<<

- 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 , 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 − p and e − p scattering. The 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 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 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 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 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 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 that would not be detected. Infrared divergence in the cross section is canceled by a similar infrared divergence in the vertex correction. Calculations including target recoil, from the target, the target vertex correction, and two- 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. Edith Borie Proton Radius Puzzle, October 2012 Radiative Corrections 15 / 22 The infrared divergent term in F1 will be canceled by a contribution from bremsstrahlung of real soft photons. F2 is not infrared divergent. At very high momentum transfers, it becomes negligibly small. At very high momentum transfers, it becomes negligibly small. For the momentum transfers of interest in this experiment, F2(t) is comparable to the other radiative corrections, and it results in a nonmultiplicative contribution to the radiative corrections. A calculation of this contribution to the radiatively corrected cross section will be given. 2 The lepton trace will be evaluated to leading order in α/π, so that F2 ≈ 0 and 2α α (F + F )2 ≈ 1+ (F (2) + F (2)), F F ≈ F (2) 1 2 π 1 2 1 2 π 2 The revised lepton trace is then given by

(F + F )2 F F 1 2 [pµpν + pµpν + g µν q2/2] − 1 2 (p + p )µ(p + p )ν m2 3 1 1 3 m2 1 3 1 3

(2) (2) Inclusion of the vacuum polarization contributions replaces F1 + F2 by (2) (2) F1 + F2 + U2m + U2e .

Dr. Edith Borie Proton Radius Puzzle, October 2012 Radiative Corrections 16 / 22 2 The term with (F1 + F2) ≈ 1 corresponds to the uncorrected cross section. The term of order α/π contributes a factor multiplying the uncorrected cross section, thus giving δel = δvertex + δVP . δvertex = 2(F1 + F2 − 1) and δVP = (2α/π)(U2m + U2e ). 2 2 In the limiting case Q /m ≫ 1, the expression for δel agrees with the corresponding expression in Eq.(3.36) of a paper by Maximon and Tjon. The contributions to δ from the vertex correction and from vacuum polarization (without the contribution from radiation of real (soft) photons) were also given in the textbook by Akhiezer and Berestetskii. The presence of an additional correction, not proportional to the uncorrected cross section, was also mentioned in that work Only the contribution of the last term in the lepton trace µ ν (proportional to (p1 + p3) (p1 + p3) ) requires additional calculation.

Dr. Edith Borie Proton Radius Puzzle, October 2012 Radiative Corrections 17 / 22 The proton trace was given before and the additional term in the 2 spin-averaged square of the matrix element |Mfi | is then (without the factor −F2)

2 2 4π α 2 2 2 GM 2p (p + p )p (p + p )+(p + p ) q /2 m2M2(q2)2 h  2 1 3 4 1 3 1 3  G 2 − G 2 + E M (p + p ) (p + p ) 2 2(1 − q2/4M2) 1 3 2 4  i

Evaluation of the scalar products gives finally

2 2 2 2 2 4π α 2 ′ 2 2 2 q GE − GM ′ 2 GM 2(E+E ) +(4m −q ) + 2(E+E ) m2(q2)2 h  2M2  (1 − q2/4M2) i

Dr. Edith Borie Proton Radius Puzzle, October 2012 Radiative Corrections 18 / 22

extra term/total vertex 1.4 extra term/total (vertex+VP)

1.2

1

0.8

0.6 radiative correction ratio 0.4

0.2

0 0 20 40 60 80 100 120 140 160 180 scattering angle Ratio of the additional contribution to the radiatively corrected cross section to the usual contribution to the radiatively corrected cross section from the noninfrared part of the vertex correction for an initial muon momentum of 153MeV/c. The dashed curve shows a similar ratio, but including the contribution due to vacuum polarization. (muon and electron loops).

Dr. Edith Borie Proton Radius Puzzle, October 2012 Radiative Corrections 19 / 22 The VP contribution from electron loops is significantly larger than the other contributions. To check whether higher order electron vacuum polarization 2 (corresponding to two loops) might be important, the value of (α/π)U2e and is shown as a comparison with other contributions (these without the overall factor (α/π))

◦ 2 2 2 θ( ) Q (GeV /c) F1 − 1 F2 U2m U2e (α/π)U2e 30 0.00611 0.0564 0.4588 0.0345 2.7582 0.0128 60 0.02133 0.1256 0.3841 0.1069 3.2146 0.0240 90 0.03907 0.1394 0.3259 0.1746 3.4164 0.0271 120 0.05394 0.1240 0.2908 0.2221 3.5239 0.0288

Only the noninfrared part of F1 is given. The last column indicates that the contribution from fourth order electron vacuum polarization is comparable to other contributions. In the limit Q2/m2 ≫ 1, this results in an additional 2 contribution to δeVP of 2(α/π) U4e where

2 2 5 1 Q U e U U e U m 4 = 2e + 2 2 − + ζ(3) + ln( 2 ) 24 4 me

Dr. Edith Borie Proton Radius Puzzle, October 2012 Radiative Corrections 20 / 22 Conclusions

◮ Since the VP contribution from the electron loops is significantly larger than the other contributions, even fourth-order electron VP should be included. ◮ For the momentum transfers of interest, the nonmultiplicative contribution to the radiative corrections is not negligible. However, the calculation of the additional correction is straightforward. ◮ The contributions from soft photon radiation, the proton vertex correction, and two-photon corrections should also be included. ◮ Since these depend on the proton structure, they should probably be given separately.

Dr. Edith Borie Proton Radius Puzzle, October 2012 Conclusions 21 / 22 I thank . Gilman for extensive email correspondence regarding this work.

Thank you for your attention

Dr. Edith Borie Proton Radius Puzzle, October 2012 Conclusions 22 / 22