ENVELOPE EQUATIONS
 Transverse envelope equations
 1-dimensional (8 Slides)
 2-dimensional => Coupled oscillations of the transverse beam
sizes (8)
 Longitudinal envelope equation
 Longitudinal envelope equation far from transition (4)
 Evolution of the phase space ellipse near transition, without
collective effects (16)
 Longitudinal envelope equation near transition (6)
 Bunch rotation with (or without) SC (and or BB impedance) (7)
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
1 /50
TRANSVERSE ENVELOPE EQUATIONS (1/16)
 Consider a particle in an ensemble of particles which obeys the
single-particle equations
x′ = px
 The total force is
Fx ( x, s )
p′x = 2
β E total
Fx ( x, s ) = Fxext + FxSC
€
 € distribution f ( x , px , s ) . Averaging over
Let’s consider a particle
the particle distribution, we obtain the equations of motion for the
centre of€the beam
< x >′ = < px €
>
< Fx ( x, s ) > < Fxext >
< px >′ =
= 2
2
β E total
β E total
as < FxSC > = 0 , because of Newton’s 3rd law
€
€
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
2 /50
TRANSVERSE ENVELOPE EQUATIONS (2/16)
 For a linear machine, one has

Fxext
= − Kx ( s ) x
2
β E total
< x >′′ + K x ( s ) < x > = 0
€
 The 2nd moments satisfy the equations
< x 2 >′ = 2 < x x′ > = €
2 < x px >
SC
F
< x px >′ = < x ′ px > + < x p′x > = < px2 > − K x ( s ) < x 2 > + < x 2 x
>
β E total
SC
F
< px2 >′ = 2 < px p′x >= − 2 K x ( s ) < x px > + 2 < px 2 x
>
β E total
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
3 /50
TRANSVERSE ENVELOPE EQUATIONS (3/16)
 To study space-charge effects, we are interested in the position and
momentum offsets of the particle from their respective averages, i.e.
Δx = x − < x >

Δpx = px − < px >
< Δx 2 >′ = 2 < Δx Δpx >
€
<Δx Δpx >′ = <Δpx2 > − K x ( s ) <Δx 2 > + <Δx
€
€
€
FxSC
>
2
β E total
SC
F
<Δpx2 >′ = − 2 K x ( s ) <Δx Δpx > + 2 <Δpx 2 x
>
β E total
 Define the rms beam emittance εx,rms =
€
and rms beam size
σx =
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
€
<Δx 2 > <Δpx2 > − <Δx Δpx > 2
<Δx 2 >
4 /50
TRANSVERSE ENVELOPE EQUATIONS (4/16)

2
2
ε
+
<Δx
Δp
>
x
<Δpx2 > = x,rms
<Δx 2 >
σ ′x =
€
σ ′x′ =
€
 €
<Δx Δpx >
<Δx 2 >
<Δx Δpx > 2
−
2 3/2
2
<Δx
>
<Δx >
<Δx Δpx >′
Finally, the transverse envelope equation can be obtained
2
εx,rms
<Δx FxSC >
σ ′x′ + K x ( s ) σ x − 3 −
=0
2
σx
σ x β E total
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
5 /50
TRANSVERSE ENVELOPE EQUATIONS (5/16)
 The SC force was derived in the previous “SC course” (assuming
for instance a uniform transverse distribution and a round beam)
FxSC
eλ
Δx
= 2
2
β E total β E total 2 π ε0 γ 2 a 2
FxSC
Δx
= K sc, x 2 2
2
β E total
a
=>
€
with
€
2 N l rp
K sc, x 2 = 2 3
β γ
Therefore,
€
a = 2σ x
Nb e
λ=
= Nl e
l
€
€
<Δx FxSC > K sc, x 2
=
2
β E total
4
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
6 /50
TRANSVERSE ENVELOPE EQUATIONS (6/16)
 The 1-dimensional envelope equation can finally be written
εx2 K sc, x 2
a′′ + K x ( s ) a − 3 −
=0
a
a
a = 2σ x
 εx = 4 εx,rms
Effect €
of space charge on the equilibrium beam size
smooth approximation
€
€
K x = ( Qx 0 / R )
a0
, in the
2
€
€
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
7 /50
TRANSVERSE ENVELOPE EQUATIONS (7/16)

The equilibrium beam size is therefore found from
2
K sc, x 2 εx2
 Qx 0 
− 3 =0

 a0 −
 R 
a0
a0
εx R
which yields a =
κ + 1+ κ 2
Qx 0
€
2
0
 (
κ=
2 εx Qx 0
The beam size is significantly perturbed by the space-charge force
when κ ≥ 1
€
 )
K sc, x 2 R
€
If the beam size becomes larger than the vacuum chamber aperture,
there will be a beam loss
€
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
8 /50
TRANSVERSE ENVELOPE EQUATIONS (8/16)
 For weak beam intensities, i.e.
κ << 1
2
2
a02 = a00
+ Δ a00
εx R
with a =
Qx 0
2
00
 €
€2
2
Δ a00 = κ a00
€ a00 describes the beam size in the absence of
The parameter
space charge. Interpreting Δ a00 as a perturbation on the singleparticle€
tune according to a02 = εx R / ( Qx 0 + Δ Qx ) , gives an expression
for the shift of the single-particle tune due to space charge
€
€
K€sc, x 2 R  It is ½ of the result found in the “SC
Δ Qx = −
course” with a = 2 σ (as expected)
2 εx
x
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
9 /50
€
TRANSVERSE ENVELOPE EQUATIONS (9/16)
 Let’s come back to the general case, i.e. consider a beam with
unequal transverse beam sizes => The envelope equations are
therefore given by (we saw in the “SC course”, that in the SC force,
a 2 must be replaced by a a + b / 2 )
(
)
2K sc, x 2
€
€
2
x
3
ε
a′′ + K x a −
−
=0
a+ b a
εy2
b′′ + K y b −
− 3 =0
a+ b b
2K sc, x 2
€
€
€
a = 2σ x
εx = 4 εx,rms
b = 2σ y
εy = 4 εy,rms
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
€
10 /50
TRANSVERSE ENVELOPE EQUATIONS (10/16)
 Both transverse planes have thus to be treated jointly for highintensity beams due to space-charge coupling
  The beam may execute some collective motion on top of equilibrium
beam sizes a 0 and b0
Let the horizontal and vertical beam sizes be
€ the perturbations
€
Δ a and
where
respect to the equilibrium sizes
a ( s ) = a0 − Δ a ( s )
b ( s ) = b0 + Δ b ( s )
Δ
€b are considered small with
€
 Linearizing yields€
K a = 4 K x€−
€
Δ a′′ + K a Δ a = K Δ b
€
Δ b′ + K b Δ b = K Δ a
2K sc, x 2 ( 2 a0 + 3b0 )
a0 ( a0 + b0 )
2
,
Kb = 4 Ky −
€
K=
2K sc, x 2
( a0 + b0 )
2
2K sc, x 2 ( 2b0 + 3a0 )
b0 ( a0 + b0 )
2
 The transverse beam sizes execute coupled oscillations
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
11 /50
TRANSVERSE ENVELOPE EQUATIONS (11/16)
 The equilibrium beam sizes a 0 and b0 are found from the following
equations
εx2
K x a0 −
− 3 =0
€ a0 + b0€ a0
2K sc, x 2
εy2
K y b0 −
− 3 =0
a0 + b0 b0
2K sc, x 2
€
 Using the smooth approximation
€
K x = ( Qx 0 / R )
2
K y = ( Qy 0 / R )
2
€shifts, yields
and assuming small tune
€
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
K a = ( Qa / R )
2
K b = ( Qb / R )
2
€
€
12 /50
TRANSVERSE ENVELOPE EQUATIONS (12/16)
Qa = 2Qx 0 + ΔQa = 2Qx 0 −
Qb = 2Qy 0 + ΔQb = 2Qy 0 −
€
 K sc, x 2 R 2 ( 2 a0 + 3b0 )
2Qx 0 a0 ( a0 + b0 )
2
K sc, x 2 R 2 ( 2b0 + 3a0 )
2Qy 0 b0 ( a0 + b0 )
2
The coupled equations can be re-written
€
2
d Δa
2
2
+
Q
Δ
a
=
K
R
Δb
a
2
dφ
€
d 2Δ b
2
2
+ Qb Δ b = K R Δ a
2
dφ
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
€
φ = Ω0 t
13 /50
TRANSVERSE ENVELOPE EQUATIONS (13/16)
 Far from the coupling resonance Qa = Qb , the solutions of the
homogeneous equations (of the coupled oscillations) are given by
Δa = Δa0 e
j Qa φ
€
 In the presence of coupling, the coupled oscillations can be solved
by searching the normal (i.e. decoupled) modes (u,v) linked by a
simple rotation
€
 Δb = Δb0 e
j Qb φ
€ Δa 
 cosα − sin α   u 
 =
 
Δb
sin
α
cos
α
 v 
  
"b
The equations of the 2 normal modes can be found
€
d 2u
2
+
Q
u=0
u
2
dφ
d 2v
2
+
Q
v =0
v
2
dφ
!
"a
!
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
€
€
14 /50
TRANSVERSE ENVELOPE EQUATIONS (14/16)
with (assuming small tune shifts)
Qu = Qa −
€
 C
tan α
2
C
tan ( 2 α ) =
€Δ
C
Qv = Qb +
tan α
2
R2 K
C =
Q0
The solutions for the decoupled modes are
€ u = Ue
Δ = Qb − Qa
Qx 0 ≈ Qy 0 ≈ Q0
j Qu φ
v = V e j Qv φ
€ (U,V) are constants
€ of motion, which
where
depend on the initial
€
conditions 
Δa = Δa0 e
j Qu φ
€
€
j Qv φ
cosα − Δb0 e
sin α
Δb = Δa0 e j Qu φ sin α + Δb0 e j Qv φ cosα
€
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
15 /50
TRANSVERSE ENVELOPE EQUATIONS (15/16)
 2
Δa = Δa Δa *
Therefore,
2
2
2
2
2
2
2
Δa = Δa0 cos2 α + Δb0 sin 2 α − Δa0 Δb0 sin ( 2 α ) cos [ ( Qv − Qu ) φ ]
2
Δb = Δa0 sin α + Δb0
2
€
cos α + Δa0 Δb0 sin ( 2 α ) cos [ ( Qv − Qu ) φ ]
2
€

€
 2
Δa + Δb = Δa0 + Δb0
2
Furthermore, using the fact that
€
2  −1/ 2

 C  
C

cos ( 2 α ) = cos arctan 
  =  1+ 2 
Δ 
 Δ  

2
€

sin 2 α =
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
C /2
2
Δ2 + C + Δ Δ2 + C
2
16 /50
TRANSVERSE ENVELOPE EQUATIONS (16/16)
and, averaging over time (i.e. Φ), it yields (when the resonance
is crossed)
2
2
2
(
2
Δa = Δa0 − Δa0 − Δb0
2
)Δ +
2
C /2
2
C + Δ Δ2 + C
2
2
2
€
2
Δb = Δb0 +
( Δa
2
0
− Δb0
2
)Δ +
2
C /2
2
C + Δ Δ2 + C
2
€
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
17 /50
LONGITUDINAL ENVELOPE EQUATION (1/33)
 Consider as canonical coordinates
ΔE = β 2 E total δ
Δφ = ω RF Δt = h Ω0 Δt
 €
€
€
The single-particle equations to be solved (far away from transition)
are
Δ€
φ˙ = p
 The total force is
€
 Δφ = φ − φ s
h Ω0 η
p= 2
ΔE
β γ E rest
p˙ = Fl
Fl = Flext + FlSC
€
Without SC one has
Fl
ext
2
s0
= − ω Δφ

2
˙
˙
Δφ + ω s0 Δφ = 0
€
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
18 /50
LONGITUDINAL ENVELOPE EQUATION (2/33)
 With SC, one can follow exactly what was done in the transverse
plane, making the following replacements
€
σ€
x =
εx =
< Δx
2
€
Δx

Δφ
px

p
K€
x

ω s02

φ˜ =
<Δx 2 >
> < Δpx2 > − < Δx
€
Δpx€
>2

E0 =
< Δφ 2 >
< Δφ 2 > < p 2 > − < Δφ p> 2
€
 € the longitudinal envelope equation can be obtained
Finally,
SC
€2
E
<Δ
φ
F
>
˙φ˜˙ + ω 2 φ˜ − 0 −
l
=0
s0
3
φ˜
φ˜
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
19 /50
LONGITUDINAL ENVELOPE EQUATION (3/33)
 Using the fact that
(see “SC course”)
K sc, l
Fl
SC
2
s0
= − ηSC ω Δφ
, with
ηSC
3 π N b rp E rest g0 h 2 Sgn ( η )
b
=
g0 = 1 + 2 ln  
2
ˆ
R€γ e VRF cos φ s
 a
K sc, l
=
Δφˆ 3
Δφˆ = π f 0 h τ b
€
and the fact that

€
< Δφ 2 > ˜
=φ
˜
φ
€
For a parabolic
distribution
€
2
s0
2
E
˙φ˜˙ + ω 2 φ˜ − 0 + ω K sc, l = 0
s0
φ˜ 3
Δφˆ 3
Half bunch length
Let’s now express E0 as a function of the usual longitudinal emittance
εl,rms [ eVs ]
€
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
20 /50
LONGITUDINAL ENVELOPE EQUATION (4/33)
Δφ = ω RF t = h Ω0 t

h Ω0 η
p= 2
ΔE
β γ E rest
h 2 Ω20 η εl,rms [ eVs ]
E 0 =€ 2
×
β γ E rest
π

 h 2 Ω20 η
p Δφ = 2
ΔE t
β γ E rest
Now, let’s convert the envelope equation to one for the half
bunch€
length
ˆ . For a parabolic distribution
Δφ
Δφˆ =
5 φ˜
€
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
€
21 /50
LONGITUDINAL ENVELOPE EQUATION (5/33)

2
2
ω
K
25
E
˙ˆ˙
s0
sc,
l
0
Δφ + ω s02 Δφˆ +
−
=0
2
3
ˆ
ˆ
Δφ
Δφ
Envelope equation for the half bunch length far from transition
€
€
 Let’s now derive the evolution of the phase space ellipse near
transition (before deriving the envelope equation near transition)
Synchrotron oscillations equation
d Δφ h η Ω0
= 2
ΔE
dt
β E total
d ΔE e VˆRF Ω0
e VˆRF Ω0
=
sin ( φ s + Δφ ) − sin φ s ] ≈
cos φ s Δφ
[
dt
2π
2π
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
22 /50
LONGITUDINAL ENVELOPE EQUATION (6/33)
d  β 2 E total d Δφ

d t  h η Ω0 d t

with
β
,
E total
,
η
,
Ω0
 e VˆRF Ω0
cos φ s Δφ = 0
−
2π

which depend on time t
€
Approximation:
We neglect the slow time variations of all the
parameters except
€
€
€

€
€
η
E total
d  E total d Δφ  h e VˆRF Ω20 cos φ s
Δφ = 0

−
2
dt η
dt 
2π β
Assumed to be time independent
€
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
23 /50
LONGITUDINAL ENVELOPE EQUATION (7/33)
Furthermore,
above transition

€

€
€
€

γ = γ t + γ˙ t
, with t < 0 below transition and t > 0
1
1 2 γ˙ t
η= 2 − 2 ≈ 3
γt γ
γt
η
2 γ˙ t
≈ 4
E total γ t E rest
and
E total = γ E rest ≈ γ t E rest
€
d  1 d Δφ 
 2
 + Δφ = 0
d t  ω s0 d t 
 e Vˆ h η cos φ  1/ 2
RF
s
ω
=
Ω
−


with
s0
0
2
 2 π β E total 
But here, this should be considered as a definition only, as the
€
beam particle do not make synchrotron
oscillations and therefore
it loses its meaning of oscillation frequency
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
24 /50
LONGITUDINAL ENVELOPE EQUATION (8/33)
t
2
4

β
E
γ
One can write ω = 3 , with
rest t
T
=

Tc
c
 4 π f 2 γ˙ h e Vˆ cos φ

0
RF
s
2
s0
 1/ 3


It is called the nonadiabatic time
€
€
Physical meaning of the nonadiabatic time: When the time is close
enough to transition, the particle will not be able to catch up with the
rapid changing of the bucket shape

One has to solve
d  Tc3 d Δφ 

 + Δφ = 0
dt t dt 
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
25 /50
LONGITUDINAL ENVELOPE EQUATION (9/33)
x
Defining a new time variable,
and considering only t > 0 for
the moment

€
0
x=
t
Tc

 + Δφ€= 0

Δφ = ϕ y 2 / 3
€
€
∫
d  1 d Δφ

dx x dx
One has to solve
Let’s call
y=
2 3/2
u du = x
3

  2 2 
   
2
d ϕ 1 dϕ   3  
+
+ 1− 2 ϕ = 0
2
dy
y dy 
y 




Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
26 /50
LONGITUDINAL ENVELOPE EQUATION (10/33)
The solution of this equation can be written
ϕ = C1 J 2 / 3 ( y ) + C2 N 2 / 3 ( y ) = C3 [ J 2 / 3 ( y ) cos χ + N 2 / 3 ( y ) sin χ ]
Constants to be determined from
the initial conditions
with J2/3 the Bessel function of the 1st kind (or simply the Bessel
function), and N2/3 the Bessel function of the 2nd kind (also called
Weber or Neumann function)

€
Δφ = b x [ J 2 / 3 ( y ) cos χ + N 2 / 3 ( y ) sin χ ]
Constants to be determined from
the initial conditions
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
27 /50
€
LONGITUDINAL ENVELOPE EQUATION (11/33)
ΔE = β 2 E total δ

and
β 2 E total d Δφ
ΔE =
h Ω0 η d t
1
d Δφ
γ t3
d Δφ
δ=
=
h Ω0 η d t
2 γ˙ t h Ω0 d t
€
€
γ t3
 δ=
2 γ˙ t h Ω0
€
Δp
δ=
p0
 Δφ b x 3 / 2
+

 x Tc
Tc
 2 J
 

 2 N2 / 3

2/3
− J 5 / 3  cos χ + 
− N 5 / 3  sin χ  


 3y

 3 y
 
using the fact that
n
J n′ ( y ) = J n ( y ) − J n +1 ( y )
y
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
n
N n′ ( y ) = N n ( y ) − N n +1 ( y )
y
28 /50
LONGITUDINAL ENVELOPE EQUATION (12/33)
Now the idea is, with the 2 equations
Δφ = b x [ J 2 / 3 ( y ) cos χ + N 2 / 3 ( y ) sin χ ]
γ t3
δ=
2€γ˙ t h Ω0
to solve for
 Δφ b x 3 / 2
+

 x Tc
Tc
cos χ
and
cos2 χ + sin 2 χ = 1
€
€
 2 J
 



2
N
2/3
− N 5 / 3  sin χ  
  2 / 3 − J 5 / 3  cos χ + 

 3y

 3 y
 
sin χ
to derive an equation linking
α φφ Δφ 2 + 2 α φδ Δφ δ + αδδ δ 2 = 1
€
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
€
, and then write the equation
Δφ
and δ
€
29 /50
LONGITUDINAL ENVELOPE EQUATION (13/33)
2
 2





J
2
J
2
J
2
J
 2 3/ 3 +  2 / 3 − J 5 / 3  + 32/ /23  2 / 3 − J 5 / 3 

x
 x

 3y

 3y

1


with α φφ =
2 2
2
2
2
det b x   2 N


N
2 N2 / 3  2 N2 / 3
2/3
2/3
− N5 / 3  + 3 + 3 / 2 
− N 5 / 3 
+ 
x
x

 3y

  3y
€
€
€
1
αδδ =
det 2 b 2 x 3
α φδ
 2 γ˙ t h Ω0 Tc  2 2
2
J
+
N


( 2/3 2/3 )
γ t3


 2 J 2 T h Ω γ˙ t


J 2 / 3 Tc  2 J 2 / 3
2/3 c
0
− 2 5/2 3 
− J 5 / 3  2 γ˙ t h Ω0 
−
2
4 3
b x γt
b x γt  3 y


1 
=


2
2


det  2 N 2 / 3 Tc h Ω0 γ˙ t
N 2 / 3 Tc
2 N2 / 3

˙
−
−
−
N
2
γ
t
h
Ω
5/3 
0 
2
4 3
2
5/2 3 

b
x
γ
b
x
γ
3
y



t
t 
 2 N2 / 3

 2 J2 / 3

det = J 2 / 3 
− N5 / 3  − N2 / 3 
− J5 / 3 
 3y

 3y

Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
30 /50
LONGITUDINAL ENVELOPE EQUATION (14/33)
Using the fact that

Jα ( y ) Nα′ ( y ) − Jα′ ( y ) Nα ( y ) =
2
πy
 2 N2 / 3

 2 J2 / 3

2
det = J 2 / 3 
− N5 / 3  − N2 / 3 
− J 5 / 3  = J 2 / 3 N 2′ / 3 − N 2 / 3 J 2′ / 3 =
πy
 3y

 3y

€
€

π2
α φφ = 2 2
9b x
2 2

˙
π x 2 γ h Ω0 Tc
2
2
αδδ =
J
+
N


(
2
/
3
2
/3 )
9 b2 
γ t3

2
€
€

2 
2 
3y
3y
J5 / 3  +  2 N 2 / 3 −
N5 / 3  
  2 J2 / 3 −
 
 
2
2

α φδ
π2
=
9 b2
2
 2 γ˙ h Ω0 Tc2  
3y



3y
N 5 / 3 − 2 N 2 / 3  − J2 / 3  2 J2 / 3 −
J5 / 3  

  N2 / 3 
3
 2



γt
2


Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
31 /50
LONGITUDINAL ENVELOPE EQUATION (15/33)
Let’s now compute the emittance of the tilted ellipse
δ
ψ
€
Δφ
€
€
Method: Let’s find (u , v) where the ellipse is upright
 Δφ   cosψ − sinψ   u 

=
 
 δ   sinψ cosψ   v 
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
32 /50
LONGITUDINAL ENVELOPE EQUATION (16/33)
=> The ellipse is upright if
 2α
1
φδ
ψ = arctan
2
 α φφ − αδδ



It is the tilt angle of the ellipse
€
Using the fact that
cos ( arctan x ) =
1
1+ x2
the equation of the ellipse can be written
€
€
2
1
1
2  2
α φφ + αδδ + ( α φφ − αδδ ) + 4 α φδ  u +  α φφ + αδδ −


2
2
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
sin ( arctan x ) =
(α
φφ
− αδδ
)
2
x
1+ x2
+ 4α
2
φδ
 2
 v = 1
33 /50
LONGITUDINAL ENVELOPE EQUATION (17/33)
The area of the ellipse is thus
2
Α =π
α φφ + αδδ +
(α
φφ
2
− αδδ
)
2

and
+ 4α
A=
2
φδ
α φφ + αδδ −
(α
φφ
− αδδ
)
2
2
+ 4 α φδ
π
α φφ αδδ − α φδ2
h Ω0
A = εl [ eVs ] 2
β E total
€
 3 y N5 / 3


3 y J5 / 3 
3
Using the fact that J 2 / 3 
− 2 N 2 / 3  + N 2 / 3  2 J2 / 3 −
=−



2
2 
π
€
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
34 /50
LONGITUDINAL ENVELOPE EQUATION (18/33)
=> The remaining unknown b
can be found and is given by
b=
2 εl h 2 Ω20 γ˙ Tc2
3 m0 c 2 β 2 γ t4
One now has all the parameters of the ellipse which can be plotted at
any time t => The evolution of the bunch length, the energy spread
and the phase space ellipse€can be studied near transition
Δφ max =
€
αδδ
α φφ αδδ − α
2
φδ
δmax =
α φφ
α φφ αδδ − α φδ2
Properties in an ellipse
(see “Introduction”)
€
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
35 /50
LONGITUDINAL ENVELOPE EQUATION (19/33)
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
36 /50
LONGITUDINAL ENVELOPE EQUATION (20/33)
Case of the CERN PS nTOF bunch without SC
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
37 /50
LONGITUDINAL ENVELOPE EQUATION (21/33)
 Let’s now concentrate on the derivation of the envelope equation
NEAR TRANSITION. Far from transition, the envelope equation was
found to be
2
2
ω
K
25
E
˙˙
s0
sc, l
0
Δφˆ + ω s02 Δφˆ +
−
=0
2
3
Δφˆ
Δφˆ
which can also be written
€
with
€
K sc, l1 S1
1 d2 τb
+ τb + 2 − 3 = 0
2
2
ω s0 d t
τb
τb
3 N b rp E rest g0 Sgn ( η )
K sc, l1 = 3 3 3 = 2
π h f 0 π h f 03 R γ t2 e VˆRF cos φ s
€
K sc, l
16 η 2 εl2
S1 = 2 4 2 2 2
π β γ E rest ω s0
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
εl = 5 εl,rms
38 /50
LONGITUDINAL ENVELOPE EQUATION (22/33)
t
Close to transition x =
Tc
x
ω = 2
Tc
2
s0

t
ω = 3
Tc
2
s0
€
€

K sc, l1 S1
1 d2 τb
+ τb + 2 − 3 = 0
€
2
x dx
τb
τb
Knowing that, close to transition, one has to make the substitution (as
seen in the previous slides)
€
1 d2 τb
x d x2
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009

d  1 d τb 


dx x dx 
39 /50
LONGITUDINAL ENVELOPE EQUATION (23/33)

One has to solve
K sc, l1 S1
d  1 d τb 

 + τb + 2 − 3 = 0
dx x dx 
τb
τb
which can also be written
€
27
36


10
K
10 x S2
d 1 d τ bns
sc, l1
−
=0

 + τ bns +
2
3
dx x dx 
τ bns
τ bns
with
64 εl2 γ˙ 2 Tc4
S2 = 2 4 8 2
π β γ t E rest
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
Full (4σ) bunch length in ns
40 /50
LONGITUDINAL ENVELOPE EQUATION (24/33)
Case of the CERN PS nTOF bunch
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
41 /50
LONGITUDINAL ENVELOPE EQUATION (25/33)
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
42 /50
LONGITUDINAL ENVELOPE EQUATION (26/33)
Full bunch length [ns]
See “Wake fields and impedances”
=> Constant with opposite sign as SC
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
43 /50
€
LONGITUDINAL ENVELOPE EQUATION (27/33)
  Bunch rotation with SC (and or constant inductive impedance)
=> The envelope equation can also be used to estimate, for a given
compression factor, the needed RF voltage, the compression time,
the evolution of the bunch length and the evolution of the momentum
spread
The envelope equation can also be (sometimes) written
K sc, l 2 εl22
d 2 rz
+ K RF rz − 2 − 3 =0
2
dz
rz
rz
K RF
e VˆRF h η cos φ s
=
2 π R 2 γ β 2 m0 c 2
Depends on the
source (was 1 before)
€
Half bunch length (in meters)
3 gˆ rp N b η
K sc, l 2 =
2 γ 3β 2
 rpipe 
1
gˆ = + 2ln 

2€
 rbeam 
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
Δp
εl 2 = η rz 0
= cte
p0
44 /50
LONGITUDINAL ENVELOPE EQUATION (28/33)
  2
The bunch is matched
(at constant energy) when
Before, we used
Δφˆ
€
rz = rz0
and we had
€
€
€
rz
=>
Δφˆ = π h f 0 τ b
Full bunch length [in s]
τb
rz = β c
2
 Here, we use
 It can be checked that this equation is the same as the one we used
before (with the small difference on the g-factor), remembering that
€
Used before
and we have
d rz d rz
=
=0
2
dz
dz
€
π 2
Δp
εl = β γ E rest τ b
2
p0
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
45 /50
LONGITUDINAL ENVELOPE EQUATION (29/33)
 Example of bunch rotation estimated in the past in a 2 GeV protonaccumulator-compressor for a neutrino factory
VˆRF = 7 MV
13
N b = 10 p/b
( Δp / p0 ) max = 2 ( σ p / p0 ) = 1.5 ×10
−3
€
€
gˆ ≈ 4
€
€
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
46 /50
LONGITUDINAL ENVELOPE EQUATION (30/33)
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
47 /50
LONGITUDINAL ENVELOPE EQUATION (31/33)
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
48 /50
LONGITUDINAL ENVELOPE EQUATION (32/33)
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
49 /50
LONGITUDINAL ENVELOPE EQUATION (33/33)
Elias Métral, USPAS2009 course, Albuquerque, USA, June 22-26, 2009
50 /50