r−rθ˙2=evsBz
γm =±v2sBz
Bρ , z¨=∓vs2Bx Bρ ,
which can be transformed into Eq. (2.22) by changing the time variable to the coor-dinate of orbital distances, i.e. x��= ¨r/v2s.
I.4 Particle Motion in Dipole and Quadrupole Magnets
We consider a on-momentum particle with p=p0, expand the magnetic field up to first order inxandz, i.e.
Bz=∓B0+∂Bz
∂x x=∓B0+B1x, Bx=∂Bz
∂x z=B1z, (2.23) where B0/Bρ = 1/ρ signifies the dipole field in defining a closed orbit, and the quadrupole gradient function B1 = ∂Bz/∂x is evaluated at the closed orbit. The betatron equations of motion, Eq. (2.22), become Hill’s equation:
x��+Kx(s)x= 0, z��+Kz(s)z= 0, (2.24) Kx= 1/ρ2+K1(s), Kz=−K1(s),
whereK1(s) =∓B1(s)/Bρis the effective focusing function with dimension [m−2].
Here the upper and lower signs correspond respectively to the positive and negative charged particles. The sign-convention isK1 > 0 for horizontal focusing, and thus vertical defocusing. The focusing index is defined as n(s) = ±ρ2K1(s), or Kx =
1
ρ2(1−n) and Kz = ρ12n. A weak focusing accelerator requires 0 ≤ n(s) ≤ 1, while a strong-focusing accelerator, |n| �1, e.g. n(s) ≈ ±350 for the AGS. Some observations about the linearized betatron equations (2.24) are given below.
• In a quadrupole, where 1/ρ= 0, we haveKx=−Kz, i.e. a horizontally focusing quadrupole is also a vertically defocusing quadrupole and vice versa.
• A horizontal bending dipole has a focusing functionKx= 1/ρ2, andKz= 0. A dipole with entrance and exit angles perpendicular to the edge of the dipole field is called asector dipole(see Fig. 2.2a). The entrance and exit angles of particle trajectories in non-sector type dipoles are not perpendicular to the dipole edge.
There is an edge focusing/defocusing effect (see Exercise 2.2.2) on all dipoles.
• The focusing functions Kx, Kz is periodic functions of the longitudinal coor-dinate s in one revolution. One can design an accelerator lattice with many identical focusing periods. The number of identical building blocks is called thesuperperiodP. The solution of periodic Hill’s equation satisfies theFloquet theorem.
Figure 2.2:Schematic drawing of the particle trajectory in asector dipole and in a rectangular dipole. Note that the particle orbit is perpendic-ular to the pole-faces of the sector dipole magnet, and makes an angle θ/2 with the pole-faces in the rectan-gular dipole.
Exercise 2.1
1. In the Frenet-Serret coordinate system (ˆx,ˆs,ˆz), transverse magnetic fields are Bx=− 1
1 +x/ρ
∂As
∂z , Bz= 1 1 +x/ρ
∂As
∂x . Derive Eq. (2.22) from the Hamiltonian of Eq. (2.15).
2. Derive Eq. (2.22) through the following geometric argument. Let (ˆx,s,ˆz) be localˆ polar coordinates inside a dipole. The particle coordinate is
�r= (ρ+x)ˆx+zˆz,
whereρis the bending radius. The momentum of the particle is�p=γm�r, where˙ γ is constant in the static magnetic field, and the overdot corresponds to the derivative with respect to timet. Similarly,d�p/dt=γm�r.¨
(a) Using Eq.(2.7), show that
�r˙= ˙xˆx+ (ρ+x) ˙θˆs+ ˙zz,ˆ
�r¨= [¨x−(ρ+x) ˙θ2]ˆx+ [2 ˙xθ˙+ (ρ+x)¨θ]ˆs+ ¨zz,ˆ
whereθ=s/ρis the angle associated with the reference orbit, i.e. ds=ρdθ.
(b) Usingd�p/dt=e�v×B, with� B�=Bxxˆ+Bzz, show thatˆ
¨
x−(ρ+x) ˙θ2=v2sBz
Bρ , z¨=−v2sBx Bρ ,
whereBρ=γmvs/eis the momentum rigidity andvsis the longitudinal velocity.
EXERCISE 2.1 41 (c) Transform the time coordinate to the longitudinal distance swith ds =ρdθ,
wheredθ=vsdt/(ρ+x),and show that where the prime is the derivative with respect tos.
3. Inside the vacuum chamber of an accelerator, we have∇ ×B� = 0 and∇ ×E� = 0.
Thus the electric field and magnetic field can be expanded by scalar potentials with B� =−∇Φm, �E=−∇Φe, where both scalar potentials satisfy the Laplace equation whereh= 1/ρ, andρis the radius of curvature. Expressing the scalar potential in power series of particle coordinates, show thatAij satisfies the following iteration relation:
Ai,j+2 = −A��i,j−ihA��i−1,j+ih�A�i−1,j−Ai+2,j−(3i+ 1)hAi+1,j
−3ihAi−1,j+2−i(3i−1)h2Ai,j−3i(i−1)h2Ai−2,j+2
−i(i−1)2h3Ai−1,j−i(i−1)(i−2)h3Ai−3,j+2,
where the prime is the derivative with respect tos. Assuming A00 = 0, A10 = 0, andA01=−B00 in a rectangular coordinate system withh=h�= 0, show that the magnetic potential, up to the fourth order withi+j≤4, is4
Φ = −B00z+1
4. The field components in the current-free region of an axial symmetric solenoid are Bx=x
4A word of caution: the magnetic potential obtained here can not be used as the potential in the Hamiltonian of Eq. (2.14). In particular, the potential for a quadrupole is given by theA11term and the skew quadrupole arises from theA20term, etc. However, this serves as a general method for deriving the magnetic field map.
(a) Show that the coefficients are b2k+1= −1
2(k+ 1)b�2k, b2k+2= 1
2(k+ 1)b�2k+1,
where the prime is the derivative with respect to s. Show that the vector potential is
Ax=z ∞ k=0
b2k
2(k+ 1)(x2+z2)k, Az=−x ∞ k=0
b2k
2(k+ 1)(x2+z2)k, As= 0.
In a cylindrical coordinate system, where�r = xxˆ+zˆz, r = √
x2+z2, and φˆ= (−zxˆ+xz)/r, show that the vector potential can be expressed asˆ
A�=− 1
2r b0(s)− 1
16r3b��0(s) +· · ·
φ.ˆ
(b) The Hamiltonian of Eq. (2.15) for the particle motion in the solenoid is H=−p+ 1
2p[(px−eAx)2+ (pz−eAz)2].
Show that the linearized equation of motion is (see also Exercise 2.6.2) x��+ 2gz�+g�z= 0, z��−2gx�−g�x= 0,
whereg = eb0/2p = eB�/2p is the strength of the solenoid. The linearized equation can be solved analytically. Lettingy=x+jz, show that the coupled equation of motion becomes
y��−j2gy�−jg�y= 0.
Transforming the coordinates into the rotating frame, show that the system is decoupled, i.e.
¯
y=ye−jθ(s), where θ= s
0
gds,
¯
y��+g2y¯= 0.
Thus the solenoidal field, in the rotating frame, provides both horizontal and vertical focusing, independent of the direction of the solenoidal field. Note also that the effects of the ends of a solenoid, included in theg�terms, have been included to obtain this Hill’s equation in the rotating frame.
(c) Up to third order, show that the equation of motion is x��+ 2gz�+g�z = g��
2z�(x2+z2) +g���
8z(x2+z2), z��−2gx�−g�x = −g��
2x�(x2+z2)−g���
8x(x2+z2).
EXERCISE 2.1 43 5. Consider the transverse magnetic field in the Frenet-Serret coordinate system.5 For
normal multipoles with mid-plane symmetry with
Bz(z) =Bz(−z), Bx(z) =−Bx(−z), Bs(z) =−Bs(−z), the most general form of expansion is
Bz=
Show that Maxwell’s equations give the following relations:
ai,k= i+ 1 where the prime is the derivative with respect tos. Assuming that we can measure theBzat the mid-plane as a function ofx, s, i.e.
Bz(z= 0) =B0,0+B1,0x+B2,0x2+B3,0x3+· · ·, whereBi,0are functions ofs, show that the field map is
Bz = B0,0+B1,0x+B2,0x2−(B2,0+B0,0��
Show that in a pure multipole magnet, where ρ → ∞, the magnetic field can be expanded as
wherej is the complex number. Thus for a finite length quadrupole withB1,0� �= 0, the end field has an octupole-like magnetic multipole field.
5See K. Steffen, CERN85-19, p. 25 (1985).
II Linear Betatron Motion
Transverse particle motion around a closed orbit is called betatron motion. Since the amplitude of betatron motion is normally small, we study the linearized Hill’s equa-tion: Eq. (2.24). The focusing functions are normally arranged to be periodic with Kx,z(s+L) =Kx,z(s), whereLis the length of a periodic structure in an accelerator.
For example, Fig. 2.3 shows a schematic drawing of the Fermilab booster lattice, where four combined function magnets are arranged to form a basic focusing-defocusing pe-riodic (FODO) cell. Exploiting the pepe-riodic nature, we apply the Floquet theorem (see Appendix A, Sec. I.5) to facilitate the design of an accelerator lattice, In this section we study linear betatron motion. betatron tune, envelope equation. Floquet transformation, the action and Courant–Snyder invariant, σ-matrix, beam distribu-tion and emittance.
Figure 2.3: A schematic drawing of the Fermi-lab booster lattice, made of 24 FODO cells with cell-length 19.7588 m. Each period consists of four combined-function magnets of length 2.8896 m and focusing function KF = 0.02448 m−2 and KD =
−0.02082 m−2. A small trim focusing quadrupole is used to change the betatron tune. The nominal be-tatron tunes areνx= 6.7 andνz= 6.8.