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

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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

2πmcλuB0

= 0.934λu[cm]B0[T]

Sλu ≈ λu(

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

Undulator review

2πmcλuB0

= 0.934λu[cm]B0[T]

Sλu ≈ λu(

Undulator review

2πmcλuB0

= 0.934λu[cm]B0[T]

Sλu ≈ λu(

Undulator review

2πmcλuB0

= 0.934λu[cm]B0[T]

Sλu ≈ λu(

Undulator review

2πmcλuB0

= 0.934λu[cm]B0[T]

Sλu ≈ λu(

Undulator review

2πmcλuB0

= 0.934λu[cm]B0[T]

Sλu ≈ λu(

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

Undulator review

2πmcλuB0

= 0.934λu[cm]B0[T]

Sλu ≈ λu(

Undulator wavelength

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

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

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.

= cT ′ − λu

The observer sees radiation with acompressed wavelength,

along withharmonics which satisfy the samecondition.

λ1 = cT ′ − λu

The fundamental wavelength

The fundamental wavelength must becorrected for the observer angle θ

λ1 = cT ′ − λu cos θ

v− cos θ

β− cos θ

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

T ′ =Sλu

S ≈ 1 +K 2

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

λ1 ≈ λu(

4γ2β− 1 +

λu2γ2

2β− 2γ2 + γ2θ2

v− cos θ

β− cos θ

T ′ =Sλu

S ≈ 1 +K 2

λ1 ≈ λu(

4γ2β− 1 +

λu2γ2

2β− 2γ2 + γ2θ2

v− cos θ

β− cos θ

T ′ =Sλu

S ≈ 1 +K 2

λ1 ≈ λu(

4γ2β− 1 +

λu2γ2

2β− 2γ2 + γ2θ2

v− cos θ

β− cos θ

T ′ =Sλu

S ≈ 1 +K 2

λ1 ≈ λu(

4γ2β− 1 +

λu2γ2

2β− 2γ2 + γ2θ2

v− cos θ

β− cos θ

T ′ =Sλu

S ≈ 1 +K 2

λ1 ≈ λu(

4γ2β− 1 +

λu2γ2

2β− 2γ2 + γ2θ2

v− cos θ

β− cos θ

T ′ =Sλu

S ≈ 1 +K 2

λ1 ≈ λu(

4γ2β− 1 +

λu2γ2

2β− 2γ2 + γ2θ2

v− cos θ

β− cos θ

T ′ =Sλu

S ≈ 1 +K 2

λ1 ≈ λu(

4γ2β− 1 +

=λu2γ2

2β− 2γ2 + γ2θ2

v− cos θ

β− cos θ

T ′ =Sλu

S ≈ 1 +K 2

λ1 ≈ λu(

4γ2β− 1 +

λu2γ2

2β− 2γ2 + γ2θ2

λ1 ≈λu2γ2

2β− 2γ2 + γ2θ2

≈ λu2γ2

β− 1

2β− (γθ)2

≈ λu2γ2

1− β2

[1− ββ

2β− (γθ)2

≈ λu2γ2

β(1 + β)+

2β− [γθ]2

regrouping terms

1− β2

1− β2 = (1 + β)(1− β)

λ1 ≈λu2γ2

2β− 2γ2 + γ2θ2

≈ λu2γ2

β− 1

2β− (γθ)2

≈ λu2γ2

1− β2

[1− ββ

2β− (γθ)2

≈ λu2γ2

β(1 + β)+

2β− [γθ]2

regrouping terms

1− β2

1− β2 = (1 + β)(1− β)

λ1 ≈λu2γ2

2β− 2γ2 + γ2θ2

≈ λu2γ2

β− 1

2β− (γθ)2

≈ λu2γ2

1− β2

[1− ββ

2β− (γθ)2

≈ λu2γ2

β(1 + β)+

2β− [γθ]2

regrouping terms

1− β2

1− β2 = (1 + β)(1− β)

λ1 ≈λu2γ2

2β− 2γ2 + γ2θ2

≈ λu2γ2

β− 1

2β− (γθ)2

≈ λu2γ2

1− β2

[1− ββ

2β− (γθ)2

≈ λu2γ2

β(1 + β)+

2β− [γθ]2

regrouping terms

1− β2

1− β2 = (1 + β)(1− β)

λ1 ≈λu2γ2

2β− 2γ2 + γ2θ2

≈ λu2γ2

β− 1

2β− (γθ)2

≈ λu2γ2

1− β2

[1− ββ

2β− (γθ)2

≈ λu2γ2

β(1 + β)+

2β− [γθ]2

regrouping terms

1− β2

1− β2 = (1 + β)(1− β)

λ1 ≈λu2γ2

2β− 2γ2 + γ2θ2

≈ λu2γ2

β− 1

2β− (γθ)2

≈ λu2γ2

1− β2

[1− ββ

2β− (γθ)2

≈ λu2γ2

β(1 + β)+

2β− [γθ]2

regrouping terms

1− β2

1− β2 = (1 + β)(1− β)

λ1 ≈λu2γ2

2β− 2γ2 + γ2θ2

≈ λu2γ2

β− 1

2β− (γθ)2

≈ λu2γ2

1− β2

[1− ββ

2β− (γθ)2

≈ λu2γ2

β(1 + β)+

2β− [γθ]2

regrouping terms

1− β2

1− β2 = (1 + β)(1− β)

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

λ1 ≈λu2γ2

β(1 + β)+

2β− (γθ)2

≈ λu2γ2

2β− (γθ)2

)and directly on axis

λ1 ≈λu2γ2

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

λ1 ≈2× 10−2

2 (104)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

β(1 + β)+

2β− (γθ)2

≈ λu2γ2

2β− (γθ)2

λ1 ≈λu2γ2

λ1 ≈2× 10−2

2 (104)2

)= 1.5× 10−10m = 1.5A

λ1 ≈λu2γ2

β(1 + β)+

2β− (γθ)2

)≈ λu

2β− (γθ)2

and directly on axis

λ1 ≈λu2γ2

λ1 ≈2× 10−2

2 (104)2

)= 1.5× 10−10m = 1.5A

λ1 ≈λu2γ2

β(1 + β)+

2β− (γθ)2

)≈ λu

2β− (γθ)2

λ1 ≈λu2γ2

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

λ1 ≈2× 10−2

2 (104)2

)= 1.5× 10−10m = 1.5A

λ1 ≈λu2γ2

β(1 + β)+

2β− (γθ)2

)≈ λu

2β− (γθ)2

λ1 ≈λu2γ2

λ1 ≈2× 10−2

2 (104)2

= 1.5× 10−10m = 1.5A

λ1 ≈λu2γ2

β(1 + β)+

2β− (γθ)2

)≈ λu

2β− (γθ)2

λ1 ≈λu2γ2

λ1 ≈2× 10−2

2 (104)2

)= 1.5× 10−10m = 1.5A

λ1 ≈λu2γ2

β(1 + β)+

2β− (γθ)2

)≈ λu

2β− (γθ)2

λ1 ≈λu2γ2

λ1 ≈2× 10−2

2 (104)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

β(1 + β)+

2β− (γθ)2

)≈ λu

2β− (γθ)2

λ1 ≈λu2γ2

λ1 ≈2× 10−2

2 (104)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

β(1 + β)+

2β− (γθ)2

)≈ λu

2β− (γθ)2

λ1 ≈λu2γ2

λ1 ≈2× 10−2

2 (104)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

β(1 + β)+

2β− (γθ)2

)≈ λu

2β− (γθ)2

λ1 ≈λu2γ2

λ1 ≈2× 10−2

2 (104)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

β(1 + β)+

2β− (γθ)2

)≈ λu

2β− (γθ)2

λ1 ≈λu2γ2

λ1 ≈2× 10−2

2 (104)2

)= 1.5× 10−10m = 1.5A

Higher harmonics

dt ′= 1− ~n · ~β(t ′)

≈ 1− β[αφ+

(1− θ2

2− α2

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 ′≈ 1−

(1− 1

)(1 + αφ− θ2

2− α2

Higher harmonics

dt ′= 1− ~n · ~β(t ′)

≈ 1− β[αφ+

(1− θ2

2− α2

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 ′≈ 1−

(1− 1

)(1 + αφ− θ2

2− α2

Higher harmonics

dt ′= 1− ~n · ~β(t ′)

≈ 1− β[αφ+

(1− θ2

2− α2

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:

{φ, ψ,

√1− θ2

}~β = β

{α, 0,

√1− α2

dt ′≈ 1−

(1− 1

)(1 + αφ− θ2

2− α2

Higher harmonics

dt ′= 1− ~n · ~β(t ′)

≈ 1− β[αφ+

(1− θ2

2− α2

~n ≈{φ, ψ, (1− θ2/2)

}~β ≈ β

{α, 0, (1− α2/2)

dt ′≈ 1−

(1− 1

)(1 + αφ− θ2

2− α2

Higher harmonics

dt ′= 1− ~n · ~β(t ′)

≈ 1− β[αφ+

(1− θ2

2− α2

~n ≈{φ, ψ, (1− θ2/2)

}~β ≈ β

{α, 0, (1− α2/2)

dt ′≈ 1−

(1− 1

)(1 + αφ− θ2

2− α2

Higher harmonics

dt ′= 1− ~n · ~β(t ′)

≈ 1− β[αφ+

(1− θ2

2− α2

~n ≈{φ, ψ, (1− θ2/2)

}~β ≈ β

{α, 0, (1− α2/2)

dt ′≈ 1−

(1− 1

)(1 + αφ− θ2

2− α2

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

Higher harmonics

dt ′≈ 1−

(1− 1

)(1 + αφ− θ2

2− α2

≈ 1− 1− αφ+θ2

(θ2 + α2 +

)− 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 ′≈ 1−

(1− 1

)(1 + αφ− θ2

2− α2

)≈ 1− 1− αφ+

(θ2 + α2 +

)− 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 ′≈ 1−

(1− 1

)(1 + αφ− θ2

2− α2

)≈ 1− 1− αφ+

(θ2 + α2 +

)− 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 ′≈ 1−

(1− 1

)(1 + αφ− θ2

2− α2

)≈ 1− 1− αφ+

(θ2 + α2 +

)− 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 ′≈ 1−

(1− 1

)(1 + αφ− θ2

2− α2

)≈ 1− 1− αφ+

(θ2 + α2 +

)− 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 ′≈ 1−

(1− 1

)(1 + αφ− θ2

2− α2

)≈ 1− 1− αφ+

(θ2 + α2 +

)− 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 ′≈ 1−

(1− 1

)(1 + αφ− θ2

2− α2

)≈ 1− 1− αφ+

(θ2 + α2 +

)− 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 ′≈ 1−

(1− 1

)(1 + αφ− θ2

2− α2

)≈ 1− 1− αφ+

(θ2 + α2 +

)− 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)

pure sine

on axis

ω1t = ωut ′ − K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)

ω1t = ωut ′ +1

6sin (2ωut ′)

0 π/2 π

pure sine

on axis

ω1t = ωut ′ − K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)

ω1t = ωut ′ +1

6sin (2ωut ′)

0 π/2 π

pure sine

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

pure sine

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

pure sine

on axis

ω1t = ωut ′ − K 2/4

1 + (γθ)2 + K 2/2sin (2ωut ′)

ω1t = ωut ′ +1

6sin (2ωut ′)

E 3E 5E

Energy

nsity (

. units)

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

ctional D

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

ctional D

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

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

nsity (

. 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

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.

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.

N−1∑m=0

~k·~rN−1∑m=0

e i2πmε

Diffraction grating

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

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

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

N−1∑m=0

e i(~k·~r+2πmε)

= e i~k·~r

N−1∑m=0

e i2πmε

Diffraction grating

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

= 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

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

(sin (πNε)

sin (πε)

)e iπ(N−1)ε

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

∣∣∣∣∣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

(sin (πNε)

sin (πε)

)e iπ(N−1)ε

∣∣∣∣∣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

(sin (πNε)

sin (πε)

)e iπ(N−1)ε

∣∣∣∣∣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

(sin (πNε)

sin (πε)

)e iπ(N−1)ε

∣∣∣∣∣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

(sin (πNε)

sin (πε)

)e iπ(N−1)ε

∣∣∣∣∣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

(sin (πNε)

sin (πε)

)e iπ(N−1)ε

∣∣∣∣∣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

(sin (πNε)

sin (πε)

)e iπ(N−1)ε

∣∣∣∣∣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

(sin (πNε)

sin (πε)

)e iπ(N−1)ε

∣∣∣∣∣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

(sin (πNε)

sin (πε)

)e iπ(N−1)ε

∣∣∣∣∣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

(sin (πNε)

sin (πε)

)e iπ(N−1)ε

∣∣∣∣∣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]

nsity (

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]

nsity (

Beam coherence

2πε=5o

0 10 20 30 40

2πε [degrees]

nsity (

Beam coherence

2πε=10o

0 10 20 30 40

2πε [degrees]

nsity (

Beam coherence

2πε=15o

0 10 20 30 40

2πε [degrees]

nsity (

Beam coherence

2πε=20o

0 10 20 30 40

2πε [degrees]

nsity (

Beam coherence

2πε=25o

0 10 20 30 40

2πε [degrees]

nsity (

Beam coherence

2πε=30o

0 10 20 30 40

2πε [degrees]

nsity (

Beam coherence

2πε=35o

0 10 20 30 40

2πε [degrees]

nsity (

Beam coherence

2πε=40o

0 10 20 30 40

2πε [degrees]

nsity (

Beam coherence

2πε=45o

0 10 20 30 40

2πε [degrees]

nsity (

Beam coherence

2πε=45o

0 10 20 30 40

2πε [degrees]

nsity (

Undulator monochromaticity

nsity (

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

nsity (

32 poles

16 poles

96 poles

nsity (

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

Emittance

divergence[mrad2

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

Emittance

divergence[mrad2

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

Emittance

divergence[mrad2

APS emittance

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

10µrad

-10µrad

10µm-10µm

σradiation = 9.1µm

σ′radiation = 7.7µrad

√Lλ

4π= 1.3µm

σ′γ=

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

10µrad

-10µrad

10µm-10µm

√Lλ

4π= 1.3µm

σ′γ=

L= 7.1µrad

σy= 9.1µm

σ′y= 3.0µrad

APS emittance

10µrad

-10µrad

10µm-10µm

√Lλ

4π= 1.3µm

σ′γ=

L= 7.1µrad

σy= 9.1µm

σ′y= 3.0µrad

APS emittance

10µrad

-10µrad

10µm-10µm

√Lλ

4π= 1.3µm

σ′γ=

L= 7.1µrad

σy= 9.1µm

σ′y= 3.0µrad

APS emittance

10µrad

-10µrad

10µm-10µm

√Lλ

4π= 1.3µm

σ′γ=

L= 7.1µrad

σy= 9.1µm

σ′y= 3.0µrad

APS emittance

10µrad

-10µrad

10µm-10µm

√Lλ

4π= 1.3µm

σ′γ=

L= 7.1µrad

σy= 9.1µm

σ′y= 3.0µrad

APS emittance

10µrad

-10µrad

10µm-10µm

√Lλ

4π= 1.3µm

σ′γ=

L= 7.1µrad

σy= 9.1µm

σ′y= 3.0µrad

APS emittance

10µrad

-10µrad

10µm-10µm

√Lλ

4π= 1.3µm

σ′γ=

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.

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

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