today’s outline - august 26, 2013csrri.iit.edu/~segre/phys570/13f/lecture_03.pdfa plane wave has...

Today’s Outline - August 26, 2013

• Coherence of x-ray sources

• The x-ray tube

• The synchrotron

• The bending magnet source

• Segmented arc approximation• Off-axis emission• Curved arc emission• Characteristic energy• Power and flux• Polarization

• Wiggler & undulator introduction

C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 1 / 29



• The x-ray tube

• The synchrotron

• The bending magnet source• Segmented arc approximation

• Off-axis emission• Curved arc emission• Characteristic energy• Power and flux• Polarization





• The x-ray tube

• The synchrotron

• The bending magnet source• Segmented arc approximation• Off-axis emission

• Curved arc emission• Characteristic energy• Power and flux• Polarization





• The x-ray tube

• The synchrotron

• The bending magnet source• Segmented arc approximation• Off-axis emission• Curved arc emission

• Characteristic energy• Power and flux• Polarization





• The x-ray tube

• The synchrotron

• The bending magnet source• Segmented arc approximation• Off-axis emission• Curved arc emission• Characteristic energy

• Power and flux• Polarization





• The x-ray tube

• The synchrotron

• The bending magnet source• Segmented arc approximation• Off-axis emission• Curved arc emission• Characteristic energy• Power and flux

• Polarization





• The x-ray tube

• The synchrotron

• The bending magnet source• Segmented arc approximation• Off-axis emission• Curved arc emission• Characteristic energy• Power and flux• Polarization



What is coherence?

So far, in our discussion, we have assumed that x-rays are“plane waves”. What does this really mean?

A plane wave has perfect coherence (like a laser).

Real x-rays are not perfect plane waves in two ways:

• they are not perfectly monochromatic

• they do not travel in a perfectly co-linear direction

Because of these imperfections the “coherence length” of anx-ray beam is finite and we can calculate it.


Longitudinal coherence

Definition: Distance over which two waves from the same source pointwith slightly different wavelengths will completely dephase.

λ

λ−∆λ

P

2LL

Two waves of slightly different wavelengthsλ and λ−∆λ are emitted from the samepoint in space simultaneously.

After a distance LL, the two waves will beexactly out of phase and after 2LL they willonce again be in phase.

2LL = Nλ2LL = (N + 1)(λ−∆λ)

Nλ = Nλ+ λ− N∆λ−∆λ

0 = λ−N∆λ−∆λ −→ λ = (N + 1)∆λ −→ N ≈ λ

∆λ−→ LL =

λ2

2∆λ




λ

λ−∆λ

P

2LL



2LL = Nλ

2LL = (N + 1)(λ−∆λ)


0 = λ−N∆λ−∆λ −→ λ = (N + 1)∆λ −→ N ≈ λ

∆λ−→ LL =

λ2

2∆λ




λ

λ−∆λ

P

2LL



2LL = Nλ2LL = (N + 1)(λ−∆λ)


0 = λ−N∆λ−∆λ −→ λ = (N + 1)∆λ −→ N ≈ λ

∆λ−→ LL =

λ2

2∆λ




λ

λ−∆λ

P

2LL



2LL = Nλ2LL = (N + 1)(λ−∆λ)


0 = λ−N∆λ−∆λ

−→ λ = (N + 1)∆λ −→ N ≈ λ

∆λ−→ LL =

λ2

2∆λ




λ

λ−∆λ

P

2LL



2LL = Nλ2LL = (N + 1)(λ−∆λ)


0 = λ−N∆λ−∆λ −→ λ = (N + 1)∆λ

−→ N ≈ λ

∆λ−→ LL =

λ2

2∆λ




λ

λ−∆λ

P

2LL



2LL = Nλ2LL = (N + 1)(λ−∆λ)


0 = λ−N∆λ−∆λ −→ λ = (N + 1)∆λ −→ N ≈ λ

∆λ

−→ LL =λ2

2∆λ




λ

λ−∆λ

P

2LL



2LL = Nλ2LL = (N + 1)(λ−∆λ)


0 = λ−N∆λ−∆λ −→ λ = (N + 1)∆λ −→ N ≈ λ

∆λ−→ LL =

λ2

2∆λ


Transverse coherence

Definition: The lateral distance along a wavefront over which there is acomplete dephasing between two waves, of the same wavelength, whichoriginate from two separate points in space.

λ

D

∆θ

2LT

P

R

∆θ

If we assume that the two wavesoriginate from points with a smallangular separation ∆θ, Thetransverse coherence length is givenby:

λ

2LT= tan ∆θ ≈ ∆θ

D

R= tan ∆θ ≈ ∆θ

LT =λR

2D




λ

D

∆θ

2LT

P

R

∆θ


λ

2LT= tan ∆θ

≈ ∆θ

D

R= tan ∆θ

≈ ∆θ

LT =λR

2D




λ

D

∆θ

2LT

P

R

∆θ


λ

2LT= tan ∆θ ≈ ∆θ

D

R= tan ∆θ ≈ ∆θ

LT =λR

2D


Coherence lengths at the APS

For a typical 3rd generation undulator source, such as at the AdvancedPhoton Source the vertical source size is D = 100µm and we are typicallyR = 50m away with our experiment. If we assume a typical wavelength ofλ = 1A, and a monochromator resolution of ∆λ/λ = 10−5 we have for thevertical direction:

LL =λ2

2∆λ

=λ

2· λ

∆λ=

1× 10−10

2 · 10−5= 5µm

LT =λR

2D

=(1× 10−10) · 50

2 · (100× 10−6)= 25µm




LL =λ2

2∆λ=λ

2· λ

∆λ

=1× 10−10

2 · 10−5= 5µm

LT =λR

2D

=(1× 10−10) · 50

2 · (100× 10−6)= 25µm




LL =λ2

2∆λ=λ

2· λ

∆λ=

1× 10−10

2 · 10−5

= 5µm

LT =λR

2D

=(1× 10−10) · 50

2 · (100× 10−6)= 25µm




LL =λ2

2∆λ=λ

2· λ

∆λ=

1× 10−10

2 · 10−5= 5µm

LT =λR

2D

=(1× 10−10) · 50

2 · (100× 10−6)= 25µm




LL =λ2

2∆λ=λ

2· λ

∆λ=

1× 10−10

2 · 10−5= 5µm

LT =λR

2D=

(1× 10−10) · 50

2 · (100× 10−6)

= 25µm




LL =λ2

2∆λ=λ

2· λ

∆λ=

1× 10−10

2 · 10−5= 5µm

LT =λR

2D=

(1× 10−10) · 50

2 · (100× 10−6)= 25µm


X-ray tube schematics

Fixed anode tube

Rotating anode tube

• low power

• low maintenance

• high power

• high maintenance


X-ray tube schematics

Fixed anode tube Rotating anode tube

• low power

• low maintenance

• high power

• high maintenance


X-ray tube spectrum

• Minimum wavelength(maximum energy) setby acceleratingpotential

• Bremßtrahlungradiation providessmooth background(charged particledeceleration)

• Highest intensity at the characteristic fluorescence emission energy ofthe anode material

• Unpolarized, incoherent x-rays emitted in all directions from anodesurface, must be collimated with slits


Synchrotron sources

Bending magnet

• Wide horizontal beam

• Broad spectrum to highenergies

Undulator

• Highly collimated beam

• Highly peaked spectrumwith harmonics


Bending magnet spectra

0 20000 40000 60000 80000 1e+050

1e+13

2e+13

3e+13

APS

NSLS

ALS

ESRF

Lower energy sources, such as NSLS have lower peak energy and higherintensity at the peak.Higher energy sources, such as APS have higher energy spectrum and areonly off by a factor of 2 intensity at low energy.



0 20000 40000 60000 80000 1e+050

1e+13

2e+13

3e+13

APS

NSLS

ALS

ESRF

Lower energy sources, such as NSLS have lower peak energy and higherintensity at the peak.

Higher energy sources, such as APS have higher energy spectrum and areonly off by a factor of 2 intensity at low energy.



0 20000 40000 60000 80000 1e+050

1e+13

2e+13

3e+13

APS

NSLS

ALS

ESRF

Lower energy sources, such as NSLS have lower peak energy and higherintensity at the peak.Higher energy sources, such as APS have higher energy spectrum and areonly off by a factor of 2 intensity at low energy.



100 1000 10000 1e+051e+05

1e+10

1e+15

APS

NSLS

ALS

ESRF

Logarithmic scale shows clearly how much more energetic and intense thebending magnet sources at the 6 GeV and 7 GeV sources are.


Review of special relativity

v

β =v

cγ =

√1

1− β2

E = γmc2

β =

√1− 1

γ2−→ β ≈ 1− 1

2

1

γ2

use binomial expansion since 1/γ2 << 1

Let’s calculate these quantitiesfor an electron at NSLS andAPS

me = 0.511 MeV/c2

NSLS: E = 1.5 GeV

γ =1.5× 109

0.511× 106= 2935

APS: E = 7.0 GeV

γ =7.0× 109

0.511× 106= 13700



v

β =v

c

γ =

√1

1− β2

E = γmc2

β =

√1− 1

γ2−→ β ≈ 1− 1

2

1

γ2



me = 0.511 MeV/c2

NSLS: E = 1.5 GeV

γ =1.5× 109

0.511× 106= 2935

APS: E = 7.0 GeV

γ =7.0× 109

0.511× 106= 13700



v

β =v

cγ =

√1

1− β2

E = γmc2

β =

√1− 1

γ2−→ β ≈ 1− 1

2

1

γ2



me = 0.511 MeV/c2

NSLS: E = 1.5 GeV

γ =1.5× 109

0.511× 106= 2935

APS: E = 7.0 GeV

γ =7.0× 109

0.511× 106= 13700



v

β =v

cγ =

√1

1− β2

E = γmc2

β =

√1− 1

γ2

−→ β ≈ 1− 1

2

1

γ2



me = 0.511 MeV/c2

NSLS: E = 1.5 GeV

γ =1.5× 109

0.511× 106= 2935

APS: E = 7.0 GeV

γ =7.0× 109

0.511× 106= 13700



v

β =v

cγ =

√1

1− β2

E = γmc2

β =

√1− 1

γ2−→ β ≈ 1− 1

2

1

γ2



me = 0.511 MeV/c2

NSLS: E = 1.5 GeV

γ =1.5× 109

0.511× 106= 2935

APS: E = 7.0 GeV

γ =7.0× 109

0.511× 106= 13700


“Headlight” effect

In electron rest frame:

emission is symmetric about theaxis of the acceleration vector

In lab frame:

emission is pushed into the direc-tion of motion of the electron


Relativistic emission

1/γv

a

the electron is in constant trans-verse acceleration due to theLorentz force from the magneticfield of the bending magnet

~F = e~v × ~B = me~a

the aperture angle of the radiationcone is 1/γ

the angular frequency of the elec-tron in the ring is ωo ≈ 106 andthe cutoff energy for emission is

Emax ≈ γ3ωo

for the APS, with γ ≈ 104 we have

Emax ≈ (104)3 · 106 = 1018



1/γv

a


~F = e~v × ~B = me~a


the angular frequency of the elec-tron in the ring is ωo ≈ 106

andthe cutoff energy for emission is

Emax ≈ γ3ωo


Emax ≈ (104)3 · 106 = 1018



1/γv

a


~F = e~v × ~B = me~a


the angular frequency of the elec-tron in the ring is ωo ≈ 106 andthe cutoff energy for emission is

Emax ≈ γ3ωo


Emax ≈ (104)3 · 106 = 1018


Flux and brilliance

There are a number of important quantities which are relevant to thequality of an x-ray source:

photon fluxphoton densitybeam divergenceenergy resolution

source type opticssource type opticssource type opticssource type optics

All these quantities are conveniently taken into account in a measurecalled brilliance

brilliance

=flux [photons/s]

divergence[mrad2

]· source size [mm2] [0.1% bandwidth]


Flux and brilliance


photon flux

photon densitybeam divergenceenergy resolution



brilliance

=flux [photons/s]

divergence[mrad2



Flux and brilliance


photon fluxphoton density

beam divergenceenergy resolution



brilliance

=flux [photons/s]

divergence[mrad2



Flux and brilliance


photon fluxphoton densitybeam divergence

energy resolution



brilliance

=flux [photons/s]

divergence[mrad2



Flux and brilliance





brilliance

=flux [photons/s]

divergence[mrad2



Flux and brilliance



source type

optics

source type

optics

source type

opticssource type optics


brilliance

=flux [photons/s]

divergence[mrad2



Flux and brilliance



source type opticssource type opticssource type optics

source type

optics


brilliance

=flux [photons/s]

divergence[mrad2



Flux and brilliance




source type

optics


brilliance =flux [photons/s]

divergence[mrad2



Flux and brilliance




source type

optics



divergence[mrad2

]

· source size [mm2] [0.1% bandwidth]


Flux and brilliance




source type

optics



divergence[mrad2

]· source size [mm2]

[0.1% bandwidth]


Flux and brilliance




source type

optics



divergence[mrad2



Computing brilliance


divergence[mrad2

]· source size [mm2] · [0.1% bandwidth]

Energy

Flu

x

hν

∆hν

The source size depends on the elec-tron beam size, its excursion, andany slits which define how much ofthe source is visible by the observer.

The divergence is the angular spreadthe x-ray beam in the x and y direc-tions.

α ≈ x/z β ≈ y/z ,where z is the distance from thesource over which there is a lateralspread x and y in each direction




divergence[mrad2


Energy

Flu

x

hν

∆hν

For a specific photon flux distribu-tion, we would normally integrate toget the total flux.

But this ignoresthat most experiments are only in-terested in a specific energy hν.

Take a bandwidth ∆hν = hν/1000,which is about 10 times widerthan the bandwidth of the typicalmonochromator.

Compute the integrated photon fluxin that bandwidth.




divergence[mrad2


Energy

Flu

x

hν

∆hν

For a specific photon flux distribu-tion, we would normally integrate toget the total flux. But this ignoresthat most experiments are only in-terested in a specific energy hν.

Take a bandwidth ∆hν = hν/1000,which is about 10 times widerthan the bandwidth of the typicalmonochromator.

Compute the integrated photon fluxin that bandwidth.




divergence[mrad2


Energy

Flu

x

hν

∆hν

The source size depends on the elec-tron beam size, its excursion, andany slits which define how much ofthe source is visible by the observer.

The divergence is the angular spreadthe x-ray beam in the x and y direc-tions.

α ≈ x/z β ≈ y/z ,where z is the distance from thesource over which there is a lateralspread x and y in each direction


Segmented arc approximation

A

B C

v∆t’

c∆t’

(c-v)∆t’

• Approximate the electron’s pathas a series of segments

• At each corner the electron isaccelerated and emits radiation

• Consider the emissions at pointsB and C

The electron travels the distance from B to C in ∆t ′ while the light pulseemitted at B travels further, c∆t ′, in the same time.The light pulse emitted at C is therefore, a distance (c − v)∆t ′ behind thepulse emitted at B.The observer will measure a time between the two pulses:

∆t =(c − v)∆t ′

c=(

1− v

c

)∆t ′ = (1− β)∆t ′



A

B C

v∆t’

c∆t’

(c-v)∆t’




The electron travels the distance from B to C in ∆t ′

while the light pulseemitted at B travels further, c∆t ′, in the same time.The light pulse emitted at C is therefore, a distance (c − v)∆t ′ behind thepulse emitted at B.The observer will measure a time between the two pulses:

∆t =(c − v)∆t ′

c=(

1− v

c

)∆t ′ = (1− β)∆t ′



A

B C

v∆t’

c∆t’

(c-v)∆t’




The electron travels the distance from B to C in ∆t ′ while the light pulseemitted at B travels further, c∆t ′, in the same time.

The light pulse emitted at C is therefore, a distance (c − v)∆t ′ behind thepulse emitted at B.The observer will measure a time between the two pulses:

∆t =(c − v)∆t ′

c=(

1− v

c

)∆t ′ = (1− β)∆t ′



A

B C

v∆t’

c∆t’

(c-v)∆t’




The electron travels the distance from B to C in ∆t ′ while the light pulseemitted at B travels further, c∆t ′, in the same time.The light pulse emitted at C is therefore, a distance (c − v)∆t ′ behind thepulse emitted at B.

The observer will measure a time between the two pulses:

∆t =(c − v)∆t ′

c=(

1− v

c

)∆t ′ = (1− β)∆t ′



A

B C

v∆t’

c∆t’

(c-v)∆t’





∆t =(c − v)∆t ′

c=(

1− v

c

)∆t ′ = (1− β)∆t ′



A

B C

v∆t’

c∆t’

(c-v)∆t’





∆t =(c − v)∆t ′

c

=(

1− v

c

)∆t ′ = (1− β)∆t ′



A

B C

v∆t’

c∆t’

(c-v)∆t’





∆t =(c − v)∆t ′

c=(

1− v

c

)∆t ′

= (1− β)∆t ′



A

B C

v∆t’

c∆t’

(c-v)∆t’





∆t =(c − v)∆t ′

c=(

1− v

c

)∆t ′ = (1− β)∆t ′


Doppler compression

A

B C

v∆t’

c∆t’

(c-v)∆t’

∆t = (1− β)∆t ′

Since 0 < β < 1 this translatesto a Doppler compression of theemitted wavelength.

Recall that

β =

√1− 1

γ2,

but for synchrotron radiation, γ > 1000, so 1/γ � 1 and we can,therefore, approximate

β =

(1− 1

γ2

)1/2

= 1− 1

2

1

γ2+

1

2

1

2

1

2!

1

γ4+ · · · ≈ 1− 1

2γ2


Doppler compression

A

B C

v∆t’

c∆t’

(c-v)∆t’

∆t = (1− β)∆t ′


Recall that

β =

√1− 1

γ2,

but for synchrotron radiation, γ > 1000,

so 1/γ � 1 and we can,therefore, approximate

β =

(1− 1

γ2

)1/2

= 1− 1

2

1

γ2+

1

2

1

2

1

2!

1

γ4+ · · · ≈ 1− 1

2γ2


Doppler compression

A

B C

v∆t’

c∆t’

(c-v)∆t’

∆t = (1− β)∆t ′


Recall that

β =

√1− 1

γ2,

but for synchrotron radiation, γ > 1000, so 1/γ � 1

and we can,therefore, approximate

β =

(1− 1

γ2

)1/2

= 1− 1

2

1

γ2+

1

2

1

2

1

2!

1

γ4+ · · · ≈ 1− 1

2γ2


Doppler compression

A

B C

v∆t’

c∆t’

(c-v)∆t’

∆t = (1− β)∆t ′


Recall that

β =

√1− 1

γ2,


β =

(1− 1

γ2

)1/2

= 1− 1

2

1

γ2+

1

2

1

2

1

2!

1

γ4+ · · · ≈ 1− 1

2γ2


Doppler compression

A

B C

v∆t’

c∆t’

(c-v)∆t’

∆t = (1− β)∆t ′


Recall that

β =

√1− 1

γ2,


β =

(1− 1

γ2

)1/2

= 1− 1

2

1

γ2+

1

2

1

2

1

2!

1

γ4+ · · ·

≈ 1− 1

2γ2


Doppler compression

A

B C

v∆t’

c∆t’

(c-v)∆t’

∆t = (1− β)∆t ′


Recall that

β =

√1− 1

γ2,


β =

(1− 1

γ2

)1/2

= 1− 1

2

1

γ2+

1

2

1

2

1

2!

1

γ4+ · · · ≈ 1− 1

2γ2


Off-axis emission

A

B C

v cosα ∆t’

c∆t’

α

(c-v cosα)∆t’

Consider the emission from seg-ment AB, which is not alongthe line toward the observer.

While on the AB segment, theelectron moves only a distancev cosα∆t ′ in the direction ofthe BC segment.

The light pulse emitted at A still travels c∆t ′, in the same time.The light pulse emitted at B is therefore, a distance (c − v cosα)∆t ′

behind the pulse emitted at A. The observer will measure a time betweenthe two pulses:

∆t =(c − v cosα)∆t ′

c=(

1− v

ccosα

)∆t ′ = (1− β cosα)∆t ′


Off-axis emission

A

B C

v cosα ∆t’

c∆t’

α

(c-v cosα)∆t’

Consider the emission from seg-ment AB, which is not alongthe line toward the observer.While on the AB segment, theelectron moves only a distancev cosα∆t ′ in the direction ofthe BC segment.




c=(

1− v

ccosα

)∆t ′ = (1− β cosα)∆t ′


Off-axis emission

A

B C

v cosα ∆t’

c∆t’

α

(c-v cosα)∆t’


The light pulse emitted at A still travels c∆t ′, in the same time.

The light pulse emitted at B is therefore, a distance (c − v cosα)∆t ′



c=(

1− v

ccosα

)∆t ′ = (1− β cosα)∆t ′


Off-axis emission

A

B C

v cosα ∆t’

c∆t’

α

(c-v cosα)∆t’



behind the pulse emitted at A.

The observer will measure a time betweenthe two pulses:


c=(

1− v

ccosα

)∆t ′ = (1− β cosα)∆t ′


Off-axis emission

A

B C

v cosα ∆t’

c∆t’

α

(c-v cosα)∆t’





c=(

1− v

ccosα

)∆t ′ = (1− β cosα)∆t ′


Off-axis emission

A

B C

v cosα ∆t’

c∆t’

α

(c-v cosα)∆t’





c

=(

1− v

ccosα

)∆t ′ = (1− β cosα)∆t ′


Off-axis emission

A

B C

v cosα ∆t’

c∆t’

α

(c-v cosα)∆t’





c=(

1− v

ccosα

)∆t ′

= (1− β cosα)∆t ′


Off-axis emission

A

B C

v cosα ∆t’

c∆t’

α

(c-v cosα)∆t’





c=(

1− v

ccosα

)∆t ′ = (1− β cosα)∆t ′


Corrected Doppler shift

A

B C

v cosα ∆t’

c∆t’

α

(c-v cosα)∆t’

∆t = (1− βcosα)∆t ′

Since α is very small:

cosα ≈ 1− α2

2

and γ is very large, we have

∆t

∆t ′≈ 1−

(1− 1

2γ2

)(1− α2

2

)= 1− 1 +

α2

2+

1

2γ2− α2

2γ2

∆t

∆t ′≈ α2

2+

1

2γ2=

1 + α2γ2

2γ2

called the time compression ratio.



A

B C

v cosα ∆t’

c∆t’

α

(c-v cosα)∆t’

∆t = (1− βcosα)∆t ′


cosα ≈ 1− α2

2


∆t

∆t ′≈ 1−

(1− 1

2γ2

)(1− α2

2

)

= 1− 1 +α2

2+

1

2γ2− α2

2γ2

∆t

∆t ′≈ α2

2+

1

2γ2=

1 + α2γ2

2γ2




A

B C

v cosα ∆t’

c∆t’

α

(c-v cosα)∆t’

∆t = (1− βcosα)∆t ′


cosα ≈ 1− α2

2


∆t

∆t ′≈ 1−

(1− 1

2γ2

)(1− α2

2

)= 1− 1 +

α2

2+

1

2γ2− α2

2γ2

∆t

∆t ′≈ α2

2+

1

2γ2=

1 + α2γ2

2γ2




A

B C

v cosα ∆t’

c∆t’

α

(c-v cosα)∆t’

∆t = (1− βcosα)∆t ′


cosα ≈ 1− α2

2


∆t

∆t ′≈ 1−

(1− 1

2γ2

)(1− α2

2

)= 1− 1 +

α2

2+

1

2γ2− α2

2γ2

∆t

∆t ′≈ α2

2+

1

2γ2

=1 + α2γ2

2γ2




A

B C

v cosα ∆t’

c∆t’

α

(c-v cosα)∆t’

∆t = (1− βcosα)∆t ′


cosα ≈ 1− α2

2


∆t

∆t ′≈ 1−

(1− 1

2γ2

)(1− α2

2

)= 1− 1 +

α2

2+

1

2γ2− α2

2γ2

∆t

∆t ′≈ α2

2+

1

2γ2=

1 + α2γ2

2γ2



Radiation opening angle

The Doppler shift is defined in terms of the time compression ratio

f

f ′=

∆t ′

∆t=

2γ2

1 + α2γ2

-0.001 -0.0005 0 0.0005 0.001

α (radians)

1e+07

1e+08

f / f’

γ=14000

γ=3000

• For APS and NSLSparameters the Doppler blueshift is between 107 and 109

• The intesection of thehorizontal and vertical dashedlines indicate whereα = ±1/γ and f /f ′ is onehalf of it’s maximum value

• The highest energy emittedradiation appears within acone of half angle 1/γ


Curved arc emission

1/γ

ωo

B

ρ

But in the limit, the compression ra-tio:

∆t

∆t ′

∣∣∣∆t→0

=dt

dt ′= 1− β cosα

so we need to treat the electron pathas a continuous arc.An electron moving in a constantmagnetic field describes a circularpath

FLorentz = evB a =dp

dt=

v 2

ρ

evB = mv 2

ρ−→ mv = p = ρeB


Curved arc emission

1/γ

ωo

B

ρ


∆t

∆t ′

∣∣∣∆t→0

=dt


so we need to treat the electron pathas a continuous arc.

An electron moving in a constantmagnetic field describes a circularpath


dt=

v 2

ρ

evB = mv 2



Curved arc emission

1/γ

ωo

B

ρ


∆t

∆t ′

∣∣∣∆t→0

=dt




dt=

v 2

ρ

evB = mv 2



Curved arc emission

1/γ

ωo

B

ρ


∆t

∆t ′

∣∣∣∆t→0

=dt



FLorentz = evB

a =dp

dt=

v 2

ρ

evB = mv 2



Curved arc emission

1/γ

ωo

B

ρ


∆t

∆t ′

∣∣∣∆t→0

=dt




dt=

v 2

ρ

evB = mv 2



Curved arc emission

1/γ

ωo

B

ρ


∆t

∆t ′

∣∣∣∆t→0

=dt




dt=

v 2

ρ

evB = mv 2

ρ

−→ mv = p = ρeB


Curved arc emission

1/γ

ωo

B

ρ


∆t

∆t ′

∣∣∣∆t→0

=dt




dt=

v 2

ρ

evB = mv 2



Electron bending radius

1/γ

ωo

B

ρ

mv = p = ρeB

but the electron is relativistic so wemust correct the momentum to retainconsistent laws of physics p → γmv

γmv = ρeB

at a synchrotron γ � 1 so v ≈ c

γmc ≈ ρeB −→ γmc2 ≈ ρecB

since E = mc2 and c = 2.998× 108m/s2 we have

ρ =E [J]

ecB[T]=E [eV]

cB[T]= 3.3

E [GeV]

B[T]



1/γ

ωo

B

ρ

mv = p = ρeB


γmv = ρeB


γmc ≈ ρeB

−→ γmc2 ≈ ρecB


ρ =E [J]

ecB[T]=E [eV]

cB[T]= 3.3

E [GeV]

B[T]



1/γ

ωo

B

ρ

mv = p = ρeB


γmv = ρeB




ρ =E [J]

ecB[T]=E [eV]

cB[T]= 3.3

E [GeV]

B[T]



1/γ

ωo

B

ρ

mv = p = ρeB


γmv = ρeB




ρ =E [J]

ecB[T]

=E [eV]

cB[T]= 3.3

E [GeV]

B[T]



1/γ

ωo

B

ρ

mv = p = ρeB


γmv = ρeB




ρ =E [J]

ecB[T]=E [eV]

cB[T]

= 3.3E [GeV]

B[T]



1/γ

ωo

B

ρ

mv = p = ρeB


γmv = ρeB




ρ =E [J]

ecB[T]=E [eV]

cB[T]= 3.3

E [GeV]

B[T]


Curved arc emission

1/γ

ωo

B

ρ

The observer, looking in the plane of the circulartrajectory,

“sees” the electron oscillate over a halfperiod in a time ∆t (observer’s frame).The electron, in the laboratory frame, travels thisarc in:

∆t ′ =(1/γ)ρ

v=

1

γωo

Because of the Doppler shift, the observer sees theelectron emitting a pulse of radiation of length

∆t ∝ ∆t ′

γ2=

1

γ3ωo

The Fourier transform of this pulse is the spectrumof the radiation from the bending magnet.


Curved arc emission

1/γ

ωo

B

ρ

The observer, looking in the plane of the circulartrajectory, “sees” the electron oscillate over a halfperiod

in a time ∆t (observer’s frame).The electron, in the laboratory frame, travels thisarc in:

∆t ′ =(1/γ)ρ

v=

1

γωo


∆t ∝ ∆t ′

γ2=

1

γ3ωo



Curved arc emission

1/γ

ωo

B

ρ

The observer, looking in the plane of the circulartrajectory, “sees” the electron oscillate over a halfperiod in a time ∆t (observer’s frame).

The electron, in the laboratory frame, travels thisarc in:

∆t ′ =(1/γ)ρ

v=

1

γωo


∆t ∝ ∆t ′

γ2=

1

γ3ωo



Curved arc emission

1/γ

ωo

B

ρ

The observer, looking in the plane of the circulartrajectory, “sees” the electron oscillate over a halfperiod in a time ∆t (observer’s frame).The electron, in the laboratory frame, travels thisarc in:

∆t ′ =(1/γ)ρ

v

=1

γωo


∆t ∝ ∆t ′

γ2=

1

γ3ωo



Curved arc emission

1/γ

ωo

B

ρ


∆t ′ =(1/γ)ρ

v=

1

γωo


∆t ∝ ∆t ′

γ2=

1

γ3ωo



Curved arc emission

1/γ

ωo

B

ρ


∆t ′ =(1/γ)ρ

v=

1

γωo


∆t ∝ ∆t ′

γ2

=1

γ3ωo



Curved arc emission

1/γ

ωo

B

ρ


∆t ′ =(1/γ)ρ

v=

1

γωo


∆t ∝ ∆t ′

γ2=

1

γ3ωo



Characteristic Energy of a Bending Magnet

The radiation from a bending magnet is defined by it’s characteristicfrequency, ωc which, when the calculation is performed rigorously is:

ωc =3

2γ3ωo

but since T is the period of the rotation through the full circle of radius ρ

ωo =2π

T= 2π

c

2πρ=

c

ρ=

ceB

γmc

we can therefore calculate the characteristic energy Ec

Ec = ~ωc =3

2γ3 ceB

γmc=

3

2ceB

γ2

mc=

3eB

2m

E2

(mc2)2

converting to storage ring units

Ec [keV] = 0.665E2[GeV]B[T]




ωc =3

2γ3ωo


ωo =2π

T

= 2πc

2πρ=

c

ρ=

ceB

γmc


Ec = ~ωc =3

2γ3 ceB

γmc=

3

2ceB

γ2

mc=

3eB

2m

E2

(mc2)2






ωc =3

2γ3ωo


ωo =2π

T= 2π

c

2πρ

=c

ρ=

ceB

γmc


Ec = ~ωc =3

2γ3 ceB

γmc=

3

2ceB

γ2

mc=

3eB

2m

E2

(mc2)2






ωc =3

2γ3ωo


ωo =2π

T= 2π

c

2πρ=

c

ρ=

ceB

γmc


Ec = ~ωc =3

2γ3 ceB

γmc=

3

2ceB

γ2

mc=

3eB

2m

E2

(mc2)2






ωc =3

2γ3ωo


ωo =2π

T= 2π

c

2πρ=

c

ρ=

ceB

γmc


Ec = ~ωc =3

2γ3 ceB

γmc

=3

2ceB

γ2

mc=

3eB

2m

E2

(mc2)2






ωc =3

2γ3ωo


ωo =2π

T= 2π

c

2πρ=

c

ρ=

ceB

γmc


Ec = ~ωc =3

2γ3 ceB

γmc=

3

2ceB

γ2

mc

=3eB

2m

E2

(mc2)2






ωc =3

2γ3ωo


ωo =2π

T= 2π

c

2πρ=

c

ρ=

ceB

γmc


Ec = ~ωc =3

2γ3 ceB

γmc=

3

2ceB

γ2

mc=

3eB

2m

E2

(mc2)2




Bending magnet spectrum

When the radiation pulse time isFourier transformed, we obtainthe spectrum of a bendingmagnet.

Scaling by the characteristicenergy, gives a universal curve

1.33×1013E2 I

(ω

ωc

)2

K 22/3

(ω

2ωc

)where K2/3 is a modified Besselfunction of the second kind.

0 20 40 60 80 100Photon Energy (keV)

0.0

0.5

1.0

1.5

2.0

2.5

3.0

Photo

n F

lux (

x10

+1

3)

APS

NSLS

ALS

ESRF

7.0 GeV

6.0 GeV

2.6 GeV

1.9 GeV 1.3 T

1.2 T

0.8 T

0.6 T





1.33×1013E2 I

(ω

ωc

)2

K 22/3

(ω

2ωc

)where K2/3 is a modified Besselfunction of the second kind.

0 2 4 6 8 10E/E

c

0.0

0.5

1.0

1.5

2.0

2.5

3.0

Photo

n F

lux (

x10

+1

3)

APS

NSLS

ALS

ESRF

3.1 keV

5.4 keV

19.2 keV

19.4 keV





1.33×1013E2 I

(ω

ωc

)2

K 22/3

(ω

2ωc

)where K2/3 is a modified Besselfunction of the second kind. 0 2 4 6 8 10

E/Ec

0.0

0.5

1.0

1.5

2.0

2.5

3.0

Photo

n F

lux (

x10

+1

3)

APS

NSLS

ALS

ESRF

3.1 keV

5.4 keV

19.2 keV

19.4 keV


Power from a bending magnet

The radiated power is given in storage ring units by:

P[kW] = 1.266E2[GeV]B2[T]L[m]I [A]

where L is the length of the arc visible to the observer and I is the storagering current.We can calculate this for the ESRF where E = 6 GeV, B = 0.8 T,Ec = 19.2 keV and the bending radius ρ = 24.8 m. Assuming that theaperture is 1 mm2 at a distance of 20 m, the angular aperture is1/20 = 0.05 mrad and the flux at the characteristic energy is given by:

Flux = (1.95×1013)(0.052mrad2)(62GeV2)(0.2A) = 3.5×1011ph/s/0.1%BW

The arc length is L = (24.8m)(0.05mrad) = 1.24mm and we have:

P = 1.266(6GeV)2(0.8T)2(1.24× 10−3m)(0.2A) = 7.3W




P[kW] = 1.266E2[GeV]B2[T]L[m]I [A]

where L is the length of the arc visible to the observer and I is the storagering current.

We can calculate this for the ESRF where E = 6 GeV, B = 0.8 T,Ec = 19.2 keV and the bending radius ρ = 24.8 m. Assuming that theaperture is 1 mm2 at a distance of 20 m, the angular aperture is1/20 = 0.05 mrad and the flux at the characteristic energy is given by:



P = 1.266(6GeV)2(0.8T)2(1.24× 10−3m)(0.2A) = 7.3W




P[kW] = 1.266E2[GeV]B2[T]L[m]I [A]




P = 1.266(6GeV)2(0.8T)2(1.24× 10−3m)(0.2A) = 7.3W




P[kW] = 1.266E2[GeV]B2[T]L[m]I [A]


Flux = (1.95×1013)

(0.052mrad2)(62GeV2)(0.2A) = 3.5×1011ph/s/0.1%BW


P = 1.266(6GeV)2(0.8T)2(1.24× 10−3m)(0.2A) = 7.3W




P[kW] = 1.266E2[GeV]B2[T]L[m]I [A]


Flux = (1.95×1013)(0.052mrad2)

(62GeV2)(0.2A) = 3.5×1011ph/s/0.1%BW


P = 1.266(6GeV)2(0.8T)2(1.24× 10−3m)(0.2A) = 7.3W




P[kW] = 1.266E2[GeV]B2[T]L[m]I [A]


Flux = (1.95×1013)(0.052mrad2)(62GeV2)

(0.2A) = 3.5×1011ph/s/0.1%BW


P = 1.266(6GeV)2(0.8T)2(1.24× 10−3m)(0.2A) = 7.3W




P[kW] = 1.266E2[GeV]B2[T]L[m]I [A]


Flux = (1.95×1013)(0.052mrad2)(62GeV2)(0.2A)

= 3.5×1011ph/s/0.1%BW


P = 1.266(6GeV)2(0.8T)2(1.24× 10−3m)(0.2A) = 7.3W




P[kW] = 1.266E2[GeV]B2[T]L[m]I [A]




P = 1.266(6GeV)2(0.8T)2(1.24× 10−3m)(0.2A) = 7.3W




P[kW] = 1.266E2[GeV]B2[T]L[m]I [A]




P = 1.266(6GeV)2

(0.8T)2(1.24× 10−3m)(0.2A) = 7.3W




P[kW] = 1.266E2[GeV]B2[T]L[m]I [A]




P = 1.266(6GeV)2(0.8T)2

(1.24× 10−3m)(0.2A) = 7.3W




P[kW] = 1.266E2[GeV]B2[T]L[m]I [A]




P = 1.266(6GeV)2(0.8T)2(1.24× 10−3m)

(0.2A) = 7.3W




P[kW] = 1.266E2[GeV]B2[T]L[m]I [A]




P = 1.266(6GeV)2(0.8T)2(1.24× 10−3m)(0.2A)

= 7.3W




P[kW] = 1.266E2[GeV]B2[T]L[m]I [A]




P = 1.266(6GeV)2(0.8T)2(1.24× 10−3m)(0.2A) = 7.3W


Polarization

A bending magnet also produces circularly polarized radiation

• If the observer is in the plane of theelectron orbit, the electron motionlooks like a half period of linearsinusoidal motion

• From above, the motion looks like anarc in the clockwise direction

• From below, the motion looks like anarc in the counterclockwise direction

The result is circularly polarized radiation above and below the on-axisradiation.


Wigglers and undulators

Wiggler

Like bending magnet except:

• larger ~B → higher Ec

• more bends → higher power

Undulator

Different from bending magnet:

• shallow bends → smaller source

• interference → peaked spectrum


Wiggler radiation

• The electron’s trajectory through a wiggler can be considered as aseries of short circular arcs, each like a bending magnet

• If there are N poles to the wiggler, there are 2N arcs

• Each arc contributes as might a single bending magnet but the morelinear path means that the effective length, L, is much longer.

• The magnetic field varies along the length of the wiggler and is higherthan that in a bending magnet, having an average value ofBrms = Bo/

√2

• This results in a significantly higher power load on all downstreamcomponents

Power [kW] = 1.266E2e [GeV]B2[T]L[m]I [A]


Wiggler radiation

• The electron’s trajectory through a wiggler can be considered as aseries of short circular arcs, each like a bending magnet

• If there are N poles to the wiggler, there are 2N arcs

• Each arc contributes as might a single bending magnet but the morelinear path means that the effective length, L, is much longer.

• The magnetic field varies along the length of the wiggler and is higherthan that in a bending magnet, having an average value ofBrms = Bo/

√2

• This results in a significantly higher power load on all downstreamcomponents

Power [kW] = 0.633E2e [GeV]B2

o [T]L[m]I [A]


today’s outline - august 26, 2013csrri.iit.edu/~segre/phys570/13f/lecture_03.pdfa plane wave has...

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