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Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 1
Dejan Trbojevic and Vasily Morozov
Non-scaling Fixed Field Alternating Gradient Permanent Magnet Cancer
Therapy Accelerator
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 2
Non-scaling Fixed Field Alternating Gradient Permanent Magnet Cancer Therapy Accelerator
1. Introduction
2. Permanent magnets - Halbach
4. NS-FFAG ring with permanent magnets
5. Matching of the ring to the straight section
3. FFAG straight section solution P. Meads
6. Acceleration
Scaling – Non-Scaling FFAG (NS-FFAG)
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 3
SCALING FFAG The orbits follow non-linear magnetic field: BR~Bo(r/ro)k
Due to restrictions on the particle motion stability in the vertical plane the length of the opposite bending magnet can not be shorter than 2/3 of the positive field bend. This increases the circumference.
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 4
NON-SCALING FFAG
- Orbit offsets are proportional to the dispersion function:
Δx = Dx * δp/p
- To reduce the orbit offsets to ±4 cm range, for momentum range of δp/p ~ ± 50 % the dispersion function Dx has to be of the order of:
Dx ~ 4 cm / 0.5 = 8 cm
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 5
Muon acceleration from 10 to 20 GeV
Triplet NS-FFAG for muon acceleration
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 6
Scaling FFAG – Non scaling FFAG
Scaling FFAG properties: • Zero chromaticity. • Orbits parallel for different δp/p • Relatively large circumference. • Relatively large physical aperture (80 cm – 120 cm). • RF - large aperture • Tunes are fixed for all energies no
integer resonance crossing. • Negative momentum compaction. • B =Bo(r/ro)k non-linear field • Large acceptance • Large magnets • Very large range in Δp/p= ±90% • could be isochronous CW operation
Non-Scaling FFAG properties: • Chromaticity is changing. • Orbits are not parallel. • Relatively small circumference. • Relatively small physical
aperture (0.50 cm – 10 cm). • RF - smaller aperture. • Tunes move 0.4-0.1 in basic cell
resonance crossing for protons • Momentum compaction changes. • B = Bo+x Go linear field • Smaller acceptance • Small magnets • Large range in Δp/p=±60% • Very difficult to be isochronous
B = Bo+r Go B =Bo(r/ro)k
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 7
Permanent magnets – Halbach structure
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 8
Permanent magnet materials
Nd-Fe-B Type Rare Earth Magnets
Electron Energy Corporation - courtesy of : Jinfang Liu and Peter Dent 924 Links Ave. Landisville PA 17538 Br = 1.35 T
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 9
Permanent magnet material
N4561
1.35 T
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 10
Halbach permanent magnets
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 11
Halbach magnet–from his original publication
11
QLD=8.0 cm BL= 4.8 cm QLF=11 cm GF = 2.2 T/0.015 m = 150.0 T/m GD= -2.2 T/0.013 m=-170.0 T/m
3.5 cm
21.0 cm
Bg= Br ln(OD/ID)
Br=1.35 T OD=21.0 cm ID =3.5 cm ln(OD/ID)=1.79
Bg=2.4 T
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 12
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 13
The doublet previous ring with all elements:
r=4.278 m
24 doublets 12 cavities Three kickers
C= 26.88 m -50%<δp/p<+50% Ek_inj =31 MeV Ek_max = 250 MeV
D=8.56 m
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 14
Magnetic Properties:
LBD = 22 cm LBF = 30 cm Gd = -14.3 T/m Gf= 8.73 T/m Bdo= 0.804 T Bfo= 0.563 T
Values of the magnetic fields at the maximum orbit offsets:
Bd max-= 0.804 + (-14.3)⋅(-0.0484) = 1.496 T Bd max+= 0.804 + (-14.2)⋅(0.107) = -0.715 T
Bf max+= 0.563 + 8.73 ⋅ 0.141 = 1.794 T Bf max- = 0.563 + 8.73 ⋅ (-0.102) = -0.327 T
δp/p x0ff(m) 50 0.140638 40 0.111097 30 0.082114 20 0.053819 10 0.026376 0 0.000000 -10 -0.025024 -20 -0.048317 -30 -0.069370 -40 -0.087506 -50 -0.101838
Offsets at F
δp/p x0ff(m) 50 0.107354 40 0.083583 30 0.060737 20 0.039014 10 0.018662 0 0.000000 -10 -0.016560 -20 -0.030484 -30 -0.041077 -40 -0.047447 -50 -0.048481
Offsets at D
Minimum horizontal aperture:
Amin=0.140638+0.101838+6σ ∼ 26 cm
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 15
Arc design NS-FFAG
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 16
Arc design NS-FFAG
By = 2.2342 T Gf = 118 T/m Gd =-125 T/m θ = 0.05236
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 17
Arc design NS-FFAG: Tunes vs. momentum
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 18
Maximum orbit offsets vs. momentum
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 19
βX and βy vs. momentum
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 20
Previous solution for the FFAG straight section P.F. Meads Jr., IEEE Transactions on Nuclear Science, Vol. NS-30, No. 4. (1983) pp. 2448-2450.
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 21
φ fmin
φfο
φdο
xmax
xmin ρdo
ρdmin
ρdmax
ρfmin
ρfmax
ρfo
φdο
φdο
φdο
amin
€
ρdo =pceBD
→ pc =1.62141 MeVc
ρd max =pmaxeBD
→ pmax = 2.43213MeVc
ρd min =pmineBD
→ pmin = 0.81071MeVc
ρ fo =pceBF
ρ f max =pmaxeBF
ρ f min =pmineBD
Input parameters are: xmax and xmin from the arc NS-FFAG pmax, po, and pmin, Dx, βx, βy,
Unknowns: BD , BF , Ffo , Fdo , and lo
To be matched to the input parameters of the linac: βx, βy, αx,αy
Matching arcs to the straight section
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 22
φdmax
ρdmin
ρdmax
ρdo
Matching arcs to the straight section
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 23
Matching Cell @ entrance
pmin
pmax
po
xm
ax x
min
ρfmax
ρfo
ρfmin
Φfmax
Φfo
w
j
umax=amax+lo tan(Φfo-Φfmax)
umin=amin+lo tan(Φfmin-Φfo)
Φfo- Φfmax= Φdo- Φdmax
Φfmin- Φfo= Φdmin- Φdo
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 24
Why “SYNCH” and “PTC” are correct codes? A comparison to the analytical results:
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 25
Initial conditions are defined by: 1. THE INITIAL VECTOR includes:
1.1 Initial orbit offsets (xoff, x’off, yoff, y’off) @ Δp/p =(0.50, 0.40, …,-0.50) 1.2 -Δs longitudinal orbit offsets (Δs=CΔp/p - CΔp/p =0) 1.3 value of theΔp/p
2. Betatron functions βx , αx , βy , αy , Dx , D’x @ Δp/p =0.5, 0.4,…., -0.4, -0.5
Name vector #part xo (mm) xo’(rad) yo (m) yo’(rad) -Δs(m) Δp/p
IC5p PVEC 10 16.467 0.0 0.0 0.0 -0.09077 0.5
IC4p 12.860 0.0 0.0 0.0 -0.06976 0.4
IC3p 9.391 0.0 0.0 0.0 -0.04997 0.3
….. … … … … … …
IC4n -9.135 0.0 0.0 0.0 0.03193 -0.4
IC5n -10.553 0.0 0.0 0.0 0.02972 -0.5
name IBET μxo, μxo βxo, βyo αxo, αyo γxo, γxo Dx , Dy D’x ,D’x
B50p 0.0 0.4661 0.0 0.0 0.0550 0.0
0.0 2.1608 0.0 0.0 0.0 0.0
Matching arcs to the straight section
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 26
Matching arcs to the straight section What are the variables and what are the constraints?
OPI
BDX2 BDX1
QDX2 QFX3
QFX1 OC1
GFX1 GF1 GD8 GDX2 GFX3 OC1P OC2P OC3P OC4P OPIP θBDX1 θBDX2
QD8 QF1
VARIABLES
αX = 0 αY = 0 βX ≈ 1 βY ≈ 1 DX = 0 D’X = 0 x’off = 0 xoff = 0 νX = 0.25 νY = 0.25
@ all Δp/p
OC4
OC3 CONSTRAINTS
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 27
Initial conditions:
αx = 0,αy = 0 D’x= 0, xoff’= 0
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 28
Matching arcs to the straight section At the beginning a total number of constraints for momentum values of: -50%, -40%, -30%, -20%, -10%, +10%, +20%, +30%, +40%,+50%, 10 x 10=100 ! a total number of variables could be: 30 cells x 2 = 60 different gradients +2θ+3 straight section gradients=65!
αX = 0, αY = 0, βX ≈ 1, βY≈ 1, DX= 0, D’X = 0, x’off = 0, xoff= 0, νX = 0.25, νY = 0.25
It happen to be that if one succeeds to match it @dp/p=+50% and @dp/p=-50% It is enough.
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 29
Matching arcs to the straight section
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 30
Matching orbit offsets @ dp/p=50 % to the straight section
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 31
Matching dispersion @ dp/p=50 % to the straight section
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 32
Matching βx and βy @ dp/p=50 % to the straight section
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 33
Matching arcs to the straight section
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 34
Matching arcs to the straight section
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 35
Race track NS-FFAG
4.48 m
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 36
Racetrack dispersion-40%<dp/p<50%
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 37
Racetrack βx and βy -40%<dp/p<50%
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 38
All at once: Fixed field & fixed focusing Magnification 30 TIMES
400 MeV/u
200 MeV/u
2 mm
9 mm
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 39
Reaching the patient with parallel beams 4.
16 m
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 40
Vasily Morozov - Dejan Trbojevic NS-FFAG 10 fixed gradients
KBF1 = 212.7332 T/m
KBD1 = -179.260 T/m
KBF2 = 214.650 T/m
KBD2 = -173.543 T/m
KBF3 = 216.805 T/m
KBD3 = -171.042 T/m
KBF4 = 220.030 T/m
KBD4 = -178.477 T/m
KBD5 = -182.891 T/m
KFTRP1 = 25.5 T/m KDTRP2 = -25.5 T/m KFTRP3 = 25.5 T/m
LBFTRP = 0.20 m LBDTRP = 0.34 m LBFTRP = 0.20 m
BFtr = 1.905 T BDtr = 0.4035 T
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 41
Acceleration PHASE JUMP each turn Acceleration is performed with the phase jump after each turn. The phase jump during acceleration with the fixed frequency by M. Blaskiewicz. - The RF frequency needs to be in a high range, 370 MHz because of
required large number of RF cycles between the passages of bunches in order to achieve higher values of Q and to limit the frequency swing. The total stored energy in the cavity is related to the amplitude VRF of the RF voltage as:
where ωr is the angular resonant frequency, Q is the quality factor, and R is the resistance. The cavity voltage dependence on the klystron voltage (driven by the low level drive)
€
U =VRF2
2ωrRQ
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 42
Requires a loaded quality factor Q=50
Full horizontal aperture 28 cm
Full vertical aperture 3 cm, R/Q = 33 Ohm (circuit)
for β=0.24
ACCELERATION:
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 43
PROTON ACCELERATION 31-250 MeV
The bunch train lls half of the ring at the injection. The changes from injection to the maximum energy of 250 MeV between βinj= 0.251 to βextr.= 0.614. There is 80 ns time to change the cavity frequency when there is no beam. With Q=50 and f=374 MHz the exponential decay time for the eld is 43 ns, where R=Q 33 at =0.25. If the synchronous voltage is 22 kV, a number of turns required for acceleration of protons is:
for twelve cavities (ncav=12). It is very clear that higher effective voltage on the cavities could improve the the resonance crossing problem as well as the patient treatment time. A total power for one RF driver is 100 kW, for twelve cavities this make 1.2 MW and this is a price for the fast acceleration.
€
Nturns =250 − 31( ) MeV[ ]
20 keV /cav[ ]* ncav≈ 912
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 44
Acceleration: 26.88 meter circumference
31 MeV < proton kinetic energy < 250 MeV, 0.24 < β < 0.61
Central rf frequency = 374 MHz
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 45
Summary • Today most of the proton cancer therapy accelerators are cyclotrons or slow extraction synchrotrons.
• An example of racetrack NS-FFAG was described were fast acceleration is assumed with a total number of turns less then 900 as the integer resonance crossing can be a problem.
• To simplify the solution the permanent separated function magnets of the Halbach structure are proposed.
• The orbit offsets in the example presented are within 11mm<Δx<17 mm. This allows use of an aperture of 35 mm. With the outside diameter of 21 cm from the available Nd-Fe-B (for temperatures less than 70 C) materials a bending dipole field of 2.4 T could be obtained.
• Advantage of the accelerator is very small magnets and simplified operation, as the magnets are permanent. Acceleration is assumed to be with a fast phase jump scheme where the voltage on the cavities is changed within one turn of the circulating bunch.
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 46
Why SYNCH and PTC are correct codes? A comparison to the analytical results:
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 47
Why SYNCH and PTC are correct codes? A comparison to the analytical results:
Dejan Trbojevic, September 15, 2011 FFAG 11 workshop – Oxford Trinity College 48
Why SYNCH and PTC are correct codes? A comparison to the analytical results:
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