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Space Charge
Luigi Palumbo
Università di Roma "La Sapienza"
and LNF-INFN
CAS, 25 May 2005
EQUATION OF MOTION
The motion of charged particles is governed by the Lorentz force :
d (mγ v )
ext
= Fe.m.
= e( E + v × B )
dt
Where m is the rest mass, γ the relativistic factor and v the particle velocity
Charged particles are accelerated, guided and confined by external
electromagnetic fields.
Acceleration is provided by the electric field of the RF cavity
Magnetic fields are produced in the bending magnets for guiding the
charges on the reference trajectory (orbit), in the quadrupoles for the
transverse confinement, in the sextupoles for the chromaticity correction.
CAS, 25 May 2005
SELF FIELDS AND WAKE FIELDS
There is another important source of e.m. fields : the beam itself
Direct self fields
Space Charge
Image self fields
Wake fields
CAS, 25 May 2005
These fields depend on the current and on the charges velocity.
They are responsible of many phenomena of beam dynamics:
• energy loss
(wake-fields)
• energy spread and emittance degradation
• shift of the synchronous phase and frequency (tune)
• shift of the betatron frequencies (tunes)
• instabilities.
CAS, 25 May 2005
Fields of a point charge with uniform motion
y’
y
v
o’
o
r
E′ =
q
r
r′
4πε o r ′3
r
B′ = 0
x,x’
q
z
z’
vt is the position of the point charge in the lab. frame O.
• In the moving frame O’ the charge is at rest
• The electric field is radial with spherical symmetry
• The magnetic field is zero
E ′x =
q
x′
4πε o r ′3
E ′y =
q
y′
4πε o r ′3
E′z =
q
z′
4πε o r ′3
CAS, 25 May 2005
Relativistic transforms of the fields from O’ to O
Bx = B′x
E x = E ′x
E y = γ ( E ′y + vB′z ) B y = γ ( B′y − vE z′ / c 2 )
E z = γ ( E ′z − vB′y )
(
′2
′2
Bz = γ ( B′z + vE ′y / c 2 )
r′ = x + y + z
)
1/ 2
2
′
 x′ = γ ( x − vt )

 y′ = y
 z′ = z

[
r ′ = γ 2 ( x − vt ) 2 + y 2 + z 2
]
1/2
CAS, 25 May 2005
E x = E ′x =
q
4πε o r ′3
E y = γE ′y =
E z = γE ′z =
x′
q
y′
4πε o r ′3
q
z′
4πε o r ′3
=
=
=
q
4πε o
q
4πε o
q
4πε o
[γ
[γ
[γ
γ ( x − vt )
2
2
2 3/ 2
2
2 3/ 2
2
2 3/ 2
( x − vt ) + y + z
γy
2
2
( x − vt ) + y + z
γz
2
2
( x − vt ) + y + z
The field pattern is moving r
q
with the charge and it can E =
4πε o
be observed at t=0.
[γ
]
2
]
]
r
γr
2 2
2
x +y +z
]
2 3/ 2
The fields have lost the spherical symmetry but still keep a
symmetry with respect to the x-axis.
CAS, 25 May 2005
t=0
y
r
θ
x = r cos θ
z
x
r
E=
q
(
γ2
4πε o 1 − β 2 sin 2 θ
)
3/ 2
r
r
r3
y 2 + z 2 = r 2 sin 2 θ
γ=0
γ≠0
γ >>1
CAS, 25 May 2005
B is transverse to the motion direction
Bx = 0
B y = −vE z / c 2
Bz = vE y / c 2
r r
r
v×E
B⊥ = 2
c
γ→∞
CAS, 25 May 2005
Two charges in the rest frame O’
q
Fr′ =
q
1
qq
4πε o r 2
Two charges in the laboratory frame O
Relativistic transform
Fr =
1
γ
Fr′ =
1
qq
4πε o γr 2
Lorentz force
Fr = q ( Er − vBϕ ) = q ( Er − β 2 Er ) =
q
γ2
Er =
qq
4πε o γr 2
1
CAS, 25 May 2005
Direct Space Charge Forces
CAS, 25 May 2005
What do we mean with “space charge”?
It is the net effect of the Coulomb interactions in a multi-particle
system.
Space Charge Regime dominated by the self field produced by the
particle distribution.
Collective Effects
CAS, 25 May 2005
Debye Length λD
r C
Φ (r ) =
r
C=
e
4πεo
real
uniform
The particle distribution around a test particle will deviate from the
continuous distribution.
CAS, 25 May 2005
The effective potential of a test charge can be defined as the sum of
the potential of the uniform distribution and a “perturbed” term.
r C − r / λD
Φ p (r ) = e
r
λD =
εo k B T
e2 n
kB= Boltzman constant
T = Temperature
kB T = average kinetic energy of the particles
n = particle density (N/V)
CAS, 25 May 2005
The effective interaction range of the test charge is limited to the
Debye length
Smooth functions for the charge and field distributions can be used
as long as the Debye length remains small compared to the particle
bunch size
λD
CAS, 25 May 2005
Longitudinal Electric field of a uniform charged cylinder
<z>=1.50 [ m]Ez [ V/ m]
200000
As computed by a multiparticle tracking code
100000
1.498
1.499
1.501
z[ m]
-100000
-200000
Analytical expression
CAS, 25 May 2005
Space charge of a relativistic cylindrical distribution
Cylindrical finite bunch, uniformly charged,
With circular cross section
R
L
Transverse electric field at r = a
Er
Ez
5
1.5
1
4
0.5
3
-0.002
2
0.004
0.006
-0.5
1
-0.002
0.002
-1
0.002
0.004
0.006
-1.5
Longitudinal electric field at r = 0
CAS, 25 May 2005
Relativistic Uniform Cylinder
I
ρ= 2
πa v
Gauss’s law
2πrlε o Er = ρπr 2l
Ampere’s law
2lBϕ = µ o Jlr
a
I
J= 2
πa
∫ ε E ⋅ dS = ∫ ρdV
Linear with r
o
Er =
ρr
Ir
=
for r ≤ a
2
2ε o 2πε o a v
Bϕ =
∫ B ⋅ dl = µ ∫ J ⋅ dS
o
Bϕ =
µ o Jr
2
Ir
= µo
for r ≤ a
2
2πa
CAS, 25 May 2005
β
c
Er
λo = ρπa 2 (C / m)
for r < a
λ ( r ) = λo ( r / a ) 2
Er ( r ) =
J = β cρ ( A / m 2 )
λo r
2πε o a 2
β
λo β r
Bθ (r ) =
Er ( r ) =
2πε o c a 2
c
I = Jπa 2 = β cλ 0( A)
The Lorentz Force
(
)
Fr = e(Er − β cBϑ ) = e 1 − β Er =
2
eEr
γ2
eλo r
=
2πε oγ 2 a 2
• has only radial component
• is a linear function of the transverse coordinate
The attractive magnetic force, which becomes significant at high
energy, tends to compensate the repulsive electric force.
CAS, 25 May 2005
Longitudinal Space Charge Forces
In order do derive the relationship between the longitudinal and
transverse forces inside a beam, let us consider the case of
cylindrical symmetry and ultra-relativistic bunches. We know that a
varying magnetic field produces a rotational electric field:
∂
∫ E ⋅ dl = − ∂t ∫ B ⋅ ndS
S
z
z+∆z
We choose as path a
rectangle going through the
beam pipe and the beam,
parallel to the axis.
CAS, 25 May 2005
b
b
r
r
E z (r , z )∆z + ∫ Er (r , z +∆z )dr − E z (b, z )∆z − ∫ Er (r , z )dr =
b
∂
= −∆z ∫ Bθ (r )dr
∂t
r
b
 ∂E (r , z ) ∂Bθ (r , z ) 
dr
−
E z (r , z ) = E z (b, z ) + ∫  r

∂t
 ∂z

r
b
∂
E z (r , z ) = E z (b, z ) + ∫ [Er (r , z ) − vBθ (r , z )]dr
∂z
r
b
E z (r , z ) = E z (b, z ) +
∂
[Er (r , z ) − vBθ (r , z )]dr
∫
∂z
r
CAS, 25 May 2005
E z (r , z ) = E z (b, z ) + (1 − β 2 )
b
∂
Er (r , z )dr
∫
∂z r
where (1-β2)=1/γ2. For perfectly conducting walls Ez=0.
b
1 ∂
E z (r , z ) =
E (r , z )dr
2 ∂z ∫ r
γ
r
Uniform beam in a circular p.c. pipe.
b
r
r2
b ∂λ ( z )
Fz (r , z ) =
(
1
−
+
2
ln
)
2
2
4πε 0γ
a
a ∂z
−e
CAS, 25 May 2005
Beam motion in a linear channel
CAS, 25 May 2005
r.m.s. emittance
εrms =
γx 2 + 2αxx ′ + βx ′2 = εrms
x
x ′ − xx ′
2
2
2
x’
x’rms
x
xrms
x 2 = βεrms and
x ′2 = γεrms
α =−
β′
2
=−
1
2εrms
xx ′
d 2
x =−
dz
εrms
γβ − α 2 = 1
CAS, 25 May 2005
Emittance degradation
Longitudinal correlation along
the bunch induced by e.m
non linear e.m. fields
x’
x’
x
x
RF fields, solenoidal fields, space charge, wake fields
CAS, 25 May 2005
Equation of motion in a drift space:
d 2 r eE r
eI
γm 2 = 2 =
r
2
2
dt
γ
2πγ εo a v
2
d2r
d
r
2 2
=
β
c
dt 2
dz 2
K=
d2r
eI
K
=
r= 2 r
2
3
2 3
dz
2πmγ εo a v
a
eI
2I
=
2πmγ 3εov 3 Ioβ 3 γ 3
4πεo mc 3
Io =
e
Generalized perveance
Alfven current
CAS, 25 May 2005
γ=1
γ=5
γ = 10
L(t)
Rs(t)
∆t
CAS, 25 May 2005
Transport in a Long Solenoid
ks =
qB
2mcβγ
K (ζ ) =
2Ig(ζ )
Io (βγ )
3
0.9
0.8
g(ζ) g 0.7
0.6
0.5
0
0.0005
0.001
0.0015
metri
0.002
0.0025
ζ
K (ζ )
R′′ + k R =
R
2
s
R′′ = 0
==> Equilibrium solution ==> Req (ζ ) =
K (ζ )
ks
CAS, 25 May 2005
Small perturbations around the equilibrium solution
CAS, 25 May 2005
Emittance Oscillations are driven by space charge differential
defocusing in core and tails of the beam
px
Projected Phase Space
x
Slice Phase
Space
CAS, 25 May 2005
Envelope oscillations drive Emittance oscillations
2
1.5
R(ζ)
envelopes
1
0.5
0
δγ
=0
γ
-0.5
0
1
2
3
4
5
metri
80
60
ε(z)emi 40
qB
ks =
2mcβγ
20
0
0
1
2
3
4
5
metri
ε(z) =
R
2
R′2 − RR′
2
÷ sin
( 2k z)
s
CAS, 25 May 2005
BEAM DYNAMICS MODELING
Space charge
On Axis
∆t
RF Field + Solenoids
Off Axis
∆t
CAS, 25 May 2005
On axis
Envelope equation (linear fields)
Space charge
Solenoid field
RF field
 4ε
e
&x& + βγ 2 β&x + (k rf + k sol ) 2 x = 3 E xsc (ξ s , Axs , x) + 
γ m
 γ
th
n
 4ε
e
&y& + βγ 2 β&y + (k rf + k sol ) 2 y = 3 E xsc (ξ s , Axs , y ) + 
γ m
 γ
2
 1
 3
 x
th
n
2
 1
 3
 y
CAS, 25 May 2005
CODES used for simulations of Space Charge Effects
PARMELA, ASTRA
Multi-particle tracking code, includes space charge but not wake fields
HOMDYN
Relies on a multi-envelope model based on the time dependent evolution
of a uniform bunch
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Slice analysis through the bunch
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Space charge with image currents
CAS, 25 May 2005
Effects of conducting or magnetic screens
Let us consider a point charge q close to a conducting screen.
The electrostatic field can be derived through the "image method".
Since the metallic screen is an equi-potential plane, it can be removed
provided that a "virtual" charge is introduced such that the potential
is constant on the screen
q
q
-q
CAS, 25 May 2005
A constant current in the free space produces circular magnetic field.
If µr≈1, the material, even in the case of a good conductor, does not
affect the field lines.
I
CAS, 25 May 2005
For ferromagnetic type, with µr>>1, the very high magnetic
permeability makes the tangential magnetic field zero at the
boundary so that the magnetic field is perpendicular to the surface,
just like the electric field lines close to a conductor.
I
I
I
In analogy with the image method we get the magnetic field, in the
region outside the material, as superposition of the fields due to two
symmetric equal currents flowing in the same direction.
CAS, 25 May 2005
CAS, 25 May 2005
Time-varying fields
Static electric fields vanish inside a conductor for any finite
conductivity, while magnetic fields pass through unless of an high
permeability.
This is no longer true for time changing fields, which can penetrate
inside the material only in a region δw called skin depth. Inside the
conducting material we write the following Maxwell equations:
∂B

B = µH
∇
×
=
−
E


∂t
 D = εE

∂D

 ∇×H = J +
 J = σE
∂t

Copper σ = 5.8 107 (Ωm)-1
Aluminium σ = 3.5 107 (Ωm)-1
Stainless steel σ = 1.4 106 (Ωm)-1 .
CAS, 25 May 2005
Consider a plane wave (Hz,Ey) propagating in the material
∂2Ey
∂x
2
− εµ
∂2Ey
∂t
2
− σµ
∂E y
∂t
=0
( the same equation holds for Hz). Assuming that fields propagate
in the x-direction with the law:
Ey
i
ω
t
−
γ
x
H z = H˜ oe
Ey = E˜ oe iωt − γx
2
2
iωt− γx
(γ + εµω − i ωµσ )E˜o e
=0
x
Hz
∆w
We say that the material behaves like a conductor if σ>>ωε thus:
γ ≅ (1 + i )
σµε
2
CAS, 25 May 2005
Fields propagating along “x” are attenuated.
The attenuation constant measured in meters is called skin depth δw:
δw
δw ≅
1
2
=
ℜ(γ )
ωσµ
∆w
The skin depth depends on the material properties and the frequency.
Fields pass through the conductor wall if the skin depth is larger than
the wall thickness ∆w. This happens at relatively low frequency.
At higher frequency, for a good conductor δw<< ∆w and both
electric and magnetic fields vanish inside the wall.
For the copper
δw ≅
6.66
(cm)
f
For a pipe 2mm thick, the fields pass through the wall up to 1 kHz.
(Skin depth of Aluminium is larger by a factor 1.27)
CAS, 25 May 2005
• compare the wall thickness and the skin depth (region of
penetration of the e.m. fields) in the conductor.
• If the fields penetrate and pass through the material, they can
interact with bodies in the outer region.
• If the skin depth is very small, fields do not penetrate, the
electric filed lines are perpendicular to the wall, as in the static
case, while the magnetic field line are tangent to the surface.
I
I
-I
-I
CAS, 25 May 2005
Circular Perfectly Conducting Pipe
(Beam at Center)
In the case of charge distribution, and
γ→∞, the electric field lines are
perpendicular to the direction of motion.
The transverse fields intensity can be
computed like in the static case,
applying the Gauss and Ampere laws.
2
 r
λ(r) = λo  ;
 a
∫
S
E r (2πr)∆z =
λ (r)∆z
εo
λ(r)∆z
β
; Bθ = E r
2πεo r
c
λ r
λ oβ r
E r (r) = o
;
B
(r
)
=
θ
2π εo a 2
2πεoc a 2
Er =
F⊥ (r) = e( E r − βc Bθ ) =
e
γ
2
Er
CAS, 25 May 2005
• Due to the symmetry, the transverse fields produced by an ultrarelativistic charge inside the pipe are the same as in the free space.
• For a distribution with cylindrical symmetry, in the ultrarelativistic regime, there is a cancellation of the electric and
magnetic forces.
• The uniform beam produces exactly the same forces as in the
free space.
• This result does not depend on the longitudinal distribution of the
beam. In general one has to consider the local charge density λ(z)
CAS, 25 May 2005
Parallel Plates (Beam at Center)
q
-q
q
-q
q
2h
In some cases, the beam pipe cross
section is such that we can consider
only the surfaces closer to the beam,
which behave like two parallel
plates. In this case, we use the image
method to a charge distribution of
radius a between two conducting
plates 2h apart. By applying the
superposition principle we get the
total image field at a position y
inside the beam.
CAS, 25 May 2005
λ(z )
E (z ,y) =
2π ε o
im
y
∞
∑
λ (z ) π 2
(−1)
≅
y
2
2
2
(2nh) − y 4 π ε o h 12
n=1
n
−2y
∞
 1
λ(z )
1 
im
n
Ey (z ,y) =
(−1) 
−

2π ε o
 2nh + y 2nh − y 
n=1
∑
y
Where we have assumed h>>a>y.
For d.c. or slowly varying currents, the boundary condition imposed
by the conducting plates does not affect the magnetic field.
CAS, 25 May 2005
From the divergence equation we derive also the other transverse
component:
∂ im
∂ im
− λ(z ) π 2
im
Ex = − Ey ⇒ Ex (z,x) =
x
2
∂x
∂y
4 π ε o h 12
Including also the direct space charge force, we get:
eλ(z )x  1
π2 
Fx (z ,x) =
 2 2−
2
π ε o  2a γ
48h 
eλ(z )y  1
π2 
Fy (z ,x) =
 2 2+
2
π ε o  2a γ
48h 
There is no cancellation of the electric and magnetic forces due to
the "image" charges.
CAS, 25 May 2005
Parallel Plates (Beam at Center) a.c. currents
Usually, the frequency beam spectrum is quite rich of harmonics,
especially for bunched beams.
It is convenient to decompose the current into a d.c. component, I,
for which δw>>∆w, and an a.c. component, Î, for which δw<< ∆w.
While the d.c. component of the magnetic does not perceives the
presence of the material, its a.c. component is obliged to be
tangent at the wall. For a charge density λ we have I=λv.
We can see that this current produces a magnetic field able to
cancel the effect of the electrostatic force.
CAS, 25 May 2005
˜ (z)y π 2
λ
;
E˜ y (z, x) =
π ε o 48h2
I
-I
I
-I
I
β
B˜ x (z,x) = E˜ y (z,x)
c
2
˜
˜Fy (z,x) = e λ (z)y π
π ε oγ 2 48h2
2 
˜ ( z) x  1
π
λ
e


F˜x (z, x ) =
2
2 2 −
24 h 
2π ε oγ  a
2 
˜ ( z)y  1
π
λ
e


F˜y ( z, x ) =
2
2 2 +
24h 
2π ε oγ  a
There is cancellation of the electric and magnetic forces !!
CAS, 25 May 2005
Parallel Plates - General expression of the force
Taking into account all the boundary conditions for d.c. and a.c.
currents, we can write the expression of the force as:
2
2
1  1

 
π2 
π
π
2
λ m β 
λ u
Fu =
+
 2  2 m
2 
2
2  
2π ε o  γ  a
24h 
 24h 12 g  
u = x, y
e
where λ is the total current, and λ its d.c. part. We take the sign (+) if
u=y, and the sign (–) if u=x.
CAS, 25 May 2005
Space charge effects in storage rings
CAS, 25 May 2005
Self Fields and betatron motion
Consider a perfectly circular accelerator with radius ρx. The beam
circulates inside the beam pipe. The transverse single particle
motion in the linear regime, is derived from the equation of
motion. Including the self field forces in the motion equation, we
have
O
y
ρ
x
z
x
d (mγ v )
r
r
ext
self
= F (r ) + F (r )
dt
dv
=
dt
r
r
self
F (r ) + F (r )
ext
mγ
CAS, 25 May 2005
For the single particle "transverse dynamics" we write:
r
r = (ρ x+ x )eˆx + yeˆ y
r
v = x&eˆx + y& eˆ y + ωo (ρ x + x )eˆz
r
a = &x& − ωo2 (ρ x+ x ) eˆx + &y&eˆ y + [ω& o (ρ x + x ) + 2ωo x& ]eˆz
[
]
For the motion along x:
&x& − ωo2 (ρ x + x ) =
(
1
Fxext + Fxself
mγ
)
Which, with respect to the azimuthal position s=vzt becomes:
&x& = v z2 x′′ = ωo2 (ρ x + x )2 x′′
x′′ −
(
1
1
Fxext + Fxself
=
ρ x + x mv z2γ
)
CAS, 25 May 2005
We assume small transverse displacements x with respect to the
closed orbit, and only dipoles for bending and quadrupole to keep
the beam around the closed orbit:
ext
x
F
≈F
ext
o
 ∂Fxext
+ 
 ∂x

 x
 x =0
x <<ρx
Around the closed orbit, putting vz= βc, we get
 1
 ∂Fxext
1

−
x′′ + 
2
2
 ρ x β Eo  ∂x




 x = 1 Fxself ( x)
2


Eo
β
 x =0 
where Eo is the particle energy. This equation expressed as function of
“s” reads:
 1

1
x ′′(s)+  2 + K x (s) x(s) = 2 Fxself (x,s)
β Eo
 ρ x (s)

CAS, 25 May 2005
•In the analysis of the motion of the particles in presence of the
self field, we will adopt a simplified model where particles
execute simple harmonic oscillations around the reference orbit.
•This is the case where the focussing term is constant. Although
this condition in never fulfilled in a real accelerator, it provides a
reliable model for the description of the beam instabilities
1
x ′′(s)+ K x x(s) = 2 Fxself (x)
β Eo
 Q x 2
1
self
x ′′( s) + 
x(
s)
=
F
( x , s)

x
2
β Eo
 ρx 
x(s) = Ax cos
[ K s− ϕ ]
x
x
ax βx = Ax ⇒ βx const.
K x (s)βx2 = 1
1
1
βx = ' =
µx
Kx
µ x (s) = K x s
ω
1
Qx = x =
ω o 2π
L
∫
o
 Q 2
ds′
= ρx K x ⇒ K x =  x 
β( s′)
 ρx 
CAS, 25 May 2005
Transverse Incoherent Effects
We take the linear term of the transverse force in the betatron
equation:
 ∂Fxs.c. 
s.c.

Fx ( x, z ) ≅ 


 ∂x  x = 0
x
2
 Qx 
1  ∂Fxs.c. 
 x =
x′′ + 
x
2


β Eo  ∂x  x = 0
 ρx 
(Qx + ∆Qx )
2
≅ Qx2
 ∂ Fxs.c. 


+ 2Qx ∆Qx ⇒ ∆Qx = −
2 β 2 EoQx  ∂x 
ρ x2
The betatron shift is negative since the space charge forces are
defocusing on both planes. Notice that the tune shift is in general
function of “z”, therefore there is a tune spread inside the beam.
CAS, 25 May 2005
Consequences of the space charge tune shifts
In circular accelerators the values of the betatron tunes should not be
close to rational numbers in order to avoid the crossing of linear and
non-linear resonances where the beam becomes unstable.
The tune spread induced by the space charge force can make hard to
satisfy this basic requirement. Typically, in order to avoid major
resonances the stability requires
∆Qu < 0.3
CAS, 25 May 2005
Example: Incoherent betatron tune shift for an uniform
electron beam of radius a, length lo, inside circular perfectly
conducting Pipe
 ∂ Fxs.c.  ∂
eλo x
eλo
 =

=
2 2
2 2
∂
x
∂
x
2
πε
o
γ
a
2
πε
o
γ
a


re , p =
e2
4πεomoc 2
ρ x2 Ne 2
∆Qx = −
4πεoa 2 β 2γ 2 EoQxolo
(electrons : 2.82 10 −15 m, protons :1.53 10 −18 m)
ρ x2 Nre, p
∆Qx = − 2 2 3
a β γ Qxolo
For a real bunched beams the space charge forces, and the tune shift
depend on the longitudinal and radial position of the charge.
CAS, 25 May 2005
PS Booster, accelerate proton bunches
From 50 to 800 MeV in about 0.6 s.
The tunes occupied by the particle are
indicated in the diagram by the shaded
area. As time goes on, the energy
increase and the space charge tune
spread gets smaller covering at t=100
ms the tune area shown by the darker
area. The point of highest tune
correspond to the particles which are
least affected by the space charge. This
point moves in the Q diagram since the
external focusing is adjusted such that
the reduced tune spread lies in a region
free of harmful resonances.
Finally, the small dark area shows the situation at t=600 ms when the beam has
Reached the energy of 800 MeV. The tune spread reduction is lower than
expected with the energy increase (1/γ3) dependence since the bunch dimensions
also decrease during the acceleration.
CAS, 25 May 2005
END
CAS, 25 May 2005