- Email: [email protected]
Degraders and ionization cooling
Nuclear Physics B (Proc. Suppl.) 149 (2005) 289–294 www.elsevierphysics.com
Degraders and ionization cooling∗ Francis J. M. Farley Yale University, New Haven, Connecticut 06520
†
Procedures are given for minimizing the increase in phase space when a beam is degraded, with and without an axial magnetic field and in a lithium lens. Optimum focusing of the beam onto the degrader is recommended. The transverse diffusion of the beam is as important as the spread in angle, so high density, low Z materials such as diamond and boron carbide are the best. For chaotic beams, focusing by increasing axial fields is suitable and the corresponding acceptance is calculated. Ionization cooling of muon beams by various materials is compared for typical cases.
In ionization cooling we degrade the beam, reducing the forward and transverse momenta together, then accelerate to restore the forward momentum; so the angles in the beam are reduced. But multiple scattering in the degrader increases the transverse angles. We need to degrade with minimum increase in transverse phase space. I first discuss how to do this for a monoenergetic beam and then consider a beam with large ∆p/p which cannot be brought to a unique focus. For more detail see my paper ”Optimum strategy for degraders and ionization cooling” [1] With a perfect pencil beam going in, the output phase space for a degrader of length t is an inclined ellipse. There is an angle-position correlation with σθ ∝ t1/2 and σx ∝ t3/2 . At the output the emittance ∝ t2 . The beam appears to diverge from a point near the centre of the degrader and we can refer back to the virtual source at this point which is an upright ellipse. To calculate the ellipse we can use the well known transport formalism [2] . A particle, represented by the vector X, is transported by the matrix R
X2
x 2 R11 R12 x1 = = = R X1 2 R21 R22 1
∗ Combines
talks to WG3 and WG4 plus discussions for correspondence: 8 Chemin de Saint Pierre, 06620 le Bar sur Loup, France [email protected]
† Address
0920-5632/$ – see front matter © 2005 Published by Elsevier B.V. doi:10.1016/j.nuclphysbps.2005.05.050
The ellipse in phase space, which for example can define the one standard deviation contour, is represented by the symmetric sigma matrix σ using the equation X T σ −1 X = 1
(1)
which written out reads 2 (2) x2 σ22 − 2xθσ12 + θ2 σ11 = ∆ = σ11 σ22 − σ12
The variances in position and angle are σx2 = σ11 and σθ2 = σ22 . ∆ is the determinant of the sigma matrix √ and the area of the ellipse in phase space is π ∆. It will be convenient to √ omit the factor π and refer to = σx .σθ = ∆ as the emittance of the beam. The rule for transporting the sigma matrix is
σ2 = Rσ1 RT
(3)
and applying this for a simple drift space of length t gives 1 0 1 t σ11 σ12 (4) σ2 = σ12 σ22 t 1 0 1 σ11 + 2tσ12 + t2 σ22 σ12 + tσ22 (5) = σ12 + tσ22 σ22 For a small distance dt this reduces to 2σ12 σ22 dt dσ = σ22 0
(6)
and one can integrate term by term to get back to the previous equation. One sees that σ22 is independent of the others (zero if there is no scattering), so this can be integrated first. Then one can
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integrate for σ12 which depends on σ22 . Finally one can integrate for σ11 . If there is scattering σ22 changes and the differential becomes 2σ12 σ22 dσ = dt (7) σ22 K(p) with the variance in θ being σ22 = K(p) = 200/X0 p2 β 2
(8)
where X0 is the radiation length in the degrader material. Integrating this term by term, as described above, one can split the resulting matrix into two parts σ out = σ beam + σ degrad
(9)
in which σ beam is the matrix the beam would have at the end of the degrader if there had been no scattering nor loss of energy, 0 0 0 0 0 + t2 σ22 σ12 + tσ22 σ11 + 2tσ12 σ beam = (10) 0 0 0 σ22 σ12 + tσ22 and σ degrad is a matrix characterizing the degrader, which is independent of the input beam, A B degrad σ = (11) B C
the input beam with the degrader characteristic. Equation (9) shows that the final sigma matrix is the sum of the two component sigma matrices. This is the convolution theorem. It can be proved for any two sigma matrices that represent Gaussian beam distributions; it may be true more generally. The convolution of two beams is obtained by adding the corresponding sigma matrices. This is rather useful. It is much easier to add two matrices than to carry out the convolution integral in four dimensions. It applies in any situation, for example if the degrader is in a magnetic field, or lithium lens, etc. Minimum output emittance To minimize the output emittance for a given degrader, one must adjust the input beam shape in eqn (9) so that the determinant of σ out is a minimum. det(σ out )
= 2beam + (AC − B 2 ) beam beam beam + Aσ22 − 2Bσ12 ) +(Cσ11
The first two terms in this equation are invariant but one can change the components of σ beam (keeping the emitance fixed) to minimize the third term in brackets. One then finds 2out = 2beam + 2degrad + 2beam degrad
(15)
that is in which the three terms are given by t K(p)ds C(t) = 0 t C(s)ds B(t) = 0 t B(s)ds A(t) = 2
out = beam + degrad (12) (13) (14)
0
The degrader matrix is the sigma matrix of the beam coming out of the degrader when the input is a perfect pencil beam. This characterizes the degrader and the determinant gives the emittance degrad called the degrader emittance. Convolution theorem A more general input beam can be split up conceptually into a bundle of pencil beams. The output in this case is the convolution integral of
(16)
In the optimum case the degrader adds its characteristic emittance to the beam: one can do worse, but one cannot do better. The optimum input beam is beam A B beam (17) σ = degrad B C
This defines the input beam as it should be at the end of the degrader, if the degrader was not there. It should match the shape of a pencil beam coming out of the degrader. One can transpose back a distance −d to get an upright ellipse. One finds that the beam should be focused to an image point distance d = B/C before the end of the degrader; for thin degraders this corresponds to the centre of the slab.
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Table 1 Degrader materials Material
diamond boron carbide beryllium graphite water lithium liquid H2
Density g/cm3 3.510 2.520 1.848 2.265 1.000 0.534 0.071
W = dE/dx M eV /cm 6.125 4.253 2.946 3.952 1.991 0.875 0.286
X0 cm 12 20 35 19 36 155 865
W 2 X0 M eV 2 /cm 450 361 306 294 143 119 71
degrad mm.mR 123 154 182 189 389 467 782
The final column gives degrad for 300 M eV muons losing 50 M eV with optimum focusing.
For thin degraders the degrader emittance is √ (18) degrad = Kt2 / 12
with K given by eqn (8). For a given loss of momentum the thickness t is proportional to 1/W where W = dp/dx, so degrad ∝ 1/W 2 X0 . The best material will be the one with the highest value of W 2 X0 . This is true for any particle at any energy for any decrement in momentum. Because the lateral spreading of the beam contributes to degrad , high specific energy loss is favoured above high radiation length; this implies that high density, low Z materials are the best. Various useful materials are compared in Table 1. The ratio of image size to convergence angle for the optimum beam is √ degrad beam β⊥ (19) = σx /σθ = β⊥ = t/ 12
Choice of material One sees from Table 1 that the best material is diamond (not usually available in sufficient sizes) followed by boron carbide, graphite and beryllium. In comparison liquid hydrogen appears unattractive. The final colum shows the degrader emittances for 300 M eV muons losing 50 M eV with optimum focusing.
Axial magnetic field One might do better by putting the degrader in an axial magnetic field B to wrap up the transverse momenta. Putting k = eB/2βγm0 c2 one can solve the transport equations for this case [1]. At every turn in the field the degrader ellipse becomes vertical K t/k 2 0 (20) σ degrad = 0 t 2
with the degrader emittance degrad = Kt2 /2nπ for n spiral turns. Comparing with eqn (18) the improvement is a factor 1.814n. But with no field, liquid hydrogen is 5.085 times worse than boron carbide (see Table 1), so it requires 2.8 spiral turns in the field for it to become better. In a field of 7T the spiral wavelength for 315 M eV /c muons is 94 cm, so we would need 2.6 m of hydrogen to improve on a simple slab of boron carbide. Lithium lens In a lithium lens of radius a with surface field B0 , similar formulae apply [1] with k = eB0 /βγm0 c2 a and the improvement factor over zero field for n turns is 3.628n. With radius a = 2 cm, and current 500 kA, the surface field is 5 T and the spiral wavelength would be 41 cm.
Ionization cooling Using unnormalized emittances, the change in emittance due to degrading and re-acceleration
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Table 2 Cooling fraction with liquid hydrogen and boron carbide Material thickness β⊥ in = 10 mm.R cm cm % B4 C 21.1 6 (opt) 32.0 ” ” 10 31.8 ” ” 20 31.0 ” ” 30 30.1 ” ” 90 24.9 LH2 , no field 315 90 26.2 LH2 , 7 T 315 30 39.1
in = 3 mm.R % 29.7 29.0 26.5 23.8 8.1 1.2 29.8
Percent cooling for 300 M eV /c muons with 90 M eV energy loss and two initial emittances in
back to the original energy is δa = δp/p = −W t/p
(21)
whereas the increase in emittance due to the scattering in the degrader with optimum matching is √ (22) δdegrad = Kt2 / 12
if the degrader is too thick there is no cooling. The optimum thickness is such that the cooling effect is twice the scattering effect √ 3W (23) topt = Kp
In liquid hydrogen in a field of 7 T , the equilibrium emittance is 386 mm.mR, but one recalls that this is not really effective before several turns in the field, the spiral wavelength being 94 cm. Hydrogen gives the smallest equilibrium emittance in the field, better than lithium (701), beryllium (926) and boron carbide (1121), all figures in units of mm.mR in a field of 7 T . For this reason liquid hydrogen has been favoured for detailed study; but boron carbide in zero field might be equally effective. The equilibrium emittance in a lithium lens [1] with the parameters cited above is rather better at 152 mm.mR.
and the net cooling is
√ −δ/ = W topt /2p = 3W 2 X0 /400
(24)
We see that the fractional cooling decreases as the beam is cooled, but never goes to zero. The optimum material is the one with the highest W 2 X0 , see Table 1. In an axial magnetic field, eqn (20) shows that the increase of emittance due to scattering is proportional to t rather than t2 . So in this case there is no optimum thickness. But if the emittance is smaller than the equilibrium value, there is no cooling: equilib =
β2W X
66.8 200c (25) = 2 β W X0 B 0 (eB/m0 c)m0 c
where B is in Tesla and the other symbols have the units of Table 1.
Influence of beam focusing To realize the optimum performance with boron carbide requires the beam to be concentrated onto a small spot with large angles, the transverse beta values being given by eqn (19). But the initial muon beams have a spread in momentum and arise from pi-mu decay at various momenta and various places along the beam transport system. Therefore they cannot be brought to a fine focus. For an energy loss of 90 M eV the degrader length for boron carbide is 21.1 cm, so the ideal input beam has β⊥ = 6.1 cm. In contrast a liquid hydrogen degrader would be 315 cm long, requiring β⊥ = 91cm, which could be much easier to achieve. Does this mean that hydrogen is better? It is instructive to compare hydrogen and boron carbide for the
F.J.M. Farley / Nuclear Physics B (Proc. Suppl.) 149 (2005) 289–294
same values of β⊥ . We can calculate the final emittance with boron carbide for any initial beam focused to a waist at the centre of the degrader by using the convolution theorem, which means adding in quadrature the corresponding values of σx and σθ . As an example we compare in table 2 the overall cooling of one stage for an input momentum of 300 M eV /c with energy loss 90 M eV for various values of β⊥ and two initial emittances. Hydrogen in a magnetic field of 7 T requires β⊥ = 30 cm, without field β⊥ = 90 cm. Initial emittances are un-normalized. It is seen that boron carbide is competitive in most cases and the value of β⊥ is not critical. As a refractory material requiring no cooling, it is much easier to use. Graphite and beryllium will give almost the same performance.
293
be brought to a precise focus. They can nevertheless be squeezed to small values of β⊥ by means of a solenoid with gradually increasing magnetic field. Using the (well-known) equations for a particle of momentum p spiraling at angle φ to an axial field B, we compute the available phase space. The particle wraps itself round a cylinder of radius r (see fig. 1) with p sin φ = QBr
(26)
where Q is the constant of proportionality. If B is increasing with axial distance z the radial field Br at radius r is obtained using Gauss’ theorem Br = −
r dB 2 dz
(27)
In distance ds = dz/ cos φ along the orbit this radial field changes the spiral angle φ by dφ =
sin φ dB sin φ Br .dz QBr ds = =− cos φ 2B cos φ Br p
(28)
so
dB d(sin φ) = 2B sin φ
(29)
implying sin2 φ = B/B0
(30)
where B0 is the field at which φ becomes 90 deg and the spiraling particle is reflected (magnetic mirror). Substituting sin φ from (26) one finds that Br2 is constant, implying that the flux through the orbit is a constant of the motion. For small spiral angles φ2 = B/B0 = (r0 /r)2 . Figure 1. Section of spiral orbit of radius r in magnetic field B
Focusing with solenoid fields High intensity muon beams come from pions of various momenta, decaying at various distances, and may include a wide range of muon momenta. Therefore they are ”chaotic” and cannot
(31)
For a particular particle B0 and r0 are constant but depend on the starting conditions; φr is invariant. Now consider a solenoid of radius R. For a particle starting at radius b the maximum spiral radius for which the orbit cannot hit the walls is (R − b)/2. This estimate is correct when b = 0 and when b = R but underestimates the available orbits at intermediate values of b; so our estimate of the available phase space will be conservative. The corresponding maximum spiral angle is φ =
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QB(R − b)/2p and the solid angle available for particles in the annulus b to (b+db) is πφ2 , making the corresponding 4-dimensional phase space 2 QB 2 (R − b)2 (32) d4 = 2π b.db 2p
Integrating over b from 0 to R gives 4 and we take the square root to get the corresponding 2-dimensional phase space 2 with the factor π omitted following the convention used throughout this paper, 3 BR2 QBR2 (33) = 2 = √ 8 p p 24
in which we have set Q = 3 corresponding to the convenient choice of units B in T , p in M eV /c and r in cm. We see that if the field is increased progressively to squeeze the orbits as per eqn(31), keeping the flux through the solenoid constant, the phase space available for the particles remains unchanged. (This will break down when φ becomes larger than about 0.5 radians; sin φ then increases more slowly than φ, our approximation is no longer valid and the particles will eventually be reflected; but we are not straying into this region). Using eqn(26) β⊥ = r/φ = p/QB = p/3B
(34)
Putting in numbers, for muons of momentum 200 M eV /c squeezed by a solenoid of radius 9 cm with field 10 T , the available normalized 2-dimensional phase space is 67π mm.R, which would comfortably accept the beams of 20π mm.R requiring to be cooled [3]. The corresponding value of β⊥ is 7 cm so these muons would be matched rather well into a boron carbide degrader. After the degrader the beam could be expanded for the acceleration stage by reducing the field, following eqn(31): for example with the field reduced to 1 T the solenoid radius would be increased to 28 cm. The high field regions can be quite short.
REFERENCES 1. Francis J.M. Farley, http://arxiv.org/abs/physics/0404072, submitted to Nuclear Instruments and Methods A. 2. K.L. Brown, D.C. Carey, Ch. Iselin and F. Rothacker, CERN 80-04 (1980) 3. M.M. Alsharo’a et al., Physical Review Special Topics, Accelerators and Beams, 6, 081001 (2003)
Degraders and ionization cooling∗ Francis J. M. Farley Yale University, New Haven, Connecticut 06520
†
Procedures are given for minimizing the increase in phase space when a beam is degraded, with and without an axial magnetic field and in a lithium lens. Optimum focusing of the beam onto the degrader is recommended. The transverse diffusion of the beam is as important as the spread in angle, so high density, low Z materials such as diamond and boron carbide are the best. For chaotic beams, focusing by increasing axial fields is suitable and the corresponding acceptance is calculated. Ionization cooling of muon beams by various materials is compared for typical cases.
In ionization cooling we degrade the beam, reducing the forward and transverse momenta together, then accelerate to restore the forward momentum; so the angles in the beam are reduced. But multiple scattering in the degrader increases the transverse angles. We need to degrade with minimum increase in transverse phase space. I first discuss how to do this for a monoenergetic beam and then consider a beam with large ∆p/p which cannot be brought to a unique focus. For more detail see my paper ”Optimum strategy for degraders and ionization cooling” [1] With a perfect pencil beam going in, the output phase space for a degrader of length t is an inclined ellipse. There is an angle-position correlation with σθ ∝ t1/2 and σx ∝ t3/2 . At the output the emittance ∝ t2 . The beam appears to diverge from a point near the centre of the degrader and we can refer back to the virtual source at this point which is an upright ellipse. To calculate the ellipse we can use the well known transport formalism [2] . A particle, represented by the vector X, is transported by the matrix R
X2
x 2 R11 R12 x1 = = = R X1 2 R21 R22 1
∗ Combines
talks to WG3 and WG4 plus discussions for correspondence: 8 Chemin de Saint Pierre, 06620 le Bar sur Loup, France [email protected]
† Address
0920-5632/$ – see front matter © 2005 Published by Elsevier B.V. doi:10.1016/j.nuclphysbps.2005.05.050
The ellipse in phase space, which for example can define the one standard deviation contour, is represented by the symmetric sigma matrix σ using the equation X T σ −1 X = 1
(1)
which written out reads 2 (2) x2 σ22 − 2xθσ12 + θ2 σ11 = ∆ = σ11 σ22 − σ12
The variances in position and angle are σx2 = σ11 and σθ2 = σ22 . ∆ is the determinant of the sigma matrix √ and the area of the ellipse in phase space is π ∆. It will be convenient to √ omit the factor π and refer to = σx .σθ = ∆ as the emittance of the beam. The rule for transporting the sigma matrix is
σ2 = Rσ1 RT
(3)
and applying this for a simple drift space of length t gives 1 0 1 t σ11 σ12 (4) σ2 = σ12 σ22 t 1 0 1 σ11 + 2tσ12 + t2 σ22 σ12 + tσ22 (5) = σ12 + tσ22 σ22 For a small distance dt this reduces to 2σ12 σ22 dt dσ = σ22 0
(6)
and one can integrate term by term to get back to the previous equation. One sees that σ22 is independent of the others (zero if there is no scattering), so this can be integrated first. Then one can
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F.J.M. Farley / Nuclear Physics B (Proc. Suppl.) 149 (2005) 289–294
integrate for σ12 which depends on σ22 . Finally one can integrate for σ11 . If there is scattering σ22 changes and the differential becomes 2σ12 σ22 dσ = dt (7) σ22 K(p) with the variance in θ being σ22 = K(p) = 200/X0 p2 β 2
(8)
where X0 is the radiation length in the degrader material. Integrating this term by term, as described above, one can split the resulting matrix into two parts σ out = σ beam + σ degrad
(9)
in which σ beam is the matrix the beam would have at the end of the degrader if there had been no scattering nor loss of energy, 0 0 0 0 0 + t2 σ22 σ12 + tσ22 σ11 + 2tσ12 σ beam = (10) 0 0 0 σ22 σ12 + tσ22 and σ degrad is a matrix characterizing the degrader, which is independent of the input beam, A B degrad σ = (11) B C
the input beam with the degrader characteristic. Equation (9) shows that the final sigma matrix is the sum of the two component sigma matrices. This is the convolution theorem. It can be proved for any two sigma matrices that represent Gaussian beam distributions; it may be true more generally. The convolution of two beams is obtained by adding the corresponding sigma matrices. This is rather useful. It is much easier to add two matrices than to carry out the convolution integral in four dimensions. It applies in any situation, for example if the degrader is in a magnetic field, or lithium lens, etc. Minimum output emittance To minimize the output emittance for a given degrader, one must adjust the input beam shape in eqn (9) so that the determinant of σ out is a minimum. det(σ out )
= 2beam + (AC − B 2 ) beam beam beam + Aσ22 − 2Bσ12 ) +(Cσ11
The first two terms in this equation are invariant but one can change the components of σ beam (keeping the emitance fixed) to minimize the third term in brackets. One then finds 2out = 2beam + 2degrad + 2beam degrad
(15)
that is in which the three terms are given by t K(p)ds C(t) = 0 t C(s)ds B(t) = 0 t B(s)ds A(t) = 2
out = beam + degrad (12) (13) (14)
0
The degrader matrix is the sigma matrix of the beam coming out of the degrader when the input is a perfect pencil beam. This characterizes the degrader and the determinant gives the emittance degrad called the degrader emittance. Convolution theorem A more general input beam can be split up conceptually into a bundle of pencil beams. The output in this case is the convolution integral of
(16)
In the optimum case the degrader adds its characteristic emittance to the beam: one can do worse, but one cannot do better. The optimum input beam is beam A B beam (17) σ = degrad B C
This defines the input beam as it should be at the end of the degrader, if the degrader was not there. It should match the shape of a pencil beam coming out of the degrader. One can transpose back a distance −d to get an upright ellipse. One finds that the beam should be focused to an image point distance d = B/C before the end of the degrader; for thin degraders this corresponds to the centre of the slab.
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F.J.M. Farley / Nuclear Physics B (Proc. Suppl.) 149 (2005) 289–294
Table 1 Degrader materials Material
diamond boron carbide beryllium graphite water lithium liquid H2
Density g/cm3 3.510 2.520 1.848 2.265 1.000 0.534 0.071
W = dE/dx M eV /cm 6.125 4.253 2.946 3.952 1.991 0.875 0.286
X0 cm 12 20 35 19 36 155 865
W 2 X0 M eV 2 /cm 450 361 306 294 143 119 71
degrad mm.mR 123 154 182 189 389 467 782
The final column gives degrad for 300 M eV muons losing 50 M eV with optimum focusing.
For thin degraders the degrader emittance is √ (18) degrad = Kt2 / 12
with K given by eqn (8). For a given loss of momentum the thickness t is proportional to 1/W where W = dp/dx, so degrad ∝ 1/W 2 X0 . The best material will be the one with the highest value of W 2 X0 . This is true for any particle at any energy for any decrement in momentum. Because the lateral spreading of the beam contributes to degrad , high specific energy loss is favoured above high radiation length; this implies that high density, low Z materials are the best. Various useful materials are compared in Table 1. The ratio of image size to convergence angle for the optimum beam is √ degrad beam β⊥ (19) = σx /σθ = β⊥ = t/ 12
Choice of material One sees from Table 1 that the best material is diamond (not usually available in sufficient sizes) followed by boron carbide, graphite and beryllium. In comparison liquid hydrogen appears unattractive. The final colum shows the degrader emittances for 300 M eV muons losing 50 M eV with optimum focusing.
Axial magnetic field One might do better by putting the degrader in an axial magnetic field B to wrap up the transverse momenta. Putting k = eB/2βγm0 c2 one can solve the transport equations for this case [1]. At every turn in the field the degrader ellipse becomes vertical K t/k 2 0 (20) σ degrad = 0 t 2
with the degrader emittance degrad = Kt2 /2nπ for n spiral turns. Comparing with eqn (18) the improvement is a factor 1.814n. But with no field, liquid hydrogen is 5.085 times worse than boron carbide (see Table 1), so it requires 2.8 spiral turns in the field for it to become better. In a field of 7T the spiral wavelength for 315 M eV /c muons is 94 cm, so we would need 2.6 m of hydrogen to improve on a simple slab of boron carbide. Lithium lens In a lithium lens of radius a with surface field B0 , similar formulae apply [1] with k = eB0 /βγm0 c2 a and the improvement factor over zero field for n turns is 3.628n. With radius a = 2 cm, and current 500 kA, the surface field is 5 T and the spiral wavelength would be 41 cm.
Ionization cooling Using unnormalized emittances, the change in emittance due to degrading and re-acceleration
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Table 2 Cooling fraction with liquid hydrogen and boron carbide Material thickness β⊥ in = 10 mm.R cm cm % B4 C 21.1 6 (opt) 32.0 ” ” 10 31.8 ” ” 20 31.0 ” ” 30 30.1 ” ” 90 24.9 LH2 , no field 315 90 26.2 LH2 , 7 T 315 30 39.1
in = 3 mm.R % 29.7 29.0 26.5 23.8 8.1 1.2 29.8
Percent cooling for 300 M eV /c muons with 90 M eV energy loss and two initial emittances in
back to the original energy is δa = δp/p = −W t/p
(21)
whereas the increase in emittance due to the scattering in the degrader with optimum matching is √ (22) δdegrad = Kt2 / 12
if the degrader is too thick there is no cooling. The optimum thickness is such that the cooling effect is twice the scattering effect √ 3W (23) topt = Kp
In liquid hydrogen in a field of 7 T , the equilibrium emittance is 386 mm.mR, but one recalls that this is not really effective before several turns in the field, the spiral wavelength being 94 cm. Hydrogen gives the smallest equilibrium emittance in the field, better than lithium (701), beryllium (926) and boron carbide (1121), all figures in units of mm.mR in a field of 7 T . For this reason liquid hydrogen has been favoured for detailed study; but boron carbide in zero field might be equally effective. The equilibrium emittance in a lithium lens [1] with the parameters cited above is rather better at 152 mm.mR.
and the net cooling is
√ −δ/ = W topt /2p = 3W 2 X0 /400
(24)
We see that the fractional cooling decreases as the beam is cooled, but never goes to zero. The optimum material is the one with the highest W 2 X0 , see Table 1. In an axial magnetic field, eqn (20) shows that the increase of emittance due to scattering is proportional to t rather than t2 . So in this case there is no optimum thickness. But if the emittance is smaller than the equilibrium value, there is no cooling: equilib =
β2W X
66.8 200c (25) = 2 β W X0 B 0 (eB/m0 c)m0 c
where B is in Tesla and the other symbols have the units of Table 1.
Influence of beam focusing To realize the optimum performance with boron carbide requires the beam to be concentrated onto a small spot with large angles, the transverse beta values being given by eqn (19). But the initial muon beams have a spread in momentum and arise from pi-mu decay at various momenta and various places along the beam transport system. Therefore they cannot be brought to a fine focus. For an energy loss of 90 M eV the degrader length for boron carbide is 21.1 cm, so the ideal input beam has β⊥ = 6.1 cm. In contrast a liquid hydrogen degrader would be 315 cm long, requiring β⊥ = 91cm, which could be much easier to achieve. Does this mean that hydrogen is better? It is instructive to compare hydrogen and boron carbide for the
F.J.M. Farley / Nuclear Physics B (Proc. Suppl.) 149 (2005) 289–294
same values of β⊥ . We can calculate the final emittance with boron carbide for any initial beam focused to a waist at the centre of the degrader by using the convolution theorem, which means adding in quadrature the corresponding values of σx and σθ . As an example we compare in table 2 the overall cooling of one stage for an input momentum of 300 M eV /c with energy loss 90 M eV for various values of β⊥ and two initial emittances. Hydrogen in a magnetic field of 7 T requires β⊥ = 30 cm, without field β⊥ = 90 cm. Initial emittances are un-normalized. It is seen that boron carbide is competitive in most cases and the value of β⊥ is not critical. As a refractory material requiring no cooling, it is much easier to use. Graphite and beryllium will give almost the same performance.
293
be brought to a precise focus. They can nevertheless be squeezed to small values of β⊥ by means of a solenoid with gradually increasing magnetic field. Using the (well-known) equations for a particle of momentum p spiraling at angle φ to an axial field B, we compute the available phase space. The particle wraps itself round a cylinder of radius r (see fig. 1) with p sin φ = QBr
(26)
where Q is the constant of proportionality. If B is increasing with axial distance z the radial field Br at radius r is obtained using Gauss’ theorem Br = −
r dB 2 dz
(27)
In distance ds = dz/ cos φ along the orbit this radial field changes the spiral angle φ by dφ =
sin φ dB sin φ Br .dz QBr ds = =− cos φ 2B cos φ Br p
(28)
so
dB d(sin φ) = 2B sin φ
(29)
implying sin2 φ = B/B0
(30)
where B0 is the field at which φ becomes 90 deg and the spiraling particle is reflected (magnetic mirror). Substituting sin φ from (26) one finds that Br2 is constant, implying that the flux through the orbit is a constant of the motion. For small spiral angles φ2 = B/B0 = (r0 /r)2 . Figure 1. Section of spiral orbit of radius r in magnetic field B
Focusing with solenoid fields High intensity muon beams come from pions of various momenta, decaying at various distances, and may include a wide range of muon momenta. Therefore they are ”chaotic” and cannot
(31)
For a particular particle B0 and r0 are constant but depend on the starting conditions; φr is invariant. Now consider a solenoid of radius R. For a particle starting at radius b the maximum spiral radius for which the orbit cannot hit the walls is (R − b)/2. This estimate is correct when b = 0 and when b = R but underestimates the available orbits at intermediate values of b; so our estimate of the available phase space will be conservative. The corresponding maximum spiral angle is φ =
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QB(R − b)/2p and the solid angle available for particles in the annulus b to (b+db) is πφ2 , making the corresponding 4-dimensional phase space 2 QB 2 (R − b)2 (32) d4 = 2π b.db 2p
Integrating over b from 0 to R gives 4 and we take the square root to get the corresponding 2-dimensional phase space 2 with the factor π omitted following the convention used throughout this paper, 3 BR2 QBR2 (33) = 2 = √ 8 p p 24
in which we have set Q = 3 corresponding to the convenient choice of units B in T , p in M eV /c and r in cm. We see that if the field is increased progressively to squeeze the orbits as per eqn(31), keeping the flux through the solenoid constant, the phase space available for the particles remains unchanged. (This will break down when φ becomes larger than about 0.5 radians; sin φ then increases more slowly than φ, our approximation is no longer valid and the particles will eventually be reflected; but we are not straying into this region). Using eqn(26) β⊥ = r/φ = p/QB = p/3B
(34)
Putting in numbers, for muons of momentum 200 M eV /c squeezed by a solenoid of radius 9 cm with field 10 T , the available normalized 2-dimensional phase space is 67π mm.R, which would comfortably accept the beams of 20π mm.R requiring to be cooled [3]. The corresponding value of β⊥ is 7 cm so these muons would be matched rather well into a boron carbide degrader. After the degrader the beam could be expanded for the acceleration stage by reducing the field, following eqn(31): for example with the field reduced to 1 T the solenoid radius would be increased to 28 cm. The high field regions can be quite short.
REFERENCES 1. Francis J.M. Farley, http://arxiv.org/abs/physics/0404072, submitted to Nuclear Instruments and Methods A. 2. K.L. Brown, D.C. Carey, Ch. Iselin and F. Rothacker, CERN 80-04 (1980) 3. M.M. Alsharo’a et al., Physical Review Special Topics, Accelerators and Beams, 6, 081001 (2003)