initial consideration on the low- section of muon linac
DESCRIPTION
Initial consideration on the low- section of muon linac. 4th meeting on muon g-2/EDM experiment Jan. 30, 2009 Masanori Ikegami KEK. Outline. Assumptions Space-charge effect? Two possible options Summary. Assumptions. Muon generation with 40,000 muons per bunch - PowerPoint PPT PresentationTRANSCRIPT
Initial consideration on the low- section of muon linac
4th meeting on muon g-2/EDM experiment
Jan. 30, 2009
Masanori Ikegami
KEK
Outline
• Assumptions
• Space-charge effect?
• Two possible options
• Summary
Assumptions
• Muon generation with – 40,000 muons per bunch– Isotropic momentum spread of 3 keV/c – RMS size of 2.5 mm – RMS pulse length of 3 psec
• Ideal initial acceleration to =0.08– Simply adding the longitudinal momentum keeping
the RMS beam size, RMS pulse length, and the momentum spread.
=0.08 corresponds to 3 MeV for proton.
Kinetic parameters
MUON PROTON
Mass [MeV/c2] 105.7 938.3 0.08 0.08 1.0032 1.0032 0.0803 0.0803p0 [MeV/c] 8.48 75.3Ek [MeV] 0.340 3.02
Transverse distribution
€
r⊥= x2 = y2 = 2.5 mm( )
€
′ x 2 ≡dxds
⎛ ⎝ ⎜
⎞ ⎠ ⎟2
≈px
2
p0
= 0.355 mrad( )
Analogously,
€
′ y 2 ≡dyds
⎛ ⎝ ⎜
⎞ ⎠ ⎟2
≈py
2
p0
= 0.355 mrad( )
assuming
€
px2 = py
2 = pz − p0( )2
= 3 keV /c( )
Longitudinal distribution
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rz = Δz2 = βc Δt 2 = 0.07 mm( )
€
Δt 2 = 3 psec( )assuming
€
Δz ≡ z − z0
of the design particle.
€
z0 ≡ βct0
€
Δ ′ z 2 ≡dΔzds
⎛ ⎝ ⎜
⎞ ⎠ ⎟2
≈pz − p0( )
2
p0
= 0.355 mrad( )
with being the longitudinal position
€
Δt ≡ t − t0 with
€
t0 being the arrival time of the design
particle to a certain position.
Longitudinal distribution (cont.)
€
Δφ2 = 2πf Δt 2 = 0.35 deg( )
being the kinetic energy of the design particle.
€
fcoordinate with
€
Δφ≡2πfΔt
€
ΔE2 ≈ 2E0
pz − p0( )2
p0
= 0.24 keV( )
We here introduce for the longitudinal
We assume
€
f = 324MHz
the fundamental frequency.
We also introduce
€
ΔE = E − E0 with
€
E0 = 0.34MeV
Comparison with J-PARC MEBT
x [mm]
x’ [mrad]
2.5-2.5
0.36
-0.36
0.89 mm•mrad
Δ [deg]
ΔE [keV]
0.35-0.35
0.24
-0.24
0.084 keV•deg 0.025 mm•mrad
x [mm]
x’ [mrad]
1.1-1.1
2.6
-2.6 2.9 mm•mrad
Δ [deg]
ΔE [keV]
5.8-5.8
16
-16 93 keV•deg 3.15 mm•mrad
MUON MUON
J-PARC J-PARC
€
E0 = 0.34MeV
€
E0 = 3MeV€
ε⊥ =
€
ε⊥ =
€
εz =
€
εz =
Envelope equation
€
d2r⊥ds2 = −kβ 0
2r⊥+K3D 1− f α( )( )
2r⊥rz
+ε⊥
2
r⊥3
€
d2rz
ds2 = −ks02rz +
K3D f α( )r⊥
2 +ε z
2
rz3
€
r⊥≡ x2 = y2
€
rz ≡ z2
€
s
€
kβ 0 , ks0
; distance along the beam line
; external focusing strength
Envelope equation for an axial symmetric bunched beam is;
€
ε⊥,ε z ; unnormalized rms emittance
Envelope equation (cont.)
; generalized perveance for a bunched beam
; classical radius of the particle€
K3D ≡3Nrc
β 2γ 3
€
N ; number of particle per bunch
€
rc ≡q2
4πmc2
€
rc =1.53×10−18 m for proton
€
rc =1.36 ×10−17m for muon
Envelope equation (cont.)
with ; bunch frequency,
€
I ; peak current,
€
f
For J-PARC linac 30 mA operation,
€
N = 4 ×104
For muon linac103 difference in
€
K3D
€
N =I
qf
€
q ; elementary electric charge
€
N =30 ×10−3
1.602 ×10−19 × 324 ×106 = 5.8 ×108
€
K3D =3× 5.8 ×108 ×1.53×10−18
0.082 = 4.17 ×10−7 m( )
€
K3D =3× 4 ×104 ×1.36 ×10−17
0.082 = 2.56 ×10−10 m( )
Envelope equation (cont.)
€
f α( )
€
α ≡ r⊥γrz
; aspect ratio in the beam frame
€
f α( ) ; form factor
€
α ≡ r⊥γrz
For α > 1,
€
f α( ) ≡α 2
α 2 −11−
1
α 2 −1tan−1 α 2 −1( )
⎡ ⎣ ⎢
⎤ ⎦ ⎥
α ~ 40 in muon linac@=0.08
Beam envelope for drift (=0.08)
The beam envelope for drift (without external focusing)Obtained by integrating the envelope equationNo accelerationDrift for 12 m or 0.5 sec
€
r⊥ m( )
€
rz m( )
€
s m( )
€
s m( )
Red: w/ space chargeBlue: w/o space charge
Red: w/ space chargeBlue: w/o space charge
Two options
• Option 1: No longitudinal focusing with low frequency s is set to 0 deg (on crest).
– The frequency is chosen to be 324 MHz or lower.
3 ps 0.36 deg @324 MHz 210-5 deviation in energy gain – Optional transverse focusing
• Option 2: With longitudinal focusing matched to the injected beam– The focusing strength is chosen to be matched with the pulse wi
dth and momentum spread to avoid emittance growth from the nonlinear nature of the RF force.
– The transverse focusing is introduced, at least, to cancel the RF defocusing force.
Transverse focusing
• In both options, transverse beam manipulation before and after the accelerating section may be effective to ease the tolerance for the good field region.– The increased transverse momentum spread in the
accelerating section should be addressed.
x
x’
Accelerating section
Quad’s Quad’sbeam
Option 1: No longitudinal focusing with low frequency
Instability
€
Δp / p
Might be feasible with small number of cells.
Num. of cells Num. of cells
€
Δp / p
Non-relativistic approximation.Start with 0.36 deg deviation from the crest.
10-4
10-4
Tentative plan for Option 1
€
E0 = 2.5 MV / m( )
€
φs = 0 deg( )
€
T = 0.8
€
f = 324 MHz( )
Plan 1-A: DTL without DTQ
= 0.08 to 0.7Energy gain: 42 MeVTotal length: 21 mNumber of cells: 58Time of travel: 0.18 s
Comparable to J-PARC DTL
€
Es max ≈ 0.8EKilpatrick =14.3 MV / m( )
It may be possible to increase assuming a short pulse.
€
E0
Tentative plan for Option 1 (cont.)
€
E0 = 5.0 MV / m( )
€
φs = 0 deg( )
€
T = 0.8
€
f = 324 MHz( )
Plan 1-B: Higher gradient DTL without DTQ
= 0.08 to 0.7Energy gain: 42 MeVTotal length: 10.5 mNumber of cells: 29Time of travel: 0.09 s
€
Es max ≈ 1.6EKilpatrick = 28.6 MV / m( )
is increased assuming a short pulse.
€
E0
Issues for Option 1
• Instability– Small number of cells is required.– Effect from abrupt cell variation?
• Small margin for the momentum spread specification– Tolerance for RF phase and amplitude– Transverse field uniformity
• Low shunt impedance at high- side– Unable to introduce frequency jump
ZT2 ~ 8M/m @=0.7 extrapolated from J-PARC SDTL
• Long total length
• RF defocusing from RF errors
Option 2: With longitudinal focusing matched to the injected beam
Matching condition
• We adopt the smooth approximation (or assume continuous focusing in both transverse and longitudinal).
• We find adequate external focusing strength to make the initial beam widths the equilibrium (matched beam widths).
• The external focusing strength can be found by solving the following equations.
€
kβ 02r⊥=
K3D 1− f α( )( )2r⊥rz
+ε⊥
2
r⊥3
€
ks02rz =
K3D f α( )r⊥
2 +ε z
2
rz3
Matching condition (cont.)
€
kβ 02 =
ε⊥2
r⊥4 =
′ x 2
x2Neglecting the space-charge term,
Then,
€
kβ 0 =′ x 2
x2= 0.142 rad / m( )
Similarly,
€
ks0 =′ z 2
z2= 5.07 rad / m( )
€
kβ 0
€
ks0
; Phase advance per meter for betatron oscillation
; Phase advance per meter for synchrotron oscillation
Matching condition (cont.)
€
kβ 0 ∝1
βγ
We here assume acceleration up to = 0.7.
As , we should
satisfy the following condition,€
kβ 0 rad / m( )
€
At = 0.7,
€
kβ 0 = 0.0116 rad / m( )
€
x2 = const€
x2 px2 = const
To keep ,
€
′ x 2 ∝1
βγ
€
px2
Matching condition (cont.)
€
ks0 ∝1
βγ
As ,
we should satisfy the following
condition,
€
ks0 rad / m( )
€
At = 0.7,
€
ks0 = 0.415 rad / m( )€
z2 = const
To keep ,
€
′ z 2 ∝1
βγ
€
pz − p0( )2
or
€
Δt 2 ∝1β
€
z2 pz − p0( )2
= const
Longitudinal focusing force
€
ks0 =2πqE0Tsin −φs( )
β 3γ 3λmc2
; Average accelerating field
€
E0
; Synchronize phase
€
φs
; RF wave length
€
λ
; Rest mass
€
mc2
; Transit time factor
€
T
RF defocusing
€
kRFD2 = −
12
ks02
; Focusing strength from quadrupole magnets
€
kquad
; Strength of RF defocusing
€
kRFD
€
kβ 02 = kquad
2 + kRFD2
We need to introduce transverse focusing force to cancel the RF defocusing.
Case study 1
€
ks0 =11.14 rad / m( )
€
E0 = 2.5 MV / m( )
€
φs = −30 deg( )
€
λ =0.926 m( )
€
mc2 =105.7 MeV( )€
T = 0.8
or
€
f = 324 MHz( )
Comparable to J-PARC DTL
Conventional DTL can provide sufficient longitudinal focusing at low- side, but not at high- side.
€
ks0 = 0.261 rad / m( )
for = 0.08
for = 0.7
Case study 1 (cont.)
€
ks0 rad / m( )
€
Red: RequirementBlue: Conventional DTL
€
€
Ratio
€
E0 MV / m( )Required
€
Es max ≈ 0.8EKilpatrick =14.3 MV / m( )
It may be possible to increaseassuming a short pulse.
€
E0
Case study 1 (cont.)
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φs requireddeg( )
€
Another possible way to vary is to adjust .
€
ks0
€
cos φs required( )
€
€
φs
Tentative plan for Option 2
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E0 = 0.5 − 6.3 MV / m( )
€
φs ≈ −30 deg( )
€
T = 0.8
€
f = 324 MHz( )
Plan 2-A: 324 MHz E0-ramped DTL
The longitudinal focusing is adjusted by ramping .
Transverse focusing can be provided with DTQ’s.
may be too high for high- side.
= 0.08 to 0.7Energy gain: 42 MeVTotal length: ~14 mNumber of cells: ~40Time of travel: 0.12 s
€
E0
€
E0
Case study 2
€
ks0 =18.11 rad / m( )
€
E0 = 3.3 MV / m( )
€
φs = −30 deg( )
€
λ =0.463 m( )
€
mc2 =105.7 MeV( )€
T = 0.8
or
€
f = 648 MHz( )
Higher frequency option
648 MHz DTL can provide sufficient longitudinal focusing at high- side with reasonable .
€
ks0 = 0.424 rad / m( )
for = 0.08
for = 0.7
€
E0
Case study 2 (cont.)
€
ks0 rad / m( )
€
Red: RequirementBlue: Conventional DTL
€
€
Ratio
€
E0 MV / m( )Required
€
Es max ≈ 0.8EKilpatrick =19.0 MV / m( )
It may be possible to increaseassuming a short pulse.
€
E0
Case study 2 (cont.)
€
φs requireddeg( )
€
Another possible way to vary is to adjust .
€
ks0
€
cos φs required( )
€
€
φs
Tentative plan for Option 2 (cont.)
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E0 = 3.3 MV / m( )
€
φs ≈ −2 → −30 deg( )
€
T = 0.8
€
f = 648 MHz( )
Plan 2-B: 648 MHz s-ramped DTL
It is disadvantageous to ramp E0 to obtain required ks0 with higher frequency.
It may be possible to adopt higher gradient to shorten total length.
= 0.08 to 0.7Energy gain: 42 MeVTotal length: ~17 mNumber of cells: ~94Time of travel: 0.14 s
Issues for Option 2
• Feasibility of precise s- or E0- ramping
• Susceptible to the longitudinal parameter of the injected beam
• Nonlinearity of the RF force in s-ramping scheme
• Low shunt impedance at high- side– Frequency jump?
• Long total length
• Transverse focusing to cancel RF defocusing – DTQ or external quad?
Other possibilities
• To match the longitudinal focusing at the injection, but gradually introduce mismatch in the downstream portion without s- or E0- ramping– Does the beam distribution adiabatically change?
• To manipulate the longitudinal beam parameter with buncher and debuncher before and after the accelerating section.
z
z’
Accelerating section
buncher debuncherbeam
Summary
• Low- section for muon linac are considered assuming ideal acceleration up to =0.08.
• Space-charge effects seem to be negligible.
• Two options are proposed:Option 1
No longitudinal acceleration
Option 2
With longitudinal acceleration matched to the injected beam