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Muon-Proton Scattering

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Muon-Proton Scattering Muon-proton Scattering Edith Borie Karlsruhe Institute of Technology (KIT), Association Euratom-KIT Institut f¨ur Hochleistungsimpuls- und Mikrowellentechnik (IHM) Kaiserstrasse 12, 76131 Karlsruhe, Germany [email protected] Dr. Edith Borie Proton Radius Puzzle, October 2012 1 / 22 Introduction The MUon proton Scattering Experiment (MUSE) at the Paul Scherrer Institut (PSI) intends to study the proton radius puzzle through simultaneous measurements of µ − p and e − p scattering. The muons will be neither ultrarelativistic nor nonrelativistic. (v/c is not approximately equal to 1 and is not extremely small.) Hence the need for equations describing the scattering cross section that do not use any approximations about the lepton energy. Expressions for the radiative corrections that do not use the 2 2 approximation Q /mµ ≫ 1 are also required. ◮ Proposal’s Kinematic Parameters ◮ Definitions ◮ The Cross Section ◮ Radiative Corrections Dr. Edith Borie Proton Radius Puzzle, October 2012 Introduction 2 / 22 Kinematic Parameters The muon momenta will be in the range (115-210) MeV/c with scattering angles in the range 20◦ to 100◦, corresponding to Q2 in the range (0.01-0.1) (GeV/c)2. For comparison, 2 2 2 mµc = 0.01116 (GeV/c) . The muon energy (E = p2 + m2) and velocity p (β = |p/E| = v/c) corresponding to the incoming momenta in the proposal are: p (MeV/c) E(MeV) β 115 156.17 0.7364 153 185.94 0.8229 210 235.08 0.8933 Dr. Edith Borie Proton Radius Puzzle, October 2012 Kinematic Parameters 3 / 22 Definitions ◮ m=lepton mass and M=target mass ◮ β = v/c (all components of velocity) ◮ α = e2/4π ◮ p1 and p3 are the incoming and outgoing lepton four-momenta ◮ p2 and p4 are the incoming and outgoing proton four-momenta ◮ q = p1 − p3 = p4 − p2 ′ ′ ◮ Lab system, p1 =(E,p) p3 =(E , p ), p2 =(M, 0), p4 =(M + q0,q) 2 2 2 ′ q = 2m − 2p1 · p3 = 2M − 2p2 · p4 = −2M(E − E ) Dr. Edith Borie Proton Radius Puzzle, October 2012 Definitions 4 / 22 The proton current has the usual on-shell form: iσ qν Γ = F (q2)γ + κF (q2) µν µ 1 µ 2 2M κ is the anomalous magnetic moment of the proton. Q2 Sachs form factors: GM = F + κF , GE = F − κF . 1 2 1 4M2 2 Q2 Q2 Or, F = (GE + GM )/(1 + ) and 1 4M2 4M2 Q2 κF = (GM − GE )/(1 + ) 2 4M2 The Gordon decomposition iσ qν (p + p ) u¯(p ) µν u(p )=¯u(p ) γ − 2 4 µ u(p ) 4 2M 2 4 µ 2M 2 is useful: (p + p ) u¯(p )Γ u(p )=u ¯(p ) (F + κF )γ − κF 2 4 µ u(p ) 4 µ 2 4 1 2 µ 2 2M 2 Dr. Edith Borie Proton Radius Puzzle, October 2012 Definitions 5 / 22 The Cross Section The invariant matrix element for scattering of a charged lepton from a proton in Born approximation is 2 µ e Mfi =u ¯(p )γ u(p ) u¯(p )Γ u(p ) 3 1 q2 4 µ 2 The cross section for scattering of the charged lepton into a given solid angle dΩ′ about an angle θ is 2 2 ′ dσ m M p /p 2 = |Mfi | dΩ′ 4π2 M + E − pE ′/p′cosθ 2 The spin sum and average of |Mfi | is 4 2 e p3 + m µ p1 + m ν |Mfi | = · Tr γ γ · 4(q2)2 h 2m 2m i p + M p + M Tr 4 Γ 2 Γ h 2M µ 2M νi Dr. Edith Borie Proton Radius Puzzle, October 2012 The Cross Section 6 / 22 The lepton trace is 1 [pµpν + pµpν + g µνq2/2] m2 3 1 1 3 The trace for the proton is 2 p4 + M p2 + M GM Tr γ γ h 2M µ 2M νi κF 2 p + M p + M + 2 (p + p ) (p + p ) Tr 4 2 2M 2 4 µ 2 4 ν 2M 2M GM κF p + M p + M − 2 (p + p ) Tr 4 2 γ 2M h 2 4 µ 2M 2M ν p + M p + M +(p + p ) Tr 4 γ 2 2 4 ν 2M µ 2M i Dr. Edith Borie Proton Radius Puzzle, October 2012 The Cross Section 7 / 22 Evaluation of the traces gives 2 G GM κF M [p p + p p + g µνq2/2] − 2 (p + p ) (p + p ) M2 2µ 4ν 2ν 4µ M2 2 4 µ 2 4 ν κF 2 1 + 2 (p + p ) (p + p ) (p · p + M2) 2M 2 4 µ 2 4 ν M2 2 4 In terms of the Sachs form factors G 2 M [p p + p p + g µνq2/2] M2 2µ 4ν 2ν 4µ G 2 − G 2 1 + E M (p + p ) (p + p ) (p · p + M2) 2M2(1 − q2/4M2) 2 4 µ 2 4 ν M2 2 4 Dr. Edith Borie Proton Radius Puzzle, October 2012 The Cross Section 8 / 22 The square of the matrix element is then 2 2 2 4π α 2 2 ′ 2 2 2 4 |Mfi | = GM (4M EE +(M + m )q + q /2) m2M2(q2)2 h 2 2 G − G ′ + E M (4M2EE + M2q2) 1 − q2/4M2 i Dr. Edith Borie Proton Radius Puzzle, October 2012 The Cross Section 9 / 22 The final result for the cross section is 2 ′ ′ 2 dσ α p /p 2 (4EE + q ) = GE dΩ′ q4 1 + (E − pE ′/p′ cos θ)/M h 1 − q2/4M2 4 2 2 2 ′ 2 1 q q m + GM (4EE + q ) 1 − + + 1 − q2/4M2 2M2 M2 i Recall that ′ ′ −q2 = Q2 = 2(EE − pp cos θ − m2) In the limit of very high lepton energies, ′ ′ ′ p ≈ E, p ≈ E , q2 ≈−4EE sin2(θ/2) and the cross section reduces to dσ α2 cos2(θ/2) 1 = · dΩ′ 4E 2 sin4(θ/2) 1+2E sin2(θ/2)/M 2 2 Q 2 1 2 GE GM + 2tan (θ/2) + h2M2 1+ Q2/4M2 1+ Q2/4M2 i Dr. Edith Borie Proton Radius Puzzle, October 2012 The Cross Section 10 / 22 12 pin=115 MeV/c pin=153 MeV/c pin=210 MeV/c 10 8 6 cross section ratio 4 2 0 0 20 40 60 80 100 120 140 160 180 scattering angle Ratio of the cross section calculated with the exact formula to that calculated with the high energy approximation, but with exact muon kinematics. The form factors were taken from a parametrization given by Kelly. Dr. Edith Borie Proton Radius Puzzle, October 2012 The Cross Section 11 / 22 Radiative Corrections Radiative corrections to the electromagnetic properties of a muon or an electron modify the lepton scattering cross sections. Standard theory: the cross section for scattering is altered by a factor 1 − δ. For potential scattering at high momentum transfer it is (Schwinger in 1949): 2α 2 2 δ ≈ (ln(Q /m ) − 1) ln(E/∆E) + δvertex + δVP π ∆E is the energy acceptance of the detector. It is related to the fact that the lepton can emit very low-energy photons that would not be detected. Infrared divergence in the Bremsstrahlung cross section is canceled by a similar infrared divergence in the vertex correction. Calculations including target recoil, radiation from the target, the target vertex correction, and two-photon exchange usually assume that Q2/m2 is very large. Not the case for the proposed experiment. Dr. Edith Borie Proton Radius Puzzle, October 2012 Radiative Corrections 12 / 22 Energy of the recoiling proton is less than 50-60MeV. Contributions from radiation by the proton will be small. Two photon exchange is discussed elsewhere. Consider here only the (purely elastic) leptonic contributions δvertex and δVP . δVP includes contributions from muon and electron loops. The lepton vertex correction gives rise to two form factors F1 and F2. 2 2 If Q /m is of order unity, F2 is not negligible. There will be an extra contribution to the radiatively corrected cross section that does not simply multiply the uncorrected cross section. The muon current including (vertex) radiative corrections of order α has the form (p1 + p3)µ u¯(p ) (F + F )γµ − F u(p ) 3 1 2 2 2m 1 Here α α F (t) ≈ 1+ F (2)(t), F (t) ≈ F (2)(t) 1 π 1 2 π 2 2 2 As usual, t = q = (p1 − p3) . Dr. Edith Borie Proton Radius Puzzle, October 2012 Radiative Corrections 13 / 22 The radiative corrections also include vacuum polarization In the matrix element for scattering the contributions from muon and electron loops must be added to the term proportional to γµ. They are: α 1 2 α + (coth (φ) − 3)(1 − φ · coth(φ)) = U m 3π h3 i π 2 for muon loops and α Q2 5 α U e ln( 2 ) − = 2 3π h me 3i π for electron loops. φ is related to the momentum transfer by sinh(φ)= Q/(2m). For the contribution due to electron loops, one may use the usual high momentum transfer approximation. Dr. Edith Borie Proton Radius Puzzle, October 2012 Radiative Corrections 14 / 22 For spacelike q2 = −Q2 the form factors can be expressed in terms of a variable Θ: 1 − tanh(φ) Θ = 1 + tanh(φ) Useful: ln(Θ) = −2φ and Q2/m2 = (1 − Θ)2/Θ. α λ 1+Θ2 F (t) − 1 = ln 1+ ln(Θ) 1 π h m 1 − Θ2 3Θ2 +2Θ+3 − 1 − ln(Θ) 4(1 − Θ2) 1+Θ2 π2 1 + + L (−Θ) − ln2(Θ) + ln(Θ)ln(1 + Θ) 1 − Θ2 12 2 4 i α Θ F (t) = − ln(Θ) 2 π 1 − Θ2 Dr.
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