hysics 598ACC Accelerators: Theory and Instructors: Fred Mills, Deborah Errede Lecture 3: Equations of Motion in Accelerator Coordinates 1
Summary A. Curvilinear (Frenet-Serret) coordinate system B. The linearized equations of betatron and closed orbit motion C. Orbit lengths, periods, transition energy D. Transformation to accelerator coordinates E. Longitudinal-Transverse coupling, effect of energy jumps
Equations of Motion in Accelerator Coordinates We will describe the motion of a particle relative to an arbitrary closed curve in space, specified by r o (s), where the parameter s is chosen to be the arc length along the curve, measured with respect to some arbitrary starting point. γ β α s increases in α direction Reference Orbit r o (s) Figure 3.1 Reference orbit and basis vectors 3
Define a tangent vector α =, where the prime means total differentiation with respect to s. Next define an outward normal unit vector. This vector has the opposite direction to the usual curvature vector. Now define a third vector γ = α β. The triad form a right-handed set of basis vectors. It can be shown that, 3.1 β =Ωα + ωγ, Here Ω and ω are the curvature (or first curvature) and torsion (or second curvature) of the curve r o (s) respectively. We now define new coordinates (sxz) in the following way: find the value of s where the point in question lies in the βγ plane; x and z are the projections on the β and γ directions respectively. Then any point r can be expressed as 3. r 0 γ = ωβ r(s,x.z) = r 0 ( s) + xβ + zγ β = α Ω 4
We can carry out this transformation by a generating function, ( ) 3.3 F 3 = p r s, x,z *see notes The new momenta are 3.4 p s = p α( 1+ xω)+ω xγ zβ p x = p β p z = p γ ( ) [ ] We note that the canonical momentum p s is not simply the projection of p on α (the same will hold true for A s ). With some exceptions, accelerators are normally built on planar surfaces, where ω = 0. We will treat only situations where ω = 0. We will only include s components of vector potential, and set φ = 0. Other fields can be easily inserted later. 5
for F F F F(Qpt): =,q =,K= H 3 3 3 3 i j Qi pj t F3 r r' ps = = p see homework s r F3 r r px = = p = p βˆ x x r F3 r r pz = = p = p γˆ z z then add in electromagnetic vector potential (with ω=0). r r A = A = A (1+ x Ω) s α A α = A s (1+ x Ω) and the new canonical momentum becomes p e ps A c 1+ xω s 6
The new Hamiltonian is 3.5 *see notes previous page (no solenoidal B field here) Now we change from t to s as the independent variable. Then the correct Hamiltonian is We note that H = cm c + p s ea s c 1 +Ωx ( ) + p x G = p s = ea s c ± 1 +Ωx c ( ) H + p z 3.6 (see eqn 1.16) 1 ( ) p x p z mc 1 3.7 H c ( ) mc 1 = H ( ) is the total mechanical momentum of the particle. 7
Since (p x,p z ) << in general, that is, the angles between the orbits and the reference curve are small, we can expand the square root to obtain an approximate Hamiltonian, 3.8 G ea s ( 1+Ωx)1 p x + p z *see notes c For this set of coordinates the differential arc length ds of a general path is given (for ω = 0) [ ] + dx 3.9 ( dσ) = ( 1+Ωx)ds ( ) + ( dz) *see notes 8
see eqn 3.6 1 1 x + z x z x z H 1 (p p ) mc p p = p p 1 c Using (1+x) 1/ =1+x/ for x<<1. ------------------------------------------------------------------------------------------------- r r see homework: use (d σ ) = dr dr and ω = 0. 9
The curvature of the central orbit is caused by bending magnetic fields B z. Let the vector potential be 3.10 A s = B x + Ωx + A 1 This gives a magnetic field B in the z direction on the reference orbit. The quadratic term is necessary to satisfy [ B] s = 0 in the curvilinear coordinates. The equations of motion are, to first order, ( ) eb 1+ Ωx 3.11 p x = +Ω + e A 1 c c x, x = ( 1 +Ωx) p x *see notes 3.1 p z = e A 1, c z z = 1+Ωx ( ) p z 10
( px + pz) e Ωx G B x+ + A1 ( 1+ xω) 1 c ( p ) x + pz ' G e e A 1 px = = B( 1+Ω x) + +Ω 1 x c c x ' eb( 1+ xω) e A1 px = +Ω + c c x to first order, and x ' G ' = z p x G = p z 11
Choose 0 to be the momentum which follows the orbit r o (s) (x = z = 0), and on r o (s). 3.13 0 Ω= eb c This is the relationship between field and curvature. Now introduce a transverse gradient in the B field by 3.14 A 1 = ( x z ) (quadrupole term) *see eqn 3.10 and let e B 3.15, c =Κ = 0 0 + Δ These lead to the linearized equations 3.16 x + 0 (, *see notes Κ+Ω )x =Ω Δ 0 Kz = 0 We note that the x equation has a free oscillation (betatron oscillation) and a forced solution (particular integral). Further the frequencies of the free oscillations will depend on momentum. This is called "chromaticity". A 1 x = 0 1
' px x = (1+Ωx) ( + xω) ' eb 1 px = +Ω K0x c p p x =Ω x + (1+Ωx) ' " ' x x nd order x p (1+Ωx) =Ω (1+Ω x) + 0Ω (1+Ω x) +Ω K0x Δ = +Ω Ω +Ω 0 0 (1 x) x K x Δ 0 Δ = (K +Ω ) x +Ω +Ω x + +Ω =Ω " 0 x (K )x [ ] drop h.o.t Δ 13
similarly ' pz z = (1+Ωx) p e A = = K z c z ' 1 z 0 p z = (1+Ω x) +Ωx ' " z ' pz drop h.o.t. " K0 z px pz z = (1+Ω x) +Ω (1+Ωx) K " 0 z z= 0 14
It is customary to separate the x motion into the betatron (free) oscillations and the part proportional to momentum difference Δ (closed orbit). To first order we can write 3.17 x = X + x p () s Δ, x p + ( Κ +Ω )x p =Ω, X + ( Κ+Ω )X = 0 Now let us suppose that the reference curve is closed with length C 0. It is of interest at this point to inquire about orbit lengths and orbit periods or revolution frequencies. The length of the closed orbit of a particle with momentum deviation Δ is 3.18 C = dσ 1 +Ωx p ds = C (see eqn 3.9) 0 +ΔC Δ 15
ΔC 3.19 =α Δ C 0 = Ωx Δ p ave The approximation is again that angles are small. α is called the "momentum compaction factor". The revolution frequency is, and its variation with momentum is f = V C Δf 3.0 = ΔV f V ΔC C = 1 γ α Δ *see notes When, all particles near the reference momentum have the same frequency, γ= 1 α independent of momentum. This has important consequences for acceleration systems. This value of γ is called the "transition γ" or γ t. We can effect transformation 3.17 by a generating function 3.1 ( ) Δ x x p F = x + x p Δ + zz Ht 16
df dv dc f v c = ( ) p = m β 1 β 1 = β β β +β β β β 1 pdp m d (1 ) ( 1)(1 ) ( d ) divide through by p and collect terms dp p dβ = γ β dβ = β 1 dp γ p df 1 dp dp 1 dp = α = α f γ p p γ p 17
The transformation equations are Δ X = x x p p x = x + x p Δ 3. Z = z *see notes p z = z X x + x p Δ T = t x p + x V p 0 0 V 0 H = H ( ) (energy variable, not Hamiltonian) Here we remark that 3.3 = 1 H 0 V 0 where V 0 is the velocity of a particle with momentum 0. 18
see eqns 1.30 F F H = = H T ( t) x = ( H) ΔH or substituting Δ = then differentiating V ( ' ) x + xpδ ' F p Δ T = = t x xp + xp ( H) V0 0 0V0 x ( ' ) x + xpδ ' F p T = = t X+ xp ( H) V0 V 0 0 0 using 1 = H V 0 19
Thus we have separated the x motion into two parts, due to momentum motion, and due to betatron oscillations around the momentum motion. On the other hand we have not removed the essential coupling between these two motions. The coupling shows up in the definition of the new time variable, now canonical to H. The new Hamiltonian is F 3.4 G* = G + s erforming the indicated differentiations and substitutions, and collecting terms in powers of Δ, we find 3.5 G* = H ( )+ x + z + ( Δ) 0 Ωx p ( x p ) + 0 [( Ω + K)X KZ ]+ ΔX x p + ( Κ+Ω )x p Ω ( ) x p + Ω + K [ ] 0
According to 3.17, the coefficient of Δ is zero. Since (x p x p' ) ' = x p x '' p +(x p' ), the coefficient of (Δ) Ωx becomes p ( x p x p ). The momentum motion is always slow compared to the betatron motion so we can average the momentum terms over one or more revolutions. x p is a periodic function of s, so (x p x p' ) ' is periodic with zero mean, and on the average it contributes zero. The average of Ωx p as we saw above, is. Now we can expand (H) to second order around H 0, and obtain for the new Hamiltonian (ignoring constants and functions of s alone) ( H H0) 3.6 G* = + ( H H0) 1 0 V 0 γ α A se includes fields not yet specified; for example accelerating fields or higher order magnetic field terms (higher order multipoles than the dipole and quadrupole terms included so far). + x + z 0 V 0 + 0 Ω + K ( )X KZ [ ] ea se c α= 1 γ t 1
Let us look at the T and H equations 3.7 T = G* H η= 1 γ α = 1 1 η H H 0 V 0 0 V 0 = 1 V 0 1 η Δ CT' is the revolution period for an off momentum particle; the added arc length and velocity change included. A se may depend on t and hence T. 3.8 H = G* *see 1.6 and notes T = ea se c T = ee s where E s is the electric field (recall we are not including a scalar potential φ, so this is derived from a magnetic field). Now, however, A se depends on X and x as well, because of 3.. Then there is an added increment to x which is using ΔH Δ = V 0 3.9 e c A se t t X = ee s x p V 0
r B= 0 Since, we can write B r = A r r r 1 A and from Faraday s law E+ = 0 c t r r 1 A Therefore we can write E + = φ c t φ = 0 We are considering cases where hence r E = r 1 A c t 3
as well as an added increment to X', which is 3.30 e A se t = ee x s p *see 3. c t x 0 V 0 Thus an accelerating electric field located where there is momentum dispersion (x p,x p' ) excites betatron oscillations. A change in energy of the particle changes the definition of its equilibrium orbit, without changing the position or angle of the particle in space. The changes in these quantities are 3.31 using Accelerators: Theory and *see 3.8 and 3. so while the position and direction haven t changed in space, the position and angle have changed with respect to the new closed orbit due to the shift in energy. δh = e Eds δx = x p δh 0 V 0 δ = x δh δ X = x p = x δ V 0 ΔH Δ = V 0 4
These are the same results obtained from the transformation equations 3. for X,x, x,p x, holding x and p x constant while changing Δ. In addition to those from accelerating fields, energy jumps due to synchrotron radiation or intrabeam scattering or cooling mechanisms can cause phase space growth and place severe restraints on accelerator design for e+e- and hadron colliders, and for ion cooling systems. End of Lecture 5
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