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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
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
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
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
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
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
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
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
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
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
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
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
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 2 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 2 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 2 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 2 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 2 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 2 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 3 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 3 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 3 / 29
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 = Nλ
2LL = (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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 3 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 3 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 3 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 3 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 3 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 3 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 3 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 4 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 4 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 4 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 4 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 4 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 5 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 5 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 5 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 5 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 5 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 5 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 5 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 5 / 29
X-ray tube schematics
Fixed anode tube
Rotating anode tube
• low power
• low maintenance
• high power
• high maintenance
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 6 / 29
X-ray tube schematics
Fixed anode tube Rotating anode tube
• low power
• low maintenance
• high power
• high maintenance
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 6 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 7 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 7 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 7 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 7 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 7 / 29
Synchrotron sources
Bending magnet
• Wide horizontal beam
• Broad spectrum to highenergies
Undulator
• Highly collimated beam
• Highly peaked spectrumwith harmonics
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 8 / 29
Synchrotron sources
Bending magnet
• Wide horizontal beam
• Broad spectrum to highenergies
Undulator
• Highly collimated beam
• Highly peaked spectrumwith harmonics
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 8 / 29
Synchrotron sources
Bending magnet
• Wide horizontal beam
• Broad spectrum to highenergies
Undulator
• Highly collimated beam
• Highly peaked spectrumwith harmonics
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 8 / 29
Synchrotron sources
Bending magnet
• Wide horizontal beam
• Broad spectrum to highenergies
Undulator
• Highly collimated beam
• Highly peaked spectrumwith harmonics
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 8 / 29
Synchrotron sources
Bending magnet
• Wide horizontal beam
• Broad spectrum to highenergies
Undulator
• Highly collimated beam
• Highly peaked spectrumwith harmonics
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 8 / 29
Synchrotron sources
Bending magnet
• Wide horizontal beam
• Broad spectrum to highenergies
Undulator
• Highly collimated beam
• Highly peaked spectrumwith harmonics
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 8 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 9 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 9 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 9 / 29
Bending magnet spectra
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 10 / 29
Bending magnet spectra
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 10 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 11 / 29
Review of special relativity
v
β =v
c
γ =
√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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 11 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 11 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 11 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 11 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 11 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 11 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 11 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 11 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 11 / 29
“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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 12 / 29
“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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 12 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 13 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 13 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 13 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 13 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 13 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 14 / 29
Flux and brilliance
There are a number of important quantities which are relevant to thequality of an x-ray source:
photon flux
photon 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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 14 / 29
Flux and brilliance
There are a number of important quantities which are relevant to thequality of an x-ray source:
photon fluxphoton density
beam 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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 14 / 29
Flux and brilliance
There are a number of important quantities which are relevant to thequality of an x-ray source:
photon fluxphoton densitybeam divergence
energy 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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 14 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 14 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 14 / 29
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
optics
source type
optics
source 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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 14 / 29
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 optics
source 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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 14 / 29
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 optics
source 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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 14 / 29
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 optics
source 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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 14 / 29
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 optics
source 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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 14 / 29
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 optics
source 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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 14 / 29
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 optics
source 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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 14 / 29
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 optics
source 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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 14 / 29
Computing brilliance
brilliance =flux [photons/s]
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 15 / 29
Computing brilliance
brilliance =flux [photons/s]
divergence[mrad2
]· source size [mm2] · [0.1% bandwidth]
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 15 / 29
Computing brilliance
brilliance =flux [photons/s]
divergence[mrad2
]· source size [mm2] · [0.1% bandwidth]
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 15 / 29
Computing brilliance
brilliance =flux [photons/s]
divergence[mrad2
]· source size [mm2] · [0.1% bandwidth]
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 15 / 29
Computing brilliance
brilliance =flux [photons/s]
divergence[mrad2
]· source size [mm2] · [0.1% bandwidth]
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 15 / 29
Computing brilliance
brilliance =flux [photons/s]
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 15 / 29
Computing brilliance
brilliance =flux [photons/s]
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 15 / 29
Computing brilliance
brilliance =flux [photons/s]
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 15 / 29
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 ′
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 16 / 29
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 ′
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 16 / 29
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 ′
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 16 / 29
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 ′
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 16 / 29
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 ′
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 16 / 29
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 ′
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 16 / 29
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 ′
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 16 / 29
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 ′
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 16 / 29
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 ′
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 16 / 29
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 ′
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 16 / 29
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 ′
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 16 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 17 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 17 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 17 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 17 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 17 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 17 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 17 / 29
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 ′
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 18 / 29
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 ′
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 18 / 29
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 ′
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 18 / 29
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 ′
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 18 / 29
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 ′
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 18 / 29
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 ′
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 18 / 29
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 ′
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 18 / 29
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 ′
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 18 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 19 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 19 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 19 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 19 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 19 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 19 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 19 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 19 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 20 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 20 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 20 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 21 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 21 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 21 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 21 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 21 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 21 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 21 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 22 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 22 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 22 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 22 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 22 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 22 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 22 / 29
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]
= 3.3E [GeV]
B[T]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 22 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 22 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 23 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 23 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 23 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 23 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 23 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 23 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 23 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 23 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 23 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 24 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 24 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 24 / 29
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
= 2πc
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 24 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 24 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 24 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 24 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 24 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 24 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 24 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 24 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 24 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 25 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 25 / 29
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 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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 25 / 29
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 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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 25 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 26 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 26 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 26 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 26 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 26 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 26 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 26 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 26 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 26 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 26 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 26 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 26 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 26 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 26 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 27 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 27 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 27 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 27 / 29
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.
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 27 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 28 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 28 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 28 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 28 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 28 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 28 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 28 / 29
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
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 28 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 29 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 29 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 29 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 29 / 29
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]
C. Segre (IIT) PHYS 570 - Fall 2013 August 26, 2013 29 / 29