−1∑m=0

km = 1 + k + k2 + · · ·+ kN−2 + kN−1


SN−1 =N−2∑m=0

km = 1 + k + k2 + · · ·+ kN−2

so we can write that SN−1 = SN − kN−1

and



SN − kSN = 1− kN −→ SN =1− kN

1− k


Geometric series


SN =N−1∑m=0

km = 1 + k + k2 + · · ·+ kN−2 + kN−1


SN−1 =N−2∑m=0

km = 1 + k + k2 + · · ·+ kN−2


SN = 1 + kSN−1

= 1 + k(SN − kN−1) = 1 + kSN − kN


SN − kSN = 1− kN −→ SN =1− kN

1− k


Geometric series


SN =N−1∑m=0

km = 1 + k + k2 + · · ·+ kN−2 + kN−1


SN−1 =N−2∑m=0

km = 1 + k + k2 + · · ·+ kN−2


SN = 1 + kSN−1 = 1 + k(SN − kN−1)

= 1 + kSN − kN


SN − kSN = 1− kN −→ SN =1− kN

1− k


Geometric series


SN =N−1∑m=0

km = 1 + k + k2 + · · ·+ kN−2 + kN−1


SN−1 =N−2∑m=0

km = 1 + k + k2 + · · ·+ kN−2




SN − kSN = 1− kN −→ SN =1− kN

1− k


Geometric series


SN =N−1∑m=0

km = 1 + k + k2 + · · ·+ kN−2 + kN−1


SN−1 =N−2∑m=0

km = 1 + k + k2 + · · ·+ kN−2




SN − kSN = 1− kN −→ SN =1− kN

1− k


Geometric series


SN =N−1∑m=0

km = 1 + k + k2 + · · ·+ kN−2 + kN−1


SN−1 =N−2∑m=0

km = 1 + k + k2 + · · ·+ kN−2




SN − kSN = 1− kN

−→ SN =1− kN

1− k


Geometric series


SN =N−1∑m=0

km = 1 + k + k2 + · · ·+ kN−2 + kN−1


SN−1 =N−2∑m=0

km = 1 + k + k2 + · · ·+ kN−2




SN − kSN = 1− kN −→ SN =1− kN

1− k


Intensity from a diffraction grating

Restoring the expression for k = e i2πε, we have:

N−1∑m=0

e i2πmε = SN =1− e i2πNε

1− e i2πε=

(e−iπNε − e iπNε

e−iπε − e iπε

)e iπNε

e iπε

SN =

(sin (πNε)

sin (πε)

)e iπ(N−1)ε

Therefore, for the diffraction grating we can calculate the intensity at thedetector as

I =

∣∣∣∣∣e i~k·~rN−1∑m=0

e i2πmε

∣∣∣∣∣2

=∣∣∣e i~k·~rSN

∣∣∣2 =

∣∣∣∣e i~k·~r sin (πNε)

sin (πε)e iπ(N−1)ε

∣∣∣∣2

I =sin2 (πNε)

sin2 (πε)




N−1∑m=0

e i2πmε = SN

=1− e i2πNε

1− e i2πε=



)e iπNε

e iπε

SN =

(sin (πNε)

sin (πε)

)e iπ(N−1)ε


I =

∣∣∣∣∣e i~k·~rN−1∑m=0

e i2πmε

∣∣∣∣∣2


∣∣∣2 =



∣∣∣∣2

I =sin2 (πNε)

sin2 (πε)




N−1∑m=0


1− e i2πε

=



)e iπNε

e iπε

SN =

(sin (πNε)

sin (πε)

)e iπ(N−1)ε


I =

∣∣∣∣∣e i~k·~rN−1∑m=0

e i2πmε

∣∣∣∣∣2


∣∣∣2 =



∣∣∣∣2

I =sin2 (πNε)

sin2 (πε)




N−1∑m=0


1− e i2πε=



)e iπNε

e iπε

SN =

(sin (πNε)

sin (πε)

)e iπ(N−1)ε


I =

∣∣∣∣∣e i~k·~rN−1∑m=0

e i2πmε

∣∣∣∣∣2


∣∣∣2 =



∣∣∣∣2

I =sin2 (πNε)

sin2 (πε)




N−1∑m=0


1− e i2πε=



)e iπNε

e iπε

SN =

(sin (πNε)

sin (πε)

)e iπ(N−1)ε


I =

∣∣∣∣∣e i~k·~rN−1∑m=0

e i2πmε

∣∣∣∣∣2


∣∣∣2 =



∣∣∣∣2

I =sin2 (πNε)

sin2 (πε)




N−1∑m=0


1− e i2πε=



)e iπNε

e iπε

SN =

(sin (πNε)

sin (πε)

)e iπ(N−1)ε


I =

∣∣∣∣∣e i~k·~rN−1∑m=0

e i2πmε

∣∣∣∣∣2


∣∣∣2 =



∣∣∣∣2

I =sin2 (πNε)

sin2 (πε)




N−1∑m=0


1− e i2πε=



)e iπNε

e iπε

SN =

(sin (πNε)

sin (πε)

)e iπ(N−1)ε


I =

∣∣∣∣∣e i~k·~rN−1∑m=0

e i2πmε

∣∣∣∣∣2


∣∣∣2 =



∣∣∣∣2

I =sin2 (πNε)

sin2 (πε)




N−1∑m=0


1− e i2πε=



)e iπNε

e iπε

SN =

(sin (πNε)

sin (πε)

)e iπ(N−1)ε


I =

∣∣∣∣∣e i~k·~rN−1∑m=0

e i2πmε

∣∣∣∣∣2


∣∣∣2

=



∣∣∣∣2

I =sin2 (πNε)

sin2 (πε)




N−1∑m=0


1− e i2πε=



)e iπNε

e iπε

SN =

(sin (πNε)

sin (πε)

)e iπ(N−1)ε


I =

∣∣∣∣∣e i~k·~rN−1∑m=0

e i2πmε

∣∣∣∣∣2


∣∣∣2 =



∣∣∣∣2

I =sin2 (πNε)

sin2 (πε)




N−1∑m=0


1− e i2πε=



)e iπNε

e iπε

SN =

(sin (πNε)

sin (πε)

)e iπ(N−1)ε


I =

∣∣∣∣∣e i~k·~rN−1∑m=0

e i2πmε

∣∣∣∣∣2


∣∣∣2 =



∣∣∣∣2

I =sin2 (πNε)

sin2 (πε)


Beam coherence


2πε=0

0 10 20 30 40

2πε [degrees]

Inte

nsity (

arb

units)

With the height and width of the peak dependent on the number of poles.


Beam coherence


2πε=0

0 10 20 30 40

2πε [degrees]

Inte

nsity (

arb

un

its)



Beam coherence


2πε=5o

0 10 20 30 40

2πε [degrees]

Inte

nsity (

arb

un

its)



Beam coherence


2πε=10o

0 10 20 30 40

2πε [degrees]

Inte

nsity (

arb

un

its)



Beam coherence


2πε=15o

0 10 20 30 40

2πε [degrees]

Inte

nsity (

arb

un

its)



Beam coherence


2πε=20o

0 10 20 30 40

2πε [degrees]

Inte

nsity (

arb

un

its)



Beam coherence


2πε=25o

0 10 20 30 40

2πε [degrees]

Inte

nsity (

arb

un

its)



Beam coherence


2πε=30o

0 10 20 30 40

2πε [degrees]

Inte

nsity (

arb

un

its)



Beam coherence


2πε=35o

0 10 20 30 40

2πε [degrees]

Inte

nsity (

arb

un

its)



Beam coherence


2πε=40o

0 10 20 30 40

2πε [degrees]

Inte

nsity (

arb

un

its)



Beam coherence


2πε=45o

0 10 20 30 40

2πε [degrees]

Inte

nsity (

arb

un

its)



Beam coherence


2πε=45o

0 10 20 30 40

2πε [degrees]

Inte

nsity (

arb

un

its)



Undulator monochromaticity

2πε

Inte

nsity (

arb

un

its)

32 poles

16 poles

96 poles

The more poles in the undulator,the more monochromatic the beam

The APS has a 72 pole undulatorof 3.3 cm period

Higher order harmonics have nar-rower energy bandwidth but lowerpeak intensity as seen from the pha-sor diagram representation



2πε

Inte

nsity (

arb

un

its)

32 poles

16 poles

96 poles






2πε

Inte

nsity (

arb

un

its)

3rd

harmonic

5th

harmonic

1st harmonic





Synchrotron time structure

There are two importanttime scales for a storagering such as the APS: pulselength and interpulse spacing

The APS pulse length in 24-bunch mode is 90 ps whilethe pulses come every 154 ns

Other modes include single-bunch mode for timing ex-periments and 324-bunchmode (inter pulse timing of11.7 ns) for a more constantx-ray flux












Emittance

Is there a limit to the brilliance of an undulator source at a synchrotron?

the brilliance is inversely proportional to the square of the product of thelinear source size and the angular divergence

brilliance =flux [photons/s]

divergence[mrad2

]· source size [mm2] [0.1% bandwidth]

the product of the source size and divergence iscalled the emittance, ε and the brilliance is thuslimited by the product of the emittance of theradiation in the horizontal and vertical directionsεxεy

this emittance cannot be changed but it can berotated or deformed by magnetic fields as theelectron beam travels around the storage ringas long as the area is kept constant

y

y

σ

σ

y

y


Emittance




divergence[mrad2




y

y

σ

σ

y

y


Emittance




divergence[mrad2


the product of the source size and divergence iscalled the emittance, ε

and the brilliance is thuslimited by the product of the emittance of theradiation in the horizontal and vertical directionsεxεy


y

y

σ

σ

y

y


Emittance




divergence[mrad2




y

y

σ

σ

y

y


Emittance




divergence[mrad2



this emittance cannot be changed but it can berotated

or deformed by magnetic fields as theelectron beam travels around the storage ringas long as the area is kept constant

y

y

σ

σ

y

y


Emittance




divergence[mrad2




y

y

σ

σ

y

y


APS emittance

For photon emission from a single electron in a 2m undulator at 1A

y

10µrad

-10µrad

10µm-10µm

y

σradiation = 9.1µm

σ′radiation = 7.7µrad

σγ=

√Lλ

4π= 1.3µm

σ′γ=

√λ

L= 7.1µrad

current APS electron beamparameters are

σy= 9.1µm

σ′y= 3.0µrad

must convolute to get pho-ton emission from entirebeam (in vertical direction)


APS emittance


y

10µrad

-10µrad

10µm-10µm

y



σγ=

√Lλ

4π= 1.3µm

σ′γ=

√λ

L= 7.1µrad


σy= 9.1µm

σ′y= 3.0µrad



APS emittance


y

10µrad

-10µrad

10µm-10µm

y



σγ=

√Lλ

4π= 1.3µm

σ′γ=

√λ

L= 7.1µrad


σy= 9.1µm

σ′y= 3.0µrad



APS emittance


y

10µrad

-10µrad

10µm-10µm

y



σγ=

√Lλ

4π= 1.3µm

σ′γ=

√λ

L= 7.1µrad


σy= 9.1µm

σ′y= 3.0µrad



APS emittance


y

10µrad

-10µrad

10µm-10µm

y



σγ=

√Lλ

4π= 1.3µm

σ′γ=

√λ

L= 7.1µrad


σy= 9.1µm

σ′y= 3.0µrad



APS emittance


y

10µrad

-10µrad

10µm-10µm

y



σγ=

√Lλ

4π= 1.3µm

σ′γ=

√λ

L= 7.1µrad


σy= 9.1µm

σ′y= 3.0µrad



APS emittance


y

10µrad

-10µrad

10µm-10µm

y



σγ=

√Lλ

4π= 1.3µm

σ′γ=

√λ

L= 7.1µrad


σy= 9.1µm

σ′y= 3.0µrad



APS emittance


y

10µrad

-10µrad

10µm-10µm

y



σγ=

√Lλ

4π= 1.3µm

σ′γ=

√λ

L= 7.1µrad


σy= 9.1µm

σ′y= 3.0µrad



Evolution of APS parameters

Parameter 1995 2001 2005σx 334 µm 352 µm 280 µmσ′x 24 µrad 22 µrad 11.6 µrad

σy 89 µm 18.4 µm 9.1 µmσ′y 8.9 µrad 4.2 µrad 3.0 µrad

When first commissioned in 1995, the APS electron beam size anddivergence was relatively large, particularly in the horizontal, x direction

By the end of the first decade of operation, the horizontal source sizedecreased by about 16% and its horizontal divergence by more than 50%

At the same time the vertical source size decreased by over 90% and thevertical divergence by nearly 67%

The next big upgrade (slated for 2022) will make the beam more square inspace and by choosing the undulator correctly, a higher performanceinsertion device.


































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