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Introduction to Particle Accelerators
Synchrotron Radiation
What is it ?
 Rate of energy loss
 Longitudinal damping
 Transverse damping
 Quantum fluctuations
 Wigglers

Rende Steerenberg (BE/OP)
13 January 2011
Acceleration and Electro-Magnetic Radiation
An accelerating charge emits Electro-Magnetic waves.
Example:
An antenna is fed by an oscillating current and it emits
electro magnetic waves.
In our accelerator we know to types of acceleration:
Longitudinal – RF system
Transverse – Magnetic fields, dipoles, quadrupoles, etc..
Force due to magnetic field
gives change of direction

dp
d
(
m

v
)  m  a
F 
dt
dt
Momentum change
So:

m  v  constant
R. Steerenberg, 13-Jan-2011
Newton’s law
Direction changes but not magnitude
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Rate of EM radiation
The rate at which a relativistic lepton radiates EM
energy is :
Force // velocity
Longitudinal  square of energy (E2)
Force  velocity
Transverse  square of magnetic field (B2)
PSR  E2 B2
In our accelerators:
Transverse force > Longitudinal force
Therefore we only consider radiation due to ‘transverse
acceleration’ (thus magnetic forces)
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Rate of energy loss (1)
This EM radiation generates an energy loss of the
particle concerned, which can be calculated using:
constant
 2 rc
P  
 3 m c
2
0

3

E F

2
2
Electron radius
Velocity of light
Total energy
‘Accelerating’ force
Lepton rest mass
Our force can be written as: F = evB = ecB
 2 e rc
Thus: P  
 3 m c 
2
3
2
0
3

E B

2
p E
but ( B )  
e ec
2
 2 rc
Which gives us: P  
 3 m c
0
R. Steerenberg, 13-Jan-2011
2

3
E


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v
1
c
2
4
Rate of energy loss (2)
 2 rc

P

We have:
 3 m c
0
2

3
E


4
2
,which gives the energy loss
We are interested in the energy loss per revolution
for which we need to integrate the above over 1 turn
ds
Bending radius
Thus:  Pdt   P
inside the magnets
c
ds
d
However:   2 
c
c
Lepton
energy
4
r
4 1
E  2 d
Finally this gives: u 
3
2
3 m0 c 

Gets very large if

E is large !!!
C
R. Steerenberg, 13-Jan-2011
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
CE 4

5
What about the synchrotron oscillations ?
The RF system, besides increasing the energy has to
make up for this energy loss u.
All the particles with the same phase, , w.r.t. RF
waveform will have the same energy gain E = Vsin
However,
Lower energy particles lose less energy per turn
Higher energy particles lose more energy per turn
u 
CE

What will happen…???
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4
Synchrotron motion for leptons
E
u 
CE
4

t (or )
All three particles will gain the same energy from the
RF system
The black particle will lose more energy than the red
one.
This leads to a reduction in the energy spread,
since u varies with E4.
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Longitudinal damping in numbers (1)
Remember how we calculated the synchrotron
frequency.
It was based on the change in energy: dE  V sin 
Now we have to add an extra term, the energy loss du
dE  V sin   du
becomes
dE
 frevV sin   frev du
dt
Our equation for the synchrotron oscillation becomes
then:
Extra term for
d   2 h
2
2
 2 h


f

V


f


rev
rev du  0
2
E
dt
 E

2
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energy loss
8
Longitudinal damping in numbers (2)
This term: 2 h f rev 2 du
E
Can be written as: 2 h f
This now becomes:
du du dE

E dE E
2
rev
du dE
but
dE E
dE
1 df

E
 f
rev
rev
du
 2 h df f
dE
rev
du 1 d

dE T dt
rev
1
d
T
dt
The synchrotron oscillation differential
equation becomes now:
Damped SHM,
rev
rev
d  du 1 d  2 h

2


 f rev V   0
2
dt
dE Trev dt  E

2
R. Steerenberg, 13-Jan-2011
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as expected
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Longitudinal damping in numbers (3)
So, we have:
d 2 du 1 d  2 h

2



f

V

  0
rev
2
dt
dE Trev dt  E

The damping coefficient  
du 1
dE T
rev
This confirms that the variation of u as a function
of E leads to damping of the synchrotron
oscillations as we already expected from our
reasoning on the 3 particles in the longitudinal
phase space.
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Longitudinal damping time
du 1
dE T
CE
du
4CE

We know that u  
and thus

dE

du
4u
CE
Not totally
So approximately:

u 
correct since
dE
E

E
The damping coefficient is given by:  
rev
4
3
4
For the damping time we have then:
Energy
1
ET
Damping time = 
 4u
rev

Revolution time
CE
4
Energy loss/turn

The damping time decreases rapidly (E3) as we
increase the beam energy.
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Damping & Longitudinal emittance
Damping of the energy spread leads to shortening
of the bunches and hence a reduction of the
longitudinal emittance.
E
Initial
Later…
E

d
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Some LHC numbers
Energy loss per turn at:
injection at 450 GeV = 1.15 x 10-1 eV
Collision at 7 TeV = 6.71 x 103 eV
Power loss per meter in the main dipoles at 7 TeV is
0.2 W/m
Longitudinal damping time at:
Injection at 450 GeV = 48489.1 hours
Collision at 7 TeV = 13 hours
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What about the betatron oscillations ? (1)
Each photon emission reduces the transverse and longitudinal
energy or momentum.
Lets have a look in the vertical plane:
Emitted photon (dp)
total momentum (p)
momentum lost dp
ideal trajectory
particle
R. Steerenberg, 13-Jan-2011
particle trajectory
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What about the betatron oscillations ? (2)
The RF system must make up for the loss in longitudinal
energy dE or momentum dp.
However, the cavity only supplies energy parallel to ideal
trajectory.
ideal trajectory
new particle trajectory
old particle trajectory
Each passage in the cavity increases only the longitudinal
energy.
This leads to a direct reduction of the amplitude of the
betatron oscillation.
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Vertical damping in numbers (1)
The RF system increases the momentum p by dp or
energy E by dE
Tan(α)= α
pt = transverse momentum
pT= total momentum
If α <<
pt
y' 
p
p = longitudinal momentum
dp is small
pt
pt  dp 
 dp 
new( y ' ) 
 1    y ' 1  
p  dp p 
p
p

The change in transverse angle is thus given by:
dp
dE
dy'   y '
  y'
p
E
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Vertical damping in numbers (2)
A change in the transverse angle alters the betatron
oscillation amplitude
y’

a.sin 
da   .dy'.sin 
dE
da    . y ' .sin 
E
dE
da     . y ' .sin 
E
dE
da  a
 sin 
E
2
y
dy’
 0
2
2
 0
a
da
da
1 dE

a
2 E
Summing over many
photon emissions
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Vertical damping in numbers (3)
We found:
dE is just the change in
energy per turn u
(energy given back by RF)
da
1 dE

a
2 E
da  a
The change in amplitude/turn is thus:
u
a
Which is also: a  
2E
da
u
Thus:

a
dt
2 ET
Change in amplitude/second
Revolution time
This shows exponential damping with coefficient:
2 ET
Damping time =
u
R. Steerenberg, 13-Jan-2011
 u
2ET
(similar to longitudinal case)

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CE
4

18
Horizontal damping in numbers
da
1u
Vertically we found:

a
2E
This is still valid horizontally
However, in the horizontal plane, when a particle
changes energy (dE) its horizontal position changes
too
OK since =1
dr
dp
dE
u
p
p
p
r
p
E
E
 is related to D(s) in the
bending magnets
da
u
 1  2 
horizontally we get:
a
2E
2 ET  1 


Horizontal damping time:
u  1  2 
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Some intermediate remarks….
Transverse damping for LHC time at:
Injection at 450 GeV = 48489.1 hours
Collision at 7 TeV = 26 hours
Longitudinal and transverse emittances all shrink as a function of
time.
For leptons damping times are typically a few milliseconds up to a
few seconds.
Advantages:
Reduction in losses
Injection oscillations are damped out
Allows easy accumulation
Instabilities are damped
Inconvenience:
Lepton machines need lots of RF power, therefore LEP was stopped
All damping is due to the energy gain from the RF system an not
due to the emission of synchrotron radiation
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Is there a limit to this damping ? (1)
Can the bunch shrink to microscopic dimensions ?
No ! , Why not ?
For the horizontal emittance h there is heating term
due to the horizontal dispersion.
What would stop dE and v of damping to zero?
For v there is no heating term. So v can get very
small. Coupling with motion in the horizontal plane
finally limits the vertical beam size
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Is there a limit to this damping ? (2)
In the transverse plane the damping seems to be
limited.
What about the longitudinal plane ?
Whenever a photon is emitted the particle energy
changes.
This leads to small changes in the synchrotron
oscillations.
This is a random process.
Adding many such random changes (quantum
fluctuations), causes the amplitude of the
synchrotron oscillation to grow.
When growth rate = damping rate then damping
stops, which give a finite equilibrium energy spread.
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Quantum fluctuations (1)
Quantum fluctuation is defined as:
Fluctuation in number of photons emitted in one damping
time
Let Ep be the average energy of one emitted photon
Revolution time
ET
E
seconds  turns
Damping time 
u
u
Energy loss/turn
u
Number of photons emitted/turn = E
Number of emitted photons in one
damping time can then be given by:
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p
u E E

Ep u Ep
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Quantum fluctuations (2)
E
Number of emitted photons in one damping time = E
Random
E
process
r.m.s. deviation = E
Energy of one
emitted photon
E
E  EE
The r.m.s. energy deviation =
E
p
p
p
p
p
The average photon energy Ep  E3
The r.m.s. energy spread  E2
The damping time  E3
Higher energy  faster longitudinal damping,
but also larger energy spread
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Wigglers (1)
The damping time in all planes  ET
u
If the loss of energy, u, increases, the damping time
decreases and the beam size reduces.
To be able to control the beam size we add ‘wigglers’
N S N
S N
S N
S N S N
S
beam
S N
S N
S N
S N S N
S N
It is like adding extra dipoles, however the wiggles
does not give an overall trajectory change, but
increases the photon emission
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Wigglers (2)
What does the wiggler in the different planes?
Vertically:
We do not really need it (no heating term), but the vertical
emittance would be reduced
Horizontally:
The emittance will reduce.
A change in energy gives a change in radial position
We know the dispersion function: dr  D ( s )
dE
E
In order to reduce the excitation of horizontal oscillations
we should put our wiggler in a dispersion free area (D(s)=0)
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Wigglers (3)
Longitudinally:
The wiggler will increase the number of photons emitted
It will increase the quantum fluctuations
It will increase the energy spread
Conclusion:
Wigglers increase longitudinal emittance and
decrease transverse emittance
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Summary
Damping due to addition of longitudinal
momentum !
Longitudinal:
4
Energy loss per turn: u  
Damping time:
Transverse:
1


CE

ET
4u
rev
1 2ETrev
Vertical damping time: 
u

Horizontal damping time: 1  2 ET  1 

u  1  2 
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Questions….,Remarks…?
Synchrotron
radiation
Damping
Quantum
fluctuations
Wigglers
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