Problem 3 : Solution/marking scheme Large Hadron Collider (10 points)

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1 Problem 3 : Solution/marking sheme Large Hadron Collider 10 points) Part A. LHC Aelerator 6 points) A1 0.7 pt) Find the exat expression for the final veloity v of the protons as a funtion of the aelerating voltage V, and fundamental onstants. Solution A1: [0.7] Conservation of energy: m p 2 + = m p 2 γ = m p 2 1 v 2 / Penalties No or inorret total energy -0.3 Missing rest mass - Solve for veloity: m p v = 1 2 ) 2 m p 2 + without proton rest mass: [0.5] m p 2 γ = m p 2 1 v 2 / Solve for veloity: mp v = 1 2 ) 2 Classial solution: [] 2 e V v = m p 1

2 A2 0.8 pt) For partiles with high energy and low rest mass the relative deviation = v)/ of the final veloity v from the speed of light is very small. Find a suitable approximation for and alulate for eletrons with an energy of 60.0 GeV. Solution A2: [0.8] veloity from previous question): m e v = 1 2 ) 2 me m e 2 or 1 2 ) 2 + relative differene: relative differene 1 2 = v m e 2 m e 2 + = 1 v ) 2 or 1 2 me 2 ) = lassial solution gives no points 0.0 2

3 A3 1.0 pt) Derive an expression for the uniform magneti flux density B neessary to keep the proton beam on a irular trak. The expression should only ontain the energy of the protons E, the irumferene L, fundamental onstants and numbers. You may use suitable approximations if their effet is smaller than the preision given by the least number of signifiant digits. Calulate the magneti flux density B for a proton energy of E = 7.00 TeV. Solution A3: [1.0] Balane of fores: γ m p v 2 r = m p v 2 = e v B 0.3 r 1 v2 2 In ase of a mistake, partial points an be given for intermediate steps up to max ). Examples: Energy: Therefore: With Example: Lorentz fore Example: γ m p v 2 E = γ 1) m p 2 γ m p 2 γ = r E m p 2 E v 2 r = e B 0.3 follows: Solution: v and r = L 2π B = 2π E e L B = 5.50T Penalty for < 2 or > 4 signifiant digits - Calulation without approximations is also orret but does not give more points B = 2π m p e L ) E 2 m p m ) E Penalty for eah algebrai mistake - Classial alulation gives ompletely wrong result and maximum 0.3 pt [0.3] m p v 2 = e v B r B = 2π 2 mp E L e 3

4 B = T Penalty for < 2 or > 4 signifiant digits - 4

5 A4 1.0 pt) An aelerated harged partile radiates energy in the form of eletromagneti waves. The radiated power P rad of a harged partile that irulates with a onstant angular veloity depends only on its aeleration a, its harge q, the speed of light and the permittivity of free spae ɛ 0. Use a dimensional analysis to find an expression for the radiated power P rad. Solution A4: [1.0] Ansatz: P rad = a α q β γ ɛ δ 0 Dimensions: [a]=ms 2, [q]=c=as, []=ms 1,[ɛ 0 ]=AsVm) 1 =A 2 s 2 Nm 2 ) 1 =A 2 s 4 kgm 3 ) 1 All dimensions orret 0.3 Three dimensions orret Two dimensions orret if dimensions: N and Coulomb [ɛ 0 ]= C 2 Nm 2 ) 1 From this follows: m α s 2α Cβ mγ s γ C 2δ N δ m 2δ = N m s N : δ = 1, C : β + 2 δ = 0, m : α + γ 2δ = 1, s : 2 α + γ = 1 And therefore: Two equations orret α = 2, β = 2, γ = 3, δ = 1 if dimensions: N and As [ɛ 0 ]=A 2 s 2 Nm 2 ) 1 From this follows: m α s 2α Aβ s β mγ s γ A2δ s 2δ N δ m 2δ = N m s N : δ = 1, A : β +2 δ = 0, m : α+γ 2δ = 1, s : 2 α+β γ +2δ = 1 And therefore: Two equations orret α = 2, β = 2, γ = 3, δ = 1 if dimensions: kg and As [ɛ 0 ]=A 2 s 4 kg m 3 ) 1 m α s 2α Aβ s β mγ s γ A2δ s 4δ kg m2 kg δ = m3δ s 3 5

6 From this follows: kg : δ = 1, A : β +2 δ = 0, m : α+γ 3δ = 2, s : 2 α+β γ +4δ = 3 Two equations orret And therefore: α = 2, β = 2, γ = 3, δ = 1 Radiated Power: P rad a2 q 2 3 ɛ 0 Other solutions with other units are possible and are aepted No solution but realise that unit of harge must vanish β = 2δ 6

7 A5 1.0 pt) Calulate the total radiated power P tot of the LHC for a proton energy of E = 7.00 TeV Note table 1). You may use appropriate approximations. Solution A5: [1.0] Radiated Power: Energy: Aeleration: Therefore: Total radiated power: P rad = γ4 a 2 e 2 6π 3 ɛ 0 E = γ 1)m p 2 or equally valid E γ m p 2 a 2 r with r = L 2π E P rad = m p 2 + e 2 E e 2 1)4 6πɛ 0 r 2 or m p 2 )4 6πɛ 0 r not required P rad = W) P tot = P rad = 5.13kW penalty for missing fator 2 for the two beams): - - penalty for wrong numbers 2808 and/or numbers ome from table 1): - - 7

8 A6 1.5 pt) Determine the time T that the protons need to pass through this field. Solution A6: [1.5] 2nd Newton s law Conservation of energy: Sine d F = dp dt = p f p i T leads to with p i = E tot = m 2 + e V Etot 2 = m 2 ) 2 + p f ) 2 p f = 1 ) m 2 + e V ) 2 e V 2 m 2 ) 2 = 2e m V + T = d p f = d ) e V 2 2e m p V T = 218ns Alternative solution [1.5] 2nd Newton s law F = dp leads to dt = p f p i with p i = d T veloity from A1 or from onservation of energy m p v = 1 2 ) 2 m p 2 + and hene for γ p f = γ m p v = γ = 1/ 1 + e V m p 2 1 v2 2 = 1 + ) m p T = d p f = d m mp p 2 ) + e V 2 m p 2 1 = d e V m p 2 m p 1 2 ) 2 m p 2 + ) e V 2 2e m p V T = 218ns Alternative solution: integrate time [1.5] 8

9 Energy inreases linearly with distane x Ex) = e V x d d dx t = dt = 0 vx) ) 2 mp vx) = m p = m p 2 + m p 2 + e V x d ) e V x 2 d mp 2 ) 2 e V x d t = 1 = Substitution : ξ = b 0 t = m p d e V ξ 1 + ξ) ξ := oshs) 1 + e V x d m p 2 e V x d m p 2 ) 2 1 e V x d m p 2 d m p 2 e V oshs) sinhs)ds osh 2 s) 1 dξ dx = with b 1 = osh 1 1), b 2 = osh e V ) m p 2 e V d m p 2 dξ b = e V m p 2 dξ = sinhs) ds = m p d [sinhs)] b 2 e V b1 T = 218ns Alternative: differential equation [1.5] d = d dt m v 1 v2 2 F = dp dt m a 1 v2 = a = s = d m 2 ) ) 3 1 v ṡ2 2 ) m a v2 2 = γ 3 m a Ansatz : st) = i 2 t 2 + k l with boundary onditions s0) = 0, v0) = 0 st) = e2 V 2 t m 2 d 2 m d) V s = d T = d ) e m T = 218ns 9

10 lassial solution: [0.4] F = d aeleration a = F m p = m p d And hene for the time d = 1 2 a T 2 T = T = d 2 mp 2d a T = 194ns 10

11 Part B. Partile identifiation 4 points) B1 0.8 pt) Express the partile rest mass m in terms of the momentum p, the flight length l and the flight time t assuming that the partiles with elementary harge e travel with veloity lose to on straight traks in the ToF detetor and that it travels perpendiular to the two detetion planes see Figure 2). Solution B1: [0.8] with veloity v = l t relativisti momentum gets mass m = p t l p = p = t m v 1 v2 2 1 l2 t 2 2 = m l 1 l2 t 2 2 p l t 2 2 l Alternative [0.8] with flight distane: l, flight time t gets: relativisti momentum t = l β) p = m β 1 β 2 therefore the veloity: β = p m2 2 + p 2 insert into the expression for t: mass: m = p t l t = l m2 2 + p 2 p ) 2 p ) 2 p = t ) 2 l) l non-relativisti solution: [0.0] flight time: t = l/v veloity: v = p m t = l m p and m = p t l this solution gives no points

12 B2 0.7 pt) Calulate the minimal length of a ToF detetor that allows to safely distinguish a harged kaon from a harged pion given both their momenta are measured to be 1.00 GeV/. For a good separation it is required that the differene in the time-of-flight is larger than three times the time resolution of the detetor. The typial resolution of a ToF detetor is 150 ps 1 ps = s). Solution B2: [0.7] Flight time differene between kaon and pion Flight time differene between kaon and pion t = 450ps = s t = l p m 2 π 2 + p 2 m 2 K 2 + p 2 ) = 450ps = s l = t p m 2K + p2 / 2 m 2 π + p 2 / 2 m 2 K + p2 / 2 = GeV/ 2 and m 2 π + p 2 / 2 = GeV/ 2 l = s GeV2 /GeV) l = s = m = 1.28m Penalty for < 2 or > 4 signifiant digits - Non-relativisti solution: [0.3] Flight time differene between kaon and pion t = l p m K m π ) = 450ps = s length: l = tp m K m Π = s 1GeV/ )GeV/ 2 l = /0.363 s = / m l = m = 0.372m Penalty for < 2 or > 4 signifiant digits - 12

13 B3 1.7 pt) Express the partile mass as a funtion of the magneti flux density B, the radius R of the ToF tube, fundamental onstants and the measured quantities: radius r of the trak and time-of-flight t. Solution B3: [1.7] Partile is travelling perpendiular to the beam line hene the trak length is given by the length of the ar Lorentz fore transverse momentum, sine there is no longitudinal momentum, the momentum is the same as the transverse momentum Use formula from B1 to alulate the mass trak length: length of ar l = 2 r asin R 2 r penalty for just taking a straight trak l = R) -0.4 partial points for intermediate steps, maximum 0.4 Lorentz fore γ m vt 2 = e v t B p T = r e B 0.4 r partial points for intermediate steps, maximum 0.3 longitudinal momentum=0 p = p T 0.5 momentum m = p t l p = e r B ) 2 p ) 2 t = e r B 2r asin R 2r ) ) 2 ) partial points for intermediate steps, maximum 0.5 Non-relativisti: trak length: length of ar [0.9] l = 2 r asin R 2 r 0.5 penalty for just taking a straight trak l = R) -0.4 partial points for intermediate steps, maximum 0.4 m = p t l = e r B t 2r asin R 2r partial points for intermediate steps, maximum 0.3 = e B t 2 asin R 2r

14 B4 0.8 pt) Identify the four partiles by alulating their mass. Partile Radius r [m] Time of flight [ns] A B C D Solution B4: [0.8] Partile ar p p pt/l pt/l pt/l Mass Mass [m] [ MeV ] [ mkg s ] [ MeV s m ] [ MeV ] [kg] [ MeV ] [kg] A B C D Partiles A and C are protons, B is a Pion and D a Kaon orret mass and identifiation: per partile penalty for orret mass but no or wrong identifiation for 1 or 2 partiles - penalty for orret mass but no or wrong identifiation for 3 or 4 partiles - wrong mass, orret momentum:per partile wrong momentum, orret ar for 3 or 4 partiles wrong momentum, orret ar for 1 or 2 partiles non relativisti solution m = pt/l Partile identifiation is not possible [0.4] Partile ar p p m = p t/l m = p t/l m = p t/l [m] [ MeV ] [ mkg s ] [ MeV s m ] [ MeV ] [kg] A B C D orret mass or orret momentum: per partile wrong momentum, orret ar for 3 or 4 partiles wrong momentum, orret ar for 1 or 2 partiles 14