Problems of Chapter 1: Introduction

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1 Chapter 1 Problems of Chapter 1: Introduction 1.1 Problem 1 1: Luminosity of Gaussian bunches a) If the bunches can be described by Gaussian ellipsoids with ( ( )) x 2 ρ exp 2σx 2 + y2 2σy 2 + z2 2σz 2, (1.1) show that the luminosity reduces to N + N L = f 0 N b π(4σ x σ y ), (1.2) where it is assumed that the beams move along the z-axis. The number of particles per bunch in the electron and positron beams are N and N +, respectively. The number of bunches in each beam is given by N b. (Assume the bunches do not change shape either due to the accelerator optics or the interaction of the beams passing through each other.) b) An e + e storage ring (CESR) operates at 5.3 GeV with 7 bunches of e + and 7 bunches of e orbiting in opposite directions. Assume the current per bunch is initially 8 ma, and the ring circumference is 768 m. For σ x = m, σ y = m, and σ z = 2.2 cm, what is the initial luminosity in one of the experiments? What is the integrated luminosity of this experiment for a 3 hr run if the beam lifetimes are both 2 hr? (Assume that the beam currents decay exponentially.) a) First we need to normalize the Gaussian distribution. For a single variable x, a Gaussian density function is normalized such that 1 2πσx e x2 2σx dx = 1. (1.3) 1

2 2 Accelerator Physics: Example Problems with Solutions So the density function for the ± beam must be ( N ± ρ ± = exp 1 ( x 2 (2π) 3/2 σ x σ y σ z 2 σ 2 x + y2 σy 2 + z2 σz 2 )). (1.4) The luminosity for a single bunch crossing is from Eq. (CM: 1.4) with a slight modification: L = v + v ρ + (x, y, z ct)ρ (x, y, z + ct) dx dy dz dt = 2cN +N (2π) 3 σxσ 2 yσ 2 z 2 e x2 /σ 2 x e y 2 /σ 2 y = N +N 4π 3 σxσ 2 yσ 2 z 2 e 1 2 [(z ct)2 +(z+ct) 2 ]/σ 2 z dx dy dz dt e x2 /σ 2 x e y 2 /σ 2 y e z 2 /σ 2 z e ξ 2 /σ 2 z dx dy dz dξ, (1.5) with ξ = ct where we have assumed that the velocity of each beam is essentially the speed of light in the high energy limit. Remembering our normalization, the integrals are all of the form So we must have e ζ2 σ 2 dζ = πσ. (1.6) L = N +N 4πσ x σ y, (1.7) and for N b bunches per revolution with a revolution frequency f 0, the instantaneous luminosity for one experiment must be b) For CESR with the given parameters, so f 0 = m/s 768 m N ± = I 0 ef 0 = L = f 0 N b L = f 0 N b N + N 4πσ x σ y. (1.8) = Hz, (1.9) 8 ma ( C) ( s 1 ) = , (1.10) L 0 = ( Hz) 7 ( ) 2 4π ( m) ( m) cm 2 s 1 (1.11) for the initial luminosity of one experiment. If we assume that each beam decays exponentially, then I ± (t) = I 0 e t/τ with τ = 2 hr. Since L

3 Problems of Chapter 1: Introduction 3 I + (t)i (t) e 2t/τ, we find the total integrated luminosity over a 3 hr run would be 3hr L = L 0 e 2t τ τ dt = L0 [1 ] e 2(3hr)/τ 0 2 ( ) = ( cm 2 s 1 ) s (1 e 3 ) = cm 2 2 = 62 nb 1, where 1 b = cm 2. (1.12)

4 4 Accelerator Physics: Example Problems with Solutions 1.2 Problem 1 2: Brightness of laser beam Calculate the brightness of a NdYAG laser with the following parameters: λ = 1.064µ m (1.13) Power = 20 W (1.14) Bandwidth ω = 120 GHz 2π (1.15) Beam divergence = 10 mrad. (1.16) The energy of a single photon is u = 2π c λ = 2π (197 MeV fm) m = ev, (1.17) so the flux of photons is dn/dt = P/u = s 1. In laser physics, the divergence angle θ d is generally specifies a narrow beam with an opening cone of ±θ d /2 and in this case yields a solid angle of dω = 2π θd /2 0 0 sin θ dθ dϕ π (0.005) 2 = sr. (1.18) The bandwidth of 120 GHz can be converted into wavelength units by and λ λ u = h ν = ω Using Eq. (CM 1.5) we then get = ω ω = u u, (1.19) = ( evs) 2π ( Hz) = ev. (1.20) d 4 n dφ Ω = 1000 dt dω ( du /u) = 1000 ( s) ( ) ( /1.165) = sr 1 s 1. (1.21)

5 Problems of Chapter 1: Introduction Problem 1 3: Equations of motion In a fixed Cartesian coordinates system, show that the equations of motion for a charged particle moving in a magnetic field may be written as x = q p (1 + x 2 + y 2 ) 1 2 [y B z (1 + x 2 )B y + x y B x ], and (1.22) y = q p (1 + x 2 + y 2 ) 1 2 [x B z (1 + y 2 )B x + x y B y ], (1.23) where the primes denote derivatives with respect to z (i. e., x = dx/dz). Here it has been assumed that the electric field is zero and that dz/dt 0. Solution[64]: In these conditions, the motion equation is [ ] 1 with γ = 1 v2 2 c 2 Due to the trivial relation q( v B) = d p dt = d (γm v) = γmd v dt dt + mdγ v (1.24) dt, v = ˆx dx dt + ŷ dy dt + ẑ dz dt and B = ˆxB x + ŷb y + ẑb z. we can write d dt = dz d dt dz = ż d dz (1.25) v = ż(ˆxx + ŷy + ẑ 1) or v = v = ż 1 + x 2 + y 2 (1.26) where the prime represents the derivative with respect to z. The vector product in the left side of Eq. (1.24) becomes v B ˆx ŷ ẑ ż(y B z B y ) = żx ży ż B x B y B = ż(b x x B z ) (1.27) z ż(x B y y B x ) Besides, from Eq. (1.26) we have d v dt = ż2 (ˆxx + ŷy ) and dv dt = ż2 x x + y y 1 + x 2 + y 2 (1.28)

6 6 Accelerator Physics: Example Problems with Solutions Therefore the components of Eq. (1.24) can be written as or γmż 2 x + m dγ dt żx = qż(y B z B y ) (1.29) γmż 2 y + m dγ dt ży = qż(b x x B z ) (1.30) m dγ dt ż = qż(x B y y B x ) (1.31) γmżx + m c 2 ż3 γ 3 (x x + y y )x = q(y B z B y ) (1.32) having considered that γmży + m c 2 ż3 γ 3 (x x + y y )y = q(b x x B z ) (1.33) m c 2 ż3 γ 3 (x x + y y ) = q(x B y y B x ) (1.34) dγ dt = v d v γ3 c 2 dt = ż3 γ 3 c 2 (x x + y y ) (1.35) Inserting the Eq into Eqs. (1.32 and 1.33) we obtain which will yield γmżx + q(x B y y B x )x = q(y B z B y ) (1.36) γmży + q(x B y y B x )y = q(b x x B z ) (1.37) or x = q γmż [y B z (1 + x 2 )B y + x y B x ] (1.38) y = q γmż [x B z (1 + x 2 )B x + x y B y ] (1.39) x = q p 1 + x 2 + y 2 [y B z (1 + x 2 )B y + x y B x ] (1.40) y = q p 1 + x 2 + y 2 [x B z (1 + x 2 )B x + x y B y ] (1.41) having considered Eq. (1.26) and that p = p = γmv.

7 Problems of Chapter 1: Introduction Problem 1 4: Decay of a pion to a muon Consider a charged pion decaying into a muon plus an antineutrino: π µ + ν µ. (1.42) Use M π ± = 140 ev/c 2, m µ = 106 MeV/c 2, and m ν = 0. a) In the rest system of the pion, what are the energies and momenta of the muon and antineutrino? b) For a moving pion with total energy U π = γm π c 2 find an expression for the direction, θ µ of the muon relative to the pion in the lab in terms of the angle θ µ in the in the pion s rest system. µ µ ν θ µ π π θ µ Pion at rest Pion in lab ν In the pion rest frame the total energy is just the rest-mass energy which, after the decay, transforms into the sum of the muon and neutrino total energies; namely M π c 2 = U µ + U ν = m µ c 2 + W µ + W ν (1.43) since m ν = 0. Besides, the pion momentum in its rest frame is obviously zero and this implies that p µ + p ν = 0 p µ = p ν p µ = p ν p µ = p ν (1.44) or p µc = (U µ) 2 (m µ c 2 ) 2 = p νc = W ν (1.45) which combined with Eq. (1.43) gives rise to

8 8 Accelerator Physics: Example Problems with Solutions Hence we have M π c 2 U µ = (U µ) 2 (m µ c 2 ) 2, or (1.46) Uµ = (M πc 2 ) 2 + (m µ c 2 ) 2 2(M π c 2. (1.47) ) W µ = (M πc 2 ) 2 + (m µ c 2 ) 2 2(M π c 2 ) = (M πc 2 m µ c 2 ) 2 2(M π c 2 ) m µ c 2, = MeV (1.48) W ν = M π c 2 m µ c 2 W µ = 4.13 MeV. (1.49) In this case, the Lorentz transformation of the momentum longitudinal component is where ( p µ cos θ µ = γ p µ cos θµ + β U µ ) = γp µ cos θµ + βγ U µ c c, (1.50) γ = U π M π c 2, and βγ = γ 2 1 = U π 2 (M π c 2 ) 2 M π c 2 = p π M π c, (1.51) which altogether yield cos θ µ = U πp µ cos θ µ + U µp π p µ M π c 2. (1.52)

9 Problems of Chapter 1: Introduction Problem 1 5: Collider vs fixed-target energies The Tevatron collides protons (m p = GeV) at 1 TeV per beam. What is the equivalent proton beam energy required to produce the same centerof-mass energy with a stationary hydrogen target? How fast would you have to drive your new 1.3 ton VW Beetle to have the same kinetic energy as a bunch of protons with this energy? (The speed of sound in air is 330 m/s.) In symmetric colliders the center of mass is standing, meaning that laboratory and center-of-mass frames coincide; namely Then consider the invariant U CM = 2 U Tevatron = 2 U T = 2 TeV. (1.53) p µ p µ c 2 = U 2 p 2 c 2 = 1 c 2 (U CM) 2 P 2 CM = U 2 CM, (1.54) or for the beam (x) hitting a fixed target: p µ p µ c 2 = (U x + m p c 2 ) 2 p 2 xc 2 = (U x + m p c 2 ) 2 (U 2 x m 2 pc 4 ) 2U x m p c 2, (1.55) since U x m p c 2. Solving for U x yields U x = 2 U 2 T m p c 2 = = GeV. (1.56) Assuming that the VW Beetle is nonrelativistic, it would have a kinetic energy of W VW = 1 2 M VWv 2 = N Bunches U x or = [ev] [J/eV] = J, (1.57) v = 2WVW M VW = 2290 m/s Mach 7. (1.58)

10 10 Accelerator Physics: Example Problems with Solutions 1.6 Problem 1 6: HERA center-of-mass system HERA collides 920 GeV protons with 27.5 GeV electrons with zero crossing angle. a) What is the center-of-mass energy? b) What is the velocity of the center of mass in the lab system? Let us recall the expression of the momentum energy four-vector ( ) U p µ = c, p, with U = (mc 2 ) 2 + (pc) 2, (1.59) whose squared modulus is an invariant. In fact: p µ p µ = U 2 c 2 (p2 x + p 2 y + p 2 z) = m2 c 4 c 2 + p2 c 2 c 2 p 2 = (mc) 2. (1.60) Therefore the following identities subsist U 2 (pc) 2 = U 2 (p c) 2 = U 2. (1.61) Where the asterisk, as well as the subscript cm, indicates the centerof-mass frame, whose momentum is obviously null. Labeling with L the laboratory frame, we have with U 2 cm = U 2 L (p L c) 2, (1.62) U L = U p + U e, and p L = p p p e, (1.63) where U p = 920 GeV and U e = 27.5 GeV are the given data of the problem. Moreover, the momenta and energies coincide since we are in an ultrarelativistic regime. Hence the center-of-mass energy results in U cm = (U p + U e ) 2 + p p p e 2 = 318 GeV. (1.64) As far as the center-of-mass velocity is concerned, we simply have v cm = β cm c = p L U L c = c = ms 1. (1.65)

11 Problems of Chapter 1: Introduction Problem 1 7: SLAC rf voltage and power The Stanford Linear Accelerator is 3.05 km long and can accelerate electrons up to 50 GeV. a) What is the average accelerating gradient of the rf cavities? b) For bunches of electrons per bunch and a duty cycle of 100 Hz, what is the power transferred to the beam? a) The average accelerating gradient is 50 GeV G = = 16.4 MeV/m. (1.66) 3.05 km b) The power transferred to the beam is given by the duty cycle f times the total energy N e U e of the electron beam: P = 100 [Hz] [ev] [C/e] = 32 kw. (1.67)

12 12 Accelerator Physics: Example Problems with Solutions 1.8 Problem 1 8: Fixed-target interaction rate An experiment has a 10 cm long liquid hydrogen target with a density of ρ =.063 g/cm 3. (1.68) Estimate the interaction rate for p+p collisions for a beam of protons every two minutes. Assume the total cross section is 40 mb. Note: 1 barn = cm 2 and Avogadro s number is Assuming that the diameter of the beam is smaller than the radius of the target so that the whole beam passes through the target: dn dt = 0.063[g/cm3 ] 10[cm] [g] 120[s] [b] [cm 2 /b] = [s 1 ]. (1.69)

CERN Accelerator School. Intermediate Accelerator Physics Course Chios, Greece, September Low Emittance Rings

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