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Transcript
```Radiation effects in electron storage rings
Average radiated power restored by RF
U ≅ 10 – 3 of E0
0
• Electron loses energy each turn
• RF cavities provide voltage to accelerate electrons
back to the nominal energy
Electron Dynamics
VRF > U0
• Average rate of energy loss produces DAMPING of electron
oscillations in all three degrees of freedom (if properly arranged!)
Quantum fluctuations
• Statistical fluctuations in energy loss (from quantised emission
of radiation) produce RANDOM EXCITATION of these
oscillations
L. Rivkin, PSI
Equilibrium distributions
Introduction to Accelerator Physics Course
CERN Accelerator School, Baden bei Wien, September 2004
• The balance between the damping and the excitation of the
electron oscillations determines the equilibrium distribution of
particles in the beam
BD 2
Average energy loss per turn
Energy gain in the RF cavities
 Every turn electron radiates small amount of energy
 Only the longitudinal component of the momentum
is increased in the RF cavity
E1 = E0 – U 0 = E0 1 –
U0
E0
RF Cavity
 Since the radiation is emitted along the tangent to
the trajectory, only the amplitude of the momentum
changes
P
P⊥
P||
U0
c
eVRF = U 0
P⊥
U0
c
P
P||
U
U
P1 = P0 – c0 = P0 1 – 0
E0
 The transverse momentum, or the amplitude of the
betatron oscillation remains small
BD 3
BD 4
Energy of betatron oscillation
Exponential damping
 Transverse momentum corresponds to the
energy of the betatron oscillation
Eβ ∝ A 2
 But this is just the exponential decay law!
A12 = A02 1 –
U0
E0
or
A1 ≅ A 0 1 –
∆A = – U0
A
2E
 The amplitudes are exponentially
U0
2E0
damped
−t
A= A ⋅e τ
 The relative change in the betatron oscillation
amplitude that occurs in one turn (time T0)
0
with the damping decrement
∆A = – U0
A
2E
1
τ
=
U0
2 E T0
BD 5
In a linear accelerator:
p⊥
The electron beam “emittance”:
p
x′ = p⊥ decreases ∝ 1
E
p
p⊥
p
In a storage ring beam passes many
times through same RF cavity
BD 6
Source
area, S
Angular
divergence, Ω
The brightness
depends on the
geometry of the
source, i.e., on the
electron beam
emittance
RF
Emittance = S x Ω
 Clean loss of energy every turn (no change in x’)
 Every turn is re-accelerated by RF (x’ is reduced)
 Particle energy on average remains constant
BD 7
BD 8
Damping time
Emittance damping in linacs:
 the time it would take particle to lose all of its energy
ε
Ω
ε
4
ε
2
Ω
2
τε =
ε ∝ −1γ
Ω
4
 or in terms of radiated power
or
γ
2γ
E T0
U0
remember that
τε =
Pγ ∝ E 4
τε ∝
γε = const.
4γ
E T0 E
=
U0
Pγ
1
E3
BD 9
Longitudinal motion:
Longitudinal motion:
phase stability
RF
 RF cavity provides accelerating field
with frequency
• h – harmonic number
U0
 The energy gain:
BD 10
VRF
U0
τ
f RF = h ⋅ f 0
 Particle ahead of synchronous one
VRF
• gets too much energy from the RF
• goes on a longer orbit (not enough B)
>> takes longer to go around
• comes back to the RF cavity closer to synchronous part.
τ
U RF = eVRF (τ )
 Synchronous particle:
 Particle behind the synchronous one
• has design energy
• gains from the RF on the average as
much as it loses per turn U0
• gets too little energy from the RF
• goes on a shorter orbit (too much B)
• catches-up with the synchronous particle
BD 11
BD 12
Longitudinal motion:
damping of synchrotron oscillations
Pγ ∝ E 2 B 2
During one period of synchrotron oscillation:
 when the particle is in the upper half-plane, it loses more
energy per turn, its energy gradually reduces
ε
U > U0
U < U0
τ
 when the particle is in the lower half-plane, it loses less
energy per turn, but receives U0 on the average, so its
Pγ∝E2B2
Displaced off the design orbit particle sees fields that
are different from design values
 betatron oscillations: zero on average
• linear term in B2 - averages to zero
- small
 energy deviation
Pγ ∝ E 2
• different energy:
• different magnetic field
particle moves on a different orbit, defined by the
off-energy or dispersion function Dx
The synchrotron motion is damped
⇒ both contribute to linear term in
 the phase space trajectory is spiraling towards the origin
Pγ (ε )
BD 13
Pγ∝E2B2
BD 14
Energy balance
Energy gain from the RF system: URF = eVRF τ = U0 + eVRF ⋅τ
To first order in ε
Urad = U0 + U ′ ⋅ ε
electron energy changes slowly, at any instant it is
moving on an orbit defined by Dx
dU
′
U ≡

synchronous particle (τ = 0) will get exactly the energy loss per
turn

we consider only linear oscillations

Each turn electron gets energy from RF and loses energy to
radiation within one revolution time T0
dE
E0
after some algebra one can write
U′ =
∆ε = U0 + eVRF ⋅τ – U0 + U′ ⋅ε
U0
2 +D
E0

D ≠ 0 only when ρk ≠ 0
VRF =
dVRF
dτ
τ= 0
dε = 1 eV ⋅τ – U′⋅ε
dt T0 RF
An electron with an energy deviation will arrive after one turn
at a different time with respect to the synchronous particle
dτ = –α ε
E0
dt
BD 15
BD 16
Synchrotron (time - energy) oscillations
Synchrotron oscillations: damped harmonic oscillator
d 2ε + 2α dε + Ω 2ε = 0
ε dt
dt 2
αeVRF
where the oscillation frequency Ω 2 ≡
T0E0
U′
typically αε <<Ω
αε ≡
the damping is slow:
2T0
Combining the two equations



The ratio of amplitudes at any instant
τ= α ε
ΩE0
Oscillations are 90 degrees out of phase
θε = θτ + π
2
The motion can be viewed in the phase space of conjugate
variables
the solution is then:
ε t =ε0e –αεtcos Ωt + θε

ε, τ
εˆ
similarly, we can get for the time delay:
τt
=τ0e –αεtcos
αε , Ωτ
E0
ε
E0
τ
τˆ
Ωt + θτ
αε
Ωτ
BD 17
Orbit Length
dl = 1 + ρx ds
dl
ds
Like the tunes Qx, Qy - α depends on the whole optics
ρ
x

Horizontal displacement has two parts:
To first order xβ does not change L
xε – has the same sign around the ring
Length of the off-energy orbit
∆L = δ ⋅
L ε = dl =
Ds
∆p
ds where δ = p = ∆E
E
ρs
A quick estimate for separated function guide field:
α=
x = xβ + xε

Ds
ds
α≡ 1
L ρs
Momentum compaction factor
Length element depends on x

BD 18
x
1 + ρε ds = L 0 + ∆L
∆L = α ⋅ δ
L
1
L 0ρ0
mag
D s ds =
ρ = ρ0 in dipoles
1 D ⋅L
mag ρ = ∞ elsewhere
L 0ρ 0

But

Since dispersion is approximately
D ≈ R2 ⇒ α ≈ 12 typically < 1%
Q
Q
and the orbit change for ~ 1% energy deviation
L mag = 2πρ 0
α=
D
R
∆L = 1 ⋅ δ ≈ 10 – 4
L
Q2
BD 19
BD 20
Something funny happens on the way around the ring...
Revolution time changes with energy

T0 =
L0
cβ
∆T = ∆L – ∆β
T
L
β
dβ 1 dp
Particle goes faster (not much!) β = γ 2 ⋅ p

while the orbit length increases (more!)

The “slip factor” η ≅ α
since
∆T = α – 1 ⋅ dp = η ⋅ dp
p
T
γ2 p

Ring is above “transition energy”
Not only accelerators work above transition
(relativity)
∆L = α ⋅ dp
p
L
α >> 12
γ
α ≡ 12
γ tr
isochronous ring: η = 0 or γ = γ tr
BD 21
Robinson theorem
Damping partition numbers
 Transverse betatron oscillations
are damped with
 Synchrotron oscillations
are damped twice as fast
Damping only
• If damping was the whole story, the beam emittance (size)
would shrink to microscopic dimensions!*
• Lots of problems! (e.g. coherent radiation)
U0
1 1
τx = τ z = 2ET0
U0
1
τε = ET0
Quantum fluctuations
takes care of the problem!
• It is sufficient to use quasi-classical picture:
 The total amount of damping (Robinson theorem)
depends only on energy and loss per turn
» Emission time is very short
» Emission times are statistically independent
U0
1
1
1 2U0
τx + τ y + τ ε = ET0 = 2ET0 J x + J y + J ε
the sum of the partition numbers
BD 22
(each emission - only a small change in electron energy)
Purely stochastic (Poisson) process
Jx + Jz + Jε = 4
BD 23
BD 24
Visible quantum effects
Damping only
I have always been somewhat amazed that a purely quantum
effect can have gross macroscopic effects in large machines;
• If damping was the whole story, the beam emittance (size)
would shrink to microscopic dimensions!*
• Lots of problems! (e.g. coherent radiation)
and, even more,
that Planck’s constant has just the right magnitude needed to
make practical the construction of large electron storage rings.
* How small? On the order of electron wavelength
h = λC
E =γmc 2 = hν = hc ⇒ λ e = 1γ mc
γ
λe
h
A significantly larger or smaller value of
λ C = 2.4⋅ 10 –12m – Compton wavelength
Diffraction limited electron emittance
ε≥
λC
1
×N 3
4πγ
–
would have posed serious -- perhaps insurmountable -problems for the realization of large rings.
fermions
Mathew Sands
BD 25
Quantum excitation of energy oscillations
Photons are emitted with typical energy u ph ≈ h ω typ
at the rate (photons/second)
N
γ3
= hc ρ
P
= γ
u ph
Fluctuations in this rate excite oscillations
During a small interval ∆t electron emits photons
losing energy of
Actually, because of fluctuations, the number is
resulting in spread in energy loss
N = N
⋅ ∆t
N ⋅ u ph
N ±
N
± N ⋅ u ph
For large time intervals RF compensates the energy loss, providing
damping towards the design energy E0
Steady state: typical deviations from E0
≈ typical fluctuations in energy during a damping time τε
BD 26
We then expect the rms energy spread to be
τε ≈
and since
σε ≈
E0 ⋅ u ph
E0
Pγ
and
σ ε ≈ N ⋅τ ε ⋅ u ph
Pγ = N ⋅ u ph
geometric mean of the electron and photon energies!
Relative energy spread can be written then as:
σε
≈γ
E0
λ– e
ρ
λ– e = mh c ∼ 4 ⋅ 10 – 13m
it is roughly constant for all rings
• typically
E ∝ ρ2
e
σε
~ const ~ 10 – 3
E0
BD 28
Equilibrium bunch length
Bunch length is related to the energy spread ε
More detailed calculations give

ρ
• for the case of an ‘isomagnetic’ lattice ρ s = ∞0
σε
E
with
2
=
Cq E
in dipoles
elsewhere
Energy deviation and time of arrival
(or position along the bunch)
are conjugate variables (synchrotron oscillations)
σ
στ = α ε
Ωs E
2

Jερ0
recall that Ωs ∝ VRF
τ
τ= α ε
Ωs E
Two ways to obtain short bunches:
hc = 1.468 ⋅ 10 – 6 m
Cq = 55
32 3 m ec 2 3
GeV 2
It is difficult to obtain energy spread < 0.1%
στ ∝ 1

RF voltage (power!)

Momentum compaction factor in the limit of α = 0
isochronous ring: particle position along the bunch is
frozen
VRF
στ ∝ α
• limit on undulator brightness!
BD 29
Horizontal oscillations: equilibrium
After an electron emits a photon
u ph
 its energy decreases:
= E0 1 + δ
E = E0 1 –
E0
E = E0 - uph
 Neither its position nor angle change after emission
 its reference orbit has smaller radius (Dispersion)
xref = D ⋅ δ
It will start a betatron oscillation
around this new reference orbit
xβ = D ⋅ δ
Horizontal oscillations excitation
Emission of photons is a random process
 Again we have random walk, now in x. How far particle
will wander away is limited by the radiation damping
The balance is achieved on the time scale of the damping
time τx = 2 τε
σ xβ ≈
In
N ⋅
τx ⋅ D ⋅ δ = 2 ⋅ D ⋅
smooth approximation for D
or, typically 10-3 of R,
reduced further by Q2 focusing!
In large rings Q2 ~ R, so D ~ 1m
σ
⋅ E
σ xβ ≈ 2 R
2
E
Q
Typical horizontal beam size ~ 1 mm
Quantum effect visible to the naked eye!
Vertical size - determined by coupling
σε
E
Equilibrium horizontal emittance
Beam emittance
Detailed calculations
for isomagnetic lattice
Betatron oscillations
εx0 ≡
2
σxβ
CqE 2 H mag
=
⋅
ρ
Jx
β
• Particles in the beam execute betatron oscillations with
x’
different amplitudes.
and
= 1 D 2 + βD′ + αD
β
H
mag
• Gaussian (electrons)
• “Typical” particle: 1 - σ ellipse
(in a place where α = β’ = 0)
2
is average value in the bending magnets
Emittance ≡
2
H ~ Dβ ~ R3
Q
For simple lattices
(smooth approximation)
σx’
Transverse beam distribution
H = γD 2 + 2αDD′ + βD′ 2
where
Area = π ⋅ ε
Units of ε m ⋅ rad
σ x2
β
σx = ε β
σ x′ = ε / β
2
εx 0 ≈
x
σx
CqE R 1
⋅ρ⋅ 3
Jx
Q
BD 33
1 e –x 2 / 2σ 2dx
2π σ
µ
x
σx′
Area = π εx

Probability to be inside n-σ ellipse
P1 = 1 – e – 1 2 = 0.39
Pn = 1 – e – n
2
2
2
E
ε∝
θ 3FFODO µ
Jx
Emittance
x’
Probability to be inside 1-σ ellipse
x′
BD 34
FODO Lattice
emittance
σx
Electron rings emittance definition
 1 - σ ellipse

σ
β = σx
100
2-D Gaussian distribution
n x dx =
ε = σ x ⋅ σ x′
ε ∝ 13
Q
10
1
0
BD 35
20
40 60 80 100 120 140 160 180
BD 36
Ionization cooling
E
p⊥
p||
absorber
acceleration
σ0
σ′0
σ′ =
σ′ 0 >> σ′ MS
2
σ′ 02 + σ′ MS
damping, but there is
multiple scattering
in the absorber that
blows up the
emittance
to minimize the
blow up due to
multiple
scattering in the
absorber we can
focus the beam
Momentum compaction factor
α=
I1
2πR
I1 =
D ds
ρ
I2 =
ρ2
ds
ds
Energy loss per turn
I3 =
U 0 = 1 Cγ E 4 ⋅ I 2
2π
I4 =
D 2k + 1 ds
2
ρ
I5 =
H ds
3
re
Cγ = 4π
= 8.858 ⋅ 10 – 5 m 3
3 m ec 2 3
GeV
ρ3
ρ
ρ
BD 37
Damping parameter
D =
I4
I2
I1 =
Damping times, partition numbers
I2 =
J ε = 2 + D , Jx = 1 – D , J y = 1
I3 =
τ0
2ET0
τ0 =
Ji
U0
τi=
σε
E
2
Cq E 2 I 3
=
⋅
Jε
I2
Equilibrium emittance
σ2 C E2 I
εx 0 = xβ = q ⋅ 5
β
Jx
I2
BD 38
Smooth approximation
ds
ρ2
x ≈ a β n cos s – ϕ0
βn
ds
ρ
x s = a β s cos ϕ s – ϕ0
Betatron oscillation
approximated by
harmonic oscillation
D ds
ρ
3
I4 =
D 2k + 1 ds
2
ρ
I5 =
H ds
3
ρ
ρ
hc = 1.468 ⋅ 10 – 6 m
Cq = 55
32 3 m ec 2 3
GeV 2
H = γD 2 + 2αDD ′ + βD′ 2
BD 39
ϕs =
s
0
⇐ x′′ + k eff ⋅ x = 0, k eff = 12
βn

around the ring

Dispersion
obeys the equation

Momentum compaction
factor α
2π Q =
ds
βs
β s = β n = const
ds = 1 ⋅2 π R
βn βn
D′′ + k eff D = 1
R
α=
⇒ βn = R
Q
⇒ Dn =
D
β2
= n2
R
R
⇒
β n2 R
=
R Q2
α ≈ 12
Qx
BD 40
```
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