Download Radiation from accelerated charged particles

Electron beam dynamics in storage rings
Lecture 10 - Melbourne
Synchrotron radiation
and its effect on electron dynamics
Lecture 10: Synchrotron radiation
Lecture 11: Radiation damping
Lecture 12: Radiation excitation
Lecture 10 - E. Wilson - 3/20/2008 - Slide 1
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Summary of last lecture
– Cavities II
• Conducting surfaces
• Quality Factor, Filling Time and Shunt
Impedance
• Corrugated structures
• Dispersion in a waveguide
• Iris loaded wavequide
• Standing-wave and travelling wave structures.
• Feeding, coupling and tuning structures.
• Different modes
Lecture 10 - E. Wilson - 3/20/2008 - Slide 2
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Contents
Introduction: synchrotron light sources
Lienard-Wiechert potentials
Angular distribution of power radiated by accelerated particles
non-relativistic motion: Larmor’s formula
relativistic motion
velocity || acceleration
velocity ⊥ acceleration: synchrotron radiation
Angular and frequency distribution of energy radiated:
the radiation integral
radiation integral for bending magnet radiation
radiation integral for undulator and wiggler radiation
Lecture 10 - E. Wilson - 3/20/2008 - Slide 3
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What is synchrotron radiation
Electromagnetic radiation is emitted by charged particles when accelerated
The electromagnetic radiation emitted when the charged particles are
accelerated radially (v ⊥ a) is called synchrotron radiation
It is produced in the synchrotron radiation sources using bending magnets
undulators and wigglers
Lecture 10 - E. Wilson - 3/20/2008 - Slide 4
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Why are synchrotron radiation sources important
Broad Spectrum which covers
from microwaves to hard X-rays:
the user can select the wavelength
required for experiment
synchrotron light
High Flux and High Brightness: highly collimated photon beam generated by a
small divergence and small size source (partial coherence)
High Stability: submicron source stability
Polarisation: both linear and circular (with IDs)
Pulsed Time Structure: pulsed length down to tens of picoseconds allows the
resolution of processes on the same time scale
Flux = Photons / ( s • BW)
Brightness = Photons / ( s • mm2 • mrad2 • BW )
Lecture 10 - E. Wilson - 3/20/2008 - Slide 5
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diamond
1.E+20
1.E+18
1.E+16
2
2
Brightness (Photons/sec/mm /mrad /0.1%)
Brightness
1.E+14
1.E+12
1.E+10
1.E+08
1.E+06
X-rays from Diamond
will be 1012 times
brighter than from
an X-ray tube,
105 times brighter
than the SRS !
X-ray
tube
60W bulb
Candle
1.E+04
1.E+02
Lecture 10 - E. Wilson - 3/20/2008 - Slide 6
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Applications
Medicine, Biology, Chemistry, Material Science, Environmental Science and more
Biology
Reconstruction of the 3D structure of a nucleosome
with a resolution of 0.2 nm
The collection of precise information on the molecular
structure of chromosomes and their components can
improve the knowledge of how the genetic code of
DNA is maintained and reproduced
Lecture 10 - E. Wilson - 3/20/2008 - Slide 7
Archeology
A synchrotron X-ray beam at the SSRL facility
illuminated an obscured work erased, written over
and even painted over of the ancient
mathematical genius Archimedes, born 287 B.C.
in Sicily.
X-ray fluorescence imaging revealed the
hidden text by revealing the iron contained in
the ink used by a 10th century scribe. This xray image shows the lower left corner of the
page.
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Layout of a synchrotron radiation source
Electrons are generated and
accelerated in a linac, further
accelerated to the required energy in
a booster and injected and stored in
the storage ring
The circulating electrons emit an
intense beam of synchrotron radiation
which is sent down the beamline
Lecture 10 - E. Wilson - 3/20/2008 - Slide 8
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Synchrotron Light Sources
Lecture 10 - E. Wilson - 3/20/2008 - Slide 9
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Lienard-Wiechert Potentials (I)
The equations for vector potential and scalar potential
1 ∂ 2Φ
ρ
∇Φ− 2
=
−
c ∂t 2
ε0
2
J
1 ∂2 A
∇ A− 2 2 =− 2
c ∂t
c ε0
2
with the current and charge densities of a single charged particle, i.e.
ρ ( x , t ) = eδ (3) ( x − r (t ))
J ( x , t ) = ev (t )δ (3) ( x − r (t ))
have as solution the Lienard-Wiechert potentials
Φ ( x ,t ) =
⎤
1 ⎡
e
4πε 0 ⎢⎣ ( 1 − β ⋅ n )R ⎥⎦ ret
A ( x ,t ) =
⎡
⎤
eβ
4πε 0 c ⎢⎣ ( 1 − β ⋅ n )R ⎥⎦ ret
1
[ ]ret means computed at time t’
t = t '+
R (t ' )
c
Lecture 10 - E. Wilson - 3/20/2008 - Slide 10
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Lineard-Wiechert Potentials (II)
The electric and magnetic fields generated by the moving charge are
computed from the potentials
∂A
E = −∇V −
B = ∇× A
∂t
and are called Lineard-Wiechert fields
⎤
e ⎡
n −β
e ⎡ n × (n − β ) × β& ⎤
E (x, t) =
+
⎥
⎢
4πε 0 ⎢⎣ γ 2 (1 − β ⋅ n ) 3 R 2 ⎥⎦ ret 4πε 0 c ⎣⎢ (1 − β ⋅ n ) 3 R ⎦⎥
ret
velocity field
acceleration field
B ( x ,t ) =
1
∝
R
1
[
n×E
c
]
rit
r
E ⊥ B ⊥ nˆ
Power radiated by a particle on a surface is the flux of the Poynting vector
S=
1
μ0
E×B
Φ Σ ( S )(t ) = ∫∫ S ( x , t ) ⋅ n dΣ
Σ
Angular distribution of radiated power
d 2P
= ( S ⋅ n)(1 − n ⋅ β ) R 2
dΩ
Lecture 10 - E. Wilson - 3/20/2008 - Slide 11
radiation emitted by the particle
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Angular distribution of radiated power:
non relativistic motion
Assuming
β ≈ 0 and substituting the acceleration field
⎡ n × (n × β& ) ⎤
Eacc ( x , t ) =
⎢
⎥
4πε 0 c ⎢⎣
R
⎥⎦ ret
e
2
1
e2
d 2P
&)
(
β
=
REacc =
n
×
n
×
dΩ μ 0 c
(4π )2 ε 0 c
2
2
d 2P
e2
&
β sin 2 θ
=
2
dΩ (4π ) ε 0 c
θ is the angle between the acceleration and the observation direction
Integrating over the angles gives the total radiated power
P=
e2
6 πε 0 c
β&
2
Larmor’s formula
Lecture 10 - E. Wilson - 3/20/2008 - Slide 12
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Angular distribution of radiated power:
relativistic motion
Substituting the acceleration field
2
1
e2
d 2P
=
REacc =
dΩ μ o c
(4π )2 ε 0 c
[
n × (n − β ) × β&
(1 − n ⋅ β ) 5
]
2
emission is peaked in the
direction of the velocity
The pattern depends on the details of velocity and acceleration but it is
dominated by the denominator
Total radiated power: computed either by integration over the angles or by
relativistic transformation of the 4-acceleration in Larmor’s formula
P=
e2
6πε 0 c
[
γ 6 ( β& ) 2 − ( β × β& ) 2
]
Lecture 10 - E. Wilson - 3/20/2008 - Slide 13
Relativistic generalization of
Larmor’s formula
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velocity || acceleration
Assuming
β | | β&
dP
e2
=
dΩ (4π )2 ε 0 c
and substituting the acceleration field
[
n × ( n − β ) × β&
( 1 − n ⋅ β )5
]
2
e β&
2
2
sin 2 θ
=
(4π )2 ε 0 c ( 1 − β cos θ )5
⎡ 1
⎣ 3β
θ max = arccos⎢
Total radiated power
P=
Lecture 10 - E. Wilson - 3/20/2008 - Slide 14
e2
6 πε 0 c
2
β& γ 6
( 1 + 15β − 1)⎤⎥⎦ → 21γ
e2
dp
P=
6 πε 0 m 2 c 3 dt
2
2
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velocity ⊥ acceleration: synchrotron radiation
Assuming
β ⊥ β&
dP
e2
=
dΩ (4π )2 ε 0 c 2
and substituting the acceleration field
[
n × ( n − β ) × β&
]
( 1 − n ⋅ β )5
2
e β&
2
2
1
=
(4π )2 ε 0 c ( 1 − β cos θ )3
⎡
sin 2 θ cos 2 φ ⎤
⎢1 − 2
2⎥
⎣ γ ( 1 − β cos θ ) ⎦
cone aperture
∼ 1/γ
When the electron velocity approaches the speed of light the emission
pattern is sharply collimated forward
Lecture 10 - E. Wilson - 3/20/2008 - Slide 15
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Total radiated power via synchrotron radiation
Integrating over the whole solid angle we obtain the total instantaneous
power radiated by one electron
P=
e2
6πε 0 c
2
&
β γ4 =
e2
6πε 0 c
β&
2
E
e2
dp 2 e 2 c γ 4
e4
2 2
γ
=
=
=
E
B
4
2 3
2
4 5
6πε 0 m c dt
6πε 0 ρ
6πε 0 m c
Eo
4
2
• Strong dependence 1/m4 on the rest mass
• P(v ⊥ a) ≈ γ2 P(v || a)
• proportional to 1/ρ2 (ρ is the bending radius)
• proportional to B2 (B is the magnetic field of the bending dipole)
The radiation power emitted by an electron beam in a storage ring is very high.
The surface of the vacuum chamber hit by synchrotron radiation must be
cooled.
Lecture 10 - E. Wilson - 3/20/2008 - Slide 16
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Energy loss via synchrotron radiation emission
in a storage ring
In the time Tb spent in the bendings
the particle loses the energy U0
2πρ e 2 γ 4
U 0 = ∫ Pdt = PTb = P
=
c
3ε 0 ρ
i.e. Energy Loss per turn (per electron)
e 2γ 4
E (GeV ) 4
= 88.46
U 0 (keV ) =
3ε 0 ρ
ρ ( m)
Power radiated by a beam of average
current Ib: this power loss has to be
compensated by the RF system
I b ⋅ Trev
e
eγ 4
E (GeV ) 4 I ( A)
P ( kW ) =
I b = 88.46
3ε 0 ρ
ρ ( m)
N tot =
Power radiated by a beam of average
eγ 4
L(m) I ( A) E (GeV ) 4
P ( kW ) =
LI b = 14.08
current Ib in a dipole of length L
6πε 0 ρ 2
ρ ( m) 2
(energy loss per second)
Lecture 10 - E. Wilson - 3/20/2008 - Slide 17
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The radiation integral (I)
The energy received by an observer (per unit solid angle at the source) is
∞
∞
d 2W
d 2P
= ∫
dt = cε 0 ∫ | RE (t ) |2 dt
dΩ −∞ dΩ
−∞
Using the Fourier Transform we move to the frequency space
∞
d 2W
= 2cε 0 ∫ | RE (ω ) |2 dω
dΩ
0
Angular and frequency distribution of the energy received by an observer
2
d 3I
2 ˆ
= 2ε 0 cR E (ω )
dΩdω
Neglecting the velocity fields and assuming the observer in the far field:
n constant, R constant
2
∞
&
3
2
d I
e
n × ( n − β ) × β iω ( t − n ⋅ r ( t ) / c )
Radiation Integral
=
e
dt
2
2
∫
dΩdω 4πε 0 4π c −∞ (1 − n ⋅ β )
[
Lecture 10 - E. Wilson - 3/20/2008 - Slide 18
]
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The radiation integral (II)
The radiation integral can be simplified to [see Jackson]
d I
eω
=
dΩdω 4πε 0 4π 2 c
3
2
2
∞
2
iω ( t − n ⋅ r ( t ) / c )
n
×
(
n
×
β
)
e
dt
∫
−∞
How to solve it?
9 determine the particle motion
r (t ); β (t ); β& (t )
9 compute the cross products and the phase factor
9 integrate each component and take the vector square modulus
Calculations are generally quite lengthy: even for simple cases as for the
radiation emitted by an electron in a bending magnet they require Airy
integrals or the modified Bessel functions (available in MATLAB)
Lecture 10 - E. Wilson - 3/20/2008 - Slide 19
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Radiation integral for synchrotron radiation
Trajectory of the arc of circumference
⎛ ⎛
⎞
βc ⎞
βc
⎜
r (t ) = ⎜ ρ ⎜⎜1 − cos t ⎟⎟, sin t , 0 ⎟⎟
ρ ⎠
ρ
⎝ ⎝
⎠
In the limit of small angles we compute
⎡
⎤
⎛ βct ⎞
⎛ βct ⎞
⎟⎟ sin θ ⎥
⎟⎟ + ε ⊥ cos⎜⎜
n × (n × β ) = β ⎢− ε || sin⎜⎜
ρ
ρ
⎠
⎠
⎝
⎝
⎣
⎦
⎡ ρ ⎛ βct ⎞
⎤
⎛ n ⋅ r (t ) ⎞
⎟⎟ cos θ ⎥
ω ⎜t −
⎟ = ω ⎢t − sin⎜⎜
c ⎠
⎝
⎣ c ⎝ ρ ⎠
⎦
Substituting into the radiation integral and introducing
2
ξ=
ρω
2 2 3/ 2
(
1
+
γ
θ )
3
3cγ
⎤
⎛ 2ωρ ⎞
d 3I
e2
γ 2θ 2
2 2 2⎡
2
2
⎜
⎟
=
1
+
K
(
)
+
K
(
)
ξ
γ
θ
ξ
1/ 3
⎢ 2/3
⎥
2 2
dΩ dω 16 π 3ε 0 c ⎜⎝ 3cγ 2 ⎟⎠
1
+
γ
θ
⎦
⎣
(
Lecture 10 - E. Wilson - 3/20/2008 - Slide 20
)
20/29
Critical frequency and critical angle
2
⎤
⎛ 2ωρ ⎞
γ 2θ 2
d I
e
2 2 2⎡
2
2
⎟
⎜
+
+
ξ
γ
θ
ξ
=
1
K
(
)
K
(
)
1/ 3
⎢ 2/3
⎥
2 2
+
γ
θ
dΩ dω 16 π 3ε 0 c ⎜⎝ 3cγ 2 ⎟⎠
1
⎣
⎦
3
2
(
)
Using the properties of the modified Bessel function we observe that the
radiation intensity is negligible for ξ >> 1
ξ=
ωρ
2 2 3/ 2
(
1
+
γ
θ ) >> 1
3cγ 3
Critical frequency
Critical angle
3c 3
γ
2ρ
1/ 3
1 ⎛ ωc ⎞
θc = ⎜ ⎟
γ⎝ω ⎠
ωc =
Higher frequencies
have smaller critical
angle
For frequencies much larger than the critical frequency and angles much
larger than the critical angle the synchrotron radiation emission is negligible
Lecture 10 - E. Wilson - 3/20/2008 - Slide 21
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Frequency distribution of radiated energy
Integrating on all angles we get the frequency distribution of the energy
radiated
dI
d 3I
3e 2 ω
γ
=
dΩ =
ω
πε
ωC
dω ∫∫
d
d
Ω
4
c
0
4π
dI
e 2 ⎛ ωρ ⎞
≈
⎜
⎟
dω 4πε 0 c ⎝ c ⎠
∞
K
∫
ω ω
/
5/3
( x)dx
C
1/ 3
ω << ωc
⎛ω ⎞
dI
3π e
≈
γ ⎜⎜ ⎟⎟
dω
2 4πε 0 c ⎝ ωc ⎠
2
1/ 2
e −ω / ωc
ω >> ωc
often expressed in terms of the
function S(ξ) with ξ = ω/ωc
∞
9 3
S (ξ ) =
ξ ∫ K 5 / 3 ( x)dx
8π ξ
dI
3e 2γ ω
=
dω 4πε 0 c ωc
∞
∫ S (ξ )dξ = 1
0
∞
2 e 2γ
K 5 / 3 ( x)dx =
S (ξ )
∫
9ε 0 c
ω / ωc
Lecture 10 - E. Wilson - 3/20/2008 - Slide 22
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Frequency distribution of radiated energy
It is possible to verify that the integral over the frequencies agrees with
the previous expression for the total power radiated [Hubner]
ω
ω
∞
U 0 1 dI
1 2e 2γ
e 2c γ 4
= ∫
P=
dω =
ωc ∫ ξ dξ ∫ K 5 / 3 ( x)dx =
Tb Tb 0 dω
Tb 9ε 0 c 0
6ε 0 c ρ 2
ξ
The frequency integral extended up to the critical frequency contains half
of the total energy radiated, the peak occurs approximately at 0.3ωc
It is also convenient to define the critical
photon energy as
ε c = hω c =
3 hc 3
γ
2 ρ
For electrons, the critical energy in practical
units reads
50% 50%
E [GeV ]3
= 0.665 ⋅ E[GeV ]2 ⋅ B [T ]
ε c [keV ] = 2.218
ρ [ m]
Lecture 10 - E. Wilson - 3/20/2008 - Slide 23
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Synchrotron radiation emission as a
function of beam the energy
Dependence of the frequency distribution of the energy radiated via synchrotron
emission on the electron beam energy
No dependence on
the energy at longer
wavelengths
Lecture 10 - E. Wilson - 3/20/2008 - Slide 24
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Polarisation of synchrotron radiation
2
⎤
⎛ 2ωρ ⎞
γ 2θ 2
d 3I
e2
2 2 2⎡
2
2
⎜
⎟
γ
θ
ξ
ξ
=
1
+
K
(
)
+
K
(
)
1/ 3
⎢ 2/3
⎥
2 2
γ
θ
dΩ dω 16 π 3ε 0 c ⎜⎝ 3cγ 2 ⎟⎠
1
+
⎣
⎦
(
)
Polarisation in
the orbit plane
Polarisation orthogonal
to the orbit plane
In the orbit plane θ = 0, the polarisation is purely horizontal
Integrating on all frequencies we get the angular distribution of the energy
radiated
d 2I ∞ d 3I
7 e 2γ 5
1
=∫
dω =
dΩ 0 dω dΩ
64πε 0 ρ ( 1 + γ 2θ 2 )5 / 2
⎡ 5 γ 2θ 2 ⎤
⎢1 +
2 2⎥
⎣ 7 1+γ θ ⎦
Integrating on all the angles we get a polarization on the plan of the orbit 7
times larger than on the plan perpendicular to the orbit
Lecture 10 - E. Wilson - 3/20/2008 - Slide 25
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Synchrotron radiation from Undulators and wigglers
Periodic array of magnetic
poles providing a sinusoidal
magnetic field on axis:
B = (0, B0 sin( ku z ), 0,)
Solution of equation of motions:
⎛
⎞
λu K
λu K 2
⎟ ˆ
r (t ) = −
ω
t
sin ωu t ⋅ xˆ + ⎜⎜ β z ct +
cos(
2
)
u ⎟⋅ z
2
2πγ
16πγ
⎝
⎠
eB0 λu
Undulator
parameter
2π mc
1 ⎛ K2 ⎞
⎟
β z = 1 − 2 ⎜⎜1 +
2 ⎟⎠
2γ ⎝
K=
Constructive interference of radiation emitted at different poles
λu
− λu cos θ = nλ
β
⎞
λu ⎛ K 2
+ γ 2θ 2 ⎟⎟
λn = 2 ⎜⎜1 +
2γ n ⎝
2
⎠
d=
Lecture 10 - E. Wilson - 3/20/2008 - Slide 26
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Synchrotron radiation from undulators and wigglers
t1
t2
t3
t4
t5
Continuous spectrum characterized
by εc = critical energy
γ –1
Bending Magnet — A “Sweeping Searchlight”
bending magnet - a “sweeping searchlight”
εc(keV) = 0.665 B(T)E2(GeV)
eg: for B = 1.4T E = 3GeV εc = 8.4 keV
Dipoles
(10-100) γ –1
(bending magnet fields are usually
lower ~ 1 – 1.5T)
Wiggler — Incoherent Superposition
wiggler - incoherent superposition K > 1
Quasi-monochromatic spectrum with
peaks at lower energy than a wiggler
(γ N)–1
Undulator — Coherent Interference
undulator - coherent interference K < 1
λu ⎛ K 2 ⎞ λu
1+
λn =
⎟≈ 2
2 ⎜
2nγ ⎝
2 ⎠ nγ
nE[GeV ]2
ε n (eV ) = 9.496
⎛ K2 ⎞
λu [m] ⎜1 +
⎟
2
⎝
⎠
Lecture 10 - E. Wilson - 3/20/2008 - Slide 27
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Synchrotron radiation from Undulators and wigglers
Radiated intensity vs K: as K increases the higher harmonic becomes stronger
For large K the wiggler spectrum becomes
similar to the bending magnet spectrum,
2Nu times larger.
Fixed B0, to reach the bending magnet
critical wavelength we need the harmonic
number n:
K
1
2
10
20
n
1
5 383 3015
Lecture 10 - E. Wilson - 3/20/2008 - Slide 28
28/29
Summary
Accelerated charged particles emit electromagnetic radiation
Radiation is stronger for circular orbit
Synchrotron radiation is stronger for light particles
Undulators and wigglers enhance the synchrotron radiation emission
Synchrotron radiation has unique characteristics
Lecture 10 - E. Wilson - 3/20/2008 - Slide 29
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Bibliography
J. D. Jackson, Classical Electrodynamics, John Wiley & sons.
E. Wilson, An Introduction to Particle Accelerators, OUP, (2001)
M. Sands, SLAC-121, (1970)
R. P. Walker, CAS CERN 94-01 and CAS CERN 98-04
K. Hubner, CAS CERN 90-03
J. Schwinger, Phys. Rev. 75, pg. 1912, (1949)
B. M. Kincaid, Jour. Appl. Phys., 48, pp. 2684, (1977).
Lecture 10 - E. Wilson - 3/20/2008 - Slide 30
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